SOLUTIONS MANUAL for Physics for Scientists & Engineers with Modern Physics 5th Edition by Douglas C by StudyGuide

Physics for Scientists & Engineers with Modern Physics 5th Edition by Douglas C. Giancoli.

TABLE OF CONTENTS CHAPTER 1 Introduction, Measurement, Estimating CHAPTER 2 Describing Motion: Kinematics in One Dimension CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors CHAPTER 4 Dynamics: Newton’s Laws of Motion CHAPTER 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces CHAPTER 6 Gravitation and Newton’s Synthesis CHAPTER 7 Work and Energy CHAPTER 8 Conservation of Energy CHAPTER 9 Linear Momentum CHAPTER 10 Rotational Motion CHAPTER 11 Angular Momentum; General Rotation CHAPTER 12 Static Equilibrium; Elasticity and Fracture CHAPTER 13 Fluids CHAPTER 14 Oscillations CHAPTER 15 Wave Motion CHAPTER 16 Sound CHAPTER 17 Temperature, Thermal Expansion, and the Ideal Gas Law CHAPTER 18 Kinetic Theory of Gases CHAPTER 19 Heat and the First Law of Thermodynamics CHAPTER 20 Second Law of Thermodynamics

CHAPTER 21 Electric Charge and Electric Field CHAPTER 22 Gauss’s Law CHAPTER 23 Electric Potential CHAPTER 24 Capacitance, Dielectrics, Electric Energy Storage CHAPTER 25 Electric Current and Resistance CHAPTER 26 DC Circuits CHAPTER 27 Magnetism CHAPTER 28 Sources of Magnetic Field CHAPTER 29 Electromagnetic Induction and Faraday’s Law CHAPTER 30 Inductance, Electromagnetic Oscillations, and AC Circuits CHAPTER 31 Maxwell’s Equations and Electromagnetic Waves CHAPTER 32 Light: Reflection and Refraction CHAPTER 33 Lenses and Optical Instruments CHAPTER 34 The Wave Nature of Light: Interference and Polarization CHAPTER 35 Diffraction CHAPTER 36 The Special Theory of Relativity CHAPTER 37 Early Quantum Theory and Models of the Atom CHAPTER 38 Quantum Mechanics CHAPTER 39 Quantum Mechanics of Atoms CHAPTER 40 Molecules and Solids CHAPTER 41 Nuclear Physics and Radioactivity CHAPTER 42 Nuclear Energy; Effects and Uses of Radiation CHAPTER 43 Elementary Particles CHAPTER 44 Astrophysics and Cosmology

CHAPTER 1: Introduction, Measurement, Estimating Responses to Questions 1.

(a) A particular person’s foot. Merits: reproducible. Drawbacks: not accessible to the general public; not invariable (could change size with age, time of day, etc.); not indestructible. (b) Any person’s foot. Merits: accessible. Drawbacks: not reproducible (different people have different size feet); not invariable (could change size with age, time of day, etc.); not indestructible. Neither of these options would make a good standard.

The distance in miles is given to one significant figure, and the distance in kilometers is given to five significant figures! The value in kilometers indicates more precision than really exists or than is meaningful. The last digit represents a distance on the same order of magnitude as a car’s length! The sign should perhaps read “7.0 mi (11 km),” where each value has the same number of significant figures, or “7 mi (11 km),” where each value has about the same % uncertainty.

The number of digits you present in your answer should represent the precision with which you know a measurement; it says very little about the accuracy of the measurement. For example, if you measure the length of a table to great precision, but with a measuring instrument that is not calibrated correctly, you will not measure accurately. Accuracy is a measure of how close a measurement is to the true value.

If you measure the length of an object and you report that it is “4,” you haven’t given enough information for your answer to be useful. There is a large difference between an object that is 4 meters long and one that is 4 km long. Units are necessary to give meaning to a numerical answer.

You should report a result of 8.32 cm. Your measurement had three significant figures. When you multiply by 2, you are really multiplying by the integer 2, which is an exact value. The number of significant figures is determined by the measurement.

The correct number of significant figures is three: sin 30.0º = 0.500.

Useful assumptions include the population of the city, the fraction of people who own cars, the average number of visits to a mechanic that each car makes in a year, the average number of weeks a mechanic works in a year, and the average number of cars each mechanic can see in a week. (a) There are about 800,000 people in San Francisco. Assume that half of them have cars. If each of these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a year. If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be seen each week. Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a week. The final estimate, then, is 800 car mechanics in San Francisco. (b) Answers will vary. But following the same reasoning, the estimate is 1/1000 of the population.

© 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Responses to MisConceptual Questions 1.

(c) As stated in the text, scientific laws are descriptive – they are meant to describe how nature behaves. Since our understanding of nature evolves, so do the laws of physics, when evidence can convince the community of physicists. The laws of physics are not permanent, and are not subject to political treaties. In fact, there have been major changes in the laws of physics since 1900 – particularly due to relativity and quantum mechanics. The laws of physics apply in chemistry and other scientific fields, since those areas of study are based on physics. Finally, the laws of physics are man-made, not a part of nature. They are our “best description” of nature as we currently understand it. As stated in the text, “Laws are not lying there in nature, waiting to be discovered.”

(e) The first product is 142.08 m, which is only accurate to the 10’s place, since 37 m/s has only two significant figures. The second product is 74.73 m, which is only accurate to the 1’s place, since 5.3 s has only two significant figures. Thus the sum of the two terms can only be accurate to the 10’s place. 142.08 + 74.73 = 216.81, which to the 10’s place is 220 m.

(a) The total number of digits present does not determine the precision, as the leading zeros in (c) and (d) are only place holders. Rewriting all the measurements in units of meters shows that (a) implies a precision of 0.0001m, (b) and (c) both imply a precision of 0.001 m, and (d) implies a precision of 0.01 m. Note that since the period is shown, the trailing zeros are significant. If all the measurements are expressed in meters, (a) has 4 significant figures, (b) and (c) each have 3 significant figures, and (d) has 2 significant figures.

(b) The leading zeros are not significant. Rewriting this number in scientific notation, 7.8  10−3 , shows that it only has two significant digits.

(b) When you add or subtract numbers, the final answer should contain no more decimal places than the number with the fewest decimal places. Since 25.2 has one decimal place, the answer must be rounded to one decimal place, or to 26.6. Thus the answer has 3 significant figures.

(b) The word “accuracy” is often misused. If a student repeats a measurement multiple times and obtains the same answer each time, it is often assumed to be accurate. In fact, students are frequently given an “ideal” number of times to repeat the experiment for “accuracy.” However, systematic errors may cause each measurement to be inaccurate. A poorly working instrument may also limit the accuracy of your measurement.

(a) Quoting the textbook, “precision” refers to the repeatability of the measurement using a given instrument. Precision and accuracy are often confused. “Accuracy” is defined by answer (b).

(d) This addresses misconceptions about squared units and about which factor should be in the numerator of the conversion. This error can be avoided by treating the units as algebraic symbols that must be cancelled out.

(e) When making estimates, the estimator may frequently believe that their answers are more significant than they actually are. This question helps the estimator realize what an order-ofmagnitude estimation is NOT supposed to accomplish.

10. (d) This addresses the fact that the generic unit symbol, like [L], does not indicate a specific system of units. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 1

Introduction, Measurement, Estimating

Solutions to Problems 1.

(a) 777

3 significant figures

(b) 81.60

4 significant figures

(c)

7.03

3 significant figures

(d) 0.03

1 significant figure

(e)

0.0086

2 significant figures

(f)

6465

4 significant figures

(g) 8700

2 significant figures

(a) 5.859 = 5.859  100 (b) 21.8 = 2.18  101 (c)

0.0068 = 6.8  10−3

(d) 328.65 = 3.2865  102

(e)

0.219 = 2.19  10−1

(f)

444 = 4.44  102

(a) 8.69  105 = 869, 000 (b) 9.1103 = 9100 (c)

2.5  10−1 = 0.25

(d) 4.76 102 = 476 (e) 3.62 10−5 = 0.0000362 0.35 m

% uncertainty =

(a) % uncertainty =

3.25 m

(b) % uncertainty =

 100% = 11%

0.2 s 4.5s

 100% = 4.444%  4%

0.2 s

 100% = 0.4444%  0.4% 45s (c) The time of 4.5 minutes is 270 seconds. 0.2 s % uncertainty =  100% = 0.0741%  0.07% 270 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

To add values with significant figures, adjust all values to be added so that their exponents are all the same. ( 9.2  103 s ) + ( 6.3  104 s ) + ( 0.008  106 s ) = ( 9.2  103 s ) + ( 63  103 s ) + (8  103 s ) = ( 9.2 + 63 + 8 )  103 s = 80.2  103 s = 8.0  104 s

When adding, keep the least accurate value, and so keep to the “ones” place in the last set of parentheses. 7.

When you multiply, the result should have as many digits as the number with the least number of significant digits used in the calculation.

( 4.079  10 m )(0.057  10 m ) = 2.325m  2.3 m 2

−1

The “best value” uncertainty is taken to be 0.01 m. 0.01m 2 % uncertainty =  100% = 0.787%  0.8% 1.27 m 2

 (radians) 0 0.10 0.12

sin( ) 0.00 0.10 0.12

tan( ) 0.00 0.10 0.12

0.20 0.24 0.25

0.20 0.24 0.26

Keeping 2 significant figures in the angle, and expressing the angle in radians, the largest angle that has the same sine and tangent is 0.24 radians . In degrees, the largest angle (keeping 2 significant figure) is 12 .

10. (a) To find the number of students, multiply the sample size times the percentage. ( 215 students )( 0.372 ) = 79.98 students But the number of students must be an integer, so the number is 80 students . (b) The original statement gave the percentage with too many significant figures. The calculation is 80 / 215 = 0.372093023… , but that result should only have 2 significant figures, since the value of 80 (as an integer, and not an estimate) has only 2 significant figures. The percentage should have been quoted as 37%. That would still give the correct number of students. ( 215 students )( 0.37 ) = 79.55 students, which rounds to 80 students. 11. We calculate what fraction 9 seconds is of an entire year. 1yr  9s   −5 −5  1yr   3.156  107s   100% = 2.85  10 %  3  10 %    12. In order to find the approximate uncertainty in the area, calculate the area for the specified radius, the minimum radius, and the maximum radius. Subtract the extreme areas. The uncertainty in the area is then half this variation in area. The uncertainty in the radius is assumed to be 0.1  104 cm .

(

2 Aspecified =  rspecified =  5.1  104 cm

) = 8.17  10 cm 2

( ) = 7.85  10 cm =  ( 5.2  10 cm ) = 8.49  10 cm

2 Amin =  rmin =  5.0  104 cm

2 Amax =  rmax

Chapter 1

Introduction, Measurement, Estimating

(

)

A = 12 ( Amax − Amin ) = 12 8.49  109 cm 2 − 7.85  109 cm 2 = 0.32  109 cm 2

Note that the last value above has only 1 significant figure. Thus the area should be quoted as

A = (8.2  0.3)  109 cm2 . 13. In order to find the approximate uncertainty in the volume, calculate the volume for the minimum radius and the volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the volume is then half of this variation in volume. 3 3 Vspecified = 43  rspecified = 43  ( 0.64 m ) = 1.098 m3 3 Vmin = 43  rmin = 43  ( 0.60 m ) = 0.905 m 3 3

3 Vmax = 43  rmax = 43  ( 0.68 m ) = 1.317 m 3 3

(

)

V = 12 (Vmax − Vmin ) = 12 1.317 m3 − 0.905 m3 = 0.206 m3

Note that the last value above has only 1 significant figure. Thus the percent uncertainty is V 0.206 m 3 =  100 = 18.6  20 % . Vspecified 1.098 m 3 14. (a) 286.6 mm

286.6  10−3 m

0.286 6 m

(b) 74  V

74  10−6 V

0.000 074 V

(c)

430 mg

430  10 −3 g

0.43g (if last zero is not significant)

(d) 47.2 ps

47.2  10−12 s

0.000 000 000 047 2 s

(e)

22.5 nm

22.5  10−9 m

0.000 000 022 5 m

(f) 2.50 gigavolts

2.5  109 volts

2, 500, 000, 000 volts

Note that in part (f) in particular, the correct number of significant digits cannot be determined when you write the number in this format. 15. (a) 3  106 volts

3megavolt = 3 Mvolt

(b) 2  10−6 meters

2 micrometers = 2  m

5 kilodays = 5 kdays

(d) 18  102 bucks

18 hectobucks = 18 hbucks or 1.8 kilobucks

(e)

−7

9  10 seconds

900 nanoseconds = 900 ns or 0.9 s

16. Assuming a height of 5 feet 10 inches, then 5'10" = ( 70 in )(1 m 39.37 in ) = 1.8 m . Assuming a weight of 165 lbs, then (165 lbs )( 0.456 kg 1 lb ) = 75.2 kg . Technically, pounds and mass measure two separate properties. To make this conversion, we have to assume that we are at a location where the acceleration due to gravity is 9.80 m/s2.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

17. (a)

(b)

SA Earth SA Moon

2 4 REarth 2 4 RMoon

4 3

( 6.38  10 km ) = 13.4 = (1.74  10 km ) ( 6.38  10 km ) = 49.3 = (1.74 10 km )

2 REarth

2 RMoon

3 3  REarth REarth = 4 = 3 3 RMoon VMoon 3  RMoon

VEarth

Instructor Solutions Manual

18. To answer this question, convert 15 m/s to mi/h, and compare to the speed limit.  1 mi   3600s  15 m s    = 33.56 mi h  34 mi h  1609 m   1h  Since the conversion only gives about 34 mi/h, the driver is not exceeding the speed limit . We could also change 35 mi/h to m/s, and find that 35 mi/h is more than 15 m/s. 19. (a) 14 billion years = 1.4 1010 years (b)

 3.156  107s  = 4.4  1017s )  1yr  

(

1.4  1010 yr 

(

)

20. (a) 93 million miles = 93 106 miles (1610 m 1 mile ) = 1.5  1011 m (b) 1.5  1011 m = (1.5  1011 m )(1km 103 m ) = 1.5  108 km 21. To add values with significant figures, adjust all values to be added so that their units are all the same. 1.90 m + 142.5cm + 6.27  105  m = 1.90 m + 1.425 m + 0.627 m = 3.952 m  3.95 m When you add, the final result is to be no more accurate than the least accurate number used. In this case, that is the first measurement, which is accurate to the hundredths place when expressed in meters.

(

)

22. (a) 1.0 10−10 m = 1.0 10−10 m ( 39.37 in 1 m ) = 3.9 10 −9 in (b)

(1.0 cm ) 

1 m   1 atom  8   = 1.0  10 atoms −10  100 cm   1.0  10 m 

23. (a)

(1km h ) 

(b)

(1m s ) 

(c)

(1km h ) 

0.621 mi 

 = 0.621mi h , so the conversion factor is  1 km  3.28 ft 

 = 3.28 ft s, so the conversion factor is  1m 

0.621mi h 1km h

3.28 ft s 1m s

1000 m   1 h

0.278 m s  .   = 0.278 m s, so the conversion factor is 1km h  1 km   3600 s 

Note that if more significant figures were used in the original factors, such as 0.6214 miles per kilometer, then more significant figures could have been included in the answers.

Chapter 1

Introduction, Measurement, Estimating

(

)

(

24. (a) 1 ft 2 = 1 ft 2 (1 yd 3 ft ) = 0.111 yd 2, and so the conversion factor is (b) 1 m = 1 m 2

) ( 3.28 ft 1 m ) = 10.8 ft , and so the conversion factor is 2

0.111 yd 2 1 ft 2

10.8ft 2 1m 2

25. (a) Find the distance by multiplying the speed by the time.

(

)(

)

1.00 ly = 2.998  108 m s 3.156  107 s = 9.462  1015 m  9.46  1015 m (b) Do a unit conversion from ly to AU.  9.462  1015 m   1 AU  4 (1.00 ly )    = 6.31 10 AU  11  1.00 ly   1.50  10 m  26. One mile is 1609 m, according to the unit conversions in the front of the textbook. Thus it is 109 m longer than a 1500-m race. The percentage difference is calculated here. 109 m  100% = 7.3% 1500 m 27. From Example 1–6, the thickness of a page of this book is about 6  10−5 m. The wavelength of orange krypton-86 light is found from the fact that 1,650,763.73 wavelengths of that light is the definition of the meter.  6  10−5 m   1, 650, 763.73 wavelengths  1page    = 99  100 1m  1page    28. The original definition of the meter was that 1 meter was one ten-millionth of the distance from the Earth’s equator to either pole. The distance from the equator to the pole would be one-fourth of the circumference of a perfectly spherical Earth. Thus the circumference would be 40 million meters:

C = 4  107 m . We use the circumference to find the radius. C = 2 r → r =

4  107 m

= 6.37  106 m 2 2 The value in the front of the textbook is 6.38  106 m. Since the circumference and the radius are proportional to each other, the % error would be the same in both cases. We calculate it for the radius. R 0.01  106 m =  100 = 0.1567  0.2 % R 6.38  106 m

29. The value of 12 liters per km can also be expressed as 1 km per 12 liters. We convert the units of km per liter to miles per gallon.

1km  3.785 L   0.6214 mi 

  12 L  1gal  

1km

 = 0.196 gal  0.20 gal 

30. Since the meter is longer than the yard, the soccer field is longer than the football field. 1.094 yd Lsoccer − Lfootball = 100.0 m  − 100.0 yd = 9.4 yd 1m © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Lsoccer − Lfootball = 100.0 m − 100.0 yd 

= 8.6 m 1.094 yd Since the soccer field is 109.4 yd compared to the 100.0-yd football field, the soccer field is

9.4 % longer than the football field.

 3.156  107s  7  = 3.16  10 s 1 yr   7  3.156  10 s   1  109 ns  (b) # of nanoseconds in 1.00 yr: 1.00 yr = (1.00 yr )  = 3.16  1016 ns    1yr   1 s  1 yr   = 3.17  10−8 yr 1.00 s = (1.00 s )  (c) # of years in 1.00 s: 7   3.156  10 s  1.00 yr = (1.00 yr ) 

31. (a) # of seconds in 1.00 yr:

 10−15 kg  1 proton or neutron  = 1012 protons or neutrons   −27 10 kg   1 bacterium   −17   1 proton or neutron  10 kg (b)  = 1010 protons or neutrons   −27 10 kg   1 DNA molecule   2  10 kg  1 proton or neutron  (c)  = 1029 protons or neutrons   −27 10 kg   1 human   41  10 kg  1 proton or neutron  68 (d)    = 10 protons or neutrons −27 1 galaxy 10 kg   

32. (a) 

33. The surface area of a sphere is found by A = 4 r 2 = 4 ( d 2 ) =  d 2 . 2

(

) = 7.478  10 m 2

(a)

2 =  4.879  106 m AMercury =  DMercury

(b)

2  DEarth   REarth   6.38  106 m   DEarth = =  = D  =R  = 6.84 2 6 AMercury  DMercury  Moon   Mercury   2.4395  10 m 

AEarth

34. The radius of the ball can be found from the circumference (represented by “c” in the equations below), and then the volume can be found from the radius. Finally, the mass is found from the volume of the baseball multiplied by the density ( = mass/volume) of a nucleon.

cball = 2 rball → rball =

c  3 ; Vball = 43  rball = 43   ball  2  2 

cball

   mnucleon   m  mnucleon V = Vball  nucleon =   ball   3 3 4 4 1  Vnucleon   3  rnucleon   3  ( 2 d nucleon ) 

mball = Vball  nucleon = Vball 

3  0.23m    c  mnucleon c   =   ball   = mnucleon  ball  = (10−27 kg )   3   (10−15 m )   2   43  ( 12 d nucleon )    d nucleon    3

4 3

= 3.9  1014 kg  4  1014 kg

Chapter 1

Introduction, Measurement, Estimating

35. (a) 3200 = 3.28  103  1  103 = 103 (b) 86.30  103 = 8.630  104  10  104 = 105 (c)

0.076 = 7.6  10−2  10  10−2 = 10−1

(d) 15.0  108 = 1.5  109  1 109 = 109 36. The textbook is slightly smaller than 25 cm deep and 5 cm wide. With books on both sides of a shelf, the shelf would then need to be about 50 cm deep. If the aisle is 1.5 m wide, then about 1/4 of the floor space is covered by shelving. The number of books on a single shelf level is then 1 4









= 1.3  10 books. With 8 shelves of books, the total number of ( 6500 m )  ( 0.25 1mbook )( 0.05 m )  2

books stored is as follows:  1.3  105 books  8 shelves = 1.04  106 books  1  106 books )  ( shelf level   37. The distance across the U.S. is about 3000 miles.

( 3000 mi )(1 km 0.621 mi )(1 hr 10 km )  500 hr

Of course, it would take more time on the clock for a runner to run across the U.S. The runner obviously could not run for 500 hours non-stop. If he or she could run for 5 hours a day, then it would take about 100 days to cross the country. 38. A commonly accepted measure is that a person should drink eight 8-oz. glasses of water each day. That is about 2 quarts, or 2 liters of water per day. Approximate the lifetime as 70 years.

( 70 y )( 365 d 1 y )( 2 L 1 d )  5  104 L 39. To determine an estimation for the number of cells in the human body we will make several assumptions. First, we assume that the cell is spherical, so we can find its volume using the formula for the volume of a sphere, V = 43  r 3. And then we will assume that the mass of an “average” human is 65 kg, which corresponds to a weight of about 140 lbs. We can use the given density to find the volume of our typical human, and then compare that to the volume of a cell. kg 1m3 1cell cells 65   = 1.24  1014 3 6 − human 1000 kg 43  5  10 m human

(

)

So for an order-of-magnitude estimate, we would say there are about 1 1014 cells in a human . 40. An NCAA-regulation football field is 360 feet long (including the end zones) and 160 feet wide, which is about 110 meters by 50 meters, or 5500 m2. We assume the mower has a cutting width of 0.5 meters and that a person mowing can walk at about 4.5 km/h, which is about 3 mi/h. Thus the distance to be walked is as follows: area 5500 m 2 d= = = 11000 m = 11 km width 0.5 m 1h  2.5 h to mow the field. At a speed of 4.5 km/h, it will take about 11km  4.5 km

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

41. There are about 3  108 people in the U.S. Assume that half of them have cars, that they drive an average of 12,000 miles per year, and that their cars get an average of 20 miles per gallon of gasoline.  1 automobile   12, 000 mi auto   1 gallon  11 3  108 people     20 mi   1  10 gal y 2 people 1 y    

(

)

42. In estimating the number of dentists, the assumptions and estimates needed are: • the population of the city • the number of patients that a dentist sees in a day • the number of days that a dentist works in a year • the number of times that each person visits the dentist each year We estimate that a dentist can see 10 patients a day, that a dentist works 225 days a year, and that each person visits the dentist twice per year. (a) For San Francisco, the population is approximately 800,000 (according to the U.S. Census Bureau). The number of dentists is found by the following calculation:  2 visits        1 yr 1 dentist   year 8  105 people      700 dentists   1 person   225 workdays   10 visits     workday  (b) For Marion, Indiana, the population is about 30,000. The number of dentists is found by a similar calculation to that in part (a), and would be about 30 dentists . There were about 40 dentists listed in a recent “yellow pages” phone book.

(

)

43. Make the estimate that each person has 1.5 loads of laundry per week, and that there are 300 million people in the United States. loads week 52 weeks 0.1kg kg kg   = 2.34  10  2  10 ( 300  10 people )  1.5 1person 1 yr 1load yr yr 6

1/ 3

 3V  . For a 1000-kg rock, 44. The volume of a sphere is given by V =  r , so the radius is r =    4  3

4 3

which weighs about 2200 lb, the volume is calculated from the density, and then the diameter from the volume.  2200 lb   1ft 3  3 V = (1T )    186 lb  = 11.8ft 1T    3 1/3  3V  = 2  3 (11.8ft )  = 2.82 ft  3ft d = 2r = 2     4  4    1/3

45. The covering of 49 km in two and a half days can be used to convert km into days. They have 270 km – 49 km = 221 km remaining. 2.5days 221km  = 11.3days  11days 49 km

Chapter 1

Introduction, Measurement, Estimating

46. The person walks 4 km h , 12 hours each day. The radius of the Earth is about 6380 km, and the distance around the Earth at the equator is the circumference, 2 REarth. We assume that the person can “walk on water,” so we ignore the existence of the oceans.  1 h   1 day  = 835days  800 days 2 ( 6380 km )     4 km   12 h  47. We approximate the jar as a cylinder with a uniform cross-sectional area. In counting the jelly beans in the top layer, we find about 25 jelly beans. Thus we estimate that one layer contains about 25 jelly beans. In counting vertically, we see that there are about 20 rows. Thus we estimate that there are 25  20 = 500 jellybeans in the jar. 48. The maximum number of buses would be needed during rush hour. We assume that a bus can hold 50 passengers. (a) The current population of Washington, D.C. is about 660,000 as of 2014 statistics. We estimate that 10% of them ride the bus during rush hour. 1bus 1driver 66, 000 passengers   = 1320 drivers  1000 drivers 50 passengers 1bus (b) For Marion, Indiana, the population is about 30,000. Because the town is so much smaller geographically, we estimate that only 5% of the current population rides the bus during rush hour. 1bus 1driver   30 drivers 1500 passengers  50 passengers 1bus 49. Consider the diagram shown (not to scale). The balloon is a distance h = 300 m above the surface of the Earth, and the tangent line from the balloon height to the surface of the Earth indicates the location of the horizon, a distance d away from the balloon. Use the Pythagorean theorem. ( r + h )2 = r 2 + d 2 → r 2 + 2rh + h 2 = r 2 + d 2

2 rh + h 2 = d 2 → d = 2 rh + h 2

(

)

d = 2 6.4  106 m ( 300 m ) + ( 300 m ) = 6.20  10 4 m  6  10 4 m (  40 mi ) 2

50. At $1000 per day, you would earn $30,000 in 30 days. With the other pay method, you would get $0.01 2t −1 on the tth day. On the first day, you get $0.01 21−1 = $0.01. On the second day, you get

( ) ( ) $0.01( 2 ) = $0.02. On the third day, you get $0.01( 2 ) = $0.04. On the 30th day, you get $0.01( 2 ) = $5.4  10 , which is over 5 million dollars. Get paid by the second method. 2 −1

30 −1

3 −1

51. Assume that the tires last for 5 years, and so there is a tread wearing of 0.2 cm/year. Assume the average tire has a radius of 40 cm, and a width of 10 cm. Thus the volume of rubber that is becoming pollution each year from one tire is the surface area of the tire times the thickness per year that is wearing. Also assume that there are 1.5  108 automobiles in the country – approximately one automobile for every two people. And there are 4 tires per automobile. The mass wear per year is given by the following calculation. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

 mass   surface area   thickness wear    year  =   ( density of rubber )( # of tires ) tire year       2 ( 0.4 m )( 0.1m )   0.002 m   1200 kg  8 8  1yr   1m3  ( 6.0  10 tires ) = 4  10 kg y  1 tire    

=

52. In the figure in the textbook, the distance d is perpendicular to the radius that is drawn approximately vertically. Thus there is a right triangle, with legs of d and R, and a hypotenuse of R + h. Since h R, h 2 2 Rh. d 2 + R 2 = ( R + h ) = R 2 + 2 Rh + h 2 → d 2 = 2 Rh + h 2 → d 2  2 Rh → 2

d2 2h

( 4400 m ) = 6.5  106 m 2 (1.5 m ) 2

A better measurement gives R = 6.38  106 m. 53. For you to see the Sun “disappear,” your line of sight to the top of the Sun must be tangent to the Earth’s surface. Initially, your eyes are a distance h1 = 20 cm above the sand and you see the first sunset at a (very small) angle of 1 relative to perfectly horizontal. Then you stand up, elevating your eyes to the height h2 = 130 cm. When you stand, your line of sight is tangent to the Earth’s surface at a slightly different angle  2 , so that is the direction to the second sunset. The change in angle  2 − 1 is the angle through which the Sun appears to move relative to the Earth during the time to be measured. The distance d is the distance from your eyes to a point on the horizon where you “see” the sun set. d1

To 2nd sunset

To 1st sunset

1

2

Earth center

Use the Pythagorean theorem for the following relationship: 2 d 2 + R 2 = ( R + h ) = R 2 + 2 Rh + h 2 → d 2 = 2 Rh + h 2 The distance h is much smaller than the distance R, so h 2 2 Rh which leads to d 2  2 Rh. We also have from the same triangle that d R = tan  , so d = R tan  . Combining these two relationships gives d 2  2 Rh = R 2 tan 2  , so R =

2h tan 2 

The change in angle,  =  2 − 1 , can be found from the two heights and the radius of the Earth. The elapsed time between the two sightings can then be found from the change in angle, because we know that a full revolution takes 24 hours. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 1

Introduction, Measurement, Estimating

2h tan  2

 2 = tan −1

2h1

→ 1 = tan −1 2h2

= tan −1

R t sec  = o 3600 s 360 24 h  1h

2 ( 0.20 m )

= tan −1

2 (1.5 m ) 6.38  106 m

6.38  10 m 6

(

= 3.93  10 −2

(

= 1.43  10 −2

)

) ;  =  −  = ( 2.50  10 ) o

−2

→

−2 o 3600 s   ( 2.50  10 )   3600 s        24 h  24 h  t = =  = 6.0 s  o  o  1h   360 1h   360    

54. Density units =

M  =  3  . SI units for density are kg m 3. volume units  L  mass units

55. (a) For the equation v = At 3 − Bt , the units of At 3 must be the same as the units of v. So the units 4 of A must be the same as the units of v t 3 , which would be L T . Also, the units of Bt must

be the same as the units of v. So the units of B must be the same as the units of v t , which would be L T 2 . (b) For A, the SI units would be m s4 , and for B the SI units would be m s2 . 56. (a) The quantity vt 2 has units of ( m s ) ( s 2 ) = m  s, which do not match with the units of meters for x. The quantity 2at has units ( m s 2 ) ( s ) = m s , which also do not match with the units of meters for x. Thus this equation cannot be correct .

(b) The quantity v 0 t has units of ( m s )( s ) = m, and 12 at 2 has units of ( m s 2 )( s 2 ) = m. Thus, since each term has units of meters, this equation can be correct .

(c) The quantity v 0 t has units of ( m s )( s ) = m, and 2at 2 has units of ( m s 2 )( s 2 ) = m. Thus, since each term has units of meters, this equation can be correct . 57. We consider the “Planck time.”

(a) tP =

(b) tP =

Gh c5

 L3   ML2  3 2 5 5  MT 2   T     =  L L T M  =  T  = T 2  = T  5    MT 3 L5  T 3      L  T 

→ units of t P =

( 6.67  10 m kg  s )( 6.63  10 kg  m s ) = 1.35  10 s  10 s = c ( 3.00  10 m s )

Gh 5

−11

−34

−43

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

58. The percentage accuracy is

Instructor Solutions Manual

 100% = 1  10−5% . The distance of 20,000,000 m needs to

2  10 m be distinguishable from 20,000,002 m, which means that 8 significant figures are needed in the distance measurements. 7

59. Divide the number of atoms by the Earth’s surface area.

# atoms m

6.02  1023 atoms 4 R

2 Earth

6.02  1023 atoms

(

4 6.38  10 m 6

)

= 1.18  109

atoms m2

This is more than one billion atoms per square meter. 60. Multiply the number of chips per wafer by the number of wafers that can be made from a cylinder. We assume the number of chips per wafer is more accurate than 1 significant figure.

 750 chips   1 wafer   250 mm  = 6.25  105  6.3  105 chips     wafer   0.300 mm   1 cylinder  cylinder  61. We assume that there are 40 hours of work per week and that you type 50 weeks out of the year. (Everybody deserves a vacation!) 1char 1min 1hour 1week 1year 1.0  1012 bytes      = 5.556  104 years 1byte 150 char 60 min 40 hour 50 weeks

(

)

 56, 000 years 62. The volume of water used by the people can be calculated as follows: 3

L day   365day   1000 cm   1km  3 −3 ( 4  10 people )  1200   1 y   1L   105cm  = 4.38  10 km y 4 people      3

The depth of water is found by dividing the volume by the area. 5 V 4.38  10−3 km3 yr  −5 km   10 cm  7.3 10 d= = =     = 7.3cm yr  7 cm yr 60 km 2 yr   1 km  A  63. We do a “units conversion” from bytes to minutes, using the given CD reading rate. 8 bits 1sec 1min 783.216  106 bytes    = 74.592 min  75 min 6 1byte 1.4  10 bits 60 sec

(

)

 10−10 m   1nm    −9  = 0.10 nm  1Å   10 m 

64. (a) 1.0Å = (1.0Å ) 

 10−10 m   1fm  5   10−15 m  = 1.0  10 fm 1Å   

(b) 1.0Å = (1.0Å ) 

 1Å  = 1.0  1010 Å  −10  10 m   9.46  1015 m   1Å  25 (d) 1.0 ly = (1.0 ly )   −10  = 9.5  10 Å  1 ly    10 m  (c) 1.0 m = (1.0 m ) 

Chapter 1

Introduction, Measurement, Estimating

65. Assume that the alveoli are spherical and that the volume of a typical human lung is about 2 liters, which is 0.002 m3. The diameter can be found from the volume of a sphere, 43  r 3. 4 3

 r =  ( d 2) = 3

4 3

d3 6

( 3 10 )  6 = 2 10 m 8

−3

 6 ( 2  10−3 ) 3  → d= m  8  3  10  

1/ 3

= 2  10 −4 m

66. A pencil has a diameter of about 0.7 cm. If held about 0.75 m from the eye, it can just block out the Moon. The ratio of pencil diameter to arm length is the same as the ratio of Moon diameter to Moon distance. From the diagram, we have the following ratios. Pencil

Moon

Pencil Distance

Moon Distance

Pencil diameter

Pencil distance

Moon diameter =

Moon diameter

→

Moon distance Pencil diameter

Pencil distance The actual value is 3480 km.

( Moon distance ) =

7  10−3 m

( 3.8  10 km )  3500 km 0.75 m 5

67. To calculate the mass of water, we need to find the volume of water and then convert the volume to 2 mass. The volume of water is the area of the city 35 km times the depth of the water (1.0 cm).

(

)

2 5    10−3 kg   1 metric ton  2  10 cm  5 5 35 km 1.0 cm ( ) )  1 km   (  1 cm3   103 kg  = 3.5  10  4  10 metric tons       

To find the number of gallons, convert the volume to gallons. 2 5    1 L   1 gal  = 9.26  107 gal  9  107 gal 2  10 cm  ( 35 km )   (1.0 cm )    3 3   1  10 cm   3.78 L   1 km   

68. We assume that the amount the ocean rises will be very small compared to the radius of the Earth. With this assumption, we calculate the volume of the water melted and set it equal to the volume of a thin “shell” of water, whose volume is 2/3 of the Earth’s surface area, times the water thickness. 2 Volume of Greenland ice = 23 4 REarth  water thickness, 3

2 m = 23 4 ( 6.38  106 m ) t → (1.7 10 km ) ( 2.5 km )  1000   1km  6

m (1.7 10 km ) ( 2.5 km )  1000   1km  = 12.46 m  12 m t= 2 2 4 ( 6.38  106 m ) 3 6

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

69. A cubit is about a half of a meter, by measuring several people’s forearms. Thus the dimensions of Noah’s ark would be 150 m long, 25 m wide, 15 m high . The volume of the ark is found by multiplying the three dimensions.

V = (150 m )( 25 m )(15 m ) = 5.625  104 m 3  6  104 m 3 70. The volume of the oil will be the area times the thickness. The area is  r =  ( d 2 ) . 2

V =  ( d 2) t → d = 2 2

t

 1m    100 cm  = 3  103 m −10

1000 cm 3 

 ( 2  10 m )

This is approximately 2 miles. 71. Utilize the fact that walking totally around the Earth along the meridian would trace out a circle whose full 360° would equal the circumference of the Earth. 3 1o    2 6.38  10 km   0.621mi  (1minute of arc )      = 1.15mi 360o  60 minutes of arc     1km 

(

)

72. (a) Note that sin15.0o = 0.259 and sin15.5o = 0.267, so  sin  = 0.267 − 0.259 = 0.008.

 0.5o     100 =    15.0o 100 = 3%     

 8  10−3    sin   100 =    0.259 100 = 3%  sin    

(b) Note that sin 75.0o = 0.966 and sin 75.5o = 0.968, so  sin  = 0.968 − 0.966 = 0.002.

 2  10−3    sin   100 =    0.966 100 = 0.2%  sin    

 0.5o     100 =    75.0o 100 = 0.7%     

A consequence of this result is that when you use a protractor, and you have a fixed uncertainty in the angle ( 0.5o in this case), you should measure the angles from a reference line that gives a large angle measurement rather than a small one. Note above that the angles around 75° had only a 0.2% error in sin , while the angles around 15° had a 3% error in sin . 73. Consider the diagram as shown in the text. Let l represent 85 strides that he walks upstream. Then from the diagram find the distance d across the river. d → tan 60o = l  0.8 m  o d = l tan 60o = ( 85 strides )   tan 60 = 117.8m  120 m  stride  But since the problem asks for an estimate, we might just keep 1 significant figure, and call the distance 100 meters. 74. Consider the body to be a cylinder, about 170 cm tall (  57 ) , and about 12 cm in cross-sectional radius (which corresponds to a 30-inch waist). The volume of a cylinder is given by the area of the cross section times the height.

V =  r 2 h =  ( 0.12 m ) (1.7 m ) = 7.69  10−2 m 3  8  10 −2 m 3 2

Chapter 1

Introduction, Measurement, Estimating

75. According to census data, there are about 800,000 people in San Francisco. We then approximate each household as having 4 people, that each household needs 1 plumber visit per year, that a plumber can make 4 visits per day, and that a plumber works 250 days each calendar year.  1 visit        1 household 1 plumber 1year    year 8  105 people       =  4 people   1 household   4 visits   250 work days     work day    

(

)

200 plumbers 76. We use values from Table 1–3. mhuman 102 kg = −17 = 1019 mDNA 10 kg molecule

77. The units for each term must be in liters, since the volume is in liters.

 units of 4.1 m =  L →  units of 4.1 =

L m

 units of 0.018 year  =  L →  units of 0.018 =

L year

 units of 2.7 = L 78. The density is the mass divided by volume. There will be only 1 significant figure in the answer. mass 6g density = = = 2.118 g cm 3  2 g cm 3 3 volume 2.8325cm 79. Multiply the volume of a spherical universe times the density of matter, adjusted to ordinary matter. The volume of a sphere is 43  r 3 .

(

m = V = 1  10−26 kg m 3

)  (13.0  10 m ) ( 0.04 ) 25

4 3

= 3.68  1051 kg  4  1051 kg

CHAPTER 2: Describing Motion: Kinematics in One Dimension Responses to Questions 1.

A car speedometer measures only speed. It does not give any information about the direction of motion, and so does not measure velocity.

If the velocity of an object is constant, the speed must also be constant. (A constant velocity means that the speed and direction are both constant.) If the speed of an object is constant, the velocity CAN vary. For example, a car traveling around a curve at constant speed has a varying velocity, since the direction of the velocity vector is changing.

If the velocity of an object is constant, the speed and the direction of travel must also be constant. If that is the case, then the average velocity is the same as the instantaneous velocity, because nothing about its velocity is changing. The ratio of displacement to elapsed time will not be changing, no matter the actual displacement or time interval used for the measurement.

There is no general relationship between the magnitude of speed and the magnitude of acceleration. For example, one object may have a large but constant speed. The acceleration of that object is then 0. Another object may have a small speed but be gaining speed, and therefore have a positive acceleration. So in this case the object with the greater speed has the lesser acceleration. Or consider two objects that are dropped from rest at different times. If we ignore air resistance, then the object dropped first will always have a greater speed than the object dropped second, but both will have the same acceleration of 9.80 m/s2.

The accelerations of the motorcycle and the bicycle are the same, assuming that both objects travel in a straight line. Acceleration is the change in velocity divided by the change in time. The magnitude of the change in velocity in each case is the same, 10 km/h, so over the same time interval the accelerations will be equal.

Yes, for example, a car that is traveling northward and slowing down has a northward velocity and a southward acceleration.

The velocity of an object can be negative when its acceleration is positive. If we define the positive direction to be to the right, then an object traveling to the left that is slowing down will have a negative velocity with a positive acceleration. The velocity of an object can also be positive when its acceleration is negative. If again we define the positive direction to be to the right, then an object traveling to the right that is slowing down will have a positive velocity and a negative acceleration.

If north is defined as the positive direction, then an object traveling to the south and increasing in speed has both a negative velocity and a negative acceleration. Or, if “up” is defined as the positive direction, then an object falling due to gravity has both a negative velocity and a negative acceleration (if air resistance is ignored).

If the two cars emerge side by side, then the one moving faster is passing the other one. Thus car A is passing car B. With the acceleration data given for the problem, the ensuing motion would be that car A would pull away from car B for a time, but eventually car B would catch up to and pass car A.

Chapter 2

Describing Motion: Kinematics in One Dimension

10. Yes. Remember that acceleration is a change in velocity per unit time, or a rate of change in velocity. So, velocity can be increasing while the rate of increase goes down. For example, suppose a car is traveling at 40 km/h and one second later is going 50 km/h. One second after that, the car’s speed is 55 km/h. The car’s speed was increasing the entire time, but its acceleration in the second time interval was lower than in the first time interval. Thus its acceleration was decreasing even as the speed was increasing. Another example would be an object falling WITH air resistance, but with a speed less than its terminal velocity. Let the downward direction be positive. As the object falls, it gains speed, and the air resistance increases. As the air resistance increases, the acceleration of the falling object decreases, and so it gains speed less quickly the longer it falls. 11. If there were no air resistance, the ball’s only acceleration during flight would be the acceleration due to gravity, so the ball would land in the catcher’s mitt with the same speed it had when it left the bat, 120 km/h. Since the acceleration is the same through the entire flight, the time for the ball’s speed to change from 120 km/h to 0 on the way up is the same as the time for its speed to change from 0 to 120 km/h on the way down. In both cases the ball has the same magnitude of displacement. 12. (a) If air resistance is negligible, the acceleration of a freely falling object stays the same as the object falls. The object’s speed increases, but since it increases at a constant rate, the acceleration is constant. (b) In the presence of air resistance, the acceleration decreases. Air resistance increases as the speed increases. If the object falls far enough, the acceleration will become zero and the velocity will become constant. That velocity is often called the terminal velocity. 13. Average speed is the displacement divided by the time. Since the distances from A to B and from B to C are equal, you spend more time traveling at 70 km/h than at 90 km/h, so your average speed should be less than 80 km/h. If the distance from A to B (or B to C) is x km, then the total distance traveled is 2x. The total time required to travel this distance is x/70 h plus x/90 h. Then 2x 2(90)(70) d v= = = = 78.75 km/h  79 km/h. t x 70 + x 90 90 + 70 14. Yes. Consider an object thrown straight up in the air. It has a nonzero acceleration due to gravity for its entire flight (neglecting air resistance). However, at the highest point it momentarily has a zero velocity. A car, at the moment it starts moving from rest, has zero velocity and nonzero acceleration. 15. Yes. Any time the velocity is constant, the acceleration is zero. For example, a car traveling at a constant 90 km/h in a straight line has nonzero velocity and zero acceleration. 16. A rock falling from a cliff has a constant acceleration if air resistance is neglected. An elevator moving from the second floor to the fifth floor making stops along the way does NOT have a constant acceleration. Its acceleration will change in magnitude and direction as the elevator starts and stops. The dish resting on a table has a constant (zero) acceleration. 17. The two conditions are that the motion needs to be near the surface of the Earth, and that there is no air resistance. An example where the second condition is not even a reasonable approximation is that of parachuting. The air resistance caused by the parachute results in the acceleration not being constant, and having values much different than 9.8 m s 2. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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18. The slope of the position versus time curve is the velocity. The object starts at the origin with a constant velocity (and therefore zero acceleration), which it maintains for about 20 s. For the next 10 s, the positive curvature of the graph indicates the object has a positive acceleration; its speed is increasing. From 30 s to 45 s, the graph has a negative curvature; the object uniformly slows to a stop, changes direction, and then moves backwards with increasing speed. During this time interval its acceleration is negative, since the object is slowing down while traveling in the positive direction and then speeding up while traveling in the negative direction. For the final 5 s shown, the object continues moving in the negative direction but slows down, which gives it a positive acceleration. During the 50 s shown, the object travels from the origin to a point 20 m away, and then back 10 m to end up 10 m from the starting position. 19. Initially, the object begins with a speed of 14 m/s, moving in the positive direction with a constant acceleration, until t = 45 s, when it has a velocity of about 37 m/s in the positive direction. The acceleration then decreases, reaching an instantaneous acceleration of 0 at t = 50 s, when the object has its maximum speed of about 38 m/s. The object then begins to slow down, but continues to move in the positive direction. The object stops moving at t = 90 s and stays at rest until t = 108 s. Then the object begins to move in the positive direction again, at first with a larger acceleration, and then a lesser acceleration. At the end of the recorded motion, the object is still moving to the right and gaining speed.

Responses to MisConceptual Questions 1.

(d) It is a common misconception that a positive acceleration always increases the speed, as in (b) and (c). However, when the velocity and acceleration are in opposite directions the speed will decrease.

(d) Since the velocity and acceleration are in opposite directions, the object will slow to a stop. However, since the acceleration remains constant, the object will only momentarily stop before moving toward the left.

(c) Since the distances are the same, a common error is to assume that the average speed will be halfway between the two speeds, or 40 km/h. However, it takes the car longer to travel the 4 km at 30 km/h than to travel the other 4 km at 50 km/h. Since more time is spent at 30 km/h the average speed will be closer to 30 km/h than to 50 km/h.

(d) From Eq. 2–12(c), for an object with constant acceleration starting from rest, the final speed is given by v = 2ad . Thus if aB = 4aA, then v B = 2v A.

(a) A common misconception is that the acceleration of an object in free-fall depends upon the motion of the object. If there is no air resistance, the accelerations is the same throughout the flight, even at the top. If the acceleration were 0 at the top, the object would not move since it also has a velocity of 0 at the top. The velocity is constantly changing throughout the flight.

(c) A common misconception is that the acceleration of an object in free-fall depends upon the motion of the object. If there is no air resistance, the accelerations for the two balls are the same magnitude and direction throughout both of their flights.

(a) Since the distance between the rocks increases with time, a common misconception is that the velocities are increasing at different rates. However, both rocks fall under the influence of gravity, so their velocities increase at the same rate.

Chapter 2

Describing Motion: Kinematics in One Dimension

(e) Both objects experience the same acceleration, so the rock that is dropped first always has a higher speed than the second rock. Since the first rock is always moving faster, it is always getting farther away from the second rock.

(b, c) Each of the given equations are based on Eq. 2–12(a–d). Answer (a) has the acceleration replaced properly with –g, but the initial velocity is downward and as such should be negative. Answer (d) is incorrect because the initial velocity has been inserted for the average velocity. Answers (b) and (c) have the correct signs for each variable and the known values are inserted properly.

10. (a) Increasing speed means that the slope must be getting steeper over time. In graphs (b) and (e) the slope remains constant, so these are cars moving at constant speed. In graph (c), as time increases x decreases. However the rate at which it decreases is also decreasing. This is a car slowing down. In graph (d) the car is moving away from the origin, but again it is slowing down. The only graph in which the slope is increasing with time is graph (a). 11. (c, e) The two lines have different slopes at point A, so they don’t have the same instantaneous velocity or speed. The object represented by the dashed line has traveled more distance, since it has traveled “forward and back” to get to the point represented by A. But they are both at the same position after the same amount of time, and so they have the same average velocity.

Solutions to Problems 1.

The distance of travel (displacement) can be found by rearranging Eq. 2–2 for the average velocity. Also note that the units of the velocity and the time are not the same, so the speed units will be converted.  1h  v = x t → x = vt = ( 85 km h )   ( 2.0s ) = 0.0472 km  47 m  3600 s 

The average speed is given by Eq. 2–1, using d to represent distance traveled.

v = d t = 235 km 2.85 h = 82.5 km h 3.

The average velocity is given by Eq. 2–2. x 8.5 cm − 5.2 cm 3.3cm v= = = = 0.61cm s 3.4 s − ( −2.0 s ) 5.4 s t The average speed cannot be calculated. To calculate the average speed, we would need to know the actual distance traveled, and it is not given. We only have the displacement.

(a) The speed of sound is intimated in the problem as 1 mile per 5 seconds. The speed is calculated as follows. distance  1mi   1610 m  speed = =  = 322 m s  300 m s  time  5s   1 mi  (b) The speed of 322 m s would imply the sound traveling a distance of 966 meters (which is approximately 1 km) in 3 seconds. So the rule could be approximated as 1 km every 3 s .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The time for the first part of the trip is calculated from the initial speed and the first distance, using d to represent distance. d d 210 km v1 = 1 → t1 = 1 = = 2.211h t1 v1 95 km h The time for the second part of the trip is now calculated. t2 = t total − t1 = 4.5 h − 2.211h = 2.289 h The distance for the second part of the trip is calculated from the average speed for that part of the trip and the time for that part of the trip. d v 2 = 2 → d 2 = v2 t2 = ( 65 km h )( 2.289 h ) = 148.8 km  150 km t2 (a) The total distance is then d total = d1 + d 2 = 210 km + 148.8 km = 358.8 km  360 km . (b) The average speed is NOT the average of the two speeds, and so is NOT 85 km/h. Use the definition of average speed, Eq. 2–1. d 358.8 km v = total = = 79.73 km h  80 km h t total 4.5 h Note that the result has 2 significant figures, and so could be expressed as 8.0  101 km h.

The distance traveled is 38 m + 12 ( 38 m ) = 57 m, and the displacement is 38 m − 12 ( 38 m ) = 19 m. The total time is 7.4 s + 1.8 s = 9.2 s. distance 57 m (a) Average speed = = = 6.2 m s time elapsed 9.2 s (b) Average velocity = vavg =

displacement time elapsed

19 m 9.2 s

= 2.1m s

The distance traveled is 3200 m ( 8 laps  400 m lap ). That distance probably has either 3 or 4 significant figures, since the track distance is probably known to at least the nearest meter for competition purposes. The displacement is 0 because the ending point is the same as the starting point. 3200 m  1 min  d (a) Average speed = =   = 3.68 m s  t 14.5 min  60 s  (b) Average velocity = v = x t = 0 m s

The average speed is the distance divided by the time. d  1  109 km   1yr  1 d  5 5 v = =    = 1.141 10 km h  1 10 km h  t  1yr   365.25 d  24 h 

Both objects will have the same time of travel. If the truck travels a distance xtruck , then the distance the car travels will be xcar = xtruck + 310 m. Use Eq. 2–2 for average speed, v = x t , solve for time, and equate the two times. x x xtruck x + 310 m t = truck = car = truck 75 km h 95 km h v truck vcar

Chapter 2

Describing Motion: Kinematics in One Dimension

Solving for xtruck gives xtruck = ( 310 m )

( 75 km h )

( 95 km h − 75 km h )

= 1163m.

 1163 m  60 min    = 0.9304 min = 55.82 s  56 s . v truck  75000 m h  1h  x  1163 m + 310 m  60 min  Also note that t = car =    = 0.9303 min = 56 s. vcar  95000 m h  1h  The time of travel is t =

xtruck

=

ALTERNATE SOLUTION: The speed of the car relative to the truck is 95 km h − 75 km h = 20 km h. In the reference frame of the truck, the car must travel 310 m to catch it. 0.31 km  3600 s  t =   = 56s 20 km h  1 h  10. The distance traveled is 560 km (280 km outgoing, 280 km return). The displacement ( x ) is 0 because the ending point is the same as the starting point. (a) To find the average speed, we need the distance traveled (560 km) and the total time x x 280 km elapsed. During the outgoing portion, v1 = 1 and so t1 = 1 = = 2.947 h. 95 km h v1 t1 During the return portion, v 2 =

x 2 t 2

, and so t2 =

x2 v2

280 km 55 km h

= 5.091h. Thus the

total time, including lunch, is ttotal = t1 + tlunch + t2 = 2.947 h + 1h + 5.091h = 9.038 h. v=

xtotal

560 km

= 61.96 km h  62 km h t total 9.038 h (b) To find the average velocity, use the displacement and the elapsed time. The displacement is 0 since the person returns to their starting point.

v = x t = 0

11. Since the locomotives have the same speed, they each travel half the distance, 4.75 km. Find the time of travel from the average speed. x x 4.75 km  60 min  v= → t = = = 0.03065 h   = 1.839 min  1.8 min v 155 km h t  1h  12. (a) The area between the concentric circles is equal to the length times the width of the spiral path.  R22 −  R12 = w l →

l =

 ( R22 − R12 ) w

 ( 0.058 m ) − ( 0.025 m )  2

−6

1.6  10 m

= 5.378  103 m  5400 m

 1s  1min    = 74.69 min  75 min  1.2 m  60 s 

(b) 5.378  103 m 

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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13. (a)

(b) The average velocity is the displacement divided by the elapsed time.

x ( 3.0 ) − x ( 0.0 )

 27 + 10 ( 3.0 ) − 2 ( 3.0 )3  m − ( 27 m )  = = −8.0 m s

3.0s − 0.0s 3.0s (c) The instantaneous velocity is given by the derivative of the position function.

(

)

= 10 − 6t 2 m s

10 − 6t 2 = 0 → t =

s = 1.3s dt 3 This can be seen from the graph as the “highest” point on the graph. 14. The average speed for each segment of the trip is given by v = For the first segment, t1 =

For the second segment, t2 =

1900 km 720 km h

d2 v2

d t

, so t =

d v

for each segment.

= 2.639 h.

2700 km 990 km h

= 2.727 h.

Thus the total time is ttot = t1 + t2 = 2.639 h + 2.727 h = 5.366 h  5.4 h . The average speed of the plane for the entire trip is d 1900 km + 2700 km v = tot = = 857.2 km h  860 km h . t tot 5.366 h Note that Eq. 2–12d does NOT apply in this situation. The numeric average of the two speeds is 855 km/h. 15. In order for the two ships to meet “in the middle,” they must both have the same average velocity (ignoring their different directions). We can find the time for each segment of the trip using d d , so t = for each segment. Both ships travel 600 km + 800 km + 600 km = 2000 km. v= t v d d d 600 km 800 km 600 km + + = 80 h First ship: t = 1 + 2 + 3 = v1 v 2 v3 20 km h 40 km h 20 km h Second ship: v =

d t

2000 km 80 h

= 25 km h

Chapter 2

Describing Motion: Kinematics in One Dimension

16. Slightly different answers may be obtained since the data comes from reading the graph. (a) The instantaneous velocity is given by the slope of the tangent line to the curve. At t = 10.0s, 3m − 0 the slope is approximately v (10 )  = 0.3 m s . 10.0 s − 0 (b) At t = 30.0s, the slope of the tangent line to the curve, and thus the instantaneous velocity, is 20 m − 8 m = 1.2 m s . approximately v ( 30 )  35 s − 25 s (c) The average velocity is given by v = (d) The average velocity is given by v = (e) The average velocity is given by v =

x ( 5.0 ) − x ( 0 ) 5.0 s − 0 s

1.5 m − 0

x ( 30.0 ) − x ( 25.0 ) 30.0 s − 25.0 s x ( 50.0 ) − x ( 40.0 ) 50.0 s − 40.0 s

5.0 s = =

= 0.30 m s .

16 m − 9 m 5.0 s

= 1.4 m s .

10 m − 19.5 m 10.0 s

= −0.95 m s .

17. Slightly different answers may be obtained since the data comes from reading the graph. (a) The indication of a constant velocity on a position-time graph is a constant slope, which occurs from t = 0 s to t  18s . (b) The greatest velocity will occur when the slope is the highest positive value, which occurs at

t  27 s . (c) The indication of a 0 velocity on a position-time graph is a slope of 0, which occurs at

t  38s . (d)

The object moves in both directions . When the slope is positive, from t = 0 s to t  38 s,

the object is moving in the positive direction. When the slope is negative, from t  38 s to t = 50 s, the object is moving in the negative direction. 18. The v vs. t graph is found by taking the slope of the x vs. t graph. Both graphs are shown here.

19. The average speed of sound is given by vsound = x t , and so the time for the sound to travel from the end of the lane back to the bowler is tsound =

x vsound

16.5 m 340 m s

= 4.85  10 −2 s. Thus the time for

the ball to travel from the bowler to the end of the lane is given by tball = ttotal − tsound = © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

2.75s − 4.85  10 −2 s = 2.7015s. And so the speed of the ball is as follows.

v ball =

x tball

16.5 m 2.7015 s

= 6.1077 m s  6.11m s

20. For the car to pass the train, the car must travel the length of the train AND the distance the train travels. The distance the car travels can thus be written as either d car = vcar t = ( 95 km h ) t or

d car = l train + v train t = 1.50 km + ( 75 km h ) t. To solve for the time, equate these two expressions for

the distance the car travels.

( 95 km h ) t = 1.50 km + ( 75 km h ) t → t =

1.50 km 20 km h

 60 min   = 4.5 min  1h 

= 0.075 h 

Note that this is the same as calculating from the reference frame of the train, in which the car is moving at 20 km/h and must travel the length of the train. The distance the car travels during this time is d = ( 95 km h )( 0.075 h ) = 7.125 km  7.1km . If the train is traveling in the opposite direction from the car, then the car must travel the length of the train MINUS the distance the train travels. Thus the distance the car travels can be written as either d car = ( 95 km h ) t or d car = 1.50 km − ( 75 km h ) t. To solve for the time, equate these two expressions for the distance the car travels. ( 95 km h ) t = 1.50 km − ( 75 km h ) t →

1.50 km 170 km h

 3600s   = 31.8s  32 s  1h 

= 8.824  10−3 h 

The distance the car travels during this time is d = ( 95 km h ) ( 8.824  10−3 h ) = 0.84 km . 21. (a) The average acceleration of the sprinter is a =

v t

9.00 m s − 0.00 m s 1.48 s

= 6.08 m s 2 .

(b) We change the units for the acceleration. 2

 1km   3600s  a = ( 6.08 m s )  = 7.88  104 km h 2     1000 m   1h  2

22. The initial velocity is v0 = 15 km h, the final velocity is v = 65 km h, and the displacement is x − x0 = 4.0 km = 4000 m. Find the average acceleration from Eq. 2–12c.

v 2 = v02 + 2a ( x − x0 ) → 2

v 2 − v02

2 ( x − x0 )

 1m s  ( 65 km h ) 2 − (15 km h ) 2     3.6 km h    = 3.9  10−2 m s 2 = 2 ( 4000 m )

23. The initial velocity of the car is the average velocity of the car before it accelerates. x 120 m v= = = 24 m s = v0 5.0 s t The final velocity is v = 0, and the time to stop is 3.7 s. Use Eq. 2–12a to find the acceleration. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

v = v0 + at → a =

v − v0 t

0 − 24 m s 3.7 s

= −6.486 m s 2

(

)  9.801 mg s  = 0.66 g’s .

Thus the magnitude of the acceleration is 6.5m s2 , or 6.486 m s 2 





24. We assume that the speedometer can read to the nearest km/h, and so the value of 120 km/h has three significant digits. The time can be found from the average acceleration, a = v t.

t =

v a

120 km h − 65 km h 1.8 m s 2

 1m s    3.6 km h  = 8.488s  8.5s

( 55 km h ) 

1.8 m s 2

25. (a) The average velocity is the displacement divided by the elapsed time. x 385 m − 25 m v= = = 21.2 m s t 20.0 s − 3.00 s (b) a =

v t

45.0 m s − 11.0 m s 20.0 s − 3.00 s

= 2.00 m s 2

26. The acceleration is the second derivative of the position function. dx d 2x dv 2 x = 4.8t + 7.3t → v = = 4.8 + 14.6t → a = 2 = = 14.6 m s 2 dt dt dt 27. (a) Since the units of A times the units of t must equal meters, the units of A must be m s . Since the units of B times the units of t 2 must equal meters, the units of B must be

m s2 . (b) The acceleration is the second derivative of the position function. dx d 2x dv x = At + Bt 2 → v = = A + 2 Bt → a = 2 = = 2B m s2 dt dt dt (c)

v = A + 2 Bt → v ( 5 ) = ( A + 12 B ) m s

a = 2B m s2

(d) The velocity is the derivative of the position function. dx x = At + Bt −3 → v = = A − 3Bt −4 dt

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

28. To estimate the velocity, find the average velocity over each time interval, and assume that the car had that velocity at the midpoint of the time interval. To estimate the acceleration, find the average acceleration over each time interval, and assume that the car had that acceleration at the midpoint of the time interval. A sample of each calculation is shown. From 2.00 s to 2.50 s, for average velocity: 2.50 s + 2.00 s tmid = = 2.25 s 2 x 13.79 m − 8.55 m 5.24 m vavg = = = = 10.48 m s 2.50 s − 2.00 s 0.50 s t From 2.25 s to 2.75 s, for average acceleration: 2.25 s + 2.75 s = 2.50 s tmid = 2 v 13.14 m s − 10.48 m s 2.66 m s = = aavg = t 2.75 s − 2.25 s 0.50 s

= 5.32 m s 2

v (m/s)

a (m/s )

t (s)

29. We call the location where the cars are initially next to each other x0 = 0. The position of the first

car can be written as x = x0 + v0t + 12 at 2 = 25.0t − 12 ( 2.0 ) t 2 . The position of the second car can be written as x = x0 + v0t + 12 at 2 = 12 ( 2.0 ) t 2 . We equate the two position equations and solve for the time.

25.0t − 12 ( 2.0) t 2 = 12 ( 2.0) t 2 → 25.0t = 2.0t 2 → t = 0 or 12.5 s

The value of 0 is the time when the cars were initially next to each other, so the “next” time they are next to each other is t = 12.5s  13s .

Chapter 2

Describing Motion: Kinematics in One Dimension

30. Slightly different answers may be obtained since the data comes from reading the graph. (a) The greatest velocity is found at the highest point on the graph, which is at t  48 s . (b) The indication of a constant velocity on a velocity–time graph is a slope of 0, which occurs from t = 90 s to t  108 s . (c) The indication of a constant acceleration on a velocity–time graph is a constant slope, which occurs from t = 0 s to t  42 s , again from t  65 s to t  83 s , and again from t = 90 s to t  108 s . (d) The magnitude of the acceleration is greatest when the magnitude of the slope is greatest, which occurs from t  65 s to t  83 s .

31. Slightly different answers may be obtained since the data comes from reading the graph. We assume that the short, nearly horizontal portions of the graph are the times that shifting is occurring, and those times are not counted as being “in” a certain gear. v 2 24 m s − 14 m s = = 2.5 m s 2 . (a) The average acceleration in 2nd gear is given by a2 = 8 s−4 s t2 (b) The average acceleration in 4th gear is given by a4 =

v 4 t4

44 m s − 37 m s 27 s − 16 s

= 0.6 m s 2 .

32. (a) The train’s constant speed is v train = 5.0 m s, and the location of the empty box car as a

function of time is given by xtrain = vtrain t = ( 5.0 m s ) t. The fugitive has v0 = 0 m s and a = 1.4 m s 2 until his final speed is 6.0 m s . The elapsed time during the acceleration is

tacc =

v − v0

6.0 m s

= 4.286 s. Let the origin be the location of the fugitive when he starts a 1.4 m s 2 to run. The first possibility to consider is, “Can the fugitive catch the empty box car before he reaches his maximum speed?” During the fugitive’s acceleration, his location as a function of time is given by Eq. 2–12b, xfugitive = x0 + v0t + 12 at 2 = 0 + 0 + 12 1.4 m s 2 t 2 . For him to catch

(

)

( 5.0 m s ) t = 12 (1.4 m s 2 ) t 2 . The solutions of this

the train, we must have xtrain = xfugitive →

are t = 0s, 7.1s. Thus the fugitive cannot catch the car during his 4.286 s of acceleration. Now the equation of motion of the fugitive changes. After the 4.286 s of acceleration, he runs with a constant speed of 6.0 m s . Thus his location is now given (for times t  4.286s ) by the following. 2 xfugitive = 12 1.4 m s 2 ( 4.286s ) + ( 6.0 m s )( t − 4.286s ) = ( 6.0 m s ) t − 12.86 m

(

)

So now, for the fugitive to catch the train, we again set the locations equal. xtrain = xfugitive → ( 5.0 m s ) t = ( 6.0 m s ) t − 12.86 m → t = 12.86 s  13s (b) The distance traveled to reach the box car is given by the following. xfugitive ( t = 12.86s ) = ( 6.0 m s )(12.86s ) − 12.86 m = 64 m 33. The acceleration can be found from Eq. 2–12c.

v 2 = v02 + 2a ( x − x0 ) → a =

v 2 − v02

2 ( x − x0 )

0 − ( 26 m s ) 2 ( 88 m )

= −3.8 m s 2

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

34. The acceleration can be found from Eq. 2–12a. v − v0 0 − 28 m s v = v0 + at → a = = = −4.4 m s 2 t 6.3s 35. By definition, the acceleration is a =

v − v0

22 m s − 13 m s

= 1.38 m s 2  1.4 m s 2 . t 6.5 s The distance of travel can be found from Eqs. 2–8 and 2–9. v +v 13 m s + 22 m s x − x0 = vt = 0 t = ( 6.5s ) = 114 m  110 m 2 2 It can also be found from Eq. 2–12b.

(

)

x − x0 = v0t + 12 at 2 = (13 m s )( 6.5s ) + 12 1.38 m s 2 ( 6.5s ) = 113.7 m  110 m 2

It can also be found from Eq. 2–12c.

v = v + 2a ( x − x0 ) → x − x0 = 2

2 0

v 2 − v02 2a

( 22 m s ) − (13 m s ) 2

(

2 1.38 m s 2

)

= 114.1m  110 m

There are slight differences in the answers because of rounding the acceleration. 36. The sprinter starts from rest. The average acceleration is found from Eq. 2–12c.

(11.5 m s ) − 0 = = 3.674 m s 2  3.67 m s 2 v = v + 2a ( x − x0 ) → a = 2 ( x − x0 ) 2 (18.0 m ) 2

2 0

v 2 − v02

Her elapsed time is found by solving Eq. 2–12a for time. v − v0 11.5 m s − 0 v = v0 + at → t = = = 3.13s a 3.674 m s 2 37. The words “slowing down uniformly” imply that the car has a constant acceleration. The distance of travel is found from combining Eqs. 2–8 and 2–9. v +v  28.0 m s + 0 m s  2 x − x0 = 0 t =  ( 8.60 sec ) = 1.20  10 m 2 2   38. The final velocity of the truck is zero. The initial velocity is found from Eq. 2–12c with v = 0 and solving for v 0. Note that the acceleration is negative.

v 2 = v02 + 2a ( x − x0 ) → v0 = v 2 − 2a ( x − x0 ) = 0 − 2 ( −6.00 m s 2 ) ( 45 m ) = 23m s 39. For the baseball, v0 = 0, x − x0 = 3.5 m, and the final speed of the baseball (during the throwing motion ) is v = 43 m s. The acceleration is found from Eq. 2–12c.

v 2 = v02 + 2a ( x − x0 ) → a =

( 43 m s ) − 0 = 264 m s 2  260 m s 2 2 ( x − x0 ) 2 ( 3.5 m ) v 2 − v02

Chapter 2

Describing Motion: Kinematics in One Dimension

40. The final velocity of the driver is zero. The acceleration is found from Eq. 2–12c with v = 0 and solving for a. 2

v 2 − v02

2 ( x − x0 )

  1000 m  1h   0 − ( 95 km h )    1km  3600 s     = = −435.2 m s 2 2 ( 0.80 m )

Converting to “g’s”: a =

435.2 m s 2

( 9.80 m s ) g 2

= 44 g’s .

41. (a) The final velocity of the car is 0. The distance is found from Eq. 2–12c with an acceleration of a = −0.50 m s 2 and an initial velocity of 85 km h . 2

  1000 m  1h   0 − ( 85 km h )    2 2 1km  3600 s   v − v0   x − x0 = = = 557 m  560 m 2

(

2 −0.50 m s

)

(b) The time to stop is found from Eq. 2–12a.



v − v0 a

 1000 m  1h      1km  3600 s   = 47.22 s  47 s

0 − ( 85 km h )  =



( −0.50 m s ) 2

(c) Take x0 = x ( t = 0 ) = 0 m. Use Eq. 2–12b, with a = −0.50 m s 2 and an initial velocity of 85 km h . The first second is from t = 0 s to t = 1s, and the fifth second is from t = 4 s to t = 5s.

 1m s  2 (1s ) + 12 ( −0.50 m s 2 ) (1s ) = 23.36 m →   3.6 km h 

x ( 0 ) = 0 ; x (1) = 0 + ( 85 km h )  x (1) − x ( 0 ) = 23 m

 1m s  2 ( 4 s ) + 12 ( −0.50 m s 2 ) ( 4 s ) = 90.44 m   3.6 km h 

x ( 4 ) = 0 + ( 85 km h ) 

 1m s  2 ( 5s ) + 12 ( −0.50 m s 2 ) ( 5s ) = 111.81m   3.6 km h 

x ( 5 ) = 0 + ( 85 km h ) 

x ( 5 ) − x ( 4 ) = 111.81m − 90.44 m = 21.37m  21m

42. The origin is the location of the car at the beginning of the reaction time. The initial speed of the car

 1000 m   1h    = 26.39 m s . The location where the brakes are applied is  1km   3600s 

is ( 95 km h ) 

found from the equation for motion at constant velocity. x0 = v0 t R = ( 26.39 m s )( 0.40 s ) = 10.56 m This is now the starting location for the application of the brakes. In each case, the final speed is 0.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(a) Solve Eq. 2–12c for the final location. v 2 = v02 + 2a ( x − x0 ) → x = x0 +

v 2 − v02 2a

= 10.56 m +

0 − ( 26.39 m s )

(

2 −2.5 m s 2

= 149.84 m  150 m

)

(b) Solve Eq. 2–12c for the final location with the second acceleration.

x = x0 +

v 2 − v02 2a

= 10.56 m +

0 − ( 26.39 m s )

(

2 −5.5 m s 2

)

= 74 m

43. Calculate the distance that the car travels during the reaction time and the deceleration. x1 = v0 t = (18.0 m s )( 0.380 s ) = 6.84 m

v = v + 2ax2 → x2 = 2

2 0

v 2 − v02 2a

0 − (18.0 m s )

(

2 −3.65 m s 2

)

= 44.38 m

x = 6.84 m + 44.4 m = 51.22 m ( only 3 significant figures ) Since she is only 24.0 m from the intersection, she will NOT be able to stop in time. She will be 27.2 m past the intersection. 44. The average velocity is defined by Eq. 2–2, v =

x

x − x0

. Compare this expression to t t Eq. 2–12d, v = 12 ( v + v0 ) . A relation for the velocity is found by integrating the expression for the acceleration, since the acceleration is the derivative of the velocity. Assume the velocity is v0 at time t = 0. v t dv a = A + Bt = → d v = ( A + Bt ) dt →  d v =  ( A + Bt ) dt → v = v0 + At + 12 Bt 2 dt v 0 0

Find an expression for the position by integrating the velocity, assuming that x = x0 at time t = 0.

v = v0 + At + 12 Bt 2 = x

(

)

→ dx = v0 + At + 12 Bt 2 dt →

(

)

x − x0

to 12 ( v + v0 ) .

2 2 3  dx =  v0 + At + 12 Bt dt → x − x0 = v0t + 12 At + 16 Bt

Compare v= 1 2

t x − x0 t

( v + v0 ) =

v0 t + 12 At 2 + 16 Bt 3

t v0 + v0 + At + 12 Bt 2 2

= v0 + 12 At + 16 Bt 2 = v0 + 12 At + 14 Bt 2

They are different, so v  12 ( v + v0 ) . 45. Use the information for the first 180 m to find the acceleration, and the information for the full motion to find the final velocity. For the first segment, the train has v0 = 0 m s, v1 = 18 m s, and a displacement of x1 − x0 = 180 m. Find the acceleration from Eq. 2–12c. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

v12 = v02 + 2a ( x1 − x0 ) → a =

(18 m s ) − 0 = 0.90 m s 2 2 ( x1 − x0 ) 2 (180 m ) 2

v12 − v02

Find the speed of the train after it has traveled the total distance (total displacement of x2 − x0 = 180 m + 85 m = 265 m ) using Eq. 2–12c.

v22 = v02 + 2a ( x2 − x0 ) → v2 = v02 + 2a ( x2 − x0 ) = 2 ( 0.90 m s 2 ) ( 265 m ) = 22 m s 46. Calculate the acceleration from the velocity–time data using Eq. 2–12a, and then use Eq. 2–12b to calculate the displacement at t = 2.0s and t = 6.0s. The initial velocity is v0 = 85 m s. a=

v − v0 t

162 m s − 85 m s 10.0 s

= 7.7 m s 2

x = x0 + v0t + 12 at 2 →

(

x ( 7.0 s ) − x ( 2.0 s ) =  x0 + v0 ( 7.0 s ) + 12 a ( 7.0 s )

) − ( x + v ( 2.0 s ) + a ( 2.0 s ) ) 0

1 2

(

= v0 ( 7.0 s − 2.0 s ) + 12 a ( 7.0 s ) − ( 2.0 s )  = ( 85 m s )( 5.0 s ) + 12 7.7 m s 2 2

)( 45s ) 2

= 598.25 m  6.0  102 m 47. During the final part of the race, the runner must have a displacement of 1200 m in a time of 180 s (3.0 min). Assume that the starting speed for the final part is the same as the average speed thus far. 8800 m x v= = = 5.432 m s = v0 t ( 27  60 ) s The runner will accomplish this by accelerating from speed v0 to speed v for t seconds, covering a

distance d1 , and then running at a constant speed of v for (180 − t ) seconds, covering a distance d 2 . We have these relationships from Eq. 2–12a and Eq. 2–12b. v = v0 + at d1 = v0t + 12 at 2 d 2 = v (180 − t ) = ( v0 + at )(180 − t )

1200 m = d1 + d 2 = v0t + 12 at 2 + ( v0 + at )(180 − t ) → 1200 m = 180v0 + 180at − 12 at 2 →

(

)

(

)

1200 m = (180 s )( 5.432 m s ) + (180 s ) 0.20 m s 2 t − 12 0.20 m s 2 t 2 → 0.10t − 36t + 222.24 = 0 → t = 2

36  362 − 4 ( 0.10 )( 222.24 ) 2 ( 0.10 )

= 353.7 s, 6.28s

Since we must have t  180 s, the solution is t = 6.3s . 48. For the runners to cross the finish line side-by-side means they must both reach the finish line in the same amount of time from their current positions. Take Mary’s current location as the origin. Use Eq. 2–12b. For Sally:

22 = 5.0 + 5.0t + 12 ( −0.40 ) t 2 → t 2 − 25t + 85 = 0 → t=

25  252 − 4 ( 85 ) 2

= 4.059 s, 20.94 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The first time is the time she first crosses the finish line, and so it is the time to be used for the problem. Now find Mary’s acceleration so that she crosses the finish line in that same amount of time. 22 − 4t 22 − 4 ( 4.059 ) 22 = 0 + 4t + 12 at 2 → a = 1 2 = = 0.70 m s 2 For Mary: 2 1 t 4.059 ) 2 2( 49. Define the origin to be the location where the speeder passes the police car. Start a timer at the instant that the speeder passes the police car, and find another time that both cars have the same displacement from the origin. For the speeder, traveling with a constant speed, the displacement is given by the following.  1m s  xs = vs t = (135 km h )   ( t ) = ( 37.5 t ) m  3.6 km h  For the police car, the displacement is given by two components. The first part is the distance traveled at the initially constant speed during the 1 second of reaction time.  1m s  xp1 = v0p (1.00 s ) = ( 95 km h )   (1.00 s ) = 26.39 m  3.6 km h  The second part of the police car’s displacement is that which occurs during the accelerated motion, which lasts for ( t − 1.00 ) s. So this second part of the police car displacement, using Eq. 2–12b, is given as follows. Note that this part of the displacement is NOT VALID for t < 1.00 s. 2 2 xp2 = v0p ( t − 1.00 ) + 12 ap ( t − 1.00 ) = ( 26.39 m s )( t − 1.00 ) + 12 ( 2.60 m s 2 ) ( t − 1.00 )  m   So the total police car displacement is:

(

xp = xp1 + xp 2 = 26.39 + 26.39 ( t − 1.00 ) + 1.30 ( t − 1.00 )

Now set the two displacements equal, and solve for the time. 26.39 + 26.39 ( t − 1.00 ) + 1.30 ( t − 1.00 ) = 37.5 t 2

10.55 

(10.55 ) − 4.00

→ t 2 − 10.55t + 1.00 = 0

= 9.57  10 −2 s, 10.5s

The first answer from the quadratic formula is a “fake” answer, because the equation for the police car’s displacement while accelerating, which was used to find that solution, was not valid for times t < 1.00 s. Thus we may ignore that answer. The answer of 10.5 s is the time for the police car to overtake the speeder. As a check on the answer, the speeder travels xs = ( 37.5 m s ) (10.5 s ) = 394 m, and the police car travels xp =  26.39 + 26.39 ( 9.5) + 1.30 ( 9.5 )  m = 394 m.





50. Define the origin to be the location where the speeder passes the police car. Start a timer at the instant that the speeder passes the police car. Both cars have the same displacement 8.00 s after the initial passing by the speeder. But the police car only is accelerating for 7.00 s. For the speeder, traveling with a constant speed, the displacement is given by xs = vs t = ( 8.00vs ) m. For the police car, the displacement is given by two components. The first part is the distance traveled at the initially constant speed during the 1.00 s of reaction time.  1m s  xp1 = v0p (1.00 s ) = ( 95 km h )   (1.00 s ) = 26.39 m  3.6 km h  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

The second part of the police car displacement is that during the accelerated motion, which lasts for 7.00 s. So this second part of the police car displacement, using Eq. 2–12b, is given by the following. 2 2 xp 2 = v0p ( 7.00 s ) + 12 ap ( 7.00 s ) = ( 26.39 m s )( 7.00 s ) + 12 ( 2.60 m s 2 ) ( 7.00 s ) = 248.43 m Thus the total police car displacement is xp = xp1 + xp2 = ( 26.39 + 248.43) m = 274.82 m. Now set the two displacements equal, and solve for the speeder’s velocity.  3.6 km h  ( 8.00vs ) m = 274.82 m → vs = 34.35 m s = ( 34.35 m s )   = 124 km h  1m s  51. From the given information, we can break the race up into three segments: Segment # 1: Time of 0.15 seconds, no displacement. Segment # 2: Starting speed of 0, final speed v, constant acceleration a, displacement 30.0 m, elapsed time t2 . Thus v = at 2 and 30 m = 12 at 22 . Segment # 3: Constant velocity v, displacement of 370 m, elapsed time t3 . Thus 370 m = v t3 . We also know that t2 + t3 = 54.85s. So we have 4 relationships defining the 4 unknowns of

v , a, t2 , and t3. We solve that system of 4 equations. v = at 2 ; 30 = 12 at 22 ; 370 = vt3 ; t2 + t3 = 54.85

Substitute the 1st equation into the 3rd.

30 = 12 at 22 ; 370 = at 2t3 ; t2 + t3 = 54.85

Divide the 2nd equation by the 1st.

at t 2t = 1 2 32 = 3 ; t2 + t3 = 54.85 30 at 2 t2 2

Substitute the 1st equation into the 2nd.

370 370 30

2t3 t2

→ t2 =

60 370

t3 ;

60 370

t3 + t3 = 54.85

54.85

= 47.20 s → t2 = 54.85 − t3 = 7.65s 1.162 v 7.839 m s 370 m 370 m v= = = 7.839 m s ; a = = = 1.025 m s 2 t3 t2 47.20 s 7.65s 370

t3 + t3 = 1.162t3 = 54.85 → t3 =

Solve for t3 .

The requested results are v = 7.8 m s ; a = 1.0 m s 2 . 52. Choose downward to be the positive direction, and take y0 = 0 at the top of the cliff. The initial velocity is v0 = 0, and the acceleration is a = 9.80 m s 2. The displacement is found from Eq. 2–12b, with x replaced by y. 2 y = y0 + v0t + 12 at 2 → y − 0 = 0 + 12 ( 9.80 m s 2 ) ( 3.25s ) → y = 51.8 m 53. Choose downward to be the positive direction, and take y0 = 0 to be at the top of the Empire State Building. The initial velocity is v0 = 0, and the acceleration is a = 9.80 m s . (a) The elapsed time can be found from Eq. 2–12b, with x replaced by y. 2

y − y0 = v0t + 12 at 2

→

2y a

2 ( 380 m ) 9.80 m s 2

= 8.806 s  8.8 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The final velocity can be found from Eq. 2–12a. v = v0 + at = 0 + ( 9.80 m s 2 ) ( 8.806 s ) = 86 m s 54. Choose downward to be the positive direction. The initial velocity is v0 = 0, the final velocity is

 1m s  2 v = ( 55 km h )  = 15.28 m s , and the acceleration is a = 9.80 m s . Solve Eq. 2–12a   3.6 km h  for the time.

v = v0 + at → t =

v − v0 a

15.28 m s − 0 9.80 m s 2

= 1.6 s

55. Choose upward to be the positive direction, and take y0 = 0 to be the height from which the ball was thrown. The acceleration is a = −9.80 m s 2. The displacement upon catching the ball is 0, assuming it was caught at the same height from which it was thrown. The starting speed can be found from Eq. 2–12b, with x replaced by y. y = y0 + v0t + 12 at 2 = 0 →

v0 =

y − y0 − 12 at 2

(

)

= − 12 at = − 12 −9.80 m s 2 ( 2.6 s ) = 12.74 m s  13 m s

t The height can be calculated from Eq. 2–12c, with a final velocity of v = 0 at the top of the path.

v = v + 2 a ( y − y0 ) → y = y 0 + 2

2 0

v 2 − v02 2a

= 0+

0 − (12.74 m s )

(

2 −9.80 m s 2

)

= 8.3 m

56. Choose upward to be the positive direction, and take y0 = 0 to be at the height where the ball was hit. For the upward path, v0 = 22 m s, v = 0 at the top of the path, and a = −9.80 m s 2. (a) The displacement can be found from Eq. 2–12c, with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → y = y0 +

v 2 − v02 2a

=0+

0 − ( 22 m s )

(

2 −9.80 m s 2

)

= 25 m

(b) The time of flight can be found from Eq. 2–12b, with x replaced by y, using a displacement of 0 for the displacement of the ball returning to the height from which it was hit. y = y0 + v0t + 12 at 2 = 0 → t ( v0 + 12 at ) = 0 →

t = 0, t =

2v0

2 ( 22 m s )

= 4.5s − a −9.80 m s 2 The result of t = 0 s is the time for the original displacement of zero (when the ball was hit), and the result of t = 4.5 s is the time to return to the original displacement. Thus the answer is t = 4.5 seconds. (c) This is an estimate primarily because the effects of the air have been ignored. There is a nontrivial amount of air effect on a baseball as it moves through the air – that’s why pitches like the “curve ball” work, for example. So ignoring the effects of air make this an estimate. Another effect is that the problem says “almost” straight up, but the problem was solved as if the initial velocity was perfectly upwards. Finally, we assume that the ball was caught at the same height as which it was hit. That was not stated in the problem either, so that is an estimate.

Chapter 2

Describing Motion: Kinematics in One Dimension

57. Choose downward to be the positive direction, and take y0 = 0 to be at the maximum height of the kangaroo. Consider just the downward motion of the kangaroo. Then the displacement is y = 1.45 m, the acceleration is a = 9.80 m s 2 , and the initial velocity is v0 = 0 m s . Use Eq. 2–12b to calculate the time for the kangaroo to fall back to the ground. The total time is then twice the falling time. y = y0 + v0t + 12 at 2 = 0 ttotal = 2

2y a

→

y = 12 at 2 → tfall =

2 (1.45 m )

( 9.80 m s ) 2

2y a

→

= 1.09 s

58. Choose upward to be the positive direction, and take y0 = 0 to be at the floor level, where the jump starts. For the upward path, y = 1.2 m, v = 0 at the top of the path, and a = −9.80 m s 2. (a) The initial speed can be found from Eq. 2–12c, with x replaced by y. v 2 = v02 + 2a ( y − y0 ) →

v0 = v 2 − 2a ( y − y0 ) = −2ay = −2 ( −9.80 m s 2 ) (1.2 m ) = 4.8497 m s  4.8 m s (b) The time of flight can be found from Eq. 2–12b, with x replaced by y, using a displacement of 0 for the displacement of the jumper returning to the original height. y = y0 + v0t + 12 at 2 = 0 → t ( v0 + 12 at ) = 0 → t = 0, t =

2v0

2 ( 4.897 m s )

= 0.99 s −a 9.80 m s 2 The result of t = 0 s is the time for the original displacement of zero (when the jumper started to jump), and the result of t = 0.99 s is the time to return to the original displacement. Thus the answer is t = 0.99 seconds.

59. Choose upward to be the positive direction, and y0 = 0 to be the height from which the stone is thrown. We have v0 = 18.0 m s , a = −9.80 m s 2 , and y − y0 = 11.0 m. (a) The velocity can be found from Eq, 2–11c, with x replaced by y. v 2 = v02 + 2a ( y − y0 ) = 0 →

v =  v02 + 2ay = 

(18.0 m s ) + 2 ( −9.80 m s 2 ) (11.0 m ) = 10.4 m s 2

Thus the speed is v = 10.4 m s . (b) The time to reach that height can be found from Eq. 2–12b. 2 (18.0 m s ) 2 ( −11.0 m ) y = y0 + v0 t + 12 at 2 → t 2 + t+ =0 2 −9.80 m s −9.80 m s 2

t − 3.6735t + 2.245 = 0 → t =

3.6735 

→

( 3.6735 ) − 4 ( 2.2449 ) 2

= t = 2.90 s, 0.774 s 2 (c) There are two times at which the object reaches that height – once on the way up ( t = 0.774 s ) , 2

and once on the way down ( t = 2.90 s ).

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

60. Choose downward to be the positive direction, and take y0 = 0 to be the height from which the object is released. The initial velocity is v0 = 0, and the acceleration is a = g . Then we can calculate the position as a function of time from Eq. 2–12b, with x replaced by y, as y ( t ) = 12 gt 2 . At the end of each second, the position would be as follows. 2 2 y ( 0 ) = 0 ; y (1) = 12 g ; y ( 2 ) = 12 g ( 2 ) = 4 y (1) ; y ( 3 ) = 12 g ( 3 ) = 9 y (1) The distance traveled during each second can be found by subtracting two adjacent position values from the above list. d (1) = y (1) − y ( 0 ) = y (1) ; d ( 2 ) = y ( 2 ) − y (1) = 3 y (1) ; d ( 3 ) = y ( 3 ) − y ( 2 ) = 5 y (1) We could do this in general. Let n be a positive integer, starting with 0. y ( n ) = 12 gn 2

y ( n + 1) = 12 g ( n + 1)

(

d ( n + 1) = y ( n + 1) − y ( n ) = 12 g ( n + 1) − 12 gn 2 = 12 g ( n + 1) − n 2 2

(

)

= 12 g n 2 + 2n + 1 − n 2 = 12 g ( 2n + 1)

The value of ( 2n + 1) is always odd, in the sequence 1, 3, 5, 7, …. 61. Choose downward to be the positive direction, and the origin to be at the location of the plane. The 2 parachutist has v0 = 0, a = g = 9.80 m s , and will have y − y0 = 3800 m − 450 m = 3350 m when she pulls the ripcord. Eq. 2–12b, with x replaced by y, is used to find the time when she pulls the ripcord.

2 ( y − y0 ) a =

y = y0 + v0t + 12 at 2 → t =

(

)

2 ( 3350 m ) 9.80 m s 2 = 26.147 s  26.1s

The speed is found from Eq. 2–12a.

v = v0 + at = 0 + ( 9.80 m s 2 ) ( 26.147 s ) = 256.24 m s  256 m s  3.6 km h    922 km h  1m s 

( 256.24 m s ) 

This is well over 550 miles per hour! 62. Choose downward to be the positive direction, and y0 = 0 to be at the start of the pelican’s dive. The pelican has an initial velocity is v0 = 0, an acceleration of a = g , and a final location of y = 16.0 m. Find the total time of the pelican’s dive from Eq. 2–12b, with x replaced by y. y = y0 + v0 t + 12 at 2 → y = 0 + 0 + 12 at 2 → tdive =

2y a

2 (16.0 m ) 9.80 m s 2

= 1.81 s.

The fish can take evasive action if he sees the pelican at a time of 1.81 s – 0.20 s = 1.61 s into the dive. Find the location of the pelican at that time from Eq. 2–12b. 2 y = y0 + v0t + 12 at = 0 + 0 + 12 9.80 m s 2 (1.61 s ) = 12.7 m

(

)

Thus the fish must spot the pelican at a minimum height from the surface of the water of 16.0 m − 12.7 m = 3.3m .

Chapter 2

Describing Motion: Kinematics in One Dimension

63. Choose upward to be the positive direction, and y0 = 0 to be at the throwing location of the stone. The initial velocity is v0 = 15.5 m s , the acceleration is a = −9.80 m s 2 , and the final location is y = −75 m . (a) Using Eq. 2–12b and substituting y for x, we have the following. y = y0 + v0 t + 12 at 2 → 4.9 m s 2 t 2 − (15.5 m s ) t − 75 m = 0 →

(

15.5 

)

(15.5 ) − 4 ( 4.9 )( −75 ) = 5.802 s, − 2.638s 2 ( 4.9 ) 2

The positive answer is the physical answer: t = 5.80 s . (b) Use Eq. 2–12a to find the velocity just before hitting. v = v0 + at = 15.5 m s + ( −9.80 m s 2 ) ( 5.802 s ) = −41.4 m s →

v = 41.4 m s .

(c) The total distance traveled will be the distance up plus the distance down. The distance down will be 75 m more than the distance up. To find the distance up, use the fact that the speed at the top of the path will be 0. Then using Eq. 2–12c we have the following.

v = v + 2 a ( y − y0 ) → y = y 0 + 2

2 0

v 2 − v02 2a

= 0+

0 − (15.5 m s )

(

2 −9.80 m s 2

)

= 12.26 m.

Thus the distance up is 12.26 m, the distance down is 87.26 m, and the total distance traveled is 99.5 m . 64. (a) Choose upward to be the positive direction, and y0 = 0 at the ground. The rocket has v0 = 0, a = 3.2 m s 2 , and y = 725m when it runs out of fuel. Find the velocity of the rocket when it runs out of fuel from Eq. 2–12c, with x replaced by y. 2 v725 = v02 + 2a ( y − y0 ) → m

v725 m =  v02 + 2a ( y − y0 ) =  0 + 2 ( 3.2 m s 2 ) ( 725 m ) = 68.12 m s  68 m s The positive root is chosen since the rocket is moving upwards when it runs out of fuel. (b) The time to reach the 725 m location can be found from Eq. 2–12a. v725 m − v0 68.12 m s − 0 v725 m = v0 + at725 m → t725 m = = = 21.29 s  21s a 3.2 m s 2 (c) For this part of the problem, the rocket will have an initial velocity v0 = v725m = 68.12 m s , an acceleration of a = −9.80 m s 2, and a final velocity of v = 0 at its maximum altitude. The altitude reached from the out-of-fuel point can be found from Eq. 2–12c. 2 v 2 = v725 + 2a ( y − 725 m ) → m

ymax = 725 m +

2 0 − v725 m

= 725 m +

− ( 68.12 m s )

(

2 −9.80 m s 2

)

= 725 m + 237 m = 962 m  960 m

(d) The time for the “coasting” portion of the flight can be found from Eq. 2–12a. v − v725 m 0 − 68.12 m s = = 6.95s v = v725 m + atcoast → tcoast = a −9.80 m s 2 Thus the total time to reach the maximum altitude is t = 21.29 s + 6.95s = 28.24 s  28s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(e) For this part of the problem, the rocket has v0 = 0 m s, a = −9.80 m s 2, and a displacement of −962 m (it falls from a height of 962 m to the ground). Find the velocity upon reaching the Earth from Eq. 2–12c. v 2 = v02 + 2a ( y − y0 ) →

v =  v02 + 2a ( y − y0 ) =  0 + 2 ( −9.80 m s 2 ) ( −962 m ) = −137.3 m s  −137 m s (f)

The negative root was chosen because the rocket is moving downward, which is the negative direction. The time for the rocket to fall back to the Earth is found from Eq. 2–12a. v − v0 −137.3 m s − 0 = = 14.01s v = v0 + at → tfall = a −9.80 m s 2 Thus the total time for the entire flight is t = 28.24 s + 14.01s = 42.25s  42 s .

65. Choose upward to be the positive direction, and y0 = 0 to be the location of the nozzle. The initial velocity is v0 , the acceleration is a = −9.80 m s 2 , the final location is y = −1.8 m, and the time of flight is t = 2.5s. Using Eq. 2–12b and substituting y for x gives the following.

y = y0 + v0t + 12 at 2 →

v0 =

y − 12 at 2 t

(

)

−1.8 m − 12 −9.80 m s 2 ( 2.5s )

= 11.53 m s  12 m s

2.5s

66. Choose downward to be the positive direction, and take y0 = 0 to be the height where the package was released. The initial velocity is v0 = −6.40 m s , the acceleration is a = 9.80 m s 2 , and the displacement of the package will be y = 105 m. The time to reach the ground can be found from Eq. 2–12b, with x replaced by y. 2 ( −6.40 m s ) 2 (105 m ) 2v 2y y = y0 + v0t + 12 at 2 → t 2 + 0 t − = 0 → t2 + t− =0 → 2 9.80 m s 9.80 m s 2 a a

t − 1.306t − 21.43 = 0 → t = 2

1.306 

(1.306 ) − 4 ( −21.43) 2

= 5.33s, − 4.02 s

The correct time is the positive answer, t = 5.33s . 67. (a) Choose y = 0 to be the ground level, and positive to be upward. Then y = 0 m, y0 = 15 m, a = − g , and t = 0.83s describe the motion of the balloon. Use Eq. 2–12b.

y = y0 + v0 t + 12 at 2 →

v0 =

y − y0 − 12 at 2 t

(

)

0 − 15 m − 12 −9.80 m s 2 ( 0.83s )

( 0.83s )

= −14.01m s  −14 m s

So the speed is 14 m s .

Chapter 2

Describing Motion: Kinematics in One Dimension

(b) Consider the change in velocity from being released to being at Roger’s room, using Eq. 2–12c.

v 2 = v02 + 2ay → y =

v 2 − v02 2a

− ( −14.01m s )

(

2 −9.80 m s 2

)

= 10.01m

Thus the balloons are coming from 25 m above the ground. 68. Choose upward to be the positive direction, and y0 = 0 to be the level from which the ball was thrown. The initial velocity is v 0 , the instantaneous velocity is v = 13 m s, the acceleration is a = −9.80 m s 2, and the location of the window is y = 18 m . (a) Using Eq. 2–12c and substituting y for x, we have v 2 = v02 + 2a ( y − y0 ) →

v 0 =  v 2 − 2 a ( y − y0 ) = 

(13 m s ) − 2 ( −9.80 m s 2 ) (18 m ) = 22.84 m s  23 m s 2

Choose the positive value because the initial direction is upward. (b) At the top of its path, the velocity will be 0, and so we can use the initial velocity as found above, along with Eq. 2–12c. v 2 − v02

v = v + 2 a ( y − y0 ) → y = y 0 + 2

2 0

= 0+

0 − ( 22.84 m s )

(

2 −9.80 m s 2

)

= 26.62 m  27 m

(c) We want the time elapsed from throwing (velocity v0 = 22.84 m s) to reaching the window (velocity v = 13 m s ). Use Eq. 2–12a. v − v0 13 m s − 22.84 m s = = 1.004 s  1.0 s v = v0 + at → t = a −9.80 m s 2 (d) We want the time elapsed from the window ( v0 = 13m s ) to reaching the street

( v = −22.84 m s ). Use Eq. 2–12a. v = v0 + at → t =

v − v0

−22.84 m s − 13 m s

= 3.657 s  3.7 s . a −9.80 m s 2 The total time from throwing to reaching the street again is 1.004s + 3.657 s = 4.661s  4.7 s. 69. Choose downward to be the positive direction, and y0 = 0 to be the height from which the stone is dropped. Call the location of the top of the window ytop . From the distance and time given you can use Eq. 2–12b to find the velocity at the top of the window, vtop . That speed can then be used to find the height from which the stone was dropped. ( ybottom − ytop ) = vtopt + 12 at 2 → v top

(y =

bottom

− ytop ) − 12 at 2 t

(

)

2.2 m − 12 9.80 m s 2 ( 0.28 s )

2 vtop = v02 + 2a ( ytop − y0 ) → ytop =

( 0.28 s )

2 v top − v02

− y0 =

= 6.485 m s

( 6.485 m s )

(

2 9.80 m s 2

)

= 2.146 m  2.1m

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

70. Choose up to be the positive direction, so a = − g . Let the ground be the y = 0 location. As an intermediate result, the velocity at the bottom of the window can be found from the data given. Assume the rocket is at the bottom of the window at t = 0, and use Eq. 2–12b. 2 ytop of = ybottom of + vbottom of tpass + 12 atpass → window

window

(

)

10.0 m = 8.0 m + vbottom of ( 0.15s ) + 12 −9.80 m s 2 ( 0.15s ) window

→ vbottom of = 14.07 m s window

Now use the velocity at the bottom of the window with Eq. 2–12c to find the launch velocity, assuming the launch velocity was achieved at the ground level. 2 2 vbottom = vlaunch + 2 a ( y − y0 ) → of window 2 vlaunch = vbottom − 2 a ( y − y0 ) = of window

(14.07 m s ) − 2 ( −9.80 m s 2 ) ( 8.0 m ) = 18.84 m s 2

 18.8 m s

The maximum height can also be found from Eq. 2–12c, using the launch velocity and a velocity of 0 at the maximum height. 2 2 vmaximum = vlaunch + 2a ( ymax − y0 ) → height 2 2 v maximum − vlaunch

ymax = y0 +

height

− (18.84 m s )

(

2 −9.80 m s 2

)

= 18.1m

71. Choose up to be the positive direction. Let the bottom of the cliff be the y = 0 location. The equation of motion for the dropped ball is yball = y0 + v0 t + 12 at 2 = 55.0 m + 12 ( −9.80 m s 2 ) t 2 . The equation of motion for the thrown stone is ystone = y0 + v0t + 12 at 2 = ( 24.0 m s ) t + 12 ( −9.80 m s 2 ) t 2 . Set the two equations equal and solve for the time of the collision. Then use that time to find the location of either object. yball = ystone → 55.0 m + 12 −9.80 m s 2 t 2 = ( 24.0 m s ) t + 12 −9.80 m s 2 t 2 →

(

55.0 m = ( 24.0 m s ) t → t =

)

55.0 m 24.0 m s

(

)

= 2.292 s

(

)

yball = y0 + v0t + 12 at 2 = 55.0 m + 0 + 12 −9.80 m s 2 ( 2.292 s ) = 29.3 m 2

(

)

ystone = y0 + v0t + 12 at 2 = 0 + ( 24.0 m s ) ( 2.292 s ) + 12 −9.80 m s 2 ( 2.292 s ) = 29.3 m 2

72. The displacement is found from the integral of the velocity, over the given time interval. t2

x =  v dt = t1

t = 3.6 s

 ( 25 + 18t ) dt = ( 25t + 9t ) 2

t =1.3s

t = 3.6 s t =1.3s

=  25 ( 3.6 ) + 9 ( 3.6 )  −  25 (1.3 ) + 9 (1.3 ) 



 



= 158.93 m  1.6  10 2 m

Chapter 2

Describing Motion: Kinematics in One Dimension

73. (a) The velocity is the integral of the acceleration. dv a= → d v = adt → d v = A tdt → dt

 d v = A tdt →

v − v0 = 23 At 3/ 2 → v = v0 + 23 At 3/ 2 → v = 7.5 m s + 23 ( 3.0 m s 5/2 ) t 3/ 2 (b) The displacement is the integral of the velocity. dx v= → dx = v dt → dx = v0 + 23 At 3/ 2 dt → dt

(

)

(

)

3/ 2 5/ 2 5/2 5/ 2  dx =  v0 + 23 At dt → x = v0t + 23 52 At = ( 7.5 m s ) t + 154 3.0 m s t

(c)

(

) 5.0s = 6.7 m s v ( t = 5.0 s ) = 7.5 m s + ( 3.0 m s ) ( 5.0 s ) = 29.86 m s  3.0  10 m s x ( t = 5.0 s ) = ( 7.5 m s )( 5.0 s ) + ( 3.0 m s ) ( 5.0 s ) = 82.22 m  82 m a ( t = 5.0s ) = 3.0 m s5/2

3/ 2

5/2

2 3

5/ 2

5/2

4 15

74. (a) The velocity is found by integrating the acceleration with respect to time. Note that with the substitution given in the hint, the initial value of u is u0 = g − k v0 = g . a=

→ d v = adt → d v = ( g − k v ) dt →

dt Now make the substitution that u  g − k v.

u  g − kv → d v = − u

g − kv

u

= − k  dt → ln u g = − kt → ln

(1 − e ) k

= dt →

u g

dv g − kv

−

du 1 k u

= dt

= dt →

du u

= − kdt

= − kt → u = ge − kt = g − k v →

− kt

(b) As t goes to infinity, the value of the velocity is vterm = lim t →

g 1− e ) = . We also note that ( k k

− kt

if the acceleration is zero (which happens at terminal velocity), then a = g − k v = 0 →

vterm =

g . k

75. Choose the upward direction to be positive, and y0 = 0 to be the level from which the object is thrown. The initial velocity is v0 and the velocity at the top of the path is v = 0 m s. The height at the top of the path can be found from Eq. 2–12c with x replaced by y. −v 2 v 2 = v02 + 2a ( y − y0 ) → y − y0 = 0 . 2a From this we see that the displacement is inversely proportional to the acceleration, and so if the acceleration is reduced by a factor of 6 by going to the Moon, and the initial velocity is unchanged, then the displacement increases by a factor of 6 . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

76. (a) For the free-falling part of the motion, choose downward to be the positive direction, and y0 = 0 to be the height from which the person jumped. The initial velocity is v0 = 0, the acceleration is a = 9.80 m s 2 , and the location of the net is y = 18.0 m. Find the speed upon reaching the net from Eq. 2–12c with x replaced by y. v 2 = v02 + 2a ( y − y0 ) →

v =  0 + 2a ( y − 0 ) =  2 ( 9.80 m s 2 ) (18.0 m ) = 18.78 m s The positive root is selected since the person is moving downward. For the net-stretching part of the motion, choose downward to be the positive direction, and y0 = 18.0 m to be the height at which the person first contacts the net. The initial velocity is

v0 = 18.78 m s, the final velocity is v = 0, and the location at the stretched position is y = 19.0 m. Find the acceleration from Eq. 2–12c with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → a=

v 2 − v02

02 − (18.78 m s )

= −176 m s 2  −180 m s 2

2 ( y − y0 ) 2 (1.0 m ) This is about 18 g’s. (b) For the acceleration to be smaller, in the above equation we see that the displacement would have to be larger. This means that the net should be “loosened.” 77. We treat the value of 30 g’s as if it had 2 significant figures. The initial velocity of the car is  1m s  v0 = ( 95 km h )   = 26.39 m s. Choose x0 = 0 to be location at which the deceleration  3.6 km h  begins. We have v = 0 m s and a = −30 g = −294 m s 2 . Find the displacement from Eq. 2–12c.

v = v + 2a ( x − x0 ) → x = x0 + 2

2 0

v 2 − v02 2a

= 0+

0 − ( 26.39 m s )

(

2 −294 m s 2

)

= 1.18 m  1.2 m

78. We are given that x ( t ) = 2.0 − 3.6t + 1.7t 2 , where x is in meters and t is in seconds. (a) The value of 2.0 must have units of meters, and is the initial position of the ball. The value of 3.6 must have units of m/s, and is the initial speed of the ball – the speed at t = 0. Note that the ball is initially moving in the negative direction, since – 3.6 is used. The value of 1.7 must have units of m s 2 , and based on Eq. 2–12b is ½ the acceleration of the ball. (b) As given above, the units of 2.0 are meters, the units of 3.6 are m/s, and the units of 1.7 are m s 2. (c)

( ) x ( 2.0s ) = 2.0 m − ( 3.6 m s )( 2.0s ) + (1.7 m s ) ( 2.0s ) = 1.6 m x ( 3.0s ) = 2.0 m − ( 3.6 m s )( 3.0s ) + (1.7 m s ) ( 3.0s ) = 6.5 m

x (1.0s ) = 2.0 m − ( 3.6 m s )(1.0s ) + 1.7 m s 2 (1.0s ) = 0.1m

(d) v =

x t

6.5m − 0.1m 2.0 s

= 3.2 m s

Chapter 2

Describing Motion: Kinematics in One Dimension

79. (a) The sounds will not occur at equal time intervals because the longer any particular bolt falls, the higher its speed. The bolts all start from rest and all have the same acceleration, so at any moment in time, they will all have the same speed. However, they have different distances to travel in reaching the floor and therefore will be falling for different lengths of time. The farther a bolt falls, and thus the later a bolt hits the tin plate, the longer it has been accelerating and therefore the faster it is moving. (b) With equal distances between bolts, each successive bolt, having fallen a longer time when its predecessor reaches the plate, will have a higher average velocity and thus travel the inter-bolt distance in shorter periods of time. Thus the sounds will occur with smaller and smaller intervals between sounds. Thus the time between “clinks” will decrease as the string falls. (c) To hear the sounds at equal intervals, the bolts would have to be tied at distances corresponding to equal time intervals. The first bolt (call it bolt # 0) is touching the plate. Since each bolt has an initial speed of 0, the distance of fall and time of fall for each bolt are related to each other by di = 12 gti2. For bolt # 1, d1 = 12 gt12. For bolt # 2, we want t 2 = 2t1, so d 2 = 12 gt22 = 12 g ( 2t1 )

(

)

= 4 12 gt12 = 4d1 . Likewise t3 = 3t1, which leads to d 3 = 9d1 ; t 4 = 4t1, which leads to d 4 = 16d1 , and so

on. If the distance from the bolt initially on the pan to the next bolt is d1, then the distance from that bolt to the next one is 3d1, the distance to the next bolt is 5d1, and so on. The accompanying table shows these

relationships in a simpler format. 80. Choose downward to be the positive direction, and the origin to be at the top of the building. The 2 barometer has y0 = 0, v0 = 0, and a = g = 9.8 m s . Use Eq. 2–12b to find the height of the building, with x replaced by y. y = y0 + v0 t + 12 at 2 = 0 + 0 + 12 9.8 m s 2 t 2

(

)

yt = 2.0 = 12 9.8 m s 2 ( 2.0 s ) = 20 m 2

)

(

)

yt = 2.3 = 12 9.8 m s 2 ( 2.3 s ) = 26 m 2

The difference in the estimates is 6 m. If we assume the height of the building is the average of the 6m two measurements, then the % difference in the two values is  100 = 26% . 23m

 1m s   = 11.11m s.  3.6 km h 

81. The speed limit is 40 km h 

(a) For your motion, you would need to travel (10 + 15 + 50 + 15 + 70 ) m = 160 m to get the front of the car to the third stoplight. The time to travel the 160 m is found using the distance and the speed limit. 160 m x x = v  t → t = = = 14.40 s 11.11m s v No , you cannot make it to the third light without stopping, since it takes you longer than 13.0 seconds to reach the third light.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The second car needs to travel 165 m before the third light turns red. This car accelerates from v0 = 0 m s to a maximum of v = 11.11m s with a = 2.00 m s 2. Use Eq. 2–12a to determine the duration of that acceleration. v − v0 11.11m s − 0 m s v = v0 + at → tacc = = = 5.556 s a 2.00 m s 2 The distance traveled during that time is found from Eq. 2–12b. 2 ( x − x0 )acc = v0tacc + 12 atacc2 = 0 + 12 ( 2.00 m s 2 ) ( 5.556 s ) = 30.87 m Since 5.556 s have elapsed, there are 13.0 – 5.556 = 7.444 s remaining to clear the intersection. The car travels another 7.444 s at a speed of 11.11 m/s, covering a distance of xconstant = vavg t = speed

(11.11m s )( 7.444 s ) = 82.70 m. Thus the total distance is 30.87 m + 82.70 m = 113.57 m. No , the car cannot make it through all three lights without stopping. The car has to travel another 51.43 m to clear the third intersection, and is traveling at a speed of 11.11 m/s. Thus the front of the car would clear the intersection a time 51.43 m x t= = = 4.6 s after the light turns red. 11.11m s v 82. (a) Choose downward to be the positive direction, and y0 = 0 to be the level from which the car was dropped. The initial velocity is v0 = 0, the final location is y = H , and the acceleration is a = g . Find the final velocity from Eq. 2–12c, replacing x with y.

v 2 = v02 + 2a ( y − y0 ) → v =  v02 + 2a ( y − y0 ) =  2 gH . The speed is the magnitude of the velocity, v =

2gH .

(b) Solving the above equation for the height, we have that H =

v2 2g

. Thus for a collision of

v = 35 km h , the corresponding height is as follows.

v2 2g

  1m s   ( 35 km h )    3.6 km h   

(

2 9.80 m s 2

= 4.823 m  4.8 m

)

  1m s   ( 95 km h )   2 v  3.6 km h    H= = = 35.53 m  36 m 2 2g

(

2 9.80 m s

)

83. Choose downward to be the positive direction, and y0 = 0 to be at the roof from which the stones are dropped. The first stone has an initial velocity of v 0 = 0 and an acceleration of a = g . Eqs. 2–12a and 2–12b (with x replaced by y) give the velocity and location, respectively, of the first stone as a function of time. v = v0 + at → v1 = gt1 y = y0 + v0t + 12 at 2 → y1 = 12 gt12 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

The second stone has the same initial conditions, but its elapsed time is t2 = t1 − 1.50 s, and so has velocity and location equations as follows. 2 v2 = g ( t1 − 1.50s ) y2 = 12 g ( t1 − 1.50s ) The second stone reaches a speed of v 2 = 12.0 m s at a time given by the following.

t2 =

v2 g

= t1 − 1.50 s → t1 = 1.50 s +

v2 g

= 1.50 s +

12.0 m s 9.80 m s

(

The location of the first stone at that time is y1 = 12 gt12 = 12 9.80 m s

= 2.724 s 2

) ( 2.724 s) = 36.36 m. 2

The location of the second stone at that time is 2 2 2 y2 = 12 g ( t1 − 1.50s ) = 12 9.80 m s ( 2.724 − 1.50s ) = 7.34 m.

(

)

Thus the distance between the two stones is y1 − y2 = 36.36 m − 7.34 m = 29.0 m . 84. For the motion in the air, choose downward to be the positive direction, and y0 = 0 to be at the height of the diving board. Then diver has v0 = 0, (assuming the diver does not jump upward or downward), a = g = 9.80 m s 2, and y = 4.0 m when reaching the surface of the water. Find the diver’s speed at the water’s surface from Eq. 2–12c, with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → v =  v02 + 2a ( y − y0 ) = 0 + 2 ( 9.80 m s 2 ) ( 4.0 m ) = 8.85 m s

For the motion in the water, again choose down to be positive, but redefine y0 = 0 to be at the surface of the water. For this motion, v0 = 8.85 m s , v = 0, and y − y0 = 1.8 m . Find the acceleration from Eq. 2–12c, with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → a =

v 2 − v02

0 − ( 8.85 m s )

= −22 m s 2

2 ( y − y0 ) 2 (1.8 m ) The negative sign indicates that the acceleration is directed upwards.

85. Take the origin to be the location where the speeder passes the police car. The speeder’s constant  1m s  speed is vspeeder = (140 km h )   = 38.89 m s , and the location of the speeder as a function  3.6 km h  of time is given by xspeeder = vspeeder tspeeder = ( 38.89 m s ) tspeeder. The police car has an initial velocity of v 0 = 0 m s and a constant acceleration of apolice. The location of the police car as a function of time is given by Eq. 2–12b. 2 xpolice = v0t + 12 at 2 = 12 apolicetpolice . (a) The position vs. time graphs would qualitatively look like the graph shown here.

Speeder Police car

t1 t (b) The time to overtake the speeder occurs when the speeder has gone a distance of 850 m. The time is found using the speeder’s equation from above. 850 m 850 m = ( 38.89 m s ) tspeeder → tspeeder = = 21.86 s  22 s 38.89 m s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(c) The police car’s acceleration can be calculated knowing that the police car also had gone a distance of 850 m in a time of 21.86 s. 850 m = 12 a p ( 21.86 s )

→ ap =

2 ( 850 m )

( 21.86 s )

= 3.558 m s 2  3.6 m s 2

(d) The speed of the police car at the overtaking point can be found from Eq. 2–12a. v = v0 + at = 0 + ( 3.558 m s 2 ) ( 21.86 s ) = 77.78 m s  78 m s Note that this is exactly twice the speed of the speeder, and so is 280 km/h. (e) The police officer would not want to pass the car at that high speed, because then he would go past the car for a long distance before being able to pull the speeder over. The police officer would not have a constant acceleration, but instead, once he gets close to the speeder, will start to slow down so as to match the speeder’s velocity and pull in close behind the speeder. 86. Choose downward to be the positive direction, and y0 = 0 to be at the height of the bridge. Agent Bond has an initial velocity of v0 = 0, an acceleration of a = g , and will have a displacement of y = 15 m − 3.5 m = 11.5 m. Find the time of fall from Eq. 2–12b with x replaced by y.

y = y0 + v0t + 12 at 2 → t =

2y a

2 (11.5 m ) 9.80 m s 2

= 1.532 s

If the truck is approaching with v = 25 m s , then he needs to jump when the truck is a distance away given by d = v t = ( 25 m s )(1.532 s ) = 38.3 m. Convert this distance into “poles.” d = ( 38.3 m )(1 pole 25 m ) = 1.53 poles

So he should jump when the truck is about 1.5 poles away from the bridge. 87. Assume that y0 = 0 for each child is the level at which the child loses contact with the trampoline surface. Choose upward to be the positive direction. (a) The second child has v02 = 4.0 m s, a = − g = −9.80 m s 2 , and v = 0 m s at the maximum height position. Find the child’s maximum height from Eq. 2–12c, with x replaced by y. v 2 = v022 + 2a ( y2 − y0 ) → y 2 = y0 +

v 2 − v022 2a

= 0+

0 − ( 4.0 m s )

(

2 −9.80 m s 2

)

= 0.8163 m  0.82 m

(b) Since the first child can bounce up to one-and-a-half times higher than the second child, the first child can bounce up to a height of 1.5 ( 0.8163 m ) = 1.224 m = y1 − y0 . Eq. 2–12c is again used to find the initial speed of the first child. v 2 = v012 + 2a ( y1 − y0 ) →

v01 =  v 2 − 2a ( y1 − y0 ) = 0 − 2 ( −9.80 m s 2 ) (1.224 m ) = 4.898 m s  4.9 m s The positive root was chosen since the child was initially moving upward. (c) To find the time that the first child was in the air, use Eq. 2–12b with a total displacement of 0, since the child returns to the original position. y = y0 + v01t1 + 12 at12 → 0 = ( 4.898 m s ) t1 + 12 ( −9.80 m s 2 ) t12 → t1 = 0 s, 0.9996 s The time of 0 s corresponds to the time the child started the jump, so the correct answer is 1.0 s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

88. First consider the “uphill lie,” in which the ball is being putted down the hill. Choose x0 = 0 to be the ball’s original location, and the direction of the ball’s travel as the positive direction. The final velocity of the ball is v = 0 m s , the acceleration of the ball is a = −1.8 m s 2 , and the displacement of the ball will be x − x0 = 6.0 m for the first case and x − x0 = 8.0 m for the second case. Find the initial velocity of the ball from Eq. 2–12c.

v = v + 2a ( x − x0 ) 2

2 0

 0 − 2 ( −1.8 m s 2 ) ( 6.0 m ) = 4.648 m s  → v0 = v − 2a ( x − x0 ) =  2   0 − 2 ( −1.8 m s ) ( 8.0 m ) = 5.367 m s 2

The range of acceptable velocities for the uphill lie is 4.6 m s to 5.4 m s , a spread of 0.719 m/s if more digits are kept. Now consider the “downhill lie,” in which the ball is being putted up the hill. Use a very similar setup for the problem, with the basic difference being that the acceleration of the ball is now a = −2.6 m s 2. Find the initial velocity of the ball from Eq. 2–12c.

v = v + 2a ( x − x0 ) 2

2 0

 0 − 2 ( −2.6 m s 2 ) ( 6.0 m ) = 5.586 m s  → v0 = v − 2a ( x − x0 ) =  2   0 − 2 ( −2.6 m s ) ( 8.0 m ) = 6.450 m s 2

The range of acceptable velocities for the downhill lie is 5.6 m s to 6.5 m s , a spread of 0.864 m/s if more digits are kept. Because the range of acceptable velocities is smaller for putting down the hill, more control in putting is necessary, and so putting the ball downhill (the “uphill lie”) is more difficult.

 1m s   = 9.722 m s.  3.6 km h 

89. The car’s initial speed is v0 = ( 35 km h ) 

Case I: trying to stop. The constraint is, with the braking deceleration of the car ( a = −5.8 m s 2 ) , can the car stop in a 28 m displacement? The 2.0 seconds has no relation to this part of the problem. Using Eq. 2–12c, the distance traveled during braking is as follows.

( x − x0 ) =

v 2 − v02 2a

0 − ( 9.722 m s )

(

2 −5.8 m s 2

)

= 8.14 m She can stop the car in time.

Case II: crossing the intersection. The constraint is, with the acceleration of the car

  65 km h − 45 km h   1m s   = 0.9259 m s 2  , can she get through the intersection   a =  6.0 s   3.6 km h     (travel 43 meters) in the 2.0 seconds before the light turns red? Using Eq. 2–12b, the distance traveled during the 2.0 sec is 2 ( x − x0 ) = v0t + 12 at 2 = ( 9.722 m s )( 2.0 s ) + 12 0.9259 m s 2 ( 2.0 s ) = 21.3 m.

(

)

She should stop.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

90. The critical condition is that the total distance covered by the passing car and the approaching car must be less than 500 m so that they do not collide. The passing car has a total displacement composed of several individual parts. These are: i) the 10 m of clear room at the rear of the truck; ii) the 20 m length of the truck; iii) the 10 m of clear room at the front of the truck; and iv) the distance the truck travels. Since the truck travels at a speed of v = 18 m s , the truck will have a displacement of xtruck = (18 m s ) t. Thus the total displacement of the car during passing is xpassing = 40 m + (18 m s ) t. car

To express the motion of the car, we choose the origin to be at the location of the passing car when the decision to pass is made. For the passing car, we have an initial velocity of v0 = 18 m s and an acceleration of a = 0.60 m s 2 . Find xpassing from Eq. 2–12b. car

(

)

xpassing = xc − x0 = v0t + at = (18 m s ) t + 12 0.60 m s 2 t 2 1 2

car

Set the two expressions for xpassing equal to each other in order to find the time required to pass. car

(

)

(

)

2 2 40 m + (18 m s ) tpass = (18 m s ) tpass + 12 0.60 m s 2 tpass → 40 m = 12 0.60 m s 2 tpass →

tpass =

80 2 s = 11.55s 0.60

Calculate the displacements of the two cars during this time. xpassing = 40 m + (18 m s )(11.55s ) = 247.9 m car

xapproaching = vapproaching t = ( 25 m s )(11.55s ) = 288.75 m car

car

Thus the two cars together have covered a total distance of 247.9 m + 288.75 m = 536.65 m, which is more than allowed. The car should not pass. 91. For the falling rock, choose downward to be the positive direction, and y0 = 0 to be the height from which the stone is dropped. The initial velocity is v0 = 0 m s, the acceleration is a = g , the final position is y = H , and the time of fall is t1 . Using Eq. 2–12b with y substituting for x, we have H = y0 + v0t + 12 t 2 = 0 + 0 + 12 gt12 . For the sound wave, use the constant speed equation that

vsnd =

x t

H T − t1

, which can be rearranged to give t1 = T −

H vsnd

, where T = 4.1s is the total time

elapsed from dropping the rock to hearing the sound. Insert this expression for t1 into the equation for H from the stone, and solve for H.

 H  H = g T −  vsnd   1 2

−5

→

g 2v

2 snd

 gT

H2 −

 vsnd



+ 1 H + 12 gT 2 = 0 →



4.2388  10 H − 1.1182 H + 82.369 = 0 → H = 73.9 m, 2.63 10 4 m 2

Chapter 2

Describing Motion: Kinematics in One Dimension

If the larger answer is used in t1 = T −

H vs

, a negative time of fall results, and so the physically

correct answer is the smaller one: H = 74 m . 92. The speed of the conveyor belt is found from Eq. 2–2 for average velocity. 1.2 m x x = vt → v = = = 0.4286 m min  0.43 m min t 2.8 min The rate of burger production, assuming the spacing given is center to center, can be found as follows.

 1 burger   0.4286 m  burgers  0.25 m   1 min  = 1.7 min    93. (a) Choose up to be the positive direction. Let the throwing height of both objects be the y = 0 location, and so y0 = 0 for both objects. The acceleration of both objects is a = − g . The equation of motion for the rock, using Eq. 2–12b, is yrock = y0 + v0 rock t + 12 at 2 = v0 rock t − 12 gt 2 , where t is the time elapsed from the throwing of the rock. The equation of motion for the ball, 2 being thrown 1.00 s later, is yball = y0 + v0 ball ( t − 1.00s ) + 12 a ( t − 1.00s ) =

v0 ball ( t − 1.00s ) − 12 g ( t − 1.00s ) . Set the two equations equal (meaning the two objects are at the same place) and solve for the time of the collision. 2

yrock = yball → v0 rock t − 12 gt 2 = v0 ball ( t − 1.00 s ) − 12 g ( t − 1.00 s )

→

(15.0 m s ) t − 12 ( 9.80 m s 2 ) t 2 = ( 22.0 m s )( t − 1.00 s ) − 12 ( 9.80 m s 2 ) ( t − 1.00 s ) ( −16.8 ) t = −26.9 (16.8 m s ) t = ( 26.9 m ) → t = 1.60 s after the rock was thrown

→

(b) Use the time for the collision to find the position of either object.

(

)

yrock = v0 rock t − 12 gt 2 = (15.0 m s )(1.60 s ) − 12 9.80 m s 2 (1.60 s ) = 11.4 6 m  11.5 m 2

(

)

yball = v0 ball ( t − 1.00 s ) − 12 g ( t − 1.00 s ) = ( 22.0 m s )( 0.60 s ) − 12 9.80 m s 2 ( 0.60 s ) 2

= 11.44 m (c) Now the ball is thrown first, and so yball = v0 ballt − 12 gt 2 and yrock = v0 rock ( t − 1.00s ) − 12 g ( t − 1.00s ) . Again set the two equations equal to find the time of collision. 2

yball = yrock → v0 ball t − 12 gt 2 = v0 rock ( t − 1.00 s ) − 12 g ( t − 1.00 s )

→

( 22.0 m s ) t − 12 ( 9.80 m s 2 ) t 2 = (15.0 m s )( t − 1.00 s ) − 12 ( 9.80 m s 2 ) ( t − 1.00 s ) ( 22.0 m s ) t = (15.0 m s ) t − 15 + 9.8t − 4.9 ( 2.80 m s ) t = 19.9 m → t = 7.11s after the ball was thrown

→

But this answer can be deceptive. Where do the objects collide?

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

Instructor Solutions Manual

)

yball = v0 ball t − 12 gt 2 = ( 22.0 m s )( 7.11s ) − 12 9.80 m s 2 ( 7.11s ) = −91.3 m 2

(

)

yrock = v0 rock ( t − 1.00 s ) − 12 g ( t − 1.00 s ) = (15.0 m s )( 6.11s ) − 12 9.80 m s 2 ( 6.11s ) 2

= −91.3 m Thus, assuming they were thrown from ground level, they collide below ground level, which cannot happen. Thus they never collide . Or, they had to be more than 91.3 m above the

ground at their launch point. 94. (a) During the interval from A to B, it is moving in the negative direction , because its displacement is negative. (b) During the interval from A to B, it is speeding up , because the magnitude of its slope is increasing (changing from less steep to more steep). (c) During the interval from A to B, the acceleration is negative , because the graph is concave downward, indicating that the slope is getting more negative, and thus the acceleration is negative. (d) During the interval from D to E, it is moving in the positive direction , because the displacement is positive. (e) During the interval from D to E, it is speeding up , because the magnitude of its slope is (f)

increasing (changing from less steep to more steep). During the interval from D to E, the acceleration is positive , because the graph is

concave upward, indicating the slope is getting more positive, and thus the acceleration is positive. (g) During the interval from C to D, the object is not moving in either direction . The velocity and acceleration are both 0 .

95. This problem can be analyzed as a series of three one-dimensional motions; the acceleration phase, the constant speed phase, and the deceleration phase. The maximum speed of the train is:  1m s  ( 95 km h )   = 26.39 m s  3.6 km h  In the acceleration phase, the initial velocity is v0 = 0 m s , the acceleration is a = 1.1m s 2 , and the final velocity is v = 26.39 m s. Find the elapsed time for the acceleration phase from Eq. 2–12a. v − v0 26.39 m s − 0 v = v0 + at → tacc = = = 23.99 s a 1.1m s 2 Find the displacement during the acceleration phase from Eq. 2–12b. 2 ( x − x0 )acc = v0t + 12 at 2 = 0 + 12 (1.1m s 2 ) ( 23.99 s ) = 316.5 m In the deceleration phase, the initial velocity is v0 = 26.39 m s , the acceleration is a = −2.0 m s 2 , and the final velocity is v = 0 m s. Find the elapsed time for the deceleration phase from Eq. 2–12a. v − v0 0 − 26.39 m s = = 13.20 s v = v0 + at → tdec = a −2.0 m s 2 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

Find the distance traveled during the deceleration phase from Eq. 2–12b. 2 ( x − x0 )dec = v0t + 12 at 2 = ( 26.39 m s )(13.20 s ) + 12 −2.0 m s2 (13.20 s ) = 174.1m

(

)

The total elapsed time and distance traveled for the acceleration / deceleration phases are: tacc + tdec = 23.99 s + 13.20s = 37.19 s

( x − x0 )acc + ( x − x0 )dec = 316.5 m + 174.1m = 491m 15000 m

= 5 inter3000 m station segments. A train making the entire trip would thus have a total of 5 inter-station segments and 4 stops of 22 s each at the intermediate stations. Since 491 m is traveled during acceleration and deceleration, 3000 m − 491m = 2509 m of each segment is traveled at an

(a) If the stations are spaced 3.0 km = 3000 m apart, then there is a total of

average speed of v = 26.39 m s . The time for that 2509 m is given by x = vt → 2509 m x tconstant = = = 95.07 s. Thus a total inter-station segment will take 37.19 s + 26.39 m s v speed 95.07 s = 132.26 s. With 5 inter-station segments of 132.26 s each, and 4 stops of 22 s each, the total time is given by t3.0 km = 5 (132.26 s ) + 4 ( 22 s ) = 749 s = 12.5 min . 15000 m

= 3 inter5000 m station segments. A train making the entire trip would thus have a total of 3 inter-station segments and 2 stops of 22 s each at the intermediate stations. Since 491 m is traveled during acceleration and deceleration, 5000 m − 491m = 4509 m of each segment is traveled at an

(b) If the stations are spaced 5.0 km = 5000 m apart, then there is a total of

average speed of v = 26.39 m s. The time for that 4509 m is given by d = v t → 4509 m d t= = = 170.86 s. Thus a total inter-station segment will take 37.19 s + 170.86 s = v 26.39 m s 208.05 s. With 3 inter-station segments of 208.05 s each, and 2 stops of 22 s each, the total time is given by t5.0 km = 3 ( 208.05s ) + 2 ( 22 s ) = 668s = 11.1min . 96. To find the average speed for the entire race, we must take the total distance divided by the total time. If one lap is a distance of L, then the total distance will be 10 L. The time elapsed at a given 9L constant speed is given by t = d v , so the time for the first 9 laps would be t1 = , and 196.0 km h the time for the last lap would be t2 = L v2 , where v 2 is the average speed for the last lap. Write an expression for the average speed for the entire race, and then solve for v 2. v=

d total t1 + t2

10 L 9L

196.0 km h

v2 =

= 200.0 km h →

1 10 200.0 km h

−

= 245.0 km h

196.0 km h

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

97. Since all of the motion and most of the acceleration is downward, we will call downward the positive direction. Let the parachutist’s initial position be y 0 = 0 m, and their initial speed is v0 = 0 m s. We break the motion up into two constant-acceleration segments. The final position, duration, velocity, and acceleration for the first segment are represented by y1 , t1 , v1 , and a1, respectively. Note that the final velocity for the first segment is the initial velocity for the second segment. First segment:

2 y1

y 1 = y 0 + v0t1 + 21 a1t12 → t1 =

(

)

2 ( 75 m )

9.8 m s 2

= 3.912 s

v1 = a1t1 = 9.8 m s 2 ( 3.912 s ) = 38.34 m s

v 22 − v12

v22 = v12 + 2a2 y2 → y2 = Second segment:

v2 = v1 + a2t2 → t2 =

2 a2

v 2 − v1 a2

( 1.5 m s ) − ( 38.34 m s ) 2

(

2 −1.5 m s 2

1.5 m s − 38.34 m s −1.5 m s 2

)

= 489.2 m

= 24.56 s

(a) Total height: 75 m + 489.2 m = 564.2 m  560 m (b) Total time:

3.912 s + 24.56 s = 28.472 s  28s

98. As shown in Example 2–18, the speed with which the ball was thrown upward is the same as its speed on returning to the ground. From the symmetry of the two motions (both motions have speed = 0 at top, have same distance traveled, and have same acceleration), the time for the ball to rise is the same as the time for the ball to fall, 1.4 s. Choose upward to be the positive direction, and the origin to be at the level where the ball was thrown. For the ball, v = 0 at the top of the motion, and a = − g . Find the initial velocity from Eq. 2–12a.

v = v0 + at → v0 = v − at = 0 − ( −9.80 m s 2 ) (1.4 s ) = 13.72 m s  14 m s 99. To find the distance, we divide the motion of the robot into three segments. First, the initial acceleration from rest; second, motion at constant speed; and third, deceleration back to rest.

(

)

d1 = v0 t + 12 a1t12 = 0 + 12 0.20 m s 2 ( 4.5s ) = 2.025 m 2

v1 = a1t1 = ( 0.20 m s 2 ) ( 4.5s ) = 0.90 m s d 2 = v1t2 = ( 0.90 m s )( 68s ) = 61.2 m

v2 = v1 = 0.90 m s

(

)

d3 = v2t3 + 12 a1t12 = ( 0.90 m s )( 2.5s ) + 12 −0.40 m s 2 ( 2.5s ) = 1.0 m 2

d = d1 + d 2 + d 3 = 2.025 m + 61.2 m + 1.0 m = 64.225 m  64 m

100. Choose upward to be the positive direction, and the origin to be at the level where the ball was thrown. The velocity at the top is v = 0, and the ball will have an acceleration of a = − g . If the maximum height that the ball reaches is y = H , then the relationship between the initial velocity and the maximum height can be found from Eq. 2–12c, with x replaced by y. v 2 = v02 + 2a ( y − y0 ) → 0 = v02 + 2 ( − g ) H → H = v02 2 g It is given that v0 Bill = 1.5v0 Joe

( v ) 2 g ( v0 Bill ) , so = 0 Bill 2 = = 1.52 = 2.25  2.3 . 2 H Joe ( v0 Joe ) 2 g ( v0 Joe ) 2

H Bill

Chapter 2

Describing Motion: Kinematics in One Dimension

101. (a) Multiply the reading rate times the bit density to find the bit reading rate. 1.2 m 1bit N=  = 4.3  106 bits s −6 1s 0.28  10 m (b) The number of excess bits is N − N 0 . N − N 0 = 4.3  106 bits s − 1.4  106 bits s = 2.9  106 bits s

N − N0 N

2.9  106 bits s 4.3  106 bits s

= 0.67 = 67%

102. (a) The two bicycles will have the same velocity at any A time when the instantaneous slopes of their x vs. t B graphs are the same. That occurs near the time t1 x as marked on the graph. (b) Bicycle A has the larger acceleration, because its graph is concave upward, indicating a positive acceleration. Bicycle B has no acceleration because t1 t its graph has a constant slope. (c) The bicycles are passing each other at the times when the two graphs cross, because they both have the same position at that time. The graph with the steepest slope is the faster bicycle, and so is the one that is passing at that instant. So at the first crossing, bicycle B is passing bicycle A. At the second crossing, bicycle A is passing bicycle B. (d) Bicycle B has the highest instantaneous velocity at all times until the time t1, where both graphs have the same slope. For all times after t1 , bicycle A has the highest instantaneous velocity. The largest instantaneous velocity is for bicycle A at the latest time shown on the graph. (e) The bicycles appear to have the same average velocity. If the starting point of the graph for a particular bicycle is connected to the ending point with a straight line, the slope of that line is the average velocity. Both appear to have the same slope for that “average” line. 103. In this problem, note that a  0 and x  0. Take your starting position as 0. Then your position is given by Eq. 2–12b, x1 = vM t + 12 at 2, and the other car’s position is given by x2 = x + v At. Set the two positions equal to each other and solve for the time of collision. If this time is negative or imaginary, then there will be no collision. x1 = x2 → v M t + 12 at 2 = x + v At → 12 at 2 + ( v M − v A ) t − x = 0

( v A − vM )  ( vM − v A ) − 4 12 a ( − x ) 2

2 12 a

No collision: ( v M − v A ) − 4 a ( − x )  0 → 2

1 2

x

( vM − v A )

−2a

CHAPTER 3: Kinematics in Two or Three Dimensions; Vectors Responses to Questions 1.

No, the two velocities are not equal. Velocity is a vector quantity, with a magnitude and direction. If two vectors have different directions, they cannot be equal.

No. The car may be traveling at a constant speed of 60 km/h and going around a curve, in which case it would be accelerating.

(i)

During one year, the Earth travels a distance equal to the circumference of its orbit, but has a displacement of 0 relative to the Sun. (ii) Any kind of cross country “round trip” air travel would result in a large distance traveled, but a displacement of 0. (iii) The displacement for a race car from the start to the finish of the Indianapolis 500 auto race is 0, even though the car has traveled 500 miles.

The length of the displacement vector is the straight-line distance between the beginning point and the ending point of the trip and therefore the shortest distance between the two points. If the path is a straight line, then the length of the displacement vector is the same as the length of the path. If the path is curved or consists of different straight line segments, then the distance from beginning to end will be less than the path length. Therefore, the displacement vector can never be longer than the length of the path traveled, but it can be shorter.

Since both the batter and the ball started their motion at the same location (where the ball was hit) and ended their motion at the same location (where the ball was caught), the displacement of both was the same.

V is the magnitude of the vector V ; it is not necessarily larger than the magnitudes V1 and/or V2. For instance, if V1 and V2 have the same magnitude as each other and are in opposite directions, then V is zero. The magnitude of the sum is dependent on not just the magnitudes V1 and V2 , but also depends on the angle between V1 and V2.

If the two vectors are in the same direction, the magnitude of their sum will be a maximum, and will be 9.5 km. If the two vectors are in the opposite direction, the magnitude of their sum will be a minimum, and will be 0.5 km. If the two vectors are oriented in any other configuration, the magnitude of their sum will be between 0.5 km and 9.5 km.

Two vectors of unequal magnitude can never add to give the zero vector. The only way that two vectors can add up to give the zero vector is if they have the same magnitude and point in exactly opposite directions. However, three vectors of unequal magnitude can add to give the zero vector. As a one-dimensional example, a vector 10 units long in the positive x direction added to two vectors of 4 and 6 units each in the negative x direction will result in the zero vector. In two dimensions, if their sum using the tail-to-tip method gives a closed triangle, then the vector sum will be zero. See the diagram, in which A + B + C = 0

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(a) The magnitude of a vector can equal the length of one of its components if the other components of the vector are all 0; i.e., if the vector lies along one of the coordinate axes. (b) The magnitude of a vector can never be less than one of its components, because each component contributes a positive amount to the overall magnitude, through the Pythagorean relationship. The square root of a sum of squares is never less than the absolute value of any individual term.

10. The odometer and the speedometer of a car both measure scalar quantities (distance and speed, respectively). 11. To find the initial speed, use the slingshot to shoot the rock directly horizontally (no initial vertical speed) from a height of 1 meter (measured with the meter stick). The vertical displacement of the rock can be related to the time of flight by Eq. 2–12b. Take downward to be positive.

(

)

y = y0 + v y 0t + 12 at 2 → 1 m = 12 gt 2 → t = 2 (1 m ) 9.8 m s 2 = 0.45 s. Measure the horizontal range R of the rock with the meter stick. Then, if we measure the horizontal range R, we know that R = v x t = v x ( 0.45 s ) , and so v x = R 0.45 s, which is the speed the slingshot imparts to the rock. The only measurements are the height of fall and the range, both of which can be measured by a meter stick. 12. The arrow should be aimed above the target, because gravity will deflect the arrow downward from a horizontal or “line of sight” flight path. The angle of aim (above the horizontal) should increase as the distance from the target increases, because gravity will have more time to act in deflecting the arrow from a straight-line path. If we assume that the arrow was shot from the same height as the target, then the “level horizontal range” formula is applicable: R = v02 sin 2 0 g →  = 12 sin −1 ( Rg v02 ). As the range and hence the argument of the inverse sine function increases, the angle increases. 13. (a) In Fig. 3–22, the object is always moving. It never stops, not even instantaneously. Thus, there is no location where v = 0. (b) v y = 0 at the highest point of the motion. (c) The object has a constant horizontal velocity. Thus, there is no location where v x = 0. 14. If the bullet was fired from the ground, then the y-component of its velocity slowed considerably by the time it reached an altitude of 2.0 km, because of both the downward acceleration due to gravity and air resistance. The x-component of its velocity would have slowed due to air resistance as well. Therefore, the bullet could have been traveling slowly enough to be caught. 15. The balloons will hit each other, although not along the line of sight from you to your friend. If there were no acceleration due to gravity, the balloons would hit each other along the line of sight. Gravity causes each balloon to accelerate downward from that “line of sight” path, each with the same acceleration. Thus each balloon falls below the line of sight by the same amount at every instant along their flight, and so they collide. This situation is similar to Conceptual Example 3–9, and to Problem 58. 16. The horizontal component of the velocity stays constant in projectile motion, assuming that air resistance is negligible. Thus the horizontal component of velocity 1.0 seconds after launch will be the same as the horizontal component of velocity 2.0 seconds after launch. In both cases the horizontal velocity will be given by v x = v0 cos  = ( 30 m s )( cos 30 ) = 26 m s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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17. A projectile has the least speed at the top of its path. At that point the vertical speed is zero. The horizontal speed remains constant throughout the flight, if we neglect the effects of air resistance. 18. (a) Cannonball A, with the larger angle, will reach a higher elevation. It has a larger initial vertical velocity, and so by Eq. 2–12c it will rise higher before the vertical component of velocity is 0. (b) Cannonball A, with the larger angle, will stay in the air longer. It has a larger initial vertical velocity, and so takes more time to decelerate to 0 and start to fall. (c) The cannonball with a launch angle closest to 45 will travel the farthest. The range is a maximum for a launch angle of 45, and decreases for angles either larger or smaller than 45. 19. (a) The ball lands at the same point from which it was thrown inside the train car – back in the thrower’s hand. (b) If the car accelerates, the ball will land behind the thrower’s hand. (c) If the car decelerates, the ball will land in front of the thrower’s hand. (d) If the car rounds a curve (assume it curves to the right), then the ball will land to the left of the thrower’s hand. (e) The ball will be slowed by air resistance, and so will land behind the thrower’s hand. 20. Your reference frame is that of the train you are riding. If you are traveling with a relatively constant velocity (not over a hill or around a curve or drastically changing speed), then you will interpret your reference frame as being at rest. Since you are moving forward faster than the other train, the other train is moving backwards relative to you. Seeing the other train go past your window from front to rear makes it look like the other train is going backwards. 21. Both rowers need to cover the same “cross river” distance. The rower with the greatest speed in the “cross river” direction will be the one that reaches the other side first. The current has no bearing on the time to cross the river because the current doesn’t help either of the boats move across the river. Thus the rower heading straight across will reach the other side first. All of his “rowing effort” has gone into crossing the river. For the upstream rower, some of his “rowing effort” goes into battling the current, and so his “cross river” speed will be only a fraction of his rowing speed. 22. When you stand still under the umbrella in a vertical rain, you are in a cylinder-shaped volume in which there is no rain. The rain has no horizontal component of velocity, and so the rain cannot move from outside that cylinder into it. You stay dry. But as you run, you have a forward horizontal velocity relative to the rain, and so the rain has a backwards horizontal velocity relative to you. It is the same as if you were standing still under the umbrella but the rain had some horizontal component of velocity towards you. The perfectly vertical umbrella would not completely shield you.

Responses to MisConceptual Questions 1.

(c) The shortest possible resultant will be 20 units, which occurs when the vectors point in opposite directions. Since 0 units and 18 units are less than 20 units, (a) and (b) cannot be correct answers. The largest possible result will be 60 units, which occurs when the vectors point in the same direction. Since 64 units and 100 units are greater than 60 units, (d) and (e) cannot be correct answers. Answer (c) is the only choice that falls between the minimum and maximum vector lengths.

(a) The components of a vector make up the two legs of a right triangle when the vector is the hypotenuse. The legs of a right triangle cannot be longer than the hypotenuse, therefore (c) and (d) cannot be correct answers. Only when the vector is parallel to the component, is the

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

magnitude of the vector equal to the magnitude of the component, as in (b). For all other vectors the magnitude of the component is less than the magnitude of the vector. 3.

(b) If you turned 90, as in (a), your path would be that of a right triangle. The distance back would be the hypotenuse of that triangle, which would be longer than 100 m. If you turned by only 30, as in (c), your path would form an obtuse triangle, the distance back would have to be greater than if you had turned 90, and therefore it too would be greater than 100 m. If you turned 180, as in (d), you would end up back at your starting point, not 100 m away. Three equal distances of 100 m would form an equilateral triangle, so (b) is the correct answer.

(b) B and A share a common origin point, and B and C share a common terminal point. Thus B = A + C.

(c) The displacement is not 0, so the magnitude of the average velocity must be greater than 0, and so (a) is incorrect. It is impossible for the magnitude of the average velocity to be larger than the average speed of 10 m/s, and so (d) is incorrect. The only way for (b) to be correct would be if the object was traveling in a straight line. Thus (c) must be the correct answer. The displacement along a quarter of a circle is shorter than the actual distance traveled (the chord of the circle is less than the arc length), and so the average velocity is less than the average speed.

(a) The bullet falls due to the influence of gravity, not due to air resistance. Therefore (b) and (c) are incorrect. Inside the rifle the barrel prevents the bullet from falling, so the bullet does not begin to fall until it leaves the barrel.

(b) Assuming that we ignore air resistance, the ball is in free-fall after it leaves the bat. If the answer were (a) the ball would continue to accelerate forward and would not return to the ground. If the answer were (c) the ball would slow to a stop and return backwards toward the bat.

(b) If we ignore air friction, the horizontal and vertical components of the velocity are independent of each other. The vertical components of the two balls will be equal when the balls reach ground level. The ball thrown horizontally will have a horizontal component of velocity in addition to the vertical component. Therefore, it will have the greater speed.

(c) Both will hit the floor at the same time. Both were launched at the same time, neither had an initial vertical velocity, and both have the same vertical acceleration. The time of flight is determined by the vertical motion, and so is the same for both.

10. (c) Both you and the ball have the same constant horizontal velocity. Therefore in the time it takes the ball to travel up to its highest point and return to ground level, your hand and the ball have traveled the same horizontal distance, and the ball will land back in your hand. 11. (d) The time of flight and the maximum height are both determined by the vertical component of the initial velocity. Since all three kicks reach the same maximum height, they must also have the same time of flight. The horizontal components of the initial velocity are different, which accounts for them traveling different distances. 12. (c) The baseball is in projectile motion during the entire flight. In projectile motion the acceleration is downward and constant; it is never zero. Therefore, (a) is incorrect. Since the ball was hit high and far, it must have had an initial horizontal component of velocity. In projectile motion the horizontal component of velocity is constant, so at the highest point the © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

magnitude of the velocity cannot be zero, and thus (b) is incorrect. However, at the highest point, the vertical component of velocity is zero, so the magnitude of the velocity has a minimum at the highest point. So, (c) is the correct answer. 13. (b) Both the monkey and bullet fall at the same rate due to gravity. If the gun was pointed directly at the monkey and gravity did not act upon either the monkey or bullet, the bullet would hit the monkey. Since both start falling at the same time and fall at the same rate, the bullet will still hit the monkey. 14. (b, e) In projectile motion the acceleration is vertical, so the x velocity is constant throughout the motion, so (a) is valid. The acceleration is that of gravity, which, when up is positive, is a constant negative value, so (b) is not valid and (c) is valid. At the highest point in the trajectory the vertical velocity is changing from a positive to a negative value. At this point the y component of velocity is zero, so (d) is valid. However, the x component of the velocity is constant, but not necessarily zero, so (e) is not valid. 15. (a) The maximum relative speed between the two cars occurs when they are traveling in opposite directions. This maximum speed would be the sum of their speeds relative to the ground or 20 m/s. Since the two cars are traveling perpendicular to each other (not in opposite directions) their relative speed must be less than the maximum relative speed.

Solutions to Problems 1.

The resultant vector displacement of the car is given by D R = D west + Dsouth- . The westward displacement is west

245km + (118 km ) cos 45.0 = 328.4 km and the southward



Dsouth-

Dwest DR

west

displacement is (118 km ) sin 45.0 = 83.4 km. The resultant displacement has a magnitude of

328.42 + 83.42 = 339 km . The direction is  = tan −1 83.4 328.4 = 14.3 south of west . 2.

The truck has a displacement of 21 + ( −26 ) = −5 blocks north and 16 blocks east. The resultant has a magnitude of

Deast

( −5) + 16 = 16.76 blocks 2

Dnorth

Dsouth

−1

 17 blocks and a direction of tan 5 16 = 17 sorth of east . DR 3.

Given that Vx = 9.40 units and Vy = −6.80 units, the magnitude of V is given by V = V + V 2 x

2 y

direction is given by  = tan −1

9.40 + ( −6.80 ) = 11.6 units . The 2

−6.80 9.40

= −35.9 , or 35.9 below the



positive x-axis.

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

The vectors for the problem are drawn approximately to scale. The resultant has a length of 17.5 m and a direction 19 north of east. If calculations are done, the actual resultant should be 17 m at 23o north of east. Keeping one more significant figure would give 17.4 m at 22.5o.

(a) See the accompanying diagram. (b) Vx = −21.8cos 23.4 = −20.0 units (c) V = Vx2 + Vy2 =

 = tan −1 6.

8.66 20.0

Vy = 21.8sin 23.4 = 8.66 units

( −20.0)2 + (8.66)2 = 21.8 units

= 23.4above the − x axis

(a) V1 x = −6.2 units

V1 y = 0 units

V2 x = 8.1cos 55 = 4.646 units  4.6 units

V2 y = 8.1sin 55 = 6.635 units  6.6 units

(b) To find the components of the sum, add the components of the individual vectors. V1 + V2 = (V1 x + V2 x ,V1 y + V2 y ) = ( −1.554, 6.635) V1 + V2 =

( −1.554 )2 + ( 6.635)2 = 6.815 units  6.8 units  = tan −1

6.635 1.554

= 77

The sum has a magnitude of 6.8 units , and is 77o clockwise from the negative x-axis , or 103o counterclockwise from the positive x-axis. 7.

We see from the diagram that A = 6.8î and B = −5.5î. (a) C = A + B = 6.8î + ( −5.5) î = 1.3î . The magnitude is 1.3 units , and the direction is +x . (b) C = A − B = 6.8î − ( −5.5) î = 12.3î . The magnitude is 12.3 units , and the direction is +x . (c)

C = B − A = ( −5.5) î − 6.8î = −12.3î . The magnitude is 12.3 units , and the direction is –x .

(a) vnorth = ( 815 km h )( cos 41.5 ) = 610.4 km h  6.10  102 km h

vwest = ( 815 km h )( sin 41.5 ) = 540.0 km h  5.40 10 2 km h (b) d north = vnorth t = ( 610.4 km h )(1.75 h ) = 1068.2 km  1070 km d west = v west t = ( 540.0 km h )(1.75 h ) = 945 km

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The x component is negative and the y component is positive, since the summit is to the west of north. The angle measured counterclockwise from the positive x axis would be 128.4o. Thus the components are found to be as follows. x = 4580 cos128.4 = −2844.9 m y = 4580 sin128.4 = 3589.3 m z = 2450 m r = −2840 m ˆi + 3590 m ˆj + 2450 m kˆ

10. Ax = 42.0cos 28.0 = 37.08

r =

( −2844.9 ) + ( 3589.3) + ( 2450 ) = 5190 m 2

Ay = 42.0sin 28.0 = 19.72

Bx = −29.7 cos 56.0 = −16.61

B y = 29.7 sin 56.0 = 24.62

C x = 31.0 cos 270 = 0.0

C y = 31.0sin 270 = −31.0

(a)

( A + B + C ) = 37.08 + ( −16.61) + 0.0 = 20.47  20.5 ( A + B + C ) = 19.72 + 24.62 + ( −31.0) = 13.34  13.3 x

A + B + C = 20.5ˆi + 13.3ˆj (b)

A+B+C =

( 20.47 )2 + (13.34 )2 = 24.43  24.4

 = tan −1

11. Ax = 42.0cos 28.0 = 37.08

Ay = 42.0sin 28.0 = 19.72

Bx = −29.7 cos 56.0 = −16.61

By = 29.7 sin 56.0 = 24.62

(a)

( B − A ) = ( −16.61) − 37.08 = −53.69

13.34 20.47

= 33.1

( B − A ) = 24.62 − 19.72 = 4.90

Since the x component < 0 and the y component > 0, the vector is in the 2nd quadrant. 2 2 B − A = −53.7ˆi + 4.9ˆj ; B − A = ( −53.69 ) + ( 4.90) = 53.9

 B − A = tan −1

4.90

= 5.2 above − x axis or 174.8 −53.69 (b) ( A − B ) x = 37.08 − ( −16.61) = 53.69 ( A − B ) y = 19.72 − 24.62 = −4.90

Since the x component > 0 and the y component < 0, the vector is in the 4th quadrant. 2 2 A − B = 53.7ˆi − 4.9ˆj A − B = ( 53.69 ) + ( −4.90 ) = 53.9

 = tan −1

−4.90

= 5.2 below + x axis or 354.8 53.69 (c) Comparing results shows that B − A = − ( A − B ) . The two results are opposite of each other.

12. Ax = 42.0cos 28.0 = 37.08

Ay = 42.0sin 28.0 = 19.72

Cx = 31.0 cos 270 = 0.0

C y = 31.0sin 270 = −31.0

( A − C ) = 37.08 − 0.0 = 37.08 x

( A − C ) = 19.72 − ( −31.0 ) = 50.72 y

A − C = 37.1ˆi + 50.7ˆj A−C =

( 37.08)2 + ( 50.72 )2 = 62.8

 = tan −1

50.72 37.08

= 53.8

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

13. Ax = 42.0cos 28.0 = 37.08

Ay = 42.0sin 28.0 = 19.72

Bx = −29.7 cos 56.0 = −16.61

B y = 29.7 sin 56.0 = 24.62

C x = 31.0 cos 270 = 0.0

C y = 31.0sin 270 = −31.0

(a)

( B − 3A ) = −16.61 − 3 ( 37.08) = −127.85 ( B − 3A ) = 24.62 − 3 (19.72 ) = −34.54 x

Note that since both components are negative, the vector is in the 3rd quadrant. B − 3A = −127.9ˆi − 34.5ˆj B − 3A =

( −127.85)2 + ( −34.54 )2 = 132.4

 = tan −1

−34.54

−127.85 (b) ( 2A − 3B + 2C ) x = 2 ( 37.08) − 3 ( −16.61) + 2 ( 0.0 ) = 123.99

= 15.1below − x axis

( 2A − 3B + 2C ) = 2 (19.72 ) − 3 ( 24.62 ) + 2 ( −31.0) = −96.42 y

Note that since the x component is positive and the y component is negative, the vector is in the 4th quadrant. 2 A − 3B + 2C = 124.0ˆi − 96.4ˆj

(123.99 )2 + ( −96.42 )2 = 157.1

2 A − 3B + 2C =

(c)

 = tan −1

( C − A − B ) = 0.0 − 37.08 − ( −16.61) = −20.47 ( C − A − B ) = −31.0 − 19.72 − 24.62 = −75.34

−96.42 123.99

= 37.9below + x axis

Note that since both components are negative, the vector is in the 3rd quadrant. C − A − B = −20.5ˆi − 75.3ˆj C−A−B =

( −20.47 )2 + ( −75.34 )2 = 78.1

 = tan −1

−75.34 −20.47

14. (a) V1 = −6.0ˆi + 8.0ˆj

V1 =

( −6.0)2 + (8.0)2 = 10  = tan −1

(b) V2 = −4.5ˆi − 5.0ˆj

V2 =

( −4.5)2 + ( −5.0)2 = 6.7

(c)

V1 + V2 = −10.5ˆi + 3.0ˆj

 = tan −1

3.0 −10.5

(d) V1 − V2 = −1.5ˆi + 13.0ˆj

 = tan −1

(

V1 + V2 =

13.0 −1.5

= 74.8 below − x axis

8.0

= 127 −6.0 −5.0 = 228  = tan −1 −4.5

( −10.5)2 + ( 3.0)2 = 10.9  11

= 164

V1 − V2 =

( −1.5)2 + (13.0)2 = 13.1  13

= 96.6  97

) (

)

15. (a) V1 + V2 + V3 = 4.0î − 8.0ˆj + 1.0î + 1.0ˆj + −2.0î + 4.0ˆj = 3.0î − 3.0ˆj

V1 + V2 + V3 =

3.02 + ( −3.0 ) = 4.2 2

 = tan −1

−3.0 3.0

= 315

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

) (

Instructor Solutions Manual

)

(b) V1 − V2 + V3 = 4.0î − 8.0ˆj − 1.0î + 1.0ˆj + −2.0î + 4.0ˆj = 1.0î − 5.0ˆj

V1 − V2 + V3 = 1.02 + ( −5.0 ) = 5.1 2

 = tan −1

−5.0 1.0

= 280

16. If the angle is in the first quadrant, then Vx = V sin  and Vy = V cos  . See the first diagram. If the angle is in the second quadrant, then Vx = V sin  and Vy = −V cos  . See the second diagram. Vx



17. (a) For the magnitudes to add linearly, the two vectors must be parallel. V1 V2 (b) For the magnitudes to add according to the Pythagorean theorem, the two vectors must be at right angles to each other. V1 ⊥ V2 (c) The magnitude of V2 vector 2 must be 0. V2 = 0 18. (a) Use the Pythagorean theorem to find the possible x components. 2 95.02 = x 2 + ( −60.0 ) → x 2 = 5425 → x = 73.7 units (b) Express each vector in component form, with V the vector to be determined. 73.7î − 60.0ˆj + V î + V ˆj = −80.0î + 0.0ˆj →

(

) (

)

Vx = ( −80.0 − 73.7 ) = −153.7

Vy = 60.0

V = −153.7ˆi + 60.0ˆj

( −153.7 ) + ( 60.0 ) = 165.0 ; V = tan −1 2

V =

60.0 −153.7

= 159

19. Differentiate the position vector in order to determine the velocity, and differentiate the velocity in order to determine the acceleration. dr r = 9.60t ˆi + 6.45 ˆj − 1.50t 2 kˆ m → v = = 9.60 ˆi − 3.00t kˆ m s → dt dv a= = −3.00 kˆ m s 2 dt

(

)

(

)

20. The average velocity is found from the displacement at the two times. r ( t2 ) − r ( t1 ) r ( 3.00s ) − r (1.00s ) v avg = = t2 − t1 3.00s − 1.00s

(

)

  9.60 ( 3.00 ) î + 6.45 ˆj − (1.50 ) ( 3.00 ) 2 kˆ m   19.20î − 12.00kˆ    = 2.00s =  2 2.00s −  9.60 (1.00 ) î + 6.45 ˆj − (1.50 )(1.00 ) kˆ m      

(

)

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(

)

= 9.60 ˆi − 6.00 kˆ m s

The magnitude of the instantaneous velocity is found from the velocity vector. dr v= = 9.60 î − 3.00t kˆ m s dt v ( 2.00 ) = 9.60 î − ( 3.00 )( 2.00 ) kˆ m s = 9.60 î − 6.00 kˆ m s →

( (

)

(

)

( 9.60 ) + ( 6.00 ) m s = 11.3 m s 2

Note that, since the acceleration of this object is constant, the average velocity over the time interval is equal to the instantaneous velocity at the midpoint of the time interval. 21. (a) Average velocity is displacement divided by elapsed time. Since the displacement is not known, the average velocity cannot be determined . A special case exists in the case of constant acceleration, where the average velocity is the numeric average of the initial and final velocities. But this is not specified as motion with constant acceleration, and so that special case cannot be assumed. (b) Define east as the positive x-direction, and north as the positive y-direction. The average acceleration is the change in velocity divided by the elapsed time. v 25.7ˆi m s − ( −16.0ˆj m s ) = = 3.21ˆi m s 2 + 2.00ˆj m s 2 aavg = 8.00s t aavg =

( 3.21m s ) + ( 2.00 m s ) = 3.78 m s 2

 = tan −1

2.00

= 31.9 3.21 (c) Average speed is distance traveled divided by elapsed time. Since the distance traveled is not known, the average speed cannot be determined . 2

22. Note that the acceleration vector is constant, and so Eqs. 3–13a and 3–13b are applicable. Also v 0 = 0 and r0 = 0.

(

)

(a) v = v 0 + at = 4.0t ˆi + 3.0t ˆj m s → (b) v = v x2 + v 2y =

v x = 4.0t m s, v y = 3.0t m s

( 4.0t m s ) + ( 3.0t m s ) = 5.0t m s 2

(

)

(c)

r = r0 + v 0t + 12 at 2 = 2.0t 2 ˆi + 1.5t 2 ˆj m

(d)

v x ( 2.0 ) = 8.0 m s, v y ( 2.0 ) = 6.0 m s, v ( 2.0 ) = 10.0 m s, r ( 2.0 ) = 8.0 ˆi + 6.0 ˆj m

(

)

23. Choose downward to be the positive y-direction for this problem. Her acceleration is directed along the slope. (a) The vertical component of her acceleration is directed downward, and its magnitude will be

(

)

given by a y = a sin  = 1.80 m s2 sin 30.0o = 0.900 m s2 . (b) The time to reach the bottom of the hill is calculated from Eq. 2–12b, with a y displacement of 125 m, v y 0 = 0, and a y = 0.900 m s 2 .

(

)

y = y0 + v y 0t + 12 a y t 2 → 125 m = 0 + 0 + 12 0.900 m s 2 ( t ) t=

2 (125 m )

( 0.900 m s ) 2

→

= 16.7 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

24. The average speed is the distance traveled divided by the elapsed time, Eq. 2–1. distance traveled 11.5 km average speed = = = 2.1km h time elapsed 5.5 h The average velocity is the displacement divided by the elapsed time. Let north be the y-direction and up be the z-direction. 8.0 km ˆj + 0.85 km kˆ = 1.455 ˆj + 0.1545 kˆ km h v avg = 5.5 h

(

vavg =

)

(1.455 km h ) + ( 0.1545 km h ) = 1.46 km h  1.5 km h 2

0.1545

 = tan −1

1.455

= 6.1elevation above straight north

25. The three displacements for the ant are shown in the diagram, along with the net displacement. In x and y components, they are +10.0 cm ˆi , 10.0 cos 40.0 ˆi + 10.0 sin 40.0 ˆj cm, and

(

)

(10.0 cos110.0 î + 10.0 sin110.0 ˆj) cm. To find the average velocity, divide the net displacement by the elapsed time. (a) r = +10.0 cm î + 10.0 cos 40.0î + 10.0sin 40.0ˆj cm

(

)

+ 10.0 cos110 ˆi + 10.0sin110 ˆj cm

(

)

= 14.24 ˆi + 15.82 ˆj cm v avg = (b)

v avg =

r t

(14.24 ˆi + 15.82 ˆj) cm = 2.48 ˆi + 2.75ˆj cm s ( ) 2.40s + 1.80s + 1.55s

( 2.48cm s )2 + ( 2.75cm s )2 = 3.70cm s

 = tan −1

vy vx

= tan −1

2.75 2.48

= 48.0

26. (a) Differentiate the position vector, r = ( 3.0 t 2 î − 6.0t 3ˆj) m , with respect to time in order to determine the velocity, and differentiate the velocity to determine the acceleration. dr dv v= = 6.0 t î − 18.0t 2 ˆj m s a= = 6.0 î − 36.0t ˆj m s 2 dt dt 2 3 (b) r ( 3.5s ) = 3.0 ( 3.5) î − 6.0 ( 3.5) ˆj m = 37 î − 260ˆj m

(

)

(



(



)

(

)

2 v ( 3.5s ) = 6.0 ( 3.5) ˆi − 18.0 ( 3.5 ) ˆj m s = 21 ˆi − 220ˆj m s





27. Find the position at t = 5.0 s, and then subtract the initial point from that new location. 2 3 r ( 5.0) = 5.0 ( 5.0) + 6.0 ( 5.0 )  m ˆi + 7.0 − 3.0 (5.0 )  m ˆj = 175m ˆi − 368 m ˆj

(



) (





)



r = 175 m î − 368 m ˆj − 2.0 m î + 7.0 m ˆj = 173m î − 375 m ˆj

r =

(173m )2 + ( −375 m )2 = 413m  410 m

 = tan −1

−375 173

= −65.2  −65

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

 1m s   = 30.6 m s ,  3.6 km h 

28. The deceleration is along a straight line. The starting velocity is 110 km h 

and the ending velocity is 0 m/s. The acceleration is found from Eq. 2–12a. 30.6 m s v = v0 + at → 0 = 30.6 m s + a ( 7.0 s ) → a = − = −4.37 m s 2 7.0 s The horizontal acceleration is ahoriz = a cos  = −4.37 m s2 ( cos 26 ) = −3.9 m s2 . The vertical acceleration is avert = a sin  = −4.37 m s2 ( sin 26 ) = −1.9 m s 2 . The horizontal acceleration is to the left in Fig, 3–43, and the vertical acceleration is down. 29. Call east the positive x-direction and north the positive y-direction. Then this relative velocity relationship follows (see the accompanying diagram). v plane rel. = v plane + v air rel. ground

rel. air

15

v plane rel.

ground

Equate the x components of the velocity vectors. The magnitude of v plane rel.

v plane

ground

is given as

135 km 1.25 h

ground

rel. air

v air rel.

= 108 km h.

 ground

(108 km h ) sin15.0 = 0 + vwind x

→

v wind x = 27.95 km h

From the y components of the relative velocity equation, we find vwind y.

−108cos15.0 = −185 + vwind y → vwind y = 185 − 108cos15.0 = 80.68 km h The magnitude of the wind velocity is as follows. 2 2 vwind = vwind + vwind = x y

( 27.95 km h ) + (80.68 km h ) = 85.38 km h  85 km h 2

The direction of the wind is  = tan

−1

v wind-y

= tan −1

v wind-x

80.68 27.95

= 70.89  71 north of east .

30. The velocity vector can be found from Eq. 3–13a, and the position vector can be found from Eq. 3–13b, since the acceleration vector is constant. The time at which the object comes to rest is found by setting the velocity vector equal to 0. Both components of the velocity must be 0 at the same time for the object to be at rest. v = v 0 + at = −14 î − 7.0ˆj m s + 6.0t î + 3.0t ˆj m s = ( −14 + 6.0t ) î + ( −7.0 + 3.0t ) ˆj m s

(

)

(

)

v rest = 0.0 ˆi + 0.0ˆj m s =  ( −14 + 6.0t ) ˆi + ( −7.0 + 3.0t ) ˆj m s →

( v x ) rest = 0.0 = −14 + 6.0t → t =

(v ) y

= 0.0 = −7.0 + 3.0t → t =

s = 73 s

6.0 7.0

s = 73 s

3.0 Since both components of velocity are 0 at t = 73 s, the object is at rest at that time. rest

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

)

(

)

Instructor Solutions Manual

(

)

r = r0 + v 0t + 12 at 2 = 0.0 î + 0.0ˆj m + −14t î − 7.0t ˆj m + 12 6.0t 2 î + 3.0t 2 ˆj m

( ) = ( −14 ( ) + 6.0 ( ) ) ˆi m + ( −7.0 ( ) + 3.0 ( ) ) ˆj m (

)

2 2 = −14 ( 73 ) ˆi − 7.0 ( 73 ) ˆj m + 12 6.0 ( 73 ) ˆi + 3.0 ( 73 ) ˆj m

7 3

7 2 3

1 2

(

)

7 3

(

7 2 3

1 2

)

= −16.3î − 8.16ˆj m  −16.3î − 8.2ˆj m 31. Since the acceleration vector is constant, Eqs. 3–13a and 3–13b are applicable. The particle reaches its maximum x coordinate when the x velocity is 0. Note that v 0 = 5.0 m s î and r0 = 0.

(

)

v = v 0 + at = 5.0 ˆi m s + −3.0t ˆi + 4.5t ˆj m s v x = ( 5.0 − 3.0t ) m s → v x = 0 = ( 5.0 − 3.0t x − max ) m s → t x − max =

5.0 m s 3.0 m s 2

= 1.67 s

v ( t x − max ) = 5.0 ˆi m s +  −3.0 (1.67 ) ˆi + 4.5 (1.67 ) t ˆj m s = 7.5 m s ˆj

(

)

r = r0 + v 0 t + 12 at 2 = 5.0 t î m + 12 −3.0t 2 î + 4.5t 2 ˆj m 2 2 r ( t x − max ) = 5.0 (1.67 ) î m s + 12  −3.0 (1.67 ) î + 4.5 (1.67 ) ˆj m = 4.2î m + 6.3ˆj m





32. Choose downward to be the positive y-direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction, v x 0 = 3.0 m s and ax = 0. In the vertical direction,

v y 0 = 0, a y = 9.80 m s , y0 = 0, and the final location is y = 7.5 m. The time for the tiger to reach 2

the ground is found from applying Eq. 2–12b to the vertical motion.

(

)

y = y0 + v y 0 t + 12 a y t 2 → 7.5m = 0 + 0 + 12 9.80 m s 2 t 2 → t =

2 ( 7.5m ) 9.80 m s 2

= 1.237 sec

The horizontal displacement is calculated from the constant horizontal velocity. x = v x t = ( 3.0 m s )(1.237 sec ) = 3.7 m 33. Choose downward to be the positive y-direction. The origin will be at the point where the diver dives from the cliff. In the horizontal direction, v x 0 = 2.5 m s and a x = 0. In the vertical direction, v y 0 = 0, a y = 9.80 m s2 , y0 = 0, and the time of flight is t = 3.5 s. The height of the cliff is found

from applying Eq. 2–12b to the vertical motion.

(

)

y = y0 + v y 0t + 12 a y t 2 → y = 0 + 0 + 12 9.80 m s 2 ( 3.5s ) = 6.0 101 m 2

The distance from the base of the cliff to where the diver hits the water is found from the horizontal motion at constant velocity: x = v x t = ( 2.5 m s )( 3.5s ) = 8.75 m  8.8 m 34. Choose downward to be the positive y-direction. The origin will be at the point where the ball is thrown from the roof of the building. In the vertical direction, v y 0 = 0, a y = 9.80 m s 2 , y0 = 0, and the displacement is 7.5 m. The time of flight is found from applying Eq. 2–12b to the vertical motion. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(

)

y = y0 + v y 0t + 12 a y t 2 → 7.5 m = 12 9.80 m s 2 t 2 → t =

2 ( 9.0 m ) 9.80 m s 2

= 1.237 sec

The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity. x = v x t → v x = x t = 9.5 m 1.237 s = 7.7 m s 35. Choose downward to be the positive y-direction. The origin is the point where the ball is thrown from the roof of the building. In the vertical direction v y 0 = 0, y0 = 0, and a y = 9.80 m s 2. The initial horizontal velocity is 10.8 m/s and the horizontal range is 21.0 m. The time of flight is found from the horizontal motion at constant velocity. x = v x t → t = x v x = ( 21.0 m ) (10.8 m s ) = 1.944 s The vertical displacement, which is the height of the building, is found by applying Eq. 2–12b to the vertical motion. 2 y = y0 + v y 0 t + 12 a y t 2 → y = 0 + 0 + 12 ( 9.80 m s 2 ) (1.944 s ) = 18.5 m 36. Choose the point at which the football is kicked as the origin, and choose upward to be the positive y-direction. When the football reaches the ground again, the y-displacement is 0. For the football, 2 v y 0 = (18.0sin 31.0 ) m s, a y = −9.80 m s , and the final y-velocity will be the opposite of the starting y-velocity. Use Eq. 2–12a to find the time of flight. v y − v y 0 ( −18.0 sin 31.0 ) m s − (18.0 sin 31.0 ) m s = = 1.89 s v y = v y 0 + at → t = a −9.80 m s 2 37. Apply the “level horizontal range” formula. v 2 sin 2 0 R= 0 → g

( 2.5 m ) ( 9.80 m s 2 ) = 0.5799 sin 2 0 = 2 = 2 v0 ( 6.5 m s ) Rg

2 0 = sin −1 0.5799 →  0 = 18, 72

There are two angles because each angle gives the same range. If one angle is  = 45 +  , then  = 45 −  is also a solution. The two paths are shown in the graph. 38. When shooting the gun vertically, half the time of flight is spent moving upwards. Thus the upwards flight takes 1.7 seconds. Choose upward as the positive y-direction. Since at the top of the flight, the vertical velocity is zero, find the launching velocity from Eq. 2–12a. v y = v y 0 + at → v y 0 = v y − at = 0 = ( 9.80 m s 2 ) (1.7 s ) = 16.66 m s Using this initial velocity and an angle of 45o in the level horizontal range formula derived in the text will give the maximum range for the gun. R=

v02 sin 2 0 g

(16.66 m s ) sin ( 2  45o ) 2

9.80 m s 2

= 28 m

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

39. Choose the origin to be where the projectile is launched, and upwards to be the positive y-direction. The initial velocity of the projectile is v 0 , the launching angle is  0 , a y = − g, and v y 0 = v0 sin 0 . (a) The maximum height is found from Eq. 2–12c with v y = 0 at the maximum height.

ymax = 0 +

v y2 − v y20 2a y

−v02 sin 2  0 −2 g

v02 sin 2  0 2g

( 38.8 m s ) sin 2 42.2 2

(

2 9.80 m s 2

)

= 34.7 m

(b) The total time in the air is found from Eq. 2–12b, with a total vertical displacement of 0 for the ball to reach the ground. y = y0 + v y 0t + 12 a y t 2 → 0 = v0 sin  0t − 12 gt 2 →

2v0 sin  0

2 ( 38.8 m s ) sin 42.2

= 5.3189 s  5.32 s and t = 0 9.80 m s 2 g The time of 0 represents the launching of the ball. (c) The total horizontal distance covered is found from the horizontal motion at constant velocity.

x = v x t = ( v0 cos  0 ) t = ( 38.8 m s )( cos 42.2 )( 5.3189 s ) = 153 m

(d) The velocity of the projectile 1.50 s after firing is found as the vector sum of the horizontal and vertical velocities at that time. The horizontal velocity is a constant v 0 cos  0 =

( 38.8 m s )( cos 42.2 ) = 28.74 m s. The vertical velocity is found from Eq. 2–12a. v y = v y 0 + at = v0 sin  0 − gt = ( 38.8 m s ) sin 42.2 − ( 9.80 m s 2 ) (1.50 s ) = 11.36 m s

Thus the speed of the projectile is as follows.

v x2 + v y2 =

( 28.74 m s ) + (11.36 m s ) = 30.9 m s 2

40. First, we find the vertical component of the launch speed using the fact that she rises 1.2 m above the launch point, and that her vertical speed at that height is 0. Use Eq. 2–12c, and take upwards to be positive. v y2 − v y2 0 → v y 0 = v y2 − 2a y ( ymax − y0 ) = 2 9.80 m s 2 (1.2 m ) = 4.85 m s ymax − y0 = 2a y

(

)

Next, find the time of flight knowing the starting vertical velocity and the total vertical displacement, using Eq. 2–12b. Take the water surface to be y = 0. y = y0 + v y 0 t + 12 a y t 2 → 0 = 2.0 + 4.85t + 12 −9.80 m s 2 t 2 → 4.90t 2 − 4.85t − 2.0 = 0

(

4.85 

)

( 4.85 ) + 4 ( 4.90 )( 2.0 ) 2

= 1.30 s or − 0.313s 9.80 We use the positive time. Now find the horizontal component of the launch speed from the time and the horizontal distance traveled. x 2.2 m x = v x t → v x = = = 1.69 m s t 1.30 s Now that we have both components of the launch velocity, we can find the magnitude and direction.

v0 =

v x2 + v y2 0 =

 0 = tan −1

vy0 vx

(1.69 m s ) + ( 4.85 m s ) = 5.136 m s  5.1m s

= tan −1

4.85 m s 1.69 m s

= 70.7  71 above the horizontal

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

41, First use the level horizontal range formula to find the launching speed for the maximum range, using 45. R=

v02 sin 2 0 g

→ v0 =

Rmax g

sin  2 ( 45 )

Rmax g

Then use the level horizontal range formula again to find the launching angle for the speed found above, with the given range. v 2 sin 2 0 Rg Rg 68 m R= 0 → sin 2 0 = 2 = = = 0.7158 → 2 0 = 45.71, 134.29 g Rmax g 95 m v0

 0 = 22.86, 67.15  23, 67 42. Choose the origin to be at ground level, under the place where the projectile is launched, and upwards to be the positive y direction. For the projectile, v0 = 62.0 m s,  0 = 35.0, a y = − g , y0 = 125 m, and v y 0 = v0 sin 0 .

(a) The time taken to reach the ground is found from Eq. 2–12b, with a final height of 0. y = y0 + v y 0t + 12 a y t 2 → 0 = y0 + v0 sin 0t − 12 gt 2 → t=

−v0 sin  0  v02 sin 2  0 − 4 ( − 12 g ) y0 2 ( − 12 g )

−35.56 m s 

( −35.56 m s ) + 2450 m 2 s 2 2

−9.80 m s 2

= 9.848s, − 2.590 s = 9.85s

Choose the positive time since the projectile was launched at time t = 0. (b) The horizontal range is found from the horizontal motion at constant velocity. x = v x t = ( v0 cos  0 ) t = ( 62.0 m s ) ( cos 35.0 ) ( 9.848 s ) = 5.00  10 2 m

(c) At the instant just before the particle reaches the ground, the horizontal component of its velocity is the constant v x = v0 cos  0 = ( 62.0 m s ) cos 35.0 = 50.8 m s . The vertical component is found from Eq. 2–12a. v y = v y 0 + at = v0 sin  0 − gt = ( 62.0 m s ) sin 35.0 − ( 9.80 m s 2 ) ( 9.848s ) = −60.9 m s

(d) The magnitude of the velocity is found from the x and y components calculated in part (c) above.

v = v x2 + v y2 =

( 50.8 m s ) + ( −60.9 m s ) = 79.3m s 2

(e) The direction of the velocity is  = tan −1

(f)

vy vx

= tan −1

−60.9 50.8

= −50.2, and so the object is

moving 50.2 below the horizon . The maximum height above the cliff top reached by the projectile will occur when the y-velocity is 0, and is found from Eq. 2–12c. v y2 = v y2 0 + 2a y ( y − y0 ) → 0 = v02 sin 2  0 − 2 gymax ymax =

v02 sin 2  0 2g

( 62.0 m s ) sin 2 35.0 2

(

2 9.80 m s 2

)

= 64.5 m

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

43. (a) Use the level horizontal range formula derived in the text. R=

v02 sin 2 0

→ v0 =

sin 2 0

( 7.80 m ) ( 9.80 m s 2 ) sin 54.0

= 9.72 m s

(b) Now increase the speed by 5.0% and calculate the new range. The new speed would be 9.72 m s (1.05 ) = 10.21m s and the new range would be as follows.

(10.21m s ) sin 54 2

v02 sin 2 0

9.80 m s 2

= 8.606 m

This is an increase of 0.81m (10% increase ) . 44. We choose the origin at the same place as in Example 3–11. With the new definition of the coordinate axes, we have the following data: y0 = 0, y = +1.00 m, v y 0 = −12.0 m s, v x 0 = −16.0 m s, a = 9.80 m s 2 . Use Eq. 2–12b to solve for the time. y = y0 + v y 0t + 12 gt 2 → 1.00 m = 0 − (12.0 m s ) t + 4.90 m s 2 t 2 →

(

)

( 4.90 m s ) t − (12.0 m s ) t − (1.00 m ) = 0 2

This is the same equation as in Example 3–11, and so we know the appropriate solution is t = 2.53s. We use that time to calculate the horizontal distance the ball travels. x = v x 0 t = ( −16.0 m s )( 2.53s ) = −40.5 m Since the x-direction is now positive to the left, the negative value means that the ball lands 40.5 m to the right of where it departed the punter’s foot. 45. (a) Choose the origin to be the location where the car leaves the ramp, and choose upward to be the positive y-direction. At the end of its flight over the 8 cars, the car must be at y = −1.5m. Also for the car, v y 0 = 0, a y = − g , v x = v0 , and x = 22 m. The time of flight is found from the horizontal motion at constant velocity: x = v x t → t = x v0 . That expression for the time is used in Eq. 2–12b for the vertical motion. 2 y = y0 + v y 0t + 12 a y t 2 → y = 0 + 0 + 12 ( − g )( x v0 ) → v0 =

− g ( x )

(

2( y)

)

− 9.80 m s 2 ( 22 m )

2 ( −1.5 m )

= 39.76 m s  4.0  101 m s

(b) Again choose the origin to be the location where the car leaves the ramp, and choose upward to be the positive y-direction. The y-displacement of the car at the end of its flight over the 8 cars must again be y = −1.5 m. For the car, v y 0 = v0 sin 0 , a y = − g , v x = v0 cos  0 , and

x = 22 m. The launch angle is  0 = 9.0. The time of flight is found from the horizontal motion at constant velocity. x x = v x t → t = v0 cos  0 That expression for the time is used in Eq. 2–12b for the vertical motion. y = y0 + v y 0 t + a y t 1 2

 x  → y = v0 sin  0 + (−g )  v0 cos  0  v0 cos  0  x

1 2

→

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

g ( x )

v0 =

( 9.80 m s ) ( 22 m ) 2

2 ( x tan  0 − y ) cos 2  0

2 ( ( 22 m ) tan 9.0 + 1.5 m ) cos 2 9.0

= 22 m s

46. Choose the origin to be the point on the ground directly below the point where the baseball was hit. Choose upward to be the positive y-direction. Then y0 = 1.2 m, y = 12.5 m at the end of the motion, v y 0 = ( 27.0sin 45.0 ) m s = 19.09 m s , and a y = −9.80 m s 2 . Use Eq. 2–12b to find the time of flight. y = y0 + v y 0 t + 12 a y t 2 t=

→

1 2

a y t 2 + v y 0 t + ( y0 − y ) = 0

−v y 0  v y2 0 − 4 ( 12 a y ) ( y0 − y ) 2 12 a y

−19.09 

→

(19.09 ) − 2 ( −9.80 )( −11.3 ) 2

−9.80

= 0.728 s, 3.168 s The smaller time is the time the baseball reached the building’s height on the way up, and the larger time is the time the baseball reached the building’s height on the way down. We must choose the larger result, because the baseball cannot land on the roof on the way up. Now calculate the horizontal distance traveled using the horizontal motion at constant velocity. x = v x t = ( 27.0 cos 45.0 ) m s  ( 3.168s ) = 60.5 m

47. Choose the origin to be the location on the ground directly below the airplane at the time the supplies are dropped, and choose upward as the positive y-direction. For the supplies, y0 = 265 m, v y 0 = 0, a y = − g, and the final y-location is y = 0 m. The initial (and constant) x-velocity of the supplies is

 1h  1000 m    = 34.72 m s. The time for the supplies to reach the ground is  3600 s  1km 

v x = 125 km h 

found from Eq. 2–12b. y = y0 + v y 0t + 12 a y t 2

→

0 = y0 + 0 + 12 at 2 → t =

−2 y0

−2 ( 265 m )

( −9.80 m s ) 2

Then the horizontal distance of travel for the package is found from the horizontal constant velocity. x = v x t = ( 34.72 m s )

−2 ( 265 m )

( −9.80 m s ) 2

= 255 m

48. Choose the origin to be the point of launch, and upwards to be the positive y-direction. The initial velocity is v0 , the launching angle is  0 , a y = − g , y0 = 0, and v y 0 = v0 sin 0 . Eq. 2–12a is used to find the time required to reach the highest point, at which v y = 0. v y = v y 0 + atup

→

tup =

vy − vy0

0 − v0 sin  0

a −g Eq. 2–12c is used to find the height at this highest point. v y2 = v y2 0 + 2a y ( ymax − y0 )

→

ymax = y0 +

v y2 − v y2 0 2a y

v0 sin  0 g = 0+

−v02 sin 2  0 −2 g

v02 sin 2  0 2g

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Eq. 2–12b is used to find the time for the object to fall the same distance with a starting velocity of 0. v 2 sin 2  0 v sin  0 2 + 0tdown − 12 gtdown → tdown = 0 y = yo + v y 0t + 12 a y t 2 → 0 = 0 2g g A comparison shows that tup = tdown . 49. The constant acceleration of the projectile is given by a = −9.80 m s 2 ˆj. We use Eq. 3–13a with the given velocity, the acceleration, and the time to find the initial velocity. v = v 0 + at → v 0 = v − at = ( 7.8 ˆi + 5.2 ˆj) m s − ( −9.80 m s 2 ˆj) ( 3.0s ) = ( 7.8 ˆi + 34.6 ˆj) m s

( 7.8 m s ) + ( 34.6 m s ) = 35.47 m s , and the original launch direction is 2

The initial speed is v0 = given by  0 = tan −1

34.6 m s 7.8 m s

= 77.30. Use this information with the level horizontal range formula

to find the range. (a) R =

v02 sin 2 0

( 35.47 m s ) ( sin154.6 ) 2

= = 55 m g 9.80 m s 2 (b) We use the vertical information to find the maximum height. The initial vertical velocity is 34.6 m/s, and the vertical acceleration is −9.80 m s 2 . The vertical velocity at the maximum height is 0, and the initial height is 0. Use Eq. 2–12c. v y2 = v y2 0 + 2a y ( ymax − y0 ) →

ymax = y0 +

v y2 − v y2 0 2a y

−v02 y 2a y

− ( 34.6 m s )

(

2 −9.80 m s 2

)

= 61m

(c) From the information above and the symmetry of projectile motion, we know that the final speed just before the projectile hits the ground is the same as the initial speed, and the angle is the same as the launching angle, but below the horizontal. So vfinal = 35 m s and

final = 77 below the horizontal . 50. Choose the origin to be where the projectile is launched, and upwards to be the positive y-direction. The initial velocity of the projectile is v0 , the launching angle is  0 , a y = − g , and v y 0 = v0 sin 0 . The range of the projectile is given by the level horizontal range formula, R =

v02 sin 2 0

. The g maximum height of the projectile will occur when its vertical speed is 0. Apply Eq. 2–12c. v02 sin 2  0 2 2 2 2 v y = v y 0 + 2a y ( y − y0 ) → 0 = v0 sin  0 − 2 gymax → ymax = 2g Now find the angle for which R = ymax . R = ymax

→

v02 sin 2 0 g

v02 sin 2  0 2g

→

sin 2 0 = 12 sin 2  0 →

2sin 0 cos 0 = 12 sin 0 → 4 cos 0 = sin 0 → tan 0 = 4 → 0 = tan −1 4 = 76 2

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

51. The angle is in the direction of the velocity, so find the components of the velocity, and use them to define the angle. Let the positive y-direction be down. vy gt v x = v0 v y = v y 0 + a y t = gt  = tan −1 = tan −1 vx v0 52. Choose upward to be the positive y-direction. The origin is the point from which the pebbles are released. In the vertical direction, a y = −9.80 m s 2 , the velocity at the window is v y = 0, and the vertical displacement is 8.0 m. The initial y-velocity is found from Eq. 2–12c. v y2 = v y2 0 + 2a y ( y − y0 ) →

vy0 =

v y2 − 2a y ( y − y0 ) =

(

)

0 − 2 −9.80 m s 2 ( 8.0 m ) = 12.5 m s

Find the time for the pebbles to travel to the window from Eq. 2–12a. v y − v y 0 0 − 12.5 m s = = 1.28s v y = v y 0 + at → t = a −9.80 m s 2 Find the horizontal speed from the horizontal motion at constant velocity. x = v x t → v x = x t = 8.5 m 1.28 s = 6.6 m s This is the speed of the pebbles when they hit the window. 53. (a) Use the level horizontal range formula to find her takeoff speed.

v02 sin 2 0 g

→ v0 =

sin 2 0

( 9.80 m s ) (8.0 m ) = 8.854 m s  8.9 m s 2

sin 90

(b) Let the launch point be at the y = 0 level, and choose upward to be positive. Use Eq. 2–12b to solve for the time to fall to 2.5 meters below the starting height, and then calculate the horizontal distance traveled. y = y0 + v0 y t + 12 a y t 2 → − 2.5 m = ( 8.854 m s ) sin 5t + 12 ( −9.80 m s 2 ) t 2

4.9t 2 − 6.261t − 2.5 m = 0 → t=

6.261 

( 6.261) − 4 ( 4.9 )( −2.5) 6.261  9.391 = = −0.319 s, 1.597 s 2 ( 4.9 ) 2 ( 4.9 ) 2

Use the positive time to find the horizontal displacement during the jump. x = v0 x t = v0 cos 45t = ( 8.854 m s ) cos 45 (1.597 s ) = 10.0 m She will land exactly on the opposite bank, neither long nor short. 54. The horizontal component of the speed does not change during the course of the motion, and so v x = v x 0. The net vertical displacement is 0 if the firing level equals the landing level, and so y − y0 = 0. Eq. 2–12c then gives v y = v y 0 + 2a y ( y − y0 ) = v y 0 . Thus v y = v y 0 , and from the 2

horizontal v x2 = v x20. The initial speed is v0 = v x20 + v y20 . The final speed is v = v x2 + v y2

= v x20 + v y20 = v0 . Thus v = v0 .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

55. (a) Use the level horizontal range formula to find the two angles. 2 v 2 sin 2 0 Rg ( 20.0 m ) 9.80 m s R= 0 → sin 2 0 = 2 = = 0.8711 → 2 g v0 (15.0 m s )

(

)

2 0 = 60.59, 119.41 →  0 = 30.29, 59.71  30.3, 59.7

(b) The maximum height can be found from the y-component of the velocity and Eq. 2–12c, using the fact that the vertical velocity is 0 at the top of the path. Call upwards the positive direction. v y2 = v y2 0 + 2a y ( y − y0 ) →

y − y0 = y − y0 =

v y2 − v y2 0 2a y v y2 − v y2 0 2a y

v − ( v0 sin 0 )

(

2 −9.80 m s 2 v − ( v0 sin 0 )

(

2 −9.80 m s 2

) 2

)

(15.0 m s )( sin 30.29 ) = 2.92 m = 2

(

2 9.80 m s 2

)

(15.0 m s )( sin 59.71 ) = 8.56 m = 2

(

2 9.80 m s 2

)

(c) The time of flight can be found Eq. 2–12a and using the fact that the initial and final vertical velocities are opposite of each other. v y − v y 0 −2v y 0 −2v0 sin  0 2v0 sin  0 v y = v y 0 + ayt → t = = = = ay ay g −g t1 =

2 (15.0 m s )( sin 30.29 ) 9.80 m s

= 1.54 s

t2 =

2 (15.0 m s )( sin 59.71 ) 9.80 m s 2

= 2.64 s

56. Choose the origin to be the location on the ground directly underneath the ball when served, and choose upward as the positive y-direction. Then for the ball, y0 = 2.30 m, v y 0 = 0, a y = − g , and the y-location when the ball just clears the net is y = 0.90 m. The time for the ball to reach the net is calculated from Eq. 2–12b. y = y0 + v y 0 t + 12 a y t 2 → 0.90 m = 2.30 m + 0 + 12 −9.80 m s 2 t 2 →

(

t to =

)

2 ( −1.40 m )

= 0.5345s −9.80 m s 2 The x-velocity is found from the horizontal motion at constant velocity. 15.0 m x x = v x t → v x = = = 28.06  28.1m s t 0.5345 s This is the minimum speed required to clear the net. net

To find the full time of flight of the ball, set the final y-location to be y = 0, and again use Eq. 2–12b. y = y0 + v y 0t + 12 a y t 2 → 0.0 m = 2.30 m + 12 −9.80 m s 2 t 2 →

(

t total =

)

2 ( −2.30 m )

= 0.6851  0.685s −9.80 m s 2 The horizontal position where the ball lands is found from the horizontal motion at constant velocity. x = v x t = ( 28.06 m s )( 0.6851s ) = 19.22  19.2 m Since this is between 15.0 and 22.0 m, the ball lands in the “good” region .

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

57. The time of flight is found from the constant velocity relationship for horizontal motion.

x = v x t → t = x v x = 8.0 m 9.1m s = 0.88s The y-motion is symmetric in time – it takes half the time of flight to rise, and half to fall. Thus the time for the jumper to fall from his highest point to the ground is 0.44 sec. His vertical speed is zero at the highest point. From this time, starting vertical speed, and the acceleration of gravity, the maximum height can be found. Call upward the positive y-direction. The point of maximum height is the starting position y0 , the ending position is y = 0, the starting vertical speed is 0, and a = − g . Use Eq. 2–12b to find the height. 2 y = y0 + v y 0 t + 12 a y t 2 → 0 = y0 + 0 − 12 ( 9.8 m s 2 ) ( 0.44 s ) → y0 = 0.95 m . 58. Choose the origin to be the location from which the balloon is fired, and choose upward as the positive y-direction. Assume the boy in the tree is a distance H up from the point at which the balloon is fired, and that the tree is a distance d horizontally from the point at which the balloon is fired. The equations of motion for the balloon and boy are as follows, using constant acceleration relationships.

 d

xballoon = v0 cos 0t

yballoon = 0 + v0 sin 0t − 12 gt 2

yboy = H − 12 gt 2

Use the horizontal motion at constant velocity to find the elapsed time after the balloon has traveled d to the right. d d = v0 cos  0 t D tD = → v0 cos  0 Where is the balloon vertically at that time? 2

    d d 1 yballoon = v0 sin  0 t D − gt = v0 sin  0 v − g  = d tan  0 − 2 g   v0 cos  0  v0 cos  0   v0 cos  0  Where is the boy vertically at that time? Note that H = d tan  0 . d

2 D

1 2

    d d 1 yboy = H − gt = H − g   = d tan  0 − 2 g    v0 cos  0   v0 cos  0  Note that yballoon = yboy , and so the boy and the balloon are at the same height and the same 1 2

2 D

1 2

horizontal location at the same time. Thus they collide! 59. Choose upward to be the positive y-direction. The origin is the point from which the football is kicked. The initial speed of the football is v0 = 20.0 m s . We have v y 0 = v0 sin 37.0 = 12.04 m s , y0 = 0, and a y = −9.80 m s2 . In the horizontal direction, v x = v0 cos 37.0 = 15.97 m s , and x = 36.0 m. The time of flight to reach the goalposts is found from the horizontal motion at constant speed. x = v x t → t = x v x = 36.0 m 15.97 m s = 2.254 s Now use this time with the vertical motion data and Eq. 2–12b to find the height of the football when it reaches the horizontal location of the goalposts. 2 y = y0 + v y 0t + 12 a y t 2 = 0 + (12.04 m s )( 2.254 s ) + 12 −9.80 m s 2 ( 2.254 s ) = 2.24 m

(

)

Since the ball’s height is less than 3.05 m, the football does not clear the bar . It is 0.81 m too low when it reaches the horizontal location of the goalposts. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

To find the distances from which a score can be made, redo the problem (with the same initial conditions) to find the times at which the ball is exactly 3.05 m above the ground. Those times would correspond with the maximum and minimum distances for making the score. Use Eq. 2–12b. y = y0 + v y 0 t + 12 a y t 2 → 3.05 = 0 + (12.04 m s ) t + 12 −9.80 m s 2 t 2 →

(

)

(12.04 ) − 4 ( 4.90 )( 3.05) 4.90t − 12.04t + 3.05 = 0 → t = = 2.1703s, 0.2868s 2 ( 4.90 ) x = v t = 15.97 m s ( 0.2868 s ) = 4.580 m ; x = v t = 15.97 m s ( 2.1703s ) = 34.660 m 1

12.04 

So the kick must be made in the range from 4.6 m to 34.7 m . 60. Choose the origin to be at the bottom of the hill, just where the incline starts. The equation of the line describing the hill is y2 = x tan  . The equations of the motion of the object are y1 = v0 y t + 12 a y t 2 and x = v 0 x t , with v0 x = v0 cos  and v0 y = v0 sin  . Solve the horizontal

equation for the time of flight, and insert that into the vertical projectile motion equation. t=

 x  gx 2 x tan → y1 = v0 sin  − g =  −  2v02 cos 2  v0 cos   v0 cos  

= v0 x v0 cos  Equate the y-expressions for the line and the parabola to find the location where the two x-coordinates intersect. gx 2 gx → tan  − tan  = 2 → x tan  = x tan  − 2 2 2v0 cos  2v0 cos 2 

( tan  − tan  )

1 2

2v02 cos 2 

g This intersection x-coordinate is related to the desired quantity d by x = d cos . d cos  = ( tan  − tan  )

2v02 cos 2 

2v02

(

)

sin  cos  − tan  cos 2  g g cos  To maximize the distance, set the derivative of d with respect to  equal to 0, and solve for  . d (d ) 2v02 d sin  cos  − tan  cos 2  = d g cos  d → d=

(

= =

2v02 g cos 

)

sin  ( − sin  ) + cos  ( cos  ) − tan  ( 2 ) cos  ( − sin  )

2v02

2v 2

0  − sin 2  + cos 2  + 2 tan  cos  sin   = cos 2 + sin  tan   = 0 g cos   g cos 



1    tan   This expression can be confusing, because it would seem that a negative sign enters the solution. In order to get appropriate values, 180 or  radians must be added to the angle resulting from the inverse tangent operation, to have a positive angle. Thus a more appropriate expression would be the following: cos 2 + sin  tan  = 0 →  = 12 tan −1  −

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

  1    . This can be shown to be equivalent to  = + , because  = 12  + tan −1  −   2 4  tan       1   tan −1  − = tan −1 ( − cot  ) = cot −1 cot  − =  − .  2 2  tan   61. See the diagram. Solve for R, the horizontal range, which is the horizontal speed times the time of flight. R R = ( v0 cos 0 ) t → t = v0 cos   h = ( v0 sin  0 ) t − 12 gt 2 → R2 − R

2v02 cos 2  0 tan  g

R= =

1 2

gt 2 − ( v0 sin  0 ) t + h = 0 →

2hv02 cos 2  0 g

=0 2

 2v 2 cos 2  0 tan   2hv02 cos 2  0   0 − 4  g g   2

v0 cos  0 

v sin  0  v02 sin 2  0 − 2 gh 

 0

 g Which sign is to be used? We know the result if h = 0 from the level horizontal range formula. v cos  0 Substituting h = 0 gives R = 0  v0 sin  0  v0 sin  0 . To agree with the level horizontal g range formula, we must choose the + sign, and so R =

v0 cos  0  g

v sin  0 + v02 sin 2  0 − 2 gh  .

 0



We see from this result that if h > 0, the range will shorten, and if h < 0, the range will lengthen. 62. Call the direction of the boat relative to the water the positive direction. For the jogger moving towards the bow, we have the following: v =v +v = 2.5m s î + 8.8 m s î = 11.3m s î . jogger rel. water

jogger rel. boat

boat rel. water

For the jogger moving towards the stern, we have the following. v =v +v = −2.5m s î + 8.8 m s î = 6.3m s î jogger rel. water

jogger rel. boat

boat rel. water

63. Call the direction of the flow of the river the x-direction, and the direction of Huck walking relative to the raft the y-direction. v Huck = v Huck + v raft rel. = 0.70ˆj m s + 1.50ˆi m s rel. bank

rel. raft

bank

(

)

= 1.50ˆi + 0.70ˆj m s Magnitude: vHuck

= 1.502 + 0.702 = 1.66 m s

rel. bank

Direction:  = tan −1

0.70 1.50

= 25relative to river

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

64. From Fig. 3–33 in Example 3–13, it is seen that vboat rel. = vboat rel. cos  = (1.85 m s ) cos 40.4 = 1.41m s . shore

water

65. The boat is traveling directly across the stream, with a heading of  = 23.5 upstream, and speed of vboat rel. = 4.30 m s . water

(a) v water rel. = vboat rel. sin  = ( 4.30 m s ) sin 23.5 = 1.71m s water

shore

(b) vboat rel. = vboat rel. cos  = ( 4.30 m s ) cos 23.5 = 3.94 m s water

shore

66. Call the direction of the boat relative to the water the x-direction, and upward the y-direction. Also see the diagram. v passenger = v passenger + v boat rel. rel. water

rel. boat

water

( = ( 2.12ˆi + 0.42ˆj) m s

)

= 0.60 cos 45ˆi + 0.60 sin 45ˆj m s + 1.70ˆi m s

67. Call east the positive x-direction and north the positive y-direction. Then the following vector velocity relationship exists. (a) v plane rel. = v plane + v air rel. ground

rel. air

ground

(

)

= −688ˆj km h + 85.0 cos 45.0ˆi + 85.0 sin 45.0ˆj km h

(

)

= 60.1ˆi − 628ˆj km h v plane rel. =

( 60.1km h ) + ( −628 km h ) = 631km h 2

ground

 = tan −1

60.1

= −5.47 = 5.47 east of south −628 (b) The plane is away from its intended position by the distance the air has caused it to move. The wind speed is 85.0 km/h, so after 11.0 min the plane is off course by the following amount.  1h  d = vt = ( 85.0 km h )(11.0 min )   = 15.6 km .  60 min 

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

68. Call east the positive x-direction and north the positive y-direction. Then the following vector velocity relationship exists. v plane rel. = v plane + v air rel. → ground

rel. air

ground

(

)

−v plane rel. ˆj = −688sin  ˆi + 688 cos  ˆj km h ground

(

)

+ 85.0 cos 45.0ˆi + 85.0 sin 45.0ˆj km h Equate x components in the above equation. 0 = −688sin  + 85.0 cos 45.0 →

 = sin −1

85.0 cos 45.0 688

= 5.01, west of south

69. Choose the x-direction to be the direction of train travel (the direction the passenger is facing) and choose the y-direction to be up. This relationship exists among the velocities: v rain rel. = v rain rel. + v train rel. . From the diagram, find the ground

train

ground

expression for the speed of the raindrops. v train rel. vT v ground = → vrain rel. = T tan  = vrain rel. vrain rel. tan  ground ground

ground

70. From the diagram,  = tan −1 118 m 265 m = 24.0. Equate the vertical components of the velocities in the diagram to find the speed of the boat relative to the shore. v boat rel. cos  = vboat rel. sin 45.0 → shore

water

v boat rel. = ( 2.80 m s )

sin 45.0

= 2.17 m s cos 24.0 Equate the horizontal components of the velocities. v boat rel. sin  = v boat rel. cos 45.0 − v water → shore

shore

v water rel. shore

water

rel. shore

= v boat rel. cos 45.0 − v boat rel. sin  water

shore

= ( 2.80 m s ) cos 45.0 − ( 2.17 m s ) sin 24.0 = 1.10 m s 71. Call the direction of the flow of the river the x-direction, and the direction straight across the river the y-direction. Call the location of the swimmer’s starting point the origin. v swimmer = v swimmer + v water rel. = 0.60 m s ˆj + 0.50 m s ˆi rel. shore

rel. water

shore

(a) Since the swimmer starts from the origin, the distances covered in the x- and y-directions will be exactly proportional to the speeds in those directions. x v x t v x x 0.50 m s = = → = → x = 46 m y v y t v y 55 m 0.60 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The time is found from the constant velocity relationship for either the x- or y-directions. y 55 m y = v y t → t = = = 92 s v y 0.60 m s 72. (a) Call the direction of the flow of the river the x-direction, and the direction straight across the river the y-direction. v water rel. 0.50 m s 0.50 shore sin  = = →  = sin −1 = 56.44  56 0.60 vswimmer 0.60 m s rel. water

(b) From the diagram her speed with respect to the shore is found as follows. vswimmer = vswimmer cos  = ( 0.60 m s ) cos 56.44 = 0.332 m s rel. shore

rel. water

The time to cross the river can be found from the constant velocity relationship. y 55 m y = v y t → t = = = 170 s = 2.8 min v y 0.332 m s 73. Call east the positive x-direction and north the positive y-direction. From the first diagram, this relative velocity relationship is seen. v car 1 rel. = v car 1 rel. + v car 2 rel. → street

car 2

street

(

)

v car 1 rel. = v car 1 rel. − v car 2 rel. = 35ˆj km h − 55ˆi km h = −55ˆi + 35ˆj km h car 2

street

For the other relative velocity relationship: v car 2 rel. = v car 2 rel. + v car 1 rel. → street

car 1

street

(

)

v car 2 rel. = v car 2 rel. − v car 1 rel. = 55ˆi km h − 35ˆj km h = 55ˆi − 35ˆj km h car 1

street

Notice that the two relative velocities are opposites of each other: v car 2 rel. = − v car 1 rel. . car 1

car 2

74. See the diagram for the orientation of the velocities. Apply the law of sines to the triangle formed by the three vectors. vplane vair rel. vair rel. rel. air ground ground = → sin  = sin128 → vplane sin128 sin rel. air

 vair rel.   82 km h  ground    = sin  sin128 = sin −1  sin128  = 6.6   560 km h   vplane  rel. air   −1

So the plane should head in a direction of 38.0 + 6.6 = 44.6north of east .

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

75. The river is flowing to the right in the diagram. From the diagram,  = 132.0. Use the law of sines to determine the angle  . v boat

v water rel.

rel. water

sin

shore

sin

→

 v water rel.   5.60 km h  shore  sin  = sin −1  sin132.0  =13.37  = sin v   18.0 km h   boat   rel. water  −1

(a) The boat should head at the angle of 42.0 + 13.37 = 55.4 upstream . (b) The across-river components of v boat rel. and v boat rel. are equal. Use that to solve for vboat rel. . shore

v boat v boat

shore

cos ( + 42.0 ) →

cos 42.0 = v boat

rel. shore

shore

rel. water

= (18.0 km h )

rel. shore

cos 55.37 cos 42.0

= 13.76 km h  13.8 km h

vboat

cos 42.0 =

rel. shore

1.45 km t

→ t =

1.45 km vboat

cos 42.0

1.45 km 13.76 km h ( cos 42.0 )

= 0.142 h

rel. shore

The time is also equal to 8.51 min. 76. Let east be the positive x-direction, north be the positive y-direction, and up be the positive z-direction. Then the plumber’s resultant displacement in component notation is D = 55m ˆi − 38 m ˆj − 12 m kˆ . Since this is a 3-



Dxy

dimensional problem, it requires 2 angles to determine his location (similar to latitude and longitude on the surface of the Earth). For the xy (horizontal) plane, see the first figure. D −38  = tan −1 y = tan −1 = −35 = 35 south of east Dx 55

Dxy =

Dx2 + Dy2 =

one leg, and the vertical displacement D z as the other leg. See the second figure, and the following calculations. −12 m D = −12.3 = 12 below the x -axis  2 = tan −1 z = tan −1 55 m Dx

Dxy2 + Dz2 =

( 55)2 + ( −38)2 = 66.9 m  67 m

For the vertical motion, consider another right triangle, made up of D x as

Dx2 + Dy2 + Dz2 =

Dx 

( 55)2 + ( −38)2 + ( −12 )2 = 68 m

The result is that the displacement is 68 m , at an angle of 35 south of east , and 12 below the x -axis .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

77. Assume that the golf ball takes off and lands at the same height, so that the level horizontal range formula derived in the text can be applied. The only variable is to be the acceleration due to gravity. REarth = v02 sin 2 0 g Earth RMoon = v02 sin 2 0 g Moon REarth RMoon

v02 sin 2 0 g Earth v sin 2 0 g Moon 2 0

1 g Earth 1 g Moon

(

g Moon g Earth

32 m 180 m

= 0.1778

→

)

g Moon = 0.1778 g Earth = 0.1778 9.80 m s 2 = 1.742 m s 2  1.7 m s 2

78. Work in the frame of reference in which the train is at rest. Then, relative to the train, the car is moving at 20 km/h. The car has to travel 1.4 km in that frame of reference to pass the train, and so the time to pass can be found from the constant horizontal velocity relationship. x 1.40 km  3600 s  x = v x t → tsame = =   = 252 s  250 s 20 km h  1h  ( v x )same direction direction

The car travels 1.4 km in the frame of reference of the stationary train, but relative to the ground, the car is traveling at 95 km/h. Thus relative to the ground the car travels this distance:  1h  x = v x tsame = ( 95 km h )( 252 s )   = 6.65 km  6.7 km direction  3600 s  If the car and train are traveling in opposite directions, the velocity of the car relative to the train is 170 km/h. Thus the time to pass will be as follows. Note that the relative speed has 3 significant figures, due to addition. x 1.40 km  3600 s  topposite = = = 29.64 s  29.6 s ( v x )opposite 170 km h  1 h  direction direction

The distance traveled by the car relative to the ground is calculated.  1h  x = v x topposite = ( 95 km h ) ( 29.64s )   = 0.78 km direction  3600 s  79. The minimum speed will be that for which the ball just clears the fence; i.e., the ball has a height of 8.0 m when it is 98 m horizontally from home plate. The origin is at home plate, with upward as the positive y direction. For the ball, y0 = 1.0 m, y = 8.0 m, a y = − g , v y 0 = v0 sin 0 , v x = v0 cos  0 , and  0 = 36.

0 y0 = 1.0 m

y = 8.0 m

x = 98 m See the diagram (not to scale). For the constant-velocity horizontal x . For the vertical motion, apply Eq. 2–12b. motion, x = v x t = v0 cos  0 t , and so t = v0 cos  0

y = y0 + v y 0t + 12 a y t 2 = y0 + v0 ( sin 0 ) t − 12 gt 2

Substitute the value of the time of flight for the first occurrence only in the above equation, and then solve for the time. x − 12 gt 2 → y = y0 + v0 t sin  0 − 12 gt 2 → y = y0 + v0 sin  0 v0 cos  0

 1.0 m − 8.0 m + ( 98 m ) tan 36   y0 − y + x tan 0  = 2   = 3.620s g 9.80 m s 2    

2

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

Finally, use the time with the horizontal range to find the initial speed. 98 m x x = v0 cos  0 t → v0 = = = 33 m s t cos  0 ( 3.620 s ) cos 36 80. See the sketch of the geometry. We assume that the hill is sloping downward to the right. If we take the point where the child jumps as the origin, with the x-direction 1.3m positive to the right and the y-direction positive upwards, 12 then the equation for the hill is given by y = − x tan12. The path of the child (shown by the dashed line) is that of projectile motion. With the same origin and coordinate system, the horizontal motion of the child is given by x = v0 cos15 ( t ) , and the vertical motion of the child will be given by Eq. 2–12b, y = v0 sin15t − 12 gt 2. The landing point of the child is given by xlanding = 1.3cos12 and ylanding = −1.3sin12. Use the horizontal motion and landing point to find an expression for the time the child is in the air, and then use that time to find the initial speed. 1.3cos12 x , tlanding = x = v0 cos15 ( t ) → t = v0 cos15 v0 cos15 Equate the y-expressions, and use the landing time. We also use the trigonometric identity that sin12 cos15 + sin15 cos12 = sin (12 + 15 ) .

ylanding = yprojectile →

2 − 1.3sin12 = v0 sin15tlanding − 12 gtlanding →

 1.3cos12  −1.3sin12 = v0 sin15 − g  v0 cos15  v0 cos15  1.3cos12

v02 = 12 g

→

1 2

cos 2 12  1.3

   → v0 = 3.728 m s  3.7 m s sin 27  cos15 

81. Consider the downward vertical component of the motion, which will occur in half the total time. Take the starting position to be y = 0, and the positive direction to be downward. Use Eq. 2–12b with an initial vertical velocity of 0. 2

y = y0 + v 0 y t + a y t 1 2

→ h = 0 + 0 + gt 1 2

2 down

 t  9.80 2 = g  = t = 1.225t 2  1.2t 2 2 8   1 2

As can be seen from the equation, by starting the analysis from the “top” point of the motion, the initial vertical speed is 0. This eliminates the need to know the original launch speed or direction in that calculation. We then also realize that the time for the object to rise is the same as the time for it to fall, so we only have to analyze the downward motion. 82. Choose downward to be the positive y-direction. The origin is at the point from which the divers push off the cliff. In the vertical direction, the initial velocity is v y 0 = 0, the acceleration is

a y = 9.80 m s 2 , and the displacement is 35 m. The time of flight is found from Eq. 2–12b.

(

)

y = y0 + v y 0t + 12 a y t 2 → 35 m = 0 + 0 + 12 9.80 m s 2 t 2 → t =

2 ( 35 m ) 9.8 m s 2

= 2.7 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity. x = v x t → v x = x t = 5.0 m 2.7 s = 1.9 m s 83. The level horizontal range formula derived in the text can be used to find the launching velocity of the grasshopper.

v02 sin 2 0 g

→ v0 =

Rg sin 2 0

( 0.80 m ) ( 9.80 m s 2 ) sin 90

= 2.8 m s

Since there is no time between jumps, the horizontal velocity of the grasshopper is the horizontal component of the launching velocity. v x = v0 cos  0 = ( 2.8 m s ) cos 45o = 2.0 m s 84. Work in the frame of reference in which the car is at rest at ground level. In this reference frame, the  1m s  helicopter’s horizontal speed is 208 km h − 135 km h = 73km h   = 20.28 m s . For  3.6 km h  the vertical motion, choose the level of the helicopter to be the origin, and downward to be positive. Then the package’s y-displacement is y = 78.0 m, v y 0 = 0, and a y = g. The time for the package to fall is calculated from Eq. 2–12b. y = y0 + v y 0 t + 12 a y t 2

→

(

)

78.0 m = 12 9.80 m s 2 t 2

→ t=

2 ( 78.0 m ) 9.80 m s 2

= 3.99 sec

The horizontal distance that the package must move, relative to the “stationary” car, is found from the horizontal motion at constant velocity. x = v x t = ( 20.28 m s )( 3.99 s ) = 80.9 m Thus the angle under the horizontal for the package release will be as follows.  78.0 m   y   = tan −1   = tan −1   = 43.95  44  x   80.9 m  85. The proper initial speeds will be those for which the ball has traveled a horizontal distance somewhere between 10.78 m and 11.22 m while it changes height from 2.10 m to 3.05 m y = 0.95m 0 with an initial (shooting) angle of 38.0o. Choose the origin to be at the shooting location of the basketball, with upward x = 10.78 m − 11.22 m as the positive y-direction. Then the vertical displacement is 2 y = 0.95m, a y = −9.80 m s , v y 0 = v0 sin 0 , and the (constant) x-velocity is v x = v0 cos  0 . See the diagram (not to scale). For the constant-velocity horizontal motion, x = v x t = v0 cos  0 t and so t =

x v0 cos  0

. For the vertical motion, apply Eq. 2–12b.

y = y0 + v y 0t + 12 a y t 2 = ( v0 sin  ) t − 12 gt 2

Substitute the expression for the time of flight and solve for the initial velocity. g ( x )  x  y = ( v0 sin  ) t − gt = v0 sin  − g  = x tan  − 2 2v0 cos 2  0 v0 cos  0  v0 cos  0  1 2

x

1 2

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

v0 =

g ( x )

2 cos 2  0 ( − y + x tan  )

For x = 11.00 m − 0.22 m = 10.78 m, the shortest shot:

( 9.80 m s ) (10.78 m ) v = = 11.078 m s  11.1m s . 2 cos 38.0 ( −0.95 m + (10.78 m ) tan 38.0 ) 2

For x = 11.00 m + 0.22 m = 11.22 m, the longest shot:

( 9.80 m s ) (11.22 m ) v = = 11.274 m s  11.3 m s . 2 cos 38.0 ( −0.95 m + (11.22 m ) tan 38.0 ) 2

86. (a) Take the ground to be the y = 0 level, with upward as the positive direction. Use Eq. 2–12b to solve for the time, with an initial vertical velocity of 0. y = y0 + v0 y t + 12 a y t 2 → 150 m = 910 m + 12 ( −9.80 m s 2 ) t 2 →

2 (150 − 910 )

( −9.80 m s ) 2

= 12.45s  12 s

(b) The horizontal motion is at a constant speed, since air resistance is being ignored. x = v x t = ( 4.0 m s )(12.45s ) = 49.8 m  5.0  101 m

87. First we find the time of flight for the ball. From that time we can calculate the vertical speed of the ball. From that vertical speed we can calculate the total speed of the ball, and the % change in the speed. We chose the downward direction to be positive for vertical motion.  1m s  v x = v0 x = (130 km h )   = 36.11m s  3.6 km h  x = v x t

→

(

x vx

18 m 36.11m s

= 0.498 s

)

v y = v0 y + at = 9.80 m s 2 ( 0.498 s ) = 4.88 m s v=

v x2 + v y2 =

% change =

( 36.11m s ) + ( 4.88 m s ) = 36.44 m s

v − v0 v0

 100 =

36.44 − 36.11 36.11

 100 = 0.91%

88. Choose the origin to be the point from which the projectile is launched, and choose upward as the positive y-direction. The y-displacement of the projectile is 135 m and the horizontal range of the projectile is 195 m. The acceleration in the y-direction is a y = − g , and the time of flight is 6.6 s. The horizontal velocity is found from the horizontal motion at constant velocity. x 195 m x = v x t → v x = = = 29.55 m s t 6.6 s Calculate the initial y-velocity from the given data and Eq. 2–12b. 2 y = y0 + v y 0t + 12 a y t 2 → 135 m = v y 0 ( 6.6 s ) + 12 −9.80 m s 2 ( 6.6 s ) → v y 0 = 52.79 m s

(

)

Thus, the initial velocity and direction of the projectile are as follows. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

v0 = v x2 + v y2 0 =

 = tan −1

vy0 vx

Instructor Solutions Manual

( 29.55 m s ) + ( 52.79 m s ) = 60.4978 m s  6.0  101 m s 2

= tan −1

52.79 m s 29.55 m s

= 61

89. Choose the origin to be the water level directly underneath the diver when she left the board. Choose upward as the positive y-direction. For the diver, y0 = 3.5 m , the final y-position is y = 0.0 m (water level), a y = − g , the time of flight is t = 1.3 s, and the horizontal displacement is

x = 3.0 m. (a) The horizontal velocity is determined from the horizontal motion at constant velocity. x 3.0 m x = v x t → v x = = = 2.308 m s t 1.3s The initial y-velocity is found using Eq. 2–12b.

(

)

y = y0 + v y 0t + 12 a y t 2 → 0 m = 3.5 m + v y 0 (1.3s ) + 12 −9.80 m s 2 (1.3s )

→

v y 0 = 3.678 m s Thus, the velocity in both vector and magnitude / direction format are as follows. 2 2 v = v 2 + v 2 = ( 2.308 m s ) + ( 3.678 m s ) = 4.3 m s v = 2.3ˆi + 3.7 ˆj m s 0

(

 = tan −1

)

vy0 vx

= tan −1

3.678 m s 2.308 m s

= 57 above the horizontal

(b) The maximum height will be reached when v y = 0. Use Eq. 2–12c.

(

)

v y2 = v y2 0 + 2ay → 0 = ( 3.678 m s ) + 2 −9.80 m s 2 ( ymax − 3.5 m ) → 2

ymax = 4.2 m above the water. (c) To find the velocity when she enters the water, the horizontal velocity is the (constant) value of v x = 2.308 m s. The vertical velocity is found from Eq. 2–12a.

(

)

v y = v y 0 + at = 3.678 m s + −9.80 m s 2 (1.3s ) = −9.062 m s. The velocity is as follows. v = 2.3ˆi − 9.1ˆj m s f

(

)

vf = v x2 + v y2 =

 f = tan −1

vfy vfx

( 2.308 m s ) + ( −9.062 m s ) = 9.351m s  9.4 m s

= tan −1

−9.062 m s 2.308 m s

= −76 ( below the horizontal )

90. Find the time of flight from the vertical data, using Eq. 2–12b. Call the floor the y = 0 location, and choose upwards as positive. y = y0 + v0 y t + 12 a y t 2 → 3.05 m = 2.40 m + (12 m s ) sin 35t + 12 −9.80 m s 2 t 2

(

)

4.90t − 6.883t + 0.65 m = 0 → 2

6.883  6.8832 − 4 ( 4.90 )( 0.65 ) 2 ( 4.90 )

= 1.303s, 0.102 s

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(a) Use the larger time for the time of flight. The shorter time is the time for the ball to rise to the basket height on the way up, while the longer time is the time for the ball to be at the basket height on the way down. x = v x t = v0 ( cos 35 ) t = (12 m s )( cos 35 )(1.303s ) = 12.81m  13 m (b) The angle to the horizontal is determined by the components of the velocity. v x = v0 cos  0 = 12 cos 35 = 9.830 m s v y = v y 0 + at = v0 sin  0 − gt = 12 sin 35 − 9.80 (1.303) = −5.886 m/s

 = tan −1

vy vx

= tan −1

−5.886 9.830

= −30.9  −31 , 31 below the horizontal.

91. (a) Choose downward to be the positive y-direction. The origin is the point where the arrow leaves the bow. In the vertical direction, v y 0 = 0, y0 = 0, and a y = 9.80 m s 2. In the horizontal direction, x = 38.0 m and v x = 21.3 m s. The time of flight is found from the horizontal motion at constant velocity. x = v x t → t = x v x = 38.0 m 21.3 m s = 1.784 s This time can now be used in Eq. 2–12b to find the vertical drop of the arrow. 2 y = y0 + v y 0t + 12 a y t 2 → y = 0 + 0 + 12 ( 9.80 m s 2 ) (1.784 s ) = 15.6 m (b) For the arrow to hit the target at the same level, the level horizontal range formula derived in the text applies. The range is 38.0 m, and the initial velocity is 21.3 m/s. Solving for the angle of launch results in the following. ( 38.0 m ) ( 9.80 m s 2 ) v02 sin 2 0 Rg −1 1 → sin 2 0 = 2 →  0 = 2 sin = 27.6 R= 2 g v0 ( 21.3 m s ) Because of the symmetry of the range formula, there is also an answer of the complement of the above answer, which would be 62.4. The larger angle is less reasonable – an archer rarely points the bow at such a steep angle. 92. (a) Call the upward direction positive for the vertical motion. Then the velocity of the ball relative to a person on the ground is the vector sum of the horizontal and vertical motions. The horizontal velocity is v x = 12.0 m s and the vertical velocity is v y = 3.0 m s . v = 12.0 m s ˆi + 3.0 m s ˆj → v =

 = tan −1

3.0 m s 12.0 m s

(12.0 m s ) + ( 3.0 m s ) = 12.4 m s 2

= 14 above the horizontal .

(b) The only change is the initial vertical velocity, and so v y = −3.0 m s. v = 12.0 m s ˆi − 3.0 m s ˆj → v =

 = tan −1

−3.0 m s 12.0 m s

(12.0 m s ) + ( −3.0 m s ) = 12.4 m s 2

= 14 below the horizontal .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

93. Let the launch point be the origin of coordinates, with right and upwards as the positive directions. The equation of the line representing the ground is ygnd = − x. The equations representing the motion of the rock are xrock = v0 t and yrock = − 12 gt 2 , which can be combined into yrock = − 12

2 xrock . v02 Find the intersection (the landing point of the rock) by equating the two expressions for y, and so finding where the rock meets the ground. 2v02 g 2 x 2v0 2 (15 m s ) 1 yrock = ygnd → − 2 2 x = − x → x = → t= = = = 3.1s v0 v0 9.80 m s 2 g g

94. Choose the origin to be the point at ground level directly below where the ball was hit. Call upwards the positive y-direction. For the ball, we have v0 = 28 m s,  0 = 64, a y = − g, y0 = 0.9 m, and y = 0.0 m. (a) To find the horizontal displacement of the ball, the horizontal velocity and the time of flight are needed. The (constant) horizontal velocity is given by v x = v0 cos  0. The time of flight is found from Eq. 2–12b. y = y0 + v y 0t + 12 a y t 2 → 0 = y0 + v0 sin  0t − 12 gt 2 →

−v0 sin  0  v02 sin 2  0 − 4 ( − 12 g ) y0 2 ( − 12 g )

− ( 28 m s ) sin 64 

( 28 m s ) sin 2 64 − 4 ( − 12 ) ( 9.80 m s 2 ) ( 0.9 m ) 2 ( − 12 ) ( 9.80 m s 2 ) 2

= 5.171s, − 0.0355s Choose the positive time, since the ball was hit at t = 0. The horizontal displacement of the ball will be found by the constant velocity relationship for horizontal motion.

x = v x t = v0 cos  0t = ( 28 m s )( cos 64 )( 5.171s ) = 63.47 m  63 m

(b) The center fielder catches the ball right at ground level. He ran 105 m – 63.47 m = 41.53 m to catch the ball, so his average running speed would be as follows. d 41.53 m vavg = = = 8.031m s  8.0 m s t 5.171s 95. Choose the origin to be the point at the top of the building from which the ball is shot, and call upwards the positive y-direction. The initial velocity is v0 = 24 m s at an angle of  0 = 42. The acceleration due to gravity is a y = − g. (a) v x = v0 cos  0 = ( 24 m s ) cos 42 = 17.84 m s  18 m s v y 0 = v0 sin  0 = ( 24 m s ) sin 42 = 16.06 m s  16 m s

(b) Since the horizontal velocity is known and the horizontal distance is known, the time of flight can be found from the constant velocity equation for horizontal motion. x 65 m x = v x t → t = = = 3.643s v x 17.84 m s With that time of flight, calculate the vertical position of the ball using Eq. 2–12b. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(

)

y = y0 + v y 0t + 12 a y t 2 = (16.06 m s )( 3.643s ) + 12 −9.80 m s 2 ( 3.643s )

= −6.52 m  −6.5 m So the ball will strike 6.5 m below the top of the building. 96. (a) For the upstream trip, the boat will cover a distance of D 2 with a net speed of v − u , so the time is t1 =

D 2 v −u

2 (v − u )

. For the downstream trip, the boat will cover a distance of D 2

with a net speed of v + u , so the time is t2 = round trip will be t = t1 + t2 =

D 2(v − u )

D 2 v+u

D 2(v + u )

D 2 (v + u )

. Thus the total time for the

(v − u ) 2

(b) For the boat to go directly across the river, it must be angled against the current in such a way that the net velocity is straight across the river, as in the diagram. This equation must be satisfied: v boat rel. = v boat rel. + v water rel. = v + u. shore

Thus vboat rel. =

water

shore

v water rel. = u shore

v boat rel. = v



v boat rel. shore

water

v 2 − u 2 , and the time to go a distance D 2 across

shore

the river is t1 =

D 2 v −u 2

D 2 v2 − u2

. The same relationship would be in effect for crossing

back, so the time to come back is given by t2 = t1 and the total time is t = t1 + t2 =

v − u2 The speed v must be greater than the speed u. The velocity of the boat relative to the shore when going upstream is v − u. If v  u , the boat will not move upstream at all, and so the first part of the trip would be impossible. Also, in part (b), we see that v is longer than u in the triangle, since v is the hypotenuse, and so we must have v  u. 2

97. Since the ball is being caught at the same height from which it was struck, use the level horizontal range formula to find the horizontal distance the ball travels. R=

v02 sin 2 0

( 28 m s ) sin ( 2  55 ) 2

= = 75.175 m g 9.80 m s 2 Then as seen from above, the location of home plate, the point where the ball must be caught, and the initial location of the outfielder are shown in the diagram. The dark arrow shows the direction in which the outfielder must run. The length of that distance is found from the law of cosines as applied to the triangle.

x= =

a 2 + b 2 − 2ab cos  75.1752 + 852 − 2 ( 75.175) ( 85) cos 22 = 32.048 m

The angle  at which the outfielder should run is found from the law of sines.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

sin 22

sin 

Instructor Solutions Manual

 75.175 sin 22  = 61.49 or 118.51  = sin −1   32.048 m 75.175 m  32.048  2 2 2 Since 75.175  85 + 32.048 , the angle must be acute, so we choose  = 61.49. =

→

Now assume that the outfielder’s time for running is the same as the time of flight of the ball. The time of flight of the ball is found from the horizontal motion of the ball at constant velocity. R 75.175 m R = v x t = v0 cos  0 t → t = = = 4.681s v0 cos  0 ( 28 m s ) cos 55 Thus the average velocity of the outfielder must be vavg =

d t

32.048 m 4.681s

= 6.8 m s at an angle of

61 relative to the outfielder’s line of sight to home plate.

98. The lifeguard will be carried downstream at the same rate as the child. Thus only the horizontal 45 m = motion need be considered. To cover 45 meters horizontally at a rate of 2.0 m/s takes 2m s 22.5s  23s for the lifeguard to reach the child. During this time they would both be moving

downstream at 1.0 m/s, and so would travel (1.0 m s )( 22.5s ) = 22.5m  23m downstream. 99. The acceleration is the derivative of the velocity. dv a= = 4.5 m s2 ˆj dt Since the acceleration is constant, we can use Eq. 3–13b. r = r0 + v 0t + 12 at 2 = 2.5 ˆi − 3.1ˆj + −3.0 ˆi t + 12 4.5 ˆj t 2

( ) ( ) = ( 2.5 − 3.0t ) m ˆi + ( −3.1 + 2.25t ) m ˆj

(

)

The shape is parabolic , with the parabola opening in the y-direction. 100. We choose to initially point the boat downstream at an angle of  relative to straight across the river, because then all horizontal velocity components are in the same direction, and the algebraic signs might be less confusing. If the boat should in reality be pointed upstream, the solution will give a negative angle. We use v BW = 1.60 m s, the speed of the boat relative to the water (the rowing speed); v WS = 0.80 m s, the speed of the water relative to the shore (the current); and v R = 3.00 m s, his running speed. The width of the river is w = 1300 m, and the length traveled along the bank is l. The time spent in the water is tW , and the time running is tR. The actual vector velocity of the boat is v BS = v BW + v WS. That vector addition is illustrated on the diagram (not drawn to scale).

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

The distance straight across the river (w) is the velocity component across the river, times the time in the water. The distance along the bank (l) is the velocity component parallel to the river, times the time in the water. The distance along the bank (as he runs back to the point straight across the river) is also his running speed times the time running. These three distances are expressed below. w = ( v BW cos  ) t W ; l = ( v BW sin  + v WS ) t W ; l = v R tR The total time is t = t W + t R , and needs to be expressed as a function of  . Use the distance relations above to write this function.  ( v sin  + v WS )  l ( v sin  + v WS ) tW = t W + BW = t W 1 + BW t = tW + tR = t W +  vR vR vR  

vBW vR cos 

v + v R

+ vBW sin   =

w vBW vR

( v + v ) sec  + v R

To find the angle corresponding to the minimum time, we set

dt d

= =

dt d

tan  

= 0 and solve for the angle.

d 

 w ( vR + v WS ) sec  + vBW tan    d  v BW v R  w

( vR + v WS ) tan  sec  + v BW sec 2   = 0 →

v BW v R 

( v + v ) tan  + v R

sec   sec  = 0 → sec  = 0, sin  = −

v BW v R + v WS

The first answer is impossible, and so we must use the second solution. v BW 1.60 m s sin  = − =− = −0.421 →  = sin −1 ( −0.421) = −24.9 3.00 m s + 0.80 m s v R + v WS To know that this is really a minimum and not a maximum, some argument must be made. The maximum time would be infinity, if he pointed his boat either directly upstream or downstream. Thus, this angle should give a minimum. A second derivative test could be done, but that would be algebraically challenging. A graph of t vs.  could also be examined to see that the angle is a minimum. Here is a portion of such a graph, showing a minimum time of somewhat more than 900 seconds near  = −25. It is also interesting to note from the equations above that the angle for the minimum time is determined only by the velocities, not the actual distance across the river. The time he takes in getting to the final location can be calculated from the angle. 1300 m w tW = = = 895.77 s v BW cos  (1.60 m s ) cos ( −24.9 )

l = ( v BW sin  + v WS ) t W = (1.60 m s ) sin ( −24.9 ) + 0.80 m s  ( 895.77 s ) = 113.12 m tR =

l vR

113.12 m 3.00 m s

= 37.71s

t = t W + tR = 895.77 s + 37.71s = 933.48s  9.33  102 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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Thus he must point the boat 24.9 upstream, taking 895.8 seconds to cross, and landing 113 m from the point directly across from his starting point. Then he runs the 113 m from his landing point to the point directly across from his starting point, in 37.7 seconds, for a total elapsed time of 933 seconds (about 15.6 minutes). 101. We have vcar rel. = 12 m s . Use the diagram, showing the snow falling ground

straight down relative to the ground and the car moving parallel to the ground, and illustrating v snow rel. = v snow rel. + v car rel. , to calculate the ground

car

ground

v snow rel. ground

v car rel.

→ vsnow rel. = (12 m s ) tan 9.0 = 1.901 m s  1.9 m s

ground

tan 9.0 =

car

ground

other speeds.

vsnow rel.

v snow rel.

9.0

vcar rel.

ground

102. First, we find the direction of the straight-line path that the boat must take to pass 150 m to the east of the buoy. See the first diagram (not to scale). We find the net displacement of the boat in the horizontal and vertical directions, and then calculate the angle. x = ( 2800 m ) sin 22.5 + 150 m = 1221.51m y = ( 2800m ) cos 22.5 = 2586.86 m

 = tan −1

y

= tan −1

buoy 150 m

2800 m

boat path

2586.86 m

= 64.7234 x 1221.51m This angle gives the direction that the boat must actually travel, so it is the direction of the velocity of the boat with respect to the shore, v boat rel.. So

(

22.5o



shore

)

v boat rel. = vboat rel. cos  ˆi + sin  ˆj . Then, using the second diagram (also not shore

shore

to scale), we can write the relative velocity equation relating the boat’s travel and the current. The relative velocity equation gives us the following. v boat rel. = v boat rel. + v water rel. → shore

water

(

shore

)

(

v water rel. shore

v boat rel.

)

vboat rel. cos  î + sin  ˆj = 2.1m s cos  î + sin  ˆj + 0.20 m s î → shore

vboat rel. cos  = ( 2.1m s ) cos  + 0.20 m s ; vboat rel. sin  = ( 2.1m s ) sin  shore

v boat rel.

water

shore





shore

These two component equations can then be solved for v boat rel. and  . One technique is to isolate the shore

terms with  in each equation, and then square those equations and add them. That gives a quadratic equation for vboat rel. . We will represent v boat rel. by “x” and drop the units in order to make shore

shore

the equations simpler to write. Once we have a value for v boat rel., we can solve for  . shore

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

cos  =

x cos  − 0.20 2.1

; sin  =

x sin  2.1

→

 x cos  − 0.20   x sin   cos  + sin  =   +  =1 2.1    2.1  2

x 2 cos 2  − 2 ( 0.20 ) x cos  + ( 0.20 ) + x 2 sin 2  = ( 2.1) 2

x 2 − ( 0.40 cos  ) x − 4.37 = x 2 − 0.17079 x − 4.37 = 0 x=

0.17080 

( −0.17080 )2 − 4 (1)( −4.37 )

0.17080  4.1844

= 2.1776 m s , − 2.0068 m s 2 2 We must have v boat rel. > 0 since it is the magnitude of a vector, so vboat rel. = 2.1776 m s. shore

sin  =

x sin 

shore

( 2.1776 m s ) sin 64.723

= 0.93768 →  = sin −1 0.93768 = 69.7 N of E

2.1m s 2.1m s The final answer was very sensitive to the number of significant digits carried in the problem, so we kept more than usual.

CHAPTER 4: Dynamics: Newton’s Laws of Motion Responses to Questions 1.

When you give the wagon a sharp pull forward, the force of friction between the wagon and the child acts on the child to move her forward. But the force of friction acts at the contact point between the child and the wagon – we assume the child is sitting in the wagon. The lower part of the child begins to move forward, while the upper part, following Newton’s first law (the law of inertia), remains almost stationary, making it seem as if the child falls backward. The “backwards” motion is relative to the wagon, not to the ground.

(a) Francesca, standing on the ground beside the truck, will see the box remain motionless while the truck accelerates out from under it. Since there is no friction, there is no net horizontal force on the box and so the box will not gain any speed. Thus Francesca would describe the motion of the box in terms of Newton’s first law – there is no force on the box, so it does not accelerate. (b) Phil, riding on the truck, will see the box appear to accelerate backwards with respect to his frame of reference, which is not inertial. He might even say something about the box being “thrown” backwards from the truck, and try to invoke Newton’s second law to explain the motion of the box. But the source of the force would be impossible to specify. (Phil better hold on to the truck, though; if the truck bed is frictionless, he too will slide off if he is just standing!)

Yes, the net force can be zero on a moving object. If the net force is zero, then the object’s acceleration is zero, but its velocity is not necessarily zero. Instead of classifying objects as “moving” and “not moving,” Newtonian dynamics classifies them as “accelerating” and “not accelerating.” Both zero velocity and constant velocity fall in the “not accelerating” category.

If the acceleration of an object is zero, the vector sum of the forces acting on the object is zero (Newton’s second law), so there can be forces on an object that has no acceleration. For example, a book resting on a table is acted on by gravity and the normal force from the table, but it has zero acceleration, because the vector sum of the two forces on it is 0.

If only one force acts on an object, the net force cannot be zero, and so the object cannot have zero acceleration, by Newton’s second law. It is possible for the object to have zero velocity, but only for an instant. For example (if we neglect air resistance), a ball thrown up into the air has only the force of gravity acting on it. Its speed will decrease while it travels upward, stops, and then falls back to the ground. At the instant the ball is at its highest point, its velocity is zero. However, the ball has a non-zero net force and a non-zero acceleration throughout its flight, including at the top, where its velocity is zero.

(a) A force is needed to bounce the ball back up, because the ball changes direction, and so the ball accelerates. If the ball accelerates, there must be a force. (b) The pavement exerts the force on the golf ball.

As you take a step on the log, your foot exerts a force on the log in the direction opposite to the direction in which you want to move, which pushes the log “backwards.” (The log exerts an equal and opposite force forward on you, by Newton’s third law.) If the log had been on the ground, friction between the ground and the log could have kept the log from moving. However, the log is floating in water, which offers little resistance to the movement of the log as you push it backwards while walking forward.

Chapter 4

Dynamics: Newton’s Laws of Motion

(a) When you first start moving on the bicycle, you need to exert a strong force to accelerate the bike and yourself, as well as overcoming friction. Once you are moving at a constant speed, you only need to exert a force to equal the forces that oppose your motion–friction and air resistance. (b) When the bike is moving at a constant speed, the net force on it is zero. Since friction and air resistance are present, you would slow down if you didn’t pedal to keep the net force on the bike (and you) equal to zero.

When giving a sharp pull, the key is the suddenness of the application of the force. When a large, sudden force is applied to the bottom string, the bottom string will have a large tension in it. Because of the stone’s inertia, it tends to stay at rest, opposing the quick pull. If a slow and steady pull is applied, the tension in the bottom string increases. We approximate that condition as considering the stone to be in equilibrium until the string breaks. The free-body diagram for the stone would look like this diagram. While the stone is in equilibrium, Newton’s second law gives Fup = Fdown + mg. Thus the tension in the upper string is going to be larger than the tension in the lower string because of the weight of the stone, and so the upper string will break first.

Fup stone

Fdown

10. The acceleration of both rocks is found by dividing their weight (the force of gravity on them) by their mass. The 2-kg rock has twice the weight as the 1-kg rock, but also twice the mass as the 1-kg rock, so the acceleration is the same for both. 11. When you pull the rope at an angle, only the horizontal component of the pulling force will be accelerating the box across the table. This is a smaller horizontal force than originally used, and so the horizontal acceleration of the box will decrease. 12. Let us find the acceleration of the Earth, assuming the mass of the freely falling object is m = 1 kg. If the mass of the Earth is M, then the acceleration of the Earth would be found using Newton’s third law and Newton’s second law. FEarth = Fobject → MaEarth = mg → aEarth = g m M Since the Earth has a mass that is on the order of 1025 kg, then the acceleration of the Earth is on the −25 −24 2 order of 10 g, or about 10 m s . This tiny acceleration is undetectable. 13. Because the acceleration due to gravity on the Moon is less than it is on the Earth, an object with a mass of 10 kg will weigh less on the Moon than it does on the Earth. Therefore, it will be easier to lift on the Moon. (When you lift something, you exert a force to oppose its weight.) However, when throwing the object horizontally, the force needed to accelerate it to the desired horizontal speed is proportional to the object’s mass, F = ma. Therefore, you would need to exert the same force to throw the 2-kg object with a given speed on the Moon as you would on Earth. 14. In a tug of war, the team that pushes hardest against the ground wins. It is true that both teams have the same force on them due to the tension in the rope. But the winning team pushes harder against the ground and thus the ground pushes harder on the winning team, making a net unbalanced force. See the free body diagram below. The forces are FT G , the force on team 1 from the ground, FT G , 1

the force on team 2 from the ground, and FTR , the force on each team from the rope.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

) (

Instructor Solutions Manual

)

The net force on both teams FT G − FTR on the winning team , FTR − FT G on the losing team is in 1

the “winning” direction. 15. If you are at rest, the net force on you is zero. Hence the ground exerts a force on you exactly equal to your weight. The two forces acting on you sum to zero, and so you don’t accelerate. If you squat down and then push with a larger force against the ground, the ground then pushes back on you with a larger force by Newton’s third law, and you can then rise into the air. 16. The victim’s head is not actually thrown backwards. If the victim’s car was initially at rest, or even moving forward, the impact from the rear suddenly pushes the car, the seat, and the person’s body forward. The head, being attached by the somewhat flexible neck to the body, can momentarily remain where it was (inertia, Newton’s first law), thus lagging behind the body. The neck muscles must eventually pull the head forward, and that causes the whiplash. To avoid this, use the car’s headrests. Then the head is pushed forward with the rest of the person’s body and the car. 17. (a) (b) (c) (d)

The reaction force has a magnitude of 40 N. The reaction force points downward. The reaction force is exerted on Mary’s hands and arms. The reaction force is exerted by the bag of groceries.

18. The father and daughter will each have the same magnitude force acting on them as they push each other away, according to Newton’s third law. If we assume the young daughter has less mass than the father, her acceleration should be greater (a = F/m). Both forces, and therefore both accelerations, act over the same time interval (while the father and daughter are in contact), so the daughter’s final speed will be greater than her father’s. 19. Assume your weight is W. If you weighed yourself on an inclined plane that is inclined at angle , the bathroom scale would read the magnitude of the normal force between you and the plane, which would be W cos. 20. A weight of 1 N corresponds to 0.225 lb. That’s about the weight of (a), an apple. 21. When you kick a heavy desk or a wall, your foot exerts a force on the desk or wall. The desk or wall exerts a force equal in magnitude on your foot (Newton’s third law). Ouch! 22. (a) The force that causes you to stop quickly is the force of friction between your shoes and the ground. (b) If we assume the top speed of a person to be around 6 m/s (equivalent to about 12 mi/h, or a 5minute mile), and if we assume that it take 2 s to stop, then the maximum rate of deceleration is about 3 m/s². 23. The carton would collapse (a). When you jump, you accelerate upward, so there must be a net upward force on you. This net upward force can only come from the normal force exerted by the carton on you and must be greater than your weight. How can you increase the normal force of a surface on you? According to Newton’s third law, the carton pushes up on you just as hard as you push down on it. That means you push down with a force greater than your weight in order to accelerate upwards. If the carton can just barely support you, it will collapse when you exert this extra force.

Chapter 4

Dynamics: Newton’s Laws of Motion

Responses to MisConceptual Questions 1.

(a) The crate does not accelerate up or down so the net force cannot be vertical. The truck bed is frictionless and the crate is not in contact with any other surface, so there are no horizontal forces. Therefore, no net force acts on the crate. As the truck slows down the crate continues to move forward at constant speed. (How did the crate stay on the truck in the first place, to be able to travel on the truck?)

(d) For George and the truck to move forward from rest they must both experience a positive horizontal acceleration. The horizontal forces acting on George are the friction force of the ground pushing him forward and the truck pulling him backwards. The ground must push Matt forward with a stronger force than the truck is pulling him back. The horizontal forces on the truck are from George pulling the truck forward and the friction from the ground in the opposite direction. For the truck to accelerate forward, the force from George must be greater than the backward force of friction from the ground. By Newton’s third law, the force of the truck on George and the force of George on the truck are equal and opposite. Since (i) the force of the ground on George is greater than the force of the truck on George, (ii) the force of the truck on George is equal to the force of George on the truck, and (iii) the force of George on the truck is greater than the friction force of the ground on the truck, the ground exerts a greater friction on George than on the truck. Symbolically, Fon George  Fon George = Fon truck  Fon truck . by ground

by truck

by George

by ground

(d) In order to hold the backpack up, the rope must exert a vertical force equal to the backpack’s weight, so that the net vertical force on the backpack is zero. The force, F, exerted by the rope on each side of the pack is always along the length of the rope. The vertical component of this force is F sin θ, where θ is the angle the rope makes with the horizontal, and there are two vertical components – one on the left of the pack, and one on the right. The higher the pack is, the smaller θ becomes and the larger F must be to hold the pack up there. No matter how hard you pull, the rope can never be horizontal because it must exert an net upward (vertical) mg component of force to balance the pack’s weight. 2 F sin  = mg → F = 2 sin 

(c) The boat accelerates forward by horizontal forces acting on the boat. The force that the man exerts on the paddles pushes the paddles, but because he and the paddles are part of the boat this force does not accelerate the boat, so (a) is not correct. As the paddle pushes on the water it causes the water to accelerate backwards. This force acts to accelerate the water, not the boat, so (b) is incorrect. By Newton’s third law, as the paddles push the water backwards, the water pushes the paddles (and thus the boat) forward, (c). With the force of the water on the paddles pushing the boat forward, the boat would move even when the water was still, so (d) is also incorrect.

(a, b, d) We assume the road is level, and the only objects in contact are you, the car, and the road. The forces in (a), (b), and (d) are all equal to 400 N in magnitude. (a) You exert a force of 400 N on the car; by Newton’s third law the force exerted by the car on you also has a magnitude of 400 N. (b) Since the car doesn’t move and the only horizontal forces acting on the car are your pushing and the force of friction on the car from the road, Newton’s second law requires these forces to have equal magnitudes (400 N) in the opposite direction. Since the road exerts a force of 400 N on the car by friction, Newton’s third law requires that the friction force on the road from the car is also 400 N.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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(c) The normal force exerted by the road on you will be equal in magnitude to your weight (assuming you are standing vertically and have no vertical acceleration). This force is not required to be 400 N. (d) The car is exerting a 400 N horizontal force on you, and since you are not accelerating, and the only horizontal forces acting on you are the force from the car and the frictional force from the ground, Newton’s second law requires that the ground must be exerting an equal and opposite horizontal force. Therefore, the magnitude of the friction force exerted by the road on you is 400 N. 6.

(c) The weight of the skier can be broken into components parallel to and perpendicular to the slope. The normal force will be equal to the perpendicular component of the skier’s weight. For a non-zero slope this component is always less than the weight of the skier.

(b) The force of the golf club acting on the ball only acts when the two objects are in contact, not as the ball flies through the air. The force of gravity acts on the ball throughout its flight. Air resistance is to be neglected so there is no force acting on the ball due to its motion through the air.

(c) Since the net force is now zero, Newton’s first law requires that the object will move in a straight line at constant speed. A non-zero net force would be needed to bring the object to rest.

(d) By Newton’s third law the force you exert on the box must be equal in magnitude to the force the box exerts on you. All of the other statements are in conflict with Newton’s third law.

10. (c) The reading on the scale will be 5 N. If the left side of the scale were attached to a fixed wall instead of the left hanging box, the right box would still be stationary, and the scale would then read the weight of the box. The tension in the strings would be the same in that scenario as they are in the diagram. The same argument would apply for hanging the scale from the ceiling (and eliminating the left box). 11. (c) The tension in the rope would be 500 N. By Newton’s third law, the force that each person pulls on the repo is equal to the force that the rope allies to the person, and that is the tension force. This is also explained in the text, Section 4–7, under the heading “Tension in a Flexible Cord.” 12. (b, d) The normal force between the skier and the snow is a contact force preventing the skier from passing through the surface of the snow. The normal force requires contact with the surface and an external net force toward the snow. The normal force does not depend upon the speed of the skier. Any slope less than 90° will have a component of gravity that must be overcome by the normal force. 13. (a) If the two forces were pulling in the same direction the net force would be the maximum and equal to the sum of the two individual forces, or 950 N. Since the forces are not parallel, the net force will be less than this maximum.

100

Chapter 4

Dynamics: Newton’s Laws of Motion

Solutions to Problems 1.

Use Newton’s second law to calculate the net force.  F = ma = ( 45 kg ) (1.4 m s2 ) = 63 N

In all cases, W = mg, where g changes with location. (a) WEarth = mg Earth = ( 74 kg ) ( 9.80 m s 2 ) = 730 N (b) WMoon = mg Moon = ( 74 kg ) (1.7 m s 2 ) = 130 N

( ) = ( 74 kg ) ( 0 m s ) = 0 N

Use Newton’s second law to calculate the tension.

 F = F = ma = (1210 kg ) (1.35m s ) = 1633.5 N  1.63  10 N 2

Use Newton’s second law to calculate the mass. F 215 N  F = ma → m = a = 2.30 m s2 = 93.5 kg

Find the average acceleration from Eq. 2–5. The average force on the car is found from Newton’s second law. v − v0 0 − 26.39 m s  1m s  v = 0 v0 = ( 95 km h )  = 26.39 m s aavg = = = −3.299 m s 2  t 8.0s  3.6 km h 

(

)

Favg = maavg = ( 950 kg ) −3.299 m s 2 = −3134 N  −3100 N

The negative sign indicates the direction of the force, in the opposite direction to the initial velocity.

v − v0

0.35 m s − 0.25 m s

The average acceleration of the blood is given by a =

= 1.0 m s 2 .

Find the average acceleration from Eq. 2–12c, and then find the force needed from Newton’s second law. We assume the train is moving in the positive direction. v 2 − v02  1m s  v = 0 v0 = (120 km h )  33.33 m s a = = avg  2 ( x − x0 )  3.6 km h 

t 0.10 s The net force on the blood, exerted by the heart, is found from Newton’s second law. F = ma = ( 20  10−3 kg )(1.0 m s 2 ) = 0.02 N

 0 − ( 33.33 m s )2  6 6 Favg = maavg = m = ( 3.6  10 kg )   = −1.333  10 N  −1.3  10 N 2 ( x − x0 ) 2 150 m ( )   v 2 − v02

The negative sign indicates the direction of the force, in the opposite direction to the initial velocity. We compare the magnitude of this force to the weight of the train. Favg 1.333  106 N = = 0.3886 mg 3.6  105 kg 9.80 m s 2

(

)(

)

101

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Thus the force is 39% of the weight of the train. By Newton’s third law, the train exerts the same magnitude of force on Superman that Superman exerts on the train, but in the opposite direction. So the train exerts a force of 1.3  106 N in the forward direction on Superman. 8.

We assume that 30 g’s has 2 significant figures. The acceleration of a person having a 30 “g”  9.80 m s 2  = 294 m s 2 . The average force causing that acceleration deceleration is a = ( 30 “ g ” )    “g” 

(

)

is F = ma = ( 65kg ) 294 m s 2 = 1.9  104 N . Since the person is undergoing a deceleration, the acceleration and force would both be directed opposite to the direction of motion. Use Eq. 2–12c to find the distance traveled during the deceleration. Take the initial velocity to be in the positive direction, so that the acceleration will have a negative value, and the final velocity will be 0.  1m s  v0 = ( 85 km h )   = 23.61m s  3.6 km h  v 2 − v02 = 2a ( x − x0 ) →

( x − x0 ) =

v 2 − v02 2a

0 − ( 23.61m s )

(

2 −294 m s 2

)

= 0.9480 m  0.95 m

Find the average acceleration from Eq. 2–12c, and then find the force needed from Newton’s second law. v 2 − v02 aavg = → 2 ( x − x0 )

 (13 m s )2 − 0  Favg = maavg = m = ( 7.0 kg )   = 211.25 N  210 N 2 ( x − x0 )  2 ( 2.8 m )  v 2 − v02

10. The problem asks for the average force on the glove, which in a direct calculation would require knowledge about the mass of the glove and the acceleration of the glove. But no information about the glove is given. By Newton’s third law, the force exerted by the ball on the glove is equal and opposite to the force exerted by the glove on the ball. So we calculate the average force on the ball, and then take the opposite of that result to find the average force on the glove. The average force on the ball is its mass times its average acceleration. Use Eq. 2–12c to find the acceleration of the ball, with v = 0, v0 = 35.0 m s , and x − x0 = 0.110 m. The initial velocity of the ball is the positive direction. aavg =

v 2 − v02

2 ( x − x0 )

0 − ( 35.0 m s ) 2 ( 0.110 m )

(

= −5568 m s 2

)

Favg = maavg = ( 0.140 kg ) −5568 m s 2 = −7.80  102 N

Thus the average force on the glove was 780 N, in the direction of the initial velocity of the ball.

102

Chapter 4

Dynamics: Newton’s Laws of Motion

11. We assume that the fish line is pulling vertically on the fish, and that the fish is not jerking the line. A free-body diagram for the fish is shown. Write Newton’s second law for the fish in the vertical direction, assuming that up is positive. The tension is at its maximum.  F = FT − mg = ma → FT = m ( g + a ) → m=

18 N

= 1.5 kg g + a 9.80 m s 2 + 2.5 m s 2 Thus a mass of 1.5 kg is the maximum that the fish line will support with the given acceleration. Since the line broke, the fish’s mass is given by m  1.5 kg (about 3 lbs).

12. Choose up to be the positive direction. Write Newton’s second law for the vertical direction, and solve for the tension force.  F = FT − mg = ma → FT = m ( g + a )

(

)

FT = (1400 kg ) 9.80 m s 2 + 0.70 m s 2 = 1.47  10 4 N  1.5  10 4 N 13. (a) The 20.0-kg box resting on the table has the free-body diagram shown. Its weight is mg = ( 20.0 kg ) ( 9.80 m s 2 ) = 196 N . Since the box is at rest, the net force on

the box must be 0, and so the normal force must also be 196 N . (b) Free-body diagrams are shown for both boxes.

F12 is the force on box 1 (the

FN1 = F12

top box) due to box 2 (the bottom box), and is the normal force on box 1. F21 is the force on box 2 due to box 1, and has the same magnitude as F12 by

Top box (#1)

Newton’s third law. FN2 is the force of the table on box 2. That is the normal force on box 2. Since both boxes are at rest, the net force on each box must be 0. Write Newton’s second law in the vertical direction for each box, taking the upward direction to be positive. FN2  F 1 = FN1 − m1 g = 0

(

)

FN1 = m1 g = (10.0 kg ) 9.80 m s 2 = 98.0 N = F12 = F21

F = F −F −m g = 0 2

Bottom box (#2)

(

m1g

m2 g

)

FN 2 = F21 + m2 g = 98.0 N + ( 20.0 kg ) 9.80 m s 2 = 294 N

F21

14. Use Eq. 2–12b with v0 = 0 to find the acceleration. x − x0 = v0t + 12 at 2 → a =

2 ( x − x0 ) t

2 ( 402 m )

( 6.40 s )

 1 “g”  = 2.00 g’s 2   9.80 m s 

= 19.63 m s 2 

The accelerating force is found by Newton’s second law.

(

)

F = ma = ( 535kg ) 19.63m s 2 = 1.05  104 N

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15. Choose up to be the positive direction. Write Newton’s second law for the vertical direction, and solve for the acceleration.  F = FT − mg = ma

FT − mg m

(

132 N − (14.0 kg ) 9.80 m s 2 14.0 kg

) = −0.371 m s  −0.4 m s 2

Since the acceleration is negative, the bucket has a downward acceleration. 16. If the thief were to hang motionless on the sheets, or descend at a constant speed, the sheets would not support him, because they would have to support the full 75 kg. But if he descends with an acceleration, the sheets will not have to support the total mass. A freebody diagram of the thief in descent is shown. If the sheets can support a mass of 62 kg,

(

then the tension force that the sheets can exert is FT = ( 62 kg ) 9.80 m s

) = 608 N.

Assume that is the tension in the sheets. Then write Newton’s second law for the thief, taking the upward direction to be positive. 2 FT − mg 608 N − ( 75 kg ) 9.80 m s = −1.7 m s 2  F = FT − mg = ma → a = m = 75 kg The negative sign shows that the acceleration is downward.

(

)

If the thief descends with an acceleration of 1.7 m/s2 or greater, the sheets will support his descent.

17. There will be two forces on the woman – her weight, and the normal force of the scales pushing up on her. A free-body diagram for the woman is shown. Choose up to be the positive direction, and use Newton’s second law to find the acceleration.  F = FN − mg = ma → 0.75mg − mg = ma →

(

)

a = −0.25 g = −0.25 9.8 m s 2 = −2.5 m s 2

Due to the sign of the result, the direction of the acceleration is down . Thus the elevator must have started to move down since it had been motionless.

 1m s   = 9.72 m s.  3.6 km h 

18. Use Eq. 2–12c to find the acceleration. The starting speed is 35 km h  v 2 = v02 + 2a ( x − x0 )



→

v 2 − v02

2 ( x − x0 )

0 − ( 9.72 m s ) 2 ( 0.017 m )

= −2779 m s 2  −2800 m s 2

  = 284 g’s  280 g’s  9.80 m s  The acceleration is negative because the car is slowing down. The required force on the occupant is found by Newton’s second law. 2779 m s 2 

(

)

F = ma = ( 68 kg ) 2779 m s2 = 1.9  105 N This huge acceleration and force would not be possible unless the car hit some very heavy, stable object.

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19. Choose up to be the positive direction. Write Newton’s second law for the elevator.  F = FT − mg = ma → a=

FT − mg m

(

) = 0.7053m s  0.705 m s

24, 950 N − ( 2375 kg ) 9.80 m s 2

2375 kg

FT mg

20. The ratio of accelerations is the same as the ratio of the force. aoptics maoptics Foptics Foptics = = = g mg mg  43  r 3 g

(

)

10  10 −12 N 3  1.0 g 1kg 106 cm 3  4 2 −6  1.0 cm 3 1000 g 1m 3  3  ( 0.5  10 m ) ( 9.80 m s )  

= 1949 →

a  2000 g’s

21. (a) Since the rocket is exerting a downward force on the gases, the gases will exert an upward force on the rocket, typically called the thrust. The free-body diagram for the rocket shows two forces – the thrust and the weight. Newton’s second law can be used to find the acceleration of the rocket.  F = FT − mg = ma → a=

FT − mg m

(

)(

3.55  107 N − 2.45  106 kg 9.80 m s 2

(

2.45  10 kg 6

)

) = 4.69 m s

(b) The velocity can be found from Eq. 2–12a. v = v0 + at = 0 + ( 4.690 m s 2 ) ( 8.0 s ) = 37.52 m s  38 m s (c) The time to reach a displacement of 9500 m can be found from Eq. 2–12b. x − x0 = v0t + 12 at 2 → t =

2 ( x − x0 ) a

2 ( 9500 m )

( 4.690 m s ) 2

= 64 s

22. (a) There will be two forces on the skydivers – their combined weight, and the upward force of air resistance, FA . Choose up to be the positive direction. Write Newton’s second law for the skydivers.  F = FA − mg = ma → 0.25mg − mg = ma →

(

)

a = −0.75 g = −0.75 9.80 m s 2 = −7.35 m s 2

Due to the sign of the result, the direction of the acceleration is down. (b) If they are descending at constant speed, then the net force on them must be zero, and so the force of air resistance must be equal to their weight.

(

)

FA = mg = (148 kg ) 9.80 m s2 = 1.45  103 N 23. In both cases, a free-body diagram for the elevator would look like the adjacent diagram. Choose up to be the positive direction. To find the MAXIMUM tension, assume that the acceleration is up. Write Newton’s second law for the elevator.  F = ma = FT − mg →

FT mg

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(

FT = ma + mg = m ( a + g ) = m ( 0.0640 g + g ) = ( 4850 kg )(1.0640 ) 9.80 m s2

)

= 5.06  104 N To find the MINIMUM tension, assume that the acceleration is down. Then Newton’s second law for the elevator becomes the following.  F = ma = FT − mg → FT = ma + mg = m ( a + g ) = m ( −0.0640 g + g )

(

)

= ( 4850 kg )( 0.9360 ) 9.80 m s 2 = 4.45  104 N

24. The velocity that the person must have when losing contact with the ground is found from Eq. 2–12c, using the acceleration due to gravity, with the condition that their speed at the top of the jump is 0. We choose up to be the positive direction. v 2 = v02 + 2a ( x − x0 ) → v0 =

(

FP mg

)

0 − 2 −9.80 m s 2 ( 0.80 m ) = 3.960 m s

v 2 − 2a ( x − x0 ) =

This velocity is the velocity that the jumper must have as a result of pushing with their legs. Use that velocity with Eq. 2–12c again to find what acceleration the jumper must have during their push on the floor, given that their starting speed is 0. v = v + 2a ( x − x0 ) 2

2 0

( 3.960 m s ) − 0 a= = = 39.20 m s 2 2 ( x − x0 ) 2 ( 0.20 m ) 2

v 2 − v02

→

Finally, use this acceleration to find the pushing force against the ground.  F = FP − mg = ma →

(

)

FP = m ( g + a ) = ( 68 kg ) 9.80 m s 2 + 39.20 m s2 = 3300 N This is about 5 times their weight! 25. (a) To calculate the time to accelerate from rest, use Eq. 2–12a. v = v0 + at → t =

v − v0 a

9.0 m s − 0 1.2 m s 2

= 7.5s

The distance traveled during this acceleration is found from Eq. 2–12b. 2 x − x0 = v0 t + 12 at 2 = 12 (1.2 m s 2 ) ( 7.5s ) = 33.75 m To calculate the time to decelerate to rest, use Eq. 2–12a. v = v0 + at → t =

v − v0 a

0 − 9.0 m s −1.2 m s 2

= 7.5s

The distance traveled during this deceleration is found from Eq. 2–12b. 2 x − x0 = v0t + 12 at 2 = ( 9.0 m s )( 7.5s ) + 12 ( −1.2 m s 2 ) ( 7.5s ) = 33.75 m To distance traveled at constant velocity is 180 m − 2 ( 33.75 m ) = 112.5 m. To calculate the time spent at constant velocity, use Eq. 2–8. x = x0 + v t → t =

x − x0 v

112.5 m s 9.0 m s

= 12.5 s  13s

Thus the times for each stage are: Accelerating: 7.5s Constant Velocity: 13s

Decelerating: 7.5s

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(b) The normal force when at rest is mg. From the free-body diagram, if up is the positive direction, we have that FN − mg = ma. Thus the change in normal force is the difference in the normal force and the weight of the person, or ma. 1.2 m s 2 FN ma a Accelerating: = = =  100 = 12% FN mg g 9.80 m s2 FN

Constant velocity: Decelerating:

FN FN

ma mg

a g

0 9.80 m s 2

−1.2 m s 2

9.80 m s 2

 100 = 0%

 100 = −12%

(c) The normal force is not equal to the weight during the accelerating and deceleration phases. 7.5s + 7.5s = 55% 7.5s + 12.5s + 7.5s 26. We break the race up into two portions. For the acceleration phase, we call the distance d1 and the time t1. For the constant speed phase, we call the distance d 2 and the time t2 . We know that d1 = 45 m, d 2 = 55 m, and t2 = 10.0 s − t1. Eq. 2–12b is used for the acceleration phase and Eq. 2–2

is used for the constant speed phase. The speed during the constant speed phase is the final speed of the acceleration phase, found from Eq. 2–12a. x − x0 = v0t + 12 at 2 → d1 = 12 at12 ; x = vt → d 2 = vt2 = v (10.0s − t1 ) ; v = v0 + at1 This set of equations can be solved for the acceleration and the velocity. d1 = 12 at12 ; d 2 = v (10.0 s − t1 ) ; v = at1 → 2d1 = at12 ; d 2 = at1 (10.0 − t1 ) → a=

t1 =

2d1 2 1

; d2 =

20.0d1

( d 2 + 2d1 )

v = at1 =

2d1 2 1

t1 (10.0 − t1 ) =

→ a=

( d 2 + 2d1 )

2d1 t12

20.0d1

2d1 t1

(10.0 − t1 ) → d 2t1 = 2d1 (10.0 − t1 ) →

2d1

 20.0d1   d + 2d  1)  ( 2 =

( d 2 + 2d1 )

( 200 s ) d 2

( d 2 + 2d1 )

200d1 ( d 2 + 2d1 ) 10.0 s (a) The horizontal force is the mass of the sprinter times their acceleration. ( d + 2d ) 2 (145 m ) 2 F = ma = m 2 2 1 = ( 66 kg ) = 154 N  150 N ( 200 s ) d1 ( 200 s2 ) ( 45 m ) (b) The velocity for the second portion of the race was found above. ( d + 2d1 ) 145 m v= 2 = = 14.5 m s 10.0 s 10.0 s

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27. (a) Use Eq. 2–12c to find the speed of the person just before striking the ground. Take down to be the positive direction. For the person, v0 = 0, y − y0 = 2.8 m, and a = 9.80 m s 2 . v 2 − v02 = 2a ( y − y0 ) → v=

2 a ( y − y0 ) =

(

)

2 9.80 m s 2 ( 2.8 m ) = 7.408 m s  7.4 m s

(b) For the deceleration, use Eq. 2–12c to find the average deceleration, choosing down to be positive. v0 = 7.408 m s v = 0 y − y0 = 0.70 m v 2 − v02 = 2a ( y − y0 ) → a=

−v02 2y

− ( 7.408 m s ) 2 ( 0.70 m )

= −39.2 m s 2

The average force on the torso ( FTorso ) due to the legs is found from Newton’s

FTorso

second law. See the free-body diagram. Down is positive. Fnet = mg − FTorso = ma →

(

)

FTorso = mg − ma = m ( g − a ) = ( 42 kg ) 9.80 m s 2 − −39.2 m s 2  = 2100 N

The force is upward. 28. Free-body diagrams for the box and the weight are shown at right. The FT tension exerts the same magnitude of force on both objects. FT FN (a) If the weight of the hanging weight is less than the weight of the box, the objects will not move, and the tension will be the same as the m2 g weight of the hanging weight. The acceleration of the box will also m1g be zero, and so the sum of the forces on it will be zero. For the box, FN + FT − m1 g = 0 → FN = m1 g − FT = m1 g − m2 g = 66.0N − 30.0 N = 36.0 N (b) The same analysis as for part (a) applies here. FN = m1 g − m2 g = 66.0 N − 60.0 N = 6.0 N (c) Since the hanging weight has more weight than the box on the table, the box on the table will be lifted up off the table, and normal force of the table on the box will be 0 N . 29. (a) Just before the player leaves the ground, the forces on the player are his weight and the floor pushing up on the player. If the player jumps straight up, then the force of the floor will be straight up – a normal force. See the left diagram. In this case, while touching the floor, FN  mg . (b) While the player is in the air, the only force on the player is their weight. See the right diagram. 30. The two forces must be oriented so that the northerly component of the first force is exactly equal to the southerly component of the second force. Thus the second force must act southwesterly . See the diagram. 31. (a) Just as the ball is being hit, ignoring air resistance, there are two main forces on the ball: the weight of the ball, and the force of the bat on the ball. (b) As the ball flies toward the outfield, the only force on it is its weight, if air resistance is ignored.

  F1 + F2 Fbat mg mg

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32. Consider the point in the rope directly below Arlene. That point can FT FT be analyzed as having three forces on it – Arlene’s weight pushing 10o 10o down, the tension in the rope towards the right point of connection, mg and the tension in the rope towards the left point of connection. Assuming the rope is massless, those two tensions will be of the same magnitude. Since the point is not accelerating the sum of the forces must be zero. In particular, consider the sum of the vertical forces on that point, with up as the positive direction.  F = FT sin10.0o + FT sin10.0o − mg = 0 →

( 50.0 kg ) ( 9.80 m s 2 )

FT =

= 2 sin10.0o 2 sin10.0o Note that this is almost 3x her weight.

= 1410 N

33. The window washer pulls down on the rope with her hands with a tension force FT , so the rope pulls up on her hands with a tension force FT . The tension in the rope is also applied at the other end of the rope, where it attaches to the bucket. Thus there is another force FT pulling up on the bucket. The bucket-washer combination thus has a net force of 2FT upwards. See the adjacent free-body diagram, showing only forces on the bucket-washer combination, not forces exerted by the combination (the pull down on the rope by the person) or internal forces (normal force of bucket on person). (a) Write Newton’s second law in the vertical direction, with up as positive. The net force must be zero if the bucket and washer have a constant speed.  F = FT + FT − mg = 0 → 2 FT = mg →

(

)

FT = 12 mg = 12 ( 78 kg ) 9.80 m s 2 = 382.2 N  380 N

(b) Now the force is increased by 15%, so FT = 382.2 N (1.15) = 439.53 N. Again write Newton’s second law, but with a non-zero acceleration.  F = FT + FT − mg = ma → a=

2 FT − mg m

(

2 ( 439.53 N ) − ( 78 kg ) 9.80 m s 2 78 kg

) = 1.47 m s  1.5 m s 2

34. We draw free-body diagrams for each bucket. (a) Since the buckets are at rest, their acceleration is 0. Write Newton’s second law for each bucket, calling UP the positive direction.  F1 = FT1 − mg = 0 →

(

)

FT1 = mg = ( 3.2 kg ) 9.80 m s 2 = 31.36 N  31N

 F = F − F − mg = 0 → 2

(

)

FT2 = FT1 + mg = 2mg = 2 ( 3.2 kg ) 9.80 m s 2 = 62.72N  63 N

(b) Now repeat the analysis, but with a non-zero acceleration. The free-body diagrams are unchanged.  F1 = FT1 − mg = ma →

(

)

FT1 = mg + ma = ( 3.2 kg ) 9.80 m s 2 + 1.45 m s 2 = 36 N

 F = F − F − mg = ma → F = F + mg + ma = 2 F = 72 N 2

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35. See the free-body diagram for the bottom bucket, and write Newton’s second law to find the tension. Take the upward direction as positive.  F = FT1 − mbucket g = mbucketa →

FT1

bottom

(

)

= mbucket ( g + a ) = ( 3.2 kg ) 9.80 m s2 + 1.45 m s2 = 36 N

FT1 bottom

Next, see the free-body for the rope between the buckets. The mass of the cord is given W by mcord = cord . g

 F = FT1 − mcord g − FT1 top

FT1 = FT1 top

= mcord a →

mbucket g FT1

top

bottom

+ mcord ( g + a ) = mbucket ( g + a ) + mcord ( g + a )

bottom



=  mbucket +

2.0 N   11.25 m s 2 ) ( g + a ) =  3.2 kg +  2 ( g  9.80 m s  

Wcord 



FT1

bottom

mcord g

= 38.296 N  38 N Note that this is the same as saying that the tension at the top is accelerating the bucket and cord together. Now use the free-body diagram for the top bucket to find the tension at the bottom of the second cord.  F = FT2 − FT1 − mbucket g = mbucketa →

FT2

FT1

top

mbucket g

top

FT2 = FT1 + mbucket ( g + a ) = mbucket ( g + a ) + mcord ( g + a ) + mbucket ( g + a ) top



Wcord 



g 

= ( 2mbucket + mcord )( g + a ) =  2mbucket +



=  2 ( 3.2 kg ) +



(g + a)

 11.25 m s2 ) = 74.296 N  74 N (  9.80 m s  2.0 N

Note that this is the same as saying that the tension in the top cord is accelerating the two buckets and the connecting cord. 36. Choose the y-direction to be the “forward” direction for the motion of the snowcats, and the x-direction to be to the right on the diagram in the textbook. Since the housing unit moves in the forward direction on a straight line, there is no acceleration in the x-direction, and so the net force in the x-direction must be 0. Write Newton’s second law for the x-direction.  Fx = FAx + FBx = 0 → − FA sin 48 + FB sin 32 = 0 → FB =

FA sin 48

( 4200 N ) sin 48

= 5890 N  5900 N sin 32 sin 32 Since the x-components add to 0, the magnitude of the vector sum of the two forces will just be the sum of their y components.  Fy = FAy + FBy = FA cos 48 + FB cos 32 = ( 4200 N ) cos 48 + (5890 N ) cos 32

= 7805 N  7800 N

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37. Since all forces of interest in this problem are horizontal, draw the free-body diagram showing only the horizontal forces. FT1 is the tension in the coupling between the locomotive and the first car, and it pulls to the right on the first car. FT2 is the tension in the coupling between the first car an the second car. It pulls to the right on car 2, labeled FT2R and to the left on car 1, labeled FT2L . Both cars have the same mass m and the same acceleration a. Note that FT2R = FT2L = FT2 by Newton’s third law.

FT2

FT1

Write a Newton’s second law expression for each car.  F1 = FT1 − FT 2 = ma  F2 = FT 2 = ma Substitute the expression for ma from the second expression into the first one.

FT1 − FT 2 = ma = FT 2 → FT1 = 2 FT2 →

FT1 FT2 = 2

This can also be discussed in the sense that the tension between the locomotive and the first car is pulling two cars, while the tension between the cars is only pulling one car. 38. The net force in each case is found by vector addition with components. (a) FNet x = − F1 = −10.4 N FNet y = − F2 = −16.2 N

( −10.4 ) + ( −16.2 ) = 19.3 N



−16.2

= 57.3 −10.4 The actual angle from the x-axis is then 237.3. Thus the net force is 2

FNet =

 = tan

−1

Fnet

FNet = 19.3 N at 237 a=

FNet m

19.3 N

= 1.04 m s 2 at 237

18.5 kg

(b) FNet x = F1 cos 30o = 9.007 N FNet =

 = tan

FNet y = F2 − F1 sin 30o = 11.0 N

−1 11.0

9.007

= 50.7

Fnet

( 9.007 N ) + (11.0 N ) = 14.223N  14.2 N 2

FNet m

14.223 N 18.5 kg

 30o

= 0.769 m s 2 at 50.7

39. From the free-body diagram, the net force along the plane on the skater is mg sin  , and so the acceleration along the plane is g sin  . We use the kinematical data and Eq. 2–12b to write an equation for the acceleration, and then solve for the angle.

y x

x − x0 = v0t + 12 at 2 = v0t + 12 gt 2 sin  →

 2 (18 m ) − 2 ( 2.0 m s )( 3.3s )   2x − 2v0t  −1 sin  = sin  =   = 12  2 2   gt 2 9.80 m s 3.3s ( )   ( )   −1

 

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40. Since the sprinter exerts a force of 720 N on the ground at an angle of 22° below the horizontal, by Newton’s third law the ground will exert a force of 720 N on the sprinter at an angle of 22° above the horizontal. A free-body diagram for the sprinter is shown. (a) The horizontal acceleration will be found from the net horizontal force. Using Newton’s second law, we have the following.  Fx = FP cos 22 = ma x → ax =

FP cos 22

( 720 N ) cos 22

22o

= 10.27 m s 2  1.0  101 m s2

m 65 kg (b) Eq. 2–12a is used to find the final speed. The starting speed is 0. v = v0 + at → v = 0 + at = (10.27 m s 2 ) ( 0.32 s ) = 3.286 m s  3.3 m s

41. During the time while the force is F0 , the acceleration is a =

. Thus the distance traveled would

be given by Eq. 2–12b, with a 0 starting velocity, x − x0 = v0t + 12 at 2 = 12

t02 . The velocity at the

 F0   t0 . During the time while the m

v = v0 + at = 0 + 

end of that time is given by Eq. 2–12a, force is 2 F0 , the acceleration is a =

2 F0 m

. The distance traveled during this time interval would  F0  t . 0 m

again be given by Eq. 2–12b, with a starting velocity of 

2 F0  2 F0 2  F0   1   t0  t0 + 2   t0 = 2 t0 m  m   m  

x − x0 = v0 t + 12 at 2 = 

The total distance traveled is 12

F0 2 F 5 F0 2 t0 + 2 0 t02 = t0 . m m 2 m

42. Find the net force by adding the force vectors. Divide that net force by the mass to find the acceleration, and then use Eq. 3–13a to find the velocity at the given time. 6î + 34ˆj N  F = 16î + 12ˆj N + −10î + 22ˆj N = 6î + 34ˆj N = ma = ( 3.0 kg ) a → a = 3.0 kg

(

v = v 0 + at = 0 +

)

(

)

(

)

(

)

( 6î + 34ˆj) N ( 4.0 s ) = 8î + 45.33ˆj m s  8î + 45ˆj m s ( ) ( ) 3.0 kg

In magnitude and direction, the velocity is 46 m s at an angle of 80. 43. We use the free-body diagram with Newton’s first law for the stationary chandelier to find the forces in question. The angle is found from the horizontal displacement and the length of the wire. −1 0.15 m = 2.53 (a)  = sin 3.4 m Fnet = FT sin  − FH = 0 → FH = FT sin  x



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Fnet = FT cos − mg = 0 → FT = FH =

(b) FT =

mg cos 

→

cos 

(

)

sin  = mg tan  = ( 27 kg ) 9.80 m s 2 tan 2.53 = 11.69 N  12 N

( 27 kg ) ( 9.80 m s 2 ) cos 2.53

= 264.9 N  260 N

44. (a) Refer to the free-body diagrams shown. With the stipulation that the direction of the acceleration be in the direction of motion for both objects, we have aC = a E = a.

mE g − FT = mE a ; FT − mC g = mCa

a +y

(b) Add the equations together to solve them. ( mE g − FT ) + ( FT − mC g ) = mEa + mCa →

mE g

mCg

mE g − mC g = mE a + mCa → a=

mE − mC mE + mC

1150 kg − 1000 kg

( 9.80 m s ) = 0.68 m s 1150 kg + 1000 kg 

mE − mC



mE + mC

FT = mC ( g + a ) = mC  g +



2mC mE



mE + mC

g =

2 (1000 kg )(1150 kg )

( 9.80 m s ) 1150 kg + 1000 kg 2

= 10, 483 N  10,500 N

45. (a) Consider the free-body diagram for the block on the frictionless surface. There is no acceleration in the y direction. Use Newton’s second law for the x-direction to find the acceleration.  Fx = mg sin  = ma →

(

)

a = g sin  = 9.80 m s 2 sin 22.0 = 3.67 m s 2

x 

(b) Use Eq. 2–12c with v0 = 0 to find the final speed.

(

 mg

)

v 2 − v02 = 2a ( x − x0 ) → v = 2a ( x − x0 ) = 2 3.67 m s 2 (12.0 m ) = 9.39 m s 46. (a) Consider the free-body diagram for the block on the frictionless surface. There is no acceleration in the y direction. Write Newton’s second law for the x-direction.  Fx = mg sin  = ma → a = g sin  Use Eq. 2–12c with v0 = −5.2 m s and v = 0 m s to find the distance that it slides before stopping. v 2 − v02 = 2a ( x − x0 ) →

( x − x0 ) =

v 2 − v02 2a

(

0 − ( −5.2 m s )

y x



 mg

)

2 9.80 m s 2 sin 22.0

= −3.683 m  3.7 m up the plane

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(b) The time for a round trip can be found from Eq. 2–12a. The free-body diagram (and thus the acceleration) is the same whether the block is rising or falling. For the entire trip, v0 = −5.2 m s and v = +5.2 m s . v = v0 + at → t =

v − v0 a

( 5.2 m s ) − ( −5.2 m s )

( 9.80 m s ) sin 22 2

= 2.833s  2.8s

47. Consider a free-body diagram of the object. The car is moving to the right (the positive direction) and gaining speed. Thus the acceleration and the net force are to the right. The acceleration of the object is found from Eq. 2–12a. v − v0 28 m s − 0 v = v0 + ax t → ax = = = 5.6 m s 2 5.0 s t Now write Newton’s second law for both the vertical (y) and horizontal (x) directions. mg  Fy = FT cos  − mg = 0 → FT = cos   Fx = FT sin  = max Substitute the expression for the tension from the y-equation into the x-equation. mg max = FT sin  = sin  = mg tan  → ax = g tan  cos 

 = tan

−1

ax g

= tan

−1

5.6 m s 2 9.80 m s

(



)

= 29.74  30 = 3.0  101 

48. See the free-body diagram for the falling purse. Assume that down is the positive direction, and that the air resistance force Ffr is constant. Write Newton’s second law for the vertical direction.  F = mg − Ffr = ma → Ffr = m ( g − a ) Now obtain an expression for the acceleration from Eq. 2–12c with v 0 = 0, and substitute back into the friction force. v2 2 2 v − v0 = 2a ( x − x0 ) → a = 2 ( x − x0 )

Ffr

2    v2 ( 27 m s )  2 Ff = m  g −  = 6.3 N  = ( 2.0 kg )  9.80 m s − 2 ( x − x0 )  2 ( 55 m )   

49. (a) We draw a free-body diagram for the piece of 12.5m the rope that is directly above the person. That piece of rope should be in equilibrium. FT x   The person’s weight will be pulling down on that spot, and the rope tension will be pulling away from that spot towards the points of attachment. Write Newton’s mg second law for that small piece of the rope. ( 72.0 kg ) ( 9.80 m s 2 ) −1 mg −1 F F mg 2 sin 0 sin sin = − = → = = = 6.535    y T 2 FT 2 ( 3100 N ) tan  =

x 12.5 m

→ x = (12.5 m ) tan 6.535 = 1.432 m  1.4 m

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(b) Use the same equation to solve for the tension force with a sag of only ¼ that found above. 0.358 m = 1.641 x = 14 (1.432 m ) = 0.358 m ;  = tan −1 12.5 m

( 72.0 kg ) ( 9.80 m s 2 ) = = 1.232  104 N  12 kN FT = 2 sin  2 ( sin1.641 ) mg

The rope will not break , since FT  31kN, but it exceeds the recommended tension by a factor of about 4. 50. For each object, we have the free-body diagram shown, assuming that the string doesn’t break. Newton’s second law is used to get an expression for the tension. Since the string broke for the 2.10 kg mass, we know that the required tension to accelerate that mass was more than 22.2 N. Likewise, since the string didn’t break for the 2.05 kg mass, we know that the required tension to accelerate that mass was less than 22.2 N. These relationships can be used to get the range of accelerations.  F = FT − mg = ma → FT = m ( a + g ) FT  m2.10 ( a + g ) ; FT  m2.05 ( a + g ) → max

max

m2.10

−ga

max

m2.05

−g →

0.77 m s 2  a  1.03m s 2

→

22.2 N 2.10 kg

FT max

m2.10

FT −ga ;

− 9.80 m s 2  a 

max

m2.05

22.2 N 2.05 kg

−ga →

− 9.80 m s 2

→

0.8 m s 2  a  1.0 m s 2

51. (a) We assume that the maximum horizontal force occurs when the train is moving very slowly, and so the air resistance is negligible. Thus the maximum acceleration is given by the following. F 4  105 N amax = max = = 0.625m s 2  0.6 m s 2 m 6.4  105 kg (b) At top speed, we assume that the train is moving at constant velocity. Therefore the net force on the train is 0, and so the air resistance and friction forces together must be of the same magnitude as the horizontal pushing force, which is 1.5  105 N . 52. Use the free-body diagram to find the net force in the x direction, and then find the acceleration. Then Eq. 2–12c can be used to find the final speed at the bottom of the ramp.  Fx = mg sin  − FP = ma → a=

mg sin  − FP m

( 450 kg ) ( 9.80 m s ) sin15 − 1020 N

450 kg



 mg

= 0.2698 m s 2

v 2 = v02 + 2a ( x − x0 ) → v=

2a ( x − x0 ) =

(

)

2 0.2698 m s 2 ( 4.0 m ) = 1.469 m s  1m s

The subtraction rule for significant figures, when applied to the acceleration calculation, leads to only 1 significant figure in the answer. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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53. For a simple ramp, the decelerating force is the component of gravity along the ramp. See the free-body diagram, and use Eq. 2–12c to calculate the distance.  Fx = −mg sin  = ma → a = − g sin 

x − x0 =

v 2 − v02 2a

0 − v02

2 ( − g sin  )

v02 

2 g sin 

  1m s   (140 km h )  3.6 km h      = 4.0  102 m =

(

)

2 9.80 m s 2 sin11

54. The average force can be found from the average acceleration. Use Eq. 2–12c to find the acceleration. v 2 − v02 t v 2 = v02 + 2a ( x − x0 ) → a = 2 ( x − x0 )

F = ma = m

v 2 − v02

2 ( x − x0 )

= ( 60.0kg )

0 − (10.0 m s ) 2 ( 25.0 m )

= −120 N

The average retarding force is 1.20  102 N , in the direction opposite to the child’s velocity. 55. (a) See the free-body diagrams included.

(b) For block A, since there is no motion in the vertical direction, we have FNA = mA g . We write Newton’s second law for the

FNA

x-direction:  FAx = FT = mA aAx . For block B, we only need to consider vertical forces:  FBy = mB g − FT = mBaBy . Since the

FT mA g

mB g

two blocks are connected, the magnitudes of their accelerations will be the same, and so let aAx = aBy = a. Combine the two force equations from above, and solve for a by substitution. FT = mA a

mB g − FT = mBa

mA a + mBa = mB g →

a=g

→

m B g − mA a = m B a →

FT = mA a = g

mA + mB

mA mB mA + mB

56. (a) From Problem 55, we have the acceleration of each block. Both blocks have the same acceleration. 5.0 kg mB a=g = ( 9.80 m s 2 ) = 2.579 m s 2  2.6 m s 2 mA + m B ( 5.0 kg + 14.0 kg ) (b) Use Eq. 2–12b to find the time. x − x0 = v0t + 12 at 2 → t =

2 ( x − x0 ) a

2 (1.250 m )

( 2.579 m s ) 2

= 0.9846 s  0.98s

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Chapter 4

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→ mA = 99mB = 99 kg

57. This problem can be solved in the same way as Problem 55, with the modification that we increase mass mA by the mass of l A and we increase mass mB by the mass of l B . We take the result from Problem 55 for the acceleration and make modifications. We assume that the cord is uniform, and so the mass of any segment is directly proportional to the length of that segment. lB l mB + mC mB + B mC lA +lB mB l a=g → a=g = g mA + mB mA + mB + mC     lA lB  mA + l + l mC  +  mB + l + mC      A B A B Note that this acceleration is NOT constant, because the lengths l A and l B are functions of time. Thus constant acceleration kinematics would not apply to this system. 58. We draw a free-body diagram for each mass. We choose UP to be the positive direction. The tension force in the cord is found from analyzing the two hanging masses. Notice that the same tension force is applied to each mass. Write Newton’s second law for each of the masses. FT − m1 g = m1a1 FT − m2 g = m2 a2 Since the masses are joined together by the cord, their accelerations will have the same magnitude but opposite directions. Thus a1 = − a2 . Substitute this into the force expressions and solve for the tension force. m g − FT FT − m1 g = − m1a2 → FT = m1 g − m1a2 → a2 = 1 m1

 m1 g − FT  2m1m2 g → FT =  m1 + m2  m1 

FT − m2 g = m2 a2 = m2 

2.2 kg

4.2 kg

m2 g

m1g

Apply Newton’s second law to the stationary pulley. 2 4m1m2 g 4 ( 4.2 kg )( 2.2 kg ) 9.80 m s FC − 2 FT = 0 → FC = 2 FT = = = 56.6 N  57 N m1 + m2 6.4 kg

(

)

59. Please refer to the free-body diagrams given in the textbook for this problem. Initially, treat the two boxes and the rope as a single system. Then the only accelerating force on the system is FP . The mass of the system is 23.0 kg, and so using Newton’s second law, the acceleration of the system is F 40.0 N a= P = = 1.739 m s2  1.74 m s2 . This is the acceleration of each part of the system. m 23.0 kg Now consider mB alone. The only force on it is FBT , and it has the acceleration found above. Thus

FBT can be found from Newton’s second law.

(

)

FBT = mBa = (12.0 kg ) 1.739 m s 2 = 20.87 N  20.9 N

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Now consider the rope alone. The net force on it is FTA − FTB , and it also has the acceleration found above. Thus FTA can be found from Newton’s second law.

(

)

FTA − FTB = mC a → FTA = FTB + mCa = 20.87 N + (1.0 kg ) 1.739 m s 2 = 22.6 N As compared to Example 4–12, the acceleration in this problems is smaller, since the rope has added mass to the system. Since the acceleration is smaller, the tension in the rope for this problem at the left end is smaller than in Example 4–12. And since the rope has mass, the tension at the two ends are not the same. The tension in the rope for this problem at the right end is greater than in the Example. Here is a summary.

aProblem  aExample ; ( FTB ) Problem  ( FTB ) Example ; ( FTA ) Problem  ( FTA ) Example 59

4-12

60. (a) In the free-body diagrams below, FAB = force on block A exerted by block B, FBA = force on block B exerted by block A, FBC = force on block B exerted by block C, and FCB = force on block C exerted by block B. The magnitudes of FBA and FAB are equal, and the magnitudes of

FBC and FCB are equal, by Newton’s third law.

(b) All of the vertical forces on each block add up to zero, since there is no acceleration in the vertical direction. Thus for each block, FN = mg. For the horizontal direction, we have the following. F

 F = F − F + F − F + F = F = (m + m + m ) a → a = m + m + m AB

(c) For each block, the net force must be ma by Newton’s second law. Each block has the same acceleration since they are in contact with each other. FA net = F

mA mA + mB + mC

FB net = F

mB mA + mB + mC

F3 net = F

(d) From the free-body diagram, we see that for mC, FCB = FC net = F Newton’s third law, FBC = FCB = F

mC mA + mB + mC

. And by

. Of course, F23 and F32 are in opposite

directions. Also from the free-body diagram, we use the net force on mA.

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F − FAB = FA net = F FAB = F

→ FAB = F − F

mA + mB + mC

→

mA + mB + mC

mB + mC mA + mB + mC

By Newton’s third law, FBC = FAB = F (e) Using the given values, a =

mB + mC mA + mB + mC

F mA + mB + mC

96.0 N 30.0 kg

. = 3.20 m s 2 . Since all three masses

(

are the same value, the net force on each mass is Fnet = ma = (10.0 kg ) 3.20 m s

) = 32.0 N.

This is also the value of FCB and FBC . The value of FAB and FBA is found as follows.

(

)

FAB = FBA = ( mB + mC ) a = ( 20.0 kg ) 3.20 m s2 = 64.0 N To summarize:

FA net = FB net = FC net = 32.0 N

FAB = FBA = 64.0 N

FBC = FCB = 32.0 N

The values make sense in that in order of magnitude, we should have F  FBA  FCB , since F is the net force pushing the entire set of blocks, FAB is the net force pushing the right two blocks, and FBC is the net force pushing the right block only. 61. Let mA = 3.0 kg be the left-most mass, mB = 2.5 kg be the mass on the table, and mC = 5.0 kg be the right-most mass. We draw free-body diagrams for each mass. We also define the positive direction to be upwards for the left mass, to the right for the mass on the table, and downward for the right mass. + FT1 + FT2 FNB FT2 + FT1 mA g

mCg

mB g

Now write Newton’s second law for each object in the direction of motion, recognizing that each object has the same acceleration in the direction of motion.  FA = FT1 − mA g = mAa ;  FB = FT2 − FT1 = mBa ;  FC = mC g − FT2 = mCa Add the three equations together, and solve for the acceleration. ( FT1 − mA g ) + ( FT2 − FT1 ) + ( mC g − FT2 ) = mA a + mBa + mCa →

FT1 − mA g + FT2 − FT1 + mC g − FT2 mA + mB + mC

mC − mA mA + mB + mC

(

= 9.80 m s2

2.0 kg ) 10.5 kg

= 1.867 m s2  1.9 m s 2

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62. If m doesn’t move on the incline, it doesn’t move in the vertical direction, and so has no vertical component of acceleration. This suggests that we analyze  FN the forces parallel and perpendicular to the table. See the force diagram for the small block, and use Newton’s second law to find the acceleration of the small block.  mg mg  Fy = FN cos  − mg = 0 → FN = cos  F sin  mg sin   Fx = FN sin  = ma → a = N m = m cos  = g tan  Since the small block doesn’t move on the incline, the combination of both masses has the same horizontal acceleration of g tan  . That can be used to find the applied force.

Fapplied = ( m + M ) a = ( m + M ) g tan  Note that this gives the correct answer for the case of  = 0, where it would take no applied force to keep m stationary. It also gives a reasonable answer for the limiting case of  → 90, where no force would be large enough to keep the block from falling, since there would be no upward force component to counteract the force of gravity. 63. Because the pulleys are massless, the net force on them must be 0. Because the cords are massless, the tension will be the same at both ends of the cords. Use the free-body diagrams to write Newton’s second law for each mass. We take the direction of acceleration for each mass to be positive in the direction of motion of the object. We assume that mC is falling, mB is falling relative to its pulley, and

FTC

aC FTA

mA is rising relative to its pulley. Also note that if the acceleration

of mA relative to the pulley above it is aR , then aA = a R + aC . Then, the acceleration of mB is a B = a R − aC , since a C is in the opposite direction of aB.

 F = F − m g = m a = m (a + a ) m :  F = m g − F = m a = m (a − a ) m : F = m g − F = m a pulley:  F = F − 2 F = 0 → F = 2 F mA :

FTC

FTA

FTA mA mA g

mCg

FTA aA

mB g

Re-write this system as three equations in three unknowns FTA , a R , aC . FTA − mA g = mA ( aR + aC ) →

FTA − mA aC − mA aR = mA g

mB g − FTA = mB ( aR − aC ) →

FTA − mB aC + mB aR = mB g

mC g − 2 FTA = mC aC → 2 FTA + mC aC = mC g This system now needs to be solved. One method to solve a system of linear equations is by determinants. We show that for aC .

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aC =

− mA

0 − mBmC + mA ( 2mB ) − mA ( mC − 2mB ) g= g − mA − mBmC − mA ( 2mB ) − mA ( mC + 2mB )

2 mC 1 − mA 1

− mB

4mA mB − mA mC − mBmC −4mA mB − mA mC − mBmC

mA mC + mBmC − 4mA mB 4mA mB + mA mC + mBmC

Similar manipulations give the following results. 2 ( mBmC − mA mC ) 4mA mBmC g aR = g ; FTA = 4mA mB + mA mC + mBmC 4mA mB + mA mC + mBmC (a) The accelerations of the three masses are found below. 2 ( mBmC − mA mC ) m m + mBmC − 4mA mB a A = a R + aC = g+ A C g 4mA mB + mA mC + mBmC 4mA mB + mA mC + mBmC 3mBmC − mA mC − 4mA mB

a B = a R − aC = =

4mA mB + mA mC + mBmC

2 ( mBmC − mC mC ) 4mA mB + mA mC + mBmC

mBmC − 3mA mC + 4mA mB

mA mC + mBmC − 4mA mB 4mA mB + mA mC + mBmC

4mA mB + mA mC + mBmC

aC =

g−

(b) The tensions are shown below.

4mA mBmC

FTA =

4mA mB + mA mC + mBmC

g ; FTC = 2 FTA =

8mA mBmC 4mA mB + mA mC + mBmC

g 

64. The force F is accelerating the total mass, since it is the only force external to the system in the direction of the acceleration. If mass mA does not move relative to

mC , then all the blocks have the same horizontal acceleration, and none of the

blocks have vertical acceleration. We solve for the acceleration of the system and then find the magnitude of F from Newton’s second law. Start with free-body diagrams for mA and mB .

mB :

 F = F sin  = m a ;  F = F cos  − m g = 0 → F cos  = m g x

mB g

Square these two expressions and add them, to get a relationship between FT and a.

mA g

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FT2 sin 2  = mB2 a 2 ; FT2 cos2  = mB2 g 2 →

(

)

(

FT2 sin 2  + cos2  = mB2 g 2 + a 2

) → F = m (g + a ) 2 T

2 B

Now analyze mA .

F = F = m a → F = m a ; F = F − m g = 0

mA :

2 T

2 A

Equate the two expressions for F , solve for the acceleration and then finally the magnitude of the applied force. mB2 g 2 mB g FT2 = mB2 g 2 + a 2 = mA2 a 2 → a 2 = → a= → 2 2 mA − mB mA2 − mB2 2 T

(

)

F = ( mA + mB + mC ) a =

(

)

( mA + mB + mC ) mB

(m − m ) 2 A

2 B

(

)

65. First, draw a free-body diagram for each mass. Notice that the same tension force is applied to each mass. Choose UP to be the positive direction. Write Newton’s second law for each of the masses. FT − m2 g = m2 a2 FT − m1 g = m1a1 Since the masses are joined together by the cord, their accelerations will have the same magnitude but opposite directions. Thus a1 = − a2 . Substitute this into the force expressions and solve for the acceleration by subtracting the second equation from the first. FT − m1 g = − m1a2 → FT = m1 g − m1a2

1.7 kg

3.1 kg

m2 g

m1g

FT − m2 g = m2 a2 → m1 g − m1a2 − m2 g = m2 a2 → m1 g − m2 g = m1a2 + m2 a2 a2 =

m1 − m2

3.1kg − 1.7 kg

( 9.80 m s ) = 2.858 m s 3.1kg + 1.7 kg 2

m1 + m2 The lighter block starts with a speed of 0, and moves a distance of 1.8 meters with the acceleration found above. Using Eq. 2–12c, the velocity of the lighter block at the end of this accelerated motion can be found.

(

)

v 2 − v02 = 2a ( y − y0 ) → v = v02 + 2a ( y − y0 ) = 0 + 2 2.858 m s 2 (1.8 m ) = 3.208 m s After the 1.8 meters of rise, the lighter block has different conditions of motion. Once the heavier block hits the ground, the tension force disappears, and the lighter block is in free fall. It has an initial speed of 3.208 m/s upward as found above, with an acceleration of –9.80 m/s2 due to gravity. At its highest point, its speed will be 0. Eq. 2–12c can again be used to find the height to which it rises.

v − v = 2 a ( y − y0 ) → 2

2 0

( y − y0 ) =

v 2 − v02 2a

0 − ( 3.208 m s )

(

2 −9.80 m s 2

)

= 0.5251m

Thus the total height above the ground is 1.8 m + 1.8 m + 0.53 m = 4.1m .

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66. The velocity can be found by integrating the acceleration function, and the position can be found by integrating the position function. F = ma = Ct 2 → a = v=

C 3m

t3 =

dx dt

C 2 dv t = m dt

→ dx =

C 3m

→ dv =

C 2 t dt → m

t 3 dt →

C 2 C 3 t dt → v = t 3m m 0

 dx =  3m t dt → 0

 dv = 

C 12m

67. See the free-body diagram for the load. The vertical component of the tension force must be equal to the weight of the load, and the horizontal component of the tension accelerates the load. The angle is exaggerated in the picture. F sin  Fnet = FT sin  = ma x → a x = T ; Fnet = FT cos  − mg = 0 → m x y FT =

→ ax =

cos 

mg sin  cos 

(

 FT

)

= g tan  = 9.80 m s 2 tan 5.0 = 0.86 m s 2 mg

68. A free-body diagram for the person in the elevator is shown. The scale reading is the magnitude of the normal force. Choosing up to be the positive direction, Newton’s second law for the person says that  F = FN − mg = ma → FN = m ( g + a ) . The kg reading of the scale is the apparent weight, FN , divided by g, which gives FN-kg =

FN g

m ( g + a) g

(

)

(a) a = 0 → FN = mg = ( 75.0 kg ) 9.80 m s 2 = 735 N

FN-kg =

mg g

= m = 75.0 kg

(b) a = 0 → FN = 735 N , FN-kg = 75.0 kg (c)

a = 0 → FN = 735 N , FN-kg = 75.0 kg

(

)

(d) FN = m ( g + a ) = ( 75.0 kg ) 9.80 m s2 + 2.0 m s2 = 885 N FN-kg =

(e)

FN g

885 N 9.80 m s 2

= 90.3 kg

(

)

FN = m ( g + a ) = ( 75.0 kg ) 9.80 m s2 − 2.0 m s 2 = 585 N  590 N FN-kg =

FN g

585 N 9.80 m s 2

= 59.7 kg  6.0  101 kg

69. The given data can be used to calculate the force with which the road pushes against the car, which in turn is equal in magnitude to the force the car pushes against the road. The acceleration of the car on level ground is found from Eq. 2–12a. v − v0 21m s − 0 v − v0 = at → a = = = 1.45 m s 2 t 14.5 s

FN FP 



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The force pushing the car in order to have this acceleration is found from Newton’s second law. FP = ma = ( 920 kg ) (1.45 m s 2 ) = 1334 N We assume that this is the force pushing the car on the incline as well. Consider a free-body diagram for the car climbing the hill. We assume that the car will have a constant speed on the maximum incline. Write Newton’s second law for the x-direction, with a net force of zero since the car is not accelerating. F  F x = FP − mg sin  = 0 → sin  = mgP 1334 N F  = sin −1 P = sin −1 = 8.5 mg ( 920 kg ) 9.80 m s2

(

)

70. Consider a free-body diagram for the cyclist coasting downhill at a constant speed. Since there is no acceleration, the net force in each direction must be zero. Write Newton’s second law for the x-direction (down the plane).  Fx = mg sin  − Ffr = 0 → Ffr = mg sin 

Ffr

This establishes the size of the air friction force at 6.0 km/h, and so can be used in the next part.

  mg

Now consider a free-body diagram for the cyclist climbing the hill. FP is the force pushing the cyclist uphill. Again, write Newton’s second law for the x-direction, with a net force of 0.  Fx = Ffr + mg sin  − FP = 0 →

y FP Ffr

FP = Ffr + mg sin  = 2mg sin 

(

 

)

= 2 ( 65 kg ) 9.80 m s 2 ( sin 6.5 ) = 144.2N  140 N

71. Each rope must support 16 of Tom’s weight, and so must have a vertical component of tension given by Tvert = 16 mg. For the vertical ropes, their entire tension is vertical. T1 = 16 mg = 16 ( 78.0 kg ) ( 9.80 m s 2 ) = 127.4 N  127 N

30o

For the ropes displaced 30° from the vertical, see the first diagram. 127.4 N mg T2 vert = T2 cos 30 = 16 mg → T2 = = = 147 N 6 cos 30 cos 30 T3 For the ropes displaced 60o from the vertical, see the second diagram. o 60 127.4 N mg T3 vert = T3 cos 60 = 16 mg → T3 = = = 255 N 6 cos 60 cos 60 The corresponding ropes on the other side of the glider will have the same tensions as found here. 72. Consider the free-body diagram for the soap block on the frictionless surface. There is no acceleration in the y-direction. Write Newton’s second law for the x-direction.  Fx = mg sin  = ma → a = g sin  Use Eq. 2–12b with v0 = 0 to find the time of travel.

y x



 mg

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x − x0 = v0 t + 12 at 2 → t=

2 ( x − x0 ) a

2 ( x − x0 ) g sin 

2 ( 3.0 m )

( 9.80 m s ) sin (8.5 ) 2

= 2.0 s

Since the mass does not enter into the calculation, the time would be the same for the heavier bar of soap. 73. The horizontal component of the pulling force causes the horizontal acceleration. See the free-body diagram. We assume there is no frictional force pulling backwards on the wagon. Write Newton’s second law for the horizontal direction, and solve for the pulling force. ( 35kg ) 0.42 m s2 ma x = 26 N  Fx = Fpull cos  = max → Fpull = cos  = cos 55

(

74. We first draw generic free body diagrams for both blocks, with FC as the contact force. We write Newton’s second law for the horizontal direction for each block, taking to the right as the positive direction. Since the blocks are in contact, they will have the same acceleration.  FA = F1 − FC = ma ;  FB = FC − F2 = ma

)

 mg

Fpull

F1 − FC = FC − F2 → F1 + F2 = 2 FC (a) F2 = F1 → 2 F1 = 2 FC → FC = F1 (b) F2 = 12 F1 → (c)

3 2 F1 = 2 FC

→ FC = 43 F1

F2 = 0 → F1 = 2 FC → FC = 12 F1

We see that the contact force depends on the two external forces pushing on the blocks. 75. Consider the free-body diagram for the watch. Write Newton’s second law for both the x- and y-directions. Note that the net force in the y-direction is 0 because there is no acceleration in the y-direction. mg  Fy = FT cos  − mg = 0 → FT = cos  mg  Fx = FT sin  = ma → cos  sin  = ma

(



y x

)

a = g tan  = 9.80 m s 2 tan 24 = 4.36 m s 2

Use Eq. 2–12a with v 0 = 0 to find the final velocity (takeoff speed).

(

)

v − v0 = at → v = v0 + at = 0 + 4.36 m s 2 ( 21s ) = 92 m s

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76. (a) Draw a free-body diagram for each block. Write Newton’s second law for each block. Notice that the acceleration of block A in the yA-direction will be zero, since it has no motion in the yA-direction.  FyA = FN − mA g cos  = 0 → FN = mA g cos 

Instructor Solutions Manual

FT yB

mB g

 F = m g sin  − F = m a F = F − m g = m a → F = m (g + a ) xA



m Ag

Since the blocks are connected by the cord, a yB = a xA = a. Substitute the expression for the tension force from the last equation into the x-direction equation for block 1, and solve for the acceleration. mA g sin  − mB ( g + a ) = mA a → mA g sin  − mB g = mA a + mBa a= g

( mA sin  − mB ) ( mA + m B )

(b) If the acceleration is to be down the plane, it must be positive. That will happen if

mA sin   mB ( down the plane ) . The acceleration will be up the plane (negative) if

mA sin   mB ( up the plane ) . If mA sin  = mB , then the system will not accelerate. It will move with a constant speed if set in motion by a push. 77. (a) From Problem 76, we have an expression for the acceleration. (1.00 kg ) sin 38.0 − 1.00 kg  = −1.88 m s2 ( m sin  − mB ) = 9.80 m s 2 a=g A 2.00 kg ( mA + m B )

(

)

 −1.9 m s 2

The negative sign means that mA will be accelerating UP the plane. (b) If the system is at rest, then the acceleration will be 0. ( m sin  − mB ) = 0 → mB = mA sin  = (1.00 kg ) sin 38.0 = 0.6157 kg  0.616 kg a=g A ( mA + mB ) (c) Again from Problem 76, we have FT = mB ( g + a ) .

Case (a): FT = mB ( g + a ) = (1.00 kg ) ( 9.80 m s 2 − 1.88 m s 2 ) = 7.92 N  7.9 N Case (b): FT = mB ( g + a ) = ( 0.6157 kg ) ( 9.80 m s2 + 0 ) = 6.034 N  6.03 N

78. (a) A free-body diagram is shown for each block. We define the positive x-direction for mA to be up its incline, and the positive x-direction for mB to be down its incline. With that definition the masses will both have the same acceleration. Write Newton’s second law for each body in the x-direction, and combine those equations to find the acceleration. mA :  Fx = FT − mA g sin  A = mA a

FN-A y

FN-B FT

x A

A

B

mA g

mB g

B

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 F = m g sin  − F = m a

mB :

add these two equations

( FT − mA g sin  A ) + ( mB g sin  B − FT ) = mA a + mBa → a =

mB sin  B − mA sin  A mA + m B

(b) For the system to be at rest, the acceleration must be 0. m sin  B − mA sin  A a= B g = 0 → mB sin  B − mA sin  A → mA + m B m B = mA

sin  A

= ( 5.0 kg )

sin 34

= 7.8 kg sin  B sin 21 The tension can be found from one of the Newton’s second law expression from part (a).

(

)

mA : FT − mA g sin  A = 0 → FT = mA g sin  A = ( 5.0 kg ) 9.80 m s 2 sin 34 = 27 N

(c) As in part (b), the acceleration will be 0 for constant velocity in either direction. m sin  B − mA sin  A a= B g = 0 → mB sin  B − mA sin  A → mA + m B mA mB

sin  B sin  A

sin 21 sin 34

= 0.64

79. (a) To find the minimum force, assume that the piano is moving with a constant velocity. Since the piano is not accelerating, FT 4 = Mg . For the lower pulley, since the tension in a rope is the same throughout, and since the pulley is not accelerating, it is seen that FT1 + FT 2 = 2 FT1 = Mg → FT1 = FT 2 = Mg 2 .

FT4

It also can be seen that since F = FT 2 , that F = Mg 2 . (b) Draw a free-body diagram for the upper pulley. From that 3Mg diagram, we see that FT3 = FT1 + FT 2 + F = . 2 To summarize:

FT1 = FT 2 = Mg 2

FT3 = 3 Mg 2

FT 4 = Mg

FT2 FT4

FT1

Upper Pulley

Lower Pulley

FT2

FT3

FT1 80. Consider a free-body diagram for a grocery cart being pushed up an incline. Assuming that the cart is not accelerating, we write Newton’s second law for the x-direction. F  Fx = FP − mg sin  = 0 → sin  = mgP

 = sin −1

FP mg

= sin −1

23 N

( 25 kg ) ( 9.80 m s2 )

= 5.4

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Instructor Solutions Manual

81. The acceleration of the pilot will be the same as that of the plane, since the pilot is at rest with respect to the plane. Consider first a free-body diagram of the pilot, showing only the net force. By Newton’s second law, the net force MUST point in the direction of the acceleration, and its magnitude is ma. That net force is the sum of ALL forces on the pilot. If we assume that the force of gravity and the force of the cockpit seat on the pilot are the only forces on the pilot, then in terms of vectors, Fnet = mg + Fseat = ma. Solve this equation for the force of the

18o

Fnet mg

seat to find Fseat = Fnet − mg = ma − mg. A vector diagram of that equation is shown. Solve for the force of the seat on the pilot using components. Fx seat = Fx net = ma cos18 = ( 75 kg ) 3.8 m s 2 cos18 = 271.1N

(

Fnet



Fseat

)

Fy seat = mg + Fy net = mg + ma sin18

(

)

(

)

= ( 75 kg ) 9.80 m s 2 + ( 75 kg ) 3.8 m s 2 sin18 = 823.2 N

The magnitude of the cockpit seat force is as follows.

Fx2seat + Fy2seat =

( 271.1N )2 + (823.2 N )2 = 866.7 N  870 N

The angle of the cockpit seat force is as follows. F 823.2 N  = tan −1 y seat = tan −1 = 72 above the horizontal. Fx seat 271.1N 82. (a)

The helicopter and frame will both have the same acceleration, and so can be treated as one object if no information about internal forces (like the cable tension) is needed. A free-body diagram for the helicopter-frame combination is shown. Write Newton’s second law for the combination, calling UP the positive direction.  F = Flift − ( mH + mF ) g = ( mH + mF ) a →

(

Flift = ( mH + mF )( g + a ) = ( 6750 kg + 1080 kg ) 9.80 m s 2 + 0.80 m s 2 = 82998 N  8.30  104 N

Flift

)

( mH + mF ) g

(b) Now draw a free-body diagram for the frame alone, in order to find the tension in the cable. Again use Newton’s second law.  F = FT − mF g = mF a →

(

FT = mF ( g + a ) = (1080 kg ) 9.80 m s 2 + 0.80 m s 2

)

mF g

= 11448 N  1.14  10 4 N (c) The tension in the cable is the same at both ends, and so the cable exerts a force of 1.14  104 N downward on the helicopter. 83. Choose downward to be positive. The elevator’s acceleration is calculated by Eq. 2–12c. v 2 − v02 = 2a ( y − y0 ) → a =

v 2 − v02

0 − ( 3.5 m s )

= −2.356 m s 2

2 ( y − y0 ) 2 ( 2.6 m ) See the free-body diagram of the elevator/occupant combination. Write Newton’s second law for the elevator.

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Dynamics: Newton’s Laws of Motion

 F = mg − F = ma T

(

)

FT = m ( g − a ) = (1650 kg ) 9.80 m s 2 − −2.356 m s 2 = 2.01  104 N

84. See the free-body diagram for the fish being pulled upward vertically. From Newton’s second law, calling the upward direction positive, we have this relationship.  Fy = FT − mg = ma → FT = m ( g + a )

(a) If the fish has a constant speed, then its acceleration is zero, and so FT = mg . Thus

the heaviest fish that could be pulled from the water in this case is 45 N (10 lb ) . (b) If the fish has an acceleration of 2.0 m/s2, and FT is at its maximum of 45 N, then solve the equation for the mass of the fish. F 45 N m= T = = 3.8 kg → g + a 9.8 m s 2 + 2.0 m s 2

(

)

mg = ( 3.8 kg ) 9.8 m s 2 = 37 N (  8.4 lb ) (c) It is not possible to land a 15-lb fish using 10-lb line, if you have to lift the fish vertically. If the fish were reeled in while still in the water, and then a net used to remove the fish from the water, it might still be caught with the 10-lb line. 85. Use Newton’s second law. F = ma = m

v t

→ t =

mv F

( 2.0 10 kg )( 2.0 10 m s ) = 1.6  10 s  185 d = −3

86. Since the climbers are on ice, the frictional force for the lower two climbers is negligible. Consider the freebody diagram as shown. Note that all the masses are the same. Write Newton’s second law in the x-direction for the lowest climber, assuming he is at rest.  Fx = FT2 − mg sin  = 0

(

( 2.5 N )

FN1

FN3

x FT2

)

FT2 = mg sin  = ( 75 kg ) 9.80 m s sin 29 2



= 356.3N  360 N

FN2

FT2

FT1

Ffr

FT1

mg  mg 



Write Newton’s second law in the x direction for the mg middle climber, assuming he is at rest.  Fx = FT1 − FT2 − mg sin  = 0 → FT1 = FT2 + mg sin  = 2 FT2 g sin  = 712.7N  710 N 87. (a) The value of the constant c is found from the free-body diagram, knowing that the net force is 0 when coasting downhill at the given speed.  Fx = mg sin  − Fair = 0 → Fair = mg sin  = cv →

Fair

  mg

y x

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

mg sin 

( 72.0 kg ) ( 9.80 m s 2 ) sin 5.0  1m s    3.6 km h 

= 36.898

( 6.0 km h ) 

Instructor Solutions Manual

N ms

 37

= 37

(b) Now consider the cyclist with an added pushing force FP directed along the plane. The free-body diagram changes to reflect the additional force the cyclist must exert. The same axes definitions are used as in part (a).  Fx = FP + mg sin  − Fair = 0 →

kg s

  mg

FP = Fair − mg sin  = cv − mg sin 



=  36.898



N 

 1m s   (18.0 km h )     m s   3.6 km h  

(

Fair

)

− ( 72.0 kg ) 9.80 m s2 sin 5.0 = 123.0 N  120 N

88. (a) We use the free-body diagram to find the force needed to pull the masses at a constant velocity. We choose the “up the plane” direction as the positive direction for both masses. Then they both have the same acceleration even if it is non-zero. mA :  Fx = FT − mA g sin  A = mA a = 0 F x

Add the equations to eliminate the tension force and solve for F. ( FT − mA g sin  A ) + ( F − FT − mB g sin  B ) = 0 →

FT B

mB g

B

x A

A

mA g

F = g ( mA sin  A + mB sin  B )

(

FN-B

FN-A

 F = F − F − m g sin  = m a = 0

mb :

)

= 9.80 m s 2 ( 8.2 kg ) sin 59 + (11.5kg ) sin 32 = 128.6 N  129 N

(b) Since  A   B , if there were no connecting string, mA would have a larger acceleration than m B . If  A   B , there would be no tension. But, since there is a connecting string, there will be

tension in the string. Use the free-body diagram from above but ignore the applied force F. mA :  Fx = FT − mA g sin  A = mAa ; mb :  Fx = − FT − mB g sin  B = mBa Again add the two equations to eliminate the tension force. ( FT − mA g sin  A ) + ( − FT − mB g sin  B ) = mA a + mB a → a = −g

mA sin  A + mB sin  B mA + mB

(

= − 9.80 m s 2

)(

8.2 kg ) sin 59 + (11.5 kg ) sin 32 19.7 kg

= −6.528 m s 2  6.53 m s 2 , down the planes .

(c) Use one of the Newton’s second law expressions from part (b) to find the string tension. It must be positive if there is a tension. FT − mA g sin  A = mA a →

(

)

FT = mA ( g sin  A + a ) = ( 8.2 kg )  9.80 m s 2 ( sin 59 ) − 6.528 m s 2  = 15 N

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89. Consider a free-body diagram for the car on the hill. Write Newton’s second law for the x-direction.  Fx = mg sin  = ma → a = g sin 

FN y

Use Eq. 2–12c to determine the final velocity, assuming that the car starts from rest. The angle is given by sin  = 14 .



v 2 − v02 = 2a ( x − x0 ) → v = 0 + 2a ( x − x0 ) = 2 g ( x − x0 ) sin 

(



)

= 2 9.80 m s 2 ( 55 m ) 0.25 = 16 m s

 1m s   = 12.5 m s. Use Eq. 2–12a to find the deceleration  3.6 km h 

90. The initial speed is v0 = ( 45 km h )  of the child. v − v0 = at → a =

v − v0

0 − 12.5 m s

= −62.5 m s 2 .

t 0.20 s The net force on the child is given by Newton’s second law. Fnet = ma = (18 kg ) −62.5 m s 2 = −1116 N  −1100 N , opposite to the velocity

(

)

This force is about 250 lbs. We also assumed that friction between the seat and child is zero, and we assumed that the bottom of the seat is horizontal. If friction existed or if the seat was tilted back, then the force that the straps would have to apply would be less. 91. (a) If the 2-block system is taken as a whole system, then the net force on the system is just the force F, accelerating the total mass. Use Newton’s second law to find the force from the mass and acceleration. Take the direction of motion caused by the force (left for the bottom block, right for the top block) as the positive direction. Then both blocks have the same acceleration.

 F = F = (m + m x

top

bottom

) a = ( 9.0 kg ) ( 2.2 m s ) = 19.8 N  2.0  10 N 2

(b) The tension in the connecting cord is the only force acting on the top block, and so must be causing its acceleration. Again use Newton’s second law.  Fx = FT = mtopa = (1.5kg ) 2.2 m s2 = 3.3 N

(

)

This could be checked by using the bottom block.  Fx = F − FT = mbottoma → FT = F − mbottoma = 19.8 N − ( 7.5kg ) 2.2 m s2 = 3.3 N

(

)

92. (a) For this scenario, find your location at a time of 4.0 sec, using Eq. 2–12b. The acceleration is found from Newton’s second law. F 1200 N a = forward = → 860 kg m

 1200 N  2  ( 4.0 s ) = 71.2 m  65 m  860 kg 

x − x0 = v0t + 12 at 2 = (15 m s )( 4.0 s ) + 12 

Yes , you will make it completely through the intersection before the light turns red. The back of your car will be 6 m beyond the far side of the intersection.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) For this scenario, find your location when the car’s velocity is 0, using Eq. 2–12c. The acceleration is found from Newton’s second law. Fbraking 2300 N a= =− → v 2 = v02 + 2a ( x − x0 ) → m 860 kg

x − x0 =

v 2 − v02 2a

0 − (15 m s )

 2300 N  2 −   860 kg 

= 42.1m  41m

No , you will stop before entering the intersection. Your car will have moved 42.1 m, and so the front of your car will have crossed into the intersection. 93. The diver has two motions with different accelerations. We apply constant acceleration relationships to each motion separately. Choose the downward direction to be positive. First use Eq. 2–12c to find the velocity at which the diver enters the water assuming her initial speed is 0. That is her initial velocity when entering the water.

(

In air: v 2 = v02 + 2a ( y − y0 ) → v = v02 + 2a ( y − y0 ) = 0 + 2 9.8 m s

) ( 5.0 m ) = 9.899 m s

While in the water, there is an upwards force of 3.5 mg,. Use the free body diagram and Newton’s second law to determine her acceleration. Then use Eq. 2–12c again to find the depth to which she goes in the water, where her final velocity is 0. In water: mg − Fwater = mg − 3.5mg = ma → a = −2.5 g

v = v + 2 a ( y − y0 ) → y − y 0 = 2

2 0

v 2 − v02 2a

0 − 9.899 m s

(

2 ( −2.5 ) 9.80 m s 2

)

= 2.0 m

Fwater

94. (a) First calculate Jeanne’s speed from falling. Let downward be positive, and use Eq. 2–12c with v0 = 0.

(

)

v 2 − v02 = 2a ( y − y0 ) → v = 0 + 2a ( y − y0 ) = 2 9.8 m s 2 ( 2.0 m ) = 6.26 m s Now calculate the average acceleration as the rope stops Jeanne, again using Eq. 2–12c, with down as positive. v − v = 2 a ( y − y0 ) → a = 2

2 0

v 2 − v02

0 − ( 6.26 m s )

= −19.6 m s 2

2 ( y − y0 ) 2 (1.0 m ) The negative sign indicates that the acceleration is upward. Since this is her acceleration, the net force on Jeanne is given by Newton’s second law, Fnet = ma.

That net force will also be upward. Now consider the free-body diagram of Jeanne as she decelerates. Call DOWN the positive direction. Newton’s second law says that Fnet = ma = mg − Frope → Frope = mg − ma. The ratio of this force to Jeanne’s weight is

Frope mg

mg − ma g

= 1.0 −

a g

= 1.0 −

−19.6 m s2 9.8 m s 2

Frope

= 3.0. Thus the rope pulls upward on Jeanne

with an average force of 3.0 times her weight . (b) A completely analogous calculation for Paul gives the same speed after the 2.0 m fall, but since he stops over a distance of 0.30 m, his acceleration is –65 m/s2, and the rope pulls upward on Paul with an average force of 7.7 times his weight . Thus, Paul is more likely to get hurt .

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95. (a) The forces acting on the small segment of the rope where FP acts are FP in the y-direction and the two equal tensions acting along the direction of the rope at an angle  below the x-axis. By Newton’s third law, the tension in the rope is equal to the force that the rope applies to the car. We use the Newton’s second law in the y-direction to determine the force on the car.  Fy = 0 → FP − FBR sin  − FCR sin  = 0 → FP − 2 FCR sin  = 0

FCR =

300 N

= 1720 N  2000 N 2 sin  2 sin 5 (b) The mechanical advantage is the ratio of the force on the car to the force she is applying. FCR 1720 N = = 5.73  6 FP 300 N (c) This method is counterproductive when the mechanical advantage drops below one, which happens when the force on the car is equal to the force she applies to the rope, FCR = FP . FCR =

FP 2 sin 

→

 F  1  = sin −1  P  = sin −1   = 30 2  2 FCR 

133

CHAPTER 5: Using Newton’s Laws: Friction, Circular Motion, Drag Forces Responses to Questions 1.

Static friction between the crate and the truck bed causes the crate to accelerate.

During the block’s ascent, there are two forces on the block that are parallel to each other causing the deceleration – the component of weight parallel to the plane, and the force of friction on the block. Since the forces are parallel to each other, both pointing down the plane, they add, causing a larger magnitude force and a larger acceleration. During the block’s descent, those same two forces are opposite of each other, because the force of friction is now directed up the plane. With these two forces being opposite of each other, the net force is smaller, and so the acceleration is smaller.

When a skier is in motion, a small coefficient of kinetic friction lets the skis move easily on the snow with minimum effort. A large coefficient of static friction lets the skier rest on a slope without slipping and keeps the skier from sliding backward when going uphill.

The static friction force is the force that prevents two objects from moving relative to one another. The “less than” sign (<) in the static friction force tells you that the actual force may have any magnitude less than that given by the equation. Typically you use Newton’s second law to determine the value of the static friction force. You then verify that the force calculated is in the allowed range given by the static friction equation. The equals sign in the equation is used when you are searching for the maximum force of static friction. For example, if an object is “on the verge” of moving away from a static configuration, you would use the equals sign.

On an icy surface, you need to put your foot straight down onto the sidewalk, with no component of velocity parallel to the surface. If you can do that, the interaction between you and the ice is through the static frictional force. If your foot has a component of velocity parallel to the surface of the ice, any resistance to motion will be caused by the kinetic frictional force, which is smaller. You will be more likely to slip.

In a very simple analysis, the net force slowing the moving object is friction. If we consider that the moving object is on a level surface, then the normal force is equal to the weight. Combining these ideas, we get the following: Fnet = ma =  FN =  mg → a =  g From Table 5–1, the “steel on steel (unlubricated)” coefficient of friction (applicable to the train) is smaller than the “rubber on dry concrete” coefficient of friction (applicable to the truck). Thus, the acceleration of the train will be smaller than that of the truck, and therefore the truck’s stopping distance will be smaller, from Eq. 2–12c.

The centripetal acceleration will be greater when the speed is greater since centripetal acceleration is proportional to the square of the speed (when the radius is constant ): aR = v 2 r. When the speed is higher, the acceleration has a larger magnitude.

The acceleration will not be the same. The centripetal acceleration is inversely proportion to the radius (when the speed is constant): aR = v 2 r. Traveling around a sharp curve, which has a smaller radius, will require a larger centripetal acceleration than traveling around a gentle curve, which has a larger radius.

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The three main forces on the child are the downward force of gravity (the child’s weight), the normal force up on the child from the horse (because the child is touching the horse), and the static frictional force on the child from the surface of the horse. The frictional force provides the centripetal acceleration. If there are other forces, such as contact forces between the child’s hands, legs, or feet and the horse, which have a radial component, they will contribute to the centripetal acceleration.

10. On level ground, the normal force on the child would be the same magnitude as their weight. This is the “typical” situation. But as the child and sled come over the crest of the hill, they are moving in a curved path, which can at least be approximated by a circle. There must be a centripetal force, pointing inward toward the center of the arc. The combination of gravity (acting downward) and the normal force on his body (acting upward when the sled is at the top of the hill) provides this centripetal force, which must be greater than zero. At the top of the hill, if downward is the positive direction, Newton’s second law says Fy = mg − FN = m v 2 r. Thus the normal force must be less than the child’s weight. 11. The statement is not correct. The “drum” of the dryer provides a centripetal force on the clothes to keep them moving in a circular path. A water droplet on the solid surface of the drum will also experience this centripetal force and move in a circle. However, as soon as the water droplet is at the location of a hole in the drum there will be no centripetal force on it and it will therefore continue moving in a path in the direction of its tangential velocity, which will take it out of the drum. There is no centrifugal force throwing the water outward; rather, there is a lack of centripetal force to keep the water moving in a circular path. 12. When describing a centrifuge experiment, the acceleration of the object in the centrifuge should be specified. Stating the rpm (in essence, stating the period or frequency of the circular motion) does not give enough information to calculate the acceleration. To allow the acceleration of the object in the centrifuge to be calculated, you also need the distance from the axis of rotation to the spinning object. 13. She should let go of the string at the moment that the tangential velocity vector is directed exactly at the target. This would also be when the string is perpendicular to the desired direction of motion of the ball. See the “top view” diagram. Also see Figs. 5–16a and 5–16b in the textbook.

Target

14. The acceleration of the ball is inward, directly toward the pole, and is provided by the horizontal component of the tension in the string. This assumes that the speed of the ball is not changing. 15. For objects (including astronauts) on the inner surface of the cylinder, the normal force provides a centripetal force which points inward toward the center of the cylinder. This normal force simulates the normal force we feel when on the surface of Earth. (a) Falling objects are not in contact with the floor, so when released they will continue to move with constant velocity until they reach the shell. From the frame of reference of the astronaut inside the cylinder, it will appear that the object “falls” in a curve, behind the point directly below the drop point, rather than straight down. The “dropped” object would still “fall” – it would eventually hit the “floor” away from the astronaut, as the shell rotates.

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(b) The magnitude of the normal force on the astronaut’s feet will depend on the radius and speed v2 v2 = g (so that m = mg for all objects), then the of the cylinder. If these are such that r r normal force will feel just like it does on the surface of Earth. Also, that normal force would point towards the astronaut’s head, just like the normal force does on the surface of the Earth. (c) Because of the large size of Earth compared to humans, we cannot tell any difference between the gravitational force at our heads and at our feet. In a rotating space colony, the difference in the simulated gravity at different distances from the axis of rotation could be significant, perhaps producing dizziness or other adverse effects. Playing “catch” with a ball could be difficult since the normal parabolic paths as experienced on Earth would not occur in the rotating cylinder. It might be possible to “adjust” gravity for the astronauts by changing the rotation speed of the spaceship. 16. At the top of the bucket’s arc, the gravitational force and normal forces from the bucket, both pointing downward, must provide the centripetal force needed to keep the water moving in a circle. v2 In the limiting case of no normal force, Newton’s second law would give Fnet = mg = m , which r means that the bucket must be moving with a tangential speed of v  gr or the water will spill out of the bucket. If the speed is greater than that, there will be a downward normal force on the water from the bucket as well. At the top of the arc, the water has a horizontal velocity. As the bucket passes the top of the arc, the velocity of the water develops a vertical component due to the vertical force on it. But the bucket is traveling with the water, with the same velocity, and so continues to contain the water as the water (and bucket) “falls” through the rest of its path. 17. (a) The normal force on the car is largest at point C. In this case, the centripetal force keeping the car in a circular path of radius R is directed upward, so the normal force must be greater than the weight to provide this net upward force. (b) The normal force is smallest at point A, the crest of the hill. At this point the centripetal force must be downward (towards the center of the circle) so the normal force must be less than the weight. (Notice that the normal force is equal to the weight at point B.) (c) The driver will feel heaviest where the normal force is greatest, or at point C. (d) The driver will feel lightest at point A, where the normal force is the least. mv 2 (e) At point A, the centripetal force is weight minus normal force, or mg − FN = . The point R at which the car just loses contact with the road corresponds to a normal force of zero, which is 2 mvmax the maximum speed without losing contact. Setting FN = 0 gives mg = → vmax = Rg . R 18. Leaning in when rounding a curve puts the bicycle tire at an angle with respect to the ground, and thus the force of the ground on the bicycle (the normal force) is also at an angle, instead of being vertical – similar to a banked roadway. A component of the normal force is now pointing horizontally, towards the center of the circle of the motion. This component provides at least part of the centripetal force needed to make the bicycle turn, and so static friction between the bicycle and the roadway can be smaller. This means the bicycle is less likely to “skid” out of the circular path while turning.

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19. When an airplane is in level flight, the downward force of gravity is counteracted by the upward lift force, analogous to the upward normal force on a car driving on a level road. The lift on an airplane is perpendicular to the plane of the airplane’s wings, so when the airplane banks, the lift vector has both vertical and horizontal components (similar to the vertical and horizontal components of the normal force on a car on a banked turn). Assuming that the plane has no vertical acceleration, then the vertical component of the lift balances the weight and the horizontal component of the lift provides the centripetal force. If FL is the total lift and  = the banking angle, measured from the vertical (as shown in the diagram), then FL cos  = mg , FL sin  = m

v2 r

, and  = tan −1 ( v 2 gr ) .

(

)

20. Solve for b, to get b = − F v . The units for b are N ( m s ) = N  s m = kg  m s 2  s m = kg s. 21. The force proportional to v 2 will dominate at high speed, because at high speeds the numeric value of v 2 is much higher.

Responses to MisConceptual Questions 1.

(b) The maximum static friction force is 25 N. Since the applied force is less than this maximum, the crate will not accelerate, Newton’s second law can be used to show that the resulting friction force will be equal in magnitude, but opposite in direction to the applied force.

(b) We assume that each tire experiences a normal force of FN = 14 mg . Then the maximum accelerating force (due to static friction) available from each “drive” tire is Ffr =  s 14 mg = 0.2mg. (i)

For 2-wheel drive, Fmax = 0.4 mg and so amax = Fmax m = 0.4 g .

(ii) For 4-wheel drive, Fmax = 0.8mg and so amax = Fmax m = 0.8 g . Thus the answer is (b). 3.

(e) In circular motion the velocity is always perpendicular to the radius of the circle, so (b), (c), and (d) are incorrect. The net force is always in the same direction as the acceleration, so if the acceleration points toward the center, the net force must also. Therefore, (e) is a better choice than (a).

(a) The forces acting on the child are gravity (downward), the normal force (away from the wall), and the force of friction (parallel to the wall and in this case opposing gravity). In particular, there is nothing “pushing” outward on the rider, so answers (b), (d), and (e) cannot be correct. If (c) were correct, the rider would accelerate downward as well as radially.

(c) A common misconception is that the ball will continue to move in a curved path after it exits the tube [answers (d) or (e)]. However, to move in a curved path a net force must be acting on the ball. When it is inside the tube, the normal force from the tube wall provides the centripetal force. After the ball exits the tube there is no net force, so the ball must travel in a straight line path in the same direction it was traveling as it exited the tube.

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(b) A common error in this problem is to ignore the contribution of gravity in the centripetal force. At the top of the loop gravity assists the tension in providing the centripetal force so the tension is less than the centripetal force. At the bottom of the loop gravity opposes the tension, so the tension is greater than the centripetal force. At all other points in the loop the tension is between the maximum at the bottom and the minimum at the top.

(b) As you turn you feel the force between yourself and the car door. A common misconception is that a centrifugal force is pushing you into the door [answer (a)]. Actually, your inertia tries to keep you moving in a straight line. As the car (and door) turn right the door accelerates into you pushing you away from your straight line motion and toward the right.

(d) The phrase “steady speed” is not the same as “constant velocity,” as velocity also includes direction. A common misconception is that if a car moves at steady speed, the acceleration and net force are zero [answers (a) or (b)]. However, since the path is circular, a radially inward force must cause the centripetal acceleration. If this force (friction between the tires and road) were not present, the car would move in a straight line. It would not accelerate outward.

(e) The force must be on the car to accelerate the car, but answers (b) and (c) are describing forces on the road. Answer (a) is describing a force internal to the car, and so it cannot accelerate the car as a whole either. The accelerating force must come from outside of the car itself, and it has to be directed “sideways” as compared to the direction of motion of the car. Answer (e) is the only one that satisfies these conditions.

10. (a) Newton’s first law says that if no force acts on an object, it will either remain at rest or move with a constant velocity. Thus if you are in a car, and no net force acts on you, you will satisfy Newton’s first law. Nothing actually “flings” you sideways when your car travels around a sharp curve. In the frame of reference of the Earth, the car “turns out from under you,” and you (at least for a short time) continue in the original direction. From your frame of reference inside the car, you assume something has “flung” you sideways. Eventually the car pushes on you sideways (via a door or seatbelt), as discussed in MisConceptual Question 7. 11. (b) At the bottom of the circular loop, the acceleration of the cart is towards the center of the circle, which is upwards. This force is the vector sum of the normal force of the track on the car (upwards) and gravity (downwards). The normal force has to be larger than gravity. 12. (b) The banked curve provides, via the radial component of the normal force, a centripetal force which is the proper force only when the  v2  relationship between banking angle, speed, and radius  tan  =  is r   satisfied. Since the car is traveling faster than the design speed, the required centripetal force is greater than that provided by the normal force. Thus another centripetal force is required. A frictional force pointing up the bank would provide a “centrifugal” force – one component of that force would point radially outward. Instead, a frictional force pointing down the bank is needed, which has a component pointing radially inward. The free body diagram, taken from the solution to Problem 98, helps to illustrate this.

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13. (d) Because the car is traveling around a curve, there must be a component of the acceleration that is directed towards the center of the circle, which is perpendicular to the direction of motion. And because the car is slowing down, there must be a component of acceleration that is opposite the direction of motion. So none of the individual answers are correct. The correct direction would be the vector sum of the accelerations described in parts (b) and (c).

Solutions to Problems 1.

A free-body diagram for the crate is shown. The crate does not accelerate vertically, and so FN = mg . The crate does not accelerate horizontally, and so FP = Ffr .

(

Ffr

)

FP = Ffr = k FN = k mg = ( 0.30 )( 26 kg ) 9.80 m s2 = 76 N

If the coefficient of kinetic friction is zero, then the horizontal force required is 0 N , since there is no friction to counteract. Of course, it would take a force to START the crate moving, but once it was moving, no further horizontal force would be necessary to maintain the motion. 2.

A free-body diagram for the box is shown. Since the box does not accelerate vertically, FN = mg . (a) To start the box moving, the pulling force must just overcome the force of static friction, and that means the force of static friction will reach its maximum value of Ffr = s FN . Thus we have for the starting motion,  Fx = FP − Ffr = 0 →

FP = Ffr = s FN = s mg → s =

FP mg

35.0 N

( 4.0 kg ) ( 9.80 m s2 )

Ffr

= 0.89

(b) The same force diagram applies, but now the friction is kinetic friction, and the pulling force is NOT equal to the frictional force, since the box is accelerating to the right.  F = FP − Ffr = ma → FP − k FN = ma → FP − kmg = ma →

k = 3.

FP − ma mg

(

35.0 N − ( 4.0 kg ) 0.60 m s 2

( 4.0 kg ) ( 9.80 m s ) 2

) = 0.83

A free-body diagram for you as you stand on the train is shown. You do not accelerate vertically, and so FN = mg . The maximum static frictional force is s FN , and that must be greater than or equal to the force needed to accelerate you in order for you not to slip. Fmax  ma → s FN  ma → s mg  ma →

Ffr mg

static friction

s  a g = 0.20 g g = 0.20 The static coefficient of friction must be at least 0.20 for you to not slide.

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See the included free-body diagram. To find the maximum angle, assume that the car is just ready to slide, so that the force of static friction is a maximum. Write Newton’s second law for both directions. Note that for both directions, the net force must be zero since the car is not accelerating.  F y = FN − mg cos  = 0 → FN = mg cos 

FN Ffr  mg

 F = mg sin  − F = 0 → mg sin  = F =  F =  mg cos  fr

mg sin 

s = 5.

mg cos 



= tan  = 0.90 →  = tan −1 0.90 = 42

(a) Here is a free-body diagram for the box at rest on the plane. The force of friction is a STATIC frictional force, since the box is at rest. (b) If the box were sliding down the plane, the only change is that the force of friction would be a KINETIC frictional force. The free-body diagram would not change.

Ffr

y x

 mg



y FN

(c) If the box were sliding up the plane, the force of friction would be a KINETIC frictional force, and it would point down the plane, as shown in the second diagram.

Notice that the value of the angle is not used in this solution. 6.

Ffr

 mg

Start with a free-body diagram. Write Newton’s second law for each direction.  Fx = mg sin  − Ffr = max



Ffr

 F = F − mg cos  = ma = 0 y

Notice that the sum in the y-direction is 0, since there is no motion (and hence no acceleration) in the y-direction. Solve for the force of friction. mg sin  − Ffr = max →

(



)

Ffr = mg sin  − max = ( 25.0 kg )  9.80 m s 2 ( sin 27 ) − 0.36 m s 2  = 102.2 N  1.0  10 2 N

Now solve for the coefficient of kinetic friction. Note that the expression for the normal force comes from the y-direction force equation above. 102.2 N Ffr = = 0.47 Ffr = k FN = k mg cos  → k = mg cos  ( 25.0 kg ) 9.80 m s 2 ( cos 27 )

(

)

The direction of travel for the car is to the right, and that is also the positive horizontal direction. Using the free-body diagram, write Newton’s second law in the x-direction for the car on the level road. We assume that the car is just on the verge of skidding, so that the magnitude of the friction force is Ffr = s FN .

 F = − F = ma x

Ffr = −ma = − s mg → s =

a g

3.80 m s 2 9.80 m s 2

FN Ffr mg

= 0.3878

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Now put the car on an inclined plane. Newton’s second law in the x-direction for the car on the plane is used to find the acceleration. We again assume the car is on the verge of slipping, so the static frictional force is at its maximum. Ffr  Fx = − Ffr − mg sin  = ma → a=

− Ffr − mg sin  m

(

− s mg cos  − mg sin  m

= − g ( s cos  + sin  )



)

= − 9.80 m s 2 ( 0.3878 cos 9.3 + sin 9.3 ) = −5.3 m s 2

A free-body diagram for the bar of soap is shown. There is no motion in the y-direction and thus no acceleration in the y-direction. Write Newton’s second law for both directions, and use those expressions to find the acceleration of the soap.  Fy = FN − mg cos  = 0 → FN = mg cos 

 F = mg sin  − F = ma x

Ffr

y 



ma = mg sin  − k FN = mg sin  − k mg cos  a = g ( sin  − k cos  )

Now use Eq. 2–12b, with an initial velocity of 0, to find the final velocity. x = x0 + v0t + 12 at 2 → t=

2x a

2x g ( sin  −  k cos  )

2 ( 7.5 m )

( 9.80 m s ) ( sin 8.0 − ( 0.060 ) cos 8.0 ) 2

= 4.38s  4 s

The value of ( sin 8.0 − ( 0.060 ) cos8.0 ) has only 1 significant figure. 9.

As the skier travels down the slope at constant speed, her acceleration parallel to the slope must be zero. Newton’s second law can then be written in component form as (see the free-body diagrams and analysis from Example 5–6):  Fx = mg sin  − k FN = 0  Fy = FN − mg cos  = 0 The vertical equation can be solved for the normal force, which can then be inserted into the horizontal equation. The horizontal equation can then be solved for the coefficient of kinetic friction. mg sin  = tan  FN = mg cos  ; mg sin  − k ( mg cos  ) = 0 → k = mg cos 

10. A free-body diagram for the box is shown, assuming that it is moving to the right. The “push” is not shown on the free-body diagram because as soon as the box moves away from the source of the pushing force, the push is no longer applied to the box. It is apparent from the diagram that FN = mg for the vertical direction. We write Newton’s second law for the horizontal direction, with positive to the right, to find the acceleration of the box.  Fx = − Ffr = ma → ma = − k FN = − kmg →

(

FN Ffr mg

)

a = − k g = −0.15 9.80 m s 2 = −1.47 m s 2

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Eq. 2–12c can be used to find the distance that the box moves before stopping. The initial speed is 4.0 m/s, and the final speed will be 0.

v − v = 2a ( x − x0 ) → x − x0 = 2

2 0

v 2 − v02 2a

0 − ( 3.5 m s )

(

2 −1.47 m s 2

)

= 4.17 m  4.2 m

11. (a) A free-body diagram for the car is shown, assuming that it is moving to the right. It is apparent from the diagram that FN = mg for the vertical direction. Write Newton’s second law for the horizontal direction, with positive to the right, to find the acceleration of the car. Since the car is assumed to NOT be sliding, use the maximum force of static friction.  Fx = − Ffr = ma → ma = − s FN = − smg → a = − s g

FN Ffr mg

Eq. 2–12c can be used to find the distance that the car moves before stopping. The initial speed is given as v, and the final speed will be 0.

v − v = 2a ( x − x0 ) → 2

2 0

( x − x0 ) =

v 2 − v02 2a

0 − v2 2 ( − s g )

v2 2 s g

(b) Using the given values:

 1m s  v = ( 95 km h )   = 26.38 m s  3.6 km h 

( 26.38 m s ) = = 55 m ( x − x0 ) = 2 s g 2 ( 0.65 ) ( 9.80 m s 2 ) 2

12. We draw three free-body diagrams – one for the car, one for the trailer, and then “add” them for the combination of car and trailer. Note that since the car pushes against the ground, the ground will push against the car with an equal but oppositely directed force. FCG is the force on the car due to the

FCT

FTC

Ffr

From consideration of the vertical forces in the individual free-body diagrams, it is apparent that the normal force on each object is equal to its weight. This leads to the conclusion that Ffr = k FN T = k mT g =

FCG

mC g

ground, FTC is the force on the trailer due to the car, and FCT is the force on the car due to the trailer. Note that by Newton’s third law, FCT = FTC .

FNC

FNT

mT g

( 0.15 )( 350 kg ) ( 9.80 m s 2 ) = 514.5 N.

Now consider the combined free-body diagram. Write Newton’s second law for the horizontal direction, This allows the calculation of the acceleration of the system.  F = FCG − Ffr = ( mC + mT ) a →

FCG − Ffr

Ffr FNT + FNC

FCG

( mC + mT ) g

3600 N − 514.5 N

= 1.804 m s 2 mC + mT 1710 kg Finally, consider the free-body diagram for the trailer alone. Again write Newton’s second law for the horizontal direction, and solve for FTC .

F = F −F = m a → TC

(

)

FTC = Ffr + mT a = 514.5 N + ( 350 kg ) 1.804 m s 2 = 1146 N  1100 N © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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13. A free-body diagram for the car in motion is shown. There is no motion Ffr in the y-direction and thus no acceleration in the y-direction. Write Newton’s second law for both directions, and use those expressions to find the acceleration of the car. IF the car is on the verge of slipping, then the frictional force will be the normal force times the coefficient of static friction.  Fy = FN − mg cos  = 0 → FN = mg cos 

y x



 F = mg sin  − F = ma x

ma = mg sin  − s FN = mg sin  − s mg cos  a = g ( sin  − s cos  )

Now use Eq. 2–12c, with a final initial velocity of 0, to find the distance traveled. v 2 − v02 = 2a ( x − x0 ) →

( x − x0 ) =

v 2 − v02 2a

0 − v2 2 g ( sin  − s cos  )

− ( 22.0 m s )

(

)

2 9.80 m s 2 ( sin15 − 0.35 cos15 )

14. Assume that the static frictional force is the only force accelerating the racer. Then consider the free-body diagram for the racer as shown. It is apparent that the normal force is equal to the weight, since there is no vertical acceleration. It is also assumed that the static frictional force is at its maximum. Thus Ffr = ma → s mg = ma → s = a g The acceleration of the racer can be calculated from Eq. 2–12b, with an initial speed of 0. x − x0 = v0t + 12 at 2 → a = 2 ( x − x0 ) t 2

s =

a g

2 ( x − x0 ) gt

(

2 1.0  103 m

)

( 9.80 m s ) (12 sec ) 2

= 312 m

FN Ffr mg

= 1.4

15. (a) See the first free-body diagram, which applies to Fig. 5–6a. Write Newton’s second law for both horizontal and vertical directions, using an acceleration of 0, and solve for the pushing force. The pushing force is inclined 25° below the horizontal.  Fy = FN − F sin  − mg = 0 → FN = F sin  + mg

 F = F cos  − F = F cos  −  F = F cos  −  ( F sin  + mg ) = 0 ( 0.12 )( 31kg ) ( 9.80 m s )  mg F= = = 42.61N  43 N fr

cos  − k sin 

cos 25 − 0.12 sin 25

(b) See the second free-body diagram, which applies to Fig. 5–6b. Write Newton’s second law for both horizontal and vertical directions, using an acceleration of 0, and solve for the pushing force. The pushing force is inclined 25° above the horizontal.

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 F = F + F sin  − mg = 0 → F = − F sin  + mg  F = F cos  − F = F cos  −  F = F cos  −  ( − F sin  + mg ) = 0 ( 0.12 )( 31kg ) ( 9.80 m s )  mg F= = = 38.09N  38 N N

cos  + k sin 

cos 25 + 0.12 sin 25

16. Assume that kinetic friction is the net force causing the deceleration. See the free-body diagram for the truck, assuming that the right is the positive direction, and the direction of motion of the skidding truck. There is no acceleration in the vertical direction, and so FN = mg. Applying Newton’s second law to the x-direction gives the following.  F = − Ffr = ma → − k FN = − kmg = ma → a = − k g

Ffr

Use Eq. 2–12c to determine the initial speed of the truck, with the final speed of the truck being zero. v 2 − v02 = 2a ( x − x0 ) →

v0 = v 2 − 2a ( x − x0 ) = 0 − 2 ( −  k g )( x − x0 ) = 2 ( 0.80 ) ( 9.80 m s 2 ) ( 72 m ) = 34 m s 17. (a) Consider the free-body diagram for the snow on the roof. If the snow is just ready to slip, then the static frictional force is at its maximum value, Ffr = s FN . Write Newton’s second law in both directions, with the net force equal to zero since the snow is not accelerating.  Fy = FN − mg cos = 0 → FN = mg cos 

 F = mg sin  − F = 0 → x

mg sin  = Ffr = s FN = s mg cos  → s = tan  = tan 24 = 0.4452  0.45 If s  0.45 , then the snow would not be on the verge of slipping. (b) The same free-body diagram applies for the sliding snow. But now the force of friction is kinetic, so Ffr =  k FN , and the net force in the x-direction is not zero. Write Newton’s second law for the x-direction again, and solve for the acceleration.  Fx = mg sin  − Ffr = ma mg sin  − Ffr

mg sin  − k mg cos 

= g ( sin  − k cos  ) m m Use Eq. 2–12c with v0 = 0 to find the speed at the end of the roof. a=

v 2 − v02 = 2a ( x − x0 ) v = v0 + 2a ( x − x0 ) = 2 g ( sin  −  k cos  )( x − x0 )

(

)

= 2 9.80 m s 2 ( sin 24 − ( 0.20 ) cos 24 ) ( 6.0 m ) = 5.133 m s  5.1m s

(c) Now the snow is in projectile motion. The snow has an initial speed of 5.133 m/s, directed at an angle of 24° below the horizontal. The horizontal speed, v x = ( 5.133 m s ) cos 24 = 4.689 m s , is constant. The vertical speed changes due to gravity. Define down to be positive. Then v y 0 = ( 5.133m s ) sin 24 = 2.088 m s , a = 9.80 m s 2 , and the vertical displacement is 10.0 m. Use Eq. 2–12c to find the final vertical speed. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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v y2 − v y20 = 2a ( y − y0 ) v y = v y20 + 2a ( y − y0 ) =

( 2.088 m s ) + 2 ( 9.80 m s 2 ) (10.0 m ) = 14.15 m/s 2

Use Eq. 2–12a to find the time of flight, and then use Eq. 2–2 to find the horizontal distance traveled. v −v 14.15 m s − 2.088 m s v y − v0 = at → t = y 0 = = 1.231s a 9.80 m s 2

x = v x t = ( 4.689 m s )(1.231s ) = 5.772 m  5.8 m (d) To find the speed when it hits the ground, the horizontal and vertical components of velocity must be combined, according to the Pythagorean theorem.

v = v x2 + v y2 =

( 4.689 m s ) + (14.15 m/s ) = 14.91m s  15 m s 2

18. Consider a free-body diagram for the box, showing the forces on the box. When FP = 23 N, the block does not move. Thus in that case, the force of friction is static friction, and must be at its maximum value, given by Ffr = s FN . Write Newton’s second law in both the x- and y-directions. The net force in each case must be 0, since the block is at rest.  Fx = FP cos  − FN = 0 → FN = FP cos 

 F = F + F sin  − mg = 0 → F + F sin  = mg y

s FN + FP sin  = mg → s FP cos  + FP sin  = mg m=

FP g

( s cos  + sin  ) =

23 N 9.80 m s 2

( 0.40 cos 22 + sin 22 ) = 1.7 kg

19. (a) Since the two blocks are in contact, they can be treated as a single object as long as no information is needed about internal forces (like the force of one block pushing on the other block). Since there is no motion in the vertical direction, it is apparent that FN = ( m1 + m2 ) g , and so Ffr =  k FN = k ( m1 + m2 ) g . Write Newton’s second law for the horizontal direction.  Fx = FP − Ffr = ( m1 + m2 ) a →

FP − Ffr m1 + m2

FP − k ( m1 + m2 ) g m1 + m2

FP Ffr

(

m1 + m2

( m1 + m2 ) g

650 N − ( 0.18)(190 kg ) 9.80 m s2

)

190 kg

= 1.657 m s2  1.7 m s2 (b) To solve for the contact forces between the blocks, an individual block must be analyzed. Look at the free-body diagram for the second block. F21 is the force of the first block pushing on the second block. Again, it is apparent that FN 2 = m2 g and so Ffr2 = k FN2 = k m2 g . Write Newton’s second law for the horizontal direction.

F21 m2

Ffr2

FN2

m2 g

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F = F − F = m a → x

fr2

Instructor Solutions Manual

(

)

(

)

F21 = k m2 g + m2 a = ( 0.18)(125 kg ) 9.80 m s 2 + (125 kg ) 1.657 m s 2 = 430 N By Newton’s third law, there will also be a 430 N force to the left on block # 1 due to block # 2. (c) If the crates are reversed, the acceleration of the system will remain the same – the analysis from part (a) still applies. We can also repeat the analysis from part (b) to find the force of one block on the other, if we simply change m1 to m2 in the free-body diagram and the resulting equations.

a = 1.657 m s  1.7 m s 2

;  Fx = F12 − Ffr1 = m1a →

(

)

(

F12 m1

Ffr1

m1g

FN1

)

F12 = k m1 g + m1a = ( 0.18)( 65 kg ) 9.80 m s2 + ( 65 kg ) 1.657 m s2 = 220 N

20. (a) Consider the free-body diagram for the crate on the surface. There is no motion in the y-direction and thus no acceleration in the y-direction. Write Newton’s second law for both directions.  Fy = FN − mg cos  = 0 → FN = mg cos 

Ffr

 F = mg sin  − F = ma x



ma = mg sin  − k FN = mg sin  − k mg cos  a = g ( sin  − k cos  )

(



)

= 9.80 m s2 ( sin 25.0 − 0.19 cos 25.0 ) = 2.454 m s 2  2.5m s 2

(b) Now use Eq. 2–12c, with an initial velocity of 0, to find the final velocity.

v 2 − v02 = 2a ( x − x0 ) → v = 2a ( x − x0 ) = 2 ( 2.454 m s 2 ) (8.75 m ) = 6.6 m s

21. (a) Consider the free-body diagram for the crate on the surface. There is no motion in the y-direction and thus no acceleration in the y-direction. Write Newton’s second law for both directions, and find the acceleration.  Fy = FN − mg cos  = 0 → FN = mg cos 

 F = mg sin  + F = ma x

FN Ffr 



ma = mg sin  + k FN = mg sin  + k mg cos  a = g ( sin  + k cos  )

Now use Eq. 2–12c, with an initial velocity of −3.0 m s and a final velocity of 0 to find the distance the crate travels up the plane. v 2 − v02 = 2a ( x − x0 ) → x − x0 =

−v02 2a

(

)

− ( −3.0 m s )

2 9.80 m s 2 ( sin 25.0 + 0.17 cos 25.0 )

= −0.796 m  −0.80 m

The crate travels 0.80 m up the plane.

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(b) We use the acceleration found above with the initial velocity in Eq. 2–12a to find the time for the crate to travel up the plane. v ( −3.0 m s ) v = v0 + at → tup = − 0 = − = 0.5308s 2 aup 9.80 m s ( sin 25.0 + 0.17 cos 25.0 )

(

)

The total time is NOT just twice the time to travel up the plane, because the acceleration of the block is different for the two parts of the motion. The second free-body diagram applies to the block sliding down the plane. A similar analysis will give the acceleration, and then Eq. 2–12b with an initial velocity of 0 is used to find the time to move down the plane.  Fy = FN − mg cos  = 0 → FN = mg cos 

FN y

Ffr 



 F = mg sin  − F = ma x

ma = mg sin  − k FN = mg sin  − k mg cos  a = g ( sin  − k cos  )

x − x0 = v0t + 12 at 2 → 2 ( x − x0 )

tdown =

adown

2 ( 0.796 m )

( 9.80 m s ) ( sin 25.0 − 0.17 cos 25.0 ) 2

= 0.7778s

t = tup + tdown = 0.5308s + 0.7778s = 1.3s

It is worth noting that the final speed is about 2.0 m/s, significantly less than the 3.0 m/s original speed. 22. The force of static friction is what decelerates the crate if it is not sliding on the truck bed. If the crate is not to slide, but the maximum deceleration is desired, then the maximum static frictional force must be exerted, and so Ffr =  s FN .

Ffr

The direction of travel is to the right. It is apparent that FN = mg since there is no acceleration in the y-direction. Write Newton’s second law for the truck in the horizontal direction.

FN mg

 F = − F = ma → −  mg = ma → a = −  g = − ( 0.75) ( 9.80 m s ) = −7.4 m s 2

The negative sign indicates the direction of the acceleration – opposite to the direction of motion. 23. (a) For mB to not move, the tension must be equal to mB g , and so mB g = FT . For mA to not move, the tension must be equal to the force of static friction, and so FS = FT . Note that the normal force on mA is equal to its weight. Use these relationships to solve for mA . mB g = FT = Fs  s mA g → mA 

s

2.0 kg 0.40

= 5.0 kg → mA  5.0 kg

(b) For mB to move with constant velocity, the tension must be equal to mB g. For mA to move with constant velocity, the tension must be equal to the force of kinetic friction. Note that the normal force on mA is equal to its weight. Use these relationships to solve for mA. mB g = Fk =  k mA g → mA =

k

2.0 kg 0.30

= 6.7 kg

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Instructor Solutions Manual

24. First consider the free-body diagram for the snowboarder on the incline. Write Newton’s second law for both directions, and find the acceleration.  Fy = FN − mg cos  = 0 → FN = mg cos 

 F = mg sin  − F = ma x

ma = mg sin  − k1 FN = mg sin  − k1mg cos 

(

)

aslope = g ( sin  − k1 cos  ) = 9.80 m s 2 ( sin 28 − 0.18cos 28 ) = 3.043m s 2  3.0 m s 2

Now consider the free-body diagram for the snowboarder on the flat surface. Again use Newton’s second law to find the acceleration. Note that the normal force and the frictional force are different in this part of the problem.  Fy = FN − mg = 0 → FN = mg  Fx = − Ffr = ma maflat = − Ffr = − k2 FN = − k2 mg →

(

)

aflat = − k 2 g = − ( 0.15) 9.80 m s2 = −1.47 m s 2  −1.5m s 2

Use Eq. 2–12c to find the speed at the bottom of the slope. This is the speed at the start of the flat section. Eq. 2–12c can be used again to find the distance x. v 2 − v02 = 2a ( x − x0 ) →

vend of = v02 + 2aslope ( x − x0 ) = slope

( 5.0 m s ) + 2 ( 3.043 m s 2 ) (110 m ) = 26.35 m s = v0 on flat 2

Flat surface: v 2 − v02 = 2a ( x − x0 ) →

( x − x0 ) =

v 2 − v02 2aflat

0 − ( 26.35 m s )

(

2 −1.47 m s 2

)

= 236 m  240 m

25. The belt is sliding underneath the box (to the right), so there will be a force of kinetic friction on the box, until the box reaches a speed of 1.5 m/s. Use the freebody diagram to calculate the acceleration of the box. (a)  Fx = Ffr = ma = k FN = k mg → a = k g

FN mg

 F = F = ma =  F =  mg → a =  g x

v = v0 + at → t =

v − v0 a

Ffr

v −0

k g

1.5 m s

( 0.70 ) ( 9.80 m s 2 )

= 0.22 s

(1.5m s ) v 2 − v02 v2 = = = 0.16 m (b) x − x0 = 2a 2k g 2 ( 0.70 ) ( 9.80 m s 2 ) 2

y 26. We define the positive x-direction to be the direction of motion for each block. See the free-body diagrams. FT FNA Write Newton’s second law in both dimensions for both objects. Add the two x-equations to find the FfrA acceleration. A A Block A: mA g  FyA = FNA − mA g cos A = 0 → FNA = mA g cos A

FfrB

FNB

y x

B

B mB g

 F = FT − m g sin  − F = m a xA

frA

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Block B:  FyB = FNB − mB g cos  B = 0 → FNB = mB g cos  B

 F = m g sin  − F − F = m a xB

frB

Add the final equations together from both analyses and solve for the acceleration, noting that in both cases the friction force is found as Ffr =  FN . mA a = FT − mA g sin  A − A mA g cos  A ; mBa = mB g sin  B −  BmB g cos  B − FT mA a + mBa = FT − mA g sin  A − A mA g cos  A + mB g sin  B −  BmB g cos  B − FT →

 − mA ( sin  A + A cos  A ) + mB ( sin  − B cos  )   ( mA + m B )    − ( 2.0 kg )( sin 51 + 0.30 cos 51 ) + ( 5.0 kg )( sin 21 − 0.30 cos 21 )  = ( 9.80 m s 2 )   ( 7.0 kg )  

a = g

= −2.2 m s 2

27. We assume that the child starts from rest at the top of the slide, and then slides a distance x − x0 along the slide. A force diagram is shown for the child on the slide. First, ignore the frictional force and so consider the no-friction case. All of the motion is in the x-direction, so we will only consider Newton’s second law for the x-direction. We also set x0 = 0.

 F = mg sin  = ma → a = g sin 

Ffr

FN y x 



Use Eq. 2–12c to calculate the speed at the bottom of the slide.

v 2 − v02 = 2ax → v( No friction ) = v02 + 2ax = 2 g x sin  Now include kinetic friction. We must now consider Newton’s second law in both the x- and y-directions. The net force in the y-direction must be 0 since there is no acceleration in the y-direction.  Fy = FN − mg cos = 0 → FN = mg cos

 F = ma = mg sin  − F = mg sin  −  F = mg sin  −  mg cos  x

mg sin  − k mg cos 

= g ( sin  − k cos  ) m With this acceleration, we can again use Eq. 2–12c to find the speed after sliding a certain distance. a=

v 2 − v02 = 2ax ( x − x0 ) → v( friction ) = v02 + 2ax = 2 g x ( sin  − k cos  ) Now let the speed with friction be half the speed without friction, and solve for the coefficient of friction. Square the resulting equation and divide by g cos  to get the result.

v( friction ) = 12 v( No friction ) →

2 g x ( sin  −  k cos  ) = 12 2 g x sin 

2 g x ( sin  −  k cos  ) = 14 2 g x sin  → sin  −  k cos  = 14 sin  →

 k = 34 tan  = 34 tan 28 = 0.3988  0.40

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28. (a) Given that mB is moving down, mA must be moving up the incline, and so the force of kinetic friction on mA will be directed down the incline. Since the blocks are tied together, they will both have the same acceleration, and so a yB = a xA = a. Write Newton’s second law for each mass.  FyB = mB g − FT = mBa → FT = mB g − mBa

Instructor Solutions Manual

Ffr xA



mB g

 mA g

 F = F − m g sin  − F = m a  F = F − m g cos  = 0 → F = m g cos  xA

Take the information from the two y-equations and substitute into the x-equation to solve for the acceleration. mB g − mBa − mA g sin  − k mA g cos  = mA a → a=

mB g − mA g sin  − mA g k g cos 

( mA + m B )

(

= 12 g (1 − sin  − k g cos  )

)

= 12 9.80 m s 2 (1 − sin 34 − 0.15cos 34 ) = 1.6 m s 2

(b) To have an acceleration of zero, the expression for the acceleration must be zero. a = 12 g (1 − sin  −  k cos  ) = 0 → 1 − sin  −  k cos  = 0 →

k =

1 − sin  cos 

1 − sin 34 cos 34

= 0.53

29. Since the upper block has a higher coefficient of friction, that block will “drag behind” the lower block. Thus there will be tension in the cord, and the blocks will have the same acceleration. From the free-body diagrams for each block, we write Newton’s second law for both the x- and y-directions for each block, and then combine those equations to find the acceleration and tension. (a) Block A:  FyA = FNA − mA g cos  = 0 → FNA = mA g cos 

FT 

FNA

FfrA 

FfrB

FNB



mB g

mA g

 F = m g sin  − F − F = m a xA

frA

mA a = mA g sin  − A FNA − FT = mA g sin  − A mA g cos  − FT

Block B:  FyB = FNB − mB g cos  = 0 → FNB = mB g cos 

 F = m g sin  − F + F = m a xB

frA

mBa = mB g sin  − B FNB + FT = mB g sin  − BmB g cos  + FT Add the final equations together from both analyses and solve for the acceleration. mA a = mA g sin  − A mA g cos  − FT ; mBa = mB g sin  − BmB g cos  + FT

mA a + mBa = mA g sin  − A mA g cos  − FT + mB g sin  − BmB g cos  + FT →

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 mA ( sin  − A cos  ) + mB ( sin  −  B cos  )   ( mA + m B )  

a = g

) (

 4.0 kg )( sin 32 − 0.20 cos 32 ) + ( 4.0 kg )( sin 32 − 0.30 cos 32 )   ( 8.0 kg )  

(

= 9.80 m s 2 

= 3.1155 m s 2  3.1m s 2

(b) Solve one of the equations for the tension force. mA a = mA g sin  − A mA g cos  − FT → FT = mA ( g sin  − A g cos  − a )

(

)

= ( 4.0 kg )  9.80 m s 2 ( sin 32 − 0.20 cos 32 ) − 3.1155 m s 2  = 1.7 N

30. (a) If  A   B , the untethered acceleration of mA would be greater than that of m B . If there were no cord connecting the masses, mA would “run away” from m B . So if they are joined together, mA would be restrained by the tension in the cord, mB would be pulled forward by the tension

in the cord, and the two masses would have the same acceleration. This is exactly the situation for Problem 29. (b) If  A   B , the untethered acceleration of mA would be less than that of m B . So even if there is a cord between them, mB will move ever closer to mA , and there will be no tension in the cord. If the incline were long enough, eventually mB would catch up to mA and begin to push it down the plane. At that point they would have the same acceleration. (c) For  A   B , the analysis will be exactly like Problem 29. Refer to that free-body diagram and analysis. The acceleration and tension are as follows, taken from the Problem 29 analysis.  m ( sin  − A cos  ) + mB ( sin  −  B cos  )  a = g A  ( mA + mB )   mA a = mA g sin  − A mA g cos  − FT → FT = mA g sin  − A mA g cos  − mA a

 mA ( sin  − A cos  ) + mB ( sin  −  B cos  )   ( mA + mB )  

= mA g sin  − A mA g cos  − mA g  =

mA mB g cos 

( mA + mB )

( B − A )

For  A   B , we can follow the analysis of Problem 29 but not include the tension forces. Each block will have its own acceleration. Refer to the free-body diagram for Problem 29. Block A:  FyA = FNA − mA g cos  = 0 → FNA = mA g cos 

 F = m g sin  − F = m a xA

frA

mA aA = mA g sin  − A FNA = mA g sin  − A mA g cos  → aA = g ( sin  − A cos  ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

Block B:  FyB = FNB − mB g cos  = 0 → FNB = mB g cos 

 F = m g sin  − F = m a A

frA

mBaB = mB g sin  −  B FNB = mB g sin  −  BmB g cos  → aB = g ( sin  − B cos  )

Note that since  A   B , aA  aB as mentioned above. And FT = 0 . Once the blocks come in contact, then the analysis would include a contact force pushing in the positive direction on block A, and an equal but opposite force pushing in the negative direction on block B. The analysis would be the same as in Problem 29, but changing each tension force to the contact force and reversing their directions. In particular, the expression for the acceleration would be the same as that found in Problem 29. 31. Let mA = 3.0 kg be the left-most mass, mB = 2.5 kg be the mass on the table, and mC = 5.0 kg be the right-most mass. We draw free-body diagrams for each mass. We also define the positive direction to be upwards for the left mass, to the right for the mass on the table, and downward for the right mass. + FT1 + FT2 FNB FT2 FT1 + Ffr mA g

mCg

mB g

Now write Newton’s second law for each object in the direction of motion, recognizing that each object has the same acceleration in the direction of motion. Since the mass on the table is moving horizontally, the friction force is the coefficient of kinetic friction times the weight of m B .

F = F − m g = m a ; F = F − F − F = m a ; F = m g − F = m a A

Add the three equations together, and solve for the acceleration. ( FT1 − mA g ) + ( FT2 − FT1 − Ffr ) + ( mC g − FT2 ) = mA a + mBa + mCa →

FT1 − mA g + FT2 − FT1 − Ffr + mC g − FT2

mA + mB + mC mC − mA − k mB mA + mB + mC

(

= 9.80 m s 2

)

mC g − mA g − Ffr mA + mB + mC

mC g − mA g − k mB g mA + mB + mC

5.0 kg − 3.0 kg − ( 0.35)( 2.5 kg ) 10.5 kg

= 1.05 m s 2

 1.1m s 2 32. Draw a free-body diagram for each block. FNA

−Ffr AB

FT Ffr B

Ffr AB mA g Block A (top)

FNB

−FNA

F mB g Block B (bottom)

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Ffr AB is the force of friction between the two blocks, FNA is the normal force of contact between the two blocks, Ffr B is the force of friction between the bottom block and the floor, and FNB is the

normal force of contact between the bottom block and the floor. Neither block is accelerating vertically, and so the net vertical force on each block is zero. top: FNA − mA g = 0 → FNA = mA g

bottom: FNB − FNA − mB g = 0 → FNB = FNA + mB g = ( mA + mB ) g Take the positive horizontal direction to be the direction of motion of each block. Thus for the bottom block, positive is to the right, and for the top block, positive is to the left. Then, since the blocks are constrained to move together by the connecting string, both blocks will have the same acceleration. Write Newton’s second law for the horizontal direction for each block. top: FT − Ffr AB = mA a bottom: F − FT − Ffr AB − Ffr B = mB a (a) If the two blocks are just to move, then the force of static friction will be at its maximum, and so the frictions forces are as follows. Ffr AB = s FNA = s mA g ; Ffr B = s FNB = s ( mA + mB ) g Substitute into Newton’s second law for the horizontal direction with a = 0 and solve for F. top: FT − s mA g = 0 → FT = s mA g bottom: F − FT − s mA g − s ( mA + mB ) g = 0 → F = FT + s mA g + s ( mA + mB ) g = s mA g + s mA g + s ( mA + mB ) g

(

)

= s ( 3mA + mB ) g = ( 0.60 )(14 kg ) 9.80 m s 2 = 82.32 N  82 N

(b) Multiply the force by 1.1 so that F = 1.1 ( 82.32 N ) = 90.55 N. Again use Newton’s second law for the horizontal direction, but with a  0 and using the coefficient of kinetic friction. FT − k mA g = mA a top: bottom: F − FT − k mA g − k ( mA + mB ) g = mBa sum:

F −  k mA g −  k m A g −  k ( m A + m B ) g = ( m A + m B ) a →

F − k mA g − k mA g − k ( mA + mB ) g

F − k ( 3mA + mB ) g

( mA + mB ) ( mA + m B ) 2 90.55 N − ( 0.40 )(14.0 kg ) ( 9.80 m s ) = = 4.459 m s2  4.5 m s 2 (8.0 kg )

33. Free-body diagrams are shown for both blocks. There is a force of friction between the two blocks, which acts to the right on the top block, and to the left on the bottom block. They are a Newton’s third law pair of forces. (a) If the 4.0 kg block does not slide off, then it must have the same acceleration as the 12.0 kg block. That acceleration is caused by the force of static friction between the two blocks. To find the minimum coefficient, we use the maximum force of static friction. 5.2 m s 2 a Ffr = mtop a =  FN =  mtop g →  = = = 0.5306  0.53 g 9.80 m s 2 top top

top

Ffr

top

mtopg

bottom

Ffr

−FN

top

bottom

mbottomg © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) If the coefficient of friction only has half the value, then the blocks will be sliding with respect to one another, and so the friction will be kinetic.  = 12 ( 0.5306 ) = 0.2653 ; Ffr = mtopa =  FN = mtop g → top

(

top

)

a =  g = ( 0.2653) 9.80 m s2 = 2.6 m s2

(c) The bottom block is still accelerating to the right at 5.2 m s 2 . Since the top block has a smaller acceleration than that, it has a negative acceleration relative to the bottom block. a top rel = a top rel + a ground rel = a top rel − a bottom rel = 2.6 m s2 î − 5.2 m s2 î = −2.6 m s2 î bottom

ground

bottom

ground

The top block has an acceleration of 2.6 m s2 to the left relative to the bottom block. (d) No sliding: Fx = FP − Ffr bottom net

= mbottom abottom →

bottom

FP = Ffr bottom

+ mbottom abottom = Ffr + mbottom abottom = mtop a top + mbottom a bottom = ( mtop + mbottom ) a top

(

)

= (16.0 kg ) 5.2 m s 2 = 83 N

This is the same as simply assuming that the external force is accelerating the total mass. The internal friction need not be considered if the blocks are not moving relative to each other. Sliding: Fx

= FP − Ffr

bottom net

= mbottom abottom →

bottom

FP = Ffr

+ mbottom abottom = Ffr + mbottoma bottom = mtopa top + mbottoma bottom

bottom

top

(

)

(

)

= ( 4.0 kg ) 2.6 m s 2 + (12.0 kg ) 5.2 m s 2 = 73 N

Again this can be interpreted as the external force providing the acceleration for each block. The internal friction need not be considered. 34. To find the limiting value, we assume that the blocks are NOT slipping, but that the force of static friction on the smaller block is at its maximum value, so that Ffr =  FN . For the two-block system, there is no friction on the system, and so F = ( M + m ) a describes the

FN Ffr

 horizontal motion of the system. Thus the upper block has a vertical F mg acceleration of 0 and a horizontal acceleration of . Write ( M + m) Newton’s second law for the upper block, using the force diagram, and solve for the applied force F. Note that the static friction force will be DOWN the plane, since the block is on the verge of sliding UP the plane.

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 F = F cos  − F sin  − mg = F ( cos  −  sin  ) − mg = 0 → F = ( cos  −  sin  ) N

 F = F sin  + F cos  = F ( sin  +  cos  ) = ma = m M + m → N

M +m

F = FN ( sin  +  cos  )

= ( M + m) g

( cos  −  sin  )

( sin  +  cos  )

M +m m

( sin  +  cos  ) ( cos  −  sin  )

35. A free-body diagram for the car at one instant of time is shown. In the diagram, the car is coming out of the paper at the reader, and the center of the circular path is to the right of the car, in the plane of the paper. If the car has its maximum speed, it would be on the verge of slipping, and the force of static friction would be at its maximum value. The vertical forces (gravity and normal force) are of the same magnitude, because the car is not accelerating vertically. We assume that the force of friction is the force causing the circular motion. v2 = s FN = s mg → FR = Ffr → m r

v = s rg =

Ffr

( 0.65)( 90.0 m ) ( 9.80 m s 2 ) = 23.94 m s  24 m s

Notice that the result is independent of the car’s mass . 36. (a) Find the centripetal acceleration from Eq. 5–1.

aR = v 2 r = (1.30 m s ) 1.40 m = 1.207 m s 2  1.21m s 2 2

(b) The net horizontal force is causing the centripetal motion, and so will be the centripetal force. FR = maR = ( 22.5 kg ) (1.207 m s 2 ) = 27.16N  27.2 N 37. Find the speed from Eq. 5–3. FR =

mv 2 r

FR r

→ v=

( 310 N )( 0.90 m ) 2.0 kg

= 11.81m s  12 m s

38. Find the centripetal acceleration from Eq. 5–1. aR = v 2 r =

( 525 m s )

( 4.80  10 m 3

)  9.801 mg s  = 5.86 g’s

= 57.42 m s 2 





39. In the free-body diagram, the car is coming out of the paper at the reader, and the center of the circular path is to the right of the car, in the plane of the paper. The vertical forces (gravity and normal force) are of the same magnitude, because the car is not accelerating vertically. We assume that the force of friction is the force causing the circular motion. If the car has its maximum speed, it would be on the verge of slipping, and the force of static friction would be at its maximum value.

Ffr

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  1m s   95 km hr ( )  3.6 km hr    v2     = 0.84 2 → m v r = s FN = s mg → s = = 2

FR = Ffr

( 85 m ) ( 9.80 m s )

Notice that the result is independent of the car’s mass. 40. The centripetal force that the tension provides is given by Eq. 5–3. Solve that for the speed. FR =

mv 2

→ v=

FR r m

( 75 N )(1.3 m ) 0.45 kg

= 14.72 m s  15 m s

41. The radius of either skater’s motion is 0.80 m, and the period is 2.5 sec. Thus their speed is given by 2 ( 0.80 m ) v = 2 r T = = 2.01m s . Since each skater is moving in a circle, the net radial force on 2.5 s each one is given by Eq. 5–3.

FR = m

v2 r

( 55.0 kg )( 2.01m s )

0.80 m

= 277.8 N  280 N

42. The centripetal acceleration of a rotating object is given by Eq, 5–1, aR = v 2 r.

v = aR r =

(1.25 10 g ) r = (1.25 10 )(9.80 m s )(8.00 10 m ) = 3.130 10 m s. 5



−2



1 rev  60 s  = 3.74  10 rpm ( 3.130  10 m s )  2 8.00     10 m )   1 min   ( 2

−2

43. The car will start to skid when the required centripetal force is larger than the maximum force that static friction can supply. Since the car is on a level roadway, the maximum force of static friction is the coefficient of static friction times the weight of the car.

(10.0 m s )  s mg → r  = = 12.75 m  13 m FR  Ffr → m s g ( 0.80 ) ( 9.80 m s 2 ) r v2

44. A free-body diagram for the ball is shown, similar to Fig. 5–17, but with the angle exaggerated to make it easier to see. The tension in the FT  suspending cord must not only hold the ball up, but also provide the centripetal force needed to make the ball move in a circle. Write mg Newton’s second law for the vertical direction, noting that the ball is not accelerating vertically. mg  Fy = FT sin  − mg = 0 → FT = sin  The force moving the ball in a circle is the horizontal portion of the tension. Write Newton’s second law for that radial motion.  FR = FT cos  = maR = m v 2 r Substitute the expression for the tension from the first equation into the second equation, and solve for the angle. Also substitute in the fact that for a rotating object, v = 2 r T . We also see that if the string is of length l , then the radius of the circle is r = l cos  . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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FT cos  = sin  =

mg sin 

gT 2 4 2 l

cos  =

mv 2 r

→  = sin

−1

The tension is then given by FT =

4 2 mr T2

gT 2 4 2 l mg sin 

4 2 ml cos  T2

→

( 9.80 m s ) ( 0.500 s ) = 5.94 = sin 2

−1

4 2 ( 0.600 m )

( 0.150 kg ) ( 9.80 m s2 ) sin 5.94

= 14.2 N

45. For an unbanked curve, the centripetal force to move the car in a circular path must be provided by the static frictional force. Since the roadway is level, the normal force on the car is equal to its weight. Assume the static frictional force is at its maximum value, and use the force relationships to calculate the radius of the curve. See the free-body diagram, which assumes the center of the curve is to the right in the diagram. The static friction coefficient for rubber on wet concrete is 0.7. FR = Ffr → m v 2 r = s FN = s mg →

Ffr

  1m s   ( 50 km h )  3.6 km h      = 28.12 m  30 m r = v 2 s g =  2

( 0.7 ) ( 9.80 m s )

46. We assume the water is rotating in a vertical circle of radius r. When the bucket is at the top of its motion, there would be two forces on the water (considering the water as a single mass). The weight of the water would be FN directed down, and the normal force of the bottom of the bucket pushing on the mg water would also be down. See the free-body diagram. If the water is moving in a circle, then the net downward force would be a centripetal force.  v2  v2  F = FN + mg = ma = m r → FN = m  r − g    The limiting condition of the water falling out of the bucket means that the water loses contact with the bucket, and so the normal force becomes 0.  v2   v2  FN = m  − g  → m  critical − g  = 0 → vcritical = rg  r   r  From this, we see that yes , it is possible to whirl the bucket of water fast enough. The minimum speed is

rg . All you really need to know is the radius of the circle in which you will be swinging

the bucket. It would be approximately the length of your arm, plus the height of the bucket. 47. At the top of a circle, a free-body diagram for the passengers would be as shown, assuming the passengers are upside down. Then the car’s normal force would be pushing DOWN on the passengers, as shown in the diagram. We assume no safety devices are present. Choose the positive direction to be down, and write Newton’s second law for the passengers.  FR = FN + mg = maR = m v 2 r → FN = m ( v 2 r − g )

We see from this expression that for a high speed, the normal force is positive, meaning the passengers are in contact with the car. But as the © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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speed decreases, the normal force also decreases. If the normal force becomes 0, the passengers are no longer in contact with the car – they are in free fall. The limiting condition is as follows.

(9.80 m s ) ( 7.6 m ) = 8.6 m s

2 vmin r − g = 0 → vmin = rg =

48. (a) At the bottom of the motion, a free-body diagram of the bucket would be as shown. Since the bucket is moving in a circle, there must be a net force on it towards the center of the circle, and a centripetal acceleration. Write Newton’s second law for the bucket, with up as the positive direction.  FR = FT − mg = ma = m v 2 r →

r ( FT − mg )

(1.10 m )  25.0 N − ( 2.00 kg ) ( 9.80 m s 2 )

= 1.723  1.7 m s m 2.00 kg (b) A free-body diagram of the bucket at the top of the motion is shown. Since the bucket is moving in a circle, there must be a net force on it towards the center of the circle, and a centripetal acceleration. Write Newton’s second law for the FT bucket, with down as the positive direction. mg

 F = F + mg = ma = m v r → v = 2

r ( FT + mg )

If the tension is to be zero, then

r ( 0 + mg )

(

)

= rg = (1.10 m ) 9.80 m s 2 = 3.28 m s m The bucket must move faster than 3.28 m/s in order for the rope not to go slack.

49. A free-body diagram for the car is shown. Write Newton’s second law for the car in the vertical direction, assuming that up is positive. The normal force is twice the weight, so FN = 2mg.

 F = F − mg = ma → 2mg − mg = m v 2 r → v = rg = (125 m ) ( 9.80 m s ) = 35 m s N

50. The free-body diagram for passengers at the top of a Ferris wheel is as shown. FN is the normal force of the seat pushing up on the passenger. The sum of the forces on the passenger is producing the centripetal motion, and so must be a centripetal force. Call the downward direction positive, and write Newton’s second law for the passenger.  FR = mg − FN = ma = m v 2 r

Since the passenger is to feel “weightless,” they must lose contact with their seat, and so the normal force will be 0. The diameter is 22 m, so the radius is 11 m.

mg = m v 2 r → v =



gr =

( 9.80 m s ) (11m ) = 10.38 m s 2

  60s    = 9.0 rpm  2 (11m )   1min 

(10.38 m s ) 

1 rev

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51. The car moves in a horizontal circle, and so there must be a net horizontal centripetal force. The car is not accelerating vertically. Write Newton’s second law for both the x- and y-directions. mg  Fy = FN cos − mg = 0 → FN = cos  Fx =  FR = FN sin  = max



The amount of centripetal force needed for the car to round the curve is as follows. 2

  1.0 m s   ( 85 km h )  3.6 km h   2 v    = 1.025  104 N FR = m = (1250 kg )  r 68 m The actual horizontal force available from the normal force is as follows. mg FN sin  = sin  = mg tan  = (1250 kg ) 9.80 m s 2 tan14 = 3.054  103 N cos  Thus more force is necessary for the car to round the curve than can be y supplied by the normal force. That extra force will have to have a x horizontal component to the right in the free body diagram in order to provide the extra centripetal force. Accordingly, we add a frictional force pointed down the plane. That corresponds to the car not being able to make the curve without friction. mg Again write Newton’s second law for both directions, and again the y-acceleration is zero. mg + F sin   Fy = FN cos  − mg − Ffr sin  = 0 → FN = cosfr

(

 Fx = FN sin  + Ffr cos  = m

)



FN  

Ffr

r Substitute the expression for the normal force from the y-equation into the x-equation, and solve for the friction force. mg + Ffr sin  v2 v2 2 → ( mg + Ffr sin  ) sin  + Ffr cos  = m cos  sin  + Ffr cos  = m cos  r r

Ffr = m

v2 r

(

)

(

)

cos  − mg sin  = 1.025  10 4 N cos14 − (1250 kg ) 9.80 m s 2 sin14

= 6.980  103 N

So a frictional force of 7.0  103 N down the plane is needed to provide the necessary centripetal force to round the curve at the specified speed. 52. (a) See the free-body diagram for the pilot in the jet at the bottom of the loop. We have aR = v 2 r = 5.0 g.

v 2 r = 5.0 g → r =

v2 5.0 g

  1m s   mg (1200 km h )  3.6 km h      = 2268 m  2300 m = 2

(

5.0 9.80 m s

)

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(b) The net force must be centripetal, to make the pilot go in a circle. Write Newton’s second law for the vertical direction, with up as positive. The normal force is the effective weight.  FR = FN − mg = m v 2 r 2 The centripetal acceleration is to be v r = 5.0 g .

(

)

FN = mg + m v 2 r = 6.0mg = 6.0 ( 78 kg ) 9.80 m s 2 = 4586 N = 4600 N

(c) See the free-body diagram for the pilot at the top of the loop. Notice that the normal force is down, because the pilot is upside down. Write Newton’s second law in the vertical direction, with down as positive.  FR = FN + mg = m v 2 r = 5.0mg → FN = 5.0mg = 3100 N

53. A free-body diagram for the sinker weight is shown. We use the symbol l to represent the length of the string actually swinging the sinker. The radius of the circle in which the sinker is moving is r = l sin  . Write Newton’s second law for the vertical direction, noting that the sinker is not accelerating vertically. Take up to be positive. mg  Fy = FT cos  − mg = 0 → FT = cos  The radial force is the horizontal portion of the tension. Write Newton’s second law for the radial motion. v2  FR = FT sin  = maR = m r Substitute the tension from the vertical equation, and the relationships r = l sin  and v = 2 r T .

FT sin  = m

 = cos

−1

v2 r

gT 2 4 2 l

→

cos 

sin  =

4 2 ml sin  T2

→ cos  =

( 9.80 m s ) ( 0.65s ) = 76.52  77 = cos 2

−1

gT 2 4 2 l

4 2 ( 0.45 m )

54. To experience a gravity-type force, objects must be on the inside of the outer wall of the tube, so that there can be a centripetal force to move the objects in a circle. See the free-body diagram for a person on the inside of the outer wall, and a portion of the tube. The normal force of contact between the person and the wall must be maintaining the circular motion. Write Newton’s second law for the radial direction.  FR = FN = ma = m v 2 r

This is to have nearly the same effect as Earth gravity, with FN = 0.90 mg . Equate the two expressions for normal force and solve for the speed.

FN = m v 2 r = 0.90 mg → v = 0.90 gr =



( 0.90 ) (9.80 m s 2 ) ( 550 m ) = 69.65 m s

  86, 400 s    = 1741rev d  1700 rev day  2 ( 550 m )   1 day 

( 69.65 m s ) 

1 rev

This is about 1.2 rpm. Note that the person walks on the inner portion of the outer tube.

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55. When the bug starts to slip, that indicates that the static friction has reached its maximum value. Accordingly, we set the static frictional force equal to the centripetal force. Since the bug is on a level surface with no additional vertical forces, the normal force is equal to the weight. Note that the speed is found from the rotational speed.  rev   2 r  1min   13  v =  65    =   r  m s ( r given in meters )  min   1rev  60 s   6 

 r=

 13  r    v2 6  =m F = F =  F =  mg = m R

s g

 13     6 

( 0.42 )( 9.80 )  13     6 

→

= 8.9  10 −2 m

After the bug moves past that point, they can’t stay stationary on the turntable any more. It will slide outward from the center (although not radially), and eventually fall off the turntable. 56. The force of static friction is causing the circular motion – it is the centripetal force. The coin slides off when the static frictional force is not large enough to move the coin in a circle. The maximum static frictional force is the coefficient of static friction times the normal force, and the normal force is equal to the weight of the coin as seen in the free-body diagram, since there is no vertical acceleration. In the free-body diagram, the coin is coming out of the paper and the center of the circle is to the right of the coin, in the plane of the paper.

Ffr

The rotational speed must be changed into a linear speed. rev   1 min   2 ( 0.105 m )   v =  38.0    = 0.4178 m s min   60 s   1 rev  

v2 v2 ( 0.4178 m s ) → m = s FN = s mg →  s = = = 0.170 r rg ( 0.105 m ) ( 9.80 m s 2 ) 2

FR = Ffr

57. (a) A free-body diagram of the car at the instant it is on the top of the hill is shown. Since the car is moving in a circular path, there must be a net centripetal force downward. Write Newton’s second law for the car, with down as the positive direction. v2  FR = mg − FN = ma = m r →



v2 



r 

FN = m  g −

  

2  = ( 975 kg )  9.80 m s −

(14.0 m s )  2

88.0 m

 = 7383.4 N  7380 N 

(b) The free-body diagram for the passengers would be the same as the one for the car, leading to the same equation for the normal force on the passengers. 2   v2  (14.0 m s )  2 FN = m  g −  = ( 72.0 kg )  9.80 m s −  = 545.2 N  545 N  r  88.0 m   

Notice that this is significantly less than the 706-N weight of the passenger. Thus the passenger will feel “light” as they drive over the hill. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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

v2 

FN = m  g −

= 0 → g = v 2 r → v = gr =  r 



( 9.80 m s ) (88.0 m ) = 29.4 m s 2

58. If the masses are in line and both have the same frequency of rotation, then they will always stay in line. Consider a freeFNA FNB body diagram for both masses, from a side view, at the FTB FTB FTA m instant that they are to the left of the post. Note that the A mB same tension that pulls inward on mass 2 pulls outward on mB g mA g mass 1, by Newton’s third law. Also notice that since there is no vertical acceleration, the normal force on each mass is equal to its weight. Write Newton’s second law for the horizontal direction for both masses, noting that they are in uniform circular motion. vA2 vB2 F = F − F = m a = m F = F = m a = m  RA TA TB A A A r  RB TB B B B r A B

 rev   2 r    = 2 rf .  sec   1 rev 

The speeds can be expressed in terms of the frequency as follows: v =  f FTB = mB

vB2 rB

= mB ( 2 rB f ) rB = 4 2 mB rB f 2 2

FTA = FTB + mA

vA2 rA

= 4 mB rB f 2 + mA ( 2 rA f ) rA = 4 2 f 2 ( mA rA + mB rB ) 2

59. The force is a centripetal force, and is of magnitude 7.45 mg. Use Eq. 5–3 for centripetal force.

F =m

(

)

= 7.45 mg → v = 7.45 rg = 7.45 (11.0 m ) 9.80 m s 2 = 28.34 m s  28.3m s

( 28.34 m s ) 

1rev 2 (11.0 m )

= 0.410 rev s

60. A free-body diagram of Tarzan at the bottom of his swing is shown. The upward tension force is created by his pulling down on the vine. Write Newton’s second law in the vertical direction. Since he is moving in a circle, his acceleration will be centripetal, and points upward when he is at the bottom.

 F = F − mg = ma = m r T

→ v=

( FT − mg ) r

m The maximum speed will be obtained with the maximum tension.

   FT − mg  r 1350 N − ( 78 kg ) ( 9.80 m s 2 ) ( 4.8 m ) max    vmax = = = 6.0 m s m

78 kg

61. The fact that the pilot can withstand 9.0 g’s without blacking out, along with the speed of the aircraft, will determine the radius of the circle that he must fly as he pulls out of the dive. To just avoid crashing into the sea, he must begin to form that circle (pull out of the dive) at a height equal to the radius of that circle. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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aR = v 2 r = 9.0 g → r =

v2 9.0 g

( 310 m s )

(

9.0 9.80 m s

)

= 1.1 103 m

62. (a) We are given that x = ( 2.0 m ) cos ( 3.0 rad s ) t  and y = ( 2.0 m ) sin ( 3.0 rad s ) t  . Square both components and add them together. x 2 + y 2 = ( 2.0 m ) cos ( 3.0 rad s ) t  + ( 2.0 m ) sin ( 3.0 rad s ) t  2

= ( 2.0 m )  cos 2 ( 3.0 rad s ) t  + sin 2 ( 3.0 rad s ) t  = ( 2.0 m ) 2

This is the equation of a circle, x 2 + y 2 = r 2 , with a radius of 2.0 m. (b)

v = ( −6.0 m s ) sin ( 3.0 rad s ) t  ˆi + ( 6.0 m s ) cos ( 3.0 rad s ) t  ˆj

(

)

(

)

a = −18 m s2 cos ( 3.0 rad s ) t  ˆi + −18 m s 2 sin ( 3.0 rad s ) t  ˆj

(c)

v = v x2 + v y2 = ( −6.0 m s ) sin ( 3.0 rad s ) t  + ( 6.0 m s ) cos ( 3.0 rad s ) t  = 6.0 m s 2

a = a x2 + a 2y =

( −18 m s2 ) cos ( 3.0 rad s ) t  + ( −18 m s 2 ) sin ( 3.0 rad s ) t  2

= 18 m s 2

(d) (e)

v2 r

( 6.0 m s )

2.0 m

= 18 m s 2 = a

( ) ( ) = ( −9.0 s )  2.0 m cos ( 3.0 rad s ) t  ˆi + 2.0 m sin ( 3.0 rad s ) t  ˆj = ( 9.0 s ) ( −r )

a = −18 m s 2 cos ( 3.0 rad s ) t  ˆi + −18 m s 2 sin ( 3.0 rad s ) t  ˆj 2

We see that the acceleration vector is directed oppositely of the position vector. Since the position vector points outward from the center of the circle, the acceleration vector points toward the center of the circle. 63. Since the curve is designed for 65 km/h, traveling at a higher speed with the same radius means that more centripetal force will be required. That extra centripetal force will be supplied by a force of static friction, downward along the incline. See the free-body diagram for the car on the incline. Note that from Example 5–15 in the textbook, the nofriction banking angle is given by the following. 2

  1.0 m s   ( 65 km h )  3.6 km h   2 v    = 21.4  = tan −1 = tan −1  2

( 85 m ) ( 9.80 m s )

Write Newton’s second law in both the x- and y-directions. The car will have no acceleration in the y-direction, and centripetal acceleration in the x-direction. We also assume that the car is on the verge of skidding, so that the static frictional force has its maximum value of Ffr = s FN . Solve each equation for the normal force.  Fy = FN cos  − mg − Ffr sin  = 0 → FN cos  − s FN sin  = mg →

FN =

( cos  − s sin  )

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 F = F sin  + F cos  = F = mv r → F sin  +  F cos  = mv r → 2

FN =

mv 2 r

( sin  + s cos  )

Equate the two expressions for FN , and solve for the coefficient of friction. The speed of rounding

 1.0 m s   = 26.39 m s .  3.6 km h 

the curve is given by v = ( 95 km h ) 

mv 2 r

( cos  − s sin  ) ( sin  + s cos  )

→

2  v2   v2   ( 26.39 m s ) − 9.80 m s 2 tan 21.4    r cos  − g sin    r − g tan    85 m       s = = = = 0.33 2       v2 v2 26.39 m s ) ( 2 tan 21.4   g cos  + r sin    g + r tan    9.80 m s +  85 m      

(

)

64. Since the curve is designed for a speed of 85 km/h, traveling at that speed would mean no friction is needed to round the curve. From Example 5–15 in the textbook, the no-friction banking angle is given by



  1m s   ( 85 km h )  3.6 km h   v    = 37.18  = tan −1 = tan −1  2 2

( 75 m ) ( 9.80 m s )



 Driving at a higher speed with the same radius means that more centripetal force will be required than is present from the normal force alone. That extra centripetal force will be supplied by a force of static friction, downward along the incline, as shown in the first free-body diagram for the car on the incline. Write Newton’s second law in both the x- and y-directions. The car will have no acceleration in the y-direction, and centripetal acceleration in the x-direction. We also assume that the car is on the verge of skidding, so that the static frictional force has its maximum value of

Ffr

Ffr = s FN .

 F = F cos  − mg − F sin  = 0 → F cos  −  F sin  = mg → N

FN =

( cos  − s sin  )

 F = F sin  + F cos  = m v r → F sin  +  F cos  = m v r → 2

FN =

mv 2 r

( sin  + s cos  )

Equate the two expressions for the normal force, and solve for the speed. mv 2 r mg = → ( sin  + s cos  ) ( cos  − s sin  )

v = rg

( sin  + s cos  ) ( sin 37.18 + 0.30 cos 37.18 ) = ( 75 m ) ( 9.80 m s 2 ) ( cos  − s sin  ) ( cos 37.18 − 0.30 sin 37.18 )

= 31.74 m s  32 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Using Newton’s Laws: Friction, Circular Motion, Drag Forces

Now for the slowest possible speed. Driving at a slower speed with  the same radius means that less centripetal force will be required than Ffr FN that supplied by the normal force. That decline in centripetal force  will be supplied by a force of static friction, upward along the incline, as shown in the second free-body diagram for the car on the incline. mg Write Newton’s second law in both the x- and y-directions. The car will have no acceleration in the y-direction, and centripetal  acceleration in the x-direction. We also assume that the car is on the verge of skidding, so that the static frictional force has its maximum value of Ffr = s FN .

y x

 F = F cos  − mg + F sin  = 0 → N

FN cos  + s FN sin  = mg →

FN =

( cos  + s sin  )

 F = F sin  − F cos  = m v r → F sin  −  F cos  = m v r → 2

mv 2 r

FN =

( sin  − s cos  ) Equate the two expressions for the normal force, and solve for the speed. mv 2 r mg = → ( sin  − s cos  ) ( cos  + s sin  ) v = rg

( sin  − s cos  ) ( sin 37.18 − 0.30 cos 37.18 ) = ( 75 m ) ( 9.80 m s 2 ) ( cos  + s sin  ) ( cos 37.18 + 0.30 sin 37.18 )

= 16.57 m s  17 m s Thus the range is 17 m s  v  32 m s , which is 60 km h  v  110 km h . 65. From Example 5–16, we are given that the track radius is 250 m, and the tangential acceleration is 3.2 m/s2. Thus the tangential force is

(

)

Ftan = matan = ( 950 kg ) 3.2 m s2 = 3040 N  3.0  103 N . The centripetal force is given by Eq. 5–3. Use a speed of 12 m/s. v2 2 FR = m = ( 950 kg )(12 m s ) ( 250 m ) = 547.2 N  550 N . r 66. The car has constant tangential acceleration, which is the acceleration that causes the speed to change. Thus use constant acceleration equations to calculate the tangential acceleration. The initial  1.0 m s  speed is 0, the final speed is 270 km h   = 75 m s, and the distance traveled is one half of  3.6 km h  a circular arc of radius 220 m, so xtan = 220 m . Find the tangential acceleration using Eq. 2–12c.

v −v 2 tan

2 0 tan

= 2atan xtan → atan =

2 v tan − v02 tan

( 75 m s ) = = 4.069 m s 2  4.1m s 2 2 ( 220 m ) 2

2xtan With this tangential acceleration, we can find the speed that the car has halfway through the turn, using Eq. 2–12c, and then calculate the radial acceleration.

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Instructor Solutions Manual

2 vtan − v02 tan = 2atan xtan → vtan = v02 tan + 2atan xtan = 2 ( 4.069 m s 2 ) (110 m ) = 53.03 m s

aR =

( 53.03 m s )

= 12.78 m s 2  13 m s 2

r 220 m The total acceleration is given by the Pythagorean combination of the tangential and centripetal 2 accelerations. atotal = aR2 + atan . If static friction is to provide the total acceleration, then 2 Ffr = matotal = m aR2 + atan . We assume that the car is on the verge of slipping, and is on a level

surface, and so the static frictional force has its maximum value of Ffr = s FN = s mg . If we equate these two expressions for the frictional force, we can solve for the coefficient of static friction. 2 = s mg → Ffr = ma total = m a R2 + a tan

s =

(12.78 m s ) + ( 4.069 m s ) 2

2 a R2 + a tan

= = 1.37  1.4 g 9.80 m s 2 This is an exceptionally large coefficient of friction, and so the curve had better be banked.

67. In all cases we draw a view from above, and the car is moving clockwise around the circular path.

a tan

(a) In this case the car is gaining speed, so it has a tangential acceleration in the direction of its velocity, as well as a centripetal acceleration. The total acceleration vector is somewhat “forward.”

(b) In this case the car has a constant speed, so there is no tangential acceleration. The total acceleration is equal to the radial acceleration.

(c) In this case the car is slowing down, so its tangential acceleration is in the opposite direction as the velocity. It also has a centripetal acceleration. The total acceleration vector is somewhat “backward.”

a = aR

a tan = 0

a tan

68. (a) The object has a uniformly increasing speed, which means the tangential acceleration is constant, and so constant acceleration relationships can be used for the tangential motion. The object is moving in a circle of radius 2.0 meters. vtan + v0 tan 2  1 ( 2 r )  ( 2.0 m ) 2xtan xtan = − v0 tan = 4 = = m s t → vtan = 2 2.0 s t t (b) The initial location of the object is at 2.0 m ˆj , and the final location is 2.0 m ˆi. v avg =

r − r0 t

2.0 m ˆi − 2.0 m ˆj 2.0s

(

= 1.0 m s ˆi − ˆj

)

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(c) The velocity at the end of the 2.0 seconds is pointing in the − ˆj direction. v − v 0 − ( m s ) ˆj = = ( −  2 m s 2 ) ˆj aavg = t 2.0s 69. (a) The tangential acceleration is the time derivative of the speed. 2 d vtan d 3.2 + 1.8 t atan = = = 3.6t → atan ( 3.0s ) = 3.6 ( 3.0 ) = 10.8 m s 2  11m s 2 dt dt (b) The radial acceleration is given by Eq. 5–1.

(

aR =

2 v tan

)

( 3.2 + 1.8t ) = 2

( 3.2 + 1.8 ( 3.0 ) ) = 17 m s → a ( 3.0 s ) = 2

rad

70. We show a top view of the particle in circular motion, traveling clockwise. Because the particle is in circular motion, there must be a radially-inward component of the acceleration. v2 (a) aR = a sin  = → r

v = ar sin  =

22 m

a tan



(1.15 m s ) (3.80 m ) sin 38.0 = 1.64 m s 2

(b) The particle’s speed change comes from the tangential acceleration, which is given by a tan = a cos  . Since the tangential acceleration is constant, we use Eq. 2–12a. vtan − v0 tan = atan t →

(

)

vtan = v0 tan + atan t = 1.64 m s + 1.15 m s 2 ( cos 38.0 )( 3.00 s ) = 4.36 m s

71. The tangential force is simply the mass times the tangential acceleration.

(

aT = b + ct 2 → FT = maT = m b + ct 2

)

To find the radial force, we need the tangential velocity, which is the anti-derivative of the tangential acceleration. We evaluate the constant of integration (C) so that v = v0 at t = 0.

aT = b + ct 2 → vT = C + bt + 13 ct 3 → vT ( 0 ) = C = v0 → vT = v0 + bt + 13 ct 3 FR =

mvT2 r

( v + bt + ct ) r 0

1 3

72. The time constant  must have dimensions of T  . The units of m are  M  . Since the expression bv is a force, we must have the dimensions of b as force units divided by speed units. So the dimensions of b are as follows:

Force units speed units

 M   L T 2   M  =  T  . Thus to get dimensions of L T

T  , we must have  = m b . 73. (a) The terminal velocity is given by Eq. 5–9. This can be used to find the value of b. 2 −5 mg mg ( 3  10 kg )( 9.80 m s ) vT = → b= = = 3.27  10 −5 kg s  3  10 −5 kg s b vT (9 m s) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

167

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) From Example 5–17, the time required for the velocity to reach 63% of terminal velocity is the time constant,  = m b .

=

m b

3  10−5 kg

= 0.917 s  1s

3.27  10−5 kg s

74. From Example 5–17, we have that v =

mg 

−



m  1 − e  . We use this expression to find the position

b 



and acceleration expressions. a= v=

−

b − t  b  m = ge   m 

m  −e   −

b 

→ dx = v dt →

 b

mg m b b

−



e m  =

 dx =  v dt =  0

 mg

x=

mg 

mg b

0

mg 

−



m  1 − e  dt →

b 



m2 g  − m t   e − 1 b2   b

75. The net force on the falling object, taking downward as positive, will be  F = mg − cv 2 = ma. (a) The terminal velocity occurs when the acceleration is 0. mg − cv 2 = ma → mg − cvT2 = 0 →

(b) vT =

mg c

vT = mg c

( 75 kg ) ( 9.80 m s2 ) → c= 2 = = 0.2 kg m 2 vT ( 60 m s ) mg

(c) The curve would be qualitatively like Fig. 5–27b, reproduced here, because the speed would increase from 0 to the terminal velocity, asymptotically. But this curve would be ABOVE the one in Fig. 5–27b, because the friction force increases more rapidly. For Fig. 5–27b, if the speed doubles, the friction force doubles. But in this case, if the speed doubles, the friction force would increase by a factor of 4, bringing the friction force closer to the weight of the object in a shorter period of time. 76. (a) We choose downward as the positive direction. Then the force of gravity is in the positive direction, and the resistive force is upwards. We follow the analysis given in Example 5–17. dv b b mg  = g − v = − v − Fnet = mg − bv = ma → a =  → dt m m b  v

dv b dv b b  mg  = − dt →  = −  dt → ln  v − =− t →  mg mg m m0 b v m  v v − v− b b mg  v − mg  v− b b b   b b =− t → b = e − m t → v = mg  1 − e − m t  + v e − m t ln    0 mg  mg m b    v0 −  v0 −  b  b 0

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Using Newton’s Laws: Friction, Circular Motion, Drag Forces

Note that this motion has a terminal velocity of v terminal = mg b. Also note that if v0 = 0, then this expression is the same as that derived in Example 5–17. (b) We choose upwards as the positive direction. Then both the force of gravity and the resistive force are in the negative direction. dv b b mg  Fnet = − mg − bv = ma → a = = −g − v = −  v +  → dt m m b  v

dv b dv b b  mg  t → = − dt →  = −  dt → ln  v + = − mg mg m m0 b  v m  v v+ v+ b b mg  v + mg  v+ b b b − t − t   b mg  − m t  b b m m ln  =− t → =e → v=  e − 1  + v0 e mg  mg m b    v0 +  v0 +  b  b  m  bv   After the object reaches its maximum height trise = ln  1 + 0   , at which point the speed b  mg    will be 0, it will then start to fall. The equation from part (a), with an initial velocity of 0, will then describe its falling motion. 0

77. (a) See the free-body diagram for the coasting. Since the bicyclist has a constant velocity, the net force on the bicycle must be 0. Use this to find the value of the constant c.  Fx = mg sin  − FD = mg sin  − cv 2 = 0 → c=

mg sin  v2

( 80.0 kg ) ( 9.80 m s 2 ) sin 7.8   1m s   9.5 km h  3.6 km h     

  mg

= 15.28 kg m

y x

 15 kg m

(b) Now another force, FP , must be added down the plane to represent the additional force needed to descend at the higher speed. The velocity is still constant. See the new free-body diagram.  Fx = mg sin  + FP − FD = mg sin  + FP − cv 2 = 0 → FP = cv 2 − mg sin  2

  1m s  = (15.28 kg m )  25km h  − (80.0 kg ) ( 9.80 m s 2 ) sin 7.8 = 630 N    3.6 km h   78. We solve this problem by integrating the acceleration to find the velocity, and integrating the velocity to find the position. dv dv b dv b → =− v → = − dt → Fnet = −bv = ma = m dt dt m m v 1 2

1 2

v

1 2

=−

 dt → 2v − 2v0 = −

1 2

b m

t →

 

1 2

v =  v0 −

bt 



2m 

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

dx dt

 

x=− =

bt 

1 2

=  v0 −



2m 

2 m 

bt 

1 2

 v 0 −

3b 

2m 

 



3 2

3 3 2

2m  3 2



bt 2m

bt 

2m 

2m 



3b  1 2

 0

3 2

 dt →

2m 

bt  

1 2



2m  

b 2t 2

b 3t 3  

2m 

8m  

3b 

− 2

bt 

1 2

 

 v0 −  v0 −

+ 3v0

 dt →  dx =   v0 −



 − v0  =

 v0 −  v0 − 3v0

3b 

1 2

→ dx =  v0 −

Instructor Solutions Manual

= 3 

3 2

 v0 − v0 + 3v0

bt 2m

1 2

− 3v0

b 2t 2

b 3t 3 

8m 3 

+ 2



 vb b2 3  =  v0 t − 0 t 2 + t  2   2 12 m m   1 2

79. The only force accelerating the boat is the drag force, and so Newton’s second law becomes  F = −bv = ma. Use this to solve for the velocity and position expressions, and then find the distance traveled under the given conditions. dv dv b  F = −bv = ma = m dt → dt = − m v → −

v = v0 e m

v

=−

dt → ln =− t → m m v

Note that this velocity never changes sign. It asymptotically approaches 0 as time approaches infinity. Apply the condition that at t = 3.0 s the speed is v = 12 v0 . −

v ( t = 3.0 ) = v0 e m

( 3.0 )

= 12 v0 →

ln 2

m 3.0s Now solve for the position expression. The object will reach its maximum position when it stops, which is after an infinite time. b b b x t − t − t − t dx m m m = v0 e → dx = v0 e dt →  dx =  v0 e dt → v= dt 0 0 − t  m  −mt  m m 3.0 s m = 10.39 m  10 m  e − 1 = v0  1 − e  → x ( t =  ) = v0 = ( 2.4 m s ) b b b ln 2   b

x = −v0

80. A free-body diagram for the coffee cup is shown. Assume that the car is moving to the right, and so the acceleration of the car (and cup) will be to the left. The deceleration of the cup is caused by friction between the cup and the dashboard. For Ffr the cup to not slide on the dash, and to have the minimum deceleration time means the largest possible static frictional force is acting, so Ffr = s FN . The normal force on the cup is equal to its weight, since there is no vertical acceleration. The horizontal acceleration of the cup is found from Eq. 2–12a, with a final velocity of zero.  1m s  v0 = ( 45 km h )   = 12.5 m s  3.6 km h  v − v0 = at → a =

v − v0 t

0 − 12.5 m s 3.5 s

= −3.57 m s 2

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Write Newton’s second law for the horizontal forces, considering to the right to be positive. ( −3.57 m s2 ) = 0.36 a = − = → = − = − → = − = − F F ma ma F mg     x fr s N s s g 9.80 m s 2 81. Since the drawer moves with the applied force of 9.0 N, we assume that the maximum static frictional force is essentially 9.0 N. This force is equal to the coefficient of static friction times the normal force. The normal force is assumed to be equal to the weight, since the drawer is horizontal. 9.0 N F = 0.57 Ffr = s FN = s mg → s = fr = mg (1.6 kg ) 9.80 m s2

(

)

82. See the free-body diagram for the descending roller coaster. It starts its  1m s  descent with v0 = ( 6.0 km h )   = 1.667 m s . The total  3.6 km h 

FN Ffr

displacement in the x-direction is x − x0 = 45.0 m. Write Newton’s second law for both the x- and y-directions.  Fy = FN − mg cos = 0 → FN = mg cos



 F = ma = mg sin  − F = mg sin  −  F = mg sin  −  mg cos  x

mg sin  − k mg cos 

= g ( sin  − k cos  ) m Now use Eq. 2–12c to solve for the final velocity. v 2 − v02 = 2a ( x − x0 ) → a=

v = v02 + 2a ( x − x0 ) = v02 + 2 g ( sin  −  k cos  )( x − x0 )

(1.667 m s ) + 2 ( 9.80 m s 2 ) sin45 − ( 0.14 ) cos45 ( 45.0 m ) 2

= 23.21m s  23 m s  85 km h 83. Consider a free-body diagram of the box. Write Newton’s second law for both directions. The net force in the y-direction is 0 because there is no acceleration in the y-direction.  Fy = FN − mg cos = 0 → FN = mg cos 

Ffr

 F = mg sin  − F = ma x

Now solve for the force of friction and the coefficient of friction.  Fy = FN − mg cos = 0 → FN = mg cos  x



 F = mg sin  − F = ma



(

Ffr = mg sin  − ma = m ( g sin  − a ) = (18.0 kg )  9.80 m s 2

)(sin 32.0 ) − 0.220 m s  o

= 89.5 N Ffr = k FN = k mg cos  → k =

Ffr mg cos 

89.5 N

(18.0 kg ) ( 9.80 m s2 ) cos 32.0o

= 0.598

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84. Since mass m is dangling, the tension in the cord must be equal to the weight of mass m, and so FT = mg . That same tension is in the other end of the cord, maintaining the circular motion of mass M, and so FT = FR = MaR = M M

= mg → v =

v2 r

. Equate the expressions for tension and solve for the velocity.

mgR M

85. Consider the free-body diagram for a person in the “Rotor-ride.” FN is the normal force of contact between the rider and the wall, and Ffr is the static frictional force between the back of the rider and the wall. Write Newton’s second law for the vertical forces, noting that there is no vertical acceleration.  Fy = Ffr − mg = 0 → Ffr = mg

Ffr

If we assume that the static friction force is a maximum, then Ffr = s FN = mg → FN = m g s . But the normal force must be the force causing the centripetal motion – it is the v2 only force pointing to the center of rotation. Thus FR = FN = m . Using v = 2 r T , we have r 2 4 mr FN = . Equate the two expressions for the normal force and solve for the coefficient of T2 friction. Note that since there are 0.50 rev per sec, the period is 2.0 sec. 4 2 mr

gT 2

( 9.80 m s ) ( 2.0s ) = 0.18 = 2

= → s = T2 4 2 r 4 2 ( 5.5 m ) at Any larger value of the coefficient of friction would mean that the normal force could be smaller to achieve the same frictional force, and so the period could be longer or the cylinder radius smaller. FN =

There is no force pushing outward on the riders. Rather, the wall pushes against the riders, so by Newton’s third law, the riders push against the wall. This gives the sensation of being pressed into the wall. 86. (a) The rate of speed increase is the tangential acceleration. From that we can find the tangential speed using Eq. 2–12a. We use the tangential speed with Eq. 5–1 to find an expression for the centripetal acceleration.

v tan = v0 + atan t ; aR = t=

v2 r

( v0 + atan t )

=g →

( 410 m ) ( 9.80 m s 2 ) − 11m s

rg − v0

= = 238s  240s atan 0.22 m s 2 (b) Use Eq. 2–12b to find the total distance traveled, and then convert to the number of laps using the circumference of the track. 2 x = v0 t + 12 atan t 2 = (11m s )( 238s ) + 12 0.22 m s 2 ( 238s ) = 8849 m

(



)

  = 3.4 laps  2 ( 410 m ) 

8849 m 

1lap

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87. The tangential force is the mass times the tangential acceleration. The tangential acceleration is the change in tangential speed divided by the elapsed time. v ( 27 m s ) = 2823 N  2800 N FT = maT = m T = (1150 kg ) t (11s ) The centripetal (radial) force is given by Eq. 5–3. FR = m

v2 r

= (1150 kg )

( 27 m s )

450 m s

= 1863 N  1900 N

88. Since the walls are vertical, the normal forces are horizontal, away from the wall faces. We assume that the frictional forces are at their maximum values, so Ffr = s FN applies at each wall. We assume that the rope in the diagram is not under any tension and so does not exert any forces. Consider the free-body diagram for the climber. FNR is the normal force on the climber from the right

FfrL

FfrR climber

FNL

y x

FNR

wall, and FNL is the normal force on the climber from the left wall. The static frictional forces are FfrL = sL FNL and FfrR = sR FNR . Write Newton’s second law for both the x- and y-directions. The

net force in each direction must be zero if the climber is stationary.  Fx = FNL − FNR = 0 → FNL = FNR  Fy = FfrL + FfrR − mg = 0 Substitute the information from the x-equation into the y-equation. FfrL + FfrR = mg → sL FNL + sR FNR = mg → ( sL + sR ) FNL = mg

FNL =

( sL + sR )

( 70.0 kg ) ( 9.80 m s2 ) 1.40

= 4.90  102 N

And so FNL = FNR = 4.90  102 N . These normal forces arise as Newton’s third law reaction forces to the climber pushing on the walls. Thus the climber must exert a force of at least 490 N against each wall. 89. The mass would start sliding when the static frictional force was not large enough to counteract the component of gravity that will be pulling the mass along the curved surface. See the free-body diagram, and assume that the static frictional force is a maximum. We also assume the block has no speed, so the radial force must be 0.  Fradial = FN − mg cos  → FN = mg cos 

F

tangential

= mg sin  − Ffr → Ffr = mg sin 

Ffr = s FN = s mg cos  = mg sin  → s = tan  →

 = tan −1 s = tan −1 0.70 = 35

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90. The lamp must have the same speed and acceleration as the train. The forces on the lamp as the train rounds the corner are shown  in the free-body diagram included. The tension in the suspending FT cord must not only hold the lamp up, but also provide the centripetal force needed to make the lamp move in a circle. mg Write Newton’s second law for the vertical direction, noting that the lamp is not accelerating vertically. mg  Fy = FT cos  − mg = 0 → FT = cos  The force moving the lamp in a circle is the horizontal portion of the tension. Write Newton’s second law for that radial motion. v2 F F sin ma m =  = =  R T R r Substitute the expression for the tension from the first equation into the second equation, and solve for the speed. v2 mg → FT sin  = sin  = mg tan  = m cos  r v=

( 215 m ) ( 9.80 m s 2 ) tan18.5 = 26.6 m s

rg tan  =

91. (a) The centripetal acceleration is aR =

v2 REarth

  4 2 REarth  2 REarth T  orbit orbit  = = . The period is 365 2 REarth

orbit

days, converted into seconds. 4 2 REarth 4 2 1.50  1011 m orbit = = 5.95  10−3 m s 2 aR = 2 2 T ( 365d )( 24 h d )( 60 min h )( 60s min )

(

)

(b) The force (from Newton’s second law) is FR = mEarth aR.

(

)(

)

FR = ma = 5.98  1024 kg 5.95  10−3 m s2 = 3.56  1022 N The Sun exerts this force on the Earth. It is a gravitational force.

 1m s   = 44.44 m s .  3.6 km h 

92. The speed of the train is (160 km h ) 

(a) If there is no tilt, then the friction force must supply the entire centripetal force on the passenger.

( 55 kg )( 44.44 m s ) FR = m = = 204.94 N  2.0  102 N R ( 530 m ) v2

(b) For the banked case, the normal force will contribute to the radial force needed. Write Newton’s second law for both the x- and y-directions. The y-acceleration is zero, and the x-acceleration is radial.

y x



FN 

Ffr

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 F = F cos  − mg − F sin  = 0 → F = y

mg + Ffr sin 

cos 

 F = F sin  + F cos  = m r x

Substitute the expression for the normal force from the y-equation into the x-equation, and solve for the friction force. mg + Ffr sin  v2 sin  + Ffr cos  = m → cos  r

( mg + Ffr sin  ) sin  + Ffr cos 2  = m

cos  →

 v2  Ffr = m  cos  − g sin    r   ( 44.44 m s )2  cos 8.0o − ( 9.80 m s 2 ) sin 8.0o  = 127.9 N  1.3  10 2 N = ( 55 kg )   530 m  93. See the free-body diagram for the forces. The tension in the top string is found by setting its vertical component equal to gravity. The angle  is found from the lengths of the strings. 1.1m = 49.68  = cos −1 1.7 m

FT1

 FT2

(a) For the lower string to be slack (and so FT2 = 0 ), the centripetal force is all supplied by the horizontal component of the top string. mg  Fy = FT1 sin  − mg = 0 → FT1 = sin 

 F = F cos  = sin  cos  = m r → R

tan 

(1.1m ) ( 9.80 m s 2 ) tan 49.68

= 3.025 m s  3.0 m s

(b) Now the lower string also contributes to the centripetal force, and v = 2 in the vertical direction is still 0.

 Fy = FT1 sin  − mg = 0 → FT1 =

mg sin 

( 0.75 kg ) ( 9.80 m s 2 ) sin 49.68

rg tan 

. The net force

= 9.640 N  9.6 N

 F = F cos  + F = sin  cos  + F = m r → R

 4rg  2   3mg 3 ( 0.75 kg ) ( 9.80 m s ) mg tan    + FT2 = m → FT2 = = = 18.72 N  19 N tan  tan  tan 49.68 r

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94. (a) Consider the free-body diagram for the block on the surface. There is no motion in the y-direction and thus no acceleration in the y-direction. Write Newton’s second law for both directions, and find the acceleration.  Fy = FN − mg cos  = 0 → FN = mg cos 

Ffr 

 F = mg sin  + F = ma x



ma = mg sin  + k FN = mg sin  + k mg cos  a = g ( sin  + k cos  )

Now use Eq. 2–12c, with an initial velocity of v0 , a final velocity of 0, and a displacement of − d to find the coefficient of kinetic friction. v 2 − v02 = 2a ( x − x0 ) → 0 − v02 = 2 g ( sin  +  k cos  )( − d ) → v02

k =

2 gd cos 

− tan 

(b) Now consider the free-body diagram for the block at the top of its motion. We use a similar force analysis, but now the magnitude of the friction force is given by Ffr  s FN , and the acceleration is 0.

 F = F − mg cos  = 0 → F = mg cos   F = mg sin  − F = ma = 0 → F = mg sin  y



s  tan 



Consider the free-body diagrams for both objects, initially stationary. As sand is added, the tension will increase, and the force of static friction on the block will increase until it reaches its maximum of Ffr = s FN . Then the system will start to move. Write Newton’s second law for each object, when the static frictional force is at its maximum, but the objects are still stationary. y2  Fy bucket = m1 g − FT = 0 → FT = m1 g x2  Fy block = FN − m2 g = 0 → FN = m2 g Ffr FT  Fx block = FT − Ffr = 0 → FT = Ffr Equate the two expressions for tension, and substitute in the expression for the normal force to find the masses. m1 g = Ffr → m1 g = s FN = sm2 g →

m1 = s m2 = ( 0.42 )( 28.0 kg ) = 11.76 kg ( 2 sig. fig.) Thus 11.76 kg − 2.00 kg = 9.76 kg  1.0  101 kg of sand was added. (b)

Ffr

Ffr  s FN → mg sin   s mg cos  →

95. a)

The same free-body diagrams can be used, but now the objects will accelerate. Since they are tied together, a y1 = a x 2 = a. The frictional force is

m2 g

FT y1

m1g

now kinetic friction, given by Ffr =  k FN =  k m2 g . Write Newton’s second laws for the objects in the direction of their acceleration.  Fy bucket = m1 g − FT = m1a → FT = m1 g − m1a

F

x block

= FT − Ffr = m2 a → FT = Ffr + m2 a

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Equate the two expressions for tension, and solve for the acceleration. m1 g − m1a = k m2 g + m2 a → a=g

11.76 kg − ( 0.34 )( 28.0 kg ) = 0.552 m s2  0.6 m s2 ( m1 − k m2 ) = ( 9.80 m s 2 ) (11.76 kg + 28.0 kg ) ( m1 + m2 )

Note that the numerator in the above expression only has 1 significant figure, since m1 = 12 kg ( 2 sig. fig. ) . 96. The acceleration that static friction can provide can be found from the minimum stopping distance, assuming that the car is just on the verge of sliding. Use Eq. 2–12c. Then, assuming an unbanked curve, the same static frictional force is used to provide the centripetal acceleration needed to make the curve. The acceleration from the stopping distance is negative, and so the centripetal acceleration is the opposite of that expression. v02 v 2 − v02 −v02 v 2 − v02 = 2a ( x − x0 ) → astopping = = → aR = 2 ( x − x0 ) 2 ( x − x0 ) 2 ( x − x0 ) Equate the above expression to the typical expression for centripetal acceleration. v02 v2 aR = = → r = 2 ( x − x0 ) = 2 ( 66 m ) = 132 m r 2 ( x − x0 ) Notice that we didn’t need to know the mass of the car, the initial speed, or the coefficient of friction. 97. (a) The horizontal component of the lift force will produce a centripetal acceleration. Write Newton’s second law for both the horizontal and vertical directions, and combine those equations to solve for the time needed to reverse course (a half-period of the circular motion). Note that 2 r T= . v v2  Fvertical = Flift cos  = mg ;  Fhorizontal = Flift sin  = m r Divide these two equations. Flift sin  mv 2 v2 v2 2 v = → tan  = = = → Flift cos  rmg rg T v gT g 2 T 2

v g tan 

Flift a rad



  1.0 m s    ( 510 km h )   3.6 km h     = = 58s

( 9.80 m s ) tan 38 2

(b) The passengers will feel a change in the normal force that their seat exerts on them. Prior to the banking, the normal force was equal to their weight. During banking, the normal force will mg increase, so that Fnormal = = 1.27mg . Thus they will feel “pressed down” into their seats, cos  banking with about a 25% increase in their apparent weight. If the plane is banking to the left, they will feel pushed to the right by that extra 25% in their apparent weight.

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98. From Example 5–15 in the textbook, the no-friction banking angle is given by  = tan −1

v02 Rg

, or

v02 = Rg tan  . The centripetal force in this case is provided by a component of the normal force. Driving at a higher speed with the same radius requires more centripetal force than that provided by the normal force alone. The additional centripetal force is supplied by a force of static friction, downward along the incline. See the free-body diagram for the car on the incline. The center of the circle of the car’s motion is to the right of the car in the diagram. Write Newton’s second law in both the xand y-directions. The car will have no acceleration in the y-direction, and centripetal acceleration in the x-direction. Assume that the car is on the verge of skidding, so that the static frictional force has its maximum value of Ffr = s FN .

 F = F cos  − mg − F sin  = 0 → F cos  −  F sin  = mg → y

FN =

( cos  − s sin  )

 F = F = F sin  + F cos  = m R FN =

mg v2

→ FN sin  + s FN cos  = m

v2 R

→

mv 2 R

( sin  + s cos  )

Equate the two expressions for the normal force, and solve for the speed, which is the maximum speed that the car can have. mv 2 R mg = → ( sin  + s cos  ) ( cos  − s sin  )

v max =

sin  (1 + s tan  )

cos  (1 − s tan  )

= v0

(1 + Rg  v ) (1 −  v Rg ) s

2 0

Driving at a slower speed with the same radius requires less centripetal force than that provided by the normal force alone. The decrease in centripetal force is supplied by a force of static friction, upward along the incline. See the free-body diagram for the car on the incline. Write Newton’s second law in both the x- and ydirections. The car will have no acceleration in the y-direction, and centripetal acceleration in the x-direction. Assume that the car is on the verge of skidding, so that the static frictional force is given by Ffr = s FN .



Ffr



y x

mg 

 F = F cos  − mg + F sin  = 0 → y

FN cos  + s FN sin  = mg →

FN =

( cos  + s sin  )

 F = F = F sin  − F cos  = m R → F sin  −  F cos  = m R → x

FN =

mv 2 R

( sin  − s cos  )

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Equate the two expressions for the normal force, and solve for the speed. mv 2 R mg = → ( sin  − s cos  ) ( cos  + s sin  )

v min =

Thus v min = v0

sin  (1 − s tan  )

cos  (1 + s tan  )

= v0

2 0

(1 −  Rg v ) and v (1 +  v Rg ) 2 0

(1 −  Rg v ) (1 +  v Rg )

2 0

max

(1 +  Rg v ) . (1 −  v Rg ) 2 0

= v0

2 0

99. An object at the Earth’s equator is rotating in a circle with a radius equal to the radius of the Earth, and a period equal to one day. Use that data to find the centripetal acceleration and then compare it to g. 2 4 2 6.38  106 m 2 r

(

    aR v 4 2 r T  aR = = = 2 → = r

)

( 86, 400 s )

( 9.80 m s ) 2

= 0.00344 

3 1000

So, for example, if we were to calculate the normal force on an object at the Earth’s equator, we could not say  F = FN − mg = 0. Instead, we would have the following. v2

 F = F − mg = −m r N

→ FN = mg − m

If we then assumed that FN = mg eff = mg − m g eff = g −

v2 r

, then we see that the effective value of g is

= g − 0.003 g = 0.997 g .

100. (a) Because there is no friction between the bead and the hoop, the hoop can only exert a normal force on the bead. See the free-body diagram for the bead at the instant shown in the textbook figure. Note that the bead moves in a horizontal circle, parallel to the floor. Thus the centripetal force is horizontal, and the net vertical force must be 0. Write Newton’s second law for both the horizontal and vertical directions, and use those equations to determine the angle  . We also use the fact that the speed and the frequency are related to each other, by v = 2 f r sin  . mg  Fvertical = FN cos  − mg = 0 → FN = cos 

F

= FN sin  = m radial

FN sin  =

(b)  = cos −1

mg cos 

g 4 2 f 2 r

v2 r sin 

sin  = m

= cos −1

4 2 f 2 r 2 sin 2  r sin 

4 f r sin  2

2 2

r sin 

→  = cos −1

9.80 m s2 4 2 ( 2.00 Hz ) ( 0.250 m ) 2

g 4 f 2 r 2

= 75.6

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(c) No , the bead cannot ride as high as the center of the circle. If the bead were located there, the normal force of the wire on the bead would point horizontally. There would be no force to counteract the bead’s weight, and so it would have to slip back down below the horizontal to g balance the force of gravity. From a mathematical standpoint, the expression would 2 2 4 f r have to be equal to 0 and that could only happen if the frequency or the radius were infinitely large. 101. The goal is to form a quantity that has acceleration units, from the speed of the radius of an object in  L  , the units of the radius are L , and the units of circular motion. The units of speed are    T 

 L  . To get time units squared in the denominator, the speed must be squared.  T 2   L2  But the units of speed squared are  2  . This has one too many powers of length, so to reduce that, T  acceleration are

divide by the radius.  L2 





T L →   = 2 → a  L  T 

Factors such as 2 or , if needed for the final formula, cannot be determined with dimensional analysis. Another technique is to write the units of speed and radius raised to unknown powers, and then solve for the powers to make their product be acceleration units. m L L = L   2  T   T 

→ length units: 1 = m + n ; time units: − 2 = − n →

n = 2, m = −1 → a = r −1v 2 = v 2 r 102. We include friction from the start, and then for the no-friction result, set the coefficient of friction equal to 0. Consider a free-body diagram for the car on the hill. Write Newton’s second law for both directions. Note that the net force in the y-direction will be zero, since there is no acceleration in the y-direction.  Fy = FN − mg cos  = 0 → FN = mg cos 

Ffr

FN y 

 F = mg sin  − F = ma →



Ffr

k mg cos 

= g ( sin  − k cos  ) m m Use Eq. 2–12c to determine the final velocity, assuming that the car starts from rest. a = g sin  −

= g sin  −

v 2 − v02 = 2a ( x − x0 ) → v = 0 + 2a ( x − x0 ) = 2 g ( x − x0 )( sin  − k cos  )

The angle is given by sin  = 1 4 →  = sin −1 0.25 = 14.5o.

( ) (b)  = 0.10 → v = 2 ( 9.80 m s ) ( 55 m ) ( sin14.5 − 0.10 cos14.5 ) = 13m s

(a) k = 0 → v = 2 g ( x − x0 ) x sin  = 2 9.80 m s 2 ( 55 m ) sin14.5o = 16 m s 2

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Chapter 5

Using Newton’s Laws: Friction, Circular Motion, Drag Forces

103. The two positions on the cone correspond to two opposite directions of the force of static friction. In one case, the frictional force points UP the cone’s surface, and in the other case, it points DOWN the cone’s surface (as shown in the free-body diagram). In each case the net vertical force is 0, and force of static friction is assumed to be its maximum value. The net horizontal force is producing centripetal motion.  Fvertical = FN sin  − Ffr cos  − mg = FN sin  − s FN cos  − mg = 0 → FN =

Ffr mg



mg sin  − s cos 

F

horizontal

=FN cos  + Ffr sin  = FN cos  + s FN sin  = FN ( cos  + s sin  ) = m

FN =

v2 r

( 2 rf )

= 4 2 rmf 2 →

4 2 rmf 2

( cos  + s sin  )

Equate the two expressions for the normal force, and solve for the radius. FN =

mg sin  − s cos 

4 2 rmf 2

( cos  + s sin  )

→

rmax =

g ( cos  + s sin  )

4 2 f 2 ( sin  − s cos  )

A similar analysis will lead to the minimum radius.  Fvertical = FN sin  + Ffr cos  − mg = FN sin  + s FN cos  − mg = 0 → FN =

Ffr

sin  + s cos 

F

horizontal

=FN cos  − Ffr sin  = FN cos  − s FN sin  = FN ( cos  − s sin  ) = m

FN =

v2 r

( 2 rf )



= 4 2 rmf 2 → mg

4 2 rmf 2

( cos  − s sin  ) mg sin  + s cos 

4 2 rmf 2

( cos  − s sin  )

→

rmin =

g ( cos  − s sin  )

4 2 f 2 ( sin  + s cos  )

104. (a) See the free-body diagram for the skier when the tow rope is horizontal. Use Newton’s second law for both the vertical and horizontal directions in order to find the acceleration.  Fy = FN − mg = 0 → FN = mg

FN FT

Ffr

 F = F − F = F −  F = F −  mg = ma F −  mg ( 240 N ) − 0.25 ( 68 kg ) ( 9.80 m s ) = = 1.08 m s2  1.1m s a= x

( 68 kg )

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(b) Now see the free-body diagram for the skier when the tow rope has an upward component.  Fy = FN + FT sin  − mg = 0 → FN = mg − FT sin 

FN FT

 F = F cos  − F = F cos  −  F x

= FT cos  −  k ( mg − FT sin  ) = ma



Ffr

FT ( cos  + k sin  ) − k mg m ( 240 N )( cos12 + 0.25sin12 ) − 0.25 ( 68 kg ) 9.80 m s 2

(

) = 1.19 m s  1.2 m s 2

( 68 kg ) (c) The acceleration is greater in part (b) because the upward tilt of the tow rope reduces the normal force, which then reduces the friction. The reduction in friction is greater than the reduction in horizontal applied force, and so the horizontal acceleration increases. 105. The radial acceleration is aR =

( 6.0 m s )

(

)

, and so aR = = = 45 m s 2 . r r 0.80 m The tension force has no tangential component, and so the tangential force is seen from the diagram to be Ftang = mg cos  . Ftang = mg cos  = matang → atang = g cos  = 9.80 m s2 cos 30 = 8.5m s2

The tension force can be found from the net radial force. v2 FR = FT − mg sin  = m → r



v2 



r 

FT = m  g sin  +

 = (1.0 kg ) ( ( 9.80 m s ) sin 30 + 45 m s ) = 5.0  10 N 2

106. (a) The acceleration has a magnitude given by a = v 2 r . v2

( −15.7 m s ) + ( −23.2 m s ) = 28.01m s = 72.5 m →

( 28.01m s ) ( 72.5 m ) = 45.06 m s  45.1m s

(b) Since the acceleration points radially in and the position vector points radially out, the components of the position vector are in the same proportion as the components of the acceleration vector, but of opposite sign. ay a 15.7 m s 2 23.2 m s 2 40.6 m 72.5 m x = r x = ( 72.5 m ) = y = r = = 60.0 m ( ) 28.01m s 2 28.01m s 2 a a

182

CHAPTER 6: Gravitation and Newton’s Synthesis Responses to Questions 1.

Whether the apple is (a) attached to a tree or (b) falling, it exerts a gravitational force on the Earth equal to the force the Earth exerts on it. That force of the Earth on the apple, or the apple on the Earth, is the weight of the apple (Newton’s third law). That force is independent of the motion of the apple.

Quoting from Problem 78: “The gravitational force at different places on Earth due to the Sun and the Moon depends on each point’s distance from the Sun or Moon, and this variation is what causes the tides.” So, the tides are caused by the small changes in gravitational pull on the Earth due to the Sun and the Moon, as you consider the range from points on the Earth closest to the Sun or Moon, to points farthest from the Sun or Moon. The gravitational pull from the Sun on the side of the Earth closest to the Sun depends on the distance from the Sun to the close side of the Earth. The pull from the Sun on the far side of the Earth depends on this distance plus the diameter of the Earth. The diameter of the Earth is a very small fraction of the total Earth–Sun distance, so these two forces, although large, are nearly equal. The diameter of the Earth is a larger fraction of the Earth–Moon distance, and so the difference in gravitational force from the Moon to the two opposite sides of the Earth will be greater, and thus the displacement of the water due to the Moon is greater. For a mathematical consideration of this question, see Problem 78.

The object will weigh more at the poles. The value of r² at the equator is greater, due to the Earth’s equatorial bulge. Also, the object has centripetal acceleration at the equator which it does not have at the poles. The two effects do not oppose each other–they add to each other.

Since the Earth’s mass is much greater than the Moon’s mass, the point at which the net gravitational pull on the spaceship is zero is closer to the Moon. It is shown in Problem 7 that this occurs at about 90% of the way from the Earth to the Moon. So, a spaceship traveling from the Earth towards the Moon must therefore overcome the net pull backwards for 90% of the trip, requiring more fuel. Once it passes that point, the Moon will exert a stronger pull than the Earth, and accelerate the spacecraft towards the Moon. However, when the spaceship is returning to the Earth, it reaches the zero point at only 10% of the way from the Moon to the Earth. Therefore most of the trip towards the Earth, the spacecraft is “helped” by the net gravitational pull in the direction of travel, and so it uses less fuel on the return portion of the trip.

The gravitational force from the Sun provides the centripetal force to keep the Moon and the Earth going around the Sun. Since the Moon and Earth are at the same average distance from the Sun, they (on average) travel together around the Sun, and the Moon is not pulled away from the Earth.

As the Moon revolves around the Earth, its position relative to the distant background stars changes. This phenomenon is known as “parallax.” As a demonstration, hold your finger at arm’s length and look at it with one eye at a time. Notice that it “lines up” with different objects on the far wall depending on which eye is open. If you bring your finger closer to your face, the shift in its position against the background increases. Similarly, the Moon’s position against the background stars will shift as viewed (simultaneously) from different locations on Earth. The distance to the Moon can be calculated by the amount of angular shift, assuming the distance between the locations on Earth is known. It is also possible to determine the distance to the Moon using a lunar eclipse, but since those events are rare, measuring via eclipse is perhaps less convenient than measuring via parallax.

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At the very center of the Earth, all of the gravitational forces would cancel (assuming the Earth is spherically symmetric), and the net force on the object would be zero.

A satellite in a geostationary orbit stays over the same spot on the Earth at all times. If the satellite were above the North Pole, it would not move relative to the center of the Earth. But it would be impossible for the satellite to be stationary, because of the Earth’s gravitational force on it. The (temporarily!) stationary satellite would then fall to the Earth. Instead, the gravitational force must be perpendicular to the orbit, and that is only possible for a geostationary orbit if the satellite’s orbit is above the equator. This satellite then rotates with the same period as the Earth, above a fixed point on the equator. It is possible for other orbits (not above the equator) to have a 24-hour period, but they are not above a fixed point on the Earth’s surface during their orbits.

According to Newton’s third law, the force the Earth exerts on the Moon has the same magnitude as the force the Moon exerts on the Earth. The Moon has a larger acceleration, since it has a smaller mass (use Newton’s second law, F = ma).

10. The satellite needs a certain speed with respect to the center of the Earth to achieve orbit. The Earth rotates towards the east so it would require less speed (with respect to the Earth’s surface) to launch a satellite (a) towards the east. Before launch, the satellite is moving with the surface of the Earth and so it already has a “boost” in that direction. 11. If the antenna becomes detached from a satellite in orbit, the antenna will continue in orbit around the Earth with the satellite. The orbital period does not depend on the mass of the orbiting object– just the radius of the (assumed to be) circular orbit. If the antenna were given a component of velocity toward or away from the Earth (even a very small one), then its orbit would change. That new orbit might cause it to fall to the Earth. If the antenna were somehow slowed down relative to the center of the Earth (by some force other than gravity), it would also fall towards the Earth. 12. Ore normally has a greater density than the surrounding soil. A large ore deposit will have a larger mass than an equal amount of soil. The greater the mass of ore, the greater the acceleration due to gravity will be in its vicinity. Careful measurements of this slight increase in g (similar to what is described in Problem 80, for example) can therefore be used to estimate the mass of ore present. Also see Problem 26. 13. Yes, we are heavier at midnight. At noon, the gravitational force on a person due to the Sun and the gravitational force due to the Earth are in the opposite directions. At midnight, the two forces point in the same direction. Therefore, your apparent weight at midnight is greater than your apparent weight at noon. 14. Your apparent weight will be greatest in case (b), when the elevator is accelerating upward. The scale reading (your apparent weight) indicates your force on the scale, which, by Newton’s third law, is the same as the normal force of the scale on you. If the elevator is accelerating upward, then the net force must be upward, so the normal force (up) must be greater than your actual weight (down). When in an elevator accelerating upward, you “feel heavy.” Your apparent weight will be least in case (c), when the elevator is in free fall. In this situation your apparent weight is zero since you and the elevator are both accelerating downward at the same rate and the normal force from the scale is zero. Your apparent weight will be the same as when you are on the ground in case (d), when the elevator is moving upward at a constant speed. If the velocity is constant, acceleration is zero and N = mg.

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(Note that it doesn’t matter if the elevator is moving up or down or even at rest, as long as the velocity is constant.) 15. Set the gravitational force due to the Earth equal to the centripetal acceleration of the Moon, in a circular orbit. 2 2 M Earth M Moon M Moon vMoon vMoon rorbit G = → G= = constant 2 rorbit rorbit M Earth For example, if the Earth’s mass were doubled from what it actually is, the radius of the Moon’s orbit would have to double, if the Moon’s speed were to remain constant. Or, the Moon’s speed in orbit would have to increase by a factor of 2 , if the Moon’s orbital radius were to remain constant. 2 In general, if both the radius and orbital speed were free to change, then the product rorbit vMoon would have to double. 16. If the Earth were covered in water, then the entire surface would be at “sea level,” even though the water at the equator would be farther away from the center of the Earth than the water away from the equator. Water would not continually “flow” from the non-equatorial locations to the equator, so the concept of “sea level” is really the idea to use in determining river flow. The source of the Mississippi river is about 450 m above “sea level,” while the outlet at Louisiana is at “sea level,” so the river is flowing “downhill,” as determined by sea level. 17. The satellite remains in orbit because it has a velocity. The instantaneous velocity of the satellite is tangent to the orbit. The gravitational force provides the centripetal force needed to keep the satellite in orbit, acting like the tension in a string when twirling a mass on a string. A force is not needed to keep the satellite “up”; in fact, a downward force (gravity) is needed to change the velocity vector around in a circle. The satellite can’t just have any speed at any radius, though. For a perfectly circular orbit, the speed is determined by the orbit radius, or vice-versa, through the relationship vorbit = r g , where r is the radius of the orbit and g is the acceleration due to gravity at the orbit position. 18. Between steps, the runner is not touching the ground. Therefore there is no normal force up on the runner and so she has no apparent weight. She is momentarily in free fall since the only force is the force of gravity pulling her back toward the ground. 19. If you were in a satellite orbiting the Earth, you would have no apparent weight (no normal force). Walking, which depends on the normal force, would not be possible, unless there was some mechanism (like magnetic shoes) to create a normal force. Drinking would be possible, but only from a tube or pouch. Tipping up a glass to drink from it would not work. Scissors would not sit on a table (no apparent weight = no normal force). They might need to be tethered to the table so that they do not “float” away. 20. The centripetal acceleration of Mars in its orbit around the Sun is smaller than that of the Earth. For both planets, the centripetal force is provided by gravity, so the centripetal acceleration is inversely proportional to the square of the distance from the planet to the Sun: mplanet v 2 GmSun mplanet v 2 GmSun = a = = so centrip r r2 r r2 Since Mars is at a greater distance from the Sun than Earth, it has a smaller centripetal acceleration. Note that the mass of the planet does not appear in the equation for the centripetal acceleration.

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21. For Pluto’s moon, we can equate the gravitational force from Pluto on its moon to the centripetal force needed to keep the moon in orbit: mmoon v 2 GmPluto mmoon v 2 r 4 2 r 3 m = → = = Pluto r r2 G GT 2 This allows us to solve for the mass of Pluto if we know G, the radius of the moon’s orbit, and the velocity of the moon, which can be determined from the period and orbital radius. Note that the mass of the moon need not be known. 22. The Earth is closer to the Sun in January. The gravitational force between the Earth and the Sun is a centripetal force. When the distance decreases, the speed increases. (Imagine whirling a rock around your head in a horizontal circle. If you pull the string through your hand to shorten the distance between your hand and the rock, the rock speeds up.) If we imagine the orbit to be approximately circular, then we have the following. mEarth v 2

GmSun mEarth 2

GmSun

r r r Since the speed is greater in January, the distance must be less. This agrees with Kepler’s second law. Also see Fig. 6–18 (copied here) and the accompanying discussion about Kepler’s second law. The last statement from the Figure says: “Planets move fastest when closest to the Sun.” So in the (greatly exaggerated) Figure, the time between points 1 and 2 would be during January, and the time between points 3 and 4 would be July.

23. The Earth’s orbit is an ellipse, not a circle. Therefore, the force of gravity on the Earth from the Sun is not perfectly perpendicular to the Earth’s velocity at all points. A component of the force will be parallel (or anti-parallel) to the velocity vector and will thus cause the planet to speed up (or slow down). 24. Standing at rest, you feel an upward force on your feet. In free fall, you don’t feel that force. So, in that analysis, you don’t directly sense the gravitational field. You might, however, be aware of the acceleration during free fall, possibly due to your inner ear. 25. If we treat g as the acceleration due to gravity, it is the result of a force from one mass acting on another mass and causing the second mass to accelerate. This implies action at a distance, since the two masses do not have to be in contact. If we view g as a gravitational field, then we say that the presence of a mass changes the characteristics of the space around it by setting up a field, and the field then interacts with other masses that enter the space in which the field exists. Since the field is in contact with the mass, this conceptualization does not imply action at a distance.

Responses to MisConceptual Questions 1.

(d) If the net force on the Moon were zero [answer (a)], the Moon would move in a straight line and not orbit about the Earth. Gravity pulls the Moon away from the straight line motion. The large tangential velocity is what keeps the Moon from crashing into the Earth. The gravitational force of the Sun also acts on the Moon, but this force causes the Earth and Moon to orbit the Sun.

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(c) The non-zero gravitational force on the ISS is responsible for it orbiting the Earth instead of moving in a straight line through space. Astronauts aboard the ISS experience the same centripetal acceleration (free-fall toward the Earth) as the station and as a result do not experience a normal force (apparent weightlessness).

(f) A common misconception is that since the Earth is more massive than the Moon, it must exert a larger force. However, the force is an interaction between the Earth and Moon, so by Newton’s third law, the forces must be equal. Since the Moon is less massive than the Earth and the forces are equal, the Moon has the greater acceleration.

(b) A common misconception is that the mass of an object affects its orbital speed. However, as with all objects in free-fall, when calculating the acceleration the object’s mass is divided out of the gravitational force. All objects at the same radial distance from the Earth experience the same centripetal acceleration, and by Eq. 5–1 they have the same orbital speed.

(c) Each of the incorrect answers assumes the presence of an external force to change the orbital motion of the payload. When the payload is attached to the arm it is orbiting the Earth at the same distance and speed as the space vehicle. When it is released, the only force acting on the payload is the force of gravity, which due to the speed of the payload, keeps it in orbital motion. For the payload to fall straight down, or to follow a curved path that hits the Earth, a force would need to slow down the payload’s speed, but no such force is present. To drift out into deep space a force would be needed to overcome the gravity that is keeping it in orbit, but no such force is present.

(a) Your weight decreases with distance from the center of the Earth, but your mass is a fixed quantity, not dependent on location. Weight is the product of mass and the acceleration due to gravity, and that acceleration decreases with distance from the center of the Earth.

(c) The scale reads the normal force that the scale exerts on the person standing on it. If the elevator is in free fall (accelerating down at 9.8 m/s2), then so is the person. With that acceleration, the only force on the person is the force of gravity from the Earth, and so the scale is not putting a force on the person.

(a) The Moon doesn’t stay “up” in the sense that it is not stationary above the Earth. The Moon is constantly moving closer to the Earth (i.e., “falling”) compared to the motion it would have if there were no gravity force. But the fact that the Moon has a tangential velocity means that the centripetal force acting on it produces a circular orbit.

(c) The Moon does not create it’s own light–it only shines by reflected sunlight. At “full moon,” the Earth is between the Sun and the Moon, so that the reflected sunlight reflects towards the “night” side of the Earth. See Fig. 6–24, parts (a) and (e).

3 2 10. (b) As discussed in Section 6–5, the ratio s T is constant for objects orbiting the same

gravitational attractor M. In particular,

s3 T

GM 4 2

, from Eq. 6–6.

11. (c) Your weight is the magnitude of the force of gravity on you. As stated in Section 4–6, “The magnitude of the force of gravity on an object, mg, is commonly called the object’s weight.” As discussed in Section 6–8, g is the gravitational field of the Earth.

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12. (d) The Earth’s gravitational field g is calculated by g = G

Instructor Solutions Manual

M Earth , where r is the distance from the r2

center of the Earth (assumed to be larger than the radius of the Earth). Although g becomes very small at large distances from the Earth, the field extends to infinitely far from the Earth.

Solutions to Problems 1.

The spacecraft is at 3.00 Earth radii from the center of the Earth, or three times as far from the Earth’s center as when at the surface of the Earth. Therefore, since the force of gravity decreases as the square of the distance, the force of gravity on the spacecraft will be one-ninth of its weight at the Earth’s surface. (1650 kg ) 9.80 m s 2 FG = 19 mg Earth’s = = 1.80  103 N 9 surface This could also have been found using Eq. 6–1, Newton’s law of universal gravitation.

(

)

(a) Mass is independent of location and so the mass of the ball is 28.0 kg on both the Earth and the planet. (b) The weight is found by W = mg.

( ) = ( 28.0 kg ) (12.0 m s ) = 336 N

WEarth = mg Earth = ( 28.0 kg ) 9.80 m s 2 = 274 N Wplanet = mg planet

The force of gravity on an object at the surface of a planet is given by Newton’s law of universal gravitation, Eq. 6–1, using the mass and radius of the planet. If that is the only force on an object, then the acceleration of a freely falling object is acceleration due to gravity. m m FG = G Moon = mg Moon → 2 rMoon

g Moon = G

mMoon 2 rMoon

( 7.35 10 kg ) = 1.62 m s = ( 6.67  10 N m kg ) (1.74 10 m ) 22

−11

The acceleration due to gravity at any location on or above the surface of a planet is given by g planet = G mplanet r 2 , where r is the distance from the center of the planet to the location in question. g planet = G

mplanet r2

( 2.5rE )

= 2

mE 1 9.80 m s 2 G g = = = 1.6 m s 2 E 2 2 2 2 2.5 rE 2.5 2.5 1

The acceleration due to gravity at any location at or above the surface of a planet is given by g planet = G mPlanet r 2 , where r is the distance from the center of the planet to the location in question.

g planet = G

mPlanet r

2.80mE 2 E

 mE  = 2.80 g E = 2.80 ( 9.80 m s 2 ) = 27.4 m s 2 2   rE 

= 2.80  G

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The acceleration due to gravity is determined by the mass of the planet and the radius of the planet. Gm Gm G 2m0 2 Gm0 2 g0 = 2 0 g new = 2 new = = = 9 g0 r0 rnew ( 3r0 )2 9 r02 So g is multiplied by a factor of 2 9 .

For the net force to be zero means that the gravitational force on the spacecraft due to the Earth must be the same as that due to the Moon. Write the gravitational forces on the spacecraft, equate them, and solve for the distance x. We measure from the center of the bodies. mE mspacecraft mMoon mspacecraft FEarth − = G F G ; = − Moon 2 x2 spacecraft spacecraft (d − x)

mE mspacecraft x

x=d

( m

mMoon mspacecraft

(d − x)

Moon

+ mE

)

→

(

= 3.84  108 m

)

(d − x)

→

mMoon

x mE

d−x mMoon

5.97  1024 kg

( 7.35  10 kg + 5.97  10 kg ) 22

= 3.46  108 m

This is only about 22 Moon radii away from the Moon. Or, it is about 90% of the distance from the center of the Earth to the center of the Moon. The distance from the surface of the Earth would be 3.46  108 m − 6.38  106 m = 3.40  108 m. 8.

Assume that the two objects can be treated as point masses, with m1 = m and m2 = 6.50 kg − m. The gravitational force between the two masses is given by the following. 2 m ( 6.50 − m ) mm −11 2 2 6.50 m − m F = G 12 2 = G =  = 6.5  10 −9 N 6.67 10 N m kg 2 2 r r ( 0.25 m )

(

)

( 6.5 10 N ) ( 0.25 m ) = 6.09 → m − 6.50m + 6.09 = 0 6.50m − m = 2

−9

6.50 

(

6.67  10 −11 N  m 2 kg 2

( 6.50 ) − 4 (1)( 6.09 )

)

6.50  4.23 = 5.37, 1.13 2

m1 = 5.37 kg, m2 = 1.13 kg

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We compare the weight at the given altitude to the weight at the surface of the Earth. Use Eq. 6–1 for the gravitational force. mE mobject G 2 rE + 1600 m ) ( rE2 weight above Earth’s surface = = 2 mE mobject weight at Earth’s surface ( rE + 1600 m ) G rE2 2     6.38  106 m ) ( rE2   = − % lost = 1 − 100 1 100 = 5.014  10−2  2 6  ( r + 1600 m )2   ( 6.38  10 m + 1600 m )  E    

So the object has lost 0.05% of its weight. 10. The force on m due to 2m points in the î direction. The force on m due to 4m points in the ĵ y direction. The force on m due to 3m points in the direction given by  = tan −1 0 . Add the force x0 vectors together to find the net force. ( 2m ) m ˆ ( 4m ) m ˆ ( 3m ) m ( 3m ) m cos  ˆi + G 2 sin  ˆj F=G i+G j+G 2 2 2 2 x0 y0 x0 + y 0 x0 + y02

ˆi + G ( 3m ) m x02 + y02 x02 + y02

2m 2 ˆ 4m 2 ˆ 3m 2 G G + + i j x02 y02 x02 + y02

 2

= Gm 2 

+ 2

 x0

y0 x +y 2 0

2 0

ˆj

  4   3 y0 ˆi +    ˆj + 2 2 2 3/ 2  2 2 3/ 2   y x y x y + + 0 ( 0 0 )   ( 0 0 )   3 x0

11. (a) Since the value of g is less at the equator, you weigh less at the equator than in New York . (b) Use the values from Table 6–1 to compare weights at the two locations.

weight at equator weight at New York

mgequator mg New York

g equator g New York

9.780 m s 2 9.803m s 2

= 0.9977 = 99.77%

So you weigh 0.23% less at the equator . 12. From Chapter 3, Section 8, right before Example 3–10, the maximum range of a particle is given as v2 Rmax = 0 . Assuming that you can give the ball the same initial velocity on the other planet as you g did on Earth, the difference in range must be due to a different value for g. Use Eq. 6–4 for g. mplanet v2 m Rmax = 0 → v02 = g Earth Rmax = g planet Rmax → G 2Earth Rmax = G 2 Rmax → g rEarth Earth rplanet planet Earth planet

Rmax mplanet = mEarth

Earth

Rmax

(

= 5.98  1024 kg

45.0 m = 7.37  10 kg ) 36.5 m 24

planet

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13. To find the new weight of objects at the Earth’s surface, the new value of g at the Earth’s surface needs to be calculated. Since the spherical shape is being maintained, the Earth can be treated as a point mass. Find the density of the Earth using the actual values, and use that density to find g under the revised conditions. g original = G g original = G

; =

rE2

 3mE   4   

mE 4 3

 rE3

=G 2/3

1/ 3

 3m  → rE =  E   4 

3mE

4 rE3

( mE )1/ 3  3   4   

; g new = G

2/3

→

( 2mE )1/ 3  3   4   

= 21/ 3 G 2/3

( mE )1/ 3  3   4   

2/3

= 21/ 3 g

Thus g is multiplied by 21/3, and so the weight would be multiplied by 21/ 3 or ~ 1.26 . 14. The gravitational force of the Sun on the Earth is a centripetal force. Set those two expressions equal to each other to solve for the speed of the Earth orbiting the Sun. G

M Sun M Earth

= M Earth

2 Earth orbit

v2 rEarth

GM Sun

→ v=

rEarth

orbit

2 rEarth That speed is also give by v =

orbit

. Equate the two expressions for the speed and solve for

TEarth

M Sun , using data for the Earth.

GM Sun rEarth

2 rEarth orbit

→

TEarth

orbit 3 4 2 rEarth

M Sun =

orbit 2 Earth

(

4 2 1.50  1011 m

)

( 6.67  10 N m kg )( 3.16  10 sec ) −11

= 2.00  1030 kg

This is the same result obtained in Example 6–8 using Kepler’s third law. 15. With the assumption that the density of Europa is the same as Earth’s, the radius of Europa can be calculated.

 Europa =  Earth → g Europa =

GmEuropa 2 rEuropa

mEuropa 3 4/3 rEuropa

1/3

mEarth 3 4/3 rEarth

GmEuropa 1/3   mEuropa    rEarth      mEarth   

 mEuropa  → rEuropa = rEarth    mEarth 

Gm1/3 m 2/3 Europa Earth 2 rEarth

2 rEarth

1/3

 mEuropa  = g Earth   1/3 mEarth  mEarth  1/3

GmEarth mEuropa

1/3

 4.9  1022 kg  2 2 = ( 9.80 m s )   = 1.98 m s  2.0 m s 24 5.98 10 kg    2

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Instructor Solutions Manual

16. The expression for the acceleration due to gravity at the surface of a body is g body = G

mbody 2 rbody

, where

rbody is the radius of the body. For Mars, g Mars = 0.38 g Earth . G

mMars 2 Mars

mEarth

= 0.38 G

→

2 rEarth

r   3400 km  = 6.5  1023 kg mMars = 0.38mEarth  Mars  = 0.38 ( 5.98  1024 kg )    6380 km   rEarth  2

17. The distance from the Earth’s center is r = rEarth + 380 km = 6.38  106 m + 3.8  105 m = 6.76  106 m. Calculate the acceleration due to gravity at that location.

g =G

mEarth r

(

= 6.67  10

−11

N m kg

)

5.97  1024 kg

( 6.76 10 m ) 6

= 8.714 m s 2

 1 “g”  = 0.889 g’s 2   9.80 m s 

= 8.714 m s 2 

This is an 11% reduction from the value of g at the surface of the Earth. 18. We are to calculate the force on Earth, so we need the distance of each planet from Earth. rEarth − = (150 − 108 )  106 km = 4.2  1010 m rEarth − = ( 778 − 150 )  106 km = 6.28  1011 m Venus

Jupiter

rEarth − = (1430 − 150 )  10 km = 1.28  10 m 6

Saturn

Jupiter and Saturn will exert a rightward force, while Venus will exert a leftward force. Take the right direction as positive. mEarth mJupiter m m m m FEarth − = G + G Earth2 Saturn − G Earth2 Venus 2 rEarth − rEarth − rEarth − planets Jupiter



Saturn

318

2  = GmEarth

 ( 6.28  1011 m ) 

+ 2

Venus

95.1

− 2

  2 ( 4.2 1010 m )  0.815

(1.28 10 m ) = ( 6.67  10 N m kg ) ( 5.97  10 kg ) ( 4.02  10 m ) = 9.56  10 N  9.6  10 N −11

−22

−2

The force of the Sun on the Earth is as follows. FEarth − = G Sun

mEarth mSun 2 Earth − Sun

( 5.97 10 kg )(1.99 10 kg ) = 3.52 10 N = ( 6.67  10 N m kg ) (1.50 10 m ) 24

−11

And so the ratio is FEarth − FEarth − = 9.56  1017 N 3.52 10 22 N = 2.7 10 −5 , which is 27 millionths. planets

Sun

mleft

19. See the diagram. The gravitational forces on the center meteorite from the other two meteorites must be equal in magnitude.

mright

mcenter 21m

53m

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Gravitation and Newton’s Synthesis

Fcenter = Fcenter → G

mleft mcenter

( 21m )

mright mcenter

→

( 53 m ) 2 ( 21m ) 2 ( 21m ) 2 mleft = mright = 1300 kg = 204 kg  2.0  102 kg ( ) 2 2 ( 53 m ) ( 53 m ) left

right

20. In general, the acceleration due to gravity of the Earth is given by g = G M Earth r 2 , where r is the distance from the center of the Earth to the location in question. So for the location in question, we have the following. m m 2 g = 101 gsurface → G Earth = 101 G 2Earth → r 2 =10 rEarth 2 r rEarth

(

)

r = 10 rEarth = 10 6.38  106 m = 2.02  107 m

21. The acceleration due to gravity at any location at or above the surface of a star is given by gstar = G M star r 2 , where r is the distance from the center of the star to the location in question.

g star = G

M star r

4 M Sun r

(

= 6.67  10

−11

N m kg

)

(

4 1.99  1030 kg

(1  10 m ) 4

) = 5.31 10 m s 12

 5  1012 m s 2 22. (a) The acceleration due to gravity at any location at or above the surface of a star is given by gstar = G mstar r 2 , where r is the distance from the center of the star to the location in question. gstar = G

msun 2 Moon

(1.99  10 kg ) = 4.38  10 m s ) (1.74  10 m ) 30

(

= 6.67  10−11 N m 2 kg 2

(

)

(b) W = mgstar = ( 65 kg ) 4.38 107 m s 2 = 2.8 109 N (c) Use Eq. 2–12c, with an initial velocity of 0. v 2 = v02 + 2a ( x − x0 ) → v=

2a ( x − x0 ) =

(

)

2 4.38  107 m s 2 (1.0 m ) = 9.4  103 m s

23. Each mass M will exert a gravitational force on mass m. The vertical components of the two forces will sum to be 0, and so the net force on m is directed horizontally. That net force will be twice the horizontal component of either force. GMm FMm = 2 → x + R2

(

x2 + R2 x

 

x2 + R2

)

193

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

FMm x =

GMm

cos  =

(x + R ) 2

GMm

(x + R ) x + R 2

Instructor Solutions Manual

GMmx

(x + R ) 2

3/ 2

2GMmx

Fnet x = 2 FMm x =

(x + R ) 2

3/ 2

24. From the symmetry of the problem, we can examine diametrically opposite infinitesimal masses and see that only the horizontal components (along the x-axis) of the force will result in a net force. Any off-axis components of force will add to zero. The infinitesimal horizontal force on m due to an infinitesimal mass Gm dM is dFdMm = 2 dM . The horizontal x + r2

(

dM x2 + r2

 

x2 + r2

)

component of that force is given by the following. Gm Gm x Gmx dM = dM ( dFdMm ) x = 2 2 cos  dM = 2 2 3/ 2 2 2 2 x +r x +r x + r2 x +r

(

)

(

) (

(

)

The total force is then found by integration over all parts of M. dFx =

Gmx dM

(x + r ) 2

3/ 2

Gmx dM

 dF =  x + r ( )

→

→ Fx =

3/ 2

GMmx

(x + r ) 2

3/ 2

From the diagram we see that it points inward towards the center of the ring. 25. The expression for g at the surface of the Earth is g = G

mE rE2

. Let g + g be the value at a distance

of rE + r from the center of Earth, which is r above the surface. (a)

g =G

mE 2 E

→ g + g = G

( rE + r )



rE2  1 +

r 



g  −2 g

rE 

mE 

r 

−2

 r   g 1 − 2  → 1+   r  rE  rE   2 E

r rE

(b) The minus sign indicated that the change in g is in the opposite direction as the change in r . So, if r increases, g decreases, and vice-versa. (c) Using this result: 1.25  105 m r g  −2 g = −2 9.80 m s 2 = −0.384 m s2 → g = 9.42 m s2 6.38  106 m rE Direct calculation: 5.98  1024 kg mE 2 2 −11 g = G 2 = 6.67  10 N m kg = 9.43m s 2 2 6 5 r 6.38  10 m + 1.25  10 m

(

)

(

)

The difference is only about 0.1%. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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26. We can find the actual g at the surface of the Earth by taking g due to the uniform Earth, subtracting away g due to the bubble as if it contained uniform Earth matter, and adding in g due to the oil-filled bubble. In the diagram, r = 1000 m (the diameter of the bubble, and the distance from the surface to the center of the bubble). The mass of matter in the bubble is found by taking the density of the matter times the volume of the bubble. Note that the radius of the bubble is 500 m. g oil = g uniform − g bubble + g bubble → present

Earth

g = g oil

(Earth matter)

rE − r

(oil)

− g uniform = g bubble − g bubble

present

Earth

( oil )

(oil)

Gmbubble

Gmbubble =

−

(Earth matter)



 G  3 = 2 mbubble − mbubble  = 2   oil −  Earth  43  rbubble r  ( oil ) r  (Earth  matter  matter)   G

The density of oil is given, but we must calculate the density of a uniform Earth. m 5.98  1024 kg  Earth = 4 E 3 = = 5.50  103 kg m3 3 6  rE 43  6.38  10 m matter 3

(

g =

G

 3  oil −  Earth  43  rbubble  r  matter  2

( 6.67  10 N m kg ) 8.0  10 kg m − 5.50  10 kg m  5.0  10 m ( ) ( ) (1.00  10 m ) −11

)

4 3

= −1.6414  10−4 m s 2  −1.6  10−4 m s 2 Finally we calculate the percentage difference. g −1.6414  10−4 m s2  100 = −1.7  10−3% (%) = g 9.80 m s 2 The negative sign means that the value of g would decrease from the uniform Earth value.

27. For an object “at rest” on the surface of the rotating Earth, there are two force vectors that add together to form the net force: Fgrav , the force of gravity, directed towards the center of the Earth; and FN , the normal force, which is given by FN = −mg eff . The sum of these two forces must produce the centripetal force that acts on the object, causing centripetal motion. See the diagram. Notice that the component axes are parallel and perpendicular to the surface of the Earth. Write Newton’s second law in vector component form for the object, and solve for g eff . The radius of the circular motion of the object is r = rE cos  , and the speed of the circular motion is v =

2 r T



Fnet

 Fgrav



, where

T is the period of the rotation, one day.

195

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Fgrav + FN = Fnet FN =

Instructor Solutions Manual

mE m ˆ mv 2 mv 2 ˆ → − G 2 j + FN = sin  i − cos  ˆj → rE r r

 4 2 r  m m mv 2   m 4 2 r   sin  ˆi +  G E2 − cos   ˆj = m  2 sin  ˆi +  G 2E − 2 cos   ˆj r r T  rE   rE    T

mv 2

 4 2 rE cos   mE 4 2 rE cos   ˆ ˆ i = m + −   sin cos G   j 2 T2 T2  rE   

 4 2 ( 6.38  106 m ) 1  4 2 ( 6.38  106 m ) 1   2 ˆ i +  9.80 m s − = m  ˆj 2 2  2 2 86, 400 s 86, 400 s ( ) ( )    

(

) (

)

= m  1.687  10 −2 m s 2 ˆi + 9.783 m s 2 ˆj

(1.687  10 m s ) = 0.0988 From this calculation we see that F points at an angle of  = tan ( 9.783m s ) −2

−1

north of the local “upwards” direction. Now solve FN = −mg eff for g eff .

( ) ( ) g = − (1.687  10 m s ) ˆi + ( 9.783m s ) ˆj → g = (1.687  10 m s ) + ( 9.783m s ) = 9.78 m s

FN = m  1.687  10 −2 m s 2 ˆi + 9.783m s 2 ˆj = − mg eff −2

→

eff

−2

eff

g eff points 0.099 below radially inward 28. Consider a distance r from the center of the Earth that satisfies r  REarth . Calculate the force due to the mass inside the radius r. mEarth 4 3 mEarth 3 mcloser to ( r ) = V =  43  r 3 = r = 3 r 3 3 4 REarth  REarth center 3

 mEarth 3   R3 r  m  r   r  m center Fgravity = G = G  Earth 2  = G 2Earth m  = mg surface    2 r r rEarth  rEarth   rEarth  Thus for Fgravity = 0.95mg , we must have r = 0.95rEarth , and so we must drill down a distance equal to mcloser to m

5% of the Earth’s radius. 0.05REarth = 0.05 ( 6.38  106 m ) = 3.19  105 m  320 km  200 mi 29. The rocket must be moving at “orbit speed” in order for the satellite to remain in the orbit when released. The speed of a satellite in circular orbit around the Earth is given in Example 6–6. vorbit = G

mEarth

v= G

mEarth r

= G

mEarth

( REarth + 780 km )

(

( 5.98 10 kg ) 24

6.67  10−11 N m 2 kg 2

) 6.38 10 m + 7.8 10 m ( ) 6

= 7.46  103 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

196

Chapter 6

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30. The speed of a satellite in a circular orbit around a body is given in Example 6–6 as

vorbit = G mbody r , where r is the distance from the satellite to the center of the body. v= G

mbody r

= G

( 5.98 10 kg ) 24

mEarth rEarth + 5.4  106 m

( 6.67 10 N m kg ) 6.38 10 m + 5.4 10 m ( ) −11

= 5.8  103 m s

31. Consider a free-body diagram for the woman in the elevator. FN is the upwards force the spring scale exerts, providing a normal force. Write Newton’s second law for the vertical direction, with up as positive.  F = FN − mg = ma → FN = m ( g + a ) (a, b) For constant speed motion in a straight line, the acceleration is 0, and so the normal force is equal to the weight. FN = mg = ( 58.0 kg ) ( 9.80 m s 2 ) = 568 N

(c) Here a = +0.18g and so FN = 1.18 mg = 1.18 ( 58.0 kg ) ( 9.80 m s 2 ) = 671N . (d) Here a = −0.18g and so FN = 0.82 mg = 0.82 ( 58.0 kg ) ( 9.80 m s 2 ) = 470 N . (e) Here a = − g and so FN = 0 N . 32. Consider a free-body diagram of yourself in the elevator. FN is the force of the scale pushing up on you, and reads the normal force. Since the scale reads 72 kg, if it were calibrated in Newtons, the normal force would be FN = ( 72 kg ) ( 9.80 m s 2 ) = 705.6 N. Write Newton’s second law in the vertical direction, with upward as positive.

 F = FN − mg = ma → a =

FN − mg

(

705.6 N − ( 62 kg ) 9.80 m s 2

62 kg m Since the acceleration is positive, the acceleration is upward.

) = 1.6 m s upward 2

33. The speed of an orbiting object is given in Example 6–6 as v = G m r , where r is the radius of the orbit, and m is the mass around which the object is orbiting. Solve the equation for m.

( 5.7 10 m )( 7.8 10 m s ) = 5.2 10 kg = v = Gm r → m = G ( 6.67 10 N m kg ) rv 2

−11

The number of solar masses is found by dividing the result by the solar mass. mgalaxy 5.2  1039 kg # solar masses = = = 2.6  109 solar masses 30 mSun 2  10 kg

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

34. Draw a free-body diagram of the monkey. Then write Newton’s second law for the vertical direction, with up as positive. F − mg  F = FT − mg = ma → a = T m For the maximum tension of 215 N, 215 N − (18.0 kg ) 9.80 m s 2 a= = 2.14 m s 2  2.1m s 2 (18.0 kg )

(

)

Thus the elevator must have an upward acceleration greater than a = 2.1m s 2 for the cord to break. Any downward acceleration would have resulted in a tension less than the monkey’s weight. 35. (a) The speed of a satellite in circular orbit around the Earth is shown in Example 6–6 to be vorbit = G

mEarth

. Thus the velocity is inversely related to the radius, and so the closer satellite r will be orbiting faster. GmEarth vclose vfar

rclose GmEarth

rfar

rclose

rEarth + 1.7  107 m rEarth + 8.5  106 m

6.38  106 m + 1.7  107 m

6.38  106 m + 8.5  106 m

= 1.253

rfar And so the close satellite is moving about 1.3x faster than the far satellite. (b) The period of a circular orbit is found by dividing the circumference of the orbit by the speed. T=

2 r vorbit

2 r

mEarth

2 r 3/ 2 GmEarth

3/ 2 2 rclose

Tclose Tfar

GmEarth 2 rfar3/ 2

r  =  close   rfar 

3/ 2

 rEarth + 8.5  106 m  =  7  rEarth + 1.7  10 m 

3/ 2

GmEarth

 6.38  106 m + 8.5  106 m  =  6 7  6.38  10 m + 1.7  10 m 

3/ 2

= 0.5077

And so the close satellite has a period of about 0.51x smaller than the far satellite. 36. The speed of an object in a circular orbit of radius r around mass m is given in Example 6–6 by v = G m r , and is also given by v = 2 r T , where T is the period of the orbiting object. Equate the two expressions for the speed and solve for T.

m r

2 r T

→

(1.74  10 m + 9.5  10 m ) = 2 = 7.05  10 s  118 min T = 2 Gm ( 6.67  10 N m kg )( 7.35  10 kg ) 6

−11

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Chapter 6

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37. The speed of an object in an orbit of radius r around mass m is the Earth is given in Example 6–6 by v = G m r , and is also given by where T is the period of the object in orbit. Equate the two expressions for the speed and solve for T.

m r

2 r r3

T = 2

→

T Gm

= 2

(

( 0.300 m )3

)

6.67  10−11 N m 2 kg 2 ( 7.0 kg )

= 4.778  104 s  4.8  104 s = 13h

38. The speed of an object in an orbit of radius r around the Earth is given in Example 6–6 by v = G mEarth r , and is also given by v = 2 r T , where T is the period of the object in orbit. Equate the two expressions for the speed and solve for T. Also, for a “near-Earth” orbit, r = rEarth . G

mEarth rEarth

T = 2

2 rEarth T

3 rEarth

GmEarth

→ T = 2

3 rEarth

GmEarth

( 6.38  10 m ) = 2 = 5070 s = 84.5 min ( 6.67  10 N m kg )( 5.98  10 m ) 6

−11

No , the result does not depend on the mass of the satellite. 39. Consider the free-body diagram for the astronaut in the space vehicle. The Moon is below the astronaut in the figure. We assume that the astronaut is in contact with the inside of the space vehicle, and so a force will be exerted on the astronaut by the spacecraft. That force has been labeled FN , and we assume that force is upwards. The F mg N magnitude of that force is the apparent weight of the astronaut. (a) If the spacecraft is moving with constant speed in a circular orbit, then gravity is supplying the centripetal force for the astronaut, and so he is in free fall and would appear to be “weightless,” He could float around the spacecraft, and there would not be a normal force on him: FN = 0 . (b) Now the astronaut has an acceleration towards the Moon. Write Newton’s second law for the astronaut, with down as the positive direction.  F = mg − FN = ma →

 GM Moon − a   2  r 

FN = mg − ma = m ( g − a ) = m 

 ( 6.67  10−11 N m 2 kg 2 )( 7.35  1022 kg )

= ( 75 kg ) 



( 2.8  10 m ) 6



− 1.8 m s 2  = −88.10 N



Because of the negative value, the normal force points in the opposite direction from what is shown on the free-body diagram–it is pointing towards the Moon. So perhaps the astronaut is pinned against the “ceiling” of the spacecraft and the ceiling is pushing him down, or safety belts are pulling down on the astronaut. The astronaut will perceive being “pushed downwards,” and so has a downward apparent weight of 88 N, towards the Moon .

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

40. We set the gravitational force equal to the expression for centripetal force, since gravity is causing circular motion. We also have to express the radius of orbit in terms of the period, using the orbit speed expression from Example 6–6. 4 2 6.38  106 m   g v 2 4 2 r acentrip = = 2 = = 8.266 m s 2  = 0.84 g 2 2  r T ( 92 min )( 60 s 1min )  9.80 m s 

(

)

= 0.70 g → v = 0.70 gr . Also for a r rotating object, the speed is given by v = 2 r T . . Equate the two expressions for the speed and solve for the period. 2 (14 m ) 2 r 2 r v = 0.70 gr = → T= = = 9.0sec 2 T 0.70 gr 0.70 9.80 m s 14 m ( ) ( )

41. The centripetal acceleration will simulate gravity. Thus

(

)

42. (a) The speed of an object in near-surface orbit around a planet is given in Example 6–6 to be v = Gm r , where m is the planet mass and r is the planet radius. The speed is also given by v = 2 r T , where T is the period of the object in orbit. Equate the two expressions for the speed. m

2 r T

→ G

m r

4 2 r 2 T2

→

4 2

GT 2

= 3

The density of a uniform spherical planet is given by  = 3m

=

3 4 2

= 3

m Volume

m 4 3

 r3

. Thus

3

4 r 4 GT GT 2 (b) For Earth, we have the following. 3 3 = = = 5.4  103 kg m 3 2 2 2 2 −11 GT 6.67  10 N m kg ( 85 min )( 60s min )  2

(

)

43. Consider the lower left mass in the diagram. The center of the orbits is the intersection of the three dashed lines in the diagram. The net force on the lower left mass is the vector sum of the forces from the other two masses, and points to the center of the orbits. To find that net force, project each force to find the component that lies along the line towards the center. The angle is  = 30. F =G

M2 l

Fnet = 2G

→ Fcomponent = F cos  = G towards center

= 3G

→

l2 2 l2 The net force is causing centripetal motion, and so is of the form M v 2 r . Note that r cos  = l 2.

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Fnet = 3G v=

M2 l2

M v2 r

M v2 l ( 2 cos  )

M v2 l

→

M2 l2

M v2 l

→

GM l

44. The effective value of the acceleration due to gravity in the elevator is g eff = g + aelevator . We take the upwards direction to be positive. The acceleration relative to the plane is along the plane, as shown in the freebody diagram. (a) The elevator acceleration is aelevator = +0.50 g . g eff = g + 0.50 g = 1.50 g →

FN a rel 

arel = geff sin  = 1.50 g sin 38 = 9.1m s



mg eff

(b) The elevator acceleration is aelevator = −0.50 g .

geff = g − 0.50 g = 0.50 g → arel = geff sin  = 0.50 g sin 38 = 3.0 m s 2 (c) The elevator acceleration is aelevator = − g .

geff = g − g = 0 → arel = geff sin  = 0sin 38 = 0 m s2 (d) The elevator acceleration is 0.

geff = g − 0 = g → arel = geff sin  = g sin 38 = 6.0 m s2 45. Use Kepler’s third law for objects orbiting the Sun.

TEarth ) = ( rNeptune rEarth ) 2

Neptune

r  TNeptune = TEarth  Neptune   rEarth 

3/ 2

→

 4.5  109 km  = (1 year )   8  1.50  10 km 

3/ 2

= 160 years

46. Use Kepler’s third law for objects orbiting the Sun. 3

 rIcarus   TIcarus  r  =T   Earth   Earth 

T  → rIcarus = rEarth  Icarus   TEarth 

2/3

 410 d  = (1.50  10 m )    365d  11

2/3

= 1.6  1011 m

47. Use Kepler’s third law for objects orbiting the Earth. The following are given.  86, 400 s  T2 = period of Moon = ( 27.4 day )  = 2.367  106 sec   1 day  r2 = radius of Moon’s orbit = 3.84  108 m r1 = radius of near-Earth orbit = rEarth = 6.38  106 m

(T1 T2 )2 = ( r1 r2 )3 T1 = T2 ( r1 r2 )

3/ 2

→

 6.38  106 m  = ( 2.367  10 sec )   8  3.84  10 m  6

3/ 2

= 5.07  103 sec ( = 84.5 min )

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48. Knowing the period of the Moon and the distance to the Moon, we can calculate the speed of the Moon by v = 2 r T . But the speed can also be calculated for any Earth satellite by

v = G mEarth r , as derived in Example 6–6. Equate the two expressions for the speed, and solve for the mass of the Earth. G mEarth r = 2 r T mEarth =

4 2 r 3 GT

→

(

4 2 3.84  108 m

)

( 6.67  10 N m kg ) ( 27.4 d )(86, 400s d ) −11

= 5.98  1024 kg

49. (a) The relationship between satellite period T, mean satellite distance r, and planet mass M can be derived from the two expressions for satellite speed: v = G m r and v = 2 r T . Equate the two expressions and solve for m. 4 2 r 3 G m r = 2 r T → m = GT 2 Substitute the values for Io to get the mass of Jupiter. mJupiter − = Io

(

4 2 4.22  108 m 2

(

4 2 6.71  108 m

(

)

(

)

= 1.90  10 27 kg

= 1.89  10 27 kg

( 6.67 10 N m kg ) ( 7.16  24  3600 s ) 4 (1.883  10 m ) = = 1.90  10 kg ( 6.67 10 N m kg ) (16.7  24  3600 s ) −11

Callisto

= 1.90  1027 kg

4 2 1.07  109 m

Ganymede

mJupiter −

)

6.67  10 −11 N m 2 kg 2 ( 3.55  24  3600 s )

mJupiter − =

(b) For the other moons, we have the following.

Europa

s ( 6.67 10 N m kg ) 1.77d  241 dh  3600  1h  −11

mJupiter − =

)

−11

Yes , the results are consistent–only about 0.5% difference between them. 50. Use Kepler’s third law, Eq. 6–7, to find the radius of each moon of Jupiter, using Io’s data for r2 and T2 .

( r1 r2 ) = (T1 T2 ) 3

→ r1 = r2 (T1 T2 )

rEuropa = rIo (TEuropa TIo )

2/3

(

2/3

(

)

= 422 103 km ( 3.55 d 1.77 d )

)

rGanymede = 422  103 km ( 7.16 d 1.77 d )

(

)

rCallisto = 422 103 km (16.7 d 1.77 d )

2/3

= 671103 km

= 1070  103 km

= 1880 103 km

The agreement with the data in the table is excellent.

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51. We equate the gravitational force to the expression for centripetal force, assuming the planet is in a circular orbit. We also use Eq. 6–6 to express the radius in terms of the period. Note that the mass of the orbiting planet does not enter the calculation. 2 Tplanet’s 4 2 orbit = → 3 rplanet’s GM star orbit 1/3

2     1/3 1/3  G ( 2 M Sun )  1.1TEarth’s   2 2  GM starTorbit  2 1/3  GM Sun TEarth  orbit     =  2 (1.1)   rplanet’s =   =  2 2    4 2 orbit  4   4     

=  2 (1.1)  2



1/3



rEarth =  2 (1.1)  2



orbit

1/3



(1.496 10 m ) = 2.0 10 m 11

52. As found in Example 6–6, the speed for an object orbiting a distance r around a mass m is given by

v = G m r. Gmstar vA vB

Gmstar

= 0.45

5.0

rB 53. Use Kepler’s third law to relate the orbits of Earth and Halley’s comet around the Sun.

rEarth ) = ( THalley TEarth ) 3

Halley

rHalley = rEarth ( THalley TEarth )

2/3

→

(

)

= 150  106 km ( 76 y 1 y )

2/3

= 2690  106 km

This value is half the sum of the nearest and farthest distances of Halley’s comet from the Sun. Since the nearest distance is very close to the Sun, we will approximate that nearest distance as 0. Then the farthest distance is twice the value above, or 5380  106 km = 5.4  1012 m . This distance approaches the mean orbit distance of Pluto, which is 5.9  1012 m. It is still in the solar system, nearest to Pluto’s orbit. 54. (a) Use Kepler’s third law to relate the Earth and the hypothetical planet in their orbits around the Sun.

planet

TEarth ) = ( rplanet rEarth ) 2

Tplanet = TEarth ( rplanet rEarth )

3/ 2

→

= (1 y )( 3 1)

3/ 2

= 5.20 y  5 y

(b) No mass data can be calculated from this relationship, because the relationship is massindependent. Any object at the orbit radius of 3 times the Earth’s orbit radius would have a period of 5.2 years, regardless of its mass.

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55. (a) In a short time t , the planet will travel a distance v t along its orbit. That distance is essentially a straight line segment for a short time duration. The time (and distance moved) during  t have been greatly exaggerated on the diagram. Kepler’s second law states that the area swept out by a line from the Sun to the planet during the planet’s motion for the  t is the same anywhere on the orbit. Take the areas swept out at the near and far points, as shown on the diagram, and approximate them as triangles (which will be reasonable for short  t ).

( Area )N = ( Area )F

→

1 2

( vN t ) d N = 12 ( vF t ) d F

→

v N vF = d F d N

(b) Since the orbit is almost circular, an average velocity can be found by assuming a circular orbit with a radius equal to the average distance. 11 11 1 2 r 2 12 ( d N + d F ) 2 2 1.47  10 m + 1.52  10 m vavg = = = = 2.973  104 m s 7 T T 3.16  10 s From part (a) we find the ratio of near and far velocities. v N v F = d F d N = 1.52 1.47 = 1.034 For this small change in velocities (3.4% increase from smallest to largest), we assume that the minimum velocity is 1.7% lower than the average velocity and the maximum velocity is 1.7% higher than the average velocity.

(

)

v N = vavg (1 + 0.017 ) = 2.973  10 4 m s (1.017 ) = 3.02  10 4 m s vF = vavg (1 − 0.017 ) = 2.973  10 4 m s ( 0.983 ) = 2.92  10 4 m s

56. (a) Take the logarithm of both sides of the Kepler’s third law expression.

 4 2  3   4 2  3  4 2  2 r log T log r 2 log T log → = → =      Gm  + 3log r →  GmJ   J   GmJ   2  4  log T = 23 log r + 12 log    GmJ 

T2 = 

This predicts a straight line graph for log(T) vs. log(r), with a slope of 3/2 and a

 4 2  y-intercept of log  .  GmJ  1 2

(b) The data is taken from Table 6–3, and the graph is shown here, with a straightline fit to the data. The data must to be converted to seconds and meters before the logarithms are calculated in order to do calculations in metric units.

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From the graph, the slope is 1.50 (as expected), and the y-intercept is –7.76.  4 2  4 2 4 2 1 log b m = → = = = 1.97  10 27 kg   J 2 2b −11 −15.52 Gm G 10 6.67 10 10   J 

(

) (

)(

)

The actual mass of Jupiter is given in Problem 18 as 318 times the mass of the Earth, which is 1.90  10 27 kg. So the value from the graph is about 4% too large. 57. (a)

Use Kepler’s third law. Eq. 6–7, to relate the orbits of the Earth and the comet around the Sun. 3

 rcomet   Tcomet  r  =T   Earth   Earth 

T  rcomet = rEarth  comet   TEarth 

→ 2/3

 2400 y  = (1 AU )    1y 

2/3

= 179.3 AU  180 AU

(b) The mean distance is the numeric average of the closest and farthest distances. 1.00 AU + rmax 179.3AU = → rmax = 357.6 AU  360 AU 2 (c) Refer to Fig. 6–18, which illustrates Kepler’s second law. If the time for each shaded region is made much shorter, then the area of each region can be approximated as a triangle. The area of each triangle is half the “base” (speed of comet multiplied by the amount of time) times the “height” (distance from Sun). So we have the following. Area min = Area max → 12 ( vmin t ) rmin = 12 ( vmax t ) rmax → vmin vmax = rmax rmin = 360 1

This relationship between speeds and distances is also derived in Problem 55. 58. (a) The Moon is Full when the Sun and Moon are directly opposite each other relative to the Earth. In this position, a person on the Earth will only be able to see either the Sun or the Moon in the sky at any given time. Therefore, as the Sun sets, the Moon rises. (b) As the Moon orbits the Earth it moves toward the East 1/29.53 of a synodic orbit, or about 12 every day. Therefore if the Moon was just rising at 6 PM on the day of the Full Moon, it would be about 12 below the horizon at 6 PM the next day, and therefore would not be visible. (c) The red dot represents the location of a person on the Earth who sees the Full Moon rise at 6 PM on the day shown as figure (a). A day later (b) the Earth has completed one full rotation and for the person at that location it is again 6 PM. When the next Full Moon arrives 29.53 days have elapsed, as shown in figure (e). That means the red dot has revolved around the Earth about 29 and a half times. Because of the half revolution, the dot is on the other side of the Earth. To the observer it is now about 6 AM and the Full Moon is setting as the Sun rises. In part (d) the Earth will have completed about 27 and a third revolutions so the red dot should be about one third of a counter-clockwise rotation from 6 PM, or about 2 AM.

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(d) The Earth completes one full revolution, or 360, around the Sun every year, or 365.25 days. The angle of the Moon in Fig. 6–24(e) relative to the “horizontal” [the dashed line in part (a)] is equal to the angle that the Earth moves between consecutive Full Moons:  29.53 days   =  360 = 29.11  365.25 days  So, in 29.53 days the Moon has orbited 360 + 29.11 = 389.11. The angular speed of the Moon is constant and can be written as the ratio of the orbital angle to orbital period for either sidereal or synodic orbits. Setting the ratios equal, solve for the sidereal period.

=

 sidereal Tsidereal

 synoptic Tsynoptic

→

  sidereal   360  = 27.32 days  = ( 29.53 days )    389.11    synoptic 

Tsidereal = Tsynoptic 

59. We choose the line joining the Earth and Moon centers to be the x-axis. The field of the Earth will point towards the Earth, and the field of the Moon will point towards the Moon.

GmEarth

( −ˆi ) + Gm

Moon

1   2 rEarth −   Moon 

( ˆi ) = (

Moon

G mMoon − mEarth ) ˆ i 2

1   2 rEarth −   Moon 

( 6.67 10 N m kg )( 7.35 10 kg − 5.97 10 kg ) ˆi = −1.07 10 m s ˆi ( ( 384 10 m ) ) −11

Earth

−2

1 2

So the magnitude is 1.07  10−2 m s2 and the direction is towards the center of the Earth . 60. (a) The gravitational field due to a spherical mass M, at a distance r from the center of the mass, is g = GM r 2 .

gSun at = Earth

GM Sun 2 Sun to Earth

( 6.67  10 N m kg )(1.99  10 kg ) = 5.93  10 m s = (1.496  10 m ) −11

−3

(b) Compare this to the field caused by the Earth at the surface of the Earth. gSun at 5.93  10−3 m s 2 Earth = = 6.05  10−4 g Earth 9.80 m s 2 No , this is not going to affect your weight significantly. The effect is less than 0.1 %.

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61. (a) From the symmetry of the situation, the net force on the object will be down. We show that explicitly by writing the field in vector component notation.





x +y

g = g left + g right =   −G





x +y

+  G





x02 + y

=  − 2G



2 0





x +y

  sin   ˆi +  −G





cos   ˆj =  −2G 2







 

x02 + y 2

2 0

m x +y 2 0

g right

sin   ˆi +  −G



 

g left

  cos   ˆj  

   y ˆj =  −2Gm  ˆj  2 2  2 2 3/ 2   x0 + y  ( x0 + y )   y

(b) If we keep y as a positive quantity, then the magnitude of the field is g = 2Gm

(x + y ) 2 0

3/ 2

We find locations of the maximum magnitude by setting the first derivative equal to 0.

g = 2Gm

 ( x02 + y 2 )3/2 − y 23 ( x02 + y 2 )1/2 2 y  =0 → → = 2Gm  2 2 3 dy   x + y (0 )

(x + y ) 2 0

3/2

(x + y ) − y (x + y ) 2y = 0 → 2 0

3/2

 

g max = g  y =

3 2

2 1/2

2 0

y max =

x0 2

 0.71x0

x0 

4Gm

=  0.77 2  = 2Gm 3/ 2 x0 2 3 3 x02  2  x0  2   x0 +     2  

There would also be a maximum at y = − x0

2. 2

62. The centripetal acceleration is aR =

v2 rEarth

  4 2 rEarth  2 rEarth T  orbit orbit  = = . The force (from Newton’s 2 rEarth

orbit

second law) is FR = mEarth aR. The period is one year, converted into seconds.

4 2 rEarth aR =

orbit

(

4 2 1.50  1011 m

) = 5.97  10 m s −3

( 3.15  10 sec ) F = ma = ( 5.98  10 kg )( 5.97  10 m s ) = 3.57  10 N T

−3

The Sun exerts this force on the Earth. It is a gravitational force.

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63. See the diagram for the two stars. D (a) The two stars don’t crash into each other because of their circular motion. The force on them is centripetal, FG FG and maintains their circular motion. Another way to consider it is that the stars have a velocity, and the gravity force causes CHANGE in velocity, not actual velocity. If the stars were somehow brought to rest and then released under the influence of their mutual gravity, they would crash into each other. (b) Set the gravity force on one of the stars equal to the centripetal force, using the relationship that v = 2 r T =  D T , and solve for the mass.

FG = G

D/2

= FR = m 2

2 2 D 3

2 ( D T )

(

2 2 mD

2 2 8.0  1011 m

)

→ G

2 2 mD

( 6.67 10 N m kg ) 14.2 yr  3.1561yr10 s  2

→

−11

= 2

= 7.5  10 29 kg





64. The acceleration due to the Earth’s gravity at a location at or above the surface is given by g = G mEarth r 2 , where r is the distance from the center of the Earth to the location in question. Find the location where g = 12 g surface. GmEarth

1 GmEarth

2 → r 2 = 2 rEarth → r=

2 Earth

2 rEarth

2 r r The distance above the Earth’s surface is as follows.

r − rEarth =

( 2 − 1) r

Earth

( 2 − 1) (6.38  10 m ) = 2.64  10 m = 0.414 r 6

Earth

65. For an object to be apparently weightless would mean that the object would have a centripetal acceleration equal to g. This is really the same as asking what the orbital period would be for an object orbiting the Earth with an orbital radius equal to the Earth’s radius. To calculate, use v2 g = aR = , along with v = 2 rEarth T , and solve for T. rEarth g=

v2 rEarth

4 2 rEarth T

→ T = 2

rEarth g

= 2

6.38  106 m 9.80 m s

= 5.07  103 s ( 84.5 min )

66. Your weight is given by the law of universal gravitation. The derivative of the weight with respect to time is found by taking the derivative of the weight with respect to distance from the Earth’s dr center, and using the chain rule. Note that = v. dt W =G

mE m r

→

dW dt

dW dr dr dt

= −2G

mE m r3

67. The speed of an object in an orbit of radius r around a planet is given in Example 6–6 as v = G mplanet r , and is also given by v = 2 r T , where T is the period of the object in orbit. Equate the two expressions for the speed and solve for T. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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mPlanet r

2 r T

→ T = 2

r3 G mPlanet

For this problem, the inner orbit has radius rinner = 7.3  107 m, and the outer orbit has radius

router = 1.7  108 m. Use these values to calculate the periods. Tinner

( 7.3 10 m ) = 2 = 2.0  10 s ( 6.67 10 N m kg )( 5.7 10 kg )

Touter

(1.7 10 m ) = 2 = 7.1  10 s ( 6.67 10 N m kg )( 5.7 10 kg )

−11

Saturn’s rotation period (day) is 10 hr 39 min, which is about 3.8  104 sec. Thus the inner ring will appear to move across the sky “faster” than the Sun (about twice per Saturn day), while the outer ring will appear to move across the sky “slower” than the Sun (about once every two Saturn days). 68. (a) The speed of a satellite orbiting the Earth is given by v = G mEarth r . For the GPS satellites,

r = rEarth + (11,000)(1.852 km ) = 6.38  106 m + 2.037  107 m = 2.675  107 m.

( 5.97  10 kg ) = 3.858  10 m s  3.9  10 m s v = ( 6.67  10 N m kg ) 2.675  10 m 24

−11

(b) The period can be found from the speed and the radius. 7 2 r 2 2.675  10 m v = 2 r T → T = = = 4.4  10 4 sec  12 h 3 3.858  10 m s v

(

)

69. (a) If the asteroid were a sphere, then the mass would be given by m =  V = 43  r 3 . We first find the mass by multiplying the density times the volume, and then use that mass to solve for the radius as if it were a sphere. kg   m =  V =  2.3  103 3  ( 40000  6000  6000 m 3 ) = 3.312  1015 kg m   1/3

   3 ( 3.312  1015 kg )   3m  3 r=  = 7005 m  7  10 m  =  kg 4      4  2.3  103 3   m    1/3

(b) The acceleration due to gravity is found from the mass and the radius of the hypothetical sphere. 3.312  1015 kg g = G m r 2 = 6.67  10−11 N m 2 kg 2 = 4.502  10−3 m s 2 2 ( 7005 m )

(

)

(

)

 5  10−3 m s 2 (c) The speed of an object orbiting a mass m is given by v = G m r . The period of an object moving in a circle path is given by T = 2 r v. Combine these relationships to find the period.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

2 r G

(

2 2  10 4 m

( 6.67 10 N m kg ) −11

)

( 3.312  10 kg ) ( 2 10 m ) 15

Instructor Solutions Manual

= 3.781 104 sec

 4  104 sec  11h

70. We apply Eq. 6–6. 2 TLRO 4 2 3 rLRO

TLRO =

→

GM Moon 3 4 2 rLRO

GM Moon

(

4 2 1.74  106 m + 5.0  10 4 m

( 6.67 10

−11

N m 2 kg 2

)

)( 7.35 10 kg ) 22

= 6796 s  1.89 h

71. In the text, it says that Eq. 6–6 is valid if the radius r is replaced with the semi-major axis s. From Fig. 6–17, the distance of closest approach rmin is seen to be rmin = s − es = s (1 − e ) , and so the semi-major axis is given by s =

4 2

GmSgrA

= 3

rmin 1− e

→ 3

 1.5  1011 m  123 AU    1AU 2    4 3 1 − 0.87   2  rmin  4     2 3 4 s 1− e  =   mSgrA = = 2 2 2 GT GT 3.156  107 s  2 2  −11 6.67 10 N m kg 15.2 y   ( )  1y   = 7.352  1036 kg  7.4  1036 kg

mSgrA

7.352  1036 kg

mSun 1.99  10 kg our Sun. 30

= 3.7  106 and so SgrA is almost 4 million times more massive than

72. The initial force of 120 N can be represented as Fgrav =

Gmplanet msatellite r2

= 120 N.

3.0  107 m

= 1.875 times the original radius. 1.6  107 m Gmplanet msatellite Gmplanet msatellite Gmplanet msatellite 1 Fnew = = = = (120 N ) = 34 N 2 2 2 3.516 r 3.516 rnew radius (1.875 r ) (b) With the larger radius, the period is T = 7200 seconds. As found in Example 6–6, orbit speed

(a) The new radius is

can be calculated by v =

Gm r

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Gm r

2 r T

→ m=

4 2 r 3 GT

(

) = = 3.1 10 kg ( 6.67 10 N m kg ) ( 7200 s ) 4 2 3.0  107 m −11

73. The apparent weight is the normal force on the passenger. For a person at rest, the normal force is equal to the actual weight. If there is acceleration in the vertical direction, either up or down, then the normal force (and hence the apparent weight) will be different than the actual weight. The speed of the Ferris wheel is v = 2 r T = 2 (11.0 m ) 12.5s = 5.529 m s . (a) See the free-body diagram for the highest point of the motion. We assume the passengers are right-side up, so that the normal force of the Ferris wheel seat is upward. The net force must point to the center of the circle, so write Newton’s second law with downward as the positive direction. mg FN The acceleration is centripetal since the passengers are moving in a circle. v2 v2 F = F = mg − F = ma = m → F = mg − m  R N N r r The ratio of apparent weight to real weight is given by the following. v2 v2 2 mg − m g− 2 ( 5.529 m s ) r = r = 1− v = 1− = 0.716 mg g rg (11.0 m ) 9.80 m s 2

(

)

And so the passengers feel “light” at the top of the motion. (b) At the bottom, consider the free-body diagram shown. We assume the passengers are right-side up, so that the normal force of the Ferris wheel seat is upward. The net force must point to the center of the circle, so write Newton’s second law with upward as the positive direction. The acceleration is centripetal since the passengers are moving in a circle. v2 v2 F = F = F − mg = ma = m → F = mg + m  R N N r r The ratio of apparent weight to real weight is given by the following. v2 2 mg + m 2 ( 5.529 m s ) r = 1+ v = 1+ = 1.284 mg rg (11.0 m ) ( 9.80 m s 2 )

And so the passengers feel “heavy” at the bottom of the motion. 74. Equate the force of gravity on a mass m at the surface of the Earth as expressed by the acceleration due to gravity to that as expressed by Newton’s law of universal gravitation. 2 2 GmEarth m g rEarth g rEarth 3g 3g 3g → = = = = = mg = G 2 3 4 rEarth mEarth  Earth 3  rEarth 4 rEarth 4 CEarth 2  CEarth 2

(

3 10 m s 2

2 3000 kg m 3

)

)( 4  10 m ) 7

= 1.25  10

−10

m3 kg s 2

 1  10

−10

m3 kg s 2

This is roughly twice the size of the accepted value of G.

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75. From the Venus data, the mass of the Sun can be determined by the following. Set the gravitational force on Venus equal to the centripetal force acting on Venus to make it orbit. 2

 2 rVenus  orbit  mVenus  4 2 mVenus rVenus  TVenus  2 GmSun mVenus mVenus v Venus orbit   = = = 2 Venus orbit

rVenus

orbit

2 Venus

3 4 2 rVenus

→ mSun =

orbit 2 Venus

3 4 2 rCallisto

Then likewise, for Callisto orbiting Jupiter, mJupiter =

orbit 2 Callisto

, and for the Moon orbiting the Earth,

3 4 2 rMoon orbit

mEarth =

. To find the density ratios, take the mass ratios with the mass expressed as density GT times volume, and expressed as found above. 3 4 2 rCallisto 2 Moon

orbit

mJupiter mSun

3  Jupiter 43  rJupiter 3  Sun 43  rSun

2 GTCallisto = 3 4 2 rVenus

→

orbit 2 Venus

 Jupiter  Sun

3 Callisto orbit

r =

2 3 TVenus rSun

2 3 3 TCallisto rVenus rJupiter

1 ( 0.01253)3 ( 224.7 ) 2 = 0.948 2 3 (16.69 ) ( 0.724 ) ( 0.0997 )3

orbit

And likewise for the Earth–Sun combination: 3 rMoon 2 3  Earth 1 ( 0.003069 )3 ( 224.7 )2 orbit TVenus rSun = 2 = = 3.98 2 3 3 3  Sun TMoon rVenus rEarth ( 27.32 ) ( 0.724 ) ( 0.0109 )3 orbit

76. The density of the sphere is uniform, and is given by  = 4 3

make the cavity is M cavity = Vcavity = 4 3

 r3

. The mass that was removed to

(  ( r 2 ) ) = M . The net force on the point mass can r

4 3

1 8

be found by finding the force due to the entire sphere, and then subtracting the force caused by the cavity alone.

Fnet = Fsphere − Fcavity = =

GMm d2

−

G ( 18 M ) m

( d − r 2)

 1

= GMm 

d

− 2

  8 ( d − r 2)  1

GMm  d

 1 1 − 2   8 (1 − r 2d ) 

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77. For a body on the equator, the net motion is circular. Consider the freebody diagram as shown. FN is the normal force, which is the apparent weight. The net force must point to the center of the circle for the object to be moving in a circular path at constant speed. Write Newton’s second law with the inward direction as positive. v2 F mg F m = − = →  R Jupiter N rJupiter



v2 

 mJupiter v2   = mG 2 −  rJupiter    rJupiter rJupiter  Use the fact that for a rotating object, v = 2 r T FN = m  g Jupiter −



FN = m  G

mJupiter 2 rJupiter

−

4 2 rJupiter 

 = mg perceived

2 TJupiter   Thus the perceived acceleration due to gravity of the object on the surface of Jupiter is as follows. mJupiter 4 2 rJupiter g perceived = G 2 − 2 rJupiter TJupiter

(1.9  10 kg ) − 4 ( 7.1  10 m ) = ( 6.67  10 N m kg ) ( 7.1  10 m ) ( 595 min )  60 s  27

−11





 1 min  

  = 2.3 g’s  9.8 m s  1g

= 22.94 m s 2 

Based on this result, you would not be crushed at all. You would feel “heavy,” but not at all crushed. 78. (a) We use the law of universal gravitation to express the force for each mass m. One mass is “near” the Moon, and so the distance from that mass to the center of the Moon is REM − rE . The other mass is “far” from the Moon, and so the distance from that mass to the center of the Moon is REM + rE .

Fnear = Moon

GmM m

( REM − rE )

Ffar

Moon

GmM m

( REM + rE )2

GmM m

 Fnear  ( REM − rE )  REM + rE   3.84  108 m + 6.38  106 m  = =  F   =  = 1.0687 8 6 GmM m  REM − rE   3.84  10 m − 6.38  10 m   far  Moon 2

( REM + rE )2

(b) We use a similar analysis to part (a). GmSm GmSm Fnear = Ffar = 2 Sun Sun ( RES − rE ) ( RES + rE )2

GmSm

 Fnear  ( RES − rE )  RES + rE   1.496  1011 m + 6.38  106 m   F  = Gm m =  R − r  =  1.496  1011 m − 6.38  106 m  = 1.000171 S   far Sun  ES E   2

( RES + rE )2

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(c) For the average gravitational force on the large masses, we use the distance between their centers. GmSmE GmM mE FSun = FMoon = 2 2 RES REM GmSmE FSun FMoon

2 RES

GmMoon mEarth

(1.99  10 kg ) ( 3.84  10 m ) = 178 = m (1.496  10 m ) ( 7.35  10 kg )

2 mS REM

2 RES

2 REM (d) Apply the expression for F as given in the statement of the problem.

 Fnear

 Fnear  − 1  Ffar FM FMoon  Ffar    Moon = 1 (1.0687 − 1) = 2.3 Moon = = FS FSun  Fnear 178 (1.000171 − 1) F   − 1 FSun  near − 1   Ffar Sun  Ffar Sun FMoon 



− 1

79. The acceleration is found from the law of universal gravitation. Using the chain rule, a relationship between the acceleration expression and the velocity can be found, which is integrated to find the velocity as a function of distance. The outward radial direction is taken to be positive, so the acceleration is manifestly negative. mm Gm Gm d v d v dr dv dv → a = − 2E = = =v → − 2E = v → F = ma = −G E2 r r dt dr dt dr r dr −GmE

GmE rE

−

 GmE  =  vd v →  = 12 v 2f → 2  r  r  2r 2r 0

= vd v → − GmE  2

GmE 2rE

= 12 v 2f → v f = 

GmE rE

→ vf = −

GmE rE

The negative sign is chosen because the object is moving towards the center of the Earth, and the outward radial direction is positive. 80. (a) See the free-body diagram for the plumb bob. The attractive gravitational force mm on the plumb bob is FM = G 2 M . Since the bob is not accelerating, the net DM force in any direction will be zero. Write the net force for both vertical and m horizontal directions. Use g = G 2Earth . rEarth

F



FM mg

= FT cos  − mg = 0 → FT =

cos   Fhorizontal = FM − FT sin  = 0 → FM = FT sin  = mg tan  vertical

mmM DM2

= mg tan  →  = tan −1 G

2 mM rEarth

gDM

mEarth DM2

= tan −1 2

214

Chapter 6

Gravitation and Newton’s Synthesis

(b) We estimate the mass of Mt. Everest by taking its volume times its mass density. If we approximate Mt. Everest as a cone with the same size diameter as height, then its volume is 2 3 3 V = 13  r 2h = 13  ( 2000 m ) ( 4000 m ) = 1.7  1010 m3 . The density is  = 3  10 kg m . Find the mass by multiplying the volume times the density.

(

)(

)

mM =  V = 3  103 kg m3 1.7  1010 m3 = 5  1013 kg (c)

With D = 5000 m, use the relationship derived in part (a).

 = tan

−1

( 5  10 kg )( 6.38  10 m ) = 8  10 degrees = tan D ( 5.97  10 kg ) ( 5000 m ) 13

2 mM rEarth

mEarth

−1

2 M

−4

81. If the ring is to produce an apparent gravity equivalent to that of Earth, then the normal force of the ring on objects must be given by FN = mg . The Sun will also exert a force on objects on the ring. See the free-body diagram. Write Newton’s second law for the object, with the fact that the acceleration is centripetal. v2 F = F = F + F = m  R Sun N r Substitute in the relationships that v = 2 r T , FN = mg , and FSun = G period of the rotation. m m 4 2 mr v2 FSun + FN = m → G Sun2 + mg = r r T2 T=

4 2 r = m + G Sun g r2



(

→ G

mSun

4 2 1.50  1011 m

(

+g=

FSun

Sun

mSun m r2

, and solve for the

4 2 r T2

)

(1.99  10 kg ) + 9.80 m s ) (1.50  10 m ) 30

6.67  10 −11 N m 2 kg 2

  = 8.99 d  86, 400s 

= 7.77  105 s 

Note that the force of the Sun is only about 1/1600 the size of the normal force. The force of the Sun could have been ignored in the calculation with no significant change in the result given above. 82. (a) The gravitational force on the satellite is given by Fgrav = G

mEarth m

, where r is the distance of r2 the satellite from the center of the Earth. Since the satellite is moving in circular motion, then v2 the net force on the satellite can be written as Fnet = m . By substituting v = 2 r T for a r 2 4 mr circular orbit, we have Fnet = . Then, since gravity is the only force on the satellite, the T2 two expressions for force can be equated, and solved for the orbit radius.

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mEarth m r2

4 2 mr

Instructor Solutions Manual

→

1/3  ( 6.67  10−11 N m 2 kg 2 )( 6.0  1024 kg ) ( 6800 s ) 2   GmEarthT 2  r=   = 2 4 2  4   

1/3

= 7.768  106 m  7.8  106 m (b) From this value the gravitational force on the satellite can be calculated. 6.0  1024 kg ( 5200 kg ) mEarth m −11 2 2 Fgrav = G = 6.67  10 N m kg = 3.449  104 N 2 2 6 r 7.768  10 m

(

)

(

)

(

)

 3.4  104 N (c) The altitude of the satellite above the Earth’s surface is given by the following. r − rEarth = 7.768  106 m − 6.38  106 m = 1.4  106 m

83. Find the “new” Earth radius by setting the acceleration due to gravity at the Sun’s surface equal to the acceleration due to gravity at the “new” Earth’s surface. GmEarth GmSun = 2 → g Earth = gSun → 2 rEarth rSun new new

rEarth = rSun

mEarth

(

= 6.96  10 m 8

5.98  1024 kg

) 1.99  10 kg = 1.21  10 m

mSun This is about 1/5 the actual Earth radius. new

84. The speed of an object orbiting the Sun is implied from Example 6–6 as v =

GmSun

vnew = 1.5v and vnew =

rnew

GmSun

→ 1.5v =

rnew

→ 1.5

GmSun r

GmSun rnew

→

rnew =

= 0.44r 1.52 Note that the answer doesn’t depend on either of the asteroid masses. 85. (a) From Example 6–6, the speed of an object in a circular orbit of radius r about mass m is

Gm r

v= r=

(b) v =

2 r T

. Use that relationship along with the definition of density to find the speed. Gm r 3v 2 4 G 

→ v2 = =

→ T=

(

Gm r

G  43  r 3

3 ( 22 m s )

4 6.67  10

2 r v

→

r −11

N m kg

2 ( 25330 m ) 22 m s

)( 2700 kg m ) 3

= 25330 m  2.5  10 4 m

= 7234s  2.0 h

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Chapter 6

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86. The speed of rotation of the Sun about the galactic center, under the assumptions made, is given by

v= G

mgalaxy rSun orbit

mgalaxy =

r v2 and so mgalaxy = Sun orbit . Substitute in the relationship that v = 2 rSun orbit T . G

4 2 ( rSun orbit ) GT 2

(

)

4 2  ( 30, 000 ) 9.5  1015 m 

( 6.67  10

−11

  3.15  107 s   N m kg ) ( 200  106 y )   1y    2

= 3.452  1041 kg  3  1041 kg

The number of solar masses is found by dividing the result by the solar mass. mgalaxy 3.452  1041 kg # stars = = = 1.726  1011  2  1011 stars 30 mSun 2.0  10 kg

217

CHAPTER 7: Work and Energy Responses to Questions 1.

“Work” as used in everyday language generally means “energy expended,” which is similar to the way “work” is defined in physics. However, in everyday language, “work” can involve mental or physical energy expended, and is not necessarily connected with displacement, as it is in physics. So a student could say she “worked” hard carrying boxes up the stairs to her dorm room (similar in meaning to the physics usage), or that she “worked” hard on a problem set (different in meaning from the physics usage).

Yes, she is doing work. She is applying a force (via her hands) to the water, and moving the water through some distance as she applies the force. When she initially stops her swimming motion, the water accelerates her in the direction of the current and so does some work on her. But once she reaches a constant “drifting” speed, equal to the current, the net force of the water on her is 0, and she no longer accelerates. The water is both pushing her and resisting her at the same time, and thus the net force of the water on the swimmer is 0.

No, not if the object is moving in a circle. Work is the product of force and the displacement in the direction of the force. Therefore, a centripetal force, which is perpendicular to the direction of motion, cannot do work on an object moving in a circle.

You are not doing any work on the wall. However, to keep your muscles under tension, metabolic processes must happen which use energy stored in your body. Another byproduct of those metabolic processes is heat, and so you might start to sweat if you pushed on the wall for a long time. Work is not being done on the wall, but energy transfers are occurring in your body which make you feel fatigued.

No. The magnitudes of the vectors and the angle between them are the relevant quantities, and these do not depend on the choice of coordinate system.

Yes. A dot product will be negative if the angle between the two vectors is an obtuse angle, because the cosine of an obtuse angle is negative. For example, if one vector points along the positive x-axis, and the other along the negative x-axis, the angle between the vectors is 180°. The cosine of 180° is –1, and so the dot product of the two vectors will be negative.

As a car accelerates forward the engine causes the tires to rotate. The force of static friction between the tires and the road accelerates the car forward. Since the force on the car and the car’s motion are in the same direction, the force of friction does positive work on the car. A second example would be the positive work done on a crate sitting in the bed of an accelerating truck. As the truck accelerates forward, the force of friction between the crate and the bed accelerates the box forward.

No. The dot product of two vectors is always a scalar, with only a magnitude. The dot product does not have a direction.

Yes. The normal force is the force perpendicular to the surface on which the object rests. If the object moves with a component of its displacement perpendicular to this surface, the normal force will do work. For instance, when you jump, the normal force does work on you in accelerating you vertically. And it is the normal force of the elevator floor on you that accelerates you in an elevator and so does work on you.

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Work and Energy

10. (a) If the force applied to the spring is the same, then F = k1 x1 = k 2 x2 , so x2 = k1 x1 k 2 . The work done on spring 1 will be W1 = 12 k1 x12 . The work done on spring 2 will be W2 = 12 k2 x22

(

)

= 12 k2 k12 x12 k22 = W1 ( k1 k2 ). Since k1  k 2 , W2  W1 , so more work is done on spring 2.

(b) If the displacement is the same, then W1 = 12 k1 x 2 and W2 = 12 k2 x 2. Since k1  k 2, W1  W2 , so more work is done on spring 1. 11. If the speed of the particle doubles, the kinetic energy increases by a factor of 4, since the kinetic energy is proportional to the square of the speed. 12. Until it reaches the x = 0 point, the end of the spring has a positive acceleration and is accelerating (and pushing) the block, and therefore will remain in contact with it. We assume that the block has no other mechanism that can accelerate it in the forward direction, and so it has the same acceleration as the spring. At the x = 0 point, the spring exerts no force, and so the block has reached its maximum velocity. But neither the block or spring will stop there. After the x = 0 point, the end of the spring begins to slow down because it will be pulled backwards by the force of the spring, but the block will continue to move (at least initially) with its maximum speed and will therefore be moving faster than the spring and will separate from it. 13. The net work done on a particle and the change in the kinetic energy do depend on the choice of reference frame. For instance, assume that in one reference frame a 1-kg object changes velocity from 0 m/s to 2 m/s in the x-direction. The change in kinetic energy would be 2 J. But in a different reference frame, moving at 1 m/s in the negative x-direction relative to the first reference frame, the velocity would change from 1 m/s to 3 m/s. The change in kinetic energy would be 4 J. If the reference frames are in relative motion, the values for net work done on a particle, the kinetic energy, and the change in the kinetic energy all will be different in different frames. The workenergy principle will still be true in each reference frame.

Responses to MisConceptual Questions 1.

(d) The force applied to the bag to hold it up is vertical, so the greatest work is done when the displacement has the largest vertical component. Answers (a) and (b) have no vertical displacement at all, and the vertical displacement of answer (d) is less than that of (c).

(a) A common misconception is that all forces do work. However, work requires that the object on which the force is acting has a component of motion in the direction of the force. So the car has to be moving for your force to do work on the car.

(e) The displacement of the box is 0 during the two minutes, so no work is done during that time.

(a) The speed of the crate is constant so the net (total) work done on the crate is zero. The normal force is perpendicular to the direction of motion, so it does no work. Your applied force and a component of gravity are in the direction of motion, and so both do positive work. The force of friction opposes the direction of motion and does negative work. For the total work to be zero, the work done by friction must equal the sum of the work done by gravity and by you.

(d) The brick wall has no displacement in either case, and so no work is done on it in either case.

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(c) Work is done when the force acting on the object has a component in the direction of motion. Gravity, which provides the centripetal force, is not zero but is always perpendicular to the motion. A common error is the notion that an object moving in a circle has no work done on it. This is true only if the object is moving at constant speed.

(c) In portion A, the vertical force of the man on the package is parallel to the vertical displacement of the package, and so he is doing work on the package. In portion C, he exerts a horizontal force on the package in order to make it accelerate, and that force is parallel to the horizontal displacement of the package, and so he is doing work on the package. In portions B and D, he is only putting a vertical force on the package, but the displacement is horizontal, so he is not doing any work on the package.

(c) The only force that does work is gravity, and both of you have the save vertical displacement. Thus the amount of work done is mgh, where h is the height of the hill. The work done is 2 equal to the change in kinetic energy, and so since you start from rest, mgh = 12 mvbottom . For

both of you then vbottom = 2 gh , independent of mass or shape of the descent path. 9.

(b) Work can be positive or negative, depending on the angle between the net force and the displacement. But kinetic energy can only be positive since it is the product of positive quantities (mass and the square of the speed).

10. (c) We assume the force of friction is the same in both cases, and so the work done will be proportional to the distance that the truck skids. The work done is also proportional to the change in kinetic energy. That change will be 9x larger in the higher-speed case since the change in kinetic energy will be proportional to the speed squared, and so the truck should skid 9x further. Thus the answer is (c). 11. (b) Since the change in speeds are equal, many students think that the change in energy will also be equal. However, the energy is proportional to the square of the speed. It takes four times as much energy to accelerate the car from rest to 60 km/h as it takes to accelerate the car from rest to 30 km/h. Therefore it takes three times the energy to accelerate the car from 30 km/h to 60 km/h as it takes to accelerate it from rest to 30 km/h. 12. (b) Since both blocks have the same initial kinetic energy, both require the same amount of work to bring them to a stop. The heavier block will have a greater frictional force on it, since the kinetic frictional force is be proportional to the mass (assuming the coefficient of friction is the same for both). So the heavier block will stop in a shorter distance, and the lighter block will go further before it stops. 13. (d) Both stones have work done on them by the gravity force. Both fall the same distance, but the heavier stone has twice the force on it, since the gravity force is proportional to the mass ( FG = mg ) . Thus, the heavier stone will have twice as much kinetic energy, since it has twice as much mass and so twice as much force on it. 14. (c) Since kinetic energy is given by K = 12 mv 2 an object must be moving (but not necessarily falling) to have kinetic energy.

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Chapter 7

Work and Energy

Solutions to Problems 1.

The force and the displacement are both downwards, so the angle between them is 0 o. Use Eq. 7–1.

(

)

WG = mgd cos  = ( 280 kg ) 9.80 m s2 ( 3.80 m ) cos 0o = 1.0  104 J 2.

The rock will rise until gravity does –70.0 J of work on the rock. The displacement is upwards, but the force is downwards, so the angle between them is 180o. Use Eq. 7–1. WG −70.0 J WG = mgd cos  → d = = = 3.86 m mg cos (1.85 kg ) 9.80 m s 2 ( −1)

(

)

The minimum force required to lift the firefighter is equal to her weight. The force and the displacement are both upwards, so the angle between them is 0o. Use Eq. 7–1.

(

)

Wclimb = Fclimbd cos  = mgd cos  = ( 55.0 kg ) 9.80 m s2 ( 28.0 m ) cos 0o = 1.51  104 J 4.

The maximum amount of work would be the work done by gravity. Both the force and the displacement are downwards, so the angle between them is 0 o. Use Eq. 7–1.

(

)

WG = mgd cos  = (1.2 kg ) 9.80 m s 2 ( 0.65 m ) cos 0o = 7.6 J

This is a relatively small amount of energy. If the person adds a larger force to the hammer during the fall, then the hammer will have a larger amount of energy to give to the nail. 5.

Draw a free-body diagram for the crate as it is being pushed across the floor. The pushing force is FP . Since the crate is not accelerating vertically, FN = mg . Since the crate is not accelerating horizontally, FP = Ffr =

x

Ffr mg

FN  k FN =  k mg . The work done to move it across the floor is the work done by the pushing force. The angle between the pushing force and the direction of motion is 0o. Use Eq. 7–1. Wpush = Fpush d cos 0o = k mgd (1) = ( 0.40 )( 46.0 kg ) ( 9.80 m s 2 ) (10.3 m ) = 1900 J 6.

(a) See the free-body diagram for the crate as it is being pulled. Since the crate is not accelerating horizontally, FP = Ffr = 230 N. The work done to move it across the floor is the work done by the pulling force. The angle between the pulling force and the direction of motion is 0. Use Eq. 7–1. WP = FP d cos 0 = ( 230 N )( 6.0 m )(1) = 1380 J  1400 J

x Ffr

(b) See the free-body diagram for the crate as it is being lifted. Since the crate is not accelerating vertically, the pulling force is the same magnitude as the weight. The angle between the pulling force and the direction of motion is 0o. Use Eq. 7–1. WP = FP d cos 0o = mgd = (1200 N )( 6.0 m ) = 7200 J

y

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(a) Write Newton’s second law for the vertical direction, with up as positive.  Fy = FL − Mg = Ma = M ( 0.10 g ) → FL = 1.10 Mg

(b) The work done by the lifting force in lifting the helicopter a vertical distance h is given by Eq. 7–1. The lifting force and the displacement are in the same direction. WL = FL h cos 0 = 1.10 Mgh . 8.

Draw a free-body diagram of the car on the incline. The minimum work will occur when the car is moved up the incline at a constant velocity. Write Newton’s second law in the x-direction, noting that the car is unaccelerated. Only the forces parallel to the plane do work.  Fx = FP − mg sin  = 0 → FP = mg sin 

FN FP 



The work done by FP in moving the car a distance d along the plane

(parallel to FP ) is given by Eq. 7–1.

(

)

WP = FPd cos 0o = mgd sin  = ( 950 kg ) 9.80 m s2 ( 510 m ) sin 9.0 = 7.4  105 J 9.

The distance over which the force acts is the area to be mowed divided by the width of the mower. The force is parallel to the displacement, so the angle between them is 0 o. Use Eq. 7–1. 200 m 2 A = 6000 J W = Fd cos  = F cos  = (15 N ) 0.50 m w

10. The minimum work required to shelve a book is equal to the weight of the book times the vertical distance the book is moved. See the diagram. Each book that is placed on the lowest shelf has its center of mass moved upwards by 15.0 cm + 11.0 cm = 26.0 cm. So the work done to move 28 books to the lowest shelf is W1 = 28 mg ( 0.260 m ) . Each book that is placed on the second shelf has its center of mass moved upwards by 15.0 cm + 38.0 cm + 11.0 cm = 64.0 cm, so the work done to move 28 books to the second shelf is W2 = 28 mg ( 0.640 m ) . Similarly,

3rd shelf 2rd shelf 64.0 cm 26.0 cm

1st shelf floor

W3 = 28 mg (1.020 m ) , W4 = 28 mg (1.400 m ) , and W5 = 28 mg (1.780 m ) . The total work done is the sum of the five work expressions. W = 28 mg ( 0.260 m + 0.640 m + 1.020 m + 1.400 m + 1.780 m )

(

)

= 28 (1.25 kg ) 9.80 m s 2 ( 5.100 m ) = 1749 J  1750 J 11. Consider the diagram shown. If we assume that the man pushes straight down on the end of the lever, then the work done by the man (the “input” work) is given by WI = FI hI . The object moves a shorter distance, as seen from the diagram, and so WO = FO hO . Equate the two amounts of work. FO hI WO = WI → FO hO = FI hI → = FI hO © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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But by similar triangles, we see that

hI hO

lI lO

, and so

FO FI

lI lO

12. Since the acceleration of the box is constant, use Eq. 2–12b to find the distance moved. Assume that the box starts from rest. 2 d = x − x0 = v0t + 12 at 2 = 0 + 12 ( 2.0 m s 2 ) ( 8.0s ) = 64 m Then the work done in moving the crate is found using Eq. 7–1. W = Fd cos 0o = mad = ( 4.0 kg ) ( 2.0 m s 2 ) ( 64 m ) = 512 J  510 J 13. The first book is already in position, so no work is required to position it. The second book must be moved upwards by a distance d, by a force equal to its weight, mg. The force and the displacement are in the same direction, so the work is mgd. The third book will need to be moved a distance of 2d by the same size force, so the work is 2 mgd. This continues through all seven books, with each needing to be raised by an additional amount of d by a force of mg. The total work done is W = mgd + 2 mgd + 3 mgd + 4 mgd + 5 mgd + 6 mgd + 7 mgd

(

)

= 28 mgd = 28 (1.6 kg ) 9.8 m s 2 ( 0.040 m ) = 17.56 J  18 J

14. (a) The motor must exert a force equal and opposite to the force of gravity on the gondola and passengers in order to lift it. The force is in the same direction as the displacement. Use Eq. 7–1 to calculate the work.

(

)

Wmotor = Fmotor d cos 0 = mgd = ( 2250 kg ) 9.80 m s2 ( 3625m − 2150 m ) = 3.25  107 J (b) Gravity would do the exact opposite amount of work as the motor, because the force and displacement are of the same magnitude, but the angle between the gravity force and the displacement is 180.

(

)

WG = FG d cos180 = −mgd = − ( 2250 kg ) 9.80 m s2 ( 3625m − 2150 m ) = −3.25  107 J (c) If the motor is generating 10% more work, than it must be able to exert a force that is 10% larger than the force of gravity. The net force then would be as follows, with up the positive direction.

Fnet = Fmotor − FG = 1.1 mg − mg = 0.1 mg = ma → a = 0.1g = 0.98 m s 2 15. Consider a free-body diagram for the grocery cart being pushed up the ramp. If the cart is not accelerating, then the net force is 0 in all directions. This can be used to find the size of the pushing force. The angles are  = 17 and  = 12. The displacement is in the x-direction. The work done by the normal force is 0 since the normal force is perpendicular to the displacement. The angle between the force of gravity and the displacement is 90 +  = 102. The angle between the normal force and the displacement is 90. The angle between the pushing force and the displacement is  +  = 29. mg sin   Fx = FP cos ( +  ) − mg sin  = 0 → FP = cos ( +  )



FP 



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(

Instructor Solutions Manual

)

Wmg = mgd cos112 = (16 kg ) 9.80 m s 2 ( 7.5 m ) cos102 = −244.5 J  −240 J WN = FN d cos 90 = 0

 mg sin12   d cos 29 = mgd sin12  cos 29 

WP = FP d cos 29 = 

(

)

= (16 kg ) 9.80 m s 2 ( 7.5 m ) sin12 = 244.5 J  240 J 16. The piano is moving with a constant velocity down the plane. FP is the force of the man pushing on the piano. (a) Write Newton’s second law in each direction for the piano, with an acceleration of 0. We assume the angle has 2 significant figures.  Fy = FN − mg cos  = 0 → FN = mg cos 

 F = mg sin  − F = 0 → F = mg sin  = ( 380 kg ) ( 9.80 m s ) ( sin 30 ) = 1862 N  1900 N



(b) The work done by the man is the work done by FP . The angle between FP and the direction of motion is 180. Use Eq. 7–1. WP = FP d cos180 = − (1862 N )( 4.6 m ) = −8565J  −8600 J .

(c) The angle between the force of gravity and the direction of motion is 60. Calculate the work done by gravity. Use Eq. 7–1. WG = FG d cos 60 = mgd cos 60 = ( 380 kg ) ( 9.80 m s 2 ) ( 4.6 m ) cos 60 = 8565 N  8600 J

(d) Since the piano is not accelerating, the net force on the piano is 0, and so the net work done on the piano is also 0. This can also be seen by adding the two work amounts calculated. Wnet = WP + WG = −8565J + 8565J = 0 J 17. If the person pulls 2 meters of rope through his hands, the rope holding the piano will get shorter by 2 meters. But that means the rope on the right side of the pulley will get shorter by 1 meter, and the rope on the left side will also get shorter by 1 meter. Thus for each meter of the load is raised, 2 m of rope must be pulled up. In terms of energy (assuming that no work is lost to friction), the work done by the man pulling on the rope must be equal to the work done on the piano. If the piano has weight mg, and it moves upwards a distance d, then the work done on the piano is mgd. The person pulls the rope a distance 2d, and therefore must exert a force of 12 mg to do the same amount of work. Wdone by = Wdone on → Fpull ( 2d ) = mgd → Fpull = 12 mg man

piano

18. Use Eq. 7.4 to calculate the dot product. A B = Ax Bx + Ay By + Az Bz = 2.0 x 2 (11.0 ) + ( −4.0 x )( 2.5 x ) + ( 5.0 )( 0 ) = 22 x 2 − 10 x 2

(

)

= 12 x 2

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19. Use Eq. 7.4 to calculate the dot product. Note that î = 1î + 0ˆj + 0kˆ , ˆj = 0î + 1ˆj + 0kˆ , and kˆ = 0î + 0ˆj + 1kˆ . î V = (1) V + ( 0 ) V + ( 0 )V = V x y z x

ˆj V = ( 0) V + (1) V + ( 0 )V = V x y z y

kˆ V = ( 0) Vx + ( 0) Vy + (1)Vz = Vz

20. Use Eq. 7.4 and Eq. 7.2 to calculate the dot product, and then solve for the angle. A B = Ax Bx + Ay By + Az Bz = ( 5.8 )( 8.2 ) + ( −3.4 )( 4.3) + ( −6.2 )( −7.0 ) = 76.34

( 5.8 ) + ( −3.4 ) + ( −6.2 ) = 9.145

A B = AB cos  →  = cos −1

AB AB

= cos −1

(8.2 ) + ( 4.3) + ( −7.0 ) = 11.607 2

76.34

( 9.145 )(11.607 )

= 44.01  44

(

)

21. We utilize the fact that if B = Bx ˆi + B y ˆj + Bz kˆ , then − B = ( − Bx ) ˆi + − B y ˆj + ( − Bz ) kˆ .

A ( − B ) = Ax ( − Bx ) + Ay ( − B y ) + Az ( − Bz ) = ( − Ax )( Bx ) + ( − Ay )( B y ) + ( − Az )( Bz ) = −  ( Ax )( Bx ) + ( Ay )( B y ) + ( Az )( Bz )  = − A B 22. See the diagram to visualize the geometric relationship between the two vectors. The angle between the two vectors is 138.

V1 V2 = VV cos  = ( 75)( 48) cos138 = −2675  −2700 1 2

V1 x

23. If A is perpendicular to B , then A B = 0. Use this to find B.

A B = Ax Bx + Ay By = ( 3.0) Bx + (1.8) By = 0 → By = − 53 Bx Any vector B that satisfies By = − 53 Bx will be perpendicular to A. For example, B = 3.0î − 5.0ˆj . Or any scalar multiple of that result. Simply switching the components and negating one of them also will give an appropriate answer. For example, B = 1.8î − 3.0ˆj or B = −1.8î + 3.0ˆj . The most general solution would be, for any non-zero constant b, that B = bî − 53 bˆj . 24. Both vectors are in the first quadrant, so to find the angle between them, we can simply subtract the angles of each of them. 4.0 2 2 F = 2.0î + 4.0ˆj N → F = ( 2.0 N ) + ( 4.0 N ) = 20 N ; F = tan −1 = tan −1 2.0 2.0 5.0 2 2 d = 1.0î + 5.0ˆj m → d = (1.0 m ) + ( 5.0 m ) = 26 m ; d = tan −1 = tan −1 5.0 1.0

(

)

(

)

(

)

(

)

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(a) W = Fd cos  = 



Instructor Solutions Manual

( 20 ) N  ( 26 ) m cos tan 5.0 − tan 2.0 = 22 J −1

−1

(b) W = Fx d x + Fy d y = ( 2.0 N )(1.0 m ) + ( 4.0 N )( 5.0 m ) = 22 J

( ) ( ) ( ) = ( 9.0ˆi − 8.5ˆj) ( −2.2ˆi − 3.1ˆj + 4.2kˆ ) = ( 9.0 )( −2.2 ) + ( −8.5)( −3.1) + ( 0 )( 4.2 )

25. (a) A ( B + C ) = 9.0î − 8.5ˆj  −8.0î + 6.1ˆj + 4.2kˆ + 5.8î − 9.2ˆj 

= 6.55  7 (b) ( A + C ) B =  9.0î − 8.5ˆj + 5.8î − 9.2ˆj  −8.0î + 6.1ˆj + 4.2kˆ

(

) ( ) ( ) = (14.8ˆi − 17.7ˆj) ( −8.0ˆi + 6.1ˆj + 4.2kˆ ) = (14.8 )( −8.0 ) + ( −17.7 )( 6.1) + ( 0 )( 4.2 )

(c)

= −226  −230 ( B + A ) C =  −8.0î + 6.1ˆj + 4.2kˆ + 9.0î − 8.5ˆj  5.8î − 9.2ˆj

(

) (

(

= 1.0ˆi − 2.4ˆj + 4.2kˆ

) (

)

) ( 5.8ˆi − 9.2ˆj) = (1.0)( 5.8) + ( −2.4)( −9.2) + ( 4.2)( 0)

= 27.88  28 26. We find the dot product using Eq. 7–2 and Eq. 7–4.

8.5 = −43.4 or 316.6 ( 9.0 ) + ( −8.5 ) = 12.379 ;  = tan −9.0

9.2 = −57.8 or 302.2 ( 5.8 ) + ( −9.2 ) = 10.876 ;  = tan −5.8

−1

Eq. 7 − 2:

A C = AC cos  = (12.379 )(10.876 ) cos ( 316.6 − 302.2 ) = 130.4

Eq. 7 − 4:

A C = 9.0ˆi − 8.5ˆj

(

) ( 5.8ˆi − 9.2ˆj) = ( 9.0 )( 5.8) + ( −8.5 )( −9.2 ) = 130.4

27. We are given that the magnitudes of the two vectors are the same, so Ax2 + Ay2 + Az2 = Bx2 + By2 + Bz2 . If the sum and difference vectors are perpendicular, their dot product must be zero. A + B = ( A + B ) ˆi + ( A + B ) ˆj + ( A + B ) kˆ x

A − B = ( Ax − Bx ) ˆi + ( Ay − B y ) ˆj + ( Az − Bz ) kˆ

( A + B ) ( A − B ) = ( A + B )( A − B ) + ( A + B )( A − B ) + ( A + B )( A − B ) x

(

) (

)

= Ax2 − Bx2 + Ay2 − B y2 + Az2 − Bz2 = Ax2 + Ay2 + Az2 − Bx2 + B y2 + Bz2 = 0

28. If C is perpendicular to B , then C B = 0. Use this along with the value of C A to find C. We also know that C has no z-component. C = C x ˆi + C y ˆj ; C B = C x Bx + C y B y = 0 ; C A = C x Ax + C y Ay = 18.0 →

9.6C x + 6.7C y = 0 ; − 4.8C x + 6.8C y = 18.0 This set of two equations in two unknowns can be solved for the components of C.

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9.6C x + 6.7C y = 0 → C y = −

9.6 6.7

 9.6 C  = −14.54C = 18.0 → C = −1.238 x  x x  6.7 

−4.8C x + 6.8C y = −4.8C x + 6.8  − Cy = −

9.6 6.7

Cx = −

9.6

( −1.238) = 1.773 → C = −1.2ˆi + 1.8ˆj

6.7

29. Note that by Eq. 7–2, the dot product of a vector A with a unit vector B would give the magnitude of A times the cosine of the angle between the unit vector and A. Thus if the unit vector lies along one of the coordinate axes, we can find the angle between the vector and the coordinate axis. We also use Eq. 7–4 to give a second evaluation of the dot product. V ˆi = V cos  x = Vx →

 x = cos −1  y = cos −1  z = cos −1

Vx V Vy V Vz V

= cos −1

Vx V +V +V 2 x

2 z

= cos −1

20.0

( 20.0 ) + ( 26.0 )2 + ( −14.0 )2 2

26.0

= cos −1 = cos −1

2 y

( 20.0 ) + ( 26.0 ) + ( −14.0 ) 2

−14.0

( 20.0 ) 2 + ( 26.0 ) 2 + ( −14.0 ) 2

= 55.9

= 43.2 = 113

30. For the diagram shown, B + C = A, or C = A − B. Let the magnitude of each vector be represented by the corresponding lowercase letter, so C = c, for example. The angle between A and B is  . Take the dot product C C. C C = ( A − B ) ( A − B ) = A A + B B − 2A B → c 2 = a 2 + b2 − 2ab cos 



31. We can represent the vectors as A = Ax î + Ay ˆj = A cos  î + A sin  ˆj and B = Bx î + B y ˆj = B cos  î + B sin  ˆj. The angle between the two vectors is  −  . Use Eqs. 7–2 and 7–4 to express the dot product. A B = AB cos ( −  ) = Ax Bx + Ay B y = A cos  B cos  + A sin  B sin  → AB cos ( −  ) = AB cos  cos  + AB sin  sin 

→

cos ( −  ) = cos  cos  + sin  sin 

32. Use the two expressions for dot product, Eqs. 7–2 and 7–4, to find the angle between the two vectors. A B = AB cos  = Ax Bx + Ay B y + Az Bz →

 = cos −1 = cos −1

Ax Bx + Ay B y + Az Bz AB (1.0 )( −1.0 ) + (1.0 )(1.0 ) + ( −2.0 )( −2.0 ) 1/2

(1.0 ) 2 + (1.0 ) 2 + ( −2.0 ) 2  ( −1.0 ) 2 + (1.0 ) 2 + ( −2.0 ) 2     

1/2

 4.0  = 48   6.0 

= cos −1 

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The (implicit) positive sign in the argument of the inverse cosine means that the angle between the two vectors is acute. 33. From Figure 7–6, we see a graphical interpretation of the scalar product as the magnitude of one vector times the projection of the other vector onto the first vector. So to show that A B + C = A B + A C is the

(

)

same as showing that A ( B + C ) = A ( B ) + A ( C ) , where the subscript is implying the component of the vector that is parallel to vector A. From the diagram, we see that ( B + C ) = ( B ) + ( C ) . Multiplying this equation by the magnitude of vector A gives A ( B + C ) = A ( B ) + A ( C ) . But from Figure 7–6, this is the same as

(

)

A B + C = A B + A C. So we have proven the statement.

34. To be perpendicular to the given vector means that the dot product will be 0. Let the unknown vector be given as uˆ = ux ˆi + u y ˆj.

(

)

uˆ 4.0ˆi + 3.0ˆj = 4.0u x + 3.0u y = 0 → u x = − 0.75u y ; unit length → u x2 + u 2y = 1 →

u x2 + u y2 = ( −0.75u y ) + u y2 = 1.5625u y2 = 1 → u y = 

1 1.5625

= 0.8, u x = 0.6

So the two possible vectors are uˆ = 0.6î − 0.8ˆj and uˆ = −0.6î + 0.8ˆj . Note that it is very easy to get a non-unit vector perpendicular to another vector in two dimensions, simply by interchanging the coordinates and negating one of them. So a non-unit vector perpendicular to ( 4.0î + 3.0ˆj) could be either ( 3.0î − 4.0ˆj) or ( −3.0î + 4.0ˆj) . Then divide each of those vectors by its magnitude (5.0) to get the possible unit vectors given above. 35. The dot product can be used to find the angle between the vectors. d1−2 = r2 − r1 =  0.230î + 0.133ˆj  10−9 m 

d1−3 d1−2

( ) = r − r = ( 0.077î + 0.133ˆj + 0.247kˆ )  10 m  d = ( 0.230î + 0.133ˆj)  10 m   ( 0.077î + 0.133ˆj + 0.247kˆ )  10 m  −9

−9

1− 3

=  3.540  10−2   10−18 m 2 d1−2 =

( 0.230)2 + ( 0.133) 2  10−9 m = 0.2657  10−9 m

d1−3 =

( 0.077 )2 + ( 0.133)2 + ( 0.247 )2  10−9 m = 0.2909  10−9 m

d1−2 d1−3 = d1−2 d1−3 cos  →

 = cos

−1

d1−2 d1−3 d1−2 d1−3

= cos

−1

3.540  10−2   10−18 m 2

( 0.2657  10 m )( 0.2909  10 m ) −9

−9

= 62.7

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36. The downward force is 420 N, and the downward displacement would be a diameter of the pedal circle. Use Eq. 7–1. W = Fd cos  = ( 420 N )( 0.36 m ) cos 0o = 151.2 J  150 J 37. The force exerted to stretch a spring is given by Fstretch = kx (the opposite of the force exerted by the spring, which is given by F = − kx ). A graph of Fstretch vs. x will be a straight line of slope k through the origin. The stretch from x1 to x2, as shown on the graph, outlines a trapezoidal area. This area represents the work. W = 12 ( kx1 + kx2 )( x2 − x1 ) = 12 k ( x1 + x2 )( x2 − x1 ) =

1 2

F = kx kx2 Force kx1

x1 Stretch distance

( 65 N m )( 0.105 m )( 0.045 m ) = 0.1536 J

 0.15 J

38. For a non-linear path, the work is found by considering the path to be an infinite number of infinitesimal (or differential) steps, each of which can be considered to be in a specific direction, namely, the direction tangential to the path. From the diagram, for each step we have dW = F d l = Fd l cos  . But d l cos  = − d l cos (180 −  ) = − dy , which is projection of the path in the direction of the force, and F = mg , the force of gravity. Find the work done by gravity. Wg =  F d l =  mg cos  d l = mg  ( −dy ) = −mg  dy = −mgh This argument could even be extended to going part way up the hill, and then part way back down, and following any kind of path. The work done by gravity will only depend on the height change of the path.

= 3040 J  3.0  10 J 3

400 300

Fx (N)

39. See the graph of force vs. distance. The work done is the area under the graph. It can be found from the formula for a trapezoid. W = 12 (12.0 m + 4.0 m )( 380 N )

200 100 0 0

x (m)

40. The x-axis is portioned into 7 segments, so each segment is 1/7 of the full 20.0-m width. The force on each segment can be approximated by the force at the middle of the segment. Thus we are performing a simple Riemann sum to find the area under the curve. The value of the mass does not come into the calculation. 7

i =1

W =  Fi xi = x  Fi = 17 ( 20.0 m )(180 N + 200 N + 175 N + 125 N + 110 N + 100 N + 95 N ) =

1 7

( 20.0 m )( 985 N ) = 2814 J  2800 J

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Instructor Solutions Manual

Another method is to treat the area as a trapezoid, with sides of 180 N and 100 N, and a base of 20.0 m. Then the work is W = 12 ( 20.0 m )(180 N + 100 N )  2800 J . 41. The work done will be the area under the Fx vs. x graph. (a) From x = 0.0 to x = 10.0 m, the shape under the graph is trapezoidal. The area is Wa = ( 400 N ) 12 (10 m + 4 m ) = 2800 J .

(b) From x = 10.0 m to x = 15.0 m, the force is in the opposite direction from the direction of motion, and so the work will be negative. Again, since the shape is trapezoidal, we find Wa = ( −200 N ) 12 ( 5 m + 2 m ) = −700 J. Thus the total work from x = 0.0 to x = 15.0 m is 2800 J − 700 J = 2100 J . 42. Apply Eq. 7–1 to each segment of the motion. W = W1 + W2 + W3 = F1d1 cos 1 + F2 d 2 cos  2 + F3d 3 cos  3 = ( 22 N )( 5.0 m ) cos 0 + ( 38 N )( 3.0 m ) cos12 + ( 22 N )( 8.5 m ) cos 0 = 408.5J  410 J

43. Since the force only has an x-component, only the x-displacement is relevant. The object moves from x = 0 to x = d. The force to insert the sharp object into the material is the opposite of the resistance force. The angle between the pushing force and the displacement is 0. d

W =  Fx dx =  bx 4 dx = 15 bd 5 44. Integrate the force over the distance the force acts to find the work. We assume the displacement is all in the y-direction. yf

0.20 m

190 3   W =  F ( y ) dy =  (150 y − 190 y ) dx =  75 y 2 − y  3  0 0 y

= 2.493J  2.49 J

45. Integrate the force over the distance the force acts to find the work. 1.0 m

1.0 m

W =  F1 dx = 

1.0 m

(

)

= 2 3.0N m1/2 (1.0m )

1/2

= 6.0 J 0 x Note that the work done is finite, even though the force is infinite at the origin. 0

dx = 2 A x

46. Because the object moves along a straight line, we know that the x-coordinate increases linearly from 0 to 10.0 m, and the y-coordinate increases linearly from 0 to 20.0 m. Use the relationship developed near the top of page 201. xb

W =  Fx dx +  Fy dy = xa

10.0 m

 3.0 xdx + 0

20.0 m

 4.0 ydy = ( 3.0 x ) 1 2

2 10.0 0

(

+ 12 4.0 y 2

)

20.0 0

= 150 J + 800 J

= 950 J 47. The work required to stretch a spring from equilibrium is proportional to the length of stretch, squared. So if we stretch the spring to 3 times its original distance, a total of 9 times as much work is required for the total stretch. Thus it would take 45.0 J to stretch the spring to a total of 6.0 cm.

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Since 5.0 J of work was done to stretch the first 2.0 cm, 40.0 J of work is required to stretch it the additional 4.0 cm. This could also be done by calculating the spring constant from the data for the 2.0 cm stretch, and then using that spring constant to find the work done in stretching the extra distance. 48. Since the force is of constant magnitude and always directed at 30 to the displacement, we have a simple expression for the work done as the object moves. finish

finish

start

W =  F d l =  F cos 30d l = F cos 30  d l = F cos 30 R =

3 FR 2

49. For each interval, the average force for that interval was calculated as the numeric average of the forces at the beginning and end of the interval. Then this force was multiplied by 10.0 cm (0.100 m) to find the work done on that interval. The total work is the sum of those work amounts. That process is expressed in a formula below. n −1

Wapplied =  12 ( Fi + Fi +1 )x = 12 ( 26.0 N + 28.5 N )( 0.100 m ) + 12 ( 28.5 N + 28.8 N )( 0.100 m ) + ... i =1

= 102.03J  102 J

Here is a graph of the force vs. position for the particle.

50. (a) The gases exert a force on the jet in the same direction as the displacement of the jet. From the graph we see the displacement of the jet during launch is 85 m. Use Eq. 7–1 to find the work.

(

)

Wgas = Fgasd cos 0 = 130  103 N (85m ) = 1.1  107 J (b) The work done by catapult is the area underneath the graph in Fig. 7–26. That area is a trapezoid.

(

)

Wcatapult = 12 1100  103 N + 65  103 N (85m ) = 5.0  107 J 51. To lift the mass off the ground, the spring must exert an upward force of magnitude mg. That means F mg the spring must be extended from its equilibrium position an amount l = = . We then use k k Eq. 7–7 to determine the amount of work done, assuming the force is directed vertically. The variable y represents the amount the spring has been stretched from its equilibrium configuration. b

mg / k

( ) ( dyˆj) =  ky dy = ky

W =  F dy ˆj =  kyˆj

1 2

2 mg / k 0

m2 g 2k 2k 2

m2 g 2 2k

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

We could also do this problem using a relationship developed in Section 7–3, which says the work done in stretching a spring is given by W = 12 kx 2 , where x is the amount if stretch. Using that relationship we get the same result. 52. The force on the object is given by Newton’s law of universal gravitation, F = G

mmE

. The force is r2 a function of distance, so to find the work, we must integrate. The directions are tricky. To use mm Eq. 7–7, we have F = −G 2 E rˆ and d l = dr rˆ. It is tempting to put a negative sign with the d l r relationship since the object moves inward, but since r is measured outward away from the center of the Earth, we must not include that negative sign. Note that we move from a large radius to a small radius. near



W = F dl =

mmE



−

  rE + 2900 km 

far

1

= GmmE 

 rE

rˆ ( dr rˆ ) = −

−G



mmE

rE + 2900 km

dr = G

mmE r

rE + 2900 km

(

)





1 ) 6.38 110 m − ( 6.38 + 2.9 )  10 m 

(

= 6.67  10−11 N  m 2 kg 2 ( 2800 kg ) 5.97  1024 kg 





= 5.5  1010 J 53. Let y represent the length of chain hanging over the table, and let  represent the weight per unit length of the chain. Then the force of gravity (weight) of the hanging chain is FG =  y. As the next small length of chain dy comes over the table edge, gravity does an infinitesimal amount of work on the hanging chain given by the force times the distance, FG dy =  ydy. To find the total amount of work that gravity does on the chain, integrate that work expression, with the limits of integration representing the amount of chain hanging over the table. yfinal

3.0 m

W =  FG dy =   ydy = 12  y 2 yinitial

3.0m 1.0 m

(

)

= 12 ( 24 N m ) 9.0 m 2 − 1.0 m 2 = 96 J

1.0 m

54. Find the velocity from the kinetic energy, using Eq. 7–10. K = mv 1 2

→ v=

2K m

(

2 6.21  10−21 J

5.31 10

−26

) = 484 m s

55. (a) Since K = 12 mv 2 , then v = 2K m and so v  tripled, the speed will be multiplied by a factor of

K . Thus if the kinetic energy is 3.

(b) Since K = 12 mv , then K  v . Thus if the speed is halved, the kinetic energy will be 2

multiplied by a factor of 1 4 .

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56. The work done on the electron is equal to the change in its kinetic energy, Eq. 7–11.

(

)(

W = K = 12 mv22 − 12 mv12 = 0 − 12 9.11 10−31 kg 1.10 106 m s

) = −5.5110 J 2

−19

Note that the work is negative since the electron is slowing down. 57. We use Eq. 7–9. (a) Wnet = 12 mv22 − 12 mv12 = 0 − 12 (1300 kg )( 20.0 m s ) = −2.6  105 J 2

(b) Wnet = 12 mv22 − 12 mv12 = 12 (1300 kg )( 20.0 m s ) − 0 = 2.6  105 J 2

(c) Wnet = 12 mv22 − 12 mv12 = 12 (1300 kg )( 20.0 m s ) − 12 (1300 kg )( 20.0 m s ) = 0 Or, the two net works could be added to get the same result. 2

58. The kinetic energies of both bullets are the same. Bullet 1 is the heavier bullet. 1 m1 = 2m2 m2v22 = 12 m1v12 → m2v22 = 2m2v12 → v22 = 2v12 → v2 = v1 2 2 The lighter bullet has the higher speed, by a factor of the square root of 2. Both bullets can do the same amount of work. 59. The force of the ball on the glove will be the opposite of the force of the glove on the ball, by Newton’s third law. Both objects have the same displacement, and so the work done on the glove is opposite the work done on the ball. The work done on the ball is equal to the change in the kinetic energy of the ball.

Won ball = ( K 2 − K1 )ball = 12 mv22 − 12 mv12 = 0 − 12 ( 0.145 kg )( 32 m s ) = −74.24 J 2

So Won glove = 74.24 J. But Won glove = Fon gloved cos 0, because the force on the glove is in the same direction as the motion of the glove. 74.24 J = Fon glove ( 0.22 m ) → Fon glove =

74.24 J 0.22 m

= 337 N  340 N , in the direction of the original

velocity of the ball. 60. The force exerted by the bow on the arrow is in the same direction as the displacement of the arrow. Thus W = Fd cos 0o = Fd = (105 N )( 0.75m ) = 78.75J. That work changes the kinetic energy of the arrow, according to the work–energy principle, Eq. 7–11. Fd = W = K 2 − K1 = 12 mv 22 − 12 mv12 → v 2 =

2Fd m

+ v12 =

2 ( 78.75 J ) 0.085 kg

61. The work needed to stop the car is equal to the change in the car’s kinetic energy. That work comes from the force of friction on the car. Assume the maximum possible Ffr frictional force (without skidding), which results in the minimum braking distance. Thus Ffr = s FN . The normal FN force is equal to the car’s weight if it is on a level surface, and so Ffr = s mg . In the diagram, the car is traveling to the right.

W = K → Ffr d cos180 = mv − mv o

1 2

2 2

1 2

2 1

d = stopping distance

→ − s mgd = − mv 1 2

+ 0 = 43 m s

2 1

→ d=

v12 2 g s

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Since d  v12 , if v1 increases by 50%, or is multiplied by 1.5, then d will be multiplied by a factor of (1.5) , or 2.25 . 2

62. (a) The work needed to stop the car is equal to the change d = stopping distance in the car’s kinetic energy. That work comes from the force of friction on the car, which is assumed to be Ffr kinetic (sliding) friction since the driver locked the brakes. Thus Ffr =  k FN . Since the car is on a level mg FN surface, the normal force is equal to the car’s weight, and so Ffr =  k mg if it is on a level surface. See the diagram for the car. The car is traveling to the right. W = K → Ffr d cos180o = 12 mv22 − 12 mv12 → −  k mgd = 0 − 12 mv12 → (b) (c)

(

)

2 ( 0.30 ) 9.80 m s 2 ( 78 m ) = 21m s

2  k gd =

v1 =

The mass does not affect the problem, since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation. Modern cars are typically equipped with anti-lock brakes, which are supposed to keep a car from skidding. If the car is skidding, then the anti-lock brakes are not functioning correctly.

63. The work done in compressing the spring gets changed into the kinetic energy of the car (assuming the spring is massless). The work done in compressing the spring is given on page 201. W = 12 kx 2 ; W = K = 12 mv22 − 12 mv12 → 12 kx 2 = 12 mv22 − 12 mv12 →

1 2

mv22 − 12 mv12 1 2

mv22 x2

 

(1200 kg )  28 =

km   1000 m  1h  





h   1km  3600 s   = 2.2  104 N m 2 (1.8 m )

64. The work done by the net force is the change in kinetic energy. W = K = 12 mv22 − 12 mv12

= 12 ( 3.5 kg ) (15.0 m s ) + ( 30.0 m s )  − 12 ( 3.5 kg ) (10.0 m s ) + ( 20.0 m s ) 









= 1094J  1100 J 65. The frictional force does the work of slowing the box. The frictional force is the coefficient of kinetic friction times the normal force, and since the box is on the horizontal floor, the normal force is equal to the weight of the box. We use Eq. 7–1 and Eq. 7–11. W = K = 12 mv 22 − 12 mv12 = − 12 mv12 ; W = Fd cos  =  k mgd cos180 = −  k mgd 1 mv 2 v2 ( 5.0 m s ) − 12 mv12 = −  k mgd →  k = 2 1 = 1 = = 0.16 2 gd 2 9.80 m s 2 ( 8.0 m ) mgd 2

(

)

234

Chapter 7

Work and Energy

66. See the free-body diagram. (a) The angle between the pushing force and the displacement is 32. WP = FP d cos  = (150 N )( 5.0 m ) cos 32 = 636.0 J  640 J

FN y FP

(b) The angle between the force of gravity and the displacement is 90 + 32 = 122. WG = FG d cos  = mgd cos  = (18 kg ) ( 9.80 m s 2 ) ( 5.0 m ) cos122



= −467.4 J  −470 J

(c) Because the normal force is perpendicular to the displacement, the work done by the normal force is 0 . (d) The net work done is the change in kinetic energy. W = WP + Wg + WN = K = 12 mv22 − 12 mv12 → v2 =

2W m

+ v12 =

2 ( 636.0 J − 467.4 J )

(18 kg )

= 4.3 m s

67. See the free-body diagram for help in the determination of the frictional force.  Fy = FN − FP sin  − mg cos  = 0 → FN = FP sin  + mg cos  Ff = k FN = k ( FP sin  + mg cos  ) (a) The angle between the pushing force and the displacement is 32. WP = FP d cos  = (150 N )( 5.0 m ) cos 32 = 636.0 J  640 J

(b) The angle between the force of gravity and the displacement is 90 + 32 = 122.

(

FN y FP Ff



)

WG = FG d cos  = mgd cos  = (18 kg ) 9.80 m s 2 ( 5.0 m ) cos122 = −467.4 J  −470 J

(c) Because the normal force is perpendicular to the displacement, the work done by the normal force is 0 . (d) To find the net work, we need the work done by the friction force. The angle between the friction force and the displacement is 180. Wf = Ff d cos  =  k FN d cos  =  k ( FP sin  + mg cos  ) d cos 

(

)

= ( 0.10 ) (150 N ) sin 32 + (18kg ) 9.80 m s 2 cos 32  ( 5.0 m ) cos180 = −114.5 J W = WP + Wg + WN + Wf = K = 12 mv22 − 12 mv12 → v2 =

2W m

+ v12 =

2 ( 636.0 J − 467.4 J − 114.5 J )

(18 kg )

= 2.5 m s

68. Assuming the stopping force is the same in both situations, then the stopping distance is proportional to the square of the starting speed. Use Eqs. 7–1 and 7–9 to show this. The stopping force is in the opposite direction of the displacement, and the ending speed is 0. W = Fd cos  = − Fd ; W = 12 mv22 − 12 mv12 = − 12 mv12

− Fd = − 12 mv12 → d =

mv12

2F We see that doubling the initial speed will make the stopping distance 4 times larger, or 4d . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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69. (a) In the Earth frame of reference, the ball changes from a speed of v1 to a speed of v1 + v 2 .

(

)

K Earth = 12 m ( v1 + v 2 ) − 12 mv12 = 12 m v12 + 2v1v 2 + v 22 − 12 mv12 = mv1v 2 + 12 mv 22 2



v1 



v2 

= 12 mv 22  1 + 2



(b) In the train frame of reference, the ball changes from a speed of 0 to a speed of v 2 .

K train = 12 mv22 − 0 = 12 mv22 (c) The work done is the change of kinetic energy, in each case.



WEarth = 12 mv22  1 + 2



v1 

2 1  ; Wtrain = 2 mv2 v2 

(d) The difference can be seen as due to the definition of work as force exerted through a distance. In both cases, the force on the ball is the same, but relative to the Earth, the ball moves farther during the throwing process than it does relative to the train. Thus more work is done in the Earth frame of reference. Another way to say it is that kinetic energy is dependent on reference frame, and so since work is the change in kinetic energy, the amount of work done will be dependent on reference frame as well. 70. (a) The spring constant is found by the magnitudes of the initial force and displacement, and so k = F x . As the spring compresses, it will do the same amount of work on the block as was done on the spring to stretch it. The work done is positive because the force of the spring is parallel to the displacement of the block. Use the work–energy principle to determine the speed of the block.

= K block = Won spring

Won block during compression

→

1 2

mvf2 = 12 kx 2 = 12

during stretching

x 2 → vf =

Fx m

(b) Now we must find how much work was done on the spring to stretch it from x 2 to x. This will be the work done on the block as the spring pulls it back from x to x 2 . x

x 2

Won spring =  Fdx =  kxdx = 12 kx 2 during stretching

1 2

mvf2 = 83 kx 2 → vf =

= 12 kx 2 − 12 k ( x 2 ) = 83 kx 2 2

x 2

3Fx 4m

71. The first car mentioned will be called car 1. So we have these statements: 2 2 K1 = 12 K 2 → 12 m1v12 = 12 12 m2 v22 ; K1,fast = K 2,fast → 12 m1 ( v1 + 8.0 ) = 12 m2 ( v2 + 8.0 )

(

)

Now use the mass information, that m1 = 2m2 . 1 2

2m2 v12 = 12

( mv ) ; 1 2

2 2

1 2

2m2 ( v1 + 8.0 ) = 12 m2 ( v 2 + 8.0 )

2v1 = v 2 ; 2 ( v1 + 8.0 ) = ( v 2 + 8.0 ) 2

→

→ 2 ( v1 + 8.0 ) = ( 2v1 + 8.0 ) 2

→

236

Chapter 7

Work and Energy

8.0

2 ( v1 + 8.0 ) = ( 2v1 + 8.0 ) → v1 =

= 5.657 m s ; v2 = 2v1 = 11.314 m s

v1 = 5.7 m s ; v2 = 11.3 m s 72. (a) From the free-body diagram for the load being lifted, write Newton’s second law for the vertical direction, with up being positive.  F = FT − mg = ma = 0.140 mg →

(

)

FT = 1.140 mg = 1.140 ( 265 kg ) 9.80 m s 2 = 2.96  103 N

(b) The net work done on the load is found from the net force. Wnet = Fnet d cos 0o = ( 0.140 mg ) d = 0.140 ( 265 kg ) 9.80 m s 2 (18.0 m )

(

= 6.54  103 J (c) The work done by the cable on the load is as follows.

)

(

)

Wcable = FT d cos 0o = (1.140 mg ) d = 1.14 ( 265 kg ) 9.80 m s 2 (18.0 m ) = 5.33  104 J (d) The work done by gravity on the load is as follows.

(

)

WG = mgd cos180o = −mgd = − ( 265 kg ) 9.80 m s 2 (18.0 m ) = −4.67 104 J (e) Use the work-energy principle to find the final speed, with an initial speed of 0.

Wnet = K 2 − K1 = mv − mv 1 2

2 2

2 1

1 2

2Wnet

→ v2 =

+v = 2 1

(

2 6.54  103 J 265 kg

) + 0 = 7.03 m s

73. (a) The work done by gravity as the elevator falls is the weight times the displacement. They are in the same direction.

(

)

WG = mgd cos 0 = ( 975kg ) 9.80 m s2 ( 22.5m ) = 2.1499  105 J  2.15  105 J

(b) The work done by gravity on the elevator is the net work done on the elevator while falling, and so the work done by gravity is equal to the change in kinetic energy. WG = K = 12 mv 2 − 0 → v=

2WG

2 mgd

(

)

= 2 gd = 2 9.80 m s 2 ( 22.5 m ) = 21.0 m s

m m (c) The elevator starts and ends at rest. Therefore, by the work-energy principle, the net work done must be 0. Gravity does positive work as it falls a distance of d + x = ( 22.5 + x ) m (assuming that x > 0), and the spring will do negative work as the spring is compressed. The work done on the spring is 12 kx 2 , and so the work done by the spring is − 12 kx 2 . W = WG + Wspring = mg ( d + x ) − 12 kx 2 = 0 → x=

mg  m 2 g 2 − 4 ( 12 k ) ( − mgd ) 2 ( 12 k )

( 975 kg ) ( 9.80 m s )  2

1 2

kx 2 − mgx − mgd = 0 →

mg  m 2 g 2 + 2 kmgd k

( 975 kg ) ( 9.80 m s 2 ) 2

(

)

(

)

+2 8.00  10 4 N m ( 975 kg ) 9.80 m s 2 ( 22.5 m ) 8.00  10 N m 4

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

and so x = 2.44 m, − 2.20 m. The positive root must be taken since we have assumed x > 0 in calculating the work done by gravity. Using the values given in the problem gives x = 2.44 m .

74. First find the kinetic energy of the train, and then find out how much work the web must do to stop the train. Note that the web does negative work, since the force is in the OPPOSITE direction of the displacement. 2

  1m s   6 Wto stop = K = mv − mv = 0 − (10 kg ) 60 km h   = −1.39  10 J  train  3.6 km h    2 2

1 2

2 1

1 2

Wweb = − kx = −1.39  10 J → k = 2

1 2

(

2 1.39  106 J

( 500 m )

) = 11.1 N m  10 N m

Note that this is not a very stiff “spring,” but it does stretch a long distance. 75. For the first part of the motion, the net force doing work is the 225 N force. For the second part of the motion, both the 225 N force and the force of friction do work. The friction force is the coefficient of friction times the normal force, and the normal force is equal to the weight. The work-energy principle is then used to find the final speed.

Ffr FN

(

Wtotal = W1 + W2 = Fpull d1 cos 0 + Fpull d 2 cos 0 + Ff d 2 cos180 = K = 12 m vf2 − vi2

) →

2  Fpull ( d1 + d 2 ) − k mgd 2 

Fpull ( d1 + d 2 ) −  k mgd 2 = 12 mvf2 → vf = =

(

)

2 ( 225 N )( 26.0 m ) − ( 0.20 )( 36.0 kg ) 9.80 m s 2 (15.0 m ) 

( 36.0 kg )

= 16.32 m s  16 m s

76. There are two forces on the meteorite–gravity and the force from the mud. Take down to be the positive direction, and then the net force is Fnet = mg − 640 x 3 . Use this (variable) force to find the work done on the meteorite as it moves in the mud, and then use the work-energy principle to find the initial velocity of the meteorite. x = 4.5 m

 ( mg − 640 x ) dx = ( mgx − 160 x ) 3

x =0

x = 4.5 m x =0

(

)

= ( 75 kg ) 9.80 m s 2 ( 4.5 m ) − 160 ( 4.5 m )

= −6.23  104 J

(

W = K = m v − v 1 2

2 2

2 1

) → v= 1

−2W m

(

−2 −6.23  10 4 J

( 75 kg )

) = 40.76 m s  41m s

77. The force exerted by the spring will be the same magnitude as the force to compress the spring. The spring will do positive work on the ball by exerting a force in the direction of the displacement. This work is equal to the change in kinetic energy of the ball. The initial speed of the ball is 0.

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Chapter 7

Work and Energy

x = 2.0 m

W = K = mv − mv = mv 2 2

1 2

2 1

1 2

 (150x + 12 x ) dx = ( 75 x + 3x )

; W=

2 2

x =0

v2 =

2 ( 348 J )

3.0 kg

x = 2.0 x =0

= 348 J

= 15 m s

78. Consider the free-body diagram for the block as it moves up the plane. 2 (a) K1 = 12 mv12 = 12 ( 6.10 kg )( 3.25 m s ) = 32.22 J  32.2 J

(b) WP = FPd cos 37.0o = (85.0 N )( 9.25m ) cos 37.0 = 627.93J

 628 J

  mg

= −332.78 J  −333J

(d) WN = FN d cos 90o = 0 J (e) Apply the work-energy principle. Wtotal = K 2 − K1 →

K 2 = Wtotal + K1 = WP + WG + WN + K1 = ( 627.93 − 332.78 + 0 + 32.22 ) J  327 J 79. We assume the force is in the x-direction, so that the angle between the force and the displacement is 0. The work is found from Eq. 7–7. x2

x =

A A W =  Fx dx =  Ae dx = − e − kx = e −0.10 k k k x = 0.10 x1 x = 0.10 m − kx

80. Since we are compressing the spring, the force and the displacement are in the same direction. X

(

)

W =  Fx dx =  kx + ax 3 + bx 4 dx = 12 kX 2 + 14 aX 4 + 15 bX 5 81. The force is constant, and so we may calculate the force by Eq. 7–3. We may also use that to calculate the angle between the two vectors. W = F d =  10.0ˆi + 9.0ˆj + 12.0kˆ kN   5.0ˆi + 4.0ˆj m  = 86 kJ

(

)

 (

) 

1/2

F =  (10.0 ) + ( 9.0 ) + (12.0 )  kN = 18.0kN ; d =  ( 5.0 ) + ( 4.0 )  m = 6.40 m 2

W = Fd cos  →  = cos

−1

W Fd

= cos

−1

8.6  104 J

(1.80  10 N ) ( 6.40 m ) 4

= 42

82. Since the forces are constant, we may use Eq. 7–3 to calculate the work done. W = ( F + F ) d =  1.50î − 0.80ˆj + 0.70kˆ N + −0.70î + 1.20ˆj N   6.0î + 8.0ˆj + 5.0kˆ m  net

(

(

)

(

)

)  (

) 

=  0.80ˆi + 0.40ˆj + 0.70kˆ N   6.0ˆi + 8.0ˆj + 5.0kˆ m  = ( 4.8 + 3.2 + 3.5) J = 11.5J

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83. (a) The force and displacement are in the same direction. W = Fd cos  ; W = K → F=

1 2

K

(

m v 22 − v12

) = ( 0.033 kg )( 85 m s ) = 372.5 N  370 N 2

1 2

= d d 0.32 m (b) Combine Newton’s second law with Eq. 2–12c for constant acceleration. 2 m v22 − v12 ( 0.033 kg )( 85 m s ) F = ma = = = 372.5 N  370 N 2x 2 ( 0.32 m )

(

)

 1m s   = 33.33 m s . The final speed is 90% of  3.6 km h 

84. The original speed of the softball is (120 km h ) 

this, or 30 m/s. The work done by air friction causes a change in the kinetic energy of the ball, and thus the speed change. In calculating the work, notice that the force of friction is directed oppositely to the direction of motion of the ball. Wfr = Ffr d cos180o = K 2 − K1 = 12 m v22 − v12 →

(

Ffr =

(

m v22 − v12 −2d

)

) = mv ( 0.9 − 1) = ( 0.25 kg )( 33.33 m s ) ( 0.9 − 1) = 1.759 N  1.8 N 2 1

−2 (15 m )

−2d

85. The kinetic energy of the spring would be found by adding together the kinetic energy of each M infinitesimal part of the spring. The mass of an infinitesimal part is given by dM = S dx, and the l x speed of an infinitesimal part is v = v0 . Calculate the kinetic energy of the mass + spring. l l

 0 

K speed v = K mass + K spring = 12 mv02 + 12  v 2 dM = 12 mv02 + 12   v0 0

mass

= 12 mv02 + = mv + 1 2

2 0

v02 M S 1 l 1 2

x  MS



l 

 x dx 2

2 0

v MS l 3 l

= 12 v02 ( m + 13 M S )

So for a generic speed v, we have Kspeed v = 12 ( m + 13 M S ) v 2 . 86. (a) Since we are solving for the average force, we assume that the acceleration is constant. We use Eq. 2–12c to find the acceleration, and Newton’s second law to find the force. a=

v 2 − v02

2 ( x − x0 )

→ F = ma = m

v 2 − v02

2 ( x − x0 )

= ( 0.025 kg )

0 − ( 360 m s ) 2 ( 0.15 m )

= −1.08  10 4 N

The negative sign indicates the direction of the force. Its magnitude is 1.08  104 N 

1.1  104 N . (b) The work done by the gel on the projectile is equal to the change in the projectile’s kinetic energy. Note that the force exerted by the gel is in the opposite direction to the direction of motion of the projectile. Use Eqs. 7–1 and 7–11. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

240

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Work and Energy

W = Fd cos  = − Fd ; W = K = 12 mv22 − 12 mv12 = − 12 mv12

( 0.025 kg )( 360 m s ) − Fd = − mv → F = = = 1.08  10 4 N  1.1  10 4 N 2d 2 ( 0.15 m ) 1 2

mv12

2 1

87. (a) The pilot’s initial speed when he hit the snow was 45 m/s. The work done on him as he fell the 1.1 m into the snow changed his kinetic energy. Both gravity and the snow did work on the pilot during that 1.1-meter motion. Gravity did positive work (the force was in the same direction as the displacement), and the snow did negative work (the force was in the opposite direction as the displacement). Wgravity + Wsnow = K → mgd + Wsnow = − 12 mv12 →

(

)

(

)

Wsnow = − 12 mv12 − mgd = − m 12 vi2 + gd = − ( 82 kg )  12 ( 45 m s ) + 9.80 m s 2 (1.1m )  2





= −8.391 104 J  −8.4  104 J (b) The work done by the snowbank is done by an upward force, while the pilot moves down. Wsnow = Fsnowd cos180o = − Fsnowd → Fsnow = −

Wsnow

=−

−8.391  104 J

= 7.628  104 N  7.6  104 N

d 1.1 m (c) During the pilot’s fall in the air, positive work was done by gravity, and negative work by air resistance. The net work was equal to his change in kinetic energy while falling. We assume he started from rest when he jumped from the aircraft. Wgravity + Wair = K → mgh + Wair = 12 mv22 − 0 →

(

)

(

)

Wair = 12 mv22 − mgh = m 12 v22 − gh = ( 82 kg )  12 ( 45 m s ) − 9.80 m s 2 ( 370 m )  2





= −2.1 105 J

88. The (negative) work done by the bumper on the rest of the car must equal the change in the car’s kinetic energy. The work is negative because the force on the car is in the opposite direction to the car’s displacement. Wbumper = K = → − 12 kx 2 = 0 − 12 mv12 → 2

  1m s   ( 5 km h )  3.6 km h   2 v    = 9  106 N m k = m 12 = (1050 kg )  2 x

( 0.015 m )

89. (a) While the package is sliding, there is a constant force of kinetic friction on it, which accelerates it from an initial speed of 0 to a final speed of v. Use Eq. 2–12a. Fnet = Ffriction = k FN = k mg = ma = m

( v − v0 ) t

→ t=

m ( v − v0 )

k mg

k g

(b) The package’s displacement can be found from Eq. 2–12c, since it has a constant acceleration.

v 2 = v02 + 2a ( x − x0 ) → x − x0 = d =

v 2 − v02 2a

v2 2k g

241

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(c) The conveyor belt’s motor did work because there was a frictional force exerted by the belt on v2 the package. The belt moved a distance vt = while exerting a force of magnitude  k mg, k g

 v2  2  = mv .  g  k 

and so by Eq. 7–1, the work done by the belt is W = Fd =  k mg 

(d) The work done to accelerate the package is the change in kinetic energy of the package, which is 12 mv 2 . Thus the amount of work done against friction must be the remainder of the total work, also 12 mv 2 . We can also calculate the work done against friction by the force of friction

 v2  1 2 times the distance that the package moved, which is  k mg   = 2 mv . So to summarize:  2k g 

Wdone = 12 mv 2 and Waccelerate = 12 mv 2 against friction

package

90. (a) The work done by gravity is found using Eq. 7–1. WG = mgd cos ( 90 −  ) = ( 85 kg ) 9.80 m s 2 (180 m ) cos 86.0 o

(

)

= 1.046  10 4 J  1.0  10 4 J (b) The work is the change in kinetic energy. The initial kinetic energy is 0. WG = K = K 2 − K1 = 12 mv22 →

v2 =

2WG m

(

2 1.046  104 J 85 kg

FN y x  

) = 15.69 m s  16 m s

91. The spring must be compressed a distance such that the work done by the spring is equal to the change in kinetic energy of the car. The distance of compression can then be used to find the spring constant. Note that the work done by the spring will be negative, since the force exerted by the spring is in the opposite direction to the displacement of the spring. m Wspring = K = → − 12 kx 2 = 0 − 12 mvi2 → x = vi k F = ma = − kx → m ( −5.0 g ) = − k vi

m k

→

( 9.80 m s )  5.0 g  k = m = 5.8  103 N m  = (1500 kg )( 25 ) 2   vi   1m s   90 km h  3.6 km h      2

92. If the rider is riding at a constant speed, then the positive work input by the rider to the (bicycle + rider) combination must be equal to the negative work done by gravity as he moves up the incline. The net work must be 0 if there is no change in kinetic energy. (a) If the rider’s force is directed downwards, then the rider will do an amount of work equal to the force times the distance parallel to the force. The distance parallel to the downward force © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

242

Chapter 7

Work and Energy

would be the diameter of the circle in which the pedals move. Then consider that by using 2 feet, the rider does twice that amount of work when the pedals make one complete revolution. So in one revolution of the pedals, the rider does this amount of work. Wrider = 2 ( 0.90mrider g ) d pedal motion

In one revolution of the front sprocket, the rear sprocket will make 42 19 revolutions, and so the back wheel (and the entire bicycle and rider as well) will move a distance of ( 42 19 ) ( 2 rwheel ) . That is a distance along the plane, and so the height that the bicycle and rider will move is h = ( 42 19 )( 2 rwheel ) sin  . Finally, the work done by gravity in moving that height is calculated. WG = ( mrider + mbike ) gh cos180 = − ( mrider + mbike ) gh = − ( mrider + mbike ) g ( 42 19 )( 2 rwheel ) sin  Set the total work equal to 0, and solve for the angle of the incline. Wrider + WG = 0 → 2 0.90mrider g  d pedal − ( mrider + mbike ) g ( 42 19 )( 2 rwheel ) sin  = 0 → motion

 = sin

( 0.90mrider ) d pedal −1

motion

( mrider + mbike )( 42 19 )( rwheel )

= sin −1

0.90 ( 65 kg )( 0.36 m )

( 77 kg )( 42 19 )  ( 0.34 m )

= 6.7

(b) If the force is tangential to the pedal motion, then the distance that one foot moves while exerting a force is now half of the circumference of the circle in which the pedals move. The rest of the analysis is the same.





Wrider = 2 ( 0.90mrider g )   rpedal  ; Wrider + WG = 0 →



 = sin −1

motion



( 0.90mrider )  rpedal motion

( mrider + mbike )( 42 19 )( rwheel )

= sin −1

0.90 ( 65 kg )( 0.18 m )

( 77 kg )( 42 19 )( 0.34 m )

= 10.5  10

93. (a) The work done by the arms of the parent will change the kinetic energy of the child. The force is in the opposite direction of the displacement. Wparent = K child = K f − K i = 0 − 12 mvi2 ; Wparent = Fparent d cos180 → − mv = − Fparent d 1 2

2 i

(18 kg )( 25 m s ) → Fparent = = = 125N  130 N  28 lbs 2d 2 ( 45 m ) mvi2

This force is achievable by an average parent. (b) The same relationship may be used for the shorter distance.

(18 kg )( 25 m s ) Fparent = = = 469N  470 N  110 lbs 2d 2 (12 m ) mvi2

This force may not be achievable by an average parent. Many people might have difficulty with a 110-pound bench press exercise, for example. 94. Because the acceleration is essentially 0, the net force on the mass is 0. The magnitude of F is found with the help of the free-body diagram in the textbook. mg  Fy = FT cos  − mg = 0 → FT = cos  mg  Fx = F − FT sin  = 0 → F = FT sin  = cos  sin  = mg tan  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

243

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(a) A small displacement of the object along the circular path is given by dr = l d , based on the definition of radian measure. The force F is at an angle  to the direction of motion. We use the symbol dr for the infinitesimal displacement, since the symbol l is already in use as the length of the pendulum.  = 0

 = 0

 =0

WF =  F dr =  ( F cos  ) l d = l  ( mg tan  ) cos  d = mg l  sin  d = − mg l cos  0 = mg l (1 − cos  0 ) 0

(b) The angle between mg and the direction of motion is ( 90 +  ) .  = 0

 = 0

 =0

WG =  mg dr = mg l  cos ( 90 +  ) d = −mg l  sin  d = mg l cos  0 = mg l ( cos  0 − 1) 0

Alternatively, it is proven in Problem 38 that the shape of the path does not determine the work done by gravity–only the height change. Since this object is rising, gravity will do negative work. WG = mgd cos  = mg ( height ) cos180 = − mgyfinal = − mg ( l − l cos  0 ) = mg l ( cos  0 − 1)

Since FT is perpendicular to the direction of motion, it does 0 work on the bob. Note that the total work done is 0, since the object’s kinetic energy does not change. 95. (a) From the graph, the shape of the force function is roughly that of a triangle. The work can be estimated using the formula for the area of a triangle of base 20 m and height 100 N. W  12 “b” “ h” = 12 ( 20.0 m )(100 N ) = 1000 J

From the red lines on the graph, indicating the triangle used, we see that the triangular approximation is an under-estimate. (b) Integrate the force function to find the exact work done. xf

W =  Fdx = xi

20.0 m

 100 − ( x − 10)  dx 2

0.0 m

20.0 m

 ( 20 x − x ) dx = 10 x − x  2

1 3

20.0 m 0.0 m

= 1333J  1330 J

0.0 m

96. Refer to the free-body diagram. The coordinates are defined simply to help analyze the components of the force. At any angle  , since the mass is not accelerating, we have the following.  Fx = F − mg sin  = 0 → F = mg sin  Find the work done in moving the mass from  = 0 to  =  0 .



mg 

244

Chapter 7

Work and Energy

 = 0

 =0

WF =  F d s =  F cos 0l d = mg l  sin  d

= −mg l cos  0 = mg l (1 − cos  0 ) 0

See the second diagram to find the height that the mass has risen. We see that h = l − l cos  0 = l (1 − cos  0 ) , and so

WF = mg l (1 − cos 0 ) = mgh . 97. The net work done on the car must be its change in kinetic energy. By applying Newton’s third law, the negative work done on the car by the spring must be the opposite of the work done in compressing the spring. W = K = −Wspring → 12 mv22 − 12 mv12 = − 12 kx 2 → 2

  1m s   66 km h  3.6 km h   2 v    = 8.3  104 N m k = m 12 = (1200 kg )  2

( 2.2 m )

98. Let us assume the pushing force is F0 . The work done on the block will be equal to its change in kinetic energy as it moves from A to B.

F0 d = 12 mvB2 → vB =

2 F0 d

m Now do the same calculation, but for applying the force from A to C. F0 ( 2d ) = 12 mvC2 → vC =

4 F0 d m

= 2vB

We see that vC = 2vB . 99. (a) The work-energy principle says the net work done is the change in kinetic energy. The climber both begins and ends the fall at rest, so the change in kinetic energy is 0. Thus the total work done (by gravity and by the rope) must be 0. This is used to find x. Note that the force of gravity is parallel to the displacement, so the work done by gravity is positive, but the force exerted by the rope is in the opposite direction to the displacement, so the work done by the rope is negative. Wnet = Wgrav + Wrope = mg ( 2 l + x ) − 12 kx 2 = 0 → 12 kx 2 − mgx − 2 l mg = 0 → x=

mg  m 2 g 2 − 4 ( 12 k ) ( −2 l mg ) 2 ( 12 k )

mg  m 2 g 2 + 4k l mg k

mg 

4k l  1  1 +  k  mg 

We have assumed that x is positive in the expression for the work done by gravity, and so the “plus” sign must be taken in the above expression. Thus x =

mg 

4k l  1 + 1 + . k  mg 

245

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(b) Use the values given to calculate

and

kx mg

Instructor Solutions Manual

( 85 kg ) ( 9.80 m s 2 )  4 ( 850 N m )( 8.0 m )  x= 1 + 1 +  = 6.665 m 1 + 1 + = k  mg  ( 850 N m )  ( 85 kg ) ( 9.80 m s 2 )  mg 

x l

4k l 

6.665 m 8.0 m

= 0.83 ;

kx mg

( 850 N m )( 6.665 m ) = 6.8 ( 85 kg ) ( 9.80 m s 2 )

100. (a) Use Eq. 7–10. 2 K = 12 mv 2 = 12 ( 2.0 kg )( 4.9 m s ) = 24.01J  24 J (b) To find the work done by the applied force, we use Eq. 7–6. Note that the magnitude of the applied force is given in Newtons by F ( x ) = 20 x, where x is in meters. 4.0 m

(

W =  20 x cos 36.8 dx = 20 cos 36.8 12 x 2

)

4.0 m 0m

= 128.11J  130 J

(c) The friction force is found from the adjacent force diagram, consider the net force in the x- and y- directions. Note that there is no acceleration in the y-direction, and thus no net force in the y-direction.  Fy = FN − F sin 36.8 − mg = 0 → FN = mg + F sin 36.8

Ffr = k FN = k ( mg + F sin 36.8 )

(

Ffr mg

)

Ffr ( x = 0 ) = k ( mg + 0 ) = 0.50 ( 2.0 kg ) 9.80 m s 2 = 9.8 N Ffr ( x = 4.0 m ) = k  mg + ( 80 N ) sin 36.8

(

)

= 0.50 ( 2.0 kg ) 9.80 m s 2 + (80 N ) sin 36.8  = 33.76N  34 N

(d) The magnitude of the friction force is given by using the functional form of the applied force. Ffr ( x ) = k  mg + ( 20 x ) sin 36.8 = ( 9.8 + 5.99 x ) N

The magnitude of the friction force is the blue line in the adjacent graph.

(e) To find the speed of the block, we need to find how much work was done by the net force, and use Eqs. 7–6 and 7–11. The net force is the horizontal component of the pushing force, minus the frictional force. Fnet ( x ) = 20 x cos 36.8 − ( 9.8 + 5.99 x ) = (10.02 x − 9.80 ) N 4.0 m

(

W =  (10.02 x − 9.80 ) dx = 5.01x 2 − 9.80 x

)

4.0 m 0m

= 40.96 J = 12 mv 22 − 12 mv12

v 2 = v12 +

2W m

( 4.9 m s ) + 2

2 ( 40.96 J ) 2.0 kg

= 8.06 m s  8.1m s

246

CHAPTER 8: Conservation of Energy Responses to Questions 1.

Friction is not conservative; it dissipates energy in the form of heat, sound, and light. Air resistance is not conservative; it dissipates energy in the form of heat and the kinetic energy of fluids. “Human” forces, for example, the forces produced by your muscles, such as pushing a box across the floor, are also not conservative. They dissipate energy in the form of heat and also through chemical processes.

The two forces on the book are the applied force upward (nonconservative) and the downward force of gravity (conservative). If air resistance is non-negligible, it is nonconservative.

At the top of the pendulum’s swing, all of its energy is gravitational potential energy; at the bottom of the swing, all of the energy is kinetic. (a) If we can ignore friction, then energy is transformed back and forth between potential energy and kinetic energy as the pendulum swings. Each swing would bring the pendulum back to the same height, and every time it reaches its lowest point, its speed would be the same. (b) If friction is present, energy is lost to friction at the pivot point and to air resistance during each swing. During each swing the total energy decreases, and so the pendulum’s amplitude decreases, and its speed is progressively smaller at the bottom of each swing. (c) When a grandfather clock is “wound up,” the amount of energy that will eventually be lost to friction and air resistance is stored as potential energy (either elastic or gravitational, depending on the clock mechanism), and part of the workings of the clock is to put that stored energy back into the pendulum at the same rate that friction is dissipating the energy. Then the motion of the clock’s pendulum approaches that of the no-friction pendulum.

The drawing shows water falling over a waterfall and then flowing back to the top of the waterfall. Since the top of the waterfall is above the bottom of the waterfall, the water has greater gravitational potential energy at the top. The optical illusion of the diagram implies that water is flowing freely from the bottom of the waterfall back to the top. Since water won’t move uphill unless work is done on it to increase its gravitational potential energy (for example, work done by a pump), the water from the bottom of the waterfall would NOT be able to make it back to the top.

Yes, the spring can leave the table. When you push down on the spring, you do work on it and it gains elastic potential energy and loses a little gravitational potential energy, since the center of mass of the spring is lowered. When you remove your hand, the spring expands, and the elastic potential energy is converted into kinetic energy and into gravitational potential energy. If enough elastic potential energy was stored, the center of mass of the spring will rise above its original position, and the spring will leave the table.

Stepping on top of a log and jumping down the other side requires you to raise your center of mass farther than just stepping over a log does. Raising your center of mass farther requires you to do more work, or use more energy. The experienced hiker uses less energy by this method.

(a) As a car accelerates uniformly from rest, the potential energy stored in the fuel is converted into kinetic energy in the engine and transmitted through the transmission into the turning of the wheels, which causes the car to accelerate (if friction is present between the road and the tires). (b) By Newton’s third law, as the tires exert a backwards force on the road, the road exerts a forward force on the tires, accelerating the car forward. Through the frictional force, the work done by the engine is converted to the kinetic energy of the car.

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Instructor Solutions Manual

Yes. If the potential energy U is negative at some location (which it can be, depending on the position where the potential energy is defined to be 0), and the absolute value of the potential energy at that location is greater than the kinetic energy K at the same location, then the total mechanical energy E will be negative.

When a child hops around on a pogo stick, gravitational potential energy (at the top of the hop) is transformed into kinetic energy as the child moves downward, and then that kinetic energy is stored as spring potential energy as the spring in the pogo stick compresses. A small amount of gravitational potential energy is also converted to spring potential energy since the spring compresses downward. As the spring begins to expand, the energy is converted back to kinetic energy and then gravitational potential energy, and the cycle repeats. Since energy is lost due to friction, the child must add energy to the system by pushing down on the pogo stick while it is on the ground to get a greater spring compression.

10. (a) The two conservative forces are the force of gravity and the elastic force of a spring. The force of gravity is accounted for in the gravitational potential energy. The spring force is accounted for in the elastic potential energy. (b) The force of water on a swimmer is a non-conservative force. Energy is not “stored” and “recovered” in the interaction between the swimmer and the water. 11. For every meter the load is raised, two meters of rope must be pulled up at the top of the building. The work done on the piano must be equal to the work done by you. Since you are pulling with half the force (the tension in the rope is equal to half of the weight of the piano), you must pull through twice the distance to do the same amount of work as that done on the piano. 12. The faster arrow has the same mass and twice the speed of the slower arrow, so will have four times the kinetic energy K = 12 mv 2 . Therefore, four times as much work must be done on the faster

(

)

arrow to bring it to rest. If the force on the arrows is constant, the faster arrow will travel four times the distance of the slower arrow in the hay. 13. For each of the water balloons, the initial energy (kinetic plus potential) will equal the final energy (all kinetic). Since the initial energy depends only on the speed and not on the direction of the initial velocity, and all balloons have the same initial speed and height, the final speeds will all be the same.  E1 = 12 mv12 + mgh = E2 = 12 mv22  14. In order to escape the Earth’s gravitational field, the rocket needs a certain minimum speed with respect to the center of the Earth. If you launch the rocket from any location except the poles, then the rocket will have a tangential velocity due to the rotation of the Earth. This velocity is towards the east and is greatest at the equator, where the surface of the Earth is farthest from the axis of rotation. In order to use the minimum amount of fuel, you need to maximize the contribution of this tangential velocity to the needed escape velocity, so launch the rocket towards the east from a point as close as possible to the equator. (As an added bonus, the gravitational pull on the rocket will be slightly less at the equator because the Earth is not a perfect sphere and the surface is farthest from the center at the equator.)

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15. The work done on the suitcase depends only on (c) the height of the table and (d) the weight of the suitcase. This can be expressed as the change in potential energy of the suitcase. Since the work done is against the force of gravity and gravity is conservative, the work done does not depend on the path. And work does not take into account the amount of time elapsed while the work is done. 16. Power is the rate of doing work. Both (c) and (d) will affect the total amount of work needed, and hence the power. Answer (b), the time the lifting takes, will also affect the power. The length of the path (a) will only affect the power if different paths take different times to traverse. 17. When you climb a mountain by going straight up, the force needed is large (and the distance traveled is small), and the power needed (work per unit time) is also large. If you take a zigzag trail, you will use a smaller force (over a longer distance, so that the work done is the same) and less power, since the time to climb the mountain will be longer. A smaller force and smaller power output make the climb seem easier. 18. Neglecting any air resistance or friction in the pivot, the pendulum bob will have the same speed at the lowest point for both launches. In both cases, the initial energy is equal to potential energy mgh plus kinetic energy 12 mv 2 , with v = 3.0 m s. (Notice that the direction of the velocity doesn’t affect the kinetic energy.) Since the total energy at any point in the swing is constant, the pendulum will have the same energy at the lowest point, and therefore the same speed, for both launches. Another observation is that the ball that is launched upward will return to the same spot with the same speed but directed downward along the trajectory, and so at that point it will be identical to the ball that was initially launched downward. 19. (a) The force is proportional to the negative of the slope of the potential energy curve, so the magnitude of the force will be greatest where the curve is steepest, at point C. (b) The force acts to the left at points A, E, and F; to the right at point C; and is zero at points B, D, and G. (c) Equilibrium exists at points B, D, and G. B is a point of neutral equilibrium, D is a point of stable equilibrium, and G is a point of unstable equilibrium. 20. (a) If the particle has E3 at x6, then it has both potential and kinetic energy at that point. As the particle moves toward x0, it gains kinetic energy as its speed increases. Its speed will be a maximum at x0. As the particle moves to x4, its speed will decrease, but will be larger than its initial speed. As the particle moves to x5, its speed will increase, then decrease to zero. The process is reversed on the way back to x6. At each point on the return trip the speed of the particle is the same as it was on the forward trip, but the direction of the velocity is opposite. (b) The kinetic energy is greatest at point x0, and least at x5.

Responses to MisConceptual Questions 1.

(c) As the ball falls, gravitational potential energy is converted to kinetic energy. As the ball hits the trampoline, kinetic energy is converted to elastic potential energy. This energy is then transferred back to kinetic energy of the ball and finally gravitational potential energy. No energy is added to the ball during the motion, so it can’t rise higher than it started. Some energy may be lost to heat during the motion, so the ball may not rise as high as it initially started.

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(e) The term “energy” is commonly misunderstood. In this problem energy refers to the sum of the kinetic and potential energies. Initially the kinetic energy is a maximum. During the flight kinetic energy is converted to potential energy, with the potential energy a maximum at the highest point. As the ball falls back down the potential energy is converted back into kinetic energy. Since no non-conservative forces act on the ball (there is no air resistance), the total energy remains constant throughout the flight.

(c) The kinetic energy depends upon the speed, and not the position of the block. Since the block moves with constant speed, the kinetic energy remains constant. As the block moves up the incline its elevation increases, and so its potential energy also increases.

(b) While moving up the hill, her kinetic energy decreases (she slows down) and her potential energy increases (her altitude increases). If there is no friction, then she doesn’t lose any energy.

(a) The kinetic energy is proportional to the square of the ball’s speed and the potential energy is proportional to the height of the ball. As the ball rises the speed and kinetic energy decrease, while the potential energy increases. As the ball falls the speed and kinetic energy increase, as the potential energy decreases.

(b) The two balls have the same initial kinetic energy and the same initial potential energy. When they hit the ground they will have the same final potential energy, so their final kinetic energies, and therefore speeds, will be the same. Even though they have the same initial and final speeds, it is a misconception to think they will spend the same time in the air. The ball thrown directly upward travels to a higher point, as all of the kinetic energy can be converted into potential energy, and therefore will spend a longer time in the air.

(e) A common misconception is that the steeper the slope, the faster the skier will be at the bottom. Without friction, all of the skier’s initial gravitational potential energy is converted into kinetic energy. The skier starting from a higher initial position will have the greater speed at the bottom. On the steeper slope, the skier accelerates faster but over a shorter time period. On the flatter slope, the skier accelerates slower but over more time.

(c) Friction is a non-conservative force, which removes energy from the system. The work done by friction is related to the product of the force of friction and the distance traveled. For a given coefficient of friction, the force of friction on the steeper slope is smaller than on the flatter slope, as it has a smaller normal force. Also, on the steeper slope the skier travels a shorter distance along the path, and so the product of a smaller friction force and a smaller distance leads to a smaller amount of work.

(b) By swinging to the exact same angle on the other side, the potential energy at the top of the motion (where the kinetic energy is 0) will be the same as the potential energy at the start of the motion (where the kinetic energy was also 0). Thus, mechanical energy is conserved.

10. (c) Friction is a non-conservative force which removes energy from the system (the person and swing). As the swing moves lower from its initial position, some of its potential energy is dissipated by friction, and the rest is kinetic energy at the bottom. Since the kinetic energy is less than the original potential energy, as the swing moves up on the other side of equilibrium, it can’t rise as far as its original position. Furthermore, it loses even more energy as it rises, again due to friction. So the maximum angle on the other side is less than the initial angle.

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11. (b) Since the person is lower than the y = 0 reference point, their y-coordinate and their gravitational potential energy are negative. 12. (a) The wattage rating of 250 Watts says that the motor can do 250 Joules of work per second, but unless the elapsed time is given, there is no statement of how much work the motor can do. 13. (c) It is certainly possible for a system to have negative potential energy, because the definition of the zero position for potential energy is arbitrary in some cases. For example, if the top of a building is defined to be the zero location for gravitational potential energy, then any place lower than that will have negative potential energy.

Solutions to Problems 1.

Subtract the initial gravitational potential energy from the final gravitational potential energy. U grav = mgy2 − mgy1 = mg ( y2 − y1 ) = ( 58 kg ) ( 9.80 m s2 ) ( 4.0 m ) = 2274 J  2300 J

The initial energy is U grav 1 = mgy1 and the final energy is U grav 2 = mgy2 . If no energy was lost to non-conservative forces, the final height would be the same as the initial height. y “lost” energy U grav 1 − U grav 2 mgy1 − mgy2 3.0 = = = 1− 2 = 1− = 0.40 initial energy U grav 1 mgy1 y1 5.0 And so 40% of the original energy was lost.

The potential energy of the spring is given by U el = 12 kx 2 where x is the distance of stretching or compressing of the spring from its natural length. 2 ( 45.0 J ) 2U el x= = = 1.07 m k 78.0 N m

(a) The change in gravitational potential energy is given by the following.

(

)

U grav = mg ( y2 − y1 ) = ( 66.5kg ) 9.80 m s2 ( 2660 m − 1150 m ) = 9.84  105 J (b) The minimum work required by the hiker would equal the change in potential energy, which is 9.84  105 J .

(c)

Yes . The actual work may be more than this, because the climber almost certainly had to overcome some dissipative forces such as air friction. Also, as the person steps up and down, they do not get the full amount of work back from each up-down event. For example, there will be friction in their joints and muscles.

The initial stretching is from the equilibrium position, x = 0. Use that to find the spring constant. U initial = 12 kx 2 → 7.5J = 12 k ( 2.0 cm )

(

)

→ k = 3.75J cm 2

U final = 12 kx 2 = 12 3.75J cm 2 ( 6.0 cm ) = 67.5 J 2

U final − U initial = 67.5J − 7.5J = 60 J = 6.0  101 J

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(a) Relative to the ground, the potential energy is given by the following. U grav = mg ( ybook − yground ) = (1.65 kg ) ( 9.80 m s 2 ) ( 2.20 m ) = 35.6 J (b) Relative to the top of the person’s head, the potential energy is given by the following. U grav = mg ( ybook − yhead ) = (1.65 kg ) ( 9.80 m s2 ) ( 2.20 m − 1.60 m ) = 9.7 J (c) The work done by the person in lifting the book from the ground to the final height is the same as the answer to part (a), 35.6 J . In part (a), the potential energy is calculated relative to the starting location of the application of the force on the book. The work done by the person is not related to the answer to part (b), because the potential energy is not calculated relative to the starting location of the application of the force on the book. The zero level for gravitational potential energy can be set an any convenient position.

Use Eq. 8–6 to find the potential energy function. k k U ( x ) = −  F ( x ) dx + C = −  − 3 dx + C = − 2 + C x 2x U ( 2.0 m ) = −

k 2 ( 2.0 m )

+C =0 → C =

→

U ( x) = −

The force is found from the relations on page 222. U U Fx = − = − (6x + 2 y ) Fy = − = − ( 2 x + 8 yz ) x y

(

F = ˆi ( −6 x − 2 y ) + ˆj ( −2 x − 8 yz ) + k −4 y 2

Fz = −

k 2x

U z

k 8 m2

= −4 y 2

)

(a) This force is conservative, because the work done by the force on an object moving from an initial position ( x1 ) to a final position ( x2 ) depends only on the endpoints. x2

(

)

(

W =  F d l =  Fx dx =  − kx + ax 3 + bx 4 dx = − 12 kx 2 + 14 ax 4 + 15 bx 5

(

) (

= − 12 kx22 + 14 ax24 + 15 bx25 − − 12 kx12 + 14 ax14 + 15 bx15

)

x2 x1

)

The expression for the work only depends on the endpoints. (b) Since the force is conservative, there is a potential energy function U such that Fx = −

(

)

Fx = −kx + ax 3 + bx 4 = −

U

→

x

U x

U ( x ) = 12 kx 2 − 14 ax 4 − 15 bx 5 + C

10. The potential energy function is derived in Problem 9. We choose the potential energy to be 0 at x = 0, and so U ( x ) = 12 kx 2 − 14 ax 4 − 15 bx 5 . From Eq. 8–4, the work done by the force in moving an object from one location to another is the opposite of the change in potential energy, and since that is the only force doing work, that amount of work is also the change in kinetic energy, between those two locations, as given in Eq. 7–11. We also have that x1 = x0 and v1 = 0. W = U ( x1 ) − U ( x2 ) = 12 mv22 − 12 mv12 →

v2 =

2 U ( x1 ) − U ( x2 ) m

− v12 =

2 m

U ( x ) − U ( x ) 0

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(a) For x2 = x0 : 2

v2 =

U ( x ) − U ( x ) = 0 0

(b) For x2 = x0 2 : 2

v2 =

U ( x ) − U ( x 2 ) 0

( 12 kx02 − 14 ax04 − 15 bx05 ) − ( 12 k 14 x02 − 14 a 161 x04 − 15 b 321 x05 )  m

( kx − ax −

bx05

U ( x ) − U ( 0 ) =

3 8

2 0

15 64

4 0

)

31 160

v2 =

2 m

( kx − ax − bx ) 1 2

2 0

1 4

4 0

1 5

5 0

(d) If we set a = 0 and b = 0 in each answer to model an ideal spring, the speeds for the ideal spring would be larger than the speeds for the spring from Problem 9, except for the starting speed [case (a)], in which they would both give a speed of 0. Note that the constants a and b must be small enough so that the argument of the square root does not become negative. 11. The forces on the skier are gravity and the normal force. The normal force is perpendicular to the direction of motion, and so does no work. Thus, the skier’s mechanical energy is conserved. Subscript 1 represents the skier at the top of the hill, and subscript 2 represents the skier at the bottom of the hill. The ground is the zero location for gravitational potential energy ( y = 0 ) . We have



v1 = 0, y1 = 115 m, and y2 = 0 (bottom of the hill). Solve for v2 , the speed at the bottom. 1 2

mv12 + mgy1 = 12 mv22 + mgy2 → 0 + mgy1 = 12 mv22 + 0 →

v2 = 2 gy1 = 2 ( 9.80 m s 2 ) (115 m ) = 47.5 m s (  106 mi h ) Note that the angle of the incline does not enter into the problem. 12. The only forces acting on Jane are gravity and the vine tension. The tension pulls in a centripetal direction, and so can do no work–the tension force is perpendicular at all times to her motion, so Jane’s mechanical energy is conserved. Subscript 1 represents Jane at the point where she grabs the vine, and subscript 2 represents Jane at the highest point of her swing. The ground is the zero location for gravitational potential energy ( y = 0 ) . We have v1 = 5.0 m s ,

y1 = 0, and v 2 = 0 (top of swing). Solve for y 2 , the height of her swing. 1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

y2 =

v12 2g

( 5.0 m s )

(

1 2

mv12 + 0 = 0 + mgy2 →

2 9.80 m s 2

)

= 1.276 m  1.3 m

No , the length of the vine does not enter into the calculation, unless the vine is less than 0.65 m long. If that were the case, she could not rise 1.3 m high. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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13. The forces on the sled are gravity and the normal force. The normal force is perpendicular to the direction of motion, and so does no work. Thus, the sled’s mechanical energy is conserved. Subscript 1 represents the sled at the bottom of the hill, and subscript 2 represents the sled at the top of the hill. The ground is the zero location for gravitational potential energy ( y = 0 ) . We have y1 = 0,

mg 

v2 = 0, and y2 = 1.22 m. Solve for v1 , the speed at the bottom. Note that the angle is not used. 1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

1 2

mv12 + 0 = 0 + mgy2 →

v1 = 2 gy2 = 2 ( 9.80 m s 2 ) (1.22 m ) = 4.89 m s 14. We assume that all the forces on the jumper are conservative, so that the mechanical energy of the jumper is conserved. Subscript 1 represents the jumper at the bottom of the jump, and subscript 2 represents the jumper at the top of the jump. Call the ground the zero location for gravitational potential energy ( y = 0 ) . We have y1 = 0, v 2 = 0.50 m s , and y2 = 2.10 m. Solve for v1 , the speed at the bottom. 1 mv12 + mgy1 = 12 mv22 + mgy2 → 12 mv12 + 0 = 12 mv22 + mgy2 → 2

( 0.50 m s ) + 2 ( 9.80 m s 2 ) ( 2.10 m ) = 6.4351m s  6.44 m s 2

v1 = v22 + 2 gy2 =

15. The mass is allowed to fall rather than “easing” it down with the hand (which introduces another force into the problem). Under this “falling” condition, mechanical energy is conserved. The unstretched spring, corresponding to the initial position, is the location of 0 elastic potential energy. There is no kinetic energy to consider since the mass is at rest at both positions. The change in height is the same as the amount of stretch of the spring. Mechanical energy is conserved. E1 = E2 → U1

+ U1 = U 2

grav

elas

grav

→ mgy1 + 0 = mgy2 + 12 k ( y2 − y1 )

+ U2

→

elas

2mg 2 ( 2.5 kg ) ( 9.80 m s ) ( y2 − y1 )2 = y1 − y2 = = = 0.56 m mg ( y1 − y2 ) = k ( y2 − y1 ) → k 88 N m ( y1 − y2 ) 2

1 2

The final position is 56 cm lower than the initial position. If we assume that the ruler is oriented so that higher numbers are nearer the floor than the lower numbers, we add 56 cm to the original 15 cm and get 71cm . 16. (a) See the diagram for the thrown ball. The speed at the top of the path will be the horizontal component of the original velocity. v top = v0 cos  = ( 7.8 m s ) cos 36o = 6.3 m s



(b) Since there are no dissipative forces in the problem, the mechanical energy of the ball is conserved. Subscript 1 refers to the ball at the release point, and subscript 2 refers to the ball at the top of the path. The release point is the zero location for gravitational potential energy ( y = 0 ) . We have v1 = 7.8 m s , y1 = 0, and v2 = v1 cos  . Solve for y2 . E1 = E2 → y2 =

(

1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

v12 1 − cos 2  2g

1 2

mv12 + 0 = 12 mv12 cos 2  + mgy2 →

) = ( 7.8 m s ) (1 − cos 36 ) = 1.07 m  1.1m 2 ( 9.80 m s ) 2

This is the height above its throwing level. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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17. Assume that all of the kinetic energy of the car becomes potential energy of the compressed spring. 2

1 2

2 mv02 = 12 kxfinal

  1m s   (1400 kg ) ( 85 km h )   2 3.6 km h   mv0   → k= 2 = = 1.6  105 N m 2

( 2.2 m )

xfinal

18. (a) Since there are no dissipative forces present, the mechanical energy of the person–trampoline– Earth combination will be conserved. We take the level of the unstretched trampoline as the zero level for both elastic and gravitational potential energy. Call “upward” the positive direction. Subscript 1 represents the jumper at the start of the jump, and subscript 2 represents the jumper upon arriving at the trampoline. There is no elastic potential energy involved in this part of the problem. We have v1 = 4.5 m s , y1 = 2.0 m, and y2 = 0. Solve for v 2 , the speed upon arriving at the trampoline. E1 = E2 → 12 mv12 + mgy1 = 12 mv22 + mgy2 → 12 mv12 + mgy1 = 12 mv22 + 0 → v2 =  v12 + 2 gy1 = 

( 4.5 m s ) + 2 ( 9.80 m s 2 ) ( 2.0 m ) =  7.710 m s  7.7 m s 2

The speed is the absolute value of v 2 . (b) Now let subscript 3 represent the jumper at the maximum stretch of the trampoline, and x represent the amount of stretch of the trampoline. We have v 2 = −7.710 m s , y2 = 0, x2 = 0,

v3 = 0, and x3 = y3 . There is no elastic energy at position 2, but there is elastic energy at position 3. Also, the gravitational potential energy at position 3 is negative, and so y3  0. A quadratic relationship results from the conservation of energy condition. E2 = E3 → 12 mv22 + mgy2 + 12 kx22 = 12 mv32 + mgy3 + 12 kx32 → 1 2

mv22 + 0 + 0 = 0 + mgy3 + 12 ky32 →

y3 =

(

− mg  m 2 g 2 − 4 ( 12 k ) − 12 mv 22 2 ( 12 k )

(

ky32 + mgy3 − 12 mv22 = 0 →

1 2

)

− mg  m 2 g 2 + kmv 22 k

) ( 62 kg ) ( 9.80 m s ) + ( 5.8 10 N m ) ( 62 kg )( 7.71m s ) ( 5.8 10 N m )

− ( 62 kg ) 9.80 m s 2 

= −0.263 m, 0.242 m Since y3  0, y3 = −0.26 m. So he depresses the trampoline 0.26 m . The second term under the quadratic is almost 600 times larger than the first term, indicating that the problem could have been approximated by not even including gravitational potential energy for the final position. If that approximation were made, the result would have been found by taking the negative result from the following solution. E2 = E3 →

1 2

mv22 = 12 ky32 → y3 = v2

m k

= ( 7.71m s )

62 kg 5.8  104 N m

= 0.25 m

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19. Use conservation of energy. The level of the ball on the uncompressed spring is taken as the zero location for both gravitational potential energy ( y = 0 ) and elastic potential energy ( x = 0 ) . It is diagram 2 in the figure. Take “upward” to be positive for both x and y. (a) Subscript 1 represents the ball at the launch point, and subscript 2 represents the ball at the location where the ball just leaves the spring, at the uncompressed length. We have v1 = 0, x1 = y1 = −0.220 m, and

x = 0, y=0

x2 = y 2 = 0. Solve for v 2 .

E1 = E2 →

1 2

mv12 + mgy1 + 12 kx12 = 12 mv22 + mgy2 + 12 kx22 →

0 + mgy1 + kx = mv + 0 + 0 → v2 = 1 2

2 1

1 2

2 2

kx12 + 2mgy1 m

( 875 N m )( 0.220 m ) + 2 ( 0.380 kg ) ( 9.80 m s 2 ) ( −0.220 m ) = 10.4 m s ( 0.380 kg ) 2

v2 =

(b) Subscript 3 represents the ball at its highest point. We have v1 = 0, x1 = y1 = −0.160 m, v3 = 0, and x3 = 0. Solve for y3 . E1 = E3 →

1 2

mv12 + mgy1 + 12 kx12 = 12 mv32 + mgy3 + 12 kx32 →

( 875 N m )( 0.220 m ) 0 + mgy1 + kx = 0 + mgy3 + 0 → y2 − y1 = = = 5.69 m 2mg 2 ( 0.380 kg ) ( 9.80 m s 2 ) 1 2

kx12

2 1

20. Since there are no dissipative forces present, the mechanical energy of the roller coaster will be conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the zero location for gravitational potential energy. We have v1 = 0 and y1 = 32 m. Point 2:

1 2

mv12 + mgy1 = 12 mv22 + mgy2 ; y2 = 0 → mgy1 = 12 mv22 →

(

)

v2 = 2 gy1 = 2 9.80 m s 2 ( 32 m ) = 25 m s

Point 3:

1 2

mv12 + mgy1 = 12 mv32 + mgy3 ; y3 = 26 m → mgy1 = 12 mv32 + mgy3 →

(

)

v3 = 2 g ( y1 − y3 ) = 2 9.80 m s 2 ( 6 m ) = 11m s

Point 4:

1 2

mv12 + mgy1 = 12 mv42 + mgy4 ; y4 = 14 m → mgy1 = 12 mv42 + mgy4 →

(

)

v4 = 2 g ( y1 − y4 ) = 2 9.80 m s 2 (18 m ) = 19 m s

21. With the mass at rest on the spring, the upward force due to the spring must be the same as the weight of the mass.

kd = mg → d =

mg k

The distance D is found using conservation of energy. Subscript 1 represents the mass at the top of the uncompressed spring, and subscript 2 represents the mass at the bottom of its motion, where the spring is compressed by D. Take the top of the uncompressed spring to be the zero location for both gravitational and elastic potential energy ( y = 0 ) . Choose up to be the positive direction. We have v1 = v 2 = 0, y1 = 0, and y2 = − D. Solve for D. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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E1 = E2 →

mv12 + mgy1 + 12 ky12 = 12 mv22 + mgy2 + 12 ky22 →

1 2

0 + 0 + 0 = 0 − mgD + 12 kD 2 → D =

2mg k

We see that D = 2d , and so D  d . The reason that the two distances are not equal is that putting the mass at rest at the compressed position requires that other work be done in addition to the work done by gravity and the spring. That other work is not done by a conservative force, but done instead by an external agent such as your hand. 22. (a) Draw a free-body diagram for each block. Write Newton’s second law for each block. Notice that the acceleration of block A in the yA direction is 0.

 F = F − m g cos  = 0 → F = m g cos  F = F − m g sin  = m a  F = m g − F = m a → F = m ( g − a ) Since the blocks are connected by the y1

cord, a yB = a xA = a. Substitute the expression for the tension force from the last equation into the x-direction equation for block 1, and solve for the acceleration. mB ( g − a ) − mA g sin  = mA a → mB g − mA g sin  = mA a + mBa a=g

( m − m sin  ) = 9.80 m s ( 5.0 kg − 3.5 kg sin 32) = 3.6 m s ( ) 8.5 kg (m + m ) B

(b) Find the final speed of mB (which is also the final speed of mA ) using constant acceleration relationships. mB − mA sin  v 2f = v02 + 2ay → v 2f = 2 g h → ( mA + mB )

(

v f = 2 gh

)

( m − m sin  ) = 2 9.80 m s ( 0.75 m ) ( 5.0 kg − 3.5 kg sin 32) = 2.3 m s ( ) 8.5 kg (m + m ) B

(c) Since there are no dissipative forces in the problem, the mechanical energy of the system is conserved. Subscript 1 represents the blocks at the release point, and subscript 2 represents the blocks when mB reaches the floor. The ground is the zero location for gravitational potential energy for m B , and the starting location for mA is its zero location for gravitational potential energy. Since mB falls a distance h, mA moves a distance h along the plane, and so rises a distance h sin  . The starting speed is 0. E1 = E2 → 0 + mA gh = 12 ( mA + mB ) v22 + mB gh sin  →

 mA − mB sin     mA + mB 

v2 = 2 gh 

This is the same expression found in part (b), and so gives the same numeric result. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

23. The only forces doing work are conservative forces, so we are able to use conservation of energy. We choose the gravitational potential energy to be 0 at the ground for the 3.0-kg mass, and to be 0 at the initial position for the 9.0-kg mass. We choose elastic potential energy is 0 when the spring is at its original length, with the 9.0-kg mass barely in contact with it. The 9.0-kg mass falls a distance y, the 3.0-kg mass rises a distance y, and the spring compresses an amount y. E1 = E2 → 0 = ( 3.0 kg ) gy + ( 9.0 kg ) g ( − y ) + 12 ky 2 0 = 12 ky 2 − ( 6.0 kg ) gy → y =

12.0 g k

(12.0 kg )( 9.80 m s ) 450 N m

= 0.26 m

24. The only forces acting on the bungee jumper are gravity and the elastic force from the bungee cord, so the jumper’s mechanical energy is conserved. Subscript 1 represents the jumper at the bridge, and subscript 2 represents the jumper at the bottom of the jump. Let the lowest point of the jumper’s motion be the zero location for gravitational potential energy ( y = 0 ) . The zero location for elastic potential energy is the point at which the bungee cord begins to stretch. See the diagram in the textbook. We have v1 = v 2 = 0, y1 = d , y2 = 0, and the amount of stretch of the cord x2 = d − 15. Solve for d. Note that we ignore Chris’s height, since his center of mass falls further than d. E1 = E2 →

 

d 2 −  30 + 2

1 2

mv12 + mgy1 + 12 kx12 = 12 mv22 + mgy2 + 12 kx22 → mgd = 12 k ( d − 15 )

mg 

  

2  d + 225 = 0 → d −  30 + 2

k 

d 2 − 56.7 d + 225 = 0 → d =

→

( 75 kg ) ( 9.80 m s 2 )  55 N m

56.7  56.7 2 − 4 ( 225 )

2 The larger answer must be taken because d > 15 m.

 d + 225 = 0 → 

= 52.4 m, 4.29 m → d = 52 m

25. The maximum acceleration of 4.5 g occurs where the force is a maximum, at the bottom of the motion, where the spring has its maximum compression. Write Newton’s second law for the elevator at the bottom of the motion, with up as the positive direction. Fnet = Fspring − Mg = Ma = 4.5Mg → Fspring = 5.5Mg Now consider the diagram for the elevator at various positions. If there are no non-conservative forces, then mechanical energy is conserved. Subscript 1 represents the elevator at the start of its fall, and subscript 2 represents the elevator at the bottom of its fall. The bottom of the fall is the zero location for gravitational potential energy ( y = 0 ) . There is also a point at the top of the spring that is the zero location for elastic potential energy (x = 0). We have v1 = 0, y1 = x + h , x1 = 0,

Fspring

Start of fall

Contact with spring, 0 for elastic PE

x Bottom of fall, 0 for gravitational PE

v 2 = 0, y2 = 0, and x2 = x. Use conservation of energy.

E1 = E2 →

1 2

M v12 + Mgy1 + 12 kx12 = 12 M v22 + Mgy2 + 12 kx22 →

0 + Mg ( x + h ) + 0 = 0 + 0 + 12 kx 2 → Mg ( x + h ) = 12 kx 2 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Fspring = 5.5Mg = kx → x =

5.5Mg

 5.5Mg + h  = 1 k  5.5Mg  → Mg   2    k   k 

→

9.6 Mg

k h 26. (a) The skier, while in contact with the sphere, is moving in a circular path, and so must have some component of the net force towards the center of the circle. See the free-body diagram. FN v2 Fradial = mg cos  − FN = m r r mg If the skier loses contact with the sphere, the normal force is 0. Use that  relationship to find the critical angle and speed. v2 v2 mg cos  crit = m crit → cos  crit = crit r rg Using conservation of mechanical energy, the velocity can be found as a function of angle. Let subscript 1 represent the skier at the top of the sphere, and subscript 2 represent the skier at angle  . The top of the sphere is the zero location for gravitational potential energy ( y = 0 ) .

We have v1 = 0, y1 = 0, and y2 = − ( r − r cos  ) .

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 → 0 = 12 mv22 − mg ( r − r cos  ) →

v2 = 2 g ( r − r cos  ) Combine the two relationships to find the critical angle. 2 2 g ( r − r cos  crit ) vcrit = = 2 − 2 cos  crit →  crit = cos −1 23  48 cos  crit = rg rg (b) If friction is present, another force will be present, tangential to the surface of the sphere. The friction force will not affect the centripetal relationship of cos  crit =

2 vcrit

. But the friction will rg reduce the speed at any given angle, and so the skier will be at a greater angle before the critical speed is reached. 27. Use conservation of energy, where all of the kinetic energy is transformed to thermal energy. 2

Einitial = Efinal →

1 2

  1m s   mv = Ethermal = ( 2 )( 72, 000 kg ) ( 85 km h )  = 4.0 10 7 J    3.6 km h    2

1 2

28. Apply the conservation of energy to the child, considering work done by gravity and thermal energy. Subscript 1 represents the child at the top of the slide, and subscript 2 represents the child at the bottom of the slide. The ground is the zero location for potential energy ( y = 0 ) . We have v1 = 0, y1 = 2.2 m, v 2 = 1.25 m s , and y2 = 0. Solve for the work changed into thermal energy.

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 + Ethermal →

(

)

Ethermal = mgy1 − 12 mv22 = (16.0 kg ) 9.80 m s 2 ( 2.20 m ) − 12 (16.0 kg )(1.15 m s ) = 334 J 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

29. (a) See the free-body diagram for the ski. Write Newton’s second law for forces perpendicular to the direction of motion, noting that there is no acceleration perpendicular to the plane.  F⊥ = FN − mg cos  = 0 → FN = mg cos  → Ffr = k FN = k mg cos 

Now use conservation of energy, including the non-conservative friction force. Subscript 1 represents the ski at the top of the slope, and subscript 2 represents the ski at the bottom of the slope. The location of the ski at the bottom of the incline is the zero location for gravitational potential energy ( y = 0 ) . We have v1 = 0, y1 = l sin  , and y2 = 0. Write the conservation of energy condition, and solve for the final speed. 1 mv12 + mgy1 − Ffr l = 12 mv22 + mgy2 → mg l sin  −  k mg l cos  = 12 mv22 → 2

(

)

(

v2 = 2 g l ( sin  −  k cos  ) = 2 9.80 m s  ( 85 m ) sin 28o − 0.090 cos 28o

)

= 25.49 m s  25 m s

(b) Now, on the level ground, Ffr = k mg , and there is no change in potential energy. We again use conservation of energy, including the non-conservative friction force, to relate position 2 with position 3. Subscript 3 represents the ski at the end of the travel on the level, having traveled a distance l 3 on the level. We have v 2 = 25.49 m s , y2 = 0, v3 = 0, and y3 = 0. 1 2

mv22 + mgy2 − Ffr l 3 = 12 mv32 + mgy3 →

l3 =

v22 2 g k

1 2

mv22 = k mg l 3 →

( 25.49 m s ) = 368.3 m  370 m 2 ( 9.80 m s  ) ( 0.090 ) 2

30. (a) Apply energy conservation with no non-conservative work. Subscript 1 represents the ball as it is dropped, and subscript 2 as it reaches the ground. The ground is the zero location for gravitational potential energy. We have v1 = 0, y1 = 14.0 m, and y2 = 0. Solve for v 2.

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 → mgy1 = 12 mv22 →

v2 = 2 gy1 = 2 ( 9.80 m s 2 ) (14.0 m ) = 16.6 m s (b) Apply energy conservation, but with non-conservative work due to friction included. The energy dissipated will be given by Ffr d . The distance d over which the frictional force acts will be the 14.0 m distance of fall. With the same parameters as above, and v2 = 12.5 m s , solve for the force of friction. 1 mv12 + mgy1 − Ffr d = 12 mv22 + mgy2 → mgy1 − Ffr d = 12 mv 22 → 2





Ffr = m  g

−

v22 

  

2  = ( 0.145 kg )  9.80 m s −

2d 

(12.5 m s )   = 0.612 N, upwards 2 (14.0 m )  2

31. Since there is a non-conservative force, apply energy conservation with the dissipative friction term. Subscript 1 represents the roller coaster at point 1, and subscript 2 represents the roller coaster at point 2. Point 2 is taken as the zero location for gravitational potential energy. We have v1 = 1.30 m s , y1 = 32 m, and y2 = 0. Solve for v 2 . Note that the dissipated energy is given by

Ffr d = 0.23mgd . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

260

Chapter 8

Conservation of Energy

1 2

mv12 + mgy1 − 0.23mgd = 12 mv22 + mgy2 → v2 = −0.46 gd + v12 + 2 gy1

(

)

(

)

= −0.46 9.80 m s2 ( 45.0 m ) + (1.30 m s ) + 2 9.80 m s2 ( 32 m ) = 20.64 m s  21m s 2

32. A free-body diagram for the sled is shown as it moves up the hill. From this we get an expression for the friction force.  Fy = FN − mg cos  = 0 → FN = mg cos  → Ffr = kmg cos 

x Ffr (a) We apply conservation of energy with a frictional force as given in  Eq. 8–15b. Subscript 1 refers to the sled at the start of its motion,  mg and subscript 2 refers to the sled at the top of its motion. Take the starting position of the sled to be the 0 for gravitational potential energy. We have v1 = 2.3 m s , y1 = 0, and v2 = 0. The relationship between the distance traveled along the incline (l) and the height the sled rises is y2 = l sin  . Solve for l. 1 2

mv12 + mgy1 = 12 mv22 + mgy2 + Ffr l →

l =

1 2

mv12 = mg l sin  + k mg l cos  →

( 2.3 m s ) = 0.3910 m  0.39 m 2 2 g ( sin  + k cos  ) 2 ( 9.80 m s ) ( sin 28 + 0.25 cos 28 ) 2

v12

(b) For the sled to slide back down, the friction force will now point UP the hill in the free-body diagram. In order for the sled to slide down, the component of gravity along the hill must be large than the maximum force of static friction. mg sin   Ffr → mg sin   s mg cos  → s  tan 28 → s  0.53 (c) We again apply conservation of energy including work done by friction. Subscript 1 refers to the sled at the top of the incline, and subscript 2 refers to the sled at the bottom of the incline. We have v1 = 0, y1 = l sin  , and y2 = 0. 1 2

mv12 + mgy1 = 12 mv22 + mgy2 + Ffr l → mg l sin  = 12 mv22 + k mg l cos  →

v2 = 2g l ( sin  −  k cos  ) = 2 ( 9.80 m s 2 ) ( 0.3910 m )( sin 28 − 0.25 cos 28 ) = 1.381m s  1.4 m s

33. Since there are no non-conservative forces, the mechanical energy of the projectile will be conserved. Subscript 1 represents the projectile at launch and subscript 2 represents the projectile as it strikes the ground. The ground is the zero location for potential energy ( y = 0 ) . We have

v1 = 165 m s , y1 = 135 m, and y2 = 0. Solve for v 2 . E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

v2 = v12 + 2 gy1 =

1 2

mv12 + mgy1 = 12 mv22 →

(165 m s ) + 2 ( 9.80 m s 2 ) (135 m ) = 173 m s 2

Notice that the launch angle does not enter the problem, and so does not influence the final speed. 34. We apply conservation of mechanical energy. We take the surface of the Moon to be the 0 level for gravitational potential energy. Subscript 1 refers to the location where the engine is shut off, and subscript 2 refers to the surface of the Moon. Up is the positive y-direction. (a) We have v1 = 0, y1 = h, v2 = −3.0 m s , and y2 = 0.

261

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

E1 = E2 → h=

v 22 2g

1 2

Instructor Solutions Manual

mv12 + mgy1 = 12 mv22 + mgy2 → mgh = 12 mv22 →

( −3.0 m s )

(

2 1.62 m s 2

)

= 2.8 m

(b) We have the same conditions except v1 = −2.0 m s . E1 = E2 → h=

1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

( −3.0 m s ) − ( −2.0 m s ) 2

v 22 − v12

(

2 1.62 m s 2

1 2

mv12 + mgh = 12 mv22 →

)

= 1.5 m

(c) We have the same conditions except v1 = 2.0 m s . And since the speeds, not the velocities, are used in the energy conservation calculation, this is the same as part (b), and so h = 1.5 m . E1 = E2 → h=

1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

( −3.0 m s ) − ( 2.0 m s ) 2

v 22 − v12

(

2 1.62 m s 2

1 2

mv12 + mgh = 12 mv22 →

)

= 1.5 m

35. We can apply Eq. 8–15b to this situation, since the only non-conservative force is kinetic friction, acting opposite to the direction of motion. Position 1 is the starting position, 8 meters up the hill, and position 2 is the stopping position. We take the level stretch as the 0 of gravitational potential energy. We have v1 = 0, y1 = 8.0 m, v2 = 0, and y2 = 0. The force of friction is the normal force times the coefficient of friction, and the normal force is equal to the weight. 1 mv12 + mgy1 = 12 mv22 + mgy2 + Ffr l → mgy1 = Ffr l =  k mg l → 2

l =

k

8.0 m 0.30

= 26.7 m  27 m

36. (a) Use conservation of energy to equate the potential energy at the top of the circular track to the kinetic energy at the bottom of the circular track. Take the bottom of the track to the be 0 level for gravitational potential energy. 2 Etop = Ebottom → mgr = 12 mvbottom →

vbottom = 2 gr = 2 ( 9.80 m s 2 ) ( 2.0 m ) = 6.261m s  6.3m s (b) The thermal energy produced is the opposite of the work done by the friction force. In this situation, the force of friction is the weight of the object times the coefficient of kinetic friction. Ethermal = −Wfriction = − Ffriction x = − Ffriction x cos  = −  k mg x ( cos180 ) =  k mg x

(

)

= ( 0.25)(1.0 kg ) 9.80 m s2 ( 3.0m ) = 7.35J  7.4 J

(c) The work done by friction is the change in kinetic energy of the block as it moves from point B to point C. Wfriction = K = K C − K B = 12 m vC2 − vB2 →

(

vC =

2Wfriction m

+ vB2 =

2 ( −7.35 J )

(1.0 kg )

)

+ ( 6.261m s ) = 4.9498 m s  4.9 m s 2

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Chapter 8

Conservation of Energy

(d) Use conservation of energy to equate the kinetic energy when the block just contacts the spring with the potential energy when the spring is fully compressed and the block has no speed. There is no friction on the block while compressing the spring. 2 2 Einitial = Efinal → 12 mvcontact = 12 kxmax →

( 4.9498 m s ) k = m 2 = (1.0 kg ) = 612.5 N m  610 N m 2 xmax ( 0.20 m ) 2

2 vcontact

37. Use conservation of energy, including the non-conservative frictional force, as developed in Eq. 8–15b. The block is on a level surface, so there is no gravitational potential energy change to consider. The frictional force is given by Ffr = k FN = k mg , since the normal force is equal to the weight. Subscript 1 represents the block at the compressed location, and subscript 2 represents the block at the maximum stretched position. The location of the block when the spring is neither stretched nor compressed is the zero location for elastic potential energy (x = 0). Take right to be the positive direction. We have v1 = 0, x1 = −0.050 m, v2 = 0, and x2 = 0.023 m. E1 = E2 + Ffr l → 1 2

1 2

mv12 + 12 kx12 = 12 mv22 + 12 kx22 + Ffr ( x2 − x1 ) →

kx12 = 12 kx22 + k mg ( x2 − x1 ) →

k =

(

) = −k ( x + x ) = − (180 N m ) ( −0.050m ) + ( 0.023m ) = 0.36 2mg ( x − x ) 2mg 2 ( 0.680 kg ) ( 9.80 m s ) k x12 − x22

38. (a) Use conservation of energy. Subscript 1 represents the block at the compressed location, and subscript 2 represents the block at its maximum position up the slope. The initial location of the block at the bottom of the plane is taken to be the zero location for gravitational potential energy (y = 0). The variable x will represent the amount of spring compression or stretch. We have v1 = 0, x1 = 0.50 m, y1 = 0, v2 = 0, and x2 = 0. The distance the block moves up the

E1 = E2 → 1 2

, so y2 = d sin  . Solve for d. sin  1 mv12 + mgy1 + 12 kx12 = 12 mv22 + mgy2 + 12 kx22 → 2

plane is given by d =

kx = mgy2 = mgd sin  2 1

( 75 N m )( 0.50 m ) → d= = = 0.73 m 2mg sin  2 ( 2.0 kg ) ( 9.80 m s 2 ) sin 41 kx12

(b) Now the spring will be stretched at the turning point of the motion. The first half-meter of the block’s motion returns the block to the equilibrium position of the spring. After that, the block beings to stretch the spring. Accordingly, we have the same conditions as before except that x2 = d − 0.5 m.

E1 = E2 → 1 2

1 2

mv12 + mgy1 + 12 kx12 = 12 mv22 + mgy2 + 12 kx22 →

kx12 = mgd sin  + 12 k ( d − 0.5 m )

This is a quadratic relation in d. Solving it gives d = 0.66 m . (c) The block now moves d = 0.50 m, and stops at the equilibrium point of the spring. Accordingly, x2 = 0. Apply the method of Section 8–6.

263

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

)

(

Instructor Solutions Manual

)

K + U + Ffr l = 12 m v22 − v12 + 12 k x22 − x12 + mg ( y2 − y1 ) +  k mgd cos  →

k =

− 12 kx12 + mgd sin  − mgd cos 

kx12 2mgd cos 

− tan 

( 75 N m )( 0.50 m ) = − tan 41 = 0.40 2 ( 2.0 kg ) ( 9.80 m s 2 ) ( 0.50 m ) cos 41 2

39. Because friction does work, Eq. 8–15a applies. (a) The spring is initially uncompressed, so x1 = 0. The block is stopped at the maximum compression, so v2 = 0. We are given v1 = 1.6 m s. K + U = WNC = − Ffr l → 1 2

1 2

(

)

(

)

m v22 − v12 + 12 k x22 − x12 = −mg  k ( x2 − x1 ) →

kx22 + mg  k x2 − 12 mv12 = 0 →

x2 =

− mg  k 

( mg k ) − 4 ( 12 k ) ( − 12 mv12 ) −mg k  ( mg k )2 + kmv12 = k 2 ( 12 k )

  −1  1 +  2 k  ( mg k )  ( 2.0 kg ) ( 9.80 m s 2 ) ( 0.30 )  (120 N m )( 2.0 kg )(1.6 m s ) 2  = −1  1 + 2 2 2  (120 N m ) 2.0 kg ) ( 9.80 m s 2 ) ( 0.30 )  (  

mg  k 

kmv12

= 0.1633m  0.16 m

(b) To remain at the compressed position with the minimum coefficient of static friction, the magnitude of the force exerted by the spring must be the same as the magnitude of the maximum force of static friction. kx (120 N m )( 0.1633m ) = 0.9998  1.0 kx2 = s mg → s = 2 = mg ( 2.0 kg ) ( 9.80 m s2 ) (c) If static friction is not large enough to hold the block in place, the spring will push the block back towards the equilibrium position. The block will detach from the decompressing spring at the equilibrium position because at that point the spring will begin to slow down while the block continues moving. Use Eq. 8–15a to relate the block at the maximum compression position to the equilibrium position. The block is initially at rest, so v1 = 0. The spring is relaxed at the equilibrium position, so x2 = 0. K + U = WNC = − Ffr l → 1 2

1 2

(

)

(

)

m v22 − v12 + 12 k x22 − x12 = −mg  k ( x2 − x1 ) = 0 →

mv22 − 12 kx12 + mg  k x1 = 0 →

v2 =

k m

x12 − 2 g  k x1 =

(120 N m ) 2 ( 0.1633 m ) − 2 ( 9.80 m s 2 ) ( 0.30 )( 0.1633 m ) ( 2.0 kg )

= 0.7999 m s  0.80 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

264

Chapter 8

Conservation of Energy

40. (a) If there is no air resistance, then conservation of mechanical energy can be used. Subscript 1 represents the glider when at launch, and subscript 2 represents the glider at landing. The landing location is the zero location for elastic potential energy (y = 0). We have y1 = 3200 m,

 1m s   = 133.3 m s . Solve for v 2 .  3.6 km h  E1 = E2 → 12 mv12 + mgy1 = 12 mv22 + mgy2 →

y2 = 0, and v1 = 480 km h 

v2 = v12 + 2 gy1 =

 3.6 km h    1m s 

(133.3 m s ) + 2 ( 9.80 m s 2 ) ( 3200 m ) = 283.7 m s  2

= 1021km h  1.0  103 km h (b) Now include the work done by the non-conservative frictional force. Consider the diagram of the glider. The distance over which the friction acts is given by 3200 m l = . Use the same subscript representations sin12

 1m s   = 58.33 m s . Use  3.6 km h 

as above, with y1 , v1 , and y2 as before, and v2 = 210 km h 

Eq. 8–15a and solve for the frictional force. K + U = WNC = − Ffr l → 12 mv12 + mgy1 = 12 mv 22 + mgy2 + Ffr l → Ff =

(

m v12 − v 22 + 2 gy1

)

( 980 kg ) (133.3 m s ) − ( 58.33 m s ) + 2 ( 9.80 m s 2 ) ( 3200 m )  2

 3200 m    sin12 

= 2454 N  2500 N

2

41. The escape velocity is given by Eq. 8–19. 2GM A 2GM B vesc = vesc = vesc = 2vesc → rA rB A B A B 2GM A rA

2GM A rA

2GM B rB

→

 2GM B  1 r 1 1 = 4  → A =  → rA rB 4  rB   rB 

= 4

42. (a) Equate the gravitational force to the expression for centripetal force, since the orbit is circular. Let M E represent the mass of the Earth. mSvS2 rS

GM E mS 2 S

→ mSvS2 =

GM E mS rS

→

1 2

mSvS2 = K =

GM E mS 2rs

(b) The potential energy is given by Eq. 8–17, U = − GM E mS rS .

265

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(c)

Instructor Solutions Manual

GM E mS K 1 2 rS = = − GM m U − 2 E S rS

43. Let r1 = rE + y1 and r2 = rE + y2 . Note that the difference in the two distances from the center of the Earth, r2 − r1 , is the same as the height change in the two positions, y2 − y1. Also, if the two distances are both near the surface of the Earth, then r1r2  rE2 .

 GmM E   GmM E  GmM E GmM E  1 1  GmM E −− = − = GmM E  −  = ( r2 − r1 )   r2   r1  r1 r2 r1r2   r1 r2 

U =  − 

GmM E 2 E

( y2 − y1 ) = m

GM E rE2

( y2 − y1 ) = mg ( y2 − y1 )

44. The escape velocity for an object located a distance r from a mass M is given by Eq. 8–19, vesc =

2GM r

. The orbit speed for an object located a distance r from a mass M is vorb = 2GM Sun

(a) vesc at =

rSun

Sun’s surface

v Earth

rSun

Earth orbit

7.0  10 m 8

2rEarth orbit rSun

(

2 1.50  1011 m 7.0  10 m 8

) = 6.2 10 m s 5

) = 20.7  21

2GM Sun rEarth orbit

(

)(

2 2.0  1030 kg 6.67  10−11 N m 2 kg 2 1.50  10 m 11

) = 4.2  10 m s 4

2GM Sun

vesc at

orbit

)(

rEarth orbit

(b) vesc at =

v Earth

GM Sun

orbit

Earth orbit

(

2GM Sun

vesc at Sun’s surface

2 2.0  1030 kg 6.67  10 −11 N m 2 kg 2

rEarth orbit GM Sun

2 →

vesc at =

2 vEarth

Earth orbit

rEarth orbit

orbit

Since vesc at  1.4v Earth , the Earth does not escape its orbit around the Sun. Earth orbit

orbit

45. (a) The potential energy is given by Eq. 8–17.

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UA = −

GmM E

( 6.67  10 N m kg ) ( 950 kg ) ( 5.98  10 kg ) =− ( 6.38  10 m + 4.80  10 m ) −11

= −3.3893  1010 J  −3.4  1010 J UB = −

GmM E rB

( 6.67  10 N m kg ) ( 950 kg ) ( 5.98  10 kg ) =− ( 6.38  10 m + 1.26  10 m ) −11

= −1.9964  1010 J  −2.0  1010 J

(b) An expression for the kinetic energy is found by equating the gravitational force to the expression for centripetal force, since the satellites are in circular orbits. GmM E mv 2 GmM E = → 12 mv 2 = K = = − 12 U 2 r r 2r GmM E = − 12 −3.3893  1010 J = 1.6947  1010 J  1.7  1010 J KA = 2rA GmM E

KB =

(

)

(

)

= − 12 −1.9964  1010 J = 0.9982  1010 J  1.0  1010 J

2rB (c) We use the work–energy principle to calculate the work done to change the orbit. WNet = K = Worbit + Wgravity = Worbit − U gravity → Worbit = K + U gravity → change

change

Worbit = K + U gravity = ( K B − K A ) + (U B − U A ) = ( − 12 U B + 12 U A ) + (U B − U A ) change

(

)

= 12 (U B − U A ) = 12 −1.9964  1010 J − −3.3893  1010 J = 6.96  109 J  7  109 J

46. Since air friction is to be ignored, the mechanical energy will be conserved. Subscript 1 represents the asteroid at high altitude, and subscript 2 represents the asteroid at the Earth’s surface. We have v1 = 660 m s , r1 = rE + 5.0  109 m, and r2 = rE . E1 = E2 →

1 2

 GM E m  1 2  GM E m  1 1 2  = 2 mv 2 +  −  → v 2 = v1 + 2GM E  −  r1  r2     r2 r1 

mv12 +  −

 2 ( 6.67  10 −11 N m 2 kg 2 )( 5.98  10 24 kg )    2 4 = ( 660 m s ) +  1 1   = 1.12  10 m s −   6 6 9  6.38  10 m 6.38  10 m + 5.0  10 m   47. The speed of the surface of the Earth at the equator (relative to the center of the Earth) is given by the following. It is an eastward velocity. Call east the x-direction, and up the y-direction. 6 2 rE 2 ( 6.38  10 m ) v= = = 464 m s T 86, 400 s The escape velocity from the Earth (relative to the center of the Earth) is given in Eq. 8–19. vesc =

2GM E rE

(

2 6.67  10−11 N m 2 kg 2

)( 5.98 10 kg ) = 11,182 m s 24

6.38  106 m

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Instructor Solutions Manual

(a) With the surface of the Earth traveling east and the rocket velocity to the east, the rocket velocity and surface velocity will add linearly to give the escape velocity. vrocket relative + 464 m s = 11,182 m s → vrocket relative = 10, 700 m s to surface of Earth

to surface of Earth

(b) With the surface of the Earth traveling east and the rocket velocity to the west, the rocket velocity will have to be higher than the nominal escape velocity. vrocket relative + 464 m s = −11,182 m s → vrocket relative = 11, 646 m s  11, 600 m s to surface of Earth

to surface of Earth

(c) When fired vertically upward, the rocket velocity and the Earth’s velocity are at right angles to each other, and so add according to the Pythagorean theorem to give the escape velocity. 2 vrocket relative + ( 464 m s ) = (11,182 m s ) 2

→ vrocket relative = 11,172 m s  11, 200 m s

to surface of Earth

48. (a) Since air friction is to be ignored, the mechanical energy will be conserved. Subscript 1 represents the rocket at launch, and subscript 2 represents the rocket at its highest altitude. We have v1 = v0 , v 2 = 0, r1 = rE , and r2 = rE + h where we take the final altitude to be a distance h above the surface of the Earth.  GmM E  1 2  GmM E   GmM E  E1 = E2 → 12 mv02 +  −  = 2 mv2 +  −  = −  → rE    rE + h   rE + h  −1

 2GM E  1 v02  h= − − 1  − rE = rE  2  rE 2GM E   rE v0   2GM E  (b) h = rE  − 1 2  rE v0 

−1

 2 ( 6.67  10−11 N m 2 kg 2 )( 5.98  1024 kg )  = ( 6.38  10 m )  − 1  = 8.0  106 m 2 6   ( 6.38  10 m ) (8350 m s )   −1

49. (a) From Eq. 8–19, the escape velocity at a distance r  rE from the center of the Earth is vesc =

2GM E r

vesc =

(b) vesc 

d vesc dr

2GM E r

r = −

= r −1/ 2 2GM E

→

d vesc dr

= − 12 r −3/ 2 2GM E = −

GM E 2r 3

( 6.67 10 N m kg )( 5.98 10 kg ) 3.2 10 m r = − ( ) 2r 2 ( 6.38  10 m ) −11

GM E

= −280 m s

The escape velocity has decreased by 280 m/s, and so is vesc = 1.12 104 m s − 280 m s = 1.09  104 m s .

50. (a) Since air friction is to be ignored, the mechanical energy will be conserved. Subscript 1 represents the meteorite at the high altitude, and subscript 2 represents the meteorite just before it hits the sand. We have v1 = 90.0 m s , r1 = rE + h = rE + 750 km , and r2 = rE . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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E1 = E2 →

1 2

 GM E m  1 2  GM E m   = 2 mv2 +  −  → rE   rE + h  

mv12 +  −

1

v 2 = v12 + 2GM E 

 rE

−

  rE + h  1

( 6.67  10−11 N m 2 kg 2 )( 5.98  10 24 kg )     2 = ( 90.0 m s ) + 2  1 1   6.38  106 m − 6.38  106 m + 7.50  105 m     = 3627.8 m s  3630 m s

(b) We use the work–energy principle, where work is done both by gravity (over a short distance) and the sand. The initial speed is 3627.8 m/s, and the final speed is 0. Wnet = WG + Wfr = mgd + Wfr = K = 12 m v 2f − vi2 →

(

)

(

)

Wfr = − 12 mv − mgd = − 12 ( 575 kg )( 3627.8 m s ) − ( 575 kg ) 9.80 m s 2 ( 3.25 m ) 2

2 i

= −3.78  109 J (c) The average force is the magnitude of the work done, divided by the distance moved in the sand. W 3.78  109 J Fsand = sand = = 1.16  109 N 3.25 m d sand (d) The work done by the sand shows up as thermal energy, so 3.78  109 J of thermal energy is produced. 51. The external work required (Wother ) is the change in the mechanical energy of the satellite. Note the following, from the work–energy principle. Wtotal = Wgravity + Wother → K = −U + Wother → Wother = K + U =  ( K + U ) = Emech Next we show that the mechanical energy of a circular orbit is given by Emech = − 12 With the condition that U = 0 at r = , the potential energy is given by U = −

GMm r

GmM E

. The r kinetic energy is found from the fact that for a circular orbit, the gravitational force is a centripetal force. 2 GmM E mvorbit GmM E GmM E 2 2 = → mvorbit = → K = 12 mvorbit = 12 2 r r r r Emech = K + U = 12

GmM E

−

GmM E

= − 12

GmM E

r r r Now calculate the change in the mechanical energy.

269

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Emech = − 12

GMm r

 

→ Emech =  − 12

Instructor Solutions Manual

GMm 

 1 GMm   GMm  −  1 GMm  =  12  − − 2   2  r  initial  r  initial  r  final final 

 GMm   GMm  GMm −  12 =   12 rE  2 rE initial  3rE final

=  12

52. Calculate the density of the shell. Use that density to calculate the potential energy due to a full sphere of radius r1 , and then subtract the potential energy due to a mass of radius r2 .

= 4 3

M full

 (r − r ) 3 1

3 2

sphere

4 3

 (r − r ) 3 1

3 2

4 3

 r13

M inner = sphere

4 3

 (r − r ) 3 1

3 2

4 3

 r23

GM full m  GM inner m   sphere sphere  = − Gm  M U shell = U full − U inner = − −− − M inner  full    r r r  sphere sphere sphere sphere 



=− =−

Gm 



(

r  43  r13 − r23

)

4 3

r − 4 3 1



 ( r13 − r23 )

4 3

  r13 r23 GmM  r  = − −    r  ( r13 − r23 ) ( r13 − r23 )   3 2

GmM r

53. (a) With the condition that U = 0 at r = , the potential energy is given by U = −

GmM E

. The

r kinetic energy is found from the fact that for a circular orbit, the gravitational force is a centripetal force. 2 GmM E mvorbit GmM E GmM E 2 2 = → mvorbit = → K = 12 mvorbit = 12 2 r r r r

Emech = K + U = 12

GmM E r

−

GmM E r

= − 12

GmM E r 1

GmM E

K 2 r 1 (b) The ratio of kinetic energy to potential energy is: = =− . U − GmM E 2

r (c) As the value of E decreases, since E is negative, the radius r must get smaller, since it is in the denominator of the expression for E. But as the radius gets smaller, the kinetic energy 1 increases, since K  . If the total energy decreases by 1 Joule, the potential energy decreases r by 2 Joules and the kinetic energy increases by 1 Joule.

54. (a) The escape speed from the surface of the Earth is vE = 2GM E rE . The escape velocity from the gravitational field of the sun, is vS = 2GM S rSE .

In the reference frame of the Earth, if

the spacecraft leaves the surface of the Earth with speed v (assumed to be greater than the escape velocity of Earth), then the speed v at a distance far from Earth, relative to the Earth, is found from energy conservation. We assume the potential energy relative to Earth is approximately 0 at this location. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

270

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1 2

mv 2 −

GM E m 2 E

= 12 mv 2 → v 2 = v 2 −

2GM E 2 E

= v 2 − v E2 → v 2 = v E2 + v 2

The reference frame of the Earth is orbiting the Sun with speed v 0 . If the rocket is moving with speed v relative to the Earth, and the Earth is moving with speed v0 relative to the Sun, then the speed of the rocket relative to the Sun is v  + v 0 (assuming that both speeds are in the same direction). This is to be the escape velocity from the Sun, and so vS = v  + v0 , or v  = vS − v 0 . Combine this with the relationship from above.

v 2 = vE2 + v 2 = vE2 + ( vS − v0 ) vE =

vS = v0 =

2GM E

2GM S 2 rSE

v = vE2 + ( vS − v0 )

(

)( 5.98 10 kg ) = 1.118 10 m s 24

6.38  106 m

(

)(

2 6.67  10−11 N m 2 kg 2 1.99  1030 kg

rSE

→

2 6.67  10−11 N m 2 kg 2

1.496  10 m 11

(

2 1.496  1011 m

) = 4.212 10 m s 4

) = 2.978 10 m s 4

( 3.156 10 s ) v = v + ( v − v ) + = (1.118  10 m s ) + ( 4.212  10 m s − 2.978  10 m s ) TSE 2 E

= 1.665  10 4 m s  16.7 km s

(b) Calculate the kinetic energy for a 1.00 kg mass moving with a speed of 1.665  104 m s . This is the energy required per kilogram of spacecraft mass.

(

K = 12 mv 2 = 12 (1.00 kg ) 1.665 104 m s

) = 1.39 10 J 2

55. The power must exert a force equal to the weight of the elevator, moving the elevator through the vertical height, in the given time. 2 mgh ( 845 kg ) ( 9.80 m s ) ( 32.0 m ) P= = = 2.41  104 W t (11.0s ) 56. The work necessary to lift the piano is the work done by an upward force, equal in magnitude to the weight of the piano. Thus W = Fd cos 0 = mgh. The average power output required to lift the piano is the work done divided by the time to lift the piano. 2 W mgh mgh ( 335 kg ) ( 9.80 m s ) (18.0 m ) P= = → t= = = 33.8s 1750 W t t P 57. (a) K = 12 mv 2 = 12 ( 85 kg )( 5.0 m s ) = 1062.5 J  1100 J 2

(b) The power required to stop him is the change in energy of the player, divided by the time to carry out the energy change. 1062.5J P= = 1062.5 W  1100 W 1.0 s

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Instructor Solutions Manual

58. The 18 hp is the power generated by the engine in creating a force on the ground to propel the car forward. The relationship between the power and the force is Eq. 8–21 with the force and velocity in the same direction, P = F v. Thus, the force to propel the car forward is found by F = P v . If the car has a constant velocity, then the total resistive force must be of the same magnitude as the engine force, so that the net force is zero. Thus, the total resistive force is also found by F = P v . F=

P v

(18 hp )( 746 W hp )  1m s  ( 95 km h )    3.6 km h 

= 510 N

 550 ft lb s   4.45 N  1 m    = 746 N m s = 746 W   1 hp   1 lb  3.28 ft 

59. (a) 1 hp = (1 hp ) 

 1 hp   = 0.10 hp  746 W 

(b) 75 W = ( 75 W ) 

 0.746 kW   = 130 kW  1 hp 

60. The power is the force that the motor can provide times the velocity, as given in Eq. 8–21. The force provided by the motor is parallel to the velocity of the boat. The force resisting the boat will be the same magnitude as the force provided by the motor, since the boat is not accelerating, but in the opposite direction to the velocity. P ( 35 hp )( 746 W 1 hp ) P = F v = Fv → F = = = 2686 N  2700 N v  1m s  ( 35 km h )    3.6 km h  So the force resisting the boat is 2700 N, opposing the velocity . 61. The work done in accelerating the shot put is given by its change in kinetic energy: The power is the energy change per unit time. P=

K 2 − K1 t

1 2

(

m v22 − v12 t

) = ( 7.3 kg ) (14 m s ) − 0 = 476.9 W  480 W 2

1 2

1.5 s

62. The energy transfer from the engine must replace the lost kinetic energy. From the two speeds, calculate the average rate of loss in kinetic energy while in neutral.  1m s   1m s  v1 = 95 km h  v2 = 65 km h  = 26.39 m s   = 18.06 m s  3.6 km h   3.6 km h 

K = 12 mv22 − 12 mv12 = 12 ( 950 kg ) (18.06 m s ) − ( 26.39 m s )  = −1.759  105 J 2

1.759  105 J 7.0 s

(

= 2.513  10 4 W, or 2.513  10 4 W

1 hp = 33.69 hp ) 746 W

So 2.5  10 W or 34 hp is needed from the engine. 4

63. The average power is the energy transformed per unit time. The energy transformed is the change in kinetic energy of the car. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

272

Chapter 8

Conservation of Energy

  1m s   975 kg ) ( 95 km h )  ( 2 2  1 3.6 km h   energy transformed K 2 m ( v2 − v1 )   P= = = = time

2 ( 6.4 s )

= 5.3  10 4 W  71 hp 64. The minimum force needed to lift the football player vertically is equal to his weight, mg. The distance over which that force would do work would be the change in height, y = ( 83m ) sin 33. So the work done in raising the player is W = mg y and the power output required is the work done per unit time. 2 W mg y ( 88 kg ) 9.80 m s (83m ) sin 33 P= = = = 520 W = 0.70 hp 75 sec t t

(

)

65. The force to lift the water is equal to its weight, and so the work to lift the water is equal to the weight times the vertical displacement. The power is the work done per unit time. 2 W mgh ( 27.0 kg ) 9.80 m s ( 3.50 m ) P= = = = 15.4 W 60 sec t t

(

)

66. The force to lift a person is equal to the person’s weight, so the work to lift a person up a vertical distance h is equal to mgh. The work needed to lift N people is Nmgh, and so the power needed is the total work divided by the total time. We assume the mass of the average person to be 70 kg. 2 W Nmgh 45000 ( 70 kg ) 9.80 m s ( 200 m ) P= = = = 1.715  106 W  2  106 W  2300 hp. 3600 s t t

(

)

67. We represent all 35 skiers as one person on the free-body diagram. The engine must supply the pulling force. The skiers are moving with constant velocity, and so their net force must be 0.  Fy = FN − mg cos  = 0 → FN = mg cos  P

x Ffr

 F = F − mg sin  − F = 0 → x



FP = mg sin  + Ffr = mg sin  + k mg cos 



The work done by FP in pulling the skiers a distance d is FP d since the force is parallel to the displacement. Finally, the power needed is the work done divided by the time to move the skiers up the incline. W FP d mg ( sin  +  k cos  ) d = = P= t t t =

(

)

35 ( 65 kg ) 9.80 m s 2 ( sin 23 + 0.10 cos 23 )( 250 m ) 120s

68. Draw a free-body diagram for the box being dragged along the floor. The box has a constant speed, so the acceleration is 0 in all directions. Write Newton’s second law for both the x (horizontal) and y (vertical) directions.  Fy = FN − mg = 0 → FN = mg

 Fx = FP − Ffr = 0 → FP = Ffr = k FN = kmg

 1hp  = 3.0  101 hp   746 W 

= 22424 W 

Ffr

FP mg

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The work done by FP in moving the crate a distance x is given by W = FP x cos 0o = k mg x. The power required is the work done per unit time. W  k mg x x P= = =  k mg =  k mgv x = ( 0.45)( 370 kg ) 9.80 m s 2 (1.20 m s ) = 1958 W t t t 1 hp   = 2.6 hp =1958 W    746 W 

(

)

69. The net rate of work done is the power, which can be found by P = F v = mav. The velocity is given dx dv by v = = 12.0t 2 − 16.0t − 44 and a = = 24.0t − 16.0. dt dt

(

)

(a) P = mav = ( 0.28 kg )  24.0 ( 2.0 ) − 16.0 m s 2 12.0 ( 2.0 ) − 16.0 ( 2.0 ) − 44  m s 2



= −250.9 W  −250 W

(

)

(b) P = mav = ( 0.28 kg )  24.0 ( 4.0 ) − 16.0 m s 2 12.0 ( 4.0 ) − 16.0 ( 4.0 ) − 44  m s 2



= 1881.6 W  1900 W

(c) The average net power input is the work done divided by the elapsed time. The work done is the change in kinetic energy. v ( 0 ) = −44 m s , v ( 2.0 ) = −28 m s , and v ( 4.0 ) = 84 m s .

Pavg

K

0 to 2.0

Pavg 2.0 to 4.0

t K t

1 2

(

m vf2 − vi2 t 1 2

(

) = ( 0.28 kg ) ( −28 m s ) − ( −44 m s )  = −81W

m vf2 − vi2 t

1 2

2.0 s

) = ( 0.28 kg ) (84 m s ) − ( −28 m s )  = 440 W 2

1 2

2.0 s

70. First, consider a free-body diagram for the cyclist going down hill. Write Newton’s second law for the x-direction, with an acceleration of 0 since the cyclist has a constant speed.  Fx = mg sin  − Ffr = 0 → Ffr = mg sin 

Ffr

 mg

Now consider the diagram for the cyclist going up the hill. Again, write Newton’s second law for the x-direction, with an acceleration of 0.  Fx = Ffr − FP + mg sin  = 0 → FP = Ffr + mg sin 



y x FP

Ffr Assume that the friction force is the same when the speed is the same, so the friction force when going uphill is the same magnitude as when going  downhill.  mg FP = Ffr + mg sin  = 2mg sin  The power output due to this force is given by Eq. 8–21, with the force and velocity parallel. P = FP v = 2mg v sin  = 2 ( 75 kg ) 9.80 m s 2 ( 4.0 m s ) sin 6.0o = 610 W

(

)

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Chapter 8

Conservation of Energy

71. The potential energy is given by U ( x ) = 12 kx 2 and so has a parabolic shape. The total energy of the object is E = kx . The object, when released, will gain kinetic energy and lose potential energy until it reaches the equilibrium at x = 0, where it will have its maximum kinetic energy and maximum speed. Then it continues to move to the left, losing kinetic energy and gaining potential energy, until it reaches its extreme point of x = − x0 . Then the motion reverses, until the

U(x)

2 0

1 2

E K 0

object reaches its original position. Then it will continue this oscillatory motion between x = − x0 and x = x0 . 72. (a) The total energy is E = 12 kx02 = 12 (160 N m )(1.0 m ) = 8.0  101 J . 2

(b) The kinetic energy is the total energy minus the potential energy. K = E − U = E − 12 kx 2 = 80 J − 12 (160 N m )( 0.50 m ) = 60 J = 6.0  101 J The answer has 2 significant figures. 2

(c) The maximum kinetic energy is the total energy, 80 J = 8.0  101 J . (d) The maximum speed occurs at x = 0 , the equilibrium position at the center of the motion. Use the maximum kinetic energy (which is equal to the total energy) to find the maximum speed. 2 K max = 12 mvmax → vmax =

2 K max m

2 ( 80 J ) 6.0 kg

= 5.2 m s

(e) The maximum acceleration occurs at the maximum displacement, x = 1.0 m , since F = ma = − kx → a =

amax =

k xmax m

k x

. m (160 N m )(1.0 m )

6.0 kg

= 27 m s2

73. (a) The sign of the force is the opposite of the slope of the potential energy graph, since dU F =− . So when the slope is negative, the force is positive, or to the right. dx 0  x  3m ; 6 m  x  9 m (b) The magnitude of the force is a minimum when the magnitude of the slope is a minimum. So when the slope is 0, the force is 0. x = 3m, 6 m, 9 m (c) The magnitude of the force is equal to the magnitude of the slope. So the force is largest in magnitude when the slope is the steepest in magnitude. x  4m 74. (a) To find possible minima and maxima, set the first derivative of the function equal to 0 and solve for the values of r. a b 1 dU a b U ( r ) = − 6 + 12 = 12 b − ar 6 → = 6 7 − 12 13 r r r dr r r

(

)

275

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1/ 6

 2b  → rcrit =   ,   a 

=0 → = 2 13 dr r7 r The second derivative test is used to determine the actual type of critical points found. d 2U a b 1 = −42 8 + 156 14 = 14 156b − 42ar 6 2 dr r r r

(

d 2U dr

1/ 6

 2b     a 

1 14/6

 2b     a 

)

1  156b − 42a 2b  = (156b − 84b )  0   14/6 a   2b      a  1/ 6

 2b  . We also must check the endpoints of the function.   a 

Thus, there is a minimum at r =  We see from the form U ( r ) =

( b − ar ) that as r → 0, U ( r ) → , and so there is a r 6

maximum at r = 0 . (b) Solve U ( r ) = 0 for the distance.

a b 1 U ( r ) = − 6 + 12 = 12 b − ar 6 = 0 → r r r

(

)

1 r12

(

)

= 0 or b − ar 6 = 0 →

1/ 6

b r=;r=  a

1/ 6

b r=  a

(U minimum )

U (r )

1/ 6

 2b    a 

r=

(d) For E < 0, there is bound oscillatory motion between two turning points. This could represent a chemical bond type of situation. For E > 0, the motion will be unbounded, and so the atoms will not stay together.

(e) The force is the opposite of the slope of the potential energy graph. 1/6

1/ 6

 2b    a 

F  0 for r   (f) F ( r ) = −

dU dr

12b r

−

1/6

 2b   2b   ,  r   ; F = 0 for r =   , r =   a   a 

; F  0 for 

6a r7

75. The binding energy will be U (  ) − U ( rU min ) . The value of r for which U(r) has a minimum is 1/ 6

 2b  found in Problem 74 to be r =   .  a      a b   a 2 a 2b  a 2  2b    + = − U (  ) − U ( rU min ) = 0 − U  r =    = 0 −  − 0 2  − 2b + 4b 2  = 4b 2 b a     2 b           a   a   1/6

Notice that this is just the depth of the potential well. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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76. (a) The plot is included here. To find the crossing point, solve U ( r ) = 0 for r. U ( r ) = U0 2 r

−

1 r

 2 − 1 = 0 →  r 2 r 

=0 →

r=2

To find the minimum value, set

dU dr

 4 1 = U0 − 3 + 2 = 0 →  r r  dr

= 0 and solve for r.

−

4 r

1 r2

=0 →

r=4

(b) The graph is redrawn with the energy value included. The approximate turning points are indicated by the small dots. An analytic solution to the relationship U ( r ) = −0.050U 0 gives r  2.3, 17.7. The maximum kinetic energy of the particle occurs at the minimum of the potential energy, and is found from E = K + U. E = K +U → −0.050U 0 = K + U ( r = 4 ) = K + U 0 ( 162 − 14 ) → K = 81 U 0 − 0.050U 0 = 0.075U 0

77. There is nothing in the algebra that was taken from Figure 8–25 that is dependent on direction or magnitude of the initial spacecraft trajectory, as long as u = u ˆi. So it is still true in the revised

(

)

2 figure below that v0x = −v0x + 2u , and that K = 12 m 4u − 4v0x u = 2mu ( u − v0x ) . But the value of

K will be negative if v0x  u. Thus to lose kinetic energy, v0x must be in the same direction as u, and since v0x  u , the satellite will be moving in the x-direction faster than Jupiter, and thus pass in front of it.

277

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

)

Instructor Solutions Manual

78. (a) K = 12 mv 2 = 12 3.0 10−3 kg ( 3.0 m s ) = 1.35 10−2 J  1.4 10−2 J (b) K actual = 0.35Erequired → Erequired =

K actual 0.35

1.35  10−2 J 0.35

= 3.9  10−2 J

79. (a) Use conservation of mechanical energy, assuming there are no non-conservative forces. Subscript 1 represents the water at the top of the dam, and subscript 2 represents the water as it strikes the turbine blades. The level of the turbine blades is the zero location for potential energy ( y = 0 ) . Assume that the water goes over the dam with an approximate speed of 0. We have v1 = 0, y1 = 88 m, and y2 = 0. Solve for v 2 .

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 → mgy1 = 12 mv22 →

v2 = 2 gy1 = 2 ( 9.80 m s 2 ) ( 88 m ) = 41.53 m s  42 m s (b) The energy of the water at the level of the turbine blades is all kinetic energy, and so is given by 1 mv22 . 55% of that energy gets transferred to the turbine blades. The rate of energy transfer to 2 the turbine blades is the power developed by the water.

 m 2  ( 0.55 )( 320 kg s )( 41.53 m s ) = 1.5  105 W v2  = 2  t  2

P = 0.55  12

80. First, define three speeds: v0 = 12 km h = speed when coasting downhill.

v1 = 32 km h = speed when pedaling downhill.

v2 = Speed when climbing the hill. For coasting downhill at a constant speed, consider the first free-body diagram shown. The net force on the bicyclist must be 0. Write Newton’s second law for the x-direction.  Fx = Ffr 0 − mg sin  = 0 → Ffr 0 = mg sin 

Ffr0 



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Chapter 8

Conservation of Energy

Note that this occurs at v = v0 . 32 v0 = 83 v0 . Since When pumping hard downhill, the speed is v1 = 12

Ffr1

the frictional force is proportional to v 2 , the frictional force increases by a factor of ( 83 ) : Ffr1 = ( 83 ) Ffr0 = 649 mg sin  . See the second free2

FP1



 body diagram. There is a new force, FP1 , created by the bicyclist. mg Since the cyclist is moving at a constant speed, the net force in the x-direction must still be 0. Solve for FP1 , and calculate the power associated with the force.

 F = F − mg sin  − F = 0 → F = F − mg sin  = ( − 1) mg sin  = x

fr1

64 9

fr1

P1 = Ffr1v1 = 559 mg v1 sin  Now consider the cyclist going uphill. The speed of the cyclist going up the hill is v 2. Since the frictional force is proportional to v 2 , the

frictional force is given by Ffr2 = ( v2 v0 ) mg sin  . See the third 2

55 9

mg sin 

FP2

free-body diagram. There is a new force, FP2 , created by the bicyclist. Since the cyclist is moving at a constant speed, the net force in the x-direction must still be 0.  Fx = FP2 − mg sin  − Ffr 2 = 0



Ffr2 

The power output of the cyclist while pedaling uphill is the same as when pedaling going downhill. P2 = P1 = 559 mg v1 sin  → FP 2 v2 = 559 mg v1 sin  → FP 2 = 559 mg ( v1 v2 ) sin  Combine this information with Newton’s second law equation for the bicyclist going uphill. 2 FP2 − mg sin  − Ffr 2 = 559 mg ( v1 v2 ) sin  − mg sin  − ( v2 v0 ) mg sin  = 0 This simplifies to the following cubic equation: v23 + v2 v02 − 559 v1v02 = 0. Note that since every term has speed to the third power, there is no need to do unit conversions. Numerically, this equation is v23 + 144v2 − 28160 = 0, when the speed is in km/h. Solving this cubic equation (with a spreadsheet, for example) gives v2 = 28.847 km h  29 km h . 81. (a) The speed v B can be found from conservation of mechanical energy. Subscript A represents the skier at the top of the jump, and subscript B represents the skier at the end of the ramp. Point B is taken as the zero location for potential energy ( y = 0 ) . We have v1 = 0, y1 = 40.6 m, and

y2 = 0. Solve for v 2 . EA = EB →

1 2

mvA2 + mgyA = 12 mvB2 + mgyB → mgyA = 12 mvB2 →

vB = 2 g yA = 2 ( 9.80 m s 2 ) ( 40.6 m ) = 28.209 m s  28.2 m s (b) Now we use projectile motion. We take the origin of coordinates to be the point on the ground directly under the end of the ramp. Then an equation to describe the slope is yslope = − x tan 30. The equations of projectile motion can be used to find an expression for the parabolic path that the skier follows after leaving the ramp. We take “upward” to be the positive vertical direction. The initial y-velocity is 0, and the x-velocity is v B as found above.

x = vBt ; yproj = y0 − 12 gt 2 = y0 − 12 g ( x vB )

The skier lands at the intersection of the two paths, so yslope = yproj . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

yslope = yproj

 x  → − x tan 30 = y0 − g    vB 

1 2

( 2v tan 30)  ( 2v tan 30 ) + 8 gy v 2 B

2 B

Instructor Solutions Manual

(

)

→ gx 2 − x 2v B2 tan 30 − 2 y0 vB2 = 0 →

( v tan 30 )  ( v tan 30 ) + 2 gy v 2 B

2 B

= 2g g Solving this with the given values gives x = −7.09 m, 100.8 m. The positive root is taken.

Finally, s cos 30.0 = x → s =

cos 30.0 does not enter into the calculations.

100.8 m cos 30.0

= 116 m . Note that the mass of the skier

82. (a) The tension in the cord is perpendicular to the path at all times, and so the tension in the cord doesn’t do work on the ball. Only gravity does work on the ball, and so the mechanical energy of the ball is conserved. Subscript 1 represents the ball when it is horizontal, and subscript 2 represents the ball at the lowest point on its path. The lowest point on the path is the zero location for potential energy ( y = 0 ) . We have v1 = 0, y1 = l , and y2 = 0. Solve for v 2 .

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 → mg l = 12 mv22 → v2 =

2g l

(b) Use conservation of energy, to relate points 2 and 3. Point 2 is as described above. Subscript 3 represents the ball at the top of its circular path around the peg. The lowest point on the path is the zero location for potential energy ( y = 0 ) . We have v2 = 2g l , y2 = 0, and y3 = 2 ( l − h ) = 2 ( l − 0.80 l ) = 0.40 l . Solve for v3 .

E2 = E3 →

1 2

mv22 + mgy2 = 12 mv32 + mgy3 →

1 2

m ( 2 g l ) = 12 mv32 + mg ( 0.40 l ) →

v3 = 1.2 g l 83. The ball is moving in a circle of radius ( l − h ) . If the ball is to complete the circle with the string just going slack at the top of the circle, the force of gravity must supply the centripetal force at the top of the circle. This tells the critical (slowest) speed for the ball to have at the top of the circle. 2 mvcrit 2 mg = → vcrit = gr = g ( l − h ) r To find another expression for the speed, we use energy conservation. Subscript 1 refers to the ball at the launch point, and subscript 2 refers to the ball at the top of the circular path about the peg. The zero for gravitational potential energy is taken to be the lowest point of the ball’s path. Let the speed at point 2 be the critical speed found above. E1 = E2 → 12 mv12 + mgy1 = 12 mv22 + mgy2 → mg l = 12 mg ( l − h ) + 2mg ( l − h ) →

h = 0.6l If h is any smaller than this, then the ball would be moving slower than the critical speed when it reaches the top of the circular path, and would not stay in centripetal motion. 84. (a) The work done against gravity is the change in potential energy. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(

)

Wagainst = U = mg ( y2 − y1 ) = ( 75.0 kg ) 9.80 m s 2 (125m ) = 9.19  10 4 J gravity

(b) The work done by the force on the pedals in one revolution is equal to the average tangential force times the circumference of the circular path of the pedals. That work is also equal to the potential energy change of the bicycle during that revolution, assuming that the speed of the bicycle is constant. Note that a vertical rise on the incline is related to the distance along the incline by rise = distance  ( sin ) . Wpedal = Ftan 2 r = U grav = mg ( y )1 rev = mgd1 rev sin  → force

Ftan =

1 rev

mgd1 rev sin  2 r

( 75.0 kg ) ( 9.80 m s2 ) ( 5.10 m ) sin 7.50o = = 433 N 2 ( 0.180 m )

85. The spring constant for the scale can be found from the 0.60 mm compression due to the 760 N force. 760 N F k= = = 1.27  10 6 N m . −4 xc 6.0  10 m Next, use conservation of energy for the jump. Subscript 1 represents the initial location, and subscript 2 represents the location at maximum compression of the scale spring. Assume that the location of the uncompressed scale spring is the 0 location for gravitational potential energy. We have v1 = v 2 = 0 and y1 = 1.0 m. Solve for y2 , which must be negative.

E1 = E2 → y22 + 2

1 2

mv12 + mgy1 = 12 mv22 + mgy2 + 12 ky22 → mgy1 = mgy2 + 12 ky22 →

y2 − 2

y1 = y22 + 2 xc y2 − 2 xc y1 = y22 + 1.20  10−3 y2 − 1.20  10−3 = 0

k k Use the quadratic formula to solve for y2 . Use the negative root. y2 =

−1.20  10−3 

(1.20  10 ) − 4 ( −1.20  10 ) −3 2

−3

(

)(

= −3.52  10−2 m , 3.40  10−2 m

)

Fscale = k y2 = 1.27  106 N m 3.52  10−2 m = 4.5  104 N

This is almost 60 times your weight! 86. Consider the free-body diagram for the coaster at the bottom of the loop. The net force must be an upward centripetal force. 2 2 vbottom vbottom − mg = m → F = mg + m  Fbottom = Fbottom N N R R bottom Now consider the force diagram at the top of the loop. Again, the net force must be centripetal, and so must be downward. 2 2 vtop vtop F = F + mg = m → F = m − mg .  top topN N R R top

bottom

top

Assume that the speed at the top is large enough that FN  0, and so vtop  Rg . Now apply the top

conservation of mechanical energy. Subscript 1 represents the coaster at the bottom of the loop, and subscript 2 represents the coaster at the top of the loop. The level of the bottom of the loop is the zero location for potential energy ( y = 0 ) . We have y1 = 0 and y2 = 2R. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

281

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

E1 = E2 →

1 2

Instructor Solutions Manual

2 2 mv12 + mgy1 = 12 mv22 + mgy2 → vbottom = vtop + 4 gR

The difference in apparent weights is the difference in the normal forces. 2 2 2 vbottom − v top     vtop v2 − mg  = 2mg + m FN − FN =  mg + m bottom  −  m R   R R bottom top  

(

)

= 2mg + m ( 4 gR ) R = 6mg

Notice that the result does not depend on either v or R. 87. (a) The work done by the hiker against gravity is the change in gravitational potential energy.

(

)

WG = mg y = ( 65kg ) 9.80 m s2 ( 3900 m − 2800 m ) = 7.007  105 J  7.0  105 J

(b) The average power output is found by dividing the work by the time taken. W 7.007  105 J Poutput = grav = = 42.31W  42 W t ( 4.6 h )( 3600 s 1 h )

 1 hp  = 5.7  10−2 hp   746 W 

42.31W 

(c) The output power is the efficiency times the input power. Poutput 42.31W Poutput = 0.15Pinput → Pinput = = = 280 W = 0.38 hp 0.15 0.15 88. (a) We use dimensional analysis. Both sides of the equation must have units of Newtons.  N  m 3/2 → z = 3 F = Ax 3/2 →  N  =  2  m z   (b) Use Eq. 7–7, with both the force and displacement in the x-direction. 3.0 m

W =  F d l =  Ax 3/ 2 dx = 52 Ax 5/ 2 a

3.0m 0

(

)

5/ 2 = 52 10.0 N m3/ 2 ( 3.0 m ) − 0  = 62.35 J  62 J



89. (a) Draw a free-body diagram for the block at the top of the curve. Since the block is moving in a circle, the net force is centripetal. Write Newton’s second law for the block, with down as positive. If the block is to be on the verge of falling off the track, then FN = 0.

 F = F + mg = m v r → mg = m v 2

2 top

r → vtop =

Now use conservation of energy for the block. Since the track is frictionless, there are no nonconservative forces, and mechanical energy will be conserved. Subscript 1 represents the block at the release point, and subscript 2 represents the block at the top of the loop. The ground is the zero location for potential energy ( y = 0 ) . We have v1 = 0, y1 = h, v2 = g r , and y2 = 2 r. Solve for h.

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 → 0 + mgh = 12 mgr + 2mgr →

h = 2.5 r

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Chapter 8

Conservation of Energy

(b) See the free-body diagram for the block at the bottom of the loop. The net force is again centripetal, and must be upwards. 2 r  FR = FN − mg = m v 2 r → FN = mg + m vbottom

The speed at the bottom of the loop can be found from energy conservation, mg similar to what was done in part (a) above, by equating the energy at the release point (subscript 1) and the bottom of the loop (subscript 2). We now have v1 = 0, y1 = 2h = 5r , and y2 = 0. Solve for v 2 .

E1 = E2 →

1 2

2 mv12 + mgy1 = 12 mv22 + mgy2 → 0 + 5mgr = 12 mvbottom +0 →

2 2 vbottom = 10 gr → FN = mg + m vbottom r = mg + 10mg = 11mg

(c) Again we use the free body diagram for the top of the loop, but now the normal force does not vanish. We again use energy conservation, with v1 = 0, y1 = 3r , and y2 = 0. Solve for v 2 .

 F = F + mg = m v r → F = m v 2

E1 = E2 →

1 2

r − mg

2 top

2 mv12 + mgy1 = 12 mv22 + mgy2 → 0 + 3mgr = 12 mv top +0 →

2 2 vtop = 6 gr → FN = m vtop r − mg = 6mg − mg = 5mg

(d) On the flat section, there is no centripetal force, and FN = mg . 90. The energy to be stored is the power multiplied by the time: E = Pt. The energy will be stored as the gravitational potential energy increase in the water: E = U = mg y = Vg y , where  is the density of the water, and V is the volume of the water.

(180  10 W ) ( 3600 s ) = = 1.7  10 m Pt = Vg y → V =  g y (1.00  10 kg m )( 9.80 m s ) ( 380 m ) 6

91. Assume that there are no non-conservative forces doing work, so the mechanical energy of the jumper will be conserved. Subscript 1 represents the jumper at the launch point of the jump, and subscript 2 represents the jumper at the highest point. The starting height of the jump is the zero location for potential energy ( y = 0 ) . We have y1 = 0, y2 = 1.1m, and v2 = 6.5 m s . Solve for v1 .

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 →

v1 = v22 + 2 gy2 = 92. (a) F ( r ) = −

dU ( r ) dr

( 6.5 m s ) + 2 ( 9.80 m s 2 ) (1.1 m ) = 8.0 m s 2

  1 1  r  1  r  r   = − ( −U 0 )  − 02  e − r r +  −U 0 0   −  e − r r  = −U 0 0 e − r r  +  r   r0  r  r    r r0    0

 1 1  3r + r  3r0  0 0  = 2 e −2  0.03 (b) F ( 3r0 ) F ( r0 ) = 9 r0 − r r  1 1  −U 0 e + r r  r0  0 0 −U 0

− 3 r0 r0

283

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(c)

F (r) = −

dU ( r ) dr

Instructor Solutions Manual

−C

C ( 3r0 )   1  = − ( −C )  − 2   = − 2 ; F ( 3r0 ) F ( r0 ) = = 19  0.1 1 r  r   −C 2

( r0 )2

(d) The Yukawa potential is said to be “short range” because as the above examples illustrate, the Yukawa force “drops off” more quickly then the “1 over r squared” electrostatic force. The Yukawa force drops by about 97% when the distance is tripled, while the electrostatic force only drops by about 89%. 93. (a) Use conservation of energy for the swinging motion. Subscript 1 represents the student initially grabbing the rope, and subscript 2 represents the student at the top of the swing. The location where the student initially grabs the rope is the zero location for potential energy ( y = 0 ) . We have v1 = 6.0 m s , y1 = 0, and v2 = 0. Solve for y 2 .

E1 = E2 →

1 2

mv12 + mgy1 = 12 mv22 + mgy2 → v12

=h 2g Calculate the angle from the relationship in the diagram. v2 l −h h = 1− = 1− 1 → cos  = l l 2gl 1 2

mv12 = mgy2 → y2 =

2    v12  ( 6.0 m s ) −1  = cos  1 − = − = 35.28  35 cos 1    2 ( 9.80 m s 2 ) (10.0 m )   2gl    −1

(b) At the release point, the speed is 0, and so there is no radial acceleration, v2 since aR = . Thus, the centripetal force must be 0. Use the free-body r diagram to write Newton’s second law for the radial direction.  FR = FT − mg cos  = 0 →

(



FT mg 

)

FT = mg cos  = ( 56 kg ) 9.80 m s 2 cos 35.28o = 448 N  450 N (c) Write Newton’s second law for the radial direction for any angle, and solve for the tension. v2 v2 F F mg cos m F mg cos m = −  = → =  +  R T T r r As the angle decreases, the tension increases, and as the speed increases, the tension increases. Both effects are greatest at the bottom of the swing, and so the tension is greatest at the bottom. FT = mg cos 0 + m max

v12 r

(

= ( 56 kg ) 9.80 m s

( 56 kg )( 6.0 m s ) 10.0 m

= 750 N

94. Energy conservation can be used to find the speed that the water must leave the ground. We take the ground to be the 0 level for gravitational potential energy. The speed at the top will be 0. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Conservation of Energy

Eground = Etop →

1 2

(

)

2 mvground = mgytop → vground = 2 g ytop = 2 9.80 m s 2 ( 33m ) = 25.43m s

The area of the water stream times the velocity gives a volume flow rate of water. If that is multiplied by the density, then we have a mass flow rate. That is verified by dimensional analysis. Av →  m 2   m s   kg m 3  =  kg s 

Another way to think about it is that Av is the mass that flows out of the hose per second. It takes a minimum force of Av g to lift that mass, and so the work done per second to lift that mass to a height of ytop is Av gytop . That is the power required.

(

P = Av gytop =  1.5  10−2 m

) ( 25.43 m s ) (1.00 10 kg m )(9.80 m s ) (33 m ) = 5813 W 2

 5800 W or 7.8 hp 95. A free-body diagram for the car is shown. We apply conservation of energy with a frictional force as given in Eq. 8–15b. Subscript 1 refers to the car at the start of its motion, and subscript 2 refers to the car at the end of the motion. Take the ending position of the car to be the 0 for gravitational potential energy. We have v1 = 95 km h , y2 = 0, and v 2 = 35 km h . The relationship between the distance traveled along the incline and the initial height of the car is y1 = d sin  . E1 = E2 + Efr →

(

1 2

)

Ffr

FN y 



mv12 + mgy1 = 12 mv22 + mgy2 + Efr →

(

)

Efr = 12 m v12 − v22 + mgy1 = 12 m  v12 − v 22 + 2 gd sin   2  1m s   2 2   ( 95 km h ) − ( 35 km h )    = 12 (1500 kg )   3.6 km h    +2 9.80 m s 2 3.0  10 2 m sin17  )( )  ( 

(

)

= 1.7  106 J

96. (a) The work to put m1 in place is 0, because it is still infinitely distant from the other two masses. The work to put m2 in place is the potential energy of the 2-mass system, − to put m3 in place is the potential energy of the m1 − m3 combination, − potential energy of the m2 − m3 combination, − these potential energies, and so W = −

 m1m2

W = −G 

m1m3

Gm1m2 r12

−

Gm2 m3 r23 Gm1m3 r13

Gm1m2 r12

Gm1m3 r13

. The work

, and the

. The total work is the sum of all of −

Gm2 m3 r23

→

m2 m3 

 . Notice that the work is negative, which is a result of the r23  masses being gravitationally attracted towards each other. (b) This formula gives the potential energy of the entire system. Potential energy does not “belong” to a single object, but rather to the entire system of objects that interact to give the potential energy.  r12

r13

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 m1m2 m1m3 m2 m3  + +  is the binding energy of the system. It would take r13 r23   r12

that much work (a positive quantity) to separate the masses infinitely far from each other. 97. The work is the power multiplied by the time it takes to do the work, Eq. 8–20a.  746 W   3600 s   1 J/s  6 W = P t = ( 3.0 hp )  (1.0 h )    = 8.1 10 J    1hp   1hr   1 W  98. It is shown in Problem 53 that the total mechanical energy for a satellite orbiting in a circular orbit of GmM E radius r is E = − 12 . That energy must be equal to the energy of the satellite at the surface of r the Earth plus the energy required from the fuel. (a) If launched from the equator, the satellite has both kinetic and potential energy initially. The kinetic energy is from the speed of the equator of the Earth relative to the center of the Earth. In Problem 47 that speed is calculated to be 464 m/s. GmM E GmM E Esurface + Efuel = Eorbit → 12 mv02 − + Efuel = − 12 → RE r

 1

Efuel = GmM E 

−

 RE

1 

 − 2 mv0 = ( 6.67  10 2r  1

−11

)

(

N m 2 kg 2 (1225 kg ) 5.98  10 24 kg

)

   1 1 1 2 −    − 2 (1225 kg ) ( 464 m s ) 6 6 6  6.38  10 m 2 ( 6.38  10 m + 1.375  10 m )  = 4.48  1010 J

(b) If launched from the North Pole, the satellite has only potential energy initially. There is no initial velocity from the rotation of the Earth. GmM E GmM E Esurface + Efuel = Eorbit → − + Efuel = − 12 → RE r

 1

Efuel = GmM E 

 RE

−

1 

 = ( 6.67  10

−11

2r 

)

(

N m 2 kg 2 (1225 kg ) 5.98  10 24 kg

)

   1 1 10 −    = 4.51 10 J 6 6 6  6.38 10 m  +  2 6.38 10 m 1.375 10 m ( )   99. (a) Use energy conservation and equate the energies at A and B. The distance from the center of the Earth to location B is found by the Pythagorean theorem. rB =

(13, 900 km ) + ( 8230 km ) = 16,150 km

EA = EB →

1 2

 GM E m  1 2  GM E m   = 2 mvB +  −  → rA  rB   

mvA2 +  −

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2 ( 6.67  10−11 N m 2 kg 2 )( 5.98  10 24 kg )  1 1   2 vB = vA2 + 2GM E  −  = ( 8650 m s ) +   1 1  −  rB rA     7 6  1.615  10 m 8.23  10 m   = 5220 m s

(b) Use energy conservation and equate the energies at A and C. rC = 16, 460 km + 8230 km = 24, 690 km EA = EC →

1 2

 GM E m  1 2  GM E m   →  = 2 mvC +  − rA  rC   

mvA2 +  −

1 1 vC = vA2 + 2GM E  −  =  rC rA 

2 ( 6.67  10−11 N m 2 kg 2 )( 5.98  10 24 kg )    ( 8650 m s ) +   1 1  −    7 6  2.469  10 m 8.23  10 m   2

= 3190 m s

100. (a) The force is found from the potential function by Eq. 8–7. F =−

dU ( r ) dr

(

)

 r − e − r − e − r  d  GMm − r  d  e − r  = − − e  = GMm    = GMm  2 dr  r dr  r  r   

GMm

e − r (1 +  r ) r2 (b) Find the escape velocity by using conservation of energy to equate the energy at the surface of the Earth to the energy at infinity with a speed of 0. = −

ER = E → E

1 2

2 − mvesc

GMm RE

e − R = 0 + 0 → vesc = E

2GM RE

− 12  RE

Notice that this escape velocity is smaller than the Newtonian escape velocity by a factor of − 1  RE

e 2

101. A point of stable equilibrium will have equilibrium function. a U ( x ) = + bx x

dU dx

=−

a x

dU dx

= 0 and

+ b = 0 → x2 =

d 2U dx 2 a b

 0, indicating a minimum in the potential

→ x= a b

But since the problem restricts us to x  0, the point must be x = d 2U dx

2 x= a b

2a x

3 x= a b

( a b)

3/ 2

2 ab3 / 2

a b.

 0, and so the point x = a b gives a minimum in the

potential energy function. 102. (a) The work done by gravity as the elevator falls is the opposite of the change in gravitational potential energy. We take the initial position of the spring to be the 0 location for gravitational potential energy.

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(

)

Wgrav = −U grav = U1 − U 2 = mg ( y1 − y2 ) = ( 975 kg ) 9.8 m s 2 ( 28.5 m ) = 2.723  105 J  2.72  105 J Gravity is the only force doing work on the elevator as it falls (ignoring friction), so this result is also the net work done on the elevator as it falls. (b) The net work done on the elevator is equal to its change in kinetic energy. The net work done just before striking the spring is the work done by gravity found above. WG = K 2 − K1 → mg ( y1 − y2 ) = 12 mv22 − 0 →

v2 = 2 g ( y1 − y2 ) = 2 ( 9.80 m s 2 ) ( 28.5 m ) = 23.63 m s  23.6 m s (c) Use conservation of energy. Subscript 1 represents the elevator just before striking the spring, and subscript 2 represents the elevator at the bottom of its motion. The level of the elevator just before striking the spring is the zero location for both gravitational potential energy and elastic potential energy. We have v1 = 23.63 m s , y1 = 0, and v2 = 0. We assume that y2  0.

E1 = E2 → mg

1 2

mv12 + mgy1 + 12 ky12 = 12 mv22 + mgy2 + 12 ky22 →

−

2mg



4m 2 g 2

mv12

k2 k k k 2 We must choose the negative root so that y 2 is negative. Thus, y22 + 2

y2 −

v12 = 0 → y2 =

− mg  m 2 g 2 + mk v12

( 975 kg ) 2 ( 9.80 m s 2 )2  2  − ( 975 kg ) ( 9.80 m s ) −   + ( 975 kg ) ( 8.00  104 N m ) ( 23.63m s ) 2    y = 2

8.00  104 N m

k 1/2

= −2.73m

So the spring compresses by 2.73 m.

103. We approximate that the Earth and the Moon are the same distance from the Sun, and so ignore the potential energy due to the Sun. We do consider the potential energy due to both the Earth and the Moon. Before the launch, the spacecraft is at the surface of the Earth, and after landing, it is at the surface of the Moon. The symbol rE-M is being used to represent the distance between the center of the Earth and the center of the Moon. RE is the radius of the Earth, and RM is the radius of the Moon.

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U launch U land

−

U launch rel. + U launch rel. =

to Earth

to Moon

U land rel. + U land rel. to Earth

to Moon

5.98  1024 kg =

6.38  103 km

−

GmM M

rE-M − RE R rE-M − RE = E GmM M ME M − − + M rE-M − RM RM rE-M − RM RM RE GmM E

7.35  1022 kg



   7.35  1022 kg  +  from num. & denom.  3 1022 kg

( 384 − 6.38)  103 km  factor out 3 10 km

5.98  1024 kg

( 384 − 1.74 )  103 km 598

GmM E

1.74  10 km

7.35

598 7.35 + − 384 6.38 ( ) 6.38 377.62 = = = 16.196  16.2  larger 598 7.35 598 7.35 + + ( 384 − 1.74 ) 1.74 382.26 1.74 If only the two terms involving the mass of the bodies and the radii of the bodies were used (thus, approximating the Earth and Moon as an infinite distance apart), then the answer would be 22.2. So that part of the calculation was significant. 6.38

12 6    13   7    F0     104. (a) U = −  Fdr + C = −  F0  2   −    dr + C =   −    + C 6  r  r  r r 

(b) The equilibrium distance occurs at the location where the force is 0.

   13   7  F = F0  2   −    = 0 → r0 = 21/ 6  = 21/ 6 ( 3.50  10−11 m ) = 3.93  10 −11 m   r0   r0   (c) In order to draw the graphs in terms of r0 , and to scale them to the given constants, the functions have been parameterized as follows.

   13  r 13   7  r 7     13   7  F ( r ) = F0  2   −    = F0  2    0  −    0   →  r r    r0   r   r0   r   F (r) F0

   13  r  −13   7  r  −7  = 2     −       r0   r0     r0   r0 

12 −12 6 −6 6    r      F0     r   U (r) =     −      →   −    = 6  r  6  r0   r0  r  r0   r0   

F0   

12 −12 6 −6 6    r      1     r   =   −    =     −      F0 6  r   r   6  r0   r0   r0   r0  

U (r)

1   

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290

CHAPTER 9: Linear Momentum Responses to Questions 1.

For momentum to be conserved, the system under analysis must be “closed”–not have any forces on it from outside the system. A coasting car has air friction and road friction on it, for example, which are “outside” or “external” forces and thus reduce the momentum of the car. If the ground and the air were considered part of the system, and their velocities analyzed, then the momentum of the entire system would be conserved, but not necessarily the momentum of any single component, such as the car.

The heavy object will have the greater momentum. The momentum of an object can be expressed in terms of its kinetic energy, as follows.

(

)

(

)

p = mv = m2 v 2 = m mv 2 = 2m 12 mv 2 = 2mK . Thus, if two objects have the same kinetic energy, the one with more mass has the greater momentum. 3.

Consider this problem as a very light object hitting and sticking to a very heavy object. The large object–small object combination (Earth + jumper) would have some momentum after the collision, but due to the very large mass of the Earth, the velocity of the combination is so small that it is not measurable. Thus, the jumper lands on the Earth and nothing more (measurable) happens. From the view of the person alone, the momentum of the person is changed (to zero) by the force of the ground acting on the person.

When you release an inflated but untied balloon at rest, the gas inside the balloon (at high pressure) rushes out the open end of the balloon. That escaping gas and the balloon form a closed system, and so the momentum of the system is conserved. The balloon and remaining gas acquire a momentum equal and opposite to the momentum of the escaping gas, and so move in the opposite direction to the escaping gas.

As the fish swishes its tail back and forth, it moves some water backward, away from the fish. The tail doesn’t move exactly perpendicularly to the body of the fish. If we consider the system to be the fish and the water, then, from conservation of momentum, since the water moves backwards, the fish must move forward.

The air bag greatly increases the amount of time over which the stopping force acts on the driver. If a hard object like a steering wheel or windshield is what stops the driver, then a large force is exerted over a very short time. If a soft object like an air bag stops the driver, then a much smaller force is exerted over a much longer time. For instance, if the air bag is able to increase the time of stopping by a factor of 10, then the average force on the person will be decreased by a factor of 10. This can greatly reduce the possibility of serious injury or death.

Yes. In a perfectly elastic collision, kinetic energy is conserved. In the Earth–ball system, the kinetic energy of the Earth after the collision is negligible, so the ball has the same kinetic energy leaving the floor as it had hitting the floor. The height from which the ball is released determines its potential energy, which is converted to kinetic energy as the ball falls. If it leaves the floor with this same amount of kinetic energy and a velocity upward, it will rise to the same height as it originally had as the kinetic energy is converted back into potential energy.

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Consider the boy and rowboat to be a system of objects. Initially, before diving, the momentum, of the system is 0. The forces involved in the diving motion are internal to the system, and so momentum must be conserved. Thus, when the boy dives off the back of the rowboat, the boat will move forward. The boy and the boat will have equal but opposite momenta.

He could have thrown the coins in the direction opposite the shore he was trying to reach. Since the ice was frictionless, momentum would have been conserved in the throwing motion and he would have “recoiled” from the throw with a momentum equal in magnitude and opposite in direction to that of the coins. Since his mass was greater than the mass of the coins, his speed would have been less than the speed of the coins, but, since there was no friction, he would have maintained this small speed until he reached the shore.

10. When the tennis ball rebounds from a stationary racket, it reverses its component of velocity perpendicular to the racket with very little energy loss. If the ball is struck straight on, with the racket moving forward, the ball can be returned with an energy (and a speed) equal to (or even greater than) the energy it had when it was served. In an elastic collision the magnitude of the relative velocity between the ball and the racket is the same after the collision as it was before. 11

Yes. Impulse is the product of the force and the time over which it acts. A small force acting over a longer time could impart a greater impulse than a large force acting over a shorter time.

12. If the direction of the force changes during the time interval, it can give a zero impulse, because impulse is a vector quantity. For example, a spring force would give a zero impulse to an attached mass over one period of oscillation. 13. The collision in which the two cars rebound would be more damaging. In the case of the cars rebounding, the change in momentum of each car is greater than in the case in which they stick together, because each car is not only brought to rest but also sent back in the direction from which it came. A greater impulse results from a greater force, and so most likely more damage would occur. 14. (a) The momentum of the ball is not conserved during any part of the process, because there is an external force acting on the ball at all times–the downward force of gravity. And there is an upward external force on the ball during the collision. When considering the ball as the system, there are always external forces on it, and so its momentum is not conserved. (b) With this definition of the system, all of the forces are internal, and so the momentum of the Earth–ball system is conserved during the entire process. (c) For a piece of putty falling and sticking to a steel plate, if the system is the putty, the plate, and the Earth, then momentum is conserved for the entire path. 15. “Crumple zones” are similar to air bags in that they increase the time of interaction during a collision, and therefore lower the average force required for the change in momentum that the car undergoes in the collision. The safety of the passengers is increased by such devices. 16. For maximum power, the turbine blades should be designed so that the water rebounds. The water has a greater change in momentum if it rebounds than if it just stops at the turbine blade. If the water has a greater change in momentum, then, by conservation of momentum, the turbine blades also have a greater change in momentum, and will therefore spin faster. 17. (a) The direction of the change in momentum of the ball is perpendicular to the wall and away from it, or outward. It is to the left in the figure.

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(b) The direction of the force on the ball is the same as the direction of its change in momentum. Therefore, by Newton’s third law, the direction of the force on the wall will be perpendicular to the wall and towards it, or inward. It is to the right in the figure. 18. With the spring stretched, the system of two blocks and spring has elastic potential energy. When the blocks are released, the spring pulls them back together, converting the potential energy into kinetic energy. The blocks will continue past the equilibrium position and compress the spring, eventually coming to rest as the kinetic energy changes back into potential energy. If no thermal energy is lost, the blocks will continue to oscillate. The center of mass of the system will stay stationary. Since momentum is conserved and the blocks started at rest, m1v1 = − m2 v 2 at all times, if we assume a massless spring. 19. From Eq. 9–8 for a 1-D elastic collision, v A − v B = v B − v A . Let A represent the bat, and let B represent the ball. The positive direction will be the (assumed horizontal) direction that the bat is moving when the ball is hit. We assume the batter can swing the bat with equal strength in either case, so that v A is the same in both pitching situations. Because the bat is so much heavier than the ball, we assume that v A  v A –the speed of the bat doesn’t change significantly during the collision. Then the velocity of the baseball after being hit is v B = v A + v A − v B  2v A − v B . If vB = 0, the ball tossed up into the air by the batter, then v B  2v A –the ball moves away with twice the speed of the bat. But if v B  0 , the pitched ball situation, we see that the magnitude of v B  2v A , and so the ball moves away with greater speed. If, for example, the pitching speed of the ball was about twice the speed at which the batter could swing the bat, then we would have v B  4v A . Thus the ball has greater speed after being struck, and so the ball will travel farther after being hit. This is similar to the “gravitational slingshot” effect discussed in Problem 100. 20. Passengers may be told to sit in certain seats in order to have the center of mass in a specific location. If they move during the flight, they could change the position of the center of mass of the plane and affect its behavior and response in flight. And for a relatively small plane, such as a 20passenger plane, each person will affect the location of the center of mass more than they would in a very large jetliner. 21. In order to maintain balance, your CM must be located directly above your feet. If you have a heavy load in your arms, your CM will be out in front of your body and not above your feet. So you lean backwards to get your CM directly above your feet, otherwise you might fall over forwards. 22. The 1-m length of pipe is uniform–it has the same density and size throughout, and so its CM is at its geometric center, which is its midpoint. The arm and leg are not uniform–they are more dense where there is muscle, primarily in the parts that are closest to the body. Thus the CM of the arm or leg is closer to the body than its geometric center. The CM is located closer to the more massive part of the arm or leg. For instance, the thigh is typically more massive than the calf, so the CM of a leg will be higher than the midpoint of the leg. 23. Draw a line from each vertex to the midpoint of the opposite side. The center of mass will be the point at which these lines intersect. Drawing any two of those lines is sufficient to locate the center of mass. See the figure.

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24. Far out in space there are no external forces acting on the rocket, so its momentum is conserved. When a rocket expels gas in a given direction, it puts a force on that gas. The momentum of the gas– rocket system stays constant, and so if the gas is pushed to the left, the rocket will be pushed to the right due to Newton’s third law. So the rocket must carry some kind of material to be ejected (it could be exhaust from some kind of engine, or it could be compressed gas) in order to change direction. 25. Consider Bob, Jim, and the rope as a system. The center of mass of the system is closer to Bob, because he has more mass. Because there is no net external force on the system, the center of mass will stay stationary. As the two men pull hand-over-hand on the rope they will move toward each other, eventually colliding at the center of mass. Since the CM is on Bob’s side of the midline, Jim will cross the midline and lose. 26. If there were only two particles involved in the decay, then by conservation of momentum, the momenta of the particles would have to be equal in magnitude and opposite in direction (assuming the original particle was at rest), so that the momenta of the particles would be required to lie along a line. If the momenta of the recoil nucleus and the electron do not lie along a line, then some other particle (the neutrino) must have some of the momentum. 27. When you are lying flat on the floor, your CM is inside of the volume of your body, approximately half-way between the head and the feet. When you sit up on the floor with your legs extended, your CM is outside of the volume of your body. The CM is higher when you sit up, and is slightly in front of your midsection. Lying down

CM is within the body, approximately half-way between the head and feet.

Sitting up

CM is outside the body.

28. When you stand next to a door in the position described, your center of mass is approximately over your heels. If you try to stand on your toes, your center of mass will not be over your area of support, and you will go back down on your heels. 29. The engine does not directly accelerate the car. The engine puts a force on the driving wheels, making them rotate. The wheels then push backwards on the roadway as they spin. The Newton’s third law reaction to this force is the forward-pushing of the roadway on the wheels, which accelerates the car. So it is the (external) road surface that literally accelerates the car. A car suspended over the roadway could not make itself move via the engine’s internal force. 30. The motion of the center of mass of the rocket will follow the original parabolic path after the explosion. Each individual piece of the rocket will follow a separate path after the explosion, but since the explosion was internal to the system (consisting of the rocket), the center of mass of all the exploded pieces will follow the original path (assuming that air friction is ignored).

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31. The ball that rebounds off the cylinder will give the cylinder a larger impulse and will be more likely to knock it over. The ball that rebounds has a larger change of momentum, which means the cylinder will also have a larger momentum change.

Responses to MisConceptual Questions 1.

(d) The girl moves in the opposite direction at 2.0 m/s. Since there are no external forces on the pair, momentum is conserved. The initial momentum of the system (boy and girl) is zero. The final momentum of the girl must be the same in magnitude and opposite in direction to the final momentum of the boy, so that the net final momentum is also zero.

(b) A common misconception in this problem is the belief that since the sand is dropped onto the boat, it does not exert a force on the boat, and therefore does not accelerate the boat. However, when dropped, the sand has no initial horizontal velocity. For the sand to be at rest on the deck of the boat it must be accelerated from rest to the final speed of the boat. This acceleration is provided by boat on the sand. By Newton’s third law, the sand exerts an equal but opposite force on the boat, which will cause the boat to slow down to the common final speed.

(a) Since the net momentum of the astronaut and wrench is zero, the only way for the astronaut to move toward the space station is for the wrench to move away from the station. If the astronaut throws the wrench in any other direction, the astronaut will move away from the wrench, but not toward the station. If the astronaut throws the wrench toward the station, but does not let go of it, neither the wrench nor the astronaut will move.

(a) Since the asteroid ends up in the shuttle storage bay, the asteroid and shuttle have the same final speed. This is a completely inelastic collision, so only momentum is conserved.

(c) This problem requires the student to understand the vector nature of momentum. The ball initially has a momentum toward the batter. If the ball is stopped by the catcher, the change in momentum has the same magnitude as the initial momentum. If the ball is hit straight back to the pitcher, the magnitude of the change in momentum is equal to twice the initial momentum. If the ball is hit straight up at the same speed, the change in momentum has a horizontal and a vertical vector component with the magnitude of each component equal to the initial momentum. Since the two components are perpendicular to each other, the magnitude of the change in momentum will be less than the sum of their magnitudes. As such, the greatest change in momentum occurs when the ball is hit straight back toward the pitcher.

(a) Students may consider the bouncy ball and clay to have the same momentum and as such, would be equally effective. However, since the clay and ball interact with the door differently, this is incorrect. The clay sticks to the door exerting an impulse on the door equal to the clay’s momentum. The ball bounces off of the door, exerting an impulse about equal to twice its momentum. Since the ball imparts a greater impulse to the door, it will be more effective.

(d) To solve this question a student should understand the relationships between force, time, momentum, work, and kinetic energy. Impulse is the product of the force and the time over which the force acts. For an object starting at rest, the impulse is also equal to the final momentum. Since the same force acts over the same time on both vehicles, they will have the same momentum. The lighter vehicle will have the greater speed and will therefore have traveled a greater distance in the same time. Since both vehicles start from rest with the same

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force acting on them, the work-energy principle shows that the vehicle that travels the greater distance will have the greater final kinetic energy. 8.

(e) Since the same force acts on both vehicles over the same distance, the work done on both vehicles is the same. From the work–energy principle both vehicles will have the same final kinetic energy. The lighter vehicle will travel the distance in a shorter amount of time, and will therefore experience a smaller impulse and have a smaller final momentum.

(c) A common misconception is that as the milk drains from the tank car and its mass decreases, the tank car’s speed increases. For the tank car’s speed to change, a horizontal force would have to act on the car. As the milk drains it falls vertically, so no horizontal force exits, and the tank car travels at constant speed. As the mass of the tank car decreases, the momentum decreases proportionately, as the milk carries its momentum with it.

10. (d) There are two common misconceptions in this situation. One idea is that since the truck has more mass, it has more momentum, and so will have a greater momentum change Alternatively, some think that since the smaller object has a greater change in speed, it will have the greater change in momentum. In the absence of external net forces, momentum is a conserved quantity. Therefore, if we assume there are no net external forces, any momentum lost by one of the vehicles is gained by the other, and the magnitude of the change in momentum is the same for both vehicles. 11. (c) The height to which the bowling ball rises depends upon the impulse exerted on the it by the putty and by the rubber ball. The putty sticks to the bowling ball and therefore continues to move forward at the new speed of the bowling ball ( v  5.0 m s ) . The rubber bounces backwards and therefore has a greater change in velocity ( v  10.0 m s ) . Since the putty and rubber have the same mass, the rubber exerts a greater impulse onto the bowling ball, causing the bowling ball to travel higher than when it is hit by the putty. 12. (b) Elastic collisions conserve both momentum and kinetic energy; inelastic collisions only conserve momentum. Thus, the two types of collisions are similar in that momentum is conserved in both. 13. (a) Ball A appears to have a constant horizontal velocity, but its vertical velocity changes from downward to upward; an upward change. Ball B also appears to have a constant horizontal velocity, but its vertical velocity changes from upward to downward; a downward change. Thus, answer (a) is the correct answer.

Solutions to Problems 1.

Momentum is defined in Eq. 9–1. We use the magnitude.

Consider the horizontal motion of the objects. The momentum in the horizontal direction will be conserved. Let A represent the car and B represent the load. The positive direction is the direction of the original motion of the car.

p = mv = ( 0.032 kg )( 8.4 m s ) = 0.2688 kg m s  0.27 kg m s

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Instructor Solutions Manual

pinitial = pfinal → mA vA + mB vB = ( mA + mB ) v  → v =

mA v A + mB vB mA + mB

( 7150 kg )(15.0 m s ) + 0 7150 kg + 3650 kg

= 9.93 m s

The tackle will be analyzed as a one-dimensional momentum-conserving situation. Let A represent the halfback and B represent the tackler. We take the direction of the halfback to be the positive direction, so v A  0 and vB  0. pinitial = pfinal → mA vA + mB vB = ( mA + mB ) v  →

v =

mA v A + mB vB

( 82 kg )( 4.4 m s ) + (110 kg )( −2.5 m s )

= 0.4469 m s  0.4 m s mA + mB 82 kg + 110 kg They will be moving it the direction that the halfback was running before the tackle.

The force on the gas can be found from its change in momentum. The speed of 1200 kg of the gas changes from rest to 4.5  10 4 m s, over the course of one second. Use Eq. 9–2, using the magnitudes of each side of the equation. p mv m Fgas = = = v = 4.5  10 4 m s (1200 kg s ) t t t

(

)

= 5.4  107 N, in the direction of the velocity of the gas. The force on the rocket is the equal and opposite reaction force (Newton’s third law) to the force on the gas, and so is 5.4  107 N in the opposite direction of the velocity of the gas . 5.

The force is the derivative of the momentum with respect to time. 2 dp d 4.8t ˆi − 8.0 j − 9.4t k F= = = 9.6t ˆi − 9.4 k N dt dt

(

)

(

)

The throwing of the package is a momentum-conserving action, if the water resistance is ignored. Let A represent the boat and child together, and let B represent the package. Choose the direction that the package is thrown as the positive direction. Apply conservation of momentum, with the initial velocity of both objects being 0. Use Eq. 9–3 in one dimension. pinitial = pfinal → ( mA + mB ) v = mA v A + mB v B = 0 → vA = −

mB v B

=−

( 5.30 kg )(10.0 m s ) = −0.803 m s ( 24.0 kg + 42.0 kg )

mA The boat and child move in the opposite direction as the thrown package, as indicated by the negative velocity.

Consider the motion in one dimension, with the positive direction being the direction of motion of the original nucleus. Let A represent the alpha particle, with a mass of 4.0 u, and let B represent the new nucleus, with a mass of 218 u. Use Eq. 9–3 for momentum conservation. pinitial = pfinal → ( mA + mB ) v = mA vA + mB vB → v A =

( mA + mB ) v − mB vB ( 222 u )( 320 m s ) − ( 218 u )( 280 m s )

= = 2500 m s mA 4.0 u Note that the masses do not have to be converted to kg, since all masses are in the same units, and a ratio of masses is what is significant. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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The change in momentum is the integral of the force, since the force is the derivative of the momentum. F=

→ p =  Fdt =

t = 2.0 s

 ( 26ˆi − 12t j ) dt = ( 26tˆi − 4t j ) 2

t =1.0 s

t = 2.0 s t =1.0 s

(

)

= 26ˆi − 28 j kg m s

Momentum will be conserved in one dimension in the explosion. Let A represent the fragment with the larger kinetic energy. Use Eq. 9–3. m v pinitial = pfinal → 0 = mA vA + mB vB → vB = − A A mB KA = 2KB →

 m v  mA v  = 2 ( mB v  ) = mB  − A A   mB  2 A

1 2

→

2 B

2 mB The fragment with the larger kinetic energy has half the mass of the other fragment.

10. Consider the motion in one dimension with the positive direction being the direction of motion of the bullet. Let A represent the bullet and B represent the block. Since there is no net force outside of the block–bullet system (like friction with the table), the momentum of the block and bullet combination is conserved. Use Eq. 9–3, and note that vB = 0. pinitial = pfinal → mA vA + mB vB = mA vA + mB vB → v B =

mA v A − mA v A mB

( 0.022 kg )( 240 m s ) − ( 0.022 kg )(130 m s )

= 1.2 m s

2.0 kg

11. The average force is the change in momentum divided by the elapsed time. Call the direction from the batter to the pitcher the positive x-direction, and call upwards the positive y-direction. The initial momentum is in the negative x-direction, and the final momentum is in the positive y-direction. The final y-velocity can be found using the height to which the ball rises, with conservation of mechanical energy during the rising motion.

Einitial = Efinal → Favg = Favg =

p t

1 2

(

v

y x

)

mvy2 = mgh → vy = 2 gh = 2 9.80 m s 2 ( 31.5 m ) = 24.85 m s

 24.85ˆj m s − ( −35.0ˆi m s )   = ( 2030ˆi + 1441ˆj) N −3  2.5 10 s   

( v − v ) = ( 0.145 kg )  t f

( 2030 N ) + (1441N ) = 2489 N  2.5  103 N ;  = tan −1 2

1441N 2030 N

= 35

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12. There are no external forces acting on the rocket and its gases, so the momentum of the combination will be conserved. Let M be the original mass of the rocket, before expelling gas, and let m be the expelled gas. The momentum is conserved separately in the x- and y-directions. p y : M vorig = ( M − m ) vfinal cos 

Instructor Solutions Manual

 pgas = m v gas

procket = ( M − m ) v final

px : 0 = − mvgas + ( M − m ) vfinal sin 

( M − m ) vfinal sin  mvgas = = tan  → ( M − m ) vfinal cos  M vorig M vorig ( 3180 kg )(125 m s )

vgas

tan  =

1750 m s

porig = M v orig

tan 35.0

= 159 kg

13. Momentum will be conserved in two dimensions. The fuel was ejected in the y-direction as seen by an observer at rest, and so the fuel had no x-component of velocity in that reference frame. px : mrocket v0 = ( mrocket − mfuel ) v x + mfuel 0 = 32 mrocket v x → v x = 32 v0 py : 0 = mfuel vfuel + ( mrocket − mfuel ) vy = 13 mrocket ( 2v0 ) + 32 mrocket vy → vy = −v0

Thus v = 32 v0 ˆi − v0 ˆj . 14. Since the neutron is initially at rest, the total momentum of the three particles after the decay must also be zero. Thus 0 = pproton + pelectron + pneutrino . Solve for any one of the three in terms of the other two: pproton = − ( pelectron + p neutrino ) . Any two vectors are always coplanar, since they can be translated so that they share initial points. So in this case, the common initial point and the two terminal points of the electron and neutrino momenta define a plane, which contains their sum. Then, since the proton momentum is just the opposite of the sum of the other two momenta, it is in the same plane. 15. Since no outside force acts on the two masses, their total momentum is conserved. mA v A = mA vA + mB vB → vB = =

mA mB 2.0

( v A − vA ) =

2.0 kg

( 4.0ˆi + 5.0ˆj − 2.0kˆ ) m s − ( −2.0ˆi + 3.0kˆ ) m s 

3.0 kg 

( 6.0ˆi + 5.0ˆj − 5.0kˆ ) m s  = ( 4.0ˆi + 3.3ˆj − 3.3kˆ ) m s

3.0 

 1m s   = 33.33m s . Thus, in one second, a  3.6 km h 

16. The air is moving with an initial speed of 120 km h 

volume of air measuring 45 m × 75 m × 33.33 m will have been brought to rest. By Newton’s third law, the average force on the building will be equal in magnitude to the force causing the change in momentum of the air. The mass of the stopped air is its volume times its density.

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p t

mv t

V v t

( 45 m )( 75 m )( 33.33 m ) (1.3 kg m 3 ) ( 33.33 m s − 0 ) 1s

= 4.9  106 N

17. (a) For the initial projectile motion, the horizontal velocity is constant. The velocity at the highest point, immediately before the explosion, is exactly that horizontal velocity, v x = v0 cos  . The explosion is an internal force, and so the momentum is conserved during the explosion. Let v 3 represent the velocity of the third fragment. p before = p after → mv0 cos  ˆi = 13 mv0 cos  ˆi + 13 mv0 cos  −ˆj + 13 mv 3 →

( )

v 3 = 2v0 cos  î + v0 cos  ˆj = 2 (116 m s ) cos 60.0î + (116 m s ) cos 60.0ˆj = (116 m s ) î + ( 58.0 m s ) ˆj

This is 130 m/s at an angle of 26.6 above the horizontal. (b) The energy released in the explosion is K after − K before . Note that v32 = ( 2v0 cos  ) + ( v0 cos  ) 2

= 5v02 cos 2  . K after − K before =  12 ( 13 m ) ( v0 cos  ) + 12 ( 13 m ) ( v0 cos  ) + 12 ( 13 m ) v32  − 12 m ( v0 cos  ) 2



(

)

= 12 m  13 v02 cos 2  + 13 v02 cos 2  + 13 5v02 cos 2   − v02 cos 2 



= 12 43 mv02 cos 2  = 23 (195 kg )(116 m s ) cos 2 60.0 = 4.37  105 J 2

18. Choose the direction from the batter to the pitcher to be the positive direction. Calculate the average force from the change in momentum of the ball. p = Favg t = mv → Favg = m

v t

 46.0 m s − ( − 31.0 m s)   = 2230 N, towards the pitcher 5.00  10−3 s  

= ( 0.145 kg ) 

19. (a) The impulse is the change in momentum. The direction of travel of the struck ball is the positive direction. p = mv = 4.5  10−2 kg ( 38 m s − 0 ) = 1.71kg m s  1.7 kg m s

(

)

(b) The average force is the impulse divided by the interaction time. p 1.71kg m s = = 490 N F= t 3.5  10−3 s 20. The impulse given the ball is the change in the ball’s momentum. From the symmetry of the problem, the vertical momentum of the ball does not change, and so there is no vertical impulse. Call the direction AWAY from the wall the positive direction for momentum perpendicular to the wall. final

(

)

= m v sin 45o − −v sin 45o  = 2mv sin 45o

p⊥ = mv⊥ − mv⊥ initial



)

= 2 6.0  10−2 kg ( 28 m s ) sin 45o = 2.4 kg m s, to the left

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21. (a) The momentum of the astronaut–space capsule combination will be conserved since the only forces are “internal” to that system. Let A represent the astronaut and B represent the space capsule, and let the direction the astronaut moves be the positive direction. Due to the choice of reference frame, vA = vB = 0. We also have v A = 2.50 m s .

pinitial = pfinal → mA vA + mB v B = 0 = mA vA + mB vB → vB = −vA

= − ( 2.50 m s )

125 kg

= −0.1420 m s  −0.14 m s mB 2200 kg The negative sign indicates that the space capsule is moving in the opposite direction to the astronaut. The actual change in speed is an increase of 0.14 m s . (b) The average force on the astronaut is the astronaut’s change in momentum, divided by the time of interaction. p m ( vA − vA ) (125 kg )( 2.50 m s − 0 ) = = = 521N Favg = t t 0.600 s 2 2 (c) K astronaut = 12 (125 kg )( 2.50 m s ) = 391J , K capsule = 12 ( 2200 kg )( −0.142 m s ) = 22 J 22. Call east the positive direction. (a) poriginal = mvoriginal = ( 95 kg )( 3.0 m s ) = 285 kg m s  290 kg m s, to the east fullback

fullback

(b) The impulse on the fullback is the change in the fullback’s momentum.



pfullback = m  vfinal

 fullback

  = ( 95 kg )( 0 − 3.0 m s ) = −285 kg m s  −290 kg m s fullback 

− vfinal

The negative sign indicates the impulse is to the west. (c) The impulse on the tackler is the opposite of the impulse on the fullback. 290 kg m s, to the east (d) The average force on the tackler is the impulse on the tackler divided by the time of interaction. p 285 kg m s F= = = 380 N, to the east t 0.75 s 23. (a) The impulse given the ball is the area under the F vs. t graph. Approximate the area as a triangle of “height” 250 N, and “width” 0.04 sec. A width slightly smaller than the base was chosen to compensate for the “inward” concavity of the force graph. p = 12 ( 250 N )( 0.04 s ) = 5 N s We could also count “boxes” under the graph, where each “box” has an “area” of ( 50 N )( 0.01s ) = 0.5 N s. There are almost 7 whole boxes and the equivalent of about 3 whole boxes in the partial boxes. 10 boxes would be about 5 N s . (b) The velocity can be found from the change in momentum. Call the positive direction the direction of the ball’s travel after being served. p 5N s p = mv = m ( vf − vi ) → vf = vi + = 0+  80 m s m 6.0  10-2 kg

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24. (a) See the adjacent graph of the force. (b) The area is trapezoidal. We estimate values rather than calculate them. J  12 ( 750 N + 50 N )( 0.0030s )

800

F (N)

600

= 1.2N s

400 200 0 0.0000

0.0005

0.0010

0.0015

0.0020

0.0025

0.0030

0.0035

t (s)

(c)

J =  Fdt =

0.0030

 740 − ( 2.3 10 ) t  dt = 740t − (1.15 10 ) t  5

(

)

0.0030s 0

= ( 740 N )( 0.0030s ) − 1.15  105 N s ( 0.0030s ) = 1.185 N s  1.2 N s 2

(d) The impulse found above is the change in the bullet’s momentum. 1.185N s J J = p = mv → m = = = 4.558  10 −3 kg  4.6 g 260 m s v (e) The momentum of the bullet–gun combination is conserved during the firing of the bullet. Use this to find the recoil speed of the gun, calling the direction of the bullet’s motion the positive direction. The momentum before firing is 0. pinitial = pfinal → 0 = mbullet vbullet − mgun vgun →

vgun =

mbullet v bullet mgun

( 4.558 10 kg ) ( 260 m s ) = 0.26 m s = −3

4.5 kg

25. (a) Since the velocity changes direction, the momentum changes. Take the final velocity to be in the positive direction. Then the initial velocity is in the negative direction. The average force is the change in momentum divided by the time. Favg =

p

 mv − ( −mv ) 

=

= 2

t t t (b) Now, instead of the actual time of interaction, use the time between collisions in order to get the average force over a long time. Favg =

p t

 mv − ( −mv ) 

=

= 2

mv t

26. (a) The impulse is the change in momentum. Take upwards to be the positive direction. The velocity just before reaching the ground is found from conservation of mechanical energy. Einitial = Efinal → mgh = 12 mv y2 →

v y =  2 gh = 2 ( 9.80 m s 2 ) ( 2.8 m ) = 7.408 m s , down J = p = m ( v f − v 0 ) = ( 68 kg ) 0 − ( −7.408 m s )  = 504 kg m s  5.0  102 kg m s, upwards

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Instructor Solutions Manual

(b) The net force on the person is the sum of the upward force from the ground, plus the downward force of gravity. Fnet = Fground − mg = ma → 2  v 2f − v02 )   ( 0 − ( −7.408 m s )  2 Fground = m ( g + a ) = m  g +  = ( 68 kg )  ( 9.80 m s ) +   2x  2 ( −0.010 m )   

= 1.9  105 N, upwards This is about 280 times the jumper’s weight. (c) We do this the same as part (b), but for the longer distance. 2  v 2f − v02   0 − ( −7.408 m s )  = ( 68 kg )  9.80 m s 2 + Fground = m  g +     2x  2 ( −0.5 m )   

(

)

(

)

= 4398 N  4000 N, upwards

This is about 6.5 times the jumper’s weight. 27. Take the upwards direction as positive. We assume t = 0 corresponds to the first water reaching the container’s bottom surface. (a) The scale reading as a function of time will be due to two components–the weight of the (stationary) water already in the pan, and the force needed to stop the falling water. The weight of the water in the pan is just the rate of mass being added to the pan, times the acceleration due to gravity, times the elapsed time. m Wwater = ( g )( t ) = ( 0.142 kg s ) 9.80 m s2 t = (1.392 N s ) t  (1.39 N s ) t t in pan The force needed to stop the falling water is the momentum change per unit time of the water p striking the pan, Fto stop = . The speed of the falling water when it reaches the pan can be t moving

(

)

water

found from energy conservation. We assume the water leaves the faucet with a speed of 0, and that there is no appreciable friction during the fall. Ewater = Ewater → mgh = 12 mv 2 → vat = − 2 gh at faucet

at pan

pan

The negative sign is because the water is moving downwards. p mfalling Fto stop = = v = ( 0.142 kg s ) 0 − − 2 9.80 m s 2 ( 2.35 m ) t t moving

( (

(

)

))

water

= 0.9637 N  0.964 N This force is constant, as the water constantly is hitting the pan. And we assume the water level is not rising. So the scale reading is the sum of these two terms.

Fscale = Fto stop + Wwater = 0.964N + (1.39 N s ) t moving water

in pan

(b) After 11.8 s, the reading is Fscale = 0.9637 + 1.392 (11.8s )  N = 17.39 N  17.4 N .

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(c) In this case, the level of the water rises over time. The height of the water in the cylinder at time t is the volume of water divided by the area of the cylinder. And the volume is the mass divided by the density of water. 1m 3   ( 0.142t ) kg   Vwater 3  in tube  1.00  10 kg  = 0.0768 t m hin = = 18.5  10 −4 m 2 Atube tube

(

)

The height that the water falls is now h = ( 2.35 − 0.0768 t ) m. Following the same analysis as above, the speed of the water when it strikes the surface of the already-fallen water is now v = − 2 gh , and so the force to stop the falling water is given by the following.

( (

(

)

Fto stop = ( 0.142 kg s ) 0 − − 2 9.80 m s 2 ( 2.35 − .0768t ) m moving water

= 0.6287

))

( 2.35 − .0768 t ) N

The scale reading is again the sum of two terms.

Fscale = Fto stop + Wwater moving water

 0.629N

in cylinder

(

= 0.6287

( 2.35 − .0768 t ) + 1.392 t ) N

( 2.35 − ( 0.0768 s ) t ) + (1.39 N s ) t

At t = 11.8 s, the scale reading is as follows.

(

)

Fscale = 0.6287  2.35 − .0768 (11.8 )  + 1.392 (11.8 ) N = 17.18 N  17.2 N

As expected, as the water level in the tube rises, it is moving slower when it is caused to stop, requiring less force to stop it. 28. Let A represent the 0.450-kg puck, and let B represent the 0.900-kg puck. The initial direction of puck A is the positive direction. We have vA = 7.40 m s and vB = 0. Use Eq. 9–8 to obtain a relationship between the velocities. vA − vB = − ( vA − vB ) → vB = vA + vA Substitute this relationship into the momentum conservation equation for the collision. mA vA + mB vB = mA vA + mB vB → mA vA = mA vA + mB ( vA + vA ) →

vA =

−0.450 kg ( mA − mB ) vA = ( 7.40 m s ) = −2.467 m s  2.47 m s ( west ) 1.350 kg ( mA + mB )

vB = vA + vA = 7.40 m s − 2.47 m s = 4.93 m s ( east ) 29. Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B represent the ball moving at 3.60 m/s in the opposite direction. Thus v A = 2.00 m s and

vB = −3.60 m s. Use Eq. 9–8 to obtain a relationship between the velocities. vA − vB = − ( vA − vB ) → vB = 5.60 m s + vA Substitute this relationship into the momentum conservation equation for the collision, noting that mA = mB.

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mA vA + mB vB = mA vA + mB vB → vA + vB = vA + vB → −1.60 m s = v A + ( v A + 5.60 m s ) → 2v A = −7.20 m s → v A = −3.60 m s vB = 5.60 m s + vA = 2.00 m s The balls exchanged velocities. This is always true for 1-D elastic collisions of equal-mass objects. 30. (a) Momentum will be conserved in one dimension. Call the direction of the first ball the positive direction. Let A represent the first ball, and B represent the second ball. We have vB = 0 and v B = 12 v A. Use Eq. 9–8 to obtain a relationship between the velocities.

vA − vB = − ( vA − vB ) → vA = − 12 v A

Substitute this relationship into the momentum conservation equation for the collision. pinitial = pfinal → mA v A + mB v B = mA v A + mB v B → mA v A = − 12 mA v A + mB 12 v A → mB = 3mA = 3 ( 0.280 kg ) = 0.840 kg

(b) The fraction of the kinetic energy given to the second ball is as follows. K B KA

1 2

mB v B2

mA vA2

= 1

3mA ( 12 vA ) mA vA2

= 0.75

31. Since the blocks move off in opposite directions but with the same speed after the collision, we know that v1 = −v2 . We also know that v 2 = 0. Apply Eq. 9–8 for 1-D elastic collisions.

v1 − v2 = − ( v1 − v2 ) → v1 = v2 − v1 = v2 − ( −v2 ) = 2v2 → v2 = 12 v1 , v1 = − 12 v1

Now use momentum conservation (Eq. 9–3) to solve for m2 .

m1v1 + m2 v2 = m1v1 + m2 v2 → m2 = m1

( v1 − v1 ) v2

= m1

( v1 + 12 v1 ) 1 2

3 v = m1 12 1 = 3m1 v 2 1

Thus m2 = 3m1 . 32. Let A represent the moving ball, and let B represent the ball initially at rest. The initial direction of the ball is the positive direction. We have vA = 5.5 m s , v B = 0, and vA = −3.8 m s . (a) Use Eq. 9–8 to obtain a relationship between the velocities. vA − vB = − ( vA − vB ) → vB = vA − vB + vA = 5.5 m s − 0 − 3.8 m s = 1.7 m s (b) Use momentum conservation to solve for the mass of the target ball. mA vA + mB vB = mA vA + mB vB →

mB = mA

( 5.5 m s − ( −3.8 m s ) ) = 1.7506 kg  1.8 kg ( vA − vA ) = ( 0.320 kg ) 1.7 m s ( vB − vB )

33. The one-dimensional stationary target elastic collision is analyzed in Example 9–8. The kinetic energy lost by the neutron is equal to the kinetic energy gained by the target particle. The fraction of kinetic energy lost is found as follows.

  2m  2  A mB  v A    KA − KA KB 2 1 mB vB   mA + mB   inital final final 2 = = = = KA

inital

1 2

mA v

2 A

mA v

inital

2 A

4mA mB

( mA + mB )

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4mA mB

( mA + mB )

4 (1.01)(1.01)

(1.01 + 1.01)2

= 1.00

All the initial kinetic energy is lost by the neutron, as expected for the case of stationary target mass equal to the incoming mass. 4 (1.01)( 2.01) 4mA mB (b) = = 0.890 2 ( mA + mB ) (1.01 + 2.01)2 (c) (d)

4mA mB

( mA + mB )

4 (1.01)(12.00 )

= 0.286

(1.01 + 12.00 )2 4 (1.01)( 208 ) 4mA mB = = 0.0192 2 ( mA + mB ) (1.01 + 208)2 2

Since the target is quite heavy, almost no kinetic energy is lost. The incoming particle “bounces off” the heavy target, much as a rubber ball bounces off a wall with approximately no loss in speed. 34. Both momentum and kinetic energy are conserved in this one-dimensional collision. We start with Eq. 9–3 (for a one-dimensional setting) and Eq. 9–8. mA vA + mB vB = mA v A + mB v B ; vA − vB = − ( v A − v B ) → v B = vA − v B + v A Insert the last result above back into the momentum conservation equation. mA vA + mB vB = mA vA + mB ( vA − vB + vA ) = ( mA + mB ) vA + mB ( vA − vB ) → mA vA + mB vB − mB ( vA − vB ) = ( mA + mB ) vA →

( mA − mB ) vA + 2mB vB = ( mA + mB ) vA

→

 mA − mB   2mB   + vB    mA + mB   mA + mB 

v A = v A 

Do a similar derivation by solving Eq. 9–8 for v A , which gives v A = v B − vA + vB .

mA vA + mB vB = mA ( vB − vA + vB ) + mB vB = mA ( −vA + vB ) + ( mA + mB ) vB → mA v A + mB v B − mA ( −v A + v B ) = ( mA + mB ) vB → 2mA vA + ( mB − mA ) vB = ( mA + mB ) vB →

 2mA

  mB − mA   + vB    mA + mB   mA + mB 

v B = v A 

35. We are given that v B = 0 and that object A keeps ¾ of the original kinetic energy. Apply momentum and energy conservation to solve for the mass ratio. We assume that v A  0. If have v A  0 and v A  v B , and if 1 2

mA mB

 1, we must

 1 , then v A  0 . Regardless of the mass values, vB  0.

mA vA2 = ( 43 ) 12 mA vA2 → vA = 





mA vA = mA vA + mB vB = mA  

3 2

mB v B2 = ( 14 ) 12 mA vA2 → v B =

vA ;

1 2



 1 mA



 2 mB

v A  + mB 



vA  →



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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

mA =  mB mA

 1 mA   →  2 mB 

mA + mB 

(

= 2 3

(

Instructor Solutions Manual

)

mA 2  3 = mB →

) = 13.9 or 0.0718 → mm = 14 or 0.072 2

36. (a) At the maximum compression of the spring, the blocks will not be moving relative to each other, and so they both have the same forward speed. All of the interaction between the blocks is internal to the mass–spring system, and so momentum conservation can be used to find that common speed. Mechanical energy is also conserved, and so with that common speed, we can find the energy stored in the spring and then the compression of the spring. Let A represent the 3.0 kg block, let B represent the 4.5 kg block, and let x represent the amount of compression of the spring. mA vA pinitial = pfinal → mA vA = ( mA + mB ) v  → v  = mA + mB

Einitial = Efinal → x=

1 2

mA vA2 = 12 ( mA + mB ) v 2 + 12 kx 2 → 1 m m

A B  mA vA2 − ( mA + mB ) v 2  = v A2 k k mA + mB

 ( 3.0 kg )( 4.5 kg ) 2 ( 8.0 m s ) = 0.37 m  ( 7.5 kg )  850 N m  

= 

(b) This is a stationary target elastic collision in one dimension, and so the results of Example 9–8 may be used.  m − mB   −1.5 kg  vA = vA  A  = ( 8.0 m s )   = −1.6 m s  7.5 kg   mA + mB 

 2mA

  6.0 kg   = ( 8.0 m s )   = 6.4 m s  7.5 kg   mA + mB 

vB = vA 

(c) Yes, the collision is elastic. All forces involved in the collision are conservative forces. 37. From the analysis in Example 9–10, the initial projectile speed is given by v = Compare the two speeds with the same masses. m+M 2 gh2 h2 v2 h2 5.2 = m = = = = v1 m + M 2.6 h1 h 1 2 gh1 m

2 →

v2 =

m+M m

2 gh .

2 v1

38. From the analysis in Example 9–10, we know the following. 1  mv  1  ( 0.028 kg )(190 m s )  h=   =  2  2g  m + M  2 9.80 m s  0.028 kg + 3.1kg  2

(

)

= 0.1476 m  0.15 m

From the diagram we see the following. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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l 2 = ( l − h ) + x2 2

x = l 2 − ( l − h) = 2

( 2.4 m ) − ( 2.4 m − 0.1476 m ) = 0.8287 m  0.8 m 2

39. Use conservation of momentum in one dimension, since the particles will separate and travel in opposite directions. Call the direction of the heavier particle’s motion the positive direction. Let A represent the heavier particle, and B represent the lighter particle. We have mA = 1.5mB , and

vA = vB = 0. pinitial = pfinal → 0 = mA vA + mB vB → vA = −

mB v B

= − 23 vB

mA The negative sign indicates direction. Since there was no mechanical energy before the explosion, the kinetic energy of the particles after the explosion must equal the energy released.

Ereleased = K A + K B = 12 mA v A2 + 12 mB v B2 = 12 (1.5mB ) ( 32 v B ) + 12 mB v B2 = 35 2

K B = Ereleased = 3 5

3 5

( 4600 J ) = 2760 J

( m v ) = K  1 2

2 B

5 3

K A = Ereleased − K B = 4600 J − 2760 J = 1840 J

Thus K A = 1800 J K B = 2800 J . 40. The impulse on the ball is its change in momentum. Call upwards the positive direction, so that the final velocity is positive, and the initial velocity is negative. The speeds immediately before and immediately after the collision can be found from conservation of energy. Take the floor to be the zero level for gravitational potential energy. 2 Falling: K bottom = U top → 12 mvdown = mghdown → vdown = − 2 ghdown Rising: K bottom = U top →

1 2

2 mv up = mghup → v up =

J = p = mv = m ( v up − vdown ) = m

(

= ( 0.021kg ) 2 9.80 m s 2

2 ghup

( 2 gh − ( − 2 gh )) = m 2 g ( h + h ) down

down

) ( 0.85 m + 1.5 m ) = 0.20 kg m s

The direction of the impulse is upwards, so the complete specification of the impulse is 0.20 kg m s , upwards . 41. We apply momentum conservation to find the velocity of ball A after the collision, and then calculate the initial and final kinetic energy to find the kinetic energy lost. Note that mA = mB . mA v A = mA v A + mB v B → v A = v A − v B = 4.3 m s − 3.0 m s = 1.3 m s

Fraction K lost =

K initial − K final K initial

1 2

(

mA vA2 − 12 mA vA2 + 12 mB vB2 1 2

mA vA2

) = v − ( v + v ) 2 A

2 A

2 B

vA2

( 4.3 m s ) − (1.3 m s ) + ( 3.0 m s )  = = 0.42 or 42% lost 2 ( 4.3 m s ) 2

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Instructor Solutions Manual

42. (a) Momentum is conserved in the one-dimensional collision. Let A represent the baseball and let B represent the brick. mA vA = mA vA + mB vB → v A =

mA vA − mB vB

( 0.144 kg )( 28.0 m s ) − ( 4.85 kg )(1.10 m s ) 0.144 kg

= −9.05 m s

So the baseball’s speed in the reverse direction is 9.05 m s . (b) K before = 12 mA vA2 = 12 ( 0.144 kg )( 28.0 m s ) = 56.4 J 2

K after = 12 mA vA2 + 12 mA vB2 = 12 ( 0.144 kg )( 9.05 m s ) + 12 ( 4.85 kg )(1.10 m s ) = 8.83 J 2

43. We apply momentum conservation and the kinetic energy statement to get two equations relating the two final velocities. That system of two equations is then solved for the two final velocities, by solving the momentum conservation condition for one of the two unknowns and substituting that into the kinetic energy relationship. The lighter ball is designated A, and the heavier ball is designated B. Note that mA = 2.0 kg , mB = 3.0 kg, and vA = 3.5 m s . mA vA = mA v A + mB vB ; ( 43 ) 12 mA vA2 = 12 mA vA2 + 12 mB vB2

m  v B = ( vA − vA ) ; ( ) mA v = mA v  + mBv  = mA v  + mB  A ( vA − vA )  mB  mB  mA

( ) v = v + 3 4

2 A

3 4

2 A

2 B

2 A

 m  m 3 m  v A2 − 2v A v A + v A2  → 0 =  1 + A  v A2 − 2 A v A v A +  A −  v A2 mB mB  mB   mB 4 

0 = 5v A2 − 14v A − 3.0625 → 0 = v A2 − 2.8v A − 0.6125 → v A = 3.004 m s or − 0.2039 m s

If v A = 3.004 m s , then v B =

mA mB

( vA − vA ) = 0.3307 m s, which is impossible, because A cannot

overtake B. Thus v A = −0.2039 m s and v B =

mA mB

( vA − vA ) = 2.469 m s.

vA = 0.20 m s to the left, vB = 2.5 m s to the right

44. The swinging motion will conserve mechanical energy. Take the zero level for gravitational potential energy to be at the bottom of the arc. For the pendulum to swing exactly to the top of the arc, the potential energy at the top of the arc must be equal to the kinetic energy at the bottom. 2 K bottom = U top → 12 ( m + M ) vbottom = ( m + M ) g ( 2l ) → vbottom = 2 g l Momentum will be conserved in the totally inelastic collision at the bottom of the arc. We assume that the pendulum does not move during the collision process.

pinitial = pfinal → mv = ( m + M ) vbottom → v =

m+M m

vbottom = 2

m+M m

45. (a) The collision is assumed to happen fast enough that the bullet–block system does not move appreciably during the actual collision. So the totally inelastic collision is described by momentum conservation. The conservation of energy (including the non-conservative work done by friction) can be used to relate the initial kinetic energy of the bullet–block system to the spring compression and the dissipated energy. Let m represent the mass of the bullet, M © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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represent the mass of the block, and x represent the distance the combination moves after the collision. m+M Collision: mv = ( m + M ) v  → v = v m After collision: 12 ( m + M ) v 2 = 12 kx 2 +  ( m + M ) gx → v  = v=

m+M

kx 2

m+M

1.000 kg

kx 2 m+M

+ 2  gx

(140 N m )( 0.050 m )2

−3

1.0  10 kg

1.000 kg

(

)

+ 2 ( 0.50 ) 9.80 m s 2 ( 0.050 m )  920 m s

K initial − K final

, where the kinetic K initial energies are calculated immediately before and after the collision. 2 2 K initial − K final 12 mv − 12 ( m + M ) v  ( m + M ) v 2 ( m + M ) v 2 = = 1− = 1− 2 1 K initial mv 2 mv 2  m + M v  2 m   m 

(b) The fraction of kinetic energy dissipated in the collision is

= 1−

m m+M

= 1−

0.0010 kg 1.00 kg

= 0.999

46. (a) For a perfectly elastic collision, Eq. 9–8 says vA − vB = − ( vA − vB ) . Substitute that into the coefficient of restitution definition. v  − vB ( v − vB ) e= A =− A =1 vB − vA vB − vA For a completely inelastic collision, v A = v B . Substitute that into the coefficient of restitution definition. v  − v B e= A =0 vB − vA (b) Let A represent the falling object and B represent the heavy steel plate. The speeds of the steel plate are v B = 0 and vB = 0, thus, e = − v A v A . Consider energy conservation during the falling or rising path. The potential energy of body A at height h is transformed into kinetic energy just before it collides with the plate. Choose down to be the positive direction. mgh = 12 mvA2 → vA = 2 gh The kinetic energy of body A immediately after the collision is transformed into potential energy as it rises. Also, since it is moving upwards, it has a negative velocity. mgh = 12 mvA2 → vA = − 2 gh Substitute the expressions for the velocities into the definition of the coefficient of restitution. − 2 gh e = − v A v A = − → e = h h 2 gh

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47. (a) See the diagram. px : mA vA = mA vA cos  A + mB vB cos  B

vA

p y : 0 = mA vA sin  A − mBvB sin  B

(b) Solve the x equation for cos  B and the y equation for sin  B , and then find the angle from the tangent function. mA vA sin  A

tan  B =

vA mB

 A

mB vB

 B

vA sin  A sin  B mB vB = = cos  B mA ( vA − vA cos  A ) ( vA − vA cos  A ) mB vB

 B = tan −1

vA sin  A ( 2.10 m s ) sin 35.0 = tan −1 = 48.1 2.80 m s − ( 2.10 m s ) cos 35.0 v A − v A cos  A

With the value of the angle, solve the y equation for the velocity. m v  sin  A ( 0.120 kg )( 2.10 m s ) sin 35.0 vB = A A = = 1.39 m s mB sin  B ( 0.140 kg ) sin 48.1 48. Use this diagram for the momenta after the decay. Since there was no momentum before the decay, the three momenta shown must add to 0 in both the x- and y-directions. ( pnucleus ) x = pneutrino ( pnucleus ) y = pelectron pnucleus =

( pnucleus ) x + ( pnucleus ) y = ( pneutrino ) + ( pelectron ) 2

p neutrino

p nucleus 

p electron

( 6.2  10 kg m s ) + ( 9.6  10 kg m s ) = 1.14  10 kg m s (p ) ( 9.6  10 kg m s ) = 57 (p )  = tan = tan = tan (p ) (p ) ( 6.2  10 kg m s ) −23

−23

electron

−1

−22

−23

−1

nucleus

−1

−23

neutrino

The second nucleus’ momentum is 147o from the electron’s momentum , and is 123o from the neutrino’s momentum . 49. (a) Let A represent the incoming nucleus, and B represent the target particle. Take the x-direction to be in the direction of the initial velocity of mass A (to the right in the diagram), and the y-direction to be up in the diagram. Momentum is conserved in two dimensions, and gives the following relationships (with mA = 2mB ). px : mA vA = mB vB cos  → v = 2vB cos 

vA mA

vA mB

mB  vB

p y : 0 = mA vA − mB vB sin  → vA = 2vB sin 

The collision is elastic, and so kinetic energy is also conserved. K : 12 mA vA2 = 12 mA vA2 + 12 mBvB2 → v 2 = vA2 + 2vB2 → v 2 − vA2 = 2vB2 Square the two momentum equations and add them together. v = 2vB cos  ; vA = 2vB sin  → v 2 = 4vB2 cos 2  ; vA2 = 4vB2 sin 2  → v 2 + vA2 = 4vB2 Add these two results together and use them in the x momentum expression to find the angle.

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v 2 − vA2 = 2vB2 ; v 2 + vA2 = 4vB2 → 2v 2 = 6vB2 → vB = cos  =

v 2vB

3 v = v 2 2 3

v 3

→  = 30

(b) From above, we already have v B = v A = 2v B sin  = 2

. Use that in the y momentum equation.

sin 30 = v A =

v 3

(c) The fraction transferred is the final energy of the target particle divided by the original kinetic energy. 2 1 1 K target mB v B2 2 ( 2mA ) v 3 2 2 = 1 = = 2 2 1 K original 2 mA v A mA v 3 2

(

)

50. Let the subscript “n” represent the incoming neutron, and let “He” represent the helium nucleus. We take mHe = 4mn . Take the x-direction to be the direction of the initial velocity of the neutron (to the right in the diagram), and the y-direction to be up in the diagram. Momentum is conserved in two dimensions, and gives the following relationships.  cos  He  → px : mn vn = mn v n cos  n + mHe vHe

v n mn

vn mHe

 n

mHe  He = 45 vHe

 cos  He  = v n cos  n v n − 4v He  sin  He  → 4v He  sin  He  = vn sin  n p y : 0 = mn v n sin  n − mHe v He

The collision is elastic, and so kinetic energy is also conserved. 2 → vn2 = vn2 + 4vHe 2 → vn2 = vn2 − 4vHe 2 K : 12 mn vn2 = 12 mn vn2 + 12 mHe vHe  , and  n . We can eliminate  n by This is a set of three equations in the three unknowns v n , v He squaring and adding the momentum equations. That can be combined with the kinetic energy equation to solve for one of the unknown speeds. 2 2 2 2 ( vn − 4vHe cos  He ) = ( vn cos  n ) ; ( 4vHe sin  He ) = ( vn sin  n ) →

 cos  He  + 16vHe 2 cos 2  He  + 16vHe 2 sin 2  He  = vn2 cos 2  n + vn2 sin 2  n → vn2 − 8vn vHe  cos  He  + 16vHe 2 = vn2 = vn2 − 4vHe 2 → vn2 − 8vn vHe

(

)

 = 0.4vn cos  He  = 0.4 6.2  105 m s cos 45 = 1.754  105 m s vHe 2 → v n2 = vn2 − 4vHe 2 = v n = v n2 − 4v He

( 6.2  10 m s ) − 4 (1.754 10 m s ) = 5.112 10 m s 5

 sin  He  = vn sin  n → 4v He

 (1.754  105 m s )    vHe  −1   = sin  4   n = sin  4 sin  He sin 45  = 76  ( 5.112  105 m s )  vn    −1

 = 1.8 105 m s ,  n = 76 . To summarize: vn = 5.1105 m s , vHe © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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51. To do this problem with only algebraic manipulations is complicated. We use a geometric approach instead. See the diagram of the geometry. Momentum conservation: mv A = mvA + mvB → v A = vA + vB

vA m

Kinetic energy conservation: 12 mvA2 = 12 mvA2 + 12 mvB2 → vA2 = vA2 + vB2 The momentum equation can be illustrated as a vector summation diagram, and the kinetic energy equation relates the magnitudes of the vectors in that summation diagram. Examination of the energy equation shows that it is identical to the Pythagorean theorem. The only way that the Pythagorean theorem can hold true is if the angle  in the diagram is a right angle. If  is a right angle, then  +  = 90, and so the angle between the final velocity vectors must be 90. 52. Let A represent the incoming neon atom, and B represent the target atom. A momentum diagram of the collision looks like the first figure. The figure can be re-drawn as a triangle, the second figure, since mA v A = mA vA + mB vB . Write the law of sines for this triangle, relating each final momentum magnitude to the initial momentum magnitude. mA vA sin  sin  = → v A = vA mA vA sin  sin 

mB v B

sin 

mA sin 

vA m

mA v = mA v  + mB v 

mB =

1 2

2 A

2 B

1 2

mA sin 2  sin 2  − sin 2 

2 A

( 20.0 u ) sin 2 55.6o

sin 2 74.4 − sin 2 50.0o

v B





 vA

vA mA

o mA  = 55.6

vA mB

mB  = 50.0o vB

mA vA



 mA sin    sin   → mA v = mA  vA  + mB  vA   sin    mB sin   2

2 A



m  vB

vA

= → v B = vA  mA vA sin  mB sin  The collision is elastic, so write the kinetic energy conservation equation, and substitute the results from above. Also note that  = 180.0 − 55.6o − 50.0o = 74.4o. 1 2

mB vB

 mA v A

→

= 39.9 u

53. There are 4 unknowns in this interaction–the three final speeds and the direction of the incident ball. But there are only 3 constraints–kinetic energy conservation, and momentum conservation in the xand y-directions. Based on the symmetry of the diagram, we will assume that the incident ball moves along the x-axis after the collision. All of the masses are identical. px : mv3 = mv3 + mv1 cos 32 + mv2 cos 32 → v3 = v3 + v1 cos 32 + v2 cos 32 p y : 0 = mv1 sin 32 − mv 2 sin 32 → 0 = v1 sin 32 − v 2 sin 32 K:

1 2

mv32 = 12 mv32 + 12 mv12 + 12 mv22 → v32 = v32 + v12 + v22

From the y-momentum conservation equation, we see that v1 = v 2 . Use that to simplify the other two equations. px : v3 = v3 + 2v1 cos 32 → v3 = v3 − 2 v1 cos 32 K : v32 = v32 + 2v12

This is a set of two equations with two unknowns. Solve for the unknowns.

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v32 = v32 + 2v12 = ( v3 − 2 v1 cos 32 ) + 2v12 = v32 − 4v3v  cos 32 + 4v12 cos 2 32 + 2v12 2

(

)

0 = −4v3v1 cos 32 + 4 cos 2 32 + 2 v12 → v1 =

4 cos 32

( 4 cos 32 + 2 ) 2

v3 =

4 cos 32

( 4 cos 32 + 2 ) 2

( 5.5 m s ) = 3.8257 m s = v2

v3 = v3 − 2 v1 cos 32 = ( 5.5 m s ) − 2 ( 3.8257 m s ) cos 32 = −0.9888 m s

Here are the final velocities: v1 = 3.8 m s , 32 above the incident direction v 2 = 3.8 m s , 32 below the incident direction v3 = 1.0 m s , opposite the incident direction

54. Choose the carbon atom as the origin of coordinates. Use Eq. 9–10. −10 m x + mO xO (12 u )( 0 ) + (16 u ) (1.13  10 m ) xCM = C C = = 6.5  10−11 m from the C atom. 12 u + 16 u mC + mO 55. Use Eq. 9–10, extended to three particles. m x + mB xB + mC xC (1.00 kg )( 0 ) + (1.50 kg )( 0.50 m ) + (1.10 kg )( 0.75 m ) xCM = A A = mA + mB + mC 1.00 kg + 1.50 kg + 1.10 kg = 0.438 m

56. Find the CM relative to the front of the Jeep. Use Eq. 9–10. m x + mfront xfront + mback xback xCM = car car mcar + mfront + mback

(1750 kg )( 2.40 m ) + 2 ( 65.0 kg )( 2.80 m ) + 3 ( 65.0 kg )( 3.90 m ) = 2.57 m 1750 kg + 5 ( 65.0 kg )

57. By the symmetry of the problem, since the centers of the cubes are along a straight line, the vertical CM coordinate will be 0, and the depth CM coordinate will be 0. The only CM coordinate to calculate is the one along the straight line joining the centers. The mass of each cube will be the 3 3 3 volume times the density, and so m1 =  ( l 0 ) , m2 =  ( 2l 0 ) , m3 =  ( 3l 0 ) . Measuring from the left edge of the smallest block, the locations of the CM’s of the individual cubes are x1 = 12 l 0 , x2 = 2 l 0 , x3 = 4.5l 0 . Use Eq. 9–10 to calculate the CM of the system. xCM =

m1 x1 + m2 x2 + m3 x3 m1 + m2 + m3

 l 30 ( 12 l 0 ) + 8  l 30 ( 2l 0 ) + 27  l 30 ( 4.5l 0 )  l + 8  l + 27  l 3 0

3 0

138 23 l0 = l0 36 6

= 3.8 l 0 from the left edge of the smallest cube

58. From the symmetry of the hydrogen equilateral triangle, and the fact that the nitrogen atom is above the center of that triangle, the center of mass will be perpendicular to the plane of the hydrogen atoms, on a line from the center of the hydrogen triangle to the nitrogen atom. We find the height of the center of mass above the triangle from the heights of the individual atoms. The masses can be expressed in any consistent units, and so atomic mass units from the periodic table will be used. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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zCM =

3mH zH + mN zN mtotal

3 (1.008 u )( 0 ) + (14.007 u )( 0.037 nm ) 3 (1.008 u ) + (14.007 u )

Instructor Solutions Manual

= 0.030 nm

And so the center of mass is 0.030 nm above the center of the hydrogen triangle. 59. Let each crate have a mass M. A top view of the pallet is shown, with the total mass of each stack listed. Take the origin to be the back left corner of the pallet. Use Eqs. 9–10, applied to both the x- and y-directions. ( 5M )( l 2 ) + ( 3M )( 3l 2 ) + ( 2M )( 5l 2 ) xCM = = 1.2 l 10 M ( 7 M )( l 2 ) + ( 2M )( 3l 2 ) + (1M )( 5l 2 ) yCM = = 0.9 l 10 M

x 3M 2M 2M y

M M M

60. Consider the following. We start with a full circle of radius 2R, with its CM at the origin. Then we draw a circle of radius R, with its CM at the coordinates ( 0.80 R, 0 ) . The full circle can now be labeled as a “shaded” part and a “white” part. The y coordinate of the CM of the entire circle, the CM of the shaded part, and the CM of the white part are all at y = 0 by the symmetry of the system. The x coordinate of the entire circle is at m x + mwhite x white xCM = 0, and can be calculated by xCM = shaded shaded . Solve mtotal this equation to solve for the CM of the “shaded” part. + mwhite xwhite m x → xCM = shaded shaded mtotal

xshaded =

mtotal xCM − mwhite xwhite

−mwhite x white

mshaded mtotal − mwhite mtotal − mwhite This is functionally the same as treating the white part of the figure as a hole of negative mass. The mass of each part can be found by multiplying the area of the part times the uniform density of the plate. −  R 2 ( 0.80 R ) − mwhite xwhite −0.80 R xshaded = = = = −0.27 R 2 mtotal − mwhite  ( 2 R ) −  R 2 3 The negative sign indicates that the CM of the “shaded” part is to the left of the center of the circle of radius 2R. 61. Consider this diagram of the cars on the raft. We are viewing from above (even though the cars don’t look like it). Notice that the origin of coordinates is located at the CM of the raft. Reference all distances to that location. The car at the NE corner has its CM at x = 9.0 m − 2.5m = 6.5m and y = 9.0 m − 1.0 m = 8.0 m. The car at the SE corner has its CM at x = 6.5m and y = −8.0 m . The car at the SW corner has its CM at x = −6.5m and y = −8.0 m .

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(1350 kg )( 6.5 m ) + (1350 kg )( 6.5 m ) + (1350 kg )( −6.5 m ) = 0.89 m 3 (1350 kg ) + 5800 kg (1350 kg )(8.0 m ) + (1350 kg )( −8.0 m ) + (1350 kg )( −8.0 m ) yCM = = −1.096 m  −1.1m 3 (1350 kg ) + 5800 kg

xCM =

62. Let M be the full mass of the wire. From the symmetry of the wire, we know that xCM = 0. Consider an infinitesimal piece of the wire, with mass dm, and coordinates ( x , y ) = ( R cos  , R sin  ) . If the length of

y dm dm dd

that piece of wire is d l , then since the wire is uniform, M d l . And from the diagram and the we have dm = R



definition of radian angle measure, we have d l = Rd . Thus dm =

R

Rd =



d . Now apply

Eq. 9–13.

yCM =



y dm = R sin  d =  sin  d = ( − cos  )0 = M M     0



 2R  .    

Thus the coordinates of the center of mass are ( xCM , yCM ) =  0,

63. Let the tip of the cone be at the origin, and the symmetry axis of the cone be vertical. From the symmetry of the cone, we know that xCM = yCM = 0, and so the center of mass lies on the z axis. We have from Eq. 9–13 that 1 zCM = z dm. The mass can be expressed as M =  dm, and so M

zCM =

 z dm . Since the object is uniform, we can express the mass as  dm

the uniform density  times the volume, for any part of the cone. That results in the following.

zCM =

 z dm =  z  dV  dm   dV

From the diagram, a disk of radius r and thickness dz has a volume of dV =  r 2dz. Finally, the geometry of the cone is such that r z = R h , and so r = zR h . Combine these relationships and integrate over the z dimension to find the center of mass. h

zCM =

 z  dV   dV

  z r 2 dz    r 2 dz

  z ( zR h ) dz 2

  ( zR h ) dz 2

 ( R h )  z 3dz

 ( R h )  z 2 dz 2

 z dz 3

0 h

 z dz 2

h4 4 h3 3

= 43 h

Thus the center of mass is at

( 0ˆi + 0ˆj + h kˆ ) . 3 4

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Instructor Solutions Manual

64. Let the peak of the pyramid be directly above the origin, and the base edges of the pyramid be parallel to the x- and yaxes. From the symmetry of the pyramid, we know that xCM = yCM = 0, and so the center of mass lies on the z axis. We have from Eq. 9–13 that zCM =

M

z dm. The mass can

be expressed as M =  dm, and so zCM =

 z dm . Since the  dm

object is uniform, we can express the mass as the uniform density  times the volume, for any part of the pyramid. That results in the following.

zCM =

 z dm =  z  dV  dm   dV

From the diagram, for the differential volume we use a square disk of side l and thickness dz, which s has a volume of dV = l 2 dz. The geometry of the pyramid is such that l = ( h − z ) . That can be h checked from the fact that l is a linear function of z, l = s for z = 0, and l = 0 for z = h. We can relate s to h by expressing the length of an edge in terms of the coordinates of the endpoints of an edge. One endpoint of each edge is at ( x =  s 2 , y =  s 2, z = 0 ) , and the other endpoint of each edge is at ( x = 0, y = 0, z = h ) . Using the Pythagorean theorem and knowing the edge length is s gives the following relationship. 2 2 s2 = ( s 2) + ( s 2) + h2 → h = s 2 We combine these relationships and integrate over the z dimension to find the center of mass. 2

zCM =

 z dm  dm

 z  dV   dV

  z l 2 dz   l 2 dz

s    z  ( h − z )  dz h  0 h

s

    ( h − z )  dz h  0 

 z ( h − z ) dz 2

= 0h

 ( h − z ) dz 2

 ( h z − 2hz + z ) dz ( h z − hz + z ) = = = h= h z − hz + z ( )  ( h − 2hz + z ) dz 2

1 2

2 3

1 4

1 3

3 h

1 4

s 2

s 4 2

 

Thus the center of mass is at  0ˆi + 0ˆj +

s 4 2

 kˆ  . 

65. (a) Find the CM relative to the center of the Earth. 5.98  1024 kg ( 0 ) + 7.35  1022 kg 3.84  108 m mEarth xEarth + mMoon xMoon xCM = = 5.98  1024 kg + 7.35  1022 kg mEarth + mMoon

(

)

(

)(

)

= 4.66  106 m from the center of the Earth This is actually inside the volume of the Earth, since REarth = 6.38  106 m. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) It is this Earth–Moon CM location that actually traces out the orbit as discussed in an earlier chapter. The Earth and Moon will orbit about this orbit path in (approximately) circular orbits. The motion of the Moon, for example, around the Sun would then be a sum of two motions: i) the motion of the Moon about the Earth–Moon CM; and ii) the motion of the Earth–Moon CM about the Sun. To an external observer, the Moon’s motion would appear to be a small radius, higher frequency circular motion (motion about the Earth–Moon CM) combined with a large radius, lower frequency circular motion (motion about the Sun). The Earth’s motion would be similar, but since the center of mass of that Earth–Moon motion is inside the Earth, the Earth would be observed to “wobble” about that CM. 66. The point that will follow a parabolic trajectory is the center of mass of the mallet. Find the CM relative to the bottom of the mallet. Each part of the mallet (handle and head) can be treated as a point mass located at the CM of the respective piece. So the CM of the handle is 12.0 cm from the bottom of the handle, and the CM of the head is 28.0 cm from the bottom of the handle. + mhead xhead ( 0.500 kg )(12.0 cm ) + ( 2.60 kg )( 28.0 cm ) m x = = 25.4 cm xCM = handle handle 3.10 kg mhandle + mhead Note that this is inside the head of the mallet. The mallet will rotate about this point as it flies through the air, giving it a wobbling kind of motion. 67. (a) Measure all distances from the original position of the woman. m x + mM xM ( 52 kg )( 0 ) + ( 72 kg )( 8.6 m ) xCM = W W = = 4.994 m mW + mM 124 kg

 5.0 m from the woman (b) Since there is no force external to the man–woman system, the CM will not move, relative to the original position of the woman. The woman’s distance will no longer be 0, and the man’s distance has changed to 8.6 m – 2.5 m = 6.1 m. m x + mM xM ( 52 kg ) xW + ( 72 kg )( 6.1m ) xCM = W W = = 4.994 m → mW + m M 124 kg xW =

( 4.994 m )(124 kg ) − ( 72 kg )( 6.1m ) 52 kg

= 3.463m

xM − xW = 6.1m − 3.463m = 2.637 m  2.6 m

(c) When the man collides with the woman, he will be at the original location of the center of mass. xM − xM = 4.994 m − 8.6 m = −3.606 m final

initial

He has moved 3.6 m from his original position. 68. (a) As in Example 9–18, the CM of the system follows the parabolic trajectory. Part I will again fall vertically, the CM will “land” a distance d from part I (as in Fig. 9–32), and part II will land a distance x to the right of the CM. We measure horizontal distances from the point underneath the explosion. x ( m + mII ) − mI xI d ( mI + 3mI ) − mI ( 0) 4 m x + mII xII → xII = CM I = = 3d xCM = I I mI + mII mII 3mI Therefore part II lands a total distance

7 3

d from the starting point.

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Instructor Solutions Manual

(b) Use a similar analysis for this case, but with mI = 3mII .

xCM =

mI xI + mII xII mI + mII

→ xII =

xCM ( mI + mII ) − mI xI mII

d ( 3mII + mII ) − 3mII ( 0 ) mII

= 4d

Therefore part II lands a total distance 5d from the starting point. 69. From Eq. 9–15, we see that v CM =

1 M

m v . i

( 35 kg ) (14î − 16ˆj) m s + ( 25 kg ) ( −25î + 14ˆj) m s ( 35 kg + 25 kg ) ( 35)(14 ) − ( 25)( 25) î kg m s + ( 35)( −16 ) + ( 25)(14 ) ˆj kg m s = ( 60 kg )

v CM =

−135ˆi kg m s − 210ˆj kg m s

( 60 kg )

= −2.3ˆi m s − 3.5ˆj m s

Note that the value of 60 kg has 2 significant figures. 70. Call the origin of coordinates the CM of the balloon, gondola, and person at rest. Since the CM is at rest, the total momentum of the system relative to the ground is 0, and the net force on the system is 0. The man climbing the rope cannot change the total momentum of the system, and so the CM must stay at rest. Call the upward direction positive. Then the velocity of the man with respect to the balloon is −v. Call the velocity of the balloon with respect to the ground v BG . Then the velocity of the man with respect to the ground is v MG = −v + v BG . Apply conservation of linear momentum in one dimension.

0 = mvMG + M vBG = m ( −v + vBG ) + M vBG →

vBG = v

m m+M

, upward

If the passenger stops, the balloon also stops, and the CM of the system remains at rest. 71. Because the interaction between the worker and the cart is internal to the worker–cart system, their total momentum will be conserved, and the center of mass of the system will move with a constant velocity relative to the ground. The velocity of the center of mass is 6.0 m/s. Once the worker starts to move, the velocity of the cart relative to the ground will be taken as v cart and the velocity of the worker relative to the ground will then be vcart + 2.0 m s. Apply Eq. 9–15, in one dimension. Letter A represents the worker, and letter B represents the cart. m v + mB vB mA ( vcart + 2.0 m s ) + mB vcart = → vCM = A A mA + mB mA + mB

( mA + mB )

( 2.0 m s ) = 6.0 m s −

110 kg

( 2.0 m s ) = 5.532 m s 470 kg The cart moves at this speed while the worker is walking. The worker walks 15 m along the cart at a relative speed of 2.0 m/s, and so he walks for 7.5 s. xcar = vcar t = ( 5.532 m s )( 7.5s ) = 41.49 m  41m vcart = vCM −

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Chapter 9

Linear Momentum

72. Use Eq. 9–19b. Call upwards the positive direction. The external force is gravity, acting downwards. The exhaust is in the negative direction, and the rate of change of mass is negative. dv dM dM  Fext = M dt − v rel dt → − Mg = Ma + vexhaust dt → vexhaust =

− Mg − Ma dM dt

−4.5Mg dM dt

(

−4.5 ( 3500 kg ) 9.80 m s 2 −27 kg s

) = 5700 m s

73. We apply Eq. 9–19b in one dimension, with “away” from the Earth as the positive direction, and “towards” the Earth as the negative direction. The external force is the force of gravity at that particular altitude, found from Eq. 6–1. dv dM M = Fext + vrel → dt dt dM dt

dv  1  d v  GM Earth M   M  d v GM Earth  − Fext  = −− + M    = M v rel  dt r2 r2   vrel  dt    vrel  dt

( 6.67  10 N  m kg )( 5.98  10 kg )  ( 25, 000 kg )  1.5 m s 2 + 2 ( −1200 m s )   ( 6.38  106 m + 6.4  106 m )

1 

−11

= 82.13 kg s  −82 kg s

The negative sign means that the mass is being ejected rather than absorbed. 74. The thrust is, in general, given as v rel

. dt (a) The mass is ejected at a rate of 4.2 kg/s, with a relative speed of 550 m/s opposite to the direction of travel. dM fuel Fthrust = v rel = ( −550 m s )( −4.2 kg s ) = 2310 N  2300 N dt fuel (b) The mass is first added at a rate of 120 kg/s, with a relative speed of 270 m/s opposite to the direction of travel, and then ejected at a rate of 120 kg/s, with a relative speed of 550 m/s opposite to the direction of travel. dM air Fthrust = vrel = ( −270 m s )(120 kg s ) + ( −550 m s )( −120 kg s ) = 33600 N dt air  3.4  104 N (c) The power developed is the force of thrust times the velocity of the airplane.



 1hp    746 W 



P =  Fthrust + Fthrust  v = ( 2310 N + 33600 N )( 270 m s ) = 9.696  10 6 W 

 fuel

air



= 1.3  10 4 hp

75. The external force on the belt is the force supplied by the motor and the oppositely-directed force of friction. Use Eq. 9–19b in one dimension. The belt is to move at a constant speed, so the acceleration of the loaded belt is 0. dv dM dM M = Fext + vrel → M ( 0 ) = Fmotor + Ffriction + v → dt dt dt © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Fmotor = ( v )

Instructor Solutions Manual

− Ffriction = ( 2.20 m s )( 75.0 kg s ) − ( −150N ) = 315 N dt The required power output from the motor is then found as the product of the force and the velocity.  1hp  Pmotor = Fmotor v = ( 315 N )( 2.20 m s ) = 693 W   = 0.93 hp  746 W  When the gravel drops from the conveyor belt, it is not accelerated in the horizontal direction by the belt and so has no further force interaction with the belt. The “new” gravel dropping on the belt must still be accelerated, so the power required is constant.

76. Because the sand is leaking out of the hole, rather than being pushed out the hole, there is no relative velocity of the leaking sand with respect to the sled (during the leaking process). Thus there is no “thrust” in this situation, and so the problem is the same as if there were no hole in the sled. From the free-body diagram, we see that the acceleration down the plane will be a = g sin  , as analyzed several times in Chapter 4. Use the constant acceleration relationships to find the time.

x = x0 + v y 0t + 12 ax t 2 → t =

2x ax

2 (120 m )

( 9.80 m s ) ( sin 32) 2

y x



 mg

= 6.8s

77. The only forces on the astronauts are internal to the 2-astronaut system, so their CM will not change. Call the CM location the origin of coordinates. That is also the initial location of the astronauts. m x + mB xB (195 kg )(12 m ) + ( 235 kg ) xB → 0= → xB = −9.96 m xCM = A A mA + mB 195 kg + 235 kg Their distance apart is xA − xB = 12 m − ( −9.96 m ) = 21.96 m  22 m . This problem can also be worked with conservation of momentum, since the only forces on the astronauts are “internal” forces. 78. Momentum will be conserved in the horizontal direction. Let A represent the railroad car, and B represent the snow. For the horizontal motion, v B = 0 and v B = v A . Momentum conservation in the horizontal direction gives the following. pinitial = pfinal → mA vA = ( mA + mB ) vA

vA =

mA vA mA + mB

( 4800 kg )( 6.70 m s )  3.80 kg  60.0 min 4800 kg +  ) (  min 

= 6.396 m s  6.4 m s

79. Let the original direction of the cars be the positive direction. We have vA = 4.50 m s and v B = 3.70 m s .

(a) Use Eq. 9–8 to obtain a relationship between the velocities. v A − v B = − ( v A − v B ) → v B = v A − v B + v A = 0.80 m s + v A Substitute this relationship into the momentum conservation equation for the collision. mA vA + mB vB = mA vA + mB vB → mA vA + mB vB = mA vA + mB ( 0.80 m s + vA ) →

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vA =

mA vA + mB ( vB − 0.80 m s ) mA + mB

( 435 kg )( 4.50 m s ) + ( 495 kg )( 2.90 m s ) 930 kg

= 3.648 m s

 3.65 m s ; vB = 0.80 m s + vA = 4.448 m s  4.45 m s (b) Calculate p = p − p for each car. pA = mA v A − mA vA = ( 435 kg )( 3.648 m s − 4.50 m s ) = −370.62 kg m s

 −370 kg m s pB = mB vB − mB vB = ( 495 kg )( 4.448 m s − 3.70 m s ) = 370.26 kg m s  370 kg m s The two changes are equal and opposite because momentum was conserved. The slight difference is due to roundoff error on the calculations. 80. This is a ballistic “pendulum” of sorts, similar to Example 9–10 in the textbook. The only difference is that the block and bullet are moving vertically instead of horizontally. The collision is still totally inelastic and conserves momentum, and the energy is still conserved in the rising of the block and embedded bullet after the collision. So we simply quote the equation from that example. m+M 2 gh → v= m

1  mv  1  ( 0.0150 kg )( 230 m s )  h=   =  = 0.3033 m  0.30 m 2  2g  m + M  2 9.80 m s  0.0150 kg + 1.40 kg  2

(

)

81. The change is momentum is due to the change in direction. p = m ( v − v ) = ( 0.145 kg ) 30.0 m s ˆj − 30.0 m s ˆi = 4.35 kg m s ˆj − ˆi f

(

)

(

)

82. (a) The meteor striking and coming to rest in the Earth is a totally inelastic collision. Let A represent the Earth and B represent the meteor. Use the frame of reference in which the Earth is at rest before the collision, and so vA = 0. Write momentum conservation for the collision.

(

)

mB vB = m + mB v  →

v = vB

mB mA + mB

(

= 2.5  10 m s 4

1.5  108 kg

) 6.0 10 kg + 1.5 10 kg = 6.25 10 m s −13

 6.3  10−13 m s

This is about 1/1000 of an atomic diameter per second. (b) The fraction of the meteor’s kinetic energy transferred to the Earth is the final kinetic energy of the Earth divided by the initial kinetic energy of the meteor. 2 K final 24 −13 1 1 m v 2 2 6.0  10 kg 6.25  10 m s Earth 2 = 1 = = 2.5  10−17 2 2 8 4 1 K initial m v 1.5  10 kg 2.5  10 m s B B 2 2

(

meteor

(

)( )(

)

(

)(

K Earth = K final − Kinitial = 12 m v2 − 0 = 12 6.0  1024 kg 6.25  10−13 m s Earth

Earth

) = 1.2 J 2

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83. (a) The average force is the momentum change divided by the elapsed time.  1m s  (1500 kg )( 0 − 45 km h )   p mv  3.6 km h  = −1.25  105 N  −1.3  105 N Favg = = = 0.15s t t The negative sign indicates direction, so the force is in the opposite direction to the original direction of motion. (b) Use Newton’s second law. We use the absolute value of the force because the problem asked for the deceleration. Favg 1.25  105 N  1g  = = 83.33 m s 2   8.5 g’s Favg = maavg → aavg = 2  m 1500 kg  9.80 m s  84. Call the final direction of the joined objects the positive x-axis. A diagram of the collision is shown. Momentum will be conserved in both the x- and y-directions. Note that v A = v B = v and v  = v 3. py :

− mv sin 1 + mv sin  2 = 0 → sin 1 = sin  2 → 1 =  2

1 2 mv B

px : mv cos 1 + mv cos  2 = ( 2m )( v 3) → cos 1 + cos  2 = 23 cos 1 + cos  2 = 2 cos 1 = 23

mv A

→ 1 = cos −1 13 = 70.5o =  2

1 +  2 = 141o 85. In each case, use momentum conservation. Let A represent the 5.5-kg object, and let B represent the 8.0-kg object. Then we have mA = 5.5 kg, v A = 6.5 m s , mB = 8.0 kg, and v B = −4.0 m s. (a) In this case, the objects stick together (totally inelastic), so v A = v B . mA v A + mB v B = ( mA + mB ) vA →

v B = v A =

mA v A + mB vB

( 5.5 kg )( 6.5 m s ) + ( 8.0 kg )( −4.0 m s )

mA + mB 13.5 kg (b) In this case, use Eq. 9–8 to find a relationship between the velocities. v A − v B = − ( v A − v B ) → v B = v A − v B + v A

= 0.277 m s  0.3 m s

mA v A + mB v B = mA v A + mB v B = mA vA + mB ( v A − v B + v A ) → v A =

( mA − mB ) vA + 2mB vB ( −2.5 kg )( 6.5 m s ) + 2 ( 8.0 kg )( −4.0 m s ) =

mA + mB

13.5 kg

= −5.9 m s

v B = v A − v B + v A = 6.5 m s − ( −4.0 m s ) − 5.9 m s = 4.6 m s (c) In this case, vA = 0.

mA v A + mB vB = mB vB →

vB =

mA v A + mB vB

( 5.5 kg )( 6.5 m s ) + ( 8.0 kg )( −4.0 m s )

= 0.4688 m s  0.5 m s 8.0 kg mB To check for “reasonableness”, first note the final directions of motion. A has stopped, and B has reversed direction. This is reasonable. Secondly, both objects are moving slower in their final state than in their initial state, so the system has lost kinetic energy. Since the system has lost kinetic energy and the directions are possible, this interaction is “reasonable”. =

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(d) In this case, vB = 0.

mA v A + mB vB = mA v A → v A =

mA v A + mB vB mA

( 5.5 kg )( 6.5 m s ) + ( 8.0 kg )( −4.0 m s ) 5.5 kg

= 0.6818 m s  0.7 m s

This answer is not reasonable because A continues to move in its original direction while B has stopped. Thus A has somehow “passed through” B. If B has stopped, A should have rebounded and so had a negative velocity. (e) In this case, v A = −4.0 m s.

mA vA + mB vB = mA vA + mB vB → vB =

(f)

( 5.5 kg ) ( 6.5 m s − ( −4.0 m s ) ) + ( 8.0 kg )( −4.0 m s )

= 3.219 m s  3.2 m s 8.0 kg The directions are reasonable, in that each object rebounds. Since the speeds of both objects are smaller than in the perfectly elastic case (b), the system has lost kinetic energy. This interaction is reasonable . As quoted above, the results for (c) and (e) are reasonable, but (d) is not reasonable .

86. For the swinging balls, their velocity at the bottom of the swing and the height to which they rise are related by conservation of energy. If the zero of gravitational potential energy is taken to be the lowest point of the swing, then the kinetic energy at the low point is equal to the potential energy at the highest point of the swing, where the speed is zero. Thus we 2 have 12 mvbottom = mgh for any swinging ball, and so the relationship 2 between speed and height is vbottom = 2 gh. From the diagram we see that

the initial height of ball A is given by h = l (1 − cos  ) = ( 0.35 m )(1 − cos 66 ) = 0.2076 m . (a) Calculate the speed of the lighter ball (A) at the bottom of its swing.

(

)

(

)

vA = 2 ghA = 2 9.80 m s 2 ( 0.35 m ) 1 − cos 66o = 2.017 m s  2.0 m s (b) This is an elastic collision with a stationary target. Accordingly, the relationships developed in Example 9–8 are applicable. Take the direction that ball A is moving just before the collision as the positive direction. ( m − mB ) ( 0.042 kg − 0.096 kg ) vA = A vA = ( 2.017 m s ) = −0.7893 m s  −0.79 m s ( mA + mB ) ( 0.042 kg + 0.096 kg ) vB =

2mA

( mA + mB )

vA =

2 ( 0.042 kg )

( 0.042 kg + 0.096 kg )

( 2.017 m s ) = 1.228 m s  1.2 m s

Notice that ball A has rebounded backwards. (c) After each collision, use the conservation of energy relationship again. 2 v 2 ( −0.7893 m s ) hA = A = = 3.2  10 −2 m = 3.2 cm 2 2g 2 9.80 m s

(

hB =

v B2 2g

)

(1.228 m s )

(

2 9.80 m s

)

= 7.7  10 −2 m = 7.7 cm

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87. (a) Use conservation of energy to find the speed of mass m before the collision. The potential energy at the starting point is all transformed into kinetic energy just before the collision.

(

)

mghA = 12 mvA2 → vA = 2 ghA = 2 9.80 m s 2 ( 3.60 m ) = 8.40 m s Use Eq. 9–8 to obtain a relationship between the velocities, noting that vB = 0.

v A − v B = v B − v A → v B = v A + v A Apply momentum conservation for the collision, and substitute the result from Eq. 9–8. mvA = mvA + M vB = mvA + M ( vA + vA ) → vA =

m−M m+M

 2.20 kg − 7.70 kg   ( 8.40 m s ) = −4.67 m s 9.90 kg  

vA = 

vB = vA + vA = −4.67 m s + 8.40 m s = 3.73 m s (b) Again use energy conservation to find the height to which mass m rises after the collision. The kinetic energy of m immediately after the collision is all transformed into potential energy. Use the angle of the plane to change the final height into a distance along the incline. v A2 2 1    = → = v m mgh h A A A 2 2g

d A =

hA sin 30.0

v A2 2 g sin 30.0

(

( −4.67 m s )

)

2 9.80 m s 2 sin 30.0

= 2.23 m

The vertical distance risen would be 1.11 m. 88. Let A represent mass m and B represent mass M. Use Eq. 9–8 to obtain a relationship between the velocities, noting that vB = 0.

v A − v B = v B − v A → v A = v B − v A After the collision, v A  0 since m is moving in the negative direction. For there to be a second collision after m moves up the ramp and comes back down, with a positive velocity at the bottom of the incline of −v A , the speed of m must be greater than the speed of M so that m can catch M. Thus −v A  v B , or v A  −v B . Substitute the result from Eq. 9–8 into the inequality. vB − vA  −vB → vB  12 vA

Now write momentum conservation for the original collision, and substitute the result from Eq. 9–8. 2m mvA = mvA + M vB = m ( vB − vA ) + M vB → vB = vA m+M Finally, combine the above result with the inequality from above. 2m vA  12 vA → 4m  m + M → m  13 M = 13 ( 7.70 kg ) = 2.57 kg m+M 89. (a) We take the CM of the system as the origin of coordinates. Then at any time, we consider the x-axis to be along the line connecting the star and the planet. Use the definition of center of mass:

xCM =

mA rA + mB ( − rB ) mA + mB

=0 →

rA =

mB mA

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(b) rA =

rB =

1.0  10−3 mA

(8.0  10 m ) = 8.0  10 m 11

mA mA (c) The geometry of this situation is illustrated d in the adjacent diagram. For small angles in 2rA  radian measure,   tan   sin  . 2 ( 8.0  108 m ) 2rA 2rA 1 ly   → d= = = 3.30  1017 m    tan    = 35 ly 15  d 9.46  10 m    1 1 ( 1000 )( 3600 ) 180 (d) We assume that stars are distributed uniformly, with an average interstellar distance of 4 ly. If we think about each star having a spherical “volume” associated with it, that volume would have a radius of 2 ly (half the distance to an adjacent star). Each star would have a volume of 3 3 4  rstar = 43  ( 2 ly ) . If wobble can be detected from a distance of 35 ly, the volume over which 3 to star 3 wobble can be detected is 43  rdetectable = 43  ( 35ly ) . 3

wobble

4 3

# stars =

r

3 detectable wobble

4 3

3  rstar

( 35ly )3 =  5400 stars ( 2 ly )3

to star

90. Let A represent the cube of mass M and B represent the cube of mass m. Thus mA = M = 2m and mB = m. Find the speed of A immediately before the collision, v A , by using energy conservation.

(

)

Mgh = 12 M vA2 → vA = 2 gh = 2 9.8 m s 2 ( 0.45 m ) = 2.970 m s Use Eq. 9–8 for elastic collisions to obtain a relationship between the velocities in the collision. We have vB = 0. v A − v B = − ( v A − v B ) → v B = v A + v A

Substitute this relationship into the momentum conservation equation for the collision. mA vA + mB vB = mA vA + mB vB → mA vA = mA vA + mB ( v A + v A ) → 2mvA = 2mvA + m ( vA + vA ) → vA =

vA 3

2 gh 3

(

)

2 9.80 m s 2 ( 0.45 m ) 3

= 0.990 m s

vB = vA + vA = 43 vA = 3.960 m s Each mass is initially moving horizontally after the collision, and so each has a vertical velocity of 0 as they start to fall. Use constant acceleration Eq. 2–12b with “down” as positive and the table top as the vertical origin to find the time of fall.

y = y0 + v0t + 12 at 2 → H = 0 + 0 + 12 gt 2 → t = 2H g Each cube then travels a horizontal distance found by x = v x t.

xM = v A t = xm = v B t =

2 gh

4 2 gh

= 23 hH = 23

( 0.45 m )( 0.95 m ) = 0.4359 m  0.44 m

= 83 hH = 83

( 0.45 m )( 0.95 m ) = 1.744 m  1.7 m

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91. The original horizontal distance can be found from the level horizontal range formula in Chapter 3. 2 R = v02 sin 2 0 g = ( 25 m s ) ( sin 56 ) ( 9.80 m s 2 ) = 52.87 m The height at which the objects collide can be found from Eq. 2–12c for the vertical motion, with v y = 0 at the top of the path. Take up to be positive.

v = v + 2ah → h = 2 y

2 y0

v y2 − v y20 2a

0 − ( 25 m s ) sin 28

(

2 −9.80 m s 2

= 7.028 m

)

Let m represent the bullet and M the skeet. When the objects collide, the skeet is moving horizontally at v0 cos  = ( 25 m s ) cos 28 = 22.07 m s = v x , and the bullet is moving vertically at

v y = 230 m s. Write momentum conservation in both directions to find the velocities after the totally inelastic collision.

( 0.20 kg )( 22.07 m s ) = 20.53 m s M +m ( 0.20 + 0.015 ) kg mv y ( 0.015 kg )( 230 m s ) → v y = = = 16.05 m s M +m ( 0.20 + 0.015 ) kg

px : M v x = ( M + m ) v x → v x = p y : mv y = ( M + m ) v y

M v x

(a) The speed v y can be used as the starting vertical speed in Eq. 2–12c to find the “extra” height,

h, that the skeet–bullet combination rises above the point of collision.

v = v + 2ah → h = 2 y

2 y0

v y2 − v y20 2a

0 − (16.05 m s )

(

2 −9.80 m s 2

)

= 13.14 m  13 m

(b) From Eq. 2–12b applied to the vertical motion after the collision, we can find the time for the skeet–bullet combination to reach the ground. y = y0 + v y t + 12 at 2 → 0 = 7.028 m + (16.05 m s ) t + 12 −9.80 m s 2 t 2 →

(

4.9t − 16.05t − 7.028 = 0 → t = 2

16.05 

)

(16.05 ) + 4 ( 4.9 )( 7.028 ) 2

9.80

3.666 s, − 0.391s The positive time root is used to find the horizontal distance traveled by the combination after the collision. xafter = v x t = ( 20.53 m s )( 3.666 s ) = 75.26 m If the collision had not happened, the skeet would have gone 12 R horizontally.

x = xafter − 12 R = 75.26 m − 12 ( 52.87 m ) = 48.83m  49 m 92. (a) This problem is a variation on the “ballistic pendulum” problem of Example 9–10. Assume the bullet hits the block immediately after leaving the rifle. Thus the bullet’s speed is 350 m/s when it hits the block. This is a totally inelastic collision, and so the post-collision velocity can be found from momentum conservation. Let the bullet mass be m, and the block mass be M. mv ( 0.0095 kg )( 760 m s ) mv = ( M + m ) v  → v  = = = 3.593 m s ( 2.0095) kg ( M + m) After the collision, conservation of mechanical energy applies since only gravity is doing work. Let l be the distance along the incline at the block–bullet’s highest point. The initial kinetic energy all becomes potential energy. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Einitial = Efinal →

l =

v 2 2 g sin 

1 2

(

( M + m ) v2 = ( M + m ) g l sin  → 2 ( 3.593 m s )

= 1.5585 m  1.6 m

)

2 9.80 m s 2 sin 25

(b) Now the post-collision velocity is the same, but friction does some non-conservative work. Thus the energy equation has to include the non-conservative work done by friction. The force of friction is the coefficient of kinetic friction times the normal force. The friction does negative work because the force of friction is opposite the direction of motion. Wnc + Einitial = Efinal ; Wnc = Fnc l cos180 = −  k ( M + m ) g l cos 

− k ( M + m ) g l cos  + 12 ( M + m ) v 2 = ( M + m ) g l sin  → l =

v 2

2 g ( sin  +  k cos  )

( 3.593 m s )

(

)

2 9.80 m s 2 ( sin 25 + 0.33cos 25 )

= 0.9127 m  0.91m

93. (a) In the reference frame of the Earth, the final speed of the Earth–asteroid system is essentially 0, because the mass of the Earth is so much greater than the mass of the asteroid. It is like throwing a ball of mud at the wall of a large building–the smaller mass stops, and the larger mass doesn’t move appreciably. Thus all of the asteroid’s original kinetic energy can be released as destructive energy.

(

K orig = 12 mv02 = 12  3200 kg m3

)  (1.0 10 m )  (1.5 10 m s ) = 1.507 10 J 3

4 3

 1.5  1021 J

 1bomb  = 38, 000 bombs 16   4.0  10 J 

(b) 1.507  1021 J 

94. We apply Eq. 9–19b, with no external forces. We also assume that the motion is all in one dimension. dv dM 1 1 M = v rel → Md v = v rel dM → dv = dM → dt dt M v rel 1 vrel

vfinal

M final

 dv = 

1 M

dM →

(

vfinal vrel

M ejected = M 0 − M final = M 0 1 − e

= ln

vfinal vrel

M final M0

→ M final = M 0 e

vfinal vrel

→

) = ( 210 kg ) (1 − e ( ) ) = 11.66 kg  12 kg 2.0 −35

95. (a) No, there is no net external force on the system. In particular, the spring force is internal to the system. (b) Use conservation of momentum to determine the ratio of speeds. Note that the two masses will be moving in opposite directions. The initial momentum, when the masses are released, is 0. pinitial = plater → 0 = mA vA − mB vB → vA vB = mB mA

1 2

mA v A2

mA  vA 

mA  mB 

= 1 =   =   = mB mA KB mB v B2 mB  vB  mB  mA  2 (d) The center of mass was initially at rest. Since there is no net external force on the system, the center of mass does not move, and so stays at rest . (c)

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96. The initial momentum is 0, and the net external force on the puck is 0. Thus momentum will be conserved in two dimensions. =p → 0 = mv ˆi + 2m ( 2v ) ˆj + mv → v = −v ˆi − 4vˆj p initial

initial

( −v ) + ( −4v ) = 2

v3 =

3 = tan −1

17v

−4v −v

= 256

97. Let the radius of the semicircular plate be R, with the center at the origin. From the symmetry of the semicircle, we know that xCM = 0, and so the center of mass lies on the y-axis. We have from Eq. 9–13 that R 1 The mass can be expressed as and yCM = y dm . M = dm ,  M so yCM =

thickness = dr y

 y dm . Since the object is uniform, we can express the mass as a uniform density   dm

times the area, for any part of the semicircle. That results in the following.

yCM =

 y dm =  y  dA  dm   dA

From the diagram, for the differential area we use a semicircular strip of width dr and length  r, which has a differential area of dA =  rdr. And from Problem 62, the y coordinate of the center of 2r . (Note the discussion immediately before Example 9–17 mass of that thin strip (like a wire) is



which mentions using the center of mass of individual objects to find the center of mass of an extended object.) We combine these relationships and integrate over the z dimension to find the center of mass. R R 2r    rdr 2  r 2 dr 2 y dm  y  dA  4R R3  3 0 0 = = = R = = yCM = R 2 1  dm   dA    rdr   rdr 2  R 3 0

 

Thus the center of mass is at  0ˆi +

4R ˆ  j . 3 

98. In this interaction, energy is conserved (initial potential energy of mass & compressed spring system = final kinetic energy of moving blocks) and momentum is conserved, since the net external force is 0. Use these two relationships to find the final speeds. pinitial = pfinal → 0 = mvm − 3mv3 m → v m = 3v3 m Einitial = E final → U spring = K final → initial 1 2

kD = 12 mv + 12 3mv32m = 12 m ( 3v3 m ) + 12 3mv32m = 6mv32m

1 2

kD 2 = 6mv32m →

2 m

v3m = D

k 12m

; v m = 3D

k 12m

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99. (a) We consider only the horizontal direction (the direction of motion of the railroad car). There is no external force in the horizontal direction. In Eq. 9–19b, the relative velocity (in the horizontal direction) of the added mass is the opposite of the horizontal velocity of the moving mass, since the added mass is moving straight down.

dv dt vf v0

= Fext + v rel = − ln

Mf M0

dM dt

= ln

→ M

M0 Mf

dv dt

= −v

→ vf = v0

dM dt

M0 Mf

→

= v0

dv v

=−

dM M

→  v0

dv v

=−

→

M0 dM M0 + t dt

(b) Evaluate the speed at t = 60.0 min.

v ( t = 60.0 ) = v0

M0 4800 kg = ( 6.70 m s ) dM 4800 kg + ( 3.80 kg min )( 60.0 min ) M0 + t dt

= 6.396 m s  6.4 m s This agrees with the answer to Problem 78. 100. The interaction between the planet and the spacecraft is elastic, because the force of gravity is conservative. Thus, kinetic energy is conserved in the interaction. Consider the problem a 1dimensional collision, with A representing the spacecraft and B representing Saturn. Because the mass of Saturn is so much larger than the mass of the spacecraft, Saturn’s speed is not changed appreciably during the interaction. Use Eq. 9–8, with v A = 10.4 km s and v B = v B = −9.6 km s. vA − vB = −vA + vB → vA = 2vB − vA = 2 ( −9.6 km s ) − 10.4 km s = −29.6 km s

Thus, there is almost a threefold increase in the spacecraft’s speed, and it reverses direction. km  1hr  1000 m 

   = 30.56 m s. So every second, the air striking an hr  3600 s  1km  area of 1 m2 (perpendicular to the wind direction) would be the air contained in a rectangular volume of dimensions 1m × 1 m × 30.56 m. Calculate the rate of air striking that surface. 1m 2 ( 30.56 m s ) 1.29 kg m 3 = 39.42 kg s  40 kg s

101. (a) The air speed is 110

(

)

(b) The force stopping the wind is exerted by the person, so the force on the person would be equal in magnitude and opposite in direction to the force stopping the wind. Calculate the force from Eq. 9–2 (in magnitude only). mwind 39.42 kg s = (1.60 m )( 0.50 m ) = 31.54 kg s vwind = 30.56 m s t m2 pwind mwind v wind mwind v wind = ( 31.54 kg s )( 30.56 m s ) Fon person = Fon wind = = = t t t = 963.9N  1000 N

The typical maximum frictional force is Ffr = s mg = (1.0 )( 70 kg ) 9.80 m s 2 = 686 N  700 N .

(

)

Thus we see that Fon person  Ffr . The wind is literally strong enough to blow a person off their feet. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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102. It is proven in the solution to Problem 51 that in an elastic collision between two objects of equal mass, with the target object initially stationary, the angle between the final velocities of the objects is 90o. For this specific circumstance, see the diagram. We assume that the target ball is hit “correctly” so that it goes in the pocket. Find 1 from the geometry of the “left’ triangle: 1 = tan −1

1.0 3.0 −1 3.0

1.0 m

3.0 m 3.0 m

1  2

= 30. Find  2 from

= 60. Since the 3.0 balls will separate at a 90 angle, if the target ball goes in the pocket, this does appear to be a

the geometry of the “right” triangle:  2 = tan good possibility of a scratch shot .

103. Since the collision is elastic, both momentum (in two dimensions) and kinetic energy are conserved. Write the three conservation equations and use them to solve for the desired quantities. The positive x-direction in the diagram is taken to the right, and the positive y-direction is taken towards the top of the picture. → 0 = mvpin sin 75 − M vball sin  → vpin sin 75 = 5vball sin  px = px initial

final

= py

initial

→ M (14.0 m s ) = mvpin cos 75 + M vball cos  →

final

70.0 m s − vpin cos 75 = 5vball cos  K initial = K final →

1 2

2 2 2 2 M (14.0 m s ) = 12 mvpin + 12 M vball → 980 m 2 s 2 = vpin + 5vball → 2

2 2 980 m2 s2 − vpin = 5vball

Square the two momentum equations and add them to eliminate the dependence on  . 2 2 2 2 vpin sin 2 75 = 25vball sin 2  ; ( 70.0 ) − 2 ( 70.0 ) vpin cos 75 + vpin cos 2 75 = 25vball cos 2  → 2

2 2 2 2 vpin sin 2 75 + ( 70.0 ) − 2 ( 70.0 ) vpin cos 75 + vpin cos 2 75 = 25vball sin 2  + 25vball cos 2  + → 2

2 = 5 ( 5vb2all ) ( 70.0 ) − 140vpin cos 75 + vpin2 = 25vball 2

Substitute from the kinetic energy equation. 2 ( 70.0 ) − 140vpin cos 75 + vpin2 = 5 980 − vpin2

(

) → 4900 − 140v cos 75 + v = 4900 − 5v pin

2 pin

2 = 140v pin cos 75 → v pin = 6.039 m s 6v pin 2 2 = 5vball → vball = 980 − vpin

1 5

( 980 − v ) = ( 980 − ( 6.039 ) ) = 13.737 m s 2 pin

1 5

   vpin sin 75  −1 ( 6.039 ) sin 75  = 4.87  = sin   5vball   5 (13.737 ) 

v pin sin 75 = 5v ball sin  →  = sin −1 

So the final answers are as follows. (a) vpin = 6.039 m s  6.0 m s (b) vball = 13.737 m s  14 m s (c)  = 4.87  4.9

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Chapter 9

Linear Momentum

104. The speed of each ball as it reaches the floor is calculated from energy conservation. We assume the radii of the balls are small compared with the height, so that they fall the full distance h before reaching the floor, and before interacting. 2 Einitial = E final → U initial = K final → mgh = 12 mvfloor → vfloor = 2 gh The larger ball is moving upwards and the smaller ball is moving downwards at the moment of collision. Call upwards the positive direction. Thus vM = 2 gh and vm = − 2 gh . Since M we assume that v M = v M . We use Eq. 9–8.

vM − vm = vm − vM → vm = 2vM − vm = 2

( 2 gh ) − ( − 2 gh ) = 3 2 gh

And now we use energy conservation again to find the final height of mass m, as its kinetic energy after the collision becomes potential energy.  = K final  → mgh = 12 m ( vm ) Uinitial

→ h =

( vm )

( 3 2 gh ) = 9 ( 2 gh ) = 9h =

2g 2g And so we see that the smaller ball rebounds to a height of 9h.

331

CHAPTER 10: Rotational Motion Responses to Questions 1.

The odometer designed for 27-inch wheels increases its reading by the circumference of a 27-inch wheel ( 27 ") for every revolution of the wheel. If a 24-inch wheel is used, the odometer will still register 27 " for every revolution, but only 24 " of linear distance will have been traveled. Thus the odometer will read a distance that is further than you actually traveled, by a factor of 27 24 = 1.125. The odometer will read 12.5% too high.

(a) A point on the rim of a disk rotating with constant angular velocity has no tangential acceleration since the tangential speed is constant. It does have radial acceleration. Although the point’s speed is not changing, its velocity is, since the velocity vector is changing direction. The point has a centripetal acceleration, which is directed radially inward. (b) If the disk’s angular velocity increases uniformly, the point on the rim will have both radial and tangential acceleration, since it is both moving in a circle and speeding up. (c) The magnitude of the radial component of acceleration will increase in case (b), but the tangential component will be constant. In case (a), neither component of linear acceleration will change.

No. The distance between the parts of a non-rigid object can change. If the distance between parts changes, then those parts can have different values of ω.

The radian is defined as the ratio of the distance traveled along an arc divided by the radius of the arc. When an angle in radians is multiplied by the radius the result is a distance. Therefore, when angular speed (which is angular displacement divided by time) is multiplied by the radius the result is the displacement along the arc divided by time, which is a linear speed. Degrees and revolutions are not defined in terms of arc lengths and cannot be used in the same way.

See the diagram. To the left is west, the direction of the angular velocity. The direction of the linear velocity of a point on the top of the wheel would be into the page, which is north. If the angular acceleration is east, which is opposite the angular velocity, the wheel is slowing down– its angular speed is decreasing. The tangential linear acceleration of the point on top will be in the opposite direction to its linear velocity– it will point south.

Using the right-hand rule, point the fingers in the direction of the Earth’s rotation, from west to east. Then the thumb points north. Thus the Earth’s angular velocity points along its axis of rotation, towards the North Star.

Yes. The magnitude of the torque exerted depends not only on the magnitude of the force but also on the lever arm, which involves both the distance from the force to the axis of rotation and the angle at which the force is applied. A small force applied with a large lever arm could create a greater torque than a larger force with a smaller lever arm.

When you do a sit-up, torque from your abdomen muscles must lift the upper half of the body from a laying-down position to a sitting-up position. The larger the moment of inertia of the upper half of the body, the more torque is needed, and thus the harder the sit-up is to do. With the hands behind the head, the moment of inertia of the upper half of the body is larger than with the hands outstretched in front.

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Just because the net force on a system is zero, the net torque need not be zero. Consider a uniform object with two equal forces on it, as shown in the first diagram. The net force on the object is zero (it would not start to translate under the action of these forces), but there is a net counterclockwise torque about the center of the rod (it would start to rotate under the action of these forces). Just because the net torque on a system is zero, the net force need not be zero. Consider an object with two equal forces on it, as shown in the second diagram. The net torque on the object is zero (it would not start to rotate under the action of these forces), but there is a net downward force on the rod (it would start to translate under the action of these forces).

10. Running involves rotating the leg about the point where it is attached to the rest of the body. Therefore, running fast requires the ability to change the leg’s rotation easily. The smaller the moment of inertia of an object, the smaller the resistance to a change in its rotational motion. The closer the mass is to the axis of rotation, the smaller the moment of inertia. Concentrating flesh and muscle high and close to the body minimizes the moment of inertia and increases the angular acceleration possible for a given torque, improving the ability to run fast. 11. Refer to the diagram of the book laying on a table. The moment of inertia about the “starred” axis (the axis parallel to the longest dimension of the book) will be the smallest. Relative to this axis, more of the mass is concentrated close to the axis.

PHYSICS

***

12. No, the mass cannot be considered as concentrated at the CM when considering rotational motion. If all of the mass were at the CM, then the object would have a rotational inertia of 0. That means it could not have any rotational kinetic energy, for example. The distribution of the mass is fundamental when describing rotational motion. 13. The moment of inertia will be larger when considering an axis through a point on the edge of the disk, because most the mass of the disk will be further from the axis of rotation than it was with the original axis position. The parallel axis theorem can be used to show that the moment of inertia about an axis at the edge will be 23 MR02 . 14. The long beam increases the rotational inertia of the walker. If the walker gets off-center from the tightrope, gravity will exert a torque on the walker causing the walker to rotate with their feet as a pivot point. With a larger rotational inertia, the angular acceleration caused by that gravitational torque will be smaller, and the walker will therefore have more time to compensate. The long length of the beam also allows the walker to make relatively small shifts in their center of mass to bring themselves back to being centered on the tightrope. It is much easier for the walker to move a long, narrow object with the precision needed for small adjustments than a short, heavy object like a barbell. 15. Applying conservation of energy at the top and bottom of the incline, assuming that there is no work done by friction, gives Etop = Ebottom → Mgh = 12 M v 2 + 12 I  2. For a solid ball, I = 52 MR 2. If the ball rolls without slipping (no work done by friction) then  = v R , and so Mgh = 12 M v 2 + 12 52 MR 2 v 2 R 2

→ v = 10 gh 7 .

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This speed is independent of the angle of the incline, and so both balls will have the same speed at the bottom. The ball on the incline with the smaller angle will take more time to reach the bottom than the ball on the incline with the larger angle. 16. The two spheres have different rotational inertias. The sphere that is hollow will have a larger rotational inertia than the solid sphere. If the two spheres are allowed to roll down an incline without slipping, the sphere with the smaller moment of inertia (the solid one) will reach the bottom of the ramp first. See the answer for Question 17 for a detailed explanation of why this happens. 17. (a) The sphere will reach the bottom first because it has a smaller rotational inertia. A detailed analysis of that is done below. (b) The sphere will have the greater speed at the bottom, and so will have more translational kinetic energy that the cylinder. (c) Both will have the same energy at the bottom, because they both started with the same potential energy at the top of the incline. (d) The cylinder will have the greater rotational kinetic energy at the bottom, because it has less translational kinetic energy than the sphere. Here is a detailed analysis of the motion. Applying conservation of energy at the top and bottom of the incline, assuming that there is no work done by friction, gives Etop = Ebottom → Mgh = 12 M v 2 + 12 I  2 . If the objects roll without slipping, then  = v R , and so Mgh = 12 M v 2 + 12 I ( v R )

→ v=

2Mgh M + I R2

. For a solid ball,

I = 52 MR 2, and for a cylinder, I = 12 MR 2 . Thus vsphere = 10gh 7 and vcyl = 4gh 3. Since

vsphere  vcyl , the sphere has the greater speed at the bottom. That is true for any amount of height change, and so the sphere is always moving faster than the cylinder after they start to move. Thus, the sphere will reach the bottom first. Since both objects started with the same potential energy, both have the same total kinetic energy at the bottom. But since both objects have the same mass and the cylinder is moving slower, the cylinder has the smaller translational kinetic energy and thus the greater rotational kinetic energy. Since rotational kinetic energy is K rot = 12 I  2 , then

K rot sphere

= 72 mgh and K rot

= 13 mgh.

cylinder

18. The two spheres will reach the bottom at the same time with the same speed–from the answer to Question 17, the speed at the bottom is not dependent on either the mass or radius. The larger, more massive sphere will have the greater total kinetic energy at the bottom, because it had more potential energy at the start, due to its larger mass.

Solutions to MisConceptual Questions 1.

(c) A common misconception is that if the riders complete the revolution at the same time, they must have the same linear velocities. The time for a rotation is the same for both riders, but Bonnie, at the outer edge, travels in a larger circle than Jill. Bonnie therefore has a greater linear velocity.

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(c) Since the disk is rigid, all of the points that rotate have the same angular velocity. The units used for angular velocity do not matter.

(b) Students may think that the rider would travel half the distance in half the time. This would be true if the object had constant angular speed. However, it is accelerating, so it will travel a shorter distance, 14  , in the first half of the time. The constant-acceleration equations can be used to calculate this result as well. When starting at rest,  = 12  t 2 . Since the angular displacement is proportion to the square of the elapsed time, if the time is cut in half, the angular displacement is only ¼ of the original amount.

(b) A common error is to think that increasing the radius of the tires would increase the speed measured by the speedometer. This is actually backwards. Increasing the size of the tires will cause the car to travel faster than it would with smaller tires, for the same angular speed. Thus the speed of the car will be greater than the speed measured by the speedometer.

(c) Torque is the product of the lever arm and the component of the force perpendicular to the arm. Although the 1000-N force has the greatest magnitude, it acts at the pivot and so the lever arm is zero, and thus the torque is also zero. The 800-N force is parallel to the lever arm and also exerts no torque. Of the three 500-N forces, (c) is both perpendicular to the lever arm and farthest from the pivot.

(a, d) The line of action of forces (a) and (d) each passes through the pivot point, and so those forces cannot exert any torque. The other three forces all exert some torque.

(c, e, f) Equations 10–10 show that there are three ways in which the torque can be written. It can be the product of the force, the lever arm, and the sine of the angle between them as in answer (c). It can be the product of the force and the component of the lever arm perpendicular to the force, as in answer (e). It can also be written as the product of the lever arm and the force perpendicular to the lever arm, as in answer (f). Doing the calculations shows that all three torques are equal.

(b) The location of the mass is very important. Imagine taking the material from the solid sphere and compressing it outward to turn the solid sphere into a hollow sphere of the same mass and radius. As you do this you would be moving mass farther away from the axis of rotation, which will increase the moment of inertia. Therefore the hollow sphere has a greater moment of inertia than the solid sphere.

(b) If you don’t consider how the location of the mass affects the moment of inertia, you may think that the two kinetic energies are nearly the same. However, a hollow cylinder (like a hoop) has twice the moment of inertia as a solid disk of the same mass and radius. The kinetic energy is proportional to the moment of inertia, so at the same angular speed the wheel with the spokes will have nearly double the kinetic energy as the solid disk. It is only “nearly double” because some of the mass is in the spokes so the moment of inertia is not exactly double.

10. (d) Because axis A passes through mass m1 , that mass has no rotation inertia relative to axis A. We 2

2  l  2 2 2 calculate I A = 2m   + m 2 l = ml + 2ml = 3ml . Because axis B passes through 2   masses m2 and m4 , those two masses have no rotational inertial relative to axis B. We calculate

( )

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 l  2  = ml . Because axis C passes through mass m1 , that mass has no rotation  2

I B = 2m 

inertia relative to axis C. We calculate I C = 2ml 2 + m

( 2l ) = 2ml + 2ml = 4ml . Thus 2

answer (d) is correct. 11. (b) It takes energy to rotate the ball. If some of the 1000 J goes into rotation, less is available for linear kinetic energy, and so the rotating ball will travel slower. 12. (b) If you do not take into account the energy of rotation, you would answer that the two objects would rise to the same height. Another common misconception is that the mass and/or diameter of the objects will affect how high they travel. When using conservation of energy to relate the total initial kinetic energy (translational and rotational) to the final potential energy, the mass and radius of the objects cancel out. The thin hoop has a larger moment of inertia (for a given mass and radius) than the solid sphere. It will therefore have a greater total initial kinetic energy, and will travel to a greater height on the ramp. 13. (b) We use conservation of energy to determine the speed of each sphere as a function of position on the incline. The sphere with the greater speed would reach the bottom of the incline first. Potential energy will be zero at the base of the incline (y = 0) and the initial height will be H. We take position 1 to be at the top of the incline, and position 2 to be at a generic location along the incline. K1 + U1 = K 2 + U 2 → 0 + mgH = 12 mv 2 + 12 I  2 + mgy

(

)



v2 



mg ( H − y ) = 12 mv 2 + 12 mr 2 2 = 12  mv 2 + 52 mr 2 v=

10 7

 = 2 ( 5 mv ) → r  1

g (H − y)

The velocity along the incline does not depend upon either the mass or the radius of the sphere. Thus, both spheres have the same speed at each point along the incline, and so both will reach the bottom of the incline at the same time. The heavier sphere had more potential energy to begin with, and so will have more kinetic energy at the bottom of the incline. 14. (a, d)

In Question 13, we found the velocity at the bottom of the incline: v =

question asks about the angular velocity, which is given by  =

10 7

R R velocity depends on the height of the incline and the radius of the sphere.

10 7

gH . This

. Thus the angular

Solutions to Problems 1.

(a)

( 45.0)( 2 rad 360 ) =  4 rad = 0.785 rad

(b)

( 60.0)( 2 rad 360 ) =  3rad = 1.05 rad

(c)

( 90.0)( 2 rad 360) =  2 rad = 1.57 rad

(d)

( 360.0) ( 2 rad 360o ) = 2 rad = 6.283rad

(e)

( 445)( 2 rad 360 ) = 89 36 rad = 7.77 rad

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Chapter 10

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The subtended angle (in radians) is the diameter of the Sun divided by the Earth–Sun distance.

=

diameter of Sun rEarth −Sun

→

  rad  1.5  1011 m = 6.545  108 m  7  108 m ) (  180 

radius of Sun = 12  rEarth −Sun = 12 ( 0.5 )  The actual radius is 6.96  108 m. 3.

We find the diameter of the spot from the definition of radian angle measure. diameter → diameter =  rEarth − Moon = 1.4  10 −5 rad 3.8  108 m = 5300 m = rEarth − Moon

(

 

The initial angular velocity is 0 =  8500

)(

)

rev   2 rad   1 min 



 = 890 rad s . Use the

min   1 rev   60 sec 

definition of angular acceleration.  0 − 890 rad s = = = −220 rad s 2 4.0 s t 5.

(a) We convert rpm to rad/s.  7200 rev   2 rad   1min  =    = 753.98 rad sec  750 rad sec  1min   1rev   60s  (b) To find the speed, we use the radius of the reading head location along with Eq. 10–4. v = R  = ( 3.00  10−2 m ) ( 753.98 rad sec ) = 22.62 m s  23 m s (c) We convert the speed of the point on the platter from m/s to bit/s, using the distance per bit. 1 bit  = 4.5  107 bit s ( 22.62 m s )   −6 0.50  10 m  

The ball rolls 2 r =  d of linear distance with each revolution.

3.1m d m  = 3.5 m → d = = 8.2  10−2 m  12.0  1rev 

12.0 rev  7.

(a) We convert rpm to rad/s.  2400 rev   2 rad   1min  =    = 251.3rad sec  250 rad sec  1min   1rev   60s  (b) To find the speed and radial acceleration, we use the full radius of the wheel, along with Eqs. 10–4 and 10–6.  0.35 m  v =  R = ( 251.3 rad sec )   = 44 m s  2 

 0.35 m  4 2  = 1.1  10 m s 2  

aR =  2 R = ( 251.3 rad sec )  2

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Instructor Solutions Manual

In each revolution, the wheel moves forward a distance equal to its circumference,  d . x 8600 m x = N rev ( d ) → N = = = 4.0  103 rev  d  ( 0.68 m )

The angular velocity is expressed in radians per second. The second hand makes 1 revolution every 60 seconds, the minute hand makes 1 revolution every 60 minutes, and the hour hand makes 1 revolution every 12 hours.

rad  1 rev   2 rad   rad sec  1.05  10−1  =  sec  60sec   1 rev  30

(a) Second hand:  = 

rad  rad  1 rev   2 rad   1min   1.75  10−3   =  sec  60 min   1 rev   60 s  1800 sec

(b) Minute hand:  =  (c) Hour hand:

rad  rad  1 rev   2 rad   1h   1.45  10−4 =   =  sec  12 h   1 rev   3600 s  21, 600 sec

(d) The average angular acceleration in each case is 0 , since the angular velocity is constant. 10. The angular speed of the merry-go-round is 2 rad 4.0s = 1.57 rad s . (a) v =  R = (1.57 rad sec )(1.2 m ) = 1.9 m s (b) The acceleration is radial. There is no tangential acceleration.

aR =  2 R = (1.57 rad sec ) (1.2 m ) = 3.0 m s 2 towards the center 2

11. The angle in radians is the diameter of the object divided by the distance to the object. 2 6.96  105 km 2 RSun Sun = = = 9.30  10 −3 rad 6 rEarth −Sun 149.6  10 km

(

 Moon =

2 RMoon

)

(

2 1.74  103 km

) = 9.06  10 rad −3

rEarth − Moon 384  10 km Since these angles are practically the same (only a 2.6% difference), solar eclipses can occur. Based on these values, the Sun would never be completely obscured. Since the orbits are not perfect circles but are ellipses, the above values are just averages. Full (total) solar eclipses do occur occasionally, but partial eclipses also occur, when the Sun is partially blocked by the Moon. 3

12. Each location will have the same angular velocity (1 revolution per day), but the radius of the circular path varies with the location. From the diagram, we see r = R cos  , where R is the radius of the Earth, and r is the radius at latitude  .

 2 rad  1day  6.38  106 m ) = 464 m s (   T  1day  86,400 s   2 rad  1day  2 (b) v =  r = r = 6.38  106 m ) cos 66.5 = 185 m s (   T  1day  86,400 s   2 rad  1day  2 (c) v =  r = r = 6.38  106 m ) cos 48.0 = 3.10  10 2 m s (   T  1day  86,400 s  (a) v =  r =

2

r =

 

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Chapter 10

Rotational Motion

13. (a) The Earth makes one orbit around the Sun in 1 year.   2 rad   1 year  −7 orbit = =  = 1.99  10 rad s  t  1year   3.16  107 s  (b) The Earth makes one revolution about its axis in 1 day.   2 rad  1 day  −5 = rotation =   = 7.27  10 rad s t  1 day  86,400 s  14. The centripetal acceleration is given by aR =  2 R. Solve for the angular velocity.

=

aR R

(100000 ) ( 9.80 m s 2 )

0.080 m

= 3500

rad  1rev   60 s 

 = 3.3  10 rpm   s  2 rad   1 min  4

15. Convert the rpm values to angular velocities. rev  2 rad  1 min   0 =  120    = 12.57 rad s min  1 rev  60 sec  

rev  2 rad  1 min    =  280    = 29.32 rad s min  1 rev  60 sec   (a) The angular acceleration is found from Eq. 10–9a.  − 0 29.32 rad s − 12.57 rad s = = = 3.35 rad s2  3.4 rad s2 t 5.0 s (b) To find the components of the acceleration, the instantaneous angular velocity is needed.  = 0 +  t = 12.57 rad s + ( 3.35 rad s 2 ) ( 2.0s ) = 19.27 rad s The instantaneous radial acceleration is given by aR =  2 R.

 0.61m  2 2  = 113 m s  110 m s  2  The tangential acceleration is given by atan =  R. aR =  2 R = (19.27 rad s )  2

(

 )  0.61m  = 1.0 m s 2

atan =  R = 3.35 rad s 2 





16. (a) The angular acceleration can be found from Eq. 10–3a. The initial angular speed is 0 and the final angular speed is 1 rpm.  1.0 rev   2 rad   1.0 min  − 0  − 0  min   1 rev   60 s  = = = 1.454  10−4 rad s2  1.5  10−4 rad s2 t 720 s (b) After 6.0 min (360 s), the angular speed is as follows.  = 0 +  t = 0 + (1.454  10−4 rad s2 ) ( 360s ) = 5.234  10−2 rad s Find the components of the acceleration of a point on the outer skin from the angular speed and the radius.

(

) a =  R = ( 5.234  10 rad s ) ( 4.25 m ) = 1.164  10 m s  1.2  10 m s

atan =  R = 1.454  10−4 rad s 2 ( 4.25 m ) = 6.180  10 −4 m s 2  6.2  10 −4 m s 2 2

−2

rad

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17. The tangential speed of the turntable must be equal to the tangential speed of the roller, if there is no slippage. v1 = v2 → 1 R1 = 2 R2 → 1 2 = R2 R1 18. (a) The direction of 1 is along the axle of the wheel, to the left. That is the −iˆ direction. The direction of  2 is also along its axis of rotation, so it is straight up. That is the +kˆ direction. That is also the angular velocity of the axis of the wheel. (b) At the instant shown in the textbook, we have the vector relationship as shown in the diagram.

( 48.0 rad s ) 2 + ( 35.0 rad s )2 = 59.4 rad s

 = 12 + 22 =  = tan −1

35.0 2 = tan −1 = 36.1 48.0 2

dω dt

. Since ω = ω1 + ω2 , and ω2 is a constant

dω1

. ω1 is rotating counterclockwise about the z axis with the angular dt velocity of 2 , and so if the figure is at t = 0, then ω1 = 1 − cos 2t ˆi − sin 2t ˆj .

(

α=

dω dt

d ( ω1 + ω2 ) dt

dω1 dt

(

)

) =   sin  t ˆi − cos  t ˆj ( )

d 1 − cos 2t ˆi − sin 2t ˆj  dt

( )

α ( t = 0 ) = 12 −ˆj = − ( 48.0 rad s )( 35.0 rad s ) ˆj = −1680 rad s 2 ˆj

19. The angular displacement can be found from Eq. 10–9d.

 =  t = 12 (0 +  ) t = 12 ( 0 + 15000 rev min )( 220s )(1min 60s ) = 2.8  104 rev 20. (a) The angular acceleration can be found from Eq. 10–9b with 0 = 0.

= 52

2 t

2 ( 26 rev )

(1.0 min )

= 52 rev min 2

rev  2 rad   1min 

   = 0.091rad s min  1rev   60 s 

(b) The final angular speed can be found from  = 12 (0 +  ) t , with 0 = 0.

=

2 t

− 0 =

2 ( 26 rev ) 1.0 min

= 52 rpm

21. (a) The angular acceleration can be found from Eq. 10–9c.

rev   2 rad   1 min  rad  2 − 02 0 − ( 780 rev min )  = =  −243 =   = −0.42 2 2  2 2 (1250 rev ) min   1 rev   60 s  s  2

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(b) The time to come to a stop can be found from  = 12 (0 +  ) t.

2

0 + 

2 (1250 rev )  60s 

 = 192.3s  190s = 3.2 min



780 rev min  1min 

22. Use Eq. 10–9d combined with Eq. 10–2a.  + 0 240 rpm + 360 rpm = = 300 rpm = 2 2 rev   1 min    =  t =  300   ( 6.8s ) = 34 rev min   60 sec   Each revolution corresponds to a circumference of travel distance.   ( 0.34 m )  = 36 m 34 rev   1 rev  23. (a) The angular acceleration can be found from  2 = 02 + 2 , with the angular velocities being found from  = v r .

 2 − 02 ( v − v0 ) = = = 2 2r 2 2

( 55 km h )2 − ( 95 km h )2   1m s     3.6 km h 

 2 rad    rev 

2 ( 0.40 m ) ( 75 rev )  2

= −3.070 rad s 2  −3.1rad s 2 (b) The time to stop can be found from  = 0 +  t , with a final angular velocity of 0.

 1m s    3.6 km h  = 12.44 s  12 s

− ( 55 km h ) 

 − 0 v − v 0 = =  r ( 0.40 m ) ( −3.070 rad s 2 )

(c) We first find the total angular displacement of the tires as they slow from 55 km/h to rest, and then convert the angular displacement to a linear displacement, assuming that the tires are rolling without slipping.  2 = 02 + 2 → 2  ( 55 km h )  1m s    v0  − 0    2 2    − 0  r  = −  0.40 m  3.6 km h   = 237.6 rad  = = 2 2 2 ( −3.070 rad s 2 ) 2

x = r  = ( 0.40 m )( 237.6 rad ) = 95 m For the total distance, add the distance moved during the time the car slows from 95 km/h to 55 km/h. The tires made 75 revolutions, so that distance is as follows.

 2 rad   = 188 m  rev 

x = r = ( 0.40 m ) ( 75rev ) 

The total stopping distance would be the sum of the two distances, 283 m.

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24. Since there is no slipping between the wheels, the tangential component of the linear acceleration of each wheel must be the same. (a) atan = atan →  small Rsmall =  large Rlarge → small

large

 large =  small

Rsmall Rlarge

2.0 cm  )  27.0  = 0.5037 rad s  0.50 rad s cm

(

= 6.8 rad s 2 





(b) Assume the pottery wheel starts from rest. Convert the speed to an angular speed, and then use Eq. 10–9a.  rev   2 rad   1 min  = 6.807 rad s  =  65     min   1 rev   60 s   − 0 6.807 rad s  = 0 +  t → t = = = 13.51s  14 s  0.5037 rad s 2 25. We are given that  = 9.5t − 13.0t 2 + 1.6t 4 . d (a)  = = 9.5 − 26.0t + 6.4t 3 , where  is in rad/sec and t is in sec.

(b)  =

d

= −26.0 + 19.2t 2 , where  is in rad sec 2 and t is in sec.

 ( 3.0) = −26.0 + 19.2 ( 3.0) = 146.8 rad s2  150 rad s2 2

(d) The average angular velocity is the angular displacement divided by the elapsed time.   ( 3.0 ) −  ( 2.0 ) avg = = 3.0 s − 2.0 s t

9.5 ( 3.0 ) − 13.0 ( 3.0 )2 + 1.6 ( 3.0 )4  − 9.5 ( 2.0 ) − 13.0 ( 2.0 )2 + 1.6 ( 2.0 )4     = 1.0 s

= 48.5 rad s  49 rad s (e) The average angular acceleration is the change in angular velocity divided by the elapsed time.   ( 3.0 ) −  ( 2.0 )  avg = = t 3.0 s − 2.0 s

9.5 − 26.0 ( 3.0 ) + 6.4 ( 3.0 )3  − 9.5 − 26.0 ( 2.0 ) + 6.4 ( 2.0 )3     = 96 rad s2 = 1.0 s

26. (a) The angular velocity is found by integrating the angular acceleration function. t t  d → d  =  dt →  d  =   dt =  ( 4.2t 2 − 9.0t ) dt →  = 1.4t 3 − 4.5t 2 = dt 0 0 0 (b) The angular position is found by integrating the angular velocity function. t t  d → d =  dt →  d =   dt =  (1.4t 3 − 4.5t 2 ) dt →  = 0.35t 4 − 1.5t 3 = dt 0 0 0

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Chapter 10

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 ( 2.0s ) = 0.35 ( 2.0 ) − 1.5 ( 2.0 ) = −6.4 rad 4

27. (a) The maximum torque will be exerted by the force of her weight, pushing tangential to the circle in which the pedal moves.  = R⊥ F = R⊥ mg = ( 0.17 m )( 58 kg ) ( 9.80 m s 2 ) = 96.6 m N  97 m N (b) She could pull upwards on one pedal (by putting her foot under it, or by using pedal straps) as she pushed down on the other pedal. She could change the pedaling mechanism to one with a larger radius. She could increase her mass, perhaps by wearing ankle weights. 28. There is a counterclockwise torque due to the force of gravity on the left block, and a clockwise torque due to the force of gravity on the right block. Call clockwise the positive direction.

 = mg l − mg l = mg ( l − l ), clockwise 2

29. The torque is calculated by  = rF sin  . See the diagram, from the top view. (a) For the first case,  = 90.

 = rF sin  = ( 0.96 m )( 37 N ) sin 90 = 36 m N (b) For the second case,  = 60.0.



 = rF sin  = ( 0.96 m )( 37 N ) sin 60.0 = 31m N

30. Each force is oriented so that it is perpendicular to its lever arm. Call counterclockwise torques positive. The torque due to the three applied forces is given by the following.  applied = ( 28 N )( 0.24 m ) − (18 N )( 0.24 m ) − ( 35 N )( 0.12 m ) = −1.8 m N forces

Since this applied torque is clockwise, we assume the wheel is rotating clockwise, and so the frictional torque is counterclockwise. Thus the net torque is as follows.  net = ( 28 N )( 0.24 m ) − (18 N )( 0.24 m ) − ( 35 N )( 0.12 m ) + 0.60 m N = −1.2 m N

= 1.2 m N, clockwise 31. (a) The force required to produce the torque can be found from  = rF sin  . The force is applied perpendicularly to the wrench, so  = 90.  95 m N = 339.3 N  340 N F= = r 0.28 m (b) The net torque still must be 95m N. This is produced by 6 forces, one at each of the 6 points. We assume for our estimate that those forces are also perpendicular to their lever arms. From the diagram, we estimate the lever arm as follows, and then calculate the force at each point.  y  + y tan 30 Lever arm = r = 12 h + x = 12    cos 30  1   + tan 30  = ( 7.5  10−3 m ) (1.15)  2 cos 30 

= y

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

 net = ( 6 Fpoint ) r → Fpoint =

 6r

95 m N

(

)

6 7.5  10−3 m (1.15)

Instructor Solutions Manual

= 1835.7 N  1800 N

32. The lever arm to the point of application of the force is along the x-axis. Thus, the perpendicular part of the force is the y component. Use Eq. 10–10b.  = RF⊥ = ( 0.120 m )( 43.4 N ) = 5.21m N, counterclockwise 33. For each torque, use Eq. 10–10c. Take counterclockwise torques to be positive. (a) Each force has a lever arm of 1.0 m.  about = − (1.0 m )( 56 N ) sin 32 + (1.0 m )( 52 N ) sin 58 = 14.42m N  14 m N C

(b) The force at C has a lever arm of 1.0 m, and the force at the top has a lever arm of 2.0 m.  about = − ( 2.0 m )( 56 N ) sin 32 + (1.0 m )( 65 N ) sin 45 = −13.39 m N  −13m N P

The negative sign indicates a clockwise torque. 34. Since all of the significant mass is located at the same distance from the axis of rotation, the moment of inertia is given by I = MR 2 .

I = MR 2 = (1.1kg )  12 ( 0.67m )  = 0.12 kg m 2 2

The hub mass can be ignored because its distance from the axis of rotation is very small, and so it has a very small rotational inertia. 35. The torque required is equal to the angular acceleration times the moment of inertia. The angular acceleration is found using Eq. 10–9a. Use the moment of inertia of a solid cylinder.  = 0 +  t →  =  t

   MR0  = ( 3.1  10 kg ) ( 7.0 m ) ( 0.72 rad s ) = 1.6  104 m N  = I  = ( 12 MR02 )   = 2t 2 ( 34 s ) t 2

36. The oxygen molecule has a “dumbbell” geometry, rotating about the dashed line, as shown in the diagram. If the total mass is M, then each atom has a mass of M/2. If the distance between them is d, then the distance from the axis of rotation to each atom is d/2. Treat each atom as a particle for calculating the moment of inertia. I = ( M 2 )( d 2 ) + ( M 2 )( d 2 ) = 2 ( M 2 )( d 2 ) = 14 Md 2 → 2

(

d = 4 I M = 4 1.9  10−46 kg m 2

) ( 5.3 10 kg ) = 1.2 10 m −26

−10

37. (a) The moment of inertia of a cylinder is 12 MR 2 . I = 12 MR 2 = 12 ( 0.380 kg )( 0.0850 m ) = 1.373  10−3 kg m 2  1.37  10−3 kg m 2 2

(b) The wheel slows down “on its own” from 1500 rpm to rest in 55.0s. This is used to calculate the frictional torque.  ( 0 − 1500 rev min )( 2 rad rev )(1 min 60 s ) = (1.373  10−3 kg m 2 )  fr = I  fr = I t 55.0 s = −3.921  10−3 m N

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The net torque causing the angular acceleration is the applied torque plus the (negative) frictional torque.   =  applied +  fr = I →  applied = I −  fr = I t −  fr (1950 rev min )( 2 rad rev )(1 min 60 s ) = (1.373  10−3 kg m 2 ) − ( −3.921  10−3 m N ) 5.00 s = 6.00  10−2 m N

38. (a) The torque gives angular acceleration to the ball only, since the arm is considered massless. The angular acceleration of the ball is found from the given tangential acceleration. The ball is treated as a point mass to find its rotational inertia about the elbow joint. a  = I = MR 2 = MR 2 tan = MRa tan = ( 3.6 kg )( 0.31m ) ( 7.0 m s 2 ) R = 7.812 m N  7.8 m N

(b) The triceps muscle must produce the torque required, but with a lever arm of only 2.5 cm, perpendicular to the triceps muscle force.  = FR⊥ → F =  R⊥ = 7.812 m N 2.5  10−2 m = 310 N

(

)

39. (a) The angular acceleration can be found from the following.   v r ( 8.5 m s ) ( 0.31m ) = = = = = 72.16 rad s 2  72 rad s 2 t t t 0.38s (b) The force required can be found from the torque, since  = Fr sin  . In this situation the force is perpendicular to the lever arm, and so  = 90. The torque is also given by  = I , where I is the moment of inertia of the arm-ball combination. Equate the two expressions for the torque, and solve for the force. Fr sin  = I F=

I r sin 

2 2 + 13 marm l arm mball d ball

r sin 90o



(1.00 kg )( 0.31m ) + 13 ( 3.7 kg )( 0.31m ) 72.16 rad s 2 ) = 619.5 N  620 N ( ( 0.025 m ) 2

40. The torque is calculated from  = I  . The rotational inertia of a rod about its end is I = 13 M l 2 .

 = I = 13 M l 2

 t

= 13 ( 0.90 kg )( 0.95 m )

( 2.4 rev s )( 2 rad rev ) 0.20 s

= 20.41m N

 2.0  101 m N

41. (a) The small ball can be treated as a particle for calculating its moment of inertia.

I = MR 2 = ( 0.35kg )(1.4 m ) = 0.686 kg m2  0.69 kg m2 2

(b) To keep a constant angular velocity, the net torque must be zero, and so the torque needed is the same magnitude as the torque caused by friction.

 = 

applied

−  fr = 0 →  applied =  fr = Ffr r = ( 0.020 N )(1.4 m ) = 2.8  10−2 m N

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42. (a) To calculate the moment of inertia about the y-axis (vertical), use the following. 2 2 2 2 I =  M i Rix2 = m ( 0.50 m ) + M ( 0.50 m ) + m (1.00 m ) + M (1.00 m ) = ( m + M ) ( 0.50 m ) + (1.00 m )  = ( 5.4 kg )  ( 0.50 m ) + (1.00 m )  = 6.75 kg m 2 2

 6.8 kg m 2

(b) To calculate the moment of inertia about the x-axis (horizontal), use the following.

I =  M i Riy2 = ( 2m + 2 M )( 0.25m ) = 2 ( m + M ) ( 0.25m ) = 0.68 kg m 2 2

(c) Because of the larger I value, it is ten times harder to accelerate the array about the y ( vertical ) axis . 43. (a) The torque exerted by the frictional force is  = rFfr sin  . The force of friction is assumed to be tangential to the clay, and so  = 90.

 total = rFfr sin  = ( ( 0.090 m ) ) (1.8 N ) sin 90 = 0.081m N 1 2

direction of rotation

(b) The time to stop is found from  = 0 +  t , with a final angular

Ffr

velocity of 0. The angular acceleration can be found from  total = I  . The net torque (and angular acceleration) is negative since the object is slowing.  − o  − o 0 − (1.6 rev s )( 2 rad rev ) = = = 13.65s  14 s t=   I ( −0.081m N ) ( 0.11kg m 2 ) 44. The torque needed is the moment of inertia of the system (merry-go-round and children) times the angular acceleration of the system. Let the subscript “mgr” represent the merry-go-round.  − 0   = I  = ( I mgr + I children ) = 12 M mgr R 2 + 2mchildren R 2 t t 2 (15 rev min )( 2 rad rev )(1min 60 s ) =  12 ( 330 kg ) + 2 ( 25 kg ) ( 2.5 m ) 10.0 s

(

)

= 211.08 m N  210 m N

The force needed is calculated from the torque and the radius. We are told that the force is directed perpendicularly to the radius (force is applied “tangentially”).  = F⊥ R sin  → F⊥ =  R = 211.08 m N 2.5 m = 84 N 45. (a) The moment of inertia of a thin rod, rotating about its end, is 13 M l 2 . There are three blades to add together.

(

)

I total = 3 13 M l 2 = M l 2 = (135 kg )( 3.75 m ) = 1898 kg m 2  1.90 103 kg m 2 2

(b) The torque required is the rotational inertia times the angular acceleration, assumed constant.  − 0 ( 6.5 rev/sec )( 2 rad rev ) = (1898 kg m 2 ) = 9700 m N  = I total = I total t 8.0s

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46. The applied force causes torque, which gives the pulley an angular acceleration. Since the applied force varies with time, so will the angular acceleration. The variable acceleration will be integrated to find the angular velocity. Finally, the speed of a point on the rim is the tangential velocity of the rim of the wheel. We use the rotational inertia as calculated in Example 10–10. t  R0 FT d  R0 FT R0 FT = = → = = → = → = R F I d dt d dt →       0 T   I dt I I 0 0

=

vT R0

R0 I

 FT dt → vT =

R02

vT ( t = 9.0 s ) =

 FT dt = 0

R02 I

(

)

2  3.00t − 0.20t dt = 0

R02

( 3.00 t 2 − 0.20 t 3 ) N s  2 3 I

( 0.330 m )  3.00 ( 2 ( 9.0 )2 − 0.203 ( 9.0 )3 ) N s  = 20.62 m s  20 m s 2 2

 0.385 kg m  Note that the answer is only “good” to one significant figure. 47. Each mass is treated as a point particle. The first mass is at the axis of rotation; the second mass is a distance l from the axis of rotation; the third mass is 2l from the axis, and the fourth mass is 3l from the axis.

(a) I = M l 2 + M ( 2l ) + M ( 3l ) = 14M l 2 (b) The torque to rotate the rod is the perpendicular component of force times the lever arm, and is also the moment of inertia times the angular acceleration. I 14 M l 2 = = 143 M l   = I = F⊥ r → F⊥ = r 3l (c) The force must be perpendicular to the rod connecting the masses, and perpendicular to the axis of rotation. An appropriate direction is shown in the diagram. 2

48. The firing force of the rockets will create a net torque, but no net force. Since each rocket fires tangentially, each force has a lever arm equal to the radius of the satellite, and each force is perpendicular to the lever arm. Thus  net = 4 FR. This torque will cause an angular acceleration according to  = I , where I = 12 MR 2 + 4mR 2 , combining a cylinder of mass M and radius R with 4 point masses of mass m and lever arm R each. The angular acceleration can be found from the  . Equating the two expressions for the torque and substituting enables us to kinematics by  = t solve for the force.  ( 1 M + 4m ) R 4 FR = I  = ( 12 M + 4m ) R 2 → F= 2 4 t  1  2 ( 3600kg ) + 4 ( 250 kg )  ( 4.0 m )( 28 rev min )( 2 rad rev )(1 min 60 s ) = = 27.37 N 4 ( 5.0 min )( 60 s min )

 27 N 49. (a) Thin hoop, radius R0

I = Mk 2 = MR02 → k = R0

(b) Thin hoop, radius R0 , width w

I = Mk 2 = 12 MR02 + 121 Mw 2 → k =

I = Mk 2 = 12 MR02 → k =

1 2

R02 + 121 w 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

Instructor Solutions Manual

) → k = (R + R )

(d) Hollow cylinder

I = Mk 2 = 12 M R 2 + R 2

(e) Uniform sphere

I = Mk 2 = 25 Mr02 → k =

2 5 r0

(f)

I = Mk 2 = 121 M l 2 → k =

1 12

(g) Long rod, through end

I = Mk 2 = 13 M l 2 → k =

1 3

(h) Rectangular thin plate

I = Mk 2 = 121 M l 2 + w2

Long rod, through center

(

1 2

) → k=

50. (a) The free-body diagrams are shown. Note that only the forces producing torque are shown on the pulley. There would also be a gravity force on the pulley (since it has mass) and a normal force from the pulley’s suspension, but they are not shown. (b) Write Newton’s second law for the two blocks, taking the positive x-direction as shown in the free-body diagrams. mA :  Fx = FTA − mA g sin  A = mA a → FTA = mA ( g sin  A + a )

(

1 12

(l + w ) 2

FTA

FNA y

x A

A

mA g

FTA

FTB

)

= ( 8.0kg )  9.80 m s 2 sin 32 + 1.00 m s 2  = 49.55 N  50 N ( 2 sig. figs.) mB :

FNB FTB

 F = m g sin  − F = m a → x

FTB = mB ( g sin  B − a )

(

)

= (10.0kg )  9.80 m s sin 61 − 1.00 m s  = 75.71N 2

B

mB g

 76 N

(c) The net torque on the pulley is caused by the two tensions. We take clockwise torques as positive.  = ( FTB − FTB ) R = ( 75.71N − 49.55 N )( 0.15m ) = 3.924m N  3.9m N Use Newton’s second law to find the rotational inertia of the pulley. The tangential acceleration of the pulley’s rim is the same as the linear acceleration of the blocks, assuming that the string doesn’t slip. a  = I = I R = ( FTB − FTB ) R →

( FTB − FTB ) R 2 a

( 75.71N − 49.55 N )( 0.15 m )2 1.00 m s

= 0.59 kg m 2

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51. (a) Since mB  mA , mB will accelerate down, mA will accelerate up, and the pulley will accelerate clockwise. Call the direction of acceleration the positive direction for each object. The masses will have the same acceleration since they are connected by a cord. The rim of the pulley has that same acceleration since the cord makes it rotate (assuming that there is no slipping), and so  pulley = a R . From the free-body diagrams for each object, we have the following.  FyA = FTA − mA g = mA a → FTA = mA g + mA a

+

FTA

FTB

FTA

FTB

mB + y

mA g

mB g

F = m g −F = m a → F = m g −m a yB

 = F r − F r = I = I R TB

We have to assume that the tensions are unequal in order to have a net torque to accelerate the pulley. Substitute the expressions for the tensions into the torque equation, and solve for the acceleration. a a FTB R − FTA R = I → ( mB g − m B a ) R − ( mA g + m A a ) R = I → R R ( m B − mA ) ( m B − mA ) a= g= g 2 mA + mB + I R mA + mB + 12 mP R 2 R 2

(

)

(

)

( 75 kg − 65 kg ) 9.80 m s 2 ) = 0.6853m s 2  0.69 m s 2 ( 1 + + 75 kg 65 kg 6.0 kg )  2 (

(b) If the moment of inertia is ignored, then from the torque equation we see that FTB = FTA , and the acceleration will be aI =0 =

( mB − mA ) ( 75 kg − 65 kg ) g= 9.80 m s 2 ) = 0.7000 m s 2 . We ( 75 kg + 65 kg ( mA + mB )

calculate the percent difference, which is small because of the relatively small mass of the pulley.  0.7000 m s2 − 0.6853m s 2  % error =    100 = 2.145%  2% 0.6853m s 2   52. A top view diagram of the hammer is shown, just at the instant of release, along with the acceleration vectors. (a) The angular acceleration is found from Eq. 10–9c.

 − (v r ) − 0 = 2 2 2

 2 = 02 + 2 →  =

2 0

a net

a tan 

a rad

( 26.5 m s ) (1.20 m ) = 9.702 rad s  9.70 rad s 2

2 ( 8 rad ) (b) The tangential acceleration is found from the angular acceleration and the radius.

(

)

atan =  R = 9.702 rad s 2 (1.20 m ) = 11.64 m s 2  11.6 m s 2 (c) The centripetal acceleration is found from the speed and the radius.

arad = v 2 R = ( 26.5 m s )

(1.20 m ) = 585.2 m s2  585 m s2

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(d) The net force is the mass times the net acceleration. It is in the same direction as the net acceleration.

(11.64 m s ) + (585.2 m s ) = 4270 N

2 2 Fnet = manet = m atan + arad = ( 7.30 kg )

(e) Find the angle from the two acceleration vectors. 11.64 m s 2 a  = tan −1 tan = tan −1 = 1.14 arad 585.2 m s 2 53. The parallel axis theorem is given in Eq. 10–17. The distance from the center of mass of the rod to the end of the rod is h = 12 l .

I = I CM + Mh 2 = 121 M l 2 + M ( 12 l ) = ( 121 + 14 ) M l 2 = 13 M l 2 2

54. We can consider the door to be made of a large number of thin horizontal rods, each of length l = 1.0 m, and rotating about one end. Two such rods are shown in the diagram. The moment of inertia of one of these rods is 1 mi l 2 , where mi is the mass of a single rod. For a collection of identical 3 rods, then, the moment of inertia would be I =  13 mi l 2 = 13 M l 2 . The i

height of the door does not enter into the calculation directly.

I = 13 M l 2 = 13 (19.0 kg )(1.0 m ) = 6.3kg m 2 2

55. (a) The parallel axis theorem (Eq. 10–17) is to be applied to each sphere. The distance from the center of mass of each sphere to the axis of rotation is h = 1.5r0 .

I for one = I CM + Mh 2 = 25 Mr02 + M (1.5r0 ) = 2.65Mr02 → I total = 5.3Mr02 2

sphere

(b) Treating each mass as a point mass, the point mass would be a distance of 1.5r0 from the axis of rotation.

I approx = 2  M (1.5r0 )  = 4.5Mr02 2

 I approx − I exact   4.5Mr02 − 5.3Mr02  4.5 − 5.3  = 100 ( ) (100 ) =  (100 )  2   I exact 5.3Mr0  5.3     

% error = 

= −15% The negative sign means that the approximation is smaller than the exact value, by about 15%. 56. (a) Treating the ball as a point mass, the moment of inertia about AB is I = MR02 . (b) The parallel axis theorem is given in Eq. 10–17. The distance from the center of mass of the ball to the axis of rotation is h = R0 .

I = I CM + Mh 2 = 25 Mr12 + MR02

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Chapter 10

(c)

Rotational Motion

 MR02 − ( 25 Mr12 + MR02 )   I approx − I exact  − 25 Mr12 100 100 = = ( ) ( ) (100)    2 2 I exact Mr12 + MR02 Mr12 + MR02     5 5

% error = 

−1

(100) = −0.3411  −0.34 2 1 + 25 ( R0 r1 ) 1 + 25 (1.0 0.0925) The negative sign means that the approximation is smaller than the exact value, by about 0.34%. −1 −1 (d) % error = 100 ) = − ( (100) = −0.2553  −0.26 2 2 1 + 25 ( R0 r1 ) 1 + 25 (1.0 0.080 ) As the radius of the ball gets smaller, the better the approximation of part (a) becomes. In this case, the approximation is smaller than the exact value, by about 0.26%. =

(100) = −

57. The 1.50-kg weight is treated as a point mass. The origin is placed at the center of the wheel, with the x-direction to the right. Let A represent the wheel and B represent the weight. m x + mB xB ( 7.0 kg )( 0 ) + (1.50 kg )( 0.24 m ) (a) xCM = A A = 8.50 kg mA + mB = 4.24  10−2 m  0.042 m from the center, on a line to the small mass (b) The moment of inertia of the wheel about the CM is found from the parallel axis theorem. 2 I = I wheel + I weight = I wheel + M wheel xCM + M weight ( x weight − xCM )

2 = 12 M wheel R 2 + M wheel xCM + M weight ( x weight − xCM )

(

= ( 7.0 kg ) 12 ( 0.34 m ) + ( 0.0424 m ) 2

) + (1.50 kg )(0.24 m − 0.0424 m ) = 0.48 kg m 2

58. We calculate the moment of inertia about one end, and then use the parallel axis theorem to find the moment of inertia about the center. Let the mass of the rod be M, and use Eq. 10–16. A small mass dM can be found as a small length dx times the mass per unit length of the rod. l M M l3 1 I end =  R 2 dM =  x 2 dx = = 3 Ml2 l l 3 0

I end = I CM + M ( 12 l )

→ I CM = I end − M ( 12 l ) = 13 M l 2 − 14 M l 2 = 121 M l 2

Alternatively we could consider the picture as measuring the distance from the center of the rod, but only showing the right half of the rod. Then we would have the following. l 2

3 M l3 l3  1 l3 M x I end =  R dM =  x dx =   M = 13 − − = = 121 M l 2 3   8 4 l l 8  l  3 −l 2 −l 2 l 2

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59. (a) We choose coordinates so that the center of the plate is at the origin. Divide the plate up into differential rectangular elements, each with an area of dA = dxdy. The mass of an

M   dxdy. The distance of that element from  l w

element is dm = 

the axis of rotation is R = x 2 + y 2 . Use Eq. 10–16 to calculate the moment of inertia. l /2 w/ 2 w/ 2 l / 2 M 4M 2 2 2 I center =  R dM =   x + y dxdy = x 2 + y 2 dxdy   lw lw 0 0 − w/ 2 − l / 2

(

)

(

w/ 2

)

 13 ( 12 l )3 + ( 12 l ) y 2  dy = 2 M   121 l 2 + y 2  dy =   lw  w =

 121 l 2 ( 12 w ) + 13 ( 12 w )3  = 121 M ( l 2 + w2 )  w 

(b) For the axis of rotation parallel to the w dimension (so the rotation axis is in the y-direction), we can consider the plate to be made of a large number of thin rods, each of length l , rotating about an axis through their center. The moment of inertia of one of these rods is 121 mi l 2 , where mi is the mass of a single rod. For a collection of identical rods, then, the moment of inertia would be I y =  121 mi l 2 = 121 M l 2 . A similar argument would give I x = 121 Mw2 . This illustrates the i

perpendicular axis theorem, Eq. 10–18, I z = I x + I y . 60. Each infinitesimally thin disk has an infinitesimal moment of inertial given by dI = 12 R 2 dm. The radius R of the disk is related to its height above the base, taken as z. R and z are related by a cone height L − z L L−z = = → R = R0 proportion: . The mass of the disk is related to its base radius R R0 L volume, which can, again, be found by a proportion, assuming the cone’s mass is uniformly distributed. The total moment of inertia is found by adding up all the disks, which in turn means integrating over the full height of the cone. Note that the volume of a right circular cone of radius r and height h is given in the Appendices as 13  r 2h . dm = ( volume of thin disk ) dI = R dm = R  R dz 1 2

1 2

I =  dI =  0

3MR02

M volume of cone M

1 2

R L 2 0

( L − z ) dz = 5

R dz = 4

2 0

2 LR

3MR02

=  R 2 dz

M 1 3

 R02 L

3M 

3MR02

 R0  dz = 2 LR  L  2 L5 2 0

 ( L − z ) dz = − 5 4

L−z

3MR02 ( L − z ) 2 L5

5 z=L

= z =0

( L − z )4 dz

3MR02 10

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Chapter 10

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61. Work can be expressed in rotational quantities as W =   , and so power can be expressed in

rotational quantities as P =

t

P = F v.

=

 

P =  = ( 245 m N )  3150



= . This is analogous to the linear motion equation

t

rev   2 rad   1min   1hp 



 = 108 hp

min   1rev   60s   746 W 

62. The energy required to bring the rotor up to speed from rest is equal to the final rotational kinetic energy of the rotor. 2

 rev  2 rad   1min   4 K rot = I  = ( 3.25  10 kg m ) 8750  1rev   60s   = 1.36  10 J min     1 2

−2

1 2

63. (a) For the daily rotation about its axis, treat the Earth as a uniform sphere, with an angular frequency of one revolution per day. 2 2 2 = 12 ( 25 MREarth K daily = 12 I daily ) daily 2

 2 rad   1day   = ( 6.0  10 kg )( 6.4  10 m )  = 2.6  1029 J      1day   86,400s   24

1 5

(b) For the yearly revolution about the Sun, treat the Earth as a particle, with an angular frequency of one revolution per year.





2 2 2 K yearly = 12 I yearly = 12  MRSun yearly



Earth



 2 rad   1day   33 = ( 6.0  10 kg )(1.5  10 m )    86,400s   = 2.7  10 J 365day    24

1 2

29 33 33 The total kinetic energy is K daily + K yearly = 2.6  10 J + 2.7  10 J = 2.7  10 J . The kinetic

energy due to the daily motion is about 10,000 times smaller than that due to the yearly motion. 64. To maintain a constant angular speed steady will require a torque  motor to oppose the frictional torque. The power required by the motor is P =  motorsteady = − frictionsteady .

  − 0   friction = I friction = 12 MR 2  f  →  t  2

  2 rad   3.8 rev s ( )     0 −  f  2   rev   = 1.054  105 W 1  = Pmotor = 12 MR 2  220 kg 5.5 m ( )( )  steady 2 18s  t   1hp   = 141.3 hp  140 hp  746 W 

= 1.054  105 W 

65. The work required is the change in rotational kinetic energy. The initial angular velocity is 0.

W = K rot = 12 I  2f − 12 I i2 = 12

(

1 2

)

 2 rad  = 1.40  104 J   7.00 s 

MR 2  2f = 14 (1240 kg )( 7.50 m ) 

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Instructor Solutions Manual

66. Mechanical energy will be conserved. The rotation is about a fixed axis, so K tot = K rot = 12 I  2 . For gravitational potential energy, we can treat the object as if all of its mass were at its center of mass. Take the lowest point of the center of mass as the zero location for gravitational potential energy. Einitial = Efinal → U initial = K final → 2 Mg 12 l (1 − cos  ) = 12 I bottom = 12

bottom =

( M l ) 2

1 3

2 bottom

(1 − cos  ) ; vbottom = bottom l =

→ 3 g l (1 − cos  )

67. The only force doing work in this system is gravity, so mechanical energy is conserved. The initial state of the system is the configuration with mA on M

the ground and all objects at rest. The final state of the system has mB just reaching the ground, and all objects in motion. Call the zero level of gravitational potential energy to be the ground level. Both masses will have the same speed since they are connected by the rope. Assuming that the rope does not slip on the pulley, the angular speed of the pulley is related to the speed of the masses by  = v R . All objects have an initial speed of 0.

1 2

mB h

→

Ei = E f

mA vi2 + 12 mBvi2 + 12 I i2 + mA g y1i + mB g y2i = 12 mA v 2f + 12 mBv 2f + 12 I  2f + mA g y1 f + mB g y2 f

 v 2f  mB gh = mA v + mB v + ( MR )  2  + mA gh R  1 2

vf =

2 f

1 2

2 ( mB − mA ) gh

( mA + mB + M ) 1 2

1 2

(

)

2 ( 41.0 kg − 35.0 kg ) 9.80 m s 2 ( 2.5 m )

( 41.0 kg + 35.0 kg + ( ) 3.1 kg ) 1 2

= 1.947 m s  1.9 m s

68. Since the lower end of the pole does not slip on the ground, there is no motion at the ground, and so any friction between the pole and the ground does no work. Thus, mechanical energy is conserved. The initial energy is the potential energy, treating all the mass as if it were at the CM. The final energy is rotational kinetic energy, for rotation about the point of contact with the ground. The angular velocity of the falling rod is equal to the linear velocity of the falling tip of the rod divided by the length. Eintial = Efinal → U initial = K final → mgh = 12 I  2 → mg l 2 = 12

(

( ml ) ( v 1 3

end

→

)

vend = 3 g l = 3 9.80 m s 2 ( 2.30 m ) = 8.22 m s

69. Apply conservation of mechanical energy. Take the bottom of the incline to be the zero location for gravitational potential energy. The energy at the top of the incline is then all gravitational potential energy, and at the bottom of the incline, there is both rotational and translational kinetic energy. Since the cylinder rolls without slipping, the angular velocity is given by  = v R .

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Chapter 10

Rotational Motion

Etop = Ebottom → Mgh = M v + I CM = M v + 2

1 2

4 3

gh =

4 3

1 2

1 1 2 2

v2 R

= 43 M v 2 →

( 9.80 m s ) ( 6.50 m ) = 9.22 m s 2

70. The total kinetic energy is the sum of the translational and rotational kinetic energies. Since the ball is rolling without slipping, the angular velocity is given by  = v R . The rotational inertia of a sphere about an axis through its center is I = 25 mR 2 . K total = K trans + K rot = mv + I  = mv + 2

1 2

1 2 2 5

v2 R

= 107 mv 2

= 0.7 ( 7.3 kg )( 3.7 m s ) = 7.0  10 J 2

71. (a) Mechanical energy is conserved as the sphere rolls without slipping down the plane. Take the zero level of gravitational potential energy to the level of the center of mass of the sphere when it is on the level surface at the bottom of the plane. All of the energy is potential energy at the top, and all is kinetic energy (of both translation and rotation) at the bottom. Eintial = Efinal → U initial = K final = K CM + K rot → 2 2 2 mgh = mg l sin  = 12 mv bottom + 12 I bottom = 12 mv bottom + 12

v bottom =

10 7

gh =

10 7

g l sin  =

10 7

(

2 5

v  ) bottom   r0 

mr02 

→

( 9.80 m s ) (10.0 m ) sin 30.0 = 8.367 m s 2

 8.37 m s

bottom = (b)

K CM K rot

v bottom r0

1 2

2 mvbottom

I

= 1

2 bottom

8.367 m s 0.245 m 1 2

= 1 2

= 34.2 rad s

2 mvbottom

 bottom    0 

( mr )  v r 2 5

2 0

5 2

(c) The translational speed at the bottom, and the ratio of kinetic energies are both independent of the radius and the mass. The rotational speed at the bottom depends on the radius. 72. (a) Since the center of mass of the spool is stationary, the net force must be 0. Thus, the force on the thread must be equal to the weight of the spool and so Fthread = Mg . (b) By the work–energy theorem, the work done is the change in kinetic energy of the spool. The spool has rotational kinetic energy.

W = Kfinal − Kinitial = 12 I  2 = 12

( MR )  = 1 2

1 4

MR 2 2

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73. (a) We work in the accelerating reference frame of the car. In the accelerating frame, we must add a fictitious force of magnitude M a train rel , in the opposite direction to the acceleration of the train. This

ground

is discussed in detail in Section 11–8 of the textbook. Since the ball is rolling without slipping,  = aball rel R . See the free-body diagram for

Ma train

train

the ball in the accelerating reference frame. Write Newton’s second law for the horizontal direction and for torques, with clockwise torques as positive. Combine these relationships to find aball rel , the acceleration of

R Mg

Ffr

train

the ball in the accelerated frame. aball rel  = − Ffr R = I = I trainR

 F = F − Ma x

train rel ground

→ Ffr = − 25 Maball rel train

= Maball rel →

− Maball rel − Matrain rel = Maball rel → 2 5

train

ground

train

aball rel = − 75 a train rel train

ground

And so as seen from inside the train, the ball is accelerating backwards. (b) Use the relative acceleration relationship. aball rel = aball rel + a train rel = − 75 atrain rel + atrain rel = 72 a train ground

train

ground

And so as seen from outside the train, the ball is accelerating forwards, but with a smaller acceleration than the train. 74. (a) Use conservation of mechanical energy. Call the zero level for gravitational potential energy to be the lowest point on which the pipe rolls. Since the pipe rolls without slipping,  = v R . See the attached diagram. Einitial = Efinal → U initial = K final = K CM + K rot 2 2 + 12 I bottom mg l sin  = 12 mv bottom

(

)

2  vbottom  2  = mvbottom → 2 R  

2 = 12 mv bottom + 12 mR 2 

vbottom =

g l sin  =

( 9.80 m s ) ( 5.60 m ) sin17.5 = 4.06 m s 2

(b) The total kinetic energy at the base of the incline is the same as the initial potential energy. K final = U initial = mg l sin  = ( 0.545 kg ) 9.80 m s 2 ( 5.60 m ) sin17.5 = 8.99 J

(

)

(c) The frictional force supplies the torque for the object to roll without slipping, and the frictional force has a maximum value. Since the object rolls without slipping,  = a R . Use Newton’s second law for the directions parallel and perpendicular to the plane, and for the torque, to solve for the coefficient of friction. a  = Ffr R = I = mR 2 R = maR → Ffr = ma  F⊥ = FN − mg cos  → FN = mg cos  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 10

Rotational Motion

 F = mg sin  − F = ma → F = mg sin  fr

Ffr  Fstatic →

1 2

mg sin   s FN = s mg cos  → s  12 tan  →

max

s = 12 tan  = 12 tan17.5 = 0.158 min

75. (a) While the ball is slipping, the acceleration of the center of mass is constant, and so constant acceleration relationships may be used. Use Eq. 2–12b with results from Example 10–21. 2

 2v 0  1  2v 0  12v02 x − x0 = v0t + at = v0  g + −  = ( ) k  2   49  k g  7 k g   7 k g  1 2

(b) Again make use of the fact that the acceleration is constant. Once the final speed is reached, the angular velocity is given by  = v r0 .

 2v  v = v0 + at = v0 + ( − k g )  0  = 75 v0 ;  = 75 v0 r0  7 k g  76. Use conservation of mechanical energy to equate the energy at points A and B. Call the zero level for gravitational potential energy to be the lowest point on which the ball rolls. Since the ball rolls without slipping,  = v r0 . EA = EB → U A = U B + K B final = U B + K B CM + K B rot → mgR0 = mgr0 + 12 mvB2 + 12 I B2

v  = mgr0 + mv + ( mr )  B   r0  1 2

2 B

1 2

2 5

2 0

→ vB =

10 7

g ( R0 − r0 )

77. (a) Use conservation of mechanical energy to equate the energy at point A to the energy at point C. Call the zero level for gravitational potential energy to be the lowest point on which the ball rolls. Since the ball rolls without slipping,  = v r0 . All locations given for the ball are for its center of mass. E A = EC → U A = U C + KC = U C + KC + KC CM

→

rot

mgR0 = mg  R0 − ( R0 − r0 ) cos   + 12 mvC2 + 12 I C2 = mg  R0 − ( R0 − r0 ) cos   + 12 mvC2 + 12

vC =

10 7

g ( R0 − r0 ) cos  =

10 7

vC2

( mr ) r 2 5

2 0

→

( 9.80 m s ) ( 0.245 m ) cos 45 = 1.557 m s  1.6 m s 2

(b) Once the ball leaves the ramp, it will move as a projectile under the influence of gravity, and the constant acceleration equations may be used to find the distance. The initial location of the ball is given by x0 = ( R0 − r0 ) sin 45 and y0 = R0 − ( R0 − r0 ) cos 45. The initial velocity of the ball © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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is given by v0 x = vC cos 45 and v0 y = vC sin 45. The ball lands when y = r0 = 0.015 m. Find the time of flight from the vertical motion, and then find D from the horizontal motion. Take the upward direction as positive for the vertical motion. y = y0 + v0 y t + 12 at 2 = R0 − ( R0 − r0 ) cos 45 + vC sin 45t − 12 gt 2 → 4.90t 2 − 1.101t − 0.07178 = 0 → t = 0.277 s, − 0.0528s We use the positive time. D = x = x0 + v0 x t = ( R0 − r0 ) sin 45 + vC cos 45t = ( 0.245 m ) sin 45 + (1.557 m s ) cos 45 ( 0.277 s ) = 0.4782 m  0.48 m

78. (a)

The friction force accelerates the center of mass of the wheel. If the wheel is spinning (and slipping) clockwise in the diagram, then the surface of the wheel that touches the ground is moving to the left, and the friction force is to the right or forward. It acts in the direction of motion of the velocity of the center of mass of the wheel. (b) Write Newton’s second law for the x-direction, the y-direction, and the rotation. Take clockwise torques (about the center of mass) as positive.  Fy = FN − Mg = 0 → FN = Mg

k FN

Ffr

k Mg

 F = F = Ma → a = M = M = M x

Ffr R

= k g

2 Mg

Ffr

2 k g

 = − F R = I →  = − MR = − MR = − R fr

1 2

Both the acceleration and angular acceleration are constant, and so constant acceleration kinematics may be used to express the velocity and angular velocity. 2 g v = v0 + at = k gt ;  = 0 +  t = 0 − k t R Note that the translational velocity starts at 0 and increases, while the angular velocity starts at 0 and decreases. Thus at some specific time T , the translational velocity and angular velocity will be related by  = v R, and the ball will then roll without slipping. Solve for the value of T needed to make that true.

 = v R → 0 −

2k g R

T =  k gT R → T =

0 R 3k g

(c) Once the ball starts rolling without slipping, there is no more frictional sliding force, and so the velocity will remain constant. R vfinal =  k gT =  k g 0 = 13 R0 3 k g

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79. (a) The total kinetic energy included the translational kinetic energy of the car’s total mass, and the rotational kinetic energy of the car’s wheels. The wheels can be treated as one cylinder. We assume the wheels are rolling without slipping, so that vCM =  Rwheels . 2 2 + 12 I wheels 2 = 12 M tot vCM + 12 K tot = K CM + K rot = 12 M tot vCM

(

1 2

2 M wheels Rwheels

2 vCM

2 wheels

1 2

tot

+ M wheels ) v 1 2

2 CM

  1m s   = (1170 kg ) ( 95 km h )  = 4.074  105 J    3.6 km h    1 2

 4.1  105 J K rot + K trans

(b) The fraction of kinetic energy in the tires and wheels is

wheels

K tot

2 ( 32 M wheels ) ( 12 M wheels + M wheels ) vCM = 1 = = 2 2 1 K tot M tot vCM M tot + 12 M wheels ) vCM + 12 I wheels 2 ( M tot + 12 M wheels ) 2 2(

K rot

1 2

2 I wheels 2 + 12 M wheels vCM

1 2

210 kg

= 0.18 1170 kg (c) A free-body diagram for the car is shown, with the frictional force of Ffr at each wheel to cause the wheels to roll. A separate diagram of one wheel is also shown. Write Newton’s second law for the horizontal motion of the car as a whole, and the rotational motion of one wheel. Take clockwise torques as positive. Since the wheels are rolling without slipping, aCM =  Rwheels .

 = 4 F R = I fr

2  = 12 M wheels Rwheels

wheels

aCM Rwheels

FN Ftow

4Ffriction

→ R

Ffr = M wheels aCM 1 8

Ffriction

 Fx = Ftow − 4 Ffr = M tot aCM →

Ftow − 4 ( 18 M wheels aCM ) = M tot aCM → aCM =

Ftow

1500 N

( M tot + M wheels ) (1170 kg ) 1 2

= 1.282 m s2  1.3m s2

(d) If the rotational inertia were ignored, we would have the following. F 1500 N  Fx = Ftow = M tot aCM → aCM = Mtow = 1100 kg = 1.364 m s2 tot % error =

aCM aCM

 100 =

1.364 m s 2 − 1.282 m s 2 1.282 m s 2

 100 = 6%

80. Write the rotational version of Newton’s second law, with counterclockwise torques as positive.  net =  N −  fr = FN l − FR = I CN CM = 52 MR 2 CM Newton’s second law for the translational motion, with left as the positive direction, gives the following. F Fnet = F = ma → a = m © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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If the sphere is rolling without slipping, we have  CM = a R . Combine these relationships to analyze the relationship between the torques. a FN l = FR + 52 MR 2 CM = FR + 52 MR 2 = FR + 52 MaR = FR + 52 FR = 75 FR → R

 N = 75  fr And since the torque due to the normal force is larger than the torque due to friction, the sphere has a counterclockwise angular acceleration, and thus the rotational velocity will decrease. 81. The torque is found from  = I  . The angular acceleration can be found from  = 0 +  t, with an initial angular velocity of 0. The rotational inertia is that of a cylinder.   − 0  = 0.5 1.6 kg 0.20 m 2 ( 22 rev s )( 2 rad rev ) = 0.74 m N  = I = 12 MR 2  ( )( )  6.0s  t  82. The linear speed is related to the angular velocity by v =  R, and the angular velocity (rad / sec) is related to the frequency (rev / sec) by  = 2 f . Combine these relationships to find values for the frequency. 1.25 m s  60 s  v v v ; f1 =  = 2 f = → f = =   = 480 rpm R 2 R 2 R1 2 ( 0.025 m )  1min  f2 =

v 2 R2

 60 s    = 210 rpm 2 ( 0.058 m )  1min  1.25 m s

83. (a) There are two forces on the yo-yo; gravity and string tension. If the top of the string is held fixed, then the tension FT does no work, and so mechanical energy is conserved. The initial gravitational potential energy is converted into rotational and translational kinetic energy. Since the yo-yo rolls without slipping at the point of contact of the string, the mg velocity of the CM is related to the angular velocity of the yo-yo by vCM = r , where r is the radius of the inner hub. Let m be the mass of the inner hub, and M and R be the mass and radius of each outer disk. Calculate the rotational inertia of the yo-yo about its CM, and then use conservation of energy to find the linear speed of the CM. We take the 0 of gravitational potential energy to be at the bottom of its fall. I CM = 12 mr 2 + 2 12 MR 2 = 12 mr 2 + MR 2

(

)

)(

) + ( 5.0  10 kg )( 3.75  10 m ) = 7.042  10 kg m = m + 2 M = 5.0  10 kg + 2 ( 5.0  10 kg ) = 0.105 kg

= 12 5.0  10−3 kg 6.5  10−3 m mtotal

−2

−3

−2

−5

−2

U initial = K final → 2 2 + 12 I CM 2 = 12 mtotal vCM + 12 mtotal gh = 12 mtotal vCM

I  2  2 vCM = 12  mtotal + CM  vCM → r r2  

I CM 2

360

Chapter 10

Rotational Motion

mtotal gh = I CM  1  mtotal + 2  2 r  

vCM =

( 0.105 kg ) ( 9.80 m s 2 ) (1.0 m )  7.042  10 −5 kg m 2 )  ( 1   + 0.105 kg ) 2 ( 2 −3   6.5 10 m ( ) 

= 1.078 m s

= 1.1m s

(b) Calculate the ratio K rot K tot . K rot K tot

K rot U initial

1 2

I CM 2

mtotal gh

1 2

I CM

2 vCM

I v2 r = 2CM CM 2r mtotal gh mtotal gh 2

( 7.042  10 kg m ) (1.078 m s ) = = 0.9412 = 94% 2 ( 6.5  10 m ) ( 0.105 kg ) ( 9.80 m s ) (1.0 m ) −5

−3

84. As discussed in the text, from the reference frame of the axle of the wheel, the points on the wheel are all moving with the same speed of v = r , where v is the speed of the axle of the wheel relative to the ground. The top of the tire has a velocity of v to the right relative to the axle, so it has a velocity of 2v to the right relative to the ground. v top rel = v top rel + v center rel = ( v to the right ) + ( v to the right ) = 2v to the right ground

center

ground

vtop rel = 2v = 2 ( v0 + at ) = 2at = 2 (1.00 m s 2 ) ( 2.75s ) = 5.50 m s ground

85. Assume that the angular acceleration is uniform. Then the torque required to whirl the rock is the moment of inertia of the rock (treated as a particle) times the angular acceleration. ( 0.60 kg )(1.5 m )2  rev   2 rad   1min   2   − 0   = I = ( mr )  =  75 min   rev   60s   = 2.4 m N 4.5s  t      That torque comes from the arm swinging the sling, and so comes from the arm muscles. More specifically, the torque comes from the non-radial component of the tension in the sling. The motion of David’s hand gives the sling a non-radial component. 86. Since the spool rolls without slipping, each point on the edge of the spool moves with a speed of v = r = vCM relative to the center of the spool, where v CM is the speed of the center of the spool relative to the ground. Since the spool is moving to the right relative to the ground, and the top of the spool is moving to the right relative to the center of the spool, the top of the spool is moving with a speed of 2v CM relative to the ground. This is the speed of the rope, assuming it is unrolling without slipping and is at the outer edge of the spool. The speed of the rope is the same as the speed of the person, since the person is holding the rope. So the person is walking with a speed of twice that of the center of the spool. Thus, if the person moves forward a distance l , in the same time the center of the spool, traveling with half the speed, moves forward a distance l 2 . The rope, to stay connected both to the person and to the spool, must therefore unwind by an amount l 2 also.

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87. (a) The linear speed of the chain must be the same as it passes over both sprockets. The linear speed is related to the angular speed by v =  R, and so R RR = F RF . If the spacing of the teeth on the sprockets is a distance d, then the number of teeth on a sprocket times the spacing distance must give the circumference of the sprocket. Nd = 2 R and so R =

(b)

R F = 52 13 = 4.0

(c)

R F = 42 28 = 1.5

Nd 2

NRd

. Thus R

2

= F

N Fd

→

2

R N F = F N R

88. (a) The free-body diagram and analysis from Problem 51 apply here, for the no-friction case. ( mB − mA ) ( mB − mA ) ( mB − mA ) a= g= g= g 2 2 2 mA + mB + I r mA + mB + 12 mP r r ( mA + mB + 12 mP )

(

)

(

)

( 3.80 kg − 3.05 kg )

( 9.80 m s ) = 1.028 m s  1.0 m s ( 3.80 kg + 3.05 kg + 0.30 kg ) 2

(b) With a frictional torque present, the torque equation from Problem 51 would be modified, and the analysis proceeds as follows. a a  = FTBr − FTA r −  fr = I = I r → ( mB g − mBa ) r − ( mA g + mAa ) r −  fr = I r → I      fr = r  ( mB − mA ) g −  mB + mA + 2  a  = r ( mB − mA ) g − ( mB + mA + 12 mp ) a  r     The acceleration can be found from the kinematical data and Eq. 2–12a. v − v0 0 − 0.20 m s v = v0 + at → a = = = −0.03226 m s 2 t 6.2 s

 fr = R ( mB − mA ) g − ( mB + mA + 12 mp ) a 

(

)

(

)

= ( 0.040 m ) ( 0.75 kg ) 9.80 m s 2 − ( 7.15 kg ) −0.03226 m s 2  = 0.3032 m N  0.30 m N

89. The mass of a hydrogen atom is 1.01 atomic mass units, and the atomic mass unit is 1.66  10 −27 kg. Since the axis passes through the oxygen atom, the oxygen atom will have no rotational inertia. (a) If the axis is perpendicular to the plane of the molecule, then each hydrogen atom is a distance l from the axis of rotation.

(

)(

I perp = 2mH l 2 = 2 (1.01) 1.66  10 −27 kg 0.096  10 −9 m

)

= 3.1  10 −47 kg m 2 (b) If the axis is in the plane of the molecule, bisecting the H-O-H bonds, each hydrogen atom is a distance of l y = l sin  = ( 9.6  10−11 m ) sin 52o = 7.564  10−11 m. Thus, the moment of inertia is as follows.

(

)(

I plane = 2mH l 2y = 2 (1.01) 1.66 10−27 kg 7.564 10−11 m

) = 1.9 10 kg m 2

−47

362

Chapter 10

Rotational Motion

90. (a) Assuming that there are no dissipative forces doing work, conservation of mechanical energy may be used to find the final height h of the hoop. Take the bottom of the incline to be the zero level of gravitational potential energy. We assume that the h hoop is rolling without sliding, so that  = v R . Relate the  conditions at the bottom of the incline to the conditions at the top by conservation of energy. The hoop has both translational and rotational kinetic energy at the bottom, and the rotational inertia of the hoop is given by I = mR 2 . 2 2 2 2 2 v 1 1 1 1 Ebottom = Etop → 2 mv + 2 I  = mgh → 2 mv + 2 mR 2 = mgh → R h=

v2 g

( 3.0 m s ) 9.80 m s 2

= 0.9184 m

The distance along the plane is given by d =

0.9184 m

= 2.972 m  3.0 m sin  sin18 (b) The time can be found from the constant acceleration of the linear motion. 2 ( 2.972 m ) 2x x = 12 ( v + v0 ) t → t = = = 1.981s v + v0 0 + 3.0 m s This is the time to go up the plane. The time to come back down the plane is the same, and so the total time is 4.0 s . 91. The wheel is rolling about the point of contact with the step, and so all torques are to be taken about that point. As soon as the wheel is off the floor, there will be only two forces that can exert torques on the wheel–the pulling force and the force of gravity. There will not be a normal force of contact between the wheel and the floor once the wheel is off the floor, and any force on the wheel from the point of the step cannot exert a torque about that very point. Calculate the net torque on the wheel, with clockwise torques positive. The minimum force occurs when the net torque is 0. (a) See the first diagram for part (a).

 = F ( 2 R − h ) − mg R − ( R − h ) = 0 2

Mg R 2 − ( R − h ) 2R − h

Mg 2 Rh − h 2 2R − h

= Mg

h 2R − h

(b) See the second diagram for part (b).

 = F ( R − h ) − mg R − ( R − h ) = 0 2

Mg R 2 − ( R − h )

Mg 2 Rh − h 2

R−h R−h This force is larger than the force for part (a), as expected due to the shorter lever arm for the pulling force in part (b).

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Instructor Solutions Manual

92. The ball has both rotational kinetic energy and translational kinetic energy. Find the fraction that is rotational. We assume the ball is a solid, uniform sphere. 1 2 1 1 mR 2 )  2 R 2 2 I 2 K rot 1 2 (5 5 2 CM = 1 = = 1 2 2 1 2 = 2 2 2 2 2 2 1 1 2 1 5vCM K tot 2 I CM + 2 mvCM 2 ( 5 mR )  + 2 mvCM 5 R  + 2 vCM 1+ 2 R 2 2 vCM = 145 K rot K tot

km  1000 m   1hr 

rev  2 rad   1min  = 40.28 m s ;  = 1200        = 40 rad s hr  1km   3600 s  min  1rev   60 s  1

= 1+

2 CM 2 2

2R 

= 1+

5 ( 40.28 m s )

2 ( 0.0375 m ) ( 40 rad s ) 2

1 1 + 182.66

= 5.4  10 −3 = 0.54%

93. Each wheel supports ¼ of the weight of the car. For rolling without slipping, there will be static friction between the wheel and the pavement. So for the wheel to be on the verge of slipping, there must be an applied torque that is equal to the torque supplied by the static frictional force. We take counterclockwise torques to the right in the diagram. The bottom of the wheel would be moving to the left relative to the pavement if it started to slip, so the frictional force is to the right. See the free-body diagram.  applied =  static = RFfr = R s FN = R s 14 mg min

 applied 1 4

Ffr

friction

(

)

= 14 ( 0.33m )( 0.65)(1250 kg ) 9.80 m s 2 = 660 m N

94. (a) The kinetic energy of the system is the kinetic energy of the two masses, since the rod is treated as massless. Let A represent the heavier mass, and B the lighter mass. K = 12 I AA2 + 12 I BB2 = 12 mA rA2A2 + 12 mB rB2A2 = 12 r 2 2 ( mA + mB ) = 12 ( 0.210 m ) ( 5.20 rad s ) ( 7.00 kg ) = 4.17 J 2

(b) The net force on each object produces centripetal motion, and so can be expressed as mr 2 . FA = mA rAA2 = ( 4.00 kg )( 0.210 m )( 5.20 rad s ) = 22.7 N 2

FB = mB rBB2 = ( 3.00 kg )( 0.210 m )( 5.20 rad s ) = 17.0 N 2

These forces are exerted by the rod. Since they are unequal, there would be a net horizontal force on the rod (and hence the axle) due to the masses. This horizontal force would have to be counteracted by the mounting for the rod and axle in order for the rod not to move horizontally. There is also a gravity force on each mass, balanced by a vertical force from the rod, so that there is no net vertical force on either mass. (c) Take the 4.00 kg mass to be the origin of coordinates for determining the center of mass. m x + mB xB ( 4.00 kg )( 0 ) + ( 3.00 kg )( 0.420 m ) = = 0.180 m from mass A xCM = A A 7.00 kg mA + mB So the distance from mass A to the axis of rotation is now 0.180 m, and the distance from mass B to the axis of rotation is now 0.240 m. Re-do the above calculations with these values. K = 12 I AA2 + 12 I BB2 = 12 mA rA2A2 + 12 mB rB2A2 = 12  2 ( mA rA2 + mB rB2 ) = 12 ( 5.20 rad s ) ( 4.00 kg )( 0.180 m ) + ( 3.00 kg )( 0.240 m )  = 4.09 J 2

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Chapter 10

Rotational Motion

FA = mA rAA2 = ( 4.00 kg )( 0.180 m )( 5.20 rad s ) = 19.5 N 2

FB = mB rBB2 = ( 3.00 kg )( 0.240 m )( 5.20 rad s ) = 19.5 N 2

Note that the horizontal forces are now equal, and so there will be no horizontal force on the rod or axle. 95. (a) See the free-body diagram. Take clockwise torques as positive. Write Newton’s second law for the rotational motion. The angular acceleration is constant, and so constant acceleration relationships can be used. We also s use the definition of radian angles,  = . R  = FR −  fr = I1 ; 1 = 0t1 + 12 1t12 = 12 1t12 ; s1 = R1

 fr

R F

Combine the relationships to find the length unrolled, s1.

(

)

s1 = R1 = R 12 1t12 = =

( 0.076 m )(1.1s )

(

2 3.3  10 −3 kg m 2

)

Rt12 2I

( FR −  fr )

( 2.5 N )( 0.076 m ) − ( 0.11m N ) = 1.115m  1.1m

(b) Now the external force is removed, but the frictional torque is still present. The analysis is very similar to that in part (a), except that the initial angular velocity is needed. That angular velocity is the final angular velocity from the motion in part (a).  FR −  fr  t = ( 2.5 N )( 0.076 m ) − ( 0.11m N )  1.1s = 26.667 rad s 1 = 0 + 1t1 =  ( ) 1 3.3  10−3 kg m 2  I 

(

)

 = − = I ;  −  = 2  = − ; s = R fr

2 2

2 1

Combine the relationships to find the length unrolled, s2 .

 − 2   − 2 I  ( 0.076 m )( 26.667 rad s ) ( 3.3  10 kg m ) s2 = R 2 = R  1  = R  1  = 2 ( 0.11m N )  2 2   −2 fr  2

−3

= 0.81m

96. (a) The disk starts from rest, and so the velocity of the center of mass is in the direction of the net t force: v = v 0 + at → v = Fnet . Thus the center of mass moves to the right. m (b) For the linear motion of the center of mass, we may apply constant acceleration equations, F where the acceleration is . m

v 2 = v02 + 2ax → v = 2

F m

x =

( 35 N )

( 21.0 kg )

( 5.2 m ) = 4.163 m s  4.2 m s

(c) The only torque is a constant torque caused by the constant string tension. That can be used to find the angular velocity.   − 0  = I  = Fr →  = Frt = Frt = 2 Ft  = I = I   1 I mr 2 mr  t  t 2 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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The time can be found from the center of mass motion under constant acceleration.

x = v0t + 12 at 2 = 12

=

2 Ft mr

2mx

F 2 t → t= m

2mx

2 F x

( 0.850 m )

2 ( 35.0 N )( 5.2 m )

( 21.0 kg )

= 9.796 rad s  9.8 rad s Note that v  r since the disk is NOT rolling without slipping. (d) The amount of string that has unwrapped is related to the angle through which the disk has turned, by the definition of radian measure, s = r  . The angular displacement is found from constant acceleration relationships. 2mx F 2 Ft Ft 2   F = 2 x  = 12 (0 +  ) t = 12 t = 12  = t = mr mr r  mr 

2 x

s = r = r

= 2 x = 10.4 m  1.0  101 m

r The length of string that has unwrapped is exactly twice the distance of motion of the center of mass. 97. Since frictional losses can be ignored, energy will be conserved for the marble. Define the 0 position of gravitational potential energy to be the bottom of the track, so that the bottom of the ball is initially a height h above the 0 position of gravitational potential energy. We also assume that the marble is rolling without slipping, so  = v r , and that the marble is released from rest. The marble has both translational and rotational kinetic energy. (a) Since r R, the marble’s CM is very close to the surface of the track. While the marble is on the loop, we approximate that its CM will be moving in a circle of radius R. When the marble is at the top of the loop, we approximate that its CM is a distance of 2R above the 0 position of gravitational potential energy. For the marble to just be on the verge of leaving the track means the normal force between the marble and the track is zero, and so the centripetal force at the top must be equal to the gravitational force on the marble. 2 mv top of loop 2 = mg → vtop = gR of R loop Use energy conservation to relate the release point to the point at the top of the loop. Erelease = Etop of → K release + U release = K top of + U top of loop

loop

loop 2 v top of

2 2 2 0 + mgh = 12 mv top + 12 I top + mg 2 R = 12 mv top + 12 of of of loop

loop

( mr ) r 2 5

2 mgh = 107 mvtop + 2mgR = 107 mgR + 2mgR = 2.7mgR → of

loop 2

+ 2mgR

h = 2.7 R

loop

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(b) Since we are not to assume that r R, then while the marble is on the loop portion of the track, it is moving in a circle of radius R − r, and when at the top of the loop, the bottom of the marble is a height of 2 ( R − r ) above the 0 position of gravitational potential energy (see the diagram). For the marble to just be on the verge of leaving the track means the normal force between the marble and the track is zero, and so the centripetal force at the top must be equal to the gravitational force on the marble. 2 mv top of loop 2 = mg → v top = g (R − r) 2 R − 2r of R−r loop Use energy conservation to equate the energy at the release point to the energy at the top of the loop. y=0 Erelease = Etop of → K release + U release = K top of + U top of loop

loop

2 2 2 + 12 I top + mg 2 ( R − r ) = 12 mv top + 12 0 + mgh = 12 mv top of of of loop

loop

2 v top of

( mr ) r 2 5

loop 2

+ 2mg ( R − r )

2 mgh = 107 mvtop + 2mg ( R − r ) = 107 mg ( R − r ) + 2mg ( R − r ) = 2.7mg ( R − r ) of loop

h = 2.7 ( R − r )

98. We calculate the moment of inertia about an axis through the geometric center of the rod. Select a differential element of the rod of length dx, a distance x from the center of the rod. Because the mass density changes

 

uniformly from 0 at x = − 12 l to 30 at x = 12 l , the mass density function is  = 20  1 +

 

mass of the differential element is then dM =  dx = 20  1 +

x

 . The

l 

x

 dx. Use Eq. 10–16 to calculate the

l 

moment of inertia. l /2

l /2   x3  x4   x I end =  R dM =  x 20  1 +  dx = 20   x 2 +  dx = 20  13 x 3 + 14  = 16 0 l 3 l l l     − l / 2 −l /2 −l /2  l /2

99. A free body diagram for the ball while the stick is in contact is shown. Write Newton’s second law for the x-direction, the y-direction, and the rotation. Take clockwise torques (about the center of mass) as positive. We assume that the ball is rolling without slipping during the time that the stick is in contact with the ball, and so the frictional force is static friction. We also assume the static friction has its maximum value of s FN . And

we have drawn the force of friction in such a direction that the ball is almost to slip in a clockwise direction–the bottom of the ball is on the verge of moving to the left relative to the table top.

 F = F − Mg = 0 → F = Mg y

F h

r Mg

Ffr

 F = F + F = F +  F = F +  Mg = Ma → a = M +  g x

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 = F ( h − r ) − F r = F ( h − r ) −  Mgr = I → fr

=

F ( h − r ) − s Mgr

I The acceleration and angular acceleration are constant, and so constant acceleration kinematics may be used to find the velocity and angular velocity as functions of time. The object starts from rest.  F ( h − r ) − s Mgr  F  +  s g  t ;  = 0 +  t =  vCM = v0 + at =  t I M    Since we have assumed that the ball is always rolling without slipping, we can say that  = vCM r . Solve for the value of h needed to make that true. The moment of inertia is I = 25 Mr 2 .

 F ( h − r ) − s Mgr  1  F  t=   = vCM r →  + s g  t →  I    rM 2   1  52 Mr  F   h=   + s g  + s Mgr + Fr  =   + s g  + s Mgr + Fr  F r  M    F  r M 

1 I  F

= 75

( F + s Mg ) = 75 r 1 + fr  F

F F   There are some things to note about this solution. Note that if there is no friction, then the ball must be hit at height h = 75 r. But as the force of friction increases, the height at which the ball must be hit increases. If the force of friction is its maximum, s Mg , and in the backwards direction as shown in the free-body diagram, then we get the MAXIMUM height at which the ball must be hit:   Mg  . And we see that the larger the striking force is, the less the maximum height hmax = 75 r  1 + s  F   is from the no-friction value of h = 75 r . But the static frictional force could perhaps be pointing in the opposite direction from that shown in the free-body diagram above, indicating that the ball is about to slip in the counter-clockwise direction, with the bottom of the ball on the verge of moving to the right on the table top. A similar   Mg  . And so the final conclusion is that analysis to that above would yield hmin = 75 r  1 − s  F   7 5

 

r 1 −

s Mg   s Mg  7   h  5 r 1 + . F  F  

The range over which the ball can be hit without slipping is centered on h = 75 r . The width of that range is proportional to the coefficient of friction, so a larger coefficient allows for a larger variance in the hitting position. The range is also inversely proportional to the force with which the ball is hit. So for a harder hit, the force must be done closer to h = 75 r than for a softer hit, for the ball to roll without slipping.

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100. The ladder is initially making a 35° angle with the building, and in its final position, will make a 90° angle with the building (it will be parallel to the ground). It’s initial angular velocity is 0. Finding it’s average angular acceleration is the same as assuming it has a constant angular acceleration, and then solving for that constant angular acceleration.   rad  2 ( 55 )   2  180  = 0.2133 rad s 2  0.21rad s 2  = 0 t + 12  t 2 →  = 2 = 2 t ( 3.0 s ) 101. (a) We assume that the front wheel is barely lifted off the ground, so that the only forces that act on the system are the normal force on the bike’s rear wheel, the static frictional force on the bike’s wheel, and the total weight of the system. We assume that the upward acceleration is zero and the angular acceleration about the center of mass is also zero. Write Newton’s second law for the x-direction, the y-direction, and rotation. Take torques about an axis through the center of mass, perpendicular to the page. Take positive torques to be clockwise.  Fy = FN − Mg = 0 → FN = Mg

 F = F = Ma → a = Mfr  CM = FN x − F y = 0 fr

Combine these equations to solve for the acceleration. FN x − Ffr y = 0 → Mgx = May → a =

x y

(b) Based on the form of the solution for the acceleration, a =

(c)

g , to minimize the acceleration y x should be as small as possible and y should be as large as possible . The rider should move upwards and towards the rear of the bicycle. 0.35 m x 9.80 m s 2 = 3.6 m s 2 a= g= 0.95 m y

(

)

102. We follow the hint given in the problem. The mass of the cutout piece is proportional by area to the mass of the entire piece. I total = 12 MR02 = I remainder + I cutout → I remainder = 12 MR02 − I cutout

I cutout = mcutout R + mcutout h 2 1

1 2

I remainder = MR − 2 0

1 2

= 12

M R02

( m 1 2

cutout

; mcutout =

R + mcutout h 2 1

 R02 2

R = M 2 1

R12

→

R02 R12

) = MR − M R ( R + h ) 1 2

2 0

1 2

2 1

( R − R − 2R h ) 4 0

4 1

2 1

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103. (a) See the free-body diagram for the falling rod. The axis of rotation would be coming out of the paper at the point of contact with the floor. There are contact forces between the rod and the table (the friction force and the normal force), but they act through the axis of rotation and so cause no torque; thus, only gravity causes torque. Write Newton’s second law for the rotation of the rod. Take counterclockwise to be the positive direction for rotational quantities. Thus in the diagram, the angle is positive, but the torque is negative. d  = I = −mg ( 12 l cos  ) = 13 ml 2 dt → 3g 3g d  d  d d   → − = = cos  = cos  d = − d  → 2l 2l dt d dt d 3g

(b)





( sin  − 1) = − 12  2 →  =

(1 − sin  ) 2l  / 2 2l l 0 The speed of the tip is the tangential speed of the tip, since the rod is rotating. At the tabletop,  = 0.

 cos  d = −   d →

v ( ) =  l = 3g l (1 − sin  ) → v ( 0 ) =

3g l

104. (a) If there is no friction, then conservation of mechanical energy can be used to find the speed of the block. We assume the cord unrolls from the cylinder without slipping, and so v block = vcord = cord R. We take the zero position of gravitational potential energy to be the bottom of the motion of the block. Since the cylinder does not move vertically, we do not have to consider its gravitational potential energy. Einitial = Efinal → U initial = K final = K block + K cylinder → mgh = mv + I  1 2

1 2

2mgD sin 

(m + M ) 1 2

 v2  → mgD sin  = mv + ( MR )  2  → R  1 2

(

1 2

)

2 ( 3.0 kg ) 9.80 m s 2 (1.80 m ) sin 27

(19.5 kg )

= 1.570 m s  1.6 m s

(b) Consider the free-body diagrams for both the block and the cylinder. We make the following observations and assumptions. Note that for the block to move down the plane from rest, FT  mg . Also note that

FN1

Ffr 1

mg  0.1Mg due to the difference in masses; thus, FT  0.1Mg .

Accordingly, we will ignore FT when finding the net vertical and horizontal forces on the cylinder, knowing that we will make less than a 10% error. Instead of trying to assign a specific direction for the force of friction between the cylinder and the depression ( Ffr 2 ) , we show a torque in the



 mg

FN2

counterclockwise direction (since the cylinder will rotate clockwise). Finally, we assume that Ffr 2 =  FN2 =  Mg . Write Newton’s second law to analyze the linear motion of the block and the rotational motion of the cylinder, and solve for the acceleration of the block. We assume the cord unrolls without slipping.

 fr 2

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Chapter 10

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 F = F − mg cos  = 0 → F = mg cos   F = mg sin  − F − F = mg sin  − F − mg cos  = ma y

 = F R −  T

fr 2

fr 1

= FT R −  FN2 R = FT R −  MgR = I  = I

a R

= 12 MRa →

FT −  Mg = Ma Add the x equation to the torque equation. mg sin  − FT −  mg cos  = ma ; FT −  Mg = 12 Ma → 1 2

mg sin  −  Mg −  mg cos  = ma + 12 Ma →

a=g =g

m ( sin  −  cos  ) −  M

(m + M ) 1 2

( 3.0 kg )( sin 27 − 0.035 cos 27 ) − ( 0.035 )( 33 kg ) = 0.057 m s 2 (19.5 kg )

Use Eq. 2–12c to find the speed after moving 1.80 m.

(

)

v 2 = v02 + 2ax → v = 2 0.057 m s 2 (1.80 m ) = 0.45 m s . 105. (a) The acceleration is found in Example 10–20 to be a constant value, a = 23 g , and so constant acceleration kinematics can be used. Take downward to be the positive direction. v y2 = v y2 0 + 2a y y → v y =

2a y y =

2 23 gh =

4 3

(b) We take the zero level for gravitational potential energy to be the starting height of the yo-yo. Then the final gravitational potential energy is negative. 2 2 Einitial = Efinal → 0 = U final + K final = − Mgh + 12 M vCM + 12 I CM → 2 Mgh = 12 M vCM + 12

(

1 2

v )  RCM   

→ vCM =

MR 2 

4 3

The answers agree, as of course they must if the analysis is correct. 106. Since there is no friction at the table, there are no horizontal forces on the rod, and so the center of mass will fall straight down. The moment of inertia of the rod about its center of mass is 121 M l 2 . Since there are no dissipative forces, energy will be conserved during the fall. Take the zero level of gravitational potential energy to be at the tabletop. The angular velocity and the center of mass v velocity are related by CM = CM . ( 12 l ) 2 2 Einitial = Efinal → U release = K final → Mg ( 12 l ) = 12 M vCM + 12 I CM →

Mg ( l ) = M v 1 2

1 2

2 CM

1 2

(

1 12

v  M l )  1CM  ( 2 l ) 2

2 2 → g l = 43 vCM → vCM =

3 4

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107. (a) The initial energy of the flywheel is used for two purposes–to give the car translational kinetic energy 30 times, and to replace the energy lost due to friction. The statement of the problem leads us to ignore any gravitational potential energy changes. 2 Wfr = K final − K initial → Ffr x cos180o = 12 M car vcar − K flywheel → 2 K flywheel = Ffr x + 12 M car vcar

  1m s   = ( 450 N ) ( 3.5  10 m ) + ( 30 ) (1200 kg ) ( 95 km h )    3.6 km h    5

1 2

= 1.7  108 J

(b) Kflywheel = 12 I  2

=

2 K flywheel I

2 K flywheel 1 2

2 flywheel

M flywheel R

(

2 1.7  108 J

1 2

)

( 270 kg )( 0.75 m )2

= 2116 rad s  2100 rad s

Work t

, where the

work done by the motor will be equal to the kinetic energy of the flywheel. 1.7  108 J W W  1min  P= → t= = = 1.519  103 s    25 min t P (150 hp )( 746 W hp )  60s 

(

)

108. (a) In order not to fall over, the net torque on the cyclist about an axis through the CM and parallel to the ground must be zero. Consider the free-body diagram shown. Sum torques about the CM, with counterclockwise as positive, and set the sum equal to zero.

Ffr

 = F x − F y = 0 → F = y = tan  → tan  = F N

(b) The cyclist is not accelerating vertically, so FN = mg . The cyclist is accelerating horizontally due to traveling in a circle. Thus the frictional force must be supplying the centripetal force, so Ffr = m v 2 r . tan  =

Ffr FN

m v2 r mg

v2 rg



→  = tan

−1

Ffr

( 8.5 m s ) = tan = 29.56  30 rg (13 m ) ( 9.80 m s 2 )

−1

Note that the answer has 2 significant figures. (c) From Ffr = m v 2 r , the smallest turning radius results in the maximum force. The maximum static frictional force is Ffr =  FN . Use this to calculate the radius.

( 8.5 m s ) = = 11.34 m  11m m v rmin = s FN = s mg → rmin = s g ( 0.65 ) ( 9.80 m s 2 ) 2

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CHAPTER 11: Angular Momentum; General Rotation Responses to Questions 1.

In order to do a somersault, the diver needs some initial angular momentum when she leaves the diving board, because angular momentum will be conserved during the free-fall motion of the dive. She cannot exert a torque about her CM on herself in isolation, and so if there is no angular momentum initially, there will be no rotation (and no somersault) during the rest of the dive.

Once the motorcycle leaves the ground, there is no net torque on it and angular momentum must be conserved. If the throttle is on, the rear wheel will spin faster as it leaves the ground because there is no torque from the ground acting on it to slow it down. The front of the motorcycle must rise up, or rotate in the direction opposite the rear wheel, in order to conserve angular momentum.

As you walk toward the center, the moment of inertia of the system of you + the turntable will decrease. If we assume that no external torque is acting on the system, then angular momentum will be conserved, and the angular speed of the turntable will increase.

Momentum and angular momentum are conserved for closed systems–systems in which there are no external forces or torques applied to the system. Probably no macroscopic systems on Earth are truly closed, and so external forces and torques (like those applied by air friction, for example) affect the systems over time, and the systems lose momentum.

The shortstop, while in mid-air, cannot exert a torque on himself, and so his angular momentum will be conserved while in the air. If the upper half of his body rotates in a certain direction during the throwing motion, then to conserve angular momentum, the lower half of his body will rotate in the opposite direction.

The cross product is unaffected–it remains the same. V1  V2 = ( − V1 )  ( − V2 )

The torque about the CM, which is the cross product between r and F, depends on x and z, but not on y, because the cross product will have no ĵ component. The torque is r  F = − zF ˆi + xF kˆ .

The angular momentum will remain constant. If the particle is moving in a straight line at constant speed, there is no net torque acting on it and therefore its angular momentum must be conserved.

Regarding forces: No. If two equal and opposite forces act on an object, the net force will be zero. If the forces are not co-linear, the two forces will produce a torque. Regarding torques: No. If an unbalanced force acts through the axis of rotation, there will be a net force on the object, but no net torque.

10. At the forward peak of the swinging motion, the child leans forward, increasing the gravitational torque about the axis of rotation by moving her center of mass forward. This increases the angular momentum of the system. At the back peak of the swinging motion, the child leans backward, increasing the gravitational torque about the axis of rotation by moving her center of mass backward. This again increases the angular momentum of the system. As the angular momentum of the system increases, the swing goes higher.

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11. The angular momentum is to be changed from the y-direction to the z-direction. As described in Conceptual Example 11–9, trying to tilt the wheel upward makes it swerve to the side. If instead the axle of the wheel was moved to tilt it “into the page,” which is the negative x-direction, a torque would point upward, and the change in the angular momentum of the wheel would also be upward, in the desired direction. 12. In both cases, angular momentum must be conserved. Assuming that the astronaut starts with zero angular momentum, she must move her limbs so that her total angular momentum remains zero. The angular momentum of her limbs must be opposite the angular momentum of the rest of her body. (a) In order to turn her body upside down, the astronaut could hold her arms straight out from her sides and rotate them from the shoulder in vertical circles. If she rotates them forward, her body will rotate backwards. (b) To turn her body about-face, she could hold her arms straight out from her sides and then pull one across the front of her body while she pulls the other behind her back. If she moves her arms counterclockwise, her body will twist clockwise. 13. Consider a helicopter in the air with the rotor spinning. To change the rotor’s angular speed, a torque must be applied to the rotor. That torque has to come from the helicopter itself, and so by Newton’s third law, and equal and opposite torque will be applied by the rotor to the helicopter. Any change in rotor speed would therefore cause the body of the helicopter to spin in a direction opposite to the change in the rotor’s angular velocity. Some large helicopters have two-rotor systems, spinning in opposite directions. That makes any change in the speed of the rotor pair require a net torque of 0, and so the helicopter body would not tend to spin. Smaller helicopters have a tail rotor which rotates in a vertical plane, causing a sideways force (thrust) on the tail of the helicopter in the opposite direction of the tendency of the tail to spin. 14. If we assume that the small parts just “come loose,” with no force (like explosions, for example) exerted to remove them, then the rotational speed of the wheel will not change. Both angular momentum and kinetic energy will be conserved. Since no net torque acts on the wheel or small parts, the angular momentum of the entire system is conserved. The small parts of the wheel that fly off will carry angular momentum with them relative to the original axis of rotation, and they will carry off kinetic energy as well. The remaining parts of the wheel will have a lower angular momentum and a lower rotational kinetic energy since it will have the same angular velocity but a smaller mass, and therefore a smaller moment of inertia. This question is similar to MisConceptional Question 3. Read that solution as well for further comments. 15. (a) Displacement, velocity, acceleration, and momentum are independent of the choice of origin, since they are all related to changes in position over some elapsed time. Moving the origin of coordinates from one location to another would not affect a change in position. But since angular momentum and torque are measured with respect to some point or axis, they would change if the origin was changed. (b) Consider a coordinate system whose origin is in a moving car. The displacement of some object in that car, as measured in that car, would be different than a displacement as measured relative to the ground. Likewise, a velocity or momentum as measured in that car would be different than as measured relative to the ground (recall the idea of relative velocity from Chapter 3). Since acceleration is related to a change in velocity, then acceleration would be independent of the velocity. And since torque is measured relative to a position, the torque would be independent of the velocity. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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16. Turning the steering wheel changes the axis of rotation of the tires, and makes the car turn. The torque is supplied by the friction between the tires and the pavement. (Notice that if the road is slippery or the tire tread is badly worn, the car may not be able to make a sharp turn.) 17. Newton’s third law is not valid in a rotating reference frame, since there is no reaction to the pseudoforce. 18. Because of the rotation of the Earth, the plumb bob will be slightly deflected by the Coriolis force, which is a “pseudoforce.” 19. In the Northern Hemisphere, the shots would be deflected to the right (similar to the deflection of winds as discussed in the text), with respect to the surface of the Earth, due to the Coriolis effect. In the Southern Hemisphere, the deflection of the shots would be to the left. The gunners had experience in the Northern Hemisphere and so miscalculated the necessary launch direction.

Solutions to MisConceptual Questions 1.

(a) Without an external torque students may think that the angular speed would remain constant. But with no external torque, the angular momentum must remain constant. The angular momentum is the product of the moment of inertia and the angular speed. As the string is shortened, the moment of inertia of the block decreases. Thus, the angular speed increases.

(a) Work is done on the object by the tension in the string, and so its kinetic energy increased. Thus the tangential velocity had to increase. Another way to consider the problem is that K = L2 2 I . As seen in Question 1, the angular momentum is constant and the rotation inertia decreases. Thus the kinetic energy (and so the speed) had to increase.

(c) Students may mistakenly reason that since no net torque acts on you and your moment of inertia decreases as the masses are released, your angular speed should increase. This reasoning is erroneous because the angular momentum of the system of you and the masses is conserved. As the masses fall they carry angular momentum with them. If you consider yourself and the masses as two separate systems, each with angular momentum from their moments of inertia and angular speed, it is easy to see that by dropping the masses, no net external torque acts on you and your moment of inertia does not change, so your angular speed will not change. The angular momentum of the masses (relative to the initial axis of rotation) also does not change until they hit the ground and friction (external torque) stops their motion.

(b) The relationship between kinetic energy, angular momentum, and rotational inertial is K = L2 2 I . We see that for the same angular momentum, the smaller kinetic energy would have the larger rotational inertia.

sin  , we see that answers (a, b, c) Since the magnitude of the cross product is given by VV 1 2 (a), (b), and (c) all guarantee that the cross product is zero.

(a) No net torque acts on the Earth so the angular momentum is conserved. As people move toward the equator their distance from the Earth’s axis increases. This increases the moment of inertia of the Earth. For angular momentum to be conserved the angular speed must decrease, and so it will take longer for the Earth to complete a full rotation.

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(c) The angular momentum will remain constant. If the particle is moving in a straight line at constant speed, there is no net torque acting on it and therefore its angular momentum must be conserved. The same results can be obtained geometrically. Assume that the coordinate system is defined so that the particle’s velocity is in the x-direction, and so is given by v = v x ˆi , and that the particle is moving in the x-y plane, so that it’s position vector is r = x ˆi + y ˆj (it is not 0

moving in the y-direction). Find the angular momentum from Eq. 11–6. ˆi ˆj kˆ L = r  v = x y 0 = −v y kˆ 0

The parameters in the expression for the angular momentum are all constants. 8.

(e) The case of an object rotating about a fixed axis is discussed in the subsection entitled “Directional Nature of Angular Momentum” on page 328. Note the footnote on that page, which says that there is always a COMPONENT of L , along the direction of ω, and its magnitude is given by I ω. Also, in Section 11–5, Eq. 11–11, we see that L = I ω when the rotation axis is both a symmetry axis, and passes through the CM. This eliminates answers (a), (b), (c), and (d).

(a) The act of extending the arms is “internal” to the person–chair system, and so no external torque is applied to the system. Thus, if we assume the conditions of Eq. 11–11, the angular momentum is conserved. Thus if the rotational inertia increase (by extending the arms), the angular velocity must decrease.

10. (e) In the configuration described, the angular momentum of the spinning bicycle wheel is towards your right, which is pointing in the positive y-direction (see the adjacent figure). When holding the wheel in front of you, your body would have a positive x coordinate. As you turn your whole body (and the wheel) to your left, that will apply a torque to the wheel that points in the positive z-direction. Thus the change in angular momentum must point in the positive z-direction also. For that to happen, the right axle must move upwards, and so you will feel the axle push your right hand upwards. 11. (d) If you are looking at the wheel and it is rotating clockwise, it’s angular momentum is pointing away from you, by applying the right-hand rule. Thus the angular momentum is in the negative x-direction.

Solutions to Problems 1.

The angular momentum is given by Eq. 11–1. The ball is treated as a point mass in order to find the rotational inertia.

L = I  = MR 2 = ( 0.210 kg )(1.25m ) (10.4 rad s ) = 3.41kg m2 s 2

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(a) The angular momentum is given by Eq. 11–1.

 1500 rev   2 rad   1 min       1 min   1 rev   60 s  

L = I  = 12 MR 2 = 12 ( 2.8 kg )( 0.18 m )  2

= 7.125 kg m 2 s  7.1kg m 2 s (b) The torque required is the change in angular momentum per unit time. The final angular momentum is zero. L − L0 0 − 7.125 kg m 2 s = = = −1.2 m N 6.0 s t The negative sign indicates that the torque is used to oppose the initial angular momentum. 3.

Since there are no external torques on the system, the angular momentum of the two-disk system is conserved. The two disks have the same final angular velocity.

Li = L f

→ I + I ( 0 ) = 2I f

→  f = 12 

There is no net torque on the diver because the only external force (gravity) passes through the center of mass of the diver. Thus the angular momentum of the diver is conserved. Subscript 1 refers to the tuck position, and subscript 2 refers to the straight position. I  2 rev  1  = 0.38 rev s L1 = L2 → I11 = I 22 → 2 = 1 1 =    I 2  1.5 sec  3.5 

The skater’s angular momentum is constant, since no external torques are applied to her. 1.0 rev 1.5s  Li = L f → I ii = I f  f → I f = I i i = 4.9 kg m2 = 1.3kg m2 2.5rev s f

(

)

She accomplishes this by starting with her arms (and possibly a leg) extended (initial angular velocity) and then pulling her limbs in to the center of her body (final angular velocity). 6.

(a) Consider the person and platform a system for angular momentum analysis. Since the force and torque to raise and/or lower the arms is internal to the system, the raising or lowering of the arms will cause no change in the total angular momentum of the system. However, the rotational inertia increases when the arms are raised. Since angular momentum is conserved, an increase in rotational inertia must be accompanied by a decrease in angular velocity.  0.80 rev s (b) Li = L f → I ii = I f  f → I f = I i i = I i = 1.333 I i  1.3 I i f 0.60 rev s The rotational inertia has increased by a factor of 1.3.

(a) For the daily rotation about its axis, treat the Earth as a uniform sphere, with an angular frequency of one revolution per day. 2 daily Ldaily = I daily = 25 MREarth

(

)

)(

= 25 5.98  1024 kg 6.38  106 m

rad   1 day   )  21day   86,400 s   = 7.08  10 kg m s 2







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Instructor Solutions Manual

(b) For the yearly revolution about the Sun, treat the Earth as a particle, with an angular frequency of one revolution per year.





2 Ldaily = I daily =  MRSun daily



Earth

(



)(

= 5.98  1024 kg 1.496  1011 m





2 rad   1 day  )  365  = 2.67  10 kg m s  day 86,400 s  2







(a) The angular momentum is the moment of inertia (modeling the skater as a cylinder) times the angular velocity. Her height does not affect the calculation. rev   2 rad  2  = 9.5kg m 2 s L = I  = 12 MR 2 = 12 ( 48 kg )( 0.15 m )  2.8   s   1rev   (b) If the rotational inertia does not change, then the change in angular momentum is strictly due to a change in angular velocity. L I final − I 0 0 − 9.5kg m 2 s = = = = −2.4 m N t t 4.0s The negative sign indicates that the torque is in the opposite direction as the initial angular momentum.

The angular momentum of the disk–rod combination will be conserved because there are no external torques on the combination. This situation is a totally inelastic collision, in which the final angular velocity is the same for both the disk and the rod. Subscript 1 represents before the collision, and subscript 2 represents after the collision. The rod has no initial angular momentum. L1 = L2 → I11 = I 22 →



I disk

I disk + I rod

2 = 1 1 = 1

  3 = ( 4.1rev s )   = 2.5rev s  1 1  5  2 MR + 12 M ( 2 R ) 

= 1 

1 2

MR 2

10. Since the person is walking radially, no torques will be exerted on the person–platform system, and so angular momentum will be conserved. The person will be treated as a point mass. Since the person is initially at the center, they have no initial rotational inertia. (a) Li = L f → I platformi = ( I platform + I person )  f

f =

I platform I platform + mR

i = 2

920 kg m 2 920 kg m 2 + ( 75 kg )( 3.0 m )

( 0.95 rad s ) = 0.548 rad s  0.55 rad s

(b) Ki = 12 I platformi2 = 12 ( 920 kg m 2 ) ( 0.95 rad s ) = 420 J 2

(

)

2  2f K f = 12 ( I platform + I person )  2f = 12 I platform + mperson Rperson

= 12 920 kg m 2 + ( 75 kg )( 3.0 m )  ( 0.548 rad s ) = 239 J  240 J





11. When the person and the platform rotate, they do so about the vertical axis. Initially there is no angular momentum pointing along the vertical axis, and so any change that the person–wheel–platform undergoes must result in no net angular momentum along the vertical axis. The first diagram shows this condition.

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(a) Now consider the next diagram. If the wheel is moved so that its angular momentum points upwards, then the person and platform must get an equal but opposite angular momentum, which will point downwards. Write the angular momentum conservation condition for the vertical direction to solve for the angular velocity of the platform.

Li = L f

→ 0 = I WW + I PP →

P = − W I W I P

The negative sign means that the platform is rotating in the opposite direction of the wheel. If the wheel is spinning counterclockwise when viewed from above, the platform is spinning clockwise. (b) Now consider the next diagram. If the wheel is pointing at a 60o angle to the vertical, then the component of its angular momentum that is along the vertical direction is I WW cos 60. Also see the simple vector diagram below the adjacent diagram. Write the angular momentum conservation condition for the vertical direction to solve for the angular velocity of the platform.

Li = L f

→ 0 = I WW cos 60 + I PP →

P = −

IW 2IP

W

Again, the negative sign means that the platform is rotating in the opposite direction of the wheel. (c) Consider this next diagram. If the wheel is moved so that its angular momentum points downwards, then the person and platform must get an equal but opposite angular momentum, which will point upwards. Write the angular momentum conservation condition for the vertical direction to solve for the angular velocity of the platform. Li = L f → 0 = I WW + I PP → P = −W I W I P

60o LW = I W  W

The platform is again rotating in the opposite direction as the wheel. If the wheel is now spinning clockwise when viewed from above, the platform is spinning counterclockwise. (d) Since the total angular momentum is 0, if the wheel is stopped from rotating, the platform will also stop. Thus P = 0 . Note that we have ignored the initial horizontal component of angular momentum. We assume that the rotating platform is not free to move in a direction corresponding to horizontal angular momentum–perhaps friction with the floor has provided an external torque to prevent the conservation of momentum along a horizontal axis. 12. Because there is no external torque applied to the wheel–clay system, angular momentum is be conserved. We assume the clay is thrown with no angular momentum, so the initial angular momentum is 0. This is a totally inelastic collision: the final angular velocity is the same for both the clay and the wheel. Subscript 1 represents before the clay is thrown, and subscript 2 represents after the clay is thrown. L1 = L2 → I11 = I 22 → 2    M wheel Rwheel =     1 2 2 2 2 1 I 2 I wheel + I clay  2 M wheel Rwheel + 2 M clay Rclay   M wheel Rwheel + M clay Rclay  2   5.0 kg )( 0.20 m ) (  = 1.377 rev s  1.4 rev s = (1.5 rev s )  2  ( 5.0 kg )( 0.20 m )2 + ( 2.8 kg ) (8.0 10 −2 m ) 

2 = 1 1 =

I wheel



= 1  1

1 2

2 M wheel Rwheel

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13. (a) The angular momentum of the merry-go-round (abbreviate m-g-r) and people combination will be conserved because there are no external torques on the combination. This situation is a totally inelastic collision, in which the final angular velocity is the same for both the merry-goround and the people. Subscript 1 represents before the collision, and subscript 2 represents after the collision. The people have no initial angular momentum. L1 = L2 → I11 = I 2 2 →

2 = 1

I1 I2

= 1



I m-g-r I m-g-r + I people

   I m-g-r + 4 M person R  I m-g-r

= 1 

  1630 kg m 2 = ( 0.80 rad s )   = 0.4696 rad s  0.47 rad s 2 2  1630 kg m + 4 ( 65 kg )( 2.1m )  (b) If the people jump off the merry-go-round radially, then they exert no torque on the merry-goround, and thus cannot change the angular momentum of the merry-go-round. The merry-goround would continue to rotate at 0.80 rad s . 14. (a) The angular momentum of the system will be conserved as the woman walks. The woman’s distance from the axis of rotation is r = R − vt.



Li = L f

  0 = ( I platform + I woman )  → woman 

→  I platform + I 0



( MR + mR )  = ( MR + m ( R − vt ) )  → ( MR + mR )  = ( M + m )  = 1 2

1 2

( MR + m ( R − vt ) ) 2

1 2

 vt  M + m 1 −   R

(b) Evaulate at r = R − v t = 0 → R = v t.

=

( 12 M + m ) 0 1 2

 

= 1 +

2m 

 0

M 

15. The values to be used are: M arm = 12 ( 0.13M ) ; M body = 0.75M ;

Rbody = 0.16 m ; l arm = 0.60 m . For the outstretched arms, see the first diagram. Use the parallel axis theorem to find the moment of inertia of the arms. I out = I body + I arms 2 2 = 12 M body Rbody +  121 2M arm l arm + 2M arm ( Rbody + 12 l arm )  2





For the arms held against the body, treat the arms like particles, since all of the arm mass is the same distance from the axis of rotation. 2 2 I in = I body + I arms = 12 M body Rbody + 2 M arm Rbody Assume that angular momentum is conserved as the arms change positions, and then find the ratio of the angular speeds.

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Lin = Lout → I inin = I outout → 1 M body Rbody + 2 M arm Rbody out I in 2 = = 2 2 2 in I out 1 M body Rbody +  121 2 M arm l arm + 2 M arm ( Rbody + 12 l arm )  2   2

( 0.75M )( 0.16 m ) + ( 0.13M )( 0.16 m ) = 2 2 2 1 0.75M )( 0.16 m ) +  121 ( 0.13M )( 0.60 m ) + ( 0.13M ) ( 0.16 m + 0.30 m )  2(   2

1 2

1.2928  10 −2 4.1008  10

−2

= 0.3153 =

out  0.3 in

The angular speed with outstretched arms is about 30% of the angular speed with the arms at the sides. 16. The rotational inertia of the grindstone is given by I = 12 MR 2. (a) The final rotation rate will occur at t = . L (t = ) 1.2 kg m 2 s = 125 rad s = L ( t =  ) = I final → final = 2 1 1 MR 2 3.0 kg )( 0.080 m ) 2 2 (

 60s   1rev  final = 125 rad s    = 1193rpm  1200 rpm  1min   2 rad  (b) Use Eq. 11–2.

=

dL dt

1.2 kg m 2 s (1 − e −0.050t ) = (1.2 kg m 2 s ) ( 0.050 s ) e −0.050t = ( 0.060 m N ) e −0.050t dt

17. The angular momentum is the total moment of inertia times the angular velocity. L = I  =  121 M l 2 + 2m ( 12 l )   = ( 121 M + 12 m ) l 2 2





18. (a) Since the lost mass carries away no angular momentum, the angular momentum of the remaining mass will be the same as the initial angular momentum. 2 f M i Ri2 Ii M i Ri2 5 Li = L f → I ii = I f  f → = = 2 = = 2.0  10 4 2 2 i I f 5 M f R f ( 0.5M i ) ( 0.01R f )

 2 rad   1 d  −2 −2  f = 2.0  10 4 i = 2.0  10 4    86,400 s  = 4.848  10 rad s  5  10 rad s 30 day    The period would be a factor of 20,000 smaller, which would make it about 130 seconds. (b) The ratio of angular kinetic energies of the spinning mass would be as follows.

 25 ( 0.5M i )( 0.01Ri ) 2  ( 2.0  104 i )  = =  = 2.0  104 → 2 2 2 1 1 2 K initial I M i Ri ) i 2 i i 2 (5 K final

1 2

I f  2f

1 2

K final = 2  104 K initial

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19. For our rough estimate, we model the hurricane as a rigid cylinder of air. Since the “cylinder” is rigid, each part of it has the same angular velocity. The mass of the air is the product of the density of air times the volume of the air cylinder.

(

) ( R) = M v

M = V =  R 2h = 1.3kg m3  8.5  104 m

( MR ) ( v

(a) K = 12 I  2 = 12

1 2

edge

) ( 4.5  10 m ) = 1.328  10 kg 2

2 edge

1 4

  1m s   16 16 = (1.328  10 kg ) (120 km h )    = 3.688  10 J  3.7  10 J 3.6 km h    2 1 1 (b) L = I  = ( 2 MR ) ( vedge R ) = 2 MRvedge 14

1 4

(

)(

)



 1m s   = 2.213  1020 kg m 2 s    3.6 km h  

= 12 1.328  1014 kg 8.5  104 m (120 km h ) 



 1.9  10 kg m s 20

20. Angular momentum will be conserved in the Earth–asteroid system, since all forces and torques are internal to the system. The initial angular velocity of the asteroid, just before collision, is asteroid = vasteroid REarth . Assuming the asteroid becomes imbedded in the Earth at the surface, the Earth and the asteroid will have the same angular velocity after the collision. We model the Earth as a uniform sphere, and the asteroid as a point mass, located at a distance of the radius of the Earth from the axis of rotation. We use m for the asteroid mass, and M for the Earth’s mass. First we calculate the relative (or fractional) change, and then convert it to a percentage change by multiplying the final result by 100. I  + I asteroidasteroid Li = L f → I EarthEarth + I asteroidasteroid = ( I Earth + I asteroid )  f →  f = Earth Earth ( I Earth + I asteroid )

rel. change =

( −  ) =  f

Earth

Earth

I Earth + I asteroid =

I EarthEarth + I asteroidasteroid −1 =

( I Earth + I asteroid )

−1 =

Earth

asteroid − I Earth − I asteroid Earth

( I Earth + I asteroid )

I Earth + I asteroid

 asteroid  − 1  Earth 

( I Earth + I asteroid )

−1

 asteroid  − 1  Earth 

2 mREarth 

I asteroid  =

asteroid Earth

( MR 2 5

2 Earth

2 + mREarth

)

  v  R m  asteroid Earth − 1    2   m  vasteroidTEarth − 1 m  asteroid − 1      TEarth  =   =  2 REarth  =  Earth 2 5

( =

M +m

2 5

)

 ( 3.5  104 m s ) ( 86400s )

2.0  105 kg  2 5

(

M +m



(

2 6.38  10 m 6

)

5.97  10 kg + 2.0  10 kg 24

2 5

M +m



− 1

 = 6.23  10−18

% change = 6.2  10−16 %

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21. The angular momentum of the person–turntable system will be conserved. Call the direction of the person’s motion the positive rotation direction. Relative to the ground, the person’s speed will be v + vT , where v is the person’s speed relative to the turntable, and vT is the speed of the rim of the turntable with respect to the ground. The turntable’s angular speed is T = v T R , and the person’s angular speed relative to the ground is P =

v + vT

R particle for calculation of the moment of inertia. Li = L f

T = −

 

v R

→ 0 = I TT + I PP = I TT + mR 2  T +

mRv I T + mR

=−

+ T . The person is treated as a point v

 →

R

( 65 kg )( 4.9 m )( 3.8 m s ) = −0.3549 rad s  −0.35 rad s 2 1850 kg m 2 + ( 65 kg )( 4.9 m )

22. We use the determinant rule, Eq. 11–3b. ˆi ˆj kˆ (a)

A  B = −A 0 0

0 = ˆi ( 0 )( B ) − ( 0 )( 0 )  + ˆj ( 0 )( 0 ) − ( − A)( B )  + kˆ ( − A)( 0 ) − ( 0 )( 0 )  B

= AB ˆj

So the direction of A  B is in the ĵ direction. (b) Based on Eq. 11–4b, we see that interchanging the two vectors in a cross product reverses the direction. So the direction of B  A is in the − ˆj direction. We could also show it explicitly with a determinant as done in part (a). ˆi ˆj kˆ B  A = 0 0 B = ˆi ( 0 )( 0 ) − ( B )( 0 ) + ˆj ( B )( − A ) − ( 0 )( 0 ) + kˆ ( 0 )( 0 ) − ( 0 )( − A ) −A 0

= − AB ˆj

(c) Since A and B are perpendicular, we have A  B = B  A = AB sin 90 = AB . 23. (a) For all three expressions, use the fact that A  B = AB sin  . If both vectors in the cross product point in the same direction, then the angle between them is  = 0. Since sin 0 = 0, a vector crossed into itself will always give 0. Thus î  î = ˆj  ˆj = kˆ  kˆ = 0 . (b) For all three expressions, use the fact that A  B = AB sin  . In each case the magnitude of each vector is 1, and the angle between them is  = 90. Since sin 90 = 1 , the magnitude of each of these cross products is 1. To determine the direction, we use the right-hand rule. From the adjacent figure, we see that if A = î and B = ˆj , then the right-hand rules says that A  B points in the k̂ direction. Thus î  ˆj = kˆ . In a similar fashion, using the

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right-hand rule with the diagram, we see that î  kˆ would point in the direction of − ˆj , so î  kˆ = − ˆj. And finally, again using the diagram with the right-hand rule, we see that ˆj  kˆ would point in the direction of î , and so ˆj  kˆ = î. We can also use the determinant rule (Eq. 11–3b) to evaluate these expressions. î ˆj kˆ î  ˆj = 1

0 = ˆi ( 0 )( 0 ) − ( 0 )(1)  + ˆj ( 0 )( 0 ) − (1)( 0 )  + kˆ (1)(1) − ( 0 )( 0 )  = kˆ

0 1

ˆi

ˆj

kˆ

ˆi  kˆ = 1

0 = ˆi ( 0 )(1) − ( 0 )( 0 )  + ˆj ( 0 )( 0 ) − (1)(1)  + kˆ (1)( 0 ) − ( 0 )( 0 )  = − ˆj

0 0

ˆi

kˆ

ˆj

ˆj  kˆ = 0 1 0 0

24. (a) (b) (c) (d)

0 = ˆi (1)(1) − ( 0 )( 0 )  + ˆj ( 0 )( 0 ) − ( 0 )(1)  + kˆ ( 0 )( 0 ) − ( 0 )(1)  = ˆi 1

East cross south is into the ground. East cross straight down is north. Straight up cross north is west. Straight up cross straight down is 0 (the vectors are anti-parallel).

25. Use the definitions of cross product and dot product, in terms of the angle between the two vectors. A  B = A B → AB sin  = AB cos  → sin  = cos  This is true only for angles with positive cosines, and so the angle must be in the first or fourth quadrant. Thus the solutions are  = 45,315. But the angle between two vectors is always taken to be the smallest angle possible, and so  = 45 . 26. We use the determinant rule, Eq. 11–3b, to evaluate the torque. ˆi ˆj kˆ τ = r  F = 5.0

3.5

6.0 m N

8.0

−4.0

 = ( −62ˆi + 20ˆj + 40kˆ ) m N



= ˆi ( 3.5)( −4.0 ) − ( 6.0 )( 8.0 )  + ˆj ( 6.0 )( 0 ) − ( 5.0 )( −4.0 )  + kˆ ( 5.0 )( 8.0 ) − ( 3.5)( 0 )  m N

Note that all the components have 2 significant figures. 27. We choose coordinates so that the plane in which the particle rotates is the x-y plane, and so the angular velocity is in the z direction. The object is rotating in a circle of radius r sin  , where  is the angle between the position vector and the axis of rotation. Since the object is rigid and rotates about a fixed axis, the linear and angular velocities of the particle are related by v =  r sin  . The magnitude of the tangential acceleration is

a tan aR

α 

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a tan =  r sin  . The radial acceleration is given by aR =

= v . We assume the r sin  r sin  object is gaining speed. See the diagram showing the various vectors involved. =v

The velocity and tangential acceleration are parallel to each other, and the angular velocity and angular acceleration are parallel to each other. The radial acceleration is perpendicular to the velocity, and the velocity is perpendicular to the angular velocity. We see from the diagram that, using the right-hand rule, the direction of a R is in the direction of ω  v. Also, since ω and v are perpendicular, we have ω  v =  v which from above is v  = aR . Since both the magnitude and direction check out, we have a R = ω  v .

We also see from the diagram that, using the right-hand rule, the direction of a tan is in the direction of α  r. The magnitude of α  r is α  r =  r sin  , which from above is  r sin  = a tan . Since both the magnitude and direction check out, we have a tan = α  r . 28. (i)

First we are to show that C = AB sin  . We follow the instructions given in the problem. 2

C2 = A  B

C 2 = C x2 + C y2 + C z2 = ( Ay Bz − Az By ) + ( Az Bx − Ax Bz ) + ( Ax By − Ay Bx ) 2

Square the quantities in parentheses

(

) ( +( A B − 2A B A B + A B )

C 2 = Ay2 Bz2 − 2 Ay Bz Az By + Az2 B y2 + Az2 Bx2 − 2 Az Bx Ax Bz + Ax2 Bz2 2 x

2 y

2 x

2 y

)

Do some factoring and re-arranging

(

)

(

)

(

)

C 2 = Ax2 By2 + Bz2 + Ay2 Bx2 + Bz2 + Az2 Bx2 + By2 − 2 ( Ax Bx Ay By + Ax Bx Az Bz + Ay B y Az Bz )

( The presence of terms like A B A B might make you think “dot product.” ) x

Now add and subtract these three terms,

( A B − A B ) + ( A B − A B ) + ( A B − A B ), 2 x

2 x

2 y

2 z

to the previous equation which then becomes, with some rearranging,

( ) ( − 2( A B A B + A B A B + A B A B )

)

(

)

C 2 = Ax2 Bx2 + By2 + Bz2 − Ax2 Bx2 + Ay2 Bx2 + By2 + Bz2 − Ay2 By2 + Az2 Bx2 + B y2 + Bz2 − Az2 Bz2 x

Do more factoring and re-grouping. Note the symmetry.

(

C 2 = Ax2 + Ay2 + Az2

)(

)

 Ax2 Bx2 + Ay2 By2 + Az2 Bz2 +

   2 Ax Bx Ay By + 2 Ax Bx Az Bz + 2 Ay By Az Bz 

Bx2 + By2 + Bz2 −  

The first term in the right-hand side of the above equation is A2 B 2 . Does the second term equal

( A B ) ? Yes. We prove that here. 2

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Instructor Solutions Manual

( A B ) = ( A B + A B + A B ) = ( A B + A B + A B )( A B + A B + A B ) 2

= A B + Ax Bx Ay By + Ax Bx Az Bz + Ay By Ax Bx + A B + Ay By Az Bz 2 x

2 x

2 y

+ Az Bz Ax Bx + Az Bz Ay By + Az2 Bz2 = Ax2 Bx2 + Ay2 By2 + Az2 Bz2 + 2 Ax Bx Ay By + 2 Ax Bx Az Bz + 2 Ay By Az Bz

 Ax2 Bx2 + Ay2 By2 + Az2 Bz2 +  C = ( A + A + A )( B + B + B ) −   2 A B A B + 2 A B A B + 2 A B A B  Thus x x z z y y z z   x x y y 2

2 x

2 y

2 z

= A2 B 2 − ( A B )

2 x

2 y

2 z

Substitute ( A B ) = AB cos  .

(

)

C 2 = A2 B 2 − ( A B ) = A2 B 2 − ( AB cos  ) = A2 B 2 1 − cos 2  = A2 B 2 sin 2  2

And so C = C = AB sin  , which is what was to be shown. (ii) Now we are to show C is perpendicular to both A and B by using dot products. A C = Ax C x + Ay C y + Az C z

= Ax ( Ay Bz − Az B y ) + Ay ( Az Bx − Ax Bz ) + Az ( Ax B y − Ay Bx ) = ( Ax Ay Bz − Ax Az B y ) + ( Ay Az Bx − Ay Ax Bz ) + ( Az Ax B y − Az Ay Bx ) = Ax Ay Bz − Ax Az B y + Ay Az Bx − Ay Ax Bz + Az Ax B y − Az Ay Bx = Bz ( Ax Ay − Ay Ax ) + B y ( Az Ax − Ax Az ) + Bx ( Ay Az − Az Ay ) =0 B C = B x C x + B y C y + Bz C z = Bx ( Ay Bz − Az B y ) + B y ( Az Bx − Ax Bz ) + Bz ( Ax B y − Ay Bx ) = ( Bx Ay Bz − Bx Az B y ) + ( B y Az Bx − B y Ax Bz ) + ( Bz Ax B y − Bz Ay Bx ) = Bx Ay Bz − Bx Az B y + B y Az Bx − B y Ax Bz + Bz Ax B y − Bz Ay Bx = Az ( B y Bx − Bx B y ) + Ay ( Bx Bz − Bz Bz ) + Ax ( Bz B y − B y Bz ) =0 (iii) Now consider Fig. 11–7, shown here. We assume that A = Ax ˆi and let B = Bx ˆi + By ˆj . A and

B are in the x-y plane, and we see that the right-hand rule indicated that C points in the z-direction, being perpendicular to both A and B. Apply Eq. 11–3c. C = ( A B − A B ) ˆi + ( A B − A B ) ˆj+ ( A B − A B ) kˆ y

= Ax By kˆ = A ( B sin  ) kˆ

Thus C = A  B follows the right-hand rule.

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29. Show that A  ( B + C ) = A  B + A  C , using Eq. 11–3c. ˆi

ˆj

A  B = Ax

Az = ˆi ( Ay Bz − Az By ) + ˆj ( Az Bx − Ax Bz ) + kˆ ( Ax By − Ay Bx )

ˆi

ˆj

kˆ

A  C = Ax

Az = ˆi ( Ay C z − Az C y ) + ˆj ( Az C x − Ax C z ) + kˆ ( Ax C y − Ay C x )

kˆ

ˆi

ˆj

kˆ

Bx + C x

By + C y

Bz + C z

A  ( B + C) =

= ˆi  Ay ( Bz + C z ) − Az ( By + C y )  + ˆj  Az ( Bx + C x ) − Ax ( Bz + C z ) + kˆ  Ax ( By + C y ) − Ay ( Bx + C x ) 

= ˆi ( Ay Bz + Ay C z ) − ( Az By + Az C y )  + ˆj ( Az Bx + Az C x ) − ( Ax Bz + Ax C z ) + kˆ ( Ax By + Ax C y ) − ( Ay Bx + Ay C x ) 

= ˆi ( Ay Bz − Az By ) + ˆj ( Az Bx − Ax Bz ) + kˆ ( Ax By − Ay Bx ) + ˆi ( Ay C z − Az C y ) + ˆj ( Az C x − Ax C z ) + kˆ ( Ax C y − Ay C x ) = AB + AC

30. We use the determinant rule, Eq. 11–3b, to evaluate the torque. ˆi ˆj kˆ

τ = r F =

8.0

6.0 m kN

2.4

−4.1

  = ( 24.6ˆi  14.4ˆj 19.2kˆ ) m kN  ( 2.5ˆi  1.4ˆj 1.9kˆ )  10 m N

= ˆi  − ( 6.0 )( −4.1)  + ˆj ( 6.0 )( 2.4 )  + kˆ  − ( 8.0 )( 2.4 )  m kN 4

The magnitude of this maximum torque is also found.

τ =

( 2.46)2 + (1.44 )2 + (1.92 )2  104 m N = 3.4  104 m N

31. (a) We use the determinant rule, Eq. 11–3b, to evaluate the cross product. î ˆj kˆ A  B = 7.4 −3.5 0 = −7.0î − 14.8ˆj + 4.29kˆ  −7.0î − 15ˆj + 4 kˆ −8.5

4.6

2.0

(b) Now use Eq. 11–3a to find the angle between the two vectors.

AB =

( −7.0)2 + ( −14.8)2 + ( 4.29 )2 = 16.92

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( 7.4 )2 + ( −3.5)2 = 8.186 ; B = ( −8.5)2 + ( 4.6)2 + ( 2.0)2 = 9.870

A  B = AB sin  →  = sin −1

AB

= sin −1

16.92

= 12.08 or 167.91 AB (8.186)( 9.870) Use the dot product to resolve the ambiguity. A B = ( 7.4 )( −8.5) + ( −3.5)( 4.6) + 0 ( 2.0) = −79 Since the dot product is negative, the angle between the vectors must be obtuse, and so

 = 167.91  170 . 32. We use the determinant rule, Eq. 11–3b, to evaluate the torque. ˆi ˆj kˆ

τ = rF =

0.280

0.335

215 cos 28.0

215sin 28.0

0mN 0

= kˆ ( 0.280 )( 215sin 28.0 ) − ( 0.335 )( 215 cos 28.0 ) m N = −35.3 m N kˆ = 35.3 m N in the − z direction This could also be calculated by finding the magnitude and direction of r, and then using Eq. 11–3a and the right-hand rule. 33. We choose the z axis to be the axis of rotation, and so ω = kˆ . We describe the location of the point as r = R cos tˆi + R sin tˆj + z0kˆ . In this description, the point is moving counterclockwise in a circle of radius R centered on the point ( 0, 0, z0 ) , and is located at ( R, 0, z0 ) at t = 0.

dr dt

ωr =

= − R sin  t ˆi + R cos  t ˆj ˆi

ˆj

kˆ

 = − R sin  t ˆi + R cos  t ˆj = v

R cos  t

R sin  t

And so we see that v = ω  r. If the origin is moved to a different place on the axis of rotation (the z axis) that would simply change the value of the z coordinate of the point to some other value, say z1 . Changing that value will still lead to v = ω  r. But if the origin is moved from the original point to something off the rotation axis, then the position vector will change. If the new origin is moved to ( x2 , y2 , z2 ) , then the position vector will change to r = ( R cos  t − x ) ˆi + ( R sin  t − y ) ˆj + ( z − z ) kˆ . See how that affects the relationships. 2

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Chapter 11

Angular Momentum; General Rotation

= − R sin  t ˆi + R cos  t ˆj

ˆi

ˆj

kˆ



R cos  t − x2

R sin  t − y2

z 0 − z2

ωr =

= ( − R sin  t +  y2 ) ˆi + ( R cos  t −  x2 ) ˆj = v

We see that with this new off-axis origin, v  ω  r. 34. We use the determinant rule, Eq. 11–3b, to evaluate the angular momentum. ˆi ˆj kˆ L = rp = x y z = ( yp − zp ) ˆi + ( zp − xp ) ˆj + ( xp − yp ) kˆ z

35. The position vector and velocity vectors are at right angles to each other for circular motion. The angular momentum for a particle moving in a circle is L = rp sin  = rmv sin 90 = mrv. The moment of inertia is I = mr 2 . L2 2I

( mrv ) 2mr

m2 r 2v 2 2mr

This is analogous to K =

mv 2 2

= 12 mv 2 = K

relating kinetic energy, linear momentum, and mass.

36. (a) See Fig. 11–34 in the textbook. We have that L = r⊥ p = dmv = mvd . The direction is into the plane of the page. (b) Since the velocity (and momentum) vectors pass through O , r and p are parallel, and so L = r  p = 0. Or, r⊥ = 0 , and so L = 0 . −p

37. See the diagram. Calculate the total angular momentum about the origin. L = r1  p + r2  ( −p ) = ( r1 − r2 )  p The position dependence of the total angular momentum only depends on the difference in the two position vectors. That difference is the same no matter where the origin is chosen, because it is the relative distance between the two particles. 38. Use Eq. 11–6 to calculate the angular momentum. ˆi ˆj −6.0

0 m2 s

3.2

−8.0

)

kˆ

L = r  p = m ( r  v ) = ( 0.085 kg ) 4.4

(

r1 − r2

(

)

= ( 0.085) 48ˆi + 35.2ˆj + 19.2kˆ kg m 2 s = 4.1ˆi + 3.0ˆj + 1.6kˆ kg m 2 s

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39. Use Eq. 11–6 to calculate the angular momentum. ˆi ˆj kˆ L = r  p = m ( r  v ) = ( 4.3kg ) 1.0

2.0

3.0 m 2 s

−5.0

2.8

−3.1

(

)

(

)

= ( 4.3) −14.6ˆi − 11.9ˆj + 12.8kˆ kg m 2 s = −63ˆi − 51ˆj + 55kˆ kg m 2 s

40. (a) See the diagram from Figure 11–16. Calculate the angular momentum about an axis along the axis of the pulley, through the center of the pulley. The angular momentum of the system is the sum of the angular momentum of the pulley and the two masses. Each angular momentum is clockwise, and so they are simply added. The two masses have the same speed, and that is equal to the tangential velocity of the rim of the pulley, assuming that the rope passes over the pulley without slipping. L = Lpulley + LA + LB = ( I  )pulley + mA vR0 + mB vR0

+ mA vR0 + mB vR0 = ( mA + mB + 12 mpulley ) vR0 R0 The external torque on the system is caused by the weights of the two masses.  = mB gR0 − mA gR0 Apply Eq. 11–9a. dL d dv = → mB gR0 − mA gR0 = ( mA + mB + 12 mpulley ) vR0  = ( mA + mB + 12 mpulley ) R0 dt dt dt dv dv ( mB − mA ) g mB gR0 − mA gR0 = ( mA + mB + 12 mpulley ) R0 → =a= dt dt ( mA + mB + 12 mpulley ) = 12 mpulley R02

( mB − mA )

(m + m + m ) A

1 2

(1.2 kg ) ( 9.80 m s 2 ) 15.6 kg

pulley

= 0.7538 m s 2  0.75 m s 2

(b) If the mass of the pulley is ignored, then we have the following. 2 ( mB − mA ) g (1.2 kg ) 9.80 m s a= = = 0.7737 m s 2 15.2 kg ( mA + m B )

(

)

 0.7737 m s 2 − 0.7538 m s 2  % error =    100 = 2.6%  3% 0.7538 m s 2   41. The rotational inertia of the compound object is the sum of the individual moments of inertia. I = I particles + I rod = m ( 0 ) + m ( 13 l ) + m ( 32 l ) + ml 2 + 13 M l 2 = ( 149 m + 13 M ) l 2 2

(a) K = 12 I  2 = 12 ( 149 m + 13 M ) l 2 2 = ( 97 m + 16 M ) l 2 2 (b) L = I  = ( 149 m + 13 M ) l 2 . The angular momentum is parallel to the axis of rotation.

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Chapter 11

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42. (a) We calculate the full angular momentum vector about the center of mass of the system. We take the instant shown in Fig. 11–35, with the positive x-axis to the right, the positive y-axis up along the axle, and the positive z axis out of the plane of the diagram towards the viewer. We take the upper mass as mass A and the lower mass as mass B. If we assume that the system is rotating counterclockwise when viewed from above along the rod, then the velocity of mass A is in the positive z direction, and the velocity of mass B is in the negative z direction. The speed is given by v =  r = ( 4.5 rad s )( 0.24 m ) = 1.08 m s. ˆi ˆj ˆi ˆj  kˆ kˆ    = m  −0.24 m 0.21m 0 + 0.24 m −0.21m 0   0 v 0 0 0 −v   









= m 2ˆi ( 0.21m ) v + 2ˆj ( 0.24 m ) v = 2mv ˆi ( 0.21m ) + ˆj ( 0.24 m )





= 2 ( 0.48 kg )(1.08 m s ) ˆi ( 0.21m ) + ˆj ( 0.24 m ) =  ˆi ( 0.2177 ) + ˆj ( 0.2488 )  kg m 2 s

The component along the axis is the ĵ component, 0.25kg m2 s . (b) The angular momentum vector will precess about the axle. The tip of the angular momentum vector traces out the dashed circle in the diagram. L 0.2177 kg m 2 s = 41  = tan −1 x = tan −1 Ly 0.2488 kg m 2 s



43. (a) We assume the system is moving such that mass B is moving down, mass A is moving to the left, and the pulley is rotating counterclockwise. We take those as positive directions. The angular momentum of masses A and B is the same as that of a point mass. We assume the rope is moving without slipping, so v = pulley R0 . L = LA + LB + Lpulley = M A vR0 + M B vR0 + I  = M A vR0 + M B vR0 + I



I 



R0 

=  ( M A + M B ) R0 +

v R0

v

(b) The net external torque about the axis of the pulley is that provided by gravity acting on M B , M B gR0 . Use Eq. 11–9a, which is applicable since the axis is fixed.

d 

I  



I  dv

 = dt → M gR = dt  ( M + M ) R + R  v  =  ( M + M ) R + R  dt B





I 



R0 

=  ( M A + M B ) R0 + a=

M B gR0

 I   ( M A + M B ) R0 + R   0 

  



a →

MBg MA + MB +

I R02

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Instructor Solutions Manual

44. We follow the notation and derivation of Eq. 11–9b. Start with the general definition of angular momentum, L =  ri  pi . Then express position and velocity with respect to the center of mass. i

ri = rCM + r

* * th i , where i is the position of the i particle with respect to the center of mass * i , which comes from differentiating the above relationship for position

vi = v CM + v

(

)

(

L =  ri  pi =  ri  mi v i =  rCM + ri*  mi v CM + v *i i

)

=  mi rCM  v CM +  mi rCM  v*i +  mi ri*  v CM +  mi ri*  v *i i

Note that the center of mass quantities are not dependent on the summation subscript, and so they may be taken outside the summation process.

  i

 

L = ( rCM  v CM )  mi + rCM   mi v*i +   mi ri*   v CM +  mi ri*  v*i i

In the first term,  mi = M . In the second term, we have the following. i

m v = m (v − v ) = m v − m v i

* i

=  mi v i − Mv CM = 0 i

This is true from the definition of center of mass velocity: v CM =

1 M

m v . i

Likewise, in the third term, we have the following.  miri* =  mi ( ri − rCM ) =  miri −  mirCM =  miri − MrCM = 0 i

This is true from the definition of center of mass: rCM =

1 M

m r. i i

Thus L = M ( rCM  v CM ) +  mi ri*  v*i = L* + ( rCM  Mv CM ) as desired. i

45. (a) The net torque to maintain the rotation is supplied by the forces at the bearings. From Fig. 11–18 we see that the net torque is 2Fd, where d is the distance from the bearings to the center of the axle. The net torque is derived in Example 11–10.

 net =

I 2 tan 

= 2 Fd → F =

( m r + m r )( sin  ) = 2 A A

I 2 2d tan 

2 B B

2d tan 

(b) We use the result from above, mA rA2 + mB rB2 )(  2 sin 2  ) 2mr 2 2 sin 2  ( F= = 2d tan  2d tan 

( 0.60 kg )( 0.30 m ) 2 (11.0 rad s )2 sin 2 28.0 = = 23.55 N  24 N ( 0.115 m ) tan 28.0

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46. As in Problem 45, the bearings are taken to be a distance d from point O. We choose the center of the circle in which mA moves as the origin, and label it O in the diagram. This choice of origin makes the position vector and the velocity vector both lie in the same horizontal plane, and so L points along the axis of rotation at all times. Thus L is parallel to ω. The magnitude of the angular momentum is as follows. L = mA rA ⊥ v = mA ( rA sin  )(  rA sin  ) = mA rA2 sin 2  dL

= τ net = 0. dt Be careful to take torques about the same point O used for the angular momentum. ( d − rA cos  )  net = 0 = FA ( d − rA cos  ) + FB ( d + rA cos  ) = 0 → FB = − FA ( d + rA cos  ) The mass is moving in a circle and so must have a net centripetal force pulling in on the mass (if shown, it would point to the right in the diagram). This force is given by FC = mA 2 rA sin  . By Newton’s third law, there must be an equal but opposite force (to the left) on the rod and axle due to the mass. But the axle is stationary, and so the net force on it must be 0.  ( d − rA cos  )  2 Fnet = FA − FB − FC = FA −  − FA  − mA rA sin  = 0 → ( d + rA cos  )  on axle 

L is constant in both magnitude and direction, and so

FA =

mA 2 rA sin  ( d + rA cos  ) 2d

FB = − FA

m  2 r sin  ( d − rA cos  ) ( d − rA cos  ) = − A A 2d ( d + rA cos  )

We see that FB points in the opposite direction as shown in the free-body diagram. 47. This is a variation on the ballistic pendulum problem. Angular momentum is conserved about the pivot at the upper end of the rod during the collision, and this is used to find the angular velocity of the system immediately after the collision. Mechanical energy is then conserved during the upward swing. Take the 0 position for gravitational potential energy to be the original location of the center of mass of the rod. The bottom of the rod will rise twice the distance of the center of mass of the system, since it is twice as far from the pivot. ml v → m ( 12 l ) v = ( I rod + I putty )  →  = Lbefore = Lafter 2 ( I rod + I putty ) collision collision

= Etop of → K after

Eafter collision

hCM

(I =

rod

swing

collision

+ I putty )  2

(I =

2(m + M ) g

= U top of → swing

+ I putty ) 

rod

+ I putty )  2 = ( m + M ) ghCM → 2

 m2 l 2v 2   = 8 g ( m + M ) ( I rod + I putty ) 2 ( m + M ) g  2 ( I rod + I putty )  rod

m2 l 2v 2

(

1 2

8 g ( m + M ) 13 M l 2 + m ( 12 l )

ml v

m2v 2

) 2g ( m + M ) ( M + m) 4 3

393

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Instructor Solutions Manual

m2v 2

hbottom = 2hCM =

g ( m + M ) ( 43 M + m )

48. Angular momentum about the pivot is conserved during this collision. Note that both objects have angular momentum after the collision. → Lbullet = Lstick + Lbullet → mbullet v0 ( 14 l ) = I stick + mbullet vf ( 14 l ) → Lbefore = Lafter collision

=

collision

initial

mbullet ( v0 − vf ) ( l ) 1 4

I stick

final

mbullet ( v0 − vf ) ( 14 l )

1 12

2 stick stick

3mbullet ( v0 − vf ) M stick l stick

3 ( 0.0035 kg )(110 m s )

( 0.27 kg )(1.0 m )

= 4.3 rad s 49. See the simple block diagram showing the two skaters on their approach path. The distance between them is l = 1.6 m . Each skater has mass m = 68 kg and speed v = 3.5 m s . (a) Their common angular speed is found from conservation of angular momentum. Their only interaction is “internal” to the 2-skater system, so the angular momentum is conserved. 2

l  Linitial = Lfinal → 2mv = I totalfinal = 2m   final → 2 2 l

final =

2 ( 3.5 m s )

= 4.375 rad s  4.4 rad s 1.6 m l (b) We calculate the final kinetic energy minus the initial kinetic energy. K final − K initial = I total 1 2

2 final

−2

( mv 1 2

2 initial

  l  2   2v  2 ) = 2m  2    l  − 2 ( 12 mv 2 ) = 0      1 2

(c) This action of pulling in on each other is “internal” to the system, and so angular momentum is again conserved. The “initial” subscript refers to their original circular motion, and the “final” subscript refers to their motion after reducing their radius.

  l  2   2v    l 2  Linitial = Lfinal → I initialinitial = I finalfinal → 2  m      = 2  m    final →   2   l   4  final =

8 ( 3.5 m s )

= 17.5 rad s  18 rad s l 1.6 m (d) We calculate the final kinetic energy minus the initial kinetic energy. The “initial” subscript refers to their original circular motion, and the “final” subscript refers to their motion after reducing their radius.

K final − K initial = I final 1 2

2 final

− I initial 1 2

2 initial

  l  2   8v  2 1   l  2   2v  2 =  2m      − 2  2m        4   l    2   l  1 2

= 3mv 2 = 3 ( 68 kg )( 3.5 m s ) = 2499 J  2500 J 2

This result is different from part (b) because each skater has a force applied to them in the direction of their motion as they are pulled closer to each other, and so positive work is done on each skater. Thus the kinetic energy increases. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

394

Chapter 11

Angular Momentum; General Rotation

50. The angular momentum of the Earth–meteorite system is conserved in the collision. The Earth is spinning counterclockwise as viewed in Fig. 11–38. We take that direction as the positive direction for rotation about the Earth’s axis, and so the initial angular momentum of the meteorite is negative. Linitial = Lfinal → I Earth0 − mRE v sin 45 = ( I Earth + I meteorite )  →

=

 = 0

I Earth0 − mRE v sin 45

( I Earth + I meteorite )

2 5

M E RE2 − mRE

0

2 5

M E RE20 − mRE v sin 45

( M R + mR ) 2 5

2 E



v 1

R ( M E + m) 2 E

= 5

RE2  52 M E − =



2 E

 2 mv     5 ME − 20 RE   20 RE  = mv

RE2 ( m + 52 M E )

( m + 52 M E )

2  v  mv  − + 1  5 ME −  20 RE  20 RE   − 0   = = −1 =  −1 =  2 0 0 0 1 + 2 M E  (m + 5 ME )  5  m       2.2  10 4 m s  + 1  2  2 rad s  ( 6.38  106 m )       86, 400   = −7.518  10 −13  −7.5  10 −13 =− 24  2 5.97  10 kg   1 + 5 5.2  1010 kg    So the change would be a decrease of about 7.5  10−13 rev/day. 51. (a) The linear momentum of the center of mass is conserved in the totally inelastic collision. pinitial = pfinal → mbeam v0 = ( mbeam + mman ) vfinal →

vfinal =

( 260 kg )(18 m s ) = 14.4 m s  14 m s ( 325 kg ) ( mbeam + mman ) mbeam v0

(b) Angular momentum about the center of mass of the system is conserved. First we find the center of mass, relative to the center of mass of the rod, taking down as the positive direction. See the diagram. m ( 0 ) + mman ( 12 l ) ( 65 kg )(1.35 m ) yCM = beam = ( 325 kg ) ( mbeam + mman ) = 0.27 m below center of rod We need the moment of inertia of the beam about the center of mass of the entire system. Use the parallel axis theorem. The location of the CM of the system relative to the CM of the beam is called rbeam . 2 I beam = 121 mbeam l 2 + mbeam rbeam ; I man = mman ( 12 l − rbeam )

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Linitial = Lfinal → mbeam v0 rbeam = ( I beam + I man ) final →

final = =

mbeam v0 rbeam

( I beam + I man )

mbeam v0 rbeam

1 12

2 + mman ( 12 l − rbeam ) mbeam l + mbeam rbeam 2

( 260 kg )(18 m s )( 0.27 m ) 2 2 2 1 260 kg )( 2.7 m ) + ( 230 kg )( 0.27 m ) + ( 65 kg )(1.08 m ) 12 (

= 5.044 rad s  5.0 rad s

52. Linear momentum of the center of mass is conserved in the totally inelastic collision. pinitial = pfinal → mv = ( m + M ) vCM → vCM = final

final

mv m+M

Angular momentum about the center of mass of the system is conserved. First we find the center of mass, relative to the center of mass of the rod, taking up as the positive direction. See the diagram. m ( 14 l ) + M ( 0 ) ml = yCM = 4(m + M ) (m + M ) The distance of the stuck clay ball from the system’s center of mass is found. ml Ml yclay = 14 l − yCM = 14 l − = 4(m + M ) 4(m + M ) We need the moment of inertia of the rod about the center of mass of the entire system. Use the parallel axis theorem. Treat the clay as a point mass.

 ml  I rod = M l + M    4(m + M ) 

1 12

Now express the conservation of angular momentum about the system’s center of mass. Linitial = Lfinal → mvyclay = ( I rod + I clay ) final →

final =

mvyclay

(I + I ) rod

clay

mvyclay 2    ml  2 2 1 +  12 M l + M   my clay    4(m + M )    

mv =

Ml 4(m + M )

  ml   Ml   +  121 M l 2 + M  m   4 m + M    4(m + M ) )    (  2

(

12mv ( m + M )

l 7 m 2 + 11mM + 4 M 2

)

12mv l ( 7m + 4M )

396

Chapter 11

Angular Momentum; General Rotation

53. (a) See the free-body diagram for the ball, after it has moved away from the initial point. There are three forces on the ball. FN and mg are in opposite directions and each has the same lever arm about an axis passing through point O perpendicular to the plane of the paper. Thus they cause no net torque. Ffr has a 0 lever arm about an axis through O, and so also produces no torque. Thus the net torque on the ball is 0. Since we are calculating torques about a point fixed in an inertial reference frame, dL we may say that  τ = = 0 and so L is constant. Note that the ball is initially slipping dt while it rolls, and so we may NOT say that v0 = R0 at the initial motion of the ball. (b) We follow the hint, and express the total angular momentum as a sum of two terms. We take clockwise as the positive rotational direction. L = L v + L = mRvCM − I  CM

The angular momentum is constant. We equate the angular momentum of the initial motion, with vCM = v0 and  = 0 = C , to the final angular momentum, with vCM = 0 and  = 0.

L initial = L final → mRv0 − I C = mR ( 0 ) − I ( 0 ) = 0 → C =

mRv0 I CM

mRv0 5v0 = 2 = 2 mR 2R 5

(c) Angular momentum is again conserved. In the initial motion, vCM = v0 and 0 = 0.90C . Note that in the final state,  = vCM R , and the final angular momenta add to each other. Linitial = Lfinal → mRv0 − 0.90 I C = mRvCM + I

(

)  2 R  = mRv

mRv0 − 0.90 52 mR 2 

5v0





vCM

→

R 2 v CM

( mR ) R 2 5

→

1 10

v0 = 75 vCM → vCM = 141 v0

This answer is reasonable. There is not enough “backspin” since 0  C , and so the ball’s final state is rolling forwards. (d) Angular momentum is again conserved. In the initial motion, vCM = v0 and 0 = 1.10C . Note that in the final state,  = vCM R , and the final angular momenta add to each other. Linitial = Lfinal → mRv0 − 1.10 I C = mRvCM + I

(

)  2 R  = mRv

mRv0 − 1.10 52 mR 2 



5v0



vCM R

→

vCM

( mR ) v R 2 5

→ − 101 v0 = 75 vCM →

vCM = − 141 v0

This answer is reasonable. There is more than enough “backspin” since 0  C , and so the ball’s final state is rolling backwards. 54. Use Eq. 11–13c for the precessional angular velocity. ( 0.22 kg ) ( 9.80 m s 2 ) ( 0.035 m ) Mgr Mgr = → I= = = 7.0  10 −4 kg m 2 I   1rev  2 rad    15 rev  2 rad    5.5s  rev    1s  rev      © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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55. (a) The period of precession is related to the reciprocal of the angular precessional frequency. 2 2 1 2 2 2 ( 45 rev s )( 0.055 m ) 2 2 I  2  2 Mrdisk  2 f 2 2 frdisk = = = = T=  Mgr Mgr gr ( 9.80 m s2 ) ( 0.105 m ) = 2.611s  2.6s

(b) Use the relationship T =

2 2 2 f rdisk

derived above to see the effect on the period.

2 rdisk

r  2 Tnew r grnew rnew  disk new  2 1 = 2 = = = = 2 2 2  rdisk  rnew  1  2 rdisk Toriginal 2 f rdisk   gr r new

new

So the period would double, and thus be Tnew = 2Toriginal = 2 ( 2.611s ) = 5.222 s  5.2 s . 56. Use Eq. 11–13c for the precessional angular velocity. 9.80 m s 2 ( 0.25 m ) Mgr Mg ( 12 l axle ) g l axle = = 1 = = = 8.0 rad s 2 2 I Mrwheel rwheel   ( 0.060 m )2 ( 85 rad s ) 2

(

)

(1.3 rev s )

57. The mass is placed on the axis of rotation and so does not change the moment of inertia. The addition of the mass does change the center of mass position r, and it does change the total mass, M, to 23 M . rnew =

M ( 12 l axle ) + 12 M l axle

 new  original

M+ M 1 2

M l axle 3 2

= 23 l axle

M new g rnew

3 M ( 32 l axle ) I = 2 =2 → M original g roriginal M ( 12 l axle ) I

 new = 2 original = 2 ( 8.0 rad s ) = 16 rad s

58. The spinning bicycle wheel is a gyroscope. The angular frequency of precession is given by Eq. 11–13c. 9.80 m s 2 ) ( 0.20 m ) ( mgr mgr gr = = = 2 = 2  rwheel  ( 0.325 m ) 2 ( 5.0 rad s ) I mrwheel

 1rev   60 s    = 11rev min  2 rad   1min 

= 1.181rad s 

In the figure, the torque from gravity is directed back into the paper. This gives the direction of precession. When viewed from above, the wheel will precess counterclockwise.

398

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Angular Momentum; General Rotation

59. The Sun will pull on the bulge closer to it more than it pulls on the opposite bulge, due to the inverse-square law of gravity. These forces from the Sun, and those from the Moon, create a torque which causes the precession of the axis of rotation of the Earth. The precession is about an axis perpendicular to the plane of the orbit. During the equinox, no torque exists, since the forces on the bulges lie along a line. 60. We assume that the plant grows in the direction of the local “normal” force. In the rotating frame of the platform, there is an outward fictitious force of v2 magnitude m = mr 2 . See the free body diagram for the rotating frame of r reference. Since the object is not accelerated in that frame of reference, the “normal” force must be the vector sum of the other two forces. Write Newton’s second law in this frame of reference. mg  Fvertical = FN cos  − mg = 0 → FN = cos 

F

= FN sin  − mr = 0 → FN = 2

horizontal

mg cos 

mr

sin 

→ tan =

r g

mr 2 sin 

→  = tan

−1

r 2 g

In the inertial frame of reference, the “normal” force still must point inward. The horizontal component of that force is providing the centripetal acceleration, which points inward. 61. (a) At the North Pole, the factor m 2 r is zero, and so there is no effect from the rotating reference frame.

g  = g −  2 r = g − 0 = 9.80 m s 2 , inward along a radial line (b) To find the direction relative to a radial line, we orient the coordinate system along the tangential (x) and radial (y, with inward as positive) directions. See the diagram. At a specific latitude  , the “true” gravity will point purely in the positive y-direction, g = g ˆj. We label the effect of the rotating reference frame as g rot . The effect of g rot can be found by decomposing it along the axes. Note that the radius of rotation is not the radius of the Earth, but r = RE cos  . g = r 2 sin  ˆi − r 2 cos  ˆj rot

= RE 2 cos  sin  ˆi − RE 2 cos 2  ˆj

(

)

g = g + g rot = RE 2 cos  sin  ˆi + g − RE 2 cos 2  ˆj

The angle of deflection from the vertical ( ) can be found from the components of g .

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2 g x −1 RE cos  sin  = tan  = tan g y g − RE 2 cos 2  −1

2 rad  cos 45 sin 45 ( 6.38  10 m )  86, −2 400s  m s2 −1 −1 1.687  10  = tan = tan 2 9.783m s 2  2 rad  2 6 2 9.80 m s − ( 6.38  10 m )   cos 45  86, 400s  6

= 0.0988 The magnitude of g is found from the Pythagorean theorem. g =

g x2 + g y2 =

(1.687  10 m s ) + ( 9.783m s ) = 9.78 m s −2

And so g  = 9.78 m s 2 , 0.0988 south from an inward radial line . (c) At the equator, the effect of the rotating reference frame is directly opposite to the “true” acceleration due to gravity. Thus the values simply subtract. 2

 2 rad  6.38  106 m ) (   86.400s 

g  = g −  2 r = g −  2 REarth = 9.80 m s 2 − 

= 9.77 m s 2 , inward along a radial line . 62. (a) In the inertial frame, the ball has the tangential speed of point B, v B = rB. This is greater than the tangential speed of the women at A, v A = rA , so the ball passes in front of the women. The ball deflects to the right of the intended motion. See the diagram. (b) We follow a similar derivation to that given in Section 11–9. In the inertial frame, the ball is given an inward radial velocity v by the man at B. The ball moves radially inward a distance rB − rA during a short time t, and so rB − rA = vt. During this time, the ball moves sideways a distance sB = v Bt , while the woman moves a distance sA = v A t . The ball will pass in front of the woman

a distance given by the following. s = sB − sA = ( vB − vA ) t = ( rB − rA )  t = v t 2 This is the sideways displacement as seen from the noninertial frame, and so the deflection is

v t 2 . This has the same form as motion at constant acceleration, with s = v t 2 = 12 aCor t 2 . Thus the Coriolis acceleration is aCor = 2v . 63. The footnote on page 342 gives the Coriolis acceleration as aCor = −2ω  v. The angular velocity vector is parallel to the axis of rotation of the Earth. For the Coriolis acceleration to be 0, then the velocity must be parallel to the axis of rotation of the Earth. At the equator this means moving either due north or due south.

400

Chapter 11

Angular Momentum; General Rotation

64. The Coriolis acceleration of the ball is modified to aCor = 2 v ⊥ = 2 v cos  , where v is the vertical speed of the ball. The vertical speed is not constant as the ball falls, but is given by v = v0 + gt. Assuming the ball starts from rest, then aCor = 2 g t cos  . That is not a constant acceleration, and so to find the deflection due to this acceleration, we must integrate twice. v t dv aCor = 2 gt cos  = Cor → d vCor = 2 gt cos  dt →  d vCor = 2 g cos   t dt → dt 0 0 Cor

vCor =  gt cos  = 2

dxCor dt

xCor

 dx

→ dxCor =  gt cos  dt → 2

Cor

=   gt 2 cos  dt → 0

xCor =  gt cos  So to find the Coriolis deflection, we need the time of flight. The vertical motion is just uniform acceleration, for an object dropped from rest. Use that to find the time. 3

1 3

y = y0 + v0 y t + 12 gt 2 → t =

2 ( y − y0 ) g

 2h    g 

xCor = 13  gt 3 cos  = 13  g cos  

g 1/ 2

3/ 2

 8h 3    g 

= 13  cos   1/ 2

 8 (110 m )3   2 rad  1 = 3  ( cos 44 )  9.80 m s 2   86, 400s   

= 0.018 m

The ball is deflected by about 2 cm in falling 110 meters. 65. The diagram is a view from above the wheel. The ant is moving in a curved path, and so there is a fictitious outward radial force of m 2 r ˆi. The ant is moving away from the axis of rotation, and so there is a fictitious Coriolis force of −2m  v ˆj. The ant is moving with a constant speed, and so in the rotating reference frame the net force is 0. Thus there must be forces that oppose these fictitious forces. The ant is in contact with the spoke, and so there can be components of that contact force in each of the coordinate axes. The force opposite to the local direction of motion is friction, and so is − Ffr ˆi. The spoke is also pushing in the opposite direction to the Coriolis force, and so we have F ˆj. Finally, in the vertical direction, there is gravity ( − mgkˆ ) spoke

and the usual normal force ( FN kˆ ) . These forces are not shown on the diagram, since it is viewed from above.

(

)

Frotating = m 2 r − FFr ˆi + ( Fspoke − 2m  v ) ˆj + ( FN − mg ) kˆ frame

401

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Instructor Solutions Manual

66. (a) Because the hoop is rolling without slipping, the acceleration of the center of the center of mass is related to the angular acceleration by aCM =  R. From the free-body diagram, write Newton’s second law for the vertical direction and for rotation. We call down and clockwise the positive directions. Combine those equations to find the angular acceleration.  Fvertical = Mg − FT = MaCM → FT = M ( g − aCM )

 = F R = I = MR

aCM R

= MRaCM

M ( g − aCM ) R = MRaCM → g

 = I  = MR 2 12

= 12 MRg =

( g − aCM ) = aCM → aCM = 12 g →  = 12 dL dt

→

g R

L = 12 MRgt

(b) FT = M ( g − aCM ) = 12 Mg , and is constant in time. 67. (a) Use Eq. 11–6 to find the angular momentum. ˆi ˆj kˆ

(

)

L = r  p = m ( r  v ) = (1.50 kg ) 0

2.0

4.0 kg m 2 s = −36ˆi + 42ˆj − 21kˆ kg m 2 s

7.0

6.0

ˆi

ˆj

kˆ

(b) τ = r  F = 0

2.0

4.0 m N = 16ˆj − 8.0kˆ m N

4.0

(

)

68. Angular momentum is conserved in the interaction between the child and the merry-go-round. 2 → I mgr0 = ( I mgr + I child )  = ( I mgr + mchild Rmgr Linitial = Lfinal → L0 = Lf + Lf ) → mgr

mchild =

I mgr (0 −  ) 2 Rmgr 

child

mgr

(860 kg m ) (1.70 rad s − 1.25 rad s ) = 34.4 kg  34 kg = 2

( 3.0 m )2 (1.25 rad s )

69. (a) See the free-body diagram for the vehicle, tilted up on 2 wheels, on the verge of rolling over. The center of the curve is to the left in the diagram, and so the center of mass is accelerating to the left. The force of gravity acts through the center of mass, and so causes no torque about the center of mass, but the normal force and friction cause opposing torques about the center of mass. The amount of tilt is exaggerated. Write Newton’s second laws for the horizontal and vertical directions and for torques, taking left, up, and counterclockwise as positive.

402

Chapter 11

Angular Momentum; General Rotation

F

= FN − Mg = 0 → FN = Mg

vertical

vC2

=F =M F R  = F ( w) − F h = 0 → F ( w ) = F h horizontal

1 2

Mg ( 12 w ) = M

vC2 R

h →

vC =

vC2

RSUV

g ( SSF )car

2 C

g ( SSF )SUV

w   2h 

Rg 

(b) From the above result, we see that R =

Rcar

1 2

vC2 2h

g w

vC2

g ( SSF )

( SSF )SUV 1.05 = = 0.750 ( SSF )car 1.40

The car can negotiate a “tighter” (smaller radius) curve than the SUV can. 70. (a) The acceleration is needed since F = ma. dr dv r = ( v x 0t ) î + ( v y 0t − 12 gt 2 ) ˆj ; v = = v x 0 î + ( v y 0 − gt ) ˆj ; a = = − gˆj (as expected) dt dt î ˆj kˆ τ = r  F = r  ma = mr  a = m v t v t − 1 gt 2 0 = − g v t kˆ x0

−g

(b) Find the angular momentum from L = r  p = m ( r  v ) , and then differentiate with respect to time. ˆi ˆj kˆ L = r  p = m ( r  v ) = m v t v t − 1 gt 2 0 =  v t ( v − gt ) − v v t − 1 gt 2  kˆ x0

v y 0 − gt

vx0

 x0

(

)

= − 12 v x 0 g t 2 kˆ dL

− v g t kˆ ) = −v g t kˆ ( dt 1 2

dt We see that the two expressions are the same.

71. The mass and kinetic energy are used to find the velocity, and then the velocity is used to find the angular momentum. Since the ball is “small,” we consider it to be a point relative to the radius of the circle in which it is moving. 2K K = 12 mv 2 → v = m L = I  = mr 2

v r

= mr v = mr

2K m

= r 2 Km = ( 0.50 m ) 2 ( 0.40 J )( 0.25 kg ) = 0.2236 kg m 2 s

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Since the ball is moving clockwise when viewed from above, the angular momentum is “down,” into the plane of motion of the ball. If the ball is moving in the x-y plane, then the angular momentum is in the negative z direction. L = −0.22 kg m2 s kˆ 72. The force applied by the spaceship puts a torque on the asteroid which changes its angular momentum. We assume that the rocket ship’s direction is adjusted to always be tangential to the surface. Thus the torque is always perpendicular to the angular momentum, and so will not change the magnitude of the angular momentum, but only its direction, similar to the action of a centripetal force on an object in circular motion. From the diagram, we make an approximation. dL L L =   → dt t t t =

L



I   Fr

(

= 5

)

mr 2  Fr

L

L 

2mr  5F

 4 rev  2 rad  1day     2 rad   6.0          360    1day  1rev  86,400 s   

2 2.65  1010 kg (123 m )  =

5 ( 265 N )

(

= 1.499  105 s

1hr = 42 hr ) 3600 s

Note that in the diagram in the book, the original angular momentum of the asteroid is “up” and the torque applied by the rocket is into the page. Thus the planet’s axis would actually tilt so that the northern half would tilt backwards into the plane of the paper, and the southern half would tilt forwards out of the plane of the paper. It would not rotate clockwise as shown in the figure above. 73. The velocity is the derivative of the position. dr d =  R cos (  t ) î + R sin (  t ) ˆj = − R sin (  t ) î +  R cos (  t ) ˆj v= dt dt  = v  − sin (  t ) î + cos (  t ) ˆj

From the right-hand rule, a counterclockwise rotation in the x-y plane produces an angular velocity

v in the +kˆ -direction. Thus  =   kˆ . Now take the cross product   r. R

v    r =  kˆ    R cos (  t ) ˆi + R sin (  t ) ˆj = R 

ˆi

ˆj

R cos (  t )

R sin (  t )

kˆ v R 0

= −v sin (  t ) ˆi + v cos (  t ) ˆj = v Thus we see that v =   r .

404

Chapter 11

Angular Momentum; General Rotation

74. Note that z = v z t , and so

dz dt

= v z . To find the angular momentum, use Eq. 11–6, L = r  p.

 2 v z t  ˆ  2 v z t  ˆ  2 z  ˆ  2 z  ˆ ˆ  i + R sin   j + zk = R cos   i + R sin   j + v z t kˆ  d   d   d   d  2 v z 2 v z dr  2 v z t  ˆ  2 v z t  ˆ v= = −R sin  cos  i + R  j + v z kˆ dt d d  d   d  2 v z To simplify the notation, let   . Then the kinematical expressions are as follows. r = R cos 

d ˆ ˆ r = R cos ( t ) i + R sin ( t ) j + v z t kˆ ; v = − R sin ( t ) ˆi +  R cos ( ) ˆj + v z kˆ ˆi ˆj kˆ L = r  p = mr  v = m R cos ( t )

R sin ( t )

− R sin ( t )

R cos ( t )

vzt vz

= m  Rv z sin ( t ) − R v z t cos ( t ) ˆi + m  − R v z t sin ( t ) − Rv z cos ( t ) ˆj + m  R 2 cos 2 ( t ) + R 2 sin 2 ( t )  kˆ





= mRv z sin ( t ) −  t cos ( t ) ˆi + mRv z  − t sin ( t ) − cos ( t ) ˆj + mR 2 kˆ



= mRv z sin ( t ) −  t cos ( t ) ˆi +  − t sin ( t ) − cos ( t ) ˆj +



 

R ˆ  k vz 

 2 z  2 z  2 z   ˆ  2 z  2 z   2 z   ˆ 2 R  k cos  sin  −  i + −  − cos   j+ d  d  d  d   d  d   d   

= mRv z  sin 

75. (a) From the free-body diagram, we see that the normal force will produce a torque about the center of mass. That torque, τ = r  FN, is clockwise in the diagram

and so points into the paper, and will cause a change L = τt to the tire’s L0 original angular momentum. L also points into the page, and so the angular momentum will change to have a component into the page. That means that the mg tire will turn to the right in the diagram. (b) The original angular momentum is the moment of inertia times the angular velocity. We assume the wheel is rolling without slipping. L = τt = ( rFN sin  ) t = rmg sin  t ; L0 = I  = I v r L L0

r 2 mg sin  t Iv

( 0.32 m ) ( 8.0 kg ) ( 9.80 m s 2 ) sin12 ( 0.20 s ) = = 0.19 ( 0.83 kg m 2 ) ( 2.1m s ) 2

405

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

76. (a) See the diagram. The parallel axis theorem is used to find the moment of inertia of the arms. The values to be used are: M arm = 4.0 kg ; M body = 62.0 kg ; Rbody = 0.16 m ; l arm = 0.600 m I a = I body + I arms 2 2 = 12 M body Rbody + 2  121 M arm l arm + M arm ( Rbody + 12 l arm )  2





= 12 ( 62.0 kg )( 0.16 m )

+ 2  121 ( 4.0 kg )( 0.60 m ) + ( 4.0 kg )( 0.46 m )  = 2.7264 kg m 2  2.73kg m 2 2

(b) Now the arms can be treated like particles, since all of the mass of the arms is the same distance from the axis of rotation. 2 2 I b = I body + I arms = 12 M body Rbody + 2M arm Rbody =

1 2

( 62.0 kg )( 0.16 m ) + 2 ( 4.0 kg )( 0.16 m ) = 0.9984 kg m 2

R M arm

M body

M arm

 1.00 kg m 2

(c) Angular momentum is conserved through the change in posture. 2 2 Linitial = Lfinal → I aa = I bb → I a = Ib → Ta Tb Ib

0.9984 kg m 2

(1.2 s ) = 0.4394 s  0.44 s Ia 2.7264 kg m 2 (d) The change in kinetic energy is the final kinetic energy (arms horizontal) minus the initial kinetic energy (arms at sides). Tb =

Ta =

 2  1  2  K = K a − K b = I a − I b = ( 2.7264 kg m )  − 2 ( 0.9984 kg m 2 )     1.2 s   0.4394 s  2 a

1 2

2 b

1 2

= −64.70 J  −60 J

The subtraction rule for significant figures limits us to the “tens” place in our answer. (e) Because of the decrease in kinetic energy when lifting the arms, it is easier to lift the arms when rotating than when at rest. There is no corresponding change in kinetic energy if the person is at rest. In the rotating system, the arms tend to move away from the center of rotation. Another way to express this is that it takes work to bring the arms into the sides when rotating. 77. (a) The angular momentum delivered to the waterwheel is that lost by the water. Lwheel = −Lwater = Linitial − Lfinal = mv1 R − mv2 R → water

Lwheel t

mv1 R − mv2 R

t

 820 kg m s 2

water

mR t

( v1 − v2 ) = ( 85 kg s )( 3.0 m )( 3.2 m s ) = 816 kg m 2 s 2

(b) The torque is the rate of change of angular momentum, from Eq. 11–9. L  on = wheel = 816 kg m 2 s 2 = 816m N  820 m N t wheel (c) Power is given by Eq. 10–21, P = .  2 rev  P =  = ( 816m N )   = 854.5W  850 W  6.0s  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

406

Chapter 11

Angular Momentum; General Rotation

78. (a) The work done by the hand is the work done by the centripetal force in changing the radius. Use Eq. 7–7 to calculate the work. The radius R is measured OUTWARD from the origin, but the force is INWARD. Thus the centripetal force and the displacement are in opposite directions, which introduces a –1 factor from the cosine of the angle between the force and the displacement. F=

mv 2 R

R = R2

R = R1

 F dl = −

;W =

mv 2 R

From the conservation of angular momentum, as discussed in Ex. 11–1, we can express v as a vR function of R: mv1 R1 = mvR → v = 1 1 R 2

v R  m 1 1  R R R R 2 mv 1 1 R  W = − dR =  −  dR = − mv12 R12  3 dR = mv12 R12 R R R 2R 2 R R R R 2

 1

2  2  R1 1 v = m − 1 1   2 2 2  R2 R1   R2  2  2  ( 0.80 m ) = 12 ( 0.75 kg )( 2.4 m s )  − 1  = 3.84 J  3.8 J 2  ( 0.48 m )   

= 12 mv12 R12 

1 

− 2

(b) The work-energy theorem says that the work done is equal to the change in kinetic energy. vR Note that from the conservation of angular momentum that v 2 = 1 1 . R2 2

v R   R2  W = K 2 − K1 = mv − mv = m  1 1  − 12 mv12 = 12 mv12  12 − 1  R2   R2  2 2

1 2

2 1

1 2

We have obtained the same expression for work in both parts (a) and (b). 79. From Problem 27, we have that a tan = α  r. For this object, rotating counterclockwise and gaining angular speed, the angular acceleration is α =  kˆ . a tan = α  r =

ˆi

ˆj

kˆ

 = − R sin  ˆi +  R cos  ˆj

R cos 

R sin 

(a) We need the acceleration in order to calculate τ = r  F. The force consists of two components, a radial (centripetal) component and a tangential component. There is no torque associated with the radial component since the angle between r and Fcentrip is 180. Thus τ = r  F = r  Ftan = r  ma tan = mr  a tan .

ˆi

ˆj

kˆ

(

)

τ = mr  a tan = m R cos 

R sin 

0 = m R 2 cos 2  + R 2 sin 2  kˆ = mR 2 kˆ

− R sin 

 R cos 

(b) The moment of inertia of the particle is I = mR 2 . τ = I α = mR 2 kˆ © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

407

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

2 80. The spin angular momentum of the Moon can be calculated by Lspin = I spinspin = 52 MRMoon spin. The

2 orbital angular momentum can be calculated by Lorbit = I orbitorbit = MRorbit orbit. Because the same

side of the Moon always faces the Earth, spin = orbit.

Lspin Lorbit

2 5

2 spin MRMoon 2 orbit MRorbit

R   1.74  106 m  −6 =  Moon  = 0.4   = 8.21 10 8 R 3.84 10 m     orbit  2 5

81. We calculate spin angular momentum for the Sun, and orbital angular momentum for the planets, 2 treating them as particles relative to the size of their orbits. Angular velocities is given by  = . T 2 2 2  1 day  2 LSun = I SunSun = 25 M Sun RSun = 25 1.99  1030 kg 6.96  108 m TSun ( 25 days )  86,400 s 

(

)(

)

= 1.1217  1042 kg m s 2 LJupiter = M Jupiter RJupiter orbit

2 TJupiter

(

)(

= 190  1025 kg 778  109 m

2  1y  ) 11.9   y 3.156  10 s 2





= 1.9240  1043 kg m s In a similar fashion, we calculate the other planetary orbital angular momenta. 2 2 LSaturn = M Saturn RSaturn = 7.806  1042 kg m s orbit TSaturn 2 LUranus = M Uranus RUranus orbit

2 TUranus

2 LNeptune = M Neptune RNeptune orbit

f =

Lplanets Lplanets + LSun

= 1.695  1042 kg m s

2 TNeptune

= 2.492  1042 kg m s

(19.240 + 7.806 + 1.695 + 2.492 )  1042 kg m s = 0.965 (19.240 + 7.806 + 1.695 + 2.492 + 1.122 )  1042 kg m s

82. (a) During the jump (while airborne), the only force on the skater is gravity, which acts through the skater’s center of mass. Accordingly, there is no torque about the center of mass, and so angular momentum is conserved during the jump. (b) For a single axel, the skater must have 1.5 total revolutions. The number of revolutions during each phase of the motion is the rotational frequency times the elapsed time. Note that the rate of rotation is the same for both occurrences of the “open” position. (1.2 rev s )( 0.10 s ) + fsingle ( 0.50 s ) + (1.2 rev s )( 0.10 s ) = 1.5 rev →

f single =

1.5 rev − 2 (1.2 rev s )( 0.10 s )

( 0.50 s )

= 2.52 rev s  2.5 rev s

The calculation is similar for the triple axel. (1.2 rev s )( 0.10 s ) + f triple ( 0.50 s ) + (1.2 rev s )( 0.10 s ) = 3.5 rev →

f triple =

3.5 rev − 2 (1.2 rev s )( 0.10 s )

( 0.50 s )

= 6.52 rev s  6.5 rev s

408

Chapter 11

Angular Momentum; General Rotation

(c) Apply angular momentum conservation to relate the moments of inertia. Lsingle = Lsingle → I singlesingle = I single single → open

closed

I single

single

closed

open

I single open

single

open

f single =

open

f single

closed

1.2 rev s 2.52 rev s

closed

= 0.476  12

closed

Thus the single axel moment of inertia must be reduced by a factor of about 2. For the triple axel, the calculation is similar. I triple f single 1.2 rev s closed open = = = 0.184  15 I triple f single 6.52 rev s open

closed

Thus the triple axel moment of inertia must be reduced by a factor of about 5. 83. (a) We assume that no angular momentum is in the thrown-off mass, so the final angular momentum of the neutron star is equal to the angular momentum before collapse. 2 L0 = Lf → I 00 = I f f →  25 ( 8.0 M Sun ) RSun  0 =  25 ( 0.30 )(8.0 M Sun ) Rf2   f →

f =

2  25 ( 8.0 M Sun ) RSun 

 25 ( 0.30 )( 8.0 M Sun ) Rf2 

0 =

2  RSun 

( 0.30 ) Rf2 

0 = 103

2 RSun

Rf2

0

( )  1.0 rev  = 1.246  10 rev day  1day  = 1.442  10 rev s = )  86400s    (   3 (12  10 m )  9.0 days  2

10 6.96  108 m 3

 14, 000 rev s (b) Now we assume that the final angular momentum of the neutron star is only 0.30 times the angular momentum before collapse. Since the rotation speed is directly proportional to angular momentum, the final rotation speed will be 3/10 of that found in part (a).  f = ( 0.30 ) (1.442  104 rev s ) = 4300 rev s 84. We assume that the tensions in the two unbroken cables immediately become zero, and so they have no effect on the motion. The forces on the tower are the forces at the base joint, and the weight. The axis of rotation is through the point of attachment to the ground. Since that axis is fixed in an inertial dL system, we may use Eq. 11–9 in one dimension,  = . See the free-body diagram in the text dt to express the torque. d ( I ) 1 2 d dL d  1 d 2 1 1 1 mg sin m g sin = → l = = l → = l = 3l 2    ( )  dt 2 3 2 3 dt dt dt dt This equation could be considered, but it would yield  as a function of time. Use the chain rule to eliminate the dependence on time. d  1 d  d 1 d g 1 g sin  = 13 l = 3l = 3 l → 23 sin  d =  d → 2 dt d dt d l 3 2





 sin  d =   d →

l 0

(

3 2

g l

(1 − cos  ) = 12  2

→ = 3

g l

(1 − cos  ) =

v l

→

)

v = 3 g l (1 − cos  ) = 3 9.80 m s 2 (12 m )(1 − cos  ) = 19 1 − cos  m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

409

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Note that the same result can be obtained from conservation of energy, since the forces at the ground do no work. 85. The desired motion is pure rotation about the handle grip. Since the grip is not to have any linear motion, an axis through the grip qualifies as an axis fixed in an inertial reference frame. The pure rotation condition is expressed by aCM =  bat ( d CM − d grip ) , where d grip is the 0.050 m distance from the end of the bat to the grip. Apply Newton’s second law for both the translational motion of the center of mass, and rotational motion about the handle grip.  F = F = maCM ;  = Fd = I grip → maCMd = I grip →

m ( d CM − d grip ) d = I grip → d =

I grip

m ( d CM − d grip )

So we must calculate the moment of inertia of the bat about an axis through the grip, the mass of the bat, and the location of the center of mass. An infinitesimal element of mass is given by dm =  dx, where  is the linear mass density. 0.84 m

 ( x − d grip )  dx =

I grip =  r 2 dm =

0.84 m

 ( x − 0.050 ) ( 0.61 + 3.3x ) dx

0.84 m

 ( 3.3x − 0.33x + 0.61825 x − 0.061x + 0.001525 ) dx 4

( 3.3x − 0.33x + 0.61825 x − 0.061x + 0.001525 x ) 5

1 5

1 4

m =  dm =   dx = d CM

1 3

1 2

0.84

= 0.33685 kg m 2

0.84 m

 ( 0.61 + 3.3x ) kg m  dx = ( 0.61x + 1.1x ) 2

0.84 0

= 1.1644 kg

( 0.61x + 3.3x )   0.61 x 3.3 x kg m dx =  xdm =  x dx = + = ( )  m m m   1.1644 kg 1

1 2

0.84 m

1 4

0.84 0

= 0.53757 m

0.33685 kg m 2

(1.1644 kg )( 0.53757 m − 0.050 m )

= 0.59333m  0.593m

So the distance from the end of the bat to the “sweet spot” is xs = d + 0.050 m = 0.643m  0.64 m . 86. Calculate the three “triple products” as requested. ˆi ˆj kˆ A  B = A A A = ˆi ( A B − A B ) + ˆj ( A B − A B ) + kˆ ( A B − A B ) x

ˆi

ˆj

kˆ

B  C = Bx

Bz = ˆi ( B y C z − Bz C y ) + ˆj ( Bz C x − Bx C z ) + kˆ ( Bx C y − B y C x )

ˆi

ˆj

kˆ

C  A = Cx

C z = ˆi ( C y Az − C z Ay ) + ˆj ( C z Ax − C x Az ) + kˆ ( C x Ay − C y Ax )

410

Chapter 11

Angular Momentum; General Rotation

(

) = A (B C − B C ) + A (B C − B C ) + A (B C − B C )

A ( B  C ) = Ax ˆi + Ay ˆj + Az kˆ  ˆi ( B y C z − BzC y ) + ˆj ( BzC x − BxC z ) + kˆ ( BxC y − B y C x )  x

= Ax B y C z − Ax Bz C y + Ay Bz C x − Ay Bx C z + Az Bx C y − Az B y C x

(

) = B (C A − C A ) + B (C A − C A ) + B (C A − C A )

B ( C  A ) = Bx ˆi + B y ˆj + Bz kˆ  ˆi ( C y Az − C z Ay ) + ˆj ( C z Ax − C x Az ) + kˆ ( C x Ay − C y Ax )  x

= Bx C y Az − Bx C z Ay + B y C z Ax − B yC x Az + BzC x Ay − BzC y Ax

(

) = C (A B − A B ) +C (A B − A B ) +C (A B − A B )

C ( A  B ) = C x ˆi + C y ˆj + C z kˆ  ˆi ( Ay Bz − Az B y ) + ˆj ( Az Bx − Ax Bz ) + kˆ ( Ax B y − Ay Bx )  x

= C x Ay Bz − C x Az B y + C y Az Bx − C y Ax Bz + C z Ax B y − C z Ay Bx

A comparison of three results shows that they are all the same.

411

CHAPTER 12: Static Equilibrium; Elasticity and Fracture Responses to Questions 1.

If the object has a net force on it of zero, then its center of mass does not accelerate. But since it is not in equilibrium, it must have a net torque, and therefore have an angular acceleration. Some examples are: a) A DVD disk in a player as it comes up to speed, after just being put in the player. b) A hard drive on a computer when the computer is first turned on. c) A window fan immediately after the power to it has been shut off. d) The drum of a washing machine while it is speeding up or slowing down.

The bungee jumper is not in equilibrium, because the net force on the jumper is not zero. If the jumper were at rest and the net force were zero, then the jumper would stay at rest by Newton’s first law. The jumper has a net upward force when at the bottom of the dive, and that is why the jumper is then pulled back upwards.

The meter stick is initially supported by both fingers. As you start to slide your fingers together, more of the weight of the meter stick is supported by the finger that is closest to the center of gravity, so that the torques produced by the fingers are equal and the stick is in equilibrium. The other finger exerts a smaller normal force, and therefore has a smaller frictional force on it, and so slides more easily and moves closer to the center of gravity. The roles switch back and forth between the fingers as they alternately move closer to the center of gravity. The fingers eventually meet at the center of gravity.

In a beam balance, the movable weights are connected to the fulcrum point by relatively long lever arms, while the platform on which you stand is connected to the fulcrum point by a very short lever arm. The scale “balances” when the torque provided by your weight (large mass, small lever arm) is equal to that provided by the sliding weights (small mass, large lever arm).

(a) If we assume that the pivot point of rotation is the lower left corner of the wall in the picture, the gravity force acting through the CM provides torque to keep the wall upright. The gravity force has a relatively small lever arm (~ ½ the wall’s width) and so a sideways force would not have to be particularly large to start to move the wall. There is also a normal force from the ground to the left of the wall, but its lever arm is quite small and so its torque is also small. This might be an unstable situation. (b) With the horizontal extension, there are factors that make the wall less likely to overturn. • The mass of the second wall is larger, and so the torque caused by gravity (helping to keep the wall upright) will be larger for the second wall. • The center of gravity of the second wall is further to the right of the pivot point and so gravity exerts a larger torque to counteract the torque due to F. • The weight of the ground above the new part of the wall provides a large clockwise torque that helps to counteract the torque due to F.

If the net force on an object is not 0, then its CM will accelerate in the direction of the net force. If the net torque on the object is 0, then the object has no angular acceleration. Some examples are: a) A satellite in a circular orbit around the Earth. b) A block sliding down an inclined plane. c) An object that is in projectile motion but not rotating d) The startup motion of an elevator, changing from rest to having a non-zero velocity.

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When the person stands near the top, the ladder is more likely to slip. In the accompanying diagram, the force of the person pushing down on the ladder ( Mg ) causes a clockwise torque about the contact point with the ground, with lever arm dx. The only force causing a counterclockwise torque about that same Ly point is the reaction force of the wall on the ladder, FW. While the ladder is in equilibrium, FW will be the same magnitude as the frictional force at the ground, FG x. Since FG x has a maximum

FG y

mg FG x

value, FW will have the same maximum value, and so FW will dx have a maximum counterclockwise torque that it can exert. As the person climbs the ladder, their lever arm gets longer and so the torque due to their weight gets larger. Eventually, if the torque caused by the person is larger than the maximum torque caused by FW , the ladder will start to slip–it will not stay in equilibrium. 8.

(a) The cone will be in stable equilibrium if it is placed flat on its base. If it is tilted slightly from this position and then released, it will return to the original position. (b) The cone will be in unstable equilibrium if it is balanced on its tip. A slight displacement in this case will cause the cone to topple over. (c) If the cone is placed on its side it will be in neutral equilibrium. If the cone is displaced slightly while on its side, it will remain in its new position.

stable equilib.

unstable equilib.

neutral equilib.

Your center of mass must be above your base of support on the floor to remain balanced. When you are flat footed on the floor your center of mass is above your feet–roughly above the front edge of your heels. When you go up onto your tiptoes your center of mass attempts to move forward so that it will be above your toes. However, since you cannot move part of your body inside the wall, your center of mass cannot move forward to be above your toes. Thus you cannot balance on your tiptoes next to the wall. If you turn around and place your heels against the wall, when you rise up on your tiptoes your center of mass moves away from the wall to be over your toes, and you can stand.

10. When you rise on your tiptoes, your CM shifts forward. Since you are already standing with your nose and abdomen against the door, your CM cannot shift forward. Thus gravity exerts a torque on you and you are unable to stay on your tiptoes–you will return to being flat-footed on the floor. 11. When you start to stand up from a normal sitting position, your CM is not over your point of support (your feet), and so gravity will exert a torque about your feet that rotates you back down into the chair. You must lean forward in order that your CM is over your feet so that you can stand up.

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12. In the midst of doing a sit-up, the abdomen muscles provide a torque to rotate you up away from the floor, while the force of gravity on your upper half-body is tending to pull you back down to the floor, providing the difficulty for doing sit-ups. The force of gravity on your lower half-body provides a torque that opposes the torque caused by the force of gravity on your upper half-body, making the sit-up a little easier. With the legs bent, the lever arm for the lower half-body is shorter, and so less counter-torque is available. 13. For rotating the upper half-body, the pivot point is near the waist and hips. In that position, the arms have a relatively small torque, even when extended, due to their smaller mass, and the more massive trunk–head combination has a very short lever arm, and so also has a relatively small torque. Thus the force of gravity on the upper body causes relatively little torque about the hips tending to rotate you forward, and so the back muscles need to produce little torque to keep you from rotating forward. The force on the upper half-body due to the back muscles is small, and so the (partially rightward) force at the base of the spinal column (not shown in the diagram), to keep the spine in equilibrium, will be small. When standing and bending over, the lever arm for the upper body is much larger than while sitting, and so causes a much larger torque. The CM of the arms is also further from the support point, and so causes more torque. The back muscles, assumed to act at the center of the back, do not have a very long lever arm. Thus the back muscles will have to exert a large force to cause a counter-torque that keeps you from falling over. And accordingly, there will have to be a large force (mostly to the right, and not drawn in the diagram) at the base of the spine to keep the spine in equilibrium. 14. Configuration (b) is more likely to be stable. In configuration (a), the CG of the bottom brick is at the edge of the table, and the CG of the top brick is to the right of the edge of the table. Thus the CG of the two-brick system is not above the base of support, and so gravity will exert a torque to roll the bricks clockwise off the table. Another way to see this is that more than 50% of the brick mass is not above the base of support–50% of the bottom brick and 75% of the top brick are to the right of the edge of the table. It is not in stable, neutral, or unstable equilibrium. It will fall. In configuration (b), exactly half of the mass (75% of the top brick and 25% of the bottom brick) is over the edge of the table. Thus the CG of the pair is at the edge of the table–it is in unstable equilibrium. 15. The Young’s modulus for a bungee cord is much smaller than that for ordinary rope. From its behavior, we know that the bungee cord stretches relatively easily, compared to ordinary rope. From F A Eq. 12–4, we have E = . The value of Young’s modulus is inversely proportional to the l l 0 change in length of a material under a tension. Since the change in length of a bungee cord is much larger than that of an ordinary rope if other conditions are identical (stressing force, unstretched length, cross-sectional area of rope or cord), it must have a smaller Young’s modulus. We can also say that for a given stress (force per unit area), the bungee cord will have a greater strain (change in length divided by original length) than the rope, and therefore a smaller Young’s modulus. 16. An object under shear stress has equal and opposite forces applied across its opposite faces. This is exactly what happens with a pair of scissors. One blade of the scissors pushes down on the cardboard, while the other blade pushes up with an equal and opposite force, at a slight displacement. This produces a shear stress in the cardboard, which causes it to fail. Thus the name “shears” is justified, since they are used to produce a shear stress. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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17. Concrete or stone should definitely not be used for the support on the left. The left-hand support pulls downward on the beam, so the beam must pull upward on the support. Therefore, the support will be under tension and should not be made of ordinary concrete or stone, since these materials are weak under tension. The right-hand support pushes up on the beam and so the beam pushes down on it; it will therefore be under a compression force. Making this support of concrete or stone would be acceptable.

Solutions to MisConceptual Questions 1.

(b) The mass of the meter stick is equal to the mass of the rock. Since the meter stick is uniform, its center of mass is at the 50-cm mark, and in terms of rotational motion about a pivot at the 25-cm mark, it can be treated as though its entire mass is concentrated at the center of mass. The meter stick’s mass at the 50-cm mark (25 cm from the pivot) balances the rock at the 0-cm mark (also 25 cm from the pivot) so the masses must be equal.

(d) A common misconception is that a non-rotating object has an axis of rotation. If an object is not rotating, it is not rotating about any arbitrary point. When solving an equilibrium problem with no rotation, the student can select any axis for the torques that facilitates solving the problem.

(a) Students might think that for the net force on the beam to be zero, the tension would equal the weight of the beam. However, this does not take into account the force that the wall exerts on the hinged end. Students might assume that the tension is equal to half of the beam’s weight. However, this does not take into account the vector nature of the tension. The vertical component of the tension is equal to half of the weight, but there is also a horizontal component. Adding these two components yields a tension at least half the weight of the beam.

(c) Drawing a free-body diagram for this problem will resolve student misconceptions. When the ball is pulled to the side, there are three forces acting on the ball; the vertical weight, the horizontal applied force, and the tension along the direction of the cable. Resolving the tension into horizontal and vertical parts and applying Newton’s second law in equilibrium, they can see that the applied force is equal to the horizontal component of the tension.

(a) As the child leans forward, her center of mass moves closer to the pivot point, which decreases her lever arm. The seesaw is no longer in equilibrium and since the torque on her side has decreased, she will rise.

(c) A common misconception is that each cord will support one half of the weight regardless of the angle. An analysis of the forces using Newton’s second law in equilibrium, shows that the horizontal components of the tension are equal. Since cord A makes a larger angle with the horizontal, it has a greater total tension. See Problem 11 for a numerical example.

(c) The applied force is proportional to the stress, so increasing the force will affect the stress. The strain is how the rope responds to the stress. Increasing the force will then affect the strain. Young’s modulus is the constant of proportionality between the stress and strain. It is determined by the properties of the material, and so is not affected by pulling on the rope.

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(e) Students may consider the tension equal to the woman’s weight, or half of the woman’s weight, if they do not consider the vector nature of the forces. A free-body diagram for the point at the bottom of the woman’s foot shows three forces acting; the weight of the woman, and the diagonal tensions in the wire on each side of her foot. Applying Newton’s second law in equilibrium in the vertical direction shows that the vertical component of the tension must equal half of her weight. Since vertical displacement is small compared to the horizontal length of the wire, the total tension is much greater than the vertical component of the tension.

(b) When the number of floors are all doubled, the weight of the garage increases by a factor of two. To keep the stress (force per area) on the columns unchanged, the area of the columns should also increase by a factor of two.

10. (d) The stress (applied force) is proportional to the strain (change in length). Doubling the stress will cause the strain to double also. 11. (b) Let the subscript “1” refer to the girl, and subscript “2” refer to the boy. Let the variable l represent their respective distance from the center of the seesaw. Then their original balance is described by setting the two gravitational torques, with the fulcrum of the seesaw as the axis of rotation, equal in magnitude to each other: m1 g l 1 = m2 g l 2 . If now each child moves half their

l l distance to the pivot point, is it still balanced? In other words, is m1 g 1 = m2 g 2 ? Yes it is, 2 2 since the original equation was just divided by 2. Thus the seesaw remains balanced.

Solutions to Problems 1.

If the tree is not accelerating, then the net force in all directions is 0.  Fx = FA + FB cos105 + FC x = 0 →

105 FA

FC x = − FA − FB cos105 = −385 N − ( 515 N ) cos105 = −251.7 N

 F = F sin105 + F y

C y

=0 →

 

FC y = − FB sin105 = − ( 515 N ) sin105 = −497.5 N FC =

2 C x

 = tan −1

FC y FC x

2 C y

= tan −1

( −251.7 N ) + ( −497.5 N ) = 557.55 N  558 N 2

−497.5 N −251.7 N

= 63.2 ,  = 180 − 63.2 = 117

And so FC is 558 N, at an angle of 117 clockwise from FA . 2.

Because mass m is stationary, the tension in the rope pulling up on the sling is mg, and the force of the sling on the leg is mg, upward. Calculate torques about the hip joint, with counterclockwise as positive. See the free-body diagram for the leg. The forces on the leg exerted by the hip joint are not drawn, because they do not exert a torque about the hip joint.  = mgx2 − Mgx1 = 0 →

m=M

x1 x2

= (15.0 kg )

( 35.0 cm ) = 6.73kg ( 78.0 cm )

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(a) See the free-body diagram. Calculate torques about the pivot point P labeled in the diagram. The upward force at the pivot will not cause any torque. The total torque is zero since the crane is in equilibrium.  = Mgx − mgd = 0 →

( 2800 kg )( 7.7 m ) x= = = 2.34 m  2.3m M ( 9200 kg )

(b) Again we sum torques about the pivot point. Mass m is the unknown in this case, and the counterweight is at its maximum distance from the pivot. Mx ( 9200 kg )( 3.4 m ) = 4062 kg  4100 kg  = Mgxmax − mmax gd = 0 → mmax = dmax = ( 7.7 m ) 4.

Her torque is her weight times the distance x that she is from the left support post. 1800 m N  = = 46 kg  = mgx → m = gx 9.80 m s2 ( 4.0 m )

(

)

(a) Let m = 0. Calculate the net torque about the left end of the diving board, with counterclockwise torques positive. Since the board is in equilibrium, the net torque is zero.  = FB (1.0 m ) − Mg ( 4.0 m ) = 0 →

(

)

FB = 4 Mg = 4 ( 58 kg ) 9.80 m s 2 = 2274 N

1.0 m

2.0 m

4.0 m

 2300 N, up . Use Newton’s second law in the vertical direction to find FA .

 F = F − Mg − F = 0 → y

(

)

FA = FB − Mg = 4Mg − Mg = 3Mg = 3 ( 58 kg ) 9.80 m s 2 = 1705 N  1700 N, down . (b) Repeat the basic process, but with m = 28 kg. The weight of the board will add more clockwise torque.  = FB (1.0 m ) − mg ( 2.0 m ) − Mg ( 4.0 m ) = 0 →

(

)

FB = 4 Mg + 2mg =  4 ( 58 kg ) + 2 ( 28 kg ) 9.80 m s 2 = 2822 N  2800 N, up .

 F = F − Mg − mg − F y

→

FA = FB − Mg − mg = 4Mg + 2mg − Mg − mg = 3Mg + mg

(

)

= 3 ( 58 kg ) + 28 kg  9.80 m s 2 = 1980 N  2.0  103 N, down . 6.

Since each half of the forceps is in equilibrium, the net torque on each half of the forceps is zero. Calculate torques with respect to an axis perpendicular to the plane of the forceps, through point P, counterclockwise being positive. Consider a force diagram for one half of the forceps. F1 is the force on the half-forceps due to the plastic rod, and force FP is the force on the half-forceps from the pin joint. FP does not exert any torque about point P.

 d1

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8.50 cm

 = F d cos  − F d cos  = 0 → F = F d = (11.0 N ) 2.70 cm = 34.6 N T

1 1

The force that the forceps exerts on the rod is the opposite of F1 , and so is also 34.6 N . 7.

Let m be the mass of the beam, and M be the mass of the piano. Calculate torques about the left end of the beam, with counterclockwise torques positive. The conditions of equilibrium for the beam are used to find the forces that the support exerts on the beam.  = FR l − mg ( 12 l ) − Mg ( 14 l ) = 0

(

)

FR = ( 12 m + 14 M ) g =  12 (110 kg ) + 14 ( 320 kg )  9.80 m s 2 = 1320 N

 F = F + F − mg − Mg = 0 y

(

)

FL = ( m + M ) g − FR = ( 430 kg ) 9.80 m s 2 − 1320 N = 2890 N

The forces on the supports are equal in magnitude and opposite in direction to the above two results. FR = 1300 N down FL = 2900 N down 8.

Since the backpack is midway between the two trees, the angles in the diagram are equal. Write Newton’s second law for the vertical direction for the point at which the backpack is attached to the cord, with the weight of the backpack being the downward vertical force. The angle is determined by the distance between the trees and the amount of sag at the midpoint, as illustrated in the second diagram. 1.5 m y = tan −1 = 25.8 (a)  = tan −1 l 2 3.1m

 F = 2 F sin  − mg = 0 → (19 kg ) ( 9.80 m s ) mg = = 213.9 N  210 N F = T

2 sin 1

(b)  = tan −1 FT =

y l 2

mg 2 sin 1

2 sin 25.8

= tan −1

0.15 m 3.1m

= 2.77

(19 kg ) ( 9.80 m s2 ) 2 sin 2.77

= 1926 N  1900 N

The pivot should be placed so that the net torque on the board is zero. We calculate torques about the pivot point, with counterclockwise torques positive. The upward force FP at the pivot point is shown, but it exerts no torque about the pivot point. The mass of the child is m, the mass of the adult is M, the mass of the board is mB , and the center of gravity is at the middle of the board.

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(a) Ignore the force mB g .

 = Mgx − mg ( l − x ) = 0 → x=

m+M

( 25 kg )

( 25 kg + 75 kg )

( 9.0 m ) = 2.25 m  2.3 m from adult .

(b) Include the force mB g .

 = Mgx − mg ( l − x ) − m g ( l 2 − x ) = 0 B

( m + mB 2 )

( M + m + mB )

( 25 kg + 9 kg )

( 75 kg + 25 kg + 18 kg )

( 9.0 m ) = 2.59 m  2.6 m from adult .

10. Using the free-body diagram, write Newton’s second law for both the horizontal and vertical directions, with net forces of zero.  Fx = FT2 − FT1 cos  = 0 → FT2 = FT1 cos 

 Fy = FT1 sin  − mg = 0 → FT1 = FT2 = FT1 cos  = FT1 =

mg sin 

cos  =

mg tan 

(170 kg ) ( 9.80 m s )

 mg

mg sin  (170 kg ) 9.80 m s2

(

tan 33

FT2

FT1

) = 2565N  2600 N

sin 33

= 3059 N  3100 N

11. Draw a free-body diagram of the junction of the three wires. 53o The tensions can be found from the conditions for force FT2 equilibrium. cos 37  Fx = FT1 cos 37 − FT2 cos 53 = 0 → FT2 = cos 53 FT1

37 o

FT1 mg

 F = F sin 37 + F sin 53 − mg = 0 y

FT1 sin 37 +

cos 37

FT1 sin 53 − mg = 0 → cos 53 ( 33 kg ) 9.80 m s 2 FT1 = = 194.6 N  190 N cos 37 sin 37 + sin 53 cos 53 cos 37 cos 37 FT2 = FT1 = 1.946  102 N = 258.3 N  260 N cos 53 cos 53

(

)

(

)

12. The table is symmetric, so the person can sit near either edge 0.60 m x and the same distance will result. We assume that the person (mass M) is on the right side of the table, and that the table mg Mg (mass m) is on the verge of tipping, so that the left leg is on the verge of lifting off the floor. There will then be no normal force between the left leg of the table and the floor. Calculate torques FN about the right leg of the table, so that the normal force between the table and the floor causes no torque. Counterclockwise torques are taken to be positive. The conditions of equilibrium for the table are used to find the person’s location. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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27.0 kg

 = mg ( 0.60 m ) − Mgx = 0 → x = ( 0.60 m ) M = ( 0.60 m ) 66.0 kg = 0.24545 m Thus the distance from the edge of the table is 0.50 m − 0.24545 m = 0.25455 m  0.25 m . 13. The cork screw will pull upward on the cork with a force of magnitude Fcork , and so there is a downward force on the opener of magnitude Fcork . We assume that there is no net torque on the opener, so that it does not have an angular acceleration. Calculate torques about the rim of the bottle where the opener is resting on the rim.  = F ( 79 mm ) − Fcork ( 9 mm ) = 0 →

9 70

Fcork =

9 79

( 200 N ) to

( 400 N ) = 22.8 N to 45.6 N  20 N to 50 N

14. The beam is in equilibrium. Use the conditions of equilibrium to calculate the tension in the wire and the forces at the hinge. Calculate torques about the hinge, and take counterclockwise torques to be positive.  = ( FT sin  ) l 2 − m1 g l 1 2 − m2 g l 1 = 0 → 1

FT = 2

m1 g l 1 + m2 g l 1 l 2 sin 

= 2

(135 N )(1.70 m ) + ( 215 N )(1.70 m ) (1.35 m )( sin 35.0 )

= 620.2 N  6.20  10 2 N

 F = F − F cos  = 0 → F = F cos  = ( 620.2 N ) cos 35.0 = 508.0 N  508 N  F = F + F sin  − m g − m g = 0 → x

FH y = m1 g + m2 g − FT sin  = 135 N + 215 N − ( 620.2 N ) sin 35.0 = −5.73 N  −6 N

So the force from the wall is actually pushing slightly downward. 15. From the free-body diagram, the conditions of equilibrium are used to find the location of the girl (mass mC ). The 45-kg boy is represented by mA , and the 35-kg boy by m B . Calculate torques about the center of the seesaw, and take counterclockwise torques to be positive. The upward force of the fulcrum on the seesaw ( F ) causes no torque about the center.

 = m g ( l ) − m gx − m g ( l ) = 0 A

1 2

( mA − mB ) 1 ( 45 kg − 35 kg ) 1 3.2 m ) = 0.64 m (2 l )= 2( mC

25 kg

16. The center of gravity of each beam is at its geometric center. Calculate torques about the left end of the beam, and take counterclockwise torques to be positive. The conditions of equilibrium for the beam are used to find the forces that the support exerts on the beam. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 = F l − Mg ( l 2 ) − Mg ( l 4 ) = 0 → F = Mg = ( 980 kg ) ( 9.80 m s ) = 6002.5 N  6.0  103 N  F = F + F − Mg − Mg = 0 → F = Mg − F = Mg = ( 980 kg ) ( 9.80 m s ) = 8403.5 N  8400 N 1 2

5 8

3 2

1 2

7 8

17. (a) The pole is in equilibrium, and so the net torque on it must be zero. From the free-body diagram, calculate the net torque about the lower end of the pole, with counterclockwise torques as positive. Use that calculation to find the tension in the cable. The length of the pole is l.  = FT h − mg ( l 2 ) cos  − Mg l cos  = 0 FT =

( m 2 + M ) g l cos 

h ( 6.0 kg + 21.5 kg ) 9.80 m s2 ( 7.20 m ) cos 37

(

)

= 407.8 N  410 N 3.80 m (b) The net force on the pole is also zero since it is in equilibrium. Write Newton’s second law in both the x- and y-directions to solve for the forces at the pivot.  Fx = FP x − FT = 0 → FP x = FT = 410 N

 F = F − mg − Mg = 0 → F = ( m + M ) g = ( 33.5 kg ) ( 9.80 m s ) = 328 N 2

18. Consider the free-body diagram for the board. There are four forces on the board; an upward force at the pivot, the downward force due to the heavy object, the weight of the board itself at its center of gravity, and an upward force from the scale at the right end. Let m be the mass of the board, and M be the mass of the heavy object. Take torques about the pivot end of the board, with clockwise torques being positive.  = Mg (1.2 m ) + mg (1.5 m ) − Fscale ( 3.0 m ) = 0

Mg =

Fscale ( 3.0 m ) − mg (1.5 m ) 1.2 m

1.2 m

Fpivot

3.0 m

Fscale

( 210 N )( 3.0 m ) − ( 3.4 kg ) ( 9.80 m s2 ) (1.5 m ) = = 483 N (1.2 m )

 480 N 19. To find the normal force exerted on the road by the trailer tires, take the torques about point B, with counterclockwise torques as positive.  = mg ( 5.5 m ) − FA (8.0 m ) = 0 →

 5.5 m   5.5 m  = ( 2300 kg ) ( 9.80 m s 2 )    = 15, 496 N  8.0 m   8.0 m 

FA = mg 

 1.5  104 N

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The net force in the vertical direction must be zero.  Fy = FB + FA − mg = 0 →

(

)

FB = mg − FA = ( 2300 kg ) 9.80 m s 2 − 15, 496 N = 7044 N  7.0  103 N

20. The person is in equilibrium, and so both the net torque and net force must be zero. From the free-body diagram, calculate the net torque about the center of gravity, with clockwise torques as positive. Use that calculation to find the location of the center of gravity, a distance x from the top of the head.  = FA x − FB ( l − x ) = 0

mB g

l =

31.6 kg

FA + FB mA g + mB g mA + mB 35.1 kg + 31.6 kg The center of gravity is about 81.5 cm from the top of the head.

(1.72 m ) = 0.8149 m  0.815 m

21. (a) The man is in equilibrium, so the net force and the net torque on him must be zero. We use half of his weight, and then consider the force just on one hand and one foot, considering him to be symmetric. Take torques about the point where the foot touches the ground, with counterclockwise as positive.  = 12 mgd 2 − Fh ( d1 + d 2 ) = 0

1 2

( 68 kg ) ( 9.80 m s2 ) ( 0.95 m ) Fh = = = 231N  230 N 2 ( d1 + d 2 ) 2 (1.37 m ) mgd 2

(b) Use Newton’s second law for vertical forces to find the force on the feet.  Fy = 2 Fh + 2 Ff − mg = 0

(

)

Ff = 12 mg − Fh = 12 ( 68 kg ) 9.80 m s 2 − 231N = 103 N  100 N = 1.0  102 N

22. See the free-body diagram. Take torques about the pivot point, with clockwise torques as positive. The plank is in equilibrium. Let m represent the mass of the plank, and M represent the mass of the person. The minimum nail force would occur if there was no normal force pushing up on the left end of the board.  = mg ( 0.75 m ) cos  + Mg ( 2.25 m ) cos  − Fnails ( 0.75 m ) cos  = 0 →

Fnails =

mg ( 0.75 m ) + Mg ( 2.25 m )

( 0.75 m )

(

= mg + 3Mg

)

=  45kg + 3 ( 65 kg )  9.80 m s 2 = 2352 N  2400 N

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23. (a) See the free-body diagram. We assume that the board is at the edge of the door opposite the hinges, and that you are pushing at that same edge of the door. Then the width of the door does not enter into the problem. Force Fpush is the force of the door on the board, and is the same as the force the person exerts on the door. Take torques about the point A in the free-body diagram, where the board rests on the ground. The board is of length l. Counterclockwise torques are taken to be positive.  = Fpush l sin  − mg ( 12 l ) cos  = 0 →

FG y mg



FG x

( 66.0 kg ) ( 9.80 m s 2 )

Fpush =

Fpush

= = 323.4 N  323 N 2 tan  2 tan 45.0 (b) Because the board is firmly set against the ground, the top of the board would move upwards as the door opened. Thus the frictional force on the board at the door must be down. We also assume that the static frictional force is a maximum, and so is given by Ffr = s FN = s Fpush . Take torques about the point A in the freebody diagram, where the board rests on the ground. The board is of length l. Counterclockwise torques are taken to be positive.  = Fpush l sin  − mg ( 12 l ) cos  − Ffr l cos  = 0 →

Fpush

Ffr FG y mg



FG x

Fpush l sin  − mg ( 12 l ) cos  − s Fpush l cos  = 0 →

Fpush =

( 66.0 kg ) ( 9.80 m s2 ) = 497.5 N  5.0  102 N 2 ( tan  − s ) 2 ( tan  − s ) 2 ( tan 45.0 − 0.35) mg

24. The forces on the door are due to gravity and the hinges. Since the door is in equilibrium, the net torque and net force must be zero. Write the three equations of equilibrium. Calculate torques about the bottom hinge, with counterclockwise torques as positive. From the statement of the problem, FA y = FB y = 12 mg.

 = mg FAx =

w 2

− FAx ( h − 2d ) = 0

w d

y x

FA x

FB y

(13.0 kg ) ( 9.80 m s 2 ) (1.30 m ) = 55.2 N 2 ( h − 2d ) 2 ( 2.30 m − 0.80 m ) mgw

FA y

FB x

 F = F − F = 0 → F = F = 55.2 N  F = F + F − mg = 0 → F = F = mg = (13.0 kg ) ( 9.80 m s ) = 63.7 N x

1 2

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25. (a) Choose the coordinates as shown in the freebody diagram. (b) Write the equilibrium conditions for the horizontal and vertical forces.  Fx = Frope sin  − Fhinge = 0 → horiz

Fhinge = Frope sin  = ( 85 N ) sin 37 = 51N horiz

F = F y

rope

cos  + Fhinge − mg − W = 0 → vert

Fhinge = mg + W − Frope cos  vert

(

)

= ( 3.8 kg ) 9.80 m s 2 + 22 N − ( 85 N ) cos 37 = −8.6 N  −9 N And so the vertical hinge force actually points downward. (c) We take torques about the hinge point, with clockwise torques as positive.  = Wd sin  + mg ( 12 l ) sin  − Frope l sin ( −  ) = 0 →

d= =

Frope l sin ( −  ) − mg ( 12 l ) sin 

W sin  ( 85 N )( 5.0 m ) sin16 − ( 3.8 kg ) 9.80 m s 2 ( 2.5 m ) sin 53

(

( 22 N ) sin 53

)

= 2.436 m  2.4 m

26. First consider the triangle made by the pole and one of the wires (first diagram). It has a vertical leg of 2.6 m, and a horizontal leg of 2.0 m. The angle that the tension (along the wire) makes with the vertical is 2.0 = 37.6o . The part of the tension that is parallel to the ground is  = tan −1 2.6 therefore FT h = FT sin  .

 2.6 m

Now consider a top view of the pole, showing force parallels to the ground (second diagram). The components of the tension in each wire that are parallel to the ground each make a 30° angle with the tension force of the net. Write the equilibrium equation for the forces along the direction of the tension in the net.  F = Fnet − 2 FT h cos 30 = 0 →

Fnet = 2 FT sin  cos 30 = 2 (125 N ) sin 37.6 cos 30 = 132.1N  130 N

2.0 m

Fnet

FT h

FT h 30o 30o

27. The ladder is in equilibrium, so the net force and net torque must both be zero. By stating that the ladder is on the verge of slipping, the static frictional force at the ground, FC x is at its maximum value and so FC x = s FC y . Torques are taken about the point of contact of the ladder

with the ground, and counterclockwise torques are taken as positive. The three conditions of equilibrium are as follows.  Fx = FC x − FW = 0 → FC x = FW

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 F = F − Mg − mg = 0 → y

(

)

FC y = ( M + m ) g = ( 67.0 kg ) 9.80 m s 2 = 656.6 N

 = F ( 4.0 m ) − mg ( ) ( 3.0 m ) − Mg ( 2.1m ) = 0 1 2

Solve the torque equation for FW .

 1 (12.0 kg )( 3.0 m ) + ( 55.0 kg )( 2.1 m )  9.80 m s2 ) = 327.1 N (  4.0 m  

FW =  2

The coefficient of friction then is then found from the components of FC .

FCx

s =

FCy

FW FCy

327.1 N 656.6 N

= 0.50

28. Write the conditions of equilibrium for the ladder, with torques taken about the bottom of the ladder, and counterclockwise torques as positive. mg  = FW l sin  − mg ( 12 l cos  ) = 0 → FW = 12 tan  mg  Fx = FG x − FW = 0 → FG x = FW = 12 tan   Fy = FG y − mg = 0 → FG y = mg For the ladder to not slip, the force at the ground FG x must be less than or equal to the maximum force of static friction. mg  s mg → FG x  s FN = s FG y → 12 tan 

1 2 s

 1    2 s 

 tan  →   tan −1 

Thus the minimum angle is  min = tan −1 (1 2 s ) . 29. We assume the truck is accelerating to the right. We want the refrigerator to not tip in the non-inertial reference frame of the truck. Accordingly, to analyze the refrigerator in the non-inertial reference frame, we must add a pseudoforce in the opposite direction of the actual acceleration. The free-body diagram is for a side view of the refrigerator, just ready to tip so that the normal force and frictional force are at the lower back corner of the refrigerator. The center of mass is in the geometric center of the refrigerator. Write the conditions for equilibrium, taking torques about an axis through the center of mass, perpendicular to the plane of the paper. The normal force and frictional force cause no torque about that axis.  Fhoriz = Ffr − matruck = 0 → Ffr = matruck

F

vert

ma truck h

Ffr

= FN − mg = 0 → FN = mg FN

 = F ( w ) − F ( h ) = 0 → F = w N

1 2

FN Ffr

h w

mg ma truck

→ a truck = g

w h

(

= 9.80 m s 2

1.0 m = 5.2 m s ) 1.9 m

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30. (a) Consider the free-body diagram for each side of the ladder. Because the two sides are not identical, we must have both horizontal and vertical components to the hinge force of one side of the ladder on the other. 1 d d First determine the angle from cos  = 2 = . 2l l 1 d 0.9 m = 68.9  = cos −1 2 = cos −1 l 2.5m Write equilibrium equations for the following conditions: Vertical forces on total ladder:  Fvert = FN − mg + Fhinge − Fhinge + FN = 0 → left

vert

right

FN + FN = mg left

right

Torques on left side, about top, clockwise positive.  = FN ( l cos  ) − mg ( 0.2l ) cos  − FT ( 12 l ) sin  = 0 left

Torques on right side, about top, clockwise positive.  = − FN ( l cos  ) + FT ( 12 l ) sin  = 0 right

Subtract the second torque equation from the first.

  1  FN + FN  ( l cos  ) − mg ( 0.2l ) cos  − 2 FT ( 2 l ) sin  = 0 left right   Substitute in from the vertical forces equation, and solve for the tension. mg ( l cos  ) − mg ( 0.2 l ) cos  − 2 FT ( 12 l ) sin  = 0 →

FT =

( 0.8 cos  ) =

0.8mg

(

0.8 ( 56.0 kg ) 9.80 m s 2

) = 169.4 N  170 N

sin  tan  tan 68.9 (b) To find the normal force on the right side, use the torque equation for the right side. − FN ( l cos  ) + FT ( 12 l ) sin  = 0 → right

FN = 12 FT tan  = 12 (169.4 N ) tan 68.9 = 219.5 N  220 N right

To find the normal force on the left side, use the vertical force equation for the entire ladder. FN + FN = mg → left

right

(

)

FN = mg − FN = ( 56.0 kg ) 9.80 m s2 − 219.5 N = 329.3 N  330 N left

right

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(c) We find the hinge force components from the free-body diagram for the right side.  Fvert = FN − Fhinge = 0 → Fhinge = FN = 219.5 N right

F

horiz

vert

right

= Fhinge − FT = 0 → Fhinge = FT = 169.4 N horiz

Fhinge =

vert

horiz

2 2 + Fhinge = Fhinge horiz

hinge = tan

vert

Fhinge −1

(169.4 N ) 2 + ( 219.5 N ) 2 = 277.3 N  280 N

vert

Fhinge

= tan −1

219.5 N 169.4 N

= 52

horiz

31. Calculate the torques about the elbow joint (the dot in the free body diagram). The arm is in equilibrium. Counterclockwise torques are positive.

 = F d − mgD − Mg l = 0 M

FM =

mD + M l d

 ( 2.3 kg )( 0.12 m ) + ( 7.3 kg )( 0.300 m )  2  ( 9.80 m s ) = 970 N 0.025 m  

=

32. Figures 12–13 (b) and (c) are redrawn here with the person 45o from the horizontal, instead of the original 30o. The distances are all the same as in the original problem. We still assume that the back muscles pull at a 12o angle to the spine. The 18o angle from the original problem becomes 33o. Torques are taken about the same point at the base of the spine, with counterclockwise torques as positive.  = ( 0.48 m ) FM sin12 − ( 0.72 m )( wH ) sin 45

45o

− ( 0.48 m )( wA ) sin 45 − ( 0.36 m )( wT ) sin 45 = 0

As in the original problem, wH = 0.07 w, wA = 0.12w, wT = 0.46w. With this, the torque equation gives the following result. ( 0.72 m )( 0.07 ) + ( 0.48 m )( 0.12 ) + ( 0.36 m )( 0.46) w sin 45o = 1.94w FM = ( 0.48 m ) sin12

Take the sum of the forces in the vertical direction, set equal to zero.  Fy = FV y − FM sin 33 − 0.07w − 0.12w − 0.46w = 0 → FV y = 1.71w FM

33o

45o 12o

Take the sum of the forces in the horizontal direction, set equal to zero.  Fx = FV x − FM cos 33 = 0 → FV y = 1.63w The final result is

FV = FV2x + FV2y = 2.4w This compares to 2.5w for the more bent position.

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33. (a) Calculate the torques about the elbow joint (the dot in the freebody diagram). The arm is in equilibrium. Take counterclockwise torques as positive.  = ( FM sin  ) d − mgD = 0 →



FM mg

( 3.6 kg ) ( 9.80 m s2 ) ( 0.24 m ) = = 272.6 N FM = d sin  ( 0.12 m ) sin15 mgD

d D

 270 N (b) To find the components of FJ , write Newton’s second law for both the x- and y-directions. Then combine them to find the magnitude.  Fx = FJ x − FM cos  = 0 → FJ x = FM cos  = ( 272.6 N ) cos15 = 263.3 N

 F = F sin  − mg − F = 0 → y

(

)

FJ y = FM sin  − mg = ( 272.6 N ) sin15 − ( 3.6 kg ) 9.80 m s 2 = 35.3 N FJ =

FJ2x + FJ2y =

( 263.3 N )2 + ( 35.3 N )2 = 265.7 N  270 N

34. Calculate the torques about the shoulder joint, which is at the left end of the free-body diagram of the arm. Since the arm is in equilibrium, the sum of the torques will be zero. Take counterclockwise torques to be positive. The force due to the shoulder joint is drawn, but it does not exert any torque about the shoulder joint.

 = F d sin  − mgD − Mg l = 0 m

Fm =

mD + M l d sin 

( 3.6 kg )( 0.24 cm ) + (8.5 kg )( 0.52 m ) 9.80 m s 2 ) = 1667 N  1700 N ( ( 0.12 m ) sin15

35. There will be a normal force upwards at the ball of the one foot on the floor, equal to the person’s weight ( FN = mg ) . Calculate torques about a point on the floor directly below the leg bone (and so in line with the leg bone force, FB ). Since the foot is in equilibrium, the sum of the torques will be zero. Take counterclockwise torques as positive.  = FN ( 2d ) − FT d = 0 →

(

D = 2d

)

FT = 2 FN = 2mg = 2 ( 72 kg ) 9.80 m s 2 = 1400 N The net force in the y-direction must be zero. Use that to find FB .

 F = F + F − F = 0 → F = F + F = 2mg + mg = 3mg = 2100 N y

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36. From Section 12–4: “An object whose center of gravity is above its base of support will be stable if a vertical line projected downward from the CG falls within the base of support.” For the tower, the base of support is a circle of radius 7.7 m. If the top is 4.5 m off center, then the CG will be 2.25 m off center, and a vertical line downward from the CG will be 2.25 m from the center of the base. As long as that vertical line is less than 7.7 m from the center of the base, the tower will be in stable equilibrium . To be unstable, the CG has to be more than 7.7 m

Vertical

Ready to fall

off center, and thus the top must be more than 2 × (7.7 m) = 15.4 m off center. Thus the top will have to lean 15.4 m–4.5 m = 10.9 m further to reach the verge of instability. 37. The truck will not tip as long as a vertical line down from the CG is between the wheels. When that vertical line is at the wheel, it is in unstable equilibrium and will tip if the road is inclined any more. See the diagram for the truck at the tipping angle, showing the truck’s weight vector. 1.2 m x x tan  = →  = tan −1 = tan −1 = 29o h h 2.2 m



38. (a) The maximum distance for brick #1 to remain on brick #2 will be reached when the CM of brick #1 is directly over the edge of brick #2. Thus brick #1 will overhang brick #2 by x1 = l 2 . The maximum distance for the top two bricks to remain on brick #3 will be reached when the center of mass of the top two bricks is directly over the edge of brick #3. The CM of the top two bricks is (obviously) at the point labeled x on brick #2, a distance of l 4 from the right edge of brick #2. Thus x2 = l 4. The maximum distance for the top three bricks to remain on brick #4 will be reached when the center of mass of the top three bricks is directly over the edge of brick #4. The CM of the top three bricks is at the point labeled x on brick #3, and is found relative to the center of brick # 3 by m ( 0 ) + 2m ( l 2 ) CM = = l 3, or l 6 from the right edge of brick #3. Thus x3 = l 6. 3m The maximum distance for the four bricks to remain on a tabletop will be reached when the center of mass of the four bricks is directly over the edge of the table. The CM of all four bricks is at the point labeled x on brick #4, and is found relative to the center of brick #4 by m ( 0 ) + 3m ( l 2 ) CM = = 3l 8 , or l 8 from the right 4m edge of brick #4. Thus x4 = l 8 .

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(b) From the last diagram, the distance from the edge of the tabletop to the right edge of brick #1 is x4 + x3 + x2 + x1 = ( l 8 ) + ( l 6 ) + ( l 4 ) + ( l 2 ) = 25l 24  l Since this distance is greater than l , the answer is yes , the first brick is completely beyond the edge of the table. (c) From the work in part (a), we see that the general formula for the total distance spanned by n bricks is xn = ( l 2 ) + ( l 4 ) + ( l 6 ) +

x1 + x2 + x3 +

i =1

+ ( l 2n ) = 

(d) The arch is to span 1.0 m, so the span from one side will be 0.50 m. Thus we must solve n 0.30 m  2i  0.50 m. Evaluation of this expression for various values of n shows that 15 bricks i =1 will span a distance of 0.498 m, and that 16 bricks will span a distance of 0.507 m. Thus, it takes 16 bricks for each half-span, plus 1 brick on top and 1 brick as the base on each side [as in Fig. 12–87(b)], for a total of 35 bricks . 39. The amount of stretch can be found using the elastic modulus in Eq. 12–4. 1 F 1 315 N l = l0 = ( 0.300 m ) = 4.01 10−2 m 2 9 2 E A 3  10 N m  5.00  10 −4

(

40. (a) Stress =

F A

mg A

( 22, 000 kg ) ( 9.80 m s2 ) 1.4 m

Stress

(b) Strain =

)

= 154, 000 N m 2  1.5  105 N m 2

1.54  105 N m 2

= 3.08  10−6  3.1 10−6

Young’s Modulus 50  10 N m (c) The change in length is found from the strain. l Strain = → l = l 0 ( Strain ) = ( 8.6 m ) ( 3.08  10 −6 ) = 2.6  10 −5 m l0

41. (a) Stress = (b) Strain = (c)

F A

mg A

(1700 kg ) ( 9.80 m s2 ) 0.012 m

Stress Young’s Modulus

(

= 1.388  106 N m 2  1.4  106 N m 2

1.388  106 N m 2 200  10 N m 9

= 6.94  10−6  6.9  10−6

)

l = ( Strain )( l 0 ) = 6.94  10−6 ( 9.50 m ) = 6.593  10−5 m  6.6  10−5 m

42. The Young’s Modulus is the stress divided by the strain.

(13.4 N )  ( 12  8.5  10−3 m )  2

Young’s Modulus =

Stress Strain

F A l l 0

(

3.7  10−3 m

)(

15  10−2 m

)

= 9.6  106 N m 2

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43. The relationship between pressure change and volume change is given by Eq. 12–7. P V V = −V0 → P = − B = − 0.10  10 −2 90  109 N m 2 = 9.0  107 N m 2 B V0

(

P Patm

9.0  107 N m 2 1.0  10 N m 5

)(

)

= 9.0  10 2 , or 900 atmospheres.

44. The percentage change in volume is found by multiplying the relative change in volume by 100. The change in pressure is 199 times atmospheric pressure, since it increases from atmospheric pressure to 200 times atmospheric pressure. Use Eq. 12–7. 199 1.0  105 N m 2 V P 100 = −100 = −100 = −2  10−2 % V0 B 90  109 N m 2 The negative sign indicates that the interior space got smaller.

(

)

45. The mass can be calculated from the equation for the relationship between stress and strain. The force causing the strain is the weight of the mass suspended from the wire. Use Eq. 12–4. l 1 F mg = = → l 0 E A EA

EA l g l0

(

= 200  10 N m 9

)

 (1.3  10−3 m ) 0.030 2

( 9.80 m s ) 2

100

= 32.51kg  33 kg

46. Set the compressive strength of the bone equal to the stress of the bone. F Compressive Strength = max → Fmax = 170  106 N m 2 3.0  10−4 m2 = 5.1 104 N A

(

)(

)

47. The tensile breaking force of a wire is equal to the tensile strength of the material times the area of the wire. Set the breaking force for the aluminum wire equal to the breaking force for the steel wire. We abbreviate “tensile strength” as “TS” in the equations below. We assume that the tensile strengths are correct to two significant figures. d2 d2 Fmax = Fmax → ( TS ) Al AAl = ( TS )steel Asteel → ( TS ) Al  Al = ( TS )steel  steel → 4 4 Al steel d Al = d steel

( TS )steel = ( 2.0 mm ) ( TS) Al

500  106 N m 2 200  106 N m 2

= 3.162 mm  3.2 mm

48. (a) The maximum tension can be found from the ultimate tensile strength of the material. F Tensile Strength = max → A

(

) (

Fmax = ( Tensile Strength ) A = 500  106 N m 2  5.00  10 −4 m

) = 393 N 2

(b) To prevent breakage, thicker strings should be used, which will increase the cross-sectional area of the strings, and thus increase the maximum force. Breakage occurs because when the strings are hit by the ball, they stretch, increasing the tension. The strings are reasonably tight in the normal racket configuration, so when the tension is increased by a particularly hard hit, the tension may exceed the maximum force. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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49. (a) Compare the stress on the bone to the compressive strength to see if the bone breaks. F 3.3  104 N Stress = = A 3.6  10−4 m 2 = 9.167  107 N m 2 <1.7  108 N m 2 ( Compressive Strength of bone )

The bone will not break. (b) The change in length is calculated from Eq. 12–4. l F  0.22 m  9.167  107 N m 2 = 1.3  10 −3 m l = 0 = 9 2  E A  15  10 N m 

(

)

50. (a) The area can be found from the ultimate tensile strength of the material. Tensile Strength F  Safety Factor  → A= F =  → A Safety Factor  Tensile Strength  7.0

A = ( 240 kg ) ( 9.80 m s 2 )

= 3.293  10−5 m 2  3.3  10 −5 m 2 500  106 N m 2 (b) The change in length can be found from the stress-strain relationship, Eq. 12–5. ( 7.5 m )( 240 kg ) 9.80 m s 2 l 0F l F =E → l = = = 2.7  10−3 m −5 2 9 2 l0 A AE 3.293  10 m 200  10 N m

(

)(

(

)

51. For each support, to find the minimum cross-sectional area with a F1 F2 Strength F , where either the tensile or = safety factor means that A Safety Factor 20.0 m mg compressive strength is used, as appropriate for each force. To find the 25.0 m force on each support, use the conditions of equilibrium for the beam. Take torques about the left end of the beam, calling counterclockwise torques positive, and also sum the vertical forces, taking upward forces as positive. 25.0  = F2 ( 20.0 m ) − mg ( 25.0 m ) = 0 → F2 = 20.0 mg = 1.25mg  Fy = F1 + F2 − mg = 0 → F1 = mg − F2 = mg − 1.25mg = −0.25mg Notice that the forces on the supports are the opposite of F1 and F2 . So the force on support #1 is directed upwards, which means that support #1 is in tension. The force on support #2 is directed downwards, so support #2 is in compression. F1 Tensile Strength = → A1 9.0

A1 = 9.0

F2 A2

( 0.25mg )

= 9.0

Tensile Strength Compressive Strength

A2 = 9.0

9.0

( 0.25) ( 2.9  103 kg )( 9.80 m s2 ) 40  10 N m 6

= 1.6  10−3 m 2

→

(1.25mg ) Compressive Strength

= 9.0

(1.25) ( 2.9  103 kg )( 9.80 m s2 ) 35  10 N m 6

= 9.1  10−3 m 2

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52. The maximum shear stress is to be 1/7th of the shear strength for iron. The maximum stress will occur for the minimum area, and thus the minimum diameter. shear strength 7.0 F F 2 stress max = = → A1 =  ( 12 d ) = → 7.0 shear strength Amin

4 ( 7.0 ) F

 ( shear strength )

28 ( 3600 N )

 (170  10 N m ) 6

= 1.37  10 −2 m = 1.4 cm

53. From the free-body diagram, write Newton’s second law for the vertical direction. Solve for the maximum tension required in the cable, which will occur for an upwards acceleration.  Fy = FT − mg = ma → FT = m ( g + a )

The maximum stress is to be 1/8th of the tensile strength for steel. The maximum stress will occur for the minimum area, and thus the minimum diameter. tensile strength 8.0 FT F 2 stress max = T = → A1 =  ( 12 d ) = → 8.0 tensile strength Amin

4 ( 8.0 ) m ( g + a )

 ( tensile strength )

(

32 ( 3100 kg ) 11.6 m s 2

 ( 500  10 N m ) 6

) = 2.71  10 m  2.7 cm −2

54. There are upward forces at each support (points A and D) and a downward applied force at point C. Write the conditions for equilibrium for the entire truss by considering vertical forces and the torques about point A. Let clockwise torques be positive. Let each side of the equilateral triangle be of length l .  Fvert = FA + FD − F = 0

 = F ( l ) − F l = 0 → F = F = (1.65 10 N ) = 8250 N 1 2

1 2

FA = F − FD = 1.65  104 N − 8250 N = 8250 N (a) Analyze the forces on the pin at point A. See the second free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FA − FAB sin 60 = 0 →

FAB =

8250 N

= = 9526 N  9530 N, compression sin 60 sin 60  Fhoriz = FAC − FAB cos 60 = 0 →

FA FAC

A 60

FAB

FAC = FAB cos 60 = ( 9526 N ) cos 60 = 4763 N  4760 N, tension By the symmetry of the structure, we also know that FDB = 9526 N  9530 N, compression and FDC = 4763 N  4760 N, tension . Finally, from consideration of the vertical forces on pin C, we

see that FBC = 1.65  104 N, tension . (b) As listed above, we have struts AB and DB under compression, and struts AC, DC, and BC under tension. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

433

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

55. (a) The three forces on the truss as a whole are the tension force at point B, the load at point E, and the force at point A. Since the truss is in equilibrium, these three forces must add to be 0 and must cause no net torque. Take torques about point A, calling clockwise torques positive. Each member is 3.0 m in length.  = FT ( 3.0 m) sin 60 − Mg (6.0 m ) = 0 → FT =

Mg ( 6.0 m )

( 3.0 m ) sin 60

( 62.0 kN )( 6.0 m ) = 143kN  140 kN ( 3.0 m ) sin 60

The components of FA are found from the force equilibrium equations, and then the magnitude and direction can be found.  Fhoriz = FT − FA horiz = 0 → FA horiz = FT = 143kN

F

= FA vert − Mg = 0 → FA vert = Mg = 62.0 kN

FA =

FA2 horiz + FA2 vert =

vert

 A = tan −1

FA vert

= tan −1

(143kN ) 2 + ( 62.0 kN ) 2 = 155.9 kN  160 kN 62.0 kN

= 23.44  23 above AC FA horiz 143kN (b) Analyze the forces on the pin at point E. See the second free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FDE sin 60 − Mg = 0 →

FDE =

FDE FCE

60

E Mg

62.0 kN

= = 71.59 kN  72 kN, in tension sin 60 sin 60  Fhoriz = FDE cos 60 − FCE = 0 → FCE = FDE cos 60 = ( 71.59 kN ) cos 60 = 35.80kN  36 kN, in compression

Analyze the forces on the pin at point D. See the third free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FDC sin 60 − FDE sin 0 = 0 →

FDC 60 60

FDC = FDE = 71.59 kN  72 kN, in compression

F

horiz

D FDE

= FDB − FDE cos 60 − FDC cos 60 = 0 →

FDB

FDB = ( FDE + FDC ) cos 60 = 2 ( 71.59 kN ) cos 60 = 71.59 kN  72 kN, in tension

Analyze the forces on the pin at point C. See the fourth free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FBC sin 60 − FDC sin 0 = 0 → FBC = FDC = 71.59 kN  72 kN, in tension

F

horiz

FCA

C 60BC 60

FCE

FDC

= FCE + FBC cos 60 + FDC cos 60 − FCA = 0 →

FCA = FCE + ( FBC + FDC ) cos 60 = 35.80 kN + 2 ( 71.59 kN ) cos 60 = 107.4 kN  107 kN, in compression

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Chapter 12

Static Equilibrium; Elasticity and Fracture

Analyze the forces on the pin at point B. See the fifth free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FAB sin 60 − FBC sin 0 = 0 →

FAB FDB

FAB = FBC = 71.59 kN  72 kN, in compression

F

horiz

60 60

FBC

= FT − FBC cos 60 − FAB cos 60 − FDB = 0 →

FT = ( FBC + FAB ) cos 60 + FDB = 2 ( 71.59 kN ) cos 60 + 71.59 kN = 143.2 kN

This final result confirms the earlier calculation, so the results are consistent. We could also analyze point A to check for consistency. 56. (a) See the free-body diagram from Figure 12–30(b), as modified to B D indicate the changes in the roadway mass distribution. There is F3 F2 F1 now a third support, in the center, labeled as F3 . This free-body diagram only shows the “external” forces on the truss–the force of 60 60 gravity pulling down on the structure (due to the weight of the E A C roadway) and the upward forces from the supports, acting on the 1 1 “pins.” As in Example 12–12, if the roadway mass is 4 Mg 4 Mg 6 5 1.40  10 kg, then for one truss we should use M = 7.0  10 kg. 1 2 Mg From the symmetry of the diagram, we can see that each end support carries ¼ of the weight, and the center support carries ½ of the weight. Alternatively, the sums of forces and torques can be used to prove the same thing. So F1 = 14 Mg = F2 and F3 = 12 Mg . Note that the problem is still symmetric about a vertical line through pin C.

Analyze the forces on the pin at point A. There is the upward force on the pin from the support, the force on the pin due to strut AC, and the force on the pin due to strut AB. See the second free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = 14 Mg − FAB sin 60 = 0 → FAB =

1 4

sin 60

1 4

1 2

Mg 2 3

Mg FAC

60

FAB

( 7.0 10 kg )( 9.80 m s ) 5

1 4

2 3

= 2.0  106 N, in compression .

F

horiz

= FAC − FAB cos 60 = 0 →

FAC = FAB cos 60 =

Mg 1

( 7.0 10 kg )( 9.80 m s )  9.9  10 N, in tension . = 5

2 32 4 3 4 3 Analyze the forces on the pin at point B. See the third free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FAB sin 60 − FBC sin 0 = 0 →

FBC = FAB =

F

horiz

Mg 2 3

= 2.0  106 N, in tension .

B FDB

FAB 60 60

FBC

= FAB cos 60 + FBC cos 60 − FDB = 0 → Mg  Mg  = 2.0  106 N, in compression .  cos 60 = 2 3 2 3

FDB = ( FAB + FBC ) cos 60 = 2 

435

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

By the symmetry of the geometry, we can determine the other forces. FDE = FAB =

FDC = FBC = FCE = FAC =

Mg 2 3

Mg 2 3 Mg

= 2.0  106 N, in compression ,

= 2.0  106 N, in tension , = 9.9  105 N, in tension .

4 3 Note that each force is reduced by a factor of 2 from the original solution Example 12–12. As a check on the calculation, the net force on pin C could be calculated, and it turns out to be 0, as it must. B D (b) See the free-body diagram from Figure 12–30(b), as modified to F2 F 1 indicate these new changes in the roadway mass distribution. This free body diagram only shows the “external” forces on the truss–the force of gravity pulling down on the bridge (due to the weight of the 60 60 roadway) and the upward forces on the “pins” from the supports. As E A C in Example 12–12, if the roadway mass is 1.40  106 kg, then for one

truss we should use M = 7.0  10 kg. Write the conditions for 1 2 Mg equilibrium for the entire truss by considering the vertical forces, and by considering the torques about point A. Let clockwise torques be positive.  Fvert = F1 + F2 − Mg = 0 5

1 2

 A = Mg ( 14 l ) + Mg ( 34 l ) − F ( l ) = 0 → F = Mg ; F = Mg − F2 = Mg 1 2

1 2

Note that the problem is still symmetric about a vertical line through pin C. Also note that the forces at the ends each bear half of the weight of that single-truss-portion of the structure. From the discussion in the textbook after Example 12–12, we FCy FAy now have a force acting downward on the middle of strut AC– FAx FCx part of the weight of the roadway. This means that there must A C also be some upward forces on that strut, exerted by the pins. So we cannot analyze pin A yet, since there will be two unknown 1 2 Mg forces on pin A exerting vertical force–the vertical force on pin A from strut AC, and the vertical force in pin A from strut AB. Thus we must analyze the forces on strut AC first, to determine the vertical forces on strut AB, and then we can analyze the forces on pin A. See the second free-body diagram for the forces on strut AC. Write equilibrium equations for the horizontal and vertical directions, and for torques about point A.  Fvert = FAy + FCy − 12 Mg = 0 ;  Fhoriz = − FAx + FCx = 0 → FCx = FAx

 = Mg ( l ) − F ( 12 l ) = → F = Mg ; F = Mg − F = Mg 1 2

1 4

1 2

1 4

Since their x components are equal and their y components are equal, FA = FC = FAC , but they do NOT lie exactly along the strut, and they are not parallel to each other. Now we can analyze the forces on the pin at point A. The components FAx and FAy found above are forces of the pin on the strut, so we put in the opposite forces, FACx and FACy , which are the forces of the strut on the pin. See the third free-body diagram. We know that FACy is downward and is smaller in magnitude than F1 , so FAB must point as shown. And since

F1 = 12 M g

FACx

60

FAB

FACy = 14 M g

436

Chapter 12

Static Equilibrium; Elasticity and Fracture

FAB has a leftward component, FACx must point in the direction shown, as we assumed when analyzing strut AC.  Fvert = F1 − FACy − FAB sin 60 = 0 → FAB =

F1 − FACy

= 2

sin 60

Mg − 14 Mg 3

1 2

= 4

Mg 2 3

( 7.0 10 kg )( 9.80 m s ) 5

2 3

= 2.0  106 N, in compression .

F

= FACx − FAB cos 60 = 0 →

horiz

FACx = FAB = 1 2

Mg 4 3

( 7.0 10 kg )( 9.80 m s )  9.9 10 N, in tension . = 5

4 3

(

)(

)

FACy = 14 Mg = 14 7.0  10 kg 9.80 m s 2  1.7  106 N ( producing shear ) . 5

Next, analyze the forces on the pin at point B. See the fourth free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FAB sin 60 − FBC sin 0 = 0 → FBC = FAB =

F

horiz

Mg 2 3

B FDB

FAB 60 60

FBC

 2.0  10 N, in tension . 6

= FAB cos 60 + FBC cos 60 − FDB = 0 → Mg  Mg   2.0  106 N, in compression .  cos 60 = 2 3 2 3

FDB = ( FAB + FBC ) cos 60 = 2 

By the symmetry of the geometry, we can state the other forces: FCE = FAC for both x and y components, FDE = FAB , and FDC = FBC . As a check on the calculations, we analyze the forces on the pin at point C. See the fifth free-body diagram. Write equilibrium equations for the horizontal and vertical directions. All forces are known. Note that, for ease of viewing, we have drawn FACy and FCEy as slightly separated from each other, but they are both acting on point “C” and should overlap each other. Sum the forces in the horizontal and vertical directions.  Fhoriz = FCEx − FACx + FDC cos 60 − FBC cos 60

−

cos 60 −

60

FDC 60

FACx

FCEx FACy

FCEy

cos 60 = 0 as expected 4 3 4 3 2 3 2 3  Fvert = − FACy − FCEy + FDC sin 60 + FBC sin 60

=−

−

FBC

4 4 2 3 2 2 3 2 Here is a summary of the forces on each strut.

= 0 as expected

FACx = FCEx  9.9  105 N, in tension . FACy = FCEy  1.7  106 N ( producing shear ) . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

437

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

FAB = FDE  2.0  106 N, in compression ; FDB  2.0  106 N, in compression ; FBC = FCD  2.0  106 N, in tension .

(c) The approach of part (b) is more valid, because it is more realistic to assume that the struts have some mass. Part (b) starts to model this analysis. A similar analysis is done in Problem 57. 57. See the free-body diagram from Figure 12–31. M represents the mass of the locomotive, and each member has a length of l . Write the conditions for equilibrium for the entire truss by considering vertical forces and the torques about point A. Let clockwise torques be positive. The only gravity force is that of half the locomotive, since we are ignoring the mass of the trusses and rails.  Fvert = F1 + F2 − 12 Mg = 0

 = Mg ( l ) − F ( 2l ) = 0 → F = Mg 1 2

1 2

1 8

F1 = 12 Mg − F2 = 83 Mg

Analyze the forces on strut AC, using the free-body diagram given in Figure 12–31(b). Note that the forces at the pins are broken up into components. See the second free-body diagram. Write equilibrium equations for the horizontal and vertical directions, and for torques about point A.  Fvert = FAy + FCy − 12 Mg = 0

FCy

FAy

FAx

FCx

C 1 2

 F = −F + F = 0 → F = F  = Mg ( l ) − F ( l ) = → F = Mg horiz

1 2

1 4

FAy = 12 Mg − FCy = 14 Mg

Since their x components are equal and their y components are equal, FA = FC = FAC . Analyze the forces on the pin at point A. The components FAx and FAy found

F1 = 83 M g

above are forces of the pin on the strut, so we put in the opposite forces, FACx and

FACy , which are the forces of the strut on the pin. FAB is the force that member AB puts on the pint. See the third free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = 83 Mg − FACy − FAB sin 60 = 0 → FAB =

3 8

Mg − FACy sin 60

3 8

Mg − 14 Mg 1 2

Mg 4 3

FACx

A 60

FAB

FACy

( 53  10 kg )( 9.80 m s ) = 3

4 3

= 7.497 104 N  7.5 104 N, in compression .

F

horiz

= FACx − FAB cos 60 = 0 →

FACx = 12 FAB =

= 3.7485  104 N  3.7  104 N, in tension .

8 3 The actual force FAC has both a tension component FACx and a shearing component FACy . Since the

problem asks for just the compressive or tension force, only FACx is included in the answer. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

438

Chapter 12

Static Equilibrium; Elasticity and Fracture

Analyze the forces on the pin at point B. See the fourth free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FAB sin 60 − FBC sin 0 = 0 → FBC = FAB =

FAB

60 60

FDB

= 7.5  10 N, in tension . 4 3  Fhoriz = FAB cos 60 + FBC cos 60 − FDB = 0 → 4

FBC

Mg  Mg  = 7.5  10 4 N, in compression .  cos 60 = 4 3 4 3

FDB = ( FAB + FBC ) cos 60 = 2 

Analyze the forces on the pin at point C. See the fifth free-body diagram. Write equilibrium equations for the horizontal and vertical directions.  Fvert = FBC sin 60 + FDC sin 0 − FACy = 0 → FDC = =

FACy sin 60 Mg

− FBC =

1 4

1 2

−

Mg 4 3

1 4

1 2

−

FBC

FDC

60

FACx

Mg 4 3

60

FCE

FACy

 7.5  10 4 N, in tension .

4 3  Fhoriz = FCE + FDC cos 60 − FBC cos 60 − FACx = 0 → FCE = FACx + ( FBC − FDC ) cos 60 =

+ 0  3.7  104 N, in tension .

8 3 Analyze the forces on the pin at point D. See the sixth free-body diagram. Write the equilibrium equation for the vertical direction.  Fvert = FDE sin 60 − FDC sin 0 = 0 → FDE = FDC =

FDE 60 60

 7.5  104 N, in compression .

FDB FDC

4 3 This could be checked by considering the forces on pin E.

58. Write Newton’s second law for the horizontal direction.  Fx = F2 cos  − F1 cos  = 0 → F2 = F1



Thus the two forces are the same size. Now write Newton’s second law for the vertical direction. Fbutress 4.2  105 N F F F F F sin sin 0   = + − = → = = = 2.4  106 N  y 1 butress 1 1 o 2sin 2 sin 5

(



FButress

)

59. Draw free-body diagrams similar to Figures 12–36(a) and 12–36(b) for the forces on the right half of a round arch and a pointed arch. The load force is placed at the same horizontal position on each arch. For each half-arch, take torques about the lower right hand corner, with counterclockwise as positive.

FLoad

round

For the round arch:

 = F ( R − x ) − F Load

H round

R = 0 → FH

= FLoad

round

R−x R

439

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

For the pointed arch:

 = F ( R − x ) − F Load

H pointed

Solve for y, given that FH

y = 0 → FH

= 13 FH

= FH

pointed

→ FLoad

round

R−x

pointed

1 3

pointed

= FLoad

round

R−x y

= 13 FLoad

R−x

→

y = 3R = 3 ( 12 8.0 m ) = 12 m 60. (a) The pole will exert a downward force and a FLeft clockwise torque about the woman’s right hand. Thus there must be an upward force exerted by the 0.32 m FRight left hand to cause a counterclockwise torque for the mg pole to have a net torque of zero about the right 0.90 m hand. The force exerted by the right hand is then of such a magnitude and direction for the net vertical force on the pole to be zero.  = FLeft ( 0.32 m ) − mg ( 0.90 m ) = 0 →

 0.90 m  (10.0 kg ) ( 9.80 m s ) ( 0.90 ) = 275.6 N  280 N, upward . = 0.32  0.32 m  2

FLeft = mg 

F = F y

Left

− FRight − mg = 0 →

(

)

FRight = FLeft − mg = 275.6 N − (10.0 kg ) 9.80 m s 2 = 177.6 N  180 N, downward . (b) We see that the force due to the left hand is larger than the force due to the right hand, since both the FLeft right hand and gravity are downward. Set the left hand force equal to 150 N and calculate the location x FRight of the left hand, by setting the net torque equal to mg 0.90 m zero. 98.0 N mg  = FLeft x − mg ( 0.90 m ) = 0 → x = F ( 0.90 m ) = 150 N ( 0.90 m ) = 0.59 m Left As a check, calculate the force due to the right hand. FRight = FLeft − mg = 150 N − 98.0 N = 52 N OK (c) Follow the same procedure, setting the left hand force equal to 85 N.  = FLeft x − mg ( 0.90 m ) = 0 →

mg FLeft

( 0.90 m ) =

98.0 N 85 N

( 0.90 m ) = 1.038 m  1.0 m

FRight = FLeft − mg = 85 N − 98.0 N = −13 N OK Note that now the force due to the right hand must be pulling upwards, because the left hand is on the opposite side of the center of the pole.

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Chapter 12

Static Equilibrium; Elasticity and Fracture

61. If the block is on the verge of tipping, the normal force will be acting at the lower right corner of the block, as shown in the free-body diagram. The block will begin to rotate when the torque caused by the pulling force is larger than the torque caused by gravity, as calculated about the lower right corner of the block. For the block to be able to slide, the pulling force must be as large as the maximum static frictional force. Write the equations of equilibrium for forces in the x- and y-directions and for torque with the conditions as stated above.  Fy = FN − mg = 0 → FN = mg

 F = F − F = 0 → F = F =  F =  mg x

 = mg 2 − Fh = 0 →

mg l 2

= Fh = s mgh

Solve for the coefficient of friction in this limiting case, to find s =

l 2h

(a) If s  l 2h , then sliding will happen before tipping. (b) If s  l 2h , then tipping will happen before sliding. 62. Assume that the building has just begun to tip, so that it is essentially vertical, but that all of the force on the building due to contact with the Earth is at the lower left corner, as shown in the figure. Take torques about that corner, with counterclockwise torques as positive.  = FA ( 90.0 m ) − mg ( 23.0 m )

(

)

=  950 N m 2 (180.0 m )( 76.0 m )  ( 90.0 m )

(

)(

− 1.8  10 kg 9.80 m s 7

) ( 23.0 m ) = −2.9 10 m N 9

23.0 m

mg 90.0 m

FE y

FE x

Since this is a negative torque, the building will tend to rotate clockwise, which means it will rotate back down to the ground. Thus the building will not topple . 63. (a) The weight of the shelf exerts a downward force and a FLeft 32.0 cm clockwise torque about the point where the shelf touches the wall. Thus, there must be an upward force and a counterclockwise torque exerted by the slot for the shelf to be in equilibrium. Since any force exerted mg FRight by the slot will have a short lever arm relative to the 2.0 cm point where the shelf touches the wall, the upward force must be larger than the gravity force. Accordingly, there then must be a downward force exerted by the slot at its left edge, exerting no torque, but balancing the vertical forces. (b) Calculate the values of the three forces by first taking torques about the left end of the shelf, with the net torque being zero, and then sum the vertical forces, with the sum being zero.  = FRight 2.0  10−2 m − mg 17.0  10−2 m = 0 →

(

)

(

)

 10 m  = 549.8 N  550 N )  17.0 2.0  10 m 

FRight = ( 6.6 kg ) 9.80 m s 2 



−2



441

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

F = F y

Right

Instructor Solutions Manual

− FLeft − mg →

(

)

FLeft = FRight − mg = 549.8 N − ( 6.6 kg ) 9.80 m s 2 = 490 N

(

)

mg = ( 6.6 kg ) 9.80 m s 2 = 65 N (c) The torque exerted by the support about the left end of the rod is  = FRight ( 2.0  10−2 m ) = ( 549.8 N ) ( 2.0  10−2 m ) = 11m N 64. (a) The torque due to the sign is the product of the weight of the sign and the distance of the sign from the wall.  = mgd = ( 6.1kg ) 9.80 m s 2 ( 2.2 m ) = 130 m N, clockwise .

(

)

Fwall



wall mg (b) Since the wall is the only other object that can put force on the pole (other than the weight of the pole), then the wall must put a torque on the pole. The torque due to the hanging sign is clockwise, so the torque due to the wall must be counterclockwise. See the diagram. Also note that the wall must put a net upward force on the pole as well, so that the net force on the pole will be zero. (c) The torque on the rod can be considered as the wall pulling horizontally to the left on the top left corner of the rod and pushing horizontally to the right at the bottom left corner of the rod. The reaction forces to these put a shear on the wall at the point of contact. Also, since the wall is pulling upwards on the rod, the rod is pulling down on the wall at the top surface of contact, causing tension. Likewise the rod is pushing down on the wall at the bottom surface of contact, causing compression. Thus all three are present.

65. (a) The meter stick is in equilibrium, so the net torque and the net force are both zero. From the force diagram, write an expression for the net torque about the 90-cm mark, with counterclockwise torques as positive.  = mg ( 0.40 m ) − FT0 ( 0.90 m ) = 0 → 0.40

(

)

FT0

0.90 m

0.50 m

FT90

0.40

= ( 0.180 kg ) 9.80 m s 2 = 0.78 N 0.90 0.90 (b) Write Newton’s second law for the vertical direction with a net force of 0 to find the other tension.  Fy = FT0 + FT90 − mg = 0 → FT0 = mg

(

)

FT90 = mg − FT0 = ( 0.180 kg ) 9.80 m s 2 − 0.78 N = 0.98 N 66. The maximum compressive force in a column will occur at the bottom. The bottom layer supports the entire weight of the column, and so the compressive force on that layer is mg . For the column to be on the verge of buckling, the weight divided by the area of the column will be the compressive strength of the material. The mass of the column is its volume (area × height) times its density. mg hA g Compressive Strength = Compressive Strength = → h= A A g Note that the area of the column cancels out of the expression, and so the height does not depend on the cross-sectional area of the column. Compressive Strength 500  106 N m 2 = = 6500 m (a) hsteel = g ( 7.8  103 kg m3 )( 9.80 m s2 ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

442

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(b) hgranite =

Compressive Strength

g

170  106 N m 2

( 2.7  10 kg m )( 9.80 m s ) 3

= 6400 m

67. The radius of the wire can be determined from the relationship between stress and strain, expressed by Eq. 12–5. F

l

→ A=

Fl 0

=  r2 → r =



1 F l0

l0 A E l  E l Use the free-body diagram for the point of connection of the mass to the wire to determine the tension force in the wire. ( 25 kg ) 9.80 m s2 mg = 589.2 N  Fy = 2 FT sin  − mg = 0 → FT = 2sin  = 2sin12 o The fractional change in the length of the wire can be found from the geometry of the problem, as seen in the second diagram. l0 2 1 1 l cos  = → = −1 = − 1 = 2.234  10 −2 o cos12 l 0 cos  ( l 0 + l ) 2 Thus the radius is

(

1 FT l 0

 E l

)

1

589.2 N  1 = 3.5  10−4 m  9 2  −2   70  10 N m  ( 2.234  10 )

68. Each crossbar is in equilibrium, and so the net torque about the suspension point for each crossbar is 0. Counterclockwise torques are taken as positive. The suspension point is used so that the tension in the suspension string need not be known initially. The net vertical force must also be 0. The bottom bar:  = mD gxD − mC gxC = 0 → mC = mD

= mD

17.50 cm 5.00 cm

FCD

= 3.50mD

 F = F − m g − m g = 0 → F = ( m + m ) g = 4.50m g y

mC g

mD g

The middle bar: xB

 = F x − m gx = 0 → F = m g x CD

→ 4.50mD g = mB g

( 0.748 kg )( 5.00 cm ) m x = 0.05541kg  5.54  10 −2 kg mD = B B = 4.50 xCD ( 4.50 )(15.00 cm ) mC = 3.50mD = ( 3.50 )( 0.05541kg ) = 0.194 kg

F = F y

BCD

FBCD

xCD

mB g

FCD

− FCD c − mB g = 0 → FBCD = FCD + mB g = ( 4.50mD + mB ) g

The top bar:  = mA gxA − FBCD xBCD = 0 → mA =

( 4.50mD + mB ) gxBCD gxA

FABCD

= ( 4.50mD + mB )

= ( 4.50 )( 0.05541 kg ) + 0.748 kg 

xBCD xA

7.50 cm 30.00 cm

= 0.249 kg

xA mA g

xBCD

FBCD

443

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

69. The limiting condition for the safety of the painter is the tension in the ropes. The ropes can only exert an upward tension on the scaffold. The tension will be least in the rope that is farther from the painter. The mass of the pail is mp , the mass of the scaffold is m, and the mass of the

Instructor Solutions Manual

Fleft = 0

Fright 1.0 m

1.0 m 1.0 m 1.0 m

mp g

painter is M .

2.0 m

Find the distance to the right that the painter can walk before the tension in the left rope becomes zero. Take torques about the point where the right-side rope is attached to the scaffold, so that its value need not be known. Take counterclockwise torques as positive.  = mg ( 2.0 m ) + mp g ( 3.0 m ) − Mgx = 0 →

m ( 2.0 m ) + mp ( 3.0 m )

( 25 kg )( 2.0 m ) + ( 4.0 kg )( 3.0 m )

65.0 kg M The painter can walk to within 5 cm of the right edge of the scaffold.

m ( 2.0 m ) + mp (1.0 m )

Fright = 0

Fleft

Now find the distance to the left that the painter can walk before the tension in the right rope becomes zero. Take torques about the point where the left-side tension is attached to the scaffold, so that its value need not be known. Take counterclockwise torques as positive.  = Mgx − mp g (1.0 m ) − mg ( 2.0 m ) = 0 →

= 0.9538 m  0.95 m

1.0 m

x 1.0 m 1.0 m Mg

2.0 m

1.0 m

mp g

( 25kg )( 2.0 m ) + ( 4.0 kg )(1.0 m )

= 0.8308 m  0.83m M 65.0 kg The painter can walk to within 17 cm of the left edge of the scaffold. Thus both ends are dangerous. 70. The number of supports can be found from the compressive strength of the wood. Since the wood will be oriented longitudinally, the stress will be parallel to the grain. Compressive Strength Load force on supports Weight of roof = = Safety Factor Area of supports ( # supports )( area per support )

( # supports ) =

Weight of roof

Safety Factor

( area per support ) Compressive Strength

(1.24  10 kg )( 9.80 m s ) = 4

( 0.040 m )( 0.090 m )

( 35  10 N m ) 6

= 11.57 supports

Since there are to be more than 11 supports, and to have the same number of supports on each side, there will be 12 supports, or 6 supports on each side . That means there will be 5 support-to-support spans, each of which would be given by spacing =

10.0 m 5 gaps

= 2.00 m gap .

71. The tension in the string when it breaks is found from the ultimate strength of nylon under tension, from Table 12–2.



 FT mg

444

Chapter 12

Static Equilibrium; Elasticity and Fracture

= Tensile Strength → A FT = A ( Tensile Strength )

(

) (

)

=   12 1.15  10 −3 m  500  106 N m 2 = 519.3 N

From the force diagram for the box, we calculate the angle of the rope relative to the horizontal from Newton’s second law in the vertical direction. Note that since the tension is the same throughout the string, the angles must be the same so that the object does not accelerate horizontally.  Fy = 2 FT sin  − mg = 0 → 3− h  2  25 kg 9.80 m s ( ) −1 mg −1 2.25m = sin = 13.64  = sin h 2 FT 2 ( 519.3 N ) To find the height above the ground, consider the second diagram. 3.00 m − h tan  = → h = 3.00 m − 2.25 m ( tan  ) = 3.00 m − 2.25 m ( tan13.64 ) = 2.45 m 2.25 m

(

)

72. See the free-body diagram for the crate on the verge of tipping. From the textbook Figure 12–15 and the associated discussion, if a vertical line projected downward from the center of gravity falls outside the base of support, then the object will topple. So the limiting case is for the vertical line to intersect the edge of the base of support. Any more tilting and the gravity force would cause the block to tip over, with the axis of rotation through the lower corner of the crate. 1.00 1.00 tan  = →  = tan −1 = 40 ( 2 sig. figs.) 1.18 1.18 The other forces on the block, the normal force and the frictional force, would be acting at the lower corner and so would not cause any torque about the lower corner. The gravity force causes the tipping. 73. Consider the “partial” free-body diagram. We have not drawn in the forces where the pole is attached to the ground, because we take torques about an axis through that point, and so those forces exert no torque. The weight acts at the midpoint, and the tension is applied 2/3 of the way along the pole. Let clockwise torques be positive.  = mg ( 12 l ) cos 35 − FT ( 23 l ) sin17 = 0 → FT = mg ( 34 )

cos 35 sin17

(

)

= ( 95 kg ) 9.80 m s 2 ( 34 )

cos 35 sin17

17

FT 18

35

= 1956 N

 2.0  103 N

445

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

74. Draw a free-body diagram for one of the beams, each of length l . By Newton’s third law, if the right beam pushes down on the left beam, then the left beam pushes up on the right beam. But the geometry is symmetric for the two beams, and so the beam contact force must be horizontal. For the beam to be in equilibrium, FN = mg and so Ffr = s FN =  mg is the maximum friction force. Take torques about the top of the beam, so that Fbeam exerts no torque. Let clockwise torques be positive.

 = F l cos  − mg ( l ) cos  − F l sin  = 0 → 1 2

 = tan −1

1 2 s

= tan −1

1 2 ( 0.45 )

= 48

75. (a) The fractional decrease in the rod’s length is the strain. Use Eq. 12–4. The force applied is the weight of the man. ( 78 kg ) 9.80 m s 2 F mg l = = 2 = = 5.407  10−8 = 5.4  10−6 % 2 9 2 l 0 AE  r E  ( 0.15 ) 200  10 N m

(

)

(

)

(b) The fractional change is the same for the atoms as for the macroscopic material. Let d represent the interatomic spacing. d l = = 5.407  10−8 → d0 l0

(

)

(

d = 5.407  10−8 d 0 = 4.506  10−8

)( 2.0 10 m ) = 1.08110 m  1.110 m −10

−17

76. (a) See the free-body diagram for the system, showing forces on the engine and the forces at the point on the rope where the mechanic is pulling (the point of analysis). Let m represent the mass of the engine. The fact that the engine was raised a half-meter means that the part of the rope from the tree branch to the mechanic is 3.25 m, as well as the part from the mechanic to the bumper. From the free-body diagram for the engine, we know that the tension in the rope is equal to the weight of the engine. Use this, along with the equations of equilibrium at the point where the mechanic is pulling, to find the pulling force by the mechanic. 3.0 m = 22.62 Angle:  = cos −1 3.25 m Engine:  Fy = FT − mg = 0 → FT = mg

 F = F − 2 F sin  = 0 → F = 2mg sin  = 2 ( 310 kg ) ( 9.80 m s ) sin 22.62 = 2337 N  2300 N Load force mg ( 310 kg ) ( 9.80 m s ) (b) Mechanical advantage = = = = 1.3 Point:

Applied force

2337 N

446

Chapter 12

Static Equilibrium; Elasticity and Fracture

77. Consider the free-body diagram for the box. The box is assumed to be in equilibrium, but just on the verge of both sliding and tipping. Since it is on the verge of sliding, the static frictional force is at its maximum value. Use the equations of equilibrium. Take torques about the lower right corner where the box touches the floor, and take clockwise torques as positive. We also assume that the box is just barely tipped up on its corner, so that the forces are still parallel and perpendicular to the edges of the box.  Fy = FN − W = 0 → FN = W

 F = F − F = 0 → F = F = W = ( 0.60)( 250 N ) = 150 N x

250 N

 = Fh − W ( 0.5 m ) = 0 → h = ( 0.5 m ) F = ( 0.5 m ) 150 N = 0.83 m 78. There are three forces on the board–the two spring forces, and the weight of the board. Consider the free-body diagram shown. As discussed in the text with Fig. 12–8, the force due to the left spring must pull down, and the force due to the right spring must push up. Thus, the left spring is stretched by an amount ( 0.080 m − l left ) , and the right

Fright

Fleft

spring is compressed by an amount ( l right − 0.080 m ) . The weight acts at the center of gravity. To begin with, take torques about the left end of the board. The magnitude of the force exerted by the right spring is found from Hooke’s Law: Fright = k ( l right − 0.08cm ) . The center of the board is 31.5/2 = 15.75 cm from the left spring. Take clockwise torques as positive, and take them about an axis through the location where the left spring is attached to the board.  = mg ( 0.1575 m ) − k ( l right − 0.080 m ) ( 0.105 m ) = 0 l right =

( 0.500 kg ) ( 9.80 m s 2 ) ( 0.1575 m ) + 0.080 m = + 0.080 m k ( 0.105 m ) ( 225 N m )( 0.105 m )

mg ( 0.1575 m )

= 0.1127 m = 11.3cm

Now take torques about an axis through the location of the attachment of the right spring. The magnitude of the force exerted by the left spring is found from Hooke’s Law: Fleft = k ( 0.080 cm − l left ) .

 = mg ( 0.1575 m − 0.105m ) − k ( 0.080 m − l )( 0.105 m ) = 0 ( 0.500 kg ) ( 9.80 m s ) ( 0.0525 m ) mg ( 0.0525 m ) l = 0.080 m − = 0.080 m − left

left

( 225 N m )( 0.105 m )

k ( 0.105 m )

= 0.0691m = 6.9 cm 79. From the free-body diagram (not to scale), write the force equilibrium condition for the vertical direction.  Fy = 2FT sin  − mg = 0

18 m



0.90 m 

mg © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

447

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

FT =

mg 2 sin 



mg 2 tan 

Instructor Solutions Manual

( 56.0 kg ) ( 9.80 m s2 )  0.90 m    18 m 

2

= 5488 N  5500 N

Note that the angle is small enough (about 3o) that we have simplified the calculation by making the substitution of sin   tan  . It is not possible to increase the tension so that there is no sag. There must always be a vertical

component of the tension to balance the gravity force. The larger the tension gets, the smaller the sag angle will be, however. 80. Assume a constant acceleration as the person is brought to rest, with up as the positive direction. Use Eq. 2–12c to find the acceleration. From the acceleration, find the average force of the snow on the person, and compare the force per area to the strength of body tissue. From the free body diagram, we have Fsnow − mg = ma → Fsnow = m ( a + g ) . v = v + 2a ( x − x0 ) → a = 2

Fsnow A Fsnow

2 0

m (a + h) A

v 2 − v02

2 ( x − x0 )

0 − ( 55 m s ) 2 ( −1.0 m )

( 75 kg ) (1513 m s 2 + 9.80 m s 2 ) 0.30m

Fsnow

= 1513 m s 2

= 3.81  105 N m 2

 Tissue strength = 5  105 N m 2

A Since the average force on the person is less than the strength of body tissue, the person may escape serious injury. Certain parts of the body, such as the legs if landing feet first, may get more than the average force though, and so still sustain injury.

81. Draw a free-body diagram for half of the cable. Write Newton’s second law for both the vertical and horizontal directions, with the net force equal to 0 in each direction. Note that FT1 is the force of the wall on the left end of the rope, and that FT2 is the force of the right side of the rope on the left side of the rope. mg  Fy = FT1 sin 62 − 12 mg = 0 → FT1 = 12 sin 62 = 0.566 mg  Fx = FT2 − FT1 cos 62 = 0 →

FT1 62

FT2 1 2

FT2 = 0.566 mg ( cos 62 ) = 0.266 mg So the results are: (a) FT2 = 0.27mg

(b) FT1 = 0.57mg (c) The direction of the tension force is tangent to the cable at all points on the cable. Thus the direction of the tension force is horizontal at the lowest point , and is 62 below the horizontal at the attachment points .

448

Chapter 12

Static Equilibrium; Elasticity and Fracture

82. We are given that rod AB is under a compressive force F, so the rod pushes outward at its endpoints. Analyze the forces on the pin at point A. See the free-body diagram. Write equilibrium equations for the horizontal and vertical directions. F  Fhoriz = FAD cos 45 − FAB = 0 → FAD = cosAB45 = 2 F , in tension .  Fvert = FAC − FAD sin 45 = 0 → FAC = FAD sin 45 = 2 F

FAC

A FAB

45

FAD

= F , in compression . r By symmetry, the other outer forces must all be the same magnitude as FAB , and the other diagonal

force must be the same magnitude as FAB . FAC = FAB = FBD = FCD = F , in compression ; FAD = FBC =

83. (a)

2 F , in tension .

See the free-body diagram. To find the tension in the wire, take torques about the left edge of the beam, with counterclockwise as positive. The net torque must be 0 for the beam to be in equilibrium.  = mgx + Mg ( 12 l ) − FT sin  l = 0 → g ( 2mx + M l )

FT =

l sin  2 l sin  2 sin  We see that the tension force is linear in x. (b) Write the equilibrium condition for vertical and horizontal forces.

F = F

− FT cos  = 0 → Fhinge = FT cos  =

F = F

+ FT sin  − ( m + M ) g = 0 →

hinge horiz hinge vert

g ( 2mx + M l )

horiz

Fhinge = ( m + M ) g − FT sin  = ( m + M ) g − vert

2 l sin 

g ( 2mx + M l ) 2 l sin 

cos  =

g ( 2mx + M l ) 2 l tan 

 x sin  = mg  1 −  + 12 Mg  l

84. Take torques about the elbow joint. Let clockwise torques be positive. Since the arm is in equilibrium, the total torque will be 0.  = ( 2.0 kg ) g ( 0.15 m ) + ( 25 kg ) g ( 0.35 m ) − Fmax ( 0.050 m ) sin105 = 0 →

Fmax =

( 2.0 kg ) g ( 0.15 m ) + ( 25 kg ) g ( 0.35 m ) = 1836 N  1800 N ( 0.050 m ) sin105

449

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

85. See the free-body diagram. Assume that the ladder is just ready to slip, so the force of static friction is Ffr =  FN . The ladder is of length l, and so d1 = 12 l sin  , d 2 = 34 l sin  , and d 3 = l cos  . The ladder is in equilibrium, so the net vertical and horizontal forces are 0, and the net torque is 0. We express those three equilibrium conditions, along with the friction condition. Take torques about the point where the ladder rests on the ground, calling clockwise torques positive.  Fvert = FGy − mg − Mg = 0 → FGy = ( m + M ) g

F

horiz

= FGx − FW = 0 → FGx = FW

 = mgd + Mgd − F d = 0 → F = 1

mgd1 + Mgd 2

Ffr =  FN → FGx =  FGy These four equations may be solved for the coefficient of friction. mgd1 + Mgd 2 F FW d3 md1 + Md 2 m ( 12 l sin  ) + M ( 34 l sin ) = = =  = Gx = FGy ( m + M ) g ( m + M ) g d3 ( m + M ) ( l cos  )( m + M )

( 12 m + 34 M ) tan  12 (16.0 kg ) + 34 ( 76.0 kg ) tan 20.0 = = 0.257 (m + M ) ( 92.0 kg )

86. Draw a force diagram for the cable that is supporting the right-hand section. The forces will be the tension at the left end, FT2 , the tension at the right end, FT1 , and the weight of the section, mg. The weight acts at the midpoint of the horizontal span of the cable. The system is in equilibrium. Write Newton’s second law in both the x- and y-directions to find the tensions. 60o o o d1  Fx = FT1 cos19 − FT2 sin 60 = 0 → F T2

FT2 = FT1

cos19o sin 60

 F = F cos 60 − F sin19 − mg = 0 → o

FT2 cos 60 − mg o

FT1 =

FT1 = mg FT2 = FT1

sin19o

FT1

cos 60o − mg sin 60o sin19o

( cos19 cos 60 − sin19 sin 60 ) o

cos19o

sin 60

sin 60o

= 4.539 o

19o

cos19o

sin 60o

FT1

→

= 4.539 mg  4.5 mg

mg = 4.956 mg  5.0 mg

To find the height of the tower, take torques about the point where the roadway meets the ground, at the right side of the roadway. Note that then FT1 will exert no torque. Take counterclockwise torques as positive. For purposes of © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

450

Chapter 12

Static Equilibrium; Elasticity and Fracture

calculating the torque due to FT2 , split it into x and y components.

 = mg ( d ) + F h − F d = 0 → ( F − mg ) d = ( F cos 60 − mg ) d = ( 4.956 mg cos 60 − 0.50 mg ) 343 m h= 1 2

T2 x

T2 y

1 2

T2 y

FT2 x

1 2

FT2 sin 60o

(

4.956 mg sin 60 o

)

= 158 m 87. We consider the right half of the bridge in the diagram in the book. We divide it into two segments of length d1 and 12 d 2 , and let the mass of those two segments be M. Since the roadway is uniform, the mass of each segment will be in proportion to the length of the 3  2 section, as follows. FT2 FT3 m2

1 d = 2 2 m1 d1

→

d2 d1

m2 m1

The net horizontal force on the right tower is to be 0. From the force FN diagram for the tower, we write this. FT3 sin  3 = FT2 sin  2 From the force diagram for each segment of the cable, write Newton’s second law for both the vertical and horizontal directions. 2 Right segment: FT2 right segment, d1  Fx = FT1 cos 1 − FT2 sin  2 = 0 →

FT1 cos 1 = FT2 sin  2

 F = F cos  − F sin  − m g = 0 → y

m1 g = FT2 cos  2 − FT1 sin 1

3

Left segment:  Fx = FT3 sin 3 − FT4 = 0 → FT3 sin 3 = FT4

left segment, d 2 2

 F = F cos  − m g = 0 → y

FT3

FT4

m2 g = FT3 cos  3 We manipulate the relationships to solve for the ratio of the masses, which will give the ratio of the lengths. sin  2 FT1 cos 1 = FT2 sin  2 → FT1 = FT2 cos 1

m1 g = FT2 cos  2 − FT1 sin 1 = FT2 cos  2 − FT2 FT3 sin  3 = FT2 sin  2 → FT3 = FT2

1

FT1

m1g

sin  2 sin  3

sin  2 cos 1

m2 g



sin  2



cos 1

sin 1 = FT2  cos  2 −

→ m2 g = FT3 cos  3 = FT2

sin  2 sin  3



sin 1 



cos  3

451

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

d2 d1

m2 m1

m2 g m1 g

2 FT2 =

sin  2 sin  3



sin  2



cos 1

FT2  cos  2 −

2 sin  2 cos 1

cos (1 +  2 ) tan  3

cos  3



sin 1 

Instructor Solutions Manual

2 sin  2 cos  3 cos 1

( cos  2 cos 1 − sin  2 sin 1 ) sin  3



2 sin 60 cos19 cos 79 tan 66

= 3.821  3.8

88. (a) Note that since the friction is static friction, we may NOT use Ffr =  FN . It could be that Ffr   FN . So, we must determine Ffr by the equilibrium equations. Take an axis of rotation to be out of the paper, through the point of contact of the rope with the wall. Then neither FT nor Ffr can cause any torque. The torque equilibrium equation is as follows. mgr0 FN h = mgr0 → FN = h Take the sum of the forces in the horizontal direction. F mgr0 FN = FT sin  → FT = N = sin  h sin  Take the sum of the forces in the vertical direction. FT cos  + Ffr = mg → Ffr = mg − FT cos  = mg −

mgr0 cos 

 r  = mg  1 − 0 cot    h 

h sin 

(b) Since the sphere is on the verge of slipping, we know that Ffr =  FN .

h

mgr0  r  Ffr =  FN → mg  1 − 0 cot   =  h  h 

→ 

 r0





− cot   =  =

− cot 

89. (a) See the free-body diagram. We write the equilibrium conditions for horizontal and vertical forces, and for rotation. We also assume that both static frictional forces are at their maximum values. Take clockwise torques as positive. We solve for the smallest angle that makes the ladder be in equilibrium.  Fhoriz = FGx − FWx = 0 → FGx = FWx

 F = F + F − mg = 0 → F + F = mg  = mg ( l cos  ) − F l sin  − F l cos  = 0 vert

Gy 1 2

FGx = G FGy ; FWy =  W FWx

Substitute the first equation above into the fourth equation, and simplify the third equation, to give this set of equations. FGy + FWy = mg ; mg = 2 FWx tan  + FWy ; FWx = G FGy ; FWy = W FWx

(

)

Substitute the third equation into the second and fourth equations. FGy + FWy = mg ; mg = 2 G FGy tan  + FWy ; FWy = W G FGy

(

)

452

Chapter 12

Static Equilibrium; Elasticity and Fracture

Substitute the third equation into the first two equations. FGy + W G FGy = mg ; mg = 2 G FGy tan  + W G FGy

(

)

Now equate the two expressions for mg, and simplify. FGy + W G FGy = 2 ( G FGy tan  + W G FGy ) →

(b) For a frictional wall:  min = tan

−1

For a frictionless wall:  min = tan

1 −  W G 2 G −1

= tan

1 −  W G 2 G

−1

= tan

tan  min =

1 − ( 0.40 ) 1 − ( 0)

2 G

2 ( 0.40 )

−1

1 −  W G

= 46.4  46

2 ( 0.40 )

= 51.3  51

 51.3 − 46.4  100 = 10.6%  11%  46.4  

% diff = 

90. The mass is to be placed symmetrically between two legs of the table. When enough mass is added, the table will rise up off of the third leg, and then the normal force on the table will all be on just two legs. Since the table legs are equally spaced, the angle marked in the Mg diagram is 30o. Take torques about a line connecting the two legs ( down ) that remain on the floor, so that the normal forces on those two legs cause no torque. It is seen from the second diagram (a portion of the first diagram but enlarged) that the two forces are equidistant from the line joining the two legs on the floor. Since the lever arms are equal, then the torques will be equal if the forces are equal. Thus, to be in equilibrium, the two forces must be the same. If the force on the edge of the table is any bigger than the weight of the table, it will tip. Thus M  28 kg will cause the table to tip. Mg

30o

mg ( down )

30o R

mg R 2

R 2

91. See the free-body diagram. Let M represent the mass of the train, and m represent the mass of the bridge. Write the equilibrium conditions for torques, taken about the left end, and for vertical forces. These two equations can be solved for the forces. Take counterclockwise torques as positive. Note that the position of the train is given by x = vt.  = Mgx + mg ( 12 l ) − FB l = 0 →

 

FB =  Mg

x l

  Mg v 1  t + 2 mg    l 

+ 12 mg  = 

    1m s   2  ( 95, 000 kg ) ( 9.80 m s ) ( 80.0 km h )     3.6 km h   1  2   = t + 2 ( 23, 000 kg ) ( 9.80 m s )   260 m      

(

)

(

)

= 7.957  10 4 N s t + 1.127  105 N  8.0  10 4 N s t + 1.1  105 N

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F

vert

Instructor Solutions Manual

= FA + FB − Mg − mg = 0 →

( )( ) ( ) = − ( 7.957  10 N s ) t + 1.044  10 N  − ( 8.0  10 N s ) t + 1.0  10 N

FA = ( M + m ) g − FB = 1.18  105 kg 9.80 m s 2 −  7.957  104 N s t + 1.127  105 N  4

92. The force on the sphere from each plane is a normal force, and so is perpendicular to the plane at the point of contact. Use Newton’s second law in both the L horizontal and vertical directions to determine the magnitudes of the forces. F L sin  sin 67  Fx = FL sin  L − FR sin  R = 0 → FR = FL sin  L = FL sin 32 R sin 67

 F = F cos  + F cos  − mg = 0 → F  cos 67 + sin 32 cos 32  = mg y

FL =

 cos 67 +  cos 32    sin 32   sin 67 sin 32

= (120.9 N )

( 23kg ) ( 9.80 m s2 )

 cos 67 + sin 67 cos 32    sin 32  

sin 67 sin 32

= 120.9 N  120 N

= 210.0 N  210 N

93. There are four forces on the bar, as shown in the free-body diagram. First, calculate the magnitude of FT in the steel wire using the data given and Eq. 12–4. 200  109 N m 2 0.0050 m 2 1.0  10 −4 m E A l FT = = l0 (1.0 m ) sin 33

(

mg sin 67

FR = FL

R

)(

)

= 1.836  105 N (a) Now calculate the magnitude of F by taking torques about an axis through the hinge point. Take clockwise torques as positive. Note that FH causes no torque.

F FT

 = F ( 2.0 m ) + mg (1.0 m ) sin 33 − F (1.0 m ) cos 33 = 0 → T

FT (1.0 m ) cos 33 − mg (1.0 m ) sin 33 2.0 m

(1.836  10 N ) (1.0 m ) cos 33 − ( 55 kg ) ( 9.80 m s ) (1.0 m ) sin 33 = 7.6843  10 N 5

2.0 m

 7.7  104 N (b) Use the force equilibrium condition to find the components of the force at the hinge.  Fx = FHx − FT + F cos 33 = 0 →

(

)

FHx = FT − F cos 33 = 1.836  105 N − 7.6843  10 4 N cos 33 = 1.192  105 N  1.2  105 N

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 F = F − mg − F sin 33 = 0 → F = mg + F sin 33 = ( 55 kg ) ( 9.80 m s ) + ( 7.6843  10 N ) sin 33 = 4.239  10 N y

 4.2  105 N 94. The airplane is in equilibrium, and so the net force in each direction and the net torque are all equal to zero. First write Newton’s second law for both the horizontal and vertical directions, to find the values of the forces.

 F = F − F = 0 → F = F = 5.0  10 N  F = F − mg = 0 F = mg = ( 8.5  10 kg )( 9.80 m s ) = 8.33  10 N

Calculate the torques about the CM, calling counterclockwise torques positive.  = FLd − FDh1 − FTh2 = 0

h1 =

FL d − FT h2

(8.33  10 N ) ( 3.2 m ) − ( 5.0  10 N ) (1.6 m ) = 3.7 m = ( 5.0  10 N ) 5

95. (a) Use conservation of energy to determine the speed of the person when they reach the ground. Set the potential energy of the ground as zero (y = 0). K1 + U1 = K 2 + U 2 → 0 + mgy1 = 12 mv 2 + 0 FN v = 2 gy1 = 2 9.8 m/s 2 ( 3.0 m ) = 7.668 m/s  7.7 m/s

(

)

(b) When the person reaches the ground, two forces will act on them: the force of gravity pulling down and the normal force of the ground pushing up. The sum of these two forces provides the net decelerating force. The net work done during deceleration is equal to the change in kinetic energy.  Fd = K → ( mg − FN )( d ) = 0 − 12 mv 2

( 65 kg )( 7.668 m s ) FN = mg + = ( 65 kg ) ( 9.8 m s ) + = 4459 N  4500 N 2d 2 ( 0.50 m ) 2

mv 2

( 65 kg )( 7.668 m s ) FN = mg + = ( 65 kg ) ( 9.8 m s ) + = 1.917  105 N  1.9  105 N 2d 2 ( 0.010 m ) 2

mv 2

(d) The force is evenly spread between each leg, so divide half of the force by the area of the tibia to determine the stress. Then compare this stress to the compressive strength of the tibia given in Table 12–2. 1 F ( 4459 N ) = 2 = 7.4  106 N m 2 < 170  106 N m 2 −4 2 A 3.0  10 m The stress is much less than the compressive strength so it is unlikely that the tibia will break. (e) Repeating the calculation for the distance of 0.010 m: 5 1 F 2 1.917  10 N = = 3.2  108 N m 2 > 170  106 N m 2 −4 2 A 3.0  10 m The stress is greater than the compressive strength so the tibia will likely break.

(

)

455

CHAPTER 13: Fluids Responses to Questions 1.

Density is the ratio of mass to volume. A high density may mean that lighter molecules are packed more closely together and thus a given amount of mass is occupying a smaller volume, making a higher density. An atom of gold weighs less than an atom of lead, because gold has a lower atomic mass, but the density of gold is higher than that of lead.

The cabin of an airplane is maintained at a pressure lower than sea-level atmospheric pressure, and the baggage compartment is not pressurized. Atmospheric pressure is lower at higher altitudes, so when an airplane flies up to a high altitude, the air pressure outside a cosmetics bottle drops, compared to the pressure inside. The higher pressure inside the bottle may force the contents to leak out around the cap.

For the cylindrical container, the normal force on the water from the bottom of the container is equal to the weight of the water. In the case of the two non-cylindrical containers, perpendicular forces from the sides of the containers on the fluid will influence the net force of the water on the base, and thus influence the normal force. For the middle container, the normal forces from the sides (perpendicular to the sides) will have an upward component, which helps support the water and thus lowers the normal force required at the bottom to be less than the total weight of the water. The slanted sides support some of the weight of the water, and so the force at the bottom can be less than the weight of the water. The weight of the water is equal to the vector sum of the normal forces on ALL the water surfaces. For the container on the right, the normal forces from the sides will have a downward component, increasing the force on the base and thus increasing the normal force at the bottom to be more than the weight of the water. The slanted sides push downward on the water, increasing the normal force at the bottom. Again, the weight of the water is equal to the vector sum of the normal forces on ALL the water surfaces.

The pressure is what determines whether or not your skin will be cut. You can push both the pen and the pin with the same force, but the pressure at the point of the pin will be much greater than the pressure at by the blunt end of the pen, because the area of the pin point is much smaller. Skin has a maximum tensile strength (a topic from Chapter 12) given in N/m2. Accordingly, the higher the pressure, the more likely the skin is to be cut.

As the water boils, steam displaces some of the air in the can. With the lid off, the pressure inside is the same as the outside pressure. When the lid is put on, and the water and the can cool, the steam that is trapped in the can condenses back into liquid water. This reduces the pressure in the can to less than atmospheric pressure, and the greater force from the outside air pressure crushes the can.

“Blood pressure” should measure the pressure of the blood coming out of the heart. If the cuff is below the level of the heart, the measured pressure will be the pressure from the pumping of the heart plus the pressure due to the height of blood above the cuff. This reading will be too high. Likewise, if the cuff is above heart level, the reported pressure measurement will be too low.

Ice floats in water, so ice is less dense than water. When ice floats, it displaces a volume of water that is equal to the weight of the ice. Since ice is less dense than water, the volume of water displaced is smaller than the volume of the ice, and some of the ice extends above the top of the water. When the ice melts and turns back into water, it will fill a volume exactly equal to the original

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volume of water displaced. The water will not overflow the glass as the ice melts. In other words, the ice displaces its melted volume. 8.

The density of ice ( 917 kg m 3 ) is greater than that of alcohol ( 790 kg m 3 ) , so the ice cube will not float in a glass of alcohol. The ice cube will sink in the alcohol.

Both products have gas dissolved in them (via the process of “carbonation”), making their density less than that of water. The difference is in the sweetener in each product. The “Coke” has a significant amount of sugar (or some other sweetener, like high fructose corn syrup) dissolved in it, increasing its density and making it greater than that of water. The “Diet Coke” has a different, lowcalorie sweetener which evidently has a lower density than the “Coke” sweetener. Thus, the “Diet Coke” density (including the can) remains less than that of water. Thus, the “Coke” sinks, and the “Diet Coke” floats.

10. (a) Let object 1 be the object with the greater volume, so V1  V2 . The apparent weight of each object is the difference between its actual weight and the buoyant force. Since both objects are fully submerged the buoyant force is equal to the product of the density of the water, their volumes, and the acceleration of gravity. We set the apparent weights equal and solve for the actual weight of object 1. We use the symbol W for the actual weight. → W1 −  waterV1 g = W2 −  waterV2 g W1 = W2 +  water (V1 − V2 ) g The larger object has the greater weight . (b) Since V1  V2 and their apparent weights are equal, the ratio of the apparent weight of object 1 to its volume will be less than the ratio of the apparent weight of object 2 to its volume. W1 −  waterV1 g W2 −  waterV2 g mg mg  → 1 −  water g  2 −  water g → V1 V2 V1 V2

1 −  water   2 −  water → 1   2 The smaller object has the greater density . 11. The liquid in the vertical part of the tube over the lower container will fall into the container through the action of gravity. This action reduces the pressure in the top of the tube and draws liquid through the tube and into the tube from the upper container. As noted, the tube must be full of liquid initially for this to work. The liquid must be “cohesive,” so that the fluid remains continuous throughout the tube. If the fluid is “broken” in some fashion, such as having air in between the two parts of the fluid, the siphon will not work. Here is another explanation. See the textbook diagram. The pressure at the surface of both containers is atmospheric pressure. The pressure in each tube, measured at the level of the surface of liquid in each container, would thus also be atmospheric pressure, assuming that its velocity was small. The pressure in each tube will then decrease with height as measured upward from the level of the surface of liquid in each container. Since the portion of the tube going into the lower container is longer than the portion of the tube going into the higher container, the pressure at the highest point on the right side is lower than the pressure at the highest point on the left side. This pressure difference causes liquid to flow from the left side of the tube to the right side of the tube. And as noted in the question, the tube must be filled with liquid before this argument can be made. 12. Sand must be added to the barge. If sand is removed, the barge will not need to displace as much water since its weight will be less, and it will rise up in the water, making it even less likely to fit under the bridge. If sand is added, the barge will sink lower into the water, making it more likely to © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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fit under the bridge. Of course you would have to be careful to not “pile” the sand up so high that you lose the advantage of adding more sand. This also assumes that the density of the sand is higher than that of the water, so that adding 1 m3 of sand will displace more than 1 m3 of water. 13. As the balloon rises, the air pressure outside the balloon will decrease and be lower than the pressure inside the balloon. The excess inside air pressure will cause the balloon to expand, lowering the pressure inside but stretching the balloon in the process. If, at launch, the material of the balloon were already stretched to the limit, the expansion of the balloon due to the decreasing outside air pressure would cause the balloon to burst. Thus, the balloon is only filled to a fraction of the maximum volume. This allows plenty of room for expansion as the balloon rises. 14. The water level will fall in all three cases. (a) The boat, when floating in the pool, displaces water, causing an increase in the overall level of water in the pool. Therefore, when the boat is removed, the water returns to its original (lower) level. (b) The boat and anchor together must displace an amount of water equal to their combined weight. If the anchor is removed, this water is no longer displaced and the water level in the pool will go down. (c) If the anchor is removed and dropped in the pool, so that it rests on the bottom of the pool, the water level will again go down, but not by as much as when the anchor is removed from the boat and pool altogether. When the anchor is in the boat, the combination must displace an amount of water equal to their weight because they are floating. When the anchor is dropped overboard, it can only displace an amount of water equal to its volume, which is less than the volume of water equal to its weight. Less water is displaced so the water level in the pool goes down. Also see Example 13–13. 15. No. If the balloon is inflated, then the pressure inside the balloon is slightly greater than atmospheric pressure. Thus, the air inside the balloon is more dense than the air outside the balloon. Because of the higher density, the weight of the air inside the balloon is greater than the weight of the outside air that has been displaced. This is the same as saying that the buoyant force on the balloon is less than the weight of the air inside the balloon. And so the apparent weight of the filled balloon will be slightly greater than that of the empty balloon. 16. In order to float, you must displace an amount of water equal to your own weight. Salt water is more dense than fresh water, so the volume of salt water you must displace is less than the volume of fresh water. You will float higher in the salt water because you are displacing less water. Less of your body needs to be submerged in the water. 17. The papers move toward each other. Bernoulli’s principle says that as the speed of air increases, the pressure decreases (when there is no appreciable change in height). So as the air passes between the papers, the air pressure between the papers is lowered. The air pressure on the outside of the papers is then greater than that between the papers, and so the papers are pushed together. 18. As the water falls, its vertical speed is larger when away from the faucet than when close to it, due to gravity. Since the water is essentially incompressible, Eq. 13–7b applies, which says that a faster flow has a smaller cross-sectional area. Thus the faster moving water has a narrower stream. 19. It is possible. Due to viscosity, some of the air near the train will be pulled along at a speed approximately that of the train. By Bernoulli’s principle, that air will be at a lower pressure than air further away from the train. That difference in pressure results in a force towards the train, which could push a lightweight child towards the train. The child would be pushed, not “sucked,” but the effect is the same–a net force towards the train. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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20. Assumptions in Bernoulli’s Equation Steady Flow

Laminar Flow

Incompressible Fluid Non-viscous Fluid

Fluids

Modifications if assumptions were not made If the flow rate can vary with time, then each of the terms in Bernoulli’s equation could also vary with time. Additional terms would be needed to account for the energy needed to change the flow rates. Without laminar flow, turbulence and eddy currents could exist. These would create energy losses due to heating that would need to be accounted for in the equation. Work is done on the fluid as it compresses and the fluid does work as it expands. This energy would need to be accounted for in the equation. Greater pressure differences would be needed to overcome energy lost to viscous forces (via frictional heating). Pressure and velocity would also depend upon distance from pipe walls.

21. Water will not flow from the holes when the cup and water are in free fall. The acceleration due to gravity is the same for all falling objects (ignoring friction), and so the cup and water would fall together. For the water to flow out of the holes while falling would mean that the water would have an acceleration larger than the acceleration due to gravity. Another way to consider the situation is that there will no longer be a pressure difference between the top and bottom of the cup of water, since the lower water molecules don’t need to hold up the upper water molecules. 22. The lift generated by a wind depends on the speed of the air relative to the wing. For example, an airplane model in a wind tunnel will have lift forces on it even though the model isn’t moving relative to the ground. By taking off into the wind, the speed of the air relative to the wing is the sum of the plane’s speed and the wind speed. This allows the plane to take off at a lower ground speed, requiring a shorter runway. 23. As the ships move, they drag water with them. The moving water has a lower pressure than stationary water, as shown by Bernoulli’s principle. If the ships are moving in parallel paths fairly close together, the water between them will have a lower pressure than the water to the outside of either one, since it is being dragged by both ships. The ships are in danger of colliding because the higher pressure of the water on the outsides will tend to push them towards each other. 24. As the car drives through the air, the air inside the car is stationary with respect to the top, but the outside air is moving with respect to the top. There is no appreciable change in height between the two sides of the canvas top. By Bernoulli’s principle, the outside air pressure near the canvas top will be less than the inside air pressure. That difference in pressure results in a force that makes the top bulge outward. 25. The roofs are pushed off from the inside. By Bernoulli’s principle, the fast moving winds of the storm causes the air pressure above the roof to be quite low, but the pressure inside the house is still near normal levels. There is no appreciable change in height between the two sides of the roof. This pressure difference, combined with the large surface area of the roof, gives a very large force which can push the roof off the house. That is why it is advised to open some windows if a tornado is imminent, so that the pressure inside the house can somewhat equalize with the outside pressure.

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Solutions to MisConceptual Questions 1.

(c) Students frequently think that since the wood floats it experiences a greater buoyancy force. However, both objects experience the same buoyant force since they have the same volume and are in the same fluid. The wood floats because its weight is less than the buoyant force, while the iron sinks because its weight is greater than the buoyant force.

(d) A common misconception is that B will have the greater force on the bottom since it holds the greatest weight of water. The force of the water on the bottom of the container is equal to the product of the area of the bottom and the pressure at the bottom. Since all three containers have the same base areas and the same depth of water, the forces on the bottom of each will be equal. The additional weight of the water in B is supported by the diagonal walls. The smaller weight of C is countered by the additional pressure exerted by the diagonal walls on the water.

(b) Pascal’s principle states that if an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. That is exactly what this question is describing.

(c) Students may think that since part of the wood floats above the water, Beaker B will weigh more. Since the wood is in equilibrium, the weight of the water displaced by the wood is equal to the weight of the wood. Therefore, both beakers will weigh the same.

(a) The buoyant force is the weight of water displaced by the volume of the submerged object. mobject FB = mwater g =  waterVobject g =  water g . Thus we see that the buoyant force is inversely

 object

proportional to the density of the submerged object, if all the masses are the same. The helium has the lowest density, so the 1-kg helium balloon would displace the largest volume of water and thus have the largest buoyant force. 6.

(d) A common misconception is to consider the ocean liner as a solid object. The ocean liner has an outer shell of steel, which keeps the water out of the interior of the ship. However, most of the volume occupied by the ocean liner is filled with air. This makes the average density of the ocean liner less than the density of the salt water. If a hole were introduced into the outer shell, the interior would fill with water increasing the density of the ship until it was greater than that of salt water and the ship would sink, as happened to the Titanic.

(c) Students may think that since part of the ice floats above the water, the water will overflow as the ice melts. The ice, however, displaces a mass of water equal to the mass of the ice. As the ice melts, its volume decreases until it occupies the same volume as the water that the ice initially displaced.

(c) If the volume of the balloon doesn’t change “much,” then the buoyant force doesn’t change “much” either. Assume that the volume doesn’t change at all–then the buoyant force doesn’t change at all. But the weight of the balloon + contents DOES increase some as more helium is added to the balloon. So the force needed to hold the balloon down will decrease slightly, since the gravity force on the balloon + contents has increased. Thus the answer is (c).

(a) A common misconception is that the hot air causes the balloon to rise, so it would rise on the Moon just as it would on Earth. What actually happens on the Earth is the denser cold air around the balloon on Earth is heavier than the hot air in the balloon. This denser air then

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moves downward displacing the hot air balloon upward. On the Moon there is no atmosphere to move downward around the balloon, so the balloon would not rise. 10. (c) Students may reason that since water is denser than oil, the object would experience a greater buoyant force in water. This would be true if the object was completely submerged in either fluid. Since the object is floating at the surface of the liquid, the buoyant force on the object will be equal to its weight and will therefore be the same in both fluids. The object floats with less volume submerged in the water than is submerged when it floats in the oil. 11. (d) The two blocks are being held up by the buoyant force, which is equal to the combined weight of the two blocks. Thus, an amount of water has been displaced that has the same weight as the combined weight of the two blocks. If the blocks are separated and put in the water individually, they each displace an amount of water equal to their weight. So the same amount of water is displaced in either case, and thus the water level doesn’t change, whether the blocks are stacked or placed in the water individually. 12. (a, e) A common misconception is that the pressure would be higher in the faster moving water. The continuity equation requires that the water in the wider pipe travels slower than in the narrow pipe. Bernoulli’s equation shows that the pressure will be higher in the region where the water travels slower–in the wider pipe–and so answers (a) and (e) are both correct. 13. (b) The ball accelerates to the right because the pressure on the left side of the ball is greater than the pressure on the right side. From Bernoulli’s equation, the air on the right side must then be traveling faster than the air on the left side. This difference in air speed is produced by spinning the ball when it is thrown. 14. (a) Students may believe that wind above the chimney will blow the smoke back down the chimney. Actually, the wind blowing across the top of the chimney causes the air pressure above the chimney to be lower than the air pressure inside. The greater inside pressure pushes the smoke up the chimney.

Solutions to Problems 1.

The mass is found from the density of granite (found in Table 13–1) and the volume of granite.

(

)(

)

m =  V = 2.7 103 kg m3 108 m3 = 2.7 1011 kg  3 1011 kg 2.

The mass is found from the density of air (found in Table 13–1) and the volume of air.

(

)

m =  V = 1.29 kg m3 ( 5.6 m )( 4.6 m )( 2.4 m ) = 79.75kg  8.0  101 kg 3.

The mass is found from the density of gold (found in Table 13–1) and the volume of gold. m =  V = (19.3  103 kg m3 ) ( 0.54 m )( 0.31m )( 0.22 m ) = 710 kg (  1600 lb )

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The specific gravity of the mixture is the ratio of the density of the mixture to that of water. To find the density of the mixture, the mass of antifreeze and the mass of water must be known. mantifreeze =  antifreezeVantifreeze = SG antifreeze  waterVantifreeze mwater =  waterVwater

 mixture mmixture Vmixture mantifreeze + mwater SG antifreeze  waterVantifreeze +  waterVwater = = =  waterVmixture  water  water  waterVmixture

SG mixture =

SG antifreezeVantifreeze + Vwater

Vmixture

( 0.80)( 4.0 L ) + 5.0 L 9.0 L

= 0.91

To find the specific gravity of the fluid, take the ratio of the density of the fluid to that of water, noting that the same volume is used for both liquids. ( m V )fluid mfluid 89.22 g − 32.00 g  = = = 0.8612 SG fluid = fluid =  water ( m V ) water mwater 98.44 g − 32.00 g

(a) The density from the three-part model is found from the total mass divided by the total volume. Let subscript 1 represent the inner core, subscript 2 represent the outer core, and subscript 3 represent the mantle. The radii are then the outer boundaries of the labeled region. 3 3 3 3 3 4 4 4 m + m2 + m3 1m1 +  2 m2 +  3m3 1 3  r1 +  2 3  r2 − r1 +  3 3  r3 − r2  three = 1 = = 4 V1 + V2 + V3 V1 + V2 + V3  r13 + 43  r23 − r13 + 43  r33 − r23 layers 3

(

1r13 +  2 ( r23 − r13 ) +  3 ( r33 − r23 ) r33

(

)

(

)

r13 ( 1 −  2 ) + r23 (  2 −  3 ) + r33 3 r33

(1220 km )3 (1900 kg m3 ) + ( 3480 km )3 ( 6700 kg m3 ) + ( 6380 km )3 ( 4400 kg m 3 ) = ( 6380 km )3 = 5500.6 kg m 3  5500 kg m 3

(b)  one

density

M V

M 4 3

R

= 3

5.98  1024 kg 4 3

 ( 6380  10 m ) 3

= 5497.3kg m3  5.50  103 kg m 3

  one −  three   5497 kg m 3 − 5501kg m 3  density layers  % diff = 100  = 100   = −0.073  −0.07 %   5501kg m 3  three     layers   7.

The pressure exerted by the heel is caused by the heel pushing down on the floor. That downward push is the reaction to the normal force of the floor on the shoe heel. The normal force on one heel is assumed to be half the weight of the person. 1 1 56 kg ) 9.80 m s 2 Wperson 2( 2 (a) Ppointed = = = 6.1  106 N m 2 2 2 Apointed 0.45cm ( 0.01m cm )

(

(b) Pwide = 8.

1 2

Wperson Awide

)

(

( 56 kg ) ( 9.80 m s2 ) = = 1.5  105 N m 2 2 2 (18cm ) ( 0.01m cm ) 1 2

The pressure is given by Eq. 13–3.

(

)

)(

)

P =  gh = 1.00  103 kg m3 9.80 m s2 ( 46 m ) = 4.5  105 N m2 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

462

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(a) The total force of the atmosphere on the table will be the air pressure times the area of the table.

(

)

F = PA = 1.013 105 N m2 (1.7 m )( 2.6 m ) = 4.5 105 N (b) Since the atmospheric pressure is the same on the underside of the table (the height difference is minimal), the upward force of air pressure is the same as the downward force of air on the top of the table, 4.5  105 N .

10. Use Eq. 13–3 to find the pressure difference. The density is found in Table 13–1. P =  gh → P =  g h = 1.05  103 kg m 3 9.80 m s 2 (1.75 m )

(

)(

)

 1 mm-Hg  = 135mm-Hg 2   133 N m 

= 1.801  104 N m2 

11. The height is found from Eq. 13–3, using normal atmospheric pressure. The density is found in Table 13–1. 1.013  105 N m 2 P P =  gh → h = = = 13m  g ( 0.79  103 kg m 3 )( 9.80 m s 2 ) That is so tall as to be impractical in many cases. 12. The pressure difference on the lungs is the pressure change from the depth of water. The pressure unit conversion comes from Table 13–2.  133 N m 2  ( −88 mm-Hg )   P  1 mm-Hg  = −1.194 m  −1.2 m P =  g h →  h = =  g (1.00  103 kg m 3 )( 9.80 m s 2 ) He could have been 1.2 m below the surface. 13. The pressure force from each tire must support 1/4 of the weight of the car. 2 4 2 mg (1250 kg ) 9.80 m s 2  10 cm  −2 1 1.392 10 m mg PA A = → = = =   140cm2 4   3 2 2 4P 4 220  10 N m  1m 

(

)

14. The sum of the force exerted by the pressure in each tire is equal to the weight of the car.  1 m2  4 2.40  105 N m 2 190 cm 2  4 2  4 PA  10 cm  = 1861kg  1900 kg mg = 4 PA → m = = 2 g 9.80 m s

(

)(

(

)

15. (a) The absolute pressure can be found from Eq. 13–6b, and the total force is the absolute pressure times the area of the bottom of the pool. P = P0 +  gh = 1.013  105 N m 2 + 1.00  103 kg m 3 9.80 m s 2 (1.8 m )

(

)(

)

= 1.189  105 N m 2  1.19  105 N m 2

(

)

F = PA = 1.189  105 N m 2 ( 24.0 m )( 8.5 m ) = 2.4  107 N

463

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the bottom. Pressure is not directional. P = 1.19 105 N m2 16. (a) The gauge pressure is given by Eq. 13–3. The height is the height from the bottom of the hill to the top of the water tank.

(

)(

)

PG =  gh = 1.00  103 kg m3 9.80 m s2 6.0 m + ( 75m ) sin 61 = 7.0  105 N m2 (b) The water would be able to shoot up to the top of the tank (ignoring any friction). h = 6.0 m + ( 75 m ) sin 61 = 72 m 17. The pressure at points a and b are equal since they are the same height in the same fluid. If they were unequal, the fluid would flow. Calculate the pressure at both a and b, starting with atmospheric pressure at the top surface of each liquid, and then equate those pressures. Pa = Pb → P0 +  oil ghoil = P0 +  water ghwater →  oil hoil =  water hwater →

 water hwater (1.00  10 kg m ) ( 0.272 m − 0.0862 m ) = = 683kg m3 hoil ( 0.272 m ) 3

 oil =

18. (a) The depth is found from Eq. 13–6b. Note that we use the density of sea water.  1.013  105 N m 2  ( 412 atm )   1atm P − P0   = 4154.9 m  4150 m P = P0 +  gh → h = = g (1.025  103 kg m3 )( 9.80 m s2 ) (b) The force is found from Eq. 13–2. The force due to the water is due to the absolute pressure, not just the gauge pressure.  1.013  105 N m 2  F = PA = ( 413atm )  0.0100 m 2 = 4.1837  105 N  4.18  105 N  1atm  

(

)

19. The minimum gauge pressure would cause the water to come out of the faucet with very little speed. This means the gauge pressure needed must be enough to hold the water at this elevation. Use Eq. 13–3.

(

)(

)

PG =  gh = 1.00 103 kg m3 9.80 m s 2 ( 44 m ) = 4.3 105 N m 2 20. Consider the lever (handle) of the press. The net torque on that handle is 0. Use that to find the force exerted by the hydraulic fluid upwards on the small cylinder (and the lever). Then Pascal’s principle can be used to find the upwards force on the large cylinder, which is the same as the force on the sample. We use the subscript “1” for the small cylinder, and the subscript “2” for the large cylinder.

 = F ( 2l ) − F l = 0 → F = 2 F ; P = P →  1

( 12 d1 )

 ( 12 d 2 )

→

464

Chapter 13

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F2 = F1 ( d 2 d1 ) = 2 F ( d 2 d1 ) = Fsample → 2

Psample =

Fsample Asample

2 F ( d 2 d1 ) Asample

2 ( 320 N )( 5) −4

4.0  10 m

= 4.0  107 N m 2

21. The pressure in the tank is atmospheric pressure plus the pressure difference due to the column of mercury, as given in Eq. 13–6b. (a) P = P0 +  gh = 1.04 bar + Hg gh

 1.00  105 N m 2  + (13.6  103 kg m 3 )( 9.80 m s 2 ) ( 0.185 m ) = 1.29 10 5 N m 2  1bar   5 2  1.00  10 N m  (b) P = (1.04 bar )  + (13.6  103 kg m3 )( 9.80 m s 2 ) ( −0.076 m ) = 9.4  104 N m 2  1bar   = (1.04 bar ) 

22. (a) See the diagram. In the accelerated frame of the beaker, there is a pseudoforce opposite to the direction of the acceleration, and so there is a pseudo acceleration as shown on the diagram. The effective acceleration, g , is given by g = g + a. The surface of the water will be perpendicular to the effective acceleration, and thus makes an angle

 = tan −1

a g

a 

g



(b) The left edge of the water surface, opposite to the direction of the acceleration, will be higher. (c) Constant pressure lines will be parallel to the surface. From the second diagram, we see that a vertical depth of h corresponds to a depth of h perpendicular to the surface, where h = h cos  , and so we have the following. P = P0 +  g h = P0 +  g 2 + a 2 ( h cos  )

  

= P0 +  g 2 + a 2  h

  = P0 +  hg g 2 + a 2  g

And so P = P0 +  hg , as in the unaccelerated case. 23. (a) Because the pressure varies with depth, the force on the wall will also vary with depth. So to find the total force on the wall, we will have to integrate. Measure vertical distance y downward from the top level of the water behind the dam. Then at a depth y, choose an infinitesimal area of width b and height dy. The pressure due to the water at that depth is P =  gy. P =  gy ; dF = PdA = (  gy )( b dy ) → h

F =  dF =  (  gy )( b dy ) = 12  gbh 2 0

465

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The lever arm for the force dF about the bottom of the dam is h − y , and so the torque caused by that force is d = ( h − y ) dF. Integrate to find the total torque.

 =  d =  ( h − y )(  gy )( b dy ) =  gb  ( hy − y 2 ) dy h

(

=  gb 12 hy 2 − 13 y 3

) =  gbh h

1 6

Consider that torque as caused by the total force, applied at a single distance from the bottom d.  = 16  gbh 3 = Fd = 12  gbh 2d → d = 13 h (c) To prevent overturning, the torque caused by gravity about the lower right front corner in the diagram must be at least as big as the torque caused by the water. The lever arm for gravity is half the thickness of the dam. mg ( 12 t )  16  gbh 3 →  concrete ( hbt ) g ( 12 t )  16  water gbh 3 → t h



1 3

 water =  concrete

1 3

1.00  103 kg m 3 2.3  103 kg m 3

= 0.38

So we must have t  0.38h to prevent overturning. Atmospheric pressure need not be added in because it is exerted on BOTH sides of the dam, and so causes no net force or torque. In part (a), the actual pressure at a depth y is P = P0 +  gy, and of course air pressure acts on the exposed side of the dam as well. 24. From Section 12–5, the change in volume due to pressure change is

V

=−

P

, where B is the bulk V0 B modulus of the water, given in Table 12–1. The pressure increase with depth for a fluid of constant density is given by P =  g h , where h is the depth of descent. If the density change is small, then we can use the initial value of the density to calculate the pressure change, and so P   0 g h. Finally, consider a constant mass of water. That constant mass will relate the volume and density at the two locations by M = V =  0V0 . Combine these relationships and solve for the density deep in the sea,  .

V =  0V0 → =

 0V0  0V0 = = V V0 + V

1025 kg m 3  0V0 0 = =  0 gh P   (1025 kg m3 )( 9.80 m s2 )( 5.4  103 m ) V0 +  −V0  1− 1 − B B   2.0  109 N m 2

= 1054 kg m 3  1.05  103 kg m 3

 0 =

1054

= 1.028 1025 The density at the 6-km depth is about 3% larger than the density at the surface.

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Chapter 13

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25. Consider a layer of liquid of (small) height h, and ignore the pressure variation due to height in that layer. Take a cylindrical ring of water of height h, radius r, and thickness dr. See the diagram (the height is not shown). The volume of the ring of liquid is ( 2 r h ) dr, and so has a mass of dm = ( 2 r h ) dr . That mass of water has a net centripetal force on it of magnitude dFradial =  2 r ( dm ) =  2 r  ( 2 r h ) dr. That force comes from a pressure difference across the surface area of the liquid. Let the pressure at the inside surface be P , which causes an outward force, and the pressure at the outside surface be P + dP, which causes an inward force. The surface area over which these pressures act is 2 r h, the “walls” of the cylindrical ring. Use Newton’s second law. dFradial = dFouter − dFinner →  2 r  ( 2 rh ) dr = ( P + dP ) 2 rh − ( P ) 2 rh → wall

wall

 dP =   r dr → P − P =  r

dP =  r  dr → 2

1 2

2 2

→

P = P0 + 12  2 r 2

26. If the iron is floating, then the net force on it is zero. The buoyant force on the iron must be equal to its weight. The buoyant force is equal to the weight of the mercury displaced by the submerged iron. Fbuoyant = mFe g →  Hg gVsubmerged =  Fe gVtotal →

Vsubmerged Vtotal

7.8  103 kg m 3  Fe = = 0.57  57%  Hg 13.6  103 kg m 3

27. The difference in the actual mass and the apparent mass is the mass of the water displaced by the rock. The mass of the water displaced is the volume of the rock times the density of water, and the volume of the rock is the mass of the rock divided by its density. Combining these relationships yields an expression for the density of the rock. m mactual − mapparent = m =  waterVrock =  water rock →

 rock

 rock =  water

mrock m

(

= 1.00  103 kg m 3

8.94 kg = 3240 kg m ) 8.94 kg − 6.18 kg

28. (a) When the hull is submerged, both the buoyant force and the tension force act upward on the hull, and so their sum is equal to the weight of the hull, if the hull is not accelerated as it is lifted. The buoyant force is the weight of the water displaced. T + Fbuoyant = mg →

T = mg − Fbuoyant = mhull g −  waterVsub g = mhull g −  water

(

)

)(

= 2.1  104 kg 9.80 m s 2  1 −

mhull

 hull

1.00  10 kg m  3



g = mhull g  1 −



 water   hull 

= 1.794  10 N  1.8  10 N 7.8  10 kg m  3



(b) When the hull is completely out of the water, the tension in the crane’s cable must be equal to the weight of the hull.

(

)(

)

T = mg = 2.1  104 kg 9.80 m s2 = 2.058  105 N  2.1  105 N © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

467

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

29. The buoyant force of the balloon must equal the weight of the balloon, plus the weight of the helium in the balloon, plus the weight of the load. For calculating the weight of the helium, we assume it is at 0oC and 1 atm pressure. The buoyant force is the weight of the air displaced by the volume of the balloon. Fbuoyant =  airVballoon g = mHe g + mballoon g + mcargo g → mcargo =  airVballoon − mHe − mballoon =  airVballoon −  HeVballoon − mballoon = (  air −  He )Vballoon − mballoon

(

= 1.29 kg m 3 − 0.179 kg m 3

)  ( 7.15 m ) − 890 kg = 810 kg = 7900 N 3

4 3

30. The initial buoyant force is equal to the initial volume of liquid displaced. We ignore the volume of the container–perhaps by assuming it has very thin walls. The height of that volume is not known, and is designated as “h.” The final buoyant force is calculated using a smaller height. Fbuoyant =  waterVsubstance g =  water ( r 2 h ) g ; Fbuoyant =  waterVsubstance g =  water  r 2 ( h − 0.028 )  g initial

initial

final

The difference in the two buoyant forces is equal to the weight of the substance removed. Fbuoyant =  water  r 2 ( 0.028 m )  g = msubstance g →

(

)

m substance =  water  r 2 ( 0.028 m )  = 1000 kg m 3  ( 0.026 m ) ( 0.028 m ) = 0.05946 kg  59 g 2

31. The buoyant force on the ice is equal to the weight of the ice, since it floats. Fbuoyant = Wice → mseawater g = mice g → mseawater = mice →  seawaterVseawater =  iceVice → displaced

displaced

( SG)seawater  waterVsubmerged = (SG )ice  waterVice → (SG )seawater Vsubmerged = (SG )ice Vice → ice

Vsubmerged = ice

ice

(SG )ice 0.917 Vice = Vice = 0.895Vice 1.025 ( SG )seawater

Thus, the fraction above the water is Vabove = Vice − Vsubmerged = 0.105 Vice or 10.5% 32. The difference in the actual mass and the apparent mass of the aluminum is the mass of the air displaced by the aluminum. The mass of the air displaced is the volume of the aluminum times the density of air, and the volume of the aluminum is the actual mass of the aluminum divided by the density of aluminum. Combining these relationships yields an expression for the actual mass. m mactual − mapparent =  airVAl =  air actual →  Al mactual =

mapparent

 1 − air  Al

4.0000 kg

= 1−

1.29 kg m 3

= 4.0019 kg

2.70  103 kg m 3

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Chapter 13

Fluids

33. The buoyancy force due to the submerged empty soda bottles must equal the weight of the child, and is also equal to the mass of water displaced. To get the minimum number of bottles (N), we assume that each bottle is completely submerged, and so displaces 1.0 L of water. Fbuoyant = NVbottle  water g = mchild g → N=

mchild Vbottle  water

32 kg

 10 m 3  1000 kg m 3 ) (1.0 L )  (   1L  −3

= 32 bottles

34. (a) The difference in the actual mass and the apparent mass of the aluminum ball is the mass of the liquid displaced by the ball. The mass of the liquid displaced is the volume of the ball times the density of the liquid, and the volume of the ball is the mass of the ball divided by its density. Combining these relationships yields an expression for the density of the liquid. m mactual − mapparent = m =  liquidVball =  liquid ball →

 Al

 liquid =

m mball

 Al =

( 3.80 kg − 2.10 kg ) 3.80 kg

( 2.70  10 kg m ) = 1210 kg m 3

 mobject − mapparent    object . mobject  

(b) Generalizing the relation from above, we have  liquid = 

35. There are three forces on the chamber; the weight of the chamber, the tension in the cable, and the buoyant force. See the free-body diagram. (a) The buoyant force is the weight of water displaced by the chamber. 3 Fbuoyant =  waterVchamber g =  water 43  rchamber g

(

= 1.025  103 kg m 3

)  ( 2.60 m ) ( 9.80 m s ) 3

4 3

= 7.3953  10 N  7.40  10 N (b) To find the tension, use Newton’s second law for the stationary chamber. Fbuoyant = mg + FT → 5

(

)(

)

FT = Fbuoyant − mg = 7.3953  105 N − 7.44  10 4 kg 9.80 m s 2 = 1.0  10 4 N

36. Since the wood sinks in alcohol, we know its specific gravity must be greater than that of the alcohol, but less than that of water. The difference in the actual mass and the apparent mass is the mass of the alcohol displaced by the wood. The mass of the alcohol displaced is the volume of the wood times the density of the alcohol, the volume of the wood is the mass of the wood divided by the density of the wood, and the density of the alcohol is its specific gravity times the density of water. m m mactual − mapparent =  alcVwood =  alc actual = SG alc  H O actual →  wood  wood 2

0.48 kg mactual  wood = SG wood = SG alc = ( 0.79 ) = 0.88 H O ( 0.48 kg − 0.047 kg ) ( mactual − mapparent ) 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

37. Use the definition of density and specific gravity, and then solve for the fat fraction, f. mfat = mf = Vfat  fat ; mfat = m (1 − f ) = Vfat  fat free

 body = X  water =

mtotal Vtotal

free

mfat + mfat =

free

Vfat + Vfat

 fat

free

m = m (1 − f ) f

 fat

1 (1 − f )

 fat

free

 fat  fat f =

free

  X  water   fat −  fat   free  4.95 X

−

free

( 0.90 g cm )(1.10 g cm ) − ( 0.90 g cm ) =  X (1.0 g cm )( 0.20 g cm ) ( 0.20 g cm ) −

 fat

   fat  free

→

fat

 

 4.95 − 4.5  = 495 − 450  X  X 

− 4.5 → % Body fat = 100 f = 100 

38. (a) The free-body diagram for the person shows three forces; their weight, the buoyancy force, and the upwards force of the scale. Those forces must add to give 0, and that can be used to find the volume of the athlete. Fbuoyant + Fscale − Fweight =  waterVg + mapparant g − mactual g = 0 →

mactual − mapparant

V =

 water

70.2 kg − 3.4 kg 3

Fbuoy

= 6.68  10−2 m 3

1000 kg m (b) The specific gravity is the person’s density divided by the density of water. −2 3 −3 3  person m (V − VR ) ( 70.2kg ) 6.68  10 m − 1.3  10 m SG = = = = 1.072  1.07  water  water 1000 kg m 3 (c) We use the formula given with the problem. 495 495 % Body fat = − 450 = − 450 = 12% SG 1.072

(

Fscale

)

39. The buoyant force is equal to the combined weight of the helium balloons and the person. We ignore the buoyant force due to the volume of the person, and we ignore the mass of the balloon material. mperson → FB = ( mperson + mHe ) g →  airVHe g = ( mperson +  HeVHe ) g → VHe = N 43  r 3 = ( air −  He )

3mperson

4 r (  air −  He ) 3

(

3 ( 72 kg )

4 ( 0.18 m ) 1.29 kg m 3 − 0.179 kg m 3 3

)

= 2653  2700 balloons

40. The object’s weight is balanced by two buoyant forces, one from the water and one from the oil. Each buoyant force is the density of the liquid, times the volume in the liquid, times g. The edge length of the cube is l. mg = Fbuoyancy + Fbuoyancy =  oilVoil g +  waterVwater g =  oil l 2 ( 0.28l ) g +  water l 2 ( 0.72 l ) g → oil

water

470

Chapter 13

Fluids

(

)

(

)

m = l 3 ( 0.28oil + 0.72  water ) = ( 0.100 m ) 0.28 810 kg m 2 + 0.72 1000 kg m 2  3

= 0.9468 kg  0.95 kg The buoyant force is the weight of the object, mg = ( 0.9468 kg ) ( 9.80 m s2 ) = 9.3 N 41. There will be a downward gravity force and an upward buoyant force on the fully submerged tank. The buoyant force is constant, but the gravity force will decrease as the air is removed. Take upwards to be positive. Ffull = FB − mtotal g =  waterVtank g − ( mtank + mair ) g

(

=  1025 kg m3

)( 0.0157 m ) − 17.0 kg  ( 9.80 m s ) = −8.89 N  9N downward 3

Fempty = FB − mtotal g =  waterVtank g − ( mtank + mair ) g

(

=  1025 kg m3

)( 0.0157 m ) − 14.0 kg  ( 9.80 m s ) = 20.51N  21N upward 3

42. The apparent weight is the force required to hold the system in equilibrium. In the first case, the object is held above the water. In the second case, the object is allowed to be pulled under the water. Consider the free-body diagram for each case. Case 1:  F = w1 − w + Fbuoy − wsinker = 0

Fapparent = w1

Fbuoy object

Fapparent = w2

w w

sinker

Case 2:

F = w + F 2

buoy object

Fbuoy

− w + Fbuoy − wsinker = 0

sinker

Since both add to 0, equate them. Also note that the specific gravity can be expressed in terms of the buoyancy force. m w  Fbuoy = Vobject  water g = object  water g = mobject g water =  object  object S.G. object

Fbuoy sinker

wsinker wsinker

w1 − w + Fbuoy − wsinker = 0 = w2 + Fbuoy − w + Fbuoy − wsinker → sinker

object

w1 = w2 + Fbuoy = w2 + object

w S.G.

→ S.G. =

sinker

( w1 − w2 )

43. For the combination to just barely sink, the total weight of the wood and lead must be equal to the total buoyant force on the wood and the lead. Fweight = Fbuoyant → mwood g + mPb g = Vwood  water g + VPb  water g →

      water → mPb  1 − water  = mwood  water − 1 →  wood  Pb  Pb     wood    water   1   1 − 1 − 1   SG − 1    = m  wood  = ( 3.75 kg )  0.50  = 4.11kg  4.11kg mPb = mwood  wood wood   water   1 − 1  1  1 1− −        11.3   Pb    SGPb  mwood + mPb =

mwood

 water +

mPb

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44. Use Eq. 13–7b, the equation of continuity for an incompressible fluid, to compare blood flow in the aorta and in the major arteries. ( Av )aorta = ( Av )arteries → varteries =

Aaorta Aarteries

vaorta =

 (1.2 cm ) 2.0 cm 2

( 40 cm s ) = 90.5 cm s  0.9 m s

45. We apply the equation of continuity at constant density, Eq. 13–7b. The flow rate out of the duct must be equal to the flow rate into the room. V V ( 8.2 m )( 4.5 m )( 3.5 m ) = 4.0 m s Aduct vduct =  r 2 vduct = room → vduct = 2room = 60 s  2  r t to fill t to fill   ( 0.12 m ) (12 min )   room room  1 min  46. We may apply Torricelli’s theorem., Eq. 13–9.

(

)

v1 = 2 g ( y2 − y1 ) = 2 9.80 m s 2 ( 5.1m ) = 9.998 m s  1.0 101 m s 47. The pressure head can be interpreted as an initial height for the water, with a speed of 0 and at atmospheric pressure. Apply Bernoulli’s equation to the faucet location and the pressure head location to find the speed of the water at the faucet, and then calculate the volume flow rate. Since the faucet is open, the pressure there will be atmospheric as well. 2 2 +  gyfaucet = Phead + 12  vhead +  gyhead → Pfaucet + 12  vfaucet 2 vfaucet =

vfaucet =



2 + 2 g ( yhead − yfaucet ) = 2 gyhead → ( Phead − Pfaucet ) + vhead

2 gyhead

(

)

Volume flow rate = Av =  r 2 2 gyhead =   12 1.85  10 −2 m 

(

)

2 9.80 m s 2 (12.0 m )

= 4.12  10−3 m3 s 48. The flow speed is the speed of the water in the input tube. The entire volume of the water in the tank is to be processed in 2.5 h. The volume of water passing through the input tube per unit time is the volume rate of flow, as expressed in the text immediately after Eq. 13–7b. l wh V V ( 0.36 m )(1.0 m )( 0.60 m ) = Av → v = = 2 = = 0.03395 m s  3.4 cm s t At  r t  3600 s  2  ( 0.015 m ) ( 2.5 h )    1h  49. We assume that there is no appreciable height difference between the two sides of the roof. Then the net force on the roof due to the air is the difference in pressure on the two sides of the roof, times the area of the roof. The difference in pressure can be found from Bernoulli’s equation. 2 2 Pinside + 12  vinside +  gyinside = Poutside + 12  voutside +  gyoutside → 2 Pinside − Poutside = 12  air voutside =

Fair Aroof

→

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Chapter 13

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Fair =  air v 1 2

2 outside

  1m s   Aroof = (1.29 kg m ) (180 km h )    ( 6.2 m )(12.4 m )  3.6 km h    3

1 2

= 1.2  105 N

50. The volume flow rate of water from the hose, multiplied by the time of filling, must equal the volume of the pool.

( Av )hose =

Vpool

→ t=



Vpool Ahose vhose

  12 ( 6.1m )  (1.4 m ) 2

= 5.168  105 s

  1m     12 ( 85 in )    ( 0.40 m s )  39.37 in   

  = 6.0 days  60  60  24 s  1day

5.168  105 s 

51. The air pressure inside the hurricane can be estimated using Bernoulli’s equation, Eq. 13–8. Assume the pressure outside the hurricane is air pressure, the speed of the wind outside the hurricane is 0, and that the two pressure measurements are made at the same height. 2 2 Pinside + 12  vinside +  gyinside = Poutside + 12  voutside +  gyoutside → 2 Pinside = Poutside − 12  air vinside

  1000 m   1h   = 1.013  10 Pa − (1.29 kg m ) ( 300 km h )     km   3600 s    5

1 2

= 9.7  10 4 Pa  0.96 atm

52. Use the equation of continuity (Eq. 13–7b) to relate the volume flow of water at the two locations, and use Bernoulli’s equation (Eq. 13–8) to relate the pressure conditions at the two locations. We assume that the two locations are at the same height. Express the pressures as atmospheric pressure plus gauge pressure. Use subscript “1” for the larger diameter, and “2” for the smaller diameter. A1 r12  r12 = v1 2 = v1 2 A1v1 = A2 v 2 → v 2 = v1 A2 r2  r2

P0 + P1 + 12  v12 +  gy1 = P0 + P2 + 12  v22 +  gy2 → r4 P1 + 12  v12 = P2 + 12  v 22 = P2 + 12  v12 14 r2

A1v1 =  r12

2 ( P1 − P2 )

 r4    14 − 1   r2 

(

=  3.0  10 −2 m

→ v1 =

)

2 ( P1 − P2 )

 r4    14 − 1   r2 

(

→

2 36.5  103 Pa − 22.6  103 Pa

)

 ( 3.0  10−2 m )4   − 1 (1.00 10 kg m )  4 −2   2.25 10 m ( )   3

= 1.0  10−2 m 3 s

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Instructor Solutions Manual

53. Consider the volume of fluid in the pipe. At each end of the pipe there is a force towards the contained fluid, given by F = PA . Since the area of the pipe is constant, we have that Fnet = ( P1 − P2 ) A . Then, since the power required is the force on the fluid times its velocity, and

AV = Q = volume rate of flow, we have Power = Fnet v = ( P1 − P2 ) Av = ( P1 − P2 ) Q . 54. The lift force would be the difference in pressure between the two wing surfaces, times the area of the wing surface. The difference in pressure can be found from Bernoulli’s equation, Eq. 13–8. We consider the two surfaces of the wing to be at the same height above the ground. Call the bottom surface of the wing point 1, and the top surface point 2. P1 + 12  v12 +  gy1 = P2 + 12  v22 +  gy2 → P1 − P2 = 12  v22 − v12

(

)

Flift = ( P1 − P2 )( Area of wing ) = 12  v22 − v12 A

(

)

(

)

= 12 1.29 kg m3 ( 280 m s ) − (150 m s )  82 m 2 = 3.0 106 N 2

55. Use the equation of continuity (Eq. 13–7b) to relate the volume flow of water at the two locations, and use Bernoulli’s equation (Eq. 13–8) to relate the conditions at the street to those at the top floor. Express the pressures as atmospheric pressure plus gauge pressure. Astreet vstreet = Atop v top →

v top = vstreet

Astreet Atop

( ) = ( 0.78 m s ) = 2.487 m s  2.5 m s   ( 2.8 10 m )    12 5.0 10−2 m  −2

1 2

2 2 +  gystreet = P0 + Pgauge + 12  v top +  gytop → P0 + Pgauge + 12  vstreet street

top

(

)

2 2 − v top +  gy ( ystreet − ytop ) Pgauge = Pgauge + 12  vstreet top

street

 1.013  105 Pa  1 2 2 2 + 2 (1.00  103 kg m 3 ) ( 0.78 m s ) − ( 2.487 m s )     atm  

= ( 3.8 atm ) 

(

+ 1.00  103 kg m 3

)( 9.80 m s ) ( −16 m ) 2

1atm     2.2 atm 5  1.013  10 Pa 

= 2.250  105 Pa 

56. Apply both Bernoulli’s equation and the equation of continuity between the two openings of the tank. Note that the pressure at each opening will be atmospheric pressure. A A2 v 2 = A1v1 → v 2 = v1 1 A2

P1 + 12  v12 +  gy1 = P2 + 12  v22 +  gy2 → v12 − v22 = 2 g ( y2 − y1 ) = 2 gh



v12 −  v1



A1 



A12 



2 2

2  = 2 gh → v1  1 −

A2 

 = 2 gh → v1 = A 

2 gh

(1 − A A ) 2 1

2 2

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Chapter 13

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57. (a) Relate the conditions at the top surface and at the opening by Bernoulli’s equation. Ptop + 12  v22 +  gy2 = Popening + 12  v12 +  gy1 → P2 + P0 +  g ( y2 − y1 ) = P0 + 12  v12 → 2 P2

v1 =

2 P2

(b) v1 =



+ 2 g ( y2 − y1 )



 1.013  105 Pa  2 ( 0.85 atm )   atm   + 2 9.80 m s 2

+ 2 g ( y2 − y1 ) =

(

(1.00 10 kg m ) 3

) ( 2.4 m ) = 15 m s

58. (a) Apply Bernoulli’s equation to point 1, the exit hole, and point 2, the top surface of the liquid in the tank. Note that both points are open to the air and so the pressure is atmospheric pressure. Also apply the equation of continuity ( A1v1 = A2 v2 ) to the same two points.

(

)

P1 + 12  v12 +  gy1 = P2 + 12  v 22 +  gy2 → Patm + 12  v12 − v 22 = Patm +  g ( y2 − y1 ) → 1 2

 ( v12 − v22 ) =  gh →

v2 =

2 gh

A   A − 1   2 2 2 1

(

 A22

)

v12 − v22 = 2 gh → 

2 1

A



− 1  v22 = 2 gh →



2 1 2 1

2 ghA

(A − A ) 2 2

Note that since the water level is decreasing, we have v 2 = −

dh dt

, and so

=−

2 ghA12

(A − A ) 2 2

2 1

(b) Integrate to find the height as a function of time. dh dt 2

=−

2 ghA12

→

(A − A ) 2 2

( h− h ) 0

2 1

dh h

=−

2 gA12

(A − A ) 2 2

 h

dt → 2

=−

  gA12 =− t → h = h − t  0 2 2 2 2  2 ( A2 − A1 )  ( A2 − A1 )   2 gA12

2 gA12

(A − A )  2 2

2 1

dt →

(

A1 =  0.25  10 −2 m

)

= 1.963  10 −5 m 2 ; A2 =

1.3  10 −3 m 3 0.106 m

= 1.226  10 −2 m 2

  gA12  h0 − t  =0 → 2 ( A22 − A12 )    t=

(

2h0 A22 − A12 gA12

) = 2 ( 0.106 m ) (1.226  10 m ) − (1.963  10 m )  = 92 s ( 9.80 m s )(1.963  10 m ) −2

−5

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

59. (a) Relate the conditions inside the rocket and just outside the exit orifice by means of Bernoulli’s equation and the equation of continuity. We ignore any height difference between the two locations. 2 2 Pin + 12  vin2 +  gyin = Pout + 12  vout +  gyout → P + 12  vin2 = P0 + 12  vout →

2 ( P − P0 )



  v 2  = v − v = v 1 −  in     vout   2 out

2 in

2 out

Ain vin = Aout vout → Avin = A0 vout → 2 ( P − P0 )



vin vout

A0 A

  v 2  2 = v 1 −  in    vout → vout = v =   vout   2 out

(b) Thrust is defined in Section 9–10, by Fthrust = vrel

dm dt

1 → 2 ( P − P0 )



, and is interpreted as the force on the

rocket due to the ejection of mass. d ( V ) 2 ( P − P0 ) dm dV = vout = vout  = vout  ( vout Aout ) =  v 2 A0 =  Fthrust = vrel A0  dt dt dt

= 2 ( P − P0 ) A0 60. (a) Apply the equation of continuity and Bernoulli’s equation at the same height to the wide and narrow portions of the tube. 2 ( P1 − P2 ) A = v 22 − v12 → A2 v 2 = A1v1 → v 2 = v1 1 ; P1 + 12  v12 = P2 + 12  v 22 → A2  2 2 ( P1 − P2 )  A1  A22  2 ( P1 − P2 ) 2 2  A1 − = → − = → v v v 1 1   1  2 2     A2   A2 A2  2

v = 2 1

(b) v1 = A2

2 A22 ( P1 − P2 )

(A − A ) 2 1

2 2

→

v1 = A2

2 ( P1 − P2 )

 ( A12 − A22 )

2 ( P1 − P2 )

 ( A12 − A22 )

=   12 ( 0.010 m )

 133 N m 2    mm Hg  = 0.18 m s 4 4 3 2 1 2 1  −  1000 kg m 0.035 m 0.010 m ( ) ( )     ( ) 2 2 2 (18 mm Hg ) 

(

)

61. There is a forward force on the exiting water, and so by Newton’s third law there is an equal force pushing backwards on the hose. To keep the hose stationary, you push forward on the hose, and so the hose pushes backwards on you. So the force on the exiting water is the same magnitude as the force on the person holding the hose. Use Newton’s second law and the equation of continuity to find the force. Note that the 480 L/min flow rate is the volume of water being accelerated per unit time. Also, the flow rate is the product of the cross-sectional area of the moving fluid, times the speed of the fluid, and so V t = A1v1 = A2 v2 . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

476

Chapter 13

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F =m

v t

v 2 − v1 t

V   V  A v A v  V   1 1  =    ( v 2 − v1 ) =     2 2 − 1 1  =     −  A1  t   t   A2  t   A2 A1  2

1  V   1 =    2 − 2   t    r2  r1  2

2  1min 1m 3  1  1 1  3 kg   480 L   =  1.00  10 3     − m   min 60 s 1000 L     12 ( 0.75  10−2 m ) 2  12 ( 7.0  10−2 m ) 2       

= 1432 N  1400 N

62. Apply Eq. 13–11 for the viscosity force. We use the inner radius to calculate the plate area.

     ( router − rinner ) v F l  rinner  → = = F =A l Av ( 2 rinner h )(  rinner )

 0.024 m N  −2  0.0510 m  ( 0.20  10 m )   = rev 2 rad 1min     2 ( 0.0510 m )( 0.150 m )  57  ( 0.0510 m ) rev 60 s   min = 6.432  10−2 Pa s = 6.4  10−2 Pa s 63. The relationship between velocity and the force of viscosity is given by v Eq. 13–11, Fvis =  A . The variable A is the area of contact between l the moving surface and the liquid. For a cylinder, A = 2 rh. The variable l is the thickness of the fluid layer between the two surfaces. See the diagram. If the object falls with terminal velocity, then the net force must be 0, and so the viscous force will equal the weight. Note that l = 12 (1.00 cm − 0.900 cm ) = 0.05 cm.

Fweight = Fvis → mg =  A

→

( 0.15 kg ) ( 9.80 m s 2 )( 0.050 10−2 m ) = = = 0.43 m s v=  A 2 rh 2 ( 0.450  10−2 m ) ( 0.300 m ) ( 200 10 −3 N s m 2 ) mg l

mg l

64. From Poiseuille’s equation, Eq. 13–12, the volume flow rate Q is proportional to R 4 if all other Q V 1 factors are the same. Thus 4 = is constant. If the volume of water used to water the garden R t R4 is to be the same in both cases, then tR 4 is constant. 4

 R1  3 8  = t1   = 0.13t1 5 8  R2 

t1 R14 = t 2 R24 → t 2 = t1 

Thus the time has been cut by 87% . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

477

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

65. Use Poiseuille’s equation, Eq. 13–12, to find the pressure difference.  R 4 ( P2 − P1 ) Q l Q= → ( P2 − P1 ) = 8 l  R4 −3 3  −3 L  1min   10 m   8 5.8  10 ( 0.2 Pa s )( 0.102 m ) min  60s   L    = 7700 Pa ( P2 − P1 ) = 4   12 (1.8  10−3 m ) 

66. The pressure drop per cm can be found from Poiseuille’s equation, Eq. 13–12, using a length of 1 cm. The volume flow rate is the area of the aorta times the speed of the moving blood.  R 4 ( P2 − P1 ) Q= → 8 l

( P2 − P1 ) l

8 Q

R

8 R 2 v

R

8 v R

(

)

8 4  10−3 Pa s ( 0.4 m s )

(1.2 10 m ) −2

= 88.9 Pa m = 0.89 Pa cm

67. Use Poiseuille’s equation, Eq. 13–12, to find the radius, and then double the radius to find the diameter.

 R 4 ( P2 − P1 ) 8 l

 8 l Q  → d = 2R = 2     ( P2 − P1 ) 

1/ 4

→

      8.0 m 14.0 m 4.0 m ( )( )( ) −5   8 (1.8  10 Pa s ) (18.0 m )  60 s       15 min    min     d = 2  5 −3  ( 0.710  10 atm )(1.013  10 Pa atm )          

1/ 4

= 0.098 m

68. Use Poiseuille’s equation, Eq. 13–12, to find the pressure difference.  R 4 ( P2 − P1 ) Q= → 8 l

( P2 − P1 ) =

(

)(

)

Q l

8 650 cm 3 s 10 −6 m 3 cm 3 ( 0.20 Pa s )(1600 m )

R

 ( 0.145 m )

= 4

= 1198 Pa  1200 Pa 69. (a) We calculate the Reynolds number with the given formula. −2  3 2vr  2 ( 0.35 m s ) 1.2  10 m 1.05  10 kg m = = 2205  2200 Re =  4  10−3 Pa s The flow is turbulent at this speed, although perhaps only slightly. (b) Doubling the velocity doubles the Reynolds number, to 4410. The flow is now definitely turbulent .

(

)(

)

478

Chapter 13

Fluids

70. From Poiseuille’s equation, the volume flow rate Q is proportional to R 4 if all other factors are the same. Thus Q R 4 is constant. Also, if the diameter is reduced by 25%, so is the radius.

Qfinal 4 final

Qinitial

→

4 initial

Qfinal

4 Rfinal 4 initial

= ( 0.75 ) = 0.32 4

R R Qinitial R The flow rate is 32% of the original value.

71. The fluid pressure must be 78 torr higher than air pressure as it exits the needle, so that the blood will enter the vein. The pressure at the entrance to the needle must be even higher than 78 torr, due to the viscosity of the blood. To produce that excess pressure, the blood reservoir is placed above the level of the needle. Use Poiseuille’s equation, Eq. 13–12, to calculate the excess pressure needed due to the viscosity, and then use Eq. 13–6b to find the height of the blood reservoir necessary to produce that excess pressure. We assume the flow rate is small enough that we can use the static pressure difference equation.  R 4 ( P2 − P1 ) 8 l Q 8 blood l Q  1  → P2 = P1 + blood4 =  blood g h → h = Q=  P1 +  8blood l R  blood g   R4 

   133 N m 2  78 mm-Hg ( )    1mm-Hg  +     1 3 −6  h =  2.0  10 m   = 1.04 m  1.0 m −3  1050 kg  9.80 m s 2  8 ( 4  10 Pa s ) ( 0.025 m )   ) 60 s  3 (    m   4 3 −    ( 0.4  10 m )   72. In Fig. 13–37, we have  = F 2 l . Use this to calculate the surface tension.

 =

F 2l

3.1 10−3 N 2 ( 0.070 m )

= 2.2  10−2 N m

73. As in Fig. 13–37, there are 2 surfaces being increased, so  = F 2 l . Use this to calculate the force.

 = F 2l

→ F = 2 l = 2 ( 0.025 N m )( 0.215 m ) = 1.1 10−2 N

74. From Example 13–17, we have that 2 r cos  = 16 mg . The maximum supported force will occur at

 = 0o . 2 r cos  = mg → mmax = 1 6

12 r g

(

)

12 3.0  10 −5 m ( 0.072 N m ) 9.80 m s 2

= 8.3  10 −6 kg

−5 The insect’s mass is 1.6  10 kg, which is much larger than the mass that the surface tension force could support. Thus the insect will not remain on top of the water .

75. As an estimate, we assume that the surface tension force acts vertically. We assume that the freebody diagram for the cylinder is similar to Fig. 13–39(a) in the text. The weight must equal the total surface tension force. The needle is of length l.

mg = 2 FT →  needle ( 12 d needle ) l g = 2 l 2

→

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

d needle =

8

 needle g

(

8 ( 0.072 N m )

) (

7800 kg m3  9.80 m s2

)

= 1.55  10−3 m  1.5 mm

76. Consider half of the soap bubble–a hemisphere. The forces on the hemisphere will be the surface tensions on the two circles and the net force from the excess pressure between the inside and the outside of the bubble. This net force is the sum of all the forces perpendicular to the surface of the hemisphere, but must be parallel to the surface tension. Therefore we can find it by finding the force on the circle that is the base of the hemisphere. The total force must be zero. Note that the forces FT outer and FT inner act over the entire length of the circles to which they are applied. The diagram may look like there are 4 tension forces, but there are only 2. Likewise, there is only 1 pressure force, FP, but it acts over the area of the hemisphere.

2 FT = FP → 2 ( 2 r ) =  r 2 P →

P =

Instructor Solutions Manual

FT outer FT inner

r FT inner

FT outer

4 r

77. The mass of liquid that rises in the tube will have the force of gravity acting down on it, and the force of surface tension acting upwards. The two forces must be equal for the liquid to be in equilibrium. The surface tension force is the surface tension times the circumference of the tube, since the tube circumference is the length of the “cut” in the liquid surface. The mass of the risen liquid is the density times the volume. FT = mg →  2 r =  r 2 hg → h = 2  gr 78. The difference in pressure from the heart to the calf can be derived from Eq. 13–6b.  1mm-Hg  P =  g h = 1.05  103 kg m 3 9.80 m s 2 (1m ) = 1.029  104 Pa   = 77.37 mm-Hg  133 Pa 

(

)(

)

 80 mm-Hg

This means the cuff would measure the person’s blood pressure to be 80 mm-Hg higher than it would read if attached at heart level. So a typical blood pressure of 120/70 would be measured as 200/150! 79. The pressures for parts (a) and (b) stated in this problem are gauge pressures, relative to atmospheric pressure. The pressure change due to depth in a fluid can be derived from Eq. 13–6b as P =  g h.

 133 N m 2   1mm-Hg  P  = = 0.71m (a) h = g  g 1kg 106 cm 3  2  1.00 cm 3  1000 g  1m 3  ( 9.80 m s )    9.81N m 2  ( 680 mm-H 2 O )   P  1mm-H 2 O  = = 0.68 m (b) h = g  g 1kg 106 cm 3  2  1.00 cm 3  1000 g  1m 3  ( 9.80 m s )  

( 52 mm-Hg ) 

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(c) For the fluid to just barely enter the vein, the fluid pressure must be the same as the blood pressure.  133 N m 2  ( 75 mm-Hg )   P  1mm-Hg  h = = = 1.0 m g  g 1kg 106 cm 3  2  1.00 cm3  1000 g  1m 3  9.80 m s  

(

)

80. (a) The fluid in the needle is confined, and so Pascal’s principle may be applied. 2 Fplunger Fneedle  r2 A rneedle = → Fneedle = Fplunger needle = Fplunger needle = Pplunger = Pneedle → F plunger 2 2  rplunger Aplunger Aneedle Aplunger rplunger

( 0.10 10 m ) = 7.6  10 N = ( 3.2 N ) ( 0.65 10 m ) −3

−2

−4

2  133 N m 2   ( 0.65  10−2 m ) = 1.3 N   1mm-Hg 

(b) Fplunger = Pplunger Aplunger = ( 75 mm-Hg ) 

81. The change in pressure with height can be derived from Eq. 13–6b. 1.29 kg m 3 9.80 m s 2 ( 380 m ) P  g h P =  g h → = = = 0.047 → P0 P0 1.013  105 Pa

(

)(

)

( about a 5% change )

P = 0.047 atm

82. The force can be found by multiplying the pressure times the area of the pump cylinder.

(

)

Fi = Pi A = 2.10  105 N m 2  ( 0.0125 m ) = 1.0  10 2 N

(

)

Ff = Pf A = 3.10  105 N m 2  ( 0.0125 m ) = 1.5  102 N 2

The range of forces is 100 N  F  150 N . 83. We first use the kinematical data to calculate the speed of the water as it leaves the nozzle. We have  0 = 45, a y = − g , y0 = 0.60 m, y = 0.0 m, and x = 3.0 m. Use projectile motion equations from Chapter 3.

x = v0 x t = v0 cos  0 t → t = y = y0 + v y 0 t + a y t 1 2

v0 =

x cos  0

x v0 cos  0

 x  → 0 = y0 + v0 sin  0 − g  v0 cos  0  v0 cos  0  x

2 ( y0 + x tan  0 )

3.0 m 2 2

1 2

9.80 m s 2

2 ( 0.60 m + 3.0 m )

→

= 4.9497 m s

Now that the speed of the water leaving the nozzle is known, the continuity equation can be used to calculate the speed of the water entering the nozzle. Let subscript “1” be entering the nozzle, and subscript “2” be leaving the nozzle. The speed of the water entering the nozzle is the speed of the water leaving the pump. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 ( d2 2)

Instructor Solutions Manual

d   0.953cm  = v2 A1v1 = A2 v2 → v1 = v2 = v2  2  = v2   = 0.3592v2 2 A1  ( d1 2 )  1.59 cm   d1  A2

Now we apply Bernoulli’s equation between the water leaving the pump at the ground, and the water leaving the nozzle into the air. The pressure upon leaving the nozzle is atmospheric pressure. Let subscript “1” be the pump location, and subscript “2” again be leaving the nozzle. P1 +  gy1 + 12  v12 = P2 +  gy2 + 12  v22 →

(

P1 = P2 +  g ( y2 − y1 ) + 12  v22 − v12

)

(

)

(

)

= 1.013  105 Pa + 1000 kg m 3  9.80 m s 2 ( 0.60 m ) + 12 ( 4.9497 m s ) 1 − 0.35922  2

= 1.1785  105 Pa  1.18  105 Pa = 1.16 atm 84. (a) The input pressure is equal to the output pressure. Pinput = Poutput → Finput Ainput = Foutput Aoutput → Ainput = Aoutput

Finput Foutput

(

=  9.0  10−2 m

N = 1.028  10 m ) ( 960 kg380 ) ( 9.80 m s ) 2

−3

 1.0  10−3 m 2 = 10 cm 2 (b) The work is the force needed to lift the car (its weight) times the vertical distance lifted. W = mgh = ( 960 kg ) ( 9.80 m s 2 ) ( 0.52 m ) = 4892 J  4900 J

(c) The work done by the input piston is equal to the work done in lifting the car. Winput = Woutput → Finput d input = Foutput d ouptut = mgh → h=

Finput d input mg

( 380 N )( 0.13 m ) = 5.251  10 −3 m  5.3  10 −3 m ( 960 kg ) ( 9.80 m s 2 )

(d) The number of strokes is the full distance divided by the distance per stroke. 0.52 m h hfull = Nhstroke → N = full = = 99 strokes hstroke 5.251  10 −3 m (e) The work input is the input force times the total distance moved by the input piston. Winput = NFinput d input → 99 ( 380 N )( 0.13m ) = 4891J  4900 J Since the work input is equal to the work output, energy is conserved. 85. The pressure change due to a change in height can be derived from Eq. 13–6b as P =  g h. That pressure is the excess force on the eardrum, divided by the area of the eardrum. F P =  g h = → A

(

F =  g hA = 1.29 kg m 3

)( 9.80 m s ) (1120 m ) ( 0.20  10 m ) = 0.2832 N  0.28 N

86. We use Eq. 13–6c.

P = P0e (

−  0 g P0 ) y

= (1.0atm ) e

−4

(

−4

− 1.2510 m

−1

)( 2400 m )

= 0.74 atm

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Chapter 13

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87. The change in pressure can be derived from Eq. 13–6b as P =  g h. Use the density of blood.

P =  g h →

P P0

 g h P0

(1.05 10 kg m )( 9.80 m s ) ( 6 m ) = 0.609 → = 3

1.013  105 Pa

P = 0.6 atm 88. The buoyant force on the wood must be equal to the combined weight of the wood and copper. m m ( mwood + mCu ) g = Vwood  water g = wood  water g → mwood + mCu = wood  water →  wood  wood

  water   1000 kg m3  − 1 = ( 0.40 kg )  − 1 = 0.27 kg 3  600 kg m    wood 

mCu = mwood 

89. (a) We assume that the one descending is close enough to the surface of the Earth that constant density may be assumed. Take Eq. 13–6b, modify it for rising, and differentiate it with respect to time. P = P0 −  gy → dP dt

= − g

(

)(

)

= − 1.29 kg m 3 9.80 m s 2 ( −7.0 m s ) = 88.49 Pa s  88 Pa s

dt y 350 m (b) y = v t → t = = = 50 s ( 2 sig. figs.) v 7.0 m s

90. Draw free-body diagrams for the brick (mass m) and for the tub of water (mass M). Note that the water exerts an upward buoyant force on the brick, so by Newton’s third law the brick exerts an equal downward buoyant force on the water. The normal force on the container of water is the scale reading. Then using Newton’s second law in equilibrium for the water, solve for the normal force on the water. The densities are found in Table 13–1.  m    water  Fbuoy =  waterVbrick g =  water  g =   mg

  brick 

F

water

  brick 

  water   mg + Mg   brick 

= 0 = FN − Fbuoy − Mg → FN = Fbuoy + Mg = 

 1.000  103 kg m 3  50 N + 100 N = 121.7 N  1.2  102 N 3 3   2.3  10 kg m 

FN = 

91. The buoyant force, equal to the weight of mantle displaced, must be equal to the weight of the continent. Let h represent the full height of the continent, and y represent the height of the continent above the surrounding rock. Wcontinent = Wdisplaced → Ah continent g = A ( h − y )  mantle g → mantle



y = h 1 −



 continent   2800 kg m3  = 35 km ( )   1 − 3300 kg m3  = 5.3km  mantle   

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

92. The “extra” buoyant force on the ship, due to the loaded fresh water, is the weight of “extra” displaced seawater, as indicated by the ship floating lower in the sea. This buoyant force is given by Fbuoyant = Vdisplaced  sea g. But this “extra” buoyant force is what holds up the fresh water, and so must water

also be equal to the weight of the fresh water.

(

)

(

)

Fbuoyant = Vdisplaced  sea g = mfresh g → mfresh = 2240 m 2 (8.55m ) 1025kg m 3 = 1.96  107 kg water

This can also be expressed as a volume. 1.96  107 kg m = 1.96  104 m3 = 1.96  107 L Vfresh = fresh = 3 3  fresh 1.00  10 kg m 93. We assume that the water is launched from the same level at which it lands. Then the level range v02 sin 2 0 formula, derived in Section 3–8, applies. That formula is R = . If the range has increased g by a factor of 4, then the initial speed has increased by a factor of 2. The equation of continuity is then applied to determine the change in the hose opening. The water will have the same volume rate of flow, whether the opening is large or small. vfully open 1 = ( Av )partly → Apartly = Afully = Afully   ( Av )fully open open open open v partly open  2  open

Thus 1 2 of the hose opening was blocked. 94. The buoyant force must be equal to the weight of the water displaced by the full volume of the logs, and must also be equal to the full weight of the raft plus the passengers. Let N represent the number of passengers. weight of water displaced by logs = weight of people + weight of logs

12 (Vlog  water g ) = Nmperson g + 12 (Vlog  log g ) → N=

12Vlog (  water −  log ) mperson

mperson

(

)

12 ( 0.225 m ) ( 6.5 m ) 1000 kg m 3 (1 − 0.60 ) 2

12 rlog2 llog (  water − SGlog  water )

68 kg

12 rlog2 llog  water (1 − SGlog ) mperson

= 72.97

Thus 72 people can stand on the raft without getting wet. When the 73rd person gets on, the raft will go under the surface. 95. Apply both Bernoulli’s equation and the equation of continuity at the two locations of the stream, with the faucet being location 0 and the lower position being location 1. The pressure will be air pressure at both locations. The lower location has y1 = 0 and the faucet is at height y0 = y.

A0 v0 = A1v1 → v1 = v0

A0 A1

= v0

 ( d0 2 )  ( d1 2 )

= v0 2

d 02 d12

→

484

Chapter 13

Fluids

P0 + 12  v02 +  gy0 = P1 + 12  v12 +  gy1 → v02 + 2 gy = v12 = v02

d 04 d14

→

1/ 4

 v2  d1 = d 0  2 0   v0 + 2 gy 

96. (a) The mass of water in the tube is the volume of the tube times the density of water.

(

) (

m = V =  r 2h = 1.00  103 kg m3  0.30  10−2 m

) (12 m ) = 0.3393kg  0.34 kg 2

(b) The net force exerted on the lid is the gauge pressure of the water times the area of the lid. The gauge pressure is found from Eqs. 13–2 and 13–3.

(

)(

)

F = Pgauge A =  gh R 2 = 1.00 103 kg m3 9.80 m s 2 (12 m )  ( 0.21m ) = 1.6 104 N 2

97. (a) The buoyant force on the object is equal to the weight of the fluid displaced. The force of gravity of the fluid can be considered to act at the center of gravity of the fluid (see Section 9–8). If the object were removed from the fluid and that space re-filled with an equal volume of fluid, that fluid would be in equilibrium. Since there are only two forces on that volume of fluid, gravity and the buoyant force, they must be equal in magnitude and act at the same point. Otherwise they would be a “couple” (see Fig. 12–4), exert a non-zero torque, and cause rotation of the fluid. Since the fluid does not rotate, we may conclude that the buoyant force acts at the center of gravity. (b) From the diagrams below, if the center of buoyancy (the point where the buoyancy force acts) is above the center of gravity (the point where gravity acts) of the entire ship, when the ship tilts, the net torque about the center of mass will tend to reduce the tilt. If the center of buoyancy is below the center of gravity of the entire ship, when the ship tilts, the net torque about the center of mass will tend to increase the tilt. Stability is achieved when the center of buoyancy is above the center of gravity. Fbuoy

Fbuoy

Fbuoy mg

No Torque (Stable)

CW Torque (Stabilizing)

CCW Torque (Stabilizing)

CW Torque (Unstabilizing)

98. (a) Consider the free body diagram for the horizontal orientation. There are 5 forces; the weight of each cube, the buoyancy force on each cube, and the upward normal force from your finger. Each weight force and buoyancy force act through the center of mass of the respective cube. The buoyancy force on each cube is the volume of the respective cube times the density of water times g, and the mass of each cube is the volume of the respective cube times the density of the material. Note that the volume of the aluminum cube is 8 × the volume of the iron cube, so we simply use VAl = 8VFe rather than calculating the volumes. We take © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

torques about an axis through the location of the normal force from your finger, so that we don’t need the magnitude of that force, and we take clockwise torques as positive. Counterclockwise torques are taken as a negative. We assume that the rod does not penetrate into either cube, but is affixed to the faces.



 =  F 









 Al



− mFe g  ( x + 0.005) −  Fbuoy − mAl g  ( 0.55 − x + 0.01) = 0

buoy Fe

(  V g −  V g ) ( x + 0.005) = (  V g −  V g ) ( 0.56 − x ) V (  −  ) ( x + 0.005) = V (  −  ) ( 0.56 − x ) H 2 O Fe

H 2O

Fe Fe

H 2O Al

H 2O

Al Al

(1000 − 7800 )( x + 0.005) = 8 (1000 − 2700 )( 0.56 − x ) → x = 0.3717 m  0.37 m (b) Now there are only four external forces, all acting in a vertical line–the two buoyancy forces pointing up, and the two weight forces pointing down. The vector sum of those forces will be the total mass times the vertical acceleration. Take down to be the positive direction.  F = mFe g − Fbuoy + mAl g − Fbuoy = ( mFe + mAl ) a Fe

mFe g − Fbuoy + mAl g − Fbuoy Fe

a= = =

mFe + mAl

 FeVFe g −  H OVFe g +  AlVAl g −  H OVAl g

 FeVFe +  AlVAl

(  −  )V + (  −  ) 8V g = (  −  ) + 8 (  −  ) g Fe

H 2O

 FeVFe +  Al 8VFe

H 2O

 Fe + 8 Al

( 7800 − 1000 ) + 8 ( 2700 − 1000 ) 9.80 m s 2 ) = 6.8 m s 2 ( 7800 + 8 ( 2700 )

(c) If the objects were NOT connected by the rod, the acceleration of the Fe would be found from mFe g − Fbuoy = mFe aFe , and the acceleration of the Al from mAl g − Fbuoy = mAl aAl. Fe

mFe g − Fbuoy aFe = =

aAl = =

mFe

( 7800 − 1000 ) 7800 mAl g − Fbuoy Al

mAl

( 2700 − 1000 )

 FeVFe g −  H OVFe g 2

 FeVFe

(  −  )V g = (  −  ) g Fe

 AlVAl g −  H OVAl g 2

 AlVAl

H2O

 Fe

(  −  )V g = (  −  ) g

( 9.80 m s ) = 6.2 m s 2

 FeVFe

( 9.80 m s ) = 8.5 m s 2

H2O

 AlVAl

H 2O

 Al

2700 Thus, if the mass gets the slightest bit away from the vertical, the Fe would have a larger acceleration, and the Al a smaller acceleration, meaning that the Fe block would eventually be below the Al block. The objects would exchange positions from Fig. 13–66b.

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Chapter 13

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99. We assume that the air pressure is due to the weight of the atmosphere, with the area equal to the surface area of the Earth. F → F = PA = mg → P= A

PA g

2 4 REarth P

(

4 6.38  106 m

) (1.013 10 N m ) = 5.29 10 kg 2

9.80 m s

100. The work done during each heartbeat is the force on the fluid times the distance that the fluid moves in the direction of the force. That can be converted to pressure times volume. W = F l = PAl = PV →

Power =

W t

PV t

 133 N m 2  3 −6  ( 70  10 m ) 1 mm-Hg   = 1.14 W  1W 1 60  min  s      70  min 

(105 mm-Hg ) 

101. We assume that there is no appreciable height difference to be considered between the two sides of the window. Then the net force on the window due to the air is the difference in pressure on the two sides of the window, times the area of the window. The difference in pressure can be found from Bernoulli’s equation. 2 2 Pinside + 12  vinside +  gyinside = Poutside + 12  voutside +  gyoutside → 2 = Pinside − Poutside = 12  air voutside

Fair Awindow

→ 2

Fair =  air v 1 2

2 outside

  1m s   9.0 m 2 ) = 1.5  10 4 N Awindow = (1.29 kg m ) (180 km h )  (    3.6 km h    3

1 2

102. We label three vertical levels. Level 0 is at the pump, and the supply tube has a radius of r0 at that location. Level 1 is at the nozzle, and the nozzle has a radius of r1 . Level 1 is a height h1 = 1.1m above Level 0. Level 2 is the highest point reached by the water, and is a height h2 = 0.12 m above Level 1. We may write Bernoulli’s equation relating any 2 of the levels, and we may write the equation of continuity relating any 2 of the levels. The desired result is the gauge pressure of the pump, which would be P0 − Patm . Start by using Bernoulli’s equation to relate Level 0 to Level 1.

P0 +  gh0 + 12  v02 = P1 +  gh`1 + 12  v12 Since Level 1 is open to the air, P1 = Patm . Use that in the above equation.

P0 − Patm =  gh`1 + 12  v12 − 12  v02 Use the equation of continuity to relate Level 0 to Level 1, and then use that result in the Bernoulli expression above. r2 d2 A0 v0 = A1v1 →  r02 v0 =  r12 v1 → v0 = 12 v1 = 12 v1 r0 d0 2

 d2   d4  P0 − Patm =  gh`1 +  v −   12 v1  =  gh`1 + 12  v12  1 − 14   d0   d0  1 2

2 1

1 2

487

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Use Bernoulli’s equation to relate Levels 1 and 2. Since both levels are open to the air, the pressures are the same. Also note that the speed at Level 2 is zero. Use that result in the Bernoulli expression above. P1 +  gh1 + 12  v12 = P2 +  g ( h1 + h2 ) + 12  v 22 → v12 = 2 gh2

  d14   d14   P0 − Patm =  gh`1 +  v  1 − 4  =  g  h`1 + h2  1 − 4    d0   d0   2 1

1 2

(

= 1.00  103 kg m 3

)( 9.80 m s ) 1.1m + ( 0.12 m ) (1 − 0.5 ) 2

= 11883 N m 2  1.2  104 N m 2 103. (a) We assume that the water is launched at ground level. Since it also lands at ground level, the level range formula derived in Section 3–8 may be used.

v02 sin 2

( 5.0 m ) ( 9.80 m s 2 )

→ v0 = = = 7.221m s  7.2 m s sin 2 sin 70 g (b) The volume rate of flow is the area of the flow times the speed of the flow. Multiply by 4 for the 4 heads.

(

Volume flow rate = Av = 4 r 2 v = 4 1.5  10 −3 m

) ( 7.221m s ) 2

1L    60 s   = 12.252 L min  12 L min 3  −3  1.0  10 m   1min 

= 2.042  10 −4 m 3 s 

(c)

Use the equation of continuity to calculate the fluid speed in the supply pipe. ( Av )heads 2.042 10−4 m3 s = = 0.72 m s ( Av )supply = ( Av )heads → vsupply = 2 Asupply  0.95  10−2 m

(

)

104. From Poiseuille’s equation, the viscosity can be found from the volume flow rate, the geometry of the tube, and the pressure difference. The pressure difference over the length of the tube is the same as the pressure difference due to the height of the reservoir, assuming that the open end of the needle is at atmospheric pressure.  R 4 ( P2 − P1 )  R 4 ( P2 − P1 )  R 4  blood gh = ; P2 − P1 =  blood gh →  = Q= 8 l 8Ql 8Ql

 ( 0.20  10 −3 m ) (1.05  103 kg m 3 )( 9.80 m s 2 ) (1.40 m ) 4

=

 cm3 1 min 10−6 m 3  −2   8  4.1 3  ( 3.8  10 m ) min 60 s cm  

= 3.5  10 −3 Pa s

105. The net force is 0 if the balloon is moving at terminal velocity. Therefore the upwards buoyancy force (equal to the weight of the displaced air) must be equal to the net downwards force of the weight of the balloon material, plus the weight of the helium, plus the drag force at terminal velocity. Find the terminal velocity, and use that to find the time to rise 12 m. FB = mballoon g + mHelium g + FD → 43  r 3  air g = mballoon g + 43  r 3  He g + 12 CD  air  r 2 vT2 →

vT =

2  34  r 3 (  air −  He ) − m  g CD  air r 2

h t

→ t=h

CD  air r 2

2  43  r 3 (  air −  He ) − m  g

→

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Chapter 13

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( 0.47 ) (1.29 kg m3 )  ( 0.15 m )2 t = (15 m ) 3 2  43  ( 0.15 m ) (1.29 kg m 3 − 0.179kg m 3 ) − ( 0.0028 kg )  ( 9.80 m s 2 ) = 6.174 s  6.2 s

489

CHAPTER 14: Oscillations Responses to Questions 1.

Examples are: a child’s swing (SHM, for small oscillations), stereo speakers (complicated motion, the addition of many SHMs), the blade on a jigsaw (approximately SHM), the string on a guitar (complicated motion, the addition of many SHMs).

The acceleration of a simple harmonic oscillator is instantaneously zero as the oscillating object passes through the equilibrium point. At this point, there is no force on the oscillating object and therefore no acceleration, by Newton’s second law.

Since the real spring has mass, the mass that is moving is greater than the mass at the end of the spring. Since f =

, a larger mass means a smaller frequency. Thus, the true frequency will 2 m be smaller than the “massless spring” approximation. And since the true frequency is smaller, the true period will be larger than the “massless spring” approximation. About 1/3 the mass of the spring contributes to the total mass value. 4.

The maximum speed is given by vmax = A k m . Various combinations of changing A, k, and/or m can result in a doubling of the maximum speed. For example, if k and m are kept constant, then doubling the amplitude will double the maximum speed. Or, if A and k are kept constant, then reducing the mass to 1/4 its original value will double the maximum speed. Note that changing

2 frequency (no matter how it is done) will also double the maximum speed.

either k or m will also change the frequency of the oscillator, since f =

. So doubling the

Before the trout is released, the scale reading is zero. When the trout is released, it will fall downward, stretching the spring to beyond its equilibrium point so that the scale reads something over 5.0 kg. Then the spring force will pull the trout back up, to a point above the equilibrium point, so that the scale will read something less than 5.0 kg. The spring will undergo damped oscillations about equilibrium and eventually come to rest at equilibrium. The corresponding scale readings will oscillate about the 5.0-kg mark, and eventually come to rest at 5.0 kg.

The period of a pendulum clock is inversely proportional to the square root of g, by Eq. 14–12c,

T = 2 l g . When taken to high altitude, g will decrease (by a small amount), which means the period will increase. If the period is too long, that means the clock is running slow and so will lose time. 7.

The tire swing is a good approximation of a simple pendulum. Pull the tire back a short distance and release it, so that it oscillates as a pendulum in simple harmonic motion with a small amplitude. Measure the period of the oscillations and calculate the length of the pendulum from Eq. 14–12c, gT 2 T = 2 l g → l = . The length, l, is the distance from the center of the tire to the branch. The 2

4

height of the branch is l plus the height of the center of the tire above the ground.

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The displacement and velocity vectors are in the same direction while the oscillator is moving away from its equilibrium position. The displacement and acceleration vectors are never in the same direction.

The two masses reach the equilibrium point simultaneously. The period of oscillation is independent of amplitude and will be the same for both systems.

10. The car will bounce faster when empty. The period of the oscillation of a spring increases with increasing mass, so when the car is empty the period of the harmonic motion of the springs will be shorter, and the car will bounce faster. 11. When you rise to a standing position, you raise your center of mass and effectively shorten the length of the swing. The period of the swing will decrease, and the frequency will increase. 12. The frequency will decrease. For a physical pendulum, the period is proportional to the square root of the moment of inertia divided by the mass. When the small sphere is added to the end of the rod, both the moment of inertia and the mass of the pendulum increase. However, the relative increase in the moment of inertia will be greater because the added mass is located farther from the axis of rotation than the center of mass of the rod. Therefore, the period will increase and the frequency will decrease. 13. When walking at a normal pace, the period of a walking step is about 1 second. The faster you walk, the shorter the period. The shorter your legs, the shorter the period. If you approximate your leg as a simple pendulum of length 1 m, then the period would be T = 2 l g  2 seconds. 14. When the 264-Hz fork is set into vibration, the sound waves generated are close enough in frequency to the resonance frequency of the 260-Hz fork to cause it to vibrate. The 420-Hz fork has a resonance frequency far from 264 Hz and far from the harmonic at 528 Hz (discussed in Section 15–9), so it will not begin to vibrate. 15. To make the water “slosh,” you must shake the water (and the pan) at the natural frequency for water waves in the pan. The water then is in resonance, or in a standing wave pattern, and the amplitude of oscillation gets large. That natural frequency is determined in part by the size of the pan–smaller pans will slosh at higher frequencies, corresponding to shorter wavelengths for the standing waves. The period of the shaking must be the same as the time it takes a water wave to make a “round trip” in the pan. 16. Examples of resonance are; pushing a child on a swing (if you push at one of the limits of the oscillation), blowing across the top of a bottle, producing a note from a flute or organ pipe. 17. Forced resonance and underdamping are factors in the pronounced rattle and vibration. Forced resonance becomes a factor because hitting the bumps in the road is a “forcing” action on the car. Assuming that the bumps are regularly spaced and the car has a constant speed, then that “forcing” will have a particular frequency. The forcing becomes “forced resonance” when the frequency of hitting the road bumps matches with a resonant frequency of some part in the car. For example, if the frequency at which the car hits the bumps in the road is in resonance with the natural frequency of the springs, the car will have a large amplitude of oscillation, and so shake strongly. If those springs are underdamped, the oscillation resulting from hitting one “bump” are still occurring when you hit the next bump, and so the amplitude increases with each bump. A significant oscillation amplitude can be built up. If the springs were critically damped or over-damped, the oscillation would not build up–it would dampen before the next bump was encountered © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

491

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

18. Building with lighter materials doesn’t necessarily make it easier to set up resonance vibrations, but it does shift the fundamental frequency and decrease the ability of the building to dampen oscillations. With less damping, resonance vibrations could be more noticeable and more likely to cause damage to the structure.

Solutions to MisConceptual Questions 1.

(e) At x =  A, the velocity is zero, but at these points the acceleration is a maximum. At x = 0, the acceleration is zero, but the velocity is a maximum. For all other points the velocity and acceleration are both non-zero. Thus, there are no points where both the acceleration and velocity are simultaneously zero.

(a, c, d) At the turning points in the oscillation ( x =  A) , the velocity is zero and the acceleration is a maximum so (a) is true. At the center of the oscillation ( x = 0 ) the acceleration is zero and the velocity is a maximum, so (c) is true. Since the velocity is only zero at the turning points where the acceleration is maximum, there is no point where both the velocity and acceleration are zero, so (b) is not true. At all other points besides x =  A and x = 0 the acceleration and velocity are both non-zero values, so (d) is also true.

(c) Students may believe that the period is proportional to the mass, and therefore think that doubling the mass will double the period. However, the period is proportional to the square root of the mass, as seen in Eq. 14–7b. Therefore the mass must be quadrupled (to 4M) for the period to double.

(b) A common misconception is that the amplitude of oscillation affects the frequency. Eq. 14–7a shows that the frequency can be increased by increasing k or decreasing m. The frequency does not depend upon the amplitude.

(e) The energy of the system is given by E = 12 kA2 . Since the energy depends on the square of the amplitude, when the amplitude is doubled, the energy is multiplied by a factor of 4.

(c) The period will be unchanged, so the time will be (c), 2.0 s. The period of a simple pendulum oscillating with a small amplitude does not depend on the mass.

(b, c, d) The period is given by Eq. 14–7b, T = 2

(a) The small angle approximation is only valid in units of radians because the angle in radians is equal to the ratio of the arc length to the radius. At small angles the arc length can be approximated as a straight line, being the opposite leg of a right triangle with hypotenuse equal

. Since the ratio of mass to spring constant k is the same for both springs, the period, frequency, and angular frequency are the same for both systems. The maximum speed and maximum acceleration depend on the amplitude and the angular frequency. Since the amplitude and angular frequency are the same for both, the maximum speed and maximum acceleration are the same. The total energy depends on the amplitude and the spring constant. Since the amplitudes are the same but the spring constants are different, the total energy of the two systems is different.

492

Chapter 14

Oscillations

to the radius of the circle. This ratio is equal to the sine of the angle, so for small angles the angle in radians is equal to the sine of the angle. 9.

(c) Students may erroneously believe that the mass of the student will affect the period of oscillation. However, Eq. 14–12c shows that the period is determined by the length of the swing cords and the acceleration of gravity. It does not depend upon the weight of the child. Since the swings are identical, they should oscillate with the same period.

10. (a) To speed up the pendulum, the period of the oscillation must be decreased. Eq. 14–12c shows that the period is proportional to the square root of the length, so shortening the string will decrease the period. The period does not depend upon the mass of the bob, so changing the mass will not affect the period. 11. (d) The period of a simple pendulum is given by Eq. 14–2c, T = 2

. Note that it does not g depend on the mass. Thus if the periods are the same, the lengths must be the same.

Solutions to Problems 1.

The particle would travel four times the amplitude; from x = A to x = 0 to x = − A to x = 0 to

x = A . So the total distance = 4 A = 4 ( 0.27 m ) = 1.08 m . The extra significant figure is allowable since we could calculate the distance as A + A + A + A. 2.

The spring constant is found from the ratio of applied force to displacement. 2 Fext mg ( 72 kg ) 9.80 m s k= = = = 1.411  105 N m −3 5.0  10 m x x The frequency of oscillation is found from the total mass and the spring constant.

(

f = 3. (a)

2

)

1.411  105 N m

2

1772 kg

= 1.420 Hz  1.4 Hz

The motion starts at the maximum extension, and so the equation is most simply written as a cosine. The amplitude is the displacement at the start of the motion.  2   2  x = A cos (  t ) = A cos  t  = ( 7.8 cm ) cos  t  = ( 7.8 cm ) cos ( 9.520t ) T   0.66 

 ( 7.8 cm ) cos ( 9.5 t ) (b) Evaluate the position function at t = 1.8 s. x = ( 7.8cm ) cos 9.520s −1 (1.8s )  = −1.110 cm  −1.1cm 4.

The spring constant is the ratio of external applied force to displacement. 210 N − 68 N 142 N F k = ext = = = 591.7 N m  590 N m 0.85 m − 0.61m 0.24 m x

493

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

The period is 1.6 seconds, and the mass is 32 kg. The spring constant is calculated from Eq. 14–7b. T = 2

Instructor Solutions Manual

m k

→ T 2 = 4 2

→ k = 4 2

m T

= 4 2

32 kg

= 493.5 N m  490 N m

(1.6s )2

(a) The spring constant is found from the ratio of applied force to displacement. 2 Fext mg ( 2.4 kg ) 9.80 m s k= = = = 735 N m  740 N m 0.032 m x x

(

)

(b) The amplitude is the distance pulled down from equilibrium, so A = 2.1cm . The frequency of oscillation is found from the oscillating mass and the spring constant. f =

2

735 N m

2

2.4 kg

= 2.785 Hz  2.8 Hz

The relationship between frequency, mass, and spring constant is Eq. 14–7a, f = (a)

f =

(b)

f =

2

(

0.1769 N m

2

4.6  10−4 kg

The spring constant is the same regardless of what mass is attached to the spring.

2

= 3.1Hz

2

)

The table of data is shown along with the smoothed graph. Every quarter of a period, the mass moves from an extreme point to the equilibrium. The graph resembles a cosine wave (actually, the opposite of a cosine wave).

→ k = 4 2 f 2 m = 4 2 ( 4.0 Hz ) 2.8  10−4 kg = 0.1769 N m  0.18 N m

f =

→

k 4

= mf 2 = constant → m1 f12 = m2 f12 →

( m kg )( 0.83 Hz )2 = ( m kg + 0.83kg )( 0.60 Hz )2

→ m=

( 0.83kg )( 0.60 Hz )2 = 0.91kg ( 0.83 Hz )2 − ( 0.60 Hz )2

10. (a) We find the effective spring constant from the mass and the frequency of oscillation.

f =

2

→

k = 4 2 mf 2 = 4 2 ( 0.052 kg )( 3.5 Hz ) = 25.15 N m  25 N m 2

494

Chapter 14

Oscillations

(b) Since the objects are the same size and shape, we anticipate that the spring constant is the same.

f =

2

25.15 N m

2

0.28 kg

= 1.5 Hz

11. The spring constant can be found from the stretched distance corresponding to the weight suspended on the spring. 2 Fext mg (1.45 kg ) 9.80 m s k= = = = 58.0 N m 0.245 m x x After being stretched further and released, the mass will oscillate. It takes one-quarter of a period for the mass to move from the maximum displacement to the equilibrium position.

(

1 4

T = 14 2 m k =

)

 1.45 kg = 0.248s 2 58.0 N m

12. When the object is at rest, the magnitude of the spring force is equal to the force of gravity. This determines the spring constant. The period can then be found.  Fvertical = kx0 − mg → k = mg x0 T = 2

m k

= 2

( mg x0 )

= 2

0.14m

= 0.75s

9.80 m s 2

13. (a) For A, the amplitude is AA = 2.5 m . For B, the amplitude is AB = 3.5 m . (b) For A, the frequency is 1 cycle every 4.0 seconds, so f A = 0.25 Hz . For B, the frequency is 1 cycle every 2.0 seconds, so f B = 0.50 Hz . (c) For A, the period is TA = 4.0 s . For B, the period is TB = 2.0 s . (d) Object A has a displacement of 0 when t = 0 , so it is most conveniently expressed as a sine function.

xA = AAsin ( 2 f At ) →

xA = ( 2.5m ) sin ( 12  t )

Object B has a maximum displacement when t = 0 , so it is most conveniently expressed as a cosine function.

xB = ABcos ( 2 f Bt ) →

xB = ( 3.5m ) cos ( t )

14. (a) From the graph, the period is 0.69 s. The period and the mass can be used to find the spring constant.

T = 2

m k

→ k = 4 2

m T

= 4 2

0.0077 kg

( 0.69 s )2

= 0.6385 N m  0.64 N m

(b) From the graph, the amplitude is 0.82 cm. The phase constant can be found from the initial conditions.  2 t +   = 0.82 cm cos  2 t +   x = A cos  )   (   T   0.69  x ( 0 ) = ( 0.82 cm ) cos  = 0.43cm →  = cos −1

0.43 0.82

= 1.02 rad

495

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Because the graph is shifted to the RIGHT from the 0-phase cosine, the phase constant must be subtracted.

 2 t − 1.0  or 0.82 cm cos 9.1 t − 1.0 ( ) ( )   0.69 

x = ( 0.82 cm ) cos 

15. For this motion, we have k = 265 N m , A = 0.280 m, m = 0.350 kg, and  = k m =

( 265 N m ) 0.350 kg = 27.52 rad s. (a) Since the mass has a zero displacement and a positive velocity at t = 0, the equation is a sine function.

y ( t ) = ( 0.280 m ) sin ( 27.5rad s ) t  2

(b) The period of oscillation is given by T =

2

= 0.22831s. The spring will have  27.52 rad s its maximum extension at times given by the following. T tmax = + nT = 5.71  10−2 s + n ( 0.228s ) , n = 0,1, 2, 4 The spring will have its minimum extension at times given by the following. 3T tmin = + nT = 1.71  10−1 s + n ( 0.228s ) , n = 0,1, 2, 4

16. Eq. 14–4 is x = Acos (  t +  ) . (a) If x ( 0 ) = − A , then − A = A cos  →  = cos −1 ( −1) →  =  . (b) If x ( 0 ) = 0 , then 0 = A cos  →  = cos −1 ( 0 ) →  =  12  . (c) If x ( 0 ) = A , then A = A cos  →  = cos −1 (1) →  = 0 . (d) If x ( 0 ) = 12 A , then 12 A = A cos  →  = cos −1 ( 12 ) →  =  13  . (e) If x ( 0 ) = − 12 A , then − 12 A = A cos  →  = cos−1 ( − 12 ) →  =  23  . (f)

If x ( 0) = A

2 = A cos  →  = cos −1

2 , then A

( ) → =  . 1 2

1 4

The ambiguity in the answers is due to not knowing the direction of motion at t = 0. 17. (a) The maximum speed is derived from Eq. 14–9a. vmax =  A = 2 f A = 2 ( 441Hz ) 1.8  10 −3 m = 5.0 m s .

(

)

(b) The maximum acceleration is given by Eq. 14–9b.

(

)

amax =  2 A = 4 2 f 2 A = 4 2 ( 441Hz ) 1.8 10−3 m = 1.4 104 m s 2 . 2

18. (a) The period and frequency are found from the angular frequency.

 = 2 f → f =

1 2

=

1 5 2 4

5 8

1 f

= 1.6 s

496

Chapter 14

Oscillations

(b) The velocity is the derivative of the position. dx  5    5   5   v= x = ( 3.8 m ) cos  t+  = − ( 3.8 m )   sin  t +  6 6 dt  4  4   4

   = 3.3 m 6

 5      sin   = −7.5 m s  4  6

x ( 0 ) = ( 3.8 m ) cos 

v ( 0 ) = − ( 3.8 m ) 

dv  5   5    5   5   a= = − ( 3.8 m )   sin  t +   cos  t +  6 6 dt  4   4  4   4   5   5 v ( 2.5 ) = − ( 3.8 m )   sin  ( 2.5 ) +  = 12 m s 6  4   4 v = − ( 3.8 m ) 

  5   5 2  cos  ( 2.5 ) +  = 36 m s 6  4   4

a ( 2.5 ) = − ( 3.8 m ) 

19. (a) The object starts at the maximum displacement in the positive direction, and so will be represented by a cosine function. The mass, period, and amplitude are given.  2 A = 0.16 m ;  = = = 9.24 rad s → y = ( 0.16 m ) cos ( 9.2 t ) T 0.68s (b) The time to reach the equilibrium is one-quarter of a period, so 14 ( 0.68s ) = 0.17 s . (c) The maximum speed is given by Eq. 14–9a. vmax =  A = ( 9.24 rad s )( 0.16 m ) = 1.5 m s (d) The maximum acceleration is given by Eq. 14–9b.

amax =  2 A = ( 9.24 rad s ) ( 0.16 m ) = 14 m s 2 2

The maximum acceleration occurs at the endpoints of the motion, and is first attained at the release point. 20. Each object will pass through the origin at the times when the argument of its sine function is a multiple of  . A: 4.0 tA = nA → tA = 14 nA , nA = 1, 2,3, so tA = 14  , 12  , 43  ,  , 45  , 23  , 74  , 2 , 94  , B: 3.0 tB = nB → tB = 13 nB , nB = 1, 2,3,

so tB = 13  , 23  ,  , 43  , 53  , 2 , 73  , 83  ,3 ,

Thus we see the first three times are  s, 2 s, 3 s or 3.1s, 6.3s, 9.4 s . 21. Consider the first free-body diagram for the block while it is at equilibrium, so that the net force is zero. Newton’s second law for vertical forces, with up as positive, gives this.  Fy = FA + FB − mg = 0 → FA + FB = mg

FA

FB x

Now consider the second free-body diagram, in which the mg block is displaced a distance x from the equilibrium point. mg Each upward force will have increased by an amount − kx , since x  0. Again write Newton’s second law for vertical forces.  Fy = Fnet = FA + FB − mg = FA − kx + FB − kx − mg = −2kx + ( FA + FB − mg ) = −2kx

497

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

This is the general form of a restoring force that produces SHM, with an effective spring constant of 2k . Thus the frequency of vibration is as follows.

f =

1 2

keffective m =

2

22. The period of the jumper’s motion is T =

28.0 s 5 cycles

= 5.60 s. The spring constant can then be found

from the period and the jumper’s mass.

T = 2

m k

4 2 m

→ k=

4 2 ( 65.0 kg )

( 5.60s )2

= 81.827 N m  81.8 N m

The stretch of the bungee cord needs to provide a force equal to the weight of the jumper when he is at the equilibrium point. 2 mg ( 65.0 kg ) ( 9.80 m s ) k x = mg → x = = = 7.78 m 81.827 N m k Thus the unstretched bungee cord must be 25.0 m − 7.8 m = 17.2 m . 23. We assume that the spring is stretched some distance y 0 while the rod is in equilibrium and horizontal, to hold the rod level, as in the first figure. Calculate the net torque about an axis through the hinge while the object is in equilibrium, with clockwise torques as positive. Mg  = Mg ( 12 l ) − Fs 0 l = 12 Mg l − ky0 l = 0 → y0 = 2k Now consider the end of the rod being displaced an additional distance y below the horizontal, so that the rod makes a small angle of  as shown in the free-body diagram. Take torques about the same axis. If  is small, then there has been no appreciable horizontal displacement of the rod. 2 2 d  1 1 1 Mg F Mg k y y I M   = l − l = l − + l = = l ( (2 ) s 2  0) 3 dt 2 Include the equilibrium condition, and the approximation that y = l sin   l  . 1 2

Mg l − ky l − ky0 l = M l

− k l 2 = 13 M l 2

1 3

d 2 dt

→

d 2 dt 2

d 2 dt

→ 3k M

1 2

Mg l − ky l − Mg l = M l 1 2

1 3

d 2 dt 2

3k M

→ f =



Fs Mg

→

 =0

This is the equation for simple harmonic motion, corresponding to Eq. 14–3, with  2 =

 2 = 4 2 f 2 =

Fs0

2

3k M

24. The impulse, which acts for a very short time, changes the momentum of the mass, giving it an initial velocity v 0 . Because this occurs at the equilibrium position, this is the maximum velocity of the mass. Since the motion starts at the equilibrium position, we represent the motion by a sine function. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 14

Oscillations

J = p = mv = mv0 − 0 = mv0 → v0 = J m

k m

→ A=

J m

→

 k

→ x = A sin t =

= v max = A = A



sin 

t

 m 

25. (a) If the block is displaced a distance x to the right in Figure 14–33a, then the length of spring #1 will be increased by a distance x1 and the length of spring #2 will be increased by a distance x2 , where x = x1 + x2 . The force on the block can be written F = − keff x. Because the springs are massless, they act similar to a rope under tension, and the same force F is exerted by each spring. Thus F = − keff x = − k1 x1 = − k 2 x2 .

x = x1 + x2 = − T = 2

m keff

F k1

−

F k2

1

= −F 

 k1

1

= 2 m 

 k1

1

=−k k2  eff

→

1 keff

1 k2

1



k2 

(b) The block will be in equilibrium when it is stationary, and so the net force at that location is zero. Then, if the block is displaced a distance x to the right in the diagram, then spring #1 will exert an additional force of F1 = − k1 x, in the opposite direction to x. Likewise, spring #2 will exert an additional force F2 = − k 2 x, in the same direction as F1 . Thus the net force on the

displaced block is F = F1 + F2 = − k1 x − k 2 x = − ( k1 + k 2 ) x. The effective spring constant is thus k = k1 + k 2 , and the period is given by T = 2

m k

= 2

m k1 + k 2

26. The energy can be written in terms of the spring constant and the amplitude. Both oscillations have the same amplitude. 2 E1 = E2 → 2 12 k1 A2 = 12 k2 A2 → 2k1 = k2

(

)

The second spring constant is twice the first spring constant. 27. The various values can be found from the equation of motion, x = A cos t = 0.650 cos 7.40t. (a) The amplitude is the maximum value of x, and so A = 0.650 m .

 7.40 rad s = = 1.18 Hz . 2 2 rad (c) The total energy can be found from the maximum potential energy. 2 2 E = U max = 12 kA2 = 12 m 2 A2 = 12 (1.28 kg )( 7.40 rad s ) ( 0.650 m ) = 14.807 J  14.8 J (b) The frequency is f =

(d) The potential energy can be found from U = 12 kx 2 , and the kinetic energy from E = U + K . U = 12 kx 2 = 12 m 2 x 2 = 12 (1.28 kg )( 7.40 rad s ) ( 0.260 m ) = 2.369 J  2.37 J 2

K = E − U = 14.807 J − 2.369 J = 12.438 J  12.4 J

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28. The spring constant can be found from the stretch that the mass causes when hung vertically from the spring. The magnitude of the spring force is equal to the weight of the mass. ( 5.5 kg ) 9.80 m s2 mg kxequilibrium = mg → k = = = 215.6 N m xequilibrium 0.25 m

(

)

Use conservation of energy to determine its velocity at the equilibrium position, since all the forces doing work are conservative forces. Take the lowest point of the spring (12 cm below the equilibrium position) to be the 0 location for gravitational potential energy. The unstretched position of the spring, 25 cm above the equilibrium position, has to be taken as the 0 location for elastic potential energy. Indicate the lowest position with subscript “1” and the equilibrium position with subscript “2.” At the lowest position the system only has elastic potential energy. At the equilibrium position, there is kinetic energy, elastic potential energy, and gravitational potential energy. 2 2 2 E1 = E2 → 12 kxlowest = 12 mvequilibrium + 12 kxequilibrium + mgyequilibrium → 1 2

2 2 2 mvequilibrium = 12 kxlowest − 12 kxequilibrium − mgyequilibrium

vequilibrium = =

)

2 − xequilibrium − 2 gyequilibrium

2 lowest

215.6 N m 

(

5.5 kg

0.37m ) − ( 0.25 m )  − 2 ( 9.80 m s 2 ) ( 0.12 m ) = 0.7513 m s  0.75 m s  2

Note that this is the exact same result if we simply ignore gravitational potential energy, and treat this as a spring whose “unstretched, uncompressed” position is the equilibrium position of the hanging mass. The amplitude of the oscillation is then 0.12 m. 2 → E1 = E2 → 12 kA2 = 12 mvequilibrium

vequilibrium = A

= ( 0.12 m )

215.6 N m

= 0.7513 m s  0.75 m s m 5.5 kg The effect of gravity can be removed by redefining the equilibrium position of the mass-spring system as “offset” by gravity. 29. To compare the total energies, we can compare the maximum potential energies. Since the frequencies and the masses are the same, the spring constants are the same. 2 2 1 Ehigh kAhigh Ahigh Ahigh 2 energy energy energy energy = 1 2 = 2 =3 → = 3 Elow kA A A low low low 2 energy

energy

30. (a) The total energy is the maximum potential energy.

U = 12 E →

1 2

kx 2 = 12

( kA ) → x = A 1 2

2  0.707 A

(b) Now we are given that x = 13 A.

U E

x2 1 = = 1 kA2 A2 9 2 1

= 2

kx 2

Thus the energy is divided up into

1 9

potential and 98 kinetic .

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31. The total energy can be found from the spring constant and the amplitude.

E = 12 kA2 = 12 ( 95 N m )( 0.020 m ) = 0.019 J The total energy is represented by the horizontal line on the graph. (a) From the graph, at x = 1.5cm, we have U  0.011J . 2

(b) From energy conservation, at x = 1.5cm, we have K = E − U  0.008 J . (c) Find the speed from the estimated kinetic energy. K = 12 mv 2 →

2K m

2 ( 0.008 J ) 0.075 kg

= 0.46 m s  0.5 m s

32. (a) At equilibrium, the velocity is its maximum. Use Eq. 14–9a, and realize that the object can be moving in either direction. vmax =  A = 2 fA = 2 ( 3.2 Hz )( 0.15 m ) = 3.016 m s → vequib  3.0 m s (b) From Eq. 14–11b, we can find the speed at any position.

( 0.10 m ) v = vmax 1 − 2 = ( 3.016 m s ) 1 − = 2.248 m s  2.2 m s 2 A ( 0.15 m ) 2 2 Etotal = 12 mvmax = 12 ( 0.25 kg )( 3.016 m s ) = 1.137 J  1.1J 2

(c)

(d) Since the object has a maximum displacement at t = 0, the position will be described by the cosine function.

x = ( 0.15m ) cos ( 2 ( 3.2 Hz ) t ) →

x = ( 0.15m ) cos ( 6.4 t )

33. (a) The spring constant can be found from the mass and the frequency of oscillation.

=

= 2 f

→ k = 4 2 f 2 m = 4 2 ( 2.5 Hz ) ( 0.21kg ) = 51.82 N m  52 N m 2

m (b) The energy can be found from the maximum potential energy. 2 E = 12 kA2 = 12 ( 51.82 N m )( 0.045 m ) = 0.052 J

34. (a) The total energy of an object in SHM is constant. When the position is at the amplitude, the speed is zero. Use that relationship to find the amplitude. Etot = 12 mv 2 + 12 kx 2 = 12 kA2 → A=

m k

v 2 + x2 =

2.7 kg 310 N m

( 0.60 m s ) + ( 0.020 m ) = 5.946  10−2 m  5.9  10−2 m 2

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(b) Again use conservation of energy. The energy is all kinetic energy when the object has its maximum velocity. 2 Etot = 12 mv 2 + 12 kx 2 = 12 kA2 = 12 mv max → vmax = A

k m

(

= 5.946  10 −2 m

N m = 0.6371m s  0.64 m s ) 310 2.7 kg

35. (a) The work done in compressing the spring is stored as potential energy. The compressed location corresponds to the maximum potential energy and the amplitude of the ensuing motion. 2 ( 3.2 J ) 2W W = 12 kA2 → k = 2 = = 378.7 N m  380 N m A ( 0.13m )2 (b) The maximum acceleration occurs at the compressed location, where the spring is exerting the maximum force. If the compression distance is positive, then the acceleration is negative. kx ( 378.7 N m )( 0.13m ) F = −kx = ma → m = − = − = 4.1kg a −12 m s2 36. We assume that the collision of the bullet and block is so quick that there is no significant motion of the large mass or spring during the collision. Linear momentum is conserved in this collision. The speed that the combination has right after the collision is the maximum speed of the oscillating system. Then, the kinetic energy that the combination has right after the collision is stored in the spring when it is fully compressed, at the amplitude of its motion. m v0 pbefore = pafter → mv0 = ( m + M ) vmax → vmax = m+M 1 2

2 = 12 kA2 → ( m + M ) vmax

v0 =

A m

k (m + M ) =

1 2

( m + M ) 

m+M

 

v0  = 12 kA2 →

( 9.460 10 m ) (162.7 N m ) 7.450 10 kg + 4.148 kg ( ) ( 7.450 10 kg ) −2

−3

= 330.2 m s

37. (a) Find the period and frequency from the mass and the spring constant.

T = 2 m k = 2 0.885kg (184 N m ) = 0.4358s  0.436s f = 1 T = 1 ( 0.4358s ) = 2.29 Hz (b) The initial speed is the maximum speed, and that can be used to find the amplitude. v max = A k m → A = v max m k = ( 2.12 m s ) 0.885 kg (184 N m ) = 0.1470 m  0.147 m

amax = Ak m = ( 0.147 m )(184 N m ) ( 0.885kg ) = 30.6 m s2 (d) Because the mass started at the equilibrium position of x = 0, the position function will be proportional to the sine function.

x = ( 0.147 m ) sin 2 ( 2.29 Hz ) t  →

x = ( 0.147 m ) sin ( 4.58 t )

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(e) The total energy is the kinetic energy that the object has when at the equilibrium position. 2 2 E = 12 mvmax = 12 ( 0.885 kg )( 2.12 m s ) = 1.989 J  1.99 J (f) Use the conservation of mechanical energy for the oscillator, noting that we found total energy in part (e), and that another expression for the total energy is Etotal = 12 kA2 .

E = 12 kx 2 + 12 mv 2 = 12 kA2 →

(

)

k ( 0.40 A) + K = 12 kA2 → 2

1 2

(

)

K = 12 kA2 1 − 0.402 = E 1 − 0.402 = (1.989 J )( 0.84 ) = 1.67 J 38. The hint says to integrate Eq. 14–11a, which comes from the conservation of energy. Let the initial position of the oscillator be x0 . v=

dx → A −x ) = ( m dt 2

(A − x ) 2

=

k m

dt →



(A − x ) 2

=

m 0

dt →

x x k x − cos   = − cos −1 + cos −1 0 =  t A A m  A x −1

Make these definitions: x

  ; cos −1 k

x0 A

 . Then we have the following. x

+  =  t → x = A cos (  t +  ) A A m A The phase angle definition could be changed so that the function is a sine instead of a cosine. And the  sign can be resolved if the initial velocity is known. − cos −1

+ cos−1

=

t → − cos −1

39. We solve this using conservation of energy, equating the energy of the ball / spring / Earth system at the compressed point with the system’s energy as the ball leaves the launcher. Take the 0 location for gravitational potential energy to be at the level where the ball is on the compressed spring. The 0 location for elastic potential energy is the uncompressed position of the spring. Initially, the system has only elastic potential energy. At the point where the spring is uncompressed and the ball just leaves the spring, there will be gravitational potential energy, translational kinetic energy, and rotational kinetic energy. The ball is rolling without slipping. v2 Ei = Ef → 12 kx 2 = mgh + 12 mv 2 + 12 I  2 = mgx sin  + 12 mv 2 + 12 ( 52 ) mr 2 2 → r m 0.025 kg  9.80 m s 2 ( 0.060 m ) sin 22 + 107 ( 3.0 m s )2  k = 2 2 gx sin  + 107 v 2 = 2 2   x ( 0.060 m )

(

)

(

)

= 90.56 N m  91N m 40. The period of a pendulum is given by T = 2 l g . Solve for the length using a period of 2.0 seconds.

T = 2 l g → l =

T 2g 4 2

( 2.0 s ) ( 9.80 m s 2 ) 2

4 2

= 0.99 m

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41. The period of a pendulum is given by T = 2 l g . The length is assumed to be the same for the pendulum both on Mars and on Earth. 2 l g Mars T g Earth T = 2 l g → Mars = = → TEarth 2 l g Earth g Mars

TMars = TEarth

g Earth g Mars

= ( 2.15s )

1 0.37

= 3.5s

42. (a) The period is given by T = 50s 23cycles = 2.174s  2.2s . (b) The frequency is given by f = 23cycles 50 s = 0.46 Hz . 43. With the metronome set at 104 beats per minute, that is a frequency of 52 cycles per minute, or a period of 1/52 minutes, which is 60/52 seconds. Use Eq. 14–12c for the period of a pendulum.

T = 2

l g

→ l =

T 2g 4 2

( 60 / 52 s ) ( 9.80 m s 2 ) 2

4 2

= 0.33049 m  0.330 m

44. We assume the simple pendulum is at a small enough amplitude that it behaves as a simple harmonic oscillator. For a simple harmonic oscillator, the energy is proportional to the square of the amplitude. Use this to find the new amplitude.

E  A2 →

E 2

= constant →

E1 2 1

E2 2 2

→ A2 = A1

E2 E1

= ( 8.5 ) 2 = 12.02  12

Here is an alternate solution that does not treat the pendulum as a simple harmonic oscillator, but instead uses mechanical energy conservation, since only the gravity force does any work. (The string tension is perpendicular to the direction of motion of the pendulum.) Take the 0 level for gravitational potential energy to be at the bottom of the pendulum’s swing. The figure is not drawn to scale. For the initial location, with the maximum angle  0 , the total energy

is E = mgh = mg l (1 − cos 0 ) . Find the maximum angle to double the energy. E2 = 2 E1 = 2mg l (1 − cos  0 ) = mg l (1 − cos  new ) →

 new = cos −1 ( 2 cos  0 − 1) = cos −1 ( 2 cos8.5 − 1) = 12.03  12 45. Since the motion starts at the amplitude position at t = 0, we may describe it by a cosine function with no phase angle,  =  max cos  t. The angular velocity can be written as a function of the length,

 g   =  max cos  t . l  

 9.80 m s2  0.35s )  = −6.248  −6.2 (  0.30 m   

(a)  ( t = 0.35s ) = 15 cos 

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Chapter 14

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 9.80 m s 2  3.45s )  = 9.686  9.7 (  0.30 m     9.80 m s2  (c)  ( t = 6.00s ) = 15 cos  6.00s )  = −14.48  −14 (  0.30 m    (b)  ( t = 3.45s ) = 15 cos 

46. The period of a pendulum is given by Eq. 14–12c, T = 2 l g . (a) T = 2 l g = 2

0.47 m 9.80 m s 2

= 1.4 s

(b) If the pendulum is in free fall, there is no tension in the string supporting the pendulum bob, and no restoring force to cause oscillations. Thus there will be no period–the pendulum will not oscillate and so no period can be defined. 47. There are ( 24 h )( 60 min h )( 60s min ) = 86, 400s in a day. The clock should make one cycle in exactly two seconds (a “tick” and a “tock”), and so the clock should make 43,200 cycles per day. After one day, the clock in question is 18 seconds slow, which means that it has made 9 less cycles than required for precise timekeeping. Thus the clock is only making 43,191 cycles in a day. 43,191 Accordingly, the period of the clock must be decreased by a factor of . 43, 200 Tnew =

43,191 43, 200

 43,191   2 l old g →  43, 200 

Told → 2 l new g =  2

 43,191   43,191  l new =   l old =   ( 0.9930 m ) = 0.9926 m  43, 200   43, 200  Thus the pendulum should be shortened by 0.4 mm. 48. Use energy conservation to relate the potential energy at the maximum height of the pendulum to the kinetic energy at the lowest point of the swing. Take the lowest point to be the zero location for gravitational potential energy. See the diagram. Etop = Ebottom → K top + U top = K bottom + U bottom → 2 0 + mgh = 12 mvmax → vmax = 2 gh =

2 g l (1 − cos  max )

49. The angular displacement of a pendulum, starting from a maximum position, is given by  =  max cos ( 2 f t ) . The amplitude is 12o , which is  15 radians. o cos ( 2 f t ) = ( 15) cos ( 5.0 t )  o =  max

(a)  o ( t = 0.25 s ) = ( 15 ) cos 5.0 ( 0.25 )  = −0.15 rad (b)  o ( t = 1.60 s ) = ( 15 ) cos 5.0 (1.60 )  =  15 rad  0.21rad (The time is exactly 4 periods.) (c)  o ( t = 500 s ) = ( 15 ) cos 5.0 ( 500 )  =  15 rad  0.21rad (The time is exactly 1250 periods.)

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 2 t   , where  0 T 

50. The equation for the angular position of the pendulum can be written as  =  0 cos 

is the amplitude. Solve for the time at which the angular position is 8o, assuming that the angular position was 10o at t = 0. T  −1   T  2   = 0 cos  t  → t = = cos cos −1 0.8 = 0.1024 T    0  2 2  T  Thus, 10.24% of a period is spent in the first quarter-cycle as the pendulum moves from 10o to 8o , which means 40.96% of any full period is spent moving between those two angular positions. Accordingly, 59.04% of any full period is spent moving between the angular positions of 8o and 0o.

(

)

Thus 59% of each period is spent between +8.0o and –8.0o. 51. The balance wheel of the watch is a torsion pendulum, described by  = − k . A specific torque and angular displacement are given, and so the torsional constant can be determined. The angular frequency is given by  = k I . Use these relationships to find the mass.

 1.1  10−5 m N  = − k → k = =   4 rad  = 2 f =

k I

→

mr 2

1.1  10−5 m N m=

k 4 f r 2

2 2

 4 rad

(

4 (1.55 Hz ) 0.95  10 m 2

−2

)

= 1.636  10−3 kg = 1.6 g

52. (a) In the text, we are given that  = − k . Newton’s second law for rotation, d 2 Eq. 10–14, says that  = I  = I 2 . We assume that the torque applied by the twisting of dt the wire is the only torque. d 2 d 2 k I I k = = = − → = −  = − 2     2 2 dt dt I This is the same form as Eq. 14–3, which is the differential equation for simple harmonic oscillation. We exchange variables with Eq. 14–4, and write the equation for the angular motion. x = A cos (t +  ) →  =  0 cos ( t +  ) ,  2 =

(b)

I The period of the motion is found from the angular velocity .

2 =

k I

→ =

k I

2 T

→

T = 2

I k

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Chapter 14

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53. (a) For a physical pendulum with the small angle approximation, we may apply Eq. 14–14. We need the moment of inertia and the distance from the suspension point to the center of mass. We find the center of mass relative to the stationary end of the rod. I = I bob + I rod = M l 2 + 13 ml 2 = ( M + 13 m ) l 2

h = xCM =

M l + m ( 12 l ) M +m I

T = 2

mtotal gh

 M + 12 m  l  M +m 

=

( M + 13 m ) l 2

= 2

( M + m ) g 

( M + 13 m ) l ( M + 12 m ) g

= 2

M + 12 m 

l  M +m 

(b) If we use the expression for a simple pendulum we would have Tsimple = 2 l g . Find the fractional error.

error =

T − Tsimple T

2 =

( M + 13 m ) l − 2 ( M + 12 m ) g ( M + 13 m ) l 2 ( M + 12 m ) g

l g

( M + 13 m ) −1 ( M + 12 m ) ( M + 12 m ) = 1− ( M + 13 m ) ( M + 13 m ) ( M + 12 m )

Note that this is negative, indicating that the simple pendulum approximation is too large. Also, if m = 0, this gives an error of 0, which is correct.

( M + 12 m ) = −0.1 ( M + 13 m ) ( M + 12 m ) ( M + 12 m ) = −0.1 → 1.12 = → ( M + 13 m ) ( M + 13 m )

1−

 12 − 13 (1.12 ) = 1.12 − 1 → 0.0967 = 0.21 → = 2.17 → 2 M M M M So the rod can be a little more than twice as massive as the pendulum bob and still have only about a 10% error in the period using the simple pendulum formula. 54. This is a torsion pendulum. The angular frequency is given in the text as  = k I , where k is the torsion constant (a property of the wire, and so a constant in this problem). The rotational inertia of a rod about its center is 121 M l 2 . Note that the mass changes since the length changes.

=

T T0

k I 2

= 2

2 T

→ T = 2

I k

→

I k = I0

I I0

1 12

Ml 2

1 12

M 0 l 02

( 0.700M 0 )( 0.700l 0 ) M 0 l 02

= 0.58566

T = ( 0.58566 ) T0 = ( 0.58566 )( 5.0 s ) = 2.9 s

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55. (a) We call the upper mass M and the lower mass m. Both masses have length l. The period of the physical pendulum is given by Eq. 14–14. Note that we must find both the moment of inertia of the system about the uppermost point, and the center of mass of the system. The parallel axis theorem is used to find the moment of inertia. I = I upper + I lower = 13 M l 2 +  121 ml 2 + m ( 23 l )  = ( 13 M + 73 m ) l 2 2



h = xCM =

M ( l ) + m( l ) 3 2

1 2

M +m

T = 2

mtotal gh



 M + m l  M +m 

=

3 2

1 2

( 13 M + 73 m ) l 2

= 2

( M + m ) g  2 1

= 2

M + 23 m 

l  M +m 

( 13 M + 73 m ) l ( 12 M + 32 m ) g

 ( 7.0kg ) + ( 4.0kg ) ( 0.55m ) = 1.6495s  1.6s  ( 7.0kg ) + ( 4.0kg ) ( 9.80 m s ) 7 3

1 3

= 2

3 2

1 2

(b) It took 7.2 seconds for 5 swings, which gives a period of 1.4 seconds. That is reasonable qualitative agreement. 56. The meter stick used as a pendulum is a physical pendulum. The period is given by Eq. 14–14,

T = 2

. Use the parallel axis theorem to find the moment of inertia about the pin. Express mgh the distances from the center of mass. I = I CM + mh = 2

1 12

ml + mh → T = 2 2

 1 l2  = 2 ( )  12 + h  dh  h 

−1/ 2

mgh

= 2

 1 l2   − 12 h 2 + 1 = 0 → h =  

1 2

ml 2 + mh 2

1 12

mgh 1 12

2  1 l 2

1/ 2

  12 h + h  g 

l = 0.2887 m

x = 12 l − h = 0.500 m − 0.2887 m  0.211m from the end.

Use the distance for h to calculate the period. 1/ 2

 1 (1.00 m )2   2 + = + 0.2887 m  = 1.53s T= h  12  12 h    g  9.80 m s 2  0.2887 m  2  1 l 2

1/ 2

57. This is a physical pendulum. Use the parallel axis theorem to find the moment of inertia about the pin at point A, and then use Eq. 14–14 to find the period. I pin = I CM + Mh 2 = 12 MR 2 + Mh 2 = M ( 12 R 2 + h 2 ) T = 2 = 2

I Mgh 1 2

= 2

( R + h ) = 2 ( R + h ) 1 2

1 2

Mgh

( 0.200 m )2 + ( 0.180 m )2 = 1.08s ( 9.80 m s2 ) ( 0.180 m )

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58. The torsional constant is related to the period through the relationship given in Problem 52. The rotational inertia of a disk in this configuration is I = 12 MR 2 . I

T = 2

→ k=

4 2 I

4 2 12 MR 2

 0.154 m  0.331Hz 2 = 2 MR f = 2 ( 0.375 kg )  )  (  2  2

= 4.81  10−3 m N rad 2 59. We assume that initially, the system is critically damped, so bcritical = 4mk . Then, after aging, we assume that after 3 cycles, the car’s oscillatory amplitude has dropped to 5% of its original

amplitude. That is expressed by A = A0e A = A0 e

−

→ 0.05 A0 = Ae

ln ( 0.05) = −

2m 1

2

−

bt 2m

b ( 3T )

= Ae

=−

−

b 3 2m f

2

3b 2m

2 bcritical

4m 2

  b 36 2  = 1 + bcritical  ln ( 0.05)  2     

→ ln ( 0.05) = −

4m 2

−

3b 1 2m f

→

6 b

=−

2 − b2 bcritical

→

4m 2

−1/ 2

= 0.16

And so b has decreased to about 16% of its original value, or decreased by a factor of 6. If we used 2% instead of 5%, we would have found that b decreased to about 20% of its original value. And if we used 10% instead of 5%, we would have found that b decreased to about 6% of its original value. 60. (a) The period of the motion can be found from Eq. 14–18, giving the angular frequency for the damped motion. k

 = T=

−

b2 4m 2

( 41.0 N m ) ( 0.662 N s m )2 − = 7.455 rad s ( 0.735 kg ) 4 ( 0.735 kg ) 2

2 2 = = 0.843s   7.455 rad s

(b) If the amplitude at some time is A, then one cycle later, the amplitude will be Ae − T . Use this to find the fractional change. fractional change =

Ae − T − A

−

− T

−

( 0.662N s m ) ( 0.843s ) 2( 0.735 kg )

= e −1 = e −1 = e − 1 = −0.316 A And so the amplitude decreases by 31.6 % from the previous amplitude, every cycle. (c) Since the object is at the origin at t = 0, we will use a sine function to express the equation of motion. x = Ae

− t

sin ( t ) → 0.120 m = Ae 0.120 m

A= e

( 0.662N s m ) − (1.00 s ) 2( 0.735 kg )

x = ( 0.204 m ) e

(

−

( 0.662N s m ) (1.00 s ) 2( 0.735 kg )

= 0.204 m ;  =

sin ( 7.455 rad )

− 0.450 s

−1

sin ( 7.455 rad ) → b 2m

( 0.662 N s m ) = 0.450s −1 2 ( 0.735 kg )

sin ( 7.46 rad s ) t 

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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61. (a) The energy of the oscillator is all potential energy when the cosine (or sine) factor is 1, and so −

E = 12 kA2 = 12 kA02 e m . The oscillator is losing 8.0% of its energy per cycle. Use this to find the actual frequency, and then compare to the natural frequency. E ( t + T ) = 0.92 E ( t ) →

b 2m

=−

1 2T

ln ( 0.92 ) = −

1 2

kA02 e

−

b( t + T )



−



−

= 0.92  12 kA02e m  → e m = 0.92 →





0 ln ( 0.92 ) 4

 b2 1 2 2 02 − 2 − 0  ln ( 0.92 )  ln ( 0.92 )  f − f 0 2 b2    m 4 2 1 = = 1− 2 2 −1 = 1− −1  1− 2 −1 0 f0 4m 0 16 2 16 2 2

ln ( 0.92 ) = −2.2  10 =− 2

1 2

16

−5

 f  − f0  −3  100 = ( −2.2  10 ) % f  0 

→ % diff =  −

(b) The amplitude’s decrease in time is given by A = A0e 2 m . Find the decrease at a time of nT, and solve for n. The value of A = A0e n=−

−

was found in part (a).

bt 2m

−

−1

bnT

→ A0e = A0e 2 m

2 ln ( 0.92 )

→ 1=

b 2m

nT = −

1 2T

ln ( 0.92 ) nT →

= 23.99  24 periods

62. (a) Since the angular displacement is given as  = Ae − t cos (t ) , we see that the displacement at t = 0 is the initial amplitude, so A = 15. We evaluate the amplitude 8.0 seconds later. −1  5.5  −  8.0s −1 −1 5.5 = 15e ( ) →  = ln   = 0.1254s  0.13s 8.0s  15  (b) The approximate period can be found from the damped angular frequency. The undamped angular frequency is also needed for the calculation.

0 =

mgh I

 =  −  = 2 0

mg ( 12 l ) 1 3

ml 3g 2l

− = 2

3g 2l

(

) − 0.1254 s = 4.157 rad s ( ) 2 ( 0.85 m )

3 9.80 m s 2

−1 2

2 2 rad = = 1.5s   4.157 rad s (c) We solve the equation of motion for the time when the amplitude is half the original amplitude. ln 2 ln 2 7.5 = 15e −  t → t1/ 2 = = = 5.5s 0.1254 s −1  T =

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Chapter 14

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− t

63. The amplitude of a damped oscillator decreases according to A = A0e = A0e used to find the damping constant. bt − 2m  A0  2 ( 0.058 kg )  5.0  2m ln   = ln  A = A0 e → b=  = 0.030 kg s t ( 3.5s )  2.0   A

−

bt 2m

. The data can be

64. (a) From Problem 25(b), we can calculate the frequency of the undamped motion. m m T = 2 = 2 → k1 + k 2 2k

f =

1 2

2k m

2 m 2

125 N s

2 ( 0.215 kg ) 2

= 5.427 Hz  5.43 Hz

(b) Eq. 14–16 says x = Ae − t cos  t , which says the amplitude follows the relationship xmax = Ae − t . Use the fact that xmax = 12 A after (exactly) 68 periods have elapsed, and assume that the damping is light enough that the damped frequency is the same as the natural frequency. ln 2 f 5.427 Hz −  55T 1 A = Ae ( ) →  = ln 2 = 0.05532s −1  0.0553s −1 = ln 2 = 2 68T 68 68 (c) Again use xmax = Ae − t . xmax = Ae −  t →

A = Ae − t → t =

ln4

ln 4

= 25.06s  25.1s 0.05532 s −1  This is the time for 136 oscillations, since 68 oscillations corresponds to a “half-life.” 1 4

→    0 . m 4m We also assume that the object starts to move from maximum displacement, and so bt bt bt bt − − − − dx b x = A0e 2 m cos t and v = A0 e 2 m cos  t −   A0 e 2 m sin  t  −0 A0 e 2 m sin  t . =− dt 2m 4mk →

65. (a) For the “lightly damped” harmonic oscillator, we have b 2

−

E = 12 kx 2 + 12 mv 2 = 12 kA02 e m cos 2  t + 12 m02 A02 e m sin 2  t −

−

= 12 kA02 e m cos 2  t + 12 kA02 e m sin 2  t = 12 kA02 e m = E0 e m

(b) The fractional loss of energy during one period is as follows. Note that we use the b 2 bT bT approximation that 0 = → 4 → 1. 2m T m m −

E = E ( t ) − E ( t + T ) = E0e m − E0e −

E E



−

b( t +T ) m

−



−



= E 0e m  1 − e m  →







E0 e m  1 − e m  =



−

E0 e m

bT  = 1 − e − m  1 −  1 − bT  = bT = b2 = 2   Q  m  m m0

511

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

66. (a) For the damped oscillator, the amplitude decays according to A = A0e

−

bt 2m

. We are also given the

m0

Q value, and Q =

. We use these relationships to find the time for the amplitude to b decrease to one-third of its original value. m0

t1/3 =

2m b

 → = 0 = m Q

g l

ln 3 =

0

ln 3 =

−

; A = A0 e 2 m = 13 A0 →

g l

bt1/3

ln 3 =

(

2 ( 350 ) 9.80 m s 2

)

( 0.50 m )

ln 3 = 173.7 s  170 s

(b) The energy is all potential energy when the displacement is at its maximum value, which is the amplitude. We assume that the actual angular frequency is very nearly the same as the natural angular frequency. 2

bt  − bt  mg 2 − btm dE b mg 2 − m =− → E = kA =  m  A0 e 2 m  = A0 e ; A0 e 2l dt m l   2

1 2

dE dt t = 0

1 2

=−

( 0.27 kg )( 0.020 m )  9.80 m s 2  = =    0.50 m  2l 2Q  l  2 ( 350 )  

g l mg Q

mA02  g 

3/ 2

= 1.3  10−5 W

67. We approximate that each spring of the car will effectively support one-fourth of the mass. The rotation of the improperly-balanced car tire will force the spring into oscillation. The shaking will be most prevalent at resonance, where the frequency of the tire matches the frequency of the spring. At v resonance, the angular velocity of the car tire,  = , will be the same as the angular frequency of r the spring system,  =

=

v r

k m

→ v=r

k m

= ( 0.42 m )

14, 000 N m 1 4

(1150 kg )

= 2.9 m s

68. (a) Eq. 14–24 is used to calculate 0 .

0 = tan −1

02 −  2  2 − 02 → if  =  , 0 = tan −1 0 = 0  (b m)  (b m)

(b) With  = 0 , we have Fext = F0 cos 0t and x = A0 sin 0t. The displacement and the driving

force are one-quarter cycle ( 12  rad or 90 ) out of phase with each other. The displacement is 0 when the driving force is a maximum, and the displacement is a maximum (+A or –A) when the driving force is 0.

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Chapter 14

Oscillations

69. Eq. 14–23 gives the amplitude A as a function of driving frequency  . To find the frequency for dA maximum amplitude, we set = 0 and solve for . d −1/ 2 2 F0 F = 0   2 − 02 + b 2 2 m 2  A= 2  m m  2 − 02 + b 2 2 m 2

(

dA d

F0 m

(

)

( − ) ( −  ) + b  2

1 2

2 0

)

m2 

−3/ 2



 2 ( 2 − 02 ) 2 + 2 b 2 m 2  = 0 →

2 2 2 2 F0  2 ( − 0 ) 2 + 2 b  m  1 −2 = 0 → 2 ( 2 − 02 ) 2 + 2 b 2 m 2 = 0 → 3/ 2 m   2 −  2 2 + b 2 2 m 2  0 ) ( 

 2 = 02 −

b2 2m 2

 = 02 −

→

2m 2

70. Apply the resonance condition,  = 0 , to Eq. 14–23, along with the given condition of

A = 23.7

F0 k

. F0

( −  ) + b  m 2

2 0

A (  = 0 ) =

F0 m b 2

2 0

→ 2

F0 = b0 m

F0 m

b

2 0

F0 m

m0

2 0

F0 k

= 23.7

→ Q = 23.7

71. First put Eq. 14–23 into a form that explicitly shows A as a function of Q and has the ratio  0 .

A= m

( −  ) + b  m 2

2 0

 2  b 2 2  2 m   2 02 − 02  + 2 02 m 0  0 

 2  b2  2 m   2 − 1  + 04 2 2 2 m 0 0  0  4 0

F0 k

(  − 1) + Q1   2

2 0

→ 2 0

m

2 0

A F0 k

 2  1 2 − + 1   2  Q2  2  0  0 =

( m ) 2 0

 2  1 2 − + 1   2  Q2  2  0  0

( (  ) − 1) + (  ) Q1 2

For a value of Q = 5.0, the following graph is obtained.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

6 5 4

F0 k 2 1

0 0

0.25

0.5

0.75

1.25

1.5

1.75

 0 2

 A  1 And for the squared relationship:  =  2 1  F0 k  ( 0 )2 − 1 + ( 0 )2

(

)

30 25 20 2

 A  15 F k  0  10 5

0 0

0.25

0.5

0.75

1.25

1.5

1.75

 0 Notice that the second graph is more sharply peaked around the resonance value. 72. We are to show that x = A sin (  t +  ) is a solution of m

dx 2 dt

dx dt

+ kx = F0 cos  t by direct

substitution.

x = A sin ( t +  ) ; m

dx 2 dt

dx dt

=  A cos ( t +  ) ;

d 2x dt 2

= − 2 A sin ( t +  )

+ kx = F0 cos  t →

m  − 2 A sin ( t +  )  + b  A cos ( t +  )  + k  A sin ( t +  )  = F0 cos  t Expand the trig functions. ( kA − m 2 A) sin  t cos  + cos  t sin   + b Acos  t cos  − sin  t sin   = F0 cos  t Group by function of time.

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Chapter 14

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( kA − m 2 A ) cos  − b A sin   sin  t + ( kA − m 2A ) sin  + b A cos   cos  t = F0 cos  t The equation has to be valid for all times, which means that the coefficients of the functions of time must be the same on both sides of the equation. Since there is no sin  t on the right side of the equation, the coefficient of sin  t must be 0. ( kA − m 2 A) cos  − b A sin  = 0 → sin  cos 

kA − m 2 A

b A

k − m 2 b

m02 − m 2 b

02 −  2 2 − 2 = tan  →  = tan −1 0 b m b m

( −  ) + mb

Thus we see that Eq. 14–24 is necessary for x = A sin ( t +  ) to be the solution. This can be illustrated with the right triangle shown.

2 0

Equate the coefficients of cos  t.

0

( kA − m A) sin  + b A cos  = F →

02 −  2

b m

  b 2 2   (0 −  ) m  = F0 → A  ( k − m 2 ) + b   2b2  2b2  2 2 2 2 2 2 (0 −  ) + m 2  (0 −  ) + m 2      2b2 2 2 2   (0 −  ) F0 m2  = F0 → A = + Am     2b2  2b2   2b2  2 2 2 2 2 2 2 2 2     − + − +   − + m ( ) ( ) ( )   0 0   0 m2 m2  m 2    Thus we see that Eq. 14–23 is also necessary for x = A sin ( t +  ) to be the solution. 73. The relationship that we are to prove is stated in the text to only be accurate for weak damping, so we assume that b2 4mk = 4m202 . The textbook also says that the peak of amplitude occurs quite close to 0 unless the damping is very large. Thus we will assume that the peak occurs at 0 . We take the amplitude from Eq. 14–23. F02 F02 F02 2 2 A2 = ; A ; Let A2 = 12 Amax = = max 2 2 2 2  2 2 2 2 2 2 2 2 2 2 2 b 0 m  −  + b   m  − + b  m

(

)



2 0

 m ( −  ) + b     2

2 0

2 0 2 2 0

2b 



(

)

→ m 2  2 − 02



) + b  = 2b  2

2 0

Re-arrange this equation to be a quadratic in the variable  2 .

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

 b2  4 2b 202  2 2 m ( −  ) + b  = 2b  →  +  2 − 20   +  0 − =0 m2  m   2

2 0

b2 b 2 2   b2   2 − 2   2 − 202  − 4  04 − 2 0  m m  m   2 0

2 =

= 02 −

 2

−4 4

8b 202

02 + 404 − 404 + 2 4b 202

For ease of notation, let  

+ 4

( m

4m202 . So we can say (later) that  =

 2 = 02 −

2m b2



1 2



4b  2

2 0

= 02 −

)

4m 202 =



 2

b2 2m

4b 202

+ 4

( 4m  ) since m 2

402b2 m

 2

b 2 + 4m 202 

2 2

= 02 −

2 0

20b m

; there are two roots



; 12 − 22 =  = (1 − 2 )(1 + 2 ) 2m 2 The value of (1 − 2 ) =  is the quantity for which we are solving. If we assume that 1 and

12 = 02 −

+ 2

; 22 = 02 −

−

 2 are approximately symmetric about 0 , then we have 1 + 2 = 20 . Thus we now have the following: 0b  b  = (1 − 2 )(1 + 2 ) = (  )( 20 ) →  = = m = 20 20 m Now substitute in the definition of Q, Q =

 =

b m

0 Q

→



0

m0 b

1 Q

 dx  .  dt 

74. (a) We have that Fext = F0 cos  t and x = A sin ( t +  ) . Calculate P = Fext v = F0 cos  t  Use Eq. 14–23.

 dx   = F0 ( cos  t ) A cos ( t +  )  dt 

P = Fext v = F0 cos  t 

= F0 ( cos  t ) m

( 2 − 02 ) + b2 2 m2 2

F02

= m

( −  ) + b  m 2

2 0

cos (  t +  )

cos  t  cos  t cos  − sin  t sin  

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Chapter 14

Oscillations

F02

( −  ) + b  m 2

2 0

 cos 2 t cos  − sin  t cos  t sin   2

Substitute sin  t cos  t = 12 sin 2 t.

F02

( −  ) + b  m 2

m =

2 0

cos2 t cos  − 12 sin 2 t sin  

F02 cos  cos 2 t − 12 F02 sin  sin 2 t

( −  ) + b  m 2

2 0

(b) The average power input averaged over a cycle is given by P =

T

T 0

F02

 P ( t ) dt =

( −  ) + b  m 2

2 0

P ( t ) dt.

T 0

cos2 t cos  − 12 sin 2 t sin   dt 2

T T   2 1 = cos   cos  t dt − 2 sin   sin 2 t dt   2 T m  2 −  2 + b 2 2 m 2   0 0 ( 0 )

F02

The integrals can be evaluated from integrals in the appendices. T

 sin 2 t dt = − 0

1 2

cos ( 2 t )

t =T = 2  

=− t =0

1 2

cos 4 − cos 0 = 0 T

 t sin 2 t   cos  t dt =  (1 − sin  t ) dt =  dt −  sin  t dt = T −  2 − 4 0 0 0 0 0 T

 T − sin 2T  = T + sin 4 = T  4  2 4 2 2

=T − And so P =

F02

( 2 − 02 ) + b2 2 m2 2

T m

Note that v =

cos  = 2m

( 2 − 02 ) + b2 2 m2

dx = A cos ( t +  ) and so vmax = A = dt m F02 cos 

 F02 cos 

( −  ) + b  m 2

2 0

= 2

F0 

( 2 − 02 ) + b2 2 m2 2

F0 cos 

F0

( −  ) + b  m

2 0

02 −  2 . This can be visualized with  (b m)

a right triangle as shown in the diagram. From this diagram, b m . Substitute this into the first cos  = 2  2 − 02 + b 2 2 m 2

(

)

= 12 F0 v max cos  .

( −  ) + mb 2 0

. Thus



02 −  2

b m

result from part (b) above. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

517

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

 F02 cos 

( −  ) + b  m 2

2 0

 F02

b m

( −  ) + b  m ( −  ) + b  m 2

2 0

= 2

(

bF02  2 2m 2

)

( −  ) + b  m 2

2 0

(d) To prepare the expression to plot, some algebraic manipulation must be done. We need to m0 express the variable as  0 , and we need to incorporate Q = . b

Let



0

  2   02  2  2   0   2m 

bF02 

( ) = ( −  ) + b  m    − 1 +  b  bF02  2 2m 2

2 0

 

2 0

 2  0

 

2 0

02 m 2

 x, and factor 04 from the denominator.

 02b  2   2m 

F02

F02 x 2 

2  b   ( x 2 − 1) + x 2 2 2  0 m   2

4 0

F02

(

F02

 b2  2 2   0 m 

x2 

2  2 b  2  ( x − 1) + x  2 m 2    0 2

F02 =

2bQ

(

)

x2 − 1 +

x2 Q2

Qx 2

2m0 2b = → 2 2 2 2 x −1 + x Q x2 − 1 + x2 2

)

(

)

 2m0  6.0 x 2 Qx 2 = =  2 2 2  F0  Q 2 ( x 2 − 1) + x 2 36 ( x 2 − 1) + x 2

P

 2m0  . 2  F0 

A plot is shown below. Note that the average power has been scaled by 

 0

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75. Apply the conservation of mechanical energy to the car, calling condition #1 to be before the collision and condition #2 to be after the collision. Assume that all of the kinetic energy of the car is converted to potential energy stored in the bumper. We know that x1 = 0 and v 2 = 0.

E1 = E2 → m

x2 =

1 2

1300 kg

v1 =

mv12 + 12 kx12 = 12 mv22 + 12 kx22 → 410  103 N m

1 2

mv12 = 12 kx22 →

( 2.0 m s ) = 0.1126 m  0.11m

76. Consider the conservation of energy for the person. Call the unstretched position of the fire net the zero location for both elastic potential energy and gravitational potential energy. We can measure both the amount of stretch of the fire net and the vertical displacement for gravitational potential energy by the variable y, measured positively for the upward direction. Calculate the spring constant by conserving energy between the window height ytop = 20.0 m and the lowest location of the

(

)

person ( y bottom = −1.4 m ) . The person has no kinetic energy at either location. 2 E top = Ebottom → mgy top = mgy bottom + 12 ky bottom

k = 2mg

(y − y top

bottom

2 bottom

) = 2 ( 62 kg ) 9.80 m s 20.0m − ( −1.4 m )  = 1.3268  10 N m ( ) 2

( −1.4 m )

(a) If the person were to lie on the fire net, they would stretch the net an amount such that the upward force of the net would be equal to their weight. 2 mg ( 62 kg ) ( 9.80 m s ) Fext = k y = mg → y = = = 4.6  10−2 m = 4.6 cm 4 1.3268  10 N m k (b) To find the amount of stretch given a starting height of 38 m, again use conservation of energy. Note that there is no kinetic energy at the top or bottom positions. 2mg 2mg 2 2 Etop = Ebottom → mgytop = mgybottom + 12 kybottom → ybottom + ybottom − ytop = 0

( 62 kg ) ( 9.80 m s ) 2

2 ybottom +2

1.3268  10 N m 4

( 62 kg ) ( 9.80 m s ) 2

ybottom − 2

1.3268  10 4 N m

( 38 m ) = 0 →

2 ybottom + 0.091589 ybottom − 3.4804 = 0 → ybottom = −1.9119 m, 1.8204 m This is a quadratic equation. The solution is the negative root, since the net must be below the unstretched position. The result is that it stretches 1.9 m down if the person jumps from 38m.

77. The frequency of a simple pendulum is given by Eq. 14–12b, f =

. The pendulum is 2 l accelerating vertically which is equivalent to increasing (or decreasing) the acceleration due to gravity by the acceleration of the pendulum. (a)

f new =

(b)

f new =

g+a

2

g+a

2

g + 0.35 g

2

g − 0.35 g

2

= =

1.35 g

2

0.65 g

2

= 1.35

= 0.65

2

= 1.35 f = 1.16 f

= 0.65 f = 0.81 f

519

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

78. (a) The frequency can be found from the length of the pendulum, and the acceleration due to gravity. f =

2

9.80 m s 2

2

0.72 m

= 0.5872 Hz  0.59 Hz

(b) To find the speed at the lowest point, use the conservation of energy relating the lowest point to the release point of the pendulum. Take the lowest point to be the zero level of gravitational potential energy. Etop = Ebottom → K top + U top = K bottom + U bottom 2 0 + mg ( l − l cos  ) = 12 mvbottom +0

vbottom = 2 g l (1 − cos  )

(

)

= 2 9.80 m s2 ( 0.72 m )(1 − cos12) = 0.5553m s  0.56 m s (c) The total energy can be found from the kinetic energy at the bottom of the motion, or the potential energy at the top of the motion. 2 = 12 ( 0.295 kg )( 0.5553m s ) = 4.5 10 −2 J Etotal = 12 mvbottom 2

(

)

Etotal = mgh = mg l (1 − cos  ) = ( 0.295 kg ) 9.80 m s 2 ( 0.72 m )(1 − cos12 ) = 4.5 10 −2 J 79. We want to find an expression of the form x = A cos ( t +  ) . The angular frequency is

 = 2 f = 6.0 rad s. The velocity is given by v = − A sin (  t +  ) . Use the position and velocity at the given time to solve for the amplitude and phase angle. x = A cos (  t +  ) ; v = − A sin ( t +  ) 2

x  v  2  ; sin ( t +  ) =  −   A  A 

cos 2 (  t +  ) = 

x  v  cos (  t +  ) + sin ( t +  ) = 1 =   +  −  →  A   A  2

x + 2

2

( −0.070 m ) + 2

(1.80 m s )

( 6.0 rad s )

= 0.1184 m  0.118 m

Now find the phase angle from either the position or velocity data. x = A cos ( t +  ) →

 −0.070 m  x x = cos −1   −  t = cos−1   − ( 6.0 rad s )( 4.10s ) = 2.203 − 77.283 = −75.08 rad  A  0.1184 m  This answer must be checked, however. Recall that using the inverse cosine function on a calculator always gives a value in the 1st or 2nd quadrant, while the actual phase angle can be in any quadrant. For example, cos−1 ( 0.5) =  3 , but both cos ( 3) and cos ( 5 3 ) = 0.5. In general, cos = cos ( 2 −  ) , and so we need to check the value for the phase angle to see if it is actually

correct. If not, we need to use one which reproduces the data. x ( 4.1s ) = ( 0.1184 m ) cos ( 6.0 rad s )( 4.1s ) − 75.08 rad  = −0.06998 m  −7.0 cm v ( 4.1s ) = − ( 6.0 rad s ) ( 0.1184 m ) sin ( 6.0 rad s )( 4.1s ) − 75.08 rad  = −1.80 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

520

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The velocity data does not match. So we need to use the other phase angle. Instead of the value of x 2.203 for cos −1   in the equation above for x , we must use = 2 − 2.203 = 4.080 rad.  A

 −0.070 m  x x = cos −1   −  t = cos −1   − ( 6.0 rad s )( 4.10 s ) = 4.080 − 77.283 = −73.203 rad  A  0.1184 m  x ( 4.1s ) = ( 0.1184 m ) cos ( 6.0 rad s )( 4.1s ) − 73.203 rad  = −0.06997 m

v ( 4.1s ) = − ( 6.0 rad s )( 0.1184 m ) sin ( 6.0 rad s )( 4.1s ) − 73.203 rad  = 1.8004 m s

The phase angle of −73.203rad is rather unwieldy. We can add multiples of 2 to it without changing its value. Since 73.203 2 = 11.65 , add 12  2 .

12  2 − 73.203 = 2.1952, and so x = ( 0.118 m ) cos ( 6.0 rad s ) t + 2.2

Evaluating this expression and the one for velcoity at t = 4.1 s gives the correct results. 80. Block m stays on top of block M (executing SHM relative to the ground) without slipping due to static friction. The maximum static frictional force on m is Ffr =  s mg . This frictional force causes max

block m to accelerate, so mamax =  s mg → amax =  s g . Thus for the blocks to stay in contact without slipping, the maximum acceleration of block M is also amax =  s g . But an object in SHM has a maximum acceleration given by amax =  2 A =

k M total

A . Equate these two expressions for the

maximum acceleration.

( 0.30) ( 9.80 m s ) g amax = A = s g → A = s ( M + m ) = ( 7.25 kg ) = 0.16 m 130 N m M total k 2

81. (a) The car on the end of the cable produces tension in the cable, and stretches the cable according 1 F l 0 , where E is Young’s modulus. Rearrange this equation to see that to Eq. (12–4), l = E A EA l , and so the effective the tension force is proportional to the amount of stretch, F = l0 spring constant is k =

. The period of the bouncing can be found from the spring constant l0 and the mass on the end of the cable.

T = 2

m k

= 2

ml 0 EA

= 2

(1550 kg )( 20.0 m )

( 200 10 N m )  ( 3.2 10 m ) 9

−3

= 0.436 s  0.44 s

(b) The cable will stretch some due to the load of the car, and then the amplitude of the bouncing will make it stretch even farther. The total stretch is to be used in finding the maximum amplitude relative to the unloaded cable. The tensile strength is found in Table 12–2. F k ( xstatic + xamplitude ) = = tensile strength ( abbrev T.S. ) →  r2 A

521

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

xamplitude =

( T.S.)  r 2 k

− l =

( T.S.)  r 2 E r

−

mg l 0 E r

Instructor Solutions Manual

l0 

mg ( T.S.) − 2   E r 

 (1550 kg ) ( 9.80 m s 2 )  6 2 = 500  10 N m −  2  ( 0.0032 m ) ( 200 109 N m 2 )  

( 20.0 m )

= 2.78  10 −3 m  3  10 −3 m = 3 mm

82. The equation of motion is x = 0.25sin ( 4.70 t ) = A sin t. (a) The amplitude is A = xmax = 0.25 m . (b) The frequency is found by  = 2 f = 4.70 rad s → f = (c) The period is the reciprocal of the frequency. T = 1 f = (d) The total energy is given by the following.

(

4.70 s −1 2 2

4.7 s −1

)

= 0.748 Hz .

= 1.34 s .

2 Etotal = 12 mvmax = 12 m ( A) = 12 ( 0.650 kg )  4.70s −1 ( 0.25 m ) = 0.4487 J  0.45 J 2

(e) The potential energy is given by

(

U = 12 kx 2 = 12 m 2 x 2 = 12 ( 0.650 kg ) 4.70s −1

) ( 0.15 m ) = 0.1615 J  0.16 J . 2

The kinetic energy is given by K = Etotal − U = 0.4487 J − 0.1615 J = 0.2872 J  0.29 J . 83. The maximum velocity is given by Eq. 14–9a. 2 A 2 ( 0.15 m ) vmax =  A = 2 fA = = = 0.13m s T 7.0s The maximum acceleration is given by Eq. 14–9b. 2 4 2 A 4 ( 0.15 m ) amax =  2 A = = = 0.1209 m s2  0.12 m s2 2 2 T ( 7.0s ) amax g

0.1209 m s2 9.80 m s

= 1.2  10−2 = 1.2%

84. The spring constant does not change, but the mass does, and so the frequency will change. Use Eq. 14–7a to relate the spring constant, the mass, and the frequency for oxygen (O) and sulfur (S).

f =

1 2

fS = f O

k m mO mS

→

(

k 4

= f 2 m = constant → f O2 mO = f S2 mS →

= 3.7  1013 Hz

6.0 = 2.6  10 Hz ) 32.0 13

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Chapter 14

Oscillations

85. The period of a pendulum is given by T = 2 l g , and so the length is l =

l Austin = l Paris =

4 2 4 2

4 2

T 2 g Austin

T 2 g Paris

( 2.000s ) ( 9.793m s 2 )

T 2g

4 2

( 2.000s ) ( 9.809 m s 2 )

= 0.992238 m  0.9922 m

4 2

= 0.993859 m  0.9939 m

l Paris − l Austin = 0.993859 m − 0.992238 m = 0.001621m  1.6 mm

l Moon =

T 2 g Moon 4 2

( 2.00s ) (1.62 m s 2 ) 2

4 2

= 0.164 m

86. The force of the man’s weight causes the raft to partially sink (more than it would due to it’s own weight). That “extra” lowering of the raft displaces more water, and thus the water exerts a larger (upward) buoyant force on the raft. This extra buoyant force is a restoring force, because it is in the opposite direction of the force put on the raft by the man. This is analogous to pulling down on a mass–spring system that is in equilibrium, by applying an extra force. An effective spring constant can be calculated by knowing the mass of the man and the distance the raft sinks in reaching equilibrium. Then when the man steps off, the buoyant restoring force pushes upward on the raft, and thus the raft–water system acts like a spring. First find the spring constant. 2 Fextra ( 68 kg ) 9.80 m s k= = = 1.904  104 N m −2 3.5  10 m x (a) The frequency of vibration is determined by the “spring constant” and the mass of the raft.

(

fn =

)

1.904  104 N m

(

)(

= = 1.312 Hz  1.3 Hz 2 m 2 280 kg (b) As explained in the text, shortly after Eq. 14–2, the effects of gravity on a vertical spring (including gravitational potential energy) can be ignored if the displacement is measured from the oscillator’s equilibrium position. The total energy is thus

Etotal = 12 kA2 = 12 1.904  104 N m 3.5  10−2 m

) = 11.66 J  12 J . 2

87. For the pebble to lose contact with the board means that there is no normal force of the board on the pebble. If there is no normal force on the pebble, then the only force on the pebble is the force of gravity, and the acceleration of the pebble will be g downward, the acceleration due to gravity. This is the maximum downward acceleration that the pebble can have. Thus if the board’s downward acceleration exceeds g, then the pebble will lose contact. The maximum acceleration and the amplitude are related by amax = 4 2 f 2 A . amax = 4 2 f 2 A  g → A 

g 4 f 2

 2

9.80 m s2 4 ( 3.0 Hz ) 2

 2.8  10−2 m

88. (a) From conservation of energy, the initial kinetic energy of the car will all be changed into elastic potential energy by compressing the spring. E1 = E2 → 12 mv12 + 12 kx12 = 12 mv 22 + 12 kx22 → 12 mv12 = 12 kx22 →

( 25 m s ) k = m 2 = ( 950 kg ) = 3.711  10 4 N m  3.7  10 4 N m 2 x2 ( 4.0 m ) v12

523

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The car will be in contact with the spring for half a period, as it moves from the equilibrium location to maximum displacement and back to equilibrium. 1 2

T = 12 2

=

950 kg 3.711  104 N m

= 0.50 s

89. (a) The effective spring constant is found from the final displacement caused by the additional mass on the table. The weight of the mass will equal the upward force exerted by the compressed springs. Fgrav = Fsprings → mg = k y →

( 0.60 kg ) ( 9.80 m s2 ) = = 98 N m k= y ( 0.060 m ) mg

(b) We assume the collision takes place in such a short time that the springs do not compress a significant amount during the collision. Use momentum conservation to find the speed immediately after the collision. pbefore = pafter → mclay vclay = ( mclay + mtable ) vafter →

vafter =

mclay

clay

+ mtable )

vclay =

0.60 kg 2.40 kg

(1.65 m s ) = 0.4125 m s

As discussed in the text, if we measure displacements from the new equilibrium position, we may use an energy analysis of the spring motion without including the effects of gravity. The total elastic and kinetic energy immediately after the collision will be the maximum elastic energy, at the amplitude location. 2 2 E1 = E2 → 12 mtotal vafter + 12 kxafter = 12 kA2 → mtotal

 2.40 kg  2 2 ( 0.4125 m s ) + ( 0.060 m ) = 0.08813 m = 8.8 cm   98 N m 

2 2 vafter + xafter = 

90. We must make several assumptions. Consider a static displacement of the trampoline, by someone sitting on the trampoline mat. The upward elastic force of the trampoline must equal the downward force of gravity. We estimate that a 75-kg person will depress the trampoline about 25 cm at its midpoint. Hooke’s Law (Eq. 7–8) can then be solved for the effective spring constant. 2 mg ( 75 kg ) 9.80 m s kx = mg → k = = = 2940 N m  3000 N m 0.25 m x Another way to estimate it would be to have a person of known mass just “bounce” (not jump, so that they stay in contact with the trampoline at all times) and measure their period of vibration for a few oscillations. The spring constant can then be calculated from their mass and the period.

(

T = 2

)

4 2 m

→ k= k T2 Answers will vary based on the assumptions made. 91. The resonant angular frequency of the spring is equal to the angular frequency of the tires when traveling at 90.0 km/h. v 90.0 km h  1m s  = = 1   = 86.21rad s r 2 ( 0.58 m )  3.6 km h  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

524

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The spring constant can be calculated from this angular frequency and the mass of the tire/wheel combination. k 2 2 = → k =  2 m = ( 86.21rad s ) (17.0 kg ) = 1.263  105 N m m Hooke’s law (Eq. 7–8) can then be used to determine the distance the spring is compressed when the mass is added. We use the absolute value of the force and the distance, so that there is no negative sign in the equation. 2 1 F mg 4 ( 280 kg ) 9.80 m s F = kx → x = = = = 5.432  10−3 m  5.4  10 −3 m 1.263  105 N m k k

(

)

92. The effective spring constant is determined by the frequency of vibration and the mass of the oscillator. Use Eq. 14–7a.

f =

2

→

(

k = 4 2 f 2 m = 4 2 2.83  1013 Hz

) (16.00 u )  1.66 1u10 kg  = 8.40  10 N m −27





93. (a) The relationship between the velocity and the position of a SHO is given by Eq. 14–11b. Set that expression equal to half the maximum speed, and solve for the displacement. v =  vmax 1 − x 2 x02 = 12 vmax →  1 − x 2 x02 = 12

x 2 x02 = 43 →

→ 1 − x 2 x02 = 14

→

x =  3x0 2  0.866 x0

(b) Since F = − kx = ma for an object attached to a spring, the acceleration is proportional to the displacement (although in the opposite direction), as a = − x k m. Thus, the acceleration will have half its maximum value where the displacement has half its maximum value, at  12 x0 94. (a) The rod is a physical pendulum. Use Eq. 14–14 for the period of a physical pendulum.

T = 2

I mgh

= 2

1 3

ml 2

mg ( 12 l )

= 2

2l 3g

= 2

2 (1.00 m )

(

3 9.80 m s 2

)

= 1.64 s

(b) The simple pendulum has a period given by T = 2 l g . Use this to find the length.

T = 2

l simple g

= 2

2l 3g

→ l simple = 23 l = 23 (1.00 m ) = 0.667 m

95. We quote from the next to last paragraph of Appendix E: “… we see that at points within a solid sphere, say 1000 km below the Earth’s surface, only the mass up to that radius contributes to the net force. The outer shells beyond the point in question contribute zero net gravitational effect.” So when the mass is a distance r from the center of the Earth, there will be a force toward the center, opposite to r, due only to the mass within a sphere of radius r. We call that mass mr . It is the density of the (assumed uniform) Earth, times the volume within a sphere of radius r.

525

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

mr = Vr =

F =−

M Earth

Vr =

VEarth

Gmmr

M Earth 4 3

3  REarth

GmM Earth =−

4 3

r3 3 REarth

 r = M Earth

=−

Instructor Solutions Manual

3 REarth

GmM Earth

r 3 r r REarth The force on the apple is opposite to and proportional to the displacement, and so will execute GmM Earth . The time for the apple to return simple harmonic motion, with a “spring constant” of k = 3 REarth is the period, found from the “spring constant.” 2

T = 2

m k

3 m REarth = 2 = 2 GmM Earth GM Earth 3 REarth

= 2

( 6.38  10 m ) ( 6.67  10 N m kg )(5.98  10 kg ) 6

−11

= 507 s or 84.5min 96. The kinetic energy of the spring would be found by adding together the kinetic energy of each M infinitesimal part of the spring. The mass of an infinitesimal part is given by dM = S dx, and the D x speed of an infinitesimal part is v = v0 . Calculate the kinetic energy of the mass + spring. D D

 0 

K speed v = K mass + K spring = 12 mv02 + 12  v 2 dM = 12 mv02 + 12   v0 0

mass

v02 M S 1

= 12 mv02 +

= mv + 1 2

2 0

1 2

x  MS



D D

 x dx 2

2 0

v M S D3 D

= 12 v02 ( m + 13 M S )

So for a generic speed v, we have Kspeed v = 12 ( m + 13 M S ) v 2 . 97. Consider energy conservation for the mass over the range of motion from “letting go” (the highest point) to the lowest point. The mass falls the same distance that the spring is stretched, and has no kinetic energy at either endpoint. Call the lowest point the zero of gravitational potential energy. The variable “x” represents the amount that the spring is stretched from the equilibrium position.

Etop = Ebottom →

1 2

1  = 2 2

2g H

y=H

x=H

y=0

2 2 2 2 + mgytop + 12 kxtop = 12 mvbottom + mgybottom + 12 kxbottom mv top

0 + mgH + 0 = 0 + 0 + 12 kH 2 → f =

x=0

k m

(

2 9.80 m s

2

0.270 m

2g H 2

= 2 →  =

2g H

) = 1.36 Hz

526

Chapter 14

Oscillations

98. For an underdamped oscillation the period is approximately equal to the period of an undamped oscillation. The period is related to the mass and spring constant as in Eq. 14–7b. To measure the period of oscillation you can push down on the bumpers of the car and time the oscillation period. Then solving Eq. 14–7b gives the effective spring constant.

 2  → k = m  k  T 

T = 2

This spring constant is for the four springs acting together. To determine the spring constant of each individual spring, you would divide the mass of the car by four.

k1 =

m  2 



 =

4 T 

m 2 T2

99. We may use Eq. 10–14,  = I , as long as the axis of rotation is fixed in an inertial frame. We choose the axis to be at the point of support, perpendicular to the plane of motion of the pendulum. There are two forces on the pendulum bob, but only gravity causes any torque. Note that if the pendulum is displaced in the counterclockwise direction (as shown in Fig. 14–48 in the text), then the torque caused by gravity will be in the clockwise direction, and vice versa. See the free-body diagram in order to write Newton’s second law for rotation, with counterclockwise as the positive rotational direction. d 2    = − l = = mg sin I I  dt 2 If the angular displacement is limited to about 15, then sin    .

d 2

mg l

g  =−  l dt dt I mr 2 This is the equation of simple harmonic motion, with  = g l . Thus we can write the − mg l  = I

→

=−

 =−

displacement of the pendulum as follows, imitating Eq. 14–4.

 g  t +   =  max cos ( t +  ) →  =  max cos   l 

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CHAPTER 15: Wave Motion Responses to Questions 1.

The frequency of a simple periodic wave is equal to the frequency of its source. The wave is created by the source moving the wave medium that is in contact with the source. If you have one end of a taut string in your hand, and you move your hand with a frequency of 2 Hz, then the end of the string in your hand will be moving at 2 Hz, because it is in contact with your hand. Then those parts of the medium that you are moving exert forces on adjacent parts of the medium and cause them to oscillate. Since those two portions of the medium stay in contact with each other, they also must be moving with the same frequency. That can be repeated all along the medium, and so the entire wave throughout the medium has the same frequency as the source.

The speed of the transverse wave is the speed at which the wave disturbance moves along the cord. For a uniform cord, that speed is constant, and depends on the tension in the cord and the mass density of the cord. The speed of a tiny piece of the cord is a statement of how fast the piece of cord moves perpendicularly to the traveling wave, as the wave passes by. That speed is not constant–if a sinusoidal wave is traveling on the cord, the speed of each piece of the cord will be given by the speed relationship of a simple harmonic oscillator (Eq. 14–9a), which depends on the amplitude of the wave, the frequency of the wave, and the specific time of observation.

The maximum climb distance (4.3 m) occurs when the tall boat is at a crest and the short boat is in a trough. If we define the height difference of the boats on level seas as Δh and the wave amplitude as A, then Δh + 2A = 4.3 m. The minimum climb distance (2.5 m) occurs when the tall boat is in a trough and the short boat is at a crest. Then Δh–2A = 2.5 m. Solving these two equations for A gives a wave amplitude of 0.45 m.

(a) Striking the rod vertically from above will displace particles in a direction perpendicular to the rod and will set up primarily transverse waves. (b) Striking the rod horizontally parallel to its length will give the particles an initial displacement parallel to the rod and will set up primarily longitudinal waves.

From Eq. 15–4, the speed of waves in a gas is given by v = B  . A decrease in the density due to a temperature increase therefore leads to a higher speed of sound (assuming the bulk modulus is approximately constant). We expect the speed of sound to increase as temperature increases.

First, estimate the number of wave crests that pass a given point per second. This is the frequency of the wave. Then, estimate the distance between two successive crests, which is the wavelength. The product of the frequency and the wavelength is the speed of the wave.

The speed of a longitudinal wave is given in general by v = elastic force factor inertia factor . For a solid, that relationship is v = B  . Even though the density of solids is 1000 to 10,000 times greater than that of air, the elastic force factor (bulk modulus) of most solids is at least 106 times as great as the bulk modulus of air. This difference overcomes the larger density of most solids, and accounts for the greater speed of sound in most solids than the speed of sound in air.

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(i) Similar to the discussion in Section 15–3 for spherical waves, as a circular wave expands, the circumference of the wave increases. For the energy in the wave to be conserved, as the circumference increases, the intensity has to decrease. The intensity of the wave is proportional to the square of the amplitude. (ii) The water waves will decrease in amplitude due to dissipation of energy from viscosity in the water (dissipative or frictional energy loss).

Assuming the two waves are in the same medium, then they will both have the same speed. Since v = f  , the wave with the smaller wavelength will have twice the frequency of the other wave. From Eq. 15–7a, the intensity of wave is proportional to the square of the frequency of the wave (and also proportional to the speed of the wave, and the square of the amplitude of the wave). The speed and amplitude are the same for both waves. Thus, the wave with the shorter wavelength will transmit 4 times as much energy as the other wave.

10. Yes. Any function of ( x − v t ) will represent wave motion because it will satisfy the wave equation, Eq. 15–16. 11. The frequency must stay the same because the media is continuous–the end of one section of cord is physically tied to the other section of cord. If the end of the first section of cord is vibrating up and down with a given frequency, then since it is attached to the other section of cord, the other section must vibrate at the same frequency. If the two pieces of cord did not move at the same frequency, they would not stay connected, and then the waves would not pass from one section to another. 12. The transmitted wave has a shorter wavelength. If the wave is inverted upon reflection at the boundary between the two sections of rope, then the second section of rope must be heavier. Therefore, the transmitted wave (traveling in the heavier rope) will have a lower velocity than the incident wave or the reflected wave. The frequency does not change at the boundary, so the wavelength of the transmitted wave must also be smaller. 13. Assuming that there are no dissipative processes, then yes, the energy is conserved. The particles in the medium, which are set into motion by the wave, have both kinetic and potential energy. At the instant in which two waves interfere destructively, the displacement of the medium may be zero, but the particles of the medium will have velocity, and therefore kinetic energy. 14. Yes. If you touch the string at any node you will not disturb the motion. There will be nodes at each end as well as at the points one-third and two-thirds of the distance along the length of the string. 15. For a rope with a fixed end, the reflected pulse is inverted relative to the incoming pulse. For a rope with a free end, the reflected pulse is not inverted. See Fig. 15–18 for an illustration. For the fixed end, the rope puts a force on the connecting point as the pulse reaches the connecting point. The connecting point puts an equal and opposite force on the rope. This force is what “generates” the inverted reflected pulse. The point of connection is a node–a point of no motion. For a free end, the incoming pulse “whips” the end of the rope in a direction transverse to the wave motion, stretching it upwards. As that “whipped” end is pulled back towards the equilibrium position, this “whipping” motion generates a wave much in the same way that the pulse was originally created, and so creates a wave that is not reflected. 16. Although both longitudinal and transverse waves can travel through solids, only longitudinal waves can travel through liquids. Since longitudinal waves are detected on the Earth diametrically opposite from the location of an earthquake, but no transverse waves are detected at those locations, there must be some liquid as part of the Earth’s interior. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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17. This description of waves works well initially for both descriptions, but the waves continue after the initial motion. When the center is struck, a wave does move from the center to the rim, but then reflects from the rim back to the center. Likewise, when the rim is struck, a wave does move from the rim to the center, but the wave does not “stop” at the center. Once reaching the center, it then spreads out again to the rim. The amplitude of the waves also changes at the waves travel. As their radius increases, the amplitude decreases; and as the radius decreases, the amplitude increases. 18. Yes. A standing wave is an example of a resonance phenomenon, caused by constructive interference between a traveling wave and its reflection. The wave energy is distributed around the antinodes, which exhibit large amplitude oscillations, even when the generating oscillations from the hand are small. Note Eq. 15–18, where the amplitude of the standing wave is twice the amplitude of either constituent wave. 19. The energy of a wave is not localized at one point, because the wave is not localized at one point, and so to talk about the energy “at a node” being zero is not really a meaningful statement. The energy is not uniformly distributed along the string. Due to the interference of the waves the total energy of the medium particles at the nodes points is zero, but the energy of the medium is not zero at points of the medium that are not nodes. In fact, the antinode points have more energy than they would have if only one of the two waves were present. 20. When a hand or mechanical oscillator vibrates a string, the motion of the hand or oscillator is not exactly the same for each vibration. This variation in the generation of the wave leads to nodes which are not quite “true” nodes. In addition, real cords have damping forces which tend to reduce the energy of the wave. The reflected wave will have a smaller amplitude than the incident wave, so the two waves will not completely cancel, and the node will not be a true node. 21. AM radio waves have a much longer wavelength than FM radio waves. How much waves bend, or diffract, around obstacles depends on the wavelength of the wave in comparison to the size of the obstacle. A hill is much larger than the wavelength of FM waves, and so there will be a “shadow” region behind the hill. However, the hill is not large compared to the wavelength of AM signals, so the AM radio waves will bend around the hill. 22. Waves exhibit diffraction. If a barrier is placed between the energy source and the energy receiver, and energy is still received, it is a good indication that the energy is being carried by waves. If placement of the barrier stops the energy transfer, it may be because the energy is being transferred by particles or that the energy is being transferred by waves with wavelengths significantly smaller than the barrier.

Solutions to MisConceptual Questions 1.

(e) Students frequently have trouble distinguishing between the motion of a point on the cord and the motion of the wave on the cord. As a wave travels down the cord, a point on the cord will move vertically between the lowest point of the wave and the highest point on the wave. The wave and the point have the same amplitude. The point on the cord completes one up and down oscillation as each wavelength passes that point. Therefore, the motion of the point on the cord has the same frequency as the wave. The speed of the wave on the cord is determined by the wavelength and frequency. It is constant in time. The point on the cord moves perpendicular to the wave with a speed that varies with time. The maximum speed of the point is proportional to the wave amplitude and the wave frequency. Changing the amplitude will change the

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maximum speed of the point on the cord, but it does not change the wave speed. The wave speed and cord speed therefore are not equal. 2.

(a) A common misconception is that the waves are objects that can collide. Waves obey the superposition principle such that at any point on the rope the total displacement is the sum of the displacements from each wave. The waves pass through each other unaffected.

(d) Eq. 15–2 shows that the wave speed on a cord is related to the tension in the cord and the mass per unit length of the cord. The wave speed does not depend upon the amplitude, frequency, or wavelength. Stretching the cord increases the tension and decreases the mass per unit length, both of which increase the speed of the wave on the cord.

(d) The point on the string does not move horizontally, so answers (a) and (b) cannot be correct. The string only has zero velocity at the turning points (top and bottom) so (e) cannot be correct. Examining the graph shows that as the wave moves to the right the crest is approaching point B, so the string at B is traveling upward at this instant.

(d) Students frequently confuse the medium (an object) with the wave motion, and answer that the waves will collide and bounce off of each other. The waves obey the superposition principle such that at any point in the lake the amplitude of the wave is the sum of the individual amplitudes of each wave. This produces the various patterns when they overlap. The waves, however, will pass through each other and continue in their same pattern and direction after they pass.

(c) The speed of the wave along the slinky depends upon the mass of the slinky and the tension caused by stretching it. Since neither of those factors has changed the wave speed remains constant. The wave speed can also be written as the product of the wavelength and frequency. Therefore, as the frequency is increased, the wavelength must decrease.

(b) The standing wave is created by an initial sinusoidal wave that reflects from a fixed endpoint, and then the initial wave and reflected wave interfere. Since the waves are basically identical except for direction of travel, they both have the same amplitude (assuming no dissipation) and the same frequency. None of the solutions that have the waves traveling in the same direction (a, c, and e) can be correct, because the two waves have the same speed (since they are on the same medium), and thus they could not interfere. And since one of the waves is formed by a reflection of the other wave, they both have the same frequency, which eliminates answer (d). Also, two interfering waves of different frequencies would not make a “standing” pattern, but instead might produce something similar to “beats,” which are discussed in Chapter 16.

(a, c, d) The wave speed is dependent on the linear mass density of the medium, so the wave speed changes. The frequency does NOT change–for example, if 3 pulses per second arrive at the junction on the thin cord, 3 pulses per second will leave the junction on the heavy cord. Since the wave speed changes and the frequency does NOT change, the wavelength must change. The amplitude decreases, because there is both a reflected portion and transmitted portion (see Problem 42). The amplitude is related to the energy in the wave, and since the “incoming” energy must equal the “outgoing” energy, the “outgoing” amplitudes must be smaller than the “incoming” amplitude. Also see Fig. 15–19.

(a) A common misconception is that a wave transports matter as well as energy. However, as shown by a transverse wave on a horizontal string, the wave transports the disturbance down the string, but each part of the string stays at its initial horizontal position.

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10. (e) Eq. 15–5 shows that the energy transported by a wave is proportional to the square of the amplitude and the square of the frequency. Thus, if both the frequency and amplitude are 2 2 doubled, the energy transported increases by ( 2 )  ( 2 ) = 16. 11. (e) Eq. 15–2 for a transverse wave says that the wave speed is proportional to the square root of the

tension, v =



. Thus to double the speed, the tension must increase by a factor of 4.

Solutions to Problems 1.

The wave speed is given by v =  f . The period is 3.0 seconds, and the wavelength is 7.5 m. v =  f =  T = ( 7.5 m ) ( 3.0s ) = 2.5 m s

The distance between wave crests is the wavelength of the wave.  = v f = 343m s 265 Hz = 1.29 m

The elastic and bulk moduli are taken from Table 12–1. The densities are taken from Table 13–1. v= B  =

(a) For water:

(b) For granite:

v= E  =

2.0  109 N m 2

= 1400 m s

1.00  103 kg m 3 45  109 N m 2

= 4100 m s

2.7  103 kg m 3 200  109 N m 2

= 5100 m s

7.8  103 kg m 3

To find the wavelength, use  = v f . AM:

1 =

FM:

1 =

v f1 v f1

= =

3.00  108 m s 550  103 Hz 3.00  108 m s 88  106 Hz

= 545 m

2 =

= 3.41m

2 =

f2 v f2

3.00 108 m s 1600 103 Hz 3.00  108 m s 108  106 Hz

= 188 m

AM: 190 m to 550 m

= 2.78 m

FM: 2.8 m to 3.4 m

The speed of the longitudinal wave is given by Eq. 15–3, v = E  . The speed and the frequency are used to find the wavelength. The elastic bulk modulus is found in Table 12–1, and the density is found in Table 13–1.

=

v f

100  109 N m 2

 = f

7.8  103 kg m3 6200 Hz

= 0.58 m

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Chapter 15

Wave Motion

To find the time for a pulse to travel from one end of the cord to the other, the velocity of the pulse on the cord must be known. For a cord under tension, Eq. 15–2 gives v = FT  .

x t



m L

→ t =

x

7.2 m

( 0.65 kg ) ( 7.2 m )

m L 7.

= 0.20s

120 N

For a cord under tension, we have from Eq. 15–2 that v = FT  . The speed is also the distance traveled by the wave divided by the elapsed time, v =

x . The distance traveled is the length of the t

cord. FT

x t

→ FT =  v 2 = 

( t )

= 2

m l2

l ( t )

( t )

( 0.40 kg )( 8.7 m ) = 4.8 N 2 ( 0.85s )

The speed of the water wave is given by v = B  , where B is the bulk modulus of water, from Table 12–1, and  is the density of sea water, from Table 13–1. The wave travels twice the depth of the ocean during the elapsed time. v=



2l t

→ l =

vt 2



2.4 s

2.0  109 N m 2

1.025  103 kg m 3

= 1700 m

(a) The speed of the pulse is given by x 2 ( 710 m ) v= = = 83.53m s  84 m s t 17 s (b) The tension is related to the speed of the pulse by v = FT  . The mass per unit length of the cable can be found from its volume and density. m m = = → 2 V  ( d 2) l 2

−2 d  3 3  1.5  10 m   = =    =  ( 7.8  10 kg m )   = 1.378 kg m 2 l 2   2

v = FT  → FT = v 2  = ( 83.53m s ) (1.378 kg m ) = 9600 N 2

10. (a) Both waves travel the same distance, so x = v1 t1 = v 2 t2 . We let the smaller speed be v1, and the larger speed be v 2 . The slower wave will take longer to arrive, and so t1 is greater than t2 .

t1 = t2 + 1.5 min = t2 + 90s → v1 ( t2 + 102s ) = v2t2 → t2 =

v1 v2 − v1

( 90s ) =

5.5 km s 8.5 km s − 5.5km s

( 90s ) = 165s

x = v2t2 = ( 8.5 km s )(165s ) = 1400 km

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(b) This is not enough information to determine the epicenter. All that is known is the distance of the epicenter from the seismic station. The direction is not known, so the epicenter lies on a circle of radius 1400 km from the seismic station. Readings from at least two other seismic stations are needed to determine the epicenter’s position. 11. (a) The shape will not change. The wave will move 1.10 meters to the right in 1.00 seconds. See the graph. The parts of the string that are moving up or down at the “later” time are indicated.

(b) At the instant shown, the string at point A will be moving down. As the wave moves to the right, the string at point A will move down by 1 cm in the time it takes the “valley” between 1 m and 2 m to move to the right by about 0.25 m. y −1cm v= =  −4 cm s t 0.25 m 1.10 m s This answer will vary depending on the values read from the graph. 12. We assume that the wave will be transverse. The speed is given by Eq. 15–2. The tension in the wire is equal to the weight of the hanging mass. The linear mass density is the volume mass density times the cross-sectional area of the wire. The volume mass density is found in Table 13–1.



mball g = V

mball g = Al



(

( 4.4 kg ) ( 9.80 m s 2 )

) (

7800 kg m 3  0.50  10−3 m

)

= 84 m s

13. The speed of the longitudinal waves on the cord can be estimated from Eq. 15–3 for a long solid rod, v = E  . The time for the wave to travel from one child to the other is the distance divided by the speed. The value of Young’s modulus is found in Table 12–1.

1150 kg m 3  t = = x = (18 m ) = 1.11  10 −2 s  0.01s 9 2 E v 3  10 N m x

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Chapter 15

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14. (a) The speed of the wave at a point h above the lower end depends on the tension at that point and the linear mass density of the cord. The tension must equal the mass of the lower segment if the lower segment is in equilibrium. Use Eq. 15–2 for the wave speed. h mg FT h l = = gh FT = msegment g = mg ; v = m l 

l (b) We treat h as a variable, measured from the bottom of the cord. The wave speed at that point is given above as v = gh . The distance a wave would travel up the cord during a time dt is then

dh = vdt = gh dt. To find the total time for a wave to travel up the cord, integrate over the length of the cord.

ttotal = 

g 0

= 2

t total

dh = vdt = ghdt → dt =

→

 dt =  0

→

l g

15. For spherical waves, the intensity is inversely proportional to

, as shown in Eq. 15–8a. Thus the r2 intensity times r 2 is a constant. Evaluate the constant at 12 m and 24 m from the speaker.

( 24 m )2 I12 (12 m ) = I 24 ( 24 m ) → = =4 2 I 24 (12 m ) 2

I12

The intensity heard at 12 m is 4 times greater than the intensity heard at 24 m. 16. From Eq. 15–7, if the speed, medium density, and frequency of the two waves are the same, then the intensity is proportional to the square of the amplitude.

I 2 I1 = E2 E1 = A22 A12 = 3.5 → A2 A1 = 3.5 = 1.9 The more energetic wave has the larger amplitude. 17. (a) Assume that the earthquake waves spread out spherically from the source. Under those conditions, Eqs. (15–8a,b) apply, stating that intensity is inversely proportional to the square of the distance from the source of the wave. I 55 km I15 km = (15 km )

( 55 km )2 = 7.4  10−2 or I15 km I 55 km = ( 55 km )2 (15 km )2 = 13

(b) The intensity is proportional to the square of the amplitude, and so the amplitude is inversely proportional to the distance from the source of the wave. A55 km A15 km = 15km 55km = 0.27 or A15 km A55 km = 55km 15km = 3.7 18. (a) Assuming spherically symmetric waves, the intensity will be inversely proportional to the square of the distance from the source. Thus Ir 2 will be constant. 2 = I far rfar2 → I near rnear I near = I far

rfar2 2 rnear

(

= 2.4  106 W m 2

)(

48 km )

(1.0 km )

= 5.53  109 W m 2  5.5  109 W m 2

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(b) The power passing through an area is the intensity times the area, as given in Eq. 15–7a.

(

P = I S = 5.53  109 W m2

)( 2.0 m ) = 1.1  10 W 2

19. (a) The power transmitted by the wave is assumed to be the same as the output of the oscillator. That power is given by Eq. 15–6. The wave speed is given by Eq. 15–2. Note that the mass per unit length can be expressed as the volume mass density times the cross-sectional area, S. F FT 2 2 P = 2 2  S vf 2 A2 = 2 2  S T f 2 A2 = 2 2  S f A = 2 2 f 2 A2 S  FT  S = 2 2 ( 60.0 Hz ) ( 0.0040 m ) 2

 ( 5.0  10 −4 m ) ( 7800 kg m 3 ) ( 7.5 N ) = 0.24 W 2

(b) The frequency and amplitude are both squared in the equation. Thus, if the power is constant and the frequency doubles, the amplitude must be halved, and so will be 0.20 cm . 20. Consider a wave traveling through an area S with speed v, much like Figure 15–11. Start with Eq. 15–7a, and use Eq. 15–6. P E El E l energy I= = = = = v S St S l t S l t volume 21. (a) We start with Eq. 15–6. The linear mass density is the mass of a given volume of the cord divided by the cross-sectional area of the cord. m V  S l P = 2 2  S vf 2 A2 ;  = = = =  S → P = 2 2  vf 2 A2

(b) The speed of the transverse wave is found from the given tension and mass density, according to Eq. 15–2. P = 2 2  vf 2 A2 = 2 2 f 2 A2  FT  = 2 2 f 2 A2  FT = 2 2 (120 Hz ) ( 0.024 m ) 2

( 0.10 kg m )(135 N ) = 6.0  102 W

22. We assume that all of the wave motion is outward along the surface of the water–no waves are propagated downwards. Consider two concentric circles on the surface of the water, centered on the place where the circular waves are generated. If there is no damping, then the power (energy per unit time) being transferred across the boundary of each of those circles must be the same. Or, the power associated with the wave must be the same at each circular boundary. The intensity depends on the amplitude squared, so for the power we have this. P = I ( 2 r ) = kA2 2 r = constant → A2 =

constant 2 rk

→

A

23. (a) The only difference is the direction of motion.

D ( x, t ) = 0.015sin ( 35x + 1200t ) (b) The speed is found from the wave number and the angular frequency, Eq. 15–12.  1200 rad s v= = = 34 m s k 35 rad m

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Chapter 15

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24. To represent a wave traveling to the left, we replace x by x + vt. The resulting expression can be given in various forms.

 

 x t   x vt   +  +   = A sin  2  +  +         T  

D = A sin  2 ( x + vt )  +   = A sin  2  = A sin ( kx +  t +  )

25. The traveling wave is given by D = 0.22 sin ( 5.6 x + 34t ) . (a) The wavelength is found from the coefficient of x. 2 2 5.6 m −1 = → = = 1.122 m  1.1m  5.6 m −1 (b) The frequency is found from the coefficient of t. 34 s −1 34 s −1 = 2 f → f = = 5.411Hz  5.4 Hz 2 (c) The amplitude is the coefficient of the sine function, and so is 0.22 m . (d) The wave velocity is the ratio of the coefficients of t and x. 2 34s −1 v=f = = 6.071m s  6.1m s 5.6 m −1 2 Because both coefficients are positive, the velocity is in the negative x-direction. (e) The particles on the cord move in simple harmonic motion with the same frequency as the wave. From Chapter 14, v max = D = 2 f D.

 34s −1   ( 0.22 m ) = 7.5 m s  2 

vmax = 2 f D = 2 

The minimum speed is when a particle is at a turning point of its motion, at which time the speed is 0. Thus vmin = 0 . 26. The traveling wave in Example 15–6 is given by the following expression: D ( x, t ) = ( 0.026 m ) sin ( 4.5 rad m ) x − (157 rad s ) t + 0.66 rad  . We substitute in x = 1.00 m, and look at the expression at that point. D (1.00 m, t ) = ( 0.026 m ) sin  4.5 − 157 s −1 t + 0.66 = ( 0.026 m ) sin 5.16 − 157 s −1 t 

(

(a) a =

d 2 D (1.00 m, t ) dt

(

amax = 157s−1

(

= − 157 s −1

dD (1.00 m, t )

(

)

) ( 0.026 m ) sin 5.16 − (157 s ) t  2

) (0.026 m) = 640 m s 2

)

−1

)

(

)

= − 157 s −1 ( 0.026 m ) cos 5.16 − 157 s −1 t  dt v (1.00 m, 2.15s ) = − (157 s −1 ) ( 0.026 m ) cos 5.16 − (157 s −1 ) ( 2.15s ) 

(b) v =

= −3.325 m s  −3.3m s

(

a (1.00 m, 2.15s ) = − 157s −1

) (0.026 m) sin 5.16 − (157s ) (2.15s ) 2

−1

= −371.8m s2  −370 m s2 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

537

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

27. (a) See the graph to the right.

(b) For motion to the right, replace x by x − vt.

D ( x, t ) = ( 0.45m ) cos 2.6 ( x − 2.0t ) + 1.2 (c) See the graph above. (d) For motion to the left, replace x by x + vt. Also see the graph above.

D ( x, t ) = ( 0.45m ) cos 2.6 ( x + 2.0t ) + 1.2 28. We take the wave function to be D ( x , t ) = A sin ( kx −  t ) . The wave speed is given by v =

  = , k f

while the speed of particles on the cord is given by choosing some point on the cord (x = 0 is dD . convenient) and then calculating dt dD ( 0, t ) d  dD ( 0, t )  =  A sin ( − t ) = − A cos ( − t ) →   = A dt dt  dt  max

 A = 0.10v = 0.10

 k

→ A = 0.10

1 k

= 0.10

 (10.0 cm ) = 0.10 = 0.16 cm 2 2

29. (a) The wavelength is the speed divided by the frequency. v 345 m s = = = 0.603 m f 572 Hz (b) In general, the phase change in degrees due to a time difference is given by

 360

t T

= f t → t =

1  f 360

 360

360

x



→  =

x



( 360 ) =

t T

 90  = 4.37  10−4 s   572 Hz  360  1

0.044 m 0.603m

 360

x



( 360 ) = 26

538

Chapter 15

Wave Motion

30. The amplitude is 0.020 cm, the wavelength is 0.603 m, and the frequency is 572 Hz. The displacement is at its most negative value at x = 0, t = 0, and so the wave can be represented by a cosine that is phase shifted by half of a cycle. D ( x, t ) = A cos ( kx −  t +  )

A = 0.020 cm ; k =

2



2 f

2 ( 572 Hz ) 345 m s

(

= 10.4 m −1 ;  = 2 f = 2 ( 572 Hz ) = 3590 rad s

)

D ( x, t ) = ( 0.020 cm ) cos  10.4 m −1 x − ( 3590 rad s ) t +   , x in m, t in s Other equivalent expressions include the following. D ( x, t ) = − ( 0.020 cm ) cos  10.4 m −1 x − ( 3590 rad s ) t 

(

)

(

)

D ( x, t ) = ( 0.020 cm ) sin  10.4 m −1 x − ( 3590 rad s ) t + 32  

31. (a) For the particle of string at x = 0, the displacement is not at the full amplitude at t = 0. The particle is moving upwards, and so a maximum is approaching from the right. The general form of the wave is given by D ( x, t ) = A sin ( kx +  t +  ) . At x = 0 and t = 0, D ( 0, 0 ) = A sin  and so we can find the phase angle. D ( 0,0) = A sin  → 0.80cm = (1.00cm ) sin  →  = sin −1 ( 0.80 ) = 0.93

 2 x + 0.93  , x in cm. See the graph. It matches the description   3.0 

So we have D ( x , 0 ) = A sin 

given earlier. (b) We use the given data to write the wave function. Note that the wave is moving to the right, and that the phase angle has already been determined. D ( x, t ) = A sin ( kx +  t +  ) A = 1.00 cm ; k =

2 3.00 cm

(

= 2.09 cm −1 ;  = 2 f = 2 ( 245 Hz ) = 1540 rad s

)

D ( x, t ) = (1.00 cm ) sin  2.09 cm −1 x + (1540 rad s ) t + 0.93 , x in cm, t in s

32. The displacement of a point on the cord is given by the wave, D ( x, t ) = ( 0.12 m ) sin ( 3.0 x − 15.0t ) . (a) The angular frequency  is the coefficient of t. Use that to find the period and frequency.  15.0 rad s = = 2.3873 Hz  2.39 Hz  = 15.0 rad s ; f = 2 2 rad T=

1 f

2



2 rad 15.0 rad s

= 0.41888s  0.419 s

539

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) At x = 0.60 m , D ( 0.60 m, t ) = ( 0.12 m ) sin (1.8 − 15.0t ) .

(

)

D ( 0.60 m, 0.20s ) = ( 0.12 m ) sin 1.8 − 15.0s −1 ( 0.20s )  = −0.11m

dD ( 0.60 m, t )

v ( 0.60 m, t ) =

(

)

= ( 0.12 m ) −15.0 s −1 cos (1.8 − 15.0 t )

(

)

v ( 0.60 m, 0.20 s ) = ( −1.8 m s ) cos 1.8 − 15.0 s −1 ( 0.20 s )  = −0.65 m s d v ( 0.60 m, t )

a ( 0.60 m, t ) =

(

)

= − ( 0.12 m ) −15.0 s −1 sin (1.8 − 15.0 t )

)

(

)

a ( 0.60 m, 0.20 s ) = −27 m s 2 sin 1.8 − 15.0 s −1 ( 0.20 s )  = 25 m s 2 (c) The maximum speed is the amplitude of the velocity of any point on the cord, and the maximum acceleration is the amplitude of the acceleration of any point on the cord. D ( x0 , t ) = ( 0.12 m ) sin ( 3.0 x0 − 15.0t ) dD ( x0 , t )

v ( x0 , t ) =

(

dt = ( −1.8 m s ) cos ( 3.0 x0 − 15.0t ) d v ( x0 , t )

a ( x0 , t ) =

(

)

= ( 0.12 m ) −15.0 s −1 cos ( 3.0 x0 − 15.0t )

(

)

= − ( 0.12 m ) −15.0 s −1 sin ( 3.0 x0 − 15.0t ) 2

)

= −27 m s 2 sin ( 3.0 x0 − 15.0t ) v max = 1.8 m s ; amax = 27 m s 2

(d) We must assume that the point x is fixed, and so we differentiate with respect to t only. D ( x, t ) = ( 0.12 m ) sin ( 3.0 x − 15.0t ) v ( x, t ) =

dD ( x, t ) dt

(

)

= ( 0.12 m ) −15.0 s −1 cos ( 3.0 x − 15.0t )

Now set t = 0 to get v ( x, 0 ) = ( −1.8 m s ) cos ( 3.0 x )

(e) That expression was calculated in part (b). dD ( 0.60 m, t ) v ( 0.60 m, t ) = = ( −1.8 m s ) cos (1.8 − 15.0 t ) dt 33. To be a solution of the wave equation, the function must satisfy Eq. 15–16, D = A cos ( t − kx ) D x D t

= ( − k ) A  − sin ( t − kx )  = kA sin ( t − kx ) ;

2D

=  A  − sin ( t − kx )  = − A sin ( t − kx ) ;

D

D 2

This gives

= 2

k D 2

, and since v = 2

x  2 t Yes, the function is a solution.

x 2

t

2 D x 2

1 2 D v 2 t 2

= − k 2 A cos ( t − kx )

= − 2 A cos ( t − kx )

2 D 1 2 D  from Eq. 15–12, we have = . x 2 v 2 t 2 k

540

Chapter 15

Wave Motion

34. To be a solution of the wave equation, the function must satisfy Eq. 15–16, D = A sin kx cos  t D x D t

= kA cos kx cos  t ;

2D

x 2

1 2 D v 2 t 2

= − k 2 A sin kx cos  t

x 2

2D

= − A sin kx sin  t ;

2 D

t 2

= − 2 A sin kx cos  t

2 D 1 2 D  from Eq. 15–12, we have = . x 2 v 2 t 2  2 t x k Yes, the function is a solution . Note: this is a general form for a standing wave.

This gives

2 D

k 2 2 D

, and since v = 2

= 2

35. To be a solution of the wave equation, the function must satisfy Eq. 15–16, (a) D = A ln ( x + vt ) D x

;

x + vt

This gives

x

( x + vt )

1 2 D

x

v 2 t 2

= 2

D

;

t

Av x + vt

;

2D

Av 2

t

( x + vt )

=− 2

1 2 D

x

v 2 t 2

= 2

and so yes, the function is a solution .

= 4 ( x − vt )

 D

;

This gives

=− 2

2 D

(b) D = ( x − vt ) D

2D

2 D

x 2

2 D

= 12 ( x − vt ) 2

x 1 2 D v 2 t 2

D

;

t

= −4v ( x − vt )

;

2D t

= 12v 2 ( x − vt )

and so yes, the function is a solution .

36. We find the various derivatives for the function from Eq. 15–13c. D 2D D ( x , t ) = A sin ( kx +  t ) ; = Ak cos ( kx +  t ) ; = − Ak 2 sin ( kx +  t ) ; 2 x x D t

= A cos ( kx +  t ) ;

2D t 2

= − A 2 sin ( kx +  t )

To satisfy the wave equation, we must have 2 D

= 2

1 2 D

2 D

1 2 D

x

v 2 t 2

→ − Ak 2 sin ( kx +  t ) =

= 2 1

x v 2 t 2 v2 Since v =  k , the wave equation is satisfied.

(

− A 2 sin ( kx +  t )

)

→ k2 =

2 v2

We find the various derivatives for the function from Eq. 15–15. Make the substitution that u = x + v t , and then use the chain rule.

541

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

 d dD  u d D = = = = D ( x, t ) = D ( x + v t ) = D ( u ) ; ;  = 2 x du x du x x du  dx du  x du 2 D t

dD u

du t

;

D

dD u

2D

2   dD   dD d 2D  d dD  u 2 d D v v v v v v = = = =     du 2 du 2 t  du  t du  du du  t

t

= 2

To satisfy the wave equation, we must have

2D

2 D

= 2

1 2 D

 dD

x v 2 t 2 2 D 1 2 D d 2D 1 2 d 2D d 2D = → = v = x 2 v 2 t 2 du 2 v 2 du 2 du 2 Since we have an identity, the wave equation is satisfied.

37. Find the various derivatives for the linear combination. D ( x, t ) = C1 D1 + C2 D2 = C1 f1 ( x, t ) + C2 f 2 ( x, t ) D x D t

= C1 = C1

f1

f 2

+ C2

x f1

x f 2

+ C2

t

; ;

 2 f1

 2 f2

x

x 2

= C1 2

+ C2 2

2 D

 2 f1

t

= C1 2

 2 f2

+ C2 2

t 2 2 D 1 2 D To satisfy the wave equation, we must have = . Use the fact that both f1 and f 2 satisfy x 2 v 2 t 2 the wave equation.  1  2 f1   1  2 f 2  1   2 f1  2 f1  2 f2  2 f2  1  2 D 2 D = C1 2 + C2 2 = C1  2 2  + C2  2 2  = 2 C1 2 + C2 2  = 2 2 x 2 x x t  v t  v t   v t  v  t

Thus we see that

t

2 D

2 D x 2

1 2 D v 2 t 2

, and so D satisfies the wave equation. This is the concept of

superposition. 38. To be a solution of the wave equation, the function must satisfy Eq. 15–16, 2

= −2k ( kx −  t )  −2k ( kx −  t ) e (

− kx − t )

 D 2

x D

 D x

;

x

− kx − t )



− kx − t )



1  D 2

v t 2

1 2 D

x

v 2 t 2

= 2

 + ( −2k 2 ) e −( kx − t ) = 2k 2  2 ( kx −  t ) 2 − 1 e −( kx − t )    2

= 2 ( kx −  t )  2 ( kx −  t ) e (

= 2 ( kx −  t ) e (

t 2 D

t

D

= −2k ( kx −  t ) e (

D=e (

2 D

 + ( −2 2 ) e −( kx − t ) = 2 2  2 ( kx −  t )2 − 1 e −( kx − t )    2

→ 2k 2  2 ( kx −  t ) − 1 e (





− kx − t )

1 v

2 2  2 ( kx −  t ) − 1 e (





− kx − t )

→

2 k = 2 v  Since v = , we have an identity. Yes, the function is a solution. k 2

542

Chapter 15

Wave Motion

 for the wave given by D = A sin ( kx −  t ) .

39. We assume that A

D = A sin ( kx −  t ) → v  = A  v max v

 v max

 →

A v

2 fA v

t

 = A = − A cos ( kx −  t ) → v max

  → vmax

 =

D

 = v wave →

 vmax

v wave



2 f

100 =   0.063 f 50

40. (a) The speed of the wave in a stretched cord is given by Eq. 15–2, v = FT  . The tensions must be the same in both parts of the cord. If they were not the same, then the net longitudinal force on the joint between the two parts would not be zero, and the joint would have to accelerate along the length of the cord. vH

v = FT  →

FT  H

FT  L

L H

(b) The frequency must be the same in both sections. If it were not, then the joint between the two sections would not be able to keep the two sections together. The ends could not stay in phase with each other if the frequencies were different. f =

→



H

→

L

H vH = = L vL

L H

(

) (

)

41. (a) For the wave in the lighter cord, D ( x, t ) = ( 0.050 m ) sin  7.5 m −1 x − 12.0s −1 t  .

=

2 k

2

( 7.5m ) −1

= 0.84 m

(b) The tension is found from the velocity, using Eq. 15–2. v=



12.0 s −1 ) ( 2 → FT =  v =  2 = ( 0.10 kg m ) = 0.26 N 2 k ( 7.5 m −1 ) 2

= 1

12 22 =  2 k12 k 22

→ 2 = 1

1 2 = k1 2

2 1 = 0.5 = 0.59 m 2 ( 7.5 m −1 )

42. The tension and the frequency do not change from one side of the knot to the other. (a) We force the cord to be continuous at x = 0 for all times. This is done by setting the initial wave plus the reflected wave (the displacement of a point infinitesimally to the LEFT of x = 0) equal to the transmitted wave (the displacement of a point infinitesimally to the RIGHT of x = 0) for all times. We also use the facts that sin ( − ) = − sin  and k1v1 = k 2 v 2 .

543

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

D ( 0, t ) + DR ( 0, t ) = DT ( 0, t ) → A sin ( −k1v1t ) + AR sin ( k1v1t ) = AT sin ( −k 2 v2t ) → − A sin ( k1v1t ) + AR sin ( k1v1t ) = − AT sin ( k2 v2t ) = − AT sin ( k1v1t ) → − A + AR = − AT →

A = AT + AR

While this relationship was derived at the origin, it is a relationship of the amplitudes, and so it applies in general–not just at the origin. (b) To make the slopes match for all times, we must have

  D ( x, t ) + DR ( x, t ) =   DT ( x, t ) x x

when evaluated at the origin. We also use the result of the above derivation, and the facts that cos ( − ) = cos  and k1v1 = k 2 v 2 .  x

 D ( x, t ) + D ( x, t )

x =0

 x

 D ( x, t )

→

x =0

k1 A cos ( − k1v1t ) + k1 AR cos ( k1v1t ) = k 2 AT cos ( − k 2 v2t ) → k1 A cos ( k1v1t ) + k1 AR cos ( k1v1t ) = k 2 AT cos ( k 2 v2 t ) →

 k2 − k1  A  k2 + k1 

k1 A + k1 AR = k 2 AT = k 2 ( A − AR ) → AR = 

Use k 2 = k1

v1 v2

 k v1 − k   v1 − 1   v1 − v 2  1 1     v v   k −k   v −v  v2 k v 2 A= 1 2 A= 2 A=  1 2 A AR =  2 1  A =  k1  v1  k v1 + k    v1 + v2   k2 + k1   v1 + v2  1   1v  v +1 v v     2   2 2 2  As mentioned above, this is a general result, even though it was derived from considerations at the knot (the origin). (c) Combine the results from the previous two parts.

  k 2 − k1    k 2 + k1   k 2 − k1    k2 − k1   A = A 1 −    = A  −   k2 + k1    k 2 + k1    k 2 + k1   k 2 + k1  

AT = A − AR = A − 

    2k1   2v 2  2k1   A=  =  A =  A  k v1 + k   k2 + k1   v1 + v2  1   1v   2 43. Note that the diagrams are approximations. (a)

(b)

(c) The energy is all kinetic energy at the moment the string has 0 displacement. There is no elastic potential energy then. Each piece of the string has speed but no displacement. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

544

Chapter 15

Wave Motion

44. (a) The resultant wave is the algebraic sum of the two component waves. D = D1 + D2 = A sin ( kx −  t ) + A sin ( kx −  t +  ) = A sin ( kx −  t ) + A sin ( kx −  t +  ) 

= A2 sin 12 ( kx −  t ) + ( kx −  t +  ) cos 12 ( kx −  t ) − ( kx −  t +  ) 

    = 2 Asin 12 ( 2kx − 2 t +  )cos 12 ( ) =  2 A cos  sin  kx −  t +  2  2  (b) The amplitude is the absolute value of the coefficient of the sine function, 2 A cos

 . The 2

  wave is purely sinusoidal because the dependence on x and t is sin  kx −  t +  . 2  (c) If  = 0, 2 , 4 ,

, 2n , then the amplitude is 2 A cos

 2n = 2 A cos = 2 A cos n = 2 A ( 1) 2 2 , ( 2n + 1)  , then the amplitude

= 2 A, which is constructive interference. If  =  ,3 ,5 , is 2 A cos

 ( 2n + 1)  = 2 A cos = 2 A cos ( n + 12 )   = 0 , which is destructive interference. 2 2

 , then the resultant wave is as follows. 2           D =  2 A cos  sin  kx −  t +  =  2 A cos  sin  kx −  t +  = 2 A sin  kx −  t +  2  2  4  4 4  

(d) If  =

This wave has an amplitude of 2 A , is traveling in the positive x-direction, and is shifted to the left by an eighth of a cycle. This is “halfway” between the two original waves. The displacement is 12 A at the origin at t = 0. 45. The fundamental frequency of the full string is given by f unfingered =

= 441Hz . If the length is 2l reduced to 2/3 of its current value, and the velocity of waves on the string is not changed, then the new frequency will be as follows. 3 v 3 v 3 f fingered = 2 = =   f unfingered =   ( 441Hz ) = 662 Hz 2 ( 3 l ) 2 2l  2  2

46. The frequencies of the harmonics of a string that is fixed at both ends are given by f n = nf1 , and so the first four harmonics are f1 = 294 Hz, f 2 = 588 Hz, f3 = 882 Hz, f 4 = 1176 Hz . 47. Four loops is the standing wave pattern for the 4th harmonic, with a frequency given by f 4 = 4 f1 = 320 Hz. Thus f1 = 80 Hz, f 2 = 160 Hz, f3 = 240 Hz, and f5 = 400 Hz are all other resonant frequencies. f1 is the fundamental or first harmonic, f 2 is the first overtone or second harmonic, f 3 is the second overtone or third harmonic, and f 5 is the fourth overtone or fifth harmonic.

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48. The oscillation corresponds to the fundamental. The frequency of that oscillation is 1 1 f1 = = = 0.769 Hz The bridge, with both ends fixed, is similar to a vibrating string, and so T 1.3s

f n = n f1 = Tn =

1.3s n

n 1.3

Hz, n = 1, 2, 3

, n = 1, 2, 3

. The periods are the reciprocals of the frequency, and so

49. Since f n = nf1 , two successive overtones differ by the fundamental frequency, as shown below. f = f n +1 − f n = ( n + 1) f1 − n f1 = f1 = 320 Hz − 240 Hz = 80 Hz

50. Adjacent nodes are separated by a half-wavelength, as examination of Figure 15–26b will show. 96 m s v v = → xnode = 12  = = = 0.1103 m  0.11m f 2 f 2 ( 435 Hz ) 51. The speed of the wave is given by Eq. 15–2, v = FT  . The wavelength of the fundamental is

1 = 2 l . Thus the frequency of the fundamental is f1 = vibrating string, and so f n = n f1 =



1

, n = 1, 2,3,



. Each harmonic is present in a

52. The speed of waves on the string is given by Eq. 15–2, v = string with both ends fixed are given by Eq. 15–17b, f n =

FT  . The resonant frequencies of a nv

, where l vib is the length of the 2 l vib portion that is actually vibrating. Combining these relationships allows the frequencies to be calculated. fn =

2 l vib



n 2 l vib

f 2 = 2 f1 = 600.86 Hz

FT m l

f1 =

1 2 ( 0.64 m )

520 N

( 3.2 10 kg ) ( 0.91m ) −3

= 300.43 Hz

f3 = 3 f1 = 901.29 Hz

So the three frequencies are 300 Hz, 600 Hz, 900 Hz , to 2 significant figures. 53. The string must oscillate in a standing wave pattern to have a certain number of loops. The frequency of the standing waves will all be 60.0 Hz, the same as the oscillator. That frequency is nv also expressed by Eq. 15–7b, f n = . The speed of waves on the string is given by Eq. 15–2, 2l

v = FT  . The tension in the string will be the same as the weight of the masses hung from the end of the string, FT = mg , ignoring the mass of the string itself. Combining these relationships gives an expression for the masses hung from the end of the string. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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fn =



(

4 l 2 f n2  n2 g

4 (1.50 m ) ( 60.0 Hz ) 3.9  10−4 kg m 2

m1 =

(

1 9.80 m s m1

(b) m2 = (c)

→ m=

2 m1

m5 =

1.289 kg 4 1.289 kg

)

) = 1.289 kg  1.3kg

= 0.32 kg

= 5.2  10−2 kg

54. The tension in the string is the weight of the hanging mass, FT = mg . The speed of waves on the FT

, and the frequency is given as f = 60 Hz . The wavelength   of waves created on the string will thus be given by string can be found by v =

=

v f



( 0.070 kg ) ( 9.80 m s 2 )

( 3.9 10 kg m ) −4

60.0 Hz

= 0.6990 m .

The length of the string must be an integer multiple of half of the wavelength for there to be nodes at both ends and thus form a standing wave. Thus l =  2,  , 3 2, n 2 . The number of standing wave patterns is given by the number of integers that satisfy 0.10 m  n 2  1.5 m.

0.10 m  n 2 → n 

n 2  1.5m → n 

2 ( 0.10 m )



2 (1.5m )



2 ( 0.10 m ) 0.6990 m

2 (1.5m ) 0.6990 m

= 0.286

= 4.29

Thus we see that we must have n from 1 to 4, and so there are 4 standing wave patterns that may be achieved. 55. The standing wave is given by D = ( 2.4 cm ) sin ( 0.60 x ) cos ( 42t ) . (a) The distance between nodes is half of a wavelength.  2 d = 12  = 12 = = 5.236 cm  5.2 cm k 0.60 cm −1 (b) The component waves travel in opposite directions. Each has the same frequency and speed, and each has half the amplitude of the standing wave.  42 s −1 A = 12 ( 2.4 cm ) = 1.2 cm ; f = = = 6.685 Hz  6.7 Hz ; 2 2

v =  f = 2d node f = 2 ( 5.236 cm )( 6.685 Hz ) = 70.01cm s  70 cm s Or, v =

 k

42 rad s 0.60 rad s

( 2 sig. figs.)

= 70 cm s ( 2 sig. figs.)

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

Instructor Solutions Manual

, when x is at the particle’s location.

)

D ( x = 3.80 cm ) = ( 2.4 cm ) sin  0.60 cm −1 ( 3.80 cm )  cos ( 42t ) = (1.821cm ) cos ( 42t ) dD ( x = 3.80 cm )

(1.821cm ) cos ( 42t ) dt dt = (1.821cm )( −42 rad s ) sin ( 42t )

v ( 3.80 cm, t ) =

v ( 3.80 cm, 2.5s ) = ( −76.48 cm s ) sin ( 42 rad s )( 2.5s ) = 74.226 cm s  74 cm s 56. (a) The given wave is D1 = 4.2 sin ( 0.84 x − 47t + 2.1) . To produce a standing wave, we simply need to add a wave of the same characteristics but traveling in the opposite direction. This is the appropriate wave.

D2 = 4.2sin ( 0.84 x + 47t + 2.1) (b) The standing wave is the sum of the two component waves. We use the trigonometric identity that sin 1 + sin  2 = 2 sin 12 (1 +  2 ) cos 12 (1 −  2 ) .

D = D1 + D2 = 4.2 sin ( 0.84 x − 47t + 2.1) + 4.2 sin ( 0.84 x + 47t + 2.1) = 4.2 ( 2 ) sin 12 ( 0.84 x − 47t + 2.1) + ( 0.84 x + 47t + 2.1)  

cos ( 0.84 x − 47t + 2.1) − ( 0.84 x + 47t + 2.1) 1 2

= 8.4 sin ( 0.84 x + 2.1) cos ( −47t ) = 8.4 sin ( 0.84 x + 2.1) cos ( 47t ) We note that the origin is NOT a node. 57. The frequency is given by f =





. The wavelength and the mass density do not change

when the string is tightened. f =





→

f2 f1









F2 F1

→ f 2 = f1

F2 F1

= ( 294 Hz ) 1.22 = 325 Hz

58. (a) Plotting one full wavelength means from x = 0 to 2 2 x= = = = 1.795 m k 3.5 m −1  1.8 m. The functions to be plotted are given below. Note that  t = (1.6 rad s )(1.5s ) = 2.4 rad.

(

)

(

)

D1 = ( 0.15 m ) sin  3.5 m −1 x − 2.4  and D2 = ( 0.15 m ) sin  3.5 m −1 x + 2.4  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) The sum D1 + D2 is plotted, and the nodes and antinodes are indicated. The analytic result is given below.

(

) = ( 0.30 m ) sin ( 3.5 m x ) cos ( 2.4 )

(

)

D1 + D2 = ( 0.15 m ) sin  3.5 m −1 x − 2.4  + ( 0.15 m ) sin  3.5 m −1 x + 2.4  −1

This expression should have nodes and antinodes at positions given by the following. n → x= = 0, 0.90 m, 1.80 m 3.5 m −1 xnode = n , n = 0,1, 2 3.5 ( n + 12 )  = 0.45 m, 1.35 m → x= 3.5 m −1 xantinode = ( n + 12 )  , n = 0,1, 2 3.5 The graph agrees with the calculations. 59. The standing wave formed from the two individual waves is given below. The period is given by T = 2  = 2 1.6s −1 = 3.9 s.

(

) ( ) = ( 0.30 m ) sin ( 3.5 m x ) cos (1.6s t )

(

)

(

)

D1 + D2 = ( 0.15 m ) sin  3.5 m −1 x − 1.6s −1 t  + ( 0.15 m ) sin  3.5 m −1 x + 1.6s −1 t  −1

−1

(a) For the point x = 0, we see that the sum of the two waves is identically 0. This means that the point x = 0 is a node of the standing wave.

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(b) For the point x =  4 , we see that the position factor is  2   sin     4

   = 1 , and  2

sin 

so the amplitude of that point is twice the amplitude of either wave. Thus this point is an antinode of the standing wave. Note that D1 and D2 are “on top” of each other, since they are identical. 60. (a) The maximum swing is twice the amplitude of the standing wave. Three loops is 1.5 wavelengths, and the frequency is given. A = 12 ( 8.00 cm ) = 4.00 cm ;  = 2 f = 2 (125 Hz ) = 785.40 rad s  785 rad s ; 3 2

 = 1.85 m →  = 1.2333 m ; k =

2 1.2333 m

= 5.0946 m −1  5.09 m −1

(

)

D = A sin ( kx ) cos ( t ) = ( 4.00 cm ) sin  5.09 m −1 x  cos ( 785 rad s ) t 

(b) Each component wave has the same wavelength, the same frequency, and half the amplitude of the standing wave.

( ) D = ( 2.00 cm ) sin  ( 5.09 m ) x + ( 785 rad s ) t  D1 = ( 2.00 cm ) sin  5.09 m −1 x − ( 785 rad s ) t  −1

61. Any harmonic with a node directly above the pickup will NOT be “picked up” by the pickup. The pickup location is exactly 1/4 of the string length from the end of the string, so a standing wave with a frequency corresponding to 4 (or 8 or 12, etc.) loops will not excite the pickup. So n = 4, 8, and 12 will not excite the pickup. 62. The standing wave is the sum of the two individual standing waves. We use the trigonometric identities for the cosine of a difference and a sum. cos (1 −  2 ) = cos 1 cos  2 + sin 1 sin  2 ; cos (1 +  2 ) = cos 1 cos  2 − sin 1 sin  2 D = D1 + D2 = A cos ( kx −  t ) + A cos ( kx +  t ) = A cos ( kx −  t ) + cos ( kx +  t ) = A cos kx cos  t + sin kx sin  t + cos kx cos  t − sin kx sin  t  = 2 A cos kx cos  t Thus, the standing wave is D = 2 A cos kx cos  t. The nodes occur where the position term forces

D = 2 A cos kx cos  t = 0 for all time. Thus cos kx = 0 → kx =  ( 2n + 1)

since k = 2.0 m −1 , we have x =  ( n + 12 )

 2

m, n = 0,1, 2,

 2

, n = 0,1, 2,

. Thus,

550

Chapter 15

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63. From the description of the water’s behavior, there is an antinode at each end of the tub, and a node in the middle. Thus one wavelength is twice the tub length. v =  f = ( 2l tub ) f = 2 ( 0.45 m )( 0.85 Hz ) = 0.77 m s 64. The angle of refraction can be found from the law of refraction, Eq. 15–19. sin  2 v 2 v 2.4 m s = → sin  2 = sin 1 2 = ( sin 35 ) = 0.4747 →  2 = sin −1 0.4747 = 28 sin 1 v1 v1 2.9 m s 65. The speed in the second medium is found from the law of refraction, Eq. 15–19. sin  2 v2 sin  2  sin 31  = → v2 = v1 = ( 8.0 km s )   = 5.2 km s sin 1 v1 sin 1  sin 52  66. (a) Eq. 15–19 gives the relationship between the angles and the speed of sound in the two media. For total internal reflection (for no sound to enter the water),  water = 90 or sin  water = 1 . The air is the “incident” media. Thus the incident angle is given by the following. sin  air sin  water



vair v water

;  air = i = sin −1 sin  water



vair 

v  v  → iM = sin −1  air  = sin −1  i   vwater   vr   vwater 

(b) From the angle of incidence, the distance is found. See the diagram. v 343 m s  air M = sin −1 air = sin −1 = 13.8 1440 m s v water tan  air M =

x 1.8 m

→ x = (1.8 m ) tan13.8 = 0.44 m

67. The angle of refraction can be found from the law of refraction, Eq. 15–19. The relative velocities can be found from the relationship given in the problem.  331 + 0.60 ( −15 )  sin  2 v2 331 + 0.60T2  322  = = → sin  2 = sin 28   = sin 28   sin 1 v1 331 + 0.60T1  340   331 + 0.60 (15 ) 

  322    2 = sin −1 sin 28   = 26.40  26  340    68. The angle of refraction can be found from the law of refraction, Eq. 15–19. The relative velocities can be found from Eq. 15–3. E 2 SG1  water 1 SG1 sin  2 v2 = = = = = 2 SG2  water SG2 sin 1 v1 E 1 sin  2 = sin 1

SG1 SG2

= sin 41o

3.6 2.8

= 0.7439 →  2 = sin −1 0.7439 = 48

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69. From the statement of the problem, Bgas = 0.65Bwater . We use Eq. 15–19, the law of refraction, and we use Eq. 15–4 for the speed of a wave in a fluid. We choose the water as medium 1 and the gasoline as medium 2.

 B2  2   v2   →  = sin 1   B  v1   1 1 

sin  2 = sin 1 

 Bgas  gas   B    vgas   sin   = sin  water  gas water  = water   Bwater  gas   Bwater  water   v water     

sin  gas = sin  water 

 0.65 1.000  103 kg m 3   = 0.6583  1 0.75  103 kg m 3   

= ( sin 45 ) 

 gas = sin −1 0.6583 = 41 70. The wavelength is to be 0.85 m. Use Eq. 15–1 to find the frequency corresponding to that wavelength. v 344 m s v = f → f = = = 404.71Hz  0.85 m There will be significant diffraction only for wavelengths larger than the width of the window, and so waves with frequencies lower than 400 Hz would diffract when passing through this window. 71. The frequency is 880 Hz and the phase velocity is 440 m/s, so the wavelength is v 440 m s = = = 0.50 m. f 880 Hz (a) Use the ratio of distance to wavelength to define the phase difference. x  6  0.50 m = → x= = = 0.042 m  2 12 12 (b) Use the ratio of time to period to define the phase difference. t 2 t  = → = = 2 tf = 2 1.0  10−4 s ( 880 Hz ) = 0.55 rad T 2 T

(

72. From Eq. 15–17b, f n =

nv 2L

)

, we see that the frequency is proportional to the wave speed on the

stretched string. From Eq. 15–2, v = FT  , we see that the wave speed is proportional to the square root of the tension. Thus the frequency is proportional to the square root of the tension.

FT 2 FT 1

f2 f1

f   247 Hz  → FT 2 =  2  FT 1 =   FT 1 = 0.889 FT 1  262 Hz   f1 

Thus the tension should be decreased by 11.1% . 73. (a) The distance traveled by the reflected sound wave is found from the Pythagorean theorem.

d = 2 D 2 + ( 12 x ) = vt → 2

2 v

D 2 + ( 12 x )

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Chapter 15

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(b) Solve for t 2 . 4 x2 4 2 t 2 = 2  D 2 + ( 12 x )  = 2 + 2 D 2  v v v  A plot of t 2 vs x 2 would have a slope of 1 v 2 , which can be used to determine the value of v. 4 The y intercept of that plot is 2 D 2 . Knowing the y intercept and the value of v, the value of D v can be determined. 74. The gap between resonant frequencies is the fundamental frequency (which is thus 300 Hz for this problem), and the wavelength of the fundamental is twice the string length. v =  f = ( 2 l )( f n +1 − f n ) = 2 ( 0.65 m )( 300 Hz ) = 390 m s 75. (a) The amplitude is half the peak-to-peak distance, so 0.05 m . (b) The maximum kinetic energy of a particle in simple harmonic motion is the total energy, which is given by Etotal = 12 kA2 . Compare the two kinetic energy maxima. 2

 A   0.15 m  = = 2 = = 9 2 1 K1 max 2 kA1  A1   0.05 m 

K 2 max

1 2

kA22

76. One wavelength, or one full oscillation, corresponds to 360; a full cycle of the sinusoidal oscillation that is creating the wave disturbance. Therefore, a half of a wavelength (/2) corresponds to a half oscillation, or 180. 77. We assume that the earthquake wave is moving the ground vertically, since it is a transverse wave. An object sitting on the ground will then be moving with SHM, due to the two forces on it–the normal force upwards from the ground and the weight downwards due to gravity. If the object loses contact with the ground, then the normal force will be zero, and the only force on the object will be its weight. If the only force is the weight, then the object will have an acceleration of g downwards. Thus the limiting condition for beginning to lose contact with the ground is when the maximum acceleration caused by the wave is greater than g. Any larger downward acceleration and the ground would “fall” quicker than the object. The maximum acceleration is related to the amplitude and the frequency as follows. g g 9.80 m s2 amax =  2 A  g → A  2 = 2 2 = 2 = 0.69 m 2 4 f  4 ( 0.60 Hz ) 78. The frequency at which the water is being shaken is about 1 Hz. The sloshing coffee is in a standing wave mode, with antinodes at each edge of the cup. The cup diameter is thus a half-wavelength, or  = 16 cm. The wave speed can be calculated from the frequency and the wavelength. v =  f = (16 cm )(1 Hz ) = 16 cm s

79. The speed of a longitudinal wave in a solid is given by Eq. 15–3, v = E  . Let the density of the less dense material be 1 , and the density of the more dense material be  2 . The less dense material will have the higher speed , since the speed is inversely proportional to the square root of the density. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

553

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

v1 v2

E 1 E 2

Instructor Solutions Manual

2 = 2.1  1.4 1

80. From Eq. 15–7, if the speed, medium density, and frequency of the two waves are the same, then the intensity is proportional to the square of the amplitude.

I 2 I1 = P2 P1 = A22 A12 = 2.5 → A2 A1 = 2.5 = 1.6 The more energetic wave has the larger amplitude. 81. (a) See the adjacent plot.

(b) The wave function is found by replacing x in the pulse by x − v t. D ( x, t ) =

4.0 m 3

 x − ( 2.4 m s ) t  + 2.0 m 2

(d) The wave function is found by replacing x in the pulse by x + v t. See the adjacent plot. D ( x, t ) =

4.0 m 3

 x + ( 2.4 m s ) t  + 2.0 m 2

554

Chapter 15

Wave Motion

82. (a) The frequency is related to the tension by Eqs. 15–1 and 15–2. v 1 FT 1 1 1 1 FT df f → = = = f = =    dFT 2  FT 2 FT   2 FT f FT



→

2 FT

 FT  f  FT 

f  12 

(b) The change in frequency is 12 Hz. f f FT f  12  = 0.0550 = 5.5%  → 2 = 2  FT f FT 2 FT  436  (c) The only change in the expression

as the overtone changes is the wavelength, and the   wavelength does not influence the final result. So yes , the formula still applies.

83. Each half of the cord has a single node, at the center of the cord. Thus, each half of the cord is a half of a wavelength, assuming that the ends of the cord are also nodes. The tension is the same in both halves of the cord, and the wavelengths are the same based on the location of the node. Let subscript 1 represent the lighter density, and subscript 2 represent the heavier density. 1 FT1

v1 =

FT1

1

FT2

= 1 f1 ; v2 =

2

= 2 f 2 ; 1 = 2 ; FT1 = FT 2 ;

f1 f2

1

1

FT2

2

2

2 = 1

The frequency is higher on the lighter portion. 84. The frequency for each wire must be the same, to ensure continuity of the string at its junction. FT 2l Each wire will obey these relationships:  f = v , v = ,  = . Combine these to find the n  nodes. Note that n is the number of “loops” in the string segment, and that n loops requires n + 1 nodes. FT FT 2l 2l n FT , =  f = v, v = → → f = f = 2l    n n nAl

2 l Al

Al

nFe

2 l Fe

Fe

→

nAl nFe

l Al l Fe

 Al 0.600 m 2.70 g m 2 = = 0.400 = Fe 0.882 m 7.80 g m 5

Thus there are 3 nodes on the aluminum, since nAl = 2, and 6 nodes on the steel, since nFe = 5, but one node is shared so there are 8 total nodes. Use the formula derived above to find the lower frequency. f =

nAl

FAl

2 l Al

 Al

135 N

2 ( 0.600 m )

2.70  10 −3 kg m

= 373 Hz

The same formula could be used for the values for steel, also resulting in 373 Hz.

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85. (a) The overtones are given by f n = nf1 , n = 2, 3, 4

G : f 2 = 2 ( 392 Hz ) = 784 Hz

f 3 = 3 ( 392 Hz ) = 1176 Hz  1180 Hz

B : f 2 = 2 ( 494 Hz ) = 988 Hz

f 3 = 3 ( 440 Hz ) = 1482 Hz  1480 Hz

(b) If the two strings have the same length, they have the same wavelength. The frequency difference is then due to a difference in wave speed caused by different masses for the strings. FT FT

fG fB

vG  vB 

vG vB

G FT

mG l FT

mB mG

 f   494  → = B  =  = 1.59 mB  f G   392  mG

B

mB l (c) If the two strings have the same mass per unit length and the same tension, then the wave speed on both strings is the same. The frequency difference is then due to a difference in wavelength. For the fundamental, the wavelength is twice the length of the string. f G v  G B 2l B l G f B 494 = = = → = = = 1.26 f B v  B G 2l G l B fG 392 (d) If the two strings have the same length, they have the same wavelength. They also have the same mass density, since they have the same mass and the same length. The frequency difference is then due to a difference in wave speed caused by different tensions for the strings.

FTG fG fB

vG  vB 

vG vB

 FTB

FTG FTB

 f   392  → = G  =  = 0.630 FTB  f B   494  FTG

 86. Relative to the fixed needle position, the ripples are moving with a linear velocity given by  rev  1min   2 ( 0.108 m )  v =  33    = 0.3732 m s . 1 rev  min  60 s    This speed is the speed of the ripple waves moving past the needle. The frequency of the waves is 0.3732 m s v f = = = 240.77 Hz  240 Hz .  1.55  10−3 m 87. The speed of the pulses is found from the tension and mass per unit length of the wire. v=

255 N

= 135.93 m s

 0.138 kg 10.0 m The total distance traveled by the two pulses will be the length of the wire. The second pulse has a shorter time of travel than the first pulse, by 20.0 ms. l = d1 + d 2 = vt1 + vt2 = vt1 + v ( t1 − 2.00  10−2 )

(10.0 m ) + 2.00  10−2 (135.93 m s ) t1 = = = 4.6784  10 −2 s 2v 2 (135.93 m s ) d1 = v t1 = (135.93 m s ) ( 4.6784  10 −2 s ) = 6.36 m l + 2.00  10−2 v

The two pulses meet 6.36 m from the end where the first pulse originated. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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88. (a) The speed of transverse waves at a point y below the top of a hanging cable of length l depends on the tension at that point and the linear mass density of the cord, as given by Eq. 15–2. The tension at that point equals the weight of the part of the cable below y–see the free-body diagram. Use Eq. 15–2 for the wave speed. (l − y) FT = mlower g = m g

part



m =

(l − y) l

g =

( l − y ) g = ( 2.0 m − y ) ( 9.80 m s 2 )

l (b) See the adjacent graph.

89. Because the wave is uniform, the same energy per second (power) must pass through every spherical surface centered on the source. A spherical surface of radius r has an area 4 r 2 . Thus the intensity, which is power per unit area, must be proportional to 1 r 2 in order to keep the power constant, since the area is increasing proportional to r2. Power constant 1 I= → Power = I ( Area ) = I 4 r 2 = constant → I = → I 2 2 Area 4 r r The intensity is also proportional to the square of the amplitude, and so the amplitude must be proportional to 1 r . We assume a single frequency sinusoidal time dependence, and so we have

(

)

 A  sin k r −  t . The amplitude of this displacement is A where A is a constant. , )  ( r r

D=

90. For a resonant condition, the free end of the string will be an antinode, and the fixed end of the string will be a node. The minimum distance from a node to an antinode is  4. Other wave patterns that fit the boundary conditions of a node at one end and an antinode at the other end have lengths of 3 4 , 5 4 ,

. See the diagrams. The

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general relationship is l = ( 2n − 1)  4 , n = 1, 2, 3,

=

4l 2n − 1

, n = 1, 2, 3,

Instructor Solutions Manual

. Solving for the wavelength gives

91. The addition of the support will force the bridge to have a node at the center of the span, which would be the first overtone of the fundamental frequency. If the wave speed in the bridge material remains constant, then the resonant frequency will double, to 6.0 Hz. Since earthquakes don’t do significant shaking at that frequency, the modifications would be effective at keeping the bridge from having large oscillations during an earthquake. 92. From the figure, we can see that the amplitude is 3.5 cm, and the wavelength is 20 cm. The maximum of the wave at x = 0 has moved to x = 12 cm at t = 0.80 s, which is used to find the velocity. The wave is moving to the right. Finally, since the displacement is a maximum at x = 0 and t = 0, we can use a cosine function without a phase angle.  12 cm A = 3.5 cm ;  = 20 cm → k = = 0.10 cm −1 ; v = = 15 cm s ;  = v k = 1.5 rad s  0.80 s

D ( x, t ) = A cos ( kx −  t ) = ( 3.5 cm ) cos ( 0.10 x − 1.5 t ) , x in cm, t in s 93. From the given data, A = 0.50 m and v = 2.5 m 4.0s = 0.625 m s. We use Eq. 15–6 for the average power, with the density of sea water from Table 13–1. We estimate the area of the chest as

( 0.30 m )2 . Answers may vary according to the approximation used for the area of the chest. 2 2 2 P = 2 2  S v f 2 A2 = 2 2 (1025 kg m3 ) ( 0.30 m ) ( 0.625 m s )( 0.25 Hz ) ( 0.50 m ) = 18 W 94. The unusual decrease of water corresponds to a trough in Fig. 15–3. The crest or peak of the wave is then one-half wavelength from the shore. The peak is 107.5 km away, traveling at 550 km/hr. x 12 ( 215 km )  60 min  x = v t → t = =   = 11.7 min  12 min 550 km hr  1hr  v 95. At t = 1.0 s, the leading edge of each wave is 1.0 cm from the other wave. They have not yet interfered. The leading edge of the wider wave is at 22 cm, and the leading edge of the narrower wave is at 23 cm.

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At t = 2.0 s, the waves are overlapping. The diagram uses dashed lines to show the parts of the original waves that are undergoing interference.

At t = 3.0 s, the waves have “passed through” each other, and are no longer interfering.

96. (a) We are told that the speed of the waves only depends on the acceleration due to gravity and the wavelength. 

 L  L  v = kg    →   =  2   L  T  T 

T : −1 = −2 →  = 1 2

L :1 =  +  →  = 1 −  = 1 2

v = k g

(b) Here the speed of the waves depends only on the acceleration due to gravity and the depth of the water, h. 



v = kg h



 L  L  →   =  2   L T  T 

L :1 =  +  →  = 1 −  = 1 2

T : −1 = −2 →  = 1 2 v = k gh

559

CHAPTER 16: Sound Responses to Questions 1.

Sound exhibits several phenomena that give evidence that it is a wave. The phenomenon of interference is a wave phenomenon, and sound produces interference (such as beats). The phenomenon of diffraction is a wave phenomenon, and sound can be diffracted (such as sound being heard around corners). Refraction is a wave phenomenon, and sound exhibits refraction when passing obliquely from one medium to another. Sound also requires a medium, a characteristic of mechanical waves.

Evidence that sound is a form of energy is found in the fact that sound can do work. A sound wave created in one location can cause the mechanical vibration of an object at a different location. For example, sound can set eardrums in motion, make windows rattle, or even shatter a glass. See Fig. 14-25 for a photograph of a goblet shattering from the sound of a trumpet.

The child speaking into a cup creates sound waves which cause the bottom of the cup to vibrate. Since the string is tightly attached to the bottom of the cup, the vibrations of the cup are transmitted to longitudinal waves in the string. These waves travel down the string, and cause the bottom of the receiver’s cup to vibrate back and forth. This relatively large vibrating surface moves the adjacent air, and generates sound waves from the bottom of the receiver’s cup, traveling up into the receiver’s cup. These waves are incident on the receiver’s ear, and thus the receiver can hear the sound from the child speaking.

If the frequency were to change, the two media could not stay in contact with each other. If one medium vibrates with a certain frequency, and the other medium vibrates with a different frequency, then particles from the two media initially in contact could not stay in contact with each other. But particles must be in contact in order for the wave to be transmitted from one medium to the other, and so the frequency does not change. Since the wave speed changes in passing from air into water, and the frequency does not change, we expect the wavelength to change. Sound waves travel about four times faster in water than in air, so we expect the wavelength in water to be about four times longer than it is in air.

If the speed of sound in air depended significantly on frequency, then the sounds that we hear would be separated in time according to frequency. For example, if a chord were played by an orchestra, we would hear the high notes at one time, the middle notes at another, and the lower notes at still another. This effect is not heard for a large range of distances, indicating that the speed of sound in air does not depend significantly on frequency.

The sound-production anatomy of a person includes various resonating cavities, such as the throat. The relatively fixed geometry of these cavities, along with the relatively fixed length of the vocal cords, will determine the relatively fixed wavelengths of sound that a person can produce. Those wavelengths will have associated frequencies given by f = v . The speed of sound is determined by the gas that is filling the resonant cavities. If the person has inhaled helium, then the speed of sound will be much higher than normal, since the speed of sound waves in helium is about 3 times that in air. Thus the person’s frequencies will go up about a factor of 3. This is about a 1.5 octave shift, and so the person sounds very high pitched.

The intensity of a sound wave is proportional to the square of the frequency, so the higher-frequency tuning fork will produce more intense sound.

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The basic equation determining the pitch of the organ pipe is either f closed = closed pipe, or f open =

nv 4

, n = odd integer, for a

, n = integer, for an open pipe. In each case, the frequency is proportional 2 to the speed of sound in air. Since the speed is a function of temperature, and the length of any particular pipe is very nearly constant over the relatively small range of temperatures in a room, the frequency is also a function of temperature. Thus, when the temperature changes, the resonant frequencies of the organ pipes change. Since the speed of sound increases with temperature, as the temperature increases, the pitch of the pipes increases as well.

A tube of a given length will resonate (permit standing waves) at certain frequencies. When a mix of frequencies is input to the tube, only those frequencies close to resonant frequencies will produce sound that persists, because standing waves are created for those frequencies. Frequencies far from resonant frequencies will not persist very long at all – they will “die out” quickly. If, for example, two adjacent resonances of a tube are at 100 Hz and 200 Hz, then sound input near one of those frequencies will persist and sound relatively loud. A sound input near 150 Hz would fade out quickly, and so have a reduced amplitude as compared to the resonant frequencies. The length of the tube can be chosen to thus “filter” certain frequencies, if those filtered frequencies are not close to resonant frequencies. v 10. For a string with fixed ends, the fundamental frequency is given by f = and so the length of 2 v string for a given frequency is = . For a string, if the tension is not changed while fretting, the 2f speed of sound waves will be constant. Thus for two frequencies f1  f 2 , the spacing between the frets corresponding to those frequencies is given as follows. v v v 1 1  − 2= − =  −  1 2 f1 2 f 2 2  f1 f 2  Now see Table 16-3. Each note there would correspond to one fret on the guitar neck. Notice that as the adjacent frequencies get higher, the inter-frequency spacing also increases. The change from C to C# is 15 Hz, while the change from G to G# is 23 Hz. Thus their reciprocals get closer together, and so from the above formula, the length spacing gets closer together. Consider a numeric example. 1  v 1  v v 1 v 1 −4 −4 − C# =  − − G# =  −  = 2.07  10  = 1.41 10 C G 2  262 277  2 2  392 415  2

(

)

(

)

− G#

= 0.68 − C# The G to G# spacing is only about 68% of the C to C# spacing. G

11. The lowest-frequency notes on a piano are wrapped with wire so that the mass density of the string can be increased. To obtain a low frequency, consider this relationship: f =

. To obtain a 2  low frequency, some combination of increasing the string length, decreasing the string tension, and increase the mass combination must take place. Making the strings have a low frequency by only increasing their length might make the piano case too long to be reasonable. Making the strings have a low frequency by only decreasing their tension could put unbalanced forces across the width of the instrument, particularly if the highest frequency strings were at a very high tension, and the lowest frequency strings were at a very low tension. Increasing the mass density of the strings can at © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

least reduce the need to adjust string length or string tension. On some other stringed instruments, like a guitar, most of the frequency change from one string to the next is done by changing the mass density. 12. When you first hear the truck, you cannot see it. There is no straight line path from the truck to you. The sound waves that you are hearing are therefore arriving at your location due to diffraction. Long wavelengths are diffracted more than short wavelengths, and so you are initially only hearing sound with long wavelengths, which are low-frequency sounds. After you can see the truck, you are able to receive all frequencies being emitted by the truck, not just the lower frequencies. Thus the sound “brightens” due to your hearing more high frequency components. 13. Active noise reduction devices work on the principle of destructive interference. The electronics of the devices are fast enough to detect the noise, invert it, create the opposite wave (180o out of phase with the original) and then send the inverted signal to the headphones (or ear buds) in just the right amount of time. The original noise and the created noise will be approximately in a destructive interference relationship, so if those two sounds are combined, the person wearing the headphones will then hear a net sound signal that is very low in intensity. 14. If the frequency of the speakers is lowered, then the wavelength will be increased. Each circle in the diagram will be larger, and so the points C and D will move farther apart. 15. The wave pattern created by standing waves does not “travel” from one place to another. The node locations are fixed in space. Any one point in the medium has the same amplitude at all times. Thus the interference can be described as “interference in space” – moving the observation point from one location to another changes the interference from constructive (anti-node) to destructive (node). To experience the full range from node to anti-node, the position of observation must change, but all observations could be made at the same time by a group of observers. The wave pattern created by beats does travel from one place to another. Any one point in the medium will at one time have a 0 amplitude (node) and half a beat period later, have a maximum amplitude (anti-node). Thus the interference can be described as “interference in time”. To experience the full range from constructive interference to destructive interference, the time of observation must change, but all observations could be made at the same position. 16. For the two waves shown, the frequency of beating is higher in Figure (a) – the beats occur more frequently. The beat frequency is the difference between the two component frequencies, and so since (a) has a higher beat frequency, the component frequencies are further apart in (a). 17. There is no Doppler shift if the source and observer move in the same direction, with the same velocity. Doppler shift is caused by relative motion between source and observer, and if both source and observer move in the same direction with the same velocity, there is no relative motion. 18. If the wind is blowing but the listener is at rest with respect to the source, the listener will not hear a Doppler effect. We analyze the case of the wind blowing from the source towards the listener. The moving air (wind) has the same effect as if the speed of sound had been increased by an amount equal to the wind speed. The wavelength of the sound waves (distance that a wave travels during one period of time) will be increased by the same percentage that the wind speed has increased relative to the still-air speed of sound. Since the frequency is the speed divided by the wavelength, the frequency does not change, and so there is no Doppler effect to hear. Alternatively, the wind has the same effect as if the air were not moving but the source and listener were both moving at the same speed in the opposite direction of the wind direction. See question 17 for a discussion of that © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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situation. Since there is no relative motion between the source and the listener, there is no Doppler shift. 19. The highest frequency of sound will be heard at position C, while the child is swinging forward. Assuming the child is moving with SHM, then the highest speed is at the equilibrium point, point C. And to have an increased pitch, the relative motion of the source and detector must be towards each other. The child would also hear the lowest frequency of sound at point C, while swinging backwards.

Solutions to MisConceptual Questions 1.

(a) Students may answer that the speed of sound is the same, if they do not understand that the speed of sound is not constant, but depends upon the temperature of the air. When it is hotter, the speed of sound is greater, so it takes less time for the echo to return.

(d) Sound waves are longitudinal waves, so (a) is incorrect. The sound waves can be characterized either by the longitudinal displacement of the air molecules, or the pressure differences that cause the displacements.

(b) A common misconception is to treat the sound intensity level as a linear scale instead of a logarithmic scale. If the sound intensity doubles, the intensity level increases by about 3 dB, so the correct answer should be 73 dB.

(e) Students often think that the sound intensity is the same as loudness, and as such mistakenly answer that doubling the intensity will double the loudness. However, the ear interprets loudness on a logarithmic scale. For something to sound twice as loud, it must have an intensity ten times as great.

(b) The octave is a measure of musical frequency, not loudness. Raising a note by one octave requires doubling the frequency. Therefore, raising a note by two octaves is doubling the frequency twice, which is the same as quadrupling the frequency.

(e) In a string or open tube the lowest vibration mode is equal to half of a wavelength. In a tube closed at one end the lowest vibration mode is equal to a quarter of a wavelength. Therefore, none of the listed objects have a lowest vibration mode equal to a wavelength.

(e) A common misconception is that the frequency of a sound changes as it passes from air to water. The frequency is the number of wave crests that pass a certain point per unit time. If this value were to change as it entered the water, then wave crests would build up or be depleted over time. This would make the interface an energy source or sink, which it is not. The speed of sound in water is greater than in air, so the speed of the wave changes. Since the frequency cannot change, the increase in speed results in an increase in wavelength.

(e) As the string oscillates it causes the air to vibrate at the same frequency. Therefore the sound wave will have the same frequency as the guitar string, so answers (b) and (c) are incorrect. The speed of sound in air at 20C is 343 m/s. The speed of sound in the string is the product of the wavelength and frequency, 462 m/s, so the sound waves in air have a shorter wavelength than the waves on the string.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(c) Pushing the string straight down onto a fret does not affect the tension a significant amount, due to the fret being so close to the string. The amplitude of the wave is determined by how hard the string is plucked, not by pushing the string onto the frets. By pushing down on the string, its effective length is shortened, which shortens the wavelength and thus increases the oscillation frequency (since the wave speed on the string doesn’t change due to fretting the string).

10. (a) The fundamental wavelength of an open-ended organ pipe is twice the length of the pipe. If one end is closed the fundamental wavelength is four times the length of the pipe. Since the wavelength doubles when one end of the pipe is closed off, and the speed of sound remains constant, the fundamental frequency is cut in half. 11. (b) The open tube next to your ear is a tube open at both ends. Any ambient noise will cause some standing waves in the tube, with a specific fundamental frequency related to the length of the tube. Then, when you cover the other end of the tube with your hand, you are changing the tube from one that is open at both ends, to one that is open at just one end. The fundamental of the second tube is an octave lower than the first tube, and so you hear the pitches decrease. 12. (a) “Middle C” is a specific frequency, so answer (b) is not correct. And the notes don’t change much whether they are played loudly or softly, so answer (c) is not correct. As discussed in Section 16-5, each instrument has a different sound spectrum, or different mix of harmonics. This is what gives the instrument its unique sound. 13. (c) A common misconception is that since the cars are moving there must be a Doppler shift. In this situation however, there is no relative motion between the two vehicles. Both vehicles travel in the same direction at the same speed. Since the distance between them does not change, both people will hear the horn sound at the same frequency. 14. (e) As the string vibrates, each part of the string (other than the nodes) oscillates at the same frequency, so answer (a) is true. This oscillation excites the air to vibrate at that frequency, so answer (c) is true. The wave relationship in answers (b) and (d) are true for any wave, and so are both true in this case as well. However, the speed of the wave on the string is determined by the tension and mass of the string, and the speed of sound in the air is determined by the temperature, pressure, and density of the air. The two speeds are not necessarily the same. Since the sound wave and wave on the string have the same frequencies, but not necessarily the same wave speeds, they do not necessarily have the same wavelengths. Thus (e) is not true.

Solutions to Problems In these solutions, we usually treat frequencies as if they are significant to the whole number of units. For example, 20 Hz is taken as to the nearest Hz, and 20 kHz is taken as to the nearest kHz. We also treat all decibel values as good to whole number of decibels. So 120 dB is good to the nearest decibel. In solving these problems, the authors did not always follow the rules of significant figures rigidly. We tended to take quoted frequencies as correct to the number of digits shown, especially where other values might indicate that. For example, in Problem 49, values of 350 Hz and 355 Hz are used. We took both of those values to have 3 significant figures. 1.

The round trip time for sound is 3 seconds, so the time for sound to travel the length of the lake is half of that. Use the time and the speed of sound to determine the length of the lake.

d = vt = ( 343m s ) 12 ( 3s ) = 514.5 m  500 m

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Chapter 16

Sound

We assume that the human hearing range of frequencies is given to 2 significant figures. v 343 m s v 343 m s 20 kHz = = = 17 m = 1.7  10−2 m (a) 20 Hz = = 4 f 20 Hz f 2.0  10 Hz The range is from 1.7 cm to 17 m. v 343 m s = 1.6  10−5 m (b)  = = 6 f 22  10 Hz

The distance that the sound travels is the same on both days, and is equal to the speed of sound times the elapsed time. Use the temperature-dependent relationship for the speed of sound in air. d = v1t1 = v2t2 → ( 331 + 0.6 ( 31) ) m s  ( 4.40 s ) = ( 331 + 0.6 ( T2 ) ) m s  ( 4.80 s ) →

T2 =

s ( 331 + 0.6 ( 31) ) m s   4.40  − 331m s 4.80 s



0.6



= −17.56C  −18C

C

(a) For the fish, the speed of sound in seawater must be used. 1550 m d d = vt → t = = = 0.994 s v 1560 m s (b) For the fishermen, the speed of sound in air must be used. d 1550 m d = vt → t = = = 4.52 s v 343 m s

The “5 second rule” says that for every 5 seconds between seeing a lightning strike and hearing the associated sound, the lightning is 1 mile distant. We assume that there are 5 seconds between seeing the lightning and hearing the sound. (a) At 30oC, the speed of sound is 331 + 0.60 ( 30 ) m s = 349 m s . The actual distance to the lightning is therefore d = v t = ( 349 m s )( 5s ) = 1745 m. A mile is 1610 m.

1745 − 1610

(100 )  8% 1745 (b) At 10oC, the speed of sound is 331 + 0.60 (10 ) m s = 337 m s. The actual distance to the % error =

lightning is therefore d = v t = ( 337 m s )( 5s ) = 1685 m. A mile is 1610 m.

% error = 6.

1685 − 1610 1685

(100 )  4%

The total time T is the time for the stone to fall ( tdown ) plus the time for the sound to come back to the top of the cliff ( tup ) : T = tup + tdown. Use constant acceleration relationships for an object

dropped from rest that falls a distance h in order to find tdown , with down as the positive direction. Use the constant speed of sound to find tup for the sound to travel a distance h.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

2 2 down: y = y0 + v0tdown + 12 atdown → h = 12 gtdown



2 h = 12 gtdown = 12 g ( T − tup ) = 12 g  T − 2



h 

up: h = vsnd t up → t up =

 vsnd 

Instructor Solutions Manual

 vsnd

→ h 2 − 2vsnd 

 g

h vsnd



2 + T  h + T 2 vsnd =0



This is a quadratic equation for the height. This can be solved with the quadratic formula, but be sure to keep several significant digits in the calculations.  343 m s  2 2 h 2 − 2 ( 343 m s )  + 2.5s  h + ( 2.5s ) ( 343 m s ) = 0 → 2  9.80 m s  h 2 − ( 25725 m ) h + 7.3531  105 m 2 = 0 → h =

25725  25668 = 25697 m, 29 m 2

The larger root is impossible since it takes more than 2.5 sec for the rock to fall that distance, so h = 29 m . 7.

The relationship between the pressure and displacement amplitudes is given by Eq. 16-5. PM 3.0  10−3 Pa = = 4.5  10−9 m (a) PM = 2 v Af → A = 3 2 vf 2 1.29 kg m ( 331m s )( 250 Hz )

(

(b)

PM 2 vf

)

3.0  10−3 Pa

(

)

(

2 1.29 kg m 3 ( 331m s ) 25  103 Hz

)

= 4.5  10−11 m

The pressure amplitude is found from Eq. 16-5. The density of air is 1.29 kg m 3 .

( ) ( ) (b) P = 2 vAf = 2 (1.29 kg m ) ( 331m s ) ( 3.0 10 m ) ( 7500 Hz ) = 6.0 10 Pa (a) PM = 2 vAf = 2 1.29 kg m3 ( 331m s ) 3.0 10 −10 m ( 75 Hz ) = 6.0 10 −5 Pa −10

−3

The pressure wave can be written as Eq. 16-4. (a) Eq. 16-4 says P = −PM cos ( kx − t ) . PM = 6.0  10−5 Pa ;  = 2 f = 2 ( 75 Hz ) = 150 rad s ; k =

(

)

(



150 rad s 331m s

= 0.45 m −1

)

P = − 6.0  10−5 Pa cos  0.45 m −1 x − (150 rad s ) t 

(b) Again, P = −PM cos ( kx − t ) . PM = 6.0  10 −3 Pa ;  = 2 f = 2 ( 7500 Hz ) = 1.5  10 4  rad s ; k=

 v

1.5  10 4  rad s 331m s

(

)

= 45 m −1

(

) (

)

P = − 6.0  10−3 Pa cos  45 m −1 x − 1.5  10 4  rad s t 

(

) (

)

10. The pressure wave is P = ( 0.0035 Pa ) sin  0.38 m−1 x − 1220 s−1 t  . (a)  =

2 k

2 0.38 m −1

= 5.3m

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Chapter 16

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 1220 s −1 = = 610 Hz 2 2  1220 s −1 = 3211 m s  3200 m s (c) v = = k 0.38 m −1 (b)

f =

(d) Use Eq. 16-5 to find the displacement amplitude. PM = 2 vAf →

11.  = 10log

I I0

PM

2 vf

= 10log

( 0.0035 Pa ) = 1.2  10−13 m 2 ( 2300 kg m3 ) ( 3211m s )( 610 Hz )

3.0  10−6 W m 2 1.0  10−12 W m 2

= 64.77 dB  65dB

12. From Fig. 16-6, a 100-Hz tone at 50 dB has a loudness of about 20 phons. At 5000 Hz, 20 phons corresponds to about 20 dB . Answers may vary due to estimation in the reading of the graph. 13. From Fig. 16-6, at 40 dB the low frequency threshold of hearing is about 70 − 80 Hz . There is no intersection of the threshold of hearing with the 40 dB level on the high frequency side of the chart, and so a 40 dB signal can be heard all the way up to the highest frequency that a human can hear, 20, 000 Hz . Answers may vary due to estimation in the reading of the graph. I 14. (a) 120 dB = 10 log 120 I0

I (b) 20 dB = 10 log 20 I0

(

)

→ I120 = 1012 I 0 = 1012 1.0  10 −12 W m 2 = 1.0 W m 2

(

)

→ I 20 = 102 I 0 = 102 1.0  10 −12 W m 2 = 1.0  10 −10 W m 2

The pain level is 1010 (10 billion) times more intense than the whisper. 15. (a) From Fig. 16-6, at 100 Hz, the threshold of hearing (the lowest detectable intensity by the −9 2 ear) is approximately 5  10 W m . The threshold of pain is about 5 W m 2 . The ratio of highest to lowest intensity is thus

5 W m2 −9

5  10 W m

= 109 .

(b) At 5000 Hz, the threshold of hearing is about 10 −13 W m 2 , and the threshold of pain is about 10 −1 W m 2 . The ratio of highest to lowest intensity is

10−1 W m 2

10−13 W m 2 Answers may vary due to estimation in the reading of the graph.

(I → (I

= 1012 .

16. For the 86-dB device: 86dB = 10log ( I Signal I Noise ) tape →

Signal

I Noise ) tape = 108.6 = 4.0  108 .

For the 98-dB device: 98dB = 10log ( I Signal I Noise ) tape

Signal

I Noise ) tape = 109.8 = 6.3  109 .

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

17. From Example 16-4, we see that a sound level decrease of 3 dB corresponds to a halving of intensity. Thus, if two engines are shut down, the intensity will be cut in half, and the sound level will be 137 dB. Then, if one more engine is shut down, the intensity will be cut in half again, and the sound level will drop by 3 more dB, to a final value of 134 dB . 18. (a) The energy absorbed per second is the power of the wave, which is the intensity times the area. I 45dB = 10 log → I = 104.5 I 0 = 104.5 1.0  10−12 W m 2 = 3.162  10 −8 W m 2 I0

(

P = IA = 3.162  10−8 W m 2

)

)(5.0  10 m ) = 1.581  10 W  1.6  10 J s −5

−12

1s    1 yr  = 20016 yr  2.0  104 yr −12   7   1.581  10 J   3.16  10 s 

(b) 1.0 J 

19. The first person is a distance of r1 = 100 m from the explosion, while the second person is a distance

r2 = 5 (100 m ) from the explosion. The intensity detected away from the explosion is inversely proportional to the square of the distance from the explosion. 2

 5 (100 m )  I1 = 2 =  = 5 ;  = 10 log = 10 log 5 = 6.99 dB  7 dB I 2 r1  100 m  I2 I1

r22

20. (a) Find the intensity from the 130 dB value, and then find the power output corresponding to that intensity at that distance from the speaker. I  = 130 dB = 10 log 2.8m → I 2.8m = 1013 I 0 = 1013 (1.0  10 −12 W m 2 ) = 10 W m 2 I0

(

)

P = IA = 4 r 2 I = 4 ( 2.8 m ) 10 W m 2 = 985.2 W  990 W 2

(b) Find the intensity from the 85 dB value, and then from the power output, find the distance corresponding to that intensity. I → I = 108.5 I 0 = 108.5 (1.0  10 −12 W m 2 ) = 3.162  10 −4 W m 2  = 85dB = 10 log I0 P = 4 r 2 I → r =

P 4 I

985.2 W

(

4 3.162  10 W m −4

)

= 497.9 m  500 m  5.0  102 m

21. (a) We find the intensity of the sound from the decibel value, and then calculate the displacement amplitude from Eq. 15-7a. I → I = 10  /10 I 0 = 1012 10 −12 W m 2 = 1.0 W m 2  = 10 log I0

(

)

I = 2 2 v  f 2 A2 → A=

f

2 v

1.0 W m 2

 ( 440 Hz ) 2 (1.29 kg m ) ( 343 m s ) 3

= 2.4  10 −5 m

568

Chapter 16

Sound

(b) The pressure amplitude can be found from Eq. 16-7. I=

( PM ) 2v

→

(

)(

)

PM = 2v I = 2 ( 343 m s ) 1.29 kg m 3 1.0 W m 2 = 3.0  101 Pa

22. (a) The intensity is proportional to the square of the amplitude, so if the amplitude is 3.5 times greater, the intensity will increase by a factor of 3.52 = 12.25  12 . (b)  = 10 log I I 0 = 10 log12.25 = 10.88 dB  11dB 23. (a) The pressure amplitude is seen in Eq. 16-5 to be proportional to the displacement amplitude and to the frequency. Thus the higher frequency wave has the larger pressure amplitude, by a factor of 3.5. (b) The intensity is proportional to the square of the frequency. Thus the ratio of the intensities is the square of the frequency ratio. 2 I 3.5 f ( 3.5 f ) = = 12.25  12 If f2 24. (a) We assume that there has been no appreciable absorption in this 25 meter distance. The intensity is the power divide by the area of a sphere of radius 25 meters. We express the sound level in dB. 5.0  105 W ) ( P I P I= ;  = 10 log = 10 log = 10 log 2 4 r 2 I0 4 r 2 I 0 4 ( 25 m ) (10−12 W m 2 ) = 138dB  140 dB

(b) We find the intensity level at the new distance, and subtract 7.0 dB/km due to absorption. ( 5.0  105 W ) P 10 log  = 10 log = = 106 dB 2 4 r 2 I 0 4 (1000 m ) (10 −12 W m 2 )

 with

= 106 dB − (1.00 km )( 7.0 dB km ) = 99 dB

absorption

(c) We find the intensity level at the new distance, and subtract 7.0 dB/km due to absorption. 5.0  105 W ) ( P  = 10 log = 10 log = 92 dB 2 4 r 2 I 0 4 ( 5000 m ) (10 −12 W m 2 )

 with

= 92 dB − ( 5.00 km )( 7.0 dB km ) = 57 dB

absorption

25. (a) According to Table 16-2, the intensity in normal conversation, when about 50 cm from the −6 2 speaker, is about 3  10 W m . The intensity is the power output per unit area, and so the power output can be found from the intensity and the area. The area to use is the surface area of a sphere. P 2 I= → P = IA = I ( 4 r 2 ) = ( 3  10−6 W m 2 ) 4 ( 0.50 m ) = 9.425  10 −6 W  9.4  10 −6 W A © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

 1 person  6  = 6.37  10  6 million people −6  9.425  10 W 

(b) 60 W 

26. The intensity is given by Eq. 15-7a, I = 2 2 v  f 2 A2 , using the density of air (from Table 13-1) and the speed of sound in air.

(

)

(

I = 2  v  2 f 2 A2 = 2 1.29 kg m 3 ( 343 m s )  2 ( 440 Hz ) 1.5  10 −4 m

 = 10 log

) = 38.045 W m 2

38.045 W m 2

= 10 log

= 135.8 dB  140 dB I0 1.0  10 −12 W m 2 Note that according to Figure 16-6, this is above the threshold of pain at that frequency. 27. The intensity of the sound is defined to be the power per unit area. We assume that the sound spreads out spherically from the loudspeaker. 280 W (a) I 280 = = 1.819 W m 2 2 4 ( 3.5m )

1.819 W m 2

1.0  10−12 W m 2

 280 = 10 log 280 = 10 log

= 122.6dB  120 dB

I 45 0.2923W m 2 2 0.2923W m ; 10 log 10 log = = = = 114.7 dB  110 dB  45 2 I0 1.0  10−12 W m 2 4 ( 3.5m ) (b) According to the textbook, for one sound to be perceived as twice as loud as another means that the intensities need to differ by a factor of 10, or differ by 10 dB. So calculate the power that produces a sound level of 124.7 dB at a point 3.5 m away from the loudspeaker. I  = 124.7 dB = 10 log → I = I 0 1012.47 = 1.0  10 −12 W m 2 1012.47 = 2.951W m 2 I0 45 W

I 45 =

(

P 4 r

) (

28. For a closed tube, Fig. 16-12 indicates that f1 = temperature. v f1 = 4

→

(

)

→ P = 4 r 2 I = 4 ( 3.5 m ) 2.951W m 2 = 454 W  450 W 2

)(

v 4 f1

343 m s 4 ( 69 Hz )

v 4

. We assume the bass clarinet is at room

= 1.2 m

29. (a) If the pipe is closed at one end, only the odd harmonic frequencies are present. v nv 343 m s fn = = nf1 , n = 1, 3, 5 → f1 = = = 79.4 Hz 4 4 4 (1.08 m ) f3 = 3 f1 = 238 Hz

f5 = 5 f1 = 397 Hz

f 7 = 7 f1 = 556 Hz

(b) If the pipe is open at both ends, all the harmonic frequencies are present. v nv 343 m s fn = = n f1 , n = 1, 3, 5 → f1 = = = 159 Hz 2 2 2 (1.08 m )

f 2 = 2 f1 = 318 Hz

f3 = 3 f1 = 476 Hz

f 4 = 4 f1 = 635 Hz

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Chapter 16

Sound

30. For a pipe open at both ends, the fundamental frequency is given by f1 = =

given fundamental frequency is

20 Hz

343 m s 2 ( 20 Hz )

v 2 f1

= 8.6 m

v 2

, and so the length for a

20 kHz

343 m s 2 ( 20, 000 Hz )

= 8.6  10 −3 m

31. Each octave is a doubling of frequency. The number of octaves, n, can be found from the following. 20, 000 Hz = 2 n ( 20 Hz ) → 1000 = 2 n → log1000 = n log2 → n=

log1000 log 2

= 9.97  10 octaves

32. We approximate the shell as a closed tube of length 15 cm, and calculate the fundamental frequency. v 343 m s f = = = 572 Hz  570 Hz 4 4 ( 0.15 m ) 33. For a fixed string, the frequency of the nth harmonic is given by f n = nf1 . Thus the fundamental for this string is f1 = f3 3 = 540 Hz 3 = 180 Hz. When the string is fingered, it has a new length of 70% of the original length. The fundamental frequency of the vibrating string is also given by v , and v is constant for the string, assuming its tension is not changed. f1 = 2 1 180 Hz v v f1 = = = f1 = = 260 Hz 2 fingered 2 ( 0.70 ) 0.70 0.70 fingered 34. A flute is a tube that is open at both ends, and so the fundamental frequency is given by f =

, 2 where is the distance from the mouthpiece (antinode) to the first open side hole in the flute tube (antinode). v v 343 m s f = → = = = 0.491m 2 2 f 2 ( 349 Hz )

35. (a) At T = 18C, the speed of sound is given by v = ( 331 + 0.60 (18 ) ) m s = 341.8 m s . For an open pipe, the fundamental frequency is given by f = f =

v 2

→

v 2f

341.8 m s 2 ( 262 Hz )

v 2

= 0.652 m

(b) The frequency of the standing wave in the tube is 262 Hz . The wavelength is twice the length of the pipe, 1.30 m (c) The wavelength and frequency are the same in the air, because it is air that is resonating in the organ pipe. The frequency is 262 Hz and the wavelength is 1.30 m .

571

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

36. (a) We assume that the speed of waves on the guitar string does not change when the string is v fretted. The fundamental frequency is given by f = , and so the frequency is inversely 2 proportional to the length. 1 f  → f = constant

fE E = fA A →

fE E

 330 Hz   = 0.51m  440 Hz 

= ( 0.68 m ) 

The string should be fretted a distance 0.68 m − 0.51m = 0.17 m from the nut of the guitar. (b) The string is fixed at both ends and is vibrating in its fundamental mode. Thus the wavelength is twice the length of the string (see Fig. 16-7).  = 2 = 2 ( 0.51m ) = 1.02 m (c) The frequency of the sound will be the same as that of the string, 440 Hz . The wavelength in air is given by the following. v 343 m s = = = 0.78 m f 440 Hz 37. The speed of sound will change as the temperature changes, and that will change the frequency of the organ. Assume that the length of the pipe (and thus the resonant wavelength) does not change. v v v − v 24 f = f11 − f 24 = 11 f 24 = 24 f11 = 11



f f



v11 − v 24 =



v 24



v11 v 24



−1 =

331 + 0.60 (11)

331 + 0.60 ( 24 )

− 1 = −2.26  10 −2  −2.3%

38. (a) The difference between successive overtones for this pipe is 176 Hz. The difference between successive overtones for an open pipe is the fundamental frequency, and each overtone is an integer multiple of the fundamental. Since 264 Hz is not a multiple of 176 Hz, 176 Hz cannot be the fundamental, and so the pipe cannot be open. Thus it must be a closed pipe. (b) For a closed pipe, the successive overtones differ by twice the fundamental frequency. Thus 176 Hz must be twice the fundamental, so the fundamental is 88 Hz . This is verified since 264 Hz is 3 times the fundamental, 440 Hz is 5 times the fundamental, and 616 Hz is 7 times the fundamental. 39. The difference in frequency for two successive harmonics is to be 44 Hz. For an open pipe, two successive harmonics differ by the fundamental, so the fundamental could be 44 Hz, with 308 Hz being the 7th harmonic and 352 Hz being the 8th harmonic. For a closed pipe, two successive harmonics differ by twice the fundamental, so the fundamental could be 22 Hz. But the overtones of a closed pipe are odd multiples of the fundamental, and both overtones are even multiples of 22 Hz. So the pipe must be an open pipe . 331 + 0.60 ( 21.5) m s = 3.908 m  3.9 m v v f = → = = 2 2f 2 ( 44 Hz )

572

Chapter 16

Sound

40. (a) The difference between successive overtones for an open pipe is the fundamental frequency.

f1 = 330 Hz − 275 Hz = 55 Hz (b) The fundamental frequency is given by f1 =

v 2

. Solve this for the speed of sound.

v = 2 f1 = 2 (1.50 m )( 55 Hz ) = 165 m s  170 m s nv

41. (a) The harmonics for the open pipe are f n =

. To be audible, they must be below 20 kHz. 2 2 ( 2.18 m ) 2  10 4 Hz nv  2  104 Hz → n  = 254.2 2 343 m s

(

)

Since there are 254 harmonics, there are 253 overtones . nv (b) The harmonics for the closed pipe are f n = , n odd. Again, they must be below 20 kHz. 4 4 ( 2.18 m ) 2  10 4 Hz nv 4  2  10 Hz → n  = 508.5 4 343 m s The values of n must be odd, so n = 1, 3, 5, …, 507. There are 254 harmonics, and so there are 253 overtones .

(

)

42. To operate with the first overtone, we see from the figure that the thickness must be given by d = 1.5 . The speed of sound in the quartz is v = G  , analogous to Eqs. 15-3 and 15-4.

d = 1.5 = 1.5

v f

= 1.5

G  f

( 2.95 10 N m ) ( 2650 kg m ) 10

= 1.5

24.0  10 Hz 6

= 2.09  10 −4 m

43. A tube closed at both ends will have standing waves with displacement nodes at each end, and so has the same harmonic structure as a string that is fastened at both ends. Thus the wavelength of the fundamental frequency is twice the length of the hallway, 1 = 2 l = 18 m. f1 =

1

343 m s 18 m

= 19.056 Hz  19 Hz ; f 2 = 2 f1 = 38 Hz

44. The ear canal can be modeled as a closed pipe of length 2.5 cm. The resonant frequencies are given nv by f n = , n odd. The first several frequencies are calculated here. 4 n ( 343 m s ) nv fn = = = n ( 3430 Hz ) , n odd 4 4 2.5  10 −2 m

(

f1 = 3430 Hz

)

f 3 = 10300 Hz

f 5 = 17200 Hz

In the graph, the most sensitive frequency is between 3000 and 4000 Hz. This corresponds to the fundamental resonant frequency of the ear canal. The sensitivity decreases above 4000 Hz, but is seen to “flatten out” around 10,000 Hz again, indicating higher sensitivity near 10,000 Hz than at surrounding frequencies. This 10,000 Hz relatively sensitive region corresponds to the first overtone resonant frequency of the ear canal. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

573

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

45. The tension and mass density of the string do not change, so the wave speed is constant. The v frequency ratio for two adjacent notes is to be 21/12. The frequency is given by f = . 2 v

f =

2 1st

f1st

fret

→

f unfingered

fret

= 21/12 →

1st fret

unfingered

1/12

75.0 cm

21/12

= 70.79 cm

2 unfingered

2nd fret

1st fret

unfingered

1/12

2/12

→

nth fret

unfingered

n /12

; xnth =

unfingered

− nth = fret

fret

(1 − 2 ) − n /12

unfingered

( ) ( ) x = ( 75.0 cm ) (1 − 2 ) = 11.9 cm ; x = ( 75.0 cm ) (1 − 2 ) = 15.5 cm x = ( 75.0 cm ) (1 − 2 ) = 18.8 cm ; x = ( 75.0 cm ) (1 − 2 ) = 22.0 cm x1 = ( 75.0 cm ) 1 − 2 −1/12 = 4.2 cm ; x2 = ( 75.0 cm ) 1 − 2 −2/12 = 8.2 cm −3/12

−4/12

−5/12

−6/12

46. From Eq. 15-7a, the intensity is proportional to the square of the amplitude and the square of the A A frequency. From Fig. 16-14, the relative amplitudes are 2  0.4 and 3  0.15. A1 A1 I = 2 v  f A 2

→

 f  A  2 = = 2 2 =  2   2  = 22 ( 0.4 ) = 0.64 2 2 2 I1 2 v f1 A1 f1 A1  f1   A1 

2 2 v f 22 A22

f 22 A22

 f  A  2 =  3   3  = 32 ( 0.15 ) = 0.20 I1  f1   A1 

 2 −1 = 10 log 2 = 10 log 0.64 = −2 dB ;  3−1 = 10 log 3 = 10 log 0.24 = −7 dB

I1 Answers may vary due to the reading of the figure.

47. The beat period is 2.0 seconds, so the beat frequency is the reciprocal of that, 0.50 Hz. Thus the other string is off in frequency by 0.50 Hz . The beating does not tell the tuner whether the second string is too high or too low. 48. The 3000 Hz shrill whine is the beat frequency generated by the combination of the two sounds. This means that the brand X whistle is either 3000 Hz higher or 3000 Hz lower than the knownfrequency whistle. If it were 3000 Hz lower, then it would be just at the edge of human hearing limits. Since it cannot be heard by humans, the brand X whistle must be 3000 Hz higher than the known frequency whistle. Thus the brand X frequency is 23.5 kHz + 3kHz = 26.5 kHz . Since the original frequency is quoted to the 0.1 kHz, we assume that the 3000 Hz value is 3.0 kHz. 49. Since there are 3 beats/s when sounded with the 350 Hz tuning fork, the guitar string must have a frequency of either 347 Hz or 353 Hz. Since there are 8 beats/s when sounded with the 355 Hz tuning fork, the guitar string must have a frequency of either 347 Hz or 363 Hz. The common value is 347 Hz . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

574

Chapter 16

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50. The beat frequency is the difference in the two frequencies, or 277 Hz − 262 Hz = 15 Hz . If the frequencies are both reduced by a factor of 4, then the difference between the two frequencies will also be reduced by a factor of 4, and so the beat frequency will be 14 (15 Hz ) = 3.75 Hz  3.8 Hz . 51. The fundamental frequency of the violin string is given by f =

= 294 Hz. Change the 2 2  tension to find the new frequency, and then subtract the two frequencies to find the beat frequency. f=

( 0.975) FT



= 0.975

(



= 0.975 f

)

(

)

f = f − f  = f 1 − 0.975 = ( 294 Hz ) 1 − 0.975 = 3.7 Hz 52. (a) Since the sounds are initially 180 out of phase, x d−x B another 180 of phase must be added by a path A length difference. Thus the difference of the distances from the speakers to the point of constructive interference must be half of a wavelength. See the diagram. 343 m s v = 0.562 m ( d − x ) − x = 12  → d = 2 x + 12  → d min = 12  = = 2 f 2 ( 305 Hz ) This minimum distance occurs when the observer is right at one of the speakers. If the speakers are separated by more than 0.583 m, the location of constructive interference will be moved away from the speakers, along the line between the speakers. (b) Since the sounds are already 180 out of phase, as long as the listener is equidistant from the speakers, there will be completely destructive interference. So even if the speakers have a tiny separation, the point midway between them will be a point of completely destructive interference. The minimum separation between the speakers is 0. 53. Beats will be heard because the difference in the speed of sound for the two flutes will result in two different frequencies.

f1 =

331 + 0.60 ( 5.0 ) m s = 253.0 Hz v1 331 + 0.60 ( 25 )  m s v = = 262.1Hz ; f 2 = 2 = 2 2 ( 0.66 m ) 2 2 ( 0.66 m )

f = 262.1Hz − 253.0 Hz = 9.1Hz  9 beats sec 54. (a) The microphone must be moved to the right until the difference in distances from the two sources is half a wavelength. See the diagram. We square the expression, collect terms, isolate the remaining square root, and square again. d 2 − d1 = 12  →

( d + x) +

−

( d + x) +

= 12  +

1 2

( d + x) + 1 2

2dx − 14  2 = 

( d − x) + 2

1 2

= 12  →

( d − x) + 2

1 2

→

= 14  2 + 2 ( 12  )

( d − x) +

→ 4d 2 x 2 − 2 ( 2dx ) 14  2 + 161  4 =  2 ( 12 d − x ) + 2 

1 2

+ ( 12 d − x ) + 2

→





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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

4d x − dx + 2

1 16

 = d  − dx + x  +  4

1 4



Instructor Solutions Manual

( d + − ) → x= ( 4d −  ) 1 4



1 16

The values are d = 3.00 m, = 3.20 m, and  = v f = ( 343 m s ) ( 454 Hz ) = 0.7555 m. x = ( 0.7555 m )

1 4

( 3.00 m )  + ( 3.20 m )2 − 161 ( 0.7555 m )2 = 0.448 m 2 2 4 ( 3.00 m ) − ( 0.7555 m )

(b) When the speakers are exactly out of phase, the maxima and minima will be interchanged. The intensity maxima are 0.448 m to the left or right of the midpoint, and the intensity minimum is at the midpoint. 55. (a) For destructive interference, the smallest path difference must be one-half wavelength. Thus the wavelength in this situation must be twice the path difference, or 1.00 m. v 343 m s = 343 Hz f = =  1.00 m (b) There will also be destructive interference if the path difference is 1.5 wavelengths, 2.5 wavelengths, etc. v 343 m s 0.50 m  = 1.5 →  = = 0.333 m → f = = = 1030 Hz  1.0  10 3 Hz 1.5  0.333 m v 343 m s 0.50 m  = 2.5 →  = = 0.20 m → f = = = 1715 Hz  1700 Hz 2.5  0.20 m 56. (a) To find the beat frequency, calculate the frequency of each sound, and then subtract the two frequencies. v v 1 1 − = ( 343 m s ) − = 8.936 Hz  8.9 Hz f beat = f1 − f 2 = 1 2 2.54 m 2.72 m (b) The speed of sound is 343 m/s, and the beat frequency is 8.936 Hz. The regions of maximum intensity are one “beat wavelength” apart. 343 m s v = = = 38.38 m  38 m f 8.936 Hz 57. (a) Observer moving towards stationary source.  v   25.0 m s  f  =  1 + obs  f =  1 +  (1650 Hz ) = 1770 Hz  343 m s   vsnd  (b) Observer moving away from stationary source.  v   25.0 m s  f  =  1 − obs  f =  1 −  (1650 Hz ) = 1530 Hz  343 m s   vsnd  58. The moving object can be treated as a moving “observer” for calculating the frequency it receives and reflects. The bat (the source) is stationary.  v   = f bat  1 − object  f object vsnd    , and the bat as a Then the object can be treated as a moving source emitting the frequency f object stationary observer. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 vobject  1 − v   f object ( vsnd − vobject ) snd  = f bat  = f bat f bat =  vobject   vobject  ( vsnd + vobject ) 1 + v  1 + v    snd  snd  343 m s − 22.0 m s  )  343  = 4.40  10 Hz = 44.0 kHz m s + 22.0 m s

(

= 5.00  10 4 Hz 





59. We assume that the comparison is to be made from the frame of reference of the stationary tuba. The stationary observers would observe a Doppler-shifted frequency from the moving tuba. fsource 75 Hz f obs = = = 78.19 Hz f beat = 78.19 Hz − 75 Hz = 3.19 Hz  3 Hz .  vsource   14.0 m s   1 − v   1 − 343 m s     snd  60. The Doppler shift is 4.5 Hz, the emitted frequency from each train is 508 Hz, and so the frequency heard by the conductor on the stationary train is 512.5 Hz. Use this to find the moving train’s speed. vsnd 508 Hz  f    f= f → vsource = 1 −  vsnd = 1 −  ( 343 m s ) = 3.01m s  3 m s f ( vsnd − vsource )  512.5 Hz   61. The ocean wave has  = 44 m and v = 11m s relative to the ocean floor. The frequency of the v 11m s = 0.25 Hz. ocean wave is then f = = 44 m  (a) For the boat traveling west, the boat will encounter a Doppler shifted frequency, for an observer moving towards a stationary source. The speed v = 11m s represents the speed of the waves in the stationary medium, and so corresponds to the speed of sound in the Doppler formula. The time between encountering waves is the period of the Doppler shifted frequency.  v   5.0 m s   f observer =  1 + obs  f =  1 +  ( 0.25 Hz ) = 0.3636 Hz → moving  11m s   vsnd 

= 2.750 s  2.8s f 0.3636 Hz Note that this can also be solved by looking at the relative velocity of the waves and the boat, which is 16 m/s. The distance of 44 m (one wavelength) will be traveled in a time of 44 m d t= = = 2.75s v rel 16 m s (b) For the boat traveling east, the boat will encounter a Doppler shifted frequency, for an observer moving away from a stationary source.  v   5.0 m s   f observer =  1 − obs  f =  1 −  ( 0.25 Hz ) = 0.13636 Hz → moving  11m s   vsnd 

1 f

1 0.13636 Hz

= 7.3s

And alternatively, with a relative velocity of 4 m/s: t =

d v rel

44 m 6m s

= 7.3s.

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Instructor Solutions Manual

62. (a) For the 18 m/s relative velocity: 1 1  fsource = f = (1800 Hz ) = 1899.7 Hz  1900 Hz  vsrc   18 m s  moving 1 − v   1 − 343 m s     snd 



vsrc 

 18 m s   = (1800 Hz )  1 +  = 1894.5 Hz  1900 Hz moving  343 m s   vsnd    . The two frequencies are  f observer The frequency shifts are slightly different, with fsource  = f 1 + f observer

moving

close, but they are not identical. As a means of comparison, calculate the spread in frequencies divided by the original frequency.   − f observer f source 1899.7 Hz − 1894.5 Hz moving moving = = 0.0029 = 0.29% f source 1800 Hz (b) For the 160 m/s relative velocity: 1 1  fsource = f = (1800 Hz ) = 3373 Hz  3400 Hz  vsrc   160 m s  moving 1 − v   1 − 343 m s     snd 



 f observer = f 1 + moving



vsrc 

 160 m s   = (1800 Hz )  1 +  = 2640 Hz  2600 Hz vsnd   343 m s 

  .  f observer The difference in the frequency shifts is much larger this time, still with fsource moving

  f source − f observer moving

moving

3373 Hz − 2640 Hz

= 0.4072 = 41% f source 1800 Hz (c) For the 320 m/s relative velocity: 1 1  fsource = f = (1800 Hz ) = 26,800 Hz  27, 000 Hz  vsrc   320 m s  moving 1 − v   1 − 343 m s     snd 



vsrc 

 320 m s   = (1800 Hz )  1 +  = 3479 Hz  3500 Hz moving  343 m s   vsnd    .  f observer The difference in the frequency shifts is quite large, still with fsource  f observer = f 1 +

moving

  f source − f observer moving

moving

26,800 Hz − 3479 Hz

= 12.96  1300% f source 1800 Hz (d) The Doppler formulas are asymmetric, with a larger shift for the moving source than for the moving observer, when the two are getting closer to each other. In the following derivation, assume vsrc vsnd , and use the binomial expansion. −1

 v   v    fsource = f = f  1 − src   f  1 + src  = f observer vsnd  vsnd   vsrc  moving moving   1 − v   snd  1

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63. (a) The observer is stationary, and the source is moving. First the source is approaching, then the source is receding.  1m s  120.0 km h   = 33.33 m s  3.6 km h 

 fsource = f moving towards

 fsource = f moving away

 vsrc  1 − v   snd  1

 vsrc  1 + v   snd 

= (1580 Hz )

 33.33 m s   1 − 343 m s    1

 33.33 m s   1 + 343 m s   

= 1750 Hz

= 1440 Hz

(b) Both the observer and the source are moving, and so use Eq. 16-11.  1m s  90.0 km h   = 25 m s  3.6 km h 

( vsnd + vobs ) ( 343 m s + 25 m s ) = (1580 Hz ) = 1880 Hz ( 343 m s − 33.33 m s ) ( vsnd − vsrc ) ( v − vobs ) ( 343 m s − 25 m s )  f receding = f snd = (1580 Hz ) = 1340 Hz ( vsnd + vsrc ) ( 343 m s + 33.33 m s )  f approaching = f

(c) Both the observer and the source are moving, and so again use Eq. 16-11.  1m s  80.0 km h   = 22.22 m s  3.6 km h 

 f police

= f

car approaching

 f police = f car receding

( vsnd − vobs ) ( 343 m s − 22.22 m s ) = (1580 Hz ) = 1640 Hz ( vsnd − vsrc ) ( 343 m s − 33.33 m s )

( vsnd + vobs ) ( 343 m s + 22.22 m s ) = (1580 Hz ) = 1530 Hz ( vsnd + vsrc ) ( 343 m s + 33.33 m s )

64. The maximum Doppler shift occurs when the heart has its maximum velocity. Assume that the heart is moving away from the original source of sound. The beats arise from the combining of the original 2.25 MHz frequency with the reflected signal which has been Doppler shifted. There are two Doppler shifts – one for the heart receiving the original signal (observer moving away from stationary source) and one for the detector receiving the reflected signal (source moving away from stationary observer).  v heart  1 − v    vheart  f heart ( v − vheart ) snd   = f original  1 −  = = f original  = f original snd f heart f detector   vheart   vheart  ( vsnd + vheart )  vsnd  1 + v  1 + v    snd  snd   f = f original − f detector = f original − f original

vheart = vsnd

f 2 f original − f

(

2v heart ( vsnd − vheart ) = f original ( vsnd + vheart ) ( vsnd + vheart )

= 1.54  103 m s

→

) 2 2.25 10220HzHz − 220 Hz = 0.075 m s ( ) 6

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If instead we had assumed that the heart was moving towards the original source of sound, we would f get vblood = vsnd . Since the beat frequency is much smaller than the original frequency, 2 f original + f the f term in the denominator does not significantly affect the answer, whether added or subtracted. 65. The Doppler effect occurs only when there is relative motion of the source and the observer along the line connecting them. In the first four parts of this problem, the whistle and the observer are not moving relative to each other and so there is no Doppler shift. The wind speed increases (or decreases) the velocity of the waves in the direction of the wind, as if the speed of sound were different, but the frequency of the waves doesn’t change. We do a detailed analysis of this claim in part (a). (a) The wind velocity is a movement of the medium, and so adds or subtracts from the speed of sound in the medium. Because the wind is blowing away from the observer, the effective speed of sound is vsnd − v wind . The wavelength of the waves traveling towards the observer is

a = ( vsnd − v wind ) f 0 , where f 0 is the frequency of the sound emitted by the factory whistle.

This wavelength approaches the observer at a relative speed of vsnd − v wind . Thus the observer hears the frequency calculated here. v − v wind v − v wind = snd = f 0 = 770 Hz f a = snd a  vsnd − v wind 

 

 

(b) Because the wind is blowing towards the observer, the effective speed of sound is vsnd + v wind . The same kind of analysis as applied in part (a) gives fb = 770 Hz . (c) Because the wind is blowing perpendicular to the line towards the observer, the effective speed of sound along that line is vsnd . Since there is no relative motion of the whistle and observer, there will be no change in frequency, and so f c = 770 Hz . (d) This is just like part (c), and so f d = 770 Hz . (e) Because the wind is blowing toward the cyclist, the effective speed of sound is vsnd + v wind . The wavelength traveling toward the cyclist is e = ( vsnd + v wind ) f 0 . This wavelength approaches the cyclist at a relative speed of vsnd + vwind + vcycle. The cyclist will hear the following frequency.

fe =

(v + v snd

wind

e

+ vcycle )

(v + v snd

wind

+ vcycle )

( vsnd + vwind )

f0 =

( 343 + 15.0 + 12.0 ) m s ( 770 Hz ) ( 343 + 15.0 )

= 796 Hz (f)

Since the wind is not changing the speed of the sound waves moving towards the cyclist, the speed of sound is 343 m/s. The observer is moving towards a stationary source with a speed of 12.0 m/s.  v   12.0 m s  f f = f  1 + obs  = ( 770 Hz )  1 +  = 797 Hz  343 m s   vsnd 

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66. We represent the Mach number by the symbol M. vobj (a) M = → vobj = M vsnd = ( 0.33) ( 331m s ) + 0.60 ( 24 )  = 113.98 m s  110 m s vsnd (b) M =

vobj vsnd

→ vsnd =

vobj M

3000 km h 3.1

 1m s   = 270 m s  3.6 km h 

= 967.7 km h 

67. The speed is found from Eq. 16-12. Since the wake is 12, the angle to use in the equation is 6 . v v 2.2 km h sin  = wave → vobj = wave = = 21.05 km h  20 km h sin  sin 6 vobj 68. From Eq. 16-12, sin  = (a)  = sin −1 (b)  = sin −1

vsnd vobj

= sin −1

. The speed of sound in the ocean is taken from Table 16-1.

343m s 9400 m s

1560 m s 9400 m s

= 2.1

= 9.6

69. (a) The Mach number is the ratio of the object’s speed to the speed of sound.  1m s  1.5  104 km hr   3.6 km hr  vobs  M = = = 75.76  76 vsound 55 m s (b) Use Eq. 16-12 to find the angle. v 1 1 = sin −1 = 0.76  = sin −1 snd = sin −1 vobj M 75.76

(

)

70. Find the angle of the shock wave, and then find the distance the plane has traveled when the shock wave reaches the observer. Use Eq. 16-12. v v 1 = 30  = sin −1 snd = sin −1 snd = sin −1 vobj 2.0vsnd 2.0

tan  =

9500 m D

→ D=

9500 m tan 30

D  9500 m

= 16454 m  16 km

71. Consider one particular wave as shown in the diagram, created at time t = 0, at the location of the black dot. After a time t has elapsed from the creation of that wave, the supersonic source has moved a distance vobjt , and the wave front has moved a distance vsnd t. The line from the position of the source at time t is tangent to all of the wave fronts, showing the location of the shock wave (see Fig. 16-23d in the text). A tangent to a circle at a point is perpendicular to the radius connecting that point to the center, and so a right angle is formed. From the right triangle, the angle  can be defined.

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sin  =

vsnd t vobjt

Instructor Solutions Manual

vsnd vobj

72. (a) The displacement of the plane from the time it passes overhead to the time the shock wave reaches the listener is shown, along with the shock wave front. From the displacement and height of the plane, the angle of the shock wave front relative to the direction of motion can be found. 1.45 km 1.45 tan  = →  = tan −1 = 35.94  36 2.0 km 2.0 (b) The speed and Mach number are found from Eq. 16-12. v 330 m s = 562.2 m s  560 m s vobj = snd = sin  sin 35.94 vobj 1 1 = = = 1.7 M = vsnd sin  sin 35.94

2.0 km  1.45 km

73. The minimum time between pulses would be the time for a pulse to travel from the boat to the maximum distance and back again. The total distance traveled by the pulse will be 170 m, at the speed of sound in fresh water, 1440 m/s. 170 m d d = vt → t = = = 0.12 s v 1440 m s 74. The single mosquito creates a sound intensity of I 0 = 110−12 W m2. Thus 200 mosquitoes will create a sound intensity of 200 times that of a single mosquito. 200 I 0 I = 200 I 0  = 10 log = 10 log 200 = 23dB . I0 75. (a) The ultrasonic pulse travels at the speed of sound, and the round trip distance is twice the distance d to the object. 2d min = vtmin → d min = 12 vtmin = 12 ( 343 m s ) (1.0  10−3 s ) = 0.17 m (b) The measurement must take no longer than 1/15 s. Again, the round trip distance is twice the distance to the object. 2d max = vtmax → d max = 12 vtmax = 12 ( 343 m s ) ( 151 s ) = 11m (c) The distance is proportional to the speed of sound. So the percentage error in distance is the same as the percentage error in the speed of sound. We assume the device is calibrated to work at 20C. d v v23 C − v20 C 331 + 0.60 ( 23) m s − 343 m s = = = = 0.005248  0.5% d 343 m s v v20 C

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

v walk 



vsnd 

 76. The person will hear a frequency f towards = f 1 +



 = f 1 − person will hear a frequency f away

 from the speaker that they walk towards. The

vwalk 

 from the speaker that they walk away from. The vsnd  beat frequency is the difference in those two frequencies.  v   v  v 1.6 m s   = f  1 + walk  − f  1 − walk  = 2 f walk = 2 ( 315 Hz ) f towards − f away = 2.9 Hz vsnd 343 m s  vsnd   vsnd  

77. Relative to the 1000 Hz output, the 15 kHz output is –12 dB. P P P −12 dB = 10 log 15 kHz → − 1.2 = log 15 kHz → 10 −1.2 = 15 kHz 225 W 225 W 225 W

(

→

)

P15 kHz = ( 225 W ) 10−1.2 = 14 W

78. The power output is found from the intensity, which is the power radiated per unit area. I 115 dB = 10 log → I = 1011.5 I 0 = 1011.5 (1.0  10 −12 W m 2 ) = 3.162  10 −1 W m 2 I0 I=

P A

P 4 r

(

)

→ P = 4 r 2 I = 4 ( 9.75 m ) 3.162  10−1 W m 2 = 378 W 2

79. Assume that only the fundamental frequency is heard. The fundamental frequency of an open pipe is v given by f = . 2L v 343 m s v 343 m s = = 57 Hz f 2.5 = = = 69 Hz (a) f3.0 = 2 L 2 ( 3.0 m ) 2 L 2 ( 2.5 m ) f 2.0 = f1.0 =

v 2L v

= =

343 m s 2 ( 2.0 m )

= 86 Hz

f1.5 =

v 2L

343 m s 2 (1.5 m )

= 114.3 Hz  110 Hz

343 m s

= 171.5 Hz  170 Hz 2 L 2 (1.0 m ) (b) On a noisy day, there are a large number of component frequencies to the sounds that are being made – more people walking, more people talking, etc. Thus it is more likely that the frequencies listed above will be a component of the overall sound, and then the resonance will be more prominent to the hearer. If the day is quiet, there might be very little sound at the desired frequencies, and then the tubes will not have any standing waves in them to detect.

80. The 130 dB level is used to find the intensity, and the intensity is used to find the power emitted by the engine. It is assumed that the jet airplane engine radiates equally in all directions. I  = 130 dB = 10 log → I = 1013 I 0 = 1013 1.0  10 −12 W m 2 = 1.0  101 W m 2 I0

(

)

P = IA = I 4 r 2 = 1.0  101 W m 2 4 ( 35 m ) = 1.539  105 W 2

Now use the power emitted by the engine to find the intensity at a distance of 25 m.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

1.539  105 W

Instructor Solutions Manual

= 19.60 W m 2

4 ( 25 m ) And now use that intensity to find out how much power is intercepted by an unprotected ear. Note that the area to use now is NOT the area of a sphere, but the area of the ear, modeled as a circle of radius 2.0 cm. A

(

)

P = IA = I r 2 = 19.60 W m2  ( 0.020 m ) = 2.463  10−2 W  2.5  10−2 W 2

81. (a) We use Eq. 15-7a for intensity, and assume that we are in air which determines the density.

(

)

(

I = 2 2 v  f 2 A2 = 2 2 ( 343 m s ) 1.29 kg m 3 (100Hz ) 5.29  10 −11 m 2

)

= 2.44  10 −13 W m 2

 = 10 log

2.44  10 −13 W m 2

= −6.13dB  −6 dB 1.0  10 −12 W m 2 This is well below the intensity of the threshold of hearing at 100 Hz (about 35 dB, from Fig. 16-6), and so it is not audible. (b) We repeat the calculation for 2.0 kHz.

(

I = 2 2 v  f 2 A2 = 2 2 ( 343 m s ) 1.29 kg m 3

)( 2.0 10 Hz ) ( 5.29 10 m ) 3

−11

= 9.777  10 −11 W m 2

 = 10 log

9.777  10 −11 W m 2

= 19.9 dB  20 dB 1.0  10 −12 W m 2 The threshold of hearing at 2.0 kHz from Fig. 16-6 is less than 0 dB, so this is audible.

82. (a) The gain is given by  = 10 log

Pout

= 10 log

135 W

= 51dB . Pin 1.0  10 −3 W (b) We solve the gain equation for the noise power level. P P 10 W = 93/10 = 5  10−9 W  = 10log signal → Pnoise = signal  10 Pnoise 10 10

83. As the car approaches, the frequency of the sound from the engine is Doppler shifted up, as given by Eq. 16-9a. As the car moves away, the frequency of the engine sound is Doppler shifted down, as   . given by Eq. 16-9b. Since the frequency shift is exactly an octave, we know that f toward = 2 f away We then solve for the car’s speed. f f  =  = f toward ; f away  vsource   vsource  1 − v  1 + v    snd  snd   = 2 f away  → f toward

 vsource  1 − v   snd 



 vsource  1 + v   snd 

→ 1 +



vsource 

 vsource   = 2 1 −  → vsnd  vsnd  

vsource = 13 vsnd = 13 ( 343 m s ) = 114 m s This is about 255 miles per hour. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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84. The effective length of the tube is

eff.

fn =

Corrected frequencies:

= + 13 D = 0.550 m + 13 ( 0.030 m ) = 0.560 m.

( 2n − 1) v 4 eff

f1−4 = ( 2n − 1)

, n = 1, 2, 3

→

343m s

= 153 Hz, 459 Hz, 766 Hz, 1070 Hz 4 ( 0.560 m ) Without the correction, the values would have been 156 Hz, 468 Hz, 780 Hz, and 1090 Hz. 85. The apparatus is a closed tube. The water level is the closed end, and so is a node of air displacement. As the water level lowers, the distance from one resonance level to the next corresponds to the distance between adjacent nodes, which is one-half wavelength.  = 12  →  = 2 = 2 ( 0.395 m − 0.125 m ) = 0.540 m

f =



343 m s 0.540 m

= 635 Hz

86. Since the sound is loudest at points equidistant from the two sources, the two sources must be in phase. The difference in distance from the two sources must be an odd number of half-wavelengths for destructive interference. See Fig. 16-16 and the accompanying discussion for a derivation of that relationship. 0.25 m =  2 →  = 0.50 m f = v  = 343 m s 0.50 m = 686 Hz

0.25 m = 3 2 →  = 0.1667 m f = v  = 343 m s 0.1667 m = 2060 Hz ( out of range ) Thus we see that the frequency must be about 690 Hz. 87. Call the frequencies of four strings of the violin f A , f B , f C , f D with f A the lowest pitch. The mass per unit length will be named  . All strings are the same length and have the same tension. For a string with both ends fixed, the fundamental frequency is given by f1 = f B = 1.5 f A →

B

f C = 1.5 f B = (1.5 ) f A → 2

f D = 1.5 f C = (1.5 ) f A → 3

= 1.5

A

C

D

→ B =

= (1.5 )

A

(1.5 )

A



= 0.44  A

→ C = → D =

A

(1.5)

A

(1.5)

= 0.20 A = 0.088A

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

88. We combine the expression for the frequency in a closed tube with the Doppler shift for a source moving away from a stationary observer, Eq. 16-9b. v f1 = snd 4 vsnd f1 343 m s f= = = = 951Hz 25 m s   vsource   vsource   −2  1 + v  4  1 + v  4 8.40  10 m  1 + 343 m s      snd  snd 

(

)

89. (a) The wave speed on the string can be found from the length and the fundamental frequency. v f = → v = 2 f = 2 ( 0.32 m )( 440 Hz ) = 281.6 m s  280 m s 2 The tension is found from the wave speed and the mass per unit length.

(

)

→ FT =  v 2 = 7.21 10−4 kg m ( 281.6 m s ) = 57 N



(b) The length of the pipe can be found from the fundamental frequency and the speed of sound. v v 343 m s f = → = = = 0.1949 m  0.19 m 4 4 f 4 ( 440 Hz ) (c) The first overtone for the string is twice the fundamental. 880 Hz The first overtone for the open pipe is 3 times the fundamental. 1320 Hz 90. (a) Since both speakers are moving towards the observer at the same speed, both frequencies have the same Doppler shift, and the observer hears no beats . (b) The observer will detect an increased frequency from the speaker moving towards him and a decreased frequency from the speaker moving away. The difference in those two frequencies will be the beat frequency that is heard. 1 1   = f = f f towards f away  v train   v train  1 − v  1 + v    snd  snd 

  = f f towards − f away



( 348 Hz ) 

 v train  1 − v   snd 

−f

343 m s

 ( 343 m s − 12.0 m s )



 v train  1 + v   snd  −

= f

vsnd

 ( vsnd − v train )

−

 ( vsnd + vtrain )  vsnd

 = 24.38 Hz  24 Hz ( 343 m s + 12.0 m s )  343 m s

(c) Since both speakers are moving away from the observer at the same speed, both frequencies have the same Doppler shift, and the observer hears no beats . 91. The intensity can be found from the sound level in decibels. I  = 10 log → I = 10  /10 I 0 = 1012 (10 −12 W m 2 ) = 1.0 W m 2 I0 Consider a square perpendicular to the direction of travel of the sound wave. The intensity is the energy transported by the wave across a unit area perpendicular to the direction of travel, per unit © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

586

Chapter 16

Sound

E , where S is the area of the square. Since the energy is “moving” with the wave, S t the “speed” of the energy is v , the wave speed. In a time  t , a volume equal to V = S v t would time. So I =

contain all of the energy that had been transported across the area S. Combine these relationships to find the energy in the volume. E

→ E = IS t =

S t

(1.0 W m ) ( 0.010 m ) = 2.9  10 J = 3

I V

−9

343 m s

92. The frequency received by the stationary car is higher than the frequency emitted by the stationary car, by  f = 5 Hz.

f obs = f source +  f =

 vsnd

fsource =  f 

 vsource

fsource

→

 vsource  1 − v   snd  

 343 m s



 6m s

− 1  = ( 5 Hz ) 



− 1  = 280.8 Hz  300 Hz



93. For each pipe, the fundamental frequency is given by f =

v 2

. Find the frequency of the shortest

pipe. f =

343 m s

= 71.46 Hz 2 2 ( 2.40 m ) The longer pipe has a lower frequency. Since the beat frequency is 6.0 Hz, the frequency of the longer pipe must be 65.46 Hz. Use that frequency to find the length of the longer pipe. v v 343 m s f = → = = = 2.62 m 2 2 f 2 ( 65.46 Hz )

94. The beats arise from the combining of the original 3.80 MHz frequency with the reflected signal which has been Doppler shifted. There are two Doppler shifts – one for the blood cells receiving the original frequency (observer/cells moving away from stationary source) and one for the detector receiving the reflected frequency (source/cells moving away from stationary observer).



vblood 



vsnd 

 = f original  1 − f blood



 vblood  1 − v   f blood ( v − vblood ) snd   f detector = = f original  = f original snd  vblood   vblood  ( vsnd + vblood ) 1 + v  1 + v    snd  snd 

2v blood ( vsnd − vblood ) = f original ( vsnd + vblood ) ( vsnd + vblood ) 2 ( 0.32 m s ) 3

 f = f original − f detector = f original − f original

(

= 4.80  106 Hz

) 1.54  10 m s + 0.32 m s = 1994 Hz  2.0  10 Hz ( ) 3

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

95. It is 65.0 ms from the start of one chirp to the start of the next. Since the chirp itself is 3.0 ms long, it is 62.0 ms from the end of a chirp to the start of the next. Thus the time for the pulse to travel to the moth and back again is 62.0 ms. The distance to the moth is half the distance that the sound can travel in 62.0 ms, since the sound must reach the moth and return during the 62.0 ms.

(

)

d = vsnd t = ( 343 m s ) 12 62.0  10−3 s = 10.6 m 96. We are given that = 0.32 m and f = 440 Hz. (a) We are to change the frequency by changing the length only. We know from Eq. 15-17b that v the fundamental frequency and the length are inversely proportional, f = . We don’t change 2 the wave speed. f 440 Hz v f1 1 = = f 2 2 → 2 = 1 1 = ( 32 cm ) = 23.94cm 2 588 Hz f2 Thus the finger must be placed 32 cm − 23.94 cm = 8.05cm  8cm from the end of the string. (b) During the “vibrato”, the length of the string will vary from 23.44 cm to 24.44 cm. Calculate the range of frequencies using the same relationship between frequency and length. The speed of waves on the string still doesn’t change. 32 cm v f1 1 = = f 2 2 → f 2 = f1 1 = ( 440 Hz ) = 576.1Hz 2 24.44 cm 2

f 3 = f1

= ( 440 Hz )

32 cm 23.44 cm

= 600.7 Hz

So the range of frequencies is from 580 Hz to 600 Hz , or with one more significant digit, from 576 Hz to 601 Hz. 97. For a tube open at both ends, all integer harmonics are allowed, with f n = nf1. Thus consecutive harmonics differ by the fundamental frequency. The four consecutive harmonics give the following values for the fundamental frequency. f1 = 523 Hz − 392 Hz = 131Hz, 659 Hz − 523 Hz = 136 Hz, 784 Hz − 659 Hz = 125 Hz

The average of these is f1 = 13 (131Hz + 136 Hz + 125 Hz )  131Hz. We use that for the fundamental frequency. 343 m s v v → = = = 1.31m (a) f1 = 2 2 f1 2 (131Hz ) Note that the bugle is coiled like a trumpet so that the full length fits in a smaller distance. f 392 Hz f 523 Hz = 2.99 ; nC5 = C5 = = 3.99 ; (b) f n = nf1 → nG4 = G4 = f1 131Hz f1 131Hz nE5 =

f E5 f1

659 Hz 131Hz

= 5.03 ; nG5 =

f G5 f1

784 Hz 131Hz

= 5.98

The harmonics are 3, 4, 5, and 6 .

588

Chapter 16

98. (a)

Sound

We assume that vsrc

vsnd , and use the binomial expansion. −1

 v   v    fsource = f = f  1 − src   f  1 + src  = f observer vsnd  vsnd   vsrc  moving moving   1 − v   snd  1

(b) We calculate the percent error in general, and then substitute in the given relative velocity.

  vsrc   1 −f  f 1 +    vsrc     vsnd  1 −  v    approx. − exact   snd  = % error =  100 100    1 exact     f  vsrc    1 − v     snd    2

 v  v   v   18.0 m s  = 100  1 + src  1 − src  − 1 = −100  src  = −100   = −0.28%  343 m s   vsnd   vsnd  vsnd   The negative sign indicates that the approximate value is less than the exact value. 99. The Alpenhorn can be modeled as an open tube, and so the fundamental frequency is f = the overtones are given by f n =

f1 =

v 2

343 m s 2 ( 3.4 m )

nv 2

, n = 1, 2, 3,

v 2

, and

= 50.44 Hz  50 Hz 370

f n = nf1 = f F # → n ( 50.44 Hz ) = 370 Hz → n =

= 7.34 50.44 Thus the 7th harmonic, which is the 6th overtone , is close to F#. 100. First, find the path difference in the original configuration. Then move the obstacle to the right by d so that the path difference increases by 12  . Note that the path difference change must be on the same order as the wavelength, and so d d , since  , d . We use D to represent the difference in distance along the two paths (direct and reflected).

( D )initial = 2 d 2 + ( 12 ) −

; ( D )final = 2

(

( d + d ) + ( 12 ) −

) − (2 d + ( ) − ) →

( D )final − ( D )initial = 12  = 2 ( d + d ) + ( 12 ) − 2

( d + d ) + ( 12 ) = 12  + 2 d 2 + ( 12 ) 2

1 2

Square the last equation above. 4  d 2 + 2d d + ( d ) + ( 12



)  =  + 2 (  ) 2 d + ( ) + 4 d + ( )  2

1 4

1 2

We delete terms that are second order in the small quantities d and . 8d d = 2 d 2 + ( 12

)

→

d =

 4d

d 2 + ( 12

)

589

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

101. (a) The “singing” rod is manifesting standing waves. By holding the rod at its midpoint, it has a node at its midpoint, and antinodes at its ends. Thus the length of the rod is a half wavelength. The speed of sound in aluminum is found in Table 16-1. v v 5100 m s = = 3187.5 Hz  3200 Hz f = =  2 1.6 m (b) The wavelength of sound in the rod is twice the length of the rod, 1.6 m . (c) The wavelength of the sound in air is determined by the frequency and the speed of sound in air. 343 m s v = = = 0.1076 m  0.11m f 3187.5 Hz 102. The ratio of sound intensity passing through the door to the original sound intensity is a 30 dB decrease.  = 10log I I 0 = −30 → log I I 0 = −3 → I = 10−3 I 0 Only 1/1000 of the sound intensity passes through the door. 103. There will be two Doppler shifts in this problem – first for a stationary source with a moving “observer” (the blood cells), and then for a moving source (the blood cells) and a stationary “observer” (the receiver). Note that the velocity component of the blood parallel to the sound transmission is vblood cos 45 = 12 vblood. It is that component that causes the Doppler shift.



 = f original  1 − f blood



1 2

v blood 

vsnd

 

1 v   2 blood 1 −   vsnd − 12 vblood )  ( vsnd  f blood   f detector = = f original = f original → 1 1 v v     vsnd + 12 vblood ) ( 2 blood 2 blood 1 +  1 +  vsnd  vsnd   

vblood = 2

(f ( f 

original detector

 ) − f detector

+ f original )

vsnd

Since the cells are moving away from the transmitter / receiver combination, the final frequency  received is less than the original frequency, by 780 Hz. Thus fdetector = f original − 780 Hz.

vblood = 2

(f ( f 

original detector

= 2

 ) − f detector

+ f original )

vsnd = 2

( 780 Hz )

 2 ( 5.0  106 Hz ) − 780 Hz 

( 2f

( 780 Hz ) original

− 780 Hz )

vsnd

(1540 m s ) = 0.17 m s

104. The displacement amplitude is related to the intensity by Eq. 15-7a. The intensity can be calculated from the decibel value. The medium is air. I  = 10 log → I = (10  10 ) I 0 = 1010.5 (10 −12 W m 2 ) = 0.0316 W m 2 I0

590

Chapter 16

Sound

(a) I = 2 2 v  f 2 A2 → A=

(b)

f

2v

f

2v 

0.0316 W m 2

 ( 8.0  10 Hz ) 2 ( 343 m s ) (1.29 kg m ) 3

0.0316 W m 2

 ( 35 Hz ) 2 ( 343 m s ) (1.29 kg m ) 3

= 2.4  10−7 m

= 5.4  10−5 m

591

CHAPTER 17: Temperature, Thermal Expansion, and the Ideal Gas Law Responses to Questions 1.

1 kg of copper will have more atoms. Copper has an atomic mass less than that of lead. Since each Cu atom is less massive than each Pb atom, there will be more Cu atoms than Pb atoms in 1 kg.

Properties of materials that could be exploited in making a thermometer include: a. thermal expansion, both linear and volume, as in a liquid-in-glass thermometer or a bimetallic strip; b. the proportionality between temperature and pressure (for an ideal gas when volume is held constant, as in the constant-volume gas thermometer); c. the temperature dependence of resistivity (discussed in Chapter 25); d. the frequency (color) of emitted radiation from a heated object (blackbody radiation curve, discussed in Chapter 37).

1Co is larger than 1Fo. There are 100 Co between the freezing and boiling temperatures of water, while there are 180 Fo between the same two temperatures, so the Celsius degree must be larger.

A and B have the same temperature; the temperature of C is different than that of A and B.

No. We can only infer that the temperature of C is different from that of A and B. We cannot infer anything about the relationship of the temperatures of A and B.

To be precise, l 0 is to be the initial length of the object, at the initial temperature. In practice, however, since the value of the coefficient of expansion is so small, there will be little difference in the calculation of l caused by using either the initial or final length, unless the temperature change is quite large.

When heated, the aluminum expands more than the iron, because the expansion coefficient of aluminum is larger than that of iron. Thus the aluminum will be on the outside of the curve.

The steam pipe can have a large temperature change as the steam enters or leaves the pipe. If the pipe is fixed at both ends and the temperature changes significantly, there will be large thermal stresses which might break joints. The “U” in the pipe allows for expansion and contraction which is not possible at fixed ends. This is similar to the joints placed in concrete roadway surfaces to allow expansion and contraction.

The bimetallic strip is made of two types of metal joined together. The metal of the outside strip has a higher coefficient of linear expansion than that of the inside strip, so it will expand and contract more dramatically. If the temperature goes above the thermostat setting, the outer strip will expand more than the inner, causing the spiral to wind more tightly and tilt the glass vessel so that the liquid mercury flows away from the contact wires, and the heater turns off. If the temperature goes below the thermostat setting, the vessel tilts back as the outer strip contracts more than the inner and the spiral opens, and the heater turns on. Moving the temperature setting lever changes the initial position of the glass vessel, and thus changes the temperature at which the coil has expanded or contracted enough to move the mercury. For instance, if the lever is set at 50, the vessel is tilted with the mercury far from the contact wires. The outer strip has to shrink significantly to uncurl the spiral enough to tilt the vessel back.

592

Chapter 17

Temperature, Thermal Expansion, and the Ideal Gas Law

10. The lead will float lower. Mercury has a larger coefficient of volume expansion than lead. When the temperature rises, mercury will expand more than lead. The density of mercury will decrease more than the density of lead will decrease, and so the lead will need to displace even more mercury in order to balance its weight. 11. If one part is heated or cooled more than another part, there will be more expansion or contraction of one part of the glass compared to an adjacent part. This causes internal stress forces which may exceed the maximum strength of the glass. 12. If water is added quickly to an overheated engine, it comes into contact with the very hot metal parts of the engine. Some areas of the metal parts will cool off very rapidly; others will not. Some of the water will quickly turn to steam and will expand rapidly. The net result can be a cracked engine block or radiator, due to the thermal stress, and/or the emission of high-temperature steam from the radiator. Water should always be added slowly, with the engine running. The water will mix with the hotter water already in the system, and will circulate through the engine, gradually cooling all parts at about the same rate. 13. The coefficient of expansion is derived from a ratio of lengths:  =

l 1

. The length units cancel, l 0 T and so the coefficient does not depend on the specific length unit used in its determination, as long as the same units are used for both l and l 0 .

14. When the cold thermometer is placed in the hot water, the glass part of the thermometer will expand first, as heat is transferred to it first. This will cause the alcohol level in the thermometer to decrease. As heat is transferred to the alcohol inside the thermometer, the alcohol will then expand. Since alcohol has a larger coefficient of expansion than glass, the alcohol will eventually expand more than the glass expands, and the level of alcohol in the thermometer will rise. 15. Since Pyrex glass has a smaller coefficient of linear expansion than ordinary glass, it will expand less than ordinary glass when heated, making it less likely to crack from internal stresses. Pyrex glass is therefore more suitable for applications involving heating and cooling. An ordinary glass mug may expand to the point of cracking if boiling water is poured in it, but a Pyrex mug will not. 16. On a hot day, the pendulum will be slightly longer than at 20oC, due to thermal expansion of the brass rod. Since the period of a pendulum is proportional to the square root of its length, the period will be slightly longer on the hot day, meaning that the pendulum takes more time for one oscillation. Thus the clock will run slow. 17. The soda is mostly water. As water cools below 4oC it expands. There is more expansion of the soda as it cools below 4oC and freezes than there is available room in the can (the can has actually shrunk a small amount, making the mismatch more pronounced), and so the freezing soda pushes against the can surfaces hard enough to push them outward. Evidently the top and bottom of the can are the weakest parts. 18. The coefficient of volume expansion is much greater for alcohol than for mercury. A given temperature change will therefore result in a greater change in volume for alcohol than for mercury. This means that smaller temperature changes can be measured with an alcohol thermometer, because the length changes of the alcohol will be larger.

593

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

19. The buoyant force on the aluminum sphere is the weight of the water displaced by the sphere, which is the volume of the sphere times the density of water times g. As the aluminum and water are heated, g remains the same, the volume of the sphere increases, and the density of the water decreases (because of its increased volume). Since the volume expansion coefficient of the water is almost three times larger than that of the aluminum, the fractional decrease in the water density is larger than the fractional increase in the aluminum volume. Thus, the product of the volume of the sphere times the density of water decreases, and the buoyant force gets smaller. 20. The atom is helium. If we take the atomic mass of 6.7  10−27 kg and divide by the conversion factor from kg to u (atomic mass units), which is 1.66  10−27 kg u , we get 4.03 u. This corresponds to the atomic mass of helium. 21. Not really, as long as the pressure is very low. At low pressure, most gases will behave like an ideal gas. Some practical considerations would be the volatility of the gas and its corrosive properties. Light monatomic or diatomic gases are best.

Solutions to MisConceptual Questions 1.

(c) Students may confuse thermal expansion with elasticity and surmise that the narrower rod would expand more for the same temperature change. However, in thermal expansion the change in length is independent of the rod’s diameter or cross-sectional area.

(a) Eq. 17–1a for thermal expansion shows that the change in length depends upon the initial length, the change in temperature, and the type of material (through the coefficient of thermal expansion). Since both rods are iron and have the change in temperature, the longer rod will have the greater change in length.

(d) Eq. 17–1a for thermal expansion shows that the change in length depends upon the initial length, the change in temperature, and the coefficient of thermal expansion (which depends upon the type of material).

(c) Many students have the misconception that as the plate expands, the hole will get smaller. To understand what actually happens, imagine that the hole is filled with a steel disk. As the plate and the steel disk are heated both will expand. After the plate and disk are heated, the disk is removed from the plate. Since the disk expanded, the hole that is left must also have expanded. As the steel ring is heated its circumference will expand causing its interior to also expand.

(b) The tape will expand when it is heated. Thus, the mark on the tape at the 1.0-cm position, which was originally a distance of 1.0 cm from the zero mark, might have expanded to be 1.05 cm long. But the tape will still read 1.0 cm on its markings. Thus the tape will “read” small, and so the measured length of some object will be smaller than its actual length.

(b) A common error is to treat the temperature as doubling. If the temperature doubled the pressure would also double. However, the temperature is given in degrees Celsius, not kelvins. When the temperature is converted to kelvins, it is easy to see that the temperature only increases by about 25%, from 373 K to 473 K. The pressure then also increases by about 25%. (The actual amount is 26.8%.)

594

Chapter 17

Temperature, Thermal Expansion, and the Ideal Gas Law

(b) The absolute (Kelvin) scale must be used in the ideal gas law. If the Celsius scale is used, it appears that the temperature has doubled. However, 0°C is an arbitrarily chosen point on the temperature scale and cannot be used to determine temperature ratios. When the temperatures are converted to kelvins (293 K and 313 K) it can be seen that their ratio is about 1.07.

(a) Students frequently do not understand that gauge pressure and temperature in Celsius are a comparison of pressure and temperature to an arbitrary zero point. In the ideal gas law the temperature, pressure, and volume must be measured relative to the true zero points; absolute zero pressure, absolute zero temperature, and zero volume. The volume units are not critical, since changing them only affects the value of the ideal gas constant. Gauge pressure and Celsius temperature have additive terms that change the functional form of the ideal gas law.

(e) A common misconception is that if the temperature increases both the pressure and volume will increase. However, by the ideal gas law, when the temperature increases the product of the pressure and volume must increase. This increase can occur by increasing the pressure, increasing the volume, or both.

10. (c, d) When you blow up a balloon, you are putting a significant amount of air into the balloon, so you definitely increase the number of molecules that comprise air, and the balloon gets significantly larger as you blow it up. The pressure may increase slightly, but not very much, because the outside air pressure doesn’t change. Problem 88 indicates that a balloon may maintain the inside pressure about 5% higher than the outside pressure. 11. (b) The air in airplane cabin is usually lower pressure than normal atmospheric pressure. The temperature in the cabin is very similar to normal indoor temperatures. So when the cabin pressure decreases at constant temperature, the volume of the gas inside the bag increases, according to the ideal gas equation. 12. (c) The pressure gauge reads “gauge” pressure, so the absolute pressure is 2.0 atm for tire A and 3.0 atm for tire B. Since the volumes and temperatures of both tires are the same, the amount of gas in each tire is proportional to the absolute pressure of the gas in the tire. Thus tire B contains 1.5 times as much air as tire A.

Solutions to Problems In solving these problems, the authors did not always follow the rules of significant figures rigidly. We tended to take quoted temperatures as correct to the number of digits shown, especially where other values might indicate that. 1.

The number of atoms is found by dividing the mass of the substance by the mass of a single atom. Take the atomic mass of copper from the periodic table. 3.4  10−3 kg N Cu = = 3.2  1022 atoms of Cu −27 ( 63.546 u atom ) (1.66  10 kg u )

The number of atoms in a pure substance can be found by dividing the mass of the substance by the mass of a single atom. Take the atomic masses of gold and silver from the periodic table.

595

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

N Au N Ag

3.15  10−2 kg (196.966570 u atom ) 1.66  10−27 kg u

(

−2

3.15  10 kg

Instructor Solutions Manual

) = 107.8682 = 0.5476 → N = 0.548N

(107.8682 u atom ) (1.66  10−27 kg u )

196.966570

Because a gold atom is heavier than a silver atom, there are fewer gold atoms in the given mass. 3.

(a) T ( C ) = 95 T ( F ) − 32 = 95 68 − 32 = 20C (b) T ( F ) = 95 T ( C ) + 32 = 95 (1900 ) + 32 = 3452F  3500F

High:

T ( C ) = 95 T ( F ) − 32 = 95 136 − 32 = 57.8C

Low:

T ( C ) = 95 T ( F ) − 32 = 95  −129 − 32 = −89.4C

T ( F ) = 95 T ( C ) + 32 = 95 ( 38.5C ) + 32 = 101.3F

(a) T ( F ) = 95 T ( C ) + 32 = 95 ( −22 ) + 32 = −7.6F  −8F (b) T

( C ) = T ( F) − 32 =  −22 − 32 = −30C ( two sig. figs.) o

5 9

Assume that the temperature and the length are linearly related. The change in temperature per unit length change is as follows. 100.0C − 0.0C T = = 9.823C cm 22.79 cm − 12.61cm l Then the temperature corresponding to length l is T ( l ) = 0.0C + ( l − 12.61cm )( 9.823C cm ) . (a) T (18.50 cm ) = 0.0C + (18.50 cm − 12.61 cm )( 9.823C cm ) = 57.9C (b) T (14.40 cm ) = 0.0C + (14.40cm − 12.61cm )( 9.823C cm ) = 17.6C

When the concrete cools in the winter, it will contract, and there will be no danger of buckling. Thus, the low temperature in the winter is not a factor in the design of the highway. But when the concrete warms in the summer, it will expand. A crack must be left between the slabs equal to the increase in length of the concrete as it heats from 15oC to 50oC.

(

)

l =  l 0 T = 12  10−6 C (14 m )( 50C − 15C ) = 5.9  10−3 m 9.

Take the 300 m height to be the height in January. Then the increase in the height of the tower as the temperature rises is given by Eq. 17–1a. l =  l 0 T = 12  10−6 C ( 300 m )( 25C − 2C ) = 0.0828 m  0.08 m

(

)

10. The increase in length of the table is given by Eq. 17–1a.

(

)

l =  l 0 T = 0.20 10−6 C (1.8 m )( 7.5C ) = 2.7 10−6 m

(

)

For steel, l =  l 0 T = 12 10−6 C (1.8 m )( 7.5C ) = 1.6 10 −4 m . The change for Super Invar is only

1 60

of the change for steel.

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11. The density at 4oC is  =

1.00  103 kg

. When the water is warmed, the mass will stay the V 1.00 m 3 same, but the volume will increase according to Eq. 17–2.

(

)(

)

V = V0 T = 210  10−6 C 1.00 m 3 ( 94C − 4C ) = 1.89  10 −2 m 3 The density at the higher temperature is  =

M V

1.00  103 kg −2

1.00 m + 1.89  10 m 3

= 981kg m 3

12. The increase in length of the rod is given by Eq. 17–1a. 0.015 l l l =  l 0 T → T = → T = T0 + = 22C + = 811.5C  810C 19  10 −6 C l 0 l 0 13. The rivet must be cooled so that its diameter becomes the same as the diameter of the hole. l =  l 0 T → l − l 0 =  l 0 ( T − T0 )

T = T0 +

l −l0

l 0

= 24C +

1.870 cm − 1.872 cm

(12 10 C) (1.872 cm ) −6

= −65.03C  −65C

The temperature of “dry ice” is about −80C, so this process will be successful. 14. We assume that all of the expansion of the water is in the thickness of the mixed layer. We also assume that the volume of the water can be modeled as a shell of constant radius, equal to the radius of the earth, and so the volume of the shell is the surface area of the shell times its (very small) thickness, V = 4 RE2 d . Use Eq. 17–2.

V = 4 RE2 d → V = 4 RE2 d ; V = V T =  4 RE2 d T 4 RE2 d =  4 RE2 d T →

(

)

d =  d T = 210  10−6 C ( 50 m )( 0.5C ) = 0.00525 m  5 mm 15. The change in volume of the aluminum is given by the volume expansion formula, Eq. 17–2. The percent change is found by taking the change, dividing by the original volume, and then multiplying by 100.  V T V (100) = 0 (100) = T (100 ) = ( 75  10−6 Co ) (140C − 30C ) (100 ) = 0.825 = 0.83%

16. (a) Assume that each dimension of the plate changes according to Eq. 17–1a. A = A − A0 = ( l + l )( w + w ) − l w = l w + l w + wl + l w − l w

= l w + wl + l w Neglect the very small quantity l w.

A = l w + wl = l ( wT ) + w ( l T ) = 2 l wT

597

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) Again assume that each dimension of the plate changes according to Eq. 17–1a. V = V − V0 = ( l + l )( w + w )( t + t ) − l wt

= ( l w + l w + wl + l w )( t + t ) − l wt = ( l wt + l t w + wt l + t l w ) + ( l wt + l wt + wl t + l wt ) − l wt = ( l t w + wt l + l wt ) + ( t l w + l wt + wl t + l wt ) Neglect the very small quantities that involve products of “” terms. V  l t wT + wt l T + l w t T = 3T ( l wt ) = 3V T Eq. 17–2 says V =  V t , and so we see that   3 . 17. The amount of water that can be added to the container is the final volume of the container minus the final volume of the water. Also note that the original volumes of the water and the container are the same. We assume that the density of water is constant over the temperature change involved.

( ) = ( 27  10 C − 210  10 C ) ( 450.0 mL ) ( −80.0 C ) = 6.59 mL

Vadded = (V0 + V )container − (V0 + V )H O = Vcontainer − VH O =  container −  H O V0 T 2

−6

18. Since the coefficient of volume expansion is much larger for the coolant than for the aluminum, the coolant will expand more than the aluminum, and so coolant will overflow the cooling system. Use Eq. 17–2. Voverflow = Vcoolant − Valuminum =  coolantVcoolant T −  aluminumValuminum T = (  coolantVcoolant −  aluminumValuminum ) T = (  coolant −  aluminum ) Voriginal T

(

) (

)

=  410  10−6 C − 35  10 −6 C  (14.0 L )(12C ) = 0.063 L  63mL

19. (a) The amount of water lost is the final volume of the water minus the final volume of the container. Also note that the original volumes of the water and the container are the same. Vlost = (V0 + V ) H O − (V0 + V )container = VH O − Vcontainer =  H OV0 T −  containerV0T 2

 container =  H O − 2

Vlost V0 T

= 210  10−6 C −

 1mL    0.98324 g 

( 0.35g ) 

( 55.50 mL )( 60C − 20C )

= 5.0  10−5 C

(b) From Table 17–1, the most likely material is copper . 20. The sum of the original diameter plus the expansion must be the same for both the plug and the ring. ( l 0 + l )iron = ( l 0 + l )brass → l iron +  iron l iron T = l brass +  brass l brass T

T =

l brass − l iron

 iron l iron −  brass l brass

8.756 cm − 8.742 cm

(12 10 C ) (8.742 cm ) − (19  10 C ) (8.756 cm ) −6

−6

= −228C = Tfinal − Tinitial = Tfinal − 15C → Tfinal = −213C  −210C

598

Chapter 17

Temperature, Thermal Expansion, and the Ideal Gas Law

21. We model the vessel as having a constant cross-sectional area A. Then a volume V0 of fluid will occupy a length l 0 of the tube, given that V0 = Al 0. Likewise V = Al .

V = V − V0 = Al − Al 0 = Al and V = V0 T =  Al 0 T . Equate the two expressions for V , and get Al =  Al 0 T → l =  l 0 T . But l =  l 0 T , so we see that under the conditions of the problem,  =  . 22. (a) When a substance changes temperature, its volume will change by an amount given by Eq. 17–2. This causes the density to change. M M M M M M M 1   =  f −  = − = − = − = − 1  V V0 V0 + V V0 V0 + V0 T V0 V0  1 +  T 



 −  T  =     1 +  T 1 +  T   1 +  T 

= 

−

1 +  T 

1 , then the denominator is approximately 1, so  = −   T .

If we assume that  T

(b) The fractional change in density is  −   T = = −  T = − ( 87  10−6 C ) ( −55C − 35C ) = 7.83  10−3   This is a 0.78% increase . 23. As the wine contracts or expands, its volume changes. We assume that the volume change can only occur by a corresponding change in the headspace. Note that if the volume increases, the headspace decreases, so their changes are of opposite signs. Use Eq. 17–2. (a) The temperature decreases, so the headspace should increase. V =  V0 T = − r 2 H →

 V0 T H = − =−  r2

(

 10−3 m 3   ( −10 C )  1L  = 0.0117 m

)

420  10−6 C ( 0.750 L ) 

 ( 0.00925 m )

H = 1.5cm + 1.17 cm = 2.67 cm  2.7 cm

(b) The temperature increases, so the headspace should decrease.  10−3 m3  420  10−6 C ) ( 0.750 L )  (10 C ) ( 1L   V0 T  H = − =− = −0.0117 m 2  r2  ( 0.00925 m ) H = 1.5cm − 1.17 cm = 0.33cm  0.3cm

24. (a) The original surface area of the sphere is given by A = 4 r 2. The radius will expand with temperature according to Eq. 17–1b, rnew = r (1 + T ) . The final surface area is 2 Anew = 4 rnew , and so the change in area is A = Anew − A.

599

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

A = Anew − A = 4 r 2 (1 + T ) − 4 r 2 = 4 r 2 (1 + T ) − 1 2

= 4 r 2 1 + 2T +  2 ( T ) − 1 = 4 r 2 ( 2T ) 1 + 12 T  2





2 If the temperature change is not large, T 1 , and so A = 8 r T (b) Evaluate the above expression for the solid iron sphere. 1 2

(

A = 8 r 2T = 8 60.0  10−2 m

) (12  10 C) ( 225C − 15C) = 2.28  10 m 2

−6

−2

25. Notation: we will use T for the period, and “Temp” for the temperature. The pendulum has a period of T0 = 2 l 0 g at 17C, and a period of T = 2 l g at 26C . Notice that T  T0 since l  l 0 . With every swing of the pendulum, the heated clock will indicate that a time T0 has passed, but the actual amount of time that has passed is T . Thus the heated clock is “losing time” by an amount of T T = T − T0 every swing. The fractional loss is given by , and is found as follows. T0 T T0

T − T0 T0

2 l g − 2 l 0 g 2 l 0 g

l − l0 l0

(

l 0 + l − l 0

l 0 +  l 0 ( Temp ) − l 0

)

= 1 +  ( Temp ) − 1 = 1 + 19  10 −6 C ( 9 C ) − 1 = 8.55  10 −5

(

Thus the amount of time lost in any time period  0 is T = 8.55  10 the following. T = 8.55  10−5

(

−5

) T . For one year, we have 0

)( 3.156  10 s ) = 2698s  45 min lost 7

26. From Problem 22, we have the relationship  = −   T . In the limit of very small temperature changes, this becomes a derivative expression.  d  = −   T → = −  → = −  T dT d = 0, which means The density will have its maximum value at whatever temperature makes dT  = 0.

 = a + bT + cT 2 = 0 → T= =

−b  b 2 − 4ac 2c

−1.70  10−5 

(

−5

−7

−1.70  10−5  1.53962  10−5 2 −2.02  10−7

(1.70 10 ) − 4 ( −6.43 10 )( −2.02 10 ) 2 ( −2.02  10 )

)

= 3.98C or 80.19C

We see that  = 0 at a temperature of about 4.0C . The other temperature, 80.19C, is outside of the range of validity of the coefficient of volume expansion.

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Chapter 17

Temperature, Thermal Expansion, and the Ideal Gas Law

27. The thermal stress must compensate for the thermal expansion. E is Young’s modulus for the aluminum.

(

)(

)

Stress = F A =  E T = 25  10−6 C 70  109 N m 2 ( 38C − 12C ) = 4.6  107 N m 2 28. (a) Since the beam cannot shrink while cooling, the tensile stress must compensate in order to keep the length constant.

(

)(

)

Stress = F A =  ET = 12  10−6 C 200  109 N m2 ( 50 C ) = 1.2 108 N m 2 (b) The ultimate tensile strength of steel (from Table 12–2) is 5  10 N m , and so 8

the ultimate strength is not exceeded . There would only be a safety factor of about 4.2.

(

)(

)

Stress = F A =  ET = 12  10−6 C 20  109 N m2 ( 50 C ) = 1.2 107 N m 2 The ultimate tensile strength of concrete is 2  10 N m , and so the concrete will fracture . 6

29. (a) Calculate the change in temperature needed to increase the diameter of the iron band so that it fits over the barrel. Assume that the barrel does not change in dimensions. l =  l 0 T → l − l 0 =  l 0 (T − T0 )

T = T0 +

l −l0

l 0

= 20C +

134.122 cm − 134.110 cm

(12 10 C) (134.110 cm ) −6

= 27.457C  27C

(b) Since the band cannot shrink while cooling, the thermal stress must compensate in order to keep the length at a constant 132.122 cm. E is Young’s modulus for the material. l = AET Stress = F A =  E T → F = AE l0

(

)(

)

= 9.4  10−2 m 6.5  10 −3 m 100  109 N m 2 12  10 −6 C ( 7.457 C ) = 5500 N

30. Use the relationships T ( K ) = T ( C ) + 273.15 and T ( K ) = 95 T ( F ) − 32 + 273.15. (a) T ( K ) = T ( C ) + 273.15 = 58 + 273.15 = 331K (b) T ( K ) = 95 T ( F ) − 32 + 273.15 = 95 86 − 32 + 273.15 = 303 K (c) T ( K ) = T ( C ) + 273.15 = −55 + 273.15 = 218 K (d) T ( K ) = T ( C ) + 273.15 = 5100 + 273.15 = 5373.15 K  5400 K 31. Use the relationship that T ( K ) = 95 T ( F ) − 32 + 273.15.

T ( K ) = 95 T ( F ) − 32 + 273.15 →

T ( F ) = 95 T ( K ) − 273.15 + 32 = 95 0 − 273.15 + 32 = −459.67F 32. The temperature is 10C. First, we convert that temperature to Kelvin and Fahrenheit temperatures. T ( K ) = T ( C ) + 273.15 K = 283 K

T (  F ) = 95 T ( C ) + 32F = 50 F © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

601

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If the temperature is doubled, it would be 20C , or 566 K, or 100F. We convert the “doubled” Celsius and Kelvin temperatures to Fahrenheit. T (  F ) = 95 T ( C ) + 32 F = 95 T ( 20C ) + 32 F = 68 F

T (  F ) = 95 T ( K − 273 K ) + 32F = 95 T ( 566 K − 273 K ) + 32F = 559F The doubling of the Celsius temperature is the most believable, indicating a change from 50 F to 68 F . However, only the Kelvin scale has a true “zero,” and so it is the only valid scale for an actual doubling. 33. Use the relationship that T ( K ) = T ( C ) + 273.15. (a) T ( K ) = T ( C ) + 273.15 = ( 2000 + 273.15) K = 4273 K  4300 K

(

)

T ( K ) = T ( C ) + 273.15 = 15  106 + 273.15 K  15  106 K (b) % error =

4000C:

T T (K)

273.15 4273

273.15

 100 =

T (K)

 100 15  106 C:

 100  6.4%

273.15 15  106

 100  1.8  10 −3%

34. Assume the air is an ideal gas. Since the amount of air is constant, the value of PV T is constant. PV 1 1 T1

PV 2 2

→ T2 = T1

 40 atm  1  = 1302 K = 1029C  1000C    1 atm  9 

P2 V2

= ( 293 K ) 

P1 V1

35. Assume the gas is ideal. Since the amount of gas is constant, the value of PV T is constant.

PV 1 1 T1

PV 2 2 T2

→ V2 = V1

P1 T2 P2 T1

(

( )  1.00atm  2.80atm

= 3.50 m3 





273 + 38.0 ) K 273 K

= 1.42 m 3

36. (a) Assume that the helium is an ideal gas, and then use the ideal gas law to calculate the volume. Absolute pressure must be used, even though gauge pressure is given. nRT (12.50 mol )( 8.314 J mol K )( 283.15 K ) PV = nRT → V = = = 0.2152 m 3 P (1.350 atm ) 1.013  105 Pa atm

(

)

(b) Since the amount of gas is not changed, the value of PV T is constant. PV 1 1 T1

PV 2 2 T2

→ T2 = T1

P2 V2 P1 V1

 2.00 atm  1  = 209.7 K = −63.4 o C    1.350 atm  2 

= ( 283.15 K ) 

37. Assume the nitrogen is an ideal gas. From Example 17–10, the volume of one mole of nitrogen gas at STP is 22.4  10−3 m3. The mass of one mole of nitrogen, with a molecular mass of 28.0 u, is 28.0 grams. Use these values to calculate the density of the nitrogen gas. M 28.0  10−3 kg = = = 1.25 kg m 3 3 −3 V 22.4  10 m We could also calculate this without using the standard molar volume. ( 28.0  10−3 kg )(1.013  105 Pa ) = 1.25 kg m3 M M MP = = = = nRT nRT J  V (1.0 mol )  8.314  ( 273 K ) P mol K   © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

602

Chapter 17

Temperature, Thermal Expansion, and the Ideal Gas Law

38. Assume that the nitrogen (subscript 1) and carbon dioxide (subscript 2) are ideal gases, and that the volume and temperature are constant for the two gases. From the ideal gas law, the value of P RT is constant. Also note that concerning the ideal gas law, the identity of the gas is = n V unimportant, as long as the number of moles is considered. 21.6 kg CO 2   −3  P1 P2 n 44.01  10 kg CO 2 mol  = → P2 = P1 2 = ( 3.45atm )   = 2.20 atm 21.6 kg N 2 n1 n2 n1  

 28.01  10−3 kg N mol    2

39. (a) Assume the nitrogen is an ideal gas. The number of moles of nitrogen is found from the atomic mass, and then the ideal gas law is used to calculate the volume of the gas. 1 mole N 2 = 946.1mol n = ( 26.5 kg ) 28.01  10 −3 kg

PV = nRT → V =

nRT

( 946.1mol )(8.314 J mol K )( 273 K )

 21.2 m 3

P 1.013  10 Pa (b) Hold the volume and temperature constant, and again use the ideal gas law. 1 mole N 2 n = ( 26.5 kg + 32.2 kg ) = 2096 mol 28.01  10 −3 kg 5

PV = nRT → P=

nRT V

( 2096 mol )(8.314 J mol K )( 273 K ) 21.2 m

= 2.24  105 Pa = 2.22 atm

40. We assume that the mass of air is unchanged, and the volume of air is unchanged (since the tank is rigid). Use the ideal gas law. Note that the gauge pressure is given, but we must use absolute pressure in the ideal gas law. P1 P2 P  192 atm  = → T2 = T1 2 = ( 273 + 29 ) K    = 286 K = 13C T1 T2 P1  203atm  41. Assume the argon is an ideal gas. The number of moles of argon is found from the atomic weight, and then the ideal gas law is used to find the pressure. 1 mole Ar n = (105.0 kg ) = 2628 mol 39.95  10 −3 kg

PV = nRT → P =

nRT V

( 2628 mol )( 8.314 J mol K )( 273.15 + 21.6 ) K = 1.57  108 Pa 3 −3 ( 41.0 L ) (1.00  10 m L )

This is 1550 atm. 42. Assume that the oxygen (subscript 1) and helium (subscript 2) are ideal gases, and that the volume P RT and temperature are constant for the two gases. From the ideal gas law, the value of = is n V constant. Also note that concerning the ideal gas law, the identity of the gas is unimportant, as long as the number of moles is considered. Finally, gauge pressure must be changed to absolute pressure.

603

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

P1 n1

P2 n2

→ n2 = n1

P2 P1

Instructor Solutions Manual

 1 mole O 2  ( 9.00 atm ) = 9.171  102 moles  −3  32  10 kg  ( 9.20 atm )

= ( 30.0 kg O 2 ) 

 10 kg  ( 9.171  10 moles )  4.0  = 3.67 kg He 1 mole He −3





43. We assume that the gas is ideal, that the amount of gas is constant, and that the volume of the gas is constant. P1 P2 P  2.00atm  = → T2 = T1 2 = ( 273.15 + 30.0) K    = 606.3K = 333.15C  333C T1 T2 P1  1.00atm  44. Assume that the air is an ideal gas, and that the volume of the tire remains constant. We consider two states: the original state 1, with temperature T1 , pressure P1 , and amount of gas n1 ; and the final state 2, with temperature T2 , pressure P1 , and amount of gas n2 . The ideal gas law is used. = n1RT1 ; PV = n2 RT2 → PV 1 1

n2 n1

T1 T2

( 273 + 15) K = 0.926 ( 273 + 38) K

Thus 1 − 0.926 = 0.074 = 7.4 % must be removed. 45. Assume the oxygen is an ideal gas. Since the amount of gas is constant, the value of PV T is constant. PV PV V T  59.2 L  ( 273.15 + 56.0) K = 4.23atm 1 1 = 2 2 → P2 = P1 1 2 = ( 2.45 atm )   T1 T2 V2 T1  38.8 L  ( 273.15 + 18.0) K 46. Assume the helium is an ideal gas. Since the amount of gas is constant, the value of PV T is constant. We assume that since the outside air pressure decreases to 64% of its original value, the air pressure inside the balloon will also decrease to 64% of its original value. PV PV 1 1 = 2 2 → T1 T2

V2 V1

P1 T2 P2 T1

 1.0  ( 273 + 5.0 ) K = 1.4825  1.5 times the original volume   0.64  ( 273 + 20.0 ) K

=

47. Since the container can withstand a pressure difference of 0.50 atm, we find the temperature for which the inside pressure has dropped from 1.0 atm to 0.50 atm. We assume the mass of contained gas and the volume of the container are constant. P1 P2 P  0.50atm  = → T2 = T1 2 = ( 273.15 + 18) K    = 146.6 K  −130C T1 T2 P1  1.0atm  48. The buoyant force is equal to the weight of the displaced air. The displaced air is proportional to the volume of the balloon. So we can find the percentage change in the volume, and that will be the percentage change in the buoyant force. The amount of gas in the balloon doesn’t change, so the value of PV T doesn’t change. The pressure doesn’t change, as stated in the problem.

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PV 1 1 T1

PV 2 2

→

FB2 − FB1 FB1

V2 V1

FB2 FB1

V2 V1

T2 T1

( 273 + 25) K = 1.0347 ( 273 + 15) K

− 1 = 0.0347  3% increase

49. (a) We assume the Martian atmosphere behaves like an ideal gas. PV = nRT →

V =

nRT P

(1.0 mol )( 8.314 J mol K )( 273.15 − 50 ) K 600 Pa

= 3.09 m 3  3m 3 or 3000 L

(b) We assume the atmosphere is totally CO 2 . The density is the mass per volume. The molecular mass of CO 2 is (12+16+16) = 44 g/mol.

=

(1.0 mol )( 44 g mol )( 0.001kg g ) 3

= 1.424  10−2 kg m3  0.01kg m3

V 3.09 m (c) We compare to Earth’s value.  Mars 1.424  10−2 kg m3 = = 1.148  10−2  0.01 or 1 100 3  Earth 1.24 kg m

50. The pressure inside the bag will change to the surrounding air pressure as the volume of the bag changes. We assume the amount of gas and temperature of the gas are constant. Use the ideal gas equation. P1  1.0 atm  PV = V1  1 1 = PV 2 2 → V2 = V1  = 1.3V1 P2  0.75atm  Thus the bag has expanded by 30%. 51. We assume that all of the gas in this problem is at the same temperature. Use the ideal gas equation, but we take the two values of pressure as gauge pressure values. First, find the initial number of moles in the tank at (34+1) atm, n1 . Then find the final number of moles in the tank at (204+1) atm, n2 . The difference in those two values is the number of moles needed to add to the tank. PV ( 35atm )(12 L ) PV = n1 RT → n1 = 1 1 = 1 1 RT RT PV ( 205atm )(12 L ) PV = n2 RT → n2 = 2 2 = 2 2 RT RT (170 atm )(12 L )

nadd = n2 − n1 =

RT Use this number of moles at atmospheric pressure to find the volume of air needed to add, and then find the time needed to add it.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

PaddVadd = nadd RT → Vadd =

nadd RT Padd

(170 atm )(12 L ) RT RT 1atm

Instructor Solutions Manual

= 2040 L

 1min   = 7.0 min  290 L 

2040 L 

52. We calculate the density of water vapor, with a molecular mass of 18.0 grams per mole, from the ideal gas law. n P PV = nRT → = → V RT

=

m V

Mn V

MP RT

( 0.0180 kg mol ) (1.013  105 Pa ) = 0.588 m 3 ( 8.314 J mol K )( 373 K )

The density of water vapor (steam) from Table 13–1 is 0.598 m 3. Because this gas is very “near” a phase change state (water can also exist as a liquid at this temperature and pressure), we would not expect it to act like an ideal gas. It is reasonable to expect that the molecules will have other interactions besides purely elastic collisions. That is evidenced by the fact that steam can form droplets, indicating an attractive force between the molecules. 53. We ignore the weight of the stopper. Initially there is a net force (due to air pressure) on the stopper of 0, because the pressure is the same both above and below the stopper. With the increase in temperature, the pressure inside the tube will increase, and so there will be a net upward force given by Fnet = ( Pin − Pout ) A, where A is the cross-sectional area of the stopper. The inside pressure can be expressed in terms of the inside temperature by means of the ideal gas law for a constant volume and constant mass of gas. PinVtube PV T = 0 tube → Pin = P0 in Tin T0 T0



Fnet = ( Pin − Pout ) A =  P0



 Fnet

Tin = 

 P0 A



 T  − P0  A = P0  in − 1 A → T0   T0 

Tin



 (10.0 N ) + 1  ( 273 + 18 ) K = 454 K = 181C 2 5  (1.013  10 Pa )  ( 0.00750 m ) 

+ 1  T0 = 



= nRT1 (original conditions) and PV = nRT2 (final 54. From the ideal gas equation, we have PV 1 1 2 2 conditions), since the amount of gas is constant. Use these relationships along with the given conditions to find the original pressure and temperature. = nRT1 ; PV = nRT2 → PV − PV = nRT1 − nRT2 = nR (T1 − T2 ) → PV 1 1 2 2 1 1 2 2

PV − ( P1 + 450 Pa ) V2 = nR ( T1 − T2 ) → P1 (V1 − V2 ) − V2 ( 450 Pa ) = nR ( 9.0 K ) → 1 1 P1 =

nR ( 9.0 K ) + V2 ( 450 Pa ) V1 − V2

( 3.5 mol )(8.314 J mol K )( 9.0 K ) + ( 0.018 m3 ) ( 450 Pa ) 0.002 m 3

= 1.35  105 Pa  1  105 Pa

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T1 =

PV 1 1 nR

(1.35  10 Pa )( 0.020 m ) = 93 K  90 K  −180C 5

( 3.5 mol )(8.314 J mol K )

55. The ideal gas law can be used to relate the volume at the surface to the submerged volume of the bubble. We assume the amount of gas in the bubble doesn’t change as it rises. The pressure at the submerged location is found from Eq. 13–6b. PsubmergedVsubmerged PV P V PV = nRT → = nR = constant → surface surface = → T Tsurface Tsubmerged Vsurface = Vsubmerged

Psubmerged Tsurface Psurface Tsubmerged

= Vsubmerged

Patm +  gh Tsurface Patm

Tsubmerged

1.013  105 Pa + (1.00  103 kg m 3 )( 9.80 m s 2 ) ( 32.0 m )  ( 273.15 + 18.5) K = (1.00 cm ) 5 3

(1.013  10 Pa )

( 273.15 + 5.5) K

= 4.29 cm 3

56. From Example 17.10, 1 mole of ideal gas occupies 22.4 L at STP. 1 mole  6.02  10 23 molecules   1 L  25 3    10−3 m3  = 2.69  10 molecules m 22.4 L  mole   57. We assume that the water is close enough to 4oC so that its density is 1000 kg m . 3

 1 mol  10−3m 3   1000 kg      = 55.5 mol  −3 3  1 L   1 m   (15.9994 + 2  1.00794 )  10 kg 

1.00 L 

 6.022  1023 molecules  = 3.34  1025 molecules  1 mol  

55.5 mol 

58. (a) Since the average depth of the oceans is very small compared to the radius of the Earth, the 2 ocean’s volume can be calculated as that of a spherical shell with surface area 4 REarth and a thickness y. Then use the density of sea water to find the mass, and the molecular weight of water to find the number of moles.

(

)

(

2 y = 0.75 ( 4 ) 6.38  106 m Volume = 0.75 4 REarth

) ( 3 10 m ) = 1.15 10 m 2

 1025 kg   1 mol  22 22   = 6.55  10 moles  7  10 moles 3 −3 m 18  10 kg   

1.15  1018 m3 

(

)

(b) 6.55  1022 moles 6.02  1023 molecules 1 mol  4  1046 molecules 59. We use Eq. 17–4. PV = NkT → P =

N V

 1molecule   1003 cm 3   −23 J  −17  1.38  10  ( 3 K ) = 4  10 Pa    3 3 K  1cm   m 

kT = 

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

60. We assume an ideal gas at STP. Example 17–10 shows that the molar volume of this gas is 22.4 L. We calculate the actual volume of one mole of gas particles, assuming a volume of l 30 , and then find the ratio of the actual volume of the particles to the volume of the gas.

Vmolecules

( 6.02  10 molecules ) (( 3.0  10 m ) molecule) = = 7.3  10 ( 22.4 L ) (1  10 m 1L ) −10

−3

Vgas

−4

The molecules take up less than 0.1% of the volume of the gas. 61. Assume the gas is ideal at those low pressures, and use the ideal gas law. −6 3 N P 1  10−12 N m 2  3  108 molecules   10 m  PV = NkT → = = =   3  V kT 1.38  10−23 J K ( 273 K )  m3   1 cm 

(

)

= 300 molecules cm 3

62. Use Eq. 17–5 for the constant-volume gas thermometer to relate the boiling point to the triple point. P Tbp P ( 273.15 + 100) K → bp = = = 1.3660 T = ( 273.16 K ) Ptp Ptp 273.16 K 273.16 K 63. (a) For the constant-volume gas thermometer, we use Eq. 17–5. P 273.16 K 273.16 K T = ( 273.16 K ) → Ptp = P = (187 torr ) = 71.2 torr Ptp T ( 273.15 + 444.6) K (b) We again use Eq. 17–5. P 128 torr T = ( 273.16 K ) = ( 273.16 K ) = 491K = 218C Ptp 71.2 torr 64. From Fig. 17–18, we estimate a temperature of 373.35 K from the oxygen curve at a pressure of 268 torr. The boiling point of water is 373.15 K. (a) The inaccuracy is T = 373.35 K − 373.15 K = 0.20 K (b) As a percentage, we have the following. 0.20 K T (100 ) = (100 ) = 0.054% T 373.15 K The answers may vary due to differences in reading the graph. 65. Since the volume is constant, the temperature of the gas is proportional to the pressure of the gas, and so T P is constant for each separate amount of gas. First we calculate the two temperatures of the different amounts of gas. T1 P 273.16 K 218 = → T1 = ( 273.16 K ) 1 melt = ( 273.16 K ) = 208.21K P1 melt P1 tp P1 tp 286

T2 P2 melt

273.16 K P1 tp

P 128 → T2 = ( 273.16 K ) 2 melt = ( 273.16 K ) = 214.51K P2 tp 163

We now assume that there is a linear relationship between the melting-point temperature and the triple-point pressure, as shown in Fig. 17–18. The two points ( Ptp , T ) that define the linear relationship are (163torr, 214.51K ) and ( 286 torr, 208.21K ) . The actual melting point is the © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Temperature, Thermal Expansion, and the Ideal Gas Law

y-intercept of that linear relationship. We use Excel and a best-fit straight line to find that y-intercept. The graph is shown below.

We see the melting temperature is 222.85 K  223 K . 66. Since we assume that the glass does not expand, the measuring cup will contain 375 mL of hot water. Find the volume of water after it cools. V = V0  T = ( 375 mL ) ( 210  10−6 C ) ( 20C − 95C ) = −5.91mL  −5.9 mL The volume of cool water is 5.9 mL less than the desired volume of 375 mL. 67. (a) At 37oC, the tape will expand from its calibrated length, and so will read low . l = T = 12  10 −6 C ( 37C − 14C ) = 2.76  10 −4  2.8  10 −2 % (b) l0

(

)

68. The net force on each side of the box will be the pressure difference between the inside and outside of the box, times the area of a side of the box. The outside pressure is 1 atmosphere. The ideal gas law is used to find the pressure inside the box, assuming that the mass of gas and the volume are constant. P P T P nR ( 273 + 165) K = = constant → 2 = 1 → P2 = P1 2 = (1.00 atm ) = 1.521 atm T V T2 T1 T1 ( 273 + 15) K The area of a side of the box is given by the following. 2

(

1/3 Area = l 2 = ( Volume of box )  = 8.25  10−2 m2

) = 1.895  10 m 2/3

−1

The net force on a side of the box is the pressure difference times the area.

(

)(

)

F = (  Pressure )( Area ) = ( 0.521atm ) 1.013  105 Pa 1.895  10−1 m2 = 1.0  104 N There are only two significant figures in the pressure difference. 69. Assume the helium is an ideal gas. The volume of the cylinder is constant, and we assume that the temperature of the gas is also constant in the cylinder. From the ideal gas law, PV = nRT , under these conditions the amount of gas is proportional to the absolute pressure. 5atm + 1atm 6 P RT P P n2 P2 PV = nRT → = = constant → 1 = 2 → = = = n V n1 n2 n1 P1 28atm + 1atm 29 Thus 6 29 = 0.207  21% of the original gas remains in the cylinder.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

70. When the rod has a length l, then a small (differential) change in temperature will cause a small (differential) change in length according to Eq. 17–1a, expressed as d l =  l dT . dl

(a) d l =  l dT →

=  dT →  l1

dl l

=   dT → ln T1

=  ( T2 − T1 ) → l 2 = l 1e (

 T2 −T1 )

(b)

l l1

(c)

=   dT → ln T1

d l =  l dT → ln

l2 l1

dl l

l2 l1

=   dT →

l 2 = l 1e

  dT

 l =   dT =  ( + bT ) dT →

=  dT →

(

) → l =l e (

)

 0 T2 −T1 ) + 12 b T22 −T12   

=  0 ( T2 − T1 ) + 12 b T22 − T12

71. Assume that the air in the lungs is an ideal gas, that the amount of gas is constant, and that the temperature is constant. The ideal gas law then says that the value of PV is constant. The pressure a distance h below the surface of a fluid is given by Eq. 13–6b, P = P0 +  gh, where P0 is atmospheric pressure and  is the density of the fluid. P P +  gh ( PV )surface = ( PV )submerged → Vsurface = Vsubmerged submerged = Vsubmerged atm Psurface Patm

= ( 4.8 L )

(

1.01  105 Pa + 1025 kg m 3

)( 9.80 m s ) (9.0 m ) = 9.1L 2

1.01  105 Pa

This is obviously very dangerous, to have the lungs attempt to inflate to almost twice their volume. Thus it is not advisable to quickly rise to the surface.

= nRT0 and PV = nRT . 72. (a) Assume the pressure and amount of gas are held constant, and so PV 0 0 0 From these two expressions calculate the change in volume and relate it to the change in temperature. V nRT nRT0 nR V = V0 + V → V = V − V0 = − = (T − T0 ) = 0 T P0 P0 P0 T0 But V =  V0 T , and so V = For T0 = 293 K,  =

1 T0

1 293 K

V0 T0

T =  V0 T →  =

1 T0

= 3.4  10 −3 K , which agrees well with Table 17–1.

= nRT0 = PV . From (b) Assume the temperature and amount of gas are held constant, and so PV 0 0 these two expressions calculate change in volume and relate it to the change in pressure. V = V0 + V → V = V − V0 =

nRT0 P

−

nRT0 P0

But from Eq. 12–7, V = −V0

1 B

1

= nRT0 

P

−

1

nRT0  P0 − P  1 =   = V0 ( −P )  P0  P0  P  P

P and so V = V0

1 P

( −P ) = −V0

1 B

P →

B=P

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Chapter 17

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73. To do this problem, the “molecular weight” of air is needed. If we approximate air as 70% N2 (molecular weight 28) and 30% O2 (molecular weight 32), then the average molecular weight is 0.70 ( 28 ) + 0.30 ( 32 ) = 29. (a) Treat the air as an ideal gas. Assume that the pressure is 1.00 atm. 1.013  105 Pa 1200 m3 PV PV = nRT → n = = = 5.076  104 moles RT ( 8.315 J mol k )( 288 K )

(

)(

)

m = 5.076  104 moles 29  10−3 kg mol = 1472 kg  1500 kg (b) Find the mass of air at the lower temperature, and then subtract the mass at the higher temperature. 1.013  105 Pa 1200 m 3 PV n= = = 5.601  104 moles RT ( 8.315J mol k )( 261K )

(

)(

(

)

)(

)

m = 5.601  104 moles 29  10 −3 kg mol = 1624 kg

The mass entering the house is 1624 kg − 1472 kg = 152 kg  200 kg . The values being subtracted are only good to the nearest 100 kg (2 sig. figs. each). 74. Assume the air is an ideal gas, and that the pressure is 1.0 atm. PV = NkT →

(1.013 10 Pa ) ( 6.0  3.0  2.5) m = 1.1197 10 molecules  1.110 molecules = N= kT (1.38 10 J K ) ( 273 + 22 ) K 5

−23

1 mole    = 1860 moles  1900 moles 23  6.02  10 molecules 

1.1197  1027 molecules 

75. We assume that the last breath Galileo took has been spread uniformly throughout the atmosphere since his death and that each molecule remains in the atmosphere. Calculate the number of molecules in Galileo’s last breath, and divide it by the volume of the atmosphere, to get “Galileo molecules/m3.” Multiply that factor times the size of a breath to find the number of Galileo molecules in one of our breaths. 1.01  105 Pa 2.0  10−3 m 3 PV = = 4.9  1022 molecules PV = NkT → N = −23 kT 1.38  10 J K ( 300 K )

(

)(

(

)

2 h = 4 6.38  106 m Atmospheric volume = 4 REarth

Galileo molecules m

# Galileo molecules breath 76. We are given that P 

4.9  1022 molecules 5.8  10 m 18

= 9.6  10

)

) (1.0 10 m ) = 5.110 m 2

= 9.6  103 molecules m3

molecules  2.0  10 −3 m 3  3

molecules

 1 breath  = 19 breath  

 20

molecules breath

1 for constant temperature and V  T 2 / 3 for constant pressure. We also 2 V

assume that V  n for constant pressure and temperature. Combining these relationships gives the following.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

PV 2 = n 2 RT 4 / 3 L2 atm (1.00 atm )( 22.4 L )2 (1.00 atm )( 22.4 L )2 R = 2 4/3 = = = 0.283 nT mol 2 K 4/3 (1.00 mol )2 ( 273.15 K )4 / 3 (1.00 mol )2 ( 273.15 K )4 / 3 PV 2

77. (a) The iron floats in the mercury because Hg  Fe . As the substances are heated, the density of both substances will decrease due to volume expansion. The density of the mercury decreases more upon heating than the density of the iron, because  Hg   Fe. The net effect is that the densities get closer together, and so relatively more mercury will have to be displaced to hold up the iron, and the iron will float lower in the mercury. (b) The fraction of the volume submerged is VHg VFe . Both volumes expand as heated. The displaced

subscript “displaced” is dropped for convenience. V0 Hg (1 +  Hg T )

fractional change = =

VHg VFe − V0 Hg V0 Fe V0 Hg V0 Fe

(

− V0 Hg V0 Fe

V0 Fe (1 +  Fe T )

V0 Hg V0 Fe

)( 25C ) − 1 = 1.0045 − 1 = 3.6  10 1 + ( 35  10 )( 25C ) 1.000875

1 + 180  10 −6 C o

(1 +  T ) − 1 Hg

(1 +  Fe T )

−6

−3

% change = 0.36 %

;

78. (a) Consider the adjacent diagrams. The mercury expands due to the heat, as does the bulb volume. The volume of filled glass is equal to the volume of mercury at both temperatures. The value  l is the amount the thread of mercury moves. The additional length of the mercury column in the tube multiplied by the tube cross sectional area will be equal to the expansion of the volume of mercury, minus the expansion of the volume of the glass bulb. Since the tube volume is so much smaller than the bulb volume we can ignore any changes in the tube dimensions and in the mercury initially in the tube volume. Original volume for glass bulb and Hg in bulb: V0bulb Change in glass bulb volume:

Vglass = V0bulb  glass T

Change in Hg volume in glass bulb:

VHg = V0bulb  Hg T

Now find the additional volume of Hg, and use that to find the change in length of Hg in the tube. ( l )  r02 = VHg − Vglass = V0bulb Hg T − V0bulb glass T →

l = =

V0bulb

T (  Hg −  glass ) = 2

 r0

(

V0bulb

 ( d0 2 )

T (  Hg −  glass ) = 2

4V0bulb

 d 02

T (  Hg −  glass )

) ( 33.0C − 10.5C ) (180 − 9 ) 10 C = 7.12 cm    (1.40  10 cm ) 4 0.285 cm 3 −2

−6

(b) The formula is quoted above:

l =

4V0bulb

 d 02

T (  Hg −  glass ) .

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79. Let the original amount of air in the tire be state “1,” and the final amount of air in the tire be state “2.” The volumes and temperatures are the same for both states. The original absolute pressure is 3.4 atm. We calculate the final absolute pressure based on a 20% decrease in gauge pressure, and then use the ideal gas law to find the fraction of moles of gas that leaked out. P1 = 3.4 atm ; P2 = ( 0.80)( 2.4 atm ) = 1.92 atm ; P2 = 2.92 atm abs

gauge

PV = nRT →

abs

RT V

= constant →

P2 n2

→

n2 n1

P2 P1

2.92 atm 3.4 atm

= 0.859

n2  0.86 n1 Thus 14% of the air leaked out. 80. Since the pressure is force per unit area, if the pressure is multiplied by the surface area of the Earth, the force of the air is found. If we assume that the force of the air is due to its weight, then the mass of the air can be found. The number of molecules can then be found using the molecular mass of air (calculated in Example 17–12) and Avogadro’s number. F 2 P= → F = PA → Mg = P 4 REarth → A M =

2 P 4 REarth

(

4 6.38  106 m

) (1.01  10 Pa ) = 5.27  10 kg 2

9.80 m s

  6.02  10 molecules  = 1.1  1044 molecules    −3 1 mole  29  10 kg     1 mole

N = 5.27  1018 kg 

81. One mole of gas at STP occupies 22.4 L, as found in Example 17–10. We find the volume of the gas per particle for a mole of gas at STP. We then assume that each molecule occupies a cube of side a, and then solve for a as the average distance between molecules. −3 3 1mol  22.4 L     10 m  = 3.72  10 −26 m 3 molecule = a 3    mol   6.02  1023 molecules   1L  

(

a = 3.72  10 −26 m 2

) = 3.34  10 m 1/ 3

−9

82. The density is the mass divided by the volume. Let the original volume of the mass of iron be V0 , the original density  0 = M V0 . The volume of that same mass deep in the Earth is V = V0 + V , and so the density deep in the Earth is  = M V = M (V0 + V ) . The change in volume is due to

two effects: the increase in volume due to a higher temperature, Vtemp = V0 T , and the decrease in volume due to a higher pressure, Vpressure = − V0 P B. So V = Vtemp + Vpressure . The new density is then calculated by  = M (V0 + V ) .

=M V = =

M V0 + V

M V0 + Vtemp + Vpressure

M V0 + V0 T − V0 P B

V0 (1 + T − P B )

0

(1 + T − P B )

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

= =

1 + ( 35  10 0

Instructor Solutions Manual

0

−6

)( 2000C ) − ( 5000 atm ) (1.0110 Pa atm ) (90 10 N m )

1 + 0.07 − .00561

= 0.9395  0 →

6 % decrease

83. We assume the temperature is constant. As the oxygen pressure drops to atmospheric pressure, we can find the volume that it occupies at atmospheric pressure. We assume the final pressure inside the cylinder is atmospheric pressure. The gas would quit flowing at that pressure. (1.38  107 Pa + 1.013  105 Pa ) = 1921L P1 14 L PV PV V V = → = = ( ) 1 1 2 2 2 1 1.013  105 Pa P2 14 L of that gas is not available–it is left in the container. So there is a total of 1907 L available. 1min = 953.5 min  950 min  16 h (1907 L ) 2.0 L 84. We find the number of moles of helium in the balloon from the ideal gas law. PV = nRT → n =

PV RT

(1.08) (1.013  105 Pa ) 43  ( 0.260 m )3 = 3.3065 mol  3.31mol (8.314 J mol k )( 293 K )

 4.00g  = 13.2 g   1 mol 

3.3065 mol 

85. (a) Assume that a mass M of gasoline with volume V0 at 0oC is under consideration, and so its density is  0 = M V0 . At a temperature of 35oC, the same mass has a volume V = V0 (1 +  T ) .

=

M V

M V0 (1 + T )

0 0.68  103 kg m3 = = 0.6593  103 kg m3 −6 1 + T 1 + ( 950  10 C ) ( 33C )

 660 kg m3 (b) Calculate the percentage change in the density. ( 0.6593 − 0.68)  103 kg m3 % change =  100 = −3% 0.68  103 kg m3 V 86. The gap will be the radius of the lid minus the radius of the jar. Also note that the original radii of the lid and the jar are the same. rgap = ( r0 + r )lid − ( r0 + r ) jar = rlid − rjar = ( brass −  glass ) r0 T

(

)

= 19  10−6 C − 9  10−6 C ( 3.5cm )( 40C ) = 1.4  10−3 cm 87. The change in length is to be restricted to l  1.0  10−6 m. 1.0  10−6 m l =  l 0 T  1.0  10−6 m → T   0.11C 9  10−6 C (1.0 m )

(

)

Thus the temperature would have to be controlled to within 0.11C © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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88. (a) From Eq. 13–6c, we have that the pressure of the atmosphere varies as Pair = ( Pair )0 e − cy ,

0 g , with the subscript indicating to use the value at y = 0. We assume that the P0 helium is an ideal gas, that the helium pressure is 1.05 times the atmospheric pressure, and that the helium temperature is the same as the surrounding air. ( PHe )0 (VHe )0 PHeVHe (1.05Pair )0 (VHe )0 1.05PairVHe = → = → T0 T1 T0 T1 where c =

( Pair )0 (VHe )0 ( Pair )0 e − cyVHe T0

e − cyVHe T1

→ V = V0

T1 T0

e + cy

 0 g (1.29 kg m )( 9.80 m s ) = = 1.25  10−4 m −1 5 P0 (1.013  10 Pa ) 3

(VHe )0

→ 2

(b) The buoyant force is the weight of the air displaced by the balloon, which would be the density of the air, times the volume of the balloon, times the acceleration due to gravity. The density of the air displaced by the balloon can be found from the ideal gas equation, applied at any particular location. Fbuoy =  airVballoon g ;

PairVair = nair RT =

mair

RT →

( mol. mass )air Pair ( mol. mass ) air

Fbuoy =  airVballoon g =

Pair ( mol. mass ) air RT PairVballoon

Vballoon g =

mair Vair

=  air

g ( mol. mass ) air

 PHe  V   balloon g ( mol. mass ) air g ( mol. mass )air P V 1.05  = nHe g ( mol. mass ) air = He balloon =

RT RT 1.05 1.05 The final expression is constant since the number of moles of helium in the balloon is constant. 89. The original length of the steel band is l 0 = 2 REarth. At the higher temperature, the length of the

band is l = l 0 + l = 2 R = 2 ( REarth + R ) . The change in radius, R, would be the height above the Earth. l =  l 0 T = 2R → R =

 l 0 T =  REarth T = (12  10 −6 C )( 6.38  106 m ) ( 55C − 25C ) = 2297 m  2300 m 2

90. The change in radius with heating does not cause a torque on the rotating wheel, and so the wheel’s angular momentum does not change. Also recall that for a cylindrical wheel rotating about its axis, the moment of inertia is I = 12 mr 2.

L0 = Lfinal → I 00 = I finalfinal → final =

I 00 I final

1 2

mr020 1 2

mr 2

r020 r2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition





Instructor Solutions Manual

r020

− 0 2 final − 0 r2 r02 r02 1 1 = r = 02 − 1 = − = −1 = −1 2 2 2 0 0 r ( r0 + r ) ( r0 +  r0 T ) (1 + T )

(

1 − 1 + 2T + (T )

(1 + T )

Now assume that T

(1 + T )

1 , and so

(

) = −2T − (T ) = −T 2 + T 



= −T

(1 + T )

2 + T

(1 + T )

 −2T . Evaluate at the given values.

)

−2T = −2 25  10−6 C (80.0 C ) = −4.0  10−3 The negative sign means that the frequency has decreased. This is reasonable, because the rotational inertia increased due to the thermal expansion 91. Consider this basic geometry for the problem, with the assumption that the shape of the sagging wire is an arc of a circle. The amount of sag is greatly exaggerated in the figure. A subscript of “0” will be used for the original (low temperature) configuration, and no subscript will be used for the final (high temperature) configuration. The variable “s” will be used for the amount of “sag.” Note that “L” refers to half the length of the sagging wire.

(15.0 m ) 2 + ( R0 − s0 ) 2 = R02 → (15.0 m ) 2 + s02 (15.0 m ) 2 + ( 0.540 m )2 R0 = = 2s0 2 ( 0.540 m ) = 208.60 m

 0 = sin −1

15.0 m 208.60 m

= 7.1970  10 −2 rad

(

)

L0 = R0 0 = ( 208.60 m ) 7.1970  10−2 rad = 15.013m

Now let the wire expand due to heating.

 (

L = L0 (1 + T ) = 15.013 m 1 + 17  10 −6 ( C ) sin  =

15.0

→  = sin −1

15.0

; =

−1

) (50 C) = 15.026 m

15.026

R R R R These two expressions for  cannot be solved analytically. When solved numerically, the result is R = 147.43m. Use this value to find the new “sag.” Note that we ignore s 2 since s R.

(15.0 m )2 + ( R − s )2 = R 2 → (15.0 m )2 − 2 Rs + s 2 = 0 ; (15.0 m )2 (15.0 m )2 = = 76.3cm s 2R 2 (147.43m )

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92. We will take the radius of curvature of the assembly as being the radius to the boundary between the two materials, and so is equal to the radius of the inside curve of the steel, plus the thickness of the steel. Each strip, when curved, subtends the same angle  . r r rsteel r  = steel = brass → = brass → l steel l brass l 0 + l 1 l 0 + l 2 rsteel

rbrass

rsteel

→

rbrass

1 +  steel T 1 +  brass T l 0 +  steel l 0 T l 0 +  brass l 0 T Use the relationship that the radius of the brass is equal to the radius of the steel, plus the thickness of the steel, so rbrass = rsteel + t.

rsteel 1 +  steel T rsteel =

 1 +  steel T   ( rsteel + t ) →  1 +  brass T 

rsteel + t

→ rsteel = 

1 +  brass T t

 1 +  brass T   1 +  T − 1    steel

0.16 cm

 1 + (19  10−6 C ) ( 80 C )  − 1  −6  1 + (12  10 C ) ( 80 C ) 

= 285.99 cm

r = rsteel + t = 285.99 cm + 0.20 cm = 286.19 cm  2.9 m

93. Both the glass and the liquid expand. The expansion of the liquid would cause the volume reading to increase, but the expansion of the glass would cause the volume reading to decrease. So the actual change in reading is the difference in those two volume changes. We use the subscript “l” for the liquid and “g” for the glass. We see from the data that the volume readings are increasing with temperature, and so the volume increase of the liquid is more than the volume increase of the glass. Vreading Vreading = Vl − Vg = V0 1 +  l T − V0 (1 +  g T ) = V0  l −  g T → = V0  l −  g T

(

)

(

)

(

)

(

)

From this expression, if the graph of Vreading vs. T is linear, it should have a slope of m = V0  − g . Thus we can find the coefficient of expansion of the liquid from  =

m V0

+ g .

From the graph, the slope is seen to be 0.0492 mL/C . The effective coefficient of volume expansion is m 0.0492 mL C  l − g = = V0 100.00 mL

(

)

= 4.92  10−4 C .

l =

0.0492 mL C 100.00 mL

+ 9  10 −6 C

= 5.01  10 −4 C ; glycerin

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

94. (a) We treat the air as an ideal gas. Since the amount of air and temperature of the air are the same in both cases, the ideal gas law says PV = nRT is a constant. 185atm P PV = PV → V2 = V1 1 = (11.3 L ) = 2090.5 L  2090 L 2 2 1 1 1.00 atm P2 (b) Before entering the water, the air coming out of the tank will be at 1.00 atm pressure, and so the person will be able to breathe 2090.5 L of air.

 1 breath   1 min  = 87.10 min  87 min    2.0 L   12 breaths 

t = 2090.5 L 

(c) When the person is underwater, the temperature and pressure will be different. Use the ideal gas law to relate the original tank conditions to the underwater breathing conditions. The amount of gas will be constant, so PV T = nR will be constant. The pressure a distance h below the surface of the water is given in Eq. 13–6b, P = P0 +  gh , where P0 is atmospheric pressure and  is the density of the sea water. PV PV P T 2 2 = 1 1 → V2 = V1 1 2 T2 T1 P2 T1

(185atm ) (1.013  10 Pa atm )



  283 K     1.013  10 Pa + (1.025  10 kg m )( 9.80 m s ) ( 23.0 m )   291K  5

V2 = (11.3 L )  = 619.69 L

 1 breath   1 min  = 25.82 min  26 min    2.0 L   12 breaths 

t = ( 619.69 L ) 

95. The first method for calculating the thermal expansion is given in Eq. 17–2, with the coefficient of expansion given in Table 13–1. Using this method the volume change for each temperature difference is calculated as follows.

( ) V = V T = ( 3400  10 C ) ( 750 L )(100C − 0C ) = 255 L  260 L

VA = V0 TA = 3400  10 −6 C ( 750 L ) 0C − ( −100C )  = 255 L  260 L −6

Note that the text states that this method is accurate only if the change in volume is small compared to the actual volume. This is obviously not the case here, so the values above are not particularly accurate. The second method is to use the ideal gas law of Eq. 17–3.

pV = nRT

→

pV1 nRT1 = pV2 nRT2

→ V2 = V1

T2 T1

T  → V = V2 − V1 = V1  2 − 1  T1 

The temperature must be converted to kelvins for this method. T   273 K  VA = V1  2 A − 1  = ( 750 L )  − 1  = 433.5 L  430 L  173 K   T1 A 

T   373 K  VB = V1  2 B − 1  = ( 750 L )  − 1  = 274.7 L  270 L  273 K   T1B  The answers are different because the first method assumes that the same change in volume will occur for any fixed-size temperature change, regardless of the starting temperature, and so the only dependence on temperature is through the change in the temperature. The ideal gas law has to take

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into account the actual values of the absolute temperature, not just the temperature change itself.

 T − T0  V0  = T ,  T0  T0

The relationship derived from the ideal gas law could be expressed as V = V0 

which shows that the starting value of the temperature is important in determining the volume change. 96. There are three forces to consider: the upwards buoyant force (which is the weight of the cold air displaced by the volume of the balloon), the downward weight of the hot air inside the balloon, and the downward weight of the passengers and equipment. For the balloon to rise at constant speed, the buoyant force must equal the sum of the two weights. Fbuoyant = mhot g + 3300 N → V cold g = V hot g + 3300 N The ideal gas law can be written in terms of the gas density  and the molecular mass M as follows. m PM m PV = nRT = RT → = T = T M R V The gas inside and outside the balloon is air, and so M is the same for inside and outside. Also, since the balloon is open to the atmosphere, the pressure inside the balloon is essentially the same as the pressure outside the balloon. Thus the ideal gas law reduces to T = constant = ( T )cold = ( T )hot . This equation can be used to eliminate the density of the hot gas from the force equation, and the resulting equation can be solved for the temperature of the gas inside the balloon. T V  cold g = V  hot g + 3300 N = V  cold cold g + 3300 N → Thot

(1800 m )(1.29 kg m ) ( 273 K ) (9.80 m s ) T = = (V  g − 3300 N ) (1800 m )(1.29 kg m )( 9.80 m s ) − 3300 N  V  coldTcold g

hot

cold

= 319.3K ( 2 sig. figs.) → 319.3K − 273.15 K = 46.15C  50C One factor limiting the maximum altitude would be that as the balloon rises, the density of the air decreases, and so the temperature required gets higher. Eventually the air temperature required might be too hot and the balloon fabric could be damaged. 97. The change in lung pressure will be the pressure at the 30 cm depth, minus atmospheric pressure. We use Eq. 13–6b, P = P0 +  gh, for the pressure at the 30 cm depth. In order for air to flow through the snorkel from the atmospheric air above the water’s surface (assume to be at atmospheric pressure), the diver must reduce the pressure in his lungs to atmospheric pressure or below, by increasing the volume of the lungs. We assume that the diver is in sea water. 1atm P = ( P0 +  gh ) − P0 =  gh =  1025 kg m 3 9.80 m s 2 ( 0.3m )  1.013  105 Pa

(

)(

)

= 0.0297 atm  0.03 atm This is a 3% increase.

619

CHAPTER 18: Kinetic Theory of Gases Responses to Questions 1.

One of the fundamental assumptions for the derivation of the ideal gas law is that the average separation of the gas molecules is much greater than the diameter of the molecules. This assumption eliminates the need to consider the different sizes of the molecules.

The change in temperature when a gas is compressed or when it expands against a piston is due to the increase or decrease in the average speed of the molecules. The increase or decrease in speed comes about when the gas molecules collide elastically with the moving piston. In the case of compression, the piston is moving toward the molecules. The net result is an increase in the momentum of the gas molecules. When the gas expands, the piston is moving away from the gas molecules. In this case the net result of collisions between the molecules and the piston is a decrease in the momentum of the molecules. (See Section 9–5.)

In an elastic collision, no energy would be transferred to the walls. If the walls and the gas are all at the same temperature, then the gas molecules will not lose (or gain) energy in collisions with the walls, since temperature is proportional to the average kinetic energy. Thus it is not necessary to specify that the collisions must be elastic, since there will be no average energy transfer anyway.

Charles’s law states that if pressure is held constant, volume is proportional to temperature. The average kinetic energy of the gas molecules is also proportional to temperature. According to kinetic theory, pressure is proportional to the average kinetic energy of the gas molecules per unit volume. Therefore, if the temperature increases, the average kinetic energy also increases by the same factor. In order to keep the pressure constant, the volume must also increase, again by the same factor as the kinetic energy.

Gay-Lussac’s law states that if volume is constant, the pressure in a gas is proportional to the temperature. Kinetic theory tells us that temperature is proportional to the kinetic energy, and also the product of pressure and volume is proportional to the kinetic energy of the gas molecules. Thus, the product of pressure and volume is proportional to the temperature. If volume is held constant, then pressure is proportional to temperature, which is Gay-Lussac’s law.

Near the surface of the Earth, the N2 molecules and the O2 molecules are all at the same temperature and therefore have the same average kinetic energies. Since N2 molecules are lighter than O2 molecules, the N2 molecules will have a higher average speed, which allows them to travel higher (on average) in the atmosphere than the O2 molecules.

For an absolute vacuum, no. But for most “vacuums,” there are still a few molecules in the containers, and the temperature can be determined from the (very low) pressure.

At both temperatures (310 K and 273 K), the lower limit for molecular speed is zero. However, the higher temperature gas (310 K) will have more molecules with higher speeds. Since the total number of molecules is the same (proportional to the area under the curves), the higher temperature gas must have fewer molecules at the peak speed. (Kinetic theory predicts that the relative number of molecules with higher speeds increases with increasing temperature.)

Temperature is a macroscopic variable, measured for a whole system. It is related to the average molecular kinetic energy, which is a microscopic variable.

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10. (a) Because the escape velocity for the Moon is 1/5 that of the Earth, even heavy molecules with relatively lower average speeds would still have been able to escape. The Moon may have started with an atmosphere, but over time almost all of the molecules of gas have escaped. (b) Hydrogen is the lightest gas. For a given kinetic energy (temperature) it has the highest speed and will be most likely to escape. 11. Velocity is a vector quantity. When the velocity is averaged, the direction must be taken into account. Since the molecules travel in random paths, with no net displacement (the container is at rest), the average velocity will have to be zero. Speed is a scalar quantity, so only the (positive) magnitude is considered in the averaging process. The molecules are not at rest, so the average speed will not be zero. 12. (a) If the pressure is doubled while the volume is held constant, the temperature also doubles. v rms is proportional to the square root of temperature, so v rms it will increase by a factor of (b) The average velocity is also proportional to the square root of the temperature, so it will increase by a factor of 2 .

13. Evaporation is an indication that not all molecules have the same speed. Only molecules in a liquid that are traveling fast enough will be able to escape the surface of the liquid and evaporate. If all of the liquid particles were moving the same speed, then either they all would evaporate in a short time period, or none would evaporate even over a long time period. 14. If alcohol evaporates more quickly than water at room temperature, then it must be easier for the alcohol molecules to escape from the surface of the liquid. Alcohol molecules are more massive than water molecules, and so will not be moving as fast at the same temperature. We can therefore infer that the attractive intermolecular forces between the alcohol molecules are less than the forces between the water molecules. 15. On a hot day, cooling occurs through evaporation of perspiration. If the day is hot and dry, then the partial pressure of water vapor in the air will be low and evaporation will readily occur, since the saturated vapor pressure for water will be higher than the external pressure. If the day is hot and humid, then the partial pressure of water vapor in the air will be much higher and the air will be holding all or nearly all the water vapor it can. In this case evaporation will not occur as readily, resulting in less cooling. 16. Liquids boil when their saturated vapor pressure equals the external pressure. For water, from Table 18–2, the saturated vapor pressure of water at 20oC is about 0.023 atm. So if the external pressure is lowered to that level (about 2.3% of normal air pressure), the water will boil at that low temperature. If the water and its container is placed in a vessel that can be depressurized, and the air pumped out of the vessel, then the water can boil at room temperature. 17. Boiling occurs when the saturated vapor pressure equals the external pressure. When we say the oxygen “boils” at –183ºC, we mean that the saturated vapor pressure for oxygen will be 1 atm (the same as atmospheric pressure) at a temperature of –183ºC. At this temperature and pressure, liquid oxygen will vaporize. 18. The freezing point of water decreases slightly with higher pressure. This can be seen in Fig. 18–5. The wire exerts a large pressure on the ice (due to the weights hung at each end). The ice under the wire will melt, allowing the wire to move lower into the block. Once the wire has passed a given position, the water now above the wire will have only atmospheric pressure on it and will refreeze. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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This process allows the wire to pass all the way through the block and yet leave a solid block of ice when the wire reaches the bottom. 19. The humid air will be more dense than the dry air at the same temperature because it will have more water vapor suspended in it. 20. For all parts of this question, an examination of Fig. 18–5 will be helpful. (a) A pressure cooker is sealed and so as the temperature of its contents increases, the number of particles inside and the volume are kept constant. Thus the pressure increases according to the ideal gas law. Assuming there is water inside the pressure cooker, then an increased pressure yields a higher boiling point for the water. The water in which the food is prepared will boil at a higher temperature than normal, thereby cooking the food faster. (b) At high altitudes, the atmospheric pressure is less than it is at sea level. If atmospheric pressure decreases, the boiling point of water will decrease, and so boiling occurs at a lower temperature than at sea level. Food being cooked in an open pot (including pasta and rice) will need to cook longer at this lower temperature to be properly prepared. (c) It is actually easier to boil water at higher altitude, because the water boils at a lower temperature. Thus it will take less time to add enough heat to the water to bring it to the boiling temperature. 21. Both “vapor” and “gas” refer to a substance in the gaseous state. They differ in that a vapor is below the critical temperature and a gas is above the critical temperature for the substance. 22. (a) Yes. As an example, think of ice skating. The pressure from the weight of the skater melts the ice, and the skater glides on a thin layer of water. Also see Question 18 above. (b) No. See Fig. 18–6. The solid–liquid interface has a positive slope, and so it is not possible to melt carbon dioxide simply by applying pressure. Compare to Fig. 18–5 for water, in which the solid–liquid interface has a negative slope. 23. Dry ice is carbon dioxide in the solid state. As shown in Fig. 18–6, carbon dioxide at room temperature will be a vapor unless it is at a pressure higher than 50 atm. When a sample of solid dry ice is placed in surroundings at 1 atm pressure and room temperature, it will sublimate (change directly from solid to gas) and therefore does not last long–it doesn’t have to first melt, and then evaporate. It directly evaporates as it warms. 24. Liquid CO2 can exist at temperatures between –56.6ºC and 31ºC and pressures between 5.11 atm and 73 atm. (See Fig. 18–6.) CO2 can exist as a liquid at normal room temperature, if the pressure is between 56 and 73 atm. 25. Exhaled air contains a large amount of water vapor and is initially at a temperature equal to body temperature. When the exhaled air comes into contact with the external air on a cold day it cools rapidly and reaches the dew point. At the dew point temperature, the air can no longer hold all the water vapor and water condenses into little droplets, forming a cloud. The white cloud seen is due to the condensed water vapor. 26. A sound wave can be described as a pressure wave or a displacement wave. Transmission of the wave depends on the collisions of the gas molecules with nearby “neighbor” molecules. If the wavelength of a sound wave were to be less than or equal to the mean free path of the molecules in a gas, then there would be no interaction from one particle to another on average over the distance of an entire wavelength. The net displacement of a molecule away from an equilibrium position would be “lost” in the random movement of the molecules. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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27. The mean free path is given by Eq. 18–10b, l M =

. Ways to reduce the mean free 4 2 r 2 ( N V ) path in a gas include increasing the size of the gas molecules (in other words, using a different gas with larger molecules) and increasing the gas particle concentration (number density). Gas density can be increased either by increasing the number of molecules in a fixed volume or by decreasing the volume (or both).

Solutions to MisConceptual Questions 1.

(e) Some students may erroneously relate the temperature of the gas to the velocity of the gas molecules, and surmise that the rms speeds would be equal. However, when the two gases are at the same temperature the molecules will have the same average kinetic energy. The kinetic energy is proportional to the mass of the molecule and the square of the rms speed. Since mass B is half the mass of A, the speed of molecules of mass B must be the 2  1.4  greater than the speed of molecules of mass A.

(a) In the mixture, the oxygen molecules and helium atoms will be at the same temperature, which means that their average molecular kinetic energies will be the same. Since a helium atom has less mass than an oxygen molecule, the helium atoms will be moving faster than the oxygen molecules on average.

(d) The temperature of an ideal gas is a measure of the average kinetic energy of the gas, and so increasing the temperature will increase the average kinetic energy–(a) is true. Even though the temperature is proportional to the average kinetic energy of the molecules, each molecule can have a random kinetic energy (according to the Maxwell distribution of speeds). The speeds of individual molecules will vary about this average, and so (d) is false. In the ideal gas law the product of the pressure and volume is proportional to the temperature. Therefore, if pressure or volume is held constant as the temperature increases, the other parameter (volume or pressure) must increase, and so (b) and (c) are true. For a gas to be ideal, it is assumed that the space occupied by the gas is mostly empty. For this to be true the molecules are assumed to be far apart compared with their size. Thus (e) is also true.

(d) Because the rms speed is proportional to the square root of the temperature (Eq. 18–5), to double the rms speed means the temperature needs to increase by a factor of 4.

(c) The rms speed of a gas is an average molecular speed found by taking the square root of the average of the square of the molecule speeds. Consider a gas where 2/3 of the molecules are at rest and 1/3 move at a constant speed. The most probable speed is 0, but the rms speed is greater than zero. Therefore, the most probable speed is not necessarily the rms speed, and so (a) is false. The rms speed is an average speed and therefore it will always be equal to or smaller than the maximum speed, and so (b) is false. The temperature of a gas is determined by the average molecular kinetic energy. As the temperature increases, the molecular kinetic energy increases, and therefore the rms speed must also increase, and so (c) is true.

(b) Since there is no change in temperature, and the average speed is proportional to the square root of the temperature, as shown in Ex. 18–5, the average speed does not change.

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(c, d) Because the temperatures of the two types of molecules are the same, their average kinetic energies are the same, which is answer (d). But a molecule of nitrogen weighs less than a molecule of oxygen (atomic weight 28 for nitrogen vs. 32 for oxygen), so the rms speed of the nitrogen will be larger, which is answer (c).

(d) From Example 18–2, the rms speed for nitrogen molecules at room temperature is about 510 m/s. A passenger jet flies at about 250 m/s. Highway speed limits are around 30 m/s, and the International Space Station orbits at about 8000 m/s. So the passenger jet is the closest value.

10. (b) From Fig. 18–5, the phase diagram for water, sublimation can occur for low pressures (below 0.006 atm) and low temperatures (below the triple point temperature). This is answer (b).

Solutions to Problems These solutions do not always follow the rules of significant figures rigidly. We tended to take quoted temperatures as correct to the number of digits shown, especially where other values might indicate that. 1.

The rms speed is given by Eq. 18–5, vrms = 3kT m . Helium has an atomic mass of 4.0.

(

v rms = 3kT m = 2.

)

= 6116 m s  6  103 m s

3kT2 m 3kT1 m

T2 T1

( 273 + 160 ) K

273K

= 1.26

6.02 + 2.02 + 4.02 + 6.02 + 0.02 + 4.02 + 1.02 + 8.02 + 5.02 + 3.02 + 7.02 + 8.02 12 320 12

= 5.2

The average kinetic molecular energy is 23 kT . Set this equal to the kinetic energy of the paper clip. 1 2

(a) The mean (average) speed is as follows. 6.0 + 2.0 + 4.0 + 6.0 + 0.0 + 4.0 + 1.0 + 8.0 + 5.0 + 3.0 + 7.0 + 8.0 54.0 vavg = = = 4.5 . 12 12 (b) The rms speed is the square root of the mean (average) of the squares of the speeds.

vrms =

(

4.0 1.66  10

−27

The rms speed is given by Eq. 18–5, vrms = 3kT m . The temperature must be in Kelvins.

( vrms )2 = ( vrms )1 3.

)

3 1.38  10 −23 J K ( 6000 K )

mv 2 = 32 kT → v =

3kT m

(

)

3 1.38  10−23 J K ( 273 + 22 ) K −3

1.5  10 kg

= 2.9  10−9 m s

The rms speed is given by Eq. 18–5, vrms = 3kT m . Since the rms speed is proportional to the square root of the absolute temperature, to triple the rms speed without changing the mass, the absolute temperature must be multiplied by a factor of 9.

Tfast = 9Tslow = 9 ( 273 + 30 ) K = 2727 K = 2454C  2450C © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(a) The average translational kinetic energy of a gas molecule is 32 kT , regardless of the identity of the gas.

(

)

Kavg = 23 kT = 23 1.38  10−23 J K ( 273K ) = 5.65  10−21 J (b) The total translational kinetic energy is the average kinetic energy per molecule, times the number of molecules. Again, this doesn’t depend on the identity of the molecule.  6.02  1023 molecules  3 −23 K total = N ( K avg ) = (1.0 mol )   2 1.38  10 J K ( 298 K ) 1mol  

(

)

= 3700 J

(a) The average molecular kinetic energy is 23 kT , so the total kinetic energy for a mole would be Avogadro’s number times 23 kT .

K = N 0 ( 23 kT ) = 23 RT = 23 ( 8.314 J mol K )( 273 + 18K ) K = 3629 J  3600 J (b) 8.

K = 12 mv 2 = 3629 J → v =

2 ( 3629 J ) 65 kg

= 10.57 m s  11m s

The rms speed is related to the Kelvin temperature by Eq. 18–5. The mass of a helium atom is 4.003 u and the mass of a molecule of water is 2(1.008) + 15.994 = 18.01 u. 3kTH O 3kTHe = → ( vrms )He = ( vrms )H O → mHe mH O 2

THe =

mHe mH O

TH O = 2

4.003 u 18.01u

( 273.15 + 20.0 ) = 65.16 K or − 207.99C  −208.0C or 65.2 K

From the ideal gas law, PV = nRT , if the volume and amount of gas are held constant, the nR temperature is proportional to the pressure, PV = nRT → P = T = ( constant ) T . Thus the V temperature will be tripled. Since the rms speed is proportional to the square root of the temperature, vrms = 3kT m = ( constant ) T , v rms will be multiplied by a factor of

3  1.73.

10. The rms speed is given by Eq. 18–5, vrms = 3kT m . The temperature can be found from the ideal gas law, PV = NkT → kT = PV N . The mass of the gas is the mass of a molecule times the M number of molecules; M = Nm , and the density of the gas is the mass per unit volume,  = . V Combining these relationships gives the following.

vrms = 3kT m =

3PV Nm

3PV M



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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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11. The rms speed is given by Eq. 18–5, vrms = 3kT m , where m is the mass of one particle.

( vrms )1 = ( vrms )2

3kT m1 3kT m2

m2 m1

m2 N A

m1 N A

→

( vrms )1 = ( vrms )2

M2 M1

12. The temperature of the nitrogen gas is found from the ideal gas law, and then the rms speed is found from the temperature. The molecular mass of nitrogen gas is 28 u. PV PV = nRT → T = nR

3kT

vrms =

3k PV m nR

(

) ( 2.9 atm ) (1.013 10 Pa atm )( 7.2 m ) 28 (1.66  10 kg ) ( 2100 mol )( 8.314 J mol K )

3 1.38  10−23 J K

−27

= 328.4 m s  330 m s 13. From Eq. 18–5, we have vrms = 3kT m . (a)

d vrms dT

1/ 2

vrms 

d  3kT 





dT  m  d vrms

T =

1/ 2

1  3k 

1/ 2

1  3kT  1 1 vrms =  =    1/ 2 2 m  T 2 m  T 2 T

1 vrms

T →

vrms



1 T

2 T 2 T dT vrms (b) The temperature must be calculated in Kelvin for the formula to be applicable. We calculate the percent change relative to the winter temperature. vrms 1 T 1  30 K   =   = 0.056 = 5.6% 2 T 2  268 K  vrms 14. Assume that oxygen is an ideal gas, and that each molecule occupies the same cubical volume of l 3. Find the volume per molecule from the ideal gas law, and then the side length of that cubical molecular volume will be an estimate of the average distance between molecules. 1.38  10 −23 J K ( 273 K ) V kT PV = NkT → = = = 3.73  10−26 m 3 molecule 1.01  105 Pa N P

(

)

1/3  (1.38  10−23 J K ) ( 273 K )   kT  −9 l =  = 3.34  10 m  =  5 1.01  10 Pa  P    1/3

15. The rms speed is given by Eq. 18–5, vrms = 3kT m .

( vrms )2 = 1.050 = ( vrms )1

3kT2 m 3kT1 m

T2 T1

→

T2 = T1 (1.050 ) = ( 293.15 K )(1.050 ) = 323.20 K = 50.0C 2

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16. Gas molecules will rush into the vacuum from all directions. An estimate for the time for air to refill this vacuum region is the radius of the region divided by the rms speed of the molecules. d d 0.01m t = = = = 2  10−5 s − 23 vrms 3kT m 3 1.38  10 J K ( 293 K )

(

)

(

−27

29 1.66  10 kg

)

17. (a) The rms speed is given by Eq. 18–5, vrms = 3kT m .

3kT

v rms =

(

)

3 1.38  10−23 J K ( 273 K )

(

2 (15.9994 ) 1.66  10−27 kg

)

= 461m s

(b) Assuming that the particle has no preferred direction, then we have the following: 2 vrms = v x2 + v y2 + v z2 = 3v x2 → v x = vrms

The time for one crossing of the room is then given by t = d v x = 3d vrms , and so the time for a round trip is 2 3d vrms . Thus the number of back and forth round trips per second is the reciprocal of this time,

v rms

. 2 3d v 461m s # round trips per sec = rms = = 22.18  22 round trips per sec 2 3d 2 3 ( 6.0 m )

18. To get a speed as low as the one quoted, the mass was probably wrong. Use the incorrect speed with Eq. 18–5 to calculate the mass that your friend used. −23 3kT 3kT 3 1.38  10 J K ( 293 K ) vrms = → m= 2 = = 42 kg 2 m vrms 1.7  10−11 m s

(

)

(

)

This is approximately the mass of 1000 moles of CO 2 , but the mass of 1 molecule should have been used. That mass is 12.01 + 2(15.994) = 43.998 u.

v rms =

3kT m

(

)

3 1.38  10 −23 J K ( 293 K )

( 43.998 u ) (1.66  10−27 kg u )

= 407.5 m s  410 m s

19. (a) The average time for a molecule to travel from one side of the box to the other and back again is simply the round-trip distance, say in the x-direction, divided by the average x speed of the molecule. The frequency of collisions for that molecule from a single wall is the reciprocal of that round-trip time. The overall frequency of collisions is N times the frequency for a single particle. Use the ideal gas law to relate the number of particles in the room to the gas parameters of pressure, volume, and temperature. PV P l 3 x 2 l tround = = ; PV = NkT → N = = kT kT vx vx trip Pl 3 f =N

1 tround

N vx

= kT 2l 2l

vx P 2 kT

trip

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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(b) We approximate that v x  v x2 . From section 18–1, we have that v x2 = 13 v 2 and Eq. 18–4, 1 2

mv 2 = 23 kT . Combine these results with the result from part (a).

f =

vx P

l2 

2 kT

2 x

2 kT

l2 =

1 3

P kT

l2 =

1 3

m P 2 l = kT 2

Pl 2 4mkT

(c) We assume the pressure is at one atmosphere, and we take the molecular mass of air to be 29 u, as given in Problem 16. 2 1.013  105 Pa ) ( 3 m ) ( Pl 2 f = = = 3.27  10 28 Hz −27 −23 4mkT 4 ( 29 ) (1.66  10 kg )(1.38  10 J K ) ( 293 K )  3  10 28 Hz

20. (a) We find the average by adding the speed of every particle and then dividing by the number of particles. 1 1  2 (10 m s ) + 7 (15 m s ) + 4 ( 20 m s ) + 3 ( 25 m s )  v =  ni vi =   = 23 m s 25  + 6 ( 30 m s ) + 1( 35 m s ) + 2 ( 40 m s ) N i  (b) We find the rms speed by taking the square root of the average squared speed. vrms =

1 N

1  2 (10 m s ) + 7 (15 m s ) + 4 ( 20 m s ) + 3 ( 25 m s )  2

 n v = 25  i

2 i



+ 6 ( 30 m s ) + 1( 35 m s ) + 2 ( 40 m s ) 2

 

= 24.56 m s  25 m s

(c) The most probable speed is that one that occurs most frequently, 15 m s . 21. In the Maxwell distribution, Eq. 18–6, we see that the mass and temperature always occur as a ratio. Thus if the mass has been tripled, tripling the temperature will keep the velocity distribution constant. 22. (a) We find the rms speed by taking the square root of the average squared speed. vrms =

1500 ( 220 m s )2 + 3900 ( 440 m s )2 + 4600 ( 660 m s )2    ni v = 14, 500  2 2 2 N i  +2900 ( 880 m s ) + 1200 (1100 m s ) + 400 (1320 m s )  1

2 i

= 705.9 m s  710 m s

(b) The temperature is related to the rms speed by Eq. 18–5. vrms =

3kT m

( 2.00 10 kg ) ( 705.9 m s ) = 240.7 K  240 K → T= = 3k 3 (1.38  10 J K ) 2

−26

2 mvrms

−23

(c) Find the average speed, and then use a result from Example 18–5. 1 1 1500 ( 220 m s ) + 3900 ( 440 m s ) + 4600 ( 660 m s )  v =  ni vi =   N i 14, 500  +2900 ( 880 m s ) + 1200 (1100 m s ) + 400 (1320 m s )  = 653.9 m s  650 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

628

Chapter 18

Kinetic Theory of Gases −26  mv 2  ( 2.00  10 kg ) ( 653.9 m s ) v= → T= = = 243.4 K  240 K  m 8k 8 (1.38  10−23 J K ) 2

8 kT

Yes, the temperatures are consistent to two significant figures. 

23. (a) Show that  f ( v ) d v = N . We use a change of variable, and we make use of an integral from 0



 . 16a 3

Appendix B-5: specifically,  x 2 e − ax dx = 2



3/ 2



1 mv

 m  2 − 2 kT v =  f dv N 4 ( )    2 kT  v e d v 0 0 x=

mv 2 2kT

mv 2

→ x2 =

2kT

3/ 2



→ v2 =

1 mv

2kT m



 m  2 −  m   4 N  2 kT  v e 2 kT d v =  4 N  2 kT  0 0  m  = 4 N    2 kT 

3/ 2

2kT



3kT

2kT

2 −x

x → dv =

x 2e− x

2kT

3/ 2



3/ 2

mv 2 2kT

→ x2 =

mv 2

→ v2 =

2kT

3/ 2



1 mv

 m    2 kT 

3/ 2

= 4 N 

3kT

 v f (v ) dv = N m 2

2kT m



4k 2T 2 m

2kT



With this result, we have v 2 =

3/ 2

4k 2T 2 m

4 −x

3kT

x → dv =

x 4e− x

x e dx = N m  m

2kT m

→



→

2kT

x2 → v =

 m  4 −  m   4 N  2 kT  v e 2 kT d v =  4 N  2 kT  0 0



 3  ( for a = 1) . = a 8

 m  4 − 2 kT dv  ve  2 kT 

v 2 =  v 2 f ( v ) d v =  4 N  x=



 = N 4

use of an integral from Appendix B-5: specifically,  x 4 e − ax dx = 

2kT

. We use the same change of variable as above, and we make



3/ 2

1 x e dx = 4 N    m 0  

2kT

(b) Show that  v 2 f ( v ) d v N =

2kT

x2 → v =

3kT

 v f (v ) dv N = m 2

3kT m

and so 12 mv 2 = 23 kT .

24. From Fig. 18–6, we see that CO 2 is a vapor at 35 atm and 35C.

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25. (a) From Fig. 18–6, at atmospheric pressure, CO2 can exist as solid or vapor . (b) From Fig. 18–6, for CO2 to exist as a liquid, 5.11 atm  P  73 atm and

−56.6o C  T  31o C . 26. (a) From Fig. 18–5, water is vapor when the pressure is 0.01 atm and the temperature is 90oC. (b) From Fig. 18–5, water is solid when the pressure is 0.01 atm and the temperature is –20oC. 27. (a) At the initial conditions, the water is a liquid. As the pressure is lowered, it becomes a vapor at some pressure between 1.0 atm and 0.006 atm. It would still be a vapor at 0.004 atm. (b) At the initial conditions, the water is a liquid. As the pressure is lowered, it becomes a solid at a pressure of 1.0 atm, and then becomes a vapor at some pressure lower than 0.006 atm. It would be a vapor at 0.004 atm. 28. From Table 18–2, the saturated vapor pressure at 25oC is 3170 Pa or 0.031 atm. Since the relative humidity is 75%, the partial pressure of water is as follows. Pwater = 0.75Psaturated = 0.75 ( 3170 Pa ) = 2378 Pa  2400 Pa Pwater = 0.75Psaturated = 0.75 ( 0.031atm ) = 0.02325atm  0.023atm

29. From Table 18–2, the saturated vapor pressure at 30oC is 4240 Pa or 0.042 atm. Since the relative humidity is 35%, the partial pressure of water is as follows. Pwater = 0.35Psaturated = 0.35 ( 4240 Pa ) = 1484 Pa  1500 Pa Pwater = 0.35Psaturated = 0.35 ( 0.042 atm ) = 0.0147 atm  0.015atm

30. At the boiling temperature, the external air pressure equals the saturated vapor pressure. Thus from Table 18–2, for 80C the saturated air pressure is 4.73  104 Pa or 0.47 atm . 31. At the boiling temperature, the air pressure equals the saturated vapor pressure. The pressure of 0.80 atm is equal to 8.10  104 Pa. From Table 18–2, the temperature is between 90oC and 100oC. Since there is no entry for 8.10  104 Pa, the temperature can be estimated by a linear interpolation. Between 90oC and 100oC, the temperature change per Pa is as follows: (100 − 90 ) C = 3.236  10−4 C Pa . 4 (10.1 − 7.01) 10 Pa Thus the temperature corresponding to 7.27  104 Pa is

(

)

90C + ( 8.10 − 7.01)  10 4 Pa  3.236 10 −4 C Pa = 93.53C  94C .

32. From Table 18–2, if the temperature is 25oC, the saturated vapor pressure is 3170 Pa. If the relative humidity is 55%, then the partial pressure of water is 55% of the saturated vapor pressure, or 1740 Pa. The dew point is the temperature at which the saturated vapor pressure is 1740 Pa, and from Table 18–2 that is between 15oC and 20oC. Since there is no entry for 1740 Pa, the temperature can be estimated by a linear interpolation. Between 15oC and 20oC, the temperature change per Pa is as follows:

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Chapter 18

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( 20 − 15) Co = 8.0645  10−3 Co Pa . ( 2330 − 1710 ) Pa Thus the temperature corresponding to 1740 Pa is 15C + (1740 − 1710 ) Pa  (8.0645  10 −3 C o Pa ) = 15.24C  15C . 33. For boiling to occur at 120C, the pressure inside the cooker must be the saturated vapor pressure of water at that temperature. That value can be found in Table 18–2. For the mass to stay in place and contain the steam inside the cooker, the weight of the mass must be greater than the force exerted by the gauge pressure from the gas inside the cooker. The limiting case, to hold the temperature right at 120C, would be with the mass equal to that force.

mg = Fgauge = ( Pinside − Patm ) A = ( Pinside − Patm )  r 2 → pressure

( Pinside − Patm )  r 2

(1.99  10 Pa − 1.013  10 Pa )  (1.5  10 m ) = 0.07047 kg = 5

−3

9.80 m s 2

 7.0  101 g 34. The volume, temperature, and pressure of the water vapor are known. We use the ideal gas law to calculate the mass. The pressure must be interpolated from Table 18–2. Between 20oC and 25oC, the pressure change per temperature change per Co is as follows. ( 3170 − 2330) Pa = 168 Pa Co o ( 25 − 20) C

(

Thus the saturated vapor pressure at 22oC is 2330 Pa + 168 Pa Co

) ( 2 C ) = 2666 Pa. The water o

vapor pressure is 45% of that value. PV ( 0.45)( 2666 Pa )( 5.0 m )( 6.0 m )( 2.4 m ) PV = nRT → n = = = 32.2 mol RT (8.314 J mol K )( 273.15 K + 22 K )

mH O = ( 32.2 mol )( 0.018 kg mol ) = 0.63kg 2

35. Since the water is boiling at 130oC, the saturated vapor pressure is the same as the pressure inside the pressure cooker. We will have to interpolate From Table 18–2. 130oC is 1/3 of the way from 120oC to 150oC, so the pressure is found as follows. P = 2.00 atm + 13 ( 4.70 atm − 2.00 atm ) = 2.90 atm

(

)

P = 1.99  105 Pa + 13 4.76  105 Pa − 1.99  105 Pa = 2.91  105 Pa

Note that if the answer in atm is converted to Pascals, the answer is as follows.  1.013  105 Pa  P = 2.90 atm  = 2.94  105 Pa  1atm   36. From Table 18–2, the saturated vapor pressure at 25oC is 3170 Pa. The total amount of water vapor that can be in the air can be found from the saturated vapor pressure, using the ideal gas law. ( 3170 Pa ) 420 m3 PV PV = nRT → n = = = 546.6 moles RT ( 8.314 J mol K )( 273 + 20 ) K

(

)

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Since the relative humidity is 65%, only 65% of the total possible water is in the air. Thus 35% of the total possible water can still evaporate into the air.  18  10−3 kg  mevaporate = 0.35 ( 546.6 moles )   = 3.444 kg  3.4 kg  1 mole  37. (a) The true atmospheric pressure will be greater than the reading from the barometer. In Fig. 13–11, if there is a vapor pressure at the top of the tube, then Patm −  gh = Pvapor . The reading from the barometer will be  gh = Patm − Pvapor  Patm . The true atmospheric pressure is greater than the reading from the barometer. (b) The percent error is found from the atmospheric pressure and the vapor pressure.   gh − Patm   Pvapor   0.0015 mm-Hg  % diff =   100 =  −  100 =  −     100  760 mm-Hg   Patm   Patm 

(

)

= −2.0  10−4 % (c) From Table 18–2, the saturated water vapor pressure at STP is 611 Pa.  Pvapor  611Pa    100 = 0.603% % diff =  −  100 =  −   5  1.013  10 Pa   Patm  38. The outside air is at the dew point, 5oC. Thus the vapor pressure of water is 872 Pa, as read from Table 18–2. The ideal gas law gives the following result for the change in volume of the given mass of air. T T P T PV = nRT → = = constant → 1 = 2 nR V V1 V2 Thus, the vapor pressure of a given mass of air that moves from outside to inside is as follows. nRTin nRTout Pin = = = Pout = 872 Pa Vin Vout The saturated vapor pressure at 22oC is estimated to be 2666 Pa by using linear interpolation. 3170 Pa − 2330 Pa P22  C − 2330 Pa P25 C − P20  C P − P20  C = 22  C → = → P22  C = 2666 Pa 25C − 20C 22C − 20C 5C 2C Thus, the relative humidity is as follows. 872 Pa rel. hum. = = 0.327  33% 2666 Pa 39. We use the equation that was presented in Section 18–5, which comes from Eq. 13–6c: − (1.2510 m ) y P ( y ) = P0 e . We also use Table 18–2, to connect a boiling temperature with a specific pressure. −4

−1

(

−4

−1

)( 3000 m )

= 0.6873atm. Now interpolate Table 18–2 between 80oC (0.47 (a) P = (1.00 atm ) e atm) and 90oC (0.70 atm) to find the temperature. 0.6873 − 0.47  10C = 9.448C ; T = 89.448C  89C = 193F 0.70 − 0.47 If we use Pa instead of atm, the answer is slightly different because more significant figures are given. − 1.2510 m

632

Chapter 18

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) (

(

P = 1.013  105 Pa e

−4

− 1.2510 m

−1

)(3000 m )

= 6.962  104 Pa

6.962 − 4.73

 10C = 9.79C ; T = 89.79C  89.8C = 194F 7.01 − 4.73 (b) The temperature of 80oC corresponds to 0.47 atm. Solve for the temperature. ln ( 0.47 ) −(1.2510 m ) y 0.47 atm = (1.00 atm ) e → y= = 6040 m − 1.25  10−4 m −1 −4

−1

(

)

If we use Pa instead of atm, the answer is slightly different, due to significant figures. The temperature of 80oC corresponds to 4.73  104 Pa. ln ( 0. 473 1.013) − (1.2510 m ) y 4.73  104 Pa = 1.013  105 Pa e → y= = 6090 m − 1.25  10−4 m −1

(

)

−4

−1

(

)

40. We use the equation that was presented in Section 18–5, which comes from Eq. 13–6c: − (1.2510 m ) y P ( y ) = P0 e . We also use Table 18–2, to connect a boiling temperature with a specific pressure. (a) From the table at the front of the book, a mile is 1609 m. Thus the pressure is − (1.2510 m )(1609 m ) = 0.8178atm . Now interpolate Table 18–2 between 90oC P = (1.00 atm ) e (0.70 atm) and 100oC (1.00 atm) to find the temperature. 0.8178 − 0.70  10C = 3.93C ; T = 93.9C  94C = 201F 1.00 − 0.70 If we use Pa instead of atm, the answer is slightly different. −4

−1

−4

−1

) (

(

P = 1.013  105 Pa e

−4

− 1.2510 m

−1

)(1609 m)

= 8.284  104 Pa

8.284 − 7.01

 10C = 4.08C ; T = 94.08C  94.1C = 201F 10.13 − 7.01 (b) We first use linear interpolation of Table 18–2 between 80oC and 90oC to find the pressure corresponding to 89oC. 0.70 atm − 0.47 atm  9C = 0.207 atm ; P = 0.47atm + 0.207atm = 0.677 atm 90C − 80C

P ( y ) = P0e

(

−4

− 1.2510 m

−1

→ y=

 P ( y)   P0 

ln 

(

−4

− 1.25  10 m

−1

)

(

ln 0.677

− 1.25  10−4 m −1

)

= 3121m  3100m

If we use Pa instead of atm, the answer is slightly different. We again use linear interpolation of Table 18–2 between 80oC and 90oC to find the pressure corresponding to 89oC. 7.01  104 Pa − 4.73  104 Pa  9C = 2.052  104` Pa 90C − 80C

P ( y ) = 4.73  104 Pa + 2.052  104 Pa = 6.782  10 4 Pa

P ( y ) = P0e

(

−4

− 1.2510 m

−1

→ y=

 P ( y)    P0 

ln 

(

− 1.25  10−4 m −1

)

 6.782  104 Pa  5   1.013  10 Pa  = 3210 m

ln 

(

− 1.25  10−4 m −1

)

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

41. For one mole of gas, the “lost” volume (the volume occupied by the molecules) is the value of b. We assume spherical molecules.

b = N 0 43  ( 12 d )

1/ 3

 3b  d = 2   4 N 0 

→

  3 ( 3.2  10−5 m 3 mol ) = 2  4 ( 6.02  1023 molecules mol )   

1/ 3

= 4.7  10−10 m

42. (a) Use the van der Waals equation. RT a − P= (V n ) − b (V n ) 2 =

(8.314 J mol K )( 273 K )

0.13 N m 4 mol 2

−

( 0.70  10 m mol ) − ( 3.2  10 m mol ) ( 0.70  10 m mol ) −3

−5

(b) Use the ideal gas law. PV = nRT → P =

nRT

−3

(1.0 mol )(8.314 J mol K )( 273 K ) −3

= 3.1  106 Pa

= 3.2  106 Pa

V 0.70  10 m The value from the ideal gas law is about 3% higher than the value from the van der Waals equation of state. 3

43. The van der Waals pressure can be either higher or lower than the ideal gas pressure, depending on the volume. Accordingly, we use a parameter “c,” which is the ratio of the van der Waals pressure to the ideal pressure. The parameter will range from 0.95 to 1.05. RT a nRT − =c → RT ( c − 1) V 2 + ( an − bcnRT )V − ban 2 = 0 → PV = cPI → 2 − V n b V ( ) (V n )

− ( an − bcnRT )  n ( a − bcRT ) + 4 RT ( c − 1) ba 2

2 RT ( c − 1)

For c = 0.95, the van der Waals pressure being lower than the ideal gas pressure, we get volumes of 4.16  10−5 m3 and 2.16  10−4 m3 . For c = 1.05, the van der Waals pressure being higher than the

ideal gas pressure, we get a volume of 3.46  10−5 m3 . Note that the pressures are equal for a volume of 3.72  10−5 m3 . 44. (a) We use the ideal gas law as applied to the air before it was put into the tank. 1.013  105 Pa )( 2.3m 3 ) ( PV = nRT → n = = 95.64 mol  96 mol (8.314 J mol K )( 293 K ) (b) Use the ideal gas law. PV = nRT → P=

nRT V

( 96 mol )(8.314 J mol K )( 293 K )

( 0.012 m ) 3

= 1.9488  107 Pa  1.9  107 Pa  190 atm

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Chapter 18

Kinetic Theory of Gases

( 0.1373 N m mol ) ( 96 mol ) ( 96 mol )(8.314 J mol K )( 293 K ) = − 2 3 3 −5 ( 0.012 m ) − ( 96 mol ) ( 3.72  10 m mol ) ( 0.012 m3 ) 4

= 1.8958  107 Pa  1.9  107 Pa

 Pideal − Pvan der   1.9488  107 Pa − 1.8958  107 Pa  Waals   100 =  (d) % error =    100 = 2.796%  3%  Pvan der  1.8958  107 Pa     Waals   45. We want the mean free path to be 1.0 m. Use Eq. 18–10b with the ideal gas law. N P kT 1 PV = NkT → ; lM = = = → 2 V kT 4 2r ( N V ) 4 2r 2 P P=

(1.38 10 J K ) ( 293 K ) = 0.010 Pa = 4 2 (1.5  10 m ) (1.0 m ) −23

kT 4 2r 2 l M

−10

46. The mean free path is given by Eq. 18–10b. Combine this with the ideal gas law to find the mean free path–pressure relationship. N P 1 kT kT = = → = PV = NkT → ; lM = P V kT 4 2r 2 ( N V ) 4 2r 2 P 4 2r 2 l M (a) P =

(b) P =

(1.38 10 J K ) ( 293 K ) = 3.4 10 Pa = 4 2 (1.5  10 m ) ( 0.30 m ) (1.38 10 J K ) ( 293 K ) = 3 10 Pa  300 atm = 4 2 (1.5  10 m ) ( 3  10 m ) −23

−2

4 2r l M 2

−10

−23

4 2r l M 2

−10

At this pressure, the ideal gas equation is probably not a good model for the gas, and so the answer to part (b) should be considered very approximate. 47. The mean free path is given by Eq. 18–10b. Combine this with the ideal gas law to find the mean free path–diameter relationship. PV = NkT →

(a) d =

(b) d =

kT 2 l M P kT 2 l M P

N V

P kT

; lM =

1 4 2r ( N V ) 2

kT 4 2 ( d ) P 1 2

→ d=

kT 2 l M P

(1.38 10 J K ) ( 273 K ) = 3.9 10 m 2 ( 5.6  10 m )(1.013 10 Pa ) (1.38 10 J K ) ( 273 K ) = 1.8 10 m 2 ( 25  10 m )(1.013  10 Pa ) −23

−10

−8

−23

−10

−8

635

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

48. First, we compare the rms speed of the hydrogen to the rms speed of the air, by Eq. 18–5.

( vrms )H = ( vrms )air 2

3kT mH

kT mair

mair mH

29 2

= 3.8

Since the hydrogen is moving about 4 times faster than the air, we will use a stationary target approximation, and so NOT make the adjustment in the text for relative velocity. We also assume that the inter-molecular distance for a collision would be the sum of the radii of the hydrogen and air molecules. So instead of the factor 2r in the derivation, we use the sum of the two radii. Another way to say this is that instead of the factor r in the final result, we use the average of the two radii. The size of the air molecules are given in Example 18–8, having a diameter of 0.30 nm or a radius of 0.15 nm. Also, based on Problem 75, we assume that the radius of the hydrogen molecule is the same as the diameter of the hydrogen atom, 0.1 nm. We use these assumptions to calculate the mean free path, similar to Eq. 18–10a. We also use the ideal gas law. N P PV = NkT → ; = V kT lM =

 rH + rair  4  2  (N V )  2  2

(1.38 10 J K ) ( 288 K )  (1.0 + 1.5 )  10 m  (1.013  10 Pa ) −23

 ( rH + rair ) P 2

−10

= 1.998  10 −7 m  2  10 −7 m

49. The average time between collisions can be approximated as the mean free path divided by the mean speed. Combine the ideal gas law with Eq. 18–10b for the mean free path, and Eq. 18–7b for the mean speed. N P kT 1 1 PV = NkT → ; lM = ; = = = 2 2 V kT 4 2r ( N V ) 4 2r ( P kT ) 4 2r 2 P 8kT v=

8kT

m

→ f max =

v lM

m kT

= 16 Pr 2

 mkT

4 2r P Only one significant figure was given for the mean free path, so only one significant figure should be in the answer. The diameter of air molecules was given in Example 18–8 as 3.0  10−10 m, and so the radius is half of that. In Problem 16, the atomic mass of air was assumed to be 29 u. 2

f = 16 Pr 2

(

 mkT

)(

= 16 1.013  105 Pa 1.5  10−10 m

)

(

29 1.66  10

−27

)(



)

kg 1.38  10−23 J K ( 273 K )

= 4.8  109 Hz  5  109 collisions per second 50. Use the ideal gas law to evaluate the mean free path. Then compare the mean free path to the dimensions of the box in order to estimate the collision ratio. The size of air molecules is given in Example 18–8. The number of collisions per second is the reciprocal of the average time between collisions. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

636

Chapter 18

Kinetic Theory of Gases

PV = NkT →

N V

P kT

(1.38 10 J K ) ( 273 K ) = l = =  1.013  10 Pa  4 2r ( N V ) 4 2r P 4 2 (1.5  10 m ) (1  10 atm )   1atm 1

−23

−10

−14





= 9.30  10 m 6

lM tmolecular collisions

l box twall

N wall

tmolecular

collisions

→

N molecular

collisions

t wall

collisions

lM l box

9.30  106 m 1.40 m

= 6.64  106

collisions

The wall collisions are more than 6.6 million times more frequent than the inter-molecular collisions.

(

)

So the particles make about 1 6.64  106 = 1.5  10−7 of a collision with each other for each collision with a wall. Or, they make 6.6 million collisions with the wall for every 1 collision with another molecule. 51. If only 2% of the electrons are to have a collision in 3.5 cm or less, we then approximate that 2% of the electrons will have a collision in every 3.5 cm of travel. Thus, in this approximation, 4% of the electrons would have a collision in 7.0 cm of travel, 6% of the electrons would have a collision in 10.5 cm of travel, and so on. Thus we estimate that 50% of the electrons should have a collision in a length of 25 times 3.5 cm, or 87.5 cm. So we assume the mean free path to be 87.5 cm, or 0.875 m. We also assume that the electrons are moving much faster than the air molecules, so that we model the air molecules as stationary, and use Eq. 18–10a (instead of Eq. 18–10b). Finally, a collision will occur if an electron comes within a distance of r from a gas molecule (the radius of the gas molecule), not 2r as in the derivation in Section 18–7, because the electrons are essentially points, with 0 radius. Combine this with the ideal gas law. We assume room temperature. 1 N P kT ; lM = 2 PV = NkT → = = 2 → V kT r (N V ) r P P=

 r2l M

(1.38 10 J K ) ( 300 K ) = 0.0669 Pa  6.7  10 =  (1.5  10 m ) ( 0.875 m ) −23

−10

−2

 6.6  10−7 atm

52. (a) The average time between collisions can be approximated as the mean free path divided by the mean speed. The highest frequency for a sound wave is the inverse of this average collision time. Combine the ideal gas law with Eq. 18–10b for the mean free path, and Eq. 18–7b for the mean speed. N P 1 1 kT PV = NkT → ; lM = ; = = = 2 2 V kT 4 2r ( N V ) 4 2r ( P kT ) 4 2r 2 P 8kT v=

8kT

m

→ f max =

v lM

m kT

= 16 Pr 2

 mkT

4 2r P 2

637

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) We have estimated the molecular mass of air to be 29 u in Problem 16, and the average molecular diameter is given as 3  10−10 m in Example 18–8. f max = 16 Pr 2

 mkT

(

)(

= 16 1.013  105 Pa 1.5  10−10 m

)

(

29 1.66  10

−27

)(



)

kg 1.38  10 −23 J K ( 293 K )

= 4.6  109 Hz

This frequency is about

4.6  109 Hz 2.0  104 Hz

= 2.3  105 larger than the highest frequency in the human

audio range. 53. We use the equation derived in Example 18–9. t=

C ( x ) C

(1.0 m ) 2

(

2 4  10 −5 m 2 s

)

= 12, 500 s  3.5 h

Because this time is so long, we see that convection is much more important than diffusion. 54. From Example 18–9, we have an expression for the time to diffuse a given distance. Divide the distance by the time to get the average speed. t=

C ( x ) C

vdiffuse =

x t

(1.00 + 0.50 ) mol m3 ( 25  10 m ) = = 0.9868 s  0.99 s (1.00 − 0.50 ) mol m3 ( 95  10−11 m 2 s ) −6

1 2

25  10−6 m 0.9868 s

= 2.5  10−5 m s

The rms thermal speed is given by Eq. 18–5, vrms = 3kT m .

v rms = 3kT m =

vdiffuse vrms

(

)

3 1.38  10−23 J K ( 293 K )

2.5  10−5 m s 312.1m s

(

75 1.66  10

= 8.0  10−8

−27

)

= 312.1 m s  310 m s

( about 7 orders of magnitude difference )

55. (a) Use the ideal gas law to find the concentration of the oxygen. We assume that the air pressure is 1.00 atm, and so the pressure caused by the oxygen is 0.21 atm. PV = nRT →

( 0.21atm ) (1.013  105 Pa atm ) = = = 8.732 mol m 3  8.7 mol m 3 V RT 8.315 J mol K 293 K ( )( ) n

Half of this is 4.366 mol m 3 . (b) Use Eq. 18–11 (in difference form rather than differential form) to calculate the diffusion rate. 3 3 C1 − C2 2 2  8.732 mol m − 4.366 mol m  −5 −9 J = DA = 1 10 m s 2  10 m   2  10−3 m x  

(

)(

)

= 4.366  10−11 mol s  4  10−11 mol s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

638

Chapter 18

Kinetic Theory of Gases

(8.732 mol m + 4.366 mol m ) ( 2 10 m ) = 0.6 s = D (8.732 mol m − 4.366 mol m ) 110 m s

C ( x )

C

1 2

−3

−5

56. We use the ideal gas law to find the length.

1/3  (1  106 )(1.38  10−23 J K ) ( 273 K )   NkT  −7 3 PV = P l = NkT → l =   = 3  10 m  = 5  1.013 10 Pa  P  ( )   1/3

57. The rms speed is given by Eq. 18–5, vrms = 3kT m . Hydrogen atoms have a mass of 1 atomic mass unit. v rms =

3kT m

(

)

3 1.38  10 −23 J K ( 2.7 K )

(

1 1.66  10

−27

)

= 260 m s

The pressure is found from the ideal gas law, PV = NkT . PV = NkT → P =

NkT V

(1) (1.38  10−23 J K ) ( 2.7 K )  1 10 m   3  1cm  −6

 1 atm   5  1.01  10 Pa 

= 3.726  10 −17 Pa 

1 cm3 

= 3.689  10−22 atm  3.7  10 −22 atm

58. We assume that each molecule will have an average kinetic energy of 23 kT . Find the total number of molecules from the mass of the bacterium. N = N H O + N other 2

 )  1.66 1u10 kg   1molecule  18 u

(

= 0.70 2.0  10 −15 kg 

−27







 )  1.66 1u10 kg   1molecule  10 u

(

+ 0.30 2.0  10−15 kg 

−27











  1molecule  + 0.30  )  1.66 1u10 kg  0.70  1molecule   18 u 10 u

(

= 2.0  10 −15 kg 

−27













= 4.69  10 molecules 10

(

) (

)

K = N ( 23 kT ) = 4.69  1010 molecules 23 1.38  10 −23 J K ( 310 K ) = 3  10 −10 J

59. It is stated in the text that the relationship vrms = 3kT m is applicable to molecules within living cells at body temperature (37oC). The rms speed is given by Eq. 18–5. (a) v rms =

(b) v rms =

3kT m 3kT m

(

3 1.38  10

−23

)

J K ( 310 K )

( 89 u ) (1.66  10−27 kg u )

(

3 1.38  10

−23

= 294.7 m s  290 m s

)

J K ( 310 K )

(8.5 10 u )(1.66 10 kg u ) 4

−27

= 9.537 m s  9.5 m s

639

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

60. Following the development of the kinetic molecular theory in the textbook, the tennis balls hitting the trash can lid are similar to the particles colliding with the walls of a container causing pressure. Quoting from the text, “the average force …– averaged over many collisions–will be equal to the momentum change during one collision … divided by the time between collisions.” That average force must be the weight of the trash can lid in order to suspend it. Also, the fact that the collisions are elastic means that the change in momentum of one of the balls is twice its original momentum. 2mball vball p 2mball v ball Favg = M lid g ; Favg = = → t = M lid g t t The above expression is “seconds per ball,” so its reciprocal will be “balls per second.” ( 0.40 kg ) ( 9.80 m s 2 ) M lid g 1 balls s = = = = 2.2 balls s t 2mball vball 2 ( 0.060 kg )(15 m s ) 61. The rms speed is given by Eq. 18–5, vrms = 3kT m . Set the escape velocity to vrms and solve for T. (a) For oxygen molecules:

2 mvrms

(

)(

2 (15.9994 ) 1.66  10−27 kg 1.12  10 4 m s

(

3 1.38  10−23 J K

)

) = 1.6110 K 2

(b) For helium atoms:

2 mvrms

( 4.002602 ) (1.66  10−27 kg )(1.12  104 m s )

(

3 1.38  10

−23

J K

)

= 2.01 104 K

(c) Because the “escape temperature” is so high for oxygen, very few oxygen molecules ever escape the atmosphere. But helium, with one-eighth the mass, can escape at a much lower temperature. While the temperature of the Earth is not close to 2.0  104 K today, during the Earth’s formation its temperature was possibly much hotter–presumably hot enough that helium was able to escape the atmosphere. 62. Find the volume “allotted” per molecule in the ideal gas law for a room at 1 atm and 23C, and compare this to the volume of an actual molecule, modeled by a cubical volume. V kT d3 3 4 1 ; Vmolecule = 3  ( 2 d ) = PV = NkT → Vallotted = = 6 N P

d3 3 5 −9 Vmolecule P d 3 (1.013  10 Pa )  ( 0.3  10 m ) 6 = = = = 3.45  10−4 = ( 3.45  10−2 ) % −23 kT Vallotted 6kT 6 (1.38  10 J K ) ( 300 K ) P  0.03% 63. (a) The volume of each gas is half of the tank volume. Use the ideal gas law, with an absolute pressure of 15 atm, to find the number of molecules. −3 5 3 1 PV 15 1.013  10 Pa 2 3.1 10 m PV = NkT → N = = = 5.825  1023 molecules −23 kT 1.38  10 J K ( 293 K )

(

) ( )

(

)

 5.8  1023 molecules

640

Chapter 18

Kinetic Theory of Gases

Both gases have the same number of molecules. The identity of the gas does not enter into the ideal gas law. (b) The average kinetic energy of a molecule is 23 kT . Since both gases are the same temperature, the ratio of the average kinetic energies is 1: 1 . 2 (c) The average kinetic energy of a molecule is given by 12 mvrms . Use this to find the ratio of the rms speeds.

( vrms )He

( mv ) = ( mv ) = kT → ( v ) = m 1 2

2 rms

1 2

2 rms

3 2

rms

32 4

= 8 = 2.8

64. Assume that the water is an ideal gas, and that the temperature is constant. From Table 18–2, saturated vapor pressure at 85oC is midway between the given saturated vapor pressures for 80oC and 90oC, which is 5.87  104 Pa. To have a relative humidity of 10%, the vapor pressure will be 5.87  103 Pa. Use the ideal gas law to calculate the amount of water. PV = nRT → n =

PV RT

( 5.87  10 Pa )( 7.8 m ) 3

( 8.314 J mol K )( 273 + 85) K

= 15.38 moles

 18  10−3 kg   = 0.2769 kg  0.28 kg  1 mole 

15.38 moles 

65. (a) We calculate the volume per molecule from the ideal gas law, and assume the molecular volume is spherical. V kT 4 3 PV = NkT → = = 3r → N P 1/3  3 (1.38  10 −23 J K ) ( 273 K )  3kT   −9 rinter=  = 2.07  10 m  =  5 P 4 4 1.01 10 Pa    molecular   ( )   1/3

The intermolecular distance would be twice this “radius,” so about 4  10−9 m. This is about 14 times larger than the molecular diameter.

d intermolecular

d actual



4.14  10−9 m 3  10−10 m

d inter= 13.8 →

molecular

d actual

molecule

 14

molecule

(b) Now we scale the molecular diameter up to 4 cm. The intermolecular distance would be 13.8 times that, which is about 55 cm. (c) We replace the pressure in part (a) with 3 atm. 1/3  3 (1.38  10−23 J K ) ( 273 K )  3kT   −9 = rinter = 1.44  10 m  =  5 molecular  4 P   4 ( 3 atm ) (1.01  10 Pa/atm )  1/3

d inter molecular

d actual



2(1.44)  10−9 m −10

3  10 m

= 9.58 →

d intermolecular

d actual

molecule

 9.6

molecule

We see that the intermolecular distance has decreased with the increased pressure, from about 14 times larger to only about 10 times larger. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(d) We are to calculate the volume occupied by the molecules themselves, compared to the total volume of the gas. The total volume is the volume of the molecules plus the volume related to the inter-molecular distance. Since we don’t know the actual number of gas particles, we use the volume of one gas particle and the volume associated with the inter-molecular distance. 3

      12 d actual  Vactual  d actual  molecule  molecule   molecule  = = 3 3 3 3 Vinter- + Vactual 1  4 1      4   2 d inter-  + 3   2 d actual  molecular molecule 3  d inter-  +  d actual   molecule   molecule   molecule   molecule  4 3

1 3

  d actual  + 1  dinter molecule molecule 

 1  −2  100 = 3.8  10 % 3 13.8 1 +  

P = 1 atm: % = 

1    100 = 0.11% 3  9.58 + 1 

P = 3 atm: % = 

66. The rms speed is given by Eq. 18–5, vrms = 3kT m . From Appendix G, here are the masses in atomic mass units: 235 U : 235.044 u, 238 U : 238.051u, F : 18.998 u.

( vrms ) UF = ( vrms ) UF 235

235

3kT m UF 235

3kT m UF 238

m UF 238

m UF 235

238.051 + 6 (18.998 )

235.044 + 6 (18.998 )

352.039 349.032

= 1.00430

67. The gravitational potential energy is given by U = mgh , and the average kinetic energy is 2 K = 12 mvrms = 32 kT . We find the ratio of potential energy to kinetic energy. The molecular mass of oxygen molecules is 32 u. 2 −27 U mgh ( 32.0) (1.66  10 kg )( 9.80 m s ) (1.50 m ) = = = 1.27  10−4 −23 3 K 23 kT 1.38 10 J K 273K 23K  + ( ) ) 2(

Yes, it is reasonable to neglect the gravitational potential energy. 68. Assume that the water vapor behaves like an ideal gas. At 20oC, the saturated vapor pressure is 2.33  103 Pa. Using the ideal gas law, find the number of moles of water in the air at both 95% and 45%. Subtract those mole amounts to find the amount of water that must be removed. PV PV = nRT → n = → RT n1 − n2 =

V RT

( P1 − P2 ) =

(105 m ) ( 2.4 m ) 2

( 8.314 J mol k )( 293 K )

( 2.33  10 Pa ) ( 0.95 − 0.45) = 120.5 mol 3

 18  10−3 kg   = 2.169 kg  2.2 kg  1 mol 

120.5 mol 

642

Chapter 18

Kinetic Theory of Gases

69. The temperature can be found from the rms speed by Eq. 18–5, vrms = 3kT m . The molecular mass of nitrogen molecules is 28. vrms = 3kT m → 2

  1m s   ( 28 ) (1.66  10 kg ) ( 4.2  104 km h )   2 mvrms  3.6 km h   = 1.5  105 K  T= = −23 −27

(

3 1.38  10

J K

)

70. (a) At a temperature of 30oC, the saturated vapor pressure, from Table 18–2, is 4240 Pa. If the relative humidity is 75%, then the water vapor pressure is 75% of the saturated vapor pressure. 0.75 ( 4240 Pa ) = 3180 Pa  3200 Pa (b) At a temperature of 5oC, the saturated vapor pressure, from Table 18–2, is 872 Pa. If the relative humidity is 75%, then the water vapor pressure is 75% of the saturated vapor pressure. 0.75 ( 872 Pa ) = 654 Pa  650 Pa 71. We assume that the energy required to evaporate the water is the kinetic energy of the evaporating molecules. The rms speed is given by Eq. 18–5. 2Eevap 2 ( 2450 J ) 2 Eevap = 12 mevap vevap → vevap = = = 2210 m s mevap (1.00 10−3 kg ) v rms =

3kT m

(

)

3 1.38  10 −23 J K ( 293 K )

(

18.0 1.66  10

−27

)

= 637 m s →

vevap v rms

2210 m s 637 m s

= 3.47

72. At 30.0C, the saturated vapor pressure as found in Table 18–2 is 4240 Pa. We can find the partial pressure of the water vapor by using the equation given immediately before Example 18–6. Ppartial Rel. Hum.  100 → Ppartial = Psaturated = ( 4240 Pa )( 0.65) = 2756 Pa Rel. Hum. = Psaturated 100 The dew point is the temperature at which 2756 Pa is the saturated vapor pressure. We use linear interpolation with Table 18–2. Between 20oC and 25oC, the temperature change per Pa is as follows: ( 25 − 20 ) C = 5.952  10−3 C Pa , 3 ( 3.17 − 2.33) 10 Pa Thus, the temperature corresponding to 2756 Pa is 20C + ( 2756 Pa − 2330 Pa ) ( 5.952  10−3 C Pa ) = 22.54C  23C 73. (a) The mean speed is given by Eq. 18–7b. The atomic weight of cesium is 133. v=

8kT

m

(

)

8 1.38  10 −23 J K ( 673 K )

 (133) (1.66  10−27 kg )

= 327.3 m s  330 m s

(b) The collision frequency is the mean speed divided by the mean free path as given by Eq. 18–10b. We also use the ideal gas law. N P PV = NkT → = V kT

643

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lM =

1 4 2r 2 ( N V )

Instructor Solutions Manual

(1.38 10 J K ) ( 673 K )  133 Pa  4 2 (1.65  10 m ) (17 mm-Hg )   1mm-Hg −23

kT 4 2r 2 P

−10





−6

= 8.49  10 m f =

327.3 m s

= 3.855  107 collisions s  3.9  107 collisions s l M 8.49  10−6 m (c) The total number of collisions per second in the gas is the number of collisions per second for a single atom times half the number of atoms in the gas, because each collision involves 2 of the gas atoms. PV f total = 12 Nfsingle = fsingle 2kT

3   133 Pa  3  1m 75 cm ( )   6 3   1mm-Hg   10 cm  3.855  107 collisions s

(17 mm-Hg ) 

(

)

2 1.38  10 −23 J K ( 673 K )

)

= 3.519  10 26 collisions s  3.5  10 26 collisions s

74. We combine the ideal gas law with Eq. 18–10b for the mean free path. From Problem 45, we se that the diameter of the average air molecule is 3  10−10 m. Since air is mostly nitrogen molecules, this is a good approximation for the size of a nitrogen molecule. N P = PV = NkT → ; V kT lM =

1 4 2r ( N V ) 2

1 4 2r ( P kT ) 2

kT 4 2r 2 P

(1.38 10 J K ) ( 300 K ) l = = = 1.4  10 m 4 2r P 4 2 (1.5  10 m ) 7.5 (1.013  10 Pa ) −23

−8

−10

Note that this is about 100 times the radius of the molecules. 75. The mean free path is given by Eq. 18–10b. 1 1 lM = = = 2  1013 m 2 2 3 6 3 3 −11 4 2r ( N V ) 4 2 5.0  10 m 1atom cm 10 cm m

(

)(

)

Note that the distance from the Sun to the Earth is about 1.5  1011 m, and the radius of the solar system is on the order of 1012 m. 76. First we find the pressure from the ideal gas equation. nRT ( 8.50 mol )(8.314 J mol k )( 300 K ) P= = = 9.6367  104 Pa  9.64  104 Pa 3 0.220 m V Now find the pressure from the van der Waals equation.

644

Chapter 18

Kinetic Theory of Gases

−

(V n ) − b (V n )

= 2

nRT V − nb

−

an 2 V2

( 0.36 N m mol ) (8.50 mol ) ( 8.50 mol )( 8.314 J mol k )( 300 K ) = − 2 3 3 −5 ( 0.220 m ) − (8.50 mol ) ( 4.5  10 m mol ) ( 0.220 m3 ) 4

= 9.5997  104 Pa  9.60  104 Pa

 Pideal − Pvan der   9.6367  104 Pa − 9.5997  104 Pa  Waals  % error =   100 =    100  0.39%  Pvan der  9.5997  104 Pa     Waals   77. (a) The Van der Waals equation of state is given by Eq. 18–9, P =

−

(V n ) − b (V n )2

. At the

critical point, both the first and second derivatives of P with respect to V are 0. Use those conditions to find the critical volume, and then evaluate the critical temperature and critical pressure. RT a nRT an 2 P= − = − (V n ) − b (V n ) 2 V − nb V 2 dP dV

=−

d 2P dV 2

nRT

(V − nb ) 2nRT

(V − nb )3

+ 2 −

2an 2 V3

6an 2 V4

;

= 0 → Tcrit =

dV d 2P

;

= 0 → Tcrit =

dV 2

2an (Vcrit − nb )

3 RVcrit

3an (Vcrit − nb )

4 RVcrit

Set the two expressions for the critical temperature equal to each other, and solve for the critical volume. Then use that expression to find the critical temperature, and finally the critical pressure.

2an (Vcrit − nb )

3 RVcrit

Tcrit = Pcrit =

3an (Vcrit − nb )

3 crit

RV Vcrit − nb

−

→ Vcrit = 3nb

RVcrit4

2an (Vcrit − nb ) nRTcrit

an 2 2 crit

= =

2an ( 3nb − nb ) R ( 3nb )

8a 27bR

a 27b 2

(b) To evaluate the constants, use the ratios

Tcrit2 Pcrit

and

Tcrit Pcrit

 8a  2 Tcrit  27bR  64a = = → 2 Pcrit

27 R

27b 2

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J  2  27  8.314 304 K ) ( 2 2  27 R Tcrit N m4 mol K  a= =  = 0.365

(

64 ( 72.8 ) 1.013  105 Pa

64 Pcrit

)

mol 2

 8.314 J  304 K ) ( Tcrit 27bR 8b RTcrit  mol K  = = → b= = = 4.28  10 −5 m 3 mol 5 8a

Pcrit

27b

8 Pcrit

(

8 ( 72.8 ) 1.013  10 Pa

)

78. For each temperature, a graph of pressure vs. volume was plotted, based on Eq. 18–9, RT a P= . From the graphs, it would appear that the critical temperature for oxygen − (V n ) − b (V n )2 is approximately 150 K. We note that, for very low temperatures, the van der Waals equations gives some non-physical (negative pressure) results as well. The freezing temperature of oxygen is about 54 K. The accepted value of the critical temperature of oxygen is about 155 K.

646

CHAPTER 19: Heat and the First Law of Thermodynamics Responses to Questions 1.

When a jar of orange juice is vigorously shaken, the work done on it goes into heating the juice (increasing the kinetic energy of the molecules), mixing the components of the juice (liquid and pulp), and dissolving air in the juice (froth).

When a hot object warms a cooler object, energy is transferred from the hot object to the cold object. Temperature does NOT flow. The temperature changes of the two objects are not necessarily equal in magnitude. Under certain circumstances, they can be equal in magnitude, however. In an ideal case (where no heat energy is lost to the surroundings), the amount of heat lost by the warmer object is the same as the amount of heat gained by the cooler object.

(a) Internal energy depends on both the number of molecules of the substance and the temperature of the substance. It can also depend on the material itself. Heat will flow naturally from the object with the higher temperature to the object with the lower temperature. The object with the higher temperature may or may not be the object with the higher internal energy. (b) The two objects may consist of one with a higher temperature and smaller number of molecules, and the other with a lower temperature and a larger number of molecules. In that case it is possible for both objects to have the same internal energy, but heat will still flow from the object with the higher temperature to the one with the lower temperature.

Because the specific heat of water is quite large, it can contain a relatively large amount of thermal energy per unit mass with a relatively small increase in temperature. Since the water is a liquid, it is relatively easy to transport from one location to another, and so large quantities of energy can be moved from one place to another with relative simplicity by water. Also, the water will give off a large amount of energy as it cools.

The water will coat the plants, and so the water, not the plant, is in contact with the cold air. Thus as the air cools, the water cools before the plant does–the water insulates the plant. As the water cools, it releases energy, and raises the temperature of its surroundings, which includes the plant. Particularly if the water freezes, relatively large amounts of heat are released due to the relatively large heat of fusion for water.

The water on the cloth jacket will evaporate. Evaporation is a cooling process since energy is required to change the liquid water to vapor. As water evaporates from the moist cloth, the canteen surface is cooled. If, for instance, radiant energy from the sun falls on the canteen, the energy will evaporate the water from the cloth cover instead of heating the water inside the canteen.

Steam at 100oC contains more thermal energy than water at 100oC. The difference is due to the latent heat of vaporization, which for water is quite high. As the steam touches the skin and condenses, a large amount of energy is released, causing more severe burns. And the condensed water is still at 100oC, and so more burning can occur as that water cools.

Evaporation involves water molecules escaping the intermolecular bonds that hold the water together in the liquid state. It takes energy for the molecules to break those bonds (to overcome the bonding forces). This energy is the latent heat of vaporization. The most energetic molecules (those having the highest speed) are the ones that have the most energy (from their kinetic energy) to be able to overcome the bonding forces. The slower moving molecules remain, lowering the average kinetic energy and thus lowering the internal energy and temperature of the liquid.

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The potatoes will not cook faster if the water is boiling faster. The rate at which the potatoes cook depends on the temperature at which they are cooking, and the boiling water is the same temperature whether it is boiling fast or slow.

10. Even though the temperature is high in the upper atmosphere, which means that the gas particles are moving very fast, the density of gas particles is very low. There would be relatively very few collisions of high-temperature gas particles with the animal, and so very little warming of the animal. Instead, the animal would radiate heat to the rarified atmosphere. The emissivity of the animal is much greater than that of the rarified atmosphere, and so the animal will lose much more energy by radiation than it can gain from the atmosphere. 11. When a gas is compressed, work must be done on it by some outside force, such as a person pushing a piston. This work becomes the increase in internal energy of the gas, and if no gas is allowed to escape, an increase in internal energy results in faster average molecular speeds and a higher temperature. When a gas expands, it does work on the piston. If the gas is insulated so that no heat enters from the outside, then the energy for this work comes from the internal energy of the gas. A decrease in internal energy with no change in the number of molecules translates into a decrease in average molecular speed and therefore temperature. We can also analyze this by conservation of momentum. When the massive piston head, moving downward with a relatively low speed, collides elastically with a light molecule moving upwards, the molecule rebounds with an increased speed. This increased speed is manifest as a larger temperature. When the piston head is moving upwards the opposite happens (the molecule loses speed) and the lower speed results in a lower temperature. (You can model this with the equations in Problem 34 of Chapter 9.) 12. Snow consists of crystals with tiny air pockets in between the flakes. Air is a good insulator, so when the Arctic explorers covered themselves with snow they were using its low thermal conductivity to keep heat from leaving their bodies. (In a similar fashion, down comforters keep you warm because of all the air trapped in between the feathers.) Snow would also protect the explorers from the very cold wind and thus prevent heat loss by convection. 13. Wet sand has been cooled by conduction as the water washes over it (ocean water is usually cooler than the beach) and continues to be cooled by evaporation, and so will be cooler than dry sand. Wet sand will have a higher specific heat than dry sand since water has a high specific heat, and so the same amount of thermal energy from the Sun will cause a lower temperature increase in wet sand than in dry sand. Wet sand will also feel cooler to your skin because of the thermal conductivity of water, in much the same way as a cold piece of metal “feels” colder than a piece of wood at the same temperature. 14. A hot-air furnace heats primarily by air convection. A return path (often called a “cold air return”) is necessary for the convective currents to be able to completely circulate. If the flow of air is blocked, then the convective currents and the heating process will be interrupted. Heating will be less efficient and less uniform if the convective currents are prevented from circulating. 15. An adiabatic compression is one that takes place with no exchange of heat with the surroundings. During the compression, work is done on the gas. Since no heat leaves the gas, then the work results in an increase in the gas’s internal energy by the same amount, and therefore an increase in its temperature. ΔEint = Q – W, so if Q = 0 then ΔEint = –W.

648

Chapter 19

Heat and the First Law of Thermodynamics

16. ΔEint is proportional to the change in temperature. The change in the internal energy is zero for the isothermal process, largest for the isobaric process, and least (negative) for the adiabatic process. The work done, W, is the area under the curve and is greatest for the isobaric process and least for the adiabatic process. From the first law of thermodynamics, Q is the sum of ΔEint and W and is zero for the adiabatic process and maximum for the isobaric process.

Pressure

Isobaric

Isothermal Adiabatic Volume V1

V2 = 2V1

17. A ceiling fan makes more of a “breeze” when it is set to blow the air down (usually called the “forward” direction by fan manufacturers). This is the setting for the summer, when the breeze will feel cooling since it accelerates evaporation from the skin. In the winter, the fan should be set to pull air up. This forces the warmer air at the top of the room to move out towards the walls and down. The relocation of warmer air keeps the room feeling warmer, and there is less “breeze” effect on the occupants of the room. 18. When the garment is fluffed up, it will have the most air trapped in its structure. The air has a low thermal conductivity, and the more the garment can be “fluffed,” the more air it will trap, making it a better insulator. The “loft” value is similar to the R value of insulation, since the thicker the insulation, the higher the R value. The rate of thermal conduction is inversely proportional to the thickness of the conductor, so a thick conductor (high loft value) means a lower thermal conduction rate, and so a lower rate of losing body heat. 19. For all mechanisms of cooling, the rate of heat transfer from the hot object to the cold one is dependent on surface area. The heat sink with fins provides much more surface area than just a solid piece of metal, and so there is more cooling of the microprocessor chip. A major mechanism for cooling the heat sink is that of convection. More air is in contact with the finned heat sink than would be in contact with a solid piece of metal. There is often a fan circulating air around that heat sink as well, so that heated air can continually be replaced with cool air to promote more cooling. 20. When there is a temperature difference in air, convection currents arise. Since the temperature of the land rises more rapidly than that of the water, due to the large specific heat of water, the air above the land will be warmer than the air above the water. The warm air above the land will rise, and that rising warm air will be replaced by cooler air from over the body of water. The result is a breeze from the water towards the land–a “sea breeze.” 21. The Earth cools primarily by radiation. The clouds act as “insulation” in that they absorb energy from the radiating Earth, and reradiate some of it back to the Earth, reducing the net amount of radiant energy loss. If there are no clouds, there is no “re-radiation” from the clouds. 22. A thermometer in the direct sunlight would gain thermal energy (and thus show a higher temperature) due to receiving radiation directly from the Sun. The emissivity of air is small, and so it does not gain as much energy from the Sun as the mercury and glass do. The thermometer is to reach its equilibrium temperature by heat transfer with the air, in order to measure the air temperature. 23. Premature babies have underdeveloped skin, and they can lose a lot of moisture through their skin by evaporation. For a baby in a very warm environment, like an incubator at 37oC, there will be a large evaporative effect. A significant increase in evaporation occurs at incubator temperatures, and that evaporation of moisture from the baby will cool the baby dramatically. Thus an incubator must have © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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not only a high temperature but also a high humidity. Other factors might include radiative energy loss, blood vessels being close to the skin surface so there is less insulation than in a more mature baby, and low food consumption to replace lost energy. Also, the smaller the child, the larger the surface area-to-volume ratio, which leads to more relatively more heat being lost through the body surface area than for an older baby. 24. Air is a poorer conductor of heat than water by roughly a factor of 20, and so the rate of heat loss from your body to the air is roughly 20 times less than the rate of heat loss from your body to the water. Thus you lose heat quickly in the water, and feel cold. Another contributing factor is that water has a high heat capacity, and so as heat leaves your body and enters the water, the temperature rise for the water close to your body is small. Air has a smaller heat capacity, and so the temperature rise for the air close to your body is larger. This reduces the temperature difference between your body and the air, which reduces the rate of heat loss to the air as well. 25. The south-facing windows will allow radiant heat from the Sun to enter the room to contribute to the heating, so that less heat will need to be provided internally. The Sun doesn’t directly shine in windows on the north side of a house in the Northern hemisphere. 26. We assume that the temperature in the house is higher than that under the house. Thus heat will flow through the floor out of the house. If the house sits directly on the ground or on concrete, the heat flow will warm the ground or concrete. Dirt and concrete are relatively poor conductors of heat, and so the thermal energy that goes into them will stay for a relatively long time, allowing their temperature to rise and thus reducing the heat loss through the floor. If the floor is over a crawlspace, then the thermal energy from the floor will be heating air instead of dirt or concrete. If that warmed air gets moved away by wind currents or by convection and replaced with colder air, then the temperature difference between the inside and outside will stay relatively high, and more energy will leave through the floor, making the inside of the house cooler. 27. A thermos bottle is designed to minimize heat transfer between the liquid contents and the outside air, even when the temperature difference is large. Heat transfer by radiation is minimized by the silvered lining. Shiny surfaces have very low emissivity, ε, and thus the net rate of energy flow by radiation between the contents of the thermos and the outside air will be small. Heat transfer by conduction and convection will be minimized by the vacuum between the inner and outer walls of the thermos, since both these methods require a medium to transport heat. 28. (a) (1) Cooling through the glass panes is cooling by conduction and radiation. (2) Cooling through the frame is cooling by conduction. (3) Ventilation around the edges is cooling by convection. (b) Heavy curtains can reduce all three heat losses. The curtains will prevent air circulation around the edges of the windows, thus reducing the convection cooling. The curtains are more opaque than the glass, preventing the electromagnetic waves responsible for radiation heat transfer from reaching the glass. And the curtains provide another layer of insulation between the outdoors and the warm interior of the room, lowering the rate of conduction. 29. The specific heat of water is large, and so the water is able to absorb a lot of heat energy with just a small temperature change. This helps to keep the temperature of the cup lower, preventing it from burning. Without the water, the cup’s temperature would increase quickly and then it would burn. 30. The air just outside the window will be somewhat warmed by conduction of heat energy through the window to the air, reducing the temperature difference between the inside and outside, and thus reducing the rate of heat loss. On a windy day, convection removes that warmer air from near the outside surface of the window. This keeps the outside surface of the window cold, and so increases © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

650

Chapter 19

Heat and the First Law of Thermodynamics

the rate of heat conduction through the window. This makes the window feel colder than on a day with no wind, even if the outside temperature is the same. 31. When the sun reaches the slope of the mountain early in the day, the ground is warmed by radiation. The air above the ground is also warmed and rises by convection. This rising air will move up the slope. When the slope is in shadow, the air cools and the convection currents reverse. 32. The thermal conductivity of the wood is about 2000 times less than that of aluminum, for example. Thus it takes a long time for energy from the wood to flow into your hand. Your skin temperature rises very slowly due to contact with the wood compared to contact with the metal, and so the sensation of heating is much less. 33. Shiny surfaces have low values of ε, the emissivity. Thus, the net rate of heat flow from the person to the surroundings (outside the blanket) will be low, since most of the heat is reflected by blanket back to the person, and the person will stay warmer. The blanket will also prevent energy loss due to wind (convection), and will insulate the person, reducing heat transfer by conduction. 34. Cities situated on the ocean have less temperature extremes because the oceans are a heat reservoir. Due to ocean currents, the temperature of the ocean in a locale will be fairly constant during a season. In the winter, the ocean temperature remains above freezing. Thus, if the air and land near the ocean get colder than the oceans, the oceans will release thermal energy, moderating the temperature of the nearby region. Likewise, in the warm seasons, the ocean temperatures will be cooler than the surrounding land mass, which heats up more easily than the water. Then the oceans will absorb thermal energy from the surrounding areas, again moderating the temperature.

Solutions to MisConceptual Questions 1.

(c) A common misconception is that “cold” flows from the ice into the tea. When the ice is placed in the tea, the ice has less kinetic energy per molecule than the tea and so in molecular collisions between the tea and ice, energy transfers from the tea into the ice. This energy transfer cools the tea as it melts the ice and then heats up the ice. The transfer of energy from the warmer tea to the colder ice is called “heat.”

(c) Students frequently interpret having more ice with being colder. However, whenever ice and water are mixed together and are in thermal equilibrium they will be at the melting/freezing point of the water. Therefore the two containers will be at the same temperature.

(a) When two objects are in thermal equilibrium heat does not transfer between them. This occurs when the two objects are at the same temperature. Internal energy is an extrinsic property that depends upon the amount of the substance present. Therefore, two gases in thermal equilibrium with each other would have different internal energies if one consisted of one mole of gas and the other consisted of two moles of gas. Heat is a transfer of energy between objects that are not in thermal equilibrium. Heat is not a property of an object.

(a) A common misconception is that as heat is added to water the temperature will always rise. However, Fig. 19–5 shows that heat is added to water at its melting and boiling points without the temperature changing. At these temperatures the water is undergoing a phase change.

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(d) Radiation is emitted by all objects not at absolute zero. Very hot objects, such as the Sun, emit radiation in the visible spectrum so they appear to be glowing. If the temperature of an object is less than its surroundings, it has a net gain in energy as it absorbs more radiation than it emits. However, it is still emitting radiation. The amount of radiation emitted is independent of the object’s specific heat.

(c) The problem does not specify the initial temperatures of the ice and water. If the ice and water are both initially at 0C, then none of the ice will melt, since heat will not transfer between them. Alternatively, if the ice is at 0C and the water is at 100C, then the water can provide Q = mcT = ( 0.010 kg )( 4186 J kg C )(100 C ) = 4186 J of heat as it cools to 0C. The ice only needs Q = mL = ( 0.010 kg )( 333kJ kg ) = 3330 J to completely melt, and so in this case all the ice would melt. Since the water could provide more heat than needed, you need to know the initial temperatures of the ice and water to determine just how much of the ice would melt.

(d) As the two objects are in thermal contact, the heat given off by the hot object will equal the heat absorbed by the cold object. The objects have the same specific heat so the heat transfer is proportional to the product of the mass of each object and its change in temperature. The object with the smaller mass will then have the larger temperature change.

(d) The specific heat of an object is a measure of how much heat is required to change its temperature. Water has a high specific heat (much higher than air) so its temperature remains fairly constant even though the surrounding air may experience large temperature fluctuations.

(a) The two objects absorb the same amount of heat from the stove. From Eq. 19–2, given that the masses are the same, the object with the higher specific heat will experience the smaller temperature increase and will therefore be cooler.

10. (d) In an isothermal process the internal energy remains constant ( Eint = 0 ) . In an expansion the

gas does work on the surroundings (W  0 ) . Since the internal energy is constant and the work is positive, the first law of thermodynamics requires the heat absorbed also be positive ( Q  0 ) .

11. (d) Students may misunderstand the difference between isothermal (temperature remains constant and so Eint = 0 ) and adiabatic ( Q = 0 ) . In an isothermal process heat can be absorbed, as long as an equal amount of work is done, so statement (i) is not true. For an ideal gas the temperature is proportional to the internal energy of the gas, statements (ii) and (iii) are equivalent, and both true. 12. (b) As the gas expands adiabatically, no energy enters or leaves the cylinder in the form of heat. The energy given to the piston comes from elastic collisions between the gas particles and the piston surface. The gas particles do positive work on the piston because the force they put on the piston is in the direction of motion of the piston, and so the piston gains energy. Thus the gas particles lose energy–their internal energy (kinetic energy of translation) decreases.

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13. (c) A common misconception is that the work done in moving an object between two states is independent of the path followed. In the graph shown, the work done in going from point A to B to C by the isobaric and isovolumetric processes is equal to the area under the AB line. The work done by the isothermal process is the area under the curved line. Since the AC line includes all of the area under the AB line, as well as the area between the AB and AC lines, more work is done on the gas (negative work) in the isothermal process.

Pressure

C Isothermal

Isovolumetric B

Isobaric

Volume

1 kcal is the heat needed to raise 1 kg of water by 1C. Use this relation to find the change in the temperature. 1 kcal  (1kg )(1C )  1  ( 6800 J )     = 0.54 C  4186 J  1 kcal  3.0 kg  Thus the final temperature is 10.0C + 0.54C = 10.5C

Find the mass of warmed water from the volume of water and its density of 1025 kg m 3. Then use the fact that 1 kcal of energy raises 1 kg of water by 1C, and that the water warms by 25C . m → m =  At = 1025 kg m 3 1.0 m 2 0.5  10 −3 m = 0.5125 kg V = At =

(



( 0.5125 kg )( 25 C ) 3.

)(

)

 1bar  (1kcal ) = 12.8 kcal ; 12.8 kcal   = 0.043 bars  0.04 bars (1kg )(1C )  300 kcal 

 4.186  103 J  (a) 2500 Cal  = 1.0  107 J   1 Cal   1 kWh  (b) 2500 Cal   = 2.9 kWh  860 Cal  (c) At 12 cents per day, the food energy costs $0.35 per day . It would be impossible to feed yourself in the United States on this amount of money.

Assume that we are at the surface of the Earth so that 1 lb is equivalent to 0.454 kg.  0.454 kg   5 9 C  1 kcal = 0.2522 kcal  0.252 kcal 1Btu = (1 lb )(1F )     1 lb   1F  (1 kg )(1C )

 4186 J  = 1056 J   1 kcal 

0.2522 kcal 

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The energy generated by using the brakes must equal the car’s initial kinetic energy, since its final kinetic energy is 0. 2

  1m s   Q = mv = (1.2  10 kg ) ( 95 km h )  = 4.178  105 J  4.2  105 J    3.6 km h    1 2

2 0

1 2

1 kcal  ( 4.178 10 J )  4186  = 99.81kcal  1.0  10 kcal J 5





The energy input is causing a certain rise in temperature, expressible as a number of Joules per hour per Co. Convert that to mass using the definition of kcal, which can relate mass to heat energy.  3.2  107 J h   1 kcal  (1kg )(1C )  30C   4186 J  1 kcal = 254.8 kg h  250 kg h   

The wattage rating is 375 Joules per second. Note that 1 L of water has a mass of 1 kg.   1 kcal  4186 J   1 s   1 kg  −1  2.5  10 L  1 L  ( 60 C )  (1 kg )(1C )  kcal   375 J  = 167 s  170s = 2.8 min  

(

)

The heat absorbed can be calculated from Eq. 19–2. Note that 1 L of water has a mass of 1 kg.

  1  10−3 m 3   1.0  103 kg   Q = mcT = (18 L )  ( 4186 J kg C )( 95C − 15C ) = 6.0  106 J     3  1 L  1 m   9.

The specific heat can be calculated from Eq. 19–2. 1.65  105 J Q Q = mcT → c = = = 2096 J kg C  2100 J kg C mT ( 4.1kg )( 37.2C − 18.0C )

10. The heat absorbed by all three substances is given by Eq. 19–2, Q = mcT . Thus the amount of Q mass can be found as m = . The heat and temperature change are the same for all three cT substances. 1 1 1 1 1 1 Q Q Q : : : : : : mCu : mAl : mH O = = = cCu T cAl T cH O T cCu cAl cH O 390 900 4186 2

4186 4186 4186 : : = 10.7 : 4.65 : 1  11 : 4.7 : 1 390 900 4186

11. (a) The heat must warm both the water and the pot to 100oC. The heat is also the power times the time. The temperature change is 89Co. Q = Pt = mAl cAl + mH O cH O TH O →

(

t = =

( m c + m c ) T Al Al

H 2O H 2O

)

H 2O

P ( 0.25 kg )( 900 J kg C ) + ( 0.75 kg )( 4186 J kg C ) (89 C ) 750 W

= 399 s  4.0  102 s = 6.7 min © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) The energy provided by the pot is the power multiplied by the time of heating. This is also the energy used by the lightbulb, which is the bulb’s power rating times the time it is on. Qpot = Ppot tpot = Ebulb = Pbulb tbulb →

tbulb =

Ppot tpot Pbulb

( 750 W )( 399 s ) 60 W

= 4.99  103 s  83min

12. The heat lost by the iron must be the heat gained by the aluminum and the glycerin. mFe cFe Ti Fe − Teq = mAl cAl Teq − Ti Al + mgly cgly Teq − Ti gly

(

)

(

)

(

)

( 0.290 kg )( 450 J kg C )(142C ) = ( 0.095 kg )( 900 J kg C )( 28C ) + ( 0.250 kg ) cgly ( 28C ) cgly = 2305J kg C  2300 J kg C

13. The heat lost by the copper must be equal to the heat gained by the aluminum and the water. The aluminum and water have the same temperature change.

(

)

(

)

(

)

mCu cCu Ti Cu − Teq = mAl cAl Teq − Ti Al + mH O cH O Teq − Ti H O tt 2

( 0.145 kg )( 900 J kg C )

  ( Teq − 12.0C )  + ( 0.825 kg )( 4186 J kg C ) 

( 0.265kg )( 390 J kg C ) ( 245C − Teq ) =  Teq = 18.532C  18.5C

14. The heat lost by the horseshoe must be equal to the heat gained by the iron pot and the water. Note that 1.35 L of water has a mass of 1.35 kg.

(

mshoe cFe ( Tshoe − Teq ) = mpot cFe (Teq − Tpot ) + mH OcH O Teq − TH O 2

)

( 0.40 kg )( 450 J kg C)(Tshoe − 25.0C ) = ( 0.30 kg )( 450 J kg C )( 25.0C − 20.0C ) + (1.35 kg )( 4186 J kg C )( 25.0C − 20.0C ) Tshoe = 185.7C  190C This is an estimate because some of the water that initially contacted the hot horseshoe almost certainly turned into steam at some temperature > 100C . 15. The heat released by burning the 15 grams of candy is equal to the heat absorbed by the aluminum and water. Q15 g = mAl cAl + mH 2OcH 2O T candy

(

)

= ( 0.325 kg + 0.624 kg )( 0.22 kcal kg C ) + (1.75 kg )(1.00 kcal kg C ) ( 53.5C − 15.0C ) = 75.41kcal The heat released by 65 grams of the candy would be 65/15 times that released by the 15 grams. 65 65 Q65g = 15 Q15g = 15 ( 75.41 kcal ) = 326.8 kcal  330 Cal candy

candy

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16. (a) Since Q = mcT and Q = C T , equate these two expressions for Q and solve for C. Q = mcT = C T → C = mc

( ) C = mc = ( 38 kg ) ( 4186 J kg C ) = 1.591  10 J C  1.6  10 J C

C = mc = (1.0 kg ) 4186 J kg Co = 4200 J Co

(b) For 1.0 kg of water:

17. We assume that all of the kinetic energy of the hammer goes into heating the nail. 2 K = Q → 8 12 mhammer vhammer = mnail cFe T →

(

T =

)

(

2 8 12 mhammer vhammer

) = 4 (1.20 kg )( 7.5 m s )

( 0.014 kg )( 450 J kg C )

mnail cFe

= 42.86 C  43C

18. The heat lost by the substance must be equal to the heat gained by the aluminum, water, and glass. mx cx Ti x − Teq = mAl cAl Teq − Ti Al + mH O cH O Teq − Ti H O + mglass cglass Teq − Ti glass

(

cx =

(

)

(

)

( ) m (T − T )

(

)

mAl cAl Teq − Ti Al + mH O cH O Teq − Ti H O + mglass cglass Teq − Ti glass 2

)

(

)

( 0.105 kg )( 900 J kg C ) + ( 0.185 kg )( 4186 J kg C )    ( 35.0C − 10.5C ) + ( 0.017 kg )( 840 J kg C )   =

( 0.215 kg )( 330C − 35.0C )

= 341.16 J kg C o  341J kg C o

19. 65% of the original potential energy of the aluminum goes to heating the aluminum. 0.65U grav = Q → 0.65mAl gh = mAl cAl T → T =

0.65gh cAl

(

)

0.65 9.80 m s2 ( 55 m )

( 900 J kg C)

= 0.39 C

20. The oxygen is all at the boiling point, so any heat added will cause oxygen to evaporate (as opposed to raising its temperature). We assume that all the heat goes to the oxygen, and none to the flask. Q 3.40  105 J = = 1.6 kg Q = mLvap → m = Lvap 2.1 105 J kg 21. The silver must be heated to the melting temperature of 961oC and then melted. Q = Qheat + Qmelt = mcT + mLfusion

(

)

= ( 26.50 kg )( 230 J kg C )( 961C − 25C ) + ( 26.50 kg ) 0.88  105 J kg = 8.0  106 J

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Chapter 19

Heat and the First Law of Thermodynamics

22. The heat lost by the steam condensing and then cooling to 30oC must be equal to the heat gained by the ice melting and then warming to 30oC. msteam  Lvap + cH O Ti steam − Teq  = mice  Lfus + cH O Teq − Ti ice 

(

)

(

)

 Lfus + cH O (Teq − Ti ice )  3.33  105 J kg + ( 4186 J kg C )( 30C )  msteam = mice = (1.00 kg )  Lvap + cH O (Ti steam − Teq )   22.6  105 J kg + ( 4186 J kg C )( 70C )  2

= 0.18 kg 23. Assume that the heat from the person is only used to evaporate the water. Also, we use the heat of vaporization at room temperature (585 kcal/kg), since the person’s temperature is closer to room temperature than 100oC. Q 185 kcal Q = mLvap → m = = = 0.316 kg = 316 mL Lvap 585 kcal kg 24. Assume that all of the heat lost by the ice cube in cooling to the temperature of the liquid nitrogen is used to boil the nitrogen, and so none is used to raise the temperature of the nitrogen. The boiling point of the nitrogen is 77 K = −196o C .





mice cice  Tice − Tice  = mnitrogen Lvap →

 initial

final





mnitrogen =



mice cice  Tice − Tice 

 initial

final

 = ( 2.8  10 kg ) ( 2100 J kg C ) 0C − ( −196C )  −2

200  103 J kg

Lvap

= 5.8  10−2 kg 25. (a) The energy absorbed from the body must warm the snow to the melting temperature, melt the snow, and then warm the melted snow to the final temperature. Qa = Qwarm + Qmelt + Qwarm = mcsnow T1 + mLfusion + mcliquid T2 = m csnow T1 + Lfusion + cliquid T2  snow

liquid

(

)

= (1.0 kg ) ( 2100 J kg C )(15 C ) + 3.33  105 J kg + ( 4186 J kg C )( 37 C )  = 5.2  105 J (b) The energy absorbed from the body only has to warm the melted snow to the final temperature.

Qb = Qwarm = mcliquid T2 = (1.0 kg )( 4186 J kg C )( 35C ) = 1.5  105 J liquid

Your body only has to supply about 30% as much energy when you drink the melted snow. 26. (a) The heater must heat both the boiler and the water at the same time.

(

)

Q1 = Pt1 = mFe cFe + mH O cH O T → t1 =

( m c + m c ) T = (180 kg )( 450 J kg C) + ( 710 kg )( 4186 J kg C) (82 C) Fe Fe

H2O H2O

5.8  107 J h

= 4.316 h  4.3 h © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) Assume that after the water starts to boil, all the heat energy goes into boiling the water, and none goes to raising the temperature of the iron or the steam. 5 mH O Lvap ( 710 kg ) 22.6  10 J kg Q2 = Pt2 = mH O Lvap → t2 = = = 27.666 h 5.8  107 J h P

(

)

Thus the total time is t1 + t2 = 4.316 h + 27.666 h = 31.98 h  32 h 27. We are told that 80% of the cyclist’s energy goes to evaporation. The energy needed for evaporation is equal to the mass of the water times the latent heat of vaporization for water. Note that 1 L of water has a mass of 1 kg. Also, we use the heat of vaporization at room temperature (585 kcal/kg), since the cyclist’s temperature is closer to room temperature than 100oC. mH O Lvap ( 9.0 kg )( 585 kcal kg ) = = 6581kcal  6.6  103 kcal 0.80 Qrider = mH O Lvap → Qrider = 0.80 0.8 2

28. The heat lost by the aluminum and the water must equal the heat needed to melt the mercury and to warm the mercury to the equilibrium temperature.

(

)

mAl cAl ( TAl − Teq ) + mH O cH O TH O − Teq = mHg  Lfusion + cHg (Teq − Tmelt )  2

Lfusion = =

(

mAl cAl ( TAl − Teq ) + mH O cH O TH O − Teq 2

mHg

) −c

(T − T ) eq

melt

( 0.620 kg )( 900 J kg C) + ( 0.400 kg )( 4186 J kg C) (12.80C − 5.06C ) 1.00 kg − (138 J kg C ) 5.06C − ( −39.0C ) 

= 1.12  104 J kg

29. The kinetic energy of the bullet is assumed to warm the bullet to its melting point and then melt it. 1 mv 2 = Q = mcPb (Tmelt − Tinitial ) + mLfusion → 2

(

)

v = 2  cPb ( Tmelt − Tinitial ) + Lfusion  = 2 (130 J kg C )( 327C − 20C ) + 0.25  105 J kg  = 360 m s

30. Assume that the kinetic energy of the bullet was all converted into heat which melted the ice. 1 mbullet v 2 = Q = mice Lfusion → 2 mice =

1 2

mbullet v 2 Lfusion

( 5.5 10 kg ) ( 250 m s ) = 5.2  10 kg = 5.2 g = 2

−2

1 2

−3

3.33  10 J kg 5

31. The heat lost by the aluminum and 310 g of liquid water must be equal to the heat gained by the ice in warming in the solid state, melting, and warming in the liquid state. mAl cAl Ti Al − Teq + mH O cH O Ti H O − Teq = mice cice (Tmelt − Ti ice ) + Lfusion + cH O (Teq − Tmelt ) 

(

mice =

)

(

)

( 0.085 kg )( 900 J kg C )( 3.0 C ) + ( 0.31kg )( 4186 J kg C )( 3.0 C ) = 9.8  10 −3 kg 5 ( 2100 J kg C )( 8.5C ) + 3.3  10 J kg + ( 4186 J kg C )(17 C ) 

658

Chapter 19

Heat and the First Law of Thermodynamics

32. Use the first law of thermodynamics, Eq. 19–5, and the definition of internal energy, Eq. 19–1b. Since the work is done by the gas, it is positive. (a) The temperature does not change, and so Eint = 0 . (b) Eint = Q − W → Q = Eint + W = 0 + 4.30  103 J = 4.30  103 J 33. For the drawing of the graph, the pressure is given relative to the starting pressure, which is taken to be P0 .

P P0

Segment A is the cooling at constant pressure.

0.5

Segment B is the isothermal expansion. Since the volume doubles, the pressure must reduce by a factor of 2.

V ( L)

0 0.0

0.5

1.0

34. Segment A is the compression at constant pressure. Since the process is at a constant pressure, the path on the diagram is horizontal from 2.5 L to 1.0 L. Segment B is the isothermal expansion. Since the temperature is constant, the ideal gas law says that the product PV is constant. Since the volume is increased by a factor of 2.5, the pressure must be divided by 2.5, and so the final point on this segment is at a pressure of 1 atm / 2.5 = 0.4 atm. The path is a piece of a hyperbola. Segment C is the pressure increase at constant volume. Since the process is at a constant volume, the path on the diagram is vertical from 0.4 atm to 1.0 atm. 35. (a) The work done at constant pressure is found from Eq. 19–10a. W = PV

(

)(

= (1.00 atm ) 1.01  105 Pa atm 4.1m 3 − 2.2 m 3

P (atm)

)

= 1.919  105 J  1.9  105 J (b) The change in internal energy is calculated from the first law of thermodynamics.

V (m3)

Eint = Q − W = 6.80  105 J − 1.919  105 J = 4.9  105 J (c) See the adjacent graph.

36. (a) Since the container has rigid walls, there is no change in volume. W = PV = 0 J (b) Use the first law of thermodynamics to find the change in internal energy. Eint = Q − W = ( −425 kJ ) − 0 = −425 kJ

659

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Instructor Solutions Manual

37. Assume that all of the melted ice stays at 0oC, so that all the heat is used in melting ice, and none in warming water. The available heat is half of the original kinetic energy 1 1 mskater v 2 = Q = mice Lfusion → 2 2

(

)

mice =

1 4

mskater v 2 Lfusion

1 4

( 64 kg )( 8.5 m s ) 3.33  105 J kg

= 3.5  10 −3 kg = 3.5 g

38. (a) The pressure must be converted to absolute pressure in order to use the ideal gas equation, and so the initial pressure is 4.5 atm absolute pressure, and the lower pressure is 2.0 atm absolute pressure. Segment A is the isothermal expansion. The temperature and the amount of gas are constant, so PV = nRT is constant. Since the pressure is reduced by a factor of 2.25, the volume increases by a factor of 2.25, to a final volume of 2.25 L. Segment B is the compression at constant pressure, and segment C is the pressure increase at constant volume. (b) The work done during part A is given by Eq. 19–9. The work done during part B is given by Eq. 19–10a. There is no work done during part C since the volume doesn’t change. V V2  1.01  105 Pa   1  10−3 m 3  WA = nRT ln 2 = PV ln = 4.5atm 1.0 L ( ) ( ) 1 1  1 atm   1 L  ln 2.25 V1 V1    

= 368.6 J

 1.01  105 Pa   1  10−3 m 3  − 1.25L ( )   1 L  = −252.5J  1 atm   

WB = P2 (V1 − V2 ) = ( 2.0atm ) 

Wtotal = WA + WB = 368.6 J − 252.5J = 116.1J  120 J 39. (a) No work is done during the first step, since the volume is constant. The work in the second step is given by Eq. 19–10a, W = PV .

 1  10−3 m 3   1.01  105 Pa  9.3 L − 5.9 L ( )  1 L  = 480 J   1 atm   

W = PV = (1.4 atm ) 

(b) Since there is no overall change in temperature, Eint = 0 J . (c) The heat flow can be found from the first law of thermodynamics.

Eint = Q − W → Q = Eint + W = 0 + 480J = 480J ( into the gas ) 40. (a) The work done by an ideal gas during an isothermal volume change is given by Eq. 19–9. 7.00 m3 V2 W = nRT ln = ( 3.20 mol )( 8.314 J mol K )( 295 K ) ln = 5440 J 3.50 m3 V1 (b) Since the process is isothermal, there is no internal energy change. Apply the first law of thermodynamics. Eint = Q − W = 0 → Q = W = 5440 J (c) Since the process is isothermal, there is no internal energy change, and so Eint = 0 .

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41. (a) Since the process is adiabatic, Q = 0 J (b) Use the first law of thermodynamics to find the change in internal energy. Eint = Q − W = 0 − ( −2630J ) = 2630 J (c) Since the internal energy is proportional to the temperature, a rise in internal energy means a rise in temperature. 42. A graph of the process is shown. The expansion process is the horizontal line, and the constant volume process is the vertical line. The dashed line is an isotherm starting from the original state. (a) Work is only done in the expansion at constant pressure, since there must be a volume change in order for there to be work done. W = PV

 1.01 105 Pa  1 10−3 m3 0.28 L = 71J ( )  1L  1 atm 

= ( 2.5atm ) 

(b) Use the first law of thermodynamics to find the heat flow. Notice that the temperature change over the entire process is 0, so there is no change in internal energy. Eint = Q − W = 0 → Q = W = 71J 43. (a) The initial volume of the water is found from its mass and density. The final volume is found from the ideal gas law. The work done at constant pressure is given by Eq. 19–10a. m (1.00 kg ) V1 = = = 1.00  10−3 m 3  1.00  103 kg m 3

(

V2 =

nRT

)

 1mol   ( 8.315J mol K )( 373 K )  0.018 kg  = 1.70 m 3

(1.00 kg ) 

= P2 1.013  105 Pa Note that the initial volume is negligible. We might have assumed that since the original state was liquid, that the gas volume was 0 to begin with, without significant error.

(

)(

)

W = PV = 1.013  105 Pa 1.70 m3 = 1.722  105 J  1.72  105 J (b) The heat added to the system is calculated from the latent heat of vaporization. Then the first law of thermodynamics will give the internal energy change. Q = mLV = (1.00 kg )( 2260 kJ kg ) = 2260 kJ = 2.26  106 J

Eint = Q − W = 2.26  106 J − 1.72  105 J = 2.09  106 J Note that we can’t use Eint = 23 nRT because a phase change is involved. 44. The work done by an ideal gas during an isothermal volume change is given by Eq. 19–9. W = nRT ln

VB VA

= PAVA ln

VB VA

(

)(

)

 1.80 L   = −187 J  3.20 L 

= 1.013  105 Pa 3.20  10−3 m 3  ln

The work done by an external agent is the opposite of the work done by the gas, 187 J .

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

45. The work is given by W =

 PdV . The pressure is given by the van der Waals expression, Eq. 18–9,

with n = 1.00. The temperature is held constant. We will keep the moles as n until the last step.

  V V  RT a  an 2   nRT an 2    dV =   W =  P dV =   − − 2  dV =  nRT ln (V − bn ) + 2 V V bn V  V  V − V      V V  V −b   n  n   V2



=  nRT ln (V2 − bn ) +



1 1 an 2   an 2  (V − bn ) ln nRT V bn − − + = nRT ln 2 + an 2  −  (1 )    V2   V1  (V1 − bn )  V2 V1 

We evaluate for n = 1.00 mol, to get W = RT ln

(V2 − b )  1 1  +a − . (V1 − b )  V2 V1 

Note that if the constants a and b are 0, we get Eq. 19–9 for the work done by an ideal gas, as we would expect. 46. For the path ac, use the first law of thermodynamics to find the change in internal energy. Eac = Qac − Wac = −175J − ( −35J ) = −140 J Note that, according to the rules of significant figures, the answer has 3 significant figures. It is correct to the “ones” place. Also, since internal energy only depends on the initial and final temperatures, this E applies to any path that starts at a and ends at c. And for any path that starts at c and ends at a, Eca = −Eac = 140 J. (a) Use the first law of thermodynamics to find Qabc. Eabc = Qabc − Wabc → Qabc = Eabc + Wabc = −140 J + ( −56 J ) = −196 J

(b) Since the work along path bc is 0, Wabc = Wab = Pb Vab = Pb (Vb − Va ) . Also note that the work along path da is 0. Wcda = Wcd = Pc Vcd = Pc (Vd − Vc ) = 12 Pb (Va − Vb ) = − 12 Wabc = − 12 ( −56 J ) = 28J (c) Use the first law of thermodynamics to find Qabc.

Ecda = Qcda − Wcda → Qcda = Ecda + Wcda = 140 J + 28J = 168J (d) As found above, Ea − Ec = Eca = −Eac = 140 J (e) Since Ed − Ec = Ecd = 42 J and Eac = −140 J, then Ead = Eac + Ecd = −98 J and Eda = 98 J. Use the first law of thermodynamics to find Qda .

Eda = Qda − Wda → Qda = Eda + Wda = 98J + 0 = 98J 47. We are given that Qac = −85J, Wac = −55 J, Wcda = 38 J, Eint,a − Eint,b = Eint = 15J, and Pa = 2.2 Pd . ba

(a) Use the first law of thermodynamics to find Eint,a − Eint,c = Eca. Eint = −Eint = − ( Qac − Wac ) = −  −85J − ( −55 J )  = 30 J ca

Note that the answer has 2 significant figures. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

662

Chapter 19

Heat and the First Law of Thermodynamics

(b) Use the first law of thermodynamics to find Qcda.

Eint = Qcda − Wcda → Qcda = Eint + Wcda = Eint + Wcda = 30 J + 38 J = 68 J cda

cda

Wabc = Wab = Pa Vab = Pa (Vb − Va ) = 2.2 Pd (Vc − Vd ) = −2.2Wcda = −2.2 ( 38 J ) = −84 J (d) Use the first law of thermodynamics to find Qabc .

Eint = Qabc − Wabc → Qabc = Eint + Wabc = Eint + Wabc = −30 J − 84 J = −114 J abc

abc

(e) Since Eint,a − Eint,b = 15 J → Eint,b = Eint,a − 15 J, we have the following.

Eint = Eint,c − Eint,b = Eint,c − ( Eint,a − 15 J ) = Eint + 15 J = −30 J + 15 J = −15 J. bc

Use the first law of thermodynamics to find Qbc .

Eint = Qbc − Wbc → Qbc = Eint + Wbc = −15 J + 0 = −15 J bc

48. (a) Leg ba is an isobaric expansion, and so the work done is positive. Leg ad is an isovolumetric reduction in pressure, and so the work done on that leg is 0. Leg dc is an isobaric compression, and so the work done is negative. Leg cb is an isovolumetric expansion in pressure, and so the work done on that leg is 0. (b) From Problem 47, Wcda = Wcd + Wda = 38 J, so Wadc = Wad + Wdc = −38 J. Also from Problem 47, Wabc = −84 J, and so Wcba = Wcb + Wba = 84 J. So the net work done during the cycle is as follows.

Wnet = Wba + Wad + Wdc + Wcb = 84 J − 38 J = 46 J (c) Since the process is a cycle, the initial and final states are the same, and so the internal energy does not change.

Eint = 0 (d) Use the first law of thermodynamics, applied to the entire cycle. Eint = Qnet − Wnet → Qnet = Eint + Wnet = 0 + 46 J = 46 J tot

(e) From Problem 47(b), we have Qadc = −68 J. This is the exhaust heat. So the input heat is found as follows. Qnet = Qadc + Qcba = −68 J + Qcba = 46 J → Qcba = 114 J = Qinput efficiency =

Wnet Qinput

 100 =

46 J 114 J

 100 = 40% ( 2 sig. figs.)

49. For a diatomic gas with all degrees of freedom active, the internal energy is given by Eint = 72 nRT . Eint = 72 nRT = 72 ( 5.40 mol )( 8.314 J mol K )( 2450 K ) = 3.85  105 J

50. If there are no heat losses or mass losses, then the heating occurs at constant volume, and so Eq. 19–11a applies, Q = nCV T . Air is primarily made of diatomic molecules, and for an ideal diatomic gas, C V = 25 R.

663

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Q = nCV T = 25 nRT = 25 T =

2QT 5PV

PV 0 0

T →

(

)

2 1.8  106 J ( 293 K )

(

)

5 1.013  10 Pa ( 3.5 m )( 4.6 m )( 3.0 m ) 5

Instructor Solutions Manual

= 43.12 K  43C

51. For one mole of gas, each degree of freedom has an average energy of 12 RT , so the internal energy of a mole of the gas with f degrees of freedom is as follows.

Eint = f ( 12 RT ) = CVT → CV = 12 f R

CP = CV + R =

f 2

R+R=

f 2

R+

2 2

R = 12 ( f + 2 ) R

52. Since the gas is monatomic, the molar specific heat is given by CV = 32 R. The molar specific heat is also given by CV = McV , where M is the molecular mass. Equate these two expressions to find the molecular mass and the identity of the gas. 3 ( 8.314 J mol K ) 1kcal 1000 g 3R CV = 32 R = McV → M = = = 83.7 g mol 2cV 2 ( 0.0356 kcal kg C ) 4186 J 1kg From the periodic table, we see that the gas is krypton , which has a molecular mass of 83.6 g/mol. 53. The process is adiabatic, and so the heat transfer is 0. Apply the first law of thermodynamics. Eint = Q − W = 0 − W → W = −Eint = − nCV T = nCV (T1 − T2 ) 54. If there are no heat losses, and no work being done, then the heat due to the people will increase the internal energy of the air, as given in Eq. 19–13. Note that air is basically diatomic. Use the ideal gas equation to estimate the number of moles of air, assuming the room is initially at 293 K. PV Q = Eint = nCV T ; n = → RT Q Q RT0Q RT0Q 2T Q T = = = = = 0 5 PV nCV C PV R 5PV 0 0 0 0 V 0 0 2 0 0 CV PV RT0



  70W  ( 7200s )    person   = 42.4 K  40 C

2 ( 293 K ) (1600 people ) 

(



5 1.013  105 Pa

)( 2.2  10 m ) 4

55. (a) First find the molar specific heat at constant volume, then the molar specific heat at constant pressure, and then finally the specific heat at constant pressure.  103 cal  CV = McV = ( 0.034 kg mol )( 0.182 kcal kg K )   = 6.188cal mol K  1kcal 

cP =

CP M

CV + R M

( 6.188cal mol K ) + (1.99 cal mol K )  1kcal   3  = 0.241kcal kg K ( 0.034 kg mol )  10 cal 

(b) From the value for CV = 6.188cal mol K , we see from Table 19–4 that this gas is probably triatomic. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

664

Chapter 19

Heat and the First Law of Thermodynamics

56. (a) The change in internal energy is given by Eq. 19–13. The nitrogen is diatomic. J   Eint = nCV T = n ( 25 R ) T = 25 ( 2.00 mol )  8.314  (150 K ) = 6236 J  6240 J or mol K  

 

Eint = nCV T = ( 2.00 mol )  4.96

cal   J    4.186  (150 K ) = 6229 J  6230 J mol K   cal 

(b) The work is done at constant pressure.

 

W = PV = nRT = ( 2.00 mol )  8.314

 150 K = 2494 J  2490 J ) ( mol K  J

(c) The heat is added at constant pressure, and so Eq. 19–11b applies. J   Q = nCP T = n ( 72 R ) T = 72 ( 2.00 mol )  8.314  (150 K ) = 8730 J or mol K   Eint = Q − W → Q = Eint + W = 6229 J + 2494 J = 8720 J

57. Since the pressures and moles are given to 3 significant digits, we assume the temperatures are as well. (a) Note that a process of pressure increasing linearly with temperature is NOT one of the typical lines on a P–V diagram. It is not constant pressure, constant temperature, constant volume, or adiabatic. In particular, if the gas is ideal, then V 

T P

. Since

T1 P1



T2  460 K

720 K    , P2  1.00 atm 1.60 atm 

the volume is not constant. Accordingly, we need to calculate the volume and pressure for the range of temperatures. P − P 0.60 atm atm P = P0 + aT → P1 = P0 + aT1 ; P2 = P0 + aT2 → a = 2 1 =  2.3077  10−3 T2 − T1 260 K K

 0.60 atm  ( 460 K )  −6.1538  10−2 atm   260 K 

P0 = P1 − aT1 = 1.00 atm − 

 

With these values for a and P0, the pressure ( P = P0 + aT ) and volume  V =

nRT 

 can be P  calculated as a function of the temperature. Pressure and volume values were calculated for temperatures from 460 K to 720 K, at 10-K intervals. A few values and a graph are shown.

665

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The change in internal energy is given by Eq. 19–13. We use the molar specific heat for a diatomic gas. J   3 Eint = nCV T = n ( 25 R ) T = 25 (1.00 mol )  8.314  ( 260 K ) = 5404.1J  5.40  10 J mol K   (c) The work done by the gas is found from Eq. 19–9. To do the integral, we must have a functional relationship between the pressure and the volume. Also note that the volume decreases as the temperature and pressure increase, so we anticipate the work done by the gas to be negative. nRT nR ( P − P0 ) nR  P0  nR P0 = = PV = nRT → V = dP  1 −  → dV = P aP a  P a P2 V2

W =  PdV =  P V1

nR P0 a P

dP =

 P2 

nRP0

ln   dP = a P a P 1

(1.00 mol )  8.314 

nRP0

 −6.1538  10−2 atm )  1.60 atm  ( mol K  ln   = −104.2 J  −104 J 1.00 atm −3 atm   2.3077  10 J

K If we approximate the area under the P–V curve as a trapezoid, we get this approximate value.  1.013  105 Pa   1.0  10−3 m 3  W = 12 ( P1 + P2 )(V1 − V2 ) = 12 ( 2.60 atm )( 36.95 L − 37.77 L )    1atm 1L   

= −108J (d) Use the first law of thermodynamics. Eint = Q − W → Q = Eint + W = 5404.1J − 104.2 J = 5299.9  5.30  103 J

58. For a diatomic gas with no vibrational modes excited, we find the  parameter. C C + R 25 R + R 7 = 5 =5  = P = V CV CV R 2 For an adiabatic process, we have PV  = constant. Use this to find the final pressure. 

 1 1

 2 2

PV = PV

V   1  = 0.4568atm  0.457 atm → P2 = P1  1  = (1.00 atm )    1.75   V2  1.4

Use the ideal gas law to find the final temperature.  P  V  PV PV 1 1 = 2 2 → T2 = T1  2  2  = ( 293K )( 0.4568)(1.75) = 234 K = −39C T1 T2  P1  V1  V2

59. The work is given by W =  PdV . The pressure is given by the adiabatic condition, PV  = c, where V1

  c is a constant. Note that for the two states given, c = PV = PV 1 1 2 2 V2

W =  PdV =  V1

cV 1−

1− V

dV = 

( cV

1− 2

− cV11−

1−

) = ( PV V 

2 2

1− 2

 1− V − PV 1 1 1

1−

) = ( PV − PV ) 1 1

2 2

 −1

666

Chapter 19

Heat and the First Law of Thermodynamics

60. (a) We first find the final pressure from the adiabatic relationship, and then use the ideal gas law to find the temperatures. For a diatomic gas,  = 1.4. 



PV = PV 1 1 2 2

1.4

V   0.1210 m 3  → P2 = P1  1  = (1.00 atm )  = 7.772  10 −2 atm 3   0.750 m   V2  PV

PV = nRT → T =

; T1 =

PV 1 1 nR

(1.013  10 Pa )( 0.1210 m ) = 403.9 K  404 K = 5

( 3.65 mol )  8.314 

PV 2 2

T2 =

  mol K  J

( 7.772  10 )(1.013  10 Pa )( 0.750 m ) = 194.6 K  195 K = −2

  mol K   J   4 (b) Eint = nCV T = n ( 25 R ) T = ( 3.65 mol ) 25  8.314  (194.6 K − 403.9 K ) = −1.59  10 J mol K   nR

( 3.65 mol )  8.314

(c) Since the process is adiabatic, no heat is transferred. Q = 0 (d) Use the first law of thermodynamics to find the work done by the gas. The work done ON the gas is the opposite of the work done BY the gas. Eint = Q − W → W = Q − Eint = 0 − ( −1.59  104 J ) = 1.59  10 4 J Won = −1.59  104 J gas

61. Combine the ideal gas equation with Eq. 19–16 for adiabatic processes. PV PV P V T P1 V2   1 1 = 2 2 → 1 = 2 1 ; PV = → = PV 1 1 2 2 T1 T2 P2 V1 T2 P2 V1 CP CV

V2 T1 V1 T2

CV + R

V2 V1

3 R+R 5 = 2 3 =3 R 2 1

 −1 2

→ V

 −1 1

62. (a) To plot the graph accurately, data points must be calculated. V1 is found from the ideal gas equation, and P2 and V2 are found from the fact that the first expansion is adiabatic. Note that the origin is NOT shown on this graph.

3/2

 T1   T1   −1 3  273 K + 25 K  3 → = V V 2 1 T   T  = ( 0.086 m )  273 K − 68 K  = 0.15 m    2  2 1.1 1.0 0.9

P (atm)

 =

0.8 0.7

0.6 0.5

0.4 0.3

2 40

100

V (L)

667

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

= nRT1 → V1 = PV 1 1

nRT1 P1

  = nRT2 , PV = PV PV 2 2 2 2 1 1

(1.00 mol )  8.314 

 588 K ) ( mol K  = 48.26  10 −3 m 3 = 48.26 L J

1.013  105 Pa

→ V2 −1 =

 PV 1 1

→

nRT2

  3/ 2 −3 5 3 5/3 5/3     1.013 10 Pa 48.26 10 m ( )( )  PV  V2 =  1 1  =    nRT2   (1.00 mol )  8.314 J  ( 389 K )    mol K     P2 =

nRT2

(1.00 mol )  8.314

Instructor Solutions Manual

3/ 2

= 89.68  10−3 m 3 = 89.68 L

 389 K ) ( 1atm mol K     = 3.606  10 4 Pa   = 0.356 atm 3 5 −3 89.68  10 m  1.013  10 Pa  J

= V2 (b) Both the pressure and the volume are known at the lower left corner of the graph. P3 = P2 , V3 = V1 → PV = nRT3 → 3 3 T3 =

PV 3 3 nR

PV 2 1 nR

( 3.606 10 Pa )( 48.26 10 m ) = 209 K = −3

(1.00 mol )  8.314 

  mol K  J

(c) For the adiabatic process, state 1 to state 2: J   Eint = 23 nRT = 23 (1.00 mol )  8.314  ( 389 K − 588 K ) = −2482 J  −2480 J mol K   Q = 0 ( adiabatic ) ; W = Q − Eint = 2480 J

For the constant pressure process, state 2 to state 3: J   Eint = 23 nRT = 23 (1.00 mol )  8.314  ( 209 K − 389 K ) = −2244.7 J  −2240 J mol K  

(

)(

)

W = PV = 3.606  104 Pa 48.26  10 −3 m 3 − 89.68  10 −3 m 3 = −1494 J  −1490 J Q = W + Eint = −1494 J − 2244.7 J = −3739 J  −3740 J

For the constant volume process, state 3 to state 1: J   Eint = 23 nRT = 23 (1.00 mol )  8.314  ( 588 K − 209 K ) = 4727 J  4730 J mol K   W = PV = 0 ; Q = W + Eint = 4730 J

(d) For the complete cycle, by definition Eint = 0 . If the values from above are added, we get Eint = 4727 J − 2482 J − 2245J = 0. Add the separate values for the work done and the heat

added. W = 2482 J − 1494 J = 998J  990 J ; Q = −3739 J + 4727 J = 998J  990 J Notice that the first law of thermodynamics is satisfied.

668

Chapter 19

Heat and the First Law of Thermodynamics

63. (a) Combine the ideal gas law with Eq. 19–16 for adiabatic processes. PV = nRT → V =



 nRT  = P1− T  nR  → ( )   P 

nRT

; PV  = constant = P 

constant

P1− T  =

( nR )

But since n and R are constant for a fixed amount of gas, we have P1− T  = constant . Take the derivative of the above expression with respect to y, using the product rule and chain rule. d d P1− T  = constant → P1− T  ) = ( constant ) = 0 → ( dy dy T  (1 −  ) P −

+ P1−  T  −1

dT dy

Multiply this equation by P  T  . P   dP dT  dP P dT T (1 −  ) P − + P1−  T  −1 = (1 −  ) + =0    T  dy dy  dy T dy

We also assume that

(1 −  )( −  g ) + 

= −  g , and so get the following.

P dT T dy

= 0 , or

dT dy

(1 −  ) T 

g

(b) Use the ideal gas law. We let M represent the total mass of the gas and m represent the mass of M M one molecule, so N = . The density of the gas is  = . m V T V Vm m PV = NkT → = = = → P Nk Mk  k dT

(1 −  ) T 

g =

(1 −  )  m 

 k   g =  



(1 −  ) mg 

(1 −  ) mg (1 − 75 ) 29 (1.66  10 

(

(d) T =

)(

kg 9.80 m s 2

1.38  10

7 5

−23

J K

) = −9.77  10 K m −3

 )  101kmm   1C  = −9.77 C km  −9.8C km 1K 3

= −9.77  10−3 K m  dT

−27







y = ( −9.77 C km )( −0.1km − 4.0 km ) = 40C = T f − ( −5C ) → T f = 35C  95 F

64. The heat conduction rate is given by Eq. 19–17a. Q t

= kA

T1 − T2 l

( 460 C − 22 C ) = 82 W = ( 380 J s m C )  (1.0  10 m ) 0.64 m o

−2

669

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

65. The heat conduction rate is given by Eq. 19–17a. 15.0o C − −5o C  T1 − T2 Q o 2  = kA = 0.84 J s m C 3.0 m = 1.6  10 4 W −3 3.2  10 m l t

(

)(

(

)

66. (a) The power radiated is given by Eq. 19–18. The temperature of the tungsten is 273K + 25 K = 298 K. Q 2 4 =  AT 4 = ( 0.35) 5.67  10−8 W m 2 K 4 4 ( 0.19 m ) ( 298 K ) = 71W t (b) The net flow rate of energy is given by Eq. 19–19. The temperature of the surroundings is 268 K. Q 2 4 4 =  A ( T14 − T14 ) = ( 0.35) ( 5.67  10−8 W m 2 K 4 ) 4 ( 0.19 m ) ( 298 K ) − ( 268 K )  t

(

)

= 25 W

67. Eq. 19–20 gives the heat absorption rate for an object facing the Sun. The heat required to melt the ice is the mass of the ice times the latent heat of fusion for the ice. The mass is found by multiplying the volume of ice times its density. Q Q = mL f =  VL f =  A ( x ) L f = 1000 W m 2 eA cos  → t  A ( x ) L f  ( x ) L f t = = 2 1000 W m eA cos  1000 W m 2 e cos 

(

)

( ) ( ) ( 9.17  10 kg m )(1.0  10 m )( 3.33  10 J kg ) = 7.5  10 s  21h = (1000 W m ) ( 0.050) cos 35 2

−2

68. The distance can be calculated from the heat conduction rate, given by Eq. 19–17a. The rate is given as a power (150 W = 150 J/s). T −T T −T 0.50 C Q = P = kA 1 2 → l = kA 1 2 = ( 0.2 J s m C ) 1.5 m 2 = 1.0  10 −3 m 150 W l t P

(

)

69. This is a heat transfer by conduction, and so Eq. 19–17a is applicable. T −T 30C − 10C Q = P = kA 1 2 = ( 0.84 J s m C ) 16 m 2 = 1920 W 0.14 m t l It would take 19.2 100-W light bulbs, and so 20 bulbs are needed to maintain the temperature difference.

(

)

70. (a) The rate of heat transfer due to radiation is given by Eq. 19–18. We assume that each teapot is a sphere that holds 0.55 L. The radius and then the surface area can be found from that. 1/3

V = r 4 3

Q t

 3V  → r=   4 

 3V  → S . A. = 4 r = 4    4 

(

 3V    4 

)

2/3

=  A T14 − T24 = 4 

2/3

(T − T ) 4 1

4 2

670

Chapter 19

Heat and the First Law of Thermodynamics 2 −3 4 4    3 ( 0.55  10 m )   −8 = 4 ( 0.70 )  5.67  10   ( 358 K ) − ( 293 K )    2 4  m K  4  t ceramic  

 Q 

2/3

= 11.67W  12 W

 Q   Q   0.10   0.10    =    = (11.67W )   = 1.667 W  1.7 W  t shiny  t  ceramic  0.70   0.70  (b) We assume that the heat capacity comes primarily from the water in the teapots, and ignore the heat capacity of the teapots themselves. We apply Eq. 19–2, along with the results from part (a). The mass is that of 0.55 L of water, which would be 0.55 kg. 1  Q  Q = mcT → T = telapsed   mc  t  radiation

( T )ceramic =

11.67 W J   ( 0.55 kg )  4186  kg C  

(1800s ) = 9.1C

( T )shiny = ( T )ceramic = 1.3C 7

71. For the temperature at the joint to remain constant, the heat flow in both rods must be the same. Note that the cross-sectional areas and lengths are the same. Use Eq. 19–17 for heat conduction. −T Thot − Tmiddle T Q Q = k Al A middle cool →   =   → kCu A l l  t Cu  t  Al

Tmiddle =

kCuThot + kAlTcool kCu + kAl

( 380 J s m C )( 205C ) + ( 200 J s m C )( 0.0C ) 380 J s m C + 200 J s m C

= 134.3C  130 C

72. (a) Choose a cylindrical shell of length l, radius R, and thickness dR. Apply Eq. 19–17b, modified for the radial geometry. See the figure, which is a drawing of a crossdQ section of the pipe. Note that in the pipe material dt must be a constant, so that all of the heat energy that enters a shell (moving radially) also exits that shell. dQ dT dT = − kA = − k 2 R l → dt dR dR

dR R ln

=−

R2 R1

2 k l dQ dt 2 k l dQ dt

R

=−

(T1 − T2 ) →

dT →

2 k l

dQ dt T

dT →

2 k ( T1 − T2 ) l ln ( R2 R1 )

(b) For still water, the initial heat flow outward from the water is described by Eq. 19–2. Q T 2 k (T1 − T2 ) l Q = mcT → = mc = → ln ( R2 R1 ) t t © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

671

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

T t

2 k ( T1 − T2 ) l

mc ln ( R2 R1 )

2 k (T1 − T2 ) l

 H O R l c ln ( R2 R1 )

 

2 1

Instructor Solutions Manual

2k (T1 − T2 )

 H O R12 c ln ( R2 R1 ) 2

 71C − 18C ) ( s m C  = = 4.835 C s  4.8 C s J   4.0  2  3 3 (1.0 10 kg m ) ( 0.033m )  4186 kg C  ln  3.3    Note that this will lower the value of T1 , which would lower the rate of temperature change as 2  40

time elapses. (c) The water at the entrance is losing temperature at a rate of 4.835C s . In one second, it will

 

have traveled 8 cm, and so the temperature drop per cm is  4.835

C   1s 



 = 0.60 C cm .

s   8cm 

73. The conduction rates through the two materials must be equal. If they were not, the temperatures in the materials would be changing. Call the temperature at the boundary between the materials Tx . T −T T −T Q l1 Q l2 = k1 A 1 x = k 2 A x 2 → = T1 − Tx ; = Tx − T2 t l1 l2 t k1 A t k2 A Add these two equations together, and solve for the heat conduction rate. Q l1 Q l 2 Q  l1 l 2  1 + = T1 − Tx + Tx − T2 →  +  = T1 − T2 → t k1 A t k2 A t  k1 k 2  A Q

Q t

(T1 − T2 )

 l1 l 2  k +k   1 2

(T1 − T2 ) ( R1 + R2 )

The R-value for the brick needs to be calculated, using the definition of R given on page 584.  1ft  R=

Q t Q t

l k

 12 in    = 0.69 ft 2 h F Btu J  1Btu  1m  5 C  3600 s  0.84    s m C  1055 J   3.281ft  9 F  1h  4 in

(T1 − T2 ) ( 35 F ) = (195 ft 2 ) = 346.6 Btu h  350 Btu h ( R1 + R2 ) ( 0.69 + 19 ) ft 2 h F Btu

= 346.6

1.356 J   1h  101.2 J  1.0  102 W =   −3 h  1.29  10 Btu   3600 s  s

Btu 

74. The heat needed to warm the liquid can be calculated by Eq. 19–2. Q = mcT = ( 0.35 kg )(1.00 kcal kg C )( 37C − 5C ) = 11.2 kcal  11Cal 75. (a) Use Eq. 19–18 for total power radiated, with e = 1.0 for a perfect emitter. 2 Q 4 2 T 4 = (1.0 ) 5.67  10−8 W m 2 K 4 4 7.0  108 m ( 5500 K ) = e AT 4 = e 4 RSun t

(

) (

)

= 3.195  10 26 W  3.2  10 26 W © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

672

Chapter 19

Heat and the First Law of Thermodynamics

(b) Assume that the energy from the Sun is distributed symmetrically over a spherical surface with the Sun at the center. 3.195  1026 W P Q t = = = 1.130  103 W m 2  1.1 103 W m 2 2 2 11 A 4 RSun-Earth 4 1.5  10 m

(

)

76. The heat released can be calculated by Eq. 19–2. To find the mass of the water, use the density (of pure water) times the volume.

(

)(

Q = mcT = VcT = 1.0  103 kg m3 1.0  103 m

) ( 4186J kg C)(1C) = 4  10 J 3

77. The heat gained by the ice (to melt it and warm it) must be equal to the heat lost by the steam (in condensing and cooling). mLF + mcH O (Teq − 0 ) = mLV + mcH O 100o C − Teq 2

Teq =

LV − LF 2cH O

(

+ 50o C =

)

2260 kJ kg − 333 kJ kg

(

2 4.186 kJ kg C

)

+ 50o C = 280o C

This answer is not possible. Because this answer is too high, the steam must not all condense, and none of it must cool below 100oC. Calculate the energy need to melt a kilogram of ice and warm it to 100oC.

(

)(

)

Q = mLF + mcH O (Teq − 0 ) = (1kg ) 333 kJ kg + 4.186 kJ kg Co 100Co  = 751.6 kJ Calculate the mass of steam that needs to condense in order to provide this much energy. 751.6 kJ Q Q = mLV → m = = = 0.333 kg LV 2260 kJ kg Thus, one-third of the original steam mass must condense to liquid at 100oC in order to melt the ice and warm the melted ice to 100oC. The final mixture will be at 100oC, with 1/3 of the total mass as steam, and 2/3 of the total mass as water . 2

78. The temperature rise can be calculated from Eq. 19–2. Use only 80% of the heat generated. Q ( 0.80 )( 200 kcal h )( 0.75 h ) Q = mcT → T = = = 2.065C  2 C mc ( 70 kg )( 0.83kcal kg C ) 79. Use the heat conduction rate equation, Eq. 19–17a. These are heat flows AWAY from the climber’s body, into the environment. 34C − ( −18C ) = 32.5 W  33 W T −T Q = kA 1 2 = ( 0.025 J s m C ) 0.95 m 2 (a) t 3.8  10−2 m l 34C − ( −18C ) = 5533 W  5500 W T −T Q = kA 1 2 = ( 0.56 J s m C ) 0.95 m 2 (b) t 5.0  10−3 m l

(

)

80. We assume that all of the heat provided by metabolism goes into evaporating the water. For the energy required for the evaporation of water, we use the heat of vaporization at room temperature (585 kcal/kg), since the runner’s temperature is closer to room temperature than 100oC.  950 kcal   kg H 2O  2.2 h    = 3.6 kg  1h   585 kcal 

673

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

81. (a) To calculate heat transfer by conduction, use Eq. 19–17a for all three areas–walls, roof, and windows. Each area has the same temperature difference. Qconduction  kA    kA   kA  =   +   +    (T1 − T2 ) t  l walls  l roof  l  windows 

 ( 0.023J s m C ) ( 410 m 2 ) ( 0.1J s m C ) ( 250 m 2 )  +   0.195 m 0.055 m   ( 38C ) =  ( 0.84 J s m C ) ( 33m 2 )  +    6.5  10−3 m

= 1.812  105 W  1.8  105 W (b) The energy being added must both heat the air and replace the energy lost by conduction, as considered above. The heat required to raise the temperature is given by Eq. 19–2, Qraise = mair cair ( T ) warming . The mass of the air can be found from the density of the air times temp

its volume. The conduction heat loss is proportional to the temperature difference between the inside and outside, which, if we assume the outside temperature is still −15C, varies from 30C to 38 C. We will estimate the average temperature difference as 34 C and scale the answer from part (a) accordingly. Q  Qadded = Qraise + Qconduction =  airVcair ( T ) warming +  conduction  (1800 s ) temp  t 

 

=  1.29

kcal   4186 J  3   ( 750m )  0.24  ( 8C )  m  kg C   kcal   kg  3

J  34C   8 8 +  1.812  105   (1800 s ) = 2.996  10 J  3.0  10 J  s 38 C    (c) We assume a month is 30 days. Q 0.9Qgas =   tmonth →  t conduction Qgas =

1 Q

1  24 h  3600 s  11 tmonth = 1.812  105 J s ) ( 30 d )  (     = 5.219  10 J 0.9  t conduction 0.9  1 d  1 h 

 1 kg   $0.080   = $773.12  $770 7   5.4  10 J   kg 

5.276  1011 J 

This seems like a very high heating bill. As an extra comment, almost 90% of the heat loss is through the windows. Investing in insulated window drapes and double-paned windows in order to reduce the thermal conductivity of the windows could mean major savings for this homeowner. 82. For an estimate of the heat conduction rate, use Eq. 19–17a. T −T Q ( 37C − 34C ) = kA 1 2 = ( 0.2 J s m C ) 1.5 m 2 = 22.5 W  20 W t 4.0  10−2 m l This is only about 10% of the cooling capacity that is needed for the body. Thus convection cooling is clearly necessary.

(

)

674

Chapter 19

Heat and the First Law of Thermodynamics

83. Assume that the loss of kinetic energy is all turned into heat which changes the temperature of the squash ball. 2 2 vi2 − v 2f ( 22 m s ) − (11m s ) 2 2 1 K lost = Q → 2 m vi − v f = mcT → T = = = 0.15 C 2c 2 (1200 J kg C )

(

)

84. The handle is pushed in very quickly, so we approximate the process as adiabatic, since there will be little time for heat transfer to or from the air in the pump. We combine the ideal gas law and Eq. 19–16 for adiabatic processes. We treat air as diatomic. PV PV T PV P2 V1   1 1 PV = 2 2 → 2 = 2 2 ; PV = → = 1 1 2 2 T1 T2 T1 P1 V1 P1 V2

 =

CP CV

CV + R CV

5 R+R 7 = 2 5 =5 R 2

V  = =  = 1 T1 P1 V1 V2 V1  V2 

P2 V2

V1 V2

 −1

V  → T2 = T1  1   V2 

 −1

( )

= ( 293 K ) 20.4 = 386.6 K = 114C

85. (a) The energy required to raise the temperature of the water is given by Eq. 19–2. Use this to find the power. Q = mcT →

= mc

T



= ( 0.250 kg )  4186

  80 C  = 797 W  8.0  102 W    kg C   105s  J

t t  (b) After 105 s, the water is at 100C. So for the remaining 15 s the energy input will boil the water. Use the heat of vaporization. Q ( Power ) t ( 797 W )(15s ) = = = 5.3g Q = mLV → m = LV LV 2260 J g 86. Note–we take the depth of 30 m and the solar constant of 1000 W/m2 to have 2 sig. figures each. (a) Consider just the 30 m of crust immediately below the Earth’s surface, assume that all heat from the interior is transferred to the surface, and so it all passes through this 30-m layer. This is a heat conduction problem, and so Eq. 19–17a is appropriate. The radius of the Earth is about 6.38  106 m. T −T T −T Q 1.0 C   3600 s   2 1hour  = kA 1 2 → Qinterior = kA 1 2 t = ( 0.80 J s m C ) 4 REarth   t 30 m  l l  hour   = 4.910  1016 J  4.9  1016 J

(b) The cross-sectional area of the Earth, perpendicular to the Sun, is a circle of radius REarth , and 2 so has an area of  REarth . Multiply this area times the value of 1000 W m 2 to get the amount of energy incident on the surface of the Earth from the Sun per second, and then convert to energy per day.   3600 s   = 4.604  10 20 J 2 QSun =  REarth 1000 W m 2 1 hour    hour   

(

QSun Qinterior

4.604  1020 J 4.910  1016 J

)

= 9377

And so QSun  9400 Qinterior . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

675

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

87. Consider a rectangular block of ice. The surface area of the top of the block is A, and this surface is exposed to the air. The surface area of the bottom of the block is also A, and this surface is exposed to the water. The thickness of this block is x. We assume the heat of fusion from the ice forming at the bottom surface is conducted through the ice to the air. For the conduction, from Eq. 19–17a, we dQ kice A ( Tbottom − Ttop ) = . This is the rate of heat energy leaving the bottom surface and have dt x escaping to the air, through the ice block. For the freezing process, if a layer of ice of thickness dx, with mass dm, is formed at the bottom surface, the energy released is dQ = Lf dm = Lf  ice Adx. The

rate of release of this energy is

= Lf  ice A

. The rate of energy release must be equal to the rate

of energy conduction, if the temperature at the bottom surface is to remain constant. The density of ice is given in Table 13–1. k ice ( Tbottom − Ttop ) dQ kice A ( Tbottom − Ttop ) dx dt = x dx → = = Lf  ice A → dt x dt Lf  ice t18 cm

 0

kice ( Tbottom − Ttop ) Lf  ice

kice ( Tbottom − Ttop )

0.18m

dt =

 x dx →

Lf  ice

(

)(

t18 cm = 12 ( 0.18 m )

1 ( 0.18 m ) Lf  ice 12 ( 0.18 m ) 3.33  10 J kg 917 kg m t18 cm = 2 = kice ( Tbottom − Ttop )  2 J  18C )  (  s m C  2

→

) = 1.374  10 s 5

 1.4  105 s  1.6 d

88. Assume that the final speed of the meteorite, as it completely melts, is 0, and that all of its initial kinetic energy was used in heating the iron to the melting point and then melting the iron. 1 mvi2 = mcFe (Tmelt − Ti ) + mLfusion → 2 vi = 2  cFe ( Tmelt − Ti ) + Lfusion  = 2 ( 450 J kg C ) 1538C − ( −105C )  + 2.89  105 J kg  = 1430 m s

89. We use the ideal gas equation to find the initial temperature of the gas. Kinetic molecular theory is used to find the final temperature of the gas. Then, since the heating occurs at constant pressure, we use Eq. 19–11b. The amount of gas is a constant 0.15 mol, and the pressure is a constant 1.10 atm. PV (1.10 atm )( 3.0 L ) T1 = 1 = = 267.97 K L atm  nR  ( 0.15 mol )  0.0821  mol K   2 From kinetic molecular theory, T  vrms , so T2 = T1 (1.16 ) . The molar specific heat at constant 2

pressure for an ideal monatomic gas is 25 R. Calculate the heat from Eq. 19–11b.

Q = nCP T = n ( 25 R ) T = n ( 25 R ) (1.16 ) − 1 2

PV1 nR

 1.013  10 Pa   1.0  10−3 m 3    = 288.8 J  290 J 1atm 1L    5

= ( 25 ) ( 0.3456 )(1.10 atm )( 3.0 L ) 

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90. Use the first law of thermodynamics to find the rate of change of internal energy. Heat is being removed, so the heat term is negative. Work is being done on the gas, so the work term is also negative. dEint dQ dW Eint = Q − W → = − = −1.5  103 W − −7.5  103 W = 6.0  103 W dt dt dt This internal energy increase will happen during the compression stroke of a cycle. We assume that the time for compression is half of a cycle. 1    60s  tcompression = 12    = 0.25s  120 cycles/min   1min  Assume an ideal diatomic gas and use Eq. 19–13 in order to find the temperature change. dEint  T  Eint = nCV T → = nCV  →  dt  t compression

(

Tcompression =

dEint tcompression dt

nCV

(

= 6.0  103 W

)

) (1.00 mol )(5.00 cal0.25s mol K )( 4.186 J cal )

= 71.67 C  72 C

91. (a) The pressure varies with depth according to Eq. 13–6b. Use Eq. 13–6b to find the pressure at the original depth, and then use the ideal gas equation with a constant temperature to find the size of the bubble at the surface. P1 = Patm +  gh ; P2 = Patm ; PV = PV → 1 1 2 2 1/3

( Patm +  gh ) 43  ( 12 d1 )3 = Patm 43  ( 12 d 2 )3 →

1/3

 P +  gh    gh  d 2 = d1  atm = d1  1 +   Patm   Patm  

 (1000 kg m 3 )( 9.80 m s 2 ) (14.0 m )  = ( 3.60 cm ) 1 +  = 4.789 cm  4.79 cm in diameter 1.013  105 Pa   1/3

(b) See the accompanying graph. The path is an isotherm, so the product PV is constant. 1000 kg m3 9.80 m s2 (14.0 m ) = 2.35atm P1 = Patm +  gh = 1.0atm + 1.013  105 Pa 1atm

(

V1 = 43  ( 12 d1 )

(

)(

)

 1ml  3   1cm 

= 43  (1.80 cm )  3

= 24.4 mL V2 = 43  ( 12 d 2 )

 4.789 cm   1ml    3  2    1cm 

= 43  

= 57.5 mL

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(c) Since the process is isothermal, Eq. 19–9 is used to calculate the work. V V W = nRT ln 2 = PV ln 2 2 2 V1 V1 = P2   ( d 2 )  ln 4 3

1 2

4 3 4 3

 ( 12 d 2 )  ( 12 d1 )

 d2   0.0479 m  3 = (1.013  105 Pa )  12  ( 0.0479 m )  ln    = 4.99 J  0.0360 m   d1 

= P2  12  d 23  ln 

Since the process is isothermal, Eint = 0 . Then, by the first law of thermodynamics, Eint = Q − W → Q = Eint + W = 4.99 J . The heat is added.

92. Let the subscript 1 refer to the original state, subscript 2 refer to the compressed state, and subscript 3 refer to the final state. The first process is best described by the ideal gas law for constant temperature, and the second process by the adiabatic relationship in Eq. 19–16. We use the ideal gas equation to relate the initial and final temperatures. Note that the initial volume is 3 times the molar volume at STP, and so is 67.2 L. V   1  = 4.545atm ; T = T = 273K Process 1: PV = PV → P2 = P1  1  = (1atm )   1 1 2 2 2 1  0.22   V2  

 V   1atm   0.22  PV = PV → P3 = P2  2  =    = 0.3644 atm  V3   0.22   1   P  V  PV PV 1 1 = 3 3 → T3 = T1  3  3  = ( 273K )( 0.3644 )(1) = 99.48 K T1 T3  P1  V1   2 2

Process 2:

5/ 3

 3 3

Tmax = 273 K = T1 = T2

5.0

Tmin = 99 K = T3

4.0

Pmax = 4.5atm = P2 P (atm)

Pmin = 0.36 atm = P3

3.0 2.0

1.0

0.0 0.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

V (L)

93. We assume that the light bulb emits energy by radiation, and so Eq. 19–19 applies. Use the data for the 75-W bulb to calculate the product  A for the bulb, and then calculate the temperature of the 150-W bulb. 4 → (Q t )75 W =  A T754 W − Troom

(

 A =

(Q t )75 W

4 75 W

−T

4 room

)

( 0.90)( 75 W ) = 9.006  10−9 W K 4 4 4 ( 273 + 75) K  − ( 273 + 18) K 

4 (Q t )150 W =  A (T1504 W − Troom ) →

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 ( Q t )150 W 4  + Troom T150 W =    e A 

1/ 4

 ( 0.90 )(150 W ) 4 = + ( 291 K )  −9 4  ( 9.006  10 W K ) 

1/ 4

= 386 K = 113C  110C

94. Combine the ideal gas relationship for a fixed amount of gas,

PV 1 1 T1

PV 2 2 T2

, with the adiabatic

  relationship, PV 1 1 = PV 2 2 .

PV 1 1 T1 V1 V2

PV 2 2

→

V1 V2

V  →  1  V2 

V1 T1 V2 T2

P2 T1

  = PV → ; PV 1 1 2 2

P1 T2

V1 V2

 −1

 T   −1  273 K + 560 K  → = 2 =  V2  T1  280 K  

5/ 2

= 15.3

95. Since the specific heat is a function of temperature, we need to integrate the infinitesimal version of Eq. 19–2, and use moles instead of grams. dQ = n C dT → T2

Q =  nk T1

T3 3 0

dT =

( T −T ) = ( 4T nk

3 0

4 2

4 1

2.75 mol )(1940 J mol K ) 4 ( 281K )

( 48.0 K ) 4 − ( 22.0 K ) 4  = 305J  

96. The body’s metabolism (blood circulation in particular) provides cooling by convection. If the metabolism has stopped, then heat loss will be by conduction and radiation, at a rate of 200 W, as given. The change in temperature is related to the body’s heat loss by Eq. 19–2, Q = mcT . The specific heat of the human body is found in Table 19–1. Q mcT =P= → t t mcT ( 65 kg )( 3470 J kg C )( 36.6C − 35.6C ) t= = = 1128s  1100 s = 19 min 200 W P 97. (a) The cross-sectional area of the Earth, perpendicular to the Sun, is a circle of radius REarth , and 2 so has an area of  REarth . Multiply this area times the solar constant to get the rate at which the Earth is receiving solar energy. 2 Q 2 =  REarth ( solar constant ) =  6.38 106 m 1350 W m 2 = 1.73 1017 W t (b) Use Eq. 19–19 to calculate the rate of heat output by radiation, and assume that the temperature of space is 0 K. The whole sphere is radiating heat back into space, and so we use the full 2 surface area of the Earth, 4 REarth .

(

Q t

)(

)

1/ 4

Q 1    t  A 

= e AT 4 → T = 

  1  = (1.73  1017 J s ) 2  (1.0 ) ( 5.67  10−8 W m 2 K 4 ) 4 ( 6.38 106 m ) 

1/ 4

= 278 K = 5C

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98. This is an example of heat conduction. The heat conducted is the heat absorbed by the melting ice, Q = mice Lfusion. The area through which the heat is conducted is the total area of the six surfaces of the box, and the length of the conducting material is the thickness of the Styrofoam. We assume that all of the heat conducted into the box goes into melting the ice, and none into raising the temperature inside the box. The time can then be calculated by Eq. 19–17a. m L l T −T Q = kA 1 2 → t = ice fusion l t kAT

( 8.2 kg ) ( 3.33 105 J kg )(1.5 10−2 m ) 2 ( 0.023 J s m C )  2 ( 0.25 m )( 0.35 m ) + 2 ( 0.25 m )( 0.55 m ) + 2 ( 0.35 m )( 0.55 m ) ( 34 C )

= 3.136  104 s  3.1  10 4 s  8.7 h 99. We treat the 1000 J and 100 g as having all their figures as significant figures. (a) Solve Eq. 19–2 for the final temperature of the ice. Use the specific heat of ice. Q 1000 J Q = mc ( T − T0 ) → T = T0 + = −20C + = −15C mc (0.100 kg)(2100 J/kgC) None of the ice melts. The ice warms from −20C to − 15C. (b) The heat will melt part of the ice. Solve Eq. 19–4 for the mass of ice that will melt, with L being the latent heat of fusion of water. The temperature will remain constant at 0°C. Q 1000 J Q = mL → m = = = 3.00 g of ice will melt L 333 kJ/kg (c) Solve Eq. 19–2 for the final temperature of the water. Use the specific heat of water. Q 1000 J Q = mc ( T − T0 ) → T = T0 + = 10C + = 12C mc (0.100 kg)(4186 J/kgC) The water warms to 12C . (d) The heat will convert some of the water into steam. Solve Eq. 19–4 for the mass of the water that will be converted to steam, with L being the latent heat of vaporization of water. The temperature will remain constant at 100°C. Q 1000 J Q = mL → m = = = 0.442 g of water will be converted to steam L 2260 kJ/kg (e) Solve Eq. 19–2 for the final temperature of the steam. Use the specific heat of steam. Q 1000 J Q = mc ( T − T0 ) → T = T0 + = 110C + = 115C mc (0.100 kg)(2010 J/kgC) The steam warms from 110C to 115C . 100. (a) The bullet will gain an amount of heat equal to 50% of its loss of kinetic energy. Initially assume that the phase of the bullet does not change, so that all of the heat causes a temperature increase. 2 2 1 vi2 − v 2f 220 m s ) − (160 m s ) ( 4 2 2 1 1  m vi − v f  = Q = mcPb T → T = = = 44 C 2 2 4 (130 J kg C ) cPb (b) The final temperature of the bullet would be about 64C , which is not above the melting temperature of lead, which is 327C. Thus none of the bullet will melt .

(

)

680

Chapter 19

Heat and the First Law of Thermodynamics

101. (a) The rate of absorbing heat for an object facing the Sun is given by Eq. 19–20.

Q t

(

)

(



1 m2





) 110 cm  (1) = 3.4 W

= 1000 W m 2 eA cos  = 1000 W m 2 ( 0.85 ) 40 cm 2 



(b) The rise in temperature is related to the absorbed heat by Eq. 19–2. We assume that all absorbed heat raises the temperature of the leaf.

Q = mcT → T = T t

Q → mc

 1kcal  1 Q 3.4W o =   = 2.256 C s  2.3C s −4 mc t ( 4.5  10 kg ) ( 0.80 kcal kg C )  4186 J 

(c) This temperature rise cannot continue for hours. In one hour, the leaf’s temperature would increase to over 8000C , and the leaf would combust long before that temperature was reached. (d) We assume that the rate of heat loss by radiation must equal the rate of heat absorption of solar energy. Note that the area of the leaf that radiates is twice the area that absorbs heat energy.  Q   Q  → 1000 W m 2  Aabsorb cos  =  Aradiate T14 − T24 →  Solar =    t   t  radiation

(

)

(

)

heating

 (1000 W m 2 ) cos    1000 W m 2 ) (1) ( 4 4 + T2  =  + ( 297 K )  T1 =  −8 2 4 2    2 ( 5.67  10 W m K )  1/4

1/4

= 359 K = 86C This is very hot, which indicates that the leaf must lose energy by other means than just radiation, in order to cool to a more reasonable temperature. (e) The leaf can also lose heat by conduction to the cooler air around it; by convection , as the wind continually moves cooler air over the surface of the leaf; and evaporation of water. 102. The rate of energy absorption from the Sun must be equal to the rate of losing energy by radiation plus the rate of losing energy by evaporation if the leaf is to maintain a steady temperature. The latent heat of evaporation is taken be the value at 20oC, which is 2450 kJ/kg. Also note that the area of the leaf that radiates is twice the area that absorbs heat energy.  Q   Q   Q  + →  Solar =     t   t  radiation  t evaporation heating

mH O Levaporation

(1000 W m )  A cos  =  A (T − T ) + m (1000 W m ) cos  − 2 (T − T ) =A 2

absorb

radiate

4 1

H 2O

t

absorb

4 2

→

t

4 2

Levaporation

(1000 W m ) (1) − 2 (5.67  10 W m K ) ( 308 K ) − ( 297 K )  = ( 0.85) ( 40  10 m ) ( 2.45  10 J kg ) −8

−4

 1000 g   3600s   = 4.3g h   1kg   1 h 

= 1.196  10−6 kg s 

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T → Q =  t T , where t l  describes the average heat conductivity properties of the house, such as insulation materials and surface area of the conducting surfaces. It could have units of J h C . We see that the heat loss is proportional to the product of elapsed time and the temperature difference. We assume that the proportionality constant  does not vary during the day, so that, for example, heating by direct sunlight through windows is not considered. We also assume that  is independent of temperature, and so is the same during both the day and the night. Qturning = ( J h C )(15 h )( 22C − 8C ) + ( J h C )( 9 h )(16C − 0C ) = 354 J

103. We model the heat loss as conductive, so that, using Eq. 19–17a,

down

Qnot turning = ( J h C )(15 h )( 22C − 8C ) + ( J h C )( 9 h )( 22C − 0C ) = 408 J down

Q Qturning

408 J − 354 J 354  J

= 0.1525  15%

down

To keep the thermostat “up” requires about 15% more heat in this model than turning it down. 104. (a) Multiply the power times the time times the mass per Joule relationship for the fat. ( 95J s )( 3600s h )( 24 h d ) (1.0 kg fat 3.7  107 J ) = 0.2218 kg d  0.22 kg d (b) 1.0 kg (1d 0.2218 kg ) = 4.5d

682

CHAPTER 20: Second Law of Thermodynamics Responses to Questions 1.

Mechanical energy can be transformed completely into heat. As a moving object slides across a rough level floor and eventually stops, the mechanical energy of the moving object has been transformed completely into heat. Also, if a moving object were to be used to compress a frictionless piston containing an insulated gas, the kinetic energy of the object would become internal energy of the gas. A gas that expands adiabatically (without heat transfer) transforms internal energy into mechanical energy, by doing work on its surroundings at the expense of its internal energy. Of course, that is an ideal (reversible) process. In any non-ideal process, only a fraction of the internal energy can be changed into mechanical energy. For example, in an explosion, a large amount of internal energy is converted into mechanical energy, but some internal energy is lost to heat or remains as internal energy of the explosion fragments.

It is possible to warm the kitchen in the winter by having the oven door open. The oven heating elements radiate heat energy into the oven cavity, and if the oven door is open, the oven is just heating a larger volume than usual. There is no thermodynamic cycle involved here. However, you cannot cool the kitchen by having the refrigerator door open. The refrigerator is a heat engine which (with an input of work) takes heat from the low-temperature reservoir (inside the refrigerator) and exhausts heat to the high-temperature reservoir (the room). The refrigerator exhausts more heat than it removes from the refrigerated volume, so the room actually gets warmer with the refrigerator door open, because of the work done by the refrigerator compressor. If you could have the refrigerator exhaust into some other room, then the refrigerator would be similar to an air conditioner, and it could cool the kitchen, while heating up some other space. Or you could unplug the refrigerator and open the door. That would cool the room somewhat, but would heat up the contents of the refrigerator, which was probably not a desired outcome!

No. The definition of heat engine efficiency as e = W QL does not account for QH , the energy needed to produce the work. Efficiency should relate the input energy and the output work. This definition of efficiency is also not useful, because with this definition, if the exhaust heat QL is less than the work done W (which is possible), the “efficiency” would exceed unity.

(a) In an internal-combustion engine, the high-temperature reservoir is the ignited gas–air mixture in the cylinder. The low-temperature reservoir is the “outside” air. The burned gases leave through the exhaust pipe. (b) In the steam engine, the high-temperature reservoir is the heated, high-pressure steam from the boiler. The low-temperature reservoir is the condensed water in the condenser. In the cases of both these engines, these areas are not technically heat “reservoirs,” because they are strictly not at a constant temperature.

A Carnot engine is a four-step cycle consisting of, in order; an isothermal expansion, an adiabatic expansion, an isothermal compression, and an adiabatic compression. Each step of the cycle is a reversible process so the net entropy change for the cycle is zero. Since the processes are reversible, the Carnot engine is the most efficient engine possible. A true Carnot engine is not practical to run and cannot be constructed, but it provides a theoretical limit to the maximum efficiency.

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To utilize thermal energy from the ocean, a heat engine would need to be developed that operated between two different temperatures. If surface-temperature water was to be both the source and the exhaust, then no work could be extracted. If the temperature difference between surface and deep ocean waters were to be used, there would be considerable engineering obstacles, high expense, and potential environmental difficulties involved in having a heat engine that connected surface water and deep ocean water. Likewise, if the difference in temperature between tropical water and arctic water were to be used, major difficulties would be involved because of the large distances involved.

There are several factors which can keep real engines from Carnot efficiency, here are some of them. Note that some of them overlap–for example, if an “adiabatic” process does not run infinitesimally slowly, it won’t be truly adiabatic. • Friction in the mechanical apparatus of the engine–particularly expanding and compressing of volumes. If friction is involved in a process, that process is irreversible. • Turbulence in the working substance, which increases entropy (and is thus irreversible). • Less than perfectly insulating equipment, which means that the “adiabatic” steps are not truly adiabatic (some heat will be exchanged during those parts of the cycle). • The “isothermal” processes are not perfectly isothermal, because no heat reservoir is infinite, and so the temperature of the reservoirs may change. • The Carnot engine has to run infinitesimally slowly to be irreversible, and that is an idealization. No real process can work infinitesimally slowly.

The expansion value allows the coolant gas to move from an area of higher pressure to an area of lower pressure. When the gas moves to the low pressure area, it expands rapidly (approximated as an adiabatic expansion). Expansion requires energy, which is taken from the internal energy of the gas, thus lowering its temperature.

(a) Consider a gas enclosed in a cylinder with a movable (and frictionless) piston. If the gas and cylinder are kept at a constant temperature by contact with a thermal reservoir, then heat can be added very slowly allowing the gas to do work on the piston and expand. This isothermal expansion is reversible. If work is done very slowly on the gas, by pushing down on the piston, while the gas is still in contact with the thermal reservoir, then the same amount of heat will leave during the isothermal compression. Other processes which are not isothermal are possible, as long as they are gradual and temperature change occurs for all parts of the system at once. (b) A stove burner could not be used to add heat to a system reversibly since the heat added by a burner would not be distributed uniformly throughout the system, and because the energy delivered by the burner to the system is not recoverable.

10. (a) When the lid is removed, the chlorine gas mixes with the air in the room around the bottle so that eventually both the room and the bottle contain a mixture of air and chlorine. (b) The reverse process, in which the individual chlorine particles reorganize so that they are all in the bottle, violates the second law of thermodynamics and does not occur naturally. It would require a spontaneous decrease in entropy. (c) Adding a drop of food coloring to a glass of water is another example of an irreversible process; the food coloring will eventually disperse throughout the water but will not ever gather into a drop again. The toppling of buildings during an earthquake is another example. The toppled building will not ever get “reconstructed” by another earthquake. 11. For a refrigerator, COP = QL W . That definition makes sense because we are interested in removing heat from the low-temperature reservoir (the interior of the refrigerator). The more heat that can be removed per amount of input work, the better (more efficient) the refrigerator is. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 20

Second Law of Thermodynamics

For a heat pump, COP = QH W . The objective of the heat pump is to heat (deliver QH ) rather than cool (remove QL ). It is the heat delivered to the house that is important for the heat pump. The more heat that can be delivered to the house per amount of input work, the better the heat pump is. 12. (a) If a gas expands adiabatically then Q = 0 , and so S = 0 by Eq. 20–6, S = Q T . (b) If a gas expands isothermally, there is no change in its internal energy, and the gas does work on its surroundings. Thus by the first law of thermodynamics, there must be heat flow into the gas, and so S  0 . Thus the entropy of the gas increases. 13. Here are several examples. • The erosion of soil due to water flow over the ground. • The oxidation of various metals (copper, zinc, iron, etc.) when left exposed to the air. • Fallen leaves decaying in the woods. • A pile of compost decomposing. • A landslide. The reverse of these processes is not observed. 14. 1 kg of liquid iron will have greater entropy, since it is less ordered than solid iron and its molecules have more thermal motion. In addition, heat must be added to solid iron to melt it; the addition of heat will increase the entropy of the iron. 15. (a) For Sale: Portable air conditioner. Place this air conditioner anywhere in your house and it will remove 500 J of heat per second from the air while using only 100 W of electrical power. No external exhaust vents needed. (b) For Sale: Automobile with 100% efficient engine. All of the gasoline energy goes directly into making the car move. No radiator (cooling system) or exhaust pipe needed. 16. Some processes that would obey the first law of thermodynamics but not the second, if they actually occurred, include: • a cup of tea warms itself by gaining thermal energy from the cooler air molecules around it; • a ball sitting on a soccer field gathers energy from its surroundings and begins to roll; • a bowl of popcorn placed in the refrigerator “un-pops” as it cools; • an empty perfume bottle is placed in a room containing perfume molecules, and all of the perfume molecules move into the bottle from various directions at the same time; • water on the sidewalk coalesces into droplets, are propelled upward, and rise into the air; • a house gets warmer in the winter while the outdoors gets colder, due to heat moving from the outdoors to inside the house. 17. While the state of the papers has gone from disorder to order, they did not do so spontaneously. An outside source (you) caused the increase in order. You had to provide energy to do this (through your metabolic processes), and in doing so, your entropy increased more than the entropy of the papers decreased. The overall effect is that the entropy of the universe increased, satisfying the second law of thermodynamics. 18. The first statement, “You can’t get something for nothing,” is a whimsical way of saying that energy is conserved. For instance, one way to write the 1st law is W = Q − Eint. This says that work done by a system must have a source–either heat is input to the system or the internal energy of the system is lowered. It “costs” energy–either heat energy or internal energy–to get work done. Another way © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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to say this is that no heat engine can be built which puts out more energy in the form of work than it extracts in the form of heat or internal energy. The second statement, “You can’t even break even,” reflects the fact that a consequence of the 2nd law is that there is no heat engine that is 100% efficient. Even though the 1st law is satisfied by an engine that takes in 100 J of heat and outputs 100 J of work, the 2nd law says that is impossible. If 100 J of heat were taken in, less than 100 J of work can be output from the heat engine, even if it is an ideal heat engine. Some energy will be “lost” as exhaust energy. 19. No. Even if the powdered milk is added very slowly, it cannot be re-extracted from the water without very large investments of energy. This is not a reversible process. 20. Entropy is a state variable, so the change in entropy for the system will be the same for the two different irreversible processes that take the system from state a to state b. However, the change in entropy for the environment will not necessarily be the same. The total change in entropy (system plus environment) will be positive for irreversible processes, but the amount may be different for different irreversible processes. 21. For a reversible process, ΔStotal = ΔSsystem + ΔSenvironment = 0. But since entropy is a state variable, the value of ΔSsystem will be the same no matter how the process is carried out–reversible or not. Thus, any change in ΔStotal for a process must be due to a change in the entropy of the environment. The more “irreversible” a process is, the higher the entropy change in the environment will be. 22. In an action movie, if you see a building or car change from an exploded state to an un-exploded state, or seeing a bullet that was fired go backwards into the gun and the gunpowder “unexploded,” you would know that the movie was “running backwards.” In a movie with vehicle crashes, seeing two collided vehicles separate from each other, becoming un-wrecked as they separate, would be an indication that the movie was “running backwards.” If you saw someone “un-write” something on a piece of paper–moving a pen over paper, taking away written marks as the pen moves (but not using an eraser to delete them) would be an indication that the movie was “running backwards.” 23. The synthesis of complex molecules from simple molecules does involve a decrease in entropy of the constituent molecules, since they have become more “structured” or “ordered.” However, the molecules are not a closed system. This process does not occur spontaneously or in isolation. The living organism in which the synthesis process occurs is a part of the environment that must be considered for the overall change in entropy. The “living organism and environment” combination will have an increase in entropy that is larger than the decrease in entropy of the molecules, and so overall, the second law is still satisfied, and the entropy of the entire system will increase.

Solutions to MisConceptual Questions 1.

(c) According to the second law of thermodynamics it is impossible for heat to be entirely converted into work in a cycle or a heat engine. However, the question does not specify that we must consider a complete cycle. In an isothermal process Q = W. In an isothermal compression work is entirely converted into heat, and in an isothermal expansion heat is entirely converted into work.

(d) A frequent misconception made in calculating the efficiency of an engine is to leave the temperatures in degrees Celsius, which would imply an efficiency of 50%. However, when the temperatures are properly converted to kelvins, Eq. 20–3 gives the efficiency as only about 34%.

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(d) A common misconception in this situation is not realizing that a heat cycle running in reverse, like a refrigerator, must have a high-temperature exhaust. Furthermore, that high-temp exhaust is the sum of the heat removed from the inside of the refrigerator, plus the work done by the refrigerator’s compressor.

4. (a, c) The maximum efficiency of an engine is given by Eq. 20–3, which can be written either as e = 1 − TL TH or as e = ( TH − TL ) / TH . Increasing TH and decreasing TL as in (a) results in a higher efficiency. In (b) the temperature difference remains the same, while the hot temperature increases, which results in a lower efficiency. In (c) the efficiency increases since the temperature difference remains the same, but TH decreases. In (d) we have just the opposite changes as in part (a), and so the efficiency decreases. 5.

(a) The text states that “real engines that are well-designed reach 60 to 80% of the Carnot efficiency.” The cooling system of the engine keeps the high temperature about 120C (400 K) and the exhaust is about room temperature (300 K). The maximum efficiency would then be around 25%. 80% of this maximum would be closest to 20% efficient. Any of the other choices for this question are not reasonable.

(b) Heat must be added to the ice cube to melt it. The change in entropy is the ratio of the heat added to the (constant) temperature of the ice cube. Since heat is absorbed in the process, the entropy increases.

(d) As the ice slowly melts, it gains heat at a constant temperature and so has a positive entropy change. The heat reservoir loses heat at a constant temperature and so has a negative entropy change. Thus answers (a), (b), and (c) are incorrect. The changes are the same magnitude, since this is a (quasi) reversible process, but answer (e) has the signs opposite to what actually happens.

(b) When the gas is compressed isothermally, its temperature is constant and so Eint = 0. Thus from the 1st law, Q = W . Since the work done on the gas is negative, Q is negeative–heat is released by the gas. From Eq. 20–6, we see that if heat is lost, entropy decreases.

80 J

= 80%. But for the operating QH 100 J temperatures given, the Carnot (ideal, maximum) efficiency is T 273 K + 10 K e = 1− L = 1− = 24%. The second law does NOT state answer (a). Answer (b) is 273 K + 100 K TH a statement of the first law, which will be true, but the second law puts other constraints on the system. The calculation for answer (d) is not set up correctly; it should say 100 J = 80 J + 20 J. And answer (e) is not how a heat engine works. Heat is taken in at the higher temperature and exhausted at the lower temperature.

10. (c) There are three microstates: HTT, THT, and TTH (H = heads, T = tails).

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11. (e) As discussed in Section 20–5, entropy is a state variable. Quoting from the text: S = S b − Sa depends only on the states a and b of the system. So in the diagram for this question, every process starts in state “S” and ends in state “F.” Thus the entropy change is the same for each process. 12. (f) All thermodynamic processes are subject to the second law of thermodynamics.

The efficiency of a heat engine is given by Eq. 20–1a. We also invoke energy conservation. 2600 J W W e= = = = 0.2549 = 25% QH W + QL 2600 J + 7600 J

The Carnot efficiency is given by Eq. 20–3, with temperatures in Kelvins. T T ( 230 + 273) K = 762K = 489C  490C e = 1 − L → TH = L = TH 1− e 1 − 0.34

The efficiency of a heat engine is given by Eq. 20–1a. We also invoke energy conservation. W W e= = → QL = W (1 e − 1) → QH W + QL QL t = W t (1 e − 1) = ( 680 MW )(1 0.32 − 1) = 1445 MW  1400 MW

We calculate both the energy per second (power) delivered by the gasoline, and the energy per second (power) needed to overcome the drag forces. The ratio of these is the efficiency. W  mi   1609 m  1h  = Poutput = F v = ( 350 N )  55     = 8604 W t ( to move  h   1mi  3600s  car)

QH t

= Pinput (from gasoline)

 

=  3.2  107

J   3.8 L  1gal  

mi   1h    55      = 58056 W L   1gal  32 mi   h   3600 s 

Poutput e=

W QH

( to move car)

Pinput

8604 W 58056 W

= 0.148  0.15

(from gasoline)

(a) The work done per second is found from the engine specifications. J cycles  W   4 =  180 ( 4 cylinders )  25  = 1.8  10 J s  t  cycle cylinder  s  

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(b) The efficiency is given by Eq. 20–1a. Use that to find the heat input. W W W t 1.8  104 J s e= = → QH = → QH t = = = 8.182  104 J s  8.2  104 J s 0.22 QH e e (c) Divide the energy in a gallon of gasoline by the rate at which that energy gets used, and then multiply by the speed. 120  106 J gal s = 1466.6 → 4 8.182  10 J s gal

1466.6 32.59 6.

s  80 km   1h 

= 32.59 km gal  33km gal or 20 mi gal gal  1h   3600s 

km  1gal  gal  3.785 L 

 8.6 km L or 5.4 mi gal

(a) For the net work done by the engine to be positive, the path must be carried out clockwise. Then the work done by process bc is positive, the work done by process ca is negative, and the work done by process ab is 0. From the shape of the graph, we see that Wbc  Wca .

P b Tb = Tc

a c (b) The efficiency of the engine is given by Eq. 20–1a. So we need to find the work done and the heat input. At first glance we might V 0 assume that we need to find the pressure, volume, and temperature at the three points on the graph. But as shown here, only the temperatures and the first law of thermodynamics are needed, along with ratios that are obtained from the ideal gas law. ab: Wab = PV = 0 ; Qab = Eint = nCV (Tb − Ta ) = 23 nR (Tb − Ta )  0 ab

V V T bc: Eint = nCV ( Tc − Tb ) = 0 ; Qbc = Wbc = nRTb ln c = nRTb ln c = nRTb ln c  0 Vb Va Ta bc

ca: Wca = Pa (Va − Vc ) =

nRTa Va



V 



T 



Va 



Ta 

(Va − Vc ) = nRTa  1 − c  = nRTa  1 − c  = nR (Ta − Tc )

Qca = nCP ( Ta − Tc ) = nR (Ta − Tc )  0 5 2

T T T nRTb ln c + nR ( Ta − Tc ) Tb ln c + (Ta − Tc ) Tb ln b + (Ta − Tb ) W W + Wca Ta Ta Ta = bc = = = e= T T T Qinput Qab + Qbc 3 nR T − T + nRT ln c 3 3 T − Ta ) + Tb ln c T − Ta ) + Tb ln b ( b a) 2 2( b 2( b b Ta Ta Ta

( 423 K ) ln = 3 2

423 K 273 K

+ ( 273 K − 423 K )

( 423 K − 273 K ) + ( 423 K ) ln

423 K

= 0.0859 = 8.59%

273 K Of course, individual values could have been found for the work and heat on each process, and used in the efficiency equation instead of referring everything to the temperatures.

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Instructor Solutions Manual

The maximum efficiency is the Carnot efficiency, given in Eq. 20–3, with temperatures in Kelvin. T ( 345 + 273) K = 0.230 or 23.0% e = 1− L = 1− TH ( 530 + 273) K We assume that both temperatures are measured to the same precision–the nearest degree.

(a) Calculate the Carnot efficiency for the given temperatures. T 77 K eideal = 1 − L = 1 − = 0.7372  74% 293 K TH This is a higher efficiency than we have seen in several of the other problems, due to the wide range of temperature. (b) First of all, it would be hard to keep a cold temperature reservoir at 77 K in a lab room at 293 K. The cold temperature reservoir would have to be extremely well insulated against heat simply entering it from its surroundings, and to allow people to work around it. Secondly, if the cold temperature reservoir were right at 77 K, any exhaust heat put into the cold temperature reservoir would begin to warm it. The liquid nitrogen would start to boil and then get warmer than 77 K. The idea of a “cold temperature reservoir” is that it is large enough that the addition of heat does not significantly change its temperature. So either there would have to be a very large quantity of liquid nitrogen, or only a relatively small amount of heat could be exhausted into it. Finally, if some of the liquid nitrogen were to “boil away,” capturing and re-liquefying it would add to the expense of the overall process, partially voiding the gains realized from having a high-efficiency engine.

Find the intake temperature from the original Carnot efficiency, and then recalculate the exhaust temperature for the new Carnot efficiency, using the same intake temperature. T T ( 340 + 273) K = 958 K e1 = 1 − L1 → TH = L1 = 1 − e1 1 − 0.36 TH e2 = 1 −

TL2

→ TL2 = TH (1 − e2 ) = ( 958 K )(1 − 0.45) = 527 K = 254C  250C

10. Find the exhaust temperature from the original Carnot efficiency, and then recalculate the intake temperature for the new Carnot efficiency, using the same exhaust temperature. T e1 = 1 − L → TL = TH1 (1 − e ) = ( 540 K + 273 K )(1 − 0.22 ) = 634.1K TH1 e2 = 1 −

→ TH2 =

TH2

TL 1 − e2

634.1 K

1 − 0.42

= 1093 K = 820o C

11. (a) The work done during any process is given by Eq. 19–8. So the net work done is as follows. W = Wab + Wbc + Wcd + Wda b

=  PdV +  PdV +  PdV +  PdV a

c     =   PdV +  PdV  −   PdV +  PdV  a  a  b d b

The sum of the first two terms is the area under the abc path (the “upper” path), and the sum of the last two © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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terms is the area under the adc path (the “lower” path). So the net work is the area enclosed by the cycle. (b) Any reversible cycle can be represented as a closed loop in the P–V plane. If we select the two points on the loop with the maximum and minimum volumes, we can apply the reasoning from above to find the net work. 12. This is a perfect Carnot engine, and so its efficiency is given by Eq. 20–1a and Eq. 20–3. Use these two expressions to solve for the rate of heat output. T W W ( 45 + 273) K = 0.3416 e = = → QL = W (1 e − 1) e = 1− L = 1− TH Q H W + QL ( 210 + 273) K QL t = W t (1 e − 1) = ( 960 W )(1 0.3416 − 1) = 1850 W  1900 W

13. The maximum (or Carnot) efficiency is given by Eq. 20–3, with temperatures in Kelvins. T ( 330 + 273) K = 0.3537 e = 1− L = 1− TH ( 660 + 273) K Thus the total power generated can be found as follows. Actual Power = ( Total Power )( max. eff.)( operating eff.) →

Total Power =

Actual Power

1.4 GW

( max. eff.)( operating eff.) ( 0.3537 )( 0.65)

= 6.089 GW

Exhaust Power = Total Power − Actual Power = 6.089 GW − 1.4 GW = 4.689 GW

(

)

= 4.689  109 J s ( 3600s h ) = 1.688  1013 J h  1.7  1013 J h 14. A 10ºC decrease in the low-temperature reservoir will give a greater improvement in the efficiency of a Carnot engine. By definition, TL is less than TH , so a 10ºC change will be a larger percentage change in TL than in TH , yielding a greater improvement in efficiency. As an example, we use the values from Problem 8 above. T 67 K = 0.7713 compared to 0.7372 elower = 1 − L = 1 − TH 293 K TL ehigher = 1 − TH

TL TH

= 1−

77 K 303 K

= 0.7458 compared to 0.7372

We see that the decrease in the lower temperature was more effective. Here is a more rigorous proof. Note that we never multiply by a negative value, so the original ordering of elower on the left TL

of the comparison and ehigher on the right of the comparison is preserved. We use the sign = to TH

mean “compare to.” T −T TL elower = 1 − L 0 ; ehigher = 1 − TH TH + T0 T T L

1−

TL − T0 TH

= 1 −

TL TH + T0

; −

TL − T0 TH

= −

TL TH + T0

;

TL TH + T0

=

TL − T0 TH

;

TLTH = ( TL − T0 )( TH + T0 ) ; T0TH + T0T0 = TLT0 ; TH + T0 = TL

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Since the left hand side of this last expression is larger than the right hand side, elower  ehigher. Thus in TL

general, a change in the low-temperature reservoir has a larger effect on the efficiency than the same change in the high-temperature reservoir. 15. We assume the efficiency of the person’s metabolism is that of a reversible engine. Then we take the work from that “engine” and assume that it is all used to increase the person’s potential energy by climbing higher. T W mgh e = 1− L = = → TH QH QH

 4186 J   Q  T   1kcal   1 − ( 273 + 20 ) K  = 1441m  1400 m h = H 1 − L  = mg  TH  ( 65 kg ) ( 9.80 m s2 )  ( 273 + 37 ) K  4.0  103 kcal 

16. This is a perfect Carnot engine, and so its efficiency is given by Eq. 20–1a and Eq. 20–3. Equate these two expressions for the efficiency. T W e = 1− L = → TH QH



W 



W t 





QH 



QH t 



TL = TH  1 −

 = TH  1 −

 = ( 480 + 273) K   1 −

 ( 950 kcal s )( 4186 J kcal )  5.2  105 J s

= 654 K = 382C  380C

17. The minimum value for TH would occur if the engine were a Carnot engine. We calculate the efficiency of the engine from the given data, and use this as a Carnot efficiency to calculate TH . W t

= Poutput = 7000 W ; ( to move car)

QH t

= Pinput (from gasoline)

 

=  3.2  107

J 

1L   21.8 m  = 41035W  L   17000 m   1s 

Poutput e=

W QH

( to move car)

Pinput

7000 W 41035 W

= 1−

TL TH

→ TH =

( 273 + 25) K

(1 − e )  1 − 7000 W   

(from gasoline)

= 359 K = 86C

41035 W 

18. The heat input must come during the isothermal expansion. From V Section 20–3, page 603 we have QH = nRTH ln b = nRTH ln 2. Since this Va is a Carnot cycle, we may use Eq. 20–3 combined with Eq. 20–1a. T W → e = 1− L = TH QH

T −T  T −T  W = QH  H L  = ( nRTH ln 2 )  H L  = nR ( TH − TL ) ln 2  TH   TH  The adiabatic relationship between points b and c and the ideal gas law are used to express the temperature ratio in terms of the volume ratio. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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nRTH

  = PV → PV b b c c

Vb =

(

V  Vc → TH = TL  c  Vc  Vb 

nRTL

 −1

(

= TL 6.2 2/3

)

W = nR ( TH − TL ) ln 2 = nRTL 6.2 2/3 − 1 ln 2 → TL =

(

920 J

) (1.00 mol )(8.314 J mol K ) ln 2 (6.2 − 1) T = ( 67.22 K ) ( 6.2 ) = 226.9 K  230 K nR ln 2 6.2

2/3

−1

2/3

= 67.22 K  67 K

2/3

19. (a) The pressures can be found from the ideal gas equation. nRT PV = nRT → P = → V nRTa ( 0.50 mol )( 8.314 J mol K )( 743 K ) = Pa = 7.5  10−3 m 3 Va = 4.118  105 Pa  4.1  105 Pa Pb =

nRTb Vb

( 0.50 mol )(8.314 J mol K )( 743 K ) 15.0  10−3 m 3

= 2.059  105 Pa  2.1  105 Pa (b) The volumes can be found from combining the ideal gas law and the relationship between pressure and volume for an adiabatic process. nRTc  nRTb    = PV → PV Vc = Vb → TcVc −1 = TbVb −1 → c c b b Vc Vb 1

2.5

 T   −1  743 K  Vc =  b  Vb =   (15.0 L ) = 34.4 L  533 K   Tc  1

2.5

 T   −1  743 K  Vd =  a  Va =   ( 7.5 L ) = 17.2 L  17 L  533 K   Td  (c) The work done at a constant temperature is given by Eq. 19–9. V   15.0  = 2141J  2100 J W = nRT ln  b  = ( 0.50 mol )( 8.314 J mol K )( 743K ) ln    7.5   Va  Note that this is also the heat input during that process, so Qab = 2141J. (d) Along process cd, there is no change in internal energy since the process is isothermic. Thus, by the first law of thermodynamics, the heat exhausted is equal to the work done during that process. V   17.2  = −1536 J  −1500 J Qcd = W = nRT ln  d  = ( 0.50 mol )( 8.314 J mol K )( 533K ) ln    34.4   Vc  So 1536 J of heat was exhausted during process cd. (e) From the first law of thermodynamics, for a closed cycle, the net work done is equal to the net heat input. And the net heat input occurs only along processes ab and cd, since the other two processes are adiabatic. Wnet = Qnet = 2141J − 1536 J = 605J  600 J © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(f)

W QH

605J 2141J

= 0.28 ; e = 1 −

TL TH

= 1−

533 K 743 K

Instructor Solutions Manual

= 0.28

20. (a) We use the ideal gas law and the adiabatic process relationship to find the values of the pressure and volume at each of the four points. Pa = 8.8atm ; Ta = 623 K ; Va =

nRTa

(1.00 mol )( 0.0821L atm mol K )( 623 K ) 8.8atm

= 5.81L  5.8 L

Tb = 623K ; Vb = 2Va = 2 ( 5.81L ) = 11.62 L  11.6 L Pb = Pa

= 12 Pa = 4.4 atm

  Tc = 483 K ; PV = PV → b b c c 1

nRTb Vb

 T   −1  623 K  Vc = Vb  b  = (11.62 L )    483 K   Tc  Pc =

nRTc

nRTd

To summarize: (b) Isotherm ab:

nRTc Vc

Vc →

3/ 2

= 17.02 L  17.0 L

(1.00 mol )( 0.0821L atm mol K )( 483 K ) 17.02 L

T  Td = 483 K ; Vd = Va  a   Td  Pd =

Vb =

 −1

 623 K    483 K 

= ( 5.81L ) 

3/ 2

= 8.51L  8.5 L

(1.00 mol )( 0.0821L atm mol K )( 483 K ) 8.51L

= 2.33atm  2.3atm

= 4.66 atm  4.7 atm

Pa = 8.8atm ; Va = 5.8 L ; Pb = 4.4 atm ; Vb = 11.6 L Pc = 2.3atm ; Vc = 17.0 L ; Pd = 4.7 atm ; Vd = 8.5 L Eint = 0 ; ab

Qab = Wab = nRTa ln

Adiabat bc:

Vb Va

= (1.00 mol )( 8.314 J mol K )( 623 K ) ln 2 = 3590 J  3600 J

Qbc = 0 ;

Eint = nCV ( Tc − Tb ) = 23 nR ( Tc − Tb ) = 23 (1.00 mol )( 8.314 J mol K )( −140 K ) bc

= −1746 J  −1700 J ; Wbc = Qbc − Eint = 1746 J  1700 J bc

Isotherm cd:

Eint = 0 ; cd

Qcd = Wcd = nRTc ln

Vd Vc

= (1.00 mol )( 8.314 J mol K )( 483 K ) ln 12 = −2783J  −2800 J

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Chapter 20

Second Law of Thermodynamics

Qda = 0 ;

Adiabat da:

Eint = nCV (Tc − Tb ) = 32 nR (Tc − Tb ) = 32 (1.00 mol )( 8.314 J mol K )(140 K ) bc

= 1746 J  1700 J ; Wbc = Qbc − Eint = 1746 J  −1700 J bc

ab : Eint = 0 ;

To summarize:

Q = 3600 J ; W = 3600 J

bc:

Eint = −1700 J ; Q = 0 ;

cd:

Eint = 0 ;

da:

Eint = 1700 J ; Q = 0 ;

W Qinput

e = 1−

W = 1700 J

Q = −2800 J ; W = −2800 J W = −1700 J

3590 J + 1746 J − 2783J − 1746 J 3590 J

= 1−

807 J 3590 J

= 0.2248  0.22

( 273 + 210) K = 0.2247  0.22 ( 273 + 350) K

TH The slight disagreement is due to rounding of various calculations.

21. The adiabatic compression takes place between temperatures of 25C and 430C. Use the adiabatic relationship and the ideal gas law to express the volumes in terms of the temperatures. nRTa nRTb   ; PV = nRT → Pa = , Pb = PV = PV → a a b b Va Vb 1

nRTa Va

Va =

nRTb Vb

Vb → TaVa −1 = TbVb −1

 T   −1  273K + 430 K 1.4−1 → = b =  = 8.55 Vb  Ta   273K + 25 K  Va

22. The ideal coefficient of performance is given by Eq. 20–4c. TL ( 273 + 2.5) K = = 14.13  14 COPideal = TH − TL ( 22 − 2.5) K 23. The coefficient of performance for a refrigerator is given by Eq. 20–4c, using absolute temperatures. TL ( −8 + 273) K COP = = = 6.463  6.5 TH − TL ( 33 + 273) K − ( −8 + 273) K 24. Consider the amounts of energy transferred in 1 second. The problem states that QH = 3900 J and QL = 3400 J, thus, W = 500 J. Use Eq. 20–1b and Eq. 20–3 for the efficiency, and solve for the

outside temperature, TL . e = 1−

QL QH

= 1−

TL TH

→ TL = TH

QL QH

= ( 23 + 273) K

3400 J 3900 J

= 258 K = −15C

25. The coefficient of performance for a refrigerator is given by Eq. 20–4c, using absolute temperatures. TL  COP   7.0  → TL = TH  COP =  = ( 22 + 273) K    = 258.1K = −14.9C  −15C TH − TL  1 + COP   8.0  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

695

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

26. We initially assume a COP of 3.0. For a heat pump the COP is given by Eq. 20–5. Q Q 2800 J = 933J  930 J (a) COP = H → W = H = W COP 3.0 (b) The calculation doesn’t depend on the outdoor temperature, so W = 930 J . (c) The efficiency of a perfect Carnot engine is given by Eqs. 15–4a and 15–5. Equate these two expressions to solve for the COP expression required.  T  T W T W → 1− L = → W = QH  1 − L  e = 1− L = ; e = TH QH TH QH  TH 

COP =

QH W



QH  1 −



W0C =

COP

(TH − TL )



W−15C = QH  1 −



 TL  ( TH − TL ) 1 − T   H  Q (T − T )  T  0 + 273  = H H L = QH  1 − L  = ( 2800 J )  1 −  = 210 J TH  22 + 273   TH 

TL  TH 

QH TH

TL 

 −15 + 273  = 350 J = ( 2800 J )  1 −   22 + 273  TH  

27. The COP for an ideal heat pump is given by Eq. 20–5. Q QH TH ( 24 + 273) K (a) COP = H = = = = 16.5  17 W QH − QL TH − TL 18 K (b) COP =

QH W

→ QH = (W t )( t )( COP ) = (1200 W )( 3600s )(16.5) = 7.128  10 7 J  7.1  10 7 J

28. (a) Use Eq. 20–4a. QL COPideal =

QL W

QL QH − QL

QH QH QH

−

QL QH

TL TH TH = = T TH TL TH − TL − 1− L TH TH TH

(b) Use Eq. 20–3. TL e = 1−

TL TH

TL TH − TL

→

TL TH

= 1 − e ; COPideal =

( 273 − 18) K 42 K

TH TH TH

−

1− e 1 − (1 − e )

1− e e

= 6.1

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Chapter 20

Second Law of Thermodynamics

29. (a) The total rate of adding heat to the house by the heat pump must equal the rate of heat lost by conduction. QL + W = ( 650 W C )( Tin − Tout ) t Since the heat pump is ideal, we have the following from the efficiency. T Q QL W Tin 1 − out = 1 − L = 1 − = → QL + W = W Tin QH QL + W QL + W Tin − Tout Combine these two expressions, and solve for Tout . QL + W t

= ( 650 W C )( Tin − Tout ) =

(Tin − Tout )2 = Tout = Tin −

Tin

t ( 650 W C )

Tin

t ( 650 W C )

Tin

t ( Tin − Tout )

→

= 295 K −

(1700 W )

295 K

( 650 W C )

= 267 K = −6C

(b) If the outside temperature is less than −6C , then even if the heat pump runs continuously, it won’t provide enough heat to counteract the heat loss by conduction. Thus the house will get cooler–the heat pump is not able to “keep up.” (c) If the outside temperature is 8C, then the rate of heat loss by conduction is found to be ( 650 W C )(14 C ) = 9100 W. The heat pump must provide this much power to the house in order for the house to stay at a constant temperature. That total power is ( QL + W ) t. Use this to solve for rate at which the pump must do work.  W Tinside (QL + W ) t =   = 9100 W → t  Tinside − Toutside 

T −T   14 K  = 9100 W  inside outside  = 9100 W   = 432 W Tinside t  295 K    Since the maximum power the pump can provide is 1700 W, the pump must work 432 W = 0.25 or 25% of the time. 1700 W W

30. The COP for a heat pump is COP =

QH W

and the efficiency is e =

each other. So if the efficiency is 0.33, the COP is

1 0.33

W QH

. Thus they are reciprocals of

= 3.0 .

31. The coefficient of performance is the heat removed from the low-temperature area divided by the work done to remove the heat. In this case, the heat removed is the latent heat released by the freezing ice, and the work done is 1.2 kW times the elapsed time. The mass of water frozen is its density times its volume.

697

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

COP = V=

QL W

mLf

( COP ) Pt  Lf

W =

VLf → Pt ( 6.0 )(1200 W )( 3600 s )

(1.0 10 kg m )( 3.33 10 J kg ) 3

Instructor Solutions Manual

 1000 L   78 L 3   1m 

= 0.0778 m 3 

32. (a) The ideal COP is given by Eq. 20–4c. TL TL ( 273 + 24 ) K → COPeff = ( 0.25) = ( 0.25) = 5.304  5.3 COP = TH − TL TH − TL 14 K (b) The compressor power can be found from Eq. 20–4a. Q QL COP = L → W = → COPeff W

 33, 000 Btu   1055J   1h     h   Btu   3600s   1kW  Q t  W t= L =  1000 W  = 1.823kW  1.8 kW COPeff 5.304    1hp  (c) 1823W   = 2.4 hp  746 W  33. The $2,100 worth of heat provided by the electric heater is the same amount of heat that the heat pump would need to provide, and so that amount of heat is QH. The amount of energy required to run the heat pump to deliver that same amount if heat is found from the coefficient of performance. Q Q Q COP = H → W = H = H W COP 2.9 So if the cost for QH is divided by 2.9, we get the cost of running the heat pump to deliver the needed heat. Subtract that from the total cost to get the savings. $2,100 Savings = $2,100 − = $1,375.86  $1, 400 2.9 Divide the cost of the heat pump by the annual savings to find the break-even time. $15, 000 = 10.90 years  11years $1, 375.86 year The total savings over 20 years is the savings in heating costs, minus the price of the heat pump. Total savings = ( $1,375.86 year )( 20 years ) − $15,000 = $12,517.20  $13,000 34. Heat energy is taken away from the steam in order to condense it to liquid water, so the change in entropy will be negative. The heat transfer is the mass of the steam times the latent heat of vaporization. ( 0.28 kg ) 22.6  105 J kg mLvap Q S = = − =− = −1696 J K  −1700 J K T T ( 273 + 100 ) K

(

)

698

Chapter 20

Second Law of Thermodynamics

35. The heat added to the water is found from Eq. 19–2, Q = mcT . Because we are told to “estimate” the change in entropy, we use the average temperature of 50C in Eq. 20–6 for entropy instead of integrating. Q mcT (1.0 kg )( 4186 J kg C )(100C ) S = = = = 1296 J K  1300 J K T T ( 273 + 50) K Note, if we had integrated we would have gotten 1306 J/K. So the estimate is quite good. T final

S = 

Tinitial

dQ T

T final

= mc 

Tinitial

dT T

 Tfinal   373K  = (1.0 kg )( 4186 J kg C ) ln    = 1306.47 J K  273K   Tinitial 

= mc ln 

36. Heat energy is taken away from the water, so the change in entropy will be negative. The heat taken away from the water is found from Q = mLfusion. Note that 1.00 m3 of water has a mass of 1.00  103 kg.

S =

Q T

=−

mLfusion T

(1.00 10 kg )( 3.33 10 J kg ) = −1.22  10 J K =− 3

273K

37. Energy has been made “unavailable” in the frictional stopping of the sliding box. We take that “lost” kinetic energy as the heat term of the entropy calculation. S =

1 2

mv02

1 2

( 6.8 kg )( 4.0 m s )

= = = 0.1857 J K  0.19 J K 293 K T T Since this is a decrease in “availability,” the entropy of the universe has increased.

38. (a) S water =

Qwater

mwater Lvaporization

( 0.45 kg ) ( 2.26  106 J kg )

= = = 2727 J K  2700 J K 373 K Twater Twater (b) Because the heat to vaporize the water comes from the surroundings, and we assume that the temperature of the surroundings does not change, we have Ssurroundings = −S water = −2700 J K . (c) The entropy change of the universe for a reversible process is Suniverse = 0 . (d) If the process were irreversible, we would have Ssurroundings  S water , because the temperature of the surroundings would increase, and so S universe  0 .

39. Because the temperature change is small, we can approximate any entropy integrals by S = Q Tavg. There are three terms of entropy to consider. First, there is a loss of entropy from the water for the freezing process, S1 . Second, there is a loss of entropy from that newly-formed ice as it cools to −8.0C , S 2 . That process has an “average” temperature of −4.0C . Finally, there is a gain of

entropy by the “great deal of ice,” S 3 , as the heat lost from the original mass of water in steps 1 and 2 goes into that great deal of ice. Since it is a large quantity of ice, we assume that its temperature does not change during the processes. The density of water is 1000 kg per cubic meter.

699

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

S1 = S 2 = S3 =

Q1 T1

=−

T2 Q3 T3

=−

273K

=−

−Q1 − Q2

(1.00 10 kg ) ( 2100 J kg C)(8.0C) = −6.2453  10 J K 3

mcice T2

(1.00 10 kg )( 3.33 10 J kg ) = −1.2198  10 J K 3

mLfusion

Instructor Solutions Manual

( −4 + 273) K

mLfusion + mcice T2 T3

(1.00  10 kg ) ( 3.33  10 J kg ) + ( 2100 J kg C)(8C) = 1.32  10 J K = 3

( −8 + 273) K

S = S1 + S 2 + S3 = −1.2198  106 J K − 6.2453  104 J K + 1.32  106 J K = 3.7747  104 J K  4  104 J K 40. The equilibrium temperature is found using calorimetry, from Chapter 19. The heat lost by the aluminum is equal to the heat gained by the water. We assume the Styrofoam insulates the mixture.





 Al







mAl cAl  Tinitial − Tfinal  = mH O cH O  Tfinal − Tinitial  → 2



H2O



mAl cAlTinitial + mH O cH OTinitial Tfinal =

mAl cAl + mH O cH O 2

H2O

( 2.8 kg )( 900 J kg C )( 36.5C ) + (1.0 kg )( 4186 J kg C )( 20.0C ) = 26.20C ( 2.8 kg )( 900 J kg C ) + (1.0 kg ) ( 4186 J kg C ) T final

S = SAl + S H O = 

dQAl T

Tinitial

T final

+ 

Tfinal Tinitial

Tinitial

= mAl cAl ln

dQH O

T final

= mAl cAl 

Tinitial

H2O

+ mH O cH O ln 2

dT T

T final

+ mH O c H O  2

Tinitial

dT T

H2O

Tfinal Tinitial H 2O

= ( 2.8 kg )( 900 J kg K ) ln

( 273.15 + 26.2 ) K ( 273.15 + 26.2 ) K + (1.0 kg )( 4186 J kg K ) ln ( 273.15 + 36.5) K ( 273.15 + 20.0 ) K

= 2.3594 J K  2.4 J K

41. The same amount of heat that leaves the high temperature heat source enters the low temperature body of water. The temperatures of the heat source and body of water are constant, so the entropy is calculated without integration. S = S1 + S2 = −

Q Thigh

Q Tlow

 1

= Q

 Tlow

−

1 

 →

Thigh 

700

Chapter 20

Second Law of Thermodynamics

S t

Q 1

 1  1 1  4.186 J   − −   = ( 8.25cal s )     t  Tlow Thigh   1cal   ( 22 + 273) K ( 225 + 273) K 

= 4.77  10 −2

J K s

42. Take the energy transfer to use as the initial kinetic energy of the rock, because this energy becomes “unusable” after the collision–it is transferred to the environment. We assume that the rock and the environment are both at temperature T0 . S = Q T

→

S = K T0

43. Because the process happens at a constant temperature, we have S = Q T . The heat flow can be found from the first law of thermodynamics, the work for expansion at a constant temperature, and the ideal gas equation. V P Eint = Q − W = 0 → Q = W = nRT ln 2 = PV ln 1 → V1 P2 S =

Q T

( 7.5) (1.013  105 Pa )( 2.50  10−3 m3 )

440 K

7.5atm 1.0atm

= 8.7 J K

44. (a) We find the final temperature of the system using calorimetry, and then approximate each part of the system as having stayed at an average constant temperature. We write both heat terms as positive and then set them equal to each other. Qlost = Qgained → mhot c ( Thot − Tfinal ) = mcoolc ( Tfinal − Tcool ) → Tfinal =

Thot mhot + Tcool mcool

( 38.0C )( 3.0 kg ) + (12.0C )( 2.0 kg )

= 27.6C mcool + mhot 5.0 kg To calculate the entropy we must use the correctly signed heat terms. m c ( T − T ) ( 2.0 kg )( 4186 J kg K )( 27.6C − 12.0C ) Q Scool = cool = cool final cool = = 446.0 J kg 1 1 273 K + 12.0 K ) + ( 273 K + 27.6 K )  Tcool T + Tfinal ) 2 ( cool 2 ( avg

S hot =

Qhot Thot

mhot c ( Thot − Tfinal ) 1 2

(Thot + Tfinal )

( 3.0 kg )( 4186 J kg K )( 27.6 − 38.0C ) = −427.1J kg 1 273 K + 38.0 K ) + ( 273 K + 27.6 K )  2 (

avg

S = Scool + S hot = 446.0 J kg − 427.1J kg = 18.9 J kg  19 J kg

(b) We use the final temperature as found above, and then calculate the entropy changes by integration. Scool = 

dQcool T

Tfinal

= 

Tcool

mcool c dT T

T = mcool c ln final Tcool

 273 K + 27.6 K   = 446.2 J kg  273 K + 12.0 K 

= ( 2.0 kg )( 4186 J kg K ) ln 

701

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

S hot = 

dQhot T

Tfinal

= 

mhot c dT T

Thot

Instructor Solutions Manual

T = mhot c ln final Thot

 273K + 27.6 K   = −427.1J kg  273K + 38.0 K 

= ( 3.0 kg )( 4186 J kg K ) ln 

S = Scool + S hot = 446.2 J kg − 427.1J kg = 19.1J kg  19 J kg

Using integration only changed the answer by about 1%. 45. (a) eactual = W QH = 550 J 2500 J = 0.22

eideal = 1 − TL TH = 1 − 650 K 970 K = 0.330

Thus eactual eideal = 0.220 0.330 = 0.667  67 % of ideal (b) The heat reservoirs do not change temperature during the operation of the engine. There is an entropy loss from the input reservoir, because it loses heat, and an entropy gain for the output reservoir, because it gains heat. Note that QL = QH − W = 2500 J − 550 J = 1950 J. S = Sinput + S output = −

QH TH

QL TL

=−

2500 J 1950 J + = 0.42 J K 970 K 650 K

=−

QH TL TH

=−

= 0 TH TL TH TL TH TH A numeric calculation might give a very small number due to not keeping all digits in the calculation.

46. Since the process is at a constant volume, dQ = nCV dT. For a diatomic gas in the temperature range of this problem, CV = 52 R. S = 

dQ T

=

nCV dT

= 52 nR ln

T2 T1

= 52 (1.6 mol )( 8.314 J mol K ) ln

( 273 + 55 ) K ( 273 + 25) K

= 3.19 J K  3.2 J K

47. (a) The same amount of heat that leaves the room enters the ice and water. Q mLfusion 3.33  105 J kg S melt = melt = =m = (1219.78m ) J K Tmelt Tmelt 273 K S warming =  S room =

dQwarming T

Troom

= 

Tmelt

mcdT T

− mL − mc ( Troom − Tmelt ) Troom

= mc ln

Troom Tmelt

= m ( 4186 J kg K ) ln

293 K 273 K

= ( 295.95m ) J K

 3.33  105 J kg + ( 4186 J kg K )( 20 K )   293 K  

= −m 

= ( −1422.25m ) J K

S total = S melt + S warming + S room = (1219.78m ) J K + ( 295.95m ) J K − (1422.25m ) J K = 93.48m J K  ( 93m ) J K

This process will occur naturally. Note that we are assuming that m is in kg. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

702

Chapter 20

Second Law of Thermodynamics

(b) For this situation, every heat exchange is exactly the opposite as in part (a). Thus we have Stotal  ( −93m ) J K . This will not occur naturally.

48. Entropy is a state variable, and so the entropy difference between two states is the same for any path. Since we are told that states a and b have the same temperature, we may find the entropy change by calculating the change in entropy for an isothermal process connecting the same two states. We also use the first law of thermodynamics. Eint = nCV T = 0 = Q − W → Q = Wisothermal = nRT ln (Vb Va ) S = 

dQ T

Q T

nRT ln (Vb Va )

= nR ln (Vb Va )

49. (a) To approximate, we use the average temperature of the water. J   (1.00 kg )  4186  ( 85C ) dQ Q mcTwater kg K   S =   = = = 1128J K  1100 J K T Tavg Tavg ( 273 + 12 85) K (b) The heat input is given by Q = mcT , so dQ = mcdT .

S = 

= T1

mcdT T

T J   ( 273 + 85) K   = mc ln 2 = (1.00 kg )  4186   ln  T1 kg K   ( 273) K  

= 1134.6 J K  1130 J K The approximation is only about 0.6% different if we keep several significant figures. (c) We assume that the temperature of the surroundings is constant at 85C (the water was moved from a cold environment to a hot environment). J   − (1.00 kg )  4186  ( 85C ) dQ Q − mcTwater kg K   S =  = = = −990 J K T T Tsurroundings ( 273 + 85) K If instead the heating of the water were done reversibly, the entropy of the surroundings would decrease by 1130 J/K. For a general non-reversible case, the entropy of the surroundings would decrease, but by less than 1130 J/K (as in the calculation here). 50. (a) The equilibrium temperature is found using calorimetry, from Chapter 19. The heat lost by the water is equal to the heat gained by the aluminum.













mH O cH O  Tinitial − Tfinal  = mAl cAl  Tfinal − Tinitial  → 2

 HO 2



mAl cAlTinitial + mH O cH OTinitial Tfinal =

H2 O

mAl cAl + mH O cH O 2

( 0.11kg )( 900 J kg C )( 35C ) + ( 0.15 kg )( 4186 J kg C )( 45C ) = 43.64C  44C ( 0.11kg )( 900 J kg C ) + ( 0.15 kg ) ( 4186 J kg C )

703

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

T final

(b) S = SAl + S H O = 

dQAl T

Tinitial

T final

+ 

Tfinal Tinitial

Tinitial

= mAl cAl ln

dQH O

T final

= mAl cAl 

Tinitial

H2O

+ mH O cH O ln 2

dT T

Instructor Solutions Manual

T final

+ mH O c H O  2

Tinitial

dT T

H2O

Tfinal Tinitial

H 2O

( 273.15 + 43.64 ) K ( 273.15 + 35) K ( 273.15 + 43.64 ) K + ( 0.15 kg )( 4186 J kg K ) ln = 4.774  10−2 J K  4.8  10 −2 J K + 273.15 45 K ( )

= ( 0.11kg )( 900 J kg K ) ln

51. (a) The figure shows two processes that start at the same state. The top process is adiabatic, and the bottom process is isothermic. We see from the figure that at a volume of V/2, the pressure is greater for the adiabatic process. We also prove it analytically. Isothermal: V  PV PV V T 1 1 = 2 2 → P2 = P1 1 2 = P1   (1) = 2 P1 1 T1 T2 V2 T1  2V  Adiabatic: 

 1 1

 2 2

PV = PV



V  V   → P2 = P1  1  = P1   = 2 P1 1 V V 2   2

Since   1, we see that ( P2 )adiabatic  ( P2 )isothermic. The ratio is

No heat is transferred to or from the gas, so Sadiabatic = 

(b) For the adiabatic process:

dQisothermal T

dQ T

= 0.

V  = 0 → Qisothermal = Wisothermal = nRT ln  2  isothermal  V1 

Eint

For the isothermal process: Sisothermal = 

( P2 )adiabatic 2 P1  −1 = =2 . ( P2 )isothermic 2 P1

dQ T

isothermal

Qisothermal T

nRT ln (V2 V1 ) T

= nR ln (V2 V1 ) = nR ln ( ) = −nR ln 2 (c) Since each process is reversible, the energy change of the universe is 0, and so Ssurroundings = −Ssystem . For the adiabatic process, Ssurroundings = 0 . For the isothermal process, 1 2

Ssurroundings = nR ln 2 . (d) The two processes do not give the same S because the two processes do not both start and end

in the same state. As can be seen in the P–V diagram in part (a), while the two processes start in the same state, they do not end in the same state. Thus, S will not be the same for the two processes.

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52. (a) The gases do not interact since they are ideal, and so each gas expands to twice its volume with no change in temperature. Even though the actual process is not reversible, the entropy change can be calculated for a reversible process that has the same initial and final states. This is discussed in Example 20–9. V S N = S Ar = nR ln 2 = nR ln 2 V1 2

S total = S N + S Ar = 2nR ln 2 2

= 2 (1.00 mol )( 8.314 J mol K ) ln 2 = 11.5J K

(b) Because the containers are insulated, no heat is transferred to or from the environment. Thus dQ Ssurroundings =  = 0. T (c) Let us assume that the argon container is twice the size of the nitrogen container. Then the final nitrogen volume is 3 times the original volume, and the final argon volume is 1.5 times the original volume.

 V2   V2   = nR ln 3 ; S Ar = nR ln   = nR ln1.5  V1  N 2  V1  Ar

S N = nR ln  2

S total = S N + S Ar = nR ln 3 + nR ln1.5 = nR ln 4.5 = (1.00 mol )( 8.314 J mol K ) ln 4.5 2

= 12.5J K

53. (a) Entropy is a state function, which means that its value only depends on the state of the sample under consideration, not on its history of how it arrived at that state. A cyclical process starts and ends at the same state. Since the state is the same, the entropy is the same, and thus the change in entropy for the system is 0. Then, because all of the processes involved are reversible, the entropy change for the universe is 0, and so the entropy change for the surroundings must also be 0. (b) For the two adiabatic processes, Q is constant. Thus, dS =

dQ = 0 for every infinitesimal part of an adiabatic T

path, and so S bc = Sda = 0. For the two isothermic processes, we have the following, based on the first law of thermodynamics and Eq. 19–9. V Eint = Q − W = 0 → Qab = Wab = nRTH ln b → Va state b

Sab =

state b

Qab

 T = T  dQ = T =

state a

H state a

Scycle = Sab + Scd = nR ln

nRTH ln TH

Vb Va

= nR ln

Vb Va

; Scd = nR ln

Vd Vc

V V  V + nR ln d = nR ln  b d  Va Vc  Va Vc 

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

From the discussion on pages 603–4, we see that

Vb Va

Vc Vd

Instructor Solutions Manual

→

Vb Vd Va Vc

= 1. Thus

 Vb Vd   = nR ln1 = 0.  Va Vc 

Scycle = nR ln 

54. For a system with constant volume, the heat input is given by Eq. 19–11a, Q = nCV T . At temperature T, an infinitesimal amount of heat would result in an infinitesimal temperature change, related by dQ = nCV dT . Use this with the definition of entropy. dQ

dS =

nCV dT T

→

T nCV

This is exactly the definition of the slope of a process shown on a T–S graph, so the slope is

nCV

A function with this property would be T = T0eS / nCV . 55. We assume that the process is reversible, so that the entropy change is given by Eq. 20–8. The heat transfer is given by dQ = nCV dT . T2

S= T1

dQ T

= T1

nCV dT T

(

=

(

)

n aT + bT 3 dT T

(

)

(

=  n a + bT 2 dT = n aT + 13 bT 3 T1

)

(

)

T2 T1

)



= ( 0.15mol ) 2.08 mJ mol K 2 (1.0 K − 5.0 K ) + 13 2.57 mJ mol K 4 (1.0 K ) − ( 5.0 K )  3

= −17.182 mJ K  −17 mJ K

56. (a) Express the first law of thermodynamics in differential form, as given in Section 19–6. dEint = dQ − dW For a reversible process, dQ = TdS (Eq. 20–7), and for any process, dW = PdV . Also, since Eint = nCV T , we have dEint = nCV dT . Finally, for an ideal gas, P =

dEint = dQ − dW → nCV dT = TdS − PdV = TdS − TdS = nCV dT +

nRT

dV → dS = nCV

+ nR

nRT V

dV →

V T V (b) Use the ideal gas law, the differentiation product rule, and Eq. 19–11a, with the above result. PdV VdP nRdT PV = nRT → PdV + VdP = nRdT → + = → nRT nRT nRT PdV VdP dT dT dV dP + = → = + PV PV T T V P

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dS = nCV dS = nCV

dT T

+ nR

dV V

+ nCP

dV dP dV  dV dP  + = nCV + n ( CV + R )  + nR P  V P V V

= nC V 

P V (c) Let dS = 0 in the above result. dP dV dP dV + nCP = 0 → nCV = − nCP dS = nCV P V P V

→

dP P

=−

CP dV CV V

= −

dV V

 P = −  V

→ ln P = − ln V + C = ln V −  + C → e ln P = e ln V + C = e ln V eC →

P = V − eC →

PV  = constant = eC

−

57. (a) The kinetic energy the rock loses when it hits the ground becomes a heat flow to the ground and air. That energy is then unavailable. We assume the temperature of the ground and air, TL , does not change appreciably when the rock hits it. K Q K S = = → Elost = TL S = TL   = K T TL  TL  (b) The work done in a free expansion (which is isothermic if it is insulated) becomes unavailable V as the gas expands. From Example 20–9, S = nR ln 2 . The work done in an isothermal V1 expansion is given in Eq. 19–9, as W = nRT ln Elost = TL S = TL nR ln (V2 V1 ) = W

V2 V1

. Since it is isothermal, T = TL .

(c) Assume an amount of heat QH is transferred from a high temperature reservoir TH to a lower temperature reservoir TL . The entropy change during that process is S =

 QH

energy “lost” is TL S = TL 

 TL

−

QH TL

−

QH TH

. The

QH 

 T  = QH  1 − L  = QH eCarnot = W . The work done during the  TH   TH 

process is no longer available to do any other work, and so has become unavailable to do work in some other process following the first process. 58. The total energy stored in the copper block is found from the heat flow that initially raised its temperature above the temperature of the surroundings. Q = mcT = ( 2.2 kg )( 390J kg K )( 200 K ) = 1.716  105 J We find the entropy change assuming that amount of energy leaves the copper in a reversible process, and that amount of energy enters the surroundings. The temperature of the surroundings is assumed to be constant at 290 K. 290 K dQ mcdT  290 K   290 K  SCu =  =  = mc ln  = ( 2.2 kg )( 390 J kg K ) ln    = −450 J K T T  490 K   490 K  490 K Ssurroundings =

Q Tsurroundings

mcT Tsurroundings

1.716  105 J 290 K

= 592 J K ; S = ( 592 − 450 ) J K = 142 J K

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Elost = TL S = ( 290 K )(142 J K ) = 4.118  104 J Wavailable = Q − Elost = 1.716  105 J − 4.118  104 J = 1.3  105 J 59. For four heads: For 3 heads, 1 tail: For 2 heads, 2 tails: For 1 head, 3 tails: For four tails:

( ) W = 4 → S = k ln W = (1.38  10 J K ) ln 4 = 1.91  10 J K W = 6 → S = k ln W = (1.38  10 J K ) ln 6 = 2.47  10 J K W = 4 → S = k ln W = (1.38  10 J K ) ln 4 = 1.91  10 J K W = 1 → S = k ln W = (1.38  10 J K ) ln1 = 0 W = 1 → S = k ln W = 1.38  10−23 J K ln1 = 0 −23

−23

60. From the table below, we see that there are a total of 26 = 64 microstates.

(a) The probability of obtaining three heads and three tails is 20 64 or 5 16 . (b) The probability of obtaining six heads is 1 64 . 61. (a) There is only one microstate for 4 tails: TTTT. There are 6 microstates with 2 heads and 2 tails: HHTT, HTHT, HTTH, THHT, THTH, and TTHH. Use Eq. 20–14 to calculate the entropy change. W S = k ln W2 − k ln W1 = k ln 2 = (1.38  10−23 J K ) ln 6 = 2.47  10 −23 J K W1 (b) Apply Eq. 20–14 again. There is only 1 final microstate, and about 1.0  1029 initial microstates, according to Table 20–2. 1 W   = −9.2  10−22 J K S = k ln W2 − k ln W1 = k ln 2 = 1.38  10−23 J K ln  29  W1  1.0  10  (c) These changes are much smaller than those for ordinary thermodynamic entropy changes. The changes in entropy calculated in the referenced examples are 34 J/K, 10 J/K, 5.76 J/K, and ~ 1000 J/K. For ordinary processes, there are many orders of magnitude more particles than we have considered in this problem. That implies that there are enormously many more microstates and thus larger entropy values than in our simple coin toss examples.

(

)

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Chapter 20

Second Law of Thermodynamics

62. When throwing two dice, there are 36 possible microstates. (a) The possible microstates that give a total of 7 are: (1)(6), (2)(5), (3)(4), (4)(3), (5)(2), and (6)(1). Thus the probability of getting a 7 is 6 36 = 1 6 . (b) The possible microstates that give a total of 11 are: (5)(6) and (6)(5). Thus the probability of getting an 11 is 2 36 = 1 18 . 63. The number of microstates for macrostate A is WA = macrostate B is WB =

10! 5!5!

10! 10!0!

= 1. The number of microstates for

= 252.

(a) S = k ln WB − k ln WA = k ln

WB WA

(

)

= 1.38  10 −23 J K ln 252 = 7.63  10 −23 J K

Since S  0, this can occur naturally. W (b) S = k ln WA − k ln WB = − k ln B = − (1.38  10 −23 J K ) ln 252 = −7.63  10 −23 J K WA Since S  0, this cannot occur naturally.

 103 W h   1day   1 m 2  64. The required area is  28 = 18.30 m 2  18 m 2 . A small house with      day   9 h Sun   170 W   2

 1  1 m  1000 ft of floor space and a roof tilted at 30 has a total roof area of (1000 ft )   = o   cos 30   3.28 ft  2

107 m 2 , which is ~ 6 times larger than the area needed, and so the cells would fit on the house . But not all parts of the roof would have 9 hours of sunlight (some parts of the roof might receive hardly any sunlight), and the sunlight is not directly perpendicular to the roof at all times, so more than the minimum area of cells would be needed.

65. (a) Assume that there are no dissipative forces present, and so the energy required to pump the water to the lake is just the gravitational potential energy of the water. U grav = mgh = 1.00  105 kg s (10.0 h ) 9.80 m s 2 (105 m ) = 1.029  109 W h

(

)

(

)

 1.03  106 kWh

(1.029  10 kW h ) ( 0.75) = 5.5  10 kW = 55 MW 6

(b)

14 h

66. (a) We assume that the electrical energy comes from the 100% effective conversion of the gravitational potential energy of the water, “falling” from the surface of the lake to the base of the dam. W = mgh →

W t

m t

gh = 

V t

(

)(

)

gh = 1.00  103 kg m 3 35 m 3 s 9.80 m s 2 ( 48 m ) = 1.646  107 W

 1.6  107 W = 16 MW

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) One very obvious environmental implication is the risk of the dam failing at some point. The release of the lake water would be devastating to much of the environment immediately downstream of the dam. Other negative implications include the loss of the land and environment that has been covered by the artificial lake, and the disruption of the environment of the previously undammed river. Positive implications are that the energy produced is fairly “clean”–no fossil fuels are combusted to produce the electrical power, and downstream flooding can be controlled by the dam. 67. According to the heat figures provided by the inventor, the engine is 67% efficient. W W t 2.00 MW e= = = = 0.667 QH QH t 3.00 MW The ideal engine efficiency at the operating temperatures is given by Eq. 20–3. T 215 K eideal = 1 − L = 1 − = 0.494 425 K TH Thus, his engine is not possible, even if it were ideal. So yes , there is something “fishy” about his claim. His engine is even “better” than ideal, which means the quoted values cannot be true. 68. (a) Calculate the Carnot efficiency for an engine operated between the given temperatures. T ( 273 + 4 ) K = 0.077 = 7.7% eideal = 1 − L = 1 − TH ( 273+27 ) K (b) Such an engine might be feasible in spite of the low efficiency because of the large volume of “fuel” (ocean water) available. Ocean water would appear to be an “inexhaustible” source of heat energy. And the oceans are quite accessible. (c) The pumping of water between radically different depths would probably move smaller seadwelling creatures from their natural location, perhaps killing them in the transport process. This might affect the food chain of other local sea-dwelling creatures. Mixing the water at different temperatures will also disturb the environment of sea-dwelling creatures. There is a significant dynamic of energy exchange between the ocean and the atmosphere, and so any changing of surface temperature water might affect at least the local climate, and perhaps also contribute to larger-scale climate changes. 69. The gas is diatomic, and so  = 1.4 and CV = 52 R. (a) Find the number of moles by applying the ideal gas law to state a. 1.013  105 Pa )( 0.010 m 3 ) ( PV a a PV = nRTa → n = = = 0.406 mol  0.41mol a a RTa (8.314 J mol K )( 300 K ) (b) Find Tc using the adiabatic relationship with the ideal gas law.   PV = PV → c c a a

V  Tc = Ta  a   Vc 

nRTc Vc

Vc =

nRTa Va

Va → TcVc −1 = TaVa −1 →

 −1

= ( 300 K )( 2 )

0.4

= 396 K  400 K ( 2 sig. fig.)

(c) This is a constant volume process. Qbc = nCV (Tc − Tb ) = 52 nR (Tc − Tb ) = 52 ( 0.406 mol )( 8.314 J mol K )( 96 K ) = 810.1J  810 J

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Chapter 20

Second Law of Thermodynamics

(d) The work done by an isothermal process is given by Eq. 19–9. V  0.0050  = −702 J  −700 J Wab = nRT ln b = ( 0.406 mol )(8.314 J mol K )( 300 K ) ln   Va  0.010  (e) Use the first law of thermodynamics, and the fact that Qca = 0. Eint = Qca − Wca = −Wca → ca

Wca = −Eint = − nCV T = n ( 52 R ) (Tc − Ta ) = 52 ( 0.406 mol )( 8.314 J mol K )( 396 K − 300 K ) ca

= 810.1J  810 J

(f)

Heat is input to the gas only along path bc. (Heat is exhausted along path ab.) W Wca + Wab 810.1J − 702 J e= = = = 0.13 810.1J Qin Qbc

(g) eCarnot =

TH − TL TH

396 K − 300 K 396 K

= 0.24

70. We start with Eq. 20–4a for the COP of a refrigerator. The heat involved is the latent heat of fusion for water. Q Q COP = L → W = L → COP W W t=

QL t COP

6 tons 0.18COPideal

(

6 ( 909 kg d ) 3.33  105 J kg

 273 K + 22 K   13 K  

) = 4.446  10 J d 8

0.18 

This is the amount of energy that the air conditioner could use in a day, if it ran continuously. But the air conditioner is only running for 6 hours per day to cool the house. Thus, the homeowner needs to buy ¼ of the amount of joules found above, for 1 day of cooling.  1kWh   $0.13  = $4.014 d  $4.0 d cost day = 14 4.446  108 J d   6   3.600  10 J   kWh 

(

cost day = ( $4.014 d )

)

30d 1month

= $120 month

71. (a) For each engine, the efficiency is given by e = 0.65eCarnot. Thus,



e1 = 0.65eC −1 = 0.65  1 −





e2 = 0.65eC − 2 = 0.65  1 −



 ( 440 + 273) K   = 0.197.  = 0.65 1 − 750 273 K TH1  + ( )   TL1 

 ( 240 + 273) K   = 0.165  = 0.65 1 − TH2   ( 415 + 273) K  TL2 

For the first engine, the input heat is from the coal. W1 = e1QH1 = e1Qcoal and QL1 = QH1 − W1 = (1 − e1 ) Qcoal For the second energy, the input heat is the output heat from the first engine. W2 = e2QH2 = e2QL1 = e2 (1 − e1 ) Qcoal

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Add the two work expressions together, and solve for Qcoal.

W1 + W2 = e1Qcoal + e2 (1 − e1 ) Qcoal = ( e1 + e2 − e1e2 ) Qcoal Qcoal =

W1 + W2

(W1 + W2 ) t

→ Qcoal t =

e1 + e2 − e1e2

Calculate the rate of coal use from the required rate of input energy, Qcoal t . Qcoal t =

950  106 W 0.197 + 0.165 − ( 0.197 )( 0.165)

= 2.883  109 J s

( 2.883  10 J s )  2.81kg10 J  = 103.0 kg s  1.0  10 kg s 2





(b) The heat exhausted into the water will make the water temperature rise according to Eq. 19–2. The heat exhausted into water is the heat from the coal, minus the useful work. Qexhaust Q −W Qexhaust = Qcoal − W ; Qexhaust = mH O cH O TH O → mH O = = coal → cH O TH O cH O TH O 2

mH O 2

( Qcoal t ) − (W t )

cH O TH O 2

( 2.883 10 J s ) − ( 9.50 10 ) J s = 1.027 10 kg s 9

( 4186 J kg C )( 4.5C )

s   1m 3   1L   1gal   5 kg  =  1.027  10 = 9.8  107 gal h  3600    −3 3     s  h   1000 kg   10 m   3.785 L   72. (a) The exhaust heating rate is found from the delivered power and the efficiency. Use the output energy with Eq. 19–2, Q = mcT = VcT , to calculate the volume of air that is heated. The efficiency is given by Eq. 20–1a. e = W QH = W ( QL + W ) → QL = W (1 e − 1) →

(

)

QL t = (W t )(1 e − 1) = 8.5  108 W (1 0.38 − 1) = 1.387  109 W QL = mcT → QL t =

mcT t

VcT t

→ V t=

(QL t )  cT

The change in air temperature is 7.0 C . The heated air has a density of 1.3 kg m 3.

(1.387  10 W )(8.64  10 s day ) V t= =  cT (1.3kg m )(1.0  10 J kg C ) ( 7.0C ) ( QL t )

 10−9 km 3  3 3  = 13.17 km day  13 km day 3 1 m  

= 1.317  1010 m 3 day 

This could affect the local climate around the power plant, if the temperature rises 7.0 C o. The next part of the problem addresses that. (b) If the air is 150 m thick, find the area by dividing the volume by the thickness. Volume 13.17 km 3 A= = = 88 km 2 thickness 0.15 km This would be a square of approximately 9.4 km or 5.8 miles to a side. Thus, the local climate for a few miles around the power plant might be heated significantly. However, since hot air rises, that “blanket” of hot air would not stay at ground level. It would rise some, perhaps reducing the effect on the local climate. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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73. All of the processes are either constant pressure or constant volume, and so the heat input and output can be calculated with specific heats at constant pressure or constant volume. This tells us that heat is input when the temperature increases, and heat is exhausted when the temperature decreases. The lowest temperature will be the temperature at point b. We use the ideal gas law to find the temperatures. PV → PV = nRT → T = nR P ( 2V0 ) PV ( 3P0 )V0 ( 3P0 )( 2V0 ) = 2Tb , Tc = = 3Tb , Td = = 6Tb Tb = 0 0 , Ta = 0 nR nR nR nR (a) Process ab: Wab = PV = P0 ( −V0 ) = − PV ; 0 0

PV Qab = nCP T = 52 nR (Tb − Ta ) = − 52 nR 0 0 = − 52 PV 0 0 0 nR Process bc:

 PV  Wbc = PV = 0 ; Qbc = nCV T = 32 nR (Tc − Tb ) = 32 nR ( 2Tb ) = 32 nR  2 0 0  = 3PV 0 0  nR 

Process cd:

Wbc = PV = 3PV ; 0 0

 PV  Qcd = nC P T = 25 nR ( Td − Tc ) = 25 nR ( 3Tb ) = 25 nR  3 0 0  = 152 PV 0 0  nR   PV  Wda = PV = 0 ; Qda = nCV T = 32 nR (Ta − Td ) = − 32 nR  4 0 0  = −6 PV 0 0 0  nR  3PV 2 W − PV 0 0 0 0 erectangle = = = = 0.1905  0.19 21 QH 3PV + 152 PV 2 0 0 0 0 The same result is found from this calculation. + 152 PV − ( 25 PV + 6) 212 − 172 Q − QL ( 3PV 4 0 0 0 0) 0 0 = = = = 0.1905  0.19 erectangle = H 21 + 152 PV QH 3PV 21 2 0 0 0 0 Process da:

(b) eCarnot =

TH − TL TH

6Tb − Tb 6Tb

= 0.8333 ;

erectangle eCarnot

0.1905 0.8333

= 0.23

This engine only has 23% of the Carnot efficiency. 74. (a) The heat that must be removed from the water ( QL ) is found in three parts–cooling the liquid water to the freezing point, freezing the liquid water, and then cooling the ice to the final temperatures. QL = m ( cliquid Tliquid + Lfusion + cice Tice )

( 4186 J kg C )( 25C ) + ( 3.33  105 J kg ) 5 = ( 0.65 kg )   = 3.077  10 J  + ( 2100 J kg C )(17C ) 

The Carnot efficiency can be used to find the work done by the refrigerator. T W W e = 1− L = = → TH QH W + QL

 ( 25 + 273) K  T  − 1  = 5.048  10 4 J  5.0  10 4 J W = QL  H − 1 = 3.077  105 J   TL   ( −17 + 273) K 

(

)

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(b) Use the compressor wattage to calculate the time. The compressor power can be expressed as one-fourth of the nominal power, since it only runs 25% of the time. P = W t → t = W P = 5.048  104 J (105W ) 0.25 = 1923s  32 min 75. (a) Calculate the Carnot efficiency by e = 1 − TL TH and compare it to the 15% actual efficiency. eCarnot = 1 − TL TH = 1 − ( 85 + 273) K ( 495 + 273 ) K = 0.534 = 53.4%

Thus the engine’s relative efficiency is eactual eCarnot = 0.15 0.534 = 0.281 = 28% (b) Take the stated 145 hp as the useful power obtained from the engine. Use the efficiency to calculate the exhaust heat. W  746 W  5 P = = (145hp )   = 1.08  10 W 1hp t   e=

W QH

W QL + W

→

1  1   3600 s   1 − 1  QL = W  − 1  = Pt  − 1  = 1.08  105 J s (1 h )    e  e   1 h   0.15 

(

)

1 kcal  )  4186  = 5.3  10 kcal J

(

= 2.203  109 J  2.2  109 J = 2.203  109 J 





. Also the pressure, volume, and 76. We have a monatomic gas, so  temperature for state a are known. We use the ideal gas law, the adiabatic relationship, and the first law of thermodynamics. (a) Use the ideal gas equation to relate states a and b. Use the adiabatic relationship to relate states a and c. PV PV b b = a a → Tb Ta = 53

Pb = Pa

 22.4 L   273 K    = 0.400 atm  56.0 L   273 K 

Va Tb

= (1.00 atm ) 

Vb Ta



 PV = PV a a c c

→ 

V   22.4 L  Pc = Pa  a  = (1.00 atm )    56.0 L   Vc 

5/ 3

= 0.2172 atm  0.217 atm

(b) Use the ideal gas equation to calculate the temperature at c. PV PV P V  0.2172 atm  b b = c c = → Tc = Tb c c = ( 273 K )   (1) = 148 K Tb Tc Pb Vb  0.400 atm  (c) Process ab:

Eint = nCV T = 0 ; ab

Qab = Wab = nRT ln

Vb Va

= (1.00 mol )( 8.314 J mol K )( 273 K ) ln 2.5

= 2079.7 J  2080 J Sab =

Qab Tab

2079.7 J 273 K

= 7.62 J K

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Second Law of Thermodynamics

Wbc = 0 ;

Process bc:

Eint = Qbc = nCV T = (1.00 mol ) 32 ( 8.314 J mol K )(148 K − 273 K ) bc

= −1559 J  −1560 J c

S bc = 

dQ T

=

nCV dT T

= nCV ln

Tc Tb

= (1.00 mol ) 32 ( 8.314 J mol K ) ln

148 K 273 K

= −7.64 J K

Qca = 0 ; Sbc = 0 ( adiabatic ) ;

Process ca:

Eint = −W = −Eint − Eint = −0 − ( −1560 J ) → ca

Eint = 1560 J ; Wca = −1560 J ca

(d) e =

W Qinput

2080 J − 1560 J 2080 J

= 0.25

77. (a) The exhaust heating rate can be found from the delivered power P and the Carnot efficiency. Then use the relationship between energy and temperature change, Q = mcT , to calculate the temperature change of the cooling water. T TL TL TL W W e = 1− L = = → QL = W → QL t = W t =P TH QH QL + W TH − TL TH − TL TH − TL QL = mcT → QL t =

cT = 

cT t t Equate the two expressions for QL t , and solve for T .

TL TH − TL

=

V t

cT → T =

P  TL   V   c  TH − TL  t

8.8  108 W

285 K

(1.0  10 kg m )( 37 m s ) ( 4186 J kg C) ( 625 K − 285 K ) 3

= 4.763 K = 4.8C

(b) The addition of heat per kilogram for the downstream water is QL t = cT . S m

=

dS m

=

dQ mT

=

cdT T

289.763 K



285 K

dT T

K )  289.763  = 69.4 J kg K 285 K

(

= 4186 J kg C o  ln





78. The radiant energy that enters the room is the heat to be removed at the low temperature. It can be related to the work necessary to remove it through the ideal efficiency, Eq. 20–3. We then subtract the two rates of doing work to find the savings. T  T  T W W e = 1− L = = → W = QL  H − 1 → W t = QL t  H − 1 TH QH W + QL  TL   TL 

T



 TL



(W t )4800 = ( 4800 W )  H − 1

T



 TL



(W t )500 = ( 500 W )  H − 1

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 ( 273 + 32 ) K

(W t )savings = (W t )4800 − (W t )500 = ( 4800 W − 500 W ) 

 ( 273 + 21) K



− 1  = 160.9 W  160 W



79. (a) To find the efficiency, we need the heat input and the heat output. The heat input occurs at constant pressure, and the heat output occurs at constant volume. Start with Eq. 20–1b. nC (T − T ) Q (T − T ) e = 1− L = 1− V d a = 1− d a QH nCP (Tc − Tb )  (Tc − Tb ) So we need to express the temperatures in terms of the corresponding volume, and use the ideal gas law and the adiabatic relationship between pressure and volume to get those expressions. Note that Pc = Pb and Vd = Va . PV = nRT → T =

PV nR

→

 Pd Pa  PV   PV d d − a a P −P   − PV Va ( Pd − Pa ) ( PV nR nR    c b  d d a a) = 1− = 1− = 1− e = 1− PV PV − PV  ( PV  Pb (Vc − Vb ) V V  c c b b)   c c − b b   c − b nR nR    Va Va  Use the adiabatic relationship between pressure and volume on the two adiabatic paths.  d d

 c c

PV = PV



V  V  P V    → =  c  =  c  ; PV = PV → a = b a a b b Pc  Vd   Va  Pb  Va  Pd



−



−

 Pd Pa   Vc   Vb   Va   Va  P −P  V  − V  V  − V  b  e = 1−  c = 1 −  a  −1  a  −1 = 1 −  c  −1  b  −1  Vc Vb   V   V    V   V    −    a  −  a     a  −  a    Va Va   Vc   Vb    Vc   Vb   (b) For a diatomic ideal gas,  = 75 = 1.4. −

−

 Va   Va  V  − V  ( 4.5) −1.4 − (16) −1.4   c  b  = 1− = 0.55 e = 1−  V  −1  V  −1  ( 4.5) −1 − (16 ) −1  1.4     a  −  a   V V  c   b   80. Take the energy transfer to use as the initial kinetic energy of the cars, because this energy becomes “unusable” after the collision–it is transferred to the environment. 2

  1m s   1100 kg ) ( 75 km h )  ( 2  1 Q 2 ( 2 mvi )  3.6 km h   = 1600 J K  S = = = T

( 20 + 273) K

81. Heat will enter the freezer due to conductivity, at a rate given by Eq. 19–17a. This is the heat that must be removed from the freezer to keep it at a constant temperature, and so this heat is the value of Q L in the equation for the COP, Eq. 20–4a. The “work” quantity in the COP is the work provided by the cooling motor. The motor must remove the heat in 18% of the time that it takes for the heat to © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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enter the freezer, so that it only runs 18% of the time. To find the minimum power requirement, we assume the freezer is ideal in its operation. QL T −T Q QL t TL = kA H L ; COP = L = = → t W W ( 0.18t ) TH − TL l

37 K T −T ( 0.050 W m K ) 8.0 m 2 kA H L  TH − TL  Q t  T −T  0.12 m  37 K  l W t= L  H L =  =  258 K  ( 0.18 )  TL  ( 0.18 )  TL  ( 0.18 )  

(

)

= 98.3 W  98 W  0.13 hp 82. The cyclist provides 200 watts to power the heat pump. That is used to determine the “work” value in the coefficient of performance equation, Eq. 20–5. The coefficient of performance equation is then used to calculate the heat delivered by the heat pump. Q J  COP = H → QH = W ( COP ) =  200  (1800s )( 2.0 ) = 7.2  105 J W s  83. We need to find the efficiency in terms of the given parameters, TH , TL , Va , and Vb . So we must find the net work done and the heat input to the system. The work done during an isothermal process is given by Eq. 19–9. The work done during an isovolumentric process is 0. We also use the first law of thermodynamics. Eint = 0 = Qab − Wab → ab (isothermal): ab

V Qab = Wab = nRTH ln b  0 Va

bc (isovolumetric): Eint = Qbc − 0 → Qbc = Eint = nCV (TL − TH ) = 32 nR (TL − TH )  0 bc

cd (isothermal):

V V Eint = 0 = Qcd − Wcd → Qcd = Wcd = nRTL ln a = − nRTL ln b  0 Vb Va cd

da (isovolumetric): Eint = Qda − 0 → Qda = Eint = nCV (TH − TL ) = 32 nR (TH − TL )  0 da

V V V W = Wab + Wcd = nRTH ln b − nRTL ln b = nR ( TH − TL ) ln b Va Va Va V Qin = Qab + Qda = nRTH ln b + 23 nR ( TH − TL ) Va

  V ln b   T −T  W Va Va  = =  H L  eSterling = Qin T ln Vb + 3 T − T T   − V T T 3   H  ln b + H L  2( H H L)    Va  Va 2  TH   V

(TH − TL ) ln b

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

  V ln b   Va  = eCarnot   Vb 3  TH − TL    ln V + 2  T    H   a Since the factor in [ ] above is less than 1, we see that eSterling  eCarnot . 84. Since two of the processes are adiabatic, no heat transfer occurs in those processes. Thus the heat transfer must occur along the isobaric processes. QH = Qbc = nCP ( Tc − Tb ) ; QL = Qda = nCP (Td − Ta )

e = 1−

= 1−

nCP ( Td − Ta )

= 1−

(Td − Ta ) (Tc − Tb )

QH nCP ( Tc − Tb ) Use the ideal gas relationship, which says that PV = nRT .

PV   PV d d − a a  P (V − V ) (T − T ) ( PV − PV ) nR nR  e = 1− d a = 1−  = 1− d d a a = 1− a d a PV  Pb (Vc − Vb ) − PV  PV (Tc − Tb ) ( PV c c c c b b) − b b   nR nR  1/ 

 a a

 b b

Because process ab is adiabatic, we have PV = PV

P  → Va = Vb  b  . Because process cd is  Pa 

1/ 

P  adiabatic, we have PV = PV → Vd = Vc  b  . Substitute these into the efficiency expression.  Pa  1/  1/    P 1/   Pb     P b b Pa  Vc   − Vb    Pa   (Vc − Vb )   Pa   Pa   P Pa (Vd − Va )  e = 1− = 1− = 1−  a   b c

 a d

Pb (Vc − Vb ) 1

−1

Pb (Vc − Vb )

1−

 P  P  = 1−  b  = 1−  b   Pa   Pa 

85. The coefficient of performance for an ideal refrigerator is given by Eq. 20–4c, with temperatures in Kelvins. Use that expression to find the temperature inside the refrigerator. COP 5.0 TL COP = → TL = TH = ( 32 + 273) K  = 254 K = −19C 1 + COP 6.0 TH − TL T 86. (a) For the Carnot cycle, two of the processes are reversible adiabats, which are constant entropy processes. The other two processes are isotherms, at the low and high temperatures. See T H the adjacent diagram. (b) The area underneath any path on the T–S diagram would be



written as T dS . This integral is the heat involved in the process.

c S

718

Chapter 20

Second Law of Thermodynamics

 T dS =  T T =  dQ = Qnet For a closed cycle such as the Carnot cycle shown, since there is no internal energy change, the first law of thermodynamics says that  T dS = Qnet = Wnet , the same as  P dV . 87. First we find the equilibrium temperature from calorimetry (Section 19–4), and then calculate the entropy change of the system. The heat lost by the warm water must be equal to the heat gained by the cold water. Since the two masses of water are the same, the equilibrium temperature is just the average of the two starting temperatures, 25C. Tfinal

S = Scool + S warm =  water

water

Tinitial cool

dQ T

Tfinal

+ 

Tinitial

dQ T

298 K

= 

mc dT

273 K

298 K

+ 

323 K

mc dT T

 298 K   298 K  + mc ln     273K   323K 

= mc ln 

warm

  298 K   298 K   + ln     = 13.30 J K  13J K  323K     273K 

= ( 0.45kg )( 4186 J kg K ) ln 

88. To find the mass of water removed, find the energy that is removed from the low temperature reservoir from the work input and the Carnot efficiency. Then use the latent heat of vaporization to determine the mass of water from the energy required for the condensation. Note that the heat of vaporization used is that given in Section 19–5 for evaporation at 20oC. T TL W W e = 1− L = = → QL = W = mLvapor TH QH W + QL (TH − TL ) m=

( 550 W )( 3600s ) ( 273 + 8) K = 13.36 kg  13kg Lvapor ( TH − TL ) ( 2.45  106 J kg ) ( 25 − 8) K W

89. (a) From the table below, we see that there are 10 macrostates, and a total of 27 microstates.

(b) The probability of obtaining all 3 beans red is 1 27 . (c) The probability of obtaining 2 greens and 1 orange is 3 27 or 1 9 .

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90. A macrostate is a set of 5 cards from the deck, as given in the problem. For example, four aces and a king is a macrostate. Two jacks, two queens, and an ace is a macrostate. A microstate is a specific set of cards that meets the criterion of a certain macrostate. For example, the set (ace of spades, ace of clubs, ace of hearts, ace of diamonds, king of spades) is a microstate of the macrostate of 4 aces and a king. The problem then is asking for the relative number of microstates for the 4 given macrostates. (a) There are only 4 microstates for this macrostate, corresponding to the particular suit to which the king belongs. (b) Since every card is specified, there is only 1 microstate for this macrostate. (c) There are 6 possible jack pairs, (spade/club, spade/heart, spade/diamond, club/heart, club/diamond, and heart/diamond), 6 possible queen pairs, and 4 possible aces, so there are 6 × 6 × 4 = 144 card combinations or 144 microstates for this macrostate. (d) There are 52 possibilities for the first card, 48 possibilities for the second card, and so on. It is apparent that there are many more microstates for this macrostate than for any of the other listed macrostates. The actual value is 52 × 48 × 44 × 40 × 36 which is about 158 million. However, this assumes that the order of the cards matters, which it doesn’t. To correct for that, we must divide by 5! So the number of microstates is actually 1,317,888. It’s still the largest number by far. Thus in order of increasing probability, we have (b), (a), (c), (d). 91. The energy necessary to heat the water can be obtained using Eq. 19–2. The specific heat of the water is 4186 J kg C.

Q = mcT = (1.0 kg )( 4186J kg C )( 95°C − 25C) = 2.9302  105 J The intensity of sunlight at the Earth’s surface is  1000 W m 2. The photovoltaic panel can therefore produce energy at this rate. Q = 1000 W m 2 1.5 m 2 ( 0.22 ) = 330 J s t Dividing the energy needed to heat the water by the rate at which energy is available will give the time required to heat the water using the photovoltaic cell. 2.9302  105 J Q t= = = 888s  15 minutes Q t 330 J s

(

)(

)

(

Using the curved mirror allows all of the energy in the sunlight  1000 W m 2

)(1.5 m ) = 1500 J s 2

to go into heating the water. 2.9302  105 J Q t= = = 195s  3minutes 1500 J s Q t 92. Calculate the entropy change. S = 

dQ T

= TL

nCV dT T

=

(

)

n 1880 J mol −1 K −1 ( T TD ) dT T

(1880 J mol K ) = (1.00 mol ) −1

3 D

−1

40 K



(1880 J K ) 1 ( 40 K ) − ( 4 K )  T dT = −1

( 2230 K )

3



= 3.61  10−3 J K

720

Chapter 20

Second Law of Thermodynamics

93. (a) Divide the energy consumed by the refrigerator per year by the product of the efficiency and the energy released per kilogram of coal burned to determine amount of coal burned each year to run the refrigerator. 2  109 J m= = 303kg  300 kg (only 1 significant figure is justified) 0.33 2  107 J kg

(

)

(b) Divide the mass of the carbon by the capture per hectare. Then convert the results to square meters of forest. 303kg  104 m 2  2 2 A=   = 1780 m  2000 m (only 1 sig. fig. is justified) 1700 kg hectare  hectare  For comparison, a football field or soccer field is approximately 4500 m 2.

721

CHAPTER 21: Electric Charge and Electric Field Responses to Questions 1.

Suspend a plastic ruler by a thread and then rub it with a cloth. As shown in Fig. 21–2a, the ruler is negatively charged. Now bring the charged comb close to the ruler. If the ruler is repelled by the comb, then the comb is negatively charged. If the ruler is attracted by the comb, then the comb is positively charged.

The shirt or blouse becomes charged as a result of being tossed about in the dryer and rubbing against the dryer sides and other clothes. When you put on the charged object (shirt), it causes charge separation within the molecules of your skin (see Fig. 21–9), which results in attraction between the shirt and your skin.

Fog or rain droplets tend to form around ions because water is a polar molecule, with a positive region and a negative region. The charge centers on the water molecule will be attracted to the ions or electrons, since opposite charges attract.

See also Fig. 21–9 in the text. The negatively charged electrons in the paper are attracted to the positively charged rod and move towards it within their molecules. The attraction occurs because the negative charges in the paper are closer to the positive rod than are the positive charges in the paper, and therefore the attraction between the unlike charges is greater than the repulsion between the like charges.

A plastic ruler that has been rubbed with a cloth is charged. When brought near small pieces of paper, it will cause separation of charge (polarization) in the bits of paper, which will cause the paper to be attracted to the ruler. A small amount of charge is able to create enough electric force to be stronger than gravity. Thus the paper can be lifted. On a humid day this is more difficult because the water molecules in the air are polar. Those polar water molecules will be attracted to the ruler and to the separated charge on the bits of paper, partially neutralizing the charges and thus reducing the attraction.

The net charge on a conductor is the sum of all of the positive and negative charges in the conductor. If a neutral conductor has extra electrons added to it, then the net charge is negative. If a neutral conductor has electrons removed from it, then the net charge is positive. If a neutral conductor has the same amount of positive and negative charge, then the net charge is zero. The “free charges” in a conductor are electrons that can move about freely within the material because they are only loosely bound to their atoms. The “free electrons” are also referred to as “conduction electrons.” A conductor may have a zero net charge but still have substantial free charges.

722

Chapter 21

Electric Charge and Electric Field

Most of the electrons are strongly bound to nuclei in the metal ions. Only a few electrons per atom (usually one or two) are free to move about throughout the metal. These are called the “conduction electrons.” The rest are bound more tightly to the nucleus and are not free to move. Furthermore, in the cases shown in Figs. 21–7 and 21–8, not all of the conduction electrons will move. In Fig. 21–7, electrons will move until the attractive force on the remaining conduction electrons due to the incoming charged rod is balanced by the repulsive force from electrons that have already gathered at the left end of the neutral rod. In Fig. 21–8, conduction electrons will be repelled by the incoming rod and will leave the stationary rod through the ground connection until the repulsive force on the remaining conduction electrons due to the incoming charged rod is balanced by the attractive force from the net positive charge on the stationary rod.

The electroscope leaves are connected together at the top. That connection can be modeled as a tension force. The horizontal component of this tension force balances the electric force of repulsion. The vertical component of the tension force balances the weight of the leaves.

The balloon has been charged. The excess charge on the balloon is able to polarize the water molecules in the stream of water, similar to Fig. 21–9. This polarization results in a net attraction of the water towards the balloon, so the water stream curves towards the balloon.

10. (a) When the leaves are charged by induction, no additional charge is added to the leaves. If the charged rod is near the top of the electroscope it repels charge onto the leaves causing them to separate as in Fig. 21–11(a). When the rod is removed the charge returns to its initial equilibrium position and the leaves come back together. (b) When the leaves are charged by conduction, positive charge is placed onto the electroscope from the rod causing the leaves to separate. When the rod is removed, the charge remains on the electroscope and the leaves remain separated. If not all of the excess charge leaves the rod, then when the rod is removed, the leaves might come back together slightly from their maximum deflection. (c) Yes. The electroscope has a negative charge on the top sphere and on the leaves. Therefore the electroscope has a total net negative charge, so it must have been charged by conduction. 11. Coulomb’s law and Newton’s law are very similar in form. When expressed in SI units, the magnitude of the constant in Newton’s law is very small, while the magnitude of the constant in Coulomb’s law is quite large. Newton’s law says the gravitational force is proportional to the product of the two masses, while Coulomb’s law says the electrical force is proportional to the product of the two charges. Newton’s law only produces attractive forces, since there is only one kind of gravitational mass. Coulomb’s law produces both attractive and repulsive forces, since there are two kinds of electrical charge. 12. The gravitational force between everyday objects on the surface of the Earth is extremely small. (Recall the value of G: 6.67  10 −11 N m 2 kg 2 . ) Consider two objects sitting on the floor near each other. They are attracted to each other, but the maximum force of static friction for each is much greater than the gravitational force each experiences from the other, and so they don’t move. Even in an absolutely frictionless environment, the acceleration resulting from the gravitational force would be so small that it would not be noticeable in a short time frame. We are aware of the gravitational force between objects if at least one of them is very massive, as in the case of the Earth and satellites or the Earth and you.

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The electric force between two objects is typically zero or very close to zero because ordinary objects are typically neutral or very close to neutral. We are aware of electric forces between objects when the objects are charged. An example is the electrostatic force (static cling) between pieces of clothing when you pull the clothes out of the dryer. 13. Coulomb observed experimentally that the force between two charged objects is directly proportional to the charge on each one. For example, if the charge on either object is tripled, then the force is tripled. This is not in agreement with a force that is proportional to the sum of the charges instead of to the product of the charges. Secondly, if two equal but opposite charges are placed near to each other, they produce an attractive force. But the “sum” version would say the net force is 0 in such a case. Also, a charged object is not attracted to or repelled from a neutral insulating object. If the numerator in Coulomb’s law were proportional to the sum of the charges, then there would be a force between a neutral object and a charged object, because the their total charge would not be 0. 14. Assume that the charged plastic ruler has a negative charge residing on its surface. That charge polarizes the charge in the neutral paper, producing a net attractive force. When the piece of paper then touches the ruler, the paper can get charged by contact with the ruler, gaining a net negative charge. Then, since like charges repel, the paper is repelled by the comb. 15. The test charge creates its own electric field. The measured electric field is the sum of the original electric field plus the field of the test charge. By making the test charge small, the field that it causes is small. Therefore the actual measured electric field is not much different than the original field. Also, if the test charges are large, their fields might significantly re-distribute the charges causing the original field, and then the measurement would not represent the field of the original configuration of charges. 16. When determining an electric field, it is best, but not required, to use a positive test charge. A negative test charge would be fine for determining the magnitude of the field. But the direction of the electrostatic force on a negative test charge will be opposite to the direction of the electric field. The electrostatic force on a positive test charge will be in the same direction as the electric field. In order to avoid confusion, it is better to use a positive test charge. If we used a negative test charge but wanted to have the same result for the electric field, we would have to define E = − F q , q  0. 17. See Fig. 21–35(b). A diagram of the electric field lines around two negative charges would be just like this diagram except that the arrows on the field lines would point towards the charges instead of away from them. The distance between the charges is l.

18. The electric field will be strongest to the right of the positive charge (between the two charges) and weakest to the left of the positive charge. To the right of the positive charge, the contributions to the field from the two charges point in exactly the same direction, and therefore add. To the left of the positive charge, the contributions to the field from the two charges point in exactly opposite directions, and therefore subtract. Note that this should be confirmed by the density of field lines in Fig. 21–35a. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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19. At point C, the positive test charge would experience zero net force. At points A and B, the direction of the force on the positive test charge would be the same as the direction of the field. This direction is indicated by the arrows on the field lines. The strongest field is at point A, followed (in order of decreasing field strength) by B and then C. The field strength can be inferred by the density of the field lines at each location. 20. Electric field lines can never cross because they give the direction of the electrostatic force on a positive test charge. If they were to cross, then the force on a test charge at a given location would be in more than one direction. This is not possible, because the electric force has a unique direction at each point in space. 21. From rule 1: A test charge would be either attracted directly towards or repelled directly away from a point charge, depending on the sign of the point charge. So the field lines must be directed either radially towards or radially away from the point charge. From rule 2: The magnitude of the field due to the point charge only depends on the distance from the point charge. Thus, the density of the field lines must be the same at any location around the point charge, for a given distance from the point charge. From rule 3: If the point charge is positive, the field lines will originate from the location of the point charge. If the point charge is negative, the field lines will end at the location of the point charge. Based on rules 1 and 2, the lines are radial and their density is constant for a given distance. This is equivalent to saying that the lines must be symmetrically spaced around the point charge. 22. The two charges are located as shown in the diagram. (a) If the signs of the charges are opposite then the point on the line where E = 0 will lie to the left of Q. In that region the electric fields from the two charges will point in opposite directions, and the point will be closer to the smaller charge. (b) If the two charges have the same sign, then the point on the line where E = 0 will lie between the two charges, closer to the smaller charge. In this region, the electric fields from the two charges will point in opposite directions. 23. The electric field at point P would point in the negative x-direction. The magnitude of the field would be the same as that calculated for a positive distribution of charge on the ring:

4 o ( x 2 + a 2 )3/2

24. We assume that there are no other forces (like gravity) acting on the test charge. The direction of the electric field line shows the direction of the force on the test charge. The acceleration is always parallel to the force by Newton’s second law, so the acceleration lies along the field line. If the particle is at rest initially and then released, the initial velocity will also point along the field line, and the particle will start to move along the field line. However, once the particle has a velocity, it will not follow the field line unless the line is straight. The field line gives the direction of the acceleration, or the direction of the change in velocity.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

25. The value measured will be slightly less than the electric field value at that point before the test charge was introduced. The test charge will repel charges on the surface of the conductor and these charges will move along the surface to increase their distances from the test charge. Since they will then be at greater distances from the point being tested, they will contribute a smaller amount to the field. 26. The motion of the electron in Example 21–17 is analogous to projectile motion. In the case of the gravitational force, the acceleration of the projectile is in the same direction as the field and has a value of g no matter what the projectile is. In the case of an electron in an electric field, the direction of the acceleration of the electron and the field direction are opposite, and the value of the acceleration depends on the mass of the electron, the charge of the electron, and the magnitude of the electric field. 27. If an electric dipole is placed in a nonuniform electric field, the charges of the dipole will experience forces of different magnitudes whose directions also may not be exactly opposite. The sum of these forces might not be zero, and so there can be a net force on the dipole. 28. (a) Initially, the dipole will rotate clockwise about an axis through point 0 (its center of mass), out of the page. It will “overshoot” the equilibrium position (parallel to the field lines), come momentarily to rest and then spin counterclockwise. The dipole will continue to oscillate back and forth if no damping forces are present. If there are damping forces, the amplitude will decrease with each oscillation until the dipole comes to rest aligned with the field. (b) There will be a larger force on the negative charge, since the electric field is stronger at the location of the negative charge. Thus the dipole will accelerate to the left as it rotates. Once again, the dipole will “overshoot” the horizontal equilibrium orientation. When that happens, the positive charge will have a larger force than the negative charge, causing an acceleration to the right. The dipole will slow it’s leftward motion, and eventually start to move back to the right, even as the torque causes counterclockwise acceleration and the rotation also shifts to counterclockwise. Thus, it will oscillate both horizontally and rotationally.

Solutions to MisConceptual Questions 1.

(a) The two charges have opposite signs, so the force is attractive. Since Q 2 is located on the positive x-axis relative to Q1 at the origin, the force on Q1 will be in the positive x-direction.

(d) The signs of the charges are still opposite, so the force remains attractive. However, since Q 2 is now located at the origin with Q1 on the positive x-axis, the force on Q1 will now be toward the origin, or in the negative x-direction.

(e) A common misconception is that the object with the greater charge and smaller mass has a greater force of attraction. However, Newton’s third law applies here. The force of attraction is the same for both lightning bugs.

(d) Students sometimes believe that electric fields only exist on electric field lines. This is incorrect. The field lines represent the direction of the electric field in the region of the lines. The magnitude of the field is proportional to the density of the field lines. At point 1 the field lines are closer together than they are at point 2. Therefore the field at point 1 is larger than the field at point 2.

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(b) A common misconception students have is recognizing which object is creating the electric field and which object is interacting with the field. In this question the positive point charge is creating the field. The field from a positive charge always points away from the charge. When a negative charge interacts with an electric field, the force on the negative charge is in the opposite direction as the field. The negative charge experiences a force toward the positive charge.

(c) If the third charge is positive, it will be repelled away from the charge on its left, and also will be repelled away from the charge on its right. Those two repelling forces will be in opposite directions, and so will add to zero if the third charge is placed one-third of the distance from Q to 4Q, along the line joining them. If the third charge is negative, it will be attracted toward the charge on its left and attracted towards the charge on its right. Again the two forces can add to zero if the third charge is placed one-third of the distance from Q to 4Q, along the line joining them.

(a) Students frequently think of the plates as acting like point charges with the electric field increasing as the plates are brought closer together. However, unlike point charges, the electric field lines from each plate are parallel with uniform density. Bringing the plates closer together does not affect the electric field between them.

(c) Each particle will have the same magnitude of force acting on it. Since the electron is lighter than the proton, the electron will have a larger acceleration, and so will move faster than the proton. Thus they will meet closer to the proton’s position.

(b) In a lightning storm charged particles in the clouds and ground create large electric fields that ionize the air creating lightning bolts. People are good conductors of electric charge and when located in the electric field can serve as conduits of the lightning bolt. The inside of a metal car acts like a cavity in a conductor, shielding the occupants from the external electric fields. In each of the other options the person remains in the storm’s electric field and as such may be struck by the lightning. If the car is struck by lightning the electricity will pass along the exterior of the car, leaving the people inside unharmed.

10. (a, c, e) Lightning will generally travel from clouds to the tallest conductors in the area. If the person is in the middle of a grassy field, near the tallest tree, or on a metal observation town he/she is likely to be struck by lightning. A person inside a wooden building or inside a car is somewhat shielded from the lightning. Of all the tall objects, the metal observation tower is probably the worst–it may act like a “lightning rod.” 11. (d) The electric field at the fourth corner is the vector sum of the electric fields from the charges at the other three corners. The two positive charges create equal magnitude electric fields pointing in the positive x-direction and in the negative y-direction. These add to produce an electric field in the direction of (d) with magnitude 2 times the magnitude of the electric field from one of the charges. The electric field from the negative charge points in the direction of (b). However, since the charge is 2 times further away than the positive charges, the magnitude of the electric field will be smaller than the field from the positive charges. The resulting field will then be in the direction of (d). 12. (e) A common misconception is that the metal ball must be negatively charged. While a negatively charged ball will be attracted to the positive rod, a neutral conductor will become polarized and also be attracted to the positive rod. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Solutions to Problems 1.

Use Coulomb’s law to calculate the magnitude of the force. 1.602  10−19 C 26  1.602  10 −19 C QQ F = k 1 2 2 = 8.988  109 N  m 2 C 2 = 2.7  10 −3 N 2 12 − r 1.5  10 m

(

)

Use Coulomb’s law to calculate the magnitude of the force. 25  10−6 C 2.5  10 −3 C QQ = 1.161  10 4 N  1.2  10 4 N F = k 1 2 2 = 8.988  109 N m 2 C 2 2 r ( 0.22 m )

)

(

)(

)

Use Coulomb’s law to calculate the magnitude of the force.

F =k

)(

)

(

Use the charge per electron to find the number of electrons.  1 electron  = 3.620  1014 electrons  3.62  1014 electrons −58.0  10−6 C   −19  −1.602  10 C 

(

)

Q1Q2 r2

(1.602  10 C ) = 18.83N  19 N = ( 8.988  10 N  m C ) ( 3.5  10 m ) 2

−19

−15

The charge on the plastic comb is negative, so the comb has gained electrons, and a negative charge. 1e −    9.109  10−31 kg  −2.4  10−6 C   1e − −1.602  10−19 C   m   = 1.52  10−15  1.5  10−13 % = m 0.0090 kg

(

)

(

)

Since the magnitude of the force is inversely proportional to the square of the separation distance, 1 F  2 , if the distance is multiplied by a factor of 1/8, the force will be multiplied by a factor of 64. r F = 64 F0 = 64 ( 4.6  10−2 N ) = 2.944N  2.9 N

Since the magnitude of the force is inversely proportional to the square of the separation distance, 1 F  2 , if the force is tripled, the distance has been reduced by a factor of 3. r r 6.22 cm r= 0 = = 3.59 cm 3 3

Use the charge per electron and the mass per electron.  1 electron  = 2.122  1014  2.1  1014 electrons −34  10−6 C   −19  −1.602  10 C 

(

)

 9.109  10−31 kg  −16 −16 ( 2.122  10 e )  1 e−  = 1.933  10 kg  1.9  10 kg   14

−

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Chapter 21

Electric Charge and Electric Field

To find the number of electrons, convert the mass to moles, the moles to atoms, and then multiply by the number of electrons in an atom to find the total electrons. Then convert to charge.  1mole Au   6.022  10 23 atoms   79 electrons   −1.602 10 −19 C  12 kg Au = (12 kg Au )    1atom     1 mole electron  0.197 kg      = −4.642  108 C  −4.6  108 C

The net charge of the bar is 0 C , since there are equal numbers of protons and electrons. 10. Take the ratio of the electric force divided by the gravitational force. Note that the distance is not needed for the calculation. QQ 2 k 12 2 8.988  109 N  m 2 C 2 1.602  10 −19 C FE kQ1Q2 r = = = = 2.27  1039 FG G m1m2 Gm1m2 6.67  10−11 N  m 2 kg 2 9.11  10 −31 kg 1.67  10 −27 kg r2 The electric force is about 2.27  1039 times stronger than the gravitational force for the given scenario.

(

)(

)

11. (a) Let one of the charges be q, and then the other charge is QT − q. The force between the charges is FE = k

q ( QT − q ) 2

( qQ − q ) . To find the maximum and minimum force, set the r 2

r first derivative equal to 0. Use the second derivative test as well. k dFE k 2 = 2 ( QT − 2 q ) = 0 → q = 12 QT FE = 2 ( qQT − q ) ; r dq r d 2 FE dq

=−

2k r2

 0 → q = 12 QT gives ( FE ) max

So q1 = q2 = 12 QT gives the maximum force. (b) If one of the charges has all of the charge, and the other has no charge, then the force between them will be 0, which is the minimum possible force. So q1 = 0, q2 = QT gives the minimum force. 12. Let the right be the positive direction on the line of charges. Use the fact that like charges repel and unlike charges attract to determine the direction of the forces. In the following expressions, k = 8.988  109 N  m 2 C 2 .

(82 C )( 48C ) ˆ (82 C )( 95C ) ˆ i+k i = −145.9 N î  −150 N î 2 2 ( 0.35 m ) ( 0.70 m ) (82 C )( 48C ) ˆ ( 48C )( 95C ) ˆ F+48 = k i+k i = 623.4 N î  620 N î ( 0.35 m )2 ( 0.35 m )2 ( 95C )(82 C ) ˆ ( 95C )( 48C ) ˆ F−95 = − k i−k i = −477.5 N î  −480 N î ( 0.70 m )2 ( 0.35 m )2 F+82 = − k

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

13. The forces on each charge lie along a line connecting the charges. Let d represent the length of a side of the triangle, and let Q represent the charge at each corner. Since the triangle is equilateral, each angle is 60o. Q2 Q2 Q2 F12 = k 2 → F12 x = k 2 cos 60o , F12 y = k 2 sin 60o d d d

F13 = k

→ F13 x = − k

F1x = F12 x + F13 x = 0

Q2 d

cos 60 , F13 y = k 2

F1 y = F12 y + F13 y = 2k

F12 Q1

F13

sin 60o 2

d Q3

sin 60o = 3k 2

(17.0 10 C ) = 199.96 N F = F + F = 3k = 3 ( 8.988  10 N  m C ) d 1

2 1x

−6

2 1y

( 0.150 m )

 2.00  102 N The direction of F1 is in the y-direction . Also notice that it lies along the bisector of the opposite side of the triangle. Thus the force on the lower left charge is of magnitude 2.00  102 N , and will point 30o below the − x-axis . Finally, the force on the lower right charge is of magnitude

2.00  102 N , and will point 30o below the + x-axis . 14. (a) If the force is repulsive, both charges must be positive since the total charge is positive. Call the total charge Q. kQ ( Q − Q ) kQ Q Fd 2 Q1 + Q2 = Q F = 12 2 = 1 2 1 → Q12 − QQ1 + =0 d d k

Q  Q2 − 4 Q1 =

Fd 2 k



(

)

= 12  82.0  10−6 C 



(

82.0  10−6 C

)

−4

(12.0N )( 0.280 m ) 2  

(8.988  10 N  m C )  9

= 80.7  10−6 C, 1.3  10−6 C (b) If the force is attractive, then the charges are of opposite sign. The value used for F must then be negative. Other than that, the solution method is the same as for part (a).

Q  Q2 − 4 Q1 =

Fd 2 k



(

)

= 12  82.0  10 −6 C 



(

82.0  10 −6 C

)

−4

2 ( −12.0N )( 0.280 m ) 

(8.988 10 N  m C )  9

= 83.3  10 −6 C, − 1.3  10 −6 C

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Chapter 21

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15. Determine the force on the upper right charge, and then use the symmetry of the configuration to determine the force on the other three charges. The force at the upper right corner of the square is the vector sum of the forces due to the other three charges. Let the variable d represent the 0.100 m length of a side of the square, and let the variable Q represent the 6.45 mC charge at each corner. F41 = k

F42 = k F43 = k

Q2 d2

→ F41x = k

Q2 2d 2

, F41 y = 0

→ F42 x = k

2Q 2

4d 2

cos45o = k 2

, F42 y = k

F41

F42

F43

2Q 2 4d 2

→ F43 x = 0, F43 y = k 2 d2 d Add the x and y components together to find the total force, noting that F4 x = F4 y.

F4 x = F41x + F42 x + F43 x = k F4 = F42x + F42y = k

2Q 2

+k 2

Q2 

1+ 2 

+0= k

2

Q2 

4 

 2=k

d 

Q2 

2 1 +  = F4 y d  4  2

1  2+  d  2

( 6.45  10 C )  2 + 1  = 7.158  10 N  7.16  10 N = ( 8.988  10 N  m C )   2 2

−3

( 0.100 m )

 = tan −1





F4 y

= 45o above the x-direction. F4 x For each charge, the net force will be the magnitude determined above, and will lie along the line from the center of the square out towards the charge. 16. Determine the force on the upper right charge, and then use the symmetry of the configuration to determine the force on the other charges. The force at the upper right corner of the square is the vector sum of the forces due to the other three charges. Let the variable d represent the 0.100 m length of a side of the square, and let the variable Q represent the 4.15 mC charge at each corner. Q2 Q2 F41 = k 2 → F41 x = − k 2 , F41 y = 0 d d

F42 = k F43 = k

Q2 2d 2 Q2

→ F42 x = k

Q2 2d 2

cos45 = k

→ F43 x = 0, F43 y = − k

2Q 2 4d 2

, F42 y = k

2Q 2

F41

F42 Q4

F43 d

4d 2

d2 d2 Add the x and y components together to find the total force, noting that F4 x = F4 y.

F4 x = F41x + F42 x + F43 x = − k

2Q 2

4d 2

+k 2

+0= k

Q2 

2

d 

4 

−1 + 2 

 = −0.64645k

= F4 y

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

F4 = F + F = k 2 4x

2 4y

Q2 d2

( 0.64645) 2 = k

Instructor Solutions Manual

( 0.9142 )

( 4.15  10 C ) ( 0.9142) = 1.42  10 N = ( 8.988  10 N  m C ) 2

−3

( 0.100 m )

 = tan −1

F4 y

= 225o from the x-direction, or exactly towards the center of the square.

F4 x For each charge, there are two forces that point towards the adjacent corners, and one force that points away from the center of the square. Thus for each charge, the net force will be the magnitude of 1.42  107 N and will lie along the line from the charge inwards towards the center of the square. 17. Take the lower left hand corner of the square to be the origin of coordinates. The 2Q will have a rightward force on it due to Q, an upward force on it due to 3Q, and a diagonal force on it due to 4Q. Find the components of each force, add the components, find the magnitude of the net force, and the direction of the net force. At the conclusion of the problem is a diagram showing the net force on the charge 2Q. Q2 kQ 2 ( 2Q ) Q ( 2Q )( 4Q ) 2Q : F2 Qx = k cos 45 = k 2 2 + 2 2 = 4.8284 2 +k 2l 2 l2 l l 2 Q kQ 2 ( 2Q )( 3Q ) ( 2Q )( 4Q ) F2 Qy = k k k sin 45 6 2 2 8.8284 +  = + = 2l 2 l2 l2 l2

(

)

(

F2 Q = F

2 2 Qx

2 2 Qy

= 10.1

kQ 2 l

 2 Q = tan −1

F2 x

= tan −1

8.8284 4.8284

= 61

F2Q

F2 y

)

18. The wires form two sides of an isosceles triangle, with the two charges separated by a distance l = 2 ( 88 cm ) sin 26 = 57.86 cm and are directly horizontal from each other. Thus the electric force on each charge is horizontal. From the free-body diagram for one of the spheres, write the net force in both the horizontal and vertical directions and solve for the electric force. Then write the electric force as given by Coulomb’s law, and equate the two expressions for the electric force to find the charge. mg  Fy = FT cos  − mg = 0 → FT = cos  mg  Fx = FT sin  − FE = 0 → FE = FT sin  = cos  sin  = mg tan 

FT FE



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Chapter 21

Electric Charge and Electric Field

FE = k

(Q 2) l

= mg tan  → Q = 2 l

mg tan  k

( 2110 kg )( 9.80 m s ) tan 26 = 3.867 10 C  3.9 10 C = 2 ( 0.5786 m ) (8.988 10 N  m C ) −3

−6

19. The balls will be treated as point charges, and so Coulomb’s law may be used to relate the amount of charge to the force of attraction. Each ball will have a magnitude Q of charge, since that amount was removed from one (initially neutral) ball and added to the other.

F =k

Q1Q2

Q2 r2

→ Q=r

= ( 0.21m )

1.7  10−2 N 8.988  109 N  m 2 C 2

 1 electron  12  = 1.8  10 electrons −19  1.602  10 C 

= 2.888  10−7 C 

20. The negative charges will repel each other, so the third charge must put an opposite force on each of the original charges. Consideration of the various possible configurations leads to the conclusion that the third charge must be positive and must be between the other two charges. See the diagram for the definition of variables. For each negative charge, equate the magnitudes of the two forces on the charge. Also note that 0  x  l . 2Qq 6Q 2 3Qq 6Q 2 = → k left: k 2 = k 2 right: k 2 l l2 x (l − x) k

2Qq x

3Qq

(l − x)

( 3 + 2)

2Qq x

6Q 2 l

→

(l − x) 3

→

x 2

(l − x)

→

l = 0.4495l

→ q = 3Q

x2 l

= 3Q ( 0.4495 ) = 0.606Q 2

Thus the charge should be of magnitude 0.61 Q , and a distance 0.45 l from − 2Q towards − 3Q . 21. (a) The charge will experience a force that is always pointing towards the origin. In the diagram, there is a greater force of Qq Qq to the left, and a lesser force of to 2 2 4 0 ( d − s ) 4 0 ( d + s ) the right. So the net force is towards the origin. The same would be true if the mass were to the left of the origin. Calculate the net force. Qq Qq Qq ( d − s ) 2 − ( d + s ) 2  Fnet = − = 2 2 2 2   4 0 ( d + s ) 4 0 ( d − s ) 4 0 ( d + s ) ( d − s )

−4Qqd

4 0 ( d + s ) ( d − s )

We assume that s

−Qqd

 0 ( d + s ) ( d − s ) 2

733

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Fnet =

−Qqd

 0 ( d + s ) ( d − s ) 2

s

−Qq

 0d 3

Instructor Solutions Manual

This has the form of a simple harmonic oscillator, where the “spring constant” is kelastic =

 0 d 3

The spring constant can be used to find the period. See Eq. 14–7b.

T = 2

m kelastic

m Qq

= 2

m 0d 3 Qq

 0d 3

(b) Sodium has an atomic mass of 23. T = 2

m 0 d 3 Qq

= 2

( 29 ) (1.66  10−27 kg )  (8.85 10−12 C 2 N m 2 )( 3 10 −10 m )

(1.60 10 C ) −19

 1012 ps  = 2.4  10 s   = 0.24 ps  0.2 ps  1s  −13

22. Use Eq. 21–3 to calculate the force. Take east to be the positive x-direction. F E= → F = qE = ( −1.602  10 −19 C ) ( 2260 N C ˆi ) = −3.62  10 −16 N ˆi = 3.62  10 −16 N west q 23. Use Eq. 21–3 to calculate the electric field. Take north to be the positive y-direction. F −1.68  10−14 N ˆj E= = = −1.049  105 N C ˆj = 1.05  105 N C south q 1.602  10−19 C 24. Use Eq. 21–4a to calculate the electric field due to a point charge. Q 33.0  10−6 C = 3.95  106 N C upward E = k 2 = 8.988  109 N  m 2 C2 2 r 0.274 m ( ) Note that the electric field points away from the positive charge.

(

)

25. Use the definition of the electric field, Eq. 21–3. 7.22  10−4 N ˆj F E= = = 195 N C ˆj q 3.70  10−6 C

(

)

26. Use the definition of the electric field, Eq. 21–3. 4.0ˆi − 5.2ˆj  10−3 N F E= = = −3200 ˆi + 4200 ˆj N C −1.25  10−6 C q The magnitude of the electric field is 5300 N/C, and the direction is 53o below the positive x-axis.

(

)

(

)

734

Chapter 21

Electric Charge and Electric Field

27. Assuming the electric force is the only force on the electron, then Newton’s second law may be used to find the acceleration. 1.602  10−19 C q Fnet = ma = qE → a = E = 756 N C = 1.33  1014 m s 2 −31 m 9.11 10 kg

( (

) )

Since the charge is negative, the direction of the acceleration is opposite to the field . 28. The electric field due to the positive charge ( Q1 ) will point Q1  0 away from it, and the electric field due to the negative charge ( Q2 ) will point toward it. Thus both fields point in l 2 the same direction, towards the negative charge, and the magnitudes can be added. Q Q Q1 Q2 4k E = E1 + E2 = k 21 + k 22 = k +k = 2 ( Q1 + Q2 ) 2 2 r1 r2 ( l / 2) ( l / 2) l

(

4 8.988  109 N  m 2 C 2

( 0.060 m )

Q2  0

E1 E2

) 9.2  10 C + 5.8  10 C = 149.8 N C  150 N C ( ) −12

−12

The direction is towards the negative charge . 29. Assuming the electric force is the only force on the proton, then Newton’s second law may be used to find the electric field strength. Fnet = ma = qE →

ma q

(1.673 10 kg )( 2.4 10 )(9.80 m s ) = 0.2456 N C  0.25 N C (1.602 10 C ) −27

−19

30. The field at the point in question is the vector sum of the two fields shown in Fig. 21–59. Use the results of Example 21–12 to find the field of the very long wire of charge. 1 ˆ 1 Q E wire = j ; EQ = −cos ˆi − sin  ˆj → 2 0 y 4 0 d 2

(



E = −



 4 0 d

)

 1   1 Q sin   ˆj − 2  2 0 y 4 0 d 

cos   ˆi + 



d 2 = ( 0.070 m ) + ( 0.120 m ) = 0.0193 m 2 ; y = 0.070 m ;  = tan −1 2

Ex = − Ey =

4 0 d 1 

2 0 y

−

(

cos  = − 8.988  109 N  m 2 C 2 1

4 0 d

sin  =

7.0 cm

= 59.7

( ) cos 59.7 = −4.699  10 N C ) 0.0193 m 2.0 C

1  2

Q  − 2 sin    4 0  y d 

 ( ) ( ) sin 59.7  = −1.622  10 N C − )  0.070 cm 0.0193 m

(

= 8.988  109 N  m 2 C 2 

(

12.0 cm

2 2.5 C m

2.0 C



) (



)

E = −4.7  10 N C ˆi + −1.6  10 N C ˆj 11

735

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

E wire =

1 ˆ 1 Q −cos ˆi − sin  ˆj j ; EQ = 2 0 y 4 0 d 2

(

Instructor Solutions Manual

)→

( −4.699 10 N C ) + ( −1.622 10 N C ) = 5.0 10 N C ( −1.622 10 N C ) = 199  = tan ( −4.699 10 N C ) E = Ex2 + E y2 =

−1

31. The field due to the negative charge will point towards the negative charge, and the field due to the positive charge will point towards the negative charge. Thus the magnitudes of the two fields can be added together to find the charges.

Enet = 2 EQ = 2k

( l / 2)

8kQ l

→ Q=

El 2 8k

l 2

( 326 N C )( 0.160 m )

(

E3 = k

Q 2l Q 2

→ E2 x = k

Q 2l

cos 45 = k

→ E3 x = 0, E1 y = k

E−Q 2

8 8.988  10 N  m C 9

32. The field at the upper right corner of the square is the vector sum of the fields due to the other three charges. Let l represent the 1.22 m length of a side of the square, and let Q represent the charge at each of the three occupied corners. Q Q E1 = k 2 → E1x = k 2 , E1 y = 0 l l

E2 = k

, E2 y = k

)

= 1.16  10−10 C

2Q 4l

−Q

l l2 Add the x and y components together to find the total electric field, noting that Ex = Ey . Ex = E1x + E2 x + E3 x = k

E = Ex2 + E y2 = k

Q l

2Q 4l

+0= k

Q

2 1 +  = Ey l  4  2

Q

2 Q 1 1 +  2=k 2 2+  l  4  l  2 2

( 3.25 10 C )  2 + 1  = 3.76 10 N C = ( 8.988  10 N  m C )   2 −6

(1.22 m )

 = tan −1

Ey Ex





= 45.0 from the x-direction.

33. The fields at the center due to the two −22.0C negative charges on opposite corners (lower right and upper left in the diagram) will cancel each other, so only the other two charges need to be considered. The field due to each of the other charges will point directly toward its source charge. Accordingly, the two fields are in opposite directions and can be combined algebraically. The

Q2 = −22.0 C

Q1 = −33.8C

E2 Q2

736

Chapter 21

Electric Charge and Electric Field

distance from each charge to the center is l

E = E1 − E2 = k

Q1 2

l 2

−k

Q2 2

l 2

( 33.8 − 22.0)  10−6 C = ( 8.988  10 N  m C ) ( 0.425 m )2 2 9

Q1 − Q2 l2 2

= 1.17  106 N C , towards the − 33.8C charge

34. Choose the rightward direction to be positive. Then the field due to +Q will be positive, and the field due to –Q will be negative.

E=k

( x + a)

−k

( x − a)

 1  1 −4kQxa − =  2 2  ( x + a) ( x − a)  2 2 2   (x − a )

= kQ 

The negative sign means the field points to the left . That is reasonable, since point P is closer to the negative charge. 35. For the net field to be zero at point P, the magnitudes of the fields created by Q1 and Q2 must be equal. Also, the distance x will be taken as positive to the left of Q1 . That is the only region where the total field due to the two charges can be zero. Let the variable l represent the 12 cm distance. Q Q2 → E1 = E 2 → k 21 = k 2 x (x + l )

x=l

( Q− Q) 2

= (12 cm )

32  C

( 45C − 32C )

= 64.57cm  65 cm

36. Make use of Example 21–12. From that, we see that the electric field due to the line charge along 1 ˆ i. In particular, the field due to that line of charge has no y-dependence. In the y-axis is E1 = 2 0 x 1 ˆ j. Then the 2 0 y

a similar fashion, the electric field due to the line charge along the x-axis is E2 = total field at ( x , y ) is the vector sum of the two fields. E = E1 + E2 =

  1 ˆ 1 ˆ 1 ˆ 1 ˆ i+ j=  i + y j 2 0 x 2 0 y 2 0  x  1 

 2 0

1 x

1 y

Ey  x 2 0 y = tan −1 = tan −1 x 2 + y 2 ;  = tan −1 1  2 0 xy Ex y 2 0 x

37. (a) The field due to the charge at A will point straight downward, and the field due to the charge at B will point along the line from A to the origin, 30o below the negative x-axis. Q Q EA = k 2 → EAx = 0, EAy = −k 2 l l

+Q l

B l

EB EA

737

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

EB = k

Q l

→ EBx = − k

l Q

EBy = − k Ex = EAx + EBx = − k E = E x2 + E y2 =

 = tan

Ey −1 Ex

cos 30o = − k sin 30o = − k

3Q 2l 4l

−k = tan −1 −k

2l 2 Q

2l 2 3Q

E y = EAy + EBy = − k

3k 2Q 2

9k 2Q 2 4l

12k 2Q 2

Instructor Solutions Manual

2l 2 3kQ

3Q 2 l 2 = tan −1 −3 = tan −1 3 = 240o 3Q − 3

2l 2 (b) Now reverse the direction of EB .

EA = k EB = k

Q l

→ EAx = 0, EAy = − k

Q l

→ EBx = k

 = tan

Ey −1 Ex

3k 2Q 2

= tan −1

k −k

3Q 2l

, EBy = k

E y = EAy + EBy = −k

cos 30o = k

Ex = EAx + EBx = k E = Ex2 + E y2 =

k 2Q 2 4

4k 2 Q 2 4l

Q l

sin 30o = k

Q 2l 2

kQ l2

Q 2 l 2 = tan −1 1 = 330o 3Q − 3 2l 2

38. Near the wire, the lines should come from it almost radially, because it is almost like an infinite line of charge when the observation point is close. When the observation point is far away, the field lines will look as if they originate on a point charge at the center of the plate.

39. Consider Example 21–10. We use the result from this example, but shift the center of the ring to be at x = 12 l for the ring on the right, and at x = − 12 l for the ring on the left. The fact that the original expression has a factor of x results in the interpretation that the sign of the field expression will give the direction of the field. No special consideration needs to be given to the location of the point at which the field is to be calculated. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

738

Chapter 21

Electric Charge and Electric Field

E = E right + E left Q ( x − 12 l )

Q ( x + 12 l ) ˆi + 1 ˆi / 2 / 2 4 0 ( x − 1 l )2 + R 2  4 0 ( x + 1 l )2 + R 2  1













Q  ( x − 12 l ) ( x + 12 l )  = ˆi +  / 2 / 2  2 2 4 0  ( x − 1 l ) + R 2  ( x + 1 l ) + R 2  







 

As a check on our answer, note that this expression gives a field of 0 at the origin, as it should. 40. Both charges must be of the same sign so that the electric fields created by the two charges oppose each other and add to zero. The magnitudes of the two electric fields must be equal.

E1 = E2 → k

( l 3)

( 2 l 3)

→ 9Q1 =

9Q2 4

→

1 4

(

)

41. In each case, find the vector sum of the field caused by the charge on the left E left and the field

(

caused by the charge on the right E right

)

Eright

Eleft A

Point A: From the symmetry of the geometry, in calculating the electric field at point A only the vertical components of the fields need to be considered. The horizontal components will cancel each other. 5.0  = tan −1 = 26.6o 10.0



( 0.050 m )2 + ( 0.100 m )2 = 0.1118 m

EA = 2

kQ d

(

sin = 2 8.988  10 N  m C 9

)

4.7  10−6 C

( 0.1118 m )

sin 26.6 = 3.0  106 N C

Point B: Now the point is not symmetrically placed. The horizontal and vertical components of each individual field need to be calculated to find the resultant electric field. 5.0 5.0 = 45 = 18.4  left = tan −1  left = tan −1 5.0 15.0 d left =

( 0.050 m ) 2 + ( 0.050 m ) 2 = 0.0707 m

d right =

( 0.050 m ) 2 + ( 0.150 m )2 = 0.1581m

E x = ( Eleft ) x + ( Eright ) x = k

(

Q d

= 8.988  109 N  m 2 C 2

2 left

cos left − k

Q 2 d right



)( 4.7  10 C )  −6

 A = 90

Eleft Eright

d right

d left +Q

 right

 left

cos  right cos45

 ( 0.0707 m )

−

cos18.4 

 = 4.372  10 N C ( 0.1581m )  6

739

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

E y = ( Eleft ) y + ( Eright ) y = k

(

= 8.988  109 N  m 2 C 2 EB =

2 left

sin left + k

2 d right

Instructor Solutions Manual

sin  right





sin45 sin18.4 + )( 4.7  10 C )  0.0707  = 6.509  10 N C m 0.1581m −6

(

)

(

)

 B = tan −1

E x2 + E y2 = 7.841  106 N C  7.8  106 N C

= 56.11  56 Ex The results are consistent with Fig. 21–35b. In the figure, the field at Point A points straight up, matching the calculations. The field at Point B should be between rightward and vertical, matching the calculations. Eright

42. (a) See the diagram. From the symmetry of the positions of the charges, we see that the net electric field points along the y-axis. Q sin  ˆj E = E left + E right = 2 4 0 r 2 =2

(

4 0 l 2 + y 2

)

sin  ˆj =

(

2 0 l 2 + y 2

)

 r



 l

Eleft

ˆj 3/ 2

(b) To find the position where the magnitude is a maximum, set the first derivative with respect to y equal to 0, and solve for the y value. Qy E= → 3/ 2 2 0 l 2 + y 2

(

dE dy

(

2 0 l 2 + y

(l + y )

2 3/ 2

)

2 3/ 2

+ ( − 32 )

(

2 0 l 2 + y 2

)

5/ 2

(2y) = 0 →

(l + y )

2 5/ 2

→ y 2 = 12 l 2 → y =  l

This has to be a maximum, because the magnitude of the electric field is positive for y > 0, the field is 0 for y = 0, and E → 0 as y → . 43. From Example 21–10, the electric field along the x-axis is E =

(

4 0 x + a 2 2

)

3 2

. To find the position

where the magnitude is a maximum, we differentiate and set the first derivative equal to zero.

( x + a ) − x ( x + a ) 2x = Q ( x + a ) − 3x  = dx 4 4 ( x + a ) (x + a )

3 2

1 2

2 3

5 2

 a 2 − 2 x 2  = 0 → xM =  2 2 2 4 0 ( x + a ) 5 2

Note that E = 0 at x = 0 and x = , and that E  0 for 0  x  . Thus the value of the magnitude of E at x = xM must be a maximum. We could also show that the value is a maximum by using the second derivative test. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

740

Chapter 21

Electric Charge and Electric Field

44. Because the distance from the wire is much smaller than the length of the wire, we can approximate the electric field by the field of an infinite wire, which is derived in Example 21–12.  7.45  10−6 C  2 6 1  1 2  N m 2   2.00 m  2.79  10 N C, E= = =  8.988  109 =  2 0 x 4 0 x  C2  2.40  10−2 m away from the wire

(

)

45. This is essentially Example 21–12 again, but with different limits of integration. From the diagram here, we see that the maximum l 2 angle is given by sin  = . We evaluate the results at 2 x2 + ( l 2) that angle. l 2

sin  =

2 x2 +( l 2)  E= ( sin  ) −l 2 4 0 x sin  =

x2 +( l 2)

     l  l 2 l 2 l   = = − − = 1/ 2 2 2 2 2 4 0 x  x 2 + ( l 2 )  2 0 x ( 4 x + l 2 ) x 2 + ( l 2 )   4 0 x x 2 + ( l 2 )   46. From the diagram, we see that the x components of the two fields will cancel each other at the point P. Thus the net electric field will be in the negative y-direction, and will be twice the y-component of either electric field vector. kQ Enet = 2 E sin  = 2 2 sin  x + a2 a 2kQ = 2 1/ 2 2 2 x + a x + a2

(

)

2kQa

(

x + a2 2

)

3/ 2

in the negative y direction

47. If we consider just one wire, then from the answer to Problem 45, we would have the following. Note that the distance from the wire to the point in question is x = z 2 + ( l 2 ) . 2

Ewire =

 2 0

z 2 + ( l 2)

( 4  z + ( l 2)  + l ) 2

1/ 2

But the total field is not simply four times the above expression, because the fields due to the four wires are not parallel to each other. Consider a side view of the problem. The two black dots represent two parallel wires, on opposite sides of the square. Note that only the vertical component of the field due to each wire will actually contribute to the total field. The horizontal components will cancel.

741

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Ewire = 4 ( Ewire ) cos  = 4 ( Ewire )

  Ewire = 4   2 0  =

Instructor Solutions Manual

z 2 + ( l 2)

 z  1/ 2 2  2 2 2 z 2 + ( l 2) 4  z 2 + ( l 2)  + l 2  z + ( l 2)    l

(

)

8 l z

 0 ( 4 z 2 + l 2 )( 4 z 2 + 2 l 2 )

1/ 2

The direction is vertical, perpendicular to the loop. 48. Select a differential element of the arc which makes an angle of  with the x-axis. The length of this element is Rd , and the charge on that element is dq =  Rd . The magnitude of the field produced by that element is 1  Rd . From the diagram, considering dE = 4 0 R 2 pieces of the arc that are symmetric with respect to the xaxis (the “upper” part and the “lower” part), we see that the total field will only have an x component. The vertical components of the field due to symmetric portions of the arc will cancel each other. So we have the following. 1  Rd cos  dE horizontal = 4 0 R 2 0

Ehorizontal = 

−0

1 4 0

cos 

 Rd R2

0

2 sin  0   cos  d = sin  0 − sin ( − 0 ) =  4 0 R − 4 0 R 4 0 R 0

The field points in the negative x-direction, so E = −

2 sin  0 ˆ 4 0 R

49. (a) If we follow the first steps of Example 21–12, and refer to Fig. 21–30, then the differential 1  dy electric field due to the segment of wire is still dE = . But now there is no 2 4 0 x + y 2

(

)

symmetry, and so we calculate both components of the field. 1 1  dy  x dy dE x = dE cos  = cos  = 3/ 2 2 2 2 4 0 x + y 4 0 x + y 2

(

dE y = − dE sin  = −

)

(

 dy

4 0 x + y 2

)

sin  = −

)

 y dy

(

4 0 x + y 2 2

)

3/ 2

The anti-derivatives needed are in Appendix B4.

742

Chapter 21

Electric Charge and Electric Field

 dy y x l x  = = Ex =    3/ 2 3/ 2 4 0 ( x 2 + y 2 ) 4 0 0 ( x 2 + y 2 ) 4 0  x 2 x 2 + y 2  0 0 l

 x dy

l 4 0 x x 2 + l 2 l

y dy  l   −1  Ey = − = − = −   3/ 2 3/ 2 4 0 ( x 2 + y 2 ) 4 0 0 ( x 2 + y 2 ) 4 0  x 2 + y 2  0 0 l

 y dy

(

1 1    x − x2 + l 2  2 2 − = 2 2 4 0  x + l x  4 x x + l 0

)

Note that E y  0, and so the electric field points to the right and down. (b) The angle that the electric field makes with the x-axis is given as follows.  x − x2 + l 2 2 2 E y 4 0 x x + l x − x2 + l 2 x x2 = = = − 1+ 2 tan  = l l l l Ex

(

)

4 0 x x 2 + l 2 As l → , the expression becomes tan  = −1 , and so the field makes an angle of

45 below the x-axis .

50. (a) If we follow the first steps of Example 21–12, and refer to Fig. 21–30, then the differential 1  dy electric field due to the segment of wire is still dE = . But now there is no 2 4 0 x + y 2

(

)

symmetry, and so we calculate both components of the field. 1 1  dy  x dy dE x = dE cos  = cos  = 3/ 2 2 2 4 0 x + y 4 0 x 2 + y 2

(

dE y = − dE cos  = −

)

(

 dy

4 0 x + y 2

sin  = −

)

 y dy

(

4 0 x 2 + y 2

)

3/ 2

The anti-derivatives needed are in Appendix B4. ymax

y  dy y x x  Ex =  = =   3/ 2 3/ 2 4 0 ( x 2 + y 2 ) 4 0 y ( x 2 + y 2 ) 4 0  x 2 x 2 + y 2  y y ymax

 x dy

max

min

ymax    − 4 0 x  x 2 + y 2 max 

min

  2  x 2 + ymin  ymin

 N m 2  ( 2.85  10 C ) ( 6.50 m ) =  8.99  109  C2  ( 0.250 m )  −6



2.50 m

 

( 0.250 m ) + ( 2.50 m )



−

 ( −4.00 m )  2 2 ( 0.250 m ) + ( −4.00 m ) 

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= 3.143  104 N C  3.14  104 N C ymax

y  y dy    −1 Ey = −  = − =−  2  3/ 2 3/ 2  4 0 ( x 2 + y 2 ) 4 0 y ( x 2 + y 2 ) 4 0  x + y 2  y y ymax

 y dy

max

min

1    − 2 4 0  x + y 2 max 



=  8.99  109

  2  x 2 + ymin  1

(

−6 N m 2  2.85  10 C



 

( 6.50 m )



 

( 0.250 m ) + ( 2.50 m )



min

−

)   2 2 ( 0.250 m ) + ( −4.00 m )  1

= 585.3 N C  585 N C

(b) We calculate the infinite line of charge result, and calculate the errors. 2 2.85  10 −6 C ) (  2  9 N m  = = 2  8.99  10 = 3.153  104 N m E=  2 2 0 x 4 0 x C  ( 6.50 m )( 0.250 m )  Ex − E E Ey E

( 3.143  10 N C ) − ( 3.153  10 N m ) = −0.0032 ( 3.153  10 N m ) 4

( 585.3 N C )

( 3.153  10 N m ) 4

= 0.0186

And so we see that E x is only about 0.3% away from the value obtained from the infinite line of charge, and E y is only about 2% of the value obtained from the infinite line of charge. The field of an infinite line of charge result would be a good approximation for the field due to this wire segment. 51. Choose a differential element of the rod dx a distance x from the origin, as shown in the diagram. The charge on that differential element is Q dq = dx. The variable x is treated as positive, l so that the field due to this differential element is dE =

4 0 ( x + x )

dx

4 0 l ( x + x )

. Integrate

along the rod to find the total field. l l l Q dx Q dx Q  Q 1 1  1  E =  dE =  = = − = −     2 2  4 0 l ( x + x ) 4 0 l 0 ( x + x ) 4 0 l  x + x 0 4 0 l  x x + l  0 =

Q 4 0 x ( x + l )

744

Chapter 21

Electric Charge and Electric Field

52. As suggested, we divide the plane into long narrow strips of width dy and length l. The charge on the strip is the area of the strip times the charge per unit area: dq =  l dy. The charge per dq unit length on the strip is  = =  dy. From Example 21–12, the field due to that narrow strip is l   dy dE = . From Fig. 21–70 in the textbook, we see that this field does = 2 0 y 2 + z 2 2 0 y 2 + z 2 not point vertically. From the symmetry of the plate, there is another long narrow strip a distance y on the other side of the origin, which would create the same magnitude electric field. The horizontal components of those two fields would cancel each other, and so we only need to calculate the vertical component of the field. Then we integrate along the y-direction to find the total field.  dy  zdy dE = ; dE z = dE cos  = 2 2 2 0 y 2 + z 2 2 0 y + z

(



)



dy z  z 1  −1 y  E = Ez =  = =  tan   2 2 2 2 z  − 2 0 ( y + z ) 2 0 − ( y + z ) 2 0 z  − =

 zdy

        −  −  =  tan −1 (  ) − tan −1 ( − )  =  2 0 2 0  2  2   2 0

53. (a) Since the field is uniform, the electron will experience a constant force in the direction opposite to its velocity, so the acceleration is constant and negative. Use Eq. 2–12c with a final velocity of 0. eE F = ma = qE = −eE → a = − ; v 2 = v02 + 2ax = 0 → m

(

)( )(

)

9.11  10 −31 kg 4.16  106 m s mv02 v02 x = − =− = = = 5.70  10−3 m 3 −19 2a  eE  2eE 2 1.60  10 C 8.65  10 N C 2 −   m (b) Find the elapsed time from constant acceleration relationships. Upon returning to the original position, the final velocity will be the opposite of the initial velocity. Use Eq. 2–12a. v = v0 + at → v02

v − v0 a

(

( (

)( )(

)

6 −31 −2v0 2mv0 2 9.11  10 kg 4.16  10 m s = = = = 5.48  10−9 s −19 3 eE  − eE  1.60  10 C 8.65  10 N C    m

)

54. (a) The acceleration is produced by the electric force. Fnet = ma = qE = − eE → a=−

e m

(1.60  10 C )  2.0ˆi + 7.0ˆj  10 N C = −3.513  10 ˆi − 1.229  10 ˆj m s ( ) )  ( ( 9.11  10 kg )  −19

E=−

−31

 −3.5  1015 m s 2 ˆi − 1.2  1016 m s 2 ˆj

745

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The direction is found from the components of the velocity. v = v 0 + at = 8.0  104 m s ˆj +  −3.513  1015 ˆi − 1.229  1016 ˆj m s 2  1.0  10 −9 s

(

)

(

)

(

)

= −3.513  106 ˆi − 1.221 107 ˆj m s tan −1

 −1.221 107 m s   = 254 or − 106, relative to x − axis 6  −3.513  10 m s 

= tan −1 

vx This is the direction relative to the x-axis. The direction of motion relative to the initial direction is measured from the y-axis, and so is  = 164 counter-clockwise from the initial direction. 55. (a) The electron will experience a force in the opposite direction to the electric field. Since the electron is to be brought to rest, the electric field must be in the same direction as the initial velocity of the electron, and so is to the right . (b) Since the field is uniform, the electron will experience a constant force, and therefore have a constant acceleration. Use constant acceleration relationships to find the field strength. qE qE F = qE = ma → a = v 2 = v02 + 2ax = v02 + 2 x → m m

( E=

) = −mv = − ( 9.109 10 kg )( 7.5 10 m s ) = 32 N C 2qx 2qx 2 ( −1.602  10 C ) ( 0.050 m )

m v 2 − v02

−31

2 0

−19

56. The angle is determined by the velocity. The x component of the velocity is constant. The time to pass through the plates can be found from the x motion. Then the y velocity can be found using constant acceleration relationships. x eE x x = v0 t → t = ; v y = v y 0 + ayt = − v0 m v0 tan  =

vy vx

− =

eE x m v0

(1.60 10 C )( 5.0 10 N C ) ( 0.046 m ) = −0.4040 → =− =− mv ( 9.1110 kg )(1.00 10 m s ) −19

eEx 2 0

−31

 = tan −1 ( −0.4040 ) = −22 57. Since the field is constant, the force on the electron is constant, and so the acceleration is constant. Thus constant acceleration relationships can be used. The initial conditions are x0 = 0, y0 = 0,

v x 0 = 1.90 m s, and v y 0 = 0. F = ma = qE → a =

q m

E=−

x = x0 + v x 0 t + 12 ax t 2 = v x 0t −

e m

E ; ax = −

e m

Ex , a y = −

e m

eEx 2 t 2m

(1.60 10 C )( 2.00 10 N C ) ( 3.0 s ) = −10.1m  −1.0 10 m = (1.90 m s )( 3.0 s ) − 2 ( 9.11  10 kg ) eE (1.60 10 C ) ( −1.20 10 N C ) ( 3.0 s ) = 9.48 m  9.5 m y = y +v t + a t = − t =− 2m 2 ( 9.11  10 kg ) −19

−11

−31

−19

1 2

−11

−31

746

Chapter 21

Electric Charge and Electric Field

58. (a) The field along the axis of the ring is given in Example 21–10, with the opposite sign because this ring is negatively charged. The force on the charge is the field times the charge q. Note that if x is positive, the force is to the left, and if x is negative, the force is to the right. Assume that x R. − qQx − qQx q 1 ( −Q ) x =  F = qE = 3 / 2 3 / 2 4 0 x 2 + R 2 4 0 x 2 + R 2 4 0 R 3

(

)

(

)

This has the form of a simple harmonic oscillator, where the “spring constant” is Qq . kelastic = 4 0 R 3 (b) The spring constant can be used to find the period. See Eq. 14–7b. T = 2

m kelastic

m Qq

= 2

m 4 0 R 3

= 2

= 4

m 0 R 3 Qq

4 0 R 59. (a) The dipole moment is given by the product of the positive charge and the separation distance. 3

(

)(

)

p = Q l = 1.60 10−19 C 0.68 10−9 m = 1.088 10−28 C m  1.110−28 C m (b) The torque on the dipole is given by Eq. 21–9a.

 = pE sin  = (1.088  10−28 C m )( 2.2  104 N C ) ( sin 90 ) = 2.4  10−24 m N

(

)(

)

(c)  = pE sin  = 1.088  10−28 C m 2.2  104 N C (sin 45 ) = 1.7  10−24 m N (d) The work done by an external force is the change in potential energy. Use Eq. 21–10. W = U = ( − pE cos  final ) − ( − pE cos  initial ) = pE ( cos  initial − cos  final )

(

)(

)

= 1.088  10−28 C m 2.2  104 N C 1 − ( −1) = 4.8  10 −24 J 60. (a) The dipole moment is the effective charge of each atom times the separation distance. p 3.4  10−30 C m = 3.4  10−20 C p = Ql → Q = = −10 1.0  10 m l −20 Q 3.4  10 C = = 0.21 No, the net charge on each atom is not an integer multiple of e. This (b) e 1.60  10−19 C is an indication that the H and Cl atoms are not ionized–they haven’t fully gained or lost an electron. But rather, the electrons spend more time near the Cl atom than the H atom, giving the molecule a net dipole moment. The electrons are not distributed symmetrically about the two nuclei. (c) The torque is given by Eq. 21–9a.

 = pE sin  →  max = pE = ( 3.4  10−30 C m )( 2.5  104 N C ) = 8.5  10−26 m N

(d) The energy needed from an external force is the change in potential energy. Use Eq. 21–10. W = U = ( − pE cos  final ) − ( − pE cos  initial ) = pE ( cos  initial − cos  final )

(

)(

)

= 3.4  10−30 C m 2.5  104 N C 1 − cos 45 = 2.5  10−26 J

747

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

61. (a) There will be a torque on the dipole, in a direction to decrease  . That torque will give the dipole an angular acceleration, in the opposite direction of  . d 2 pE  = − pE sin  = I →  = 2 = − sin  dt I If  is small, so that sin    , then the equation is in the same form as Eq. 14–3, the equation of motion for the simple harmonic oscillator. d 2 pE pE d 2 pE = −  − → +    =0 sin dt 2 I I dt 2 I (b) The frequency can be found from the coefficient of  in the equation of motion.

2 =

pE I

→ f =

1  = 2 2

pE I

62. (a) From the symmetry in the diagram, we see that the resultant field will be in the y-direction. The vertical components of the two fields add together, while the horizontal components cancel. Q r Enet = 2 E sin  = 2 1/ 2 2 2 4 0 r + l r2 + l 2

(

2Qr

(

4 0 r + l 2

)

2 3/ 2



)(

2Qr

( )

4 0 r

)

2Q 4 0 r 2

(b) Both charges are the same sign–there is no “cancellation” as there is for a dipole. A long distance away from the charges, they will look like a single charge 2Q q . of magnitude 2Q , and so E = k 2 = 4 0 r 2 r 63. If the dipole is of very small extent, then the potential energy is a function of position, and so Eq. 21–10 gives U ( x ) = −p E ( x ) . Since the potential energy is known, we can use Eq. 8–7. Fx = −

dU dx

=−

 − p E ( x )  = p dx dx

Note that the dipole moment is constant, so there is no

dp dx

term. Since the field does not depend on

 dE  ˆ the y or z coordinates, all other components of the force will be 0. Thus F = Fx ˆi =  p i .  dx  64. (a) Along the dipole axis the fields from the two charges are parallel so the magnitude is found as follows. Q ( −Q ) + Enet = E+ Q + E− Q = 2 2 1 4 0 ( r − 2 l ) 4 0 ( r + 12 l ) Q ( r + 12 l ) − ( r − 12 l )  2

 =  2 2 1 1 4 0 ( r + 2 l ) ( r − 2 l )

748

Chapter 21

Electric Charge and Electric Field

Q ( 2r l ) 4 0 ( r + 12 l ) ( r − 12 l ) 2



Q ( 2r l ) 4 0 r

2Ql

4 0 r

1 2p 4 0 r 3

The same result is obtained if the point is to the left of −Q. (b) The electric field points in the same direction as the dipole moment vector. 65. (a) The net force between the thymine and adenine is due to the following forces. 0.08ke 2 ( 0.4e )( 0.2e ) O – H attraction: FOH = k = 2 2 1.80  10−10 m 1.80  10−10 m

(

( 0.4e )( 0.2e )

O – N repulsion:

FON = k

N – N repulsion:

FNN = k

H – N attraction:

) ( =

)

0.08ke

( 2.80  10 m ) ( 2.80  10 m ) 2

−10

( 0.2e )( 0.2e )

= 2

−10

0.04ke 2

( 3.00  10 m ) ( 3.00  10 m ) −10

( 0.2e )( 0.2e )

FHN = k

−10

0.04ke 2

( 2.00  10 m ) ( 2.00  10 m ) −10

−10

1  0.08 0.08 0.04 0.04   ke FA-T = FOH − FON − FNN + FHN =  − − +  2 2 2 2 2  −10  1.80 2.80 3.00 2.00  1.0 10 m  d

(8.988 10 N  m C )(1.602 10 C ) = 4.623 10 N  5 10 N = ( 0.02004 ) (1.0 10 m ) 9

−19

−10

(b) The net force between the cytosine and guanine is due to the following forces. 0.08ke 2 ( 0.4e )( 0.2e ) O – H attraction: FOH = k (2 of these) = 2 2 1.90  10−10 m 1.90  10−10 m

(

) (

( 0.4e )( 0.2e )

O – N repulsion:

FON = k

H – N attraction:

FHN = k

N – N repulsion:

FNN = k

= 2

)

0.08ke 2

( 2.90  10 m ) ( 2.90  10 m ) −10

( 0.2e )( 0.2e )

= 2

−10

( 0.2e )( 0.2e )

= 2

−10

0.04ke 2

( 3.00  10 m ) ( 3.00  10 m ) −10

(2 of these)

0.04ke 2

( 2.00  10 m ) ( 2.00  10 m ) −10

−10

0.08 0.04 0.04  1  0.08  ke −2 − +  2 2 2 2  −10 2.90 3.00 2.00  1.0  10 m  d 2  1.90

FC-G = 2 FOH − 2 FON − FNN + FHN =  2

(8.988 10 N  m C )(1.602 10 C ) = 7.116  10 N  7 10 N = ( 0.03085 ) (1.0 10 m ) 9

−19

−10

(c) For the 105 pairs of molecules, we assume that half are A–T pairs and half are C–G pairs. We average the above results and multiply by 105 .

749

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

Fnet = 12 105 ( FA-T + FC-G ) = 105 4.623  10−10 N + 7.116  10−10 N

Instructor Solutions Manual

)

= 5.850  10 −5 N  6  10 −5 N 66. Set the magnitude of the electric force equal to the magnitude of the force of gravity and solve for the distance. e2 Felectric = Fgravitational → k 2 = mg → r

(8.988  10 N  m C ) = 5.08 m r=e = (1.602  10 C ) mg ( 9.11  10 kg )( 9.80 m s ) 9

−19

−31

67. Water has an atomic mass of 18, so 1 mole of water molecules has a mass of 18 grams. Each water molecule contains 10 protons.  6.02  1023 H 2O molecules   10 protons   1.60  10−19 C  9 72 kg    1 molecule   proton  = 3.9  10 C 0.018 kg       68. For the droplet to remain stationary, the magnitude of the electric force on the droplet must be the same as the weight of the droplet. The mass of the droplet is found from its volume times the density of water. Let n be the number of excess electrons on the water droplet. FE = q E = mg → neE = 43  r 3  g →

4 r 3  g 3eE

(

) (1.00 10 kg m )(9.80 m s ) = 9.96 10  1.0 10 electrons 3 (1.602  10 C ) (150 N C )

4 1.8  10−5 m

−19

69. There are four forces to calculate. Call the rightward direction the positive direction. The value of k is 8.988 109 N  m 2 C 2 and the value of e is 1.602 10−19 C.

Fnet = FCH + FCN + FOH + FON =

k ( 0.40e )( 0.20e ) 

(1  10 m ) −9

1 1 1 1  + + − − 2 2 2 2   ( 0.30 ) ( 0.40 ) ( 0.18) ( 0.28) 

= 2.445  10−10 N  2.4  10−10 N 70. Set the Coulomb electrical force equal to the Newtonian gravitational force on the Moon. M M Q2 FE = FG → k 2 = G Moon2 Earth → rorbit rorbit

GM Moon M Earth k

( 6.67 10 N m kg )( 7.35 10 kg )( 5.98 10 kg ) = 5.7110 C (8.988 10 N  m C ) −11

71. The electric force must be radial (centripetal) in order for the electron to move in a circular orbit. Q 2 mv 2 FE = Fradial → k 2 = → rorbit rorbit

750

Chapter 21

Electric Charge and Electric Field

rorbit

(1.602 10 C ) =k = ( 8.988  10 N  m C ) = 5.2  10 m mv ( 9.109 10 kg )( 2.2 10 m s ) −19

−11

−31

72. Because of the inverse square nature of the electric field, Q1 Q2 any location where the field is zero must be closer to the weaker charge ( Q2 ). Also, in between the two charges, l d the fields due to the two charges are in the same direction and cannot cancel. Thus the only place where the field can be zero is closer to the weaker charge, but not between the charges. In the diagram, this means that l must be positive. Evaluate the net field at that location. Q Q1 2 E = − k 22 + k = 0 → Q2 ( l + d ) = Q1l 2 → 2 l (l + d )

Q2 Q1 −

5.0  10−6 C

2.5  10−5 C − 5.0  10 −6 C

( 3.2 m ) =

2.6 m from Q2 , 5.8 m from Q1

73. We consider that the sock is only acted on by two forces–the force of gravity, acting downward, and the electrostatic force, acting upwards. If charge Q is on the sweater, then it will create an electric Q A  , where A is the surface area of one side of the sweater. The same magnitude = field of E = 2 0 2 0 of charge will be on the sock, and so the attractive force between the sweater and sock is Q2 FE = QE = . This must be equal to the weight of the sweater. We estimate the sweater area as 2 0 A 0.10 m2, which is roughly a square foot. Q2 = mg → FE = QE = 2 0 A

(

Q = 2 0 Amg = 2 8.85  10−12 C 2 N m 2

)( 0.10 m ) ( 0.040 kg ) ( 9.80 m s ) = 8  10 C 2

−7

74. The sphere will oscillate sinusoidally about the equilibrium point, with an amplitude of 5.0 cm. The angular frequency of the sphere is given by  = k m = 115 N m 0.650 kg = 13.3rad s. The distance of the sphere from the table is given by r = 0.150 − 0.0500 cos (13.3 t )  m . Use this distance and the charge to give the electric field value at the tabletop. That electric field will point upwards at all times, towards the negative sphere. 8.988  109 N  m 2 C 2 3.00  10 −6 C Q 2.70  104 E=k 2 = = N C 2 2 r 0.150 − 0.0500 cos (13.3 t ) m 2 0.150 − 0.0500 cos (13.3 t )

(

)(

1.08  107

3.00 − cos (13.3 t )

)

N C , upwards

751

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Instructor Solutions Manual

75. The electric field at the surface of the pea is given by Eq. 21–4a. Solve that equation for the charge.

E=k

→ Q=

Er 2

( 3 10 N C )(3.75 10 m ) = 5 10 C = −9

r k 8.988  10 N  m C This corresponds to about 3 billion electrons. 2

−3

76. There will be a rightward force on Q1 due to Q2 , given by Coulomb’s law. There will be a leftward force on Q1 due to the electric field created by the parallel plates. Let right be the positive direction.

Q1Q2

F = k x

− Q1 E

( 6.7  10 C )( 2.6  10 C ) − 6.7  10 C 5.3  10 N C = ( 8.988  10 N  m C ) ( )( ) −6

−6

( 0.47 m )

= 0.35 N, right 77. The mass of the sphere is the density times the volume. The electric force is given by Eq. 21–1, with both spheres having the same charge, and the separation distance equal to their diameter. Q2 kQ 2 3 4 mg = k →  3r g = → ( d )2 ( 2r )2

16 gr 5

(

) (

)(

16 35 kg m 3  9.80 m s 2 1.0  10 −2 m

(

3 8.99  109 N m 2 C 2

)

) = 8.0  10 C 5

−9

78. From the symmetry, we see that the resultant field will be in the y-direction. So we take the vertical component of each field. 2Q Q r − Enet = 2 E+ sin  − E− = 2 2 2 2 2 1/ 2 4 0 r 2 4 0 r + l r +l

(

= =

)(

)

2Q 

1 r   − 3/ 2 4 0  ( r 2 + l 2 ) r2    2Q

(

4 0 r 2 + l

)

2 3/ 2

 r 3 − ( r 2 + l 2 )3/ 2   r2 

  l 2 3/ 2  2Qr 1 −  1 + 2     r   3

 l2  4 0 r  1 + 2   r 

3/ 2

Use the binomial expansion, assuming r

752

Chapter 21

Electric Charge and Electric Field

  l 2 3/ 2  2Qr 1 −  1 + 2     r   3

Enet =

 l2  4 0 r  1 + 2   r 

3/ 2





2Qr 3 1 −  1 + 32

l 2 

 r 2    = l2  5 3 4 0 r  1 + 2 2  r   



2Qr 3  − 32

l2 

2  2  r  = − 3Ql 4 0 r 5 (1) 4 0 r 4

Notice that the field points toward the negative charges. 79. This is a constant acceleration situation, similar to projectile motion in a uniform gravitational field. Let the width of the plates be l, the vertical gap between the plates be h, and the initial velocity be v 0 . Notice that the vertical motion has a maximum displacement of h/2. Let upwards be the positive vertical direction. We calculate the vertical acceleration produced by the electric field as the electron crosses the region. eE Fy = ma y = qE = −eE → a y = − m Since the electron has constant horizontal velocity in the region, we use the horizontal component of the velocity and the length of the region to solve for the time of flight across the region. l l = v0 cos  0 ( t ) → t = v0 cos  0 In half this time the electron will travel vertically from the middle of the region to the highest point. At the highest point the vertical velocity must be zero. Using Eq. 2–12a with the known acceleration and time to reach the highest point, we can solve for the square of the initial speed of the electron.   l l eE   eE   2 v y = v0 y + a y ttop → 0 = v0 sin  0 +  −   12  → v0 =   2m  sin  0 cos  0  top  m   v0 cos  0  Finally, using Eq. 2–12b we solve for the initial angle.

ytop = y0 + v0 y t top + a y t 1 2

h = l tan  0 −

eE l 2

→

4m cos  0 v 2

1 2

  1 eE  1  l l h = v0 sin  0  12 − 2  2  m  v0 cos  0   v0 cos  0 

2 0

h = 12 l tan   →  0 = tan −1

= l tan  0 −

2h l

= tan −1

eE l 2

4m cos  0 eE 

 l   2m  sin  0 cos  0 

2 (1.0 cm ) 7.2 cm

→

= l tan  0 − 12 l tan  0

= 15.52  16

80. (a) The electric field from the long wire is derived in Example 21–12. 1  E= , radially away from the wire . 2 0 r (b) The force on the electron will point radially in, producing a centripetal acceleration. e  mv 2 F = qE = 2 0 r r

(1.60 10 C )( 0.14 10 C/m ) v= 2 = 2 ( 8.99  10 N  m / C ) 4 m 9.11 10 kg e

−19

−6

−31

= 21 107m/s

Note that this speed is independent of r.

753

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

81. We treat each of the plates as if it were infinite, and then use Eq. 21–7. The fields due to the first and third plates point towards their respective plates, and the fields due to the second plate point away from it. See the diagram. The directions of the fields are given by the arrows, so we calculate the magnitude of the fields from Eq. 21–7. Let the positive direction be to the right.

1

  − 2 + 3 EA = E1 − E2 + E3 = 2 0 2 0 2 0 =

(

2 8.85  10

EB = − E1 − E2 + E3 = − =

C Nm

1 2 0

(

2 8.85  10

−12

2 0

)

(

2 8.85  10

−12

C Nm

1 2 0

)

(

2 8.85  10

–

+ + +

2

E1 E2

– – – – –

E1 E2

–

3

= −2.3  103 N C = 2.3  103 N C to the left .

= 5.6  103 N C to the right .

2 3 − 2 0 2 0

( −0.50 + 0.25 − 0.35 )  10−6 C m 2 −12

–

2 3 + 2 0 2 0

( −0.50 + 0.25 + 0.35 )  10−6 C m 2

ED = − E1 + E2 − E3 = − =

C Nm +

+ + +

2 3 + 2 0 2 0

1

= 3.4  104 N C , to the right .

( −0.50 − 0.25 + 0.35 )  10−6 C m 2

EC = − E1 + E2 + E3 = − =

−

)

– – –

1

( 0.50 − 0.25 + 0.35)  10−6 C m 2 −12

Instructor Solutions Manual

C Nm

)

= −3.4  10 4 N C = 3.4  103 N C to the left .

82. Since the electric field exerts a force on the charge in the same direction as the FT  electric field, the charge is positive . Use the free-body diagram to write the equilibrium equations for both the horizontal and vertical directions, and use those FE mg equations to find the magnitude of the charge. 43  = cos −1 = 38.6 55  l = 55cm  Fx = FE − FT sin  = 0 → FE = FT sin  = QE 43cm mg

 F = F cos  − mg = 0 → F = cos  → QE = mg tan  mg tan  (1.5  10 kg )( 9.80 m s ) tan 38.6 Q= = = 1.2  10 C y

−3

−6

( 9500 N C )

754

Chapter 21

Electric Charge and Electric Field

83. A negative charge must be placed at the center of the square. Let Q = 5.8  C be the charge at each corner, let − q be the magnitude of negative charge in the center, and let d = 9.2 cm be the side length of the square. By the symmetry of the problem, if we make the net force on one of the corner charges be zero, the net force on each other corner charge will also be zero. The force on the charge in the center is also zero, from the symmetry of the problem. Q2 Q2 F41 = k 2 → F41x = k 2 , F41 y = 0 d d

F42 = k F43 = k F4 q = k

→ F42 x = k

2d 2 Q2

2Q 2

4d 2

cos45o = k 2

→ F43 x = 0, F43 y = k

d2 qQ

→ F4 qx = − k

2qQ 2

2Q 2

+k 2

+0−k

F41

F4q −q

2Q 2 4d 2

Q2 d2

cos 45o = − k

d 2 d The net force in each direction should be zero.

 Fx = k

, F42 y = k

F42

F43 Q4

2 qQ d2  1

2qQ d2

= F4 qy

=0 →

1  1 1 +  = ( 5.8  10−6 C )  +  = 5.6  10−6 C  2 4  2 4

q = Q

So the charge to be placed is −q = −5.6  10−6 C . Note that the length of the sides of the square does not enter into the calculation. This is an unstable equilibrium . If the center charge were slightly displaced, say towards the right, then it would be closer to the right charges than the left charges, and would be attracted more to the right. Likewise the positive charges on the right side of the square would be closer to it and would be attracted more to it, moving from their corner positions. The system would not have a tendency to return to the symmetric shape, but rather would have a tendency to move away from that symmetric shape if disturbed. 84. (a) The force of sphere B on sphere A is given by Coulomb’s law.

FAB =

kQ 2

, away from B . R2 (b) The result of touching sphere B to uncharged sphere C is that the charge on B is shared between the two spheres, and so the charge on B is reduced to Q 2. Again use Coulomb’s law.

FAB = k

QQ 2

kQ 2

, away from B . R2 2R2 (c) The result of touching sphere A to sphere C is that the charge on the two spheres is shared, and so the charge on A is reduced to 3Q 4. Again use Coulomb’s law. FAB = k

( 3 Q 4 )( Q 2 ) R2

3kQ 2 8R 2

, away from B .

755

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

85. (a) The weight of the mass is only about 1.4 N. Since the tension in the string is more than that, there must be a downward electric force on the positive charge, which means that the electric field must be pointed down . Use the free-body diagram to write an expression for the magnitude of the electric field.  F = FT − mg − FE = 0 → FE = QE = FT − mg →

FT − mg Q

(

5.18 N − ( 0.145 kg ) 9.80 m s −6

3.40  10 C

FT mg

) = 1.1056 10 N C

 1.11  106 N C , downward . (b) Use Eq. 21–7.  E= →  = 2 E 0 = 2 (1.1056  106 N C )( 8.854  10 −12 ) = 1.96  10 −5 C m 2 2 0 But since the field points towards the sheet, it’s charge density must be negative.

 = −1.96  10−5 C m2 86. (a) The force will be attractive. Each successive charge is another distance d farther than the previous charge. The magnitude of the charge on the electron is e. eQ eQ eQ eQ eQ  1 1 1 1  F =k +k +k +k + =k 2  2 + 2 + 2 + 2 +  2 2 2 2 d 1 2 3 4   (d ) ( 2d ) ( 3d ) ( 4d ) =k

eQ  1

1 eQ  2

 n = 4 d d 2

n =1

 eQ 24 0d 2

(b) Now the closest Q is a distance of 3d from the electron. eQ eQ eQ eQ eQ  1 1 1 1 +k +k +k + =k 2  2 + 2 + 2 + 2 + F =k 2 2 2 2 d 3 4 5 6 ( 3d ) ( 4d ) ( 5d ) ( 6d )

 2 5  = k 2  2 = k 2   2  − 2 − 2  = k 2  −  = − d n =3 n d  n =1 n  1 2  d  6 4  4 0 d 2  6 4  eQ  1

eQ   1 

1

eQ   2

5

  

87. (a) A mole of carbon weighs 12.01 grams. An atom of carbon has 6 electrons, since the atomic number of carbon is 6. 6.02  1023 atoms 6 electrons N total = 1.00 mol   = 3.61  1024 electrons mol atom (b) One sphere will have a positive charge, and the other sphere will have the same amount of negative charge. First solve for that charge by equating the electric force to the gravitational force. Then compare that charge to the total charge. N represents the number of electrons moved from one sphere to the other. We assume that the number of electrons transferred will be small compared to the total number of electrons, and so we use the mass of 1 mole of neutral C atoms for m for both spheres. m2 ( Ne ) 2 Felectric = Fgravitation → k = G → r2 r2

756

Chapter 21

Electric Charge and Electric Field

( 0.012 kg )

6.67  10 −11

−19

(1.602  10 C ) 8.988  10

= 6.453  106 electrons

6.453  106 electrons

= 1.79  10−18 3.61  10 24 electrons Our assumption about the number of electrons transferred is justified. fraction =

88. Two forces act on the mass; gravity and the electric force. Since the acceleration is smaller than the acceleration of gravity, the electric force must be upward, opposite the direction of the electric field. The charge on the object is therefore negative. Using Newton’s second law, with down as the positive direction, we can calculate the magnitude of the charge on the object.  F = ma → mg + qE = ma →

m (a − g ) E

(1.0 kg ) (8.0 m s2 − 9.8 m s2 ) 150 N C

= −1.2  10 −2 C

89.

90. (a) From Problem 88, we know that the electric field is pointed towards the Earth’s center. Thus an electron in such a field would experience an upwards force of magnitude FE = eE. The force of gravity on the electron will be negligible compared to the electric force. FE = eE = ma →

eE m

(1.602 10 C ) (150 N C ) = 2.638 10 m s  2.6 10 m s , up . ( 9.1110 kg ) −19

−31

(b) A proton in the field would experience a downwards force of magnitude FE = eE . The force of gravity on the proton will be negligible compared to the electric force. FE = eE = ma →

(1.602 10 C ) (150 N C ) = 1.439 10 m s  1.4 10 m s , down . = a= m (1.67 10 kg ) −19

−27

a g

2.638  1013 m s2 9.80 m s2

 2.7  1012 ; Proton:

a g

1.439  1010 m s 2 9.80 m s 2

 1.5  109

757

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

91. Take Fig. 21–29 and add the angle , measured from the –z axis, as indicated in the diagram. Consider an infinitesimal length of the ring ad . The charge on that infinitesimal length is dq =  ( ad  ) =

a

( ad ) =



d . The charge creates an infinitesimal electric Q

field, d E, with magnitude dE =

d  = . From the 4  x 2 + a 2

1 dq

4  r 2

symmetry of the figure, we see that the z component of d E will be cancelled by the z component due to the piece of the ring that is on the opposite side of the y-axis. The trigonometric relationships give dE x = dE cos  and dE y = −dE sin  sin . The factor of sin  can be justified by noting that dE y = 0 when  = 0, and dE y = −dE sin  when  =  2. dEx = dE cos  = Ex =

d

4   x + a 2

x 2

x +a 2



d

(

4   x + a 2 2

)

3/ 2

 d = 4 x + a 4  ( x + a ) ( ) 2

2 3/ 2



dE y = − dE sin  sin  = − Ey = −

2 3/ 2



d

4   x + a 2

a 2

x +a 2



sin  = −

(

4   x 2 + a 2 2

)

3/ 2

sin  d

 sin  d = − 4  x + a ( − cos  ) − ( − cos 0 ) 4  ( x + a ) ( ) 2

2 3/ 2

= −



2 3/ 2



2Qa

(

4 2  x 2 + a 2

)

3/ 2

We can write the electric field in vector notation. E=

(

4  x 2 + a

)

2 3/ 2

ˆi −

2Qa

(

4 2  x 2 + a

)

2 3/ 2

ˆj =

(

4 0 x 2 + a

)

2 3/ 2

 xˆi − 2a ˆj      

92. (a) Select a differential element of the arc which makes an angle of  with the x-axis. The length of this element is Rd , and the charge on that element is dq =  Rd . The magnitude of the field produced by that element is 1  Rd . From the diagram, considering dE = 4 0 R 2 pieces of the arc that are symmetric with respect to the x-axis, we see that the total field will only have a y component, because the magnitudes of the fields due to those two pieces are the same. From the diagram we see that the field will point down. The horizontal components of the field cancel.

758

Chapter 21

Electric Charge and Electric Field

dE vertical =

1 4 0

 Rd R

sin  =

0 sin 2  d 4 0 R

 /2

E vertical = =

 /2

    /2 2 2  40 R sin  d = 40 R  sin  d = 40 R ( 12  − 14 sin 2 )− / 2 0 0 0 − / 2 − / 2 0  ( 12  ) = 0 4 0 R 8 0 R

→ E= −

0 ˆ j 8 0 R

(b) The force on the electron is given by Eq. 21–5. The acceleration is found from the force. q0 ˆ F = ma = qE = − j → 8 0 R

(

)(

) )

1.60  10−19 C 1.0  10−6 C m q0 ˆ e0 ˆ ˆj a=− j= j= 8m 0 R 8m 0 R 8 9.11  10−31 kg 8.85  10−12 C 2 N m 2 ( 0.010 m )

(

)(

= 2.5  1017 m s 2 ˆj

759

CHAPTER 22: Gauss’s Law Responses to Questions 1.

No. If the net electric flux through a surface is zero, then the net charge contained in the surface is zero. However, there may be charges both inside and outside the surface that affect the electric field at the surface. The electric field could point outward from the surface at some points and inward at others. A surface surrounding an electric dipole would have a net flux of 0 through the surface, but the electric field would not be 0 on each point of the surface. Yes. If the electric field is zero for all points on the surface, then the net flux through the surface must be zero and no net charge is contained within the surface.

No. The electric field in the expression for Gauss’s law refers to the total electric field (the electric field due to all charge), not just the electric field due to any enclosed charge. Notice, though, that if the electric field is due to charge outside the gaussian surface, then the net flux through the surface due to this charge will be zero.

The net flux will be zero. An electric dipole consists of two charges that are equal in magnitude but opposite in sign, so the net charge of an electric dipole is zero. If the closed surface encloses a zero net charge, than the net flux through it will be zero.

Yes. If the electric field is zero for all points on the surface, then the integral of E dA over the surface will be zero, the flux through the surface will be zero, and no net charge will be contained in the surface. No. If a surface encloses no net charge, then the net electric flux through the surface will be zero, but the electric field is not necessarily zero for all points on the surface. The integral of E dA over the total surface must be zero, but the electric field itself is not required to be zero. There may be charges outside the surface that will affect the values of the electric field at the surface.

The electric flux through a surface is the scalar (dot) product of the electric field vector and the area vector of the surface. Thus, in magnitude,  E = EA cos  . By analogy, the gravitational flux through a surface would be the product of the gravitational field (or force per unit mass) and the area, or g = gA cos  . Any mass, such as a planet, would be a “sink” for gravitational field, because gravitational field lines point in the direction that a mass would accelerate if free to move. Since there is no “anti-gravity” there would be no sources.

No. Gauss’s law is most useful in cases of high symmetry, where a surface can be defined over which the electric field has a constant value and a constant relationship to the direction of the outward normal to the surface. Such a “simple” surface cannot be defined for an electric dipole.

When the ball is inflated and charge is distributed uniformly over its surface, the field inside is zero. Since the ball is spherically symmetric, only a radial electric field is possible. The analysis of Example 22–3 would still be applicable to this non-conducting spherically symmetric case, if all of the charge is on the surface. When the ball is collapsed, there is no symmetry to the charge distribution, and the calculation of the electric field strength and direction at points inside the ball is quite difficult (and will likely give a non-zero result for some positions).

760

Chapter 22

Gauss’s Law

For an infinitely long wire, the electric field is radially outward from the wire, resulting from contributions from all parts of the wire. The charge outside the gaussian surface contributes to the symmetry of the electric field. For an infinitely long wire, there is no preference of left or right along the wire, and so the electric field only points radially outward. This allows us to set up a gaussian surface that is cylindrical, with the cylinder axis parallel to the wire. This surface will have zero flux through the left and right ends of the cylinder, since the net electric field and the outward surface normal are both perpendicular at all points over the left and right ends. In the case of a short wire, the electric field is not radially outward from the wire near the ends; it curves away from being perfectly radial, and actually points directly outward along the axis of the wire at both the left and right ends. We cannot define a useful gaussian surface for this case, and the electric field must be computed directly. Thus the radially-symmetric field is due to the entire wire, not just the part enclosed by the cylinder.

In Example 22–6, there is no flux through the flat ends of the cylindrical gaussian surface because the field is directed radially outward from the wire. If instead the wire extended only a short distance past the ends of the cylinder, there would be a non-radial component of the field through the ends of the cylinder. The result of the example would be altered because the value of the field at a given point would now depend not only on the radial distance from the wire but also on the distance from the ends.

10. The electric flux through the sphere remains the same, since the same charge is enclosed. The electric field at the surface of the sphere is changed, because different parts of the sphere are now at different distances from the charge. The electric field will not have the same magnitude on all parts of the sphere, and the direction of the electric field will not be parallel to the outward normal for all points on the surface of the sphere. The electric field will be stronger on the parts of the sphere closer to the charge and weaker on the parts further from the charge. 11. (a) A charge of (Q – q) will be on the outer surface of the conductor. The total charge Q is placed on the conductor but since +q will reside (via induction) on the inner surface, the remainder, (Q – q), will reside on the outer surface. (b) A charge of +q will reside on the inner surface of the conductor since that amount is attracted by the charge –q in the cavity. (Note that E must be zero inside the material of the conductor.) 12. Yes. The charge q will induce a charge –q on the inside surface of the thin metal shell, leaving the outside surface with a charge +q. The charge Q outside the sphere will feel the same electric force as it would if the metal shell were not present (assuming that the presence of Q does not distort the distribution of the charge on the outside surface). 13. The total flux through the balloon’s surface will not change because the enclosed charge does not change. The flux per unit surface area will decrease, since the surface area increases while the total flux does not change.

Solutions to MisConceptual Questions 1.

(c) The electric flux will be the same. The flux is equal to the net charge enclosed by the surface divided by ε0. If the same charge is enclosed, then the flux is the same, regardless of the shape of the surface.

761

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(e) Since there is nothing inside the ball, there are no charges where electric field lines could originate or terminate. And all the charge is of the same sign, so no field lines could both originate and terminate on the ball. Example 22–3 gives details using Gauss’s law to explain why there are no electric field lines inside the ball.

(b) Gauss’s law is always true for electrostatics, so (a) is not correct. But Gauss’s law is only simple to use if the charge distribution is very symmetric (spherical, cylindrical, planar) so that a simple gaussian surface can be used which has a simple relationship to the geometry of the charged object. So answers (c) and (d) are also not correct. Gauss’s law is true but not useful for this situation.

(a) The electric field at the surface of a uniform sphere of charge is inversely proportional to the radius of the sphere–see Example 22–3. Since the same charge is being put on a sphere with ¼ of the radius, the electric field would be 16 times larger for the smaller sphere (the Moon).

(a) If a gaussian sphere is drawn with a radius smaller than the radius of the hollow sphere, the enclosed charge would be only the +Q point charge. Thus the application of Gauss’s law for this spherically-symmetric charge distribution would give the electric field expression in answer (a).

(d) If a gaussian sphere is drawn with a radius larger than the radius of the hollow sphere, the enclosed charge would be 0. Thus the application of Gauss’s law fort his spherically-symmetric charge distribution would result in an electric field of 0, answer (d).

(c) Gauss’s law says that the total electric flux passing through a closed surface is proportional to the charge enclosed by that surface. Since there is no flux passing through the surface, the total charge enclosed by the surface must be 0, and so the flies have to have equal and opposite charges.

(d) The charge is spherically symmetric, so the electric field would be spherically symmetric, but the surface through which the flux is to be calculated is NOT spherically symmetric, and so the electric field is not uniform across a side of the box. Thus, the flux integral calculation will not simply be “uniform electric field times area,” but instead involve detailed geometric considerations.

(c) The only charge enclosed by a spherical gaussian surface that passes through point A would be the –3 C at the center.

10. (b, d) This situation is discussed in Example 22–4. We assume that answer (a) means IMMEDIATELY above the surface of the sphere, so r = r0, as opposed to meaning anywhere outside the sphere. With this interpretation, answer (a) is correct. Answer (b) is incorrect because the field outside decreases a 1/r2. It is true that the charge enclosed is constant for any r > r0, but the electric field also depends on the distance from the center. Answer (c) is correct, as shown in Example 22–4. Answer (d) is incorrect, because Example 22–4 shows that the electric field inside the sphere increases linearly with distance from the center. Answer (e) is correct, again since the electric field inside the sphere increases linearly with distance from the center.

762

Chapter 22

Gauss’s Law

Solutions to Problems 1.

The electric flux of a uniform field is given by Eq. 22–1b. (a)  E = E A = EA cos  = ( 640 N C )  ( 0.13m ) cos0 = 34 N m 2 C 2

(b)  E = E A = EA cos  = ( 640 N C )  ( 0.13m ) cos 45 = 24 N m2 C 2

 E = E A = EA cos  = ( 640 N C )  ( 0.13m ) cos 90 = 0 2

Use Eq. 22–1b for the electric flux of a uniform field. Note that the surface area vector points radially outward, and the electric field vector points radially inward. Thus the angle between the two is 180.

(

 E = E A = EA cos  = (150 N C ) 4 RE2 cos180 = −4 (150 N C ) 6.38 106 m

)

= −7.7  1016 N m 2 C 3.

(a) Since the field is uniform, no lines originate or terminate inside the cube, and so the net flux is  net = 0 . (b) There are two opposite faces with field lines perpendicular to the faces. The other four faces have field lines parallel to those faces. For the faces parallel to the field lines, no field lines enter or exit the faces. Thus  parallel = 0 .

Of the two faces that are perpendicular to the field lines, one will have field lines entering the cube, and so the angle between the field lines and the face area vector is 180. The other will have field lines exiting the cube, and so the angle between the field lines and the face area vector is 0. Thus we have  entering = E A = E0 A cos180 = − E0 l 2 and

 leaving = E A = E0 A cos 0 = E0 l 2 . 4.

(a) From the diagram in the textbook, we see that the outward flux through the hemispherical surface is the same magnitude as the inward flux through the circular surface base of the hemisphere. On the circular surface base, all of the inward flux has the same orientation– exactly opposite to the outward pointing surface normal, so on the circular base,  E =  E dA = EA cos180 = − r 2 E. Thus for the hemisphere, we have  E =  r 2 E .

(b) E is vertical and perpendicular to the axis. Then every field line would both enter through the hemispherical surface at some angle and then leave through the hemispherical surface at the same angle relative to the surface, and so  E = 0 . 5.

Use Gauss’s law to determine the enclosed charge. Note that the size of the box does not enter into the calculation. The flux is outward, so it is positive. Q  E = encl → Qencl =  E  o = 1850 N  m 2 C 8.85  10 −12 C 2 N  m 2 = 1.64  10 −8 C o

(

)(

)

763

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The net flux through each closed surface is determined by the net charge inside. Refer to the figure in the textbook.

1 = ( +Q − 3Q )  0 = −2 Q  0 ;  2 = ( +Q + 2Q − 3Q )  0 = 0 ;  3 = ( +2Q − 3Q )  0 = − Q  0 ;  4 = 0 ;  5 = +2 Q  0 7.

(a) Use Gauss’s law to determine the electric flux. Q −1.0  10−6 C  E = encl = = −1.1 105 N  m 2 C 8.85  10−12 C 2 N  m 2 o (b) Since there is no charge enclosed by surface A2,  E = 0 .

The net flux is only dependent on the charge enclosed by the surface. Since both surfaces enclose the same amount of charge, the flux through both surfaces is the same. Thus the ratio is 1: 1 .

The only contributions to the flux are from the faces perpendicular to the electric field. Over each of these two surfaces, the magnitude of the field is constant, so the flux is just E A on each of these two surfaces. Q  E = ( E A ) right + ( E A )left = Eright l 2 − Eleft l 2 = encl → 0

(

)

Qencl = ( Eright − Eleft ) l 2 0 = ( 370 N C − 650 N C )( 25m ) 8.85  10−12 C2 N  m 2 = −1.5  10 −6 C 2

10. Because of the symmetry of the problem one-sixth of the total flux will pass through each face.

 face = 16  total = 16

Qencl

0

Qencl

6 0

Notice that the side length of the cube did not enter into the calculation. 11. The charge density can be found from Eq. 22–4, Gauss’s law. The charge is the charge density times the length of the rod. 7.3  105 N  m 2 C 8.85  10 −12 C 2 N  m 2 l  0 Q  = encl = = → = = = 2.6  10−5 C m 0 0 l 0.250 m

(

)(

)

12. The egg is metal, so the electric field lines should be perpendicular to the egg at its surface.

13. The electric field can be calculated by Eq. 21–4a, and that can be solved for the magnitude of the charge.

E =k

Q r

→ Q =

E r2 k

(5.85  10 N C )( 3.50  10 m ) = 7.97  10 C = −2

−11

8.988  10 N  m C 9

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Chapter 22

Gauss’s Law

This corresponds to about 5 108 electrons. Since the field points toward the ball, the charge must be negative. Thus Q = −7.97  10−11 C . 14. The charge on the spherical conductor is uniformly distributed over the surface area of the sphere, so Q = . The field at the surface of the sphere is evaluated at r = R. 4 R 2

E (r = R) =

4 0 R

= 2

1 4 R 2 4 0

 0

15. The electric field due to a long thin wire is given in Example 22–6 as E = (a) E =

1 

1 2

(

= 8.988  109 N  m 2 C 2

1  2 0 R

(

) = −3.0  10 N C

(

) = −1.0  10 N C

)

2 −8.4  10−6 C m

)

2 −8.4  10−6 C m

2 0 R 4 0 R ( 5.0 m ) The negative sign indicates the electric field is pointed towards the wire. The magnitude of the electric field is 3.0 104 N C .

(b) E =

1 

1 2

(

= 8.988  109 N  m 2 C 2

2 0 R 4 0 R (1.5 m ) The negative sign indicates the electric field is pointed towards the wire. The magnitude of the electric field is 1.0 105 N C .

16. Because the globe is a conductor, the net charge of –1.50 mC will be arranged symmetrically around the sphere.

17. See Fig. 22–30 in the text for the diagram for this problem. (a) For r less than the inner radius of the shell, the field is that of the point charge, E =

1 Q 4 0 r 2

(b) There is no field inside the conducting material; E = 0 . (c) Outside the shell, the field is that of the point charge, E =

1 Q 4 0 r 2

(d) The shell does not affect the field due to Q alone, except in the shell material, where the field is 0. (e) The charge Q does affect the shell–it polarizes it. There will be an induced charge of –Q uniformly distributed over the inside surface of the shell, and an induced charge of +Q uniformly distributed over the outside surface of the shell. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

18. See Example 22–3 for a detailed discussion related to this problem. (a) Inside a solid metal sphere the electric field is 0 . (b) Inside a solid metal sphere the electric field is 0 . (c) Outside a solid metal sphere the electric field is the same as if all the charge were concentrated at the center as a point charge. 4.80  10−6 C 1 Q 9 2 2 = 8.988  10 N  m C = 4490 N C E = 4 0 r 2 ( 3.10 m )2 The field would point towards the center of the sphere. (d) Same reasoning as in part (c). 4.80  10−6 C 1 Q 9 2 2 =   = 674 N C E = 8.988 10 N m C 4 0 r 2 (8.00 m )2 The field would point towards the center of the sphere. (e) The answers would be no different for a thin metal shell. (f) The solid sphere of charge is dealt with in Example 22–4. We see from that Example that the 1 Q field inside the sphere is given by E = r. Outside the sphere the field is no different. 4 0 r03 So we have these results for the solid non-conducting sphere. 4.80  10−6 C E ( r = 0.250 m ) = 8.988  109 N  m 2 C2 ( 0.250 m ) = 399 N C ( 3.00 m )3

(

)

(

)

(

)

(

)

(

)

(

E ( r = 2.90 m ) = 8.988  10 N  m C 9

4.80  10−6 C

) 3.00 m ( 2.90 m ) = 4630 N C (

−6

(

) 4.803.10 10m C = 4490 N C

(

 10 C = 674 N C ) 4.80 8.00 m

E ( r = 3.10 m ) = 8.988  109 N  m 2 C 2 E ( r = 8.00 m ) = 8.988  109 N  m 2 C2

(

−6

(

All point towards the center of the sphere. 19. For points inside the nonconducting spheres, the electric field will be determined by the charge inside the spherical surface of radius r. 3

 4  r3  r Qencl = Q  3 3  = Q   4  3  r0   r0  The electric field for r  r0 can be calculated from Gauss’s law. 3

−6 r  Q  1 9 2 2 3.25  10 C E ( r  r0 ) = Q r r 8.988 10 N m C = = =   ( )  r  4 r 2  4 r 3  4 0 r 2 ( 0.075m )3  0  0 0 0 

Qencl

 

N  r C m The electric field outside the sphere is calculated from Gauss’s law with Qencl = Q. =  6.92  107

E ( r  r0 ) =

Qencl 4 0 r

= 2

Q 4 0 r 2

(8.988  10 N  m C )( 3.25  10 C ) = 2.92  10 N  m C = 9

−6

766

Chapter 22

Gauss’s Law

20. (a) When close to the sheet, we approximate it as an infinite sheet, and use the result of Example 22–7. We assume the charge is over both surfaces of the aluminum. 245  10−9 C

 ( 0.25 m ) E= = = 2.2  105 N C , away from the sheet . 2 o 2 ( 8.85  10−12 C2 N  m 2 ) 2

(b) When far from the sheet, we approximate it as a point charge. −9 1 Q 9 2 2 245  10 C E= = 8.988  10 N  m C = 9.8 N C , away from the sheet . 2 4 0 r 2 (15 m )

(

)

21. (a) Consider a spherical gaussian surface at a radius of 3.00 cm. It encloses all of the charge. Q 2  E dA = E 4 r = →

(

) 

4 0 r

(

= 8.988  109 N  m 2 C 2 2

)

6.50  10 −6 C

( 3.00 10 m ) −2

= 6.49  107 N C , radially outward .

(b) A radius of 6.00 cm is inside the conducting material, and so the field must be 0. Note that there must be an induced charge of −6.50  10−6 C on the surface at r = 4.50 cm, and then an induced charge of 6.50  10−6 C on the outer surface of the sphere. (c) Consider a spherical gaussian surface at a radius of 30.0 cm. It encloses all of the charge. Q 2  E dA = E 4 r = →

(

) 

4 0 r

(

= 8.988  109 N  m 2 C 2 2

−6

) 6.50 10 C = 6.49 10 N C , radially outward . ( 30.0 10 m ) 5

−2

22. Due to the spherical symmetry of the problem, the electric field can be evaluated using Gauss’s law and the charge enclosed by a spherical gaussian surface of radius r. We assume the electric field points radially outward. If it points inward, we will get a negative value for the electric field. 1 Q Q 2  E dA = E 4 r = encl → E = 4 rencl2 0 0

(

)

767

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Since the charge densities are constant, the charge enclosed is found by multiplying the appropriate charge density times the volume of charge enclosed by the gaussian sphere. Let r1 = 6.0 cm and r2 = 12.0 cm.

(a) Negative charge is enclosed for r  r1 .

1 Qencl 4 0 r 2

1 (−) 4 0

(

(  r ) = ( )r = 4 3

( −4.0C m ) r

−

3 0

(

3 8.85  10−12 C 2 N  m 2

)

= −1.5  1011 N C m r The negative sign means it points radially inward. (b) In the region r1  r  r2 , all of the negative charge and part of the positive charge is enclosed.

1 Qencl 4 0 r

1 (−) 4 0

(  r ) +  ( )   ( r − r ) = ( ( ) − ( ) ) ( r ) + ( )r 4 3

3 1

4 3

3 1

3 0 r

)

3 0

( 9.0 C m ) r

3 8.85  10−12 C 2 N  m 2 r 2

3 1

( −4.0 C m 3 ) − ( 9.0 C m 3 )  ( 0.060 m ) = +

(

−

(

3 8.85  10 −12 C 2 N  m 2

)

( −1.1  10 N m C ) + 3.4  10 N C m r = ( ) r 8

1 Qencl 4 0 r 2

1 (−) 4 0

(  r ) + ( )   ( r − r ) = ( ( ) − ( ) ) ( r ) + ( ) ( r ) = 4 3

3 1

3 2

4 3

3 1

−

3 1

3 2

3 0 r 2

( −4.0C m3 ) − ( 9.0C m3 ) ( 0.060 m ) + ( 9.0C m3 ) ( 0.120m ) ( 4.8  108 N m2 C ) = = 3

(

)

3 8.85  10−12 C2 N  m 2 r 2

(d) See the adjacent plot. The field is continuous at the edges of the layers. The field is continuous at the boundaries because there is no surface charge density on any boundary.

23. (a) There can be no field inside the conductor, and so there must be an induced charge of −8.00C on the surface of the spherical cavity.

768

Chapter 22

Gauss’s Law

(b) Any charge on the conducting material must reside on its boundaries. If the net charge of the cube is −6.40C, and there is a charge of −8.00C on its inner surface, there must be a charge of +1.60C on the outer surface. 24. Since the charges are of opposite sign, and since the charges are free to move since they are on conductors, the charges will attract each other and move to the inside or facing edges of the plates. There will be no charge on the outside edges of the plates. And there cannot be charge in the plates themselves, since they are conductors. All of the charge must reside on surfaces. Due to the symmetry of the problem, all field lines must be perpendicular to the plates, as discussed in Example 22–7. (a) To find the field between the plates, we choose a gaussian cylinder, perpendicular to the plates, with area A for the ends of the cylinder. We place one end inside the left plate (where the field must be zero), and the other end between the plates. No flux passes through the curved surface of the cylinder. Q  E dA =  E dA +  E dA =  E dA = encl → 0 ends side right end

Ebetween A =

A → 0

Ebetween =

 0

The field lines between the plates leave the inside surface of the left plate, and terminate on the inside surface of the right plate. A similar derivation could have been done with the right end of the cylinder inside of the right plate, and the left end of the cylinder in the space between the plates. + − (b) If we now put the cylinder from above so that the right end is + – inside the conducting material, and the left end is to the left of the left plate, the only possible location for flux is through the Eoutside left end of the cylinder. Note that there is NO charge enclosed + – by the gaussian cylinder. Q + –  E dA =  E dA +  E dA =  E dA = encl0 → ends side left end

Eoutside A =

→

Eoutside = 0

–

0 (c) If the two plates were nonconductors, the results would not change. The charge would be distributed throughout the two plates in a different fashion, and the field inside of the plates would not be zero, but the charge in the empty regions of space would be the same as when the plates are conductors. Using Gauss’ law would not be as simple, because the field inside of the plates is not zero. There would be charge enclosed in the gaussian cylinders. 25. Example 22–7 gives the electric field from a positively charged plate as E =  / 2 0 with the field pointing away from the plate. The fields from the two plates will add, as shown in the figure. (a) Between the plates the fields are equal in magnitude, but point in opposite directions. Ebetween =

  − = 0 2 0 2 0

769

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) Outside the two plates the fields are equal in magnitude and point in the same direction.

Eoutside =

   + = 2 0 2 0 0

(c) When the plates are conducting the charge lies on the surface of the plates. For nonconducting plates the same charge will be spread throughout each plate. This will not affect the electric field between or outside the two plates. It will, however, allow for a non-zero field inside each plate. 26. Because 3.0 cm << 0.65 m, we can consider the plates to be infinite in size, and ignore any edge effects. We use the result from Problem 24(a).  Q A 2 E= = → Q = EA 0 = (130 N C )( 0.65 m ) ( 8.85  10 −12 C 2 N  m 2 ) = 4.9  10 −10 C 0 0 Note that the separation distance does not enter into the calculation. 27. (a) In the region 0  r  r1 , a gaussian surface would enclose no charge. Thus, due to the spherical symmetry, we have the following. Q 2  E dA = E 4 r = encl = 0 → E = 0 0

(

)

(b) In the region r1  r  r2 , only the charge on the inner shell will be enclosed.

(

) Q

2  E dA = E 4 r =

encl

 1 4 r12 0

→ E=

 1r12  0r 2

(

) Q

2  E dA = E 4 r =

encl

 1 4 r12 +  2 4 r22 0

→ E=

 1r12 +  2 r22  0r 2

(d) To make E = 0 for r2  r , we must have  1r12 +  2 r22 = 0 . This implies that the shells are of opposite charge. (e) To make E = 0 for r1  r  r2 , we must have  1 = 0 . Or, if a charge Q = −4 1r12 were placed at the center of the shells, that would also make E = 0. 28. If the radius is to increase from r0 to 2r0 linearly during an elapsed time of T, then the rate of r t   increase must be r0 T . The radius as a function of time is then r = r0 + 0 t = r0  1 +  . Since the T  T balloon is spherical, the field outside the balloon will have the same form as the field due to a point charge. (a) Here is the field just outside the balloon surface.

1 Q 4 0 r

4 0

 t r 1 +   T

2 0

770

Chapter 22

Gauss’s Law

(b) Since the balloon radius is always smaller than 3.2 r0 , the total charge enclosed in a gaussian surface at r = 3.2 r0 does not change in time.

1 Q 4 0 r

4 0 ( 3.2r0 )

29. Due to the spherical symmetry of the problem, Gauss’s law using a sphere of radius r leads to the following. Q Q 2  E dA = E 4 r = encl → E = 4enclr 2 0 0

(

)

(a) For the region 0  r  r1 , the enclosed charge is 0. E=

Qencl

= 0

4 0 r 2

(b) For the region r1  r  r0 , the enclosed charge is the product of the volume charge density times the volume of charged material enclosed. The charge density is given by Q 3Q =4 3 4 3 = .  r0 − 3  r1 4 r03 − r13 3

(

)

Vencl = = E= 2 4 0 r 4 0 r 2 Qencl

   r −  r  3

4 3

3 1

4 0 r 2

 3  r − 3  r1  4 r03 − r13 

(

)

4 0 r 2

(r − r ) = 4 r ( r − r ) Q

3 1

3 0

3 1

Q 4 0 r 2

30. By the superposition principle for electric fields (Section 21–6), we find the field for this problem by adding the field due to the point charge at the center to the field found in Problem 29. At any −q . location r  0, the field due to the point charge is E = 4 0 r 2 (a) E = E− q + EQ =

(b) E = E− q + EQ =

(c)

E = E− q + EQ =

−q 4 0 r −q 4 0 r

+ 2

4 0 r 2

( r − r ) = 1 Q ( r − r ) − q   4 r ( r − r ) 4 r  ( r − r )  Q

−q 4 0 r

−q

+0=

Q 4 0 r

3 1

3 0

3 1

3 0

3 1

Q−q 4 0 r 2

771

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

31. (a) Create a gaussian surface that just encloses the inner surface of the spherical shell. Since the electric field inside a conductor must be zero, Gauss’s law requires that the enclosed charge be zero. The enclosed charge is the sum of the charge at the center and charge on the inner surface of the conductor. Qenc = q + Qinner = 0 Therefore Qinner = − q . (b) The total charge on the conductor is the sum of the charges on the inner and outer surfaces. Q = Qouter + Qinner → Qouter = Q − Qinner = Q + q (c) A gaussian surface of radius r  r1 only encloses the center charge, q. The electric field will therefore be the field of the single charge.

E ( r  r1 ) =

q 4 0 r 2

(d) A gaussian surface of radius r1  r  r0 is inside the conductor so E = 0 . (e) A gaussian surface of radius r  r0 encloses the total charge q + Q . The electric field will then be the field from the sum of the two charges.

E ( r  r0 ) =

q+Q 4 0 r 2

32. (a) For points inside the shell, the field will be due to the point charge only.

E ( r  r0 ) =

q 4 0 r 2

(b) For points outside the shell, the field will be that of a point charge, equal to the total charge.

E ( r  r0 ) =

q+Q 4 0 r 2

Q 4 0 r

(d) If q = −Q, we have E ( r  r0 ) =

−Q 4 0 r 2

and E ( r  r0 ) =

2Q 4 0 r 2

and E ( r  r0 ) = 0 .

33. We follow the development of Example 22–6. Because of the symmetry, we expect the field to be directed radially outward (no fringing effects near the ends of the cylinder) and to depend only on the perpendicular distance, R, from the symmetry axis of the shell. Because of the cylindrical symmetry, the field will be the same at all points on a gaussian surface that is a cylinder whose axis coincides with the axis of the shell. The gaussian surface is of radius R and length l. E is perpendicular to this surface at all points. In order to apply Gauss’s law, we need a closed surface, so we include the flat ends of the cylinder. Since E is parallel to the flat ends, there is no flux through the ends. There is only flux through the curved wall of the gaussian cylinder.

772

Chapter 22

Gauss’s Law

Qencl

 E dA = E ( 2 Rl ) = 

 Aencl 0

→ E=

 Aencl 2 0 R l

(a) For R  R0 , the enclosed surface area of the charged shell is Aencl = 2 R0 l , and so the enclosed charge is Qencl =  Aencl =  2 R0 l .

 Aencl  2 R0 l  R0 = = , radially outward . 2 0 Rl 2 0 Rl 0R

(b) For R  R0 , none of the charged shell is enclosed by the gaussian surface, and so Qencl = 0 and thus E = 0 . (c) The field for R  R0 due to the shell is the same as the field due to the long line of charge, if we substitute  = 2 R0 . 34. The geometry of this problem is similar to Problem 33, and so we use the same development, following Example 22–6. See the solution of Problem 33 for details. V V Q  E dA = E ( 2 Rl ) = encl = E encl → E = 2E enclRl 0 0 0 (a) For R  R0 , the enclosed volume of the shell is

Vencl =  R02 l .

 EVencl  E R02 l  E R02 = = E= , radially outward . 2 0 Rl 2 0 Rl 2 0 R (b) For R  R0 , the enclosed volume of the shell is Vencl =  R 2 l .

 EVencl  E R 2 l  R = = E , radially outward . 2 0 Rl 2 0 Rl 2 0

35. The geometry of this problem is similar to Problem 33, and so we use the same development, following Example 22–6. See the solution of Problem 33 for details. We choose the gaussian cylinder to be the same length as the cylindrical shells. Q Q  E dA = E ( 2 Rl ) = encl → E = 2enclRl 0 0 (a) For 0  R  R1 , no charge is enclosed, and so E =

Qencl 2 0 R l

(b) For R1  R  R2 , charge +Q is enclosed, and so E =

= 0.

Q 2 0 Rl

, radially outward .

Qencl

= 0. 2 0 R l (d) The force on an electron between the cylinders points in the direction opposite to the electric field, and so the force is inward. The electric force produces the centripetal acceleration for the electron to move in the circular orbit. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

773

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Fcentrip = eE =

eQ 2 0 Rl

→ K = 12 mv 2 =

Instructor Solutions Manual

eQ 4 0 l

Note that this is independent of the actual value of the radius, as long as R1  R  R2 . 36. The geometry of this problem is similar to Problem 33, and so we use the same development, following Example 22–6. See the solution of Problem 33 for details. We choose the gaussian cylinder to be the same length as the cylindrical shells. Q Q  E dA = E ( 2 Rl ) = encl → E = 2enclRl 0 0 (a) At a distance of R = 3.0cm, no charge is enclosed, and so E =

Qencl

= 0. 2 0 R l (b) At a distance of R = 7.0cm, the charge on the inner cylinder is enclosed.

Qencl

2 Qencl

( −0.88  10 C ) = −45 N C −9

(

)

(

(1.56 − 0.88)  10−9 C ) ( 0.120 m )(5.0 m ) = 2.0  101 N C

= 2 8.988  109 N  m 2 C 2

2 0 Rl 4 0 Rl ( 0.070 m )( 5.0 m ) The negative sign indicates that the field points radially inward. (c) At a distance of R = 12.0cm, the charge on both cylinders is enclosed.

Qencl

2 Qencl

= 2 8.988  109 N  m 2 C2

2 0 Rl 4 0 Rl The field points radially outward.

37. (a) The final speed can be calculated from the work–energy theorem, where the work is the integral of the force on the electron between the two shells. W =  F dr = 12 mv 2 − 12 mv02 Setting the force equal to the electric field times the charge on the electron, and inserting the electric field from Problem 36 gives the work done on the electron. R2

2 0 l R

W =

dR =

qQ 2 0 l

 R2    R1 

ln 

( −1.60  10 C )( −0.88  10 C ) ln  9.0 cm  = 3.51  10 =   2 ( 8.85  10 C /Nm ) ( 5.0 m )  4.5cm  −19

−6

−12

−16

Solve for the velocity from the work–energy theorem, assuming the initial velocity is 0.

(

2 3.51  10−16 J −31

) = 2.8 10 m s 7

m 9.1 10 kg (b) The electric force on the proton provides its centripetal acceleration. qQ mv 2 Fc = = qE = 2 0 l R R The velocity can be solved for from the centripetal acceleration.

(1.60 10 C )( 0.88 10 C ) = 5.5  10 m s 2 ( 8.85  10 C /Nm )(1.67  10 kg ) ( 5.0 m ) −19

−6

−12

−27

Note that as long as the proton is between the two cylinders, the velocity is independent of the radius. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

774

Chapter 22

Gauss’s Law

38. Due to the spherical symmetry of the geometry, we have the following to find the electric field at any radius r. The field will point either radially out or radially in. Q Q 2  E dA = E 4 r = encl → E = 4enclr 2 0 0

(

)

(a) For 0  r  r0 , the enclosed charge is due to the part of the charged sphere that has a radius smaller than r.  E 43  r 3  r Qencl E= = = E 2 2 4 0 r 4 0 r 3 0

(

)

(b) For r0  r  r1 , the enclosed charge is due to the entire charged sphere of radius r0 .

Qencl 4 0 r 2

 E ( 43  r03 ) 4 0 r 2

 E r03 = 3 0 r 2

(c) For r1  r  r2 , r is in the interior of the conducting spherical shell, and so E = 0 . This implies that Qencl = 0, and so there must be an induced charge of magnitude − 43  E r03 on the inner surface of the conducting shell, at r1. (d) For r  r2 , the enclosed charge is the total charge of both the sphere and the shell.

Qencl 4 0 r

= 2

Q + E

( r ) =  Q +  r  1 4 3

3 0

4 0 r 2

 4  0

3 E 0



3 0  r 2

39. The geometry of this problem is similar to Problem 33, and so we use the same development, following Example 22–6. See the solution of Problem 33 for details. Q Q  E dA = E ( 2 Rl ) = encl → E = 2enclRl 0 0 (a) For 0  R  R1 , the enclosed charge is the volume of charge enclosed times the charge density.

Qencl 2 0 Rl

 E R 2 l  R = E 2 0 Rl 2 0

(b) For R1  R  R2 , the enclosed charge is all of the charge on the inner cylinder.

Qencl 2 0 Rl

 E R12 l  R2 = E 1 2 0 Rl 2 0 R

(c) For R2  R  R3 , the enclosed charge is all of the charge on the inner cylinder, and the part of the charge in the tube that is enclosed by the gaussian cylinder.

Qencl 2 0 Rl

 E R12 l +  E ( R 2 l −  R22 l ) 2 0 Rl

 E ( R12 + R 2 − R22 ) 2 0 R

(d) For R  R3 , the enclosed charge is all of the charge on both the inner cylinder and the tube.

Qencl 2 0 Rl

 E R12 l +  E ( R32 l −  R22 l ) 2 0 Rl

 E ( R12 + R32 − R22 ) 2 0 R

775

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(e) See the graph.

Instructor Solutions Manual

10.0

6.0

E (10 N/C)

8.0

4.0 2.0 0.0 0.0

2.5

5.0

7.5

10.0

12.5

15.0

17.5

20.0

R (cm)

40. The conducting outer tube is uncharged, and the electric field is 0 everywhere within the conducting material. Because there will be no electric field inside the conducting material of the outer cylinder tube, the charge on the inner nonconducting cylinder will induce an oppositely signed, equal magnitude charge on the inner surface of the conducting tube. This charge will NOT be uniformly distributed, because the inner cylinder is not in the center of the tube. Since the conducting tube has no net charge, there will be an induced charge on the OUTER surface of the conducting tube, equal in magnitude to the charge on the inner cylinder, and of the same sign. This charge will be uniformly distributed. Since there is no electric field in the conducting material of the tube, there is no way for the charges in the region interior to the tube to influence the charge distribution on the outer surface. Thus, the field outside the tube is due to a cylindrically symmetric distribution of charge. Application of Gauss’s law as in Example 22–6, for a gaussian cylinder with a radius larger Q than the conducting tube, and a length l leads to E ( 2 R l ) = encl . The enclosed charge is the

0

amount of charge on the inner cylinder.

 E R12 Qencl =  E R l → E = =  0 ( 2 Rl ) 2 0 R 2 1

Qencl

41. We treat the source charge as a disk of positive charge of radius 4 R0 , concentric with a disk of negative charge of radius R0 . In order for the net charge of the inner space to be 0, the charge per unit area of the source disks must both have the same magnitude but opposite sign. The field due to the annulus is then the sum of the fields due to both the positive and negative rings. (a) At a distance of 0.25R0 from the center of the ring, we can approximate both of the disks as infinite planes, each producing a uniform field. Since those two uniform fields will be of the same magnitude and opposite sign, the net field is 0. (b) At a distance of 75R0 from the center of the ring, it appears to be approximately a point charge, and so the field will approximate that of a point charge, E =

4 0 ( 75 R0 )

42. The conducting sphere is uncharged, and the electric field is 0 everywhere within its interior, except for in the cavities. When charge Q1 is placed in the first cavity, a charge −Q1 will be drawn from the conducting material to the inner surface of the cavity, and the electric field will remain 0 in the conductor. That charge −Q1 will NOT be distributed symmetrically on the cavity surface. Since the conductor is neutral, a compensating charge Q1 will appear on the outer surface of the conductor © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

776

Chapter 22

Gauss’s Law

(charge can only be on the surfaces of conductors in electrostatics). Likewise, when charge Q2 is placed in the second cavity, a charge −Q2 will be drawn from the conducting material, and a compensating charge Q2 will appear on the outer surface. Since there is no electric field in the conducting material, there is no way for the charges in the cavities to influence the charge distribution on the outer surface. So the distribution of charge on the outer surface is uniform, just as it would be if there were no inner charges, and instead a charge Q1 + Q2 were simply placed on the conductor. Thus the field outside the conductor is due to a spherically symmetric distribution of Q1 + Q2 . Application of Gauss’s law leads to E =

1 Q1 + Q2 4 0

. If Q1 + Q2  0, the field will point

radially outward. If Q1 + Q2  0, the field will point radially inward. 43. (a) Choose a cylindrical gaussian surface with the flat ends parallel to and equidistant from the slab–one end “above” the slab, and the other end “below” the slab, similar to Fig. 22–16 (and see the figure here). The areas of each end is taken to be A. By symmetry the electric field (shown as red vectors) must point perpendicularly away from the slab, resulting in no flux passing through the curved part of the gaussian cylinder. By symmetry the net flux through each end of the cylinder must be equal to each other, with the electric field constant on those end surfaces:  E  dA = 2 EA. The charge enclosed by the surface is the charge density of the slab multiplied by the volume of the slab enclosed by the surface. qenc =  E ( Ad ) Gauss’s law can then be solved for the electric field.

 E Ad

 E  dA = 2 EA = 

→ E=

Ed 2 0

Note that this electric field is independent of the distance from the slab. (b) When the coordinate system of this problem is changed to axes parallel ( ẑ ) and perpendicular

( r̂ ) to the slab, it can easily be seen that the particle will hit the slab if the initial perpendicular velocity is sufficient for the particle to reach the slab before the acceleration decreases its velocity to zero. In the new coordinate system the axes are rotated by 45 counterclockwise from the original axes shown. y y r0 = y0 cos 45rˆ + y0 sin 45zˆ = 0 rˆ + 0 zˆ 2 2 v v v 0 = −v0 sin 45rˆ + v0 cos 45zˆ = − 0 rˆ + 0 zˆ 2 2 a = qE / mrˆ The perpendicular components are then inserted into Eq. 2–12c, with the final velocity equal to zero, to find the limiting case. v2 q   d  y  0 = vr20 − 2a(r − r0 ) = 0 − 2  E  0 − 0  m  2 0   2 2  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

777

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Solving for the velocity gives the minimum speed that the particle can have to reach the slab.

2 q y0  E d

v0 

 0m

44. Due to the spherical symmetry of the problem, Gauss’s law using a sphere of radius r leads to the following. Q Q 2  E dA = E 4 r = encl → E = 4enclr 2 0 0

(

)

(a) For the region 0  r  r1 , the enclosed charge is 0. E=

Qencl 4 0 r 2

= 0

(b) For the region r1  r  r0 , the enclosed charge is the product of the volume charge density times

r the volume of charged material enclosed. The charge density is given by  =  0 1 . We must r integrate to find the total charge. We follow the procedure given in Example 22–5. We divide the sphere up into concentric thin shells of thickness dr, as shown in Fig. 22–14. We then integrate to find the charge. r r r  Qencl =   E dV =   0 1 4 ( r ) dr = 4 r1 0  rdr = 2 r1 0 r 2 − r12 r r r

(

Qencl 4 0 r 2

)

(

2 r1 0 r 2 − r12 4 0 r 2

) =  r (r − r ) 2

0 1

2 1

2 0 r 2

(c) For the region r  r0 , the enclosed charge is the total charge, found by integration in a similar fashion to part (b). r r r1    Qencl =   E dV =   0 4 ( r ) dr = 4 r1 0  rdr = 2 r1 0 r02 − r12 r r r

Qencl 4 0 r

= 2

(

2 r1 0 r02 − r12 4 0 r 2

0 1

2 r

2 0 0

2 0

2 1

2 0 r 2

1.00 0.75

. E /E 0

 0 r1 ( r02 − r12 )

)

) =  r (r − r )

(d) See the attached graph. We have chosen r1 = 12 r0 . Let

E0 = E ( r = r0 ) =

(

0.50 0.25 0.00 0.0

0.5

1.0

1.5

2.0

r /r 0

778

Chapter 22

Gauss’s Law

45. (a) The force felt by one plate will be the charge on that plate multiplied by the electric field caused by the other plate. The field due to one plate is found in Example 22–7. Let the positive plate be on the left, and the negative plate on the right. We find the force on the negative plate due to the positive plate.  Fon = qon Edue to = ( b Ab ) Ea = ( b Ab ) a 2 0 plate plate plate “ b”

“ b”

“a ”

( −25 10 C m )(1.0 m )( 25 10 C m ) = −35.31N 2 ( 8.85  10 C /Nm ) −6

−6

−12

 35 N, towards the other plate .

(b) Since the field due to either plate is constant, the force on the other plate is constant, and then the work is just the force times the distance. Since the plates are oppositely charged, they will attract, and so a force equal to and opposite the force above will be required to separate them. The applied force will be in the same direction as the displacement of the plates. W = F x = ( 35.31N )( cos 0 ) ( 5.0  10−3 m ) = 0.18 J 46. Because the slab is very large, and we are considering only distances from the slab much less than its height or breadth, the symmetry of the slab results in the field being perpendicular to the slab, with a constant magnitude for a constant distance from the center. We assume that  E  0 and so the electric field points away from the center of the slab. (a) To determine the field inside the slab, choose a cylindrical gaussian surface, of length 2x  d and cross-sectional area A. Place it so that it is centered in the slab. There will be no flux through the curved wall of the cylinder. The electric field is parallel to the surface area vector on both ends, and is the same magnitude on both ends. Apply Gauss’s law to find the electric field at a distance x  12 d from the center of the slab. See the first diagram.  ( 2 xA) Q  E dA =  E dA +  E dA =  E dA + 0 = encl0 → 2EA =  0 → ends side ends

Einside =

x ; x  12 d 0

(b) Use a similar arrangement to determine the field outside the slab. Now let 2 x  d . See the second diagram. Q  E dA =  E dA = encl → 0 ends

2EA =

 ( dA) d ; x  12 d → Eoutside = 2 0 0

Notice that electric field is continuous at the boundary of the slab.

779

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

47. (a) In Problem 46, it is shown that the field outside a flat slab of nonconducting material with a d . If the charge density is positive, the field points uniform charge density is given by E = 2 0 away from the slab, and if the charge density is negative, the field points towards the slab. So for this problem’s configuration, the field outside of both half-slabs is the vector sum of the fields from each half-slab. Since those fields are equal in magnitude and opposite in direction, the field outside the slab is 0. (b) To find the field in the positively charged half-slab, we use a cylindrical gaussian surface of cross-sectional area A. Place it so that its left end is in the positively charged half-slab, a distance x > 0 from the center of the slab. Its right end is external to the slab. Due to the symmetry of the configuration, there will be no flux through the curved wall of the cylinder. The electric field is parallel to the surface area vector on the left end, and is 0 on the right end. We assume that the electric field is pointing to the left. Apply Gauss’s law to find the electric field a distance 0  x  d from the center of the slab. See the diagram. Q  E dA =  E dA +  E dA =  E dA + 0 = encl → 0 ends side left end

EA =

0 ( d − x ) A  (d − x) → E x 0 = 0 0 0

Since the field is pointing to the left, we can express this as E x 0 = −

0 ( d − x ) ˆ i. 0

(c) To find the field in the negatively charged half-slab, we use a cylindrical gaussian surface of cross-sectional area A. Place it so that its right end is in the negatively charged half-slab, a distance x < 0 from the center of the slab. Its left end is external to the slab. Due to the symmetry of the configuration, there will be no flux through the curved wall of the cylinder. The electric field is parallel to the surface area vector on the left end, and is 0 on the right end. We assume that the electric field is pointing to the right. Apply Gauss’s law to find the electric field at a distance − d  x  0 from the center of the slab. See the diagram. Q  E dA =  E dA +  E dA =  E dA + 0 = encl0 → ends side right end

EA =

−0 ( d + x ) A

0

→ E x 0 =

− 0 ( d + x )

0

Since the field is pointing to the left, we can express this as E x0 = −

0 ( d + x ) ˆ i. 0

Notice that the field is continuous at all boundaries. At the left edge ( x = − d ) , E x 0 = Eoutside . At the center ( x = 0 ) , E x 0 = E >0 . And at the right edge ( x = d ) , E x 0 = Eoutside .

780

Chapter 22

Gauss’s Law

48. We follow the development of Example 22–6. Because of the symmetry, we expect the field to be directed radially outward (no fringing effects near the ends of the cylinder) and to depend only on the perpendicular distance, R, from the symmetry axis of the cylinder. Because of the cylindrical symmetry, the field will be the same at all points on a gaussian surface that is a cylinder whose axis coincides with the axis of the cylinder. The gaussian surface is of radius R and length l. E is perpendicular to the curved surface at all points. In order to apply Gauss’s law, we need a closed surface, so we include the flat ends of the cylinder. Since E is parallel to the flat ends, there is no flux through the ends. There is only flux through the curved wall of the gaussian cylinder. Q Q  E dA = E ( 2 Rl ) = encl → E = 2enclRl 0 0 To find the field inside the cylinder, we must find the charge enclosed in the gaussian cylinder. We divide the gaussian cylinder up into coaxial thin cylindrical shells of length l and thickness dR. That shell has volume dV = 2 Rl dR. The total charge in the gaussian cylinder is found by integration. 2

R R 2 0 l  0 l R 4 3 R  R0 : Qencl =   E dV =   0   2 R l dR = R dR = R02 0 2 R02  R0  0 0  0 l R 4 Qencl 2 R02  R3 E= = = 0 2 , radially out . 2 0 R l 2 0 R l 4 0 R0 R

2 0 l

R  R0 : Qencl =   E dV = 0

R02

3  R dR = 0

 0 l R02 2

 0 l R Qencl  0 R02 2 , radially out . E= = = 2 0 R l 2 0 R l 4 0 R 2 0

49. The symmetry of the charge distribution allows the electric field inside the sphere to be calculated using Gauss’s law with a concentric gaussian sphere of radius r  r0 . The enclosed charge will be found by integrating the charge density over the enclosed volume. r  r    r4 4 0 r 3 Qencl =   E dV =   0   4 r 2 dr  = r dr = 0 0 r0 0 r0  r0  The enclosed charge can be written in terms of the total charge by setting r = r0 and solving for the charge density in terms of the total charge.



 0 r0 4

=  0 r0 →  0 = → Qencl ( r ) = r0  r0 3 The electric field is then found from Gauss’s law Q=

Qencl

 E  dA = 

→ E ( 4 r ) = 2

Qr

 0 r 4

   0  r0 

→ E=

r = Q   r0 

Q r2 4 0 r0 4

The electric field points radially outward since the charge distribution is positive. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

781

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

50. (a) The flux through any closed surface containing the total charge must be the same, so the flux through the larger sphere is the same as the flux through the smaller sphere, +235 N m2 /C . (b) Use Gauss’s law to determine the enclosed charge. Q  = encl → Qencl =  0 = 8.85  10 −12 C 2 N m 2 0

(

)( +235 N m /C ) = +2.08  10 C −9

51. The flux through a gaussian surface depends only on the charge enclosed by the surface. For both of these spheres the two point charges are enclosed within the sphere. Therefore the flux is the same for both spheres. 8.20  10−9 C + −5.00  10 −9 C Qencl = = = 362 N m 2 C 2 2 −12 0 8.85  10 C N m

(

) (

)

52. (a) We use Gauss’s law for a spherically symmetric charge distribution, and assume that all the charge is on the surface of the Earth. Note that the field is pointing radially inward, and so the dot product introduces a negative sign.

 E  dA = − E ( 4 r ) = Q

0 →

encl

Qencl = −4 0 ER

2 Earth

(

− (150 N C ) 6.38  106 m

(8.988  10 N  m C ) 9

) = −6.793  10 C  −6.8  10 C 2

(b) Find the surface density of electrons. Let n be the total number of electrons. Q ne →  = =− A A n A

=−

Q eA

=−

(8.85  10 C N m ) (150 N C ) = = ) e (1.60  10 C )

2 −4 0 EREarth

(

2 e 4 REarth

−12

0E

−19

= 8.3  109 electrons m 2

53. The gravitational field a distance r from a point mass M is given by an equation in Section 6–8, GM g = − 2 rˆ , where r̂ is a unit vector pointing radially outward from mass M. Compare this to the r 1 Q rˆ . To change the electric field to the gravitational field, electric field of a point charge, E = 4 0 r 2 we would make these changes: E → g ; Q  0 → −4 GM . Make these substitutions in Gauss’s law. Q  E  dA = encl →  g  dA = −4 GM encl 0 54. (a) Find the value of b by integrating the charge density over the entire sphere. Follow the development given in Example 22–5. r0

(

)

( ) → b = Qr

Q =   E dV =  br 4 r 2 dr = 4 b 14 r04 0

4 0

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(b) To find the electric field inside the sphere, we apply Gauss’s law to an imaginary sphere of radius r, calculating the charge enclosed by that sphere. The spherical symmetry allows us to evaluate the flux integral simply. r Q Q Qr 4 2  =  E  dA = encl ; Q =   E dV =   = → 4 r r dr 0 0  r04 r04

(

1 Qr 2 4 0 r04

)

, r  r0

(c) As discussed in Example 22–4, the field outside a spherically symmetric distribution of charge is the same as that for a point charge of the same magnitude located at the center of the sphere.

4 0 r 2

, r  r0

55. The electric field is strictly in the y-direction. So, referencing the diagram, there is no flux through the top, bottom, front, or back faces of the cube. Only the “left” and “right” faces will have flux through them. And since the flux is only dependent on the y coordinate, the flux through each of those two faces is particularly simple. Calculate the flux and use Gauss’s law to find the enclosed charge.  =  E  dA =  E  d A +  E  d A left face

(

z l

l x

y l

right face

)

( )

=  bˆj  − ˆjdA +  ( a l + b ) ˆj  ˆjdA = − bl 2 + a l 3 + bl 2 left face

right face

= a l 3 = Qencl  0

→ Qencl =  0 a l 3

56. The force on a sheet is the charge on the sheet times the average electric field due to the other sheets: But the fields due to the “other” sheets are uniform, so the field is the same over the entire sheet. The force per unit area is then the charge per unit area, times the field due to the other sheets. Fon = qon Eother = qon Eother → sheet

sheet

sheets

sheet

sheets

  (+) EII

 II ( − ) EI

F = q E   on   on other =  on Eother sheets sheet sheets  A sheet  A sheet

EIII

EII

 III ( + )

The uniform fields from each of the three sheets are indicated on the diagram. Take the positive direction as upwards. We take the direction from the diagram, and so use the absolute value of each charge density. The electric field magnitude due to each sheet is given by E =  2 0 . 7.5  10−9 C m 2  F  =  E − E = I  −  = ( 6.0 − 3.0 )  10−9 C m 2  ( ) ( )   I III II III II −12 2 2  2 A  2 ( 8.85  10 C N m )  I 0 = 1.3  10−6 N m 2 ( up )

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

−3.0  10−9 C m 2  F  =  E − E =  II  −  = ( 6.0 − 7.5)  10−9 C m 2  ( ) ( )   II III I III I −12 2 2  2 0 2 ( 8.85  10 C N m )  A  II = 2.5  10−7 N m 2 ( up )

6.0  10−9 C m 2  F  =  E − E =  III  −  = ( 3.0 − 7.5)  10−9 C m 2  ( ) ( )   III III I II I −12 2 2  2 0 2 ( 8.85  10 C N m )  A  III = −1.5  10−6 N m 2 ( down ) 57. (a) There is no charge enclosed within the sphere, and so no flux lines can originate or terminate inside the sphere. All field lines enter and leave the sphere. The net flux through the sphere is 0. (b) The maximum electric field will be at the point on the sphere closest to Q, which is the top of the sphere. The minimum electric field will be at the point on the sphere farthest from Q, which is the bottom of the sphere. Emax = Emin =

4 0 r

2 closest

4 0 r

2 farthest

= =

4 0 ( r ) 1

(d)

4 0 ( r ) 5 2 0

Thus the range of values is (c)

Q 1 2 0

25 0 r

2 0

= =

1 Q

 0 r02 1

25 0 r02

 Esphere  surface

1 Q

 0 r02

E is not perpendicular at all points. It is only perpendicular at the two points already discussed; the point on the sphere closest to the point charge, and the point on the sphere farthest from the point charge. The electric field is not perpendicular or constant over the surface of the sphere. Therefore Gauss’s law is not useful for obtaining electric field at the surface of the sphere because a gaussian surface cannot be chosen that simplifies the flux integral.

58. By Gauss’s law, the total flux through the cylinder is Q  0 . We find the flux through the ends of the cylinder, and then subtract that from the total flux to find the flux through the curved sides. The electric field is that of a point charge. On the ends of the cylinder, that field will vary in both magnitude and direction. Thus, we must do a detailed integration to find the flux through the ends of the cylinder. Divide the ends into a series of concentric circular rings, of radius R and thickness dR. Each ring will have an area of 2 RdR. The angle between E and dA is  , where tan  = R R0 . See the diagram of the left half of the cylinder. R0

cos  ( 2 R ) dR 4 0 r 2 The flux integral has three variables: r, R, and  . We express r and  in terms of R in order to integrate. The anti-derivative is found in Appendix B–4.  left =  E  dA =  end

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Gauss’s Law

R 2 + R02 ; cos  = R0

 left =  end

=  total =

(

R0 r

R 2 + R02 R0

( 2 R ) dR =

) R +R

4 0 R 2 + R02

2 QR0

2 0

4 0

QR0 

 = −    R 2 + R 2 3 / 2 2 2 2 R R + 0  0 ( ) 0  0  0

RdR

Q  Q 1  1  1− ;  both = 2 left = 1−    2 0  0  2 2  ends end Q

0

=  sides +  both →  sides = ends

0

−  both = ends

0

−

Q 1  1− =  0  2 

Q 2 0

59. (a) The total negative charge, equal to the charge on an electron, is found by integrating the electron’s charge density over all space, in the manner of Example 22–5. Use an integral from Appendix B–5.   2! − e =   E dV =  − Ae −2 r a 4 r 2 dr = −4 A e −2 r a r 2dr = −4 A = − Aa03 → 3 ( 2 a0 ) 0 0

(

= 3

 a0

)(

)

1.6  10−19 C

 ( 0.53  10 m ) −10

= 3.4  1011 C m 3

(b) The net charge inside a sphere of radius a0 will be made of two parts–the positive point charge at the center of the sphere, and some fraction of the total negative charge–since the negative charge is distributed over all space, as described by the charge density. To find the negative charge inside the sphere of radius a0 , we again use an integral from Appendix B–5. We are indicating the elementary charge by ( e ) , so as to not confuse it with the base of the natural logarithms. a0

(

Qneg =  − Ae −2r a 0

=−

) ( 4 r dr ) = −  a( )  e

4 ( e )  

 a03

4 e

3 0

−2 r a0

r 2 dr

 ( 2 a0 ) r + 2 ( 2 a0 ) r + 2   = ( e ) 5e −2 − 1 − 3    ( 2 a0 )  0 e −2 r a  0

(

)

Qnet = Qneg + Qpos = ( e ) 5e −2 − 1 + ( e ) = ( e ) 5e −2 = 1.6  10−19 C 5e −2 = 1.083  10−19 C  1.1  10−19 C (c) The field at a distance r = a0 is that of a point charge of magnitude Qnet at the origin, because of the spherical symmetry and Gauss’s law. 1.083  10−19 C 1 Qencl 1 Qnet 9 2 2 E= = = 8.988  10 N  m C = 3.5  1011 N C 2 2 2 10 − 4 0 r 4 0 a0 0.53  10 m

(

)

( (

) )

 , as discussed in Example 22–7. Because the slab is very 2 0 large, and we assume that we are considering only distances from the slab much less than its height or breadth, the symmetry of the slab results in its field being perpendicular to the slab, with a

60. The field due to the plane is Eplane =

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

constant magnitude for a constant distance from its center. We also assume that  E  0 and so the electric field of the slab points away from the center of the slab. (a) To determine the field to the left of the plane, we choose a cylindrical gaussian surface, of length x  d and cross-sectional area A. Place it so that the plane is centered inside the cylinder. See the diagram. There will be no flux through the curved wall of the cylinder. From the symmetry, the electric field is parallel to the surface area vector on both ends. We already know that the field due to the plane is the same on both ends, and by the symmetry of the problem, the field due to the slab must also be the same on both ends. Thus the total field is the same magnitude on both ends. Q  A +  dA  E dA =  E dA +  E dA =  E dA + 0 = encl → 2Eoutside A =  E → 0 0 ends side ends Eoutside = Eleft

of plane

 +  Ed 2 0

(b) As argued above, the field is symmetric on the outside of the charged matter. Eright

of plane

 +  Ed 2 0

(c) To determine the field inside the slab, we choose a cylindrical gaussian surface of cross-sectional area A with one face to the left of the plane, and the other face inside the slab, a distance x from the plane. Due to symmetry, the field again is parallel to the surface area vector on both ends, has a constant value on each end, and no flux pierces the curved walls. Apply Gauss’s law.

 E dA =  E dA +  E dA +  E dA = E left end

right end

outside

A + Einside A + 0 =

Qencl

side

0

  +  Ed   A +  E xA A + Einside A = →  0  2 0 

Qencl =  A +  E xA →  Einside =

 + E (2x − d ) , 0 xd 2 0

Notice that the field is continuous from “inside” to “outside” at the right edge of the slab, but not at the left edge of the slab. That discontinuity is due to the surface charge density. 61. Consider this sphere as a combination of two spheres. Sphere 1 is a solid sphere of radius r0 and charge density  E centered at A and sphere 2 is a second sphere of radius r0 / 2 and density −  E centered at C. (a) The electric field at A will have zero contribution from sphere 1 due to its symmetry about point A. The electric field is then calculated by creating a gaussian surface centered at point C with radius r0 / 2.

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qenc

 E  dA = 

→ E  4 ( r ) = 1 2 0

( −  E ) 43  ( 12 r0 )3

→

0

E=−

 E r0 6 0

Since the electric field points into the gaussian surface (negative) the electric field at point A points to the right . (b) At point B the electric field will be the sum of the electric fields from each sphere. The electric field from sphere 1 is calculated using a gaussian surface of radius r0 centered at A.

 E1  dA =

qenc

0

→ E1  4 r0 2 = 3

 r03 (  E )  r → E1 = E 0 0 3 0

At point B the field from sphere 1 points toward the left. The electric field from sphere 2 is calculated using a gaussian surface centered at C of radius 3r0 / 2.

qenc

 E  dA =  2

→ E2  4 ( r ) = 3 2 0

( −  E ) 43  ( 12 r0 )3

0

→ E2 = −

 E r0 54 0

At point B, the electric field from sphere 2 points toward the right. The net electric field is the sum of these two fields. The net field points to the left .

E = E1 + E2 =

 E r0 −  E r0 17  E r0 . + = 3 0 54 0 54 0

62. We assume the charge is uniformly distributed, and so the field of the pea is that of a point charge. 1 Q → E (r = R) = 4 0 R 2

(

) (

)

Q = E 4 0 R 2 = 3  106 N C 4 8.85  10−12 C2 N m 2 ( 0.00375 m ) = 5  10 −9 C 2

63. (a) In an electrostatic situation, there is no electric field inside a conductor. Thus E = 0 inside the conductor. (b) The positive sheet produces an electric field, external to itself, directed away from the plate with a magnitude as

 given in Example 22–7, of E1 = 1 . The negative sheet 2 0 produces an electric field, external to itself, directed towards the plate with a magnitude of E2 =

2 . Between the left 2 0

Qnet = 0

1

3

–

and middle sheets, those two fields are parallel and so add to each other. 2 5.00  10−6 C m 2 1 +  2 = = 5.65  105 N C , to the right Eleft = E1 + E2 = −12 2 2 2 0 2 8.85  10 C N m middle

(

)

Emiddle = 5.65  105 N C , to the right right

(d) To find the charge density on the surface of the left side of the middle sheet, choose a gaussian cylinder with ends of area A. Let one end be inside the conducting sheet, where there is no electric field, and the other end be in the area between the left and middle sheets. Apply Gauss’s law in the manner of Example 22–6. Note that there is no flux through the curved sides © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

of the cylinder, and there is no flux through the right end since it is in conducting material. Also note that the field through the left end is in the opposite direction as the area vector of the left end. Q  A  E dA =  E dA +  E dA +  E dA = − Eleftmiddle A + 0 + 0 = encl = left → 0 0 left right side end

end

 left = − 0 Eleft middle

 1 +  2  −6 2  = −5.00  10 C m 2    0

= − 0 

(e) Because the middle conducting sheet has no net charge, the charge density on the right side must be the opposite of the charge density on the left side.

 right = − left = 5.00  10−6 C m2 Alternatively, we could have applied Gauss’s law on the right side in the same manner that we did on the left side. The same answer would result. 64. Because the electric field has only x and y components, there will be no flux through the top or bottom surfaces. For the other faces, we choose a horizontal strip of height dz and width a for a differential element and integrate to find the flux. The total flux is used to determine the enclosed charge. a   z z   front =  E dA =   E0  1 +  ˆi + E0   ˆj ( a dz ˆi )  a a  ( x =a ) 0  a

z2    z = E0 a   1 +  dz = E0a  z + = 23 E0a 2  2a 0 a  0  a

  z z   back =   E0  1 +  ˆi + E0   ˆj −adzˆi = − 23 E0a 2 a  ( x =0 )   a

(

)

a z  ˆ z  z2  1   z ˆ   ˆ  right =   E0  1 +  i + E0   j ( adzj) = E0a    dz = E0a   = 2 E0a 2 a  a a  ( y =a )  2a  0 0  0  a

 left

( y =a )

 0 

 

=   E0  1 +

z ˆ z  2 1  i + E0   ˆj − adzˆj = − 2 E0a a a 

(

)

 total =  front +  back +  right +  left +  top +  bottom = 23 E0 a 2 − 23 E0 a 2 + 12 E0 a 2 − 12 E0 a 2 = 0 + 0 =0=

Qencl

→

Qencl = 0

0 We could have done this without the calculations. Since the field only depends on the z-coordinate, the value of the field on each of the four “side” faces is constant, for a given value of z. Thus, the “inward” flux through the back two faces for the strip shown in the diagram is the same as the “outward” flux through the front two faces, and so the total flux for the strip shown is zero. This is true for all the strips, no matter the value of z. 65. (a) Because the shell is a conductor, there is no electric field in the conducting material, and all charge must reside on its surfaces. All of the field lines that originate from the point charge at the center must terminate on the inner surface of the shell. Therefore the inner surface must have an equal but opposite charge to the point charge at the center. Since the conductor has the same magnitude of charge as the point charge at the center, all of the charge on the conductor is on the inner surface of the shell , in a spherically symmetric distribution. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) By Gauss’s law and the spherical symmetry of the problem, the electric field can be calculated 1 Qencl by E = . 4 0 r 2

r  0.10 m: E =

1 Qencl 4 0 r 2

(8.988  10 N  m C )( 3.0  10 C ) = 2.7  10 N  m C = 9

−6

r  0.15 m: E = 0 And since there is no electric field in the shell, we could express the second answer as r  0.10 m: E = 0 . 66. (a) At a strip such as is marked in the textbook diagram, dA is perpendicular to the surface, and E is inclined at an angle  relative to dA .  /2

(

 hemisphere =  E dA =  E cos  2 R 2 sin  d

)

 /2

= 2 R 2 E  cos  sin  d = 2 R 2 E

( sin  ) 1 2

 /2 0

=  R2E

(b) Choose a closed gaussian surface consisting of the hemisphere and the disk of radius R at the base of the hemisphere. There is no charge inside that closed gaussian surface, and so the total flux through the two surfaces (hemisphere and disk) must be zero. The field lines are all perpendicular to the disk, and all of the same magnitude, and so that flux is simple to calculate.  circle =  E dA =  E ( cos180 ) dA = − EA = − E R 2  total = 0 =  circle +  hemisphere = − E R 2 +  hemisphere →  hemisphere =  R 2 E

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CHAPTER 23: Electric Potential Responses to Questions 1.

If two points are at the same potential, then no NET work is done in moving a test charge from one point to the other. Along some segments of the path, some positive work might be done, but along other segments of the path, negative work would be done. And if the object is moved strictly along an equipotential line, then no work would be done along any segment of the path. Along any segment of the path where positive or negative work is done, a force would have to be exerted. If the object is moved along an equipotential line, then no force would be exerted along that segment of the path. This is analogous to climbing up and then back down a flight of stairs to get from one point to another point on the same floor of a building. Gravitational potential increased while going up the stairs, and decreased while going down the stairs. A force was required both to go up the stairs and down the stairs. If instead you walked on the level from one point to another, then the gravitational potential was constant, and no force was need to change gravitational potential.

A negative charge will move toward a region of higher potential. A positive charge will move toward a region of lower potential. The potential energy of both charges decreases as they move, because they gain kinetic energy.

(a) Electric potential, a scalar, is the electric potential energy per unit charge at a point in space. Electrice potential has no direction. Electric field, a vector, is the electric force per unit charge at a point in space. It has a specific direction. (b) Electric potential energy is the work done against the electric force in moving a charge from a specified location of zero potential energy to some other location. Electric potential is the electric potential energy per unit charge.

The potential energy of the electron is proportional to the voltage used to accelerate it. Thus, if the voltage is multiplied by a factor of 4, then the potential energy is increased by a factor of 4 also. Then, by energy conservation, we assume that all of the potential energy is converted to kinetic energy during the acceleration process. Thus the kinetic energy has increased by a factor of 4 also. Finally, since the speed is proportional to the square root of kinetic energy, the speed must increase by a factor of 2.

The electric field is zero at the midpoint of the line segment joining the two equal positive charges. The electric field due to each charge is of the same magnitude at that location, because the location is equidistant from both charges, but the two fields are in the opposite direction. Thus the net electric field is zero there. The electric potential is never zero along that line, except at infinity. The electric potential due to each charge is positive, so the total potential, which is the algebraic sum of the two potentials, is always positive at any finite distance from the charges.

A negative particle will have its electric potential energy decrease if it moves from a region of low electric potential to one of high potential. By Eq. 23–3, if the charge is negative and the potential difference is positive, the change in potential energy will be negative and therefore decrease.

There is no general relationship between the value of V and the value of E . Instead, the magnitude of E is equal to the rate at which V decreases over a short distance. Consider the point midway between two positive charges. E is 0 there, but V is high. Or, consider the point midway between

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Chapter 23

Electric Potential

two negative charges. E is also 0 there, but V is low, because it is negative. Finally, consider the point midway between positive and negative charges of equal magnitude. There E is not 0, because it points towards the negative charge, but V is zero. 8.

Two equipotential lines cannot cross. That would indicate that a region in space had two different values for the potential. For example, if a 40-V line and a 50-V line crossed, then the potential at the point of crossing would be both 40 V and 50 V, which is impossible. As an analogy, imagine contour lines on a topographic map. They also never cross because one point on the surface of the Earth cannot have two different values for elevation above sea level. Likewise, the electric field is perpendicular to the equipotential lines. If two lines crossed, the electric field at that point would point in two different directions simultaneously, which is not possible.

The equipotential lines (in black) are perpendicular to the electric field lines (in red).

10. a) b) c) d) e) f) g) h) i)

An equipotential surface is one in which all points are at the same potential (page 680, Example 23–7; page 692). An equipotential surface must be perpendicular to the electric field at any point (page 692). Equipotential lines and surfaces are always continuous (page 693). Equipotential lines and surfaces never end (page 693). The entire volume (including the surface) of a conductor is an equipotential in electrostatics (page 693). No work is required to move a charged particle from one point on an equipotential surface to another point on that surface (page 692). Equipotential lines and surfaces cannot cross (page 706, Question 8). Electric field lines point from higher equipotential surfaces to lower equipotential surfaces (page 664, bottom of the page). Closer spacing of equipotentials indicates a larger electric field (page 693, Fig. 23–17).

11. (a) V at other points would be lower by 10 V. E would be unaffected, since E is the negative gradient of V, and a change in V by a constant value will not change the value of the gradient. (b) If V represents an absolute potential, then yes, the fact that the Earth carries a net charge would affect the value of V at the surface. If V represents a potential difference, then no, the net charge on the Earth would not affect the choice of V. 12. The Earth’s gravitational equipotential lines are roughly circular, so the orbit of the satellite would have to be roughly circular. If the Earth were exactly spherically symmetric in its mass distribution, then the equipotential lines would be perfect circles. 13. (a) Once the two spheres are placed in contact with each other, they effectively become one larger conductor. They will have the same potential because the potential everywhere on a conducting surface is constant. (b) Because the spheres are identical in size, an amount of charge Q/2 will flow from the initially charged sphere to the initially neutral sphere so that they will have equal charges. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(c) Even if the spheres do not have the same radius, they will still be at the same potential once they are brought into contact because they still create one larger conductor. However, the amount of charge that flows will not be exactly equal to half the total charge. The larger sphere will end up with the larger charge. 14. The potential at point P would be unchanged. Each bit of positive charge will contribute an amount to the potential based on its charge and its distance from point P. Moving charges to different locations on the ring does not change their distance from P, and hence does not change their contributions to the potential at P. The value of the electric field will change. The electric field is the vector sum of all the contributions to the field from the individual charges. When the charge Q is distributed uniformly about the ring, the y and z components of the field contributions cancel, leaving a net field in the x-direction only. When the charge is not distributed uniformly, the off-axis components will not cancel, and the net field will not point in exactly the x-direction. There is no discrepancy here, because electric potential is a scalar and electric field is a vector. 15. The charge density and the electric field strength will be greatest at the pointed ends of the football because the surface there has a smaller radius of curvature than the middle. 16. You cannot calculate electric potential knowing only electric field at a point, and you cannot calculate electric field knowing only electric potential at a point. As an example, consider the uniform field between two charged, conducting plates. If the potential difference between the plates is known, then the distance between the plates must also be known in order to calculate the field. If the field between the plates is known, then the distance to a point of interest between the plates must also be known in order to calculate the potential there. In general, to find V, you must know E over some range of positions, and be able to integrate it. To find E, you must know V over a range of positions and be able to take its derivative. Thus you need E or V in the region around the point, not just at the point, in order to be able to find the other quantity. 17. If the electric field in a region of space is uniform, then you can infer that the electric potential is increasing or decreasing uniformly in that region. For example, if the electric field is 10 V/m in a region of space then you can infer that the potential difference between two points 1 meter apart (measured parallel to the direction of the field) is 10 V. If the electric potential in a region of space is uniform, then you can infer that the electric field there is zero, since the derivative of a constant (like a uniform potential) is 0. 18. The electric potential energy of two unlike charges is negative, if we take the 0 location for potential energy to be when the charges are infinitely far apart. The electric potential energy of two like charges is positive. In the case of unlike charges, work must be done to separate the charges. In the case of like charges, work must be done to move the charges together. 19. If the electric field points due north, the change in the potential will be (a) greatest (positive) in the direction opposite the field, south; (b) least (negative) in the direction of the field, north; and (c) zero in a direction perpendicular to the field, which could be east, west, straight up (vertical), or straight down. 20. Yes. In regions of space where the equipotential lines are closely spaced, the electric field is stronger than in regions of space where the equipotential lines are farther apart. The units of electric field, volts / meter, help to represent this. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 23

Electric Potential

Solutions to MisConceptual Questions 1.

(b) The similarity in the names of these concepts can lead to confusion. The electric potential is determined by the electric field and is independent of the charge placed in the field. Therefore, changing the charge will not affect the electric potential. The electric potential energy is the product of the electric potential and the electric charge. Changing the sign of the charge will change the sign of the electric potential energy.

(a) It is very important to realize that the electric field is a vector, while the electric potential is a scalar. The electric field at the point halfway between the two charges is the sum of the electric fields from each charge. The magnitudes of the fields will be the same, but they will point in opposite directions. Their sum is therefore zero. The electric potential from each charge is positive. Since the potential is a scalar, the net potential at the midpoint is the sum of the two potentials, which will also be positive.

(b) The net electric field at the center is the vector sum of the electric fields due to each charge. The fields will have equal magnitudes at the center, but the fields from the charges at opposite corners point in opposite directions, so the net field will be zero. The electric potential from each charge is a non-zero scalar. At the center the magnitudes of the four potentials are equal and sum to a non-zero value.

(d) A common misconception is that the electric field and electric potential are proportional to each other. However, the electric field is proportional to the change in the electric potential. If the electric potential is constant (no change) the electric field must be zero. The electric potential midway between equal but opposite charges is zero, but the electric field is not zero at that point, so (a) is incorrect. The electric field midway between two equal positive charges is zero, but the electric potential is not zero at that point, so (b) is incorrect. The electric field between two oppositely charged parallel plates is constant, but the electric potential decreases as you move from the positive plate towards the negative plate, so (c) is incorrect.

(a) By moving in the direction opposite that of the field, the potential is increasing, and so the electric potential energy of the charge will be increasing. If no other force is doing work on the charge, then its kinetic energy would decrease as the potential energy increases. This could also be discussed in terms of force. The electric force on the charge would be in the direction of the field, so its acceleration is in the direction opposite its motion. That means it is slowing down, losing kinetic energy.

(b) The electric field points toward lower electric potential, and away from higher electric potential. Thus if you move south in a northward field, the electric potential increases. However, if the electric potential increases from –50 volts to 0 volts, the MAGNITUDE actually decreases. So perhaps answer (d) should also be considered as the correct answer.

(c) It is often thought that the electron, with its greater acceleration and greater final speed, will have the greater final kinetic energy. However, the increase in kinetic energy is equal to the decrease in potential energy, which is the product of the object’s charge and the change in potential. In this situation, the increase in kinetic energy is equal to the decrease in electric potential energy, which is the product of the object’s charge and the change in electric potential through which it passes. The change in potential for each object has the same magnitude but they have opposite signs since they move in opposite directions in the field. The magnitude of the charge of each object is the same, but they have opposite signs, thus they experience the same change in electric potential energy and therefore they have the same final kinetic energy.

793

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Instructor Solutions Manual

(a) The electric potential is always highest near positive charge, and the electric field points from regions of high potential to regions of lower potential. So in this diagram, the potential is highest at point “A.”

(c) The potential due to the large plate depends on the charge of the plate and the distance from the plate, as illustrated in Fig. 23–16. So the potential of both of the small charges is the same. But the potential energy of each test charge–large plate combination is also proportional to the magnitude of the test charge. Thus, the two test charges each have a different potential energy.

10. (b) It is commonly thought that it only takes 2W to bring together the three charges. However, it takes W to bring two charges together. When the third charge is brought in it is repelled by both of the other charges. It therefore takes an additional 2W to bring in the third charge. Adding the initial work to bring the first two charges together gives a total work of 3W. 11. (a) It takes the most work to move a charge in the opposite direction of the local electric field–that increases the potential energy, and so takes positive work. It takes the least work to move a charge in the same direction as the local electric field–that decreases the potential energy, and so takes negative work. And it takes no work to move perpendicularly to the local electric field. Thus, it takes the most work to move the charge to point 1. 12. (a, c) The potential energy of a positive charge will increase when moving in the opposite direction of an electric field, and decrease when moving in the same direction as an electric field. Thus answer (a) is true and answer (b) is false. The force due to the electric field is in the same direction as the displacement of the positive charge, and so the electric field is doing positive work on the charge, and so answer (c) is true. It is analogous to an object falling in a gravitational field. The gravitational potential energy decreases, and gravity does positive work on the falling object. 13. (c) A vector has both direction and magnitude. Electric potential and electric potential energy both have magnitudes, but no direction. Equipotential lines have a value, representing the potential at their location, but also no unique direction. The electric field has a magnitude and a direction at every point in space, and so is a vector.

Solutions to Problems 1.

The work done by the electric field can be found from Eq. 23–2b. W Vba = − ba → Wba = − qVba = − −9.7  10−6 C ( +65 V ) = 6.3  10−4 J q

(

)

The work done by the electric field can be found from Eq. 23–2b. W Vba = − ba → Wba = − qVba = − 1.60  10−19 C  −65V − 125V  = 3.04  10−17 J q

(

)

= − (1 e )( −190 V ) = 1.90  102 eV

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Chapter 23

Electric Potential

Energy is conserved, so the change in potential energy is the opposite of the change in kinetic energy. The change in potential energy is related to the change in potential. U = qV = −K →

V =

−K q

K initial − K final

( 9.1110 kg )( 5.2 10 m s ) = −0.77 V = = 2q 2 ( −1.60  10 C ) −31

mv 2

−19

The final potential should be lower than the initial potential in order to stop the electron. 4.

The kinetic energy gained by the electron is the work done by the electric force. Use Eq. 23–2b to calculate the potential difference. 4.65  10−16 J W Vba = − ba = − = 2906 V  2910 V q −1.60  10−19 C

(

)

The electron moves from low potential to high potential, so plate B is at the higher potential. 5.

(a) Since the field is uniform, the electron will have a constant acceleration. The initial velocity is v0 = 0 and the final velocity is v = 2.0  104 m s. Use Eq. 2–11. We just use the magnitude of the charge, so that we find the magnitude of the displacement. qE Fnet = ma = qE → a = m x =

v 2 − v02 2a

(

) = ( 9.1110 kg )( 2.0 10 m s ) = 1.898 10 m  1.9 10 m 2qE 2 (1.60  10 C )( 6.0  10 N C ) −31

m v 2 − v02

−7

−19

−7

(b) For a uniform field, the magnitude of the potential difference can be found from conservation of energy. We call the final position the 0-volts location. The initial energy is all electrical potential energy, and the final energy is all kinetic energy. We again use just the magnitude of the charge, so that we find the magnitude of the potential difference. qV = 12 mvf2 →

V = 6.

1 2

mvf2 q

( 9.1110 kg )( 2.0 10 m s ) = 1.139 10 V  1.110 V = 2 (1.60  10 C ) −31

−3

−19

(a) Since the field is uniform, the proton will have a constant acceleration. The initial velocity is v0 = 0 and the displacement is x = 0.36 m . Use Eq. 2–11. Fnet = ma = qE → a =

v = v0 + 2ax =

qE m

2qE x m

(

)(

)

2 1.60  10−19 C 4.5  10 4 N C ( 0.36 m )

(1.67 10 kg ) −27

= 1.762  106 m s  1.8  106 m s

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qV = 12 mvf2 → V = 7.

1 2

mvf2 q

(1.67 10 kg )(1.762 10 m s ) = 1.620 10 V  1.6 10 V = 2 (1.60  10 C ) −27

−19

By the work energy theorem, the total work done by the external force and the electric field together, is the change in kinetic energy. The work done by the electric field is given by Eq. 23–2b. Wexternal + Welectric = K final − K initial → Wexternal − q (Vb − Va ) = K final →

(Vb − Va ) =

Wexternal − K final q

15.0  10−4 J − 4.82  10−4 J −7.40  10−6C

= −138 V

Since the potential difference is negative, we see that Va  Vb . 8.

The magnitude of the voltage can be found from Eq. 23–4b, using the magnitude of the electric field. V E = ba → Vba = Ed = ( 465V m ) 11.0  10−3 m = 5.12 V d

(

)

The distance between the plates is found from Eq. 23–4b, using the magnitude of the electric field. 45 V V V E = ba → d = ba = = 2.8  10 −2 m d E 1600 V m

10. We assume that the electric field is uniform, and so use Eq. 23–4b, using the magnitude of the electric field. V 85V E = ba = = 2.1  104 V m −3 d 4.0  10 m 11. The maximum charge will produce an electric field that causes breakdown in the air. We use the same approach as in Examples 23–4 and 23–5. 1 Q Vsurface = r0 Ebreakdown and Vsurface = → 4 0 r0



 ( 0.058 m )2 ( 3  106 V m ) = 1.1  10−6 C   8.99  10 N m C 

Q = 4 0 r02 Ebreakdown = 

12. To find the limiting value, we assume that the electric field at the radius of the sphere is the minimum value that will produce breakdown in air. We use the same approach as in Examples 23–4 and 23–5. V 45, 000 V = 0.015 m Vsurface = r0 Ebreakdown → r0 = surface = Ebreakdown 3  106 V m Vsurface =

1 Q 4 0 r0



  ( 45, 000 V )( 0.015 m )  8.99  10 N m C  1

→ Q = 4 0Vsurface r0 = 

= 7.5  10−8 C

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Chapter 23

Electric Potential

13. Find the distance corresponding to the maximum electric field, using Eq. 23–4b for the magnitude of the electric field. 65 V V V E = ba → d = ba = = 2.2  10 −5 m  2  10 −5 m 6 d E 3  10 V m 14. From Example 22–7, the electric field produced by a large plate is uniform with magnitude E =

 . 2 0

The field points away from the plate, assuming that the charge is positive. Apply Eq. 23–4a. x

( )

  ˆ x x i  ( dxˆi ) = − → V ( x ) = V0 − 2 0  2 0 2 0 0  x

V ( x ) − V ( 0 ) = V ( x ) − V0 = −  E d l = −   0

15. (a) The electric field at the surface of the Earth is the same as that of a point charge, E =

Q 4 0 r02

The electric potential at the surface, relative to V () = 0 is given by Eq. 23–5. Writing this in terms of the electric field and radius of the Earth gives the electric potential. Q V = = Er0 = ( −150 V m ) ( 6.38  106 m ) = −0.96 GV 4 0 r0 (b) Part (a) demonstrated that the potential at the surface of the Earth is 0.96 GV lower than the potential at infinity. Therefore if the potential at the surface of the Earth is taken to be zero, the potential at infinity must be V () = 0.96 GV . If the charge of the ionosphere is included in the calculation, the electric field outside the ionosphere is basically zero. The electric field between the Earth and the ionosphere would remain the same. The electric potential, which would be the integral of the electric field from infinity to the surface of the Earth, would reduce to the integral of the electric field from the ionosphere to the Earth. This would result in a negative potential, but of a smaller magnitude. 16. (a) The potential at the surface of a charged sphere is derived in Example 23–4. Q V0 = → Q = 4 0 rV → 0 0 4 0 r0

=

Q Area

Q 4 r02

4 0 rV 0 0 4 r02

V0 0 r0

( 480 V ) (8.85  10−12 C2 /Nm 2 ) = ( 0.16 m )

= 2.655  10−8 C m 2  2.7  10−8 C m 2 (b) The potential away from the surface of a charged sphere is also derived in Example 23–4. 4 0 rV Q rV rV ( 0.16 m )( 480 V ) 0 0 = = 0 0 → r= 0 0 = = 3.072 m  3.1m V = 4 0 r 4 0 r r V ( 25V ) This would be 3.072 m – 0.16 m = 2.9 m from the surface of the sphere. 17. (a) After the connection, the two spheres are at the same potential . If they were at different potentials, then there would be a flow of charge in the wire until the potentials were equalized. (b) We assume the spheres are so far apart that the charge on one sphere does not influence the distribution of charge on the other sphere. The charge splits between the spheres so that their potentials (due to the charge on them only) are equal. The initial charge on sphere 1 is Q, and the final charge on sphere 1 is Q1. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

V1 =

Q1 4 0 r1

; V2 =

Q − Q1 4 0 r2

; V1 = V2 →

Charge transferred = Q − Q1 = Q − Q

Q1 4 0 r1

( r1 + r2 )

= Q

Q − Q1 4 0 r2

Instructor Solutions Manual

→ Q1 = Q

( r1 + r2 )

18. From Example 22–6, the electric field due to a long wire is radial relative to the wire, and is of 1  . If the charge density is positive, the field lines point radially away from the magnitude E = 2 0 R wire. Use Eq. 23–4a to find the potential difference, integrating along a line that is radially outward from the wire. Rb

( )

Vb − Va = −  E d l = −  Ra

1  2 0 R

dR = −

R   ln ( Rb − Ra ) = ln a 2 0 2 0 R b

19. (a) The width of the end of a finger is about 1 cm, so consider the fingertip to be a part of a sphere of diameter 1 cm. We assume that the electric field at the radius of the sphere is the minimum value that will produce breakdown in air. We use the same approach as in Examples 23–4 and 23–5. Vsurface = r0 Ebreakdown = ( 0.005m ) ( 3  106 V m ) = 15, 000 V Since this is just an estimate, we might expect anywhere from 10,000 V to 20,000 V. 1 Q 1 4 r02 (b) Vsurface = = → 4 0 r0 4 0 r0

 = Vsurface

(8.85  10−12 C2 /N m2 ) = 2.7  10−5 C m2 0 = (15, 000 V ) r0 0.005 m

Since this is an estimate, we might say the charge density is on the order of 30  C m 2 . 20. We assume the field is uniform, and so Eq. 23–4b applies. V 0.10 V E= = = 1  107 V m −9 d 10  10 m 21. Since the field is uniform, we may apply Eq. 23–4b. Note that the electric field always points from high potential to low potential. (a) VBA = 0 . The distance between the two points is exactly perpendicular to the field lines. (b) VCB = VC − VB = ( −6.30 N C )( 7.00 m ) = −44.1V (c) VCA = VC − VA = VC − VB + VB − VA = VCB + VBA = −44.1V + 0 = −44.1V 22. Because of the spherical symmetry of the problem, the electric field in each region is the same as that of a point charge equal to the net enclosed charge. (a) For r  r2 : E =

1 Qencl 4 0 r

3 2

4 0 r

3 Q 8 0 r 2

For r1  r  r2 : E = 0 , because the electric field is 0 inside of conducting material. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

798

Chapter 23

Electric Potential

For 0  r  r1 : E =

1 Qencl 4 0 r

1 2

4 0 r

1 Q

8 0 r 2

(b) For r  r2 , the potential is that of a point charge at the center of the sphere.

1 Q 4 0 r

3 2

4 0 r

3 Q 8 0 r

, r  r2

(c) For r1  r  r2 , the potential is constant and equal to its value on the outer shell, because there is no electric field inside the conducting material.

V = V ( r = r2 ) =

3 Q 8 0 r2

, r1  r  r2

(d) For 0  r  r1 , we use Eq. 23–4a. The field is radial, so we integrate along a radial line so that E d l = E dr. r

Vr − Vr = −  E d l = −  E dr = −  1

Vr = Vr + 1

1 Q 8 0 r

dr =

Q 1

1 −   8 0  r r1 

Q 1

1 Q  1 1 Q  1 1 − = + =    +  , 0  r  r1 8 0  r r1  8 0  2 r1 r  8 0  r2 r 

(e) To plot, we first calculate V0 = V ( r = r2 ) =

3Q 8 0 r2

and E0 = E ( r = r2 ) =

3Q 8 0 r22

. Then we plot

V V0 and E E0 as functions of r r2 .

For 0  r  r1 :

Q  1 1 1 Q +   V 8 0  r2 r  1 E 8 0 r 2 1 r22 1 −1 −2 = = 3 1 + ( r r2 )  ; = = 3 2 = 3 ( r r2 ) Q Q 3 3 V0 E0 r 8 0 r2 8 0 r22

For r1  r  r2 :

3 Q V 8 0 r2 E 0 = =1; = =0 Q Q 3 3 V0 E0 8 0 r2 8 0 r22

For r  r2 :

3 Q 3 Q V 8 0 r r2 E 8 0 r 2 r22 −1 −2 = = = ( r r2 ) ; = = 2 = ( r r2 ) 3Q 3Q V0 r E0 r 2 8 0 r2 8 0 r2

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Instructor Solutions Manual

5.0 4.0

V /V 0

3.0 2.0 1.0 0.0 0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

1.25

1.50

1.75

2.00

r /r 2

5.0 4.0

E /E 0

3.0 2.0 1.0 0.0 0.00

0.25

0.50

0.75

1.00

r /r 2

23. The field is found in Problem 22–33. The field inside the cylinder is 0, and the field outside the  R0 . cylinder is 0R (a) Use Eq. 23–4a to find the potential. Integrate along a radial line, so that E d l = EdR. R R R R  R0 R VR − VR = −  E d l = −  E dR = −  dR = − 0 ln → R0 0R 0 R R R 0

VR = V0 −

 R0 R ln , R  R0 R0 0

(b) The electric field inside the cylinder is 0, so the potential inside is constant and equal to the potential on the surface, V0 . (c) No, we are not able to assume that V = 0 at R = . V  0 because there would be charge at infinity for an infinite cylinder. And from the formula derived in (a), if R = , VR = −. 24. Use Eq. 23–5 to find the potential. 1 Q 3.00  10−6 C = 8.99  109 N m 2 C2 = 1.80  105 V V= 4 0 r 0.150 m

(

)

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Chapter 23

Electric Potential

25. Use Eq. 23–5 to find the charge. 1 Q 1   → Q = ( 4 0 ) rV =  V= ( 0.15 m )(165 V ) = 2.8 10−9 C 9 2 2   4 0 r 8.99 10 N m C   26. (a) Because of the inverse square nature of the electric x d field, any location where the field is zero must be q2  0 q1  0 closer to the weaker charge ( q2 ) . Also, in between the two charges, the fields due to the two charges are parallel to each other (both to the left) and cannot cancel. Thus, the only places where the field can be zero are closer to the weaker charge, but not between them. In the diagram, this is the point to the left of q2 . Take rightward as the positive direction. 1 q2 1 q1 2 − = 0 → q2 ( d + x ) = q1 x 2 → E= 2 2 4 0 x 4 0 ( d + x )

2.0  10−6 C

(8.0cm ) = 26 cm left of q2 3.4  10−6 C − 2.0  10−6 C (b) The potential due to the positive charge is positive d x1 x2 everywhere, and the potential due to the negative charge is negative everywhere. Since the negative q2  0 q1  0 charge is smaller in magnitude than the positive charge, any point where the potential is zero must be closer to the negative charge. So consider locations between the charges (position x1 ) and to the left of the negative charge (position x2 ) as shown in the diagram. −2.0  10−6 C ) ( 8.0 cm ) ( q2  q2 d 1  q1 Vlocation 1 = +  = 0 → x1 = = = 2.963cm  4 0  ( d − x1 ) x1  ( q2 − q1 ) ( −5.4  10−6 C ) x=

q1 −

Vlocation 2 =

1 

q1 q  + 2=0 →  4 0  ( d + x2 ) x2 

( −2.0  10 C ) ( 5.0 cm ) = 11.43cm x =− =− (q + q ) (1.4  10 C ) −6

q2 d

−6

So the two locations where the potential is zero are 3.0 cm from the negative charge towards the positive charge, and 11 cm from the negative charge away from the positive charge. 27. (a) The electric potential is given by Eq. 23–5. 1 Q 1.60  10−19 C = 8.99  109 N m 2 C 2 = 5.754  105 V  5.8  105 V V= −15 4 0 r 2.5  10 m (b) The potential energy of a pair of charges is derived in Conceptual Example 23–8.

(

)

(1.60  10 C ) = 9.2  10 J = 0.58 MeV = ( 8.99  10 N m C ) U =k r 2.5  10 m −19

Q1Q2

−14

−15

28. The potential at the corner is the sum of the potentials due to each of the charges, using Eq. 23–6a. V=

( 3Q )

4 0

( −2Q )

4 0

1 Q 1  1 2Q 1 + = 4 0 l  2  4 0 2 l

( 2 + 1)

801

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

29. The work required is the difference in the potential energy of the charges, calculated with the test charge at the two different locations. The potential energy of a pair of charges is given in Eq. 23–10 1 qQ as U = . So to find the work, calculate the difference in potential energy with the test 4 0 r charge at the two locations. Let Q represent the 25C charges, let q represent the 0.15C test charge, let D represent the 6.0 cm distance, and let d represent the 1.0 cm distance. Since the potential energy of the two 25C charges doesn’t change, we don’t include it in the calculation. 1 Qq 1 Qq 1 1 Qq Qq U initial = U final = + + 4 0 D 2 4 0 D 2 4 0  D 2 − d  4 0  D 2 + d 

Wexternal = U final − U initial = force

4 0  D 2 − d 

4 0  D 2 + d 

 1 Qq    4 0 D 2 

− 2

2Qq 

1 1 1  + −   4 0   D − 2d   D + 2d  D 2 

(

= 2 8.99  109 N m 2 C 2

1 1 1  + − )( 25 10 C )( 0.15 10 C )  0.040 m 0.080 m 0.030m  −6

−6





= 0.28 J An external force needs to do positive work to move the charge. 30. (a) The potential due to a point charge is given by Eq. 23–5. 1 q 1 q q 1 1 − = Vba = Vb − Va =  −  4 0 rb 4 0 ra 4 0  rb ra 

(

= 8.99  109 N m 2 C 2

)( −3.8  10 C )  0.361 m − 0.261 m  = 3.6  10 V −6





(b) The magnitude of the electric field due to a point charge is given by Eq. 21–4b. The direction of the Eb Ea electric field due to a negative charge is towards the charge, so the field at point a will point downward, and the field at point b will point to the right. See the vector diagram. 9 2 2 −6 1 q ˆ 8.99  10 N m C 3.8  10 C ˆ Eb = i i = 2.636  105 V m ˆi = 2 2 4 0 rb ( 0.36 m )

(

Ea = −

)(

(

E b − Ea

)

)(

 Eb

)

8.99  109 N m 2 C 2 3.8  10 −6 C qˆ ˆj = −5.054  105 V m ˆj j = − 2 2 4 0 ra ( 0.26 m ) 1

Eb − Ea = 2.636  105 V m ˆi + 5.054  105 V m ˆj E b − Ea =

 = tan

−

( 2.636  10 V m ) + (5.054  10 V m ) = 5.7  10 V m

− Ea Eb

= tan

−1

5.054  105 2.636  105

= 62

802

Chapter 23

Electric Potential

31. By energy conservation, all of the initial potential energy of the charges will change to kinetic energy when the charges are very far away from each other. By momentum conservation, since the initial momentum is zero and the charges have identical masses, the charges will have equal speeds in opposite directions from each other as they move. Thus each charge will have the same kinetic energy. 1 Q2 = 2 12 mv 2 → Einitial = Efinal → U initial = K final → 4 0 r

(

)

(8.99 10 N m C )( 5.5 10 C ) = 1.79 10 m s  1.8 10 m s v= = 4 mr (1.0 10 kg ) ( 0.085 m ) 9

1 Q2

−6

32. We assume that all of the energy the proton gains in being accelerated by the voltage is changed to potential energy just as the proton’s outer edge reaches the outer radius of the silicon nucleus. 1 e (14e ) U initial = U final → eVinitial = → r 4 0

(14 ) (1.60  10−19 C ) = ( 8.99  10 N m C ) = 4.2  106 V Vinitial = 4 0 r (1.2 + 3.6 )  10−15m 1 14e

33. By energy conservation, all of the initial potential energy will change to kinetic energy of the electron when the electron is far away. The other charge is fixed, and so has no kinetic energy. When the electron is far away, there is no potential energy. ( −e )( Q ) 1 2 Einitial = Efinal → U initial = K final → = 2 mv → 4 0 r

2 ( −e )( Q )

( 4 0 ) mr

(

)( −1.60 10 C )( −1.25 10 C ) ( 9.1110 kg ) ( 0.345 m ) −19

2 8.99  109 N m 2 C 2

−10

−31

= 1.07  106 m s 34. Use Eq. 23–2b and Eq. 23–5.  1 1 ( −q )   1 q 1 ( −q )  q + − + VBA = VB − VA =     4 0 d − b 4 0 b   4 0 b 4 0 d − b 

1 1 1 1  1  1 − 1  = 2 q ( 2b − d ) − − + q  =2  4 0  d − b b b d − b  4 0  d − b b  4 0b ( d − b ) q 

35. We follow the development of Example 23–9, with Fig. 23–15. The charge on a thin ring of radius R and thickness dR is dq =  dA =  ( 2 RdR ) . Use Eq. 23–6b to find the potential of a continuous charge distribution.

803

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

V =

4  r

 2 0

4 

 ( 2 RdR ) x +R 2

0 R1

 = 2 0 R

R x +R 2

Instructor Solutions Manual

dR =

1/ 2 R  x2 + R2 ) ( R 2 0

( x +R − x +R ) 2

2 2

2 1

Note that this could also be done from Example 23–10 by superimposing a disk of positive charge of radius R2 and a disk of negative charge of radius R1, both having the same charge density. The potentials from those two disks can be added together, and we arrive at the same result. 36. All of the charge is the same distance from the center of the semicircle–the radius of the semicircle. Use Eq 23–6b to calculate the potential. l =  r0 → r0 =



; V=

= dq = 4  r 4 r  0

0 0

Q 4 0



Q 4 0 l

37. The electric potential energy is the product of the point charge and the electric potential at the location of the charge. Since all points on the ring are equidistant from any point on the axis, the electric potential integral is simple. Let q1 represent the charge on the small sphere, and q2 represent the charge of the ring. dq q1 q1q2 U = q1Vdue to q2 = q1  dq = = 2 2 2 2  4 0 r + x 4 0 r + x 4 0 r 2 + x 2 Energy conservation is used to obtain a relationship between the potential and kinetic energies at the center of the loop and at a point 2.0 m along the axis from the center. K0 + U 0 = K + U

q1q2

= 12 mv 2 +

4 0 r 4 0 r 2 + x 2 This is equation is solved to obtain the velocity at x = 2.0 m.

q1q2  1

 1  − 2 2 2 0 m  r r +x 

 1  ( 3.6 10 C )(15.0 10 C ) 1   = − 2 ( 8.85  10 C / Nm )( 7.5  10 kg )  0.12 m ( 0.12 m ) + ( 2.0 m )   −6

−12

−6

−3

= 31.85 m s  32 m/s

38. Use Eq. 23–6b to find the potential of a continuous charge distribution. Choose a differential element of length dx at position Q dx, and the x along the rod. The charge on the element is dq = 2l element is a distance r = x 2 + y 2 from a point on the y-axis. Use an indefinite integral from Appendix B–4.

804

Chapter 23

Electric Potential

Vy axis =

= 4  r 4  0 −l

Q

4 0 2 l 

(

Q dx  2l x 2 + y 2

)

  l 2 + y 2 + l  ln =  − l 8 0 l   l 2 + y 2 − l     

x2 + y 2 + x  

39. Use Eq. 23–6b to find the potential of a continuous charge distribution. Choose a differential element of length dx at position x along the rod. The charge on the element is Q dq = dx, and the element is a distance x − x from a 2l point outside the rod on the x-axis. Q dx  l dq Q   x + l  1 1 1 Q l 2 l V = ln  = = − ln ( x − x ) − l =   ,xl   4 0 r 4 0 − l x − x  4 0 2 l 8 0 l   x − l   40. For both parts of the problem, use Eq. 23–6b to find the potential of a continuous charge distribution. Choose a differential element of length dx at position x along the rod. The charge on the element is dq =  dx = axdx. (a) The element is a distance r = x 2 + y 2 from a point on the y-axis. l 1 1 dq ax dx  V = = =0   4 0 r 4 0 − l x2 + y 2 The integral is equal to 0 because the region of integration is “even” with respect to the origin, while the integrand is “odd.” Alternatively, the antiderivative can be found, and the integral can be shown to be 0. This is to be expected since the potential from points symmetric about the origin would cancel on the y-axis. (b) The element is a distance x − x from a point outside the rod on the x-axis. l l 1 dq 1 a xdx  axdx V = = =    4 0 r 4 0 − l x − x 4 0 − l x − x A substitution of z = x − x can be used to do the integration.

V = =

xdx a ( x − z )( −dz ) a  x  = =  − 1 dz   4 0 − l x − x 4 0 x + l 4 0 x− l  z z  a a 4 0

x−l

x+l

( x ln z − z ) xx +− ll =

 x + l  − 2l  , x  l x ln     4 0  x−l  a 

805

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

41. We follow the development of Example 23–9, with Fig. 23–15. The charge on a thin ring of radius R and thickness dR will now be dq =  dA = aR 2 ( 2 RdR ) . Use Eq. 23–6b to find the potential of

(

)

a continuous charge distribution. R R aR 2 ( 2 RdR ) 1 dq 1 a V = = = 4 0  r 4 0 0 2 0 0 x2 + R2

(

)

R 3dR x2 + R2

A substitution of x 2 + R 2 = u 2 can be used to do the integration.

x 2 + R 2 = u 2 → R 2 = u 2 − x 2 ; 2 RdR = 2udu R0

R 3dR

V =

2 0 0

a 1

x2 + R2

(x + R ) 2  2

R = R0

2 0 R=0

3/ 2

( u − x ) udu = a  u − ux  2

2 0 

(

− x2 x2 + R2

)  1/ 2

1 3

R = R0

 R =0

R = R0 R =0

 ( x + R ) − x ( x + R )  + x  2  a  1

2 0

3/ 2

2 1/ 2 0

2 3

a 

( 0 6 0 

R2 − 2 x2

)( x + R ) 2

2 1/ 2 0

+ 2 x3  , x  0



42.

43. The electric field from a large metal plate is uniform with magnitude E =   0 , with the field pointing away from the plate on both sides. The charge density is assumed to be the given value on EACH of the two faces of the plate. Eq. 23–4(a) can be integrated between two arbitrary points to calculate the potential difference between those points. x1

  ( x0 − x1 ) dx = 0 0 x

V = −  0

Setting the change in voltage equal to 100 V and solving for x0 − x1 gives the distance between field lines. 8.85  10−12 C2 /Nm 2 (100 V )  0 V = = 1.18  10−3 m  1mm x0 − x1 = 2 −6 0.75  10 C/m 

(

)

44. The potential at the surface of the sphere is V0 =

1 Q 4 0 r0

. The potential outside the sphere is

1 Q

r = V0 0 , and decreases as you move away from the surface. The difference in potential 4 0 r r between a given location and the surface is to be a multiple of 100 V. V =

806

Chapter 23

Electric Potential

1 Q

V0 =

 10 C  = 12,843V )  0.500.35 m  −6

(

= 8.99  109 N m 2 C2 

4 0 r0





r V0 V0 − V = V0 − V0 0 = (100 V ) n → r = r r V0 − (100 V ) n  0 (a) r1 =

r0 =

V − (100 V ) 1 0

12,843V 12, 743V

( 0.35m ) = 0.3527 m  0.35m

Note that to within the appropriate number of significant figures, this location is at the surface of the sphere. That can be interpreted that we don’t know the voltage well enough to be working with a 100-V difference. V0 12,843V (b) r10 = r0 = ( 0.35 m ) = 0.3796 m  0.38 m V0 − (100 V ) 10 11,843V (c)

r100 =

V − (100 V ) 100 0

r0 =

12,843V 2,843V

( 0.35 m ) = 1.581m  1.6 m

45. The potential due to the dipole is given by Eq. 23–7. 8.99  109 N m 2 C2 4.8  10 −30 C m cos 0 1 p cos  = (a) V = 2 4 0 r 2 4.1  10−9 m

(

= 2.6  10−3 V

(b) V =

p cos 

4 0

= 1.8  10−3 V

p cos 

4 0

)(

(

)

−Q

Q r

(8.99  10 N m C )( 4.8  10 C m ) cos 45 ( 4.1  10 m ) 9

−30

−9

(8.99  10 N m C )( 4.8  10 C m ) cos135 = (1.1  10 m ) 9

−30

−9



= −1.8  10−3 V

−Q

 −Q

46. (a) We assume that p1 and p 2 are equal in magnitude, and that each makes a 52 angle with p. The magnitude of p1 is also given by p1 = qd , where q is the net charge on the hydrogen atom, and d is the distance between the H and the O. p = qd → p = 2 p1 cos 52 → p1 = 2 cos 52 q=

p 2d cos 52

(

6.1  10−30 C m −10

)

2 0.96  10 m cos 52

= 5.2  10−20 C

This is about 0.32 times the charge on an electron. (b) Since we are considering the potential far from the dipoles, we will take the potential of each dipole to be given by Eq. 23–7. See the diagram for the angles p involved. From part (a), p1 = p2 = . 2 cos 52 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

807

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

V = Vp + Vp 1

= = =

47. E = −

dV dr

=−

Instructor Solutions Manual

p1 cos ( 52 −  )

4 0

4 0 r 2 cos 52 1

4 0 r 2 cos 52

p2 cos ( 52 +  )

4 0

cos ( 52 −  ) + cos ( 52 +  ) ( cos 52 cos  + sin 52 cos  + cos 52 cos  − sin 52 cos  ) ( 2 cos 52 cos  ) =

p cos 

4 0

d  1

q q d 1 q  1 1 q =−  =− − 2  =   dr  4 0 r  4 0 dr  r  4 0  r  4 0 r 2

48. The potential gradient is the negative of the electric field. Outside of a spherically symmetric charge distribution, the field is that of a point charge at the center of the distribution. ( 92 ) 1.60  10−19 C dV 1 q 9 2 2 8.99 10 N m C = −E = − = −  = −2.4  1021 V m 2 2 − 15 dr 4 0 r 7.5  10 m

(

)

(

)

49. The electric field between the plates is obtained from the negative derivative of the potential. dV d E=− = − (8.5 V/m) x + 4.5 V  = −8.5 V/m dx dx The charge density on the conducting plates is then calculated from the electric field between two large plates, E =  /  0 .

 = E 0 = (8.5V/m ) (8.85  10−12 C2 /Nm2 ) = 7.5  10−11 C/m2 The plate at the origin has the charge −7.5  10−11 C/m 2 and the other plate, at a positive x, has charge +7.5  10−11 C/m 2 so that the electric field points in the negative direction, in agreement with the first calculation. 50. We use Eq. 23–9 to find the components of the electric field. V V Ex = − = 0 ; Ez = − =0 x z

 a 2 + y 2 ) b − by ( 2 y ) ( y 2 − a 2 ) b ( Ey = − =−  2 = =− 2 2 y y  ( a + y 2 )  (a2 + y2 ) (a2 + y2 )  

V

( y − a ) b ˆj E= (a + y ) 2

808

Chapter 23

Electric Potential

51. We use Eq. 23–9 to find the components of the electric field. V = ay 2 + bxy − cxyz

Ex = −

V x

= −by + cyz ; E y = −

V y

= −2ay − bx + cxz ; E z = −

V z

= cxy

E = ( −by + cyz ) ˆi + ( −2ay − bx + cxz ) ˆj + ( cxy ) kˆ 52. We use the potential to find the electric field, the electric field to find the force, and the force to find the acceleration. V F qE x q V q V Ex = − ; Fx = qE x ; a x = x = =− =− x m m m x m x a x ( x = 2.5 m ) = −

2.0  10 −6 C

 2 ( 2.0 V m 2 ) ( 2.5 m ) − 3 ( 3.0 V m 3 ) ( 2.5 m ) 2  = 1.85 m s 2  5.0  10 kg  −5

 1.9 m s 2

53. (a) The potential along the y-axis was derived in Problem 38.

Vy axis =

  l 2 + y 2 + l  Q  = ln  ln  2  8 0 l   l + y 2 − l   8 0 l  Q

( l + y + l ) − ln ( l + y − l ) 2

 12 ( l 2 + y 2 )−1/ 2 2 y 12 ( l 2 + y 2 ) −1/ 2 2 y  Q  = =− − Ey = − 2 2 2 2 y 8 0 l  l + y +l l + y − l  4 0 y l 2 + y 2  V

From the symmetry of the problem, this field will point along the y-axis.

4 0 y l 2 + y 2

ˆj

Note that for y l , this reduces to the field of a point charge at the origin. (b) The potential along the x-axis was derived in Problem 39. Q   x + l  Q Vx axis = ln  ln ( x + l ) − ln ( x − l )  =  8 0 l   x − l   8 0 l Ex = −

V

=−

 1 − 1 = 1  Q    8 0 l  x + l x − l  4 0  x 2 − l 2  Q

x From the symmetry of the problem, this field will point along the x-axis. 1  Q ˆ E=  i 4 0  x 2 − l 2 

Note that for x

l , this reduces to the field of a point charge at the origin.

54. Let the side length of the equilateral triangle be l . Imagine bringing the electrons in from infinity one at a time. It takes no work to bring the first electron to its final location, because there are no other charges present. Thus W1 = 0. The work done in bringing in the second electron to its final location is equal to the charge on the electron times the potential (due to the first electron) at the final location of the second electron.

−e

l l

−e

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition



Thus W2 = ( − e )  −

1 e

Instructor Solutions Manual

1 e2

=

. The work done in bringing the third electron to its final 4 0 L location is equal to the charge on the electron times the potential (due to the first two electrons).  1 e 1 e 1 2e 2 − = Thus W3 = ( −e )  − . The total work done is the sum W1 + W2 + W3 .   4 0 l 4 0 l  4 0 l

 4 0 l 

W = W1 + W2 + W3 = 0 +

1 e2 4 0 l

1 2e 2 4 0 l

1 3e 2 4 0 l

( =

)(

3 8.99  109 N m 2 C 2 1.60  10 −19 C

(1.0  10 m )

)

−10

1eV   = 43eV −19   1.60  10 J 

= 6.9  10 −18 J = 6.9  10 −18 J 

55. The gain of kinetic energy comes from a loss of potential energy due to conservation of energy, and the magnitude of the potential difference is the potential energy per unit charge. The helium nucleus has a charge of 2e. U K 165  103 eV V = =− =− = −82.5kV q q 2e The negative sign indicates that the helium nucleus had to go from a higher potential to a lower potential. 56. The kinetic energy of the particle is given in each case. Use the kinetic energy to find the speed. (a)

1 2

mv 2 = K → v =

(b)

1 2

mv = K → v = 2

2K m 2K m

= =

(

)(

2 5.0  102 eV 1.60  10 −19 J eV −31

9.11 10 kg

(

)(

2 5.0  102 eV 1.60  10 −19 J eV −27

1.67  10 kg

) = 1.3 10 m s 7

) = 3.110 m s 5

57. The potential energy of the two-charge configuration (assuming they are both point charges) is given by Eq. 23–10. 1 Q1Q2 1 e2 U= =− 4 0 r 4 0 r

U = U final − U initial =

(

e2  1

1  −   4 0  rinitial rfinal 

)(

 )  0.110 110 m − 0.100 110 m   1.601eV   10 J 2

= 8.99  109 N m 2 C 2 1.60  10−19 C 



−9



−19



= −1.31eV Thus 1.3 eV of potential energy was lost.

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58. (a) First we derive the potential for four charges at general locations, following the discussion in Section 23–8. Refer to the diagram on the right 1 charge: No work is required to move a single charge into a position, so U1 = 0. This represents the interaction between Q1 and

2 charges:

Q2 . U 2 =

1 Q1Q2 4 0 r12

This now adds the interactions between Q1 & Q3 and Q2 & Q3 .

3 charges:

U3 =

1  Q1Q2

QQ Q Q  + 1 3 + 2 3  r13 r23  4 0  r12

This now adds the interaction between Q1 & Q4 , Q2 & Q4 , and Q3 & Q4 .

4 charges:

U4 =

1  Q1Q2

QQ QQ Q Q Q Q Q Q  + 1 3+ 1 4 + 2 3+ 2 4 + 3 4  4 0  r12 r13 r14 r23 r24 r34 

Now put in the specific distances, based on the square geometry. 1  Q1Q2 Q1Q3 Q1Q4 Q2Q3 Q2Q4 Q3Q4  + + + + + U4 =   4 0  r12 r13 r14 r23 r24 r34 

Q2  1

 + 4 0  b

1 b

(

1 Q2 + = 4+ 2 2b b  4 0b

)

(b) The potential energy of the fifth charge is due to the interaction between the fifth charge and each of the other four charges. Each of those interaction terms is of the same magnitude since the fifth charge is the same distance from each of the other four charges.

U 5th charge

Q2 4 0

Q2 4 0b

(4 2 )

2 (c) If the center charge were moved away from the center, it would be moving closer to one or two of the other charges. Since the charges are all of the same sign, by moving closer, the center charge would be repelled back towards its original position. Thus it is in a place of stable equilibrium. (d) If the center charge were moved away from the center, it would be moving closer to one or two of the other charges. Since the corner charges are of the opposite sign as the center charge, the center charge would be attracted towards those closer charges, making the center charge move even farther from the center. So it is in a place of unstable equilibrium. 59. (a) The electron was accelerated through a potential difference of 4.8 kV (moving from low potential to high potential) in gaining 4.8 keV of kinetic energy. The proton is accelerated through the opposite potential difference as the electron, and has the exact opposite charge. Thus, the proton gains the same kinetic energy, 4.8 keV . (b) Both the proton and the electron have the same kinetic energy. Use that to find the ratio of the speeds. mp ve 1.67  10 −27 kg 2 2 1 1 v = v → = = = 42.8 m m p p e e 2 2 9.11  10 −31 kg vp me The electron will be moving 42.8 times as fast as the proton. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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60. We find the energy by bringing in a small amount of charge at a time, similar to the method given in Section 23–8. Consider the sphere partially charged, with charge q < Q. The potential at the 1 q , and the work to add a charge dq to that sphere will increase the surface of the sphere is V = 4 0 r0 potential energy by dU = Vdq. Integrate over the entire charge to find the total potential energy. Q

U =  dU =  0

4 0 r0

dq =

1 Q2 8 0 r0

61. We find the energy by bringing in a small amount of spherically symmetric charge at a time, similar to the method given in Section 23–8. Consider that the sphere has been partially constructed, and so has a charge q < Q, contained in a radius r  r0 . Since the sphere is made of uniformly charged material, the charge density of the sphere must be  E = 4 3

also satisfies  E = 4

r sphere can now found.

, and so 4 3 3

r

= 3

Q 4 3

 r03

→ q=

 r03

. Thus the partially constructed sphere

Qr 3 r03

. The potential at the surface of that

Qr 3 V =

r03

1 Qr 2

4 0 r 4 0 r 4 0 r03 We now add another infinitesimally thin shell to the partially constructed sphere. The charge of that shell is dq =  E 4 r 2dr. The work to add charge dq to the sphere will increase the potential energy by dU = Vdq. Integrate over the entire sphere to find the total potential energy. Q r r 2 5 4 1 Qr E Q 4  E Q r0  r03 Q r05 2 3 U =  dU =  Vdq =  r dr r dr 4 = = =   E 4 0 r03  0 r03 0  0 r03 5  0 r03 5 0 0

3Q 2 20 0 r0

62. To convert a decimal number into a binary number, we need to start with the largest power of 2 that 6 is smaller than the decimal number with which we start. That would be 2 = 64, so there is a “1” in the 64’s place. Subtracting 64 from 108 leaves 44. The next power of 2 is 2 = 32, so there is a “1” 5

in the 32’s place. Subtracting 32 from 44 leaves 12. The next power of 2 is 2 = 16, so there is a 4

“0” in the 16’s place. The next power of 2 is 2 = 8, so there is a “1” in the 8’s place. Subtracting 8 from 12 leaves 4, so there is a “1” in the 4’s place. The remainder is now 0, so there is a “0” in the 2’s place, and a “0” in the 1’s place. 64 + 32 + 0 + 8 + 4 + 0 + 0 = 108, so 108 decimal = 1101100 binary . Note that there are 7 “bits,” corresponding to the powers of 2 from 20 to 26 (right to left). 3

63. To turn the decimal 3597 into binary, we need to use the largest power of 2 that is smaller than the 11 decimal number with which we start. That would be 2 = 2048, so there is a “1” in the 2048’s place (the 12th bit, counting from the lowest order bit). Subtracting 2048 from 3597 leaves 1549. 10 The next smaller power of 2 is 2 = 1024, so there is a “1” in the 1024’s place (the 11th bit). © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Subtracting 1024 from 1549 leaves 525. The next smaller power of 2 is 2 = 512, so there is a “1” in the 512’s place (the 10th bit). Subtracting 512 from 525 leaves 13. Thus the 256’s place, the 128’s place, the 64’s place, the 32’s place, and the 16’s place are all 0 (5 0-bits in a row). The 8’s place is the next non-zero bit. There is a “1” in the 8’s place (4th bit). Subtracting 8 from 13 leaves 5. Thus there is a “1” in the 4’s place (3rd bit). Subtracting 4 from 5 leaves 1. Thus there is a “0” in the 2’s place, and a “1” in the 1’s place (1st bit). 2048 + 1024 + 512 + 8 + 4 + 1 = 3597, so 3597 decimal = 111000001101 binary . Note that there are 12 “bits,” corresponding to the powers of 2 from 20 to 211 (right to left). 9

64. The given binary number has a “1” in the 1’s place, the 4’s place, the 16’s place, and the 64’s place, so we have 1 + 4 + 16 + 64 = 85, and 01010101 binary = 85 decimal . Another way to calculate it is that 01010101 = 20 + 22 + 24 + 26 = 1 + 4 + 16 + 64 = 85. 65. Each “1” in the given binary contributes a certain power of 2 to the total value. 1010101010101010 = 21 + 2 3 + 25 + 2 7 + 2 9 + 211 + 213 + 215

= 2 + 8 + 32 + 128 + 512 + 2048 + 8192 + 32768 = 43, 690 decimal 66. The parity bit is added so that the total number of 1’s is always even. (a) There are 4 “1” bits, so the parity bit should be a “0”. (b) There are 6 “1” bits, so the parity bit should be a “0”. (c) There are 3 “1” bits, so the parity bit should be a “1”. 16 67. (a) A 16-bit sampling would be able to represent 2 = 65,536 different voltages. 24 (b) A 24-bit sampling would be able to represent 2 = 16,777,216 different voltages.

224 = 16,777,216 different colors are possible. 68 (a) A 4-bit number can represent 16 different values, from 0 to 15. Thus if 5.0 volts is to be the maximum value, then each binary number would represent 5.0 V / 15 = 1/3 V. Since 1011 binary = 1 + 2 + 8 = 11 decimal, 1011 binary = 11( 13 V ) = 3 23 V . (b) 2.0 Volts is 6 units of 1/3 volt each, so the binary representation is 0110, which is equal to 6 in the decimal system. 69. (a) The letter “C” is two letters beyond “A.” Thus we add 2 (binary 10) to the ASCII code for “A,” and get 00100011 . The letter “T” is 19 letters beyond “A.” Thus we add 19 (binary 10011) to the ASCII code for “A,” and get, so it’s ASCII code adds 19 to the “A” value, and get 00110100. (b) The ASCII code for the letter “A” has the decimal value of 1 + 32 = 33. Thus the ASCII code for B would have a decimal value of 34, and the ASCII code for H would have a decimal value of 40. (c) Note in Fig. 23–22 that the figure is putting the highest order bit on the left of the diagram.

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Instructor Solutions Manual

70. We see that they have a base 4 counting system, because the decimal number 5 is represented as 11, which would be a “1” in the 1’s place and a “1” in the 4’s place. Making an analogy to humans, we have 5 digits on each hand, and use base 10. Since the extraterrestrials use base 4, they must have 2 fingers on each hand, for a total of 4 fingers on their two hands . 71. An 8-bit binary value has 28 = 256 values, from 0 to 255. If we assume that a binary value of 0 represents 0 volts, then a binary value of 255 would represent 5.00 volts, and each bit would then represent 5.00 / 255 = 0.0196 volts, or about 0.02 volts. (a) The binary number 01110101 is 1 + 4 + 16 + 32 + 64 = 117. Thus this binary number would 117 represent ( 5.00 V ) = 2.29 V . 255 (b) The value of 3.47 would convert as a binary number by the following proportion. 3.47 V x  3.47 V  = → x = 255   = 176.97  177 5.00 V 255  5.00 V  Thus the binary representation is the equivalent to 177 decimal, which is 10110001. 177 As a check, ( 5.00 V ) = 3.4706 V. 255 72. (a) The monitor screen is re-drawn at regular intervals. Even though the mouse cursor moves in a continuous motion, its location is only drawn at those regular intervals. It is somewhat like a strobe light effect. (b) Let the interval between images be “T,” and construct a table. Position # 1 Time = 0 Position # 2 Time = T Position # 3 Time = 2T Position # 4 Time = 3T Position # 15 Time = 14T = 0.25 seconds. Thus there are 15 images in 0.25 seconds. The refresh time would be T = 0.25 / 14 = 0.018 seconds. The refresh rate is the reciprocal of this, which is 56 Hz. 73. If 0 is white and 100 is black, then there are 5 different color states. There are three various gray shades in the picture. 0 = white ; 1 = light gray ; 2 = medium gray ; 3 = dark gray ; 4 = black. The 11th row has 5 whites; 1 medium gray; 1 dark gray; 1 light gray; 2 blacks; 3 medium grays; and 3 whites, for a total of 16 squares.

000; 000; 000; 000; 000; 010; 011; 001; 100; 100; 010; 010; 010; 000; 000; 000

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74. From Section 23–10, HD TV has 1080 × 1920 pixels for an image. According to Section 23–9, each pixel receives a 24-bit signal (an 8-bit signal for the brightness of each of the three “sub-pixels” of red, green, and blue). This amount of data is transmitted 60 times per second, and we are accumulating the amount of data for 300 seconds (three significnant figures).  1byte   1Gbyte  pixels   24 bits   60 frames   Data ( Gbytes ) =  1080  1920     ( 300 sec )    30  frame   pixel   1sec    8 bits   2 bits  = 104.3Gbytes  104 Gbytes

75. (a) Each subpixel gets 8 bits per cycle. subpixels   8 bits   60 frames   1Mbit   Data rate =  3  1080  1920   = 2850 Mb/s   frame   subpixel   1sec   2 20 bits   (b) Compared to the current rate of 19 Mb/s, the HD TV would need about 150 times more data than is currently transmitted. This must mean that the data for the TV is “compressed” into a smaller number of bits. (c) If the signal is being compressed by a factor of 150, then on average, only 1/150 of the pixels would be refreshed every frame. This is less than 1%.

1 = 6.67  10=3 = 0.667%  0.7% 150 (d) We assume the TV is receiving 19 Mb/s, and calculate what fraction of a frame of data that is

  220 bits   1frame  = 0.40 frames s .    1Mbit   (1080 1920  24 ) bits 

(19 Mb s ) 

This is equivalent to about 2.5 seconds for each frame. 76. (a) In Fig. 23–30, there are 23 rows, 16 columns, and 5 possible shadings for each pixel. It would take 5 bits to encode the row number (row number can go from 0 to 22), 4 bits to encode a column number (column number can go from 0 to 15), and 3 bits to encode the shading (shading can go from 0 to 4). Thus a 12-bit binary number could encode all the data for 1 pixel–the first 5 bits for the row number, the next 4 bits for the column number, and the last 3 bits to encode the shading. There are 23 x 16 = 368 pixels, and so the total data for one frame pixels   12 bits   would be  368   = 4416 bits frame . frame   1pixel   (b) A simple compression scheme might work as follows: • Have a 4-bit code of 1111 to start a new row, followed by a 5-bit code giving the row number. • Have a stream of 16 sets of 3 bits giving the shading for each pixel in a row. Those 48 bits would comprise the total set of data for that row. Note that with our 3-bit encoding of the shading, the only possible valid 3-bit combinations are 000, 001, 010, 011, and 100. It would be impossible by stringing any of those 3-bit combinations together to ever generate a “false row start” of 1111. (c) That makes a total of 57 bits per row. There are 23 rows, and so it would require 57 x 23 = 1311 bits to describe the entire screen. That is only 30% of the original number of bits needed to describe the frame.

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Instructor Solutions Manual

(d) This scheme would work for any picture with 4 greytones (5 possible colors per pixel, counting white), since it still describes each pixel in each row. But it would require some “decomposition” of the 57-bit string of data for each row. We assume that the computer processor in the display device would be capable of doing that process very rapidly. 77. The potential of the earth will increase because the “neutral” Earth will now be charged by the removing of the electrons. The excess charge will be the elementary charge times the number of electrons removed. We approximate this change in potential by using a spherical Earth with all the excess charge at the surface.   6.02  1023 molecules   1000 kg  4 10 e −  1.602  10−19 C   3 Q=   3  ( 0.00175 m )      3 − 0.018 kg e    H 2O molecule    m 

= 1203C V =

4 0 REarth

= 1.7  10 V ) 6.381203C  10 m

(

= 8.99  109 N m 2 C2

78. The potential at the surface of a charged sphere is that of a point charge of the same magnitude, located at the center of the sphere. 1  10−8 C 1 q = 8.99  109 N m 2 C 2 = 599.3V  600 V V = 4 0 r ( 0.15 m )

(

)

(

)

79. (a) All 8 charges are the same distance from the center of the cube. Use Eq. 23–5 for the potential of a point charge.

Vcenter = 8

4 0

1 Q

3 4 0 l

 9.24

1 Q 4 0 l

2 (b) For the 7 charges that produce the potential at a corner, 3 are a distance l away from that corner, 3 are a distance 2l away from that corner, and 1 is a distance 3l away from that corner.

Vcorner = 3

1 Q 4 0 l

4 0

 

= 3+

3 2

1  1 Q

1 Q

 5.70  4 0 l 3  4 0 l

(c) The total potential energy of the system is half the energy found by multiplying each charge times the potential at a corner. The factor of half comes from the fact that if you took each charge times the potential at a corner, you would be counting each pair of charges twice.

 

U = 12 8 ( QVcorner ) = 4  3 +

3 2

1  1 Q2



3  4 0 l

 22.8

1 Q2 4 0 l

80. Let d1 represent the distance from the left charge to point B, and let d 2 represent the distance from the right charge to point B. Let Q represent the positive charges, and let q represent the negative charge that moves. The change in potential energy is given by Eq. 23–2b.

d1 = 122 + 142 cm = 18.44 cm

d 2 = 142 + 242 cm = 27.78 cm

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Chapter 23

Electric Potential

1 

  Q Q Q Q  + +  −  4 0  0.1844 m 0.2778 m   0.12 m 0.24 m  

U b − U a = q (Vb − Va ) = q



4 0

Qq 

 0.1844 m

(

= 8.99  109 N m 2 C 2

  1 1  − +   0.2778 m   0.12 m 0.24 m   1

)( 38 10 C )( −1.8 10 C )( −3.477 m ) = 2.138 J  2.1J −6

−6

−1

81. The alpha particle has twice the charge of the proton, so it will have twice the potential energy for the same voltage. Thus the alpha will have twice the kinetic energy of the proton after acceleration. 82. The electric force on the electron must be the same magnitude as the weight of the electron in order for the electron to remain stationary. The magnitude of the electric force is the charge on the electron times the magnitude of the electric field. The electric field is the potential difference per meter: E = V d .

FE = mg ; FE = q E = eV d V =

mgd

→ eV d = mg →

( 9.11  10 kg )( 9.80 m s ) ( 0.024 m ) = 1.3  10 V = −31

−12

1.60  10−19C e Since it takes such a tiny voltage to balance gravity, the thousands of volts in a television set are more than enough (by many orders of magnitude) to move electrons upward against the force of gravity. 83. (a) We assume that Q2 is held at rest. The energy of the system will then be conserved, with the initial potential energy of the system all being changed to kinetic energy after a very long time. kQ1Q2 1 Einitial = Efinal → U initial = K final → = 2 m1v12 → r

v1 =

2kQ1Q2 m1r

(

)( 6.5 10 C ) = 2756 m s  2.8 10 m s ( 2.5 10 kg )( 4.0 10 m )

−6

2 8.99  109 N m 2 C 2

−2

(b) In this case, both the energy and the momentum of the system will be conserved. Since the initial momentum is zero, the magnitudes of the momenta of the two charges will be equal. m pinitial = pfinal → 0 = m1v1 + m2 v2 → v2 = −v1 1 m2

Einitial = Efinal → U initial = K final →

kQ1Q2 r

v1 =

2  2  m1   1 m1 = m v + m2 v =  m1v1 + m2  −v1 ( m1 + m2 ) v12  =2 m m   2   2  1 2

2 1 1

2 2

1 2

2kQ1Q2 m2

m1 ( m1 + m2 ) r

1 2

(

)( 6.5 10 C ) (3.5 10 kg ) ( 2.5 10 kg )( 6.0 10 kg )( 4.0 10 m ) −6

2 8.99  109 N m 2 C 2 −6

−6

−2

= 2105 m s  2.1  103 m s

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84. Calculate VAB = VA − VB. Represent the 0.10 m distance by the variable d.

1   kq1 kq2   kq2 kq1  k  + + −  = ( q1 − q2 )  1 −  2d   d 2d  d 2  d 

VAB = VA − VB = 

(8.99 10 N m C ) 4.5  10 C 1 − 1  = 1.2  10 V = ( ) 2  0.1m 9

−6





85. The kinetic energy of the electrons (provided by the UV light) is converted completely to potential energy at the plate since they are stopped. Use energy conservation to find the emitted speed, taking the 0 of potential energy to be at the surface of the barium. K initial = U final → 12 mv 2 = qV →

2qV m

(

)

2 −1.60  10−19 C ( −3.02 V ) −31

9.11  10 kg

= 1.03  106 m s

86. To find the angle, the horizontal and vertical components of the velocity are needed. The horizontal component is found using conservation of energy for the initial acceleration of the electron. That component does not change as the electron passes through the plates. The vertical component is found using the vertical acceleration due to the potential difference of the plates, and the time the electron spends between the plates. Horizontal: x U inital = K final → qV = 12 mv x2 t= vx Vertical: ( v y − v0 y ) → v = qE y t = qE y x FE = qE y = ma = m y t m mv x Combined: qE y x  250 V  ( 0.065 m ) qE y x qE y x E y x  0.013 m  vy mv x tan  = = = = = = = 0.1786 mv x2 2qV 2V 2 ( 3500 V ) vx vx

 = tan −1 0.1786 = 10.12  10 ( two significant figures ) 87. Use Eq. 23–5 to find the potential due to each charge. Since the triangle is equilateral, the 30-60-90 triangle relationship says that the distance from a corner to the midpoint of the opposite side is 3l 2.

VA = =

VB =

( −Q ) +

4 0 l 2

1 4 0

( −3Q ) +

l 2

4 0

(Q ) +

(Q ) = 3l 2

2Q 

1 

4 0 l 

3

4 0

 −4 +



−Q A

+Q C

−3Q

Q  3

 − 2   0 l  6  1

( −Q ) +

4 0 l 2

4 0

( −3Q ) = − 3l 2

= −

2 0 l

818

Chapter 23

Electric Potential

(Q ) +

VC =

4 0 l 2

1 4 0

( −3Q ) + l 2

1 4 0

( −Q ) = − 3l 2

1  3 Q  1 +  2 + = − l  6   0 l  3

2Q 

1 4 0

88. Since the electric field is directed downward, the surface of the Earth is a lower potential than points located above the surface. Call the surface of the Earth 0 volts. Then a height of 2.00 m has a potential of 300 V. We also call the surface of the Earth the 0 location for gravitational potential energy. Write conservation of energy relating the charged spheres at 2.00 m (where their speed is 0) and at ground level (where their electrical and gravitational potential energies are 0).

 

Einitial = Efinal → mgh + qV = 12 mv 2 → v = 2  gh +

qV 



m 

 ( 6.5 10−4 C ) ( 300 V )  = 6.3174 m s v+ = 2 ( 9.80 m s 2 ) ( 2.00 m ) +  ( 0.550 kg )  

 ( −6.5 10−4 C ) ( 300 V )  = 6.2041 m s v− = 2 ( 9.80 m s 2 ) ( 2.00 m ) +  ( 0.550 kg )   v+ − v− = 6.3174 m s − 6.2041m s = 0.1133 m s  0.11m s 89. (a) The energy is related to the charge and the potential difference by Eq. 23–3. U 5.2  106 J U = qV → V = = = 1.3  106 V 4.0 C q (b) The energy (as heat energy) is used to raise the temperature of the water and boil it. Assume that room temperature is 20oC. Use Eq. 19–2 and Eq. 19–4. Q = mcT + mLf → m=

cT + Lf

5.2  106 J J    o 5 J   4186 kg o C  ( 80 C ) +  22.6  10 kg     

= 2.0 kg

90. Use Eq. 23–7 for the dipole potential, and use Eq. 23–9 to determine the electric field. x

p cos 

4 0

Ex = −

V x

=−

(x + y )

4 0

x +y

(

)

2 1/ 2

(

4 0 x + y 2

(

)

3/ 2

1/ 2 2 2 3/ 2 − x 23 x 2 + y 2 2x  p  x +y

 4 0  

(x + y ) 2

p  2x2 − y 2  =   5/ 2  4 0  ( x 2 + y 2 )    

p  2 cos 2  − sin 2  



4 0 

 

819

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Ey = −

V y

=−

px 

4 0 

(

− 32 x 2 + y 2

)

−5/ 2

2y =



Instructor Solutions Manual

p 

 3 xy   5/ 2 4 0  ( x 2 + y 2 )   

p  3cos  sin    4 0  r3

= Notice the

1 r3

dependence in both components, which is indicative of a dipole field.

91. (a) The electric field outside a charged, spherically symmetric volume is the same as that for a point charge of the same magnitude of charge. Integrating the electric field from infinity to the radius of interest will give the potential at that radius.

E ( r  r0 ) =

Q 4 0 r

; V ( r  r0 ) = − 

4 0 r



dr =

4 0 r 

Q 4 0 r

(b) Inside the sphere the electric field is obtained from Gauss’s Law using the charge enclosed by a sphere of radius r. Q 43  r 3 Qr → E ( r  r0 ) = 4 r 2 E = 3 4  0 3  r0 4 0 r03 Integrating the electric field from the surface to r  r0 gives the electric potential inside the sphere. r

V ( r  r0 ) = V ( r0 ) −  r0

Qr 4 0 r0

dr = 3

Q 4 0 r0

−

Qr 2

8 0 r03 r

Q 4 0 r0

 r2  3 −  r2  8 0 r0  0  Q

and E0 = E ( r = r0 ) =

Q 4 0 r02

. Then we plot

V V0 and E E0 as functions of r r0 .

 r2  Qr − 3  2  2 8 0 r0  r0  1  r  r 4 0 r03 = 2  3 − 2  ; E E0 = = V V0 = Q Q r0  r0  4 0 r0 4 0 r02 Q

For r  r0 :

For r  r0 :

V V0 =

4 0 r r0 4 0 r 2 r02 −1 −2 = = ( r r0 ) ; E E0 = = 2 = ( r r0 ) Q Q r r 2 4 0 r0 4 0 r0

820

Chapter 23

Electric Potential

1.50

1.0

1.25

0.8 0.6

0.75

E /E (r 0)

V /V (r 0)

1.00

0.50 0.25 0.00

0.4 0.2 0.0

0.0

0.5

1.0

r/r 0

1.5

2.0

2.5

3.0

0.0

0.5

1.0

r/r 0

1.5

2.0

2.5

3.0

92. We first need to find the electric field. Since the charge distribution is spherically symmetric, Gauss’s law tells us the electric field everywhere. 1 Q Q 2  E dA = E 4 r = encl → E = 4 rencl2 0 0

(

)

If r  r0 , calculate the charge enclosed in the manner of Example 22–5. r



r2 



r0 

Qencl =   E dV =   0 1 − 0



r4 

 r3



r0 

3

4 r 2dr = 4 0   r 2 − 2 0

dr = 4 0  2

−

r5 

5r02 

The total charge in the sphere is the above expression evaluated at r = r0 .

 r03 r05  8 0 r03 Qtotal = 4 0  − 2  = 15  3 5r0  Outside the sphere, we may treat it as a point charge, and so the potential at the surface of the sphere is given by Eq. 23–5, evaluated at the surface of the sphere. 8 0 r03 2 1 Qtotal 1 15 = 2  0 r0 V ( r = r0 ) = = 4 0 r0 4 0 r0 15 0 The potential inside is found from Eq. 23–4a. We need the field inside the sphere to use Eq. 23–4a. The field is radial, so we integrate along a radial line so that E d l = Edr.

 r3 r5  4 0  − 2  3 1 Qencl 1  3 5r0  =  0  r − r  = E ( r  r0 ) =  0  3 5r02  4 0 r 2 4 0 r2 r

0  r r3  0  r 2 r4  Vr − Vr = −  E d l = −  E dr = −   − 2  dr = −  −  0  3 5r0   0  6 20r02  r r r r r

   r2 r 4   2  0 r02  0  r 2 r 4   r02 r04   0 − = − − − −    0  6 20r02   15 0  0  6 20r02   6 20r02   r   r

Vr = Vr +  − 0

 0  r02 r 2 r4  − +    0  4 6 20r02 

93. First express each decimal number as a binary number: 14 = 1110 ; 45 = 101101 ; 103 = 1100111. Then line the values up at the right edge to add them. Note that if there are two or more “1’s” in a column, then we will “carry” one or more to the next left-most column. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Now we convert the result to decimal: 2 + 32 + 128 = 162, which is 14 + 45 + 103. 94. As an estimate, the length of the bolt would be the voltage difference of the bolt divided by the breakdown electric field of air. 5  107 V = 16.7 m  20 m 3  106 V m If the lightning strikes from a height of 750 m (roughly a half-mile), then 40 or more steps might be involved. 95. (a) The voltage at x = 0.20 m is obtained by inserting the given data directly into the voltage equation. B 120 V m 4 V ( 0.20 m ) = = = 18750 V  19 kV 2 2 x2 + R2 ( 0.20 m ) 2 + ( 0.20 m ) 2   

(

(b)

)

The electric field is the negative derivative of the potential. E( x) = −

d 

 B 4 Bx ˆi   ˆi = 2 2 2 3 dx  ( x 2 + R 2 )  x + R ( )  

Since the voltage only depends on x the electric field points in the positive x-direction. (c) Inserting the given values in the equation of part (b) gives the electric field at x = 0.20 m. 4 120 V m 4 ( 0.20 m ) ˆi E(0.20 m) = = 1.875  105 V m  1.9  105 V m ˆi 2 2 3 ( 0.20 m ) + ( 0.20 m )   

(

)

96. Use energy conservation, equating the energy of charge − q1 at its initial position to its final position at infinity. Take the speed at infinity to be 0, and take the potential of the point charges to be 0 at infinity. 2 Einitial = Efinal → K initial + U initial = K final + U final → 12 mv02 + ( −q1 ) Vinitial = 12 mvfinal + ( −q1 ) Vfinal point

1 2

mv02 + ( − q1 )

2 q2

4 0

a2 + b2

= 0 + 0 → v0 =

q1q2

m 0

a2 + b2

point

97. Use Eq. 23–4b without the negative sign in order to find the magnitude of the voltage difference. V E= → V = E x = 3  106 V m 1 10−3 m = 3000 V x No harm is done because so very little charge is actually transferred between you and the doorknob.

(

)(

)

822

Chapter 23

Electric Potential

98. (a) From Example 22–6, the electric field due to a long wire is radial relative to the wire, and is of 1  . Since the charge density is positive, the field lines point radially away magnitude E = 2 0 R from the wire. (b) Use Eq. 23–4a to find the potential difference, integrating along a line that is radially outward from the wire. Ra

( )

1 

Va − Vb = −  E d l = −  Rb

2 0 R

dR = −

R   ln ( Ra − Rb ) = ln b 2 0 2 0 R a

99. (a) The electric potential is found from Eq. 23–5.

Vinitial =

(1.60  10 C )  27 V = ( 8.99  10 N m C ) 0.53  10 m −19

−10 4 0 ratom (b) The kinetic energy can be found from the fact that the magnitude of the net force on the electron, which is the attraction by the proton, is causing circular motion. m v2 1 qp qe 1 e2 1 e2 2 2 1 1 v v → = → = = Fnet = e = m K m e e 2 2 2 4 0 ratom 4 0 ratom 4 0 ratom ratom

(1.60  10 C ) = 2.17110 J  2.2  10 J = ( 8.99  10 N m C ) ( 0.53 10 m ) −19

1 2

−18

−10

1eV

= 2.171  10 −18 J 

= 13.57 eV  14 eV 1.60  10−19 J (c) The total energy of the electron is the sum of its kinetic and potential energies. The potential energy is found from Eq. 23–2a, and is negative since the electron’s charge is negative. 1 e2 1 e2 1 e2 + 12 = − 12 Etotal = U + K = −eV + 12 me v 2 = − 4 0 ratom 4 0 ratom 4 0 ratom = −2.171 10−18 J  −2.2  10−18 J  −14 eV

(d) If the electron is taken to infinity at rest, both its potential and kinetic energies would be 0. The amount of energy needed by the electron to have a total energy of 0 is just the opposite of the answer to part (c), 2.2  10−18 J or 14 eV . 100. Since the two fragments are spherical and the charge is uniformly distributed, they can be treated as point charges for purposes of calculating their potential energy. Use Eq. 23–10 to calculate the potential energy. Using energy conservation, the potential energy is all converted to kinetic energy as the two fragments separate to a large distance. 1 q1q2 Einitial = Efinal → U initial = K final = 4 0 r

(

= 8.99  10 N m C 9

( 38 )( 54 ) (1.60  10−19 C )

 = 250  10 eV ) 5.5 10 m + 6.2 10 m  1.601eV ( ) ( )  10 J  −15

−15

−19

= 250 MeV

This is about 25% greater than the observed kinetic energy of 200 MeV. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

101. The electric field can be determined from the potential by using Eq. 23–8, differentiating with respect to x. 1/ 2 −1/ 2 dV ( x )  d  Q  2 Q 1 2 =−  − x  = − E ( x) = − x + R02 x + R02 ( 2 x ) − 1 2 2 2  2 0 R0  dx dx  2 0 R0 

(

)

(

)

  x   1 − 1/ 2 2 0 R02  ( x 2 + R02 )    Q

Express V and E in terms of x R0 . Let X = x R0 .

V ( x) =

( x 2 + R02 )1/ 2 − x  = 2Q  4 0 R0 2 0 R02 

(

= 8.99  109 N m 2 C2 E ( x) =

( X +1 − X )

)

(

2 5.0  10−6 C

)

0.10 m

( X + 1 − X ) = (8.99  10 V ) ( X + 1 − X ) 2

  2Q  x  = 1 1− − 1/ 2 2 2 0 R0  ( x 2 + R02 )  4 0 R02   

  X 2 +1

2 5.0  10 −6 C  1− ( 0.10 m )2 

  X 2 +1

(

= 8.99  10 N m C 9

(

)



(

)

  X +1 X

= 8.99  106 V m 1 −



8.0 6.0

V (10 Volts)

10.0

4.0 2.0 0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

2.5

3.0

3.5

4.0

x /R 0 10.0

E (10 V/m)

8.0 6.0 4.0 2.0 0.0 0.0

0.5

1.0

1.5

2.0

x /R 0

824

Chapter 23

Electric Potential

102. If we assume the electric field is uniform, then we can use Eq. 23–4b to estimate the magnitude of the electric field. From Problem 22–24 we have an expression for the electric field due to a pair of oppositely charged planes. We approximate the area of a shoe as 30 cm × 8 cm. Answers will vary with the shoe size approximation. V  Q → E= = = d 0 0 A −12 2 2 2 3  0 AV ( 8.85  10 C / Nm )( 0.024m )( 6.0  10 V ) = = 1.3  10 −6 C Q= −3 1.0  10 m d

103. (a) Since the reference level is given as V = 0 at r = , the potential outside the shell is that of a point charge with the same total charge. V =

1 Q 4 0 r

 E ( 43  r23 − 43  r13 )

4 0

 ( r2 − r1 ) = E , r  r2 r 3 0 3

Note that the potential at the surface of the shell is Vr = 2

 E  2 r13   r2 − r  . 3 0  2 

(b) To find the potential in the region r1  r  r2 , we need the electric field in that region. Since the charge distribution is spherically symmetric, Gauss’s law may be used to find the electric field. 3 3 3 3 4 4  E r − r1 1 Qencl 1  E 3  r − 3  r1 Qencl 2 =  E dA = E 4 r =  0 → E = 4 0 r 2 = 4 0 3 0 r2 r2 The potential in that region is found from Eq. 23–4a. The electric field is radial, so we integrate along a radial line so that E d l = E dr.

(

)

(

)

r13   ( r − r1 ) E   E  1 2 r13  = − − = − Vr − Vr = −  E d l = −  E dr = −  E dr r dr r +  2 3 0 3 0 r  3 0  r2 r 2  r r r r r r

  E  2 r13    E  2 r13    E  2 r13   E  2 r13   1 1 1 r +  = r2 −  +  − r + + r +  2 2 2 2     r  r  3 0  r2   3 0  r  3 0  r2    3 0  r

Vr = Vr +  − 2

r13   E  3 2 1 2 r13  E  2 r 1 2 r 1 2 r r r r −2r −  − − − + + 2 = 2 2 2 2 2 r2 r r2  3 0  3 0  r 

 E  1 2 1 2 1 r13  r − 6 r − 3  , r1  r  r2 2 2 r   0 

3 1

(c) Inside the cavity there is no electric field, so the potential is constant and has the value that it has on the cavity boundary at r = r1 . Vr = 1

 E  1 2 1 2 1 r13   E 2 2 r − 6 r1 − 3  = ( r2 − r1 ) , r  r1 2 2 r1  2 0  0 

The potential is continuous at both boundaries.

825

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

104. (a) The contour values on the map are in feet. There appears to be about seven contour lines between Greenstone Lake and the first Conness Lake. Each line is to be interpreted as an 80-V difference, so the voltage difference is approximately 560 V. The two lakes appear to be about 2 km apart. The magnitude of the electric field is equal to the change in voltage divided by the separation distance, Eq. 23–4b. 560 V V E = ba = = 0.28 V m  0.3 V/m 2000 m d Greenstone Lake is lower than Cecile Lake, so the direction of the electric field is basically in the +x-direction. Answers may vary due to interpreting the figure. (b) On the Conness Glacier, the darker elevation line near the “C” in the name appears to be 11806 feet, and the darker elevation line two lines to the “northeast” of the name appears to be at 10800 feet. Those two lines appear to be approximately 1.5 km apart on average, given the data about Greenstone Lake. Thus the “voltage” is decreasing as you move to the northeast, which means the electric field is pointed in that direction. The electric field points toward lower voltage. V 11806 V − 10800 V E = ab =  0.7 V/m, 45 above the + x -axis d 1500 m Again, answers may vary due to estimations in interpreting the figure. 105. (a) We may treat the sphere as a point charge located at the center of the field. Then the electric 1 Q 1 Q , and the potential at the surface is Vsurface = . field at the surface is Esurface = 2 4 0 r0 4 0 r0 Vsurface =

(b) Vsurface =

1 Q 4 0 r0

(

)

= Esurface r0 = Ebreakdown r0 = 3  106 V m ( 0.20 m ) = 6  105 V

→ Q = ( 4 0 ) rV = 0 surface

( 0.20 m ) ( 6  105 V )

(8.99  10 N m C ) 9

= 1.33  10−5 C  1  10−5 C

826

CHAPTER 24: Capacitance, Dielectrics, Electric Energy Storage Responses to Questions 1.

Yes, there can be a potential difference between them. If the conductors have different shapes, then even if they have the same charge, they will have different charge densities and therefore different electric fields near the surface. Thus, there can be a potential difference between them–because of the different E-fields, it might take work to move a small amount of charge from one conductor to the other one. No, the definition of capacitance C = Q/V cannot be used here because it is defined for the case where the charges on the two conductors of the capacitor are equal in magnitude and of opposite signs.

Eq. 24–2 would give an underestimate. If the separation between the plates is not very small compared to the plate size, then the electric field (for a given magnitude of the charges) will actually be smaller than for the infinite plate case. For example, consider the electric field between two small charges (think of point charges) when separated by a large distance. The field is strongest near the charges, and weaker between the charges. Since the electric field is reduced, the potential difference will also be reduced. The capacitance is inversely proportional to potential and, for parallel plates, also inversely proportional to the field, so the capacitance will actually be larger than that given by the formula.

Ignoring “fringing” field effects (i.e., the E-field existing outside the region where overlap occurs), the capacitance would decrease by a factor of 2, since the area of overlap decreases by a factor of 2. (Fringing effects might actually be noticeable in this configuration.)

When a capacitor is first connected to a battery, charge flows to one plate. Because the plates are separated by an insulating material, charge cannot cross the gap. An equal amount of charge is therefore repelled from the opposite plate, leaving it with a charge that is equal and opposite to the charge on the first plate. The two conductors of a capacitor will have equal and opposite charges even if they have different sizes or shapes. The battery cannot “create” charge, and so to put a certain amount of positive charge on one plate, it must remove that amount of positive charge from the other plate, leaving it with an equal magnitude but opposite sign of charge.

Charge a parallel-plate capacitor using a battery with a known voltage V. Let the capacitor discharge through a resistor with a known resistance R and measure the time constant. This will allow calculation of the capacitance C. (Or, the capacitance could be measured with a capacitance meter.) Then use C =  0 A d and solve for ε0.

The energy stored in a capacitor network can be calculated by U = 12 CV 2. Since the voltage for the capacitor network is the same in this problem for both configurations, the configuration with the highest equivalent capacitance will store the most energy. The parallel combination has the highest equivalent capacitance, and so stores the most energy. Another way to consider this is that the total stored energy is the sum of the quantity U = 12 CV 2 for each capacitor. Each capacitor has the same capacitance, but in the parallel circuit, each capacitor has a larger voltage than in the series circuit. Thus the parallel circuit stores more energy.

If a large copper sheet of thickness l is inserted between the plates of a parallel-plate capacitor, the same magnitude of charge on the capacitor will be induced on the large flat surfaces of the copper sheet, with the negative side of the copper facing the positive side of the capacitor. This arrangement

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

can be considered to be two capacitors in series, each with a thickness of 12 ( d − l ) , assuming the copper sheet is placed symmetrically between the original capacitor plates. The new net capacitance will be C  =  0 A ( d − l ) , so the capacitance of the capacitor will be increased. The same result can be derived even if the copper is not placed in symmetrically. See the solution to Problem 18. 8.

A force is required to increase the separation of the plates of an isolated capacitor because you are pulling a positive plate away from a negative plate. Since unlike charges attract, it takes a force to move the oppositely-charged plates apart. The capacitance decreases, and so the potential between the plates increases since the charge has to remain the same. The work done in increasing the separation goes into increasing the electric potential energy stored in the capacitor.

The energy stored in the capacitor is given by Eq. 24–5, U = 12 CV 2. As the plates are separated, the voltage is held constant by the battery. Since the separation distance has doubled, the capacitance has been multiplied by ½, using Eq. 24–2. Thus the stored energy has been multiplied by ½ .

10. (c) If the voltage across a capacitor is doubled, the amount of energy it can store is quadrupled: U = 12 CV 2. 11. The dielectric will be pulled into the capacitor (to the left) by the electrostatic attractive forces between the charges on the capacitor plates and the polarized charges on the dielectric’s surface. (Note that the addition of the dielectric decreases the energy of the system.) 12. The dielectric is polarized whether the capacitor is isolated, as in Question 11, or is connected to the battery, as in this question. The induced polarization always results in attraction–there is no way to polarize the dielectric so that like charges line up with each other, creating a repulsion. So in this case the dielectric is also pulled into the capacitor. 13. (a) If the capacitor is isolated, Q remains constant, and U = 12

Q2 C

becomes U ' = 12

Q2 KC

and the

stored energy decreases. (b) If the capacitor remains connected to a battery so V does not change, U = 12 CV 2 becomes U ' = 12 KCV 2, and the stored energy increases.

14. For dielectrics consisting of polar molecules, one would expect the dielectric constant to decrease with temperature. As the thermal energy increases, the molecular vibrations will increase in amplitude, and the polar molecules will be less likely to line up with the electric field of the capacitor. 15. (a) When the dielectric is removed, the capacitance decreases by a factor of K. (b) The charge decreases since Q = CV and the capacitance decreases while the potential difference remains constant. The “lost” charge returns to the battery. (c) The potential difference stays the same because it is equal to the battery voltage. (d) If the potential difference remains the same and the capacitance decreases, the energy stored in the capacitor must also decrease, since U = 12 CV 2. (e) The electric field between the plates will stay the same because the potential difference across the plates and the distance between the plates remain constant.

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16. We meant that the capacitance did not depend on the amount of charge stored on the capacitor plates or on the potential difference between the capacitor plates. Changing the amount of charge stored or the potential difference will not change the capacitance. Changing the size or position of the plates, or the dielectric material, will change the capacitance. 17. The dielectric constant is the ratio of the capacitance of a capacitor with the dielectric between the plates to the capacitance without the dielectric. If a conductor were inserted between the plates of a capacitor such that it filled the gap and touched both plates, the capacitance would drop to zero since charge would flow from one plate to the other, rather than being “stored” on the plates. So, the dielectric constant of a good conductor would be zero.

Solutions to MisConceptual Questions 1.

(d, e) The capacitance is determined by the shape of the capacitor (area of plates and separation distance) and the material between the plates (dielectric). The capacitance is the constant ratio of the charge on the plates to the potential difference between them. Changing the charge will change the potential difference, but not the capacitance, so (d) is an incorrect answer to the question. Changing the energy stored in a capacitor will change the charge on the plates and the potential difference between them, but will not change the capacitance, so (e) is also an answer to the question.

(c) Since the capacitor are in series, they will each have the same charge. Thus the charge on the capacitor plate closest to the positive side of the battery is +Q. And then since each capacitor is electrically neutral, the charge at A is –Q, and the charge at B is +Q. Also, since the part of the circuit bounded by A and B is not actually connected to the battery, it must be neutral. It is polarized with equal and opposite charges at A and B.

(d) Since the plates are not connected to anything, the charge on them must stay the same. Assuming the separation distance is small compared to the dimensions of the plates, then the electric field between the plates is constant and does not depend on the distance. So as the distance decreases, the potential must drop so that the ratio of volts / meter (the electric field) is constant. Or, as the distance decreases, the capacitance increases, and since Q = CV, if Q is constant and C increases, V must decrease.

(a) The capacitance is given by C = K  0

(c) Connecting capacitors in series effectively increases the plate separation distance, decreasing the net capacitance. Connecting capacitors in parallel effectively increases the plate area, increasing the net capacitance. Therefore when capacitors are connected in series, the effective capacitance will be less than the capacitance of the smallest capacitor; and when connected in parallel the effective capacitance will be greater than the capacitance of the largest capacitance. This can also be seen from the capacitor combination formulas.

(b) Since Q = CV, if the charges are the same, then the smallest capacitance will have the largest potential difference. The capacitance and voltage are inversely proportional if they both have the same charge.

. Only answer (a) will decrease the capacitance. d Answers (b) and (c) will increase the capacitance, and answer (d) will not affect the capacitance.

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(c) As the plates are moved together, the capacitance increases. Since Q = CV, if the capacitance increases but the voltage is constant, then the amount of charge on the plates must increase.

(b) The electric field between the plates is the potential difference divided by the distance between the plates. Since the potential difference is constant and the distance is increasing, the electric field must decrease.

(b) When the plates were connected to the battery, a charge was established on the plates. As the battery is disconnected this charge remains constant on the plates. The capacitance decreases as the plates are pulled apart, since the capacitance is inversely proportional to the separation distance. For the charge to remain constant with smaller capacitance, the voltage between the plates increases.

10. (a) Eq. 24–10 states that the electric field is reduced, so answer (a) is correct. From Eq. 24–7, the capacitance increases, so answer (b) is not correct. Example 24–11 shows that the charge increases, and so answer (c) is not correct. From Table 24–1, we see that the dielectric strength is greater than that of air in every case, and so the maximum voltage for a given capacitor plate separation distance increases, and so answer (d) is not correct. Since U = 12 CV 2 , if the voltage is held constant but the capacitance increases, then the stored energy will increase, and so answer (e) is not correct. 11. (d) The energy is given by U = 12 CV 2. If V is doubled, then the stored energy will increase by a factor of 4, to 4U. 12. (a) Since the capacitors are in series, they each have the same charge, thus, answer (a) is true and answer (b) is false. And since neither capacitor is connected in parallel with the battery, the voltage will not be V0 for either of them. 13. (b) The energy is given by U =

. Since the charge is constant and the capacitance doubles 2C when the distance between the plates is halved, the energy will be halved.

Solutions to Problems 1.

The capacitance is found from Eq. 24–1. Q 3.5  10−3 C Q = CV → C = = = 3.6  10−6 F 960 V V

The voltage is found from Eq. 24–1. 16.5  10−8 C Q Q = CV → V = = = 25 V C 6500  10−12 F

We assume the capacitor is fully charged, according to Eq. 24–1.

(

)

Q = CV = 7.50 10−6 F (12.0 V ) = 9.00 10−5 C

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Let Q1 and V1 be the initial charge and voltage on the capacitor, and let Q2 and V2 be the final charge and voltage on the capacitor. Use Eq. 24–1 to relate the charges and voltages to the capacitance. Q1 = CV1 Q2 = CV2 Q2 − Q1 = CV2 − CV1 = C (V2 − V1 ) →

C= 5.

Q2 − Q1 V2 − V1

15  10−6 C

(121 − 83) V

The total charge will be conserved, and the final potential difference across the capacitors will be the same.

Q0 = Q1 + Q2 ; V1 = V2 → Q2 = Q0 − Q1 = Q0 − Q0

V1 = V2 = 6.

= 3.947  10−7 F = 0.39  F

Q1 C1

Q0 =

Q1 C1

C1 C1 + C2

Q2 C2

Q0 − Q1



→

Q1 = Q0

C1 C1 + C2

   C1 + C2  C2

= Q2 = Q0 

C1 C1 + C2 C1

= V =

Q0 C1 + C2

After the first capacitor is disconnected from the battery, the total charge must remain constant. The voltage across each capacitor must be the same when they are connected together, since each capacitor plate is connected to a corresponding plate on the other capacitor by a constant-potential connecting wire. Use the total charge and the final potential difference to find the value of the second capacitor. QTotal = C1V1 Q1 = C1Vfinal Q2 = C2Vfinal initial

QTotal = Q1 final

final

+ Q2

final

= ( C1 + C2 )Vfinal → C1V1

final

= ( C1 + C2 )Vfinal →

initial

 V1   185 V  initial  C2 = C1 − 1  = ( 7.7  10−6 F )  − 1  = 8.7  10 −5 F = 87  F  Vfinal   15 V    7. The work to move the charge between the capacitor plates is W = qV , where V is the voltage difference between the plates, assuming that q Q so that the charge on the capacitor does not change appreciably. The charge is then found from Eq. 24–1. 15  10−6 F ( 24 J ) Q CW  = = 1.2 C W = qV = q   → Q = q 0.30  10−3 C C The assumption that q Q is justified. If it were not, then we would have to do an integral formulation of the process.

(

)

(a) The total charge on the combination of capacitors is the sum of the charges on the two individual capacitors, since there is no battery connected to them to supply additional charge, and there is no neutralization of charge by combining positive and negative charges. The voltage across each capacitor must be the same after they are connected, since each capacitor plate is connected to a corresponding plate on the other capacitor by a constant-potential

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connecting wire. Use the total charge and the fact of equal potentials to find the charge on each capacitor and the common potential difference. Q1 = C1V1 Q2 = C2V2 Q1 = C1Vfinal Q2 = C2Vfinal initial

initial

QTotal = Q1

initial

+ Q2

initial

+ Q2

= C1V1

final

+ C2V2

initial

= C1Vfinal + C2Vfinal →

initial

( 2.70  10 F ) ( 435 V ) + ( 4.00  10 F ) ( 525 V ) ( 6.70  10 F ) −6

initial

final

+ C2V2

C1V1 Vfinal =

= Q1

initial

C1 + C2

−6

= 488.73V  489 V = V1 = V2

(

)

(

)

= C1Vfinal = 2.70  10−6 F ( 488.73V ) = 1.32  10−3 C

Q1 final

= C2Vfinal = 4.00  10−6 F ( 488.73V ) = 1.95  10−3 C

Q2 final

(b) By connecting plates of opposite charge, the total charge will be the difference of the charges on the two individual capacitors. Once the charges have equalized, the two capacitors will again be at the same potential. Q1 = C1V1 Q2 = C2V2 Q1 = C1Vfinal Q2 = C2Vfinal initial

initial

QTotal = Q1

initial

− Q2

initial

− C2V2

C1V1 Vfinal =

= Q1

initial

C1 + C2

initial

final

+ Q2

→

final

C1V1

final

initial

− C2V2

= C1Vfinal + C2Vfinal →

initial

( 2.70  10 F ) ( 435 V ) − ( 4.00  10 F ) (525 V ) = ( 6.70  10 F ) −6

−6

= 138.13 V  140 V = V1 = V2 final

final

Use Eq. 24–2.

C = 0

)

(

)

= C2Vfinal = 4.00  10−6 F (138.13 V ) = 5.5  10 −4 C

(

= C1Vfinal = 2.70  10−6 F (138.13 V ) = 3.7  10 −4 C

A d

( 0.40  10 F )( 3.4  10 m ) = 153.7 m  150 m → A= =  (8.85  10 C N m ) Cd

−6

−3

−12

If the capacitor plates were square, they would be about 12.4 m on a side. 10. The capacitance per unit length of a coaxial cable is derived in Example 24–2. 2 8.85  10−12 C 2 N m 2 2 0 2 0 C = = = = 3.5  10−11 F m l ln ( Routside Rinside ) ln ( d outside d inside ) ln ( 5.0 mm 1.0 mm )

(

)

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11. Inserting the potential at the surface of a spherical conductor into Eq. 24–1 gives the capacitance of a conducting sphere. Then inserting the radius of the Earth as the radius of the capacitor, as in Example 24–3, yields the Earth’s capacitance. Q Q C= = = 4 0 r = 4 ( 8.85  10 −12 F/m )( 6.38  106 m ) = 7.10  10 −4 F V ( Q 4 0 r ) 12. Combine Eqs. 24–1, 24–2, and 23–4b to find the area.

Q = CV =

 0 AV d

=  0 AE → A =

0E

( 4.2  10 C ) = 0.26 m 1800 V   (8.85  10 C N  m )  10 m  −6

−12

−3

13. Use Eq. 24–2 for the capacitance. −12 2 2 −4 2 0 A  0 A 8.85  10 C N  m 1.0  10 m C= → d= = = 9  10−16 m d C (1F )

(

)(

)

No , this is not practically achievable. The gap would have to be smaller than the radius of a proton. 14. From the symmetry of the charge distribution, any electric field must be radial, away from the cylinder axis, and its magnitude must be independent of the location around the axis (for a given radial location). We assume the cylinders have charge of magnitude Q in a length l. Choose a Gaussian cylinder of length d and radius R, centered on the capacitor’s axis, with d l and the Gaussian cylinder far away from both ends of the capacitor. On the ends of this cylinder, E ⊥ dA and so there is no flux through the ends. On the curved side of the cylinder, the field has a constant magnitude and E dA. Thus E dA = EdA. Write Gauss’s law. Q Acurved = E ( 2 Rd ) = encl  E dA = Ecurved 0 walls walls For R  Rb , Qencl = 0 → E ( 2 Rd )  0 = 0 → E = 0. For R  Ra , Qencl =

Q −Q d+ d = 0 , and so Qencl = 0 → E ( 2 Rd )  0 = 0 → E = 0. l l

15. We assume there is a uniform electric field between the capacitor plates, so that V = Ed , and then use Eqs. 24–1 and 24–2.  AV =  0 AE = 8.85  10−12 C2 N  m 2 38.0  10−4 m 2 8.50  105 V m Q = CV = 0 d

(

)(

)

= 2.86  10−8 C 16. We assume there is a uniform electric field between the capacitor plates, so that V = Ed , and then use Eqs. 24–1 and 24–2. A Qmax = CVmax =  0 ( Emax d ) =  0 AEmax = 8.85  10−12 F/m 8.2  10−4 m 2 3.0  106 V m d

(

)(

)

= 2.2  10−8 C © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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17. We assume there is a uniform electric field between the capacitor plates, so that V = Ed , and then use Eq. 24–1. 84  10−6 C Q Q = CV = CEd → E = = = 5.3  104 V m −6 −3 0.80  10 F 2.0  10 m Cd

(

)(

)

18. (a) We assume the capacitor is charged with charge +Q on the positive plate, and –Q in the negative plate. The uncharged conducting sheet will polarize so that negative charge will be drawn towards the positive capacitor plate, and positive charge will be drawn towards the negative capacitor plate. The same charge will be on each face of the sheet as on the original capacitor plates, and there will be no electric field in the sheet. The same electric field will be in the gaps as before the plate was inserted. Use that electric field to determine the potential difference between the two original plates, and the new capacitance. Let x be the distance from one original plate to the nearest face of the sheet, and so d − l − x is the distance from the other original plate to the other face of the sheet. Q (d − l − x)  Q Qx = E= ; Vconductor = E ( d − l − x ) = ; V(+) plate = Ex =  0 A 0 A 0 A 0 relattive to relative to conductor

( − ) plate

V( + )plate =

Q (d − l − x) A 0

relative to ( − ) plate

Qx A 0

Q (d − l ) A 0

Q C

→ C = 0

(d − l )

This formula can also be derived by considering this arrangement as two capacitors in series. A (b) Cinitial =  0

A d

; Cfinal =  0

(d − l )

;

Cfinal Cinitial

0

(d − l ) 0

d d −l

d d − 0.40d

1 0.60

= 1.7

d 19. (a) The distance between plates is obtained from Eq. 24–2.  A  A C= 0 → x= 0 x C Inserting the maximum capacitance gives the minimum plate separation and the minimum capacitance gives the maximum plate separation. −6 2  0 A ( 8.85 pF/m ) 25  10 m = = 0.22  m xmin = 1000.0  10−12 F Cmax

(

xmax =

o A Cmin

)

(8.85 pF/m ) ( 25  10−6 m 2 ) 1.0 pF

= 0.22 mm = 220 m

So 0.22  m  x  220  m . (b) Differentiating the distance equation gives the approximate uncertainty in distance. 0 A dx d 0 A  x  C =  = − C . C dC dC  C  C2 The minus sign indicates that the capacitance increases as the plate separation decreases. Since only the magnitude is desired, the minus sign can be dropped. The uncertainty is finally written in terms of the plate separation using Eq. 24–2. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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x 

0 A

 0 A     x 

x 2 C

C = 2

0 A

(c) The percent uncertainty in distance is obtained by dividing the uncertainty by the separation distance. xmin x C ( 0.22  m )( 0.1pF )(100% )  100% = min  100% = = 0.01% o A xmin (8.85 pF/m ) 25 mm 2

(

xmax xmax

 100% =

xmax C

o A

 100% =

)

( 0.22 mm )( 0.1pF )(100% ) = 10% (8.85 pF/m ) ( 25 mm 2 )

20. The goal is to have an electric field of strength E S at a radial distance of 5.0 Rb from the center of the cylinder. Knowing the electric field at a specific distance allows us to calculate the linear charge density on the inner cylinder. From the linear charge density and the capacitance we can find the potential difference needed to create the field. From the cylindrically symmetric geometry and 1  Gauss’s law, the field in between the cylinders is given by E = . The capacitance of a 2 0 R cylindrical capacitor is given in Example 24–2.  Q 1 = ES →  = 2 0 ( 5.0 Rb ) ES = E ( R = 5.0 Rb ) = l 2 0 ( 5.0 Rb )

Q = CV → V =

Q C

2  l

Q ln ( Ra Rb ) l

ln ( Ra Rb )

2 

(

=  2 0 ( 5.0 Rb ) ES 

ln ( Ra Rb ) 2 

m  )  1.00.100   10 m

)(

= ( 5.0 Rb ) ES ln ( Ra Rb ) = 5.0 1.0  10 −4 m  3.0  106 N C ln 



−4



= 10362V  1.0  104 V 21. To reduce the net capacitance, another capacitor must be added in series. 1 1 1 1 1 1 C1 − Ceq = + → = − = → Ceq C1 C2 C2 Ceq C1 C1Ceq

C2 =

C1Ceq C1 − Ceq

( 3.4  10 F )(1.6  10 F ) = 3.02  10 F  3.0  10 pF ( 3.4  10 F ) − (1.6  10 F ) −9

−9

Yes, an existing connection needs to be broken in the process. One of the connections of the original capacitor to the circuit must be disconnected in order to connect the additional capacitor in series. 22. (a) Capacitors in parallel add according to Eq. 24–3.

(

)

Ceq = C1 + C2 + C3 + C4 + C5 + C6 = 6 3.2  10−6 F = 1.92  10−5 F = 19.2 F

(b) Capacitors in series add according to Eq. 24–4. −1

−1

−6 1 1 1 1 1 1  6   = 3.2  10 F = 5.3  10−7 F Ceq =  + + + + +  =  −6 6  3.2  10 F   C1 C2 C3 C4 C5 C6  = 0.53  F

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23. The capacitors are in parallel, and so the potential is the same for each capacitor, and the total charge on the capacitors is the sum of the individual charges. We use Eqs. 24–1 and 24–2. A1

Q1 = C1V =  0

V ; Q2 = C2V =  0

Qtotal = Q1 + Q2 + Q3 =  0

A1 d1

V + 0

A2 d2

V ; Q3 = C3V =  0

V + 0

A3 d3





V =  0

+ 0

A2 d2

+ 0

A3 

V

d3 

 A1 A A   0 +  0 2 +  0 3 V  d1 d2 d3   A Q A A  Cnet = total =  =   0 1 +  0 2 +  0 3  = C1 + C2 + C3 V V d2 d3   d1 24. Capacitors in parallel add linearly, and so adding a capacitor in parallel will increase the net capacitance without removing the 5.0  F capacitor. 5.0  F + C = 22  F → C = 17  F connected in parallel

25. (a) The two capacitors are in parallel . Both capacitors have their high-voltage plates at the same potential (the middle plate), and both capacitors have their low-voltage plates at the same potential (the outer plates, which are connected). (b) The capacitance of two capacitors in parallel is the sum of the individual capacitances. C = C1 + C2 =

 d + d2  1 1 0 A 0 A + = 0 A +  = 0 A 1  d1 d2  d1 d 2   d1d 2 

 0 Al  0 Al . We see that C →  as d1 → 0 or = d1d 2 d1 ( l − d1 )

d1 → l (which is d 2 → 0 ). Of course, a real capacitor would break down as the plates got too close to each other. To find the minimum capacitance, set

dC d ( d1 )

= 0 and solve for d1.

  0 Al  ( l − 2 d1 ) = − 0 Al = 0 → d1 = 12 l = d 2 2   2 2 d ( d1 )  d1 l − d1  ( d1 l − d1 ) d

   d1 + d 2   4  l 4 = 0 A = 0 A  = 0 A    1 1 l   d1d 2  d = l  d1 + d 2   ( 2 l )( 2 l ) 

Cmin =  0 A 

Cmin =

4 0 A d1 + d 2

1 2

; Cmax = 

26. (a) The two capacitors are in series . The middle plate is shared by both capacitors, and has no direct connection to the voltage. One of the capacitors has a plate at the higher voltage, and the other capacitor has a plate at the lower voltage. (b) The capacitance of two capacitors in series is given by Eq. 24–4.

1 C

1 C1

1 C2

0 A 0 A

d1 + d 2

0 A

→ C=

0 A d1 + d1

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(c) Moving the middle plate doesn’t change the sum of d1 + d 2 , and so moving the middle plate doesn’t vary the capacitance. It has no maximum or minimum other than it’s (constant) value. 27. The maximum capacitance is found by connecting the capacitors in parallel. Cmax = C1 + C2 + C3 = 3.1  10−9 F + 5.8  10 −9 F + 1.00  10 −8 F = 1.89  10 −8 F in parallel

The minimum capacitance is found by connecting the capacitors in series. −1

−1

1 1 1  1 1 1  = 1.68  10−9 F in series Cmin =  + +  = + +  −9 −9 −8  C1 C2 C3   3.1  10 F 5.8  10 F 1.00  10 F  28. (a) From the diagram, we see that C1 and C2 are in series. That combination is in parallel with C3, and then that combination is in series with C4. Use those combinations to find the equivalent capacitance. We use subscripts to indicate which capacitors have been combined. 1 1 1 = + → C12 = 12 C ; C123 = C12 + C3 = 12 C + C = 23 C ; C12 C C

1 C1234

1 C123

1 C4

2 3C

1 C

5 3C

→ C1234 = 53 C

(b) The charge on the equivalent capacitor C1234 is given by Q1234 = C1234V = 53 CV . This is the charge on both of the series components of C1234 ( C123 and C4 ) .

Q123 = 53 CV = C123V123 = 23 CV123 → V123 = 25 V Q4 = 53 CV = C4V4 → V4 = 53 V The voltage across the equivalent capacitor C123 is the voltage across both of its parallel components. Note that the sum of the charges across the two parallel components of C123 is the same as the total charge on the two components, 53 CV .

V123 = 25 V = V12 ; Q12 = C12V12 = ( 12 C ) ( 25 V ) = 15 CV V123 = 25 V = V3 ; Q3 = C3V3 = C ( 25 V ) = 25 CV

Finally, the charge on the equivalent capacitor C12 is the charge on both of the series components of C12 .

Q12 = 15 CV = Q1 = C1V1 → V1 = 15 V ; Q12 = 15 CV = Q2 = C1V2 → V2 = 15 V Here are all the results, gathered together. Q1 = Q2 = 15 CV ; Q3 = 25 CV ; Q4 = 53 CV V1 = V2 = 15 V ; V3 = 25 V ; V4 = 53 V 29. C1 and C2 are in series, so they both have the same charge. We then use that charge to find the voltage across each of C1 and C2. Then their combined voltage is the voltage across C3. The voltage across C3 is used to find the charge on C3. Q 14.8 C Q 14.8 C = 0.925 V ; V2 = 2 = = 0.925 V Q1 = Q2 = 14.8 C ; V1 = 1 = C1 16.0 F C2 16.0 F

V3 = V1 + V2 = 1.850 V ; Q3 = C3V3 = (16.0 F )(1.85 V ) = 29.6 C From the diagram, C4 must have the same charge as the sum of the charges on C1 and C3. Then the voltage across the entire combination is the sum of the voltages across C4 and C3. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

837

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Q4 = Q1 + Q3 = 14.8C + 29.6C = 44.4 C ; V4 =

Q4 C4

Instructor Solutions Manual

44.4 C 28.5 F

= 1.56 V

Vab = V4 + V3 = 1.56 V + 1.850 V = 3.41V Here is a summary of all results. Q1 = Q2 = 14.8 C ; Q3 = 29.6 C ; Q4 = 44.4  C

V1 = V2 = 0.925V ; V3 = 1.850 V ; V4 = 1.56 V ; Vab = 3.41V 30. When the capacitors are connected in series, they each have the same charge as the net capacitance. −1

−1

1 1  1 1   9.0 V + (a) Q1 = Q2 = Qeq = CeqV =  +  V =  )  ( −6 −6 C C 0.50 10 F 0.80 10 F      1 2  = 2.769  10−6 C V1 =

Q1 C1

2.769  10−6 C 0.50  10−6 F

= 5.538 V  5.5V

V2 =

Q2 C2

2.769  10−6 C

= 3.461V  3.5V

0.80  10−6 F

(b) Q1 = Q2 = Qeq = 2.769  10−6 C  2.8  10 −6 C When the capacitors are connected in parallel, they each have the full potential difference. (c) V1 = 9.0 V

(

)

Q1 = C1V1 = 0.50  10−6 F ( 9.0 V ) = 4.5  10−6 C

V2 = 9.0 V

)

Q2 = C2V2 = 0.80  10−6 F ( 9.0 V ) = 7.2  10−6 C

31. When the switch is down the initial charge on C2 is calculated from Eq. 24–1. Q2 = C2V0 When the switch is moved up, charge will flow from C2 to C1 until the voltage across the two capacitors is equal. Q Q C V = 2 = 1 → Q2 = Q1 2 C2 C1 C1 The sum of the charges on the two capacitors is equal to the initial charge on C2.  C + C1  C Q2 = Q2 + Q1 = Q1 2 + Q1 = Q1  2  C1  C1  Inserting the initial charge in terms of the initial voltage gives the final charges.

 C2 + C1   = C2V0  C1 

Q1 

→

Q1 =

C1C2 C1 + C2

V0 ; Q2 = Q1

C2 C1

C22 C1 + C2

32. We want a small voltage drop across C1. Since V = Q C , if we put the smallest capacitor in series with the battery, there will be a large voltage drop across it. Then put the two larger capacitors in parallel, so that their equivalent capacitance is large and therefore will have a small voltage drop across them. So put C1 and C3 in parallel with each other, and then put that combination in series with C2. See the diagram. To calculate the voltage across C1, find the equivalent capacitance and the net charge. That charge is used to find the

C3 V0

C1 C2

838

Chapter 24

Capacitance, Dielectrics, Electric Energy Storage

voltage drop across C2, and then that voltage is subtracted from the battery voltage to find the voltage across the parallel combination. Q C ( C + C3 ) 1 1 1 C + C 2 + C3 Q ; Qeq = CeqV0 ; V2 = 2 = eq ; = + = 1 → Ceq = 2 1 Ceq C2 C1 + C3 C2 ( C1 + C3 ) C1 + C2 + C3 C2 C2

C2 ( C1 + C3 ) V1 = V0 − V2 = V0 −

Qeq

= V0 −

CeqV0 C2

= V0 −

C1 + C2 + C3

V0 =

C2 C1 + C2 + C3

V0 =

1.5 F 6.5 F

(12 V )

= 2.8 V 33. The energy stored by a capacitor is given by Eq. 24–5, U = 12 CV 2.

U final = 5.0U initial →

1 2

2 2 CfinalVfinal = ( 5.0) 12 CinitialVinitial

One simple way to accomplish this is to have Cfinal = 5.0 Cinitial and Vfinal = Vinitial. In order to keep the voltage the same for both configurations, any additional capacitors must be connected in parallel to the original capacitor. In order to multiply the capacitance by a factor of 5.0, we recognize that capacitors added in parallel add linearly. Thus, if a capacitor of value 4.0 C = 1000 pF = 1.00  10 pF 3

were connected in parallel to the original capacitor, the final capacitance would be 5.0 times the original capacitance with the same voltage, and the stored energy would increase by a factor of 5.0. 34. (a) The voltage across C3 and C4 must be the same, since they are in parallel. 16  F Q3 Q4 C V3 = V4 → = → Q4 = Q3 4 = ( 21 C ) = 42  C 8 F C3 C 4 C3 The parallel combination of C3 and C4 is in series with the parallel combination of C1 and C2, and so Q3 + Q4 = Q1 + Q2 . That total charge then divides between C1 and C2 in such a way that

V1 = V2 . Q1 + Q2 = Q3 + Q4 = 63 C ; V1 = V2 → C1

Q4 C4

63 C − Q1 C4

→

8.0  F

( 63 C ) = 21 C ; Q2 = 63 C − 21 C = 42  C C4 + C1 24.0 F Notice the symmetry in the capacitances and the charges. Q1 =

( 63 C ) =

(b) Use Eq. 24–1. Q 21 C = 2.625 V  2.6 V ; V2 = V1 = 2.6 V V1 = 1 = C1 8.0  F

V3 =

Q3 C3

21 C 8.0 F

= 2.625 V  2.6 V ; V4 = V3 = 2.6 V

839

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

35. Since there is no voltage drop between points a and b, we can imagine a there being a connecting wire between points a and b. Then capacitors C1 C1 and C2 are in parallel, and so have the same voltage. Also capacitors C3 and Cx are in parallel, and so have the same voltage. c V Q1 Q2 Q3 Q x V1 = V2 → = = ; V3 = Vx → C1 C2 C3 C x Since no charge flows through the voltmeter, we could also remove it C2 b from the circuit and have no change in the circuit. In that case, capacitors C1 and Cx are in series and so have the same charge. V0 Likewise capacitors C2 and C3 are in series, and so have the same charge. Q1 = Q x ; Q2 = Q3 Solve this system of equations for Cx. Q3 Qx Q Q C  8.9  F  = → C x = C3 x = C3 1 = C3 1 = ( 4.8 F )   = 2.67  F  2.7  F C3 C x Q3 Q2 C2  16.0 F  36. We have CP = C1 + C2 and 1

CS 1

C1 =

1 C1

1 C2

1 C1

1 C P − C1

1 C2

Cx d

. Solve for C1 and C 2 in terms of C P and CS .

CP ( CP − C1 ) + C1 = C1 ( C P − C1 ) C1 ( C P − C1 )

→

→ C12 − CPC1 + CPCS = 0 →

C1 ( CP − C1 )

CP  CP2 − 4CPCS

2 = 25.9  F, 7.0 F

( 32.9  F ) 2 − 4 ( 32.9  F )( 5.5 F )

32.9  F 

C2 = CP − C1 = 32.9  F − 25.9  F = 7.0  F or 32.9  F − 7.0  F = 25.9  F

So the two values are 25.9  F and 7.0 F . 37. For an infinitesimal area element of the capacitance a distance y up from the small end, the distance between the plates is d + x = d + y tan   d + y  . Since the capacitor plates are square, they are of dimension

A  A , and the

area of the infinitesimal strip is dA = A dy. The infinitesimal capacitance dC of the strip is calculated, and then the total capacitance is found by adding together all of the infinitesimal capacitances, in parallel with each other.

C = 0

A d

→ dC =  0

dA d + y

= 0

A dy d + y

 A = 0 C =  dC =   0 ln ( d + y  )  d + y 0 0 A

A dy

0 A  A  d + A  0 A   A ln d +  A − ln d  = 0 ln  ln  1 + =       d d    

(

)

840

Chapter 24

Capacitance, Dielectrics, Electric Energy Storage

We use the approximation from Page A–1 that ln (1 + x )  x − 12 x 2 . 2  0 A   A   0 A  A 1   A    0 A   A  − 2 C= ln  1 +  =  = 1 −     d d  d   d  2d     

38. The stored energy is given by Eq. 24–5.

(

)

U = 12 CV 2 = 12 2.8  10−9 F ( 720 V ) = 7.3  10−4 J 2

39. The energy density is given by Eq. 24–6.

(

)

u = 12  0 E 2 = 12 8.85  10−12 C2 N m2 (150 V m ) = 1.0  10−7 J m3 2

40. Use Eq. 24–5 to find the capacitance. 2 (1200 J ) 2U U = 12 CV 2 → C = 2 = = 9.6  10−5 F 2 3 V 5.0  10 V

(

)

41. The energy stored is obtained from Eq. 24–5, with the capacitance from Eq. 24–2.

( 4.2  10 C ) ( 0.0015 m ) = = = 2.3  10 J U= 2C 2 A 2 ( 8.85  10 C N m ) ( 0.080 m ) Q2

−4

Q 2d

−12

42. (a) The charge is constant, and the halving of separation increases the capacitance by a factor of 2. Q2 A



0 U 2 2C2 C1 d = 1 = 2 = = 2 Q U1 C2  A 0 1 2C1 2d (b) The work done is the change in energy stored in the capacitor. The work done to move the plates together will be negative work, since the plates will naturally attract each other.

U 2 − U1 = 12 U1 − U1 = − 12 U1 = − 12

43. (a) Use Eqs. 24–3 and 24–5.

Q2 2C1

=−

(

Q2 4 0

= −

Q 2d 4 0 A

)

U parallel = 12 CeqV 2 = 12 ( C1 + C2 )V 2 = 12 0.65  10−6 F ( 24 V ) = 1.872  10−4 J  1.9  10−4 J 2

(b) Use Eqs. 24–4 and 24–5.

−6 −6  C1C2  2 1  ( 0.45  10 F )( 0.20  10 F )  2 U series = CeqV =  V = 2 24 V ) (   − 6   0.65  10 F  C1 + C2    1 2

1 2

= 3.988  10−5 J  4.0  10−5 J

841

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

U = 12 QV → Q = Qseries =

(

2U V −5

2 3.988  10 J 24 V

→ Qparallel =

(

2 1.872  10−4 J 24 V

Instructor Solutions Manual

) = 1.6  10 C −5

) = 3.3  10 C −6

44. The capacitance of a cylindrical capacitor is given in Example 24–2 as C = (a) If the charge is constant, the energy can be calculated by U = 12

Q2 C

2 0 l

ln ( Ra Rb )

2 0 l Q2 ln ( Ra Rb ) ln ( 3Ra Rb ) U2 C2 C1 = = = = 1 2 2 0 l ln ( Ra Rb ) U1 1 Q C2 2 ln ( 3Ra Rb ) C1 The energy comes from the work required to separate the capacitor components. (b) If the voltage is constant, the energy can be calculated by U = 12 CV 2 . 1 2

2 0 l

U2 U1

= 2

1 2

C2V

C1V

C2 C1

ln ( 3Ra Rb ) ln ( Ra Rb ) = 1 2 0 l ln ( 3Ra Rb ) ln ( Ra Rb )

Since the voltage remained constant, and the capacitance decreased, the amount of charge on the capacitor components decreased. Charge flowed back into the battery that was maintaining the constant voltage. 45. (a) Before the capacitors are connected, the only stored energy is in the initially-charged capacitor. Use Eq. 24–5.

(

)

U1 = 12 C1V02 = 12 3.70  10−6 F (16.0 V ) = 4.736  10−4 J  4.74  10−4 J 2

(b) The total charge available is the charge on the initial capacitor. The capacitance changes to the equivalent capacitance of the two capacitors in parallel.

( 3.70  10 F ) (16.0 V ) Q = Q = CV ; C = C + C ; U = = = C C +C (8.70  10 F ) 1

1 0

1 2

−6

C12V02 1

1 2

−6

= 2.014  10−4 J  2.01  10−4 J

(c)

U = U 2 − U1 = 2.014  10−4 J − 4.736  10−4 J = −2.72  10 −4 J

46. (a) With the plate inserted, the capacitance is that of two series capacitors of plate separations d1 = x and d 2 = d − l − x. −1

 x d −x−l   A + = 0 Ci =   0 A  d −l 0 A With the plate removed the capacitance is obtained directly from Eq. 24–2: C f =

0 A d

842

Chapter 24

Capacitance, Dielectrics, Electric Energy Storage

Since the voltage remains constant the energy of the capacitor will be given by Eq. 24–5 written in terms of voltage and capacitance. The work will be the change in energy as the plate is removed. W = U f − U i = 12 ( Cf − Ci ) V02

 0 Al V02  0 A 0 A  2 V − = −  0 2d ( d − l )  d d −l 

= 12 

The net work done is negative. Although the person pulling the plate out must do work, charge is returned to the battery, resulting in a net negative work done. (b) Since the charge now remains constant, the energy of the capacitor will be given by Eq. 24–5 written in terms of capacitance and charge. Q2  1 1  Q2  d d − l  Q2l − = − W=  =   2  Cf Ci  2   0 A  0 A  2 0 A 2

 0 A V  l  0 0 A  0 AV02 l d −l   V0 and so W = . = The original charge is Q = CV0 = 2 d −l 2 0 A 2 (d − l ) 47. In both configurations, the voltage across the combination of capacitors is the same. So use U = 12 CV  .

U P = 12 CPV  = 12 ( C1 + C2 ) V 2 ; U S = 12 CSV  = 12 U P = 6.0 U S →

1 2

( C1 + C2 ) V 2 = 6.0 ( 12 )

C12 − 4.0C1C2 + C22 = 0 → C1 = C1 C2

C1C2

( C1 + C2 )

C1C2

( C1 + C2 )

V 2 → ( C1 + C2 ) = 6.0C1C2 → 2

4.0C2  16C22 − 4C22 2

= C2

4.0  12 2

(

= C2 2.0  3

)→

= 2 + 3 or 2 − 3 = 3.7 or 0.27

48. (a) The energy density is given by Eq. 24–6.

(

)

u = 12  0 E 2 = 12 8.85  10−12 C2 N m2 (150 V m ) = 9.956  10−8 J m3  1.0  10−7 J m3 2

(b) The energy stored is equal to the energy density times the volume under consideration. Since the thickness of air is so much less that the radius, we can approximate the volume as the surface area of the Earth times the thickness of the air.

(

)

(

) (

2 U air = u 4 REarth r = 9.956  10−8 J m3 4 6.38  106 m

) (10 m ) = 5.092  10 J 2

 5  108 J (c) The output from the power station is the power generated times the time duration. U station = Pt = 2  109 J s 8.64  103 s = 1.728  1014 J

(

U station

)(

1.728  1014 J

)

= 3.39  105  3  105

U air 5.092  10 J The output from the power station is about 300,000 more than the electrical potential energy stored in the first 10 m of the atmosphere. 8

843

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

49. (a) The charge remains constant, so we express the stored energy as U = 12

= 12

Q2 x

, where x C 0 A is the separation of the plates. The work required to increase the separation by dx is dW = Fdx, where F is the force on one plate exerted by the other plate. That work results in an increase in potential energy, dU . dW = Fdx = dU = 12

Q 2 dx

0 A

(b) We cannot use F = QE = Q

→

1 Q2 2 0 A

 Q Q2 =Q = because the electric field is due to both plates, 0 0 A 0 A

and charge cannot put a force on itself by the field it creates. By the symmetry of the geometry, the electric field at one plate, due to just the other plate, is 12 E . See Example 24–10. 50. The capacitance per unit length is derived in Example 24–2. 2 8.85  10 −12 C 2 N m 2 2 0 C = = = 2.858  10−11 F m l ln ( Router Rinner ) ln ( 3.5 mm 0.50 mm )

(

)

(a) The capacitance of a 1-meter length is 2.858  10−11 F.

(

)

U = 12 CV  = 12 2.858 10−11 F ( 625 V ) = 5.582 10−6 J  5.6 10−6 J 2

(b) Use the capacitance per meter to find the charge per meter. Q = CV →

= =

(

)

V = 2.858  10−11 F m ( 625 V ) = 1.786  10−8 C m  1.8  10−8 C m

l l (c) Gauss’ law is used in Example 24–2 to find the electric field between the cylinders. 1.786  10−8 C m  E= = = 6.424  105 V m  6.4  105 V m 2 0 R 2 8.85  10 −12 C 2 N m 2 0.50  10 −3 m

(

)(

)

The electric field points from higher potential to lower potential, so this electric field points radially outward . 51. The energy density u is a function of the electric field, as given in Eq. 24–6, and the electric field as found in Example 24–2 is a function of the radial distance from the center of the wire. Break the volume of the cylindrical capacitor into cylindrical shells, of length l , radius R, and thickness dR. The adjacent figure is a view along the center of the capacitor, so the length is not shown. The shaded region has an infinitesimal volume of dV = 2 Rl dR, and so the infinitesimal energy in that infinitesimal volume is dU = udV . Integrate the infinitesimal volume between the two radial limits. The value of the linear charge density is taken from the solution to Problem 50.

U l

Router

l R

inner

u dV =

Router

l R

1 2

Router dR

R Rinner

 0 E 2 2 R l dR

inner

844

Chapter 24

Capacitance, Dielectrics, Electric Energy Storage

Router

= 

1 2

Rinner

   R  2 2 1 0  ln  outer    dR =  2 RdR =  4 0 R  R  4 0  Rinner   2 0 R  Router

inner

(

)(

)

 0.35 mm  = 5.579 10 −6 J  5.6 10 −6 J m   0.05 mm 

= 8.998  109 N m 2 C 2 1.786  10 −8 C m ln 

52. (a) The electric field outside the spherical conductor is that of an equivalent point charge at the 1 Q center of the sphere, so E = , r  r0 . Consider a differential volume of radius dr, and 4 0 r 2 volume dV = 4 r 2 dr , as used in Example 22–5. The energy in that volume is dU = udV . Integrate over the region outside the conductor. 2



  1 Q Q2 1 Q2 1 2  = = − U =  dU =  udV =  0  E dV =  0   4 r dr dr  4 0 r 2  8 0 r r 2 8 0 r r r  

1 2

Q2 8 0 r0

(b) Use Eq. 24–5 with the capacitance of an isolated sphere, from the text immediately after Example 24–3. U = 12

Q2 C

= 12

Q2 4 0 r0

Q2 8 0 r0

(c) When there is a charge q < Q on the sphere, the potential of the sphere is V = work required to add a charge dq to the sphere is then dW = Vdq =

4 0 r0

increases the potential energy by the same amount, so dU = dW = Vdq =

4 0 r0

dq =

 qdq =

4 0 r0 0

. The

dq. That work 1

4 0 r0 the entire charge from 0 to Q, calculating the energy as the charge increases. U =  dU =  dW =  Vdq = 

dq. Build up

Q2 8 0 r0

53. Use Eq. 24–8 to calculate the capacitance with a dielectric.

C = K 0

A d

(

= ( 7 ) 8.85  10

−12

C Nm 2

)

 ( 5.0  10−2 m )

= 2.11  10 −10 F  2.1  10 −10 F

( 2.3  10 m ) −3

54. Use Eq. 24–8 to calculate the capacitance with a dielectric.

( 4.7  10 m ) = 2.4  10 F = ( 2.2 ) ( 8.85  10 C N  m ) C = K d (1.8  10 m ) A

−2

−12

−11

−3

845

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

55. The change in energy of the capacitor is obtained from Eq. 24–5 in terms of the constant voltage and the capacitance. U = U f − U i = 12 C0V 2 − 12 KC0V 2 = − 12 ( K − 1) C0V 2 The work done by the battery in maintaining a constant voltage is equal to the voltage multiplied by the change in charge, with the charge given by Eq. 24–1. Wbattery = V (Qf − Qi ) = V ( C0V − KC0V ) = − ( K − 1) C0V 2 The work done in pulling the dielectric out of the capacitor is equal to the difference between the change in energy of the capacitor and the energy done by the battery. W = U − Wbattery = − 12 ( K − 1) C0V 2 + ( K − 1) C0V 2

(

)

= 12 ( K − 1) C0V 2 = ( 3.4 − 1) 8.8  10 −9 F (100 V ) = 1.1  10 −4 J 2

56. We assume the charge and dimensions are the same as in Problem 41. Use Eq. 24–5 with charge and capacitance.

1 2

( 420  10 C ) ( 0.0015 m ) = = = = 333.7 J  330 J C KC K A ( 7 ) (8.85  10 C N  m )( 64  10 m )

1 2

−6

Q 2d

1 2

−12

−4

57. The initial charge on the capacitor is Qinitial = CinitialV . When the mica is inserted, the capacitance changes to Cfinal = KCinitial , and the voltage is unchanged since the capacitor is connected to the same battery. The final charge on the capacitor is Qfinal = CfinalV .

(

)

Q = Qfinal − Qinitial = CfinalV − CinitialV = ( K − 1) CinitialV = ( 7 − 1) 3.5  10 −9 F (18 V ) = 3.78  10−7 C  3.8  10 −7 C 58. The potential difference is the same on each half of the capacitor, so it can be treated as two capacitors in parallel. Each parallel capacitor has half of the total area of the original capacitor. 1 1 A A A C = C1 + C2 = K1 0 2 + K 2 0 2 = 12 ( K1 + K 2 )  0 d d d

59. The intermediate potential at the boundary of the two dielectrics can be treated as the “low” potential plate of one half and the “high” potential plate of the other half, so we treat it as two capacitors in series. Each series capacitor has half of the inter-plate distance of the original capacitor.

1 C

1 C1

1 C2

1 2

K1 0 A

1 2

K 2 0 A

K1 + K 2

2 0 A K1 K 2

→ C=

2 0 A K1K 2 d

K1 + K 2

60. The capacitor can be treated as two series capacitors with the same areas, but different plate separations and dielectrics. Substituting Eq. 24–8 into Eq. 24–4 gives the effective capacitance. −1

−1

 d1  1 1  d2  A 0 K1K 2 C = + = + =   d 1 K 2 + d 2 K1  C1 C2   K1 A 0 K 2 A 0 

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61. (a) Since the capacitors each have the same charge and the same voltage in the initial situation, Q each has the same capacitance of C = 0 . When the dielectric is inserted, the total charge of V0

2Q0 will not change, but the charge will no longer be divided equally between the two

capacitors. Some charge will move from the capacitor without the dielectric ( C1 ) to the capacitor with the dielectric ( C2 ) . Since the capacitors are in parallel, their voltages will be the same.

V1 = V2 → Q1 = (b) V1 = V2 =

( K + 1) Q1 C1

Q1 C1

Q0 =

0.43Q0 Q0 V0

→

C2 2 4.6

2Q0 − Q1

→

Q0 = 0.43Q0 ; Q2 = 1.57Q0

= 0.43V0 =

Q2 C2

1.57Q0 3.6 Q0 V0

62. Since the capacitor is disconnected from the battery, the charge on it cannot change. The capacitance of the capacitor is increased by a factor of K, the dielectric constant. 1 C C Q = CinitialVinitial = CfinalVfinal → Vfinal = Vinitial initial = Vinitial initial = (18.0 V ) = 8.2 V 2.2 Cfinal KCinitial 63. The potential can be found from the field and the plate separation. Then the capacitance is found from Eq. 24–1, and the area from Eq. 24–8. V E= ; Q = CV = CEd → d

( 0.765  10 C ) = = 4.761  10 F  4.76  10 F C= Ed ( 8.24  10 V m )(1.95  10 m ) ( 4.761  10 F )(1.95  10 m ) = 0.280 m A Cd → A= = C = K d K ( 3.75) (8.85  10 C N  m ) −6

−9

−3

−9

−3

−12

64. (a) We treat this system as two capacitors, one with a dielectric, and one without a dielectric. Both capacitors have their high-voltage plates in contact and their low-voltage plates in contact, so they are in parallel. Use Eqs. 24–2 and 24–8 for the capacitance. Note that x is measured from the right edge of the capacitor, and is positive to the left in the diagram.

C = C1 + C2 =  0

l (l − x) d

+ K 0

lx d

= 0

x l2  1 + ( K − 1) d  l 

(b) Both “capacitors” have the same potential difference, so use U = 12 CV 2 .

U = 12 ( C1 + C2 )V02 =  0

x l2  1 + ( K − 1) V02  2d  l 

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(c) We must be careful here. When the voltage across a capacitor is constant and a dielectric is inserted, charge flows from the battery to the capacitor. So the battery will lose energy and the capacitor gain energy as the dielectric is inserted. As in Example 24–10, we assume that work is done by an external agent (Wnc ) in such a way that the dielectric has no kinetic energy. Then the work-energy principle (Chapter 8) can be expressed as Wnc = U or dWnc = dU . This is analogous to moving an object vertically at constant speed. To increase (decrease) the gravitational potential energy, positive (negative) work must be done by an outside, nongravitational source. In this problem, the potential energy of the voltage source and the potential energy of the capacitor both change as x changes. Also note that the change in charge stored on the capacitor is the opposite of the change in charge stored in the voltage supply. dWnc = dU = dU cap + dU battery → Fnc dx = d 12 CV02 + d ( QbatteryV0 ) →

(

Fnc = 12 V02

dC dx

+ V0

dQbattery dx

= 12 V02

dC dx

− V0

)

dQcap dx

= 12 V02

dC dx

− V02

dC dx

= − 12 V02

dC dx

l  ( K − 1)  V  l = − ( K − 1)  2d d  l Note that this force is in the opposite direction of dx, and so is to the right. Since this force is being applied to keep the dielectric from accelerating, there must be a force of equal magnitude to the left pulling on the dielectric. This force is due to the attraction of the charged plates and the induced charge on the dielectric. The magnitude and direction of this attractive force are = − 12 V02 0

V02 0 l 2d

2 0 0

( K − 1) , left .

65. (a) We consider the cylinder as two cylindrical capacitors in parallel. The two “negative plates” are the (connected) halves of the inner cylinder (half of which is in contact with liquid, and half of which is in contact with vapor). The two “positive plates” are the (connected) halves of the outer cylinder (half of which is in contact with liquid, and half of which is in contact with vapor). Schematically, it is like Fig. 24–30 in Problem 58. The capacitance of a cylindrical capacitor is given in Example 24–2. 2 0 K liq h 2 0 K V ( l − h ) 2 0 l  h  C = Cliq + CV = K liq − K V ) + K V = C → + = (   l ln ( Ra Rb ) ln ( Ra Rb ) ln ( Ra Rb ) 

h l

 C ln ( Ra Rb )  − KV   ( Kliq − K V )  2 0 l  1

(b) For the full tank, Full:

C= =

= 1, and for the empty tank,

l 2 0 l

h l

= 0.

 K − K h + K  = 2 0 l Kliq ( liq V ) l V  ln R R ln ( Ra Rb )  ( a b)  2 ( 8.85  10 −12 C 2 N  m 2 ) ( 2.0 m )(1.4 ) ln ( 5.0 mm 4.5 mm )

= 1.5  10 −9 F

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Empty:

 K − K h + K  = 2 0 l K V ( liq V ) l V  ln R R ln ( Ra Rb )  ( a b) 

2 0 l

(

)

2 8.85  10−12 C2 N  m 2 ( 2.0 m )(1.0 ) ln ( 5.0 mm 4.5 mm )

= 1.1  10−9 F

66. Find the energy in each region from the energy density and the volume. The energy density in the 2 “gap” is given by ugap = 12  0 Egap , and the energy density in the dielectric is given by uD = 12  D ED2 = 2

1 2

 Egap  1 Egap K 0   = 2  0 K , where Eq. 24–10 is used.  K  UD U total

U gap + U D l K

uD Vol D ugap Vol gap + uD Vol D

(d − l ) +

(d − l ) K + l

1 2

= 1 2

0

2 Egap

0E A(d − l ) + 0 2 gap

1 2

2 Egap

(1.00 mm ) = 0.222 (1.00 mm )( 3.50 ) + (1.00 mm )

67. Consider the dielectric as having a layer of equal and opposite charges at each side of the dielectric. Then the geometry is like three capacitors in series. One air gap is taken to be d1 , and then the other air gap is d − d1 − l . 1 C

1 C1

1 C2

1 C3

0 A

l K 0 A

d − d1 − l

0 A

1  l

   + ( d − l )  → 0 A   K 

(8.85  10 C N  m )( 2.50  10 m ) = 1.72  10 F = C=  l + d −l   1.00  10 m  ) + (1.00  10 m )  K (   3.50 −12

0 A

−2

−10

−3





68. By leaving the battery connected, the voltage will not change when the dielectric is inserted, but the amount of charge will change. That will also change the electric field. (a) Use Eq. 24–2 to find the capacitance. −2 2 A −12 2 2  2.50  10 m  = 1.106  10−10 F  1.11  10−10 F C0 =  0 = 8.85  10 C N  m   −3 d  2.00  10 m  (b) Use Eq. 24–1 to find the initial charge on each plate.

(

)

Q0 = C0V = 1.106  10−10 F (150 V ) = 1.659  10−8 C  1.66  10−8 C

In Example 24–14, the charge was constant, so it was simple to calculate the induced charge and then the electric fields from those charges. But now the voltage is constant, and so we calculate the fields first, and then calculate the charges. So we are solving the problem parts in a different order.

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(d) We follow the same process as in part (f) of Example 24–14. E V = E0 ( d − l ) + E D l = E0 ( d − l ) + 0 l → K V (150 V ) E0 = = = 1.167  105 V m −3 l 1.00  10 m −3 −3 d −l + 2.00  10 m − 1.00  10 m + K ( 3.50 )

(

) (

)

(

)

 1.17  105 V m (e)

ED =

1.167  105 V m 3.50

Q  (h) E0 = =  0 A 0

= 3.333  104 V m  3.33  104 V m

→

(

)(

Q = EA 0 = 1.167  105 V m 0.0250 m 2

)(8.85  10 C N  m ) = 2.582  10 C −12

−8

 2.58  10−8 C

 

1

1   −8 −8  = ( 2.582  10 C )  1 −  = 1.84  10 C K  3.50 

Because the battery voltage does not change, the potential difference between the plates is unchanged when the dielectric is inserted, and so is V = 150 V .

(g) C =

2.582  10−8 C

= 1.72  10−10 pF

150 V V Notice that the capacitance is the same as in Example 24–14. Since the capacitance is a constant (a function of geometry and material, not charge and voltage), it should be the same value.

69. The capacitance will be given by C = Q / V . When a charge Q is placed on one plate and a charge –Q is placed on the other plate, an electric field will be set up between the two plates. The electric Q  field in the air-filled region is just the electric field between two charged plates, E0 = . =  0 A 0 The electric field in the dielectric is equal to the electric field in the air, divided by the dielectric E Q constant: E D = 0 = . K KA 0 The voltage drop between the two plates is obtained by integrating the electric field between the two plates. One plate is set at the origin with the dielectric touching this plate. The dielectric ends at x = l . The rest of the distance to x = d is then air filled. l

V = −  E  dx = 

Qdx

+

Qdx

Q  l

  + (d − l )  A 0  K 

KA 0 l A 0 The capacitance is the ratio of the voltage to the charge. 0

Q V

Q Q  l

  + (d − l )  A 0  K 

0 A d −l +

l K

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As a check, we evaluate this using the values of Example 24–14, and get the correct value for C. (8.85  10−12 C2 N  m 2 )( 0.0250 m 2 ) = 1.72  10−10 F 0 A C= = l 0.001m d −l + 0.002 m − 0.001m + K 3.5 70. There are two uniform electric fields–one in the air, and one in the gap. They are related by Eq. 24–10. In each region, the potential difference is the field times the distance in the direction of the field over which the field exists. E V = Eair d air + Eglass d glass = Eair d air + air d glass → K glass Eair = V

K glass d air K glass + d glass

= ( 80.0 V )

5.80

( 3.00  10 m ) ( 5.80) + ( 2.00  10 m ) −3

−3

= 2.39  104 V m

Eglass =

Eair K glass

2.39  104 V m 5.80

= 4.12  103 V m

The charge on the plates can be calculated from the field at the plate, using Eq. 22–5. Use Eq. 24–11b to calculate the charge on the dielectric.  Q Eair = plate = plate → 0 0 A

(

)(

)

Qplate = Eair 0 A = 2.39  104 V m 8.85  10 −12 C 2 N  m 2 1.45 m 2 = 3.07  10 −7 C

 

Qind = Q  1 −

1

1   −7 −7  = ( 3.07  10 C )  1 −  = 2.54  10 C 5.80 K  

71. (a) We approximate the configuration as a parallel-plate capacitor, and so use Eq. 24–2 to calculate the capacitance.

C = 0

A d

= 0

 r2 d

 6  10−12 F (b) Use Eq. 24–1.

(

= 8.85  10

−12

C Nm 2

)

 ( 4.5in )( 0.0254 m in ) 

0.06 m

= 6.053  10−12 F

)

Q = CV = 6.053  10−12 F ( 9 V ) = 5.488  10−11 C  5  10−11 C

(c) The electric field is uniform, and is the voltage divided by the plate separation. 9V V E= = = 150 V m  150 V m d 0.06 m (d) The work done by the battery to charge the plates is equal to the energy stored by the capacitor. Use Eq. 24–5.

(

)

U = 12 CV 2 = 12 6.053  10−12 F ( 9 V ) = 2.451  10−10 J  2  10−10 J 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(e) The electric field will stay the same, because the voltage will stay the same (since the capacitor is still connected to the battery) and the plate separation will stay the same. The capacitance changes, and so the charge changes (by Eq. 24–1), and so the work done by the battery changes (by Eq. 24–5). 72. (a) The capacitance of a single isolated conducting sphere is given after Example 24–3. C = 4 0 r →

C r

) 

(

= 4 8.85  10 −12 C 2 N  m 2 =  1.11  10 −10



And so C = (1.11pF cm ) r →

F   1m   1012 pF 



 = 1.11pF cm

m   100 cm   1F 

C ( pF )  r ( cm ) . The rule is about a 10% under-estimate.

(b) We assume that the human body is a sphere of radius 100 cm. Thus the rule C ( pF )  r ( cm ) says that the capacitance of the human body is about 100 pF . (c) A 0.5-cm spark would require a potential difference of about 15,000 V. Use Eq. 24–1.

Q = CV = (100 pF )(15, 000 V ) = 1.5 C

73. The energy in the capacitor, given by Eq. 24–5, is the heat energy absorbed by the water, given by Eq. 19–2. U = Qheat → 12 CV 2 = mcT →

V =

2mcT C

 

2 ( 2.4 kg )  4186

  ( 95C − 22C ) kg C  = 605.5 V  610 V J

4.0 F

74. The capacitor is charged and isolated, so the amount of charge on the capacitor cannot change. (a) The stored energy is given by Eq. 24–5, U = 12 QV . The charge does not change, and the potential difference is doubled, so the stored energy is multiplied by 2 . Q2 (b) The stored energy can also be given by U = 12 . The charge does not change. The C capacitance would be reduced by a factor of 2 if the plate separation were doubled, according to Eq. 24–2, so the stored energy is again multiplied by 2 . 75. (a) The energy stored is multiplied by 4 since the potential difference across the plates doubles and the capacitance doesn’t change: U = 12 CV 2 . (b) The energy stored is multiplied by 4 since the charge doubles and the capacitance doesn’t Q2 change: U = 12 . C (c) If the separation between the plates doubles, the capacitance is divided by 2. The voltage across the capacitor doesn’t change if it remains connected to a battery, so the energy stored is also divided by 2 : U = 12 CV 2.

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76. (a) The capacitance per unit length of a cylindrical capacitor with no dielectric is derived in C 2 0 Example 24–2, as . Adding a dielectric increases C by a factor of K. = ln ( Ra Rb ) l

C l C

2 0 K

ln ( Ra Rb )

2 0 K

(

)

2 8.85  10−12 C 2 N  m 2 ( 3.2 )

= 1.4  10−10 F m ln ( Ra Rb ) ln ( 9.0 mm 2.5 mm ) l (c) Since Q = CV , the charge per unit length will just be the capacitance per unit length from part (b), times the potential difference. Q CV = 1.4  10−10 F m 1.0  103 V = 1.4  10−7 C m = = l l (b)

(

)(

)

77. The energy stored in the capacitor is given by Eq. 24–5. The final energy is half the initial energy. Find the final charge, and then subtract the final charge from the initial charge to find the charge lost. 2 2 Qfinal Qinitial 1 1 1 1 1 Efinal = 2 Einitial → 2 =22 → Qfinal = Qinitial C C 2 1  1    −6 Qlost = Qinitial − Qfinal = Qinitial  1 −  = CV  1 −  = 2.1  10 F ( 9.0 V )( 0.2929 ) 2 2  

(

)

= 5.5  10−6 C 78. Use Eq. 24–1 with Eq. 24–8 to find the charge. Q = CV = K  0

A d

(

V = ( 3.7 ) 8.85  10

79. Use Eq. 24–1.

Q = C V ; t =

Q Q t

C V Q t

−12

C Nm 2

)

 ( 0.55  10−2 m )

( 0.10 10 m )

(1200 F )( 6.0 V ) −3

1.0  10 C s

−3



(12 V ) = 3.7  10−10 C

  = 83d  86, 400s 

= 7.2  106 s 

80. The relative change in energy can be obtained by inserting Eq. 24–8 into Eq. 24–5. Q2 A 0 U C 1 = 2C2 = 0 = d = KA 0 Q U0 C 2K 1 (2 d) 2C 0 The dielectric is attracted to the capacitor. As such, the dielectric will gain kinetic energy as it enters the capacitor. An external force is necessary to stop the dielectric. The negative work done by this force results in the decrease in energy within the capacitor. Since the charge remains constant, and the magnitude of the electric field depends on the charge, and not the separation distance, the electric field is not affected by the change in distance between the plates. The electric field between the plates is reduced by the dielectric constant, as in Eq. 24–10.

E E0 / K 1 = = E0 E0 K © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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81. We treat this as N capacitors in parallel, so that the total capacitance is N times the capacitance of a single capacitor. The maximum voltage and dielectric strength are used to find the plate separation of a single capacitor. V 100 V l 6.0  10−3 m −6 d= N = = 3.33  10 m ; = = = 1800 ES 30  106 V m d 3.33  10 −6 m

Ceq = NC = N  0 K

A d

→

(1.0  10 F )( 3.33  10 m ) = = 1.244  1.2 K= N  A 1800 (8.85  10 C N  m )(12.0  10 m )(14.0  10 m ) −6

Ceq d

−12

−6

−3

82. The total charge doesn’t change when the second capacitor is connected, since the two-capacitor combination is not connected to a source of charge. The final voltage across the two capacitors must be the same. Use Eq. 24–1. Q0 = C1V0 = Q1 + Q2 = C1V1 + C2V2 = C1V1 + C2V1

C2 = C1

(V0 − V1 ) V1

 11.2 V − 5.9 V   = 3.144  F  3.1 F 5.9 V  

= ( 3.5 F ) 

83. (a) Use Eq. 24–5 to calculate the stored energy.

(

)(

U = 12 CV 2 = 12 8.0  10−8 F 2.5  104 V

) = 25J 2

(b) The power is the energy converted per unit time. Energy 0.15 ( 25J ) P= = = 9.38  105 W  940 kW time 4.0  10−6 s 84. The pressure is the force per unit area on a face of the dielectric. The force is related to the potential dU , where x is the separation of the capacitor energy stored in the capacitor by Eq. 8–7, F = − dx plates. A dU K  0 AV 2 F K  0V 2  U = 12 CV 2 = 12  K  0 V 2 → F = − P = = = → ; x dx A 2x2 2x2 

V =

) ( 40.0 Pa ) = 154 V  150 V = K ( 3.8) (8.85  10 C N  m )

2x2P

(

2 1.0  10−4 m

−12

85. (a) The charge can be determined from Eqs. 24–1 and 24–2. 2.0  10−4 m 2 A Q = CV =  0 V = 8.85  10−12 C2 N m 2 d 5.0  10−4 m

(

)

( (

) (12 V ) = 4.248  10 C ) −11

 4.2  10−11 C (b) Since the battery is disconnected, no charge can flow to or from the plates. Thus the charge is constant.

Q = 4.2  10−11 C

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(c) The capacitance has changed and the charge has stayed constant, and so the voltage has changed. A A Q = CV = constant → C1V1 = C0V0 →  0 V1 =  0 V0 → d1 d0

V1 =

V0 =

0.85 mm

(12 V ) = 20.4V  2.0  101 V

d0 0.50 mm (d) The work is the change in stored energy.

(

)

W = U = 12 QV1 − 12 QV0 = 12 Q (V1 − V0 ) = 12 4.248  10−11 C ( 8.4 V ) = 1.8  10 −10 J  2  10−10 J Only 1 significant figure is justified due to the voltage subtraction. 86. Since the two capacitors are in series, they will both have the same charge on them. 1 1 1 V Q1 = Q2 = Qseries ; = = + → Cseries Qseries C1 C2 C2 =

(125  10 C )(155  10 F ) = = 5.17  10 F (155  10 F ) ( 25.0 V ) − (125  10 C ) −12

QseriesC1

−12

C1V − Qseries

−12

87. (a) From the diagram, we see that one group of 4 plates is connected together, and the other group of 4 plates is connected together. This common grouping shows that the capacitors are connected in parallel . (b) Since they are connected in parallel, the equivalent capacitance is the sum of the individual capacitances. The variable area will change the equivalent capacitance. A Ceq = 7C = 7 0 d

Cmin

( 2.0  10 m ) = 7.7  10 F = 7 = 7 ( 8.85  10 C N m ) d (1.6  10 m ) ( 9.0  10 m ) = 3.5  10 F A = 7 = 7 ( 8.85  10 C N m ) d (1.6  10 m ) Amin

−4

−12

−3

−4

Cmax

max

−12

−11

−3

And so the range is from 7.7 pF to 35 pF . 88. The metal conducting strips connecting cylinders b and c mean that b and c are at the same potential. Due to the positive charge on the inner cylinder and the negative charge on the outer cylinder, cylinders b and c will polarize according to the first diagram, with negative charge on cylinder c, and positive charge on cylinder b. This is then two capacitors in series, as illustrated in the second diagram. The capacitance per unit length of a cylindrical capacitor is derived in Example 24–2. 2 0 l 2 0 l 1 1 1 C1 = ; C2 = ; = + → ln ( Ra Rb ) ln ( Rc Rd ) Cnet C1 C2

– + Rc

–

Ra Rb

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

 2 0 l   2 0 l   ln R R   ln R R  ( a b )  ( c d ) CC Cnet = 1 2 =  2 0 l 2 0 l C1 + C2 ln ( Ra Rb )

= C l

2 0 l

ln ( Rc Rd ) + ln ( Ra Rb )

Instructor Solutions Manual

Cyl. d

Cyl. c

Cyl. b

Cyl. a

ln ( Rc Rd )

2 0 l

→

ln ( Ra Rc Rb Rd )

2 0

ln ( Ra Rc Rb Rd )

89. (a) The force acting on one plate by the other plate is equal to the electric field produced by one charged plate multiplied by the charge on the second plate (see Example 24–10).  Q  Q2 = F = EQ =  Q  2 A 0  2 A 0  The force is attractive since the plates are oppositely charged. Since the force is constant, the work done in increasing the distance between the two by an amount x is just the force times the increase in distance.

W = Fx =

Q2 x 2 A 0

(b) The change in energy stored between the plates is obtained using Eq. 24–5.

W = U =

Q2  1 1  Q2  2x x  Q2 x − = −  =   2  C2 C1  2   0 A  0 A  2 0 A

The work done in pulling the plates apart is equal to the increase in energy between the plates. 90. (a) For a plane conducting surface, the electric field is given by Eq. 22–5. Q  E= = → Qmax = ES 0 A = 3  106 N C 8.85  10−12 C2 N  m 2 0 0 A

(

)(

)(150  10 m ) −4

= 3.98  10−7 C  4  10−7 C (b) The capacitance of an isolated sphere is derived in the text, right after Example 24–3.

(

)

C = 4 0 r = 4 8.85  10−12 C2 N  m2 (1m ) = 1.11  10−10 F  1  10−10 F

(c) Use Eq. 24–1, with the maximum charge from part (a) and the capacitance from part (b). Q 3.98  10−7 C Q = CV → V = = = 3586 V  4000 V C 1.11  10−10 F 91. (a) Use Eq. 24–2 to calculate the capacitance. 8.85  10−12 C2 N  m 2 2.0 m 2 0 A C0 = = = 5.9  10−9 F −3 d 3.0  10 m

(

)( )

(

)

Use Eq. 24–1 to calculate the charge.

(

)

Q0 = C0V0 = 5.9  10−9 F ( 45V ) = 2.655  10−7 C  2.7  10−7 C

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Capacitance, Dielectrics, Electric Energy Storage

The electric field is the potential difference divided by the plate separation. V 45V E0 = 0 = = 15000 V m d 3.0  10−3 m Use Eq. 24–5 to calculate the energy stored.

(

)

U 0 = 12 C0V02 = 12 5.9  10−9 F ( 45V ) = 6.0  10−6 J 2

(b) Now include the dielectric. The capacitance is multiplied by the dielectric constant.

(

)

C = KC0 = 3.2 5.9  10−9 F = 1.888  10−8 F  1.9  10−8 F

The voltage doesn’t change. Use Eq. 24–1 to calculate the charge.

(

)

Q = CV = KC0V = 3.2 5.9  10−9 F ( 45V ) = 8.496  10−7 C  8.5  10−7 C

Since the battery is still connected, the voltage is the same as before, and so the electric field doesn’t change.

E = E0 = 15000 V m Use Eq. 24–5 to calculate the energy stored.

(

)

U = 12 CV 2 = 12 KC0V 2 = 12 ( 3.2 ) 5.9  10−9 F ( 45V ) = 1.9  10−5 J 2

92. Since the other values in this problem manifestly have 2 significant figures, we assume that the capacitance also has 2 significant figures. (a) The number of electrons is found from the charge on the capacitor. 30  10−15 F (1.5 V ) CV Q = CV = Ne → N = = = 2.8  105 e’s −19 1.60  10 C e (b) The thickness is determined from the dielectric strength. 1.5 V V V Emax = → d min = = = 1.5  10 −9 m d min Emax 1.0  109 V m (c) The area is found from Eq. 24–8. 30  10−15 F 1.5  10−9 m A Cd → A= = = 3.4  10−13 m 2 C = K 0 −12 2 2 d K 0 15 8.85  10 C N  m

(

)

)(

(

) )

93. (a) Both of these equations for the potential energy can be obtained from Eq. 24–5 by using Eq. 24–1 first to eliminate the charge from the equation, and second to eliminate the voltage. U = 12 QV = 12 ( CV ) V = 12 CV 2

Q = 1 Q  2 C C

= 12 QV = 12 Q 

(

2 (b) The first equation U = 12 CV

) is used when the capacitance and voltage difference across the

capacitor are known, as when the capacitor is connected to a battery. The second equation Q2   1  U = 2 C  is used when the capacitance and charge are known, as when a specific charge is   placed on an isolated capacitor. (c) Since the voltage is constant, use the first equation for the potential energy, with the capacitance equal to the product of the initial capacitance and the paper’s dielectric constant.

U = 12 CV 2 = 12 ( KC0 ) V 2 = K

( C V ) = K (U ) 1 2

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Instructor Solutions Manual

The potential energy in the capacitor will increase by a factor equal to the dielectric constant. For a paper dielectric the potential increases by a factor of 3.7 . (d) In this case the charge is constant, so we use the second equation to determine the final energy. Q2 Q2 1  Q2  1 U= = =   = (U 0 ) 2C 2 KC0 K  2C0  K The potential energy in the capacitor decreases by a factor equal to the dielectric constant. For quartz, the potential energy decreases to 1/4.3 = 0.23, or 23% of the initial value . 94. See the schematic diagram for the arrangement. The two “capacitors” are in series, and so have the same charge. Thus their voltages, which must total 25kV, will be inversely proportional to their capacitances. Let C1 be the air-filled capacitor, and C2 be the vinyl capacitor. The area of the foot is approximately twice the area of the hand, and since there are two feet on the floor and only one hand on the metal case, the 4 A area ratio is foot = . Ahand 1

Q = C1V1 = C2V2 → V1 = V2

 0 K air Ahand

C1 =

; C2 =

d air

 0 K vinyl Afoot d vinyl

 0 K vinyl Afoot C2 C1

d vinyl K vinyl Afoot d air ( 3)( 4 )( 0.63) = = = 7.56  0 K air Ahand K air Ahand d vinyl (1)(1)(1.0 ) d air

V = V1 + V2 = V2

+ V2 = 8.56V2 = 25, 000 V → V2 = 2920 V  3000 V C1 Answers will vary somewhat due to the estimations involved.

95. (a) If N electrons flow onto the plate, the charge on the top plate is − Ne, and the positive charge associated with the capacitor is Q = Ne. Since Q = CV , we have Ne = CV → V = Ne C , showing that V is proportional to N. (b) Given V = 1mV and we want N = 1, solve for the capacitance. V =

→ V =

eN

→ C C 1 N C=e = 1.60  10−19 C = 1.60  10−16 F  2  10 −16 F −3 1  10 V V (c) Use Eq. 24–8. A l2 C = 0K = 0K → d d

(

)

(1.60  10 F )(100  10 m ) = 7.76  10 m  10  m  = 0.8  m l = =  1m  K (8.85  10 C N  m ) ( 3)   Cd

−16

−9

−7

−12

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Chapter 24

Capacitance, Dielectrics, Electric Energy Storage

96. (a) The absolute value of the charge on each plate is given by Eq. 24–1. The plate with electrons has a net negative charge. Q = CV → N ( − e ) = −CV → N=

( 25  10 F ) ( 0.6 V ) = 9.375  10  9  10 electrons = −15

1.60  10−19 C e (b) Since the charge is directly proportional to the potential difference, a 2.0% decrease in potential difference corresponds to a 2.0% decrease in charge. Q = 0.02Q ;

t =

Q Q t

0.02Q Q t

0.02CV

Q t

(

)

0.02 25  10−15 F ( 0.6 V ) 0.5  10

−12

= 6  10−4 s

97. (a) From Problem 96, we have C = 25  10−15 F. Use Eq. 24–8 to calculate the area. C = K 0

( 25  10 F )(10 10 m ) = 1.130  10 m  10  m  → A= =  1m  K ( 25 ) ( 8.85  10 C N  m )   −15

−9

−12

= 1.130  m 2  1.1  m 2

(b) Each cell requires twice as much area as its plate size, according to the problem, and so half of 2 the wafer area is available for capacitor plates. Thus, there is 1.5cm available for capacitance. Each capacitor is one “bit.” 2

 106  m   1bit   1byte  7 1.5cm  2   1.13 m 2   8 bits  = 1.66  10 bytes  17 Mbytes 10 cm      2

98. (a) Use Eq. 24–2.

0 A

d (b) Use Eq. 24–1.

(8.85  10 C N  m )(110  10 m ) = 6.49  10 F  6.5  10 F = 65 F = −12

−7

(1500 m )

(

)(

)

(

)

−7

Q = CV = 6.49  10−7 F 3.5  107 V = 22.715 C  23C (c) Use Eq. 24–5.

U = 12 QV = 12 ( 22.715C ) 3.5  107 V = 4.0  108J

99. (a) One horse power is equal to 746 Watts  1hp  P = 75  103W  = 100.5 hp  1.0  102 hp   746 W  (b) The energy is the product of the power and time elapsed.

(

)

(

)

E = Pt = 75  103 W ( 4.0 hr )( 3600s hr ) = 1.08  109 J  1.1  109 J

(c) Use Eq. 24–5, in terms of the capacitance and voltage, to solve for the capacitance. 9 2U 2 1.08  10 J 2 1 = 2.990  103 F  3.0  103 F U = 2 CV → C = 2 = 2 V ( 850 V )

(

)

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(d) According to the text, the capacitance of 0.1 g of activated carbon is about 1 F. We again use Eq. 24–5 to find the capacitance and then convert to a mass of carbon. 9 2U 2 (1.08  10 J )  0.1g  C= 2 = = 2.16  107 F ; 2.16  107 F  = 2.2  106 g 2  V (10 V )  1F   2  106 g = 2000 kg

(e) Cars have a mass on the order of 1000–2000 kg, so this capacitor would more than double the mass of the car. This is not practical . 100. (a) The energy stored in a capacitor is given by Eq. 24–5. 9 2U 2 10 J U = 12 CV 2 → C = 2 = = 1.389  107 F  1 107 F 2 V (12 V )

(

)

(b) We assume the thickness of the capacitor is 0.1 mm, as given in part (c). Use Eq. 24–8. The largest dielectric constant in Table 24–1 is 300 for strontium titanate. 1.389  107 F 0.1 10 −3 m A Cd → A= = = 5.23  1011 m 2  5  1011 m 2 C = K 0 −12 2 2 d K  0 ( 300 ) 8.85  10 C N  m

(

)(

(

)

(c) There would have to be 5.23  1011 sheets of thickness 0.1 10−3 m. Thus the total thickness would be about 5.23  107 m  5 107 m. This is about 8 times the radius of the Earth. This is not practical . (d) On page 712 of the text, an approximation is made that 0.1 g of activated carbon can provide about 1F of capacitance.  0.1g   1kg  1.389  107 F    = 1.389 103 kg  1000 kg    1F   1000 g  A compact car has a mass in the 1000–1500 kg range, so this is still impractical. The weight of the car would significantly increase due to the capacitor. 101. (a) The initial capacitance is obtained directly from Eq. 24–8. K  0 A 3.7 ( 8.85 pF/m )( 0.21m )( 0.14 m ) C0 = = = 8.752  103 pF  8.8 nF −3 d 0.11  10 m (b) Maximum charge will occur when the electric field between the plates is equal to the dielectric strength. The charge will be equal to the capacitance multiplied by the maximum voltage, where the maximum voltage is the electric field times the separation distance of the plates. Qmax = C0V = C0 Ed = ( 8.752 nF ) 15  106 V/m 0.11  10−3 m

(

)(

)

= 1.444  104 nV  14  C (c) See the adjacent sketch. (d) The sheets of foil would be separated by sheets of paper with alternating sheets connected together on each side. This capacitor would consist of 100 sheets of paper with 101 sheets of foil. t = 101d Al + 100d paper = 101( 0.016 mm ) + 100 ( 0.11mm ) = 12.62 mm (e) Since the capacitors are in parallel, each capacitor has the same voltage which is equal to the total voltage. Therefore, breakdown will occur when the voltage across a single capacitor provides an electric field across that capacitor equal to the dielectric strength.

(

)(

)

Vmax = Emax d = 15  106 V/m 0.11  10−3 m = 1650 V  1700 V © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

860

CHAPTER 25: Electric Currents and Resistance Responses to Questions 1.

In the circuit (not in the battery), electrons flow from high potential energy (at the negative terminal) to low potential energy (at the positive terminal). Inside the battery, the chemical reaction does work on the electrons to take them from low potential energy to high potential energy (to the negative terminal). A chemical description could say that the chemical reaction taking place at the negative electrode leaves electrons behind on the terminal, and the positive ions created at the negative electrode pull electrons off the positive electrode.

Battery energy is what is being “used up.” As charges leave the battery terminal, they have a relatively high potential energy. Then as the charges move through the flashlight bulb, they lose potential energy. The battery uses a chemical reaction to replace the potential energy of the charges, by lowering the battery’s chemical potential energy. When a battery is “used up,” it is unable to give potential energy to charges.

Ampere-hours measures charge. The ampere is a charge per unit time, and the hour is a time, so the product is charge. 1 Ampere-hour of charge is 3600 Coulombs of charge.

Resistance is given by the relationship R =  l A. If the ratio of resistivity to area is the same for both the copper wire and the aluminum wire, then the resistances will be the same. Thus if  Cu ACu =  Al AAl or AAl ACu =  Al  Cu , the resistances will be the same for equal lengths of the two materials. Since the density of copper is about 3 times greater than the density of aluminum, the aluminum would have to have about 1/3 the cross-sectional area of the copper. Also, resistance changes with temperature. By having the two wires at different temperatures, it might be possible to have their resistances the same.

That one terminal of the battery (usually the negative one) is connected to the metal chassis, frame, and engine block of the car. This means that all voltages used for electrical devices in the car are measured with respect to the car’s frame. Also, since the frame is a large mass of metal, it can supply charges for current without significantly changing its electric potential.

Generally, water is already in the faucet spout, but it will not come out until the faucet valve is opened. Opening the valve provides the pressure difference needed to force water out of the spout. The same thing is essentially true when you connect a wire to the terminals of a battery. Electrons already exist in the wires. The battery provides the potential that causes them to move, producing a current.

If the emf in a circuit remains constant while the resistance is increased, less current will flow, and the power dissipated in the circuit will decrease. Both power equations support this result. If the current in a circuit remains constant and the resistance is increased, then the emf must increase and the power dissipated in the circuit will increase. Both equations also support this result. There is no contradiction, because the voltage, current, and resistance are also related to each other by V = IR.

When a light bulb burns out, its filament burns in two. Since the filament is part of the conducting path for the electricity flowing through the bulb, once the filament is broken, current can no longer flow through the bulb. It no longer gives off any light.

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Instructor Solutions Manual

We assume that the voltage is the same in both cases. Then if the resistance increases, the power delivered to the heater will decrease according to P = V 2 R. If the power decreases, the heating process will slow down.

10. Resistance is given by the relationship R =  L A . Thus, to minimize the resistance, you should have a small length and a large cross-sectional area. Likewise, to maximize the resistance, you should have a large length and a small cross-sectional area. (a) For the least resistance, connect the wires to the faces that have dimensions of 2a by 3a, which maximizes the area (6a2) and minimizes the length (a). (b) For the greatest resistance, connect the wires to the faces that have dimensions of a by 2a, which minimizes the area (2a2) and maximizes the length (3a). 11. When a light bulb is first turned on, it will be cool and the filament will have a lower resistance than when it is hot. Since the bulb is at a constant voltage, there will be a large current initially while it has that lower resistance. Metals expand when heated, and they expand quickly when heated quickly (which is what happens when the bulb is turned on). As bulbs age, the filament becomes more brittle due to oxidation and vaporization. At some point, the stress of thermal expansion strains the filament material beyond its ability to respond, and it breaks. It’s tensile strength (see Section 12–5) has decreased. 12. Assuming that both light bulbs have the same voltage, then since P = IV , the higher power bulb will draw the most current. Likewise assuming that both light bulbs have the same voltage, since P = V 2 R , the higher power bulb will have the lower resistance. So the 18-W bulb will draw the most current, and the 12-W bulb will have the higher resistance. 13. Transmission lines have resistance, and therefore will change some electrical energy to thermal energy (heat) as the electrical energy is transmitted. We assume that the resistance of the transmission lines is constant. Then the “lost” power is given by Plost = I 2 R, where I is the current carried by the transmission lines. The transmitted power is given by Ptrans = IV , where V is the voltage across the transmission lines. For a given value of Ptrans = IV , the higher the voltage is, the lower the current has to be. As the current is decreased, Plost = I 2 R is also decreased, so there is a lower amount of power lost. 14. The 15-A fuse is blowing because the circuit is carrying more than 15 A of current. The circuit is probably designed to only carry 15 A, so there might be a “short” or some other malfunction causing the current to exceed 15 A. Replacing the 15-A fuse with a 25-A fuse will allow more current to flow and thus make the wires carrying the current get hotter. A fire might result, or damage to certain kinds of electrical equipment. The repeatedly-blowing fuse is a warning that something is wrong with the circuit. 15. The human eye and brain cannot distinguish the on-off cycle of lights when they are operated at the normal 60 Hz frequency. At much lower frequencies, such as 5 Hz, the eye and brain are able to process the on-off cycle of the lights, and they will appear to flicker. 16. There are several factors which can be considered. As the voltage reverses with each cycle of AC, the potential energy of the electrons is raised again. Thus with each “pass” through the light, the electrons lose their potential energy, and then get it back again. Secondly, the heating of the filament (which causes the light) does not depend on the direction of the current, but only that a current exists, so the light occurs as the electrons move in both directions. Also, at 60 Hz, the current peaks 120 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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times per second. The small amount of time while the magnitude of the current is small is not long enough for the hot metal filament to cool down much, and so it stays lit during the entire cycle. Finally, the human eye sees anything more rapid than about 20 Hz as continuous, even if it is not (for instance, motion pictures are played at 24 frames per second). So even if the light was to go dim during part of each cycle, our eyes will not detect it. This is called “persistence of vision.” 17. Initially, there is no current at all, so once the voltage is applied, the magnitude of the current will be increasing from 0. But consider this; when a toaster is not in use, the Nichrome wire is at room temperature. When you turn it on the wire starts to heat up immediately. Since the resistance increases with temperature, the resistance will be increasing as the wire heats. Assuming the voltage supplied is constant, then the magnitude of the current will be decreasing as the resistance increases, until it reaches a constant temperature. 18. Current is NOT used up in a resistor. The same current flows into the resistor as flows out of the resistor. If that were not the case, there would be either an increase or decrease in the charge of the resistor, but the resistor actually stays electrically neutral, indicating equal charge flow both in and out. What does get “used up” is potential energy. The charges that leave a resistor have lower potential energy than the charges that enter a resistor. The amount of energy decrease per unit time is given by I 2 R. 19. If you turn on an electric appliance when you are outside with bare feet, and the appliance “shorts out” through you (meaning the appliance’s case is at a high voltage), the current has a direct path to ground through your feet, and you will receive a severe shock. If you are inside wearing socks and shoes with thick soles, and the appliance shorts out, the current will not have a low-resistance path to ground through you, and will most likely take an alternate path. You might receive a mild shock, but not a severe one. 20. In the two wires described, the drift velocities of the electrons will be about the same. Drift velocity depends on the rate of collisions of the electrons with the atoms in the wire, which depends on the density of the atoms. The density of the atoms is similar, so the collision rates will be similar. But the current density, and therefore the current in the wire, with twice as many free electrons per atom, will be twice as large as in the other wire. This is due to the current density being directly proportional to n, the number of free electrons per unit volume. 21. (a) If the length of the wire doubles, its end-to-end resistance also doubles by Eq. 25–3, R = 

. A Thus, the current in the wire will be reduced by a factor of two, according to Ohm’s law. We see from Eq. 25–13, I = neAv d , that if all other factors are unchanged (# of free electrons per unit volume, and cross sectional area of the wire), that already used the drift speed is proportional to the current. Thus, since the current is reduced by a factor of two, the drift velocity must also be reduced by a factor of two. (b) If the wire’s radius is doubled, its cross-sectional area is increased by a factor of 4. Then, by Eq. 25–3, resistance is lowered by a factor of 4. Thus, the current will increase by a factor of 4 by Ohm’s law. We see from Eq. 25–13 that if the current and the area both increase by the same factor, then the drift velocity will remain the same. (c) If the potential difference doubles while the resistance remains constant, the current doubles by Ohm’s law. Then, by Eq. 25–13, if the current is doubled but the other factors remain the same, the drift velocity will also double.

863

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Instructor Solutions Manual

Solutions to MisConceptual Questions 1.

(c) It may be thought that the orientation of the battery is important for the bulb to work properly, but the current can flow either direction through the light bulb and it will glow equally bright.

(c) A common misconception is that the light bulb “uses up” the current, causing more current to flow in one portion of a loop than in another. For a single loop the current is the same at every point in the loop, and so the same current flows through the light bulb as through the battery.

(a) Ohm’s law is an empirical law showing that for some materials the current through the material is proportional to the voltage across the material. For other materials, such as diodes, fluorescent light bulbs, and superconductors, Ohm’s law is not valid. Since it is not valid for all objects, it cannot be a fundamental law of physics.

(e) A common misconception is that the electrons are “used up” by the light bulb; a good analogy would be water flowing across a water wheel in a flour mill. The water flows onto the wheel at the top (high potential) and causes the wheel to rotate as the water descends along the wheel. The same amount of water leaves the bottom of the wheel as entered at the top, but does so at a lower point (low potential). The change in potential energy goes into work in the wheel. In a light bulb, electrons at higher potential energy enter the light bulb and give off that energy as they pass through the bulb. As with the water on the wheel, the same number of electrons exit the bulb as enter it.

(e) A misconception commonly found when dealing with electric circuits is that electrons are “used up” by the light bulb. The current is a measure of the rate that electrons pass a given point. If the current were different at two points in the circuit, then electrons would be building up (or being depleted) between those two points. The buildup of electrons would cause the circuit to be time-dependent, and not a steady-state system. The flow of electrons (current) must be the same at all points in a loop.

(b) Ohm’s law requires that the ratio of voltage to current be constant. Since it is not constant in this case, the material does not obey Ohm’s law.

(d) A common misconception is that the charge or current is “used up” in a resistor. The resistor removes energy from the electrons, so that the electrons exiting the resistor have less potential energy than the electrons entering, but the same number of electrons (charge carriers) enter as exit. The rate of electrons entering and exiting is equal to the constant current in the circuit.

(c) Since the unit of kilowatt-hour contains the word “watt,” it is often incorrectly thought to be a unit of power. However, the kilowatt-hour is the product of the unit of power (kilowatt) with a unit of time (hour), resulting in a unit of energy.

(b) Each device added to the circuit is added in parallel. The voltage across the circuit does not change as the devices are added. However, each new device creates a path for additional current to flow, causing the current in the circuit breaker to increase. When the current becomes too high the circuit breaker will open.

10. (b) For current to flow through an object it must complete a circuit. When a bird lands on the wire it creates a loop in parallel with the segment of wire between its feet. The voltage drop across the bird is equal to the voltage drop across the wire between the bird’s feet. Since the wire has a small resistance, there will be very little voltage drop across the wire between the bird’s feet, © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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and very little voltage drop across the bird. The bird is a good conductor (similar to a human), but since there is little voltage drop, it will experience little current flow. When a ladder is placed between the ground and the wire it creates a path for the current to flow from the highvoltage wire to ground (zero voltage). This large potential difference will enable a large current to flow through the ladder. 11. (b) A common misconception is that the electrons must travel from the switch to the light bulb for the light bulb to turn on. This is incorrect because there are electrons throughout the circuit, not just at the switch. When the switch is turned on, the electric potential across the circuit creates an electric field in the wire that causes all of the conduction electrons in the wire to move. 12. (d) The power can be calculated by P =

. If the voltage is constant and the resistance is R doubled, then the power will be cut in half.

13. (a) Answer (a) is not true. The average power delivered by an ac circuit is given by the various V2 2 forms of Eq. 25–10: P = I rmsVrms = I rms = rms . R 14. (a) From Eq. 25–14, j = nev d in absolute value, we see that drift speed and current density are proportional. Thus answer (a) is correct. Current density does depend on the number of electrons per unit volume, which makes it dependent on the type of wire, and so answer (b) is incorrect. According to Example 25–14, drift speeds are on the order of 0.05 mm/s, so both answers (c) and (e) are incorrect. Fig. 25–25 illustrates that answer (d) is not correct.

Solutions to Problems 1.

Use Eq. 25–2 for the voltage.

V = IR = ( 0.25A )( 5400  ) = 1350V  1400 V 2.

Solve Eq. 25–2a for resistance. V 120 V R= = = 29  I 4.2 A

Use the definition of current, Eq. 25–1a. Q 1.75C 1 electron → 1.75A =  = 1.09  1019 electrons s I= t s 1.60  10−19C

The ampere-hour is a unit of charge, indicating how much charge can be “stored” in the battery.  1C s   3600s  5 ( 95A h )    = 3.4  10 C   1A   1 h 

Use the definition of current, Eq. 25–1a. −19 Q (1200 ions ) 1.60  10 C ion = = 6.9  10−11 A I= −6 t 2.8  10 s

(

)

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Find the potential difference from the resistance and the current. R = 2.5  10−5  m 4.0  10−2 m = 1.0  10−6 

(

)(

)

(

)

V = IR = ( 3800 A ) 1.0  10−6  = 3.8  10−3 V 7.

(a) Use Eq. 25–2 to find the current. V 240 V V = IR → I = = = 18.46 A  18 A R 13  (b) Use the definition of current, Eq. 25–1a. Q I= → Q = I t = (18.46 A )(15min )( 60s min ) = 1.66  104 C  1.7  104 C t

(a) Use Eq. 25–2 to find the resistance. 12 V V R= = = 24  I 0.50 A (b) An amount of charge Q loses a potential energy of U = ( Q ) V as it passes through potential difference V across the resistor. The amount of charge is found from Eq. 25–1a.

U = ( Q ) V = ( I t )V = ( 0.50 A )( 60s )(12 V ) = 360 J 9.

(a) Use Eq. 25–2 for resistance. V 120 V R= = = 8.889   8.9  I 13.5 A (b) Use the definition of current, Eq. 25–1a. Q I= → Q = I t = (13.5C s )(8.0 min )( 60s min ) = 6.48  103 C  6.5  103 C t

10.

Use Ohm’s Law, Eq. 25–2, to find the current. Then use the definition of current, Eq. 25–1a, to calculate the number of electrons per minute.

V R

Q t

4.5 V 2.3 

1.956 C s



1 electron −19



60 s

1.60  10 C 1min

= 7.3  1020

electrons minute

11. (a) If the voltage drops by 15% (the same as being multiplied by 0.85), and the resistance stays the same, then by Eq. 25–2, V = IR, the current will also drop by 15%, or be multiplied by 0.85.

I final = 0.85I initial = 0.85 ( 5.25 A ) = 4.4625 A  4.5A (b) The resistance drops by 7.5% (the same as being multiplied by 0.925), and the voltage drops by 15%. Recalculate the current using Eq. 25–2. V ( 0.85) Vinitial ( 0.85) V ( 0.85) ( 0.85) I final = final = initial = = I initial = 5.25A = 4.8 A Rfinal Rinitial ( 0.925) (Vinitial I initial )( 0.925) ( 0.925) ( 0.925) 2 2 12. Use Eq. 25–3 to find the diameter, with the area as A =  r =  d 4 .

R=

l A

=

d

→ d=

4l 

R

(

4 (1.00 m ) 5.6  10−8  m

 ( 0.38  )

) = 4.3  10 m −4

866

Chapter 25

Electric Currents and Resistance

13. Use Eq. 25–3 to calculate the resistance, with the area as A =  r 2 =  d 2 4 . R=

l A

=

d

(

= 1.68  10−8  m

)

4 ( 5.1m )

 (1.5  10 m ) −3

= 4.8  10−2 

14. Use Eq. 25–3 to calculate the resistances, with the area as A =  r 2 =  d 2 4 ; 4l l . R= = A d2 4l  Al Al2 2 2.65  10−8  m (10.0 m )(1.8 mm ) RAl  d Al  Al l Al d Cu2 = = = = 0.39 2  Cu l Cu d Al2 RCu  4 l Cu 1.68  10−8 m ( 27.0 m )( 2.2 mm ) Cu  d Cu2

( (

) )

15. Use Eq. 25–3 to express the resistances, with the area as A =  r 2 =  d 2 4 , and so R =  RW = RCu →  W

d

2 W

=  Cu

 d Cu2

l A

=

d2

→

W 5.6  10−8  m = (1.8 mm ) = 3.3mm  Cu 1.68  10−8  m The diameter of the tungsten should be 3.3 mm. d W = d Cu

16. Since the resistance is directly proportional to the length, the length of the long piece must be 4.0 times the length of the short piece. l = l short + l long = l short + 4.0l short = 5.0l short → l short = 0.20l , l long = 0.80l Make the cut at 20% of the length of the wire .

l short = 0.20l , l long = 0.80l → Rshort = 0.2 R = 0.2 (18.5  ) = 3.7  , Rlong = 0.8R = 14.8  17. Use Eq. 25–5 multiplied by l A so that it expresses resistance instead of resistivity.

R = R0 1 +  (T − T0 ) = 1.16 R0 → 1 +  (T − T0 ) = 1.16 → T − T0 =

0.16

= 24 C −1 0.0068 ( C ) So raise the temperature by 24 C to a final temperature of 44C .



18. The wires have the same resistance and the same resistivity.

Rlong = Rshort →

 l long Along

 l short → Ashort

d long l = 1 short2 → = 2 1 dshort 4  d short 4  d long 2l short 2

867

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

19. Use Eq. 25–5 multiplied by l A so that it expresses resistances instead of resistivity.

R = R0 1 +  (T − T0 ) → R = R0 1 +  (T − T0 ) = 12  1 + 0.0045 ( C )

−1

( 2900 K − 293K ) = 152.8   150 

20. Calculate the voltage drop by combining Ohm’s Law (Eq. 25–2) with the expression for resistance, Eq. 25–3. 4 1.68  10−8  m (18 m ) 4 l l =I = (12 A ) = 1.7 V V = IR = I 2 A d2  1.628  10−3 m

(

)

(

)

21. In each case, calculate the resistance by using Eq. 25–3 for resistance. 3.0  10−5  m 1.0  10−2 m l x = = 3.75  10−4   3.8  10−4  (a) Rx = 2.0  10−2 m 4.0  10 −2 m Ayz

(

)( ) ( )( ) l ( 3.0  10  m )( 2.0  10 m ) = 1.5  10  = (b) R = A (1.0  10 m )( 4.0  10 m ) ( 3.0  10  m )( 4.0  10 m ) = 6.0  10  l = (c) R = A (1.0  10 m )( 2.0  10 m ) −5

−2

−3

−2

−5

−2

−3

−2

22. The length of the new wire is half the length of the original wire, and the cross-sectional area of the new wire is twice that of the original wire. Use Eq. 25–3. 1 l l l l R0 =  0 l = 12 l 0 A = 2 A0 R =  =  2 0 = 14  0 = 14 R0 A0 A 2 A0 A0 The new resistance is one-fourth of the original resistance. 23. The original resistance is R0 = V I 0 , and the high temperature resistance is R = V I , where the two voltages are the same. The two resistances are related by Eq. 25–5, multiplied by l A so that it expresses resistance instead of resistivity. R = R0 1 +  ( T − T0 )  → T = T0 +

1 R

  1V I 1I  − 1  = T0 +  − 1  = T0 +  0 − 1     R0    V I0   I 

= 23.5C +

1 0.00429 ( C )

−1

 0.4212 A   0.3646 A − 1 = 59.7C  

We have ignored the change in length due to the temperature change. 24. The total resistance is to be 3600 ohms ( Rtotal ) at all temperatures. Write each resistance in terms of Eq.25–5 (with T0 = 0 C ), multiplied by l A to express resistance instead of resistivity. o

Rtotal = R0C 1 +  CT  + R0 N 1 +  NT  = R0C + R0C CT + R0 N + R0 N NT = R0C + R0 N + ( R0C C + R0 N N ) T

For the above to be true, the terms with a temperature dependence must cancel, and the terms without a temperature dependence must add to Rtotal. Thus we have two equations in two unknowns. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Electric Currents and Resistance

R0C C

0 = ( R0C C + R0 N N ) T → R0 N = − Rtotal = R0C + R0 N = R0C −

R0C C

N

R0C ( N −  C )

N

→

0.0004 ( Co ) N = ( 3600  ) = 1600  = 1.60 k R0C = Rtotal −1 −1 ( N −  C ) 0.0004 ( Co ) + 0.0005 ( Co ) −1

R0 N = Rtotal − R0C = 3600  − 1600  = 2000  = 2.00 k 25. We choose a spherical shell of radius r and thickness dr as a differential element. The area of this element is 4 r 2 . Use Eq. 25–3, but for an infinitesimal resistance. Then integrate over the radius of the sphere.

R=

→ dR = 

1 dr

 4 r 2

1  1 1 1 1 → R =  dR =  = −  =  −  2 4  r  r 4  r1 r2   r 4 r 1

26. (a) Let the values at the lower temperature be indicated by a subscript “0.” Thus R0 =  0

= 0

4l 0

 d 02

l0 A0

. The change in temperature results in new values for the resistivity, the length, and

the diameter. Let  represent the temperature coefficient for the resistivity, and  T represent the thermal coefficient of expansion, which will affect the length and diameter. 4 l 0 1 +  T ( T − T0 )  l 4l 4 l 1 +  ( T − T0 )  =  0 1 +  ( T − T0 )  =  0 02 R= = 2 2 d  d 0 1 +  T ( T − T0 )  A  d 0 1 +  T ( T − T0 ) 

= R0

1 +  (T − T ) → R 1 +  T − T = R 1 +  T − T → ( ) )   ( 1 +  (T − T ) 0

T = T0 +

( R − R0 )

( R0 − R T )

= 20C +

(140  − 12  ) (12  ) ( 0.0045C−1 ) − (140  ) ( 5.5  10−6 C−1 )

= 20C + 2405C = 2425C  2400C (b) The net effect of thermal expansion is that both the length and diameter increase, which lowers the resistance. 4l 0 2 l 1 +  T ( T − T0 )  ld2 1 R d 02 =  d = 02 = 0 = 2 l 0d l0 R0  4 l 0 d 0 1 + T (T − T0 ) 1 + T (T − T0 ) 0  d 02 =

1 + ( 5.5  10 C−1 ) ( 2405C )  −6

= 0.9869

 R − R0  R  100 =  − 1  100 = −1.31  −1.3%   R0   R0 

% change = 

The net effect of resistivity change is that the resistance increases. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

869

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

R R0

 =

0

4l 0

 d 02

4l 0

 d 02

Instructor Solutions Manual

  0 1 +  ( T − T0 ) = = 1 +  ( T − T0 ) = 1 + ( 0.0045 C−1 ) ( 2405C )  0 0

= 11.82

 R − R0  R  100 =  − 1 100 = 1082  1100%  R0   R0 

% change = 

27. (a) Divide the cylinder up into concentric cylindrical shells of radius r, thickness dr, and length l. See the diagram. The resistance of one of those shells, from Eq. 25–3, is found. Note that the “length” in Eq. 25–3 is in the direction of the current flow, so we must substitute in dr for the “length” in Eq. 25–3. The area is the surface area of the thin cylindrical shell. Then integrate over the range of radii to find the total resistance. dr “l ” R= ; → dR =  A 2 r l r2

R =  dR =   r1

2 r l

r r1 r2

r  ln 2 2 l r1

(b) Use the data given to calculate the resistance from the above formula.  r 15  10−5  m  1.8 mm  −4 ln 2 = ln  R=  = 5.8  10  2 l r1 2 ( 0.024 m )  1.0 mm  (c) For resistance along the axis, we again use Eq. 25–3, but the current is flowing in the direction of length l. The area is the cross-sectional area of the face of the hollow cylinder.

15  10−5  m ) ( 0.024 m ) ( l l R= = = = 0.51 A  ( r22 − r12 )  (1.8  10−3 m )2 − (1.0  10−3 m )2   

28. Use Eq. 25–6 to find the power from the voltage and the current.

P = IV = ( 0.24 A )( 3.0 V ) = 0.72 W

29. Use Eq. 25–7b to find the resistance from the voltage and the power. P=

V2 R

→ R=

V2 P

( 240 V )2 3100 W

= 19 

30. Use Eq. 25–7a to find the current from the power and the resistance. P = I 2R → I =

P R=

( 0.25 W ) ( 4200  ) = 7.7  10−3 A

If the current exceeds this, the resistor will over-heat and perhaps be damaged. 31. The power needed is the voltage times the current, Eq. 25–6. P 45, 000 W P = IV → I = = = 132.4A  130 A 340 V V © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

870

Chapter 25

Electric Currents and Resistance

32. The battery’s potential energy is equal to the charge that it can deliver times its voltage.  1000 W   3600s  16 kW h    U  1kW   1h  = 1.7  105 C U = QV → Q = = 340 V V 33. (a) Since P =

→ R=

says that the resistance is inversely proportional to the power for a R P constant voltage, we predict that the 1250 W setting has the higher resistance. We assume that the 120-V voltage has 3 significant figures.

(b) Rlow =

V2 P

power

(c)

Rhi

V2 P

power

= =

(120 V )

1250 W

(120 V )

= 11.52   11.5 

1875 W

= 7.68 

34. (a) Use Eq. 25–6 to find the current. P 130 W P = IV → I = = = 1.18A  1.2 A V 110 V (b) Use Eq. 25–7b to find the resistance. P=

V2 R

→ R=

V2 P

(110 V )2 130 W

 93 

35. The power (and thus the brightness) of the bulb is proportional to the square of the voltage, V2 according to Eq. 25–7b, P = . Since the resistance is assumed to be constant, if the voltage is cut R in half from 240 V to 120 V, the power will be reduced by a factor of 4. Thus the bulb will appear only about 1/4 as bright in the United States as in Europe. 36. To find the kWh of energy, multiply the kilowatts of power consumption by the number of hours in operation.  1kW   1h  Energy = P ( in kW ) t ( in h ) = ( 650 W )  ( 5.0 min )    = 0.05417 kWh  1000 W   60 min 

 0.054 kWh To find the cost of the energy used in a month, multiply the daily kWh usage times 4 days per week of usage, times 4 weeks per month, times the cost per kWh. kWh   4 d  4 week   13eurocents   Cost =  0.05417     kWh  = 11eurocents month d   1week  1month     37. To find the cost of the energy, multiply the kilowatts of power consumption by the number of hours in operation times the cost per kWh.  1kW   24 h   $0.12  Cost = ( 25W )  ( 365day )   = $26.28  $ 26    1000 W   1day   kWh  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

871

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

38. The A h rating is the amount of charge that the battery can deliver. The potential energy of the charge is the charge times the voltage.  3600s  6 6 U = QV = ( 85 A h )   (12 V ) = 3.672  10 J  3.7  10 J 1h  

3.672  106 J 

1W s 1J



1kW 1000 W



1h 3600s

= 1.02 kWh  1.0 kWh

39. (a) Calculate the resistance from Eq. 25–2 and the power from Eq. 25–6. 3.0 V V R= = = 22.22   22  P = IV = ( 0.135 A )( 3.0 V ) = 0.405 W  0.41W I 0.135 A (b) If four AA-cells are used, the voltage will be doubled to 6.0 V. Assuming that the resistance of the bulb stays the same (by ignoring heating effects in the filament), the power that the bulb V2 would need to dissipate is given by Eq. 25–7b, P = . A doubling of the voltage means the R power is increased by a factor of 4 . This should not be tried because the bulb is probably not rated for such a high wattage. The filament in the bulb would probably burn out. 40. Find the power dissipated in the cord by Eq. 25–7a, using Eq. 25–3 for the resistance. 2 ( 2.7 m ) 2l 2 = (16.0 A ) 1.68  10−8  m = 17.77 W  18 W P = I 2R = I 2 2 A  14 0.129  10−2 m

(

)

(

)

41. Find the current used to deliver the power in each case, and then find the power dissipated in the resistance at the given current. P P2 P = IV → I = Pdissipated = I 2 R = 2 R V V

Pdissipated 12,000 V

Pdissipated 50,000 V

( 7.5  10 W ) ( 3.0  ) = 11719 W = (1.2  10 V ) ( 7.5  10 W ) ( 3.0  ) = 675 W difference = 11719 W − 675 W = 1.1  10 W = ( 5  10 V ) 5

42. Use Eqs. 25–3 and 25–7b. V2 V2 4 l l l R= = 2 = P ; = = 4 l A R r d2 d2

→

(1.5 V )2  ( 5.0  10−4 m ) = = 1.461m  1.5 m l = 4 P 4 (1.68  10−8  m ) (18 W ) V 2 d 2

If the voltage increases by a factor of 6 without the resistance changing, the power will increase by a factor of 36. The blanket would theoretically be able to deliver 540 W of power, which might make the material catch on fire or burn the occupant.

872

Chapter 25

Electric Currents and Resistance

43. (a) The efficiency of the bulb is proportional to the power radiated from the bulb. The power radiated is given by Eq. 19–18. Subscript “H” is used for the halogen bulb, and “T” for the traditional incandescent bulb. (Q t ) H − (Q t )T TH4 − TT4 ( 2900 K )4 − ( 2700 K )4 = = = 0.331  33% TT4 (Q t )T ( 2900 K )4 The halogen bulb is about 33% more efficient than the traditional incandescent bulb. (b) The radiated power is proportional to the power consumed by the light bulb. Since the halogen bulb is 33% more efficient than the incandescent, the power input can be decreased by a factor of 1.33 to provide the same output intensity. 100 W PH = PT = = 75 W 1.33 44. (a) By conservation of energy and the efficiency claim, 85% of the electrical power dissipated by the heater must be the rate at which energy is absorbed by the water. Recall that 1 mL of water has a mass of 1 g. Q mcT → 0.85Pemitted by = Pabsorbed → 0.85 ( IV ) = heat water = t t heater by water I=

mcT

( 0.120 kg )( 4186 J kg )( 95C − 25C ) = 8.208 A  8.2 A ( 0.85)(12 V )( 420s )

0.85Vt (b) Use Ohm’s law to find the resistance of the heater. 12 V V V = IR → R = = = 1.46   1.5  I 8.208 A

45. For the wire to stay a constant temperature, the power generated in the resistor is to be dissipated by radiation. Use Eq. 25–7a and 19–18, both expressions of power (energy per unit time). We assume that the dimensions requested and dimensions given are those at the higher temperature, and do not take any thermal expansion effects into account. We also use Eq. 25–3 for resistance. 4 l 4 4 4 4 I 2 R =  A Thigh − Tlow → I2 =  d l Thigh − Tlow → 2 d

(

)

(

)

1/3 2     4 (18.0 A ) ( 5.6  10−8  m ) 4I 2   d = 2 =     ( T 4 − T 4 )    2 (1.0 ) ( 5.67  10−8 W m 2 K 4 ) ( 3100 K ) 4 − ( 293 K ) 4   high low     

1/3

= 1.12  10−4 m  0.11mm

46. Use Ohm’s law and the relationship between peak and rms values. 220 V V I peak = 2 I rms = 2 rms = 2 = 0.124A  0.12 A 2500  R 47. Find the peak current from Ohm’s law, and then find the rms current from Eq. 25–9a. Vpeak 180 V I peak = = = 0.5143A  0.51A I rms = I peak 2 = ( 0.5143A ) 2 = 0.36 A R 350 

873

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

48. (a) When everything electrical is turned off, no current will be flowing into the house, even though a voltage is being supplied. Since for a given voltage, the more resistance, the lower the current, a zero current corresponds to an infinite resistance. (b) Use Eq. 25–10c to calculate the resistance. P=

2 Vrms

(120 V ) = → R= = 9.6  P (1500 W ) 2 Vrms

49. The power and current can be used to find the peak voltage, and then the rms voltage can be found from the peak voltage. I 2 (1750 W ) 2P P = I rmsVrms = peak Vrms → Vrms = = = 405.7 V  410 V I peak 6.1A 2 50. Use the average power and rms voltage to calculate the peak voltage and peak current. (a) Vpeak = 2Vrms = 2 ( 660 V ) = 933.4 V  930 V (b) P = I rmsVrms =

I peak 2

Vrms → I peak =

2P Vrms

2 (1800 W ) 660 V

= 3.9 A

51. (a) We use Eq. 25–10c and 25–10a. 2 Vrms P= → Vrms = PR = (150 W )( 4  ) = 24.49V  20 V R P 150 W I rms = = = 6.12 A  6 A Vrms 24.49 V (b) We repeat the process using the lower power. 2 Vrms P= → Vrms = PR = (1.0 W )( 4  ) = 2 V R P 1.0 W I rms = = = 0.5 A Vrms 2V 52. (a) We assume that the 2.7 hp is the average power, so the maximum power is twice that, or 5.4 hp, as seen in Figure 25–22.  746 W  5.4 hp   = 4028 W  4.0 kW  1hp  (b) Use the average power and the rms voltage to find the peak current. I 2  12 ( 4028 W )  2P P = I rmsVrms = peak Vrms → I peak = = = 11.87 A  12 A 240 V Vrms 2 53. (a) Find Vrms . Use an integral from Appendix B–4. 1/ 2

T    4 t    1/ 2 sin 2    V 2  t 1 T  2 t    T   0  = − Vrms =    V0 sin dt     8 T    T 2  T 0       T 0 

1/ 2

V 2  = 0   2 

V0 2

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Chapter 25

Electric Currents and Resistance

(b) Find Vrms . 1/ 2

1 T  Vrms =   V 2 dt  T 0 

1/ 2

T  1 T /2  1 2 =   V02 dt +  ( 0 ) dt  T T /2 T 0 

1/ 2

V 2 T  = 0 + 0 T 2 

V0 2

54. (a) The average power used can be found from the resistance and the rms voltage by Eq. 25–10c. P=

( 240 V )2

2 Vrms

= = 1800 W 32  R (b) The maximum power is twice the average power, and the minimum power is 0. Pmax = 2 P = 2 (1800 W )  3600 W Pmin = 0 W

55. (a) We follow the derivation in Example 25–14. Start with Eq. 25–14, in absolute value. j I I 4I m = = = j = nevd → vd = ne neA  N (1 mole )  N  D e d 2 2 1   e d   )  D   (2  m (1 mole )  vd =

(

)( ) = 7.1  10 m s     6.02 10 8.9 10 kg m 1.60 10 C 0.65 10 m  )( ) ( ) ( )( 4 3.2  10 −6 A 63.5  10 −3 kg

−19

−10

−3

(b) Calculate the current density from Eq. 25–11. 4 ( 3.2  10−6 A ) I I 4I j= = 2 = = = 9.643A m 2  9.6 A m 2 A r  d 2  ( 6.5  10−4 m )2 (c) The electric field is calculated from Eq. 25–16. 1 j = E → E =  j = (1.68  10−8  m )( 9.643A m 2 ) = 1.6  10−7 V m  56. (a) Use Ohm’s law to find the resistance. V 0.0220 V V = IR → R = = = 0.02933   0.029  0.75 A I (b) Find the resistivity from Eq. 25–3. l → R= A

=

R r 2

( 0.02933  )  (1.0  10−3 m ) = = 1.739  10−8  m  1.7  10−8  m ( 5.30 m ) 2

= l l (c) Use Eq. 25–11 to find the current density. I I 0.75 j= = 2 = = 2.387  105 A m 2  2.4  105 A m 2 2 A r  ( 0.0010 m ) (d) Use Eq. 25–16 to find the electric field. 1 j= E → 

(

)(

)

E =  j = 1.739  10−8  m 2.387  105 A m 2 = 4.151  10 −3 V m  4.2  10 −3 V m

875

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(e) Find the number of electrons per unit volume from the absolute value of Eq. 25–14. j 2.387  105 A m 2 j = nevd → n = = = 8.8  10 28 e − m 3 vd e 1.7  10−5 m s 1.60  10 −19 C

(

)(

)

57. We are given a charge density and a speed (like the drift speed) for both types of ions. From that we can use Eq. 25–13 (without the negative sign) to determine the current per unit area. Both currents are in the same direction in terms of conventional current–positive charge moving north has the same effect as negative charge moving south–and so they can be added. I = neAvd →

I A

(

= ( nevd )He + ( nevd )O =  2.8  1012 ions m 3

)( 2 1.60 10 C ion )( 2.0 10 m s ) + −19

( 7.0  1011 ions m 3 )( −1  1.60  10 −19 C ion )( −6.2  106 m s )  = 2.486 A m 2  2.5 A m 2 , North

58. The magnitude of the electric field is the voltage change per unit meter. V 70  10−3 V = = 7  106 V m E = x 1.0  10 −8 m 59. The speed is the change in position per unit time. x 7.20  10 −2 m − 3.40  10 −2 m v= = = 35 m s 0.0063s − 0.0052 s t Two measurements are needed because there may be a time delay from the stimulation of the nerve to the generation of the action potential. 60. The power is the work done per unit time. The work done to move a charge through a potential difference is the charge times the potential difference. The charge density must be multiplied by the surface area of the cell (the surface area of an open tube, length times circumference) to find the actual charge moved. W QV Q = = V P= t t t mol  ions   −19 C  −6 =  3  10 −7 2  6.02  10 23  1.6  10  ( 0.10 m )  20  10 m ( 0.030 V ) m s  mol  ion  

(

)

= 5.4  10−9 W 61. The energy supplied by the battery is the energy consumed by the lights. Esupplied = Econsumed → QV = Pt →

QV P

( 75A h )( 3600s h )(12V ) 122 W

 1h   = 7.377 h  7.4 h  3600s 

= 26557 s 

62. Use Eq. 25–6 to calculate the current. We assume that 120 V has 3 significant figures. P 746 W P = IV → I = = = 6.22 A V 120 V © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

876

Chapter 25

Electric Currents and Resistance

63. From Eq. 25–2, if R = V I , then G = I V I 0.38 A = 0.1267 S  0.13S G= = 3.0 V V 64. Use Eq. 25–7b to express the resistance in terms of the power, and Eq. 25–3 to express the resistance in terms of the wire geometry. l l l V2 V2 → R= P= R =  =  2 = 4 r d2 R P A 4

d

= 2

V2 P

→ d=

4 l P

V

(

)

4 9.71  10 −8  m ( 4.2 m )(1500 W )

 (110 V )

= 2.5  10 −4 m

65. (a) Calculate the total kWh used per day, and then multiply by the number of days and the cost per kWh. ( 2.2 kW )( 2.0 h d ) + 4 ( 0.1kW )( 6.0 h d ) + ( 3.0 kW )(1.0 h d ) + ( 2.0 kWh d ) = 11.8 kWh d

 $0.115  = $40.71  $40.70 per month   kWh 

Cost = (11.8 kWh d )( 30 d ) 

(b) The energy required by the household is 35% of the energy that needs to be supplied by the power plant. Household energy = 0.35 ( coal mass )( coal energy per mass ) →

coal mass =

(11.8 kWh d )( 365 d ) 

Household energy

( 0.35 )( coal energy per mass )

1000 W   3600 s 

   kW   1h  kcal   4186 J   ( 0.35 )  7500   kg   1kcal  

= 1411kg  1400 kg of coal

66. The resistance of the filament when the flashlight is on is R =

3.0 V

I 0.20 A with a combination of Eqs. 25–3 and 25–5 to find the temperature. R = R0 1 +  ( T − T0 )  →

T = T0 +

= 15 . That can be used

1 R

 1  15   − 1  = 20o C + − 1  = 2020o C  2000o C  1 −    R0 0.0045 ( C o )  1.5   

67. The water temperature rises by absorbing the heat energy that the electromagnet dissipates. Express both energies in terms of power, which is energy per unit time. Q mcT Pelectric = Pto heat → IV = heat water = → t t water m t

IV cT

( 21.5 A )( 240 V ) = 0.224 kg s  0.22 kg s ( 4186 J kg C )( 5.50 C )

This is 224 mL s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

877

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

68. (a) If the wire obeys Ohm’s law, then V = IR or I =

Instructor Solutions Manual

V , showing a linear relationship between I R and V. A graph of I vs. V should give a straight line with a slope of 1 R and a y-intercept of 0. (b) From the graph and the calculated linear fit, we see that the wire obeys Ohm’s law. 1 slope = → R 1 R= A V 0.718 = 1.39  (c) Use Eq. 25–3 to find the resistivity.

 d 2 R  ( 3.2  10 m ) (1.39  ) R= → = = = = 1.0  10−6  m A 4l 4 ( 0.11m ) l l

−4

From Table 25–1, the material is nichrome. 69. We assume that all of the current that enters at a leaves at b, so that the current is the same at each end. The current density is given by Eq. 25–11. 4 ( 2.5 A ) 4I I I = = = = 5.1  105 A m 2 ja = 2 2 2 −3 1 Aa  ( 2 a ) a  2.5  10 m

(

jb =

I Ab

 ( b) 1 2

b

)

4 ( 2.5 A )

 ( 4.0  10 m ) −3

= 2.0  105 A m 2

70. Using Eq. 25–3, we find the infinitesimal resistance first of a thin vertical slice at a horizontal distance x from the center of the left side towards the center of the right side. Let the thickness of that slice be dx. That thickness corresponds to the variable l in Eq. 25–3. The x diameter of this slice is a + ( b − a ) . Then l integrate over all the slices to find the total resistance. l dx R= → dR =  → 2 A x  1   4  a + (b − a )  l   l l

R =  dR =   0

dx x  14  a + ( b − a )  l  

=−

4

x  b−a    a + (b − a )  l  0

4 l

 ab

878

Chapter 25

Electric Currents and Resistance

71. The equivalent amount of work is 3.6  106 J. Set that equal to the change in potential energy. mg y = 3.6  106 J → m =

3.6  106 J g y

3.6  106 J

( 9.8 m s ) (1m ) 2

= 3.67  105 kg  3.7  105 kg

This is about 5000 people, or 250 cars. 72. (a) Use Eq. 25–7b to relate the power to the voltage for a constant resistance. P=

V2 R

→

P105 P117

(105 V ) R (105 V ) = = 0.805 or a 19.5% decrease 2 2 (117 V ) R (117 V ) 2

(b) The lower power output means that the resistor is generating less heat, and so the resistor’s temperature would be lower. The lower temperature results in a lower value of the resistance, which would increase the power output at the lower voltages. Thus, the decrease would be smaller (in absolute value) than the value given in the first part of the problem. 73. Assume that we have a meter of wire, carrying 35 A of current, and dissipating 1.5 W of heat. The l power dissipated is PR = I 2 R, and the resistance is R = . A l l 4 l = I2 2 = I2 → PR = I 2 R = I 2 r d2 A

l = 2I = 2 ( 35 A ) d= I PR PR 2

4 l

(1.68  10  m ) (1.0 m ) = 4.2  10 m −8

(1.5 W ) 

−3

74. (a) The power delivered to the interior is 72% of the power drawn from the source. P 950 W Pinterior = 0.72 Psource → Psource = interior = = 1319 W  1300 W 0.72 0.72 (b) The current drawn is current from the source, and so the source power is used to calculate the current. 1319 W P Psource = IVsource → I = source = = 10.99 A  11A 120 V Vsource 75. The volume of wire is unchanged by the stretching. The volume is equal to the length of the wire times its cross-sectional area, and since the length was increased by a factor of 1.25, the area was decreased by a factor of 1.25. Use Eq. 25–3. l l l A 1.25l 0 2 l = 1.25l 0 A = 0 = (1.25)  0 = 1.56 R0 = 1.56  R0 =  0 R= = A A0 1.25 A A0 0 1.25

76. The long, thick conductor is labeled as conductor number 1, and the short, thin conductor is labeled 2 as number 2. The power transformed by a resistor is given by Eq. 25–7b, P = V R , and both have the same voltage applied.

879

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

R1 =  P1 P2

R2 = 

V12 R1 2 2

R2 R1

l 1 = 2l 2

Instructor Solutions Manual

A1 = 4 A2 ( diameter1 = 2  diameter2 )

 l 2 A2 l 2 A1 1 = = 4 = 2  l 1 A1 l 1 A2 2

P1 : P2 = 2 :1

77. (a) From Eq. 25–6, if power P is delivered to the transmission line at voltage V, there must be a current I = P V . As this current is carried by the transmission line, there will be power losses of I 2 R due to the resistance of the wire. This power loss can be expressed as P = I 2 R

= P 2 R V 2 . Equivalently, there is a voltage drop across the transmission lines of V  = IR. Thus the voltage available to the users is V − V , and so the power available to the users is

P = (V − V ) I = VI − V I = VI − I 2 R = P − I 2 R. The power loss is P = P − P = P − ( P − I 2 R )

= I 2 R = P2 R V 2 . 1

, V should be as large as possible to minimize P. V2 This solution is correct “as is” for DC current, voltage, and power. If, as is more common, the power plant is producing AC electricity, then this solution is still valid if we change P to P and use rms values of voltage and current.

(b) Since P 

78. (a) Use Eq. 25–7b. P=

( 240 V )2

mcT

( 0.15 L ) 

→ R= = = 20.57   21  2800 W R P (b) Only 65% of the heat from the oven is used to heat the water. Use Eq. 19–2. 0.65 ( Poven ) t = Heat absorbed by water = Q = mcT →

t= (c)

11cents kWh

0.65 ( Poven )

( 2.8 kW )( 29.33s )

 1kg   ( 4186 J kg C )(85C )  1L  = 29.33s  29 s 0.65 ( 2800 W )

1h 3600 s

= 0.25cents

( 14 of a cent )

79. (a) The horsepower required is the power dissipated by the frictional force, since we are neglecting the energy used for acceleration.  1m s  P = F v = ( 440 N )( 45 km hr )   = 5500W  3.6 km hr 

 1hp   = 7.373 hp  7.4 hp  746 W 

P = 5500W 

(b) The charge available by each battery is Q = 95A h = 95 C s 3600 s = 3.42  10 C, so the total charge available is 24 times that. The potential energy of that charge is the charge times the voltage. That energy must be delivered (batteries discharged) in a certain amount of time to produce the 5500 W necessary. The speed of the car times the discharge time is the range of the car between recharges. 5

880

Chapter 25

Electric Currents and Resistance

→ t=

d = vt = v

QV P

QV Fv

QV P =

d v

QV F

→

(

)

24 3.42  105 C (12 V ) 440 N

= 2.24  105 m  220 km

80. (a) The resistance at the operating temperature can be calculated directly from Eq. 25–7b.

2 120 V ) ( = = 190 

→ R= 75 W R P (b) The resistance at room temperature is found by converting Eq. 25–5 into an equation for resistances and solving for R0 .

R = R0 1 +  ( T − T0 ) R

R0 =

1 +  (T − T )

1 + ( 0.0045 K

192  −1

) ( 2800 K − 293 K )

= 15.63   16 

81. (a) The angular frequency is  = 210 rad s.

f =

 210 rad s = = 33.42 Hz  33 Hz 2 2

(b) The maximum current is 1.40 A. I 1.40 A I rms = max = = 0.990 A 2 2 (c) For a resistor, V = IR.

V = IR = (1.40 A )( sin 210 t )( 21.5  ) = ( 30.1sin 210 t ) V 82. Use Eq. 25–7b. (a) P = (b) P =

V2 R

(120 V )

12 

(120 V ) 2 150 

= 1200 W

= 96 W

83. To deliver 15 MW of power at 120 V requires a current of I =

15  106 W

= 1.25  105 A. The 120 V V resistance of the connecting wires is found using Eq. 25–3, and then the power loss calculated. 2 ( 4.5  104 m ) l l l −8 = 4 (1.68  10  m ) = 77.0  R =  =  2 = 4 2 A r d2  ( 5.0  10−3 m )

(

Plost to = I 2 R = 1.25  105 A heat

) ( 77.0  ) = 1.20  10 W 2

(

1kW  $0.12  (1h )  )  1000   W kWh

Cost = ( Power )( time )( rate per kWh ) = 1.20  1012 W 









= $1.44  108  $140 million © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

881

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

84. Use Eq. 25–7b for the power in each case, assuming the resistance is constant. V2 R P13.8 V 13.82 13.8 V = 2 = = 1.3225 = 32% increase P12.0 V 12.0 2 V R

( (

) )

12.0 V

85. (a) The power is given by Eq. 25–6.

P = IV = (18 A )( 220 V ) = 3960W  4.0  103 W (b) The power dissipated by the resistor is given by PR = I 2 R, and the resistance is R =

l A

l 4 l 2 4 (1.68  10  m ) 2 ( 3.5 m ) PR = I R = I =I = I2 = (18 A ) = 18.486 W 2 2 2 r d A  (1.62  10−3 m ) 2

l

−8

 18 W

(c) Recalculate using the alternate wire. −8 2 4 1.68  10  m 2 ( 3.5 m ) 2 4 l = (18 A ) = 11.543W  12 W PR = I 2 d2  2.05  10−3 m

(

)

(

)

(d) The savings is due to the power difference.  1kW   12 h   $0.12  Savings = (18.486 W − 11.543W )  ( 30 d )      1000 W   1d   1kWh  = $0.2999 / month  30 cents per month

86. The solution assumes that the 40 W values are precise to 2 significant figures. (a) The current can be found from Eq. 25–6.

I =PV

I A = PA VA = 40W 120 V = 0.33 A

I B = PB VB = 40W 12 V = 3.3 A

(b) The resistance can be found from Eq. 25–7b. R=

VA2

(120 V )

RA = = = 360  40 W P PA (c) The charge is the current times the time.

Q = It

RB =

VB2 PB

(12 V ) 40 W

= 3.6 

QA = I At = ( 0.33 A )( 3600 s ) = 1200 C QB = I B t = ( 3.3 A )( 3600 s ) = 12, 000 C

(d) The energy is the power times the time, and the power is the same for both bulbs.

E = Pt (e)

E A = E B = ( 40 W )( 3600s ) = 1.4  105 J

Bulb B requires a larger current, and so should have larger diameter connecting wires to avoid overheating the connecting wires.

882

Chapter 25

Electric Currents and Resistance

87. The volume of wire ( Volume = length  cross-sectional area ) remains constant as the wire is stretched from the original length of l 0 to the final length of 2 l 0 . Thus the cross-sectional area changes from A0 to 12 A0. Use Eq. 25–3 for the resistance and Eq. 25–7b for the power dissipated.

R0 =

 L0

; R=

L

 2 L0

 L0

A0 A A0 A0 The power is reduced by a factor of 4 . 1 2

= 4 R0 ; P0 =

V2 R0

; P=

V2 R

V2 4 R0

= 14

V2 R0

= 14 P0

88. (a) The energy stored is electrical potential energy, found by multiplying the available charge by the potential.



 1C s   3600s   6 6    (12 V ) = 3.67  10 J  3.7  10 J   1A   1 h  

U = QV = ( 85A h ) 



(b) The work done by the potential energy applies a force to overcome the retarding force, and thus does the same amount of work as the retarding force (but of opposite sign). U 3.67  106 J W = U = Fd → d = = = 1.412  104 m  1.4  104 m 260 N F 89. (a) The D-cell provides 25 mA at 1.5 V for 820 h, at a cost of $1.70.  1kW  Energy = Pt = VIt = (1.5V )( 0.025 A )(820 h )   = 0.03075 kWh  1000 W  $1.70

Cost kWh =

= $55.28 kWh  $55 kWh 0.03075 kWh (b) The AA-cell provides 25 mA at 1.5 V for 120 h, at a cost of $1.25.  1kW  Energy = Pt = VIt = (1.5 V )( 0.025 A )(120 h )   = 0.0045 kWh  1000 W 

$1.25

Cost kWh =

= $277.78 kWh  $280 kWh 0.0045 kWh (c) Compare the battery costs with a normal commercial electric rate. $55.28 kWh  5.0  10 2 times as costly D-cell: $0.11 kWh AA-cell:

$277.78 kWh $0.11 kWh

 2.5  103 times as costly

90. The wasted power is due to losses in the wire. The current in the wire can be found by I = P V . (Note that we assume that the FULL length of copper wire is 25.0 m. Since the wire has to make a complete circuit, the length of a 2-wire connecting cord would only be ~ 12.5 m.) P2 P2  l P2  l P 2 4 l 2 = 2 = (a) PR = I R = 2 R = 2 V V A V  r2 V 2  d 2 −8 (1250 W )2 4 1.68  10  m ( 25.0 m ) = = 8.65 W 2 (120 V )2  2.59  10−3 m

(

)

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(b) PR =

(1250 W )2 4 (1.68  10  m ) ( 25.0 m ) = 3.42 W 2 (120 V )2  ( 4.12  10−3 m ) −8

P 2 4 l V 2 d

Instructor Solutions Manual

= 2

91. Eq. 25–3 can be used. The area to be used is the cross-sectional area of the pipe. 1.68  10−8  m (12.0 m ) l l R= = = = 1.60  10−4  2 2 −2 −2 A  r2 − r2    2.50  10 m − 1.50  10 m

(

outside

inside

)

(

(

) (

)

)

92. When the tank is empty, the entire length of the wire is in a non-superconducting state, and so has a non-zero resistivity, which we call  . Then the resistance of the wire when the tank is empty is given by R0 = 

. When a length x of the wire is superconducting, that portion of the wire A I has 0 resistance. Then the resistance of the wire is only due to the length l − x, and so l −x l l −x l −x = = R0 R= . This resistance, combined with the constant current, gives l A A l V = IR.

 V0  l − x  x = V0  1 −  = V0 (1 − f ) → R0  l  l  R0 

V = IR = 

f = 1−

V V0

Thus, a measurement of the voltage can give the fraction of the tank that is filled with liquid helium. 93. The heater must heat 130 m 3 of air per hour from 5C to 22C, and also replace the heat being lost at a rate of 750 kcal/h. Use Eq. 19–2 to calculate the energy needed to heat the air. The density of air is found in Table 13–1. Q m m3   kg   kcal  kcal  Q = mcT → = cT =  130 (17C ) = 485  1.29 3   0.17  o  t t h  m  kg C  h 

Power required = 485

kcal h

+ 750

kcal h

kcal  4186 J   1h 

= 1235





 = 1436 W  1400 W

h  kcal   3600s 

94. The resistor code in Figure 25–12 shows that the resistance is 24  101 = 240 . Solve for the resistivity by inserting this resistance into Eq. 25–3 along with the length and diameter of the resistor.

=R

A l

d2 4l

= ( 240  )

 ( 2.15  10−3 m )

(

−3

4 9.00  10 m

)

= 9.68  10−2  m  97  10 −3  m

Using Table 25–1, the only material that has a resistivity in this range is Germanium . 95. (a) We use Eq. 25–3. RCu  Cu l A  Cu 1.68  10−8  m = = = = 0.634 RAl  Al l A  Al 2.65  10−8  m (b) Use Eq. 25–3 with the resistivity of copper to find the resistance. 4 (125 m ) 4l l RCu =  =  = 1.68  10−8  m = 1.01  2 2 A d  1.63  10−3 m

(

)

(

)

884

Chapter 25

Electric Currents and Resistance

(c) Use Eq. 25–3 with the resistivity of aluminum. 4 (125 m ) 4l l RAl =  =  = 2.65  10 −8  m = 1.59  2 2 A d  1.63  10−3 m

(

)

(

)

(d) Use Eq. 25–7a to determine the power dissipated in each wire.

PCu = I 2 RCu = (18A ) (1.01  ) = 327 W  330 W 2

PAl = I 2 RAl = (18 A ) (1.59  ) = 515 W  520 W 2

(e) Set the resistances in Eq. 25–3 equal and solve for the ratio of the wire diameters. 4l 4l =  Cu → RAl = RCu →  Al 2 2  d Al  d Cu

 Al 2.65  10−8  m = 1.63mm = 2.05 mm  Cu 1.68  10−8  m

d Al = d Cu

(f)

In Section 25–4, the text says that the resistance of a given weight of aluminum is less than that of the same weight of copper. Use the density of copper to find the mass of the copper wire in part (b):

mCu =  CuV =

 Cu d 2 l

(8.9  10 kg m )  (1.63  10 m ) (125 m ) = 2.32 kg = 3

−3

4 4 Set the mass of aluminum equal to the mass of the copper, and then calculate the diameter of the aluminum wire.

mAl =

 Al d 2 l 4

→ d=

4mAl

 M l

4 ( 2.32 kg )

( 2.70  10 kg m )  (125 m ) 3

= 2.96  10−3 m

Using the length, diameter, and resistivity of aluminum, calculate the resistance of the aluminum wire. −8 4  Al l 4 2.65  10  m (125 m ) RAl = = = 0.481  2 d2  2.96  10−3 m

(

)

(

)

In part (b) above, the copper has a resistance of 1.01 . Thus the statement about equivalent weights of aluminum vs. copper is true. The resistance of the aluminum is about 48% of the resistance of copper. 96. Model the protons as moving in a continuous beam of cross-sectional area A. Then use Eq. 25–13, I = neAv d , where we only consider the absolute value of the current. The variable n is the number of protons per unit volume, so n =

, where N is the number of protons in the beam and l is the Al circumference of the ring. The “drift” velocity in this case is the speed of light. N N I = neAvd = eAvd = evd → Al l ( 0.58 A ) 27  103 m Il N= = = 3.3  1014 protons 8 −19 evd 1.60  10 C 3.00  10 m s

(

( )(

)

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97. The electrons are assumed to be moving with simple harmonic motion. During one cycle, an object in simple harmonic motion will move a distance equal to twice its amplitude–it will move from its minimum position to its maximum position, for example (and then back to its minimum position). From Eq. 14–9a, we know that v max = A , where  is the angular frequency of oscillation and A is the amplitude. From Eq. 25–13 in absolute value, we see that I max = ne ( Area ) vmax . Finally, the maximum current can be related to the power by Eqs. 25–9 and 25–10. The charge carrier density, n, is calculated in Example 25–14. P = I rmsVrms = 12 I maxVrms

A= =

v max



I max

 n e ( Area )

 n e ( 14  d 2 ) Vrms 4 2 ( 750 W )

(

2 ( 60 Hz ) 8.4  10 m 28

−3

)(1.60 10 C )  (1.7 10 m ) (120 V ) −19

−3

= 7.69  10 −7 m

The electron will move this distance in both directions from its equilibrium point, so it’s maximum displacement is from one extreme to the other, which is twice the amplitude  1.5  10−6 m . 98. For the cylindrical wire, its (constant) volume is given by V0 = l 0 A0 = l A, and so A = this relationship with Eq. 25–3. We assume that l

R0 =  R 

l0 A0

dR dl

=

l 02 V0

l = 2 

; R= l V0

l A

=

l2 V0

l →  l =

Since we assumed that l

;

V0 R 2 l

dR dl

l 0 , we can say that

. Combine

l

l 0.

= 2

→

V0 R 2 l

l l 0 + l

= 2



R R = 12 2 l R

l l0

V0 , and so

l l0

= 12

R R0

886

CHAPTER 26: DC Circuits Responses to Questions 1.

Even though the bird’s feet are at high potential with respect to the ground, there is very little potential difference between them, because they are close together on the wire. The resistance of the bird is much greater than the resistance of the wire between the bird’s feet. These two resistances are in parallel, so very little current will pass through the bird as it perches on the wire. However, when you put a metal ladder up against a power line, you provide a direct connection between the high potential line and ground. The ladder will have a large potential difference between its top and bottom. A person standing on the ladder will also have a large potential difference between his or her hands and feet. Even if the person’s resistance is large, the potential difference will be great enough to produce a current through the person’s body large enough to cause substantial damage or death.

Series: The main disadvantage of Christmas tree lights connected in series is that when one bulb burns out, a break is created in the circuit and none of the bulbs remain lit. Finding the burned-out bulb requires replacing each individual bulb, one at a time, until all of the bulbs are lit again. As an advantage, the bulbs are slightly easier to wire in series. Only one wire needs to be attached from bulb to bulb. Also, a “blinker bulb” can make the entire string flash on and off by cutting off the current. Parallel: The main advantage of connecting the bulbs in parallel is that one burned-out bulb does not affect the rest of the strand, and is easy to identify and replace. As a disadvantage, wiring the bulbs in parallel is slightly more difficult. Two wires must be attached from bulb to bulb.

It is possible. If 20 of the 6-V lamps were connected in series and then connected to the 120 V line, there would be a voltage drop of 6 V for each of the lamps, and they would not burn out due to too much voltage. Being in series, if one of the bulbs went out for any reason, they would all turn off.

If the lightbulbs are in series, each will have the same current. The power dissipated by the bulb as heat and light is given by P = I 2 R . Thus the bulb with the higher resistance ( R2 ) will be brighter. If the bulbs are in parallel, each will have the same voltage. The power dissipated by the bulb as 2 heat and light is given by P = V R . Thus the bulb with the lower resistance ( R1 ) will be brighter.

The outlets are connected in parallel to each other, because you can use one outlet without using the other. If they were in series, both outlets would have to be used at the same time to have a completed circuit. Also, both outlets supply the same voltage to whatever device is plugged in to the outlet, which indicates that they are wired in parallel to the voltage source.

2 The power output from a resistor is given by P = V R. To maximize this value, the voltage needs to be as large as possible and the resistance as small as possible. That can be accomplished by putting the two batteries in series, which would result in the sum of their individual voltages, and then connecting the two resistors in parallel to each other (which reduces their equivalent resistance), across the full 2-battery voltage.

The battery has to supply less power when the two resistors are connected in series than it has to supply when only one resistor is connected. P = V 2 R , so if V is constant and R increases, the power decreases.

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We assume that the bulbs are in parallel with each other. The overall resistance decreases and more current is drawn from the source. A bulb rated at 60-W and 120-V has a resistance of 240 Ω. A bulb rated at 100-W and 120-V has a resistance of 144 Ω. When only the 60-W bulb is on, the total resistance is 240 Ω. When both bulbs are lit, the total resistance is the combination of the two resistances in parallel, which is only 90 Ω.

When resistors are connected in series the equivalent resistance is the sum of the individual resistances, Req,series = R1 + R2 + …. The current has to go through each additional resistance if the resistors are in series, and therefore the equivalent resistance is greater than any individual resistance. In contrast, when capacitors are in parallel the equivalent capacitance is equal to the sum of the individual capacitors, Ceq,parallel = C1 + C2 + …. Charge drawn from the battery can go down any one of the different branches and wind up on any one of the capacitors, so the overall capacitance is greater than that of each individual capacitor. When resistors are connected in parallel, the current from the battery or other source divides into the different branches and so the equivalent resistance is less than any individual resistor in the circuit. The corresponding expression is 1/Req,parallel = 1/R1 + 1/R2 + …. The formula for the equivalent capacitance of capacitors in series follows this same form; 1/Ceq,series = 1/C1 + 1/C2 + …. When capacitors are in series, the overall capacitance is less than the capacitance of any individual capacitor. Charge leaving the first capacitor lands on the second rather than going straight to the battery. Compare the expressions defining resistance (R = V/I) and capacitance (C = Q/V). Resistance is proportional to voltage, whereas capacitance is inversely proportional to voltage. The table below summarizes the equivalent component rules.

10. No, the sign of the battery’s emf does not depend on the direction of the current through the battery. The sign of the battery’s emf depends on the direction you “go through” the battery in applying the loop rule. If you go from negative pole to positive pole, the emf is added. If you go from positive pole to negative pole, the emf is subtracted. On the other hand, the terminal voltage does depend on the direction of the current through the battery. If current is flowing through the battery in the normal orientation (leaving the positive terminal, flowing through the circuit, and arriving at the negative terminal) then there is a voltage drop across the internal resistance, and the terminal voltage is less than the emf. If the current flows in the opposite sense (as in charging the battery), then there is a voltage rise across the terminal resistance, and the terminal voltage is higher than the emf. 11. (a) With the batteries in series, a greater voltage is delivered to the lamp, and the lamp will burn brighter. Thus the current is larger than either battery would provide by itself. (b) With the batteries in parallel, the voltage across the lamp is the same as for either battery alone (assuming the batteries are the same voltage).

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Chapter 26

DC Circuits

Each battery supplies only half of the current going through the lamp, so the batteries will last longer (and the bulb stay lit longer) as compared to having just one battery, or two batteries in series. 12. When batteries are connected in series, their emfs add together, producing a larger potential. For instance, if there are two 1.5-V batteries in series in a flashlight, the potential across the bulb will be 3.0 V, and there will be more current in the bulb. The batteries do not need to be identical in this case. When batteries are connected in parallel, the currents they can generate add together, producing a larger current over a longer time period. Batteries in this case need to be nearly identical, or the battery with the larger emf will end up charging the battery with the smaller emf. 13. Yes. When a battery is being charged, current is forced through it “backwards” and Vterminal = e + Ir, so Vterminal  e. 14. Refer to Fig. 26–2. We assume that we know the emf of the battery. Connect the battery to a known resistance R, and measure the terminal voltage Vab. The current in the circuit is given by Ohm’s law to be I = r=

Vab

e − Vab I

. It is also true that Vab = e − Ir and so the internal resistance can be calculated by

e − Vab Vab

15. No, the total energy supplied by the battery is not equal to the total energy stored by the capacitor. As current passes through the resistor in the RC circuit, energy is dissipated in the resistor. Therefore, the total energy supplied by the battery during the charging is the combination of the energy dissipated in the resistor and the energy stored in the capacitor. 16. (a) Stays the same; (f) Decreases;

(b) Increases; (g) Decreases;

(d) Increases; (e) Increases; (i) Remains the same.

17. See the adjacent figure. If both switches are connected to the same wire, the circuit is complete and the light is on. If they are connected to opposite wires, the light will remain off.

18. The soles of your shoes are made of a material which has a relatively high resistance, and there is relatively high resistance flooring material between your shoes and the literal ground (the Earth). With that high resistance, a malfunctioning appliance would not be able to cause a large current flow through your body. The resistance of bare feet is much less than that of shoes, as the feet are in direct contact with the ground, so the total resistance is much lower and so a larger current would flow through your body. 19. In an analog ammeter, the internal resistor, or shunt resistor, has a small value and is in parallel with the galvanometer, so that the overall resistance of the ammeter is very small. In an analog voltmeter, the internal resistor has a large value and is in series with the galvanometer, and the overall resistance of the voltmeter is very large. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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20. If you mistakenly use an ammeter where you intend to use a voltmeter, you are inserting a short in parallel with some resistance. That means that the resistance of the entire circuit has been lowered, and almost all of the current will flow through the low-resistance ammeter. Ammeters usually have a fairly small current limit, and so the ammeter might very likely get damaged in such a scenario. Also, if the ammeter is inserted across a voltage source, the source will provide a large current, and again the meter will almost certainly be damaged, or at least disabled by burning out a fuse. 21. An ammeter is placed in series with a given circuit element in order to measure the current through that element. If the ammeter did not have very low (ideally, zero) resistance, its presence in the circuit would change the current it is attempting to measure by adding more resistance in series. An ideal ammeter has zero resistance and thus does not change the current it is measuring. A voltmeter is placed in parallel with a circuit element in order to measure the voltage difference across that element. If the voltmeter does not have a very high resistance, then its presence in parallel will lower the overall resistance and affect the circuit. An ideal voltmeter has infinite resistance so that when placed in parallel with circuit elements it will not change the value of the voltage it is reading, because no current with flow through it. 22. When a voltmeter is connected across a resistor, the voltmeter is in parallel with the resistor. Even if the resistance of the voltmeter is large, the parallel combination of the resistor and the voltmeter will be slightly smaller than the resistor alone. If Req of that part of the circuit decreases, then the overall current in the circuit will increase, so that the potential drop across the rest of the circuit will increase. Thus, the potential drop across the parallel combination will be less than the original voltage drop across the resistor. Here is another way to explain it. When the voltmeter is connected across the resistor, the equivalent resistance decreases. That makes the (resistor + voltmeter) combination a smaller fraction of the total resistance of the circuit than the resistor was alone, which means that it will have a smaller fraction of the total voltage drop across it. 23. A voltmeter has a very high resistance. When it is connected to the battery very little current will flow. A small current results in a small voltage drop due to the internal resistance of the battery, and the emf and terminal voltage (measured by the voltmeter) will be very close to the same value. However, when the battery is connected to the lower-resistance flashlight bulb, the current will be higher and the voltage drop due to the internal resistance of the battery will also be higher. As a battery is used, its internal resistance increases. Therefore, the terminal voltage will be significantly lower than the emf: Vterminal = e − Ir. A lower terminal voltage will result in a dimmer bulb, and usually indicates a “used-up” (high internal resistance) battery.

Solutions to MisConceptual Questions 1.

(a, c) Resistors are connected in series when there are no junctions between the resistors. With no junctions between the resistors, any current flowing through one of the resistors must also flow through the other resistor.

(d) It might be easy to think that all three resistors are in parallel because they are on three branches of the circuit. However, following a path from the positive terminal of the battery, all of the current from the battery passes through R2 before reaching the junction at the end of that

890

Chapter 26

DC Circuits

branch. The current then splits and part of the current passes through R1 while the remainder of the current passes through R3 before meeting at the right junction. 3.

(c) A common misconception is that the smaller resistor would have the larger current. However, the two resistors are in series so they must have the same current flowing through them. It does not matter which resistor is first.

(d) Current takes the path of least resistance. Since bulb A is in parallel with the short circuit, all of the current will pass through the short circuit, causing bulb A to go out.

(b) A common misconception is to think that adding another parallel branch would have no effect on the voltage across R4. However, when the switch is closed the additional parallel resistor makes the effective resistance of the parallel resistors smaller, and therefore the resistance of the entire circuit gets smaller. With a smaller effective resistance, a greater current flows through the battery and through R1, resulting in a greater voltage drop across R1. Since the voltage from the battery is equal to the sum of the voltages across R1 and R4 , increasing the voltage across R1 results in a decrease in voltage across R4.

(a) As explained in the answer to Question 7, when the switch is closed it adds an additional resistor in parallel to R3 and R4 , making the effective resistance of the entire circuit smaller. With less effective resistance, a greater current flows through the circuit, increasing the potential difference across R1.

(b) As the capacitor charges the voltage drop across the capacitor increases, thereby diminishing the voltage drop across the resistor. As the voltage drop across the resistor decreases, the current decreases.

(c) Even though steady current cannot flow through a capacitor, charge can build up on the capacitor allowing current to initially flow in the circuit. As the charge builds on the capacitor, the voltage drop across the capacitor increases, and the current decreases. The rate of charging is determined by the time constant, which is the product of R and C.

(d) Resistor R1 and R2 are in parallel, so each has half of the current from the battery. R3 and R4 are in series and add to produce twice as much resistance as R5 . Since they are in parallel with R5 , one third of the current from the battery goes through them, while two-thirds goes through R5. The greatest current therefore goes through R5.

10. (b) It might seem that a capacitor would discharge linearly in time, losing one-fourth of the charge every second. This is incorrect, as the capacitor discharges exponentially. That is, every 2.0 seconds half of the remaining charge on the capacitor will discharge. After 2.0 seconds, half of the charge remains. After 4.0 seconds, half of the half, or one-fourth, of the charge remains. After 6.0 seconds, one-eighth of the charge remains. 11. (a) To double the heart rate, the time of discharging must be shorter, so the discharge rate must be faster. The resistance limits the rate at which the current can flow through the circuit. Decreasing the resistance will increase the current flow causing the capacitor to discharge faster. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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12. (c) In a two-prong cord, one prong is at high voltage and the other is grounded. Electricity flows through the appliance between these two wires. However, if there is a short between the high voltage wire and the casing, the casing can become charged and electrocute a person touching the case. The third prong connects the external case to ground, so that the case cannot become charged. 13. (b) The ammeter is placed in series with a component in the circuit and therefore should have a small resistance, so there is minimal voltage drop across the ammeter. The voltmeter is placed in parallel with a component in the circuit. It should have a large resistance so that minimal current from the circuit passes through the voltmeter instead of passing through the circuit.

Solutions to Problems 1.

See Fig. 26–2 for a circuit diagram for this problem. Using the same analysis as in Example 26–1, e the current in the circuit is I = . Use Eq. 26–1 to calculate the terminal voltage. We assume R+r that the 610- resistor is known to three significant figures. 61.0   e  r = e ( R + r ) − e r = e R = 6.00 V = 5.91V (a) Vab = e − Ir = e −  ( )  R+r R+r ( 61.0 + 0.900)  R+r (b) Vab = e

R R+r

= ( 6.00 V )

610 

( 610 + 0.900 ) 

= 5.99 V

See the circuit diagram below. The current in the circuit is I. The voltage Vab is given by Ohm’s law to be Vab = IR. That same voltage is the terminal voltage of the series emf. b

Vab = ( e − Ir ) + ( e − Ir ) + ( e − Ir ) + ( e − Ir ) = 4 ( e − Ir ) and Vab = IR 4 ( e − Ir ) = IR → r =

e − 14 IR I

(1.5 V ) − 14 ( 0.47 A )(12  ) 0.47 A

We take the low-resistance ammeter to have no resistance. The circuit is shown. The terminal voltage will be 0 volts. e 1.5 V Vab = e − Ir = 0 → r = = = 0.060  I 25 A

= 0.1915   0.2 

a e

I A

892

Chapter 26

DC Circuits

See Figure 26–2 for a circuit diagram for this problem. Use Eq. 26–1. e − Vab 12.0 V − 8.2 V Vab = e − Ir → r = = = 0.040  95 A I Vab = IR → R =

Vab I

8.2 V 95 A

= 0.086 

The equivalent resistance is the sum of the two resistances: Req = R1 + R2 . The current in the circuit is then the voltage divided by the equivalent resistance: I =

e Req

e R1 + R2

. The

voltage across the 2200- resistor is given by Ohm’s law. 1800  R2 e V1800 = IR2 = R2 = e = (12.0 V ) = 8.6 V 720  + 1800  R1 + R2 R1 + R2 6.

(a) For the resistors in series, use Eq. 26–3, which says the resistances add linearly. Req = 3 ( 45  ) + 3 ( 65  ) = 330  (b) For the resistors in parallel, use Eq. 26–4, which says the resistances add reciprocally. 3 ( 65  ) + 3 ( 45  ) 1 1 1 1 1 1 1 3 3 = + + + + + = + = → Req 45  45  45  65  65  65  45  65  ( 65  )( 45  )

Req = 7.

The equivalent resistance of five 100- resistors in parallel is found, and then that resistance is divided by 10  to find the number of 10- resistors needed.

1

Req = 

 R1

( 65  )( 45  ) = 8.9  3 ( 65  ) + 3 ( 45  )

1 R2

1 R3

1 R4

1 

−1

20   5  = 20  = n (10  ) → n = = 2  10   100  

 =

R5 

(a) The maximum resistance is made by combining the resistors in series. Req = R1 + R2 + R3 = 520  + 760  + 1200  = 2.48 k (b) The minimum resistance is made by combining the resistors in parallel. 1 1 1 1 = + + → Req R1 R2 R3 −1

−1

1 1 1 1 1   1 Req =  + +  = + +  = 246   250   520  760  1200    R1 R2 R3  9.

The resistance of each bulb can be found by using Eq. 25–7b, P = V 2 R . The two individual resistances are combined in parallel. We label the bulbs by their wattage. 1 P P =V2 R → = 2 R V

893

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Instructor Solutions Manual

−1

 75 W  1 1  25 W  Req =  + = + = 144   140   2 2   R75 R25   (120 V ) (120 V )  10. The resistors can all be connected in series. Req = R + R + R = 3 ( 2.20 k ) = 6.60 k The resistors can all be connected in parallel. 1

Req

1 R

−1

 3  = R = 2.20 k = 733   3 3 R

→ Req = 

Two resistors in series can be placed in parallel with the third. 1 1 1 1 1 3 2 R 2 ( 2.20 k ) = + = + = → Req = = = 1.47 k Req R R + R R 2R 2 R 3 3 Two resistors in parallel can be placed in series with the third. −1

 1 + 1  = R + R = 3 2.20 k = 3.30 k ( )  2 2 R R

Req = R + 

11. The resistance of each bulb can be found from its power rating. 2 V2 V 2 (12.0 V ) P= → R= = = 48  3.0 W R P Find the equivalent resistance of the two bulbs in parallel. R 48  1 1 1 2 = + = → Req = = = 24  Req R R R 2 2

R R

The terminal voltage is the voltage across this equivalent resistance. Use that to find the current drawn from the battery. V V 2V Vab = IReq → I = ab = ab = ab Req R 2 R Finally, use the terminal voltage and the current to find the internal resistance, as in Eq. 26–1. 12.0 V − 11.8 V e − Vab e − Vab e − Vab Vab = e − Ir → r = = =R = ( 48  ) = 0.407   0.4  I 2Vab 2 (11.8 V )  2Vab 

   R 

12. (a) Each bulb should get one-eighth of the total voltage, but let us prove that instead of assuming it. Since the bulbs are identical, the net resistance is Req = 8R . The current flowing through the bulbs is then Vtot = IReq → I =

Vtot Req

Vtot 8R

. The voltage across one bulb is found from Ohm’s

law.

V = IR = (b)

Vtot 8R

→ R=

R= Vtot 8I

Vtot 8 =

120 V

8 120 V

8 ( 0.65 A )

= 15V = 23.08   23 

P = I 2 R = ( 0.65 A ) ( 23.08  ) = 9.751W  9.8 W 2

894

Chapter 26

DC Circuits

13. We model the resistance of the long leads as a single resistor r. Since the bulbs are in parallel, the total current is the sum of the current in each bulb, and so I = 8 I R . The voltage drop across the long leads is Vleads = Ir = 8 I R r = 8 ( 0.21A )(1.4  ) = 2.352 V. Thus the voltage across each of the parallel resistors is VR = Vtot − Vleads = 120 V − 2.352 V = 117.6 V. Since we have the current through each resistor, and the voltage across each resistor, we calculate the resistance using Ohm’s law. V 117.6 V VR = I R R → R = R = = 560  IR 0.21A The total power delivered is P = Vtot I , and the “wasted” power is I 2 r. The fraction wasted is the ratio of those powers. I 2r Ir 8 ( 0.21A )(1.4  ) fraction wasted = = = = 0.0196  0.020 IVtot Vtot 120 V So about 2% of the power is wasted. 14. To fix this circuit, connect another resistor in parallel with the 480- resistor so that the equivalent resistance is the desired 330 .

1 Req

1 R1

−1

1 R2

−1

 1 1 1   1 → R2 =  −  = −  = 1056   1100   330  480    Req R1 

So solder an 1100- resistor in parallel with the 480- resistor. 15. Each bulb will get one-tenth of the total voltage, and so Vbulb =

Vtot

. That could be proven using a 10 similar argument to that used in Problem 12. Use that voltage and the power dissipated by each bulb to calculate the resistance of a bulb.

Pbulb =

2 Vbulb

→ R=

2 Vbulb

Vtot2 100 P

(120 V )2 = 21  100 ( 7.0 W )

16. (a) The equivalent resistance is found by combining the 750- and 580- resistors in parallel, and then adding the 990- resistor in series with that parallel combination. −1

1   1 Req =  +  + 990  = 327  + 990  = 1317   1320   750  580  

(b) The current delivered by the battery is I =

V Req

12.0 V 1317 

= 9.112  10−3 A. This is the

current in the 990- resistor. The voltage across that resistor can be found by Ohm’s law. V990 = IR = 9.112  10−3 A ( 990  ) = 9.021V  9.0 V

(

)

Thus the voltage across the parallel combination must be 12.0 V − 9.021V = 2.979 V  3.0 V . This is the voltage across both the 750- and 580- resistors, since parallel resistors have the same voltage across them. Note that this voltage value could also be found as follows. Vparallel = IRparallel = 9.112  10−3 A ( 327  ) = 2.980 V  3.0 V

(

)

(c) The current in the 990- resistor was already seen to be 9.11 mA . The other currents can be found from Ohm’s Law. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

895

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

I 750 =

V750 R750

2.980 V

= 3.97 mA  4.0 mA ; I 580 =

750 

V580 R580

Instructor Solutions Manual

2.980 V 580 

= 5.137 mA  5.1mA

Note that these last two currents add to be the current in the 990- resistor. 17. (a) The three resistors on the far right are in series, so their equivalent resistance is 3R. That combination is in parallel with the next resistor to the left, as shown in the dashed box in the second figure. The equivalent resistance of the dashed box is found as follows. −1

1 1  = 3R Req1 =  +  4  R 3R  This equivalent resistance of 43 R is in series with the next two resistors, as shown in the dashed box in the third figure. The equivalent resistance of that dashed box is Req2 = 2 R + 43 R = 114 R. This 114 R is in parallel with the next resistor to the left, as shown in the fourth figure. The equivalent resistance of that dashed box is found as follows. −1

1 4  11 Req2 =  +  = 15 R  R 11R  This is in series with the last two resistors, the ones connected directly to A and B. The final equivalent resistance is given below. 11 Req = 2 R + 15 R = 1541 R = 1541 ( 215  ) = 587.67   588  (b) The current flowing from the battery is found from Ohm’s law. V 50.0 V = = 0.08508 A  8.51  10−2 A I total = Req 587.67  This is the current in the top and bottom resistors. There will be less current in the next resistor because the current splits with some current passing through the resistor in question, and the rest of the current passing through the equivalent resistance of 114 R, as shown in the last figure. The voltage across R and across 114 R must be the same, since they are in parallel. Use this to find the desired current. VR = V R → I R R = I R ( 114 R ) = ( I total − I R ) ( 114 R ) → 11 4

11 4

11 11 I R = 15 I total = 15 ( 0.08508 A ) = 0.062392A  6.24  10−2 A

18. The resistors have been numbered in the accompanying diagram to help in the analysis. R1 and R2 are in series with an equivalent resistance of

R12 = R + R = 2 R . This combination is in parallel with R3 , with an

−1

1 1  2 +  = 3 R. This combination is  R 2R 

equivalent resistance of R123 = 

in series with R4 , with an equivalent resistance of R1234 = R + R = R. 2 3

5 3

This combination is in parallel with R5 , with an equivalent resistance of

A R5 R6

−1

1 3  5 +  = 8 R. Finally, this combination is in series with R 5 R  

R12345 = 

896

Chapter 26

DC Circuits

R6 , and we calculate the final equivalent resistance.

Req = 85 R + R = 138 R

A more detailed solution with more circuit diagrams is given in Problem 19. 19. We reduce the circuit to a single loop by combining series and parallel combinations. We label a combined resistance with the subscripts of the resistors used in the combination. See the successive diagrams. R1 and R2 are in series.

R1 A

R12 = R1 + R2 = R + R = 2 R

R3 R5

R3 A

−1

R1234 = R123 + R4 = 23 R + R = 53 R

R1234 and R5 are in parallel. −1

−1

 1  1 1 1  R12345 =  +  =  5 +  = 85 R  3R R  R1234 R5 

R12345 and R6 are in series, producing the equivalent resistance. Req = R12345 + R6 = 85 R + R = 138 R

R12345

C R6

e A

R1234

R123

R123 and R4 are in series.

R12 and R3 are in parallel.

 1 1   1 + 1 = 2R R123 =  +  =  3  2R R   R12 R3 

R12

C R6

Req

Now work “backwards” from the simplified circuit. Resistors in series have the same current as their equivalent resistance, and resistors in parallel have the same voltage as their equivalent resistance. To avoid rounding errors, we do not use numeric values until the end of the problem. I eq =

8e e = 13 = = I 6 = I12345 Req R 13 R 8

 8e  5 R = 5 e ; I = V5 = 135 e = 5e = I  ( 8 ) 13 5 5 R5 R 13R  13R 

V5 = V1234 = V12345 = I12345 R12345 =  I1234 =

V1234 R1234

= 135 3

3e 13R

 3e  2 R = 2 e = V = V  ( 3 ) 13 12 3  13R 

= I 4 = I123 ; V123 = I123 R123 = 

2 e e = 13 = = I1 = I 2 R3 13R R12 2 R 13R Now substitute in numeric values. 12.0 V 2e 3e e I1 = I 2 = = = 0.217 mA ; I 3 = = 0.434 mA ; I 4 = = 0.652 mA ; 13R 13 ( 4.25 k ) 13R 13R

I3 =

I5 =

5e 13R

= I 3 ; I12 =

= 1.09 mA ; I 6 =

V12

8e 13R

= 1.74 mA ; VAB = V3 = 132 e = 1.85 V

897

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

20. (a) If the terminal voltage is to be 3.0 V, then the voltage across R1 will be 9.0 V. This can be used to find the current, which then can be used to find the value of R2 . V1 = IR1 → I = R2 =

V2 I

= R1

V2 V1

V2 = IR2 →

= (16.5  )

3.0 V 9.0 V

= 5.5

(b) If the load has a resistance of 7.0  , then the parallel combination of R2 and the load must be used to analyze the circuit. The equivalent resistance of the circuit can be found and used to calculate the current in the circuit. Then the terminal voltage can be found from Ohm’s law, using the parallel combination resistance. RR ( 5.5  )( 7.0  ) R2+load = 2 load = = 3.08  Req = 3.08  + 16.5  = 19.58  12.5  R2 + Rload

V Req

12.0 V 19.58 

= 0.6129 A

VT = IR2+load = ( 0.6129 A )( 3.08  ) = 1.888 V  1.9 V

The presence of the load has affected the terminal voltage significantly. 21. Connecting 18 of the resistors in series will enable you to make a voltage divider with a 3.5-V output. To get the desired output, measure the voltage across 7 consecutive series resistors. In the diagram that would be the voltage between points a and b. e e Req = 18 (1.0  ) I = = Req 18.0  Vab = ( 7.0  ) I = ( 7.0  )

e 18.0 

= ( 7.0  )

9.0 V 18.0 

11.0

I 7.0

= 3.5 V

22. It is given that the power used when the resistors are in series is one-fourth the power used when the resistors are in parallel. The voltage is the same in both cases. Use Eq. 25–7b, along with the definitions of series and parallel equivalent resistance. V2 V2 R1R2 1 1 Pseries = 4 Pparallel → =4 → Rseries = 4 Rparallel → ( R1 + R2 ) = 4 → Rseries Rparallel ( R1 + R2 )

( R1 + R2 )2 = 4 R1R2 → R12 + 2 R1R2 + R22 − 4 R1R2 = 0 = ( R1 − R2 )2

→ R1 = R2

Thus the two resistors must be the same, and so the “other” resistor is 4.3 k . 23. Find the maximum current and resulting voltage for each resistor under the power restriction. V2 P P = I 2R = → I= , V = RP R R I1400 =

I 2500 =

1.0 W 1.4  10  3

1.0 W 2.5  10  3

= 0.02673A

V1800 =

(1.0 W ) (1.4  103  ) = 37.42 V

= 0.02000 A

V2500 =

(1.0 W ) ( 2.5  103  ) = 50.00 V

898

Chapter 26

DC Circuits

1.0 W

(

)

= 0.01644 A V3700 = (1.0 W ) 3.7  103  = 60.83V 3.7  103  The parallel resistors have to have the same voltage, and so the voltage across that combination is limited to 50.00 V. That would require a current given by Ohm’s law and the parallel combination of the two resistors. V  1 1  1   1 I parallel = parallel = Vparallel  + = ( 50.00 V )  +   = 0.03351A Rparallel  2500  3700    R2500 R3700  I 3700 =

This is more than the maximum current that can be in R1400 . Thus, the maximum current that R1400 can carry, 0.02673A, is the maximum current for the circuit. The maximum voltage that can be applied across the combination is the maximum current times the equivalent resistance. The equivalent resistance is the parallel combination of R2500 and R3700 added to R1400 . −1 −1    1 1   1    1 Vmax = I max Req = I max  R1400 +  +   = ( 0.02673A ) 1400  +  2500  + 3700        R2800 R3700    

= 77.30V  77 V

24. We label the three identical resistors from left to right as Rleft , Rmiddle , and Rright . When the switch is opened, the equivalent resistance of the circuit increases from 32 R + r to 2 R + r. Thus, the current delivered by the battery decreases, from 3 2

e R+r

e 2R + r

. Note that final current is more than half

of the original current. (a) Because the current from the battery has decreased, the voltage drop across Rleft will decrease, since it will have less current than before. The voltage drop across Rright decreases to 0, since no current is flowing in it. The voltage drop across Rmiddle will increase, because even though the total current has decreased, the current flowing through Rmiddle has increased since before the switch was opened, only half the total current was flowing through Rmiddle. Another way to consider it is that since the voltage drop across both r and Rleft have decreased, the voltage across Rmiddle must increase to account for the entire battery voltage. Vleft decreases ; Vmiddle increases ; Vright goes to 0 .

(b) By Ohm’s law, the current is proportional to the voltage for a fixed resistance. I left decreases ; I middle increases ; I right goes to 0 (c) Since the current from the battery has decreased, the voltage drop across r will decrease, and thus the terminal voltage increases. (d) With the switch closed, the equivalent resistance is 32 R + r . Thus the current in the circuit is I closed = 3 2

e R+r

, and the terminal voltage is given by Eq. 26–1.

Vterminal = e − I closed r = e − 3 closed

e R+r



   0.40  = ( 9.0 V )  1 − 3   R+r 2  2 ( 5.50  ) + 0.40  

r = e 1 − 3



= 8.584 V  8.6 V © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

899

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(e) With the switch open, the equivalent resistance is 2 R + r. Thus, the current in the circuit is e , and again the terminal voltage is given by Eq. 26–1. I closed = 2R + r   r  0.40  e  Vterminal = e − I closed r = e − r = e 1 −  = ( 9.0 V )  1 −  2R + r closed  2R + r   2 ( 5.50  ) + 0.40   = 8.684 V  8.7 V

The terminal voltage does increase, as stated earlier. 25. (a) Note that adding resistors in series always results in a larger resistance, and adding resistors in parallel always results in a smaller resistance. Closing the switch adds another resistor in parallel with R3 and R4 , which lowers the net resistance of the parallel portion of the circuit, and thus lowers the equivalent resistance of the circuit. That means that more current will be delivered by the battery. Since R1 is in series with the battery, the voltage across it will increase. Because of that increase, the voltage across R3 and R4 must decrease so that the total voltage change from the positive side of the battery to the negative side is the same, no matter which path is taken. And since there was no voltage across R2 until the switch was closed, its voltage will increase. To summarize: V1 and V2 increase ; V3 and V4 decrease . (b) By Ohm’s law, the current is proportional to the voltage for a fixed resistance. Thus, I1 and I 2 increase ; I 3 and I 4 decrease . (c) Since the battery voltage does not change and the current delivered by the battery increases, the power delivered by the battery, found by multiplying the voltage of the battery by the current delivered, increases . (d) Before the switch is closed, the equivalent resistance is R3 and R4 in parallel, combined with R1 in series. −1

−1

 1 1   2  Req = R1 +  +  = 135  +   = 202.5   135    R3 R4  The current delivered by the battery is the same as the current through R1 .

I total =

Vbattery Req

22.0 V 202.5 

= 0.1086 A = I1

The voltage across R1 is found by Ohm’s law. V1 = IR1 = ( 0.1086 A )(135  ) = 14.66 V

The voltage across the parallel resistors is the battery voltage less the voltage across R1. Vp = Vbattery − V1 = 22.0 V − 14.66 V = 7.34 V

The current through each of the parallel resistors is found from Ohm’s law. Vp 7.34 V I3 = = = 0.0544 A = I 4 R2 135  Notice that the current through each of the parallel resistors is half of the total current, within the limits of significant figures. The currents before closing the switch are as follows. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

900

Chapter 26

DC Circuits

I1 = 0.109 A ; I 2 = 0 ; I 3 = I 4 = 0.054 A

After the switch is closed, the equivalent resistance is R2 , R3 , and R4 in parallel, combined with R1 in series. Do a similar analysis. −1

−1

 1 1 1   3  Req = R1 +  + +  = 135  +   = 180   135    R2 R3 R4  I total =

Vbattery Req

22.0 V 180 

V1 = IR1 = ( 0.1222 A )(135  ) = 16.5 V

= 0.1222 A = I1

Vp = Vbattery − V1 = 22.0 V − 16.5 V = 5.5 V

5.5 V

= 0.0407 A = I 3 = I 4 R2 135  Notice that the current through each of the parallel resistors is one-third of the total current, within the limits of significant figures. The currents after closing the switch are as follows.

I1 = 0.122 A

I2 =

I 2 = I 3 = I 4 = 0.041A

Yes, the predictions made in part (b) are all confirmed. 26. The goal is to determine r so that Pheater , since Pheater 

dPheater

dPheater dR

= 0. This ensures that R produces very little change in R = R0

2 R. The power delivered to the heater can be found by Pheater = Vheater R,

dR and so we need to determine the voltage across the heater. We do this by calculating the current drawn from the voltage source, and then subtracting the voltage drop across R  from the source voltage. e ( R + R ) RR 2 RR + R2 R ( 2 R + R ) e e Req = R + ; I total = = = = = R + R R + R R + R Req R ( 2 R + R ) R ( 2 R + R ) R + R

Vheater = e − I total r = e − dPheater dR

e ( R + R ) e ( R + R ) V2 eR e 2R ; Pheater = heater = R = e − = 2 R ( 2 R + R ) R ( 2 R + R ) ( 2 R + R ) ( 2 R + R )

( 2 R0 + R ) − R0 ( 2 ) ( 2 R0 + R )( 2 ) 2 =e = 0 → ( 2 R0 + R ) − R0 ( 2 )( 2 R0 + R )( 2 ) = 0 → 4 ( 2 R0 + R ) R=R 2

4 R + 4 R0 R + R2 − 8 R02 − 4 R0 R = 0 → R2 = 4 R02 → R = 2 R0 2 0

27. All of the resistors are in series, so the equivalent resistance is just the sum of the resistors. Use Ohm’s law then to find the current, and show all voltage changes starting at the negative pole of the battery and going counterclockwise. e 9.0 V I= = = 0.3273A  0.33A Req ( 9.5 + 16.0 + 2.0 ) 

 voltages = 9.0 V − ( 9.5  )( 0.3273A ) − (16.0  )( 0.3273A ) − ( 2.0  )( 0.3273A ) = 9.0 V − 3.109 V − 5.237 V − 0.655 V = −.001 V  0.00 V

The very slight discrepancy is due to rounding of the numbers.

901

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

28. Apply Kirchhoff’s loop rule to the circuit starting at the upper left corner of the circuit diagram, in order to calculate the current. Assume that the current is flowing clockwise. 6V − I ( 2.0  ) + 18 V − I ( 5.8  ) − 12 V − I (1.0  ) = 0 → I = = 0.682A 8.8  The terminal voltage for each battery is found by summing the potential differences across the internal resistance and emf from negative to positive. Note that for the 12-V battery, there is a voltage gain going across the internal resistance from negative to positive due to the direction of the current.

18-V battery: Vterminal = − I ( 2.0  ) + 18 V = − ( 0.682 A )( 2.0  ) + 18 V = 16.64V  17 V 12 -V battery: Vterminal = I (1.0  ) + 12 V = ( 0.682 A )(1.0  ) + 12 V = 12.682 V  13V 29. To find the potential difference between points a and b, the current must be found from Kirchhoff’s loop law. Start at point a and go counterclockwise around the entire circuit, taking the current to be counterclockwise. e − IR + e − IR − IR + e − IR = 0 → I = 2R e Vab = Va − Vb = − IR + e − IR = e − 2 IR = e − 2 R = 0V 2R Notice that the actual values for the battery voltages and the resistances were not used. I1 30. There are three currents involved, and so there must be three I3 25 independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction of 3.0 V 5.8 V I2 the three branches at the top center of the circuit. I1 = I 2 + I 3 120 56 110 Another equation comes from Kirchhoff’s loop rule applied to 64  the left loop, starting at the negative terminal of the battery and progressing counterclockwise. 5.8 V − I1 (120  ) − I1 ( 64  ) − I 2 ( 56  ) = 0 → 5.8 = 184 I1 + 56 I 2 The final equation comes from Kirchhoff’s loop rule applied to the right loop, starting at the negative terminal of the battery and progressing counterclockwise. 3.0 V − I 3 ( 25  ) + I 2 ( 56  ) − I 3 (110  ) = 0 → 3.0 = −56 I 2 + 135 I 3

Substitute I1 = I 2 + I 3 into the left loop equation, so that there are two equations with two unknowns. Both loop equations are written on the next line. 5.8 = 184 ( I 2 + I 3 ) + 56 I 2 = 240 I 2 + 184 I 3 ; 3.0 = −56 I 2 + 135 I 3 Solve the right loop equation for I 2 and substitute into the left loop equation, resulting in an equation with only one unknown, which can be solved. 135 I 3 − 3.0  135 I 3 − 3.0  + 184 I → 3.0 = −56 I 2 + 135 I 3 → I 2 = ; 5.8 = 240 I 2 + 184 I 3 = 240   3 56 56   324.8 = 32400 I 3 − 720 + 10304 I 3 → I 3 = I2 =

135 I 3 − 3.0 56

1044.8 42704

= 0.02447 A

= 0.00527 A ; I1 = I 2 + I 3 = 0.02974 A

902

Chapter 26

DC Circuits

The current in each resistor is: 120  : 0.030 A 64  : 0.030 A

56  : 0.005A

25  : 0.024 A

110  : 0.024 A

31. This circuit is identical to Example 26–8 and Figure 26–12 except for the numeric values. So we may copy the same equations as developed in that Example, but using the new values. Eq. (i): I 3 = I1 + I 2 Eq. (ii): −34 I1 + 45 − 48 I 3 = 0 Eq. (iii): −34 I1 + 25 I 2 − 85 = 0 Eq. (v):

I3 =

Eq. (iv): I 2 =

85 + 34 I1 25

= 3.4 + 1.36 I1

45 − 34 I1

= 0.9375 − 0.7083I1 48 I 3 = I1 + I 2 → 0.9375 − 0.7083I1 = I1 + 3.4 + 1.36 I1 → I1 = −0.8026 A

I 2 = 3.4 + 1.36 I1 = 2.308 A ; I 3 = 0.9375 − 0.7083I1 = 1.506 A (a) To find the potential difference between points a and d, start at point a and add each individual potential difference until reaching point d. The simplest way to do this is along the top branch. Vad = Va − Vd = I1 ( 34  ) = ( −0.8026 A )( 34  ) = −27.29 V  −27 V Slight differences will be obtained in the final answer depending on the branch used, due to rounding. For example, using the bottom branch, we get the following. Vad = Va − Vd = I 2 ( 25  ) − e1 = ( 2.308 A )( 25  ) − 85 V = −27.3V  −27 V Note that this says that the voltage at point a is lower than the voltage at point d. (b) For the 85-V battery, the terminal voltage is the potential difference from point g to point e. For the 45-V battery, the terminal voltage is the potential difference from point d to point b.

85- V battery: Vterminal = e1 − I 2 r = 85 V − ( 2.308 A )(1.0  ) = 83V 45-V battery: Vterminal = e2 − I 3 r = 45 V − (1.506 A )(1.0  ) = 43V 32. Because there are no resistors in the bottom branch, it is possible to write Kirchhoff loop equations that only have one current term, making them easier to solve. To find the current through R1, go around the outer loop counterclockwise, starting at the lower left corner. V + V 6.0 V + 9.0 V V3 − I1 R1 + V1 = 0 → I1 = 3 1 = = 0.68 A, left 22  R1 To find the current through R2 , go around the lower loop counterclockwise, starting at the lower left corner. 6.0 V V V3 − I 2 R2 = 0 → I 2 = 3 = = 0.33A, left R2 18  33. There are three currents involved, and so there must be three independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction of the three branches on the left of the circuit. I1 = I 2 + I 3 Another equation comes from Kirchhoff’s loop rule applied to the outer loop, starting at the lower left corner, and progressing counterclockwise.

9.0 V

22 

1.2 I2

18 6.0 V

1.2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

− I 3 (1.2  ) + 6.0 V − I1 ( 22  ) − I1 (1.2  ) + 9.0 V = 0 → 15 = 23.2 I1 + 1.2 I 3 The final equation comes from Kirchhoff’s loop rule applied to the bottom loop, starting at the lower left corner, and progressing counterclockwise. − I 3 (1.2  ) + 6.0 V + I 2 (18  ) = 0 → 6 = −18 I 2 + 1.2 I 3 Substitute I1 = I 2 + I 3 into the top loop equation, so that there are two equations with two unknowns.

15 = 23.2 I1 + 1.2 I 3 = 23.2 ( I 2 + I 3 ) + 1.2 I 3 = 23.2 I 2 + 24.4 I 3 ; 6 = −18I 2 + 1.2 I 3

Solve the bottom loop equation for I 2 and substitute into the top loop equation, resulting in an equation with only one unknown, which can be solved. −6 + 1.2 I 3 6 = −18 I 2 + 1.2 I 3 → I 2 = 18  −6 + 1.2 I 3  + 24.4 I → 270 = −139.2 + 27.84 I + 439.2 I → 15 = 23.2 I 2 + 24.4 I 3 = 23.2   3 3 3 18   I3 =

409.2 467.04

= 0.8762 A ; I 2 =

−6 + 1.2 I 3 18

−6 + 1.2 ( 0.8762 ) 18

= −0.2749 A  0.27 A, left

I1 = I 2 + I 3 = 0.6013A  0.60 A, left

34. (a) There are three currents involved, and so there must be three independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction of the three branches on the right of the circuit. I 2 = I1 + I 3 → I1 = I 2 − I 3 Another equation comes from Kirchhoff’s loop rule applied to the top loop, starting at the negative terminal of the battery and progressing clockwise. e1 − I1R1 − I 2 R2 = 0 → 9 = 25I1 + 48I 2

R2 I2 R3

The final equation comes from Kirchhoff’s loop rule applied to the bottom loop, starting at the negative terminal of the battery and progressing counterclockwise. e 2 − I 3 R3 − I 2 R2 = 0 → 12 = 35I 3 + 48I 2

Substitute I1 = I 2 − I 3 into the top loop equation, so that there are two equations with two unknowns. 9 = 25 I1 + 48 I 2 = 25 ( I 2 − I 3 ) + 48 I 2 = 73I 2 − 25 I 3 ; 12 = 35 I 3 + 48 I 2 Solve the bottom loop equation for I 2 and substitute into the top loop equation, resulting in an equation with only one unknown, which can be solved. 12 − 35 I 3 12 = 35 I 3 + 48 I 2 → I 2 = 48  12 − 35 I 3  9 = 73I 2 − 25 I 3 = 73   − 25 I 3 → 432 = 876 − 2555 I 3 − 1200 I 3 →  48  I3 =

444 3755

= 0.1182 A  0.12 A, up ; I 2 =

12 − 35 I 3 48

= 0.1638 A  0.16 A, left

I1 = I 2 − I 3 = 0.0456 A  0.046 A, right © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 26

DC Circuits

(b) We can include the internal resistances simply by adding 1.0 to R1 and R3 . So let R1 = 26  and let R3 = 36 . Now re-work the problem exactly as in part (a). I 2 = I1 + I 3 → I1 = I 2 − I 3

e1 − I1R1 − I 2 R2 = 0 → 9 = 26 I1 + 48I 2 e 2 − I 3 R3 − I 2 R2 = 0 → 12 = 36 I 3 + 48I 2 9 = 26 I1 + 48 I 2 = 26 ( I 2 − I 3 ) + 48 I 2 = 74 I 2 − 26 I 3 ; 12 = 36 I 3 + 48 I 2 12 = 36 I 3 + 48 I 2 → I 2 =

12 − 36 I 3 48

1 − 3I 3 4

 1 − 3I 3   − 26 I 3 → 36 = 74 − 222 I 3 − 104 I 3 →  4 

9 = 74 I 2 − 26 I 3 = 74  I3 =

38 326

= 0.1166 A  0.12 A, up ; I 2 =

1 − 3I 3 4

= 0.1626 A  0.16 A, left

I1 = I 2 − I 3 = 0.046 A, right

The currents are unchanged to 2 significant figures by the inclusion of the internal resistances. 35. We are to find the ratio of the power used when the resistors are in series, to the power used when the resistors are in parallel. The voltage is the same in both cases. Use Eq. 25–7b, along with the definitions of series and parallel equivalent resistance. Rseries = R1 + R2 + Pseries Pparallel

V 2 Rseries 2

V Rparallel

1 1 Rn = nR ; Rparallel =  + +  R1 R2 =

Rparallel Rseries

R n nR

1 

−1

R n =  =  Rn  n R

1 n2

36. (a) Since there are three currents to determine, there must be three independent equations to determine those three currents. One comes from Kirchhoff’s junction rule applied to the junction near the negative terminal of the middle battery. I1 = I 2 + I 3 Another equation comes from Kirchhoff’s loop rule applied to the top loop, starting at the negative terminal of the middle battery, and progressing counterclockwise. We add series resistances. 12.0 V − I 2 ( 29  ) + 12.0 V − I1 ( 41  ) = 0 → 24.0 = 41I1 + 29 I 2 The final equation comes from Kirchhoff’s loop rule applied to the bottom loop, starting at the negative terminal of the middle battery, and progressing clockwise. 12.0 V − I 2 ( 29  ) − 6.0 V + I 3 ( 28  ) = 0 → 6.0 = 29 I 2 − 28 I 3 Substitute I1 = I 2 + I 3 into the top loop equation, so that there are two equations with two unknowns. 24.0 = 41I1 + 29 I 2 = 41 ( I 2 + I 3 ) + 29 I 2 = 70 I 2 + 41I 3 Solve the bottom loop equation for I 2 and substitute into the top loop equation, resulting in an equation with only one unknown, which can be solved for I 3 .

905

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

6.0 = 29 I 2 − 28 I 3 → I 2 = I 3 = 0.09 A ; I 2 =

6.0 + 28 I 3 29

Instructor Solutions Manual

 6.0 + 28 I 3  + 41I →  3 29  

; 24.0 = 70 I 2 + 41I 3 = 70 

6.0 + 28 I 3

= 0.29 A ; I1 = I 2 + I 3 = 0.38 A 29 (b) The terminal voltage of the 6.0-V battery is 6.0 V − I 3r = 6.0 V − ( 0.088 A )(1.0  ) = 5.912 V  5.9 V .

37. This problem is the same as Problem 36, except the total resistance in the top branch is now 29  instead of 41 . We simply reproduce the adjusted equations here without the prose. I1 = I 2 + I 3

12.0 V − I 2 ( 29  ) + 12.0 V − I1 ( 29  ) = 0 → 24.0 = 29 I1 + 29 I 2 12.0 V − I 2 ( 29  ) − 6.0 V + I 3 ( 28  ) = 0 → 6.0 = 29 I 2 − 28 I 3 24.0 = 29 I1 + 29 I 2 = 29 ( I 2 + I 3 ) + 29 I 2 = 58 I 2 + 29 I 3

6.0 = 29 I 2 − 28 I 3 → I 2 = I 3 = 0.141A ; I 2 =

6.0 + 28 I 3

6.0 + 28 I 3 29

 6.0 + 28 I 3  + 29 I →  3 29  

; 24.0 = 58 I 2 + 29 I 3 = 58 

= 0.343A ; I1 = I 2 + I 3 = 0.484 A  0.48 A

38. The circuit diagram has been labeled with six different currents. We apply the junction rule to junctions a, b, and c. We apply the loop rule to the three loops labeled in the diagram. 1) I = I1 + I 2 ; 2 ) I1 = I 3 + I 5 ; 3) I 3 + I 4 = I

4 ) − I1R1 − I 5 R5 + I 2 R2 = 0 ; 5) − I 3 R3 + I 4 R4 + I 5 R5 = 0 6 ) e − I 2 R2 − I 4 R4 = 0 Eliminate I using equations 1) and 3). 1) I 3 + I 4 = I1 + I 2 ; 2 ) I1 = I 3 + I 5 4 ) − I1R1 − I 5 R5 + I 2 R2 = 0 ; 5) − I 3 R3 + I 4 R4 + I 5 R5 = 0 6 ) e − I 2 R2 − I 4 R4 = 0 Eliminate I 1 using equation 2).

1) I 3 + I 4 = I 3 + I 5 + I 2 → I 4 = I 5 + I 2

4 ) − ( I 3 + I 5 ) R1 − I 5 R5 + I 2 R2 = 0 → − I 3 R1 − I 5 ( R1 + R5 ) + I 2 R2 = 0

5) − I 3 R3 + I 4 R4 + I 5 R5 = 0 6 ) e − I 2 R2 − I 4 R4 = 0 Eliminate I 4 using equation 1).

4 ) − I 3 R1 − I 5 ( R1 + R5 ) + I 2 R2 = 0

5) − I 3 R3 + ( I 5 + I 2 ) R4 + I 5 R5 = 0 → − I 3 R3 + I 5 ( R4 + R5 ) + I 2 R4 = 0 6 ) e − I 2 R2 − ( I 5 + I 2 ) R4 = 0 → e − I 2 ( R2 + R4 ) − I 5 R4 = 0

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Chapter 26

DC Circuits

Eliminate I 2 using equation 4): I 2 = 5) − I 3 R3 + I 5 ( R4 + R5 ) +

1 R2

 I R + I ( R + R ) . 3

 I R + I ( R + R ) R = 0 → 3

I 3 ( R1 R4 − R2 R3 ) + I 5 ( R2 R4 + R2 R5 + R1R4 + R5 R4 ) = 0 6) e −

1 R2

 I R + I ( R + R ) ( R + R ) − I R = 0 → 3

e R2 − I 3 R1 ( R2 + R4 ) − I 5 ( R1 R2 + R1 R4 + R5 R2 + R5 R4 + R2 R4 ) = 0

Eliminate I 3 using equation 5): I 3 = − I 5

( R2 R4 + R2 R5 + R1R4 + R5 R4 ) ( R1R4 − R2 R3 )

 ( R2 R4 + R2 R5 + R1R4 + R5 R4 )   R1 ( R2 + R4 ) − I 5 ( R1R2 + R1R4 + R5 R2 + R5 R4 + R2 R4 ) = 0 ( R1R4 − R2 R3 )  

e R2 +  I 5 e=−

I 5   ( R2 R4 + R2 R5 + R1 R4 + R5 R4 ) 

 R2  

  R1 ( R2 + R4 ) − ( R1R2 + R1R4 + R5 R2 + R5 R4 + R2 R4 )   

( R1R4 − R2 R3 )

  ( 25  )(14  ) + ( 25  )(15  ) + ( 22  )(14  ) + (15  )(14  )   ( 22  )( 25  + 14  )  I5   ( 22  )(14  ) − ( 25  )(12  ) =−    25     − ( 22  )( 25  ) + ( 22 )(14 ) + (15  )( 25  ) + (15  )(14  ) + ( 25  )(14  )  = − I 5 ( 5261  ) → I 5 = −

6.0 V 5261 

= −1.140 mA ( upwards )

( R2 R4 + R2 R5 + R1R4 + R5 R4 ) ( R1R4 − R2 R3 ) ( 25  )(14  ) + ( 25  )(15  ) + ( 22  )(14  ) + (15  )(14  ) = − ( −1.140 mA ) = 0.1771A ( 22  )(14  ) − ( 25  )(12  )

I3 = − I5

I2 =

1 R2

 I R + I ( R + R ) = 3

1 25 

( 0.1771A )( 22  ) + ( −0.00114 A )( 37  ) = 0.1542 A

I 4 = I 5 + I 2 = −0.00114 A + 0.1542 A = 0.1531A I1 = I 3 + I 5 = 0.1771A − 0.00114 A = 0.1760 A We keep an extra significant figure to show the slight difference in the currents.

I 22  = 0.176A

I 25  = 0.154 A

I12  = 0.177 A

I14  = 0.153A

I15 = 0.001A, upwards

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

39. The circuit diagram from Problem 38 is reproduced, with R2 = 0. This circuit can now be simplified significantly. Resistors R1 and R5 are in

parallel; call that combination R15 . That combination is in series with R3 ;

call that combination R153 . That combination is in parallel with R4 . See

the second diagram. We calculate the equivalent resistance R153 , use that to find the current through the top branch in the second diagram, and then use that current to find the current through R5 . −1

−1

+ –

1 1 1   1 R153 =  +  + R3 =  +  + 12  = 20.92   22  15    R1 R5 

Use the loop rule for the outside loop to find the current in the top branch. 6.0 V e = = 0.2868 A e − I153 R153 = 0 → I153 = R153 20.92 

This current is the sum of the currents in R1 and R5 . Since those two resistors are in parallel, the voltage across them must be the same. V1 = V5 → I1 R1 = I 5 R5 → ( I153 − I 5 ) R1 = I 5 R5 → I 5 = I153

( R5 + R1 )

= ( 0.2868 A )

22  37 

R4 e

+ –

= 0.17 A

40. (a) As shown in the diagram, we use symmetry to reduce the number of independent currents to six. Using Kirchhoff’s junction rule, we write equations for junctions a, c, and d. We then use Kirchhoff’s loop rule to write the loop equations for loops afgba, hedch, and aba (through the voltage source). I = 2 I1 + I 2 [1] ; I 3 + I 4 = I1 [2] ; I 5 = 2 I 4 [3] 0 = −2 I1R − I 3 R + I 2 R [4] ; 0 = −2 I 4 R − I 5 R + I 3 R [5] 0 = e − I 2 R [6]

We have six equations with six unknown currents. We use the method of substitution to reduce the equations to a single equation relating the emf from the power source to the current through the power source. This resulting ratio is the effective resistance between points a and b. We insert Eqs. [2], [3], and [6] into the other three equations to eliminate I1, I2, and I5. e e I = 2 ( I 3 + I 4 ) + =2 I 3 + 2 I 4 + [1*] R R e 0 = −2 ( I 3 + I 4 ) R − I 3 R + R = − 2 I 4 R − 3I 3 R + e [4*] R 0 = −2 I 4 R − 2 I 4 R + I 3 R = −4 I 4 R + I 3 R [5*] We solve Eq. [5*] for I3 and insert that into Eq. [4*]. We then insert the two results into Eq. [1*] and solve for the effective resistance. e I 3 = 4 I 4 ; 0 = −2 I 4 R − 3 ( 4 I 4 ) R + e → I 4 = 14 R e e e 10e e 24e 12e + = = → Req = = 127 R I = 2 ( 4 I 4 ) + 2 I 4 + = 10 I 4 + = R R I 14 R R 14 R 7 R © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 26

DC Circuits

(b) As shown in the diagram, we use symmetry to reduce the number of currents to four. We use Kirchhoff’s junction rule at junctions a and d and the loop rule around loops abca (through the voltage source) and afgdcha. This results in four equations with four unknowns. We solve these equations for the ratio of the voltage source to current I, to obtain the effective resistance. I = 2 I1 + I 2 [1] ; 2 I 3 = I 2 [2] 0 = −2 I1 R + e [3] ; 0 = −2 I 2 R − 2 I 3 R + 2 I1R [4] We solve Eq. [3] for I1 and Eq. [2] for I3. These results are inserted into Eq. [4] to determine I2. Using these results and Eq. [1] we solve for the effective resistance. 3 I I e e e ; I 3 = 2 ; I1 = I 2 + I 3 → I1 = = I2 + 2 = I2 → I2 = 2R 2 2R 2 2 3R 4e e  e  e ; Req = = 43 R I = 2 I1 + I 2 = 2  = + I  2 R  3R 3R (c) As shown in the diagram, we again use symmetry to reduce the number of currents to three. We use Kirchhoff’s junction rule at points a and b and the loop rule around the loop abgda (through the power source) to write three equations for the three unknown currents. We solve these equations for the ratio of the emf to the current through the emf (I) to calculate the effective resistance. I = 3I1 [1] ; I1 = 2 I 2 [2] 0 = −2 I1 R − I 2 R + e [3] We insert Eq. [2] into Eq. [3] and solve for I1. Inserting I1 into Eq. [1] enables us to solve for the effective resistance. 2e 6e e 0 = −2 I1 R − 12 I1R + e → I1 = ; I = 3I 1 = → Req = = 65 R 5R 5R I

41. (a) To find the equivalent resistance between points a and c, apply a voltage between points a and c, find the current that flows from the voltage source, and then calculate Req = e I .

There is no symmetry to exploit. ( bottom loop ) 1) e − RI 3 = 0

(a − d − b) (a − b − c) (d − b − c) ( junction a ) ( junction d ) ( junction b )

− RI1 − RI 5 + RI 2 = 0

− RI 2 − RI 6 + RI 3 = 0

− RI 5 − RI 6 + RI 4 = 0

I = I1 + I 2 + I 3

I1 = I 4 + I 5

I2 + I5 = I6

R I6

I2 I3

R

c I

From Eq. 1), substitute I 3 = e R . 2) − RI1 − RI 5 + RI 2 = 0 → I1 + I 5 = I 2

909

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

− RI 2 − RI 6 + R

= 0 → I2 + I6 =

Instructor Solutions Manual

R R  4) − RI 5 − RI 6 + R I 4 = 0 → R ( I 5 + I 6 ) = RI 4

5) I = I1 + I 2 +

; 6) I1 = I 4 + I 5 ; 7) I 2 + I 5 = I 6 R From Eq. 7), substitute I 6 = I 2 + I 5 2) I1 + I 5 = I 2 ; 3)

I 2 + I 2 + I5 =

4) R ( 2 I 5 + I 2 ) = RI 4 ; 5)

→ 2I 2 + I5 =

I = I1 + I 2 +

R → I 5 = I1 − I 4

From Eq. 6), substitute I1 = I 4 + I 5 2) 2 I1 − I 4 = I 2

3) 2 I 2 + I1 − I 4 =

;

4) R ( 2 I1 − 2 I 4 + I 2 ) = RI 4

;

; 6) I1 = I 4 + I 5

e R

I = I1 + I 2 +

e R

From Eq. 2), substitute 2 I1 − I 4 = I 2 → I 4 = 2 I1 − I 2

3) 2 I 2 + I1 − ( 2 I1 − I 2 ) =

→ 3I 2 − I1 =

R R  4) R ( 2 I1 − 2 ( 2 I1 − I 2 ) + I 2 ) = R ( 2 I1 − I 2 ) → R ( 3I 2 − 2 I1 ) = R  ( 2 I1 − I 2 )

I = I1 + I 2 +

e R

From Eq. 3), substitute 3I 2 − I1 =

 

4) R  3I 2 − 2  3I 2 −

I = 3I 2 −

e R

e e        = R  2  3I 2 −  − I 2  → R  −3I 2 + 2  = R  5I 2 − 2  R  R R R     

R R I = From Eq. 5), substitute 2 14 I

 

→ I1 = 3I 2 −

e 

+ I2 +

4) R  −3 ( 14 I ) + 2

e

→ I = 4I2

R ( 5 R  + 3R ) e e  1 = Req =  = R  5 ( 4 I ) − 2  → R R I 8 ( R + R ) 

e

(b) In this case, apply a voltage between points a and b. Now there is symmetry. In this case no current would flow through resistor R, and so that branch can be eliminated from the circuit. See the adjusted diagram. Now the upper left two resistors (from a to d to b) are in series, and the lower right two resistors (from a to c to b) are in series. These two combinations are in parallel with each other, and with the resistor between a and b. The equivalent resistance is now relatively simple to calculate. −1

−1

 1 + 1 + 1  = 4  = 1R Req =     2  2R R 2R   2R 

d R

R R

b R R

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Chapter 26

DC Circuits

42. (a) We label each of the currents as shown in the accompanying figure. Using Kirchhoff’s junction rule and the first three junctions (a–c) we write equations relating the entering and exiting currents. I = I1 + I 2

[1]

I 2 = I3 + I 4

[2]

I1 + I 4 = I 5 [3] We use Kirchhoff’s loop rule to write equations for loops abca, abda, and bdcb. 0 = − I 2 R − I 4 R + I1 R

[4]

0 = − I 2 R − I 3R + e

[5]

0 = − I 3R + I5R + I 4 R

[6]

We have six unknown currents and six equations. We solve these equations by substitution. First, insert Eq. [3] into [6] to eliminate current I5. Next insert Eq. [2] into Eqs. [1], [4], and [5] to eliminate I2.

0 = − I 3 R + ( I1 + I 4 ) R + I 4 R → 0 = − I 3 R + I1 R + 2 I 4 R

[6*]

I = I1 + I 3 + I 4

[1*]

0 = − ( I 3 + I 4 ) R − I 4 R + I1 R → 0 = − I 3 R − 2 I 4 R + I1 R

[4*]

0 = − ( I3 + I 4 ) R − I3 R + e → 0 = − I 4 R − 2I3 R + e

[5*]

Next we solve Eq. [4*] for I4 and insert the result into Eqs. [1*], [5*], and [6*].

0 = − I 3 R − 2 I 4 R + I1 R → I 4 = 12 I1 − 12 I 3 I = I1 + I 3 + 12 I1 - 12 I 3 → I = 23 I1 + 12 I 3

[1**]

0 = − I 3 R + I1 R + 2 ( 12 I1 - 12 I 3 ) R = −2 I 3 R + 2 I1 R → I1 = I 3

[6**]

0 = − ( 12 I1 − 12 I 3 ) R − 2 I 3 R + e → 0 = − 12 I1 R − 23 I 3 R + e

[5**]

Finally we substitute Eq. [6**] into Eq [5**] and solve for I1. We insert this result into Eq. [1**] to write an equation for the current through the battery in terms of the battery emf and resistance. e e 0 = − 12 I1R − 23 I1R + e → I1 = ; I = 23 I1 + 12 I1 = 2 I1 → I = 2R R (b) We divide the battery emf by the current to determine the effective resistance. e e Req = = = R I e R 43. Define I1 to be the current to the right through the 2.00-V

battery ( e1 ) , and I 2 to be the current to the right through the 3.00-V battery ( e2 ) . At the junction, they combine to give

current I = I1 + I 2 to the left through the top branch. Apply Kirchhoff’s loop rule first to the upper loop, and then to the outer loop, and solve for the currents.

I1 + I 2

r I1 r

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

e1 − I1r − ( I1 + I 2 ) R = 0 → e1 − ( R + r ) I1 − RI 2 = 0 e2 − I 2 r − ( I1 + I 2 ) R = 0 → e2 − RI1 − ( R + r ) I 2 = 0

Solve the first equation for I 2 and substitute into the second equation to solve for I 1 . e1 − ( R + r ) I1 − RI 2 = 0 → I 2 =

e1 − ( R + r ) I1

2.00 − 4.350 I1

= 0.500 − 1.0875 I1 R 4.00 e2 − RI1 − ( R + r ) I 2 = 3.00 − 4.00 I1 − ( 4.35 )( 0.500 − 1.0875 I1 ) = 0 → 0.825 = −0.7306 I1 I1 = −1.129 A ; I 2 = 0.500 − 1.0875 I1 = 1.728 A

The voltage across R is its resistance times I = I1 + I 2 . VR = R ( I1 + I 2 ) = ( 4.00  )( −1.129 A + 1.728 A ) = 2.396 V  2.40 V

Note that the top battery is being charged–the current is flowing through it from positive to negative. Also note that the voltage across R is higher than the 2.00-V emf and lower than the 3.00-V emf, but not exactly midway between them. 44. We take each of the given times as the time constant of the RC combination.   = RC → R = C   1s 2s = 1  106  ; R2s = = = 2  106  ; R1s = = −6 −6 C 1  10 F C 1  10 F   4s 8s = 4  106  ; R2s = = = 8  106  ; R4s = = −6 −6 C 1  10 F C 1  10 F  15s = 15  106  R15s = = −6 C 1  10 F So we estimate the range of resistance to be 1M − 15 M . 45. (a) From Eq. 26–7 the product RC is equal to the time constant.  24.0  10−6 s = 1.60  10−9 F  = RC → C = = 3 R 15.0  10  (b) Since the battery has an emf of 24.0 V, if the voltage across the resistor is 14.0 V, the voltage across the capacitor will be 10.0 V as it charges. Use the expression for the voltage across a charging capacitor. t  V   V  → e − t / =  1 − C  → − = ln  1 − C  → VC = e 1 − e − t / e  e    

(

)

 

t = − ln  1 −

 10.0 V  −6 −5  = − ( 24.0  10 s ) ln  1 −  = 1.29  10 s e  24.0 V  

VC 

46. Express the stored energy in terms of the charge on the capacitor, using Eq. 24–5. The charge on the capacitor is given by Eq. 26–6a. Q2 1

U=2

C e (1 − e −t  )  2 2 1  =2 = 12 C e 2 (1 − e −t  ) = U max (1 − e −t  ) ; 2

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Chapter 26

DC Circuits

(

U = 0.75U max → U max 1 − e −t 

(

)

) = 0.75U 2

→

max

(1 − e ) = 0.75 → −t 

t = − ln 1 − 0.75 = 2.01

47. The current for a capacitor-charging circuit is given by Eq. 26–8, with R the equivalent series resistance and C the equivalent series capacitance.

e Req



−

 

 Req Ceq 

→

 C C   I ( R1 + R2 )   IReq  = − ( R1 + R2 )  1 2  ln    e  e   C1 + C2    ( 3.8  10−6 F )2   (1.35  10−3 A ) ( 4400  )  −3 −3  ln  = − ( 4400  )   = 5.879  10 s  5.9  10 s −6 (12.0V )  7.6  10 F   

t = − ReqCeq ln 

48. The capacitance is given by Eq. 24–8 and the resistance by Eq. 25–3. The capacitor plate separation d is the same as the resistor length l. Calculate the time constant.   d   K  A  =  K  = 1.0  1012  m 5.0 8.85  10 −12 C 2 N m 2 = 44 s  = RC =  ( )( )( )  0  0 d  A  49. The voltage of the discharging capacitor is given by VC = V0 e−t RC . The capacitor voltage is to be 0.0025V0 .

VC = V0 e −t RC → 0.0025 V0 = V0 e −t RC → 0.0025 = e −t RC → ln ( 0.0025 ) = −

(

)(

)

t RC

→

t = − RC ln ( 0.0025 ) = − 8.7  10  3.0  10 −6 F ln ( 0.0025 ) = 0.16 s 50. (a) At t = 0, the capacitor is uncharged and so there is no voltage difference across it. The capacitor is a “short,” and so a simpler circuit can be drawn just by eliminating the capacitor. In this simpler circuit, the two resistors on the right are in parallel with each other, and then in series with the resistor by the switch. The current through the resistor by the switch splits equally when it reaches the junction of the equal, parallel resistors. −1

1 1 Req = R +  +  = 23 R → R R I1 =

e e 2e = 3 = ; I 2 = I 3 = 12 I1 = Req 2 R 3R 3R

(b) At t = , the capacitor will be fully charged and there will be no current in the branch containing the capacitor, and so a simpler circuit can be drawn by eliminating that branch. In this simpler circuit, the two resistors are in series, and they both have the same current.

Req = R + R = 2 R → I1 = I 2 =

e Req

e 2R

; I3 = 0

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(c) At t = , since there is no current through the branch containing the capacitor, there is no potential drop across that resistor. It can be treated as a short. Therefore the voltage difference across the capacitor equals the voltage difference across the resistor through which I 2 flows.

 e R = 1 e  2  2R 

VC = VR = I 2 R =  2

This is basically a voltage divider circuit. 51. (a) After the capacitor is fully charged, there is no current through it, and so it behaves like an “open” in the circuit. In the circuit diagram, this means that I1 = I3 and I2 = I4. e Write loop equations for the leftmost loop and the outer loop in order to solve for the currents. e 12.0 V e − I 2 ( R2 + R4 ) = 0 → I 2 = = = 1.20 A R2 + R4 10.0 

e − I1 ( R1 + R3 ) = 0 → I1 =

d R4

12.0 V

= 0.800 A R1 + R3 15.0  Use these currents to find the voltage at points c and d, which will give the voltage across the capacitor. Vc = e − I 2 R2 = 12.0 V − (1.20 A )(1.0  ) = 10.8 V Vd = e − I1 R1 = 12.0 V − ( 0.800 A )(10.0  ) = 4.00 V Vcd = 10.8 V − 4.00 V = 6.8 V ; Q = CV = ( 2.2  F )( 6.8 V ) = 14.96 C  15C

(b) When the switch is opened, the emf is taken out of the circuit. Then we have the capacitor discharging through an equivalent resistance. That equivalent resistance is the series combination of R1 and R2, in parallel with the series combination of R3 and R4. Use the expression for discharging a capacitor, Eq. 26–9a. −1

−1

 1 1  1   1 Req =  + = +   = 6.16   11.0  14.0    R1 + R2 R3 + R4  Q = Q0e

− t / Req C

= 0.020Q0 →

(

)

t = − ReqC ln ( 0.020 ) = − ( 6.16  ) 2.2  10−6 F ln ( 0.020 ) = 5.3  10 −5 s 52. (a) With the currents and junctions labeled as in the diagram, we use point a for the junction rule and the right and left loops for the loop rule. We set current I3 equal to the derivative of the charge on the capacitor and combine the equations to obtain a single differential equation in terms of the capacitor charge. Solving this equation yields the charging time constant. Q I1 = I 2 + I 3 [1] ; e − I1R1 − I 2 R2 = 0 [2] ; − + I 2 R2 = 0 [3] C We use Eq. [1] to eliminate I1 in Eq. [2]. Then we use Eq. [3] to eliminate I2 from Eq. [2].  Q  0 = e − ( I 2 + I 3 ) R1 − I 2 R2 ; 0 = e − I 2 ( R1 + R2 ) − I 3 R1 ; 0 = e −   ( R1 + R2 ) − I 3 R1  R2C  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 26

DC Circuits

We set I3 as the derivative of the charge on the capacitor and solve the differential equation by separation of variables.  Q  e R2C dQ dQ R1R2C = −Q → 0 = e − R1 →  ( R1 + R2 ) − dt dt ( R1 + R2 ) ( R1 + R2 )  R2C  Q t −(R + R ) − ( R1 + R2 ) dQ dQ  1 2 = = dt →  dt  → 0 0 e R2C R1R2C R1R2C  R2C e  Q− Q −   ( R1 + R2 )  R1 + R2 

  R Ce  ( R1 + R2 ) t  ln Q  −  2  = − R1R2C  R1 + R2   0  0

  R2C e   Q −   R1 + R2   ( R + R2 ) t →   → ln =− 1   R2C e   R1R2C       R1 + R2  

(R +R ) − 1 2 t R2C e   1 − e R1R2C   R1 + R2   

From the exponential term we obtain the time constant,  =

R1R2C . R1 + R2

The resistors appear in this equation as if they are in parallel. (b) We obtain the maximum charge on the capacitor by taking the limit as time goes to infinity. (R +R ) − 1 2 t R2C e  R Ce  1 − e R1R2C  = 2 Qmax = lim t → R + R   R1 + R2 1 2   This can also be seen as a voltage divider, in the limit as time goes to infinity. 53. (a) With the switch open, the resistors are in series with each other, and so have the same current. Apply the loop rule clockwise around the left loop, starting at the negative terminal of the source, to find the current. R1 is the 8.8- resistor, and R2 is the 4.4- resistor. V − IR1 − IR2 = 0 → I =

24 V

= 1.818 A R1 + R2 8.8  + 4.4  The voltage at point a is the voltage across the 4.4  -resistor.

Va = IR2 = (1.818 A )( 4.4  ) = 8.0 V This can also be seen from the “voltage divider” relationship. (b) With the switch open, the capacitors are in series with each other. Find the equivalent capacitance. The charge stored on the equivalent capacitance is the same value as the charge stored on each capacitor in series. C1 is the 0.48 F capacitor, and C 2 is the 0.36 F capacitor.

1 Ceq

1 C1

1 C2

→ Ceq =

C1C2 C1 + C2

( 0.48 F )( 0.36 F ) = 0.2057  F ( 0.48 F + 0.36 F )

Qeq = VCeq = ( 24.0 V )( 0.2057  F ) = 4.937 C = Q1 = Q2 The voltage at point b is the voltage across the 0.36 F capacitor. Q 4.937 C Vb = 2 = = 13.7 V  14 V C2 0.36  F © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

915

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(c) The switch is now closed. After equilibrium has been reached for a long time, there is no current flowing in the capacitors, and so the resistors are again in series, and the voltage of point a must be 8.0 V. Point b is connected by a conductor to point a, and so point b must be at the same potential as point a, 8.0 V . This also means that the voltage across C 2 is 8.0 V, and the voltage across C1 is 16 V. (d) Find the charge on each of the capacitors, which are no longer in series. Q1 = V1C1 = (16 V )( 0.48 F ) = 7.68C Q2 = V2C2 = ( 8.0 V )( 0.36  F ) = 2.88C

When the switch was open, point b had a net charge of 0, because the charge on the negative plate of C1 had the same magnitude as the charge on the positive plate of C2 . With the switch closed, these charges are not equal. The net charge at point b is the sum of the charge on the negative plate of C1 and the charge on the positive plate of C2. Qb = −Q1 + Q2 = −7.68C + 2.88C = −4.80 C  −4.8 C

Thus 4.8C of charge has passed through the switch, from right to left. 54. Because there are no simple series or parallel connections in this circuit, we use Kirchhoff’s rules to write equations for the currents, as labeled in our diagram. We write junction equations for the junctions c and d. We then write loop equations for each of the three loops. We set the current through the capacitor equal to the derivative of the charge on the capacitor. Q Q I = I1 + I 3 [1] ; I = I 2 + I 4 [2] ; e − 1 − 2 = 0 [3] C1 C2 Q1 Q − I 3 R3 = 0 [4] ; 2 − I 4 R4 = 0 [5] C1 C2 We differentiate Eq. [3] with respect to time and set the derivative of the charge equal to the current. I I C d e dQ1 1 dQ2 1 − − = 0 − 1 − 2 → I 2 = − I1 2 0= dt dt C1 dt C2 C1 C2 C1 We then substitute Eq. [1] into Eq. [2] to eliminate I. Then using Eqs. [4] and [5] we eliminate I3 and I4 from the resulting equation. We eliminate I2 using the derivative equation above. Q C Q I 1 + I 3 = I 2 + I 4 ; I1 + 1 = − I1 2 + 2 R3C1 C1 R4C2 Finally, we eliminate Q2 using Eq. [3].  R4 + R3   C + C2  Q C Q  1  I1 + 1 = − I1 2 +  e − 1  → e = I1 R4  1  →  + Q1  R3C1 C1 R4  C1   C1   R3C1  e = I1 R +

 R3   C + C2  Q1 where R = R4  1   and C = C1  C  C1   R4 + R3 

This final equation represents a simple RC circuit, with time constant  = RC.  C + C2   R3  R4 R3 ( C1 + C2 )  = RC = R4  1 =  C1  R4 + R3  C1   R4 + R3  =

(8.8  )( 4.4  )( 0.48  F + 0.36  F ) = 2.5 s 8.8  + 4.4 

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Chapter 26

DC Circuits

55. The full-scale current is the reciprocal of the sensitivity. 1 I full- = = 2.9  10 −5A or 29  A 35, 000  V scale 56. The resistance is the full-scale voltage multiplied by the sensitivity.

R = Vfull- ( sensitivity ) = ( 250 V )( 35, 000  V ) = 8.75  106   8.8  106 scale

57. (a) The current for full-scale deflection of the galvanometer is 1 1 IG = = = 2.222  10 −5 A. sensitivity 45, 000  V To make an ammeter, a shunt resistor must be placed in parallel with the galvanometer. The voltage across the shunt resistor must be the voltage across the galvanometer. The total current is to be 1.0 A. See Fig. 26–28 for a circuit diagram. I IG 2.222  10−5 A I G rG = I s Rs → Rs = G rG = rG = ( 20.0  ) Is I full − I G 1.0A − 2.222  10−5 A = 4.444  10−4   4.4  10 −4  in parallel

Vfull IG

− rG =

1.00V 2.222  10−5 A

− 20.0  = 44985   45 k in series

58. (a) To make an ammeter, a shunt resistor must be placed in parallel with the galvanometer. The voltage across the shunt resistor must be the voltage across the galvanometer. See Fig. 26–28 for a circuit diagram. Vshunt = VG → ( I full − I G ) Rshunt = I G RG → Rshunt =

( 48  10 A ) ( 32  ) = 6.1  10  ( I − I ) ( 25 A − 48  10 A ) −6

I G RG

full

−5

−6

(b) To make a voltmeter, a resistor must be placed in series with the galvanometer, so that the desired full-scale voltage corresponds to the full-scale current of the galvanometer. See Fig. 26–29 for a circuit diagram. 250 V V Vfull scale = I G ( Rser + RG ) → Rser = full scale − RG = − 32  = 5.2  106  −6 48  10 A IG 59. We divide the full-scale voltage of the electronic module by the module’s internal resistance to determine the current through the module that will give full-scale deflection. Since the module and R2 are in parallel they will have the same voltage drop across them (400 mV) and their currents will add to equal the current through R1. We set the voltage drop across R1 and R2 equal to the 40 volts and solve the resulting equation for R2.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

I meter =

Instructor Solutions Manual

Vmeter 250  10−3 V V V = = 2.08  10−9 A ; I 2 = meter ; I1 = I 2 + I meter = meter + I meter 6 r R2 R2 120  10 

V  V = I1R1 + Vmeter → (V − Vmeter ) =  meter + I meter  R1 →  R2  R2 =

(12  106  ) ( 0.25V ) R1Vmeter = (V − Vmeter ) − I meter R1 ( 40 V − 0.25V ) − ( 2.08  10−9 A )(12  106  )

= 7.552  104   76 k

60. To make a voltmeter, a resistor Rser must be placed in series with the existing meter so that the desired fullscale voltage corresponds to the full-scale current of the galvanometer. We know that 25 mA produces full-scale deflection of the galvanometer, so the voltage drop across the total meter must be 25 V when the current through the meter is 25 mA.

I full

scale

Rser

Rshunt

Vfull

scale

−1   1 1   Vfull = I full Req = I full  Rser +  +   → R R scale scale scale   G shunt    

Vfull

−1

 1 1  25 V 1   1 Rser = − + = − + = 999.8   1000   −3 I full  RG Rshunt  25  10 A  33  0.20   scale

scale

The sensitivity is

1000  25 V

= 40  V

61. If the voltmeter were ideal, then the only resistance in the circuit would be the series combination of the two resistors. The current can be found from the battery and the equivalent resistance, and then the voltage across each resistor can be found. 35 V V Rtot = R1 + R2 = 44 k + 27 k = 71k ; I = = = 4.930  10 −4 A 3 Rtot 71  10 

( )( ) V = IR = ( 4.930  10 A )( 27  10  ) = 13.31V

V44 = IR1 = 4.930  10−4 A 44  103  = 21.69 V −4

Now put the voltmeter in parallel with the 44 k resistor. Find its equivalent resistance, and then follow the same analysis as above.



−1

  = 28.99 k  44 k 85 k 

Req = 

Rtot = Req + R2 = 28.99 k + 27k = 55.99 k

(

)(

)

V Rtot

35 V 55.99  10  3

= 6.251  10 −4 A

V44 = Veq = IReq = 6.251  10 −4 A 28.99  103  = 18.12 V  18 V % error =

18.12 V − 21.69 V 21.69 V

 100 = −16.46%  −16% ( reading too low )

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Chapter 26

DC Circuits

And now put the voltmeter in parallel with the 27- k resistor, and repeat the process. −1

 1 1  + Req =   = 20.49 k  27 k 95 k  Rtot = Req + R1 = 20.49 k + 44 k = 64.49 k

(

−4

)(

)

V Rtot

35 V 64.49  10  3

= 5.427  10 −4 A

V27 = Veq = IReq = 5.427  10 A 20.49  10  = 11.12V  11V % error =

11.12 V − 13.31V 13.31V

 100 = −16.45%  −16% ( reading too low )

62. The total resistance with the ammeter present is Req = 760  + 480  + 53  = 1293 . The voltage

(

)

supplied by the battery, found from Ohm’s law, is Vbattery = IReq = 5.25  10 A (1293  ) = 6.788 V. −3

When the ammeter is removed, we assume that the battery voltage does not change. The equivalent resistance changes to Req = 1240  , and the new current is again found from Ohm’s law.

Vbattery Req

6.788 V 1240 

= 5.47  10−3 A = 5.47 mA

63. Find the equivalent resistance for the entire circuit, and then find the current drawn from the source. That current will be the ammeter reading. The ammeter and voltmeter symbols in the diagram are each assumed to have resistance. ( 7500  )(18000  ) Req = 1.0  + 0.50  + 7500  + ( 7500  + 18000  )

A 0.50

7.5k

V 18k

= 12795.6  I source =

e Req

12.0 V 12795.6 

= 9.378  10−4 A = 9.38  10 −4 A

The voltmeter reading will be the source current times the equivalent resistance of the resistor– voltmeter combination. ( 7500  )(18000  ) Vmeter = I source Req = 9.378  10−4 A = 4.965V  5.0 V ( 7500  + 18000  ) If we had all ideal components (no internal resistance, no ammeter resistance, infinite meter resistance) then the values would be as follows. Req = 1.0  + 0.50  + 2 ( 7500  ) = 15001.5 

(

I source =

e Req

12.0 V 15001.5 

(

)

= 7.999  10−4 A  8.0  10−4 A

)

Vmeter = I source R = 7.999  10−4 A ( 7500  ) = 5.999 V  6.0 V The current is higher with the meters present, and the voltage across the resistor is lower. We calculate the % “error” for current and voltage. 9.378  10−4 A − 7.999  10−4 A  100 = 17.24%  17% higher Current: 7.999  10−4 A © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

4.965 V − 5.999 V

Voltage:

5.999 V

Instructor Solutions Manual

 100 = −17.24%  17% lower

64. From the first diagram, write the sum of the currents at junction a; and then substitute in for those currents as shown. I1 = I1A + I1V

e − VR − I1R2 = 0 → I1 =

e − VR R2

e − VR

; I1 A =

; I1V =

V1V RV

V1V

R2 R1 RV Then do a similar procedure for the second diagram. I 2 = I 2 A + I 2V e − I 2 R1 − VR = 0 → I 2 =

e − VR R1

e − VR

; I2A =

; I 2V =

V2 V RV

Now there are two equations in the two unknowns of R1 and R2 . Solve for the reciprocal values and then find the resistances. Assume that all resistances are measured in kilohms. e − VR VR V1V 12.0 − 5.5 5.5 5.5 6.5 5.5 = + → = + → = + 0.30556 R2 R1 RV R2 R1 18.0 R2 R1 1

e − VR

R1 8.0 R1 6.5 R2

= =

4.0 R2 5.5 R1

V2 V RV

→

+ 0.22222 →

12.0 − 4.0 R1 1 R2

2 R1

2

+ 0.30556 → 6.5 

 R1

R1 = 11.25 k ;

1 R2

2 R1

4.0 R2

4.0 18.0

→

8.0 R1

4.0 R2

+ 0.22222

− 0.05556



5.5



− 0.05556  =

+ 0.30556 →

1 R1

0.66667 7.5

→

− 0.05556 → R2 = 8.18 k

So the final results are R1 = 11k ; R2 = 8.2 k . 65. The sensitivity of the voltmeter is 1000 ohms per volt on the 3.0 volt scale, so it has a resistance of 3000 ohms. The circuit is shown in the diagram. Find the equivalent resistance of the meter–resistor parallel combination and the entire circuit.

1

Rp = 

R

1 

−1

R R

( 3000  )(8400  )

V  = R + R = 3000  + 8400  = 2211  RV  V

Req = R + Rp = 2211  + 8400  = 10611 

e R

V RV

Using the meter reading of 2.3 volts, calculate the current into the parallel combination, which is the current delivered by the battery. Use that current to find the emf of the battery. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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V Rp

2.3V 2211 

= 1.040  10−3 A

(

)

e = IReq = 1.040  10−3 A (10611  ) = 11.04V  11V

66. By calling the voltmeter “high resistance,” we can assume it has no current passing through it. Write Kirchhoff’s loop rule for the circuit for both cases, starting with the negative pole of the battery and proceeding counterclockwise. V Case 1: Vmeter = V1 = I1R1 e − I1r − I1R1 = 0 → e = I 1 ( r + R1 ) = 1 ( r + R1 ) R1

Case 2: Vmeter = V2 = I 2 R2

e − I 2 r − I 2 R2 = 0 → e = I 2 ( r + R2 ) =

V2 R2

( r + R2 )

Solve these two equations for the two unknowns of e and r. V V e = 1 ( r + R1 ) = 2 ( r + R2 ) → R1 R2

   8.1V − 9.7 V = ( 45  )(14.0  )   = 4.408   4.4    V1 R2 − V2 R1   ( 9.7 V )(14.0  ) − ( 8.1V )( 45  )   V2 − V1

r = R1 R2 

V1 R1

( r + R1 ) =

9.7 V 45 

( 4.408  + 45  ) = 10.65 V  11V

67. We know from Example 26–16 that the voltage across the resistor without the voltmeter connected is 4.0 V. Thus the minimum acceptable voltmeter reading is 95% of that: (0.95)(4.0 V) = 3.8 V. The voltage across the other resistor would then be 4.2 V, which is used to find the current in the circuit. V 4.2 V I= 2 = = 2.8  10 −4 A R2 15, 000  This current is used with the voltmeter reading to find the equivalent resistance of the meter–resistor combination, from which the voltmeter resistance can be found. V 3.8 V = 13, 570  Rcomb = comb = I 2.8  10 −4 A 1 1 1 1 1 1 = + → = − → Rcomb R1 Rmeter Rmeter Rcomb R1 Rmeter =

R1 Rcomb R1 − Rcomb

(15, 000  )(13, 570  ) 15, 000  − 13, 570 

= 142, 300   140 k

68. We connect the battery in series with the body and a resistor. The current through this series circuit is the voltage supplied by the battery divided by the sum of the resistances. The voltage drop across the body is equal to the current multiplied by the body’s resistance. We set the voltage drop across the body equal to 0.25 V and solve for the necessary resistance. e I= R + RB

VB = IRB =

e  e RB  1.5 V  → R =  − 1 RB =  − 1 (1800  ) = 9000  = 9.0 k R + RB  0.25 V   VB 

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69. The voltage drop across the two wires is the 3.0-A current times their total resistance.

Vwires = IRwires = ( 3.0 A )( 0.0065  m )( 90 m ) = 1.755V  1.8 V Thus the voltage applied to the apparatus is V = Vsource − Vwires = 120 V − 1.775 V = 118.2 V  118 V . 70. The charge on the capacitor and the current in the resistor both decrease exponentially, with a time Q2 constant of  = RC . The energy stored in the capacitor is given by U = 12 , and the power C dissipated in the resistor is given by P = I 2 R. V Q Q = Q0 e − t / RC ; I = I 0 e − t / RC = 0 e − t / RC = 0 e − t / RC R RC 2

Q02 1  Q0 e −1  Q02  Q2   Q2  −2 1 1 1 U decrease = −U = U t = 0 − U t = =  − = − =  2 (1 − e ) 2 2 2   C C  C t =0  C t =  C  1 2



Q2 Q 2  RC  −2 t / RC  Q  U dissipated =  Pdt =  I Rdt =   0 e − t / RC  Rdt = 0 2  e −2 t / RC dt = 0 2  − )0  (e RC RC 0 RC  2   0 0 2

= − 12

Q02 −2 Q02 1 − = e 1 1 − e −2 ) ( ( ) 2 C C

And so we see that U decrease = U dissipated . 71. (a) If the ammeter shows no current with the closing of the switch, then points B and D must be at the same potential, because the ammeter has some small resistance. Any potential difference between points B and D would cause current to flow through the ammeter. Thus, the potential drop from A to B must be the same as the drop from A to D. Since points B and D are at the same potential, the potential drop from B to C must be the same as the drop from D to C. Use these two potential relationships to find the unknown resistance. R3 I1 VBA = VDA → I 3 R3 = I1 R1 → = R1 I 3 VCB = VCD → I 3 Rx = I1R2 → Rx = R2

(b) Rx = R2

R3 R1

I1 I3

= R2 R3 R1

 78.6    = 111.53   112  .  685  

= ( 972  ) 

72. From the solution to Problem 71, the unknown resistance is given by Rx = R2 R3 R1 . We use that with Eq. 25–3 to find the length of the wire. R 4 L L L Rx = R2 3 = = = → 2 R1 A  ( d 2) d2

( 29.2  )( 3.26  )  (1.22  10−3 m ) = = 27.6 m 4 R1 4 ( 38.0  ) (10.6  10−8  m ) 2

R2 R3 d 2

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73. Divide the power by the required voltage to determine the current drawn by the hearing aid. P 2.5 W I= = = 0.625 A V 4.0 V Use Eq. 26–1 to calculate the terminal voltage across the series combinations of three batteries for both mercury (Hg) and dry (D) cells. VHg = 3 ( e − Ir ) = 3 1.35 V − ( 0.625 A )( 0.030  ) = 3.99 V

VD = 3 ( e − Ir ) = 3 1.50 V − ( 0.625 A )( 0.35  ) = 3.84 V The terminal voltage of the mercury cell batteries is closer to the required 4.0 V than the voltage from the dry cell. 74. One way is to connect N resistors in series. If each resistor can dissipate 0.5 W, then it will take 7 resistors in series to dissipate 3.5 W. Since the resistors are in series, each resistor will be 1/7 of the total resistance. Req 3200  R= = = 457   460  7 7 So connect 7 resistors of 460 each, rated at ½ W, in series. Each resistor could have a maximum voltage of about 15 V in order to not exceed the power rating. Thus the total voltage would be limited to about 105 V. Or, the resistors could be connected in parallel. Again, if each resistor watt can dissipate 0.5 W, then it will take 7 resistors in parallel to dissipate 3.5 W. Since the resistors are in parallel, the equivalent resistance will be 1/7 of each individual resistance. 1 1 = 7   → R = 7 Req = 7 ( 3200  ) = 22.4 k Req R So connect 7 resistors of 22.4 k each, rated at ½ W, in parallel. Each resistor could have about 105 V across it in this configuration. 75. To build up a high voltage, the cells will have to be put in series. 120 V is needed from a series of 120 V 0.80-V cells. Thus = 150 cells are needed in series to provide the desired voltage. Since 0.80 V cell these cells are all in series, their current will all be the same at 350 mA. To achieve the higher current desired, banks made of 150 cells each can be connected in parallel. Then their voltage (per bank) will still be at 120 V, but the currents would add making a total of 1.3 A = 3.71 banks  4 banks. So the total number of cells is 600 cells . The panel area 350  10 −3 A bank

(

)

is 600 cells 9.0 10−4 m2 cell = 0.54 m2 . The cells should be wired in 4 banks of 150 cells in series per bank, with the banks in parallel . This will produce 1.4 A at 120 V. To optimize the output, always have the panel pointed directly at the sun . 76. The terminal voltage and current are given for two situations. Apply Eq. 26–1 to both of these situations, and solve the resulting two equations for the two unknowns. V1 = e − I1r ; V2 = e − I 2 r → e = V1 + I1r = V2 + I 2 r →

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

V2 − V1 I1 − I 2

47.3 V − 40.8 V

Instructor Solutions Manual

= 1.512   1.5 

7.10 A − 2.80 A

e = V1 + I1r = 40.8 V + ( 7.10 A )(1.512  ) = 51.5 V 77. There are two answers because it is not known which direction the given current is flowing through the 4.0-k resistor. Assume the current is to the right. The voltage across the 4.0-k resistor is

(

)

given by Ohm’s law as V = IR = 2.85  10−3 A ( 4000  ) = 11.4 V. The voltage drop across the 8.0-k resistor must be the same, and the current through it is I =

V R

11.4 V 8000 

= 1.425  10 −3 A.

The total current in the circuit is the sum of the two currents, and so I tot = 4.275  10−3 A. That current can be used to find the terminal voltage of the battery. Write a loop equation, starting at the negative terminal of the unknown battery and going clockwise. We assume the current is to the right. Vab − ( 5000  ) I tot − 11.4 V − 12.0 V − (1.0  ) I tot = 0 →

(

)

Vab = 23.4V + ( 5001  ) 4.275  10 A = 44.78 V  45 V −3

If the current is to the left, then the voltage drop across the parallel combination of resistors is still 11.4 V, but with the opposite orientation. Again write a loop equation, starting at the negative terminal of the unknown battery and going clockwise. The current is now to the left. Vab + ( 5000  ) I tot + 11.4 V − 12.0 V+ (1.0  ) I tot = 0 →

(

)

Vab = 0.6V − ( 5001  ) 4.275  10 A = −20.78 V  −21V −3

78. The current in the circuit can be found from the resistance and the power dissipated. Then the product of that current and the equivalent resistance is equal to the battery voltage.

P = I 2R → I =

P33 R33

0.80 W 33 

= 0.1557 A

−1

1   1 Req = 33  +  +  = 73.49   88  75  

V = IReq = ( 0.1557 A )( 73.49  ) = 11.44 V  11V

79. If the switches are both open, then the circuit is a simple series circuit. Use Kirchhoff’s loop rule to find the current in that case. 6.0 V − I ( 50  + 20  + 10  ) = 0 → I = 6.0 V 80  = 0.075 A If the switches are both closed, the 20- resistor is in parallel with R, and the 10- resistor is shorted. Apply Kirchhoff’s loop rule to the outer loop of the circuit, with the 20- resistor having the current found previously. 6.0 V − ( 0.075 A )( 20  ) 6.0 V − I ( 50  ) − ( 0.075 A )( 20  ) = 0 → I = = 0.090 A 50  This is the current in the parallel combination. Since 0.075 A is in the 20- resistor, 0.015 A must be in R. The voltage drops across R and the 20- resistor are the same since they are in parallel. I 0.075 A V20 = VR → I 20 R20 = I R R → R = R20 20 = ( 20  ) = 100  IR 0.015 A

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2 80. (a) Since P = V R and the voltage is the same for each combination, the power and resistance are inversely related to each other. So for the 50-W output, use the higher-resistance filament . For the 100-W output, use the lower-resistance filament . For the 150-W output, use the filaments in parallel . P =V2 R → (b)

R50 W =

(120 V )2

Rp =

R1 + R2

( 288  )(144  ) 288  + 144 

= 96 

R100 W =

(120 V )2

= 144   140  50 W 100 W P As a check, the parallel combination of the resistors gives the following. R1 R2

= 288   290 

V2 R

(120 V ) 96 

= 150 W.

81. (a) The resistance of the wire is found as two resistors in series, using Eq. 25–3 to calculate each resistance. l l l R =  Cu Cu +  Al Al = (  Cu +  Al ) ACu AAl A

    5.0 m −8 −8   = 0.14064   0.14  = (1.68  10  m + 2.65  10  m )  (1.4  10 −3 m )2      4 (b) Use Ohm’s Law. V = IR → I =

0.075 V

R 0.14064  (c) Again, use Ohm’s Law.

= 0.5333 A  0.53 A

    5.0 m −8   = 0.02919 V  0.029 V VCu = IR Cu = ( 0.5333 A ) (1.68  10  m )  (1.4  10−3 m )2      4 VAl = Vtotal − VCu = 0.075 V − 0.029 V = 0.046 V

82. From Eq. 26–7, the product RC is equal to the time constant. 3.0 s  = 1.0  106   = RC → R = = −6 C 3.0  10 F 83. The equivalent resistance is the sum of all the resistances. The current is found from Ohm’s law. e 240 V = = 8  10−3A I= 4 Req 3  10  This is 8 milliamps, and 10 milliamps is considered to be a dangerous level in that it can cause sustained muscular contraction. The 8 milliamps could certainly be felt by the patient, and could be painful.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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84. (a) The 12- and the 25- resistors are in parallel, with a net resistance R1-2 as follows. −1

1   1 R1-2 =  +  = 8.108   12  25   R1-2 is in series with the 4.5- resistor, for a net resistance R1-2-3 as follows.

R1-2-3 = 4.5  + 8.108  = 12.608 

That net resistance is in parallel with the 18- resistor, for a final equivalent resistance as follows. −1

1 1   Req =  +  = 7.415   7.4   12.608  18  

(b) Find the current in the 18- resistor by using Kirchhoff’s loop rule for the loop containing the battery and the 18- resistor. 6.0 V e = = 0.33 A e − I18 R18 = 0 → I18 = R18 18  (c) Find the current in R1-2 and the 4.5- resistor by using Kirchhoff’s loop rule for the outer loop containing the battery and the resistors R1-2 and the 4.5- resistor.

e − I1-2 R1-2 − I1-2 R4.5 = 0 → I1-2 =

e R1-2 + R4.5

6.0 V 12.608 

= 0.4759 A

This current divides to go through the 12- and 25- resistors in such a way that the voltage drop across each of them is the same. Use that to find the current in the 12- resistor. I1-2 = I12 + I 25 → I 25 = I1-2 − I12 VR = VR 12

I12 = I1-2

→ I12 R12 = I 25 R25 = ( I1-2 − I12 ) R25 → R25

( R12 + R25 )

= ( 0.4759 A )

25  37 

= 0.32 A

(d) The current in the 4.5- resistor was found above to be I1-2 = 0.4759 A. Find the power accordingly. 2 2 P4.5 = I1-2 R 4.5 = ( 0.4759 A ) ( 4.5  ) = 1.019 W  1.0 W 85. We find the current for the bulb from the bulb’s ratings. Since the bulb will have 3.0 V across it, the resistor R will have 6.0 V across it. 2.0 W 6.0 V P V Pbulb = I bulbVbulb → I bulb = bulb = = 0.667A ; R = R = = 9.0  Vbulb 3.0 V I R 0.667A 86. The equivalent resistance of the circuit is the parallel combination of the bulb and the lower portion of the variable resistor, in series with the upper portion of the variable resistor. With the slide at position x, the resistance of the lower portion is Rlower = xRvar , and the resistance of the upper portion is Rupper = (1 − x ) Rvar . From that equivalent resistance, we find the current in the loop, the voltage across the bulb, and then the power expended in the bulb. −1

 1 Rlower Rbulb xRvar Rbulb 1  Rparallel =  + =  = Rlower + Rbulb xRvar + Rbulb  Rlower Rbulb  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Req = Rupper + Rparallel = (1 − x ) Rvar + Rparallel ; I loop =

e Req

; Vbulb = I loop Rparallel ; Pbulb =

2 Vbulb

Rbulb

(a) Consider the case in which x = 1.00. In this case, the full battery potential is across the bulb, and so it is obvious that Vbulb = 120 V. Thus Pbulb =

2 Vbulb

Rbulb

(120 V )2 240 

= 60 W .

(b) Consider the case in which x = 0.70. xRvar Rbulb ( 0.70)(150  )( 240  ) = = 73.04  Rparallel = xRvar + Rbulb ( 0.70 )(150  ) + 240 

Req = (1 − x ) Rvar + Rparallel = ( 0.30 )(150  ) + 73.04  = 118.04  I loop = Pbulb =

e Req

120 V 118.04 

= 1.017 A ; Vbulb = (1.017 A )( 73.04  ) = 74.28 V

( 74.28 V )2

= 22.99 W  23W 240  (c) Consider the case in which x = 0.40. xRvar Rbulb ( 0.40 )(150  )( 240  ) = = 48  Rparallel = xRvar + Rbulb ( 0.40 )(150  ) + 240  Req = (1 − x ) Rvar + Rparallel = ( 0.6 )(150  ) + 48  = 138  I loop = Pbulb =

e Req

120 V 138 

( 41.74 V )2 240 

= 0.8696 A ; Vbulb = ( 0.8696 A )( 48  ) = 41.74 V = 7.259 W  7.3W

87. The capacitor will charge up to 75% of its maximum value, and then discharge. The charging time is the time for one heartbeat. 1min 60s tbeat =  = 0.8333s 72 beats 1min − − −      t  V = V0  1 − e RC  → 0.75V0 = V0  1 − e RC  → e RC = 0.25 →  − beat  = ln ( 0.25) →  RC      tbeat

R=−

tbeat

C ln ( 0.25)

=−

0.8333s

(5.2  10 F ) ( −1.3863) −6

tbeat

= 1.2  105 

88. (a) Apply Ohm’s law to find the current. Vbody 120 V I= = = 0.126 A  0.13 A Rbody 950  (b) The description of “alternative path to ground” is a statement that the 25- path is in parallel with the body. Thus the full 120 V is still applied across the body, and so the current is the same: 0.13A .

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(c) If the current is limited to a total of 1.5 A, then that current will get divided between the person and the parallel path. The voltage across the body and the parallel path will be the same, since they are in parallel. Vbody = Valternate → I body Rbody = I alternate Ralternate = ( I total − I body ) Ralternate → I body = I total

Ralternate

body

+ Ralternate )

= (1.5 A )

25  950  + 25 

= 0.03846 A  38 mA

This is still a very dangerous current. 89. When the galvanometer gives a null reading, no current is passing through the galvanometer or the emf that is being measured. All of the current is flowing through the slide wire resistance. Application of the loop rule to the lower loop gives e − IR = 0, since there is no current through the emf to cause a voltage drop across any internal resistance in the emf or any resistance in the galvanometer. The current flowing through the slide wire resistor is independent of the emf in the lower loop because no current is flowing through the lower loop. We solve the loop rule equation for the current and set the currents equal. The resulting equation is solved for the unknown emf.

ex − IRx = 0 ; es − IRs = 0 → ; I =

ex Rx

→

 Rx   es  Rs 

ex = 

90. (a) In normal operation, the capacitor is fully charged by the power supply, and so the capacitor voltage is the same as the power supply voltage, and there will be no current through the resistor. If there is an interruption, the capacitor voltage will decrease exponentially–it will discharge. We want the voltage across the capacitor to be at 75% of the full voltage after 0.20 s. Use Eq. 26–9b for the discharging capacitor. − 0.20s / RC − 0.20s / RC V = V0e − t / RC ; 0.75V0 = V0e ( ) → 0.75 = e ( ) →

− ( 0.20s )

C ln ( 0.75)

− ( 0.20s )

( 7.65  10 F ) ln ( 0.75) −6

= 90877   91k

(b) When the power supply is functioning normally, there is no voltage across the resistor, so the device should NOT be connected between terminals a and b. If the power supply is not functioning normally, there will be a larger voltage across the capacitor than across the capacitor–resistor combination, since some current might be present. This current would result in a voltage drop across the resistor. To have the highest voltage in case of a power supply failure, the device should be connected between terminals b and c . 91. (a) We assume that the ammeter is ideal and so has 0 resistance, but that the voltmeter has resistance RV. Then apply Ohm’s law, using the equivalent resistance. We also assume the voltmeter is accurate, and so it is reading the voltage across the battery.

V = IReq = I

1

1 1

→ V

R

1 

+ = =I → RV  R RV V

→

1 R

I V

−

1 RV

R RV (b) We now assume the voltmeter is ideal, and so has an infinite resistance, but that the ammeter has resistance RA. We also assume that the voltmeter is accurate and so is reading the voltage across the battery. V = IReq = I ( R + RA ) → R + RA =

V I

→

V I

− RA

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92. (a) The time constant of the RC circuit is given by Eq. 26–7.  = RC = ( 33.0 k )( 4.00  F ) = 132 ms During the charging cycle, the charge and the voltage on the capacitor increases as in Eq. 26–6b. We solve this equation for the time it takes the circuit to reach 90.0 V. 90.0 V   V  V = e (1 − e −t /  ) → t = − ln 1 −  = − (132 ms ) ln 1 −  = 304 ms  e  100.0 V  (b) When the neon bulb starts conducting, the voltage on the capacitor drops quickly to 65.0 V and then starts charging. We can find the recharging time by first finding the time for the capacitor to reach 65.0 V, and then subtract that time from the time required to reach 90.0 V. 65.0 V   V  t = − ln  1 −  = − (132 ms ) ln  1 −  = 139 ms  e  100.0 V  t = 304 ms − 139 ms = 165 ms ; t2 = 304 ms + 165ms = 469 ms

(c)

93. There are three loops, and so there are three loop equations necessary to solve the circuit. We identify five currents in the diagram, and so we need two junction equations to complete the analysis. Note that if we use the resistances in “kilo-ohms,” the currents will be in “milli-amps.” Lower left junction: I 4 = I1 + I 3 Lower right junction: I 2 = I 3 + I 5

Left loop, clockwise:

16 V − 15 I1 − 12 I 4 = 0

Right loop, counterclockwise:

21V − 13I 2 − 18 I 5 = 0

Bottom loop, counterclockwise: 12 V + 18 I 5 − 12 I 4 = 0 16 − 15I1 − 12 ( I1 + I 3 ) = 0

→ 16 − 27 I1 − 12 I 3 = 0 → I1 =

21 − 13 ( I 3 + I 5 ) − 18 I 5 = 0 → 21 − 13I 3 − 31I 5 = 0 → I 5 =

16 − 12 I 3 27 21 − 13I 3 31

12 + 18 I 5 − 12 ( I1 + I 3 ) = 0 → 12 − 12 I1 − 12 I 3 + 18 I 5 = 0 Substitute the results of the first two equations into the third equation.

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12 − 12

Instructor Solutions Manual

 16 − 12 I 3  − 12 I + 18  21 − 13I 3  = 0 3  27   31 

12 − 12 16 − 12 I 3 ) − 12 I 3 + 1831 ( 21 − 13I 3 ) = 0 27 ( 12 − 94 16 + 1831 21 = (12 + 1831 13 − 94 12 ) I 3 17.082 = 14.215I 3 → I 3 =

17.082

= 1.202 mA 14.215 16 − 12 I 3 16 − 12 (1.202 ) (a) I1 = = = 0.0584 mA  5.8  10−5 A, upward 27 27 (b) Reference all voltages to the lower left corner of the circuit diagram. Then Va = 16 V, Vb = 12 V + 21V = 33 V, and Va − Vb = 16 V − 33 V = −17 V .

94. The power delivered to the starter is equal to the square of the current in the circuit multiplied by the resistance of the starter. Since the resistors in each circuit are in series we calculate the currents as the battery emf divided by the sum of the resistances. 2

P I 2 RS  I   e Req   R0eq   r + RS  = =  =  =  =  P0 I 02 RS  I 0   e R0eq   Req   r + RS + RC 

  0.02  + 0.15  =  = 0.40  0.02  + 0.15  + 0.10   95. Note that, based on the significant figures of the resistors, the 1.0- resistor will not change the equivalent resistance of the circuit as determined by the resistors in the switch bank. Case 1: n = 0 switch closed. The effective resistance of the circuit is 16.0 k. The current in the 16 V = 1.0 mA. The voltage across the 1.0- resistor is V = IR circuit is I = 16.0 k = (1.0 mA )(1.0  ) = 1.0 mV .

Case 2: n = 1 switch closed. The effective resistance of the circuit is 8.0 k. The current in the circuit is I =

16 V 8.0 k

= 2.0 mA. The voltage across the 1.0- resistor is V = IR

= ( 2.0 mA )(1.0  ) = 2.0 mV .

Case 3: n = 2 switch closed. The effective resistance of the circuit is 4.0 k. The current in the 16 V = 4.0 mA. The voltage across the 1.0- resistor is V = IR circuit is I = 4.0 k = ( 4.0 mA )(1.0  ) = 4.0 mV .

Case 4: n = 3 and n = 1 switches closed. The effective resistance of the circuit is found by the parallel combination of the 2.0-k and 8.0-k resistors. −1

1   1 Req =  +  = 1.6 k  2.0 k 8.0 k 

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Chapter 26

DC Circuits

The current in the circuit is I =

16 V 1.6 k

= 10 mA. The voltage across the 1.0- resistor is

V = IR = (10 mA )(1.0  ) = 10 mV .

So in each case, the voltage across the 1.0- resistor, if taken in mV, is the expected analog value corresponding to the digital number set by the switches. 96. (a) Use Eq. 26–6b to solve for the unknown resistance, where the ratio VC e = 0.95 .

0.95 = 1 − e − t / RC → R=−

t C ln(1 − 0.95)

=−

2.0 s

(1.4 10 F ) ln(0.05) −6

= 4.769  105   5  105 

(b) The defibrillator paddles are usually moistened so that the resistance of the skin is about 1000 Ω. Use Eq. 26–7 to solve for the capacitance. We assume the resistance has 2 significant figures.  12  10−3 s  = RC → C = = = 12  10−6 F R 1000  97. We apply the junction rule to points a and b, and then apply the loop rule to loops 1, 2, and 3. This enables us to solve for all of the currents. I 5 = I 6 + I top ; I top + I 6.8 = I 12 → I 5 − I 6 = I12 − I 6.8 →

I 5 + I 6.8 = I12 + I 6

[1]

17.00 − 5.00 I 5 − 6.00 I 6 = 0

[2] ( loop 1)

14.00 − 12.00 I12 − 6.80 I 6.8 = 0

[3] ( loop 2 )

12.00 I12 − 6.00 I 6 = 0

[4] ( loop 3)

Use Eq. [4] to solve for I 6 = I12

12.00

= 2.00 I12 . Now the equation set is as follows. 6.00 I 5 + I 6.8 = 3.00 I12 [1] ; 17.00 − 5.00 I 5 − 12.00 I12 = 0 [2] ; 14.00 − 12.00 I12 − 6.80 I 6.8 = 0 [3]

Use Eq. [1] to eliminate I 6.8 by I 6.8 = 3.00 I12 − I 5 .

17.00 − 5.00 I 5 − 12.00 I12 = 0

[2]

14.00 − 12.00 I12 − 6.80 ( 3.00 I12 − I 5 ) = 0 → 14.00 − I12 (12.00 + 20.40 ) + 6.80 I 5 = 0 Use Eq. [2] to eliminate I 5 by I 5 = 14.00 − 32.40 I12 +

17.00 − 12.00 I12 5.00

[3]

, and then Eq. [3] for I12 .

 17.00 − 12.00 I12  6.80 = 0 → 70.00 − 162.00 I + 115.60 − 81.70 I = 0 → 12 12   5.00

185.60 − 243.70 I12 = 0 → I12 =

185.60

= 0.7616 A  0.762 A = I12 243.70 17.00 − 12.00 I12 17.00 − 12.00 ( 0.7616 ) I5 = = = 1.5722 A  1.57A = I 5 5.00 5.00

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I 6.8 = 3.00 I12 − I 5 . = 3.00 ( 0.7616 ) − 1.5722 = 0.7126 A  0.713A = I 6.8 I 6 = 2.00 I12 = 2.00 ( 0.7616 A ) = 1.5232  1.52 A = I 6 98. (a) The light will first flash when the voltage across the capacitor reaches 90.0 V. t −   RC V = e0  1 − e  →  



t = − RC ln  1 −



V 

 90.0  = 0.628s = − ( 2.15  106  )( 0.150  10 −6 F ) ln  1 −   e0   105 

(b) We see from the equation that t  R, and so if R increases, the time will increase . (c) The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief. (d) Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp “goes out,” the lamp resistance increases, and the capacitor starts to recharge. It charges again for 0.628 seconds, and the process will repeat. 99. We represent the 10.00-M resistor by R10 , and the resistance of the voltmeter as RV . In the first configuration, we find the equivalent resistance ReqA , the current in the circuit I A , and the voltage drop across R. ReqA = R +

R10 RV e R ; IA = ; VR = I A R = e − VA → e = e − VA R10 + RV ReqA ReqA

In the second configuration, we find the equivalent resistance ReqB , the current in the circuit I B , and the voltage drop across R10 . RRV e R ReqB = R10 + ; IB = ; VR10 = I B R10 = e − VB → e 10 = e − VB R + RV ReqB ReqB We now have two equations in the two unknowns of R and RV . We solve the second equation for RV and substitute that into the first equation. We are leaving out much of the algebra in this solution. R R e =e = e − VA ; R R ReqA R + 10 V R10 + RV e

R10 R10 VB R10 R =e = e − VB → RV = RRV ReqB ( e R − VB R10 − VB R ) R10 + R + RV

e − VA = e

R =e R10 RV R+ R10 + RV

R   VB R10 R R10  ( e R − VB R10 − VB R )   R+   VB R10 R R10 +    ( e R − VB R10 − VB R ) 

→

VB 7.317 V R10 = (10.00 M ) = 199.92 M  200 M ( 3 sig. figs.) VA 0.366 V

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DC Circuits

100. Let the internal resistance of the voltmeter be indicated by RV , and let the 15-M resistance be indicated by R15 . We calculate the current through the probe and voltmeter as the voltage across the probe divided by the equivalent resistance of the probe and the voltmeter. We then set the voltage drop across the voltmeter equal to the product of the current and the parallel combination of RV and R15 . This can be solved for the unknown resistance. V R R V R15 RV VR15 RV I= = → ; VV = I 15 V = R15 RV R R + + + + R R R R R R R R R ( ) 15 V 15 V 15 V 15 V 15 V R+ R+ R15 + RV R15 + RV

V R15 RV − R15 RV  (15M )(10 M )  50,000 V  R15 RV  V VV = − 1 R=  − 1 = ( R15 + RV ) ( R15 + RV )  VV  ( 25M )  50 V  = 5994 M  6000 M = 6G 101.

933

CHAPTER 27: Magnetism Responses to Questions 1.

The Earth’s magnetic field is not always parallel to the surface of the Earth–it may have a component perpendicular to the Earth’s surface. The compass will tend to line up with the local direction of the magnetic field, and so one end of the compass will dip downward. The angle that the Earth’s magnetic field makes with the horizontal is called the dip angle.

The pole on a magnetic compass needle that points geographically northward is defined as the north pole of the compass. This north pole is magnetically attracted to the south pole of other magnets, and so the Earth’s magnetic field must have a south pole at the geographic North pole.

The magnetic field lines are concentric circles around the wire. With the current running to the left, the field is directed I counterclockwise when looking from the left end. So, the field goes into the page above the wire and comes out of the page below the wire. The direction of a few of the field lines are indicated on the adjacent figure.

The force is downward. The field lines point from the north pole to the south pole, or left to right. Use the right-hand rule. Your right hand fingers should point in the direction of the current (away from you). Curl your fingers in the direction of the field (to the right). Your thumb points in the direction of the force (downward). See Fig. 27–11a, copied here.

   F is always perpendicular to both B and l . B and l can be at any angle with respect to each other.

Typical current in a house circuit is 60 Hz AC. Due to the mass of the compass needle, its reaction to 60 Hz (changing direction back and forth at 60 complete cycles per second) will probably not be noticeable. A DC current in a single wire could affect a compass, depending on the relative orientation of the wire and the compass, the magnitude of the current, and the distance from the wire to the compass. A DC current being carried by two very close wires in opposite directions would not have much of an effect on the compass needle, since the two currents would cause magnetic fields that tended to cancel each other.

Use the right-hand rule to determine the direction of the force on each particle. Be sure to take into account the charge of each particle. In the plane of the diagram, the magnetic field is coming out of the paper for points above the wire, and is going into the paper for points below the wire. a: force down, toward the wire b: force to the left, opposite of the direction of the current c: force up, toward the wire d: force to the left, opposite of the direction of the current

The magnetic force will be exactly perpendicular to the velocity, which means the force is perpendicular to the direction of motion. Since there is no component of force in the direction of motion, the work done by the magnetic force will be zero, and the kinetic energy of the particle will not change. The particle will change direction, but not change speed.

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(a) Near one pole of a very long bar magnet, the magnetic field is proportional to 1 r 2 , analogous to the electric field near one point charge of a dipole. (b) Far from the magnet as a whole, the magnetic field is proportional to 1 r 3 , analogous to the electric field of a dipole when far away from the dipole.

10. Charge a experiences an initial upward force. By the right-hand rule, charge a must be positive. Charge b experiences no force, and so charge b must be uncharged. Charge c experiences an initial downward force. By the right-hand rule, charge c must be negative. 11. The negative particle will curve down (toward the negative plate) if v > E/B because the magnetic force (down) will be greater than the electric force (up). If v < E/B the negative particle will curve up toward the positive plate because the electric force will be greater than the magnetic force. The motion of a positive particle would be exactly opposite that of a negative particle. 12. No, you cannot set a resting electron into motion with a magnetic field (no matter how large the field is). A magnetic field can only put a force on a moving charge. Thus, with no force (which means no acceleration), the velocity of the electron will not change–it will remain at rest. However, you can set a resting electron into motion with an electric field. An electric field will put a force on any charged particle, moving or not. Thus, the electric force can cause the electron to accelerate from rest to a higher speed. 13. The right-hand rule tells us that as the positive particle moves to the right, if it is deflected upward the magnetic field must be into the paper, and if it is deflected downward the magnetic field must be out of the paper. Also, the stronger the magnetic field, the more tightly the charged particle will turn. The magnetic field at the first bend must be relatively small and pointed into the paper. The magnetic field at the second bend must be medium in size and pointed out of the paper. The magnetic field at the third bend must be relatively very large and pointed into the paper. In terms of the indicated letters. a: no field b: relatively weak field, into the paper c: moderate strength field, out of the paper d: no field e: very strong field, into the paper 14. The particle will move in an elongating helical path in the direction of the electric field (for a positive charge). The radius of the helix will remain constant, but the pitch (see Problem 32) will increase, because the particle will accelerate along the electric field lines. 15. The beam is deflected to the right. The current in the wire creates a magnetic field into the page surrounding the beam of electrons. This results in a magnetic force on the negative particles that is to the right. 16. Yes. One possible situation is that the magnetic field is parallel or anti-parallel to the velocity of the charged particle. In this case, the magnetic force would be zero, and the particle would continue moving in a straight line. Another possible situation is that there is an electric field with a magnitude and direction (perpendicular to the magnetic field) such that the electric and magnetic forces on the particle cancel each other out. The net force would be zero and the particle would continue moving in a straight line.

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17. No, a moving charged particle can be deflected sideways with an electric field that points perpendicular to the direction of the velocity, even when the magnetic field in the region is zero. 18. If the direction of the velocity of the electrons is changing but their speed is not, then they are being deflected by a magnetic field only, and their path will be circular or helical. If the speed of the electrons is changing but their direction is not changing, then they are being accelerated by an electric field only, that is either parallel or anti-parallel to the direction of motion. If both speed and direction are changing, the particles are possibly being deflected by both magnetic and electric fields, or they are being deflected by an electric field that is not parallel to the initial velocity of the particles. In the latter case, the component of the electron velocity anti-parallel to the field direction will continue to increase, and the component of the electron velocity perpendicular to the field direction will remain constant. Therefore, the electron will asymptotically approach a straight path in the direction opposite the field direction. If the particles continue with a circular component to their path, there must be a magnetic field present. 19. You could use a small current-carrying coil or a solenoid for the compass needle. 20. Suspend the magnet in a known magnetic field so that it is aligned with the field and free to rotate. Measure the torque necessary to rotate the magnet so that it is perpendicular to the field lines. The magnetic moment will be the torque divided by the magnetic field strength. τ = μ  B so  = B when the magnetic moment and the field are perpendicular. 21. (a) If the plane of the current loop is perpendicular to the field such that the direction of A is parallel to the magnetic field lines from the external magnetic field, the loop will be in stable equilibrium. Small displacements from this position will result in a torque that tends to return the loop to this position. (b) If the plane of the current loop is perpendicular to the field such that the direction of A is antiparallel to the field lines, the loop will be in unstable equilibrium. 22. Using the configuration shown in Fig. 27–39, the downward electric field puts a force on the charged particles. Positive charges are accelerated downward, and negative particles are accelerated upwards. As the ions move vertically, perpendicularly to the magnetic field, the magnetic field now puts a force on them. Using “Right-Hand Rule 3,” both charges of ions experience a force out of the page, which is parallel to the axis of the blood vessel. This will accelerate the ions along the length of the blood vessel, in the direction of the force vector shown. 23. The charge carriers are positive. Positive particles moving to the right in the figure will experience a magnetic force into the page, or toward point a. Therefore, the positive charge carriers will tend to move toward the side containing a; this side will be at a higher potential than the side with point b. 24. The two ions will come out of the velocity selector portion of the mass spectrometer at the same speed, since v = E B. Once the charges reach the area of the mass spectrometer where there is only a magnetic field, then the difference in the ions’ charges will have an effect. Eq. 27–15 for the mass spectrometer says that m = qBBr E , which can be rearranged as r = mE qBB. Since every quantity in the equation is constant except q, the doubly-ionized ion will hit the film at half of the radius as the singly-ionized ion.

936

Chapter 27

Magnetism

Solutions to MisConceptual Questions 1.

(c) It is common to confuse the directions of the magnetic fields with electric fields and thus indicate that the magnetic field points toward or away from the current. Since the current is flowing into the page, the right-hand rule indicates that the magnetic field is in the clockwise direction around the current. At point A the field points downward.

(b) The right-hand rule indicates that the magnetic field is clockwise around the current, so at point B the current is to the left.

(e) When a magnet is cut the resulting magnets will each have 2 poles, and each of those 2 poles will be the same strength. All of the magnetic field lines that “leave” the original N pole will terminate at the new S pole after cutting–magnetic field lines have to close on themselves. So there is no way to make a magnet with one pole of a different strength than the other.

(e) It is common to confuse the direction of electric fields (which point toward or away from the charges) with magnetic fields, which always make circles around the current.

(c) Electric fields are created by charged objects whether the charges are moving or not. Magnetic fields are created by moving charged objects. Since the proton is charged and moving, it creates both an electric field and a magnetic field.

(a) A stationary charged particle does not experience a force in a magnetic field. Therefore, the particle must be moving to experience a force. The force is a maximum when the particle is moving perpendicular to the field, not parallel to the field. Since the force is perpendicular to the motion of the particle, it acts as a centripetal force, changing the particle’s direction but not its kinetic energy. That is, since the force is perpendicular to the motion it does no work on the particle. The direction of the force is always perpendicular to the direction of motion and also perpendicular to the magnetic field.

(a) The charged particle only experiences a force when it has a component of velocity perpendicular to the magnetic field. When it moves parallel to the field, it follows a straight line at constant speed.

(d) From the equation FB = q v  B, the only difference in the two magnetic forces is that the proton and electron have opposite charges. The velocity and magnetic field are the same in both cases. Thus, the only difference is that the two forces are in opposite directions.

(c) A common misconception is that a force always does work on an object. Since the magnetic force is perpendicular to the proton’s velocity, the force acts as a centripetal force, changing the proton’s direction, but not doing any work and thus not changing its kinetic energy.

10. (a) A common misconception is that a constant magnetic field can change the magnitude of the particle’s velocity. However, the magnetic force is always perpendicular to the velocity, so it can do no work on the particle. The magnetic force only serves as a centripetal force to change the particle’s direction. 11. (e) Eq. 27–5b shows that the magnetic force depends upon the particle’s charge, its velocity, and the strength of the external the magnetic field. The direction of the force is always perpendicular to the magnetic field and the velocity of the particle. Therefore, all four statements are accurate. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

12. (b) A magnetic field is created by moving charge, but an electric field is created by an unbalanced charge of one sign or another in some region. In a metallic wire, conduction charges are moving (on average) along the wire, creating a magnetic field, but the wire itself is neutral until you get to the atomic level. Even a microscopic amount of the wire is neutral, and so there is no net electric field. 13. (c) The net force on each particle points toward the center of its circular path. Thus, for particle 1 the net force is initially pointing upwards. Using the right-hand rule for a positive charge, FB = q v  B, the force would point upwards. Thus, particle 1 must be positive. The same analysis for particle 2, which is initial moving leftwards, would give a downwards force. Thus, particle 2 must be negative.

Solutions to Problems 1.

You desire to travel 35 east of north, but your compass is actually pointing 13 east of true north. After adding 35 to what you assume is true north, you are actually hiking in a line that is 48 east of true north. See the diagram, and use the law of cosines to determine how far away you are from your destination. d 2 = ( 3 km ) + ( 3 km ) − 2 ( 3 km )( 3 km ) cos13 = 0.45134 m 2 → 2

desired 3 km 13 35

actual 3 km

d = 0.679 m  0.68 m

If the field decreases as 1 r 3, then B  1 r 3 , or B = k r 3 , or Br 3 = k .

(

)

3 3 k = Bsurface rsurface = 0.50  10−4 T REarth

)

(

3 0.50  10−4 T REarth = B (1.2 REarth )

(b)

(

3 0.50  10−4 T REarth = B ( 3.0 REarth )

)

→ B=

( 0.50 10 T ) = 2.9  10 T

→ B=

( 0.50 10 T ) = 1.9  10 T

−4

(a)

−5

(1.2 )

−4

−6

( 3.0 )

(a) Use Eq. 27–1 to calculate the force with an angle of 90 and a length of 1 meter. F = I l B sin  → F l = IB sin  = ( 7.40 A )( 0.90T ) sin 90 = 6.66 N m  6.7 N m

(b) Change the angle to 35.0. F l = IB sin  = ( 7.40 A )( 0.90T ) sin 35.0 = 3.82 N m  3.8 N m

Use Eq. 27–2.

Fmax = I l B → I = 5.

Fmax lB

0.625 N

( 4.20 m ) (8.00  10−2 T )

= 1.86 A

Use Eq. 27–1 to calculate the force. F = I l B sin  = (150 A )( 240 m ) ( 5.0  10 −5 T ) sin 68 = 1.7 N

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Magnetism

The dip angle is the angle between the Earth’s magnetic field and the current in the wire. Use Eq. 27–1 to calculate the force.

(

)

F = I l B sin  = ( 4.5A )( 2.6 m ) 5.5  10−5 T sin 35 = 3.691  10 −4 N  3.7  10 −4 N 7.

To have the maximum force, the current must be perpendicular to the magnetic field, F F = 0.55 max to find the angle between the wire as shown in the first diagram. Use l l and the magnetic field, illustrated in the second diagram. F F = 0.55 max → IB sin  = 0.55IB →  = sin −1 0.55 = 33 l l

wire



wire

(a) By the right-hand rule, the magnetic field must be pointing up, and so the top pole face must be a South pole . (b) Use Eq. 27–2 to relate the maximum force to the current. The length of wire in the magnetic field is equal to the diameter of the pole faces. 7.50  10−2 N Fmax Fmax = I l B → I = = = 4.1667 A  4.17 A l B ( 0.100 m )( 0.180 T ) (c) If the wire is tipped so that it points 10.0 downward, the angle between the wire and the magnetic field is changed to 100.0. But the length of wire now in the field has increased initial rotated from l to l cos10.0. The net effect of the changes is that the force has not changed.

(

)

Finitial = I l B ; Frotated = I ( l cos10.0 ) B sin100.0 = I l B = Finitial = 7.50  10 −2 N

The magnetic force must be equal in magnitude to the force of gravity on the wire. The maximum magnetic force is applicable since the wire is perpendicular to the magnetic field. The mass of the wire is the density of copper times the volume of the wire.

FB = mg → I l B =  ( 12 d ) l g → 2

3 3 2 −3  d 2 g ( 8.9  10 kg m )  (1.00  10 m ) ( 9.80 m s ) I= = = 1400 A 4B 4 ( 5.0  10−5 T ) 2

This answer does not seem feasible. The current is very large, and the resistive heating in the thin copper wire would probably melt it. 10. Use Eq. 27–2. The length of wire in the magnetic field is the same as the diameter of the pole faces. 1.64 N F Fmax = I l B → B = max = = 0.458T Il ( 6.45 A )( 0.555 m ) 11. We find the force using Eq. 27–3, where the vector length is broken down into two parts; the portion along the z-axis, and the portion along the line y = 2x.  ˆi + 2ˆj  L1 = −0.250 m kˆ L2 = 0.250 m    5   

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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ˆi + 2ˆj   F = I L B = I L1 + L2  B = ( 20.0 A )( 0.250 m )  −kˆ +   0.285ˆi T   5    ˆ ˆ 2 ˆ ˆ = (1.425 N )  −k  i + j  i  = − 1.425 ˆj +1.275 kˆ N 5  

)

(

)

(

F = F = 1.4252 + 1.2752 N = 1.91 N  −1.275 N   = 41.8 below the negative y -axis  −1.425 N 

 = tan −1 

12. We find the force per unit length from Eq. 27–3. Note that while the length is not known, the direction is given, and so l = l ˆi. F = I l  B = I l ˆi  B → B

FB l

ˆi

ˆj

kˆ

0.20 T

−0.42 T

0.25T

= I ˆi  B = ( 3.0 A )

(

 1m  = −0.75 ˆj − 1.26 kˆ N m    100 cm 

(

)

= − 7.5 ˆj + 13 kˆ  10−3 N cm 13. We apply Eq. 27–3 to each circumstance, and solve for the magnetic field. Let B = Bx î + By ˆj + Bz kˆ . For the first circumstance, l = l î. î ˆj

kˆ

(

)

FB = I l  B = ( 7.4 A ) 2.0 m

0 = ( −14.8A m ) Bz ˆj + (14.8A m ) B y kˆ = −2.5ˆj N →

B y = 0 ; ( −14.8A m ) Bz = −2.5 N → Bz =

2.5 N 14.8A m

For the second circumstance, l = l ˆj. ˆi ˆj

kˆ

FB = I l  B = ( 7.4 A ) 0

2.0 m

0.1689 T

( −14.8A m ) Bx = −5.0 N → Bx =

(

= 0.1689 T ; Bx unknown

(

)

= ( 2.5 N ) ˆi + ( −14.8A m ) Bx kˆ = 2.5ˆi − 5.0kˆ N →

5.0 N 14.8A m

= 0.3378T

)

Thus B = 0.34 ˆi + 0.17 kˆ T . 14. We find the net force on the loop by integrating the infinitesimal force on each infinitesimal portion of the loop within the magnetic field. The infinitesimal force is found using Eq. 27–4 with the current in an infinitesimal portion of the loop given by Id L = I − cos ˆi + sin  ˆj rd .

F =  Id L B = I 

2 −0

0

(− cos ˆi + sin ˆj) rd  B kˆ = IB r  0

(

2 −0

0

)

(cos ˆj + sin ˆi ) d

940

Chapter 27

Magnetism

(

= IB0 r sin  ˆj − cos ˆi

)

2 −0

= IB0 r sin ( 2 − 0 ) ˆj − sin 0 ˆj − cos ( 2 − 0 ) ˆi + cos0 ˆi  

0



= −2 IB0 r sin 0ˆj The trigonometric identities sin ( 2 −  ) = − sin  and cos ( 2 −  ) = cos are used to simplify the solution. 15. We find the force along the wire by integrating the infinitesimal force from each path element (given by Eq. 27–4) along an arbitrary path between the points a and b. b

)

(

)

(

F =  I d L B = I  î dx + ˆj dy  B0kˆ = IB0  −ˆj dx + î dy = IB0 −x ˆj + y î a

)

The resultant magnetic force on the wire depends on the displacement between the points a and b, and not on the path taken by the wire. Therefore, the resultant force must be the same for the curved path, as for the straight line path between the points. 16. The net force on the current loop is the sum of the infinitesimal forces obtained from each current element. From the figure, we see that at each current segment, the magnetic field is perpendicular to the current. This results in a force with only outward radial and downward vertical components. By symmetry, we find that the radial force components from segments on opposite sides of the loop cancel. The net force then is purely vertical. Symmetry also shows us that each current element contributes the same magnitude of force. F =  dF =  Id L B =  dFz kˆ = −  ( sin  ) dF kˆ = − ( sin  ) IB kˆ  d l = − I ( B sin  ) kˆ ( 2 r ) = −2 IB

r2 kˆ r2 + d 2

17. The maximum magnetic force as given in Eq. 27–5b can be used since the velocity is perpendicular to the magnetic field.

(

)(

)

Fmax = qvB = 1.60  10−19 C 7.75  105 m s ( 0.65T ) = 8.1  10−14 N By the right-hand rule, the force must be directed to the North . 18. The magnetic force will cause centripetal motion, and the electron will move in a clockwise circular path if viewed in the direction of the magnetic field. The radius of the motion can be determined. v2 Fmax = qvB = m → r

( 9.1110 kg )(1.70 10 m s ) = 1.76 10 m r= = qB (1.60 10 C ) ( 0.550 T ) mv

−31

−5

−19

Assuming the magnetic field existed in a sharply-defined region, the electron would only make onehalf of a rotation, and then exit the field going in the opposite direction from which it came, a distance of 2r = 3.52  10−5 m from its entry point.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

19. The magnetic force is to be 1/10 the bird’s weight. Fmag = qvB sin 90 = 0.10 mg → v=

0.10 mg qB

(

)(

0.10 20.0  10−3 kg 9.80 m s 2

(15.0  10 C )( 0.5 10 T ) −6

−4

) = 2.6110 m s  3 10 m s 7

This is not realistic. It’s about 10% the speed of light. 20. In this scenario, the magnetic force is causing centripetal motion, and so must have the form of a centripetal force. The magnetic force is perpendicular to the velocity at all times for circular motion. 6.6  10−27 kg 2.2 106 m s v2 mv → B= = = 0.32 T Fmax = qvB = m r qr 2 1.60  10−19 C ( 0.14 m )

(

)(

(

)

21. Since the charge is negative, the answer is the OPPOSITE of the result given from the right-hand rule applied to the velocity and magnetic field. (a) no force (b) downward (c) upward (d) inward, into the paper (e) to the left (f) to the left 22. The right-hand rule applied to the velocity and magnetic field would give the direction of the force. Use this to determine the direction of the magnetic field given the velocity and the force. (a) to the right (b) downward (c) into the page 23. The kinetic energy of the proton can be used to find its velocity. The magnetic force produces centripetal acceleration, and from this the radius can be determined. mv 2 mv 2K 2 1 → r= K = 2 mv → v = qv B = m r qB

mv qB

m =

2K m = qB

2K m qB

(

)(

2 7.2  105 eV 1.60  10−19 J eV 1.67  10 −27 kg

(1.60 10 C ) ( 0.20 T ) −19

)

= 0.61m

24. The magnetic field can be found from Eq. 27–5b, and the direction is found from the right-hand rule. Remember that the charge is negative. F 7.4  10−13 N Fmax = qvB → B = max = = 1.652 T  1.7 T qv (1.60  10−19 C )( 2.8  106 m s ) The direction would have to be East for the right-hand rule, applied to the velocity and the magnetic field, to give the proper direction of force.

942

Chapter 27

Magnetism

25. The total force on the proton is given by the Lorentz equation, Eq. 27–7. ˆi ˆj   F = q ( E + v  B ) = e 3.0ˆi − 4.2ˆj  103 V m + 6.0  103 m s 3.0  103 m s

(  

)

0.56 T

0.42 T

) ( ) ( )( = (1.60  10 C ) ( 5.1î − 7.0ˆj + 0.84kˆ )   10 N C = ( 8.16  10 î − 1.12  10 ˆj + 1.344  10 kˆ ) N = ( 0.82î − 1.1ˆj + 0.1 kˆ )   10 N

kˆ

  −5.0  10 m s   0  3

= 1.60  10 −19 C  3.0ˆi − 4.2ˆj + 2.1ˆi − 2.8ˆj + 0.84kˆ   103 N C −19

−16

−15

−16

−15

The k̂ term has only 1 significant figure due to the subtraction rule, used in the cross-product. 26. The force on the electron is given by Eq. 27–5a. Set the force expression components equal and solve for the velocity components. î ˆj kˆ F = q v  B → F î + F ˆj = −e v v v = −ev B î − e ( −v B ) ˆj → x

Fx = −ev y Bz → v y = − Fy = ev x Bz → v x =

(

eBz

=−

eBz

3.8  10 −13 N

(1.60 10 C ) ( 0.74T ) −19

−2.7  10 −13 N

(1.60 10 C ) ( 0.74T ) −19

= −3.209  106 m s

= −2.280  106 m s

)

v = − 2.3 ˆi + 3.2 ˆj  106 m s

Notice that we have not been able to determine the z component of the electron’s velocity, since it would be parallel to the magnetic field. 27. The kinetic energy of the particle can be used to find its velocity. The magnetic force produces centripetal acceleration, and from this the radius can be determined. Inserting the radius and velocity into the equation for angular momentum gives the angular momentum in terms of the kinetic energy and magnetic field. 2K mv 2 mv K = 12 mv 2 → v = qv B = → r= m r qB



L = mvr = m

2K 

 m   

2K 



m  2mK = qB  qB

 

From the equation for the angular momentum, we see that doubling the magnetic field while keeping the kinetic energy constant will cut the angular momentum in half. Lfinal = 12 Linitial

943

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

28. The force on the electron is given by Eq. 27–5a. ˆi ˆj

kˆ

FB = q v  B = − e 6.0  104 m s −7.0  104 m s −0.80 T

(

0.60 T

)(

Instructor Solutions Manual

0 = − e ( 3.6 − 5.6 )  104 T m s kˆ 0

)

= − 1.60  10−19 C −2.0  104 T m s kˆ = 3.2  10−15 N kˆ 29. (a) The velocity of the ions is found using energy conservation. The electrical potential energy of the ions becomes kinetic energy as they are accelerated. Einitial = Efinal → qV = 12 mv 2 → v=

2qV m

(

)

2  2 1.60  10−19 C  ( 5200 V )

( 6.6 10 kg ) −27

= 7.101 105 m s  7.1  105 m s

(b) Since the ion is moving perpendicular to the magnetic field, the magnetic force will be a maximum. That force will cause the ion to move in a circular path. v2 Fmax = qvB = m → r

( 6.6 10 kg )( 7.10110 m s ) = 4.308 10 m  4.3 10 m r= = qB 2 (1.60  10 C ) ( 0.340 T ) −27

−2

−19

(c) The period can be found from the speed and the radius. −2 2 r 2 4.308  10 m T= = = 3.8  10 −7 s 5 v 7.101  10 m s

(

)

30. (a) From Example 27–7, we have that r = 2

mv qB

, and so v =

rqB m

. The kinetic energy is given by

 rqB  r q B and so we see that K  r 2 .  = 2m  m 

K = 12 mv 2 = 12 m 

(b) The angular momentum of a particle moving in a circular path is given by L = mv r . From rqB mv Example 27–7, we have that r = , and so v = . Combining these relationships gives qB m rqB L = mvr = m r = qBr 2 . m 31. The velocity of each charged particle can be found using energy conservation. The electrical potential energy of the particle becomes kinetic energy as it is accelerated. Then, since the particle is moving perpendicularly to the magnetic field, the magnetic force will be a maximum. That force will cause the ion to move in a circular path, and the radius can be determined in terms of the mass and charge of the particle. Einitial = Efinal → qV = 12 mv 2 → v =

2qV m

944

Chapter 27

Magnetism

Fmax = qvB = m

rd rp

r rp

→ r=

2qV 1 m = qB B

2mV q

1 B

2mdV qd

1 B

2mpV qp

qd qp

1 B

2mV q

m mp

2mpV

q qp

= 2 → rd = 2 rp

4 2

= 2 → r = 2 rp

32. The centripetal force is caused by the magnetic field, and is given by Eq. 27–5b. v2 F = qvB sin  = qv⊥ B = m ⊥ → r

( 9.1110 kg )( 2.2 10 m s ) sin 45 = 3.163  10 m  3.2  10 m r= = qB (1.60  10 C ) ( 0.28 T ) −31

mv⊥

−5

−19

The component of the velocity that is parallel to the magnetic field is unchanged, and so the pitch is that velocity component times the period of the circular motion. mv ⊥ 2 2 r 2 m qB = = T= v⊥ v⊥ qB 2 ( 9.11  10 −31 kg )  2 m  −4 6 o p = v T = v cos 45   = ( 2.2  10 m s ) cos 45 1.60  10−19 C 0.28 T = 2.0  10 m qB ) ( )(   o

33. (a) For the particle to move upward the magnetic force must point upward, and by the right-hand rule we see that the force on a positively charged particle would be downward. Therefore, the charge on the particle must be negative . (b) In the figure we have created a right triangle to relate the horizontal distance l, the displacement d, and the radius of curvature r. Using the Pythagorean theorem we can write an expression for the radius in terms of the other two distances. r2 = ( r − d ) + l 2 → r = 2

d2 + l 2 2d

Since the momentum is perpendicular to the magnetic field, we can solve for the momentum by relating the maximum force (Eq. 27–5b) to the centripetal force on the particle. Fmax = qvB0 =

mv 2 r

→ p = mv = qB0 r =

(

qB0 d 2 + l 2

)

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

34. In order for the path to be bent by 90 within a distance d, the radius of curvature must be less than or equal to d. The kinetic energy of the protons can be used to find their velocity. The magnetic force produces centripetal acceleration, and from this, the magnetic field can be determined. 2K

K = mv → v = 1 2

B

mv ed

m =

qv B =

mv 2

→ B=

mv qr

2K 1/ 2

 2 Km  m =  2 2 ed e d 

35. The magnetic force produces an acceleration that is perpendicular to the original motion. If that perpendicular acceleration is small, it will produce a small deflection, and the original velocity can be assumed to always be perpendicular to the magnetic field. This leads to a constant perpendicular acceleration. The time that this (approximately) constant acceleration acts can be found from the original velocity v and the distance traveled l. The starting speed in the perpendicular direction will be zero. qvB F⊥ = ma⊥ = qvB → a⊥ = m

(18.5 10 C )( 5.00 10 T )(1.25 10 m ) = =   m v 2mv 2 ( 3.40  10 kg ) (155 m s )

qv B  l  1

d ⊥ = v0 ⊥ t + 12 a⊥ t = 2 2

−9

qB l 2

−5

−3

= 1.37  10−6 m This small distance justifies the assumption of constant acceleration. 36. (a) In the magnetic field, the proton will move along an arc of a circle. The distance x in the textbook diagram is a chord of that circle, and so the center of the circular path lies on the perpendicular bisector of the chord. That perpendicular bisector bisects the central angle of the circle which subtends the chord. Also recall that a radius is perpendicular to a tangent. In the diagram, 1 =  2 because they are vertical

8

5 4

6 7

r   3 2

angles. Then  2 =  4 , because they are both complements of

 3 , so 1 =  4 . We have  4 =  5 since the central angle is

1

bisected by the perpendicular bisector of the chord.  5 =  7

because they are both complements of  6 , and  7 = 8 because they are vertical angles. Thus

1 =  2 =  4 =  5 =  7 = 8 , and so in the textbook diagram, the angle at which the proton leaves is  = 45o . (b) The radius of curvature is given by r =

x = 2r cos  = 2

mv qB

, and the distance x is twice the value of r cos  .

(1.67 10 kg )(1.8 10 m s ) cos 45 = 3.110 m (1.60 10 C ) ( 0.850 T ) −27

cos  = 2

−3

−19

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Chapter 27

Magnetism

37. (a) Since the velocity is perpendicular to the magnetic field, the particle will follow a circular trajectory in the x–y plane of radius r. The radius is found using the centripetal acceleration. mv 2 mv qvB = → r= r qB From the figure we see that the distance  is the chord distance, which is twice the distance r cos  . Since the velocity is perpendicular to the radial vector, the initial direction and the angle  are complementary angles. The angles  and  are also complementary angles, so  = 30. 2mv0 mv cos30 = 3 0  = 2r cos  = qB0 qB0 (b) From the diagram, we see that the particle travels a circular path, that is 2 short of a complete circle. Since the angles  and  are complementary angles, so  = 60. The trajectory distance is equal to the circumference of the circular path times the fraction of the complete circle. Dividing the distance by the particle speed gives t. l 2 r  360 − 2 ( 60 )  2 mv0  2  4 m t = = =  =  v0 qB0  3  3qB0 v0 v0  360  38. With the plane of the loop parallel to the magnetic field, the torque will be a maximum. We use Eq. 27–9. 0.185m N  = = 3.67 T  = NIAB sin  → B = 2 NIAB sin  (1)( 3.80 A )  ( 0.0650 m ) sin 90 39. If the face of the loop of wire is parallel to the magnetic field, the angle between the perpendicular to the loop and the magnetic field is 90o. Use Eq. 27–9 to calculate the torque.

 = NIAB sin  = (1)( 5.70 A )( 0.220 m ) (1.30T ) sin 90 = 0.3586 m N  0.359 m N 2

40. The work required by an external agent is equal to the change in potential energy. The potential energy is given by Eq. 27–12, U = −μ B. (a) W = U = ( −μ B )final − ( −μ B )initial = ( μ B )initial − ( μ B )final = NIAB ( cos initial − cos  final )

= NIAB ( cos 0 − cos180 ) = 2 NIAB (b) W = NIAB ( cos initial − cos final ) = NIAB ( cos 90 − cos ( −90 ) ) = 0 41. (a) The torque is given by Eq. 27–9. The angle is the angle between the magnetic field and the perpendicular to the coil face.

  0.180 m 2  −5 −5  = NIAB sin  = 12 ( 7.10 A )     ( 5.50  10 T ) sin 36 = 7.01  10 m N 2     (b) In Example 27–11 it is stated that if the coil is free to turn, it will rotate toward the orientation so that the angle is 0. In this case, that means the north edge of the coil will rise, so that a perpendicular to its face will be parallel with the Earth’s magnetic field.

947

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

42. The magnetic dipole moment is defined in Eq. 27–10 as  = NIA. The number of turns, N, is 1. The current is the charge per unit time passing a given point, which on the average is the charge on the electron divided by the period of the circular motion, I = e T . If we assume the electron is moving in a circular orbit of radius r, then the area is  r 2. The period of the motion is the circumference of the orbit divided by the speed, T = 2 r v . Finally, the angular momentum of an object moving in a circle is given by L = mr v. Combine these relationships to find the magnetic moment. e 2 e e r 2 v er v emrv e e 2 = = = mrv = L  = NIA =  r = r = 2 r v 2 r 2 2m 2m 2m T 43. (a) The magnetic moment of the coil is given by Eq. 27–10. Since the current flows in the clockwise direction, the right-hand rule shows that the magnetic moment is down, or in the negative z-direction. 2  0.22 m  2 2 ˆ ˆ ˆ μ = NIA = 24 ( 7.6 A )    −k = −6.934 k A m  −6.9 k A m  2  (b) We use Eq. 27–11 to find the torque on the coil.  = μ  B = −6.934 kˆ A m2  0.55ˆi + 0.60ˆj − 0.65kˆ T = 4.2 ˆi − 3.8ˆj m N

( )

)

) (

(

)

(c) We use Eq. 27–12 to find the potential energy of the coil. U = −μ B = − −6.934 kˆ A m2 0.55ˆi + 0.60ˆj − 0.65kˆ T = − ( 6.934A m 2 ) ( 0.65 T )

(

)

)(

= −4.5 J x 44. To find the total magnetic moment, we divide the rod into infinitesimal pieces of thickness dx, along the length of the rod. Consider a piece a distance x from the axis. The charge on that small piece is ( Q d ) dx. As the rod rotates on its axis, the charge on that small piece creates a current loop around the axis of rotation. The area of that loop is  x2. The magnitude of the current is the charge divided by the period of rotation, or the charge times the frequency of rotation,  2 . By integrating the infinitesimal magnetic moments from each piece, we find the total magnetic moment. d Q d 2   Q  Q d 2 dx  = x dx μ =  dμ =  AdI =  ( x 2 )  = .  0 6  2 d  2d 0

45. From Section 27–5, we see that the torque is proportional to the current, so if the current drops by 18%, the output torque will also drop by 18%. Thus, the final torque is 0.82 times the initial torque. 46. In Section 27–6, it is shown that the deflection of the galvanometer needle is proportional to the product of the current and the magnetic field. Thus, if the magnetic field is decreased to 0.700 times its original value, the current must be increased by dividing the original value by 0.700 to obtain the same deflection. I B ( 63.0  A ) Binitial = 90.0  A ( IB )initial = ( IB )final → I final = initial initial = Bfinal 0.700 Binitial

948

Chapter 27

Magnetism

47. From the galvanometer discussion in Section 27–6, the amount of deflection is proportional to the I ratio of the current and the spring constant:   . Thus if the spring constant decreases by 15%, k the current can decrease by 15% to produce the same deflection. The new current will be 85% of the original current. I final = 0.85I initial = 0.85 ( 25 A ) = 21.25 A  21 A 48. Use Eq. 27–13. ( 320V m ) q E = 2 = = 1.9  105 C kg m B r ( 0.46T )2 ( 0.0080m ) 49. (a) The force on the electron due to the electric force must be the same magnitude as the force on the electron due to the magnetic force. E 1.52  10 4 V m = 5.846  106 m s  5.85  10 6 m s FE = FB → qE = qvB → v = = −3 B 2.60  10 T (b) If the electric field is turned off, the magnetic force will cause circular motion. 9.11 10−31 kg 5.846 106 m s v2 mv → r= = = 1.28  10−2 m FB = qvB = m −19 −3 r qB 1.60  10 C 2.60 10 T

(

)( )(

(

)

50. The force from the electric field must be equal to the weight. V  qE = ( ne )   = mg → d 

2 −15 mgd ( 3.8  10 kg )( 9.80 m s ) ( 0.010 m ) n= = = 6.84  7 electrons eV (1.60  10−19 ) ( 340 V )

51. (a) Eq. 27–14 shows that the Hall emf is proportional to the magnetic field perpendicular to the conductor’s surface. We can use this proportionality to determine the unknown resistance. Since the new magnetic field is oriented 90 to the surface, the full magnetic field will be used to create the Hall potential. eH  B⊥ e  63mV = → B⊥ = H B⊥ = ( 0.10T ) = 0.53T eH B⊥ eH 12 mV (b) When the field is oriented at 60 to the surface, the magnetic field, B sin60, is used to create the Hall potential. e  63 mV ( 0.10T ) B⊥ sin 60 = H B⊥ → B⊥ = = 0.61T eH 12 mV sin 60 52. (a) We use Eq. 27–14 for the Hall potential and Eq. 25–13 to write the current in terms of the drift velocity. vd Bd e 1 KH = H = = IB en ( td ) vd  B ent   (b) We set the magnetic sensitivities equal and solve for the metal thickness. n 1 1 3 1022 m−3 = → tm = s ts = (0.15 10−3 m ) = 5 10−11 m 11029 m−3 ens ts enmtm nm This is less than one-sixth the size of a typical metal atom. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

949

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(c) Use the magnetic sensitivity to calculate the Hall potential. (100 mA )( 0.1T ) IB eH = K H IB = = = 14 mV  10 mV ent (1.6 10−19 C )( 3 1022 m −3 )( 0.15 10−3 m ) 53. (a) We find the Hall field by dividing the Hall emf by the width of the metal. e 6.5  V EH = H = = 2.167 10−4 V/m  2.2 10−4 V/m d 0.030 m (b) Since the forces from the electric and magnetic fields are balanced, we can use Eq. 27–14 to calculate the drift speed. E 2.167 10−4 V/m vd = H = = 2.709 10−4 m s  2.7 10−4 m/s B 0.80T (c) We now find the density using Eq. 25–13. I 42 A n= = eAvd (1.6 10−19 C )(1.0 10−3 m ) ( 0.03m ) ( 2.709 10−4 m/s )

= 3.2 1028 electrons/m3 54.

We find the magnetic field using Eq. 27–14, with the drift velocity given by Eq. 25–13. To determine the electron density we divide the density of sodium by its atomic weight. This gives the number of moles of sodium per cubic meter. Multiplying the result by Avogadro’s number gives the number of sodium atoms per cubic meter. Since there is one free electron per atom, this is also the density of free electrons. The problem says e eH e net eH et   N A  = H = B= H =   vd d  I  I I  mA   d  ne(td ) 

(1.86 10 V )(1.60 10 C)(1.20 10 m ) ( 0.971) (1000 kg/m )( 6.022 10 e/mole ) −6

−19

−3

12.0 A

0.02299 kg/mole

= 0.757 T . 55. (a) The sign of the ions will not change the magnitude of the Hall emf, but will determine the polarity of the emf . The magnitude of the Hall emf depends on the drift velocity of the ions, the magnetic field, and the distance across which the emf is measured. (b) The flow speed corresponds to the drift speed in Eq. 27–14. 0.13  10−3 V ) ( eH eH = vBd → v = = = 0.56 m s Bd ( 0.070T )( 0.0033m ) 56. The magnetic force on the ions causes them to move in a circular path, so the magnetic force is a centripetal force. This results in the ion mass being proportional to the path’s radius of curvature. qv B = m v 2 r → m = qBr v → m r = qB v = constant = 76 u 22.8 cm

m21.0 21.0 cm m21.9

= =

76 u 22.8 cm 76 u

→ m21.0 = 70 u

m21.6 21.6 cm m22.2

→ m21.9 = 73 u

21.9 cm 22.8 cm 22.2 cm The other masses are 70 u, 72 u, 73 u, and 74 u .

= =

76 u 22.8 cm 76 u 22.8 cm

→ m21.6 = 72 u → m22.2 = 74 u

950

Chapter 27

Magnetism

57. The velocity of the ions is found using energy conservation. The electrical potential energy of the ions becomes kinetic energy as they are accelerated. Then, since the ions move perpendicularly to the magnetic field, the magnetic force will be a maximum. That force will cause the ions to move in a circular path. qv B =

mv 2 R

→ v=

 qBR  q B R qV = mv = m   = 2m  m 

qBR

1 2

→ m=

qB 2 R 2 2V

58. (a) The location of each line on the film is twice the radius of curvature of the ion. The radius of curvature can be found from Eq. 27–15. qBBr mE 2mE m= → r= → 2r = E qBB qBB

2 r12 =

(

)( (1.60  10 C ) ( 0.58T )

2 (12 ) 1.67  10−27 kg 2.84  104 V m −19

) = 2.1148  10 m −2

2 r13 = 2.2910  10−2 m 2r14 = 2.4673  10−2 m The distances between the lines are as follows. 2 r13 − 2 r12 = 2.2910  10−2 m − 2.1148  10 −2 m = 1.762  10 −3 m  2  10 −3 m 2 r14 − 2 r13 = 2.4673  10−2 m − 2.2910  10 −2 m = 1.763  10 −3 m  2  10 −3 m (b) If the ions are doubly charged, the value of q in the denominator of the expression would double, and so the actual distances on the film would be halved. Thus, the distances between the lines would also be halved.

2 r13 − 2 r12 = 1.1455  10−2 m − 1.0574  10 −2 m = 8.81  10 −4 m  9  10 −4 m 2 r14 − 2 r13 = 1.2337  10−2 m − 1.1455  10 −2 m = 8.82  10 −4 m  9  10 −4 m (c) The distances between the marks will have the same value regardless of the signs of the ions, but the curvature of the ions will be one direction (perhaps clockwise as in Fig. 27–34) for positive ions, and the opposite direction for negative ions. To keep the curvature direction the same, the magnetic field B could be reversed in direction depending on which sign of ion was being measured.

59. Since the particle is undeflected in the crossed fields, its speed is given by Eq. 27–8. Without the electric field, the particle will travel in a circle due to the magnetic force. Using the centripetal acceleration, we can calculate the mass of the particle. Also, the charge must be an integer multiple of the fundamental charge. mv 2 qv B = → r 2 −19 qBr qBr neB 2 r n (1.60 10 C ) ( 0.036T ) ( 0.024 m ) m= = = = = n ( 3.32 10−27 kg ) E v 1.5 103 V/m ( E / B)  n ( 2.0 u ) The particle has an atomic mass of a multiple of 2.0 u. The simplest two cases are that it could be a hydrogen-2 nucleus (mass of 2 u, called a deuteron), or a helium-4 nucleus (mass of 4 u, called an alpha particle): 21 H, 24 He .

951

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

60. The radius and magnetic field values can be used to find the speed of the protons. The electric field is then found from the fact that the magnetic force must be the same magnitude as the electric force for the protons to have straight paths. qvB = m v 2 r → v = qBr m FE = FB → qE = qvB →

(1.60 10 C ) ( 0.585 T ) ( 5.10 10 m ) = 1.67 10 V m E = vB = qB r m = 2

−19

−2

−27

1.67  10 kg The direction of the electric field must be perpendicular to both the velocity and the magnetic field, and must be in the opposite direction to the magnetic force on the protons.

61. The particles in the mass spectrometer follow a semicircular path as shown in Fig. 27–34. A particle has a displacement of 2r from the point of entering the semicircular region to where it strikes the film. So if the separation of the two molecules on the film is 0.50 mm, the difference in radii of the two molecules is 0.25 mm. The mass to radius ratio is the same for the two molecules. qvB = m v 2 r → m = qBr v → m r = constant

m m   =   r CO  r  N

28.0106 u

→

28.0134 u r + 2.5  10−4 m

→ r = 2.5 m

62. The kinetic energy is used to determine the speed of the particles, and then the speed can be used to determine the radius of the circular path, since the magnetic force is causing centripetal acceleration. 2K m 2 2m K 2K mv mv m K = 12 mv 2 → v = qv B = → r= = = m r qB qB qB 2m p K rp re

qB 2me K

mp me

1.67  10 −27 kg

9.11  10 −31kg

= 42.8

63. The maximum torque is found using Eq. 27–9 with sin  = 1 . Set the current equal to the voltage divided by resistance and the area as the square of the side length. 2 V   9.0V   = NIAB = N   l 2 B = 20  0.050m ) ( 0.020T ) = 3.8 10−4 m N (  R  24   64. (a) There will be one force on the rod, due to the magnetic force on the charge carriers in the rod. That force is of magnitude FB = IdB , and by Newton’s second law is equal to the mass of the rod times its acceleration. That force is constant, so the acceleration will be constant, and constant acceleration kinematics can be used.

Fnet = FB = IdB = ma → a =

IdB

v − v0

→ v=

IdB

t m t t m (b) Now the net force is the vector sum of the magnetic force and the force of kinetic friction. Fnet = FB − Ffr = IdB − k FN = IdB − k mg = ma → a=

IdB m

− k g =

v − v0 t

v t

 IdB  − k g  t  m 

→ v= 

(c) Using the right-hand rule, we find that the force on the rod is to the east, and the rod moves east. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

952

Chapter 27

Magnetism

65. Assume that the magnetic field makes an angle  with respect to the vertical. The rod will begin to slide when the horizontal magnetic force  B ( IBl cos ) is equal to the maximum static friction ( s FN ) . Find the FB normal force by setting the sum of the vertical forces equal to zero. See  the free-body diagram. FN Ffr FB sin  + FN − mg = 0 → FN = mg − FB sin  = mg − I l B sin  mg s mg IBl cos = s FN =s ( mg − I l B sin  ) → B = I l ( s sin  + cos ) We find the angle for the minimum magnetic field by setting the derivative of the magnetic field with respect to the angle equal to zero and solving for the angle. −  mg ( s cos − sin  ) dB =0= s →  = tan −1s = tan −1 0.5 = 26.6 2 d I l ( s sin  + cos )

0.5 ( 0.40 kg ) ( 9.80 m/s ) s mg = = 0.17T I l ( s sin  + cos ) ( 36 A )( 0.28m ) ( 0.5sin 26.6 + cos26.6) 2

The minimum magnetic field that will cause the rod to move is 0.17 T at 27, westward from the vertical. 66. The magnetic force must be equal in magnitude to the weight of the electron. 9.11 10−31kg 9.80 m s 2 mg = = 1.1 10−6 m s mg = qvB → v = qB 1.60  10−19 C 0.50  10−4 T

( (

)( )(

) )

The magnetic force must point upwards, and so by the right-hand rule and the negative charge of the electron, the electron must be moving west . 67. Since the magnetic and gravitational force along the entire rod is uniform, we consider the two forces acting at the center of mass of the rod. To be balanced, the net torque about the fulcrum must be zero. Using the usual sign convention for torques and Eq. 10–10a, we solve for the magnetic force on the rod.  = 0 = Mg ( 14 l ) − mg ( 14 l ) − FB ( 14 l ) → FB = ( M − m) g

l 4

Mg FB We solve for the current using Eq. 27–2. ( M − m ) g = ( 7.0m − m ) g = 6.0mg F I= B = lB lB lB lB The right-hand rule indicates that the current must flow toward the left since the magnetic field is into the page and the magnetic force is downward.

68. The magnetic force produces centripetal acceleration. 4.8  10−21 kg m s p qvB = mv 2 r → mv = p = qBr → B = = = 2.2  10−2 T −19 qr (1.60  10 C ) (1.35 m ) If the protons are rotating in the plane of a piece of paper on the desk, then the magnetic field must point upward or out of the paper to cause an inwardpointing (centripetal) force that steers the protons clockwise. It is illustrated in the adjacent figure. It is also the opposite of the case illustrated in Fig. 27–18.

953

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

69. (a) The frequency of the voltage must match the frequency of circular motion of the particles, so that the electric field is synchronized with the circular motion. The radius of each circular orbit mv is given in Example 27–7 as r = . For an object moving in circular motion, the period is qB 2 r given by T = , and the frequency is the reciprocal of the period. v T=

2 r v

→ f =

v 2 r

= 2

v Bq = mv 2 m

qB In particular we note that this frequency is independent of the radius, and so the same frequency can be used throughout the acceleration. (b) For a small gap, the electric field across the gap will be approximately constant and uniform as the particles cross the gap. If the motion and the voltage are synchronized so that the maximum voltage occurs when the particles are at the gap, the particles receive an energy increase of K = qV0 as they pass each gap. The energy gain from one revolution will include the passing of 2 gaps, so the total kinetic energy increase is 2qV0 . (c) The maximum kinetic energy will occur at the outside of the cyclotron.

( 2.0 m ) (1.60  10 C ) ( 0.50 T ) r q B  r qB  2 K max = 12 mvmax = 12 m  max  = 12 max = 12 1.67  10−27 kg m  m  2

−19

 1eV  1MeV   = 47.91MeV  48 MeV 6 −19   1.60  10 J  10 eV 

= 7.665  10−12 J 

70. The protons will follow a circular path as they move through the region of magnetic field, with a radius of curvature given in Example mv 27–7 as r = . Fast-moving protons will have a radius of curvature qB that is too large and so they will exit above the second tube. Likewise, slow-moving protons will have a radius of curvature that is too small and so they will exit below the second tube. Since the exit velocity is perpendicular to the radius line from the center of curvature, the bending angle can be calculated. l sin  = → r

r 

( 3.9 10 m )(1.60 10 C ) ( 0.32 T ) = sin 0.6643 = 41.6  42 = sin = sin  = sin r mv (1.67 10 kg )(1.8 10 m s ) −1

−1

l qB

−2

−19

−1

−27

71. With one turn of wire, we have B =

0 I

. Use the radius of the Earth for r. 2r 6 −4 I 2 rB 2 ( 6.38  10 m )(1  10 T ) = = 1.015  109 A  1  109 A B= 0 → I = −7 0 2r 4  10 T m A

954

Chapter 27

Magnetism

72. (a) The force on each of the vertical wires in the loop is perpendicular to the magnetic field and is given by Eq. 27–1, with  = 90. When the face of the loop is parallel to the magnetic field, the forces point radially away from the axis. This provides a tension in the two horizontal sides. When the face of the loop is perpendicular to the magnetic field, the force on opposite vertical wires creates a shear force in the horizontal wires. From Table 12–2, we see that the tensile and shear strengths of aluminum are the same, so either can be used to determine the minimum strength. We set tensile strength multiplied by the cross-sectional area of the wire equal to the magnetic force multiplied by the safety factor and solve for the wire diameter. 2 8( I l B) 8 (15.0 A )( 0.200 m )(1.35 T ) F d  =2    = 8( I l B) → d = 2 A 2  ( F A)  200 106 N/m2

(

)

= 4.5416 10−4 m  0.454 mm (b) The resistance is found from the resistivity using Eq. 25–3. 4 ( 0.200 m ) l R =  = 2.65 10−8  m = 0.131 2 A  4.5416 10−4  m    2  

(

)

73. The accelerating force on the bar is due to the magnetic force on the current. If the current is constant, the magnetic force will be constant, and so constant acceleration kinematics (Eq. 2–11c) can be used. v 2 − v02 v2 v 2 = v02 + 2ax → a = = 2x 2x  v2  m 2  45  10−3 kg ( 28 m s ) 2x  ma mv 2  Fnet = ma = I l B → I = = = = = 32.67A  33 A 2 l Bx 2 ( 0.36 m )(1.5 T )(1.0 m ) lB lB Using the right-hand rule, for the force on the bar to be in the direction of the acceleration shown in Fig. 27–58, the magnetic field must be down .

(

)

74. (a) For the beam of electrons to be undeflected, the magnitude of the magnetic force must equal the magnitude of the electric force. We assume that the magnetic field will be perpendicular to the velocity of the electrons so that the maximum magnetic force is obtained. E 12, 000 V m FB = FE → qv B = qE → B = = = 2.857  10 −3 T  2.9  10 −3 T 6 v 4.2  10 m s (b) Since the electric field is pointing south, the electric force is to the north. Thus the magnetic force must be to the south. Using the right-hand rule with the negative electrons, the magnetic field must be vertically upward . (c) If the electric field is turned off, then the magnetic field will cause a centripetal force, moving the electrons in a circular path. The frequency is the cyclotron frequency, Eq. 27–6. 1.60  10−19 C (12, 000 V m ) qB qE f = = = = 8.0  107 Hz −31 6 2 m 2 mv 2 9.11 10 kg 4.2  10 m s

(

)

)(

)

955

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

75. The speed of the proton can be calculated based on the radius of curvature of the (almost) circular motion. From that the kinetic energy can be calculated. qv B =

mv 2 r

→ v=

2 2

 qBr  q B r  = 2m  m 

qBr

K = 12 mv 2 = 12 m 

(1.60 10 C ) ( 0.010 T )  8.5  10 m − 10.0  10 m  K = r −r )= ( ) ( ) ( 2m 2 (1.67  10 kg ) −19

q2 B2

2 2

−3

2 1

−3

−27

 1eV  = −0.1329 eV  −0.13eV −19   1.60  10 J 

= −2.127  10 −20 J 

76. The magnetic force will produce centripetal acceleration. Use that relationship to calculate the speed. The radius of the Earth is 6.38  106 km , and the altitude is added to that.

FB = qvB = m

v2 r

(1.60 10 C )( 6.385 10 m )( 0.50 10 T ) = 1.3 10 m s → v= = m 238 (1.66 10 kg ) −19

qrB

−4

−27

Compare the size of the magnetic force to the force of gravity on the ion. 1.60  10−19 C 1.3  108 m s 0.50  10−4 T FB qvB = = = 2.7  108 Fg mg 238 1.66  10−27 kg 9.80 m s2

(

)(

)

Yes, we may ignore gravity. The magnetic force is almost 300 million times larger than gravity. 77. We assume that the horizontal component of the Earth’s magnetic field is pointing due North. The Earth’s magnetic field also has a dip angle of 22o. The angle between the magnetic field and the eastward current is 90o. Use Eq. 27–1 to calculate the magnitude of the force. F = I l B sin  = ( 330 A )(15 m ) 5.0  10 −5 T sin 90

(

)

= 0.2475 N  0.25 N

68o 22o

B Earth

Using the right-hand rule with the eastward current and the Earth’s magnetic field, the force on the wire is to the North and 68o above the horizontal . 78. (a) As the electron orbits the nucleus in the absence of the magnetic field, its centripetal acceleration is caused solely by the electrical attraction between the electron and the nucleus. Writing the velocity of the electron as the circumference of its orbit times its frequency enables us to obtain an equation for the frequency of the electron’s orbit. 2 2 r f 0 ) ( ke2 ke2 v2 2 m m f = = → = 0 r r r2 4 2 mr 3 When the magnetic field is added, the magnetic force adds or subtracts from the centripetal acceleration (depending on the direction of the field) resulting in the change in frequency. 2 ( 2 r f ) → f 2 qB f − f 2 = 0 ke2   = q 2 r f B m ( ) 0 r 2 m r2 We can solve for the frequency shift by setting f = f 0 + f , and only keeping the lowest order terms in f , since f f0.

956

Chapter 27

Magnetism

qB ( f + f ) − f 02 = 0 2 m 0 qB qB qB f 02 + 2 f 0 f + f 2 f0 f − f 02 = 0 → f =  2 m 2 m 4 m The “” indicates whether the magnetic force adds to or subtracts from the centripetal acceleration. If the magnetic force adds to the centripetal acceleration, the frequency increases. If the magnetic force is opposite in direction to the acceleration, the frequency decreases.

( f 0 + f )

(b)

79. The magnetic field points in the positive z-direction. The right-hand rule says that the forces on each piece of the loop will point outwards, away from the loop. The net force on each infinitesimal segment of the two sides of the loop parallel to the x-axis will have the same magnitude, but opposite directions (due to the opposite direction of the current in these two sides, but equal magnitude of magnetic field on each segment), so these forces cancel each other. They are represented by the two opposing forces in the diagram, one in the positive y-direction, and one in the negative y-direction. b b   Left side: x = b → Bleft = B0 1 +  = 2 B0 → Fleft = I l Bleft −î = IaB0 1 +  −î = −2aIB0î b b   

( )

Right side:



x = b + a → Bright = B0 1 + 

()

( )

b+a   a =  2 +  B0 →  b   b

()

a a   Fright = I l Bright î = IaB0  2 +  î =  2 +  aIB0î b b   Net force:

a a2 I ˆ  Fright + Fleft =  2 +  aIB0ˆi − 2aIB0ˆi = B0 i , which is to the right. b b 

80. (a) The electron, a negative particle, accelerates opposite the direction of the electric field. Therefore, for the electron to accelerate east, the electric field should point to the west. (b) By the right-hand rule, the magnetic field must point upward for the magnetic force on a northmoving electron to point westward. (c) For the electron to not accelerate, the magnetic force must be equal in magnitude, but opposite in direction to the electric force. Set the magnitudes of the forces from Eq. 21–5 and Eq. 27–5b equal and solve for the magnetic field. The direction of the field is upward. 380 V m E qE = qv B → B = = = 0.01267 T  0.013T v 3.0  10 4 m s (d) If the electron is moving faster than 3.0  10 4 m s , the magnetic force will be greater than the electric force and the electron will accelerate westward. If it is moving slower, the magnetic force will be smaller than the electric force and the electron will accelerate eastward. (e) The two forces must be equal and opposite for the particle to travel undeflected. The ratio of the electric to magnetic field can be calculated by setting the forces equal. E qE = qvB → = v = 5.5  104 m s B The value of B cannot be determined unless the value of E is known.

957

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

81. From Fig. 27–23 we see that when the angle  is positive (counterclockwise as measured from the horizontal axis), the torque is negative (clockwise). The magnitude of the torque is given by Eq. 27–9. For small angles we use the approximation sin    . Using Eq. 10–14, we can write the torque in terms of the angular acceleration, showing that it is a harmonic oscillator.  IabB  2  = − NIAB sin   − IabB = I M  →  = −    = −  I  M 

We obtain the period of motion from the angular frequency, using T = 2  . First we determine the moment of inertia of the loop, as two wires rotating about their centers of mass and two wires rotating about an axis parallel to their lengths. The mass of each wire is proportional to it’s length, as compared to the perimeter of the loop. 2 1  ( 3a + b ) mb2 b a     b  IM = 2   m  b2  + 2  m  = 12 ( a + b )  2a + 2b   2   12  2a + 2b  

2



mb2 ( 3a + b ) mb ( 3a + b ) IM = 2 = NIabB 12 ( a + b ) NIabB 3 ( a + b ) NIaB

= 2

82. From Example 20–6, we have that r =

mv qB

. The quantity mv is the momentum p , and so r =

p qB

Thus, p = qBr . 83. The airplane is a charge moving in a magnetic field. Since it is flying perpendicular to the magnetic field, Eq. 27–5b applies.

(

)

(

)

Fmax = qvB = 1160  10−6 C (120 m s ) 5.0  10−5 T = 7.0  10−6 N 84. We find the speed of the muons using conservation of energy. The accelerating potential energy becomes the kinetic energy of the electron. eV = 12 mv 2 →

(

)(

2 65 103 eV 1.60 10−19 J eV 2eV v= = m 207 9.1110−31 kg

(

)

= 1.050 10 m s Upon entering the magnetic field the muons are traveling horizontally. The magnetic field will cause the path of the muons to be an arc of a circle of radius r, and deflect an angle  from the horizontal. While in the field, the muons will travel a horizontal distance d and a vertical distance h0 . Our approximation will be to ignore the distance h0 as compared to h. We have these relationships. h d mv tan = ; sin  = ; r= → l −d r eB 11cm 0.038m h d = tan −1 = 31.15 ; r = = = 0.0735m  = tan −1 18.2cm sin  sin 31.15 l −d 7

(

)( )

)

−31 7 mv 207 9.11  10 kg 1.050  10 m s = = 0.1684T  0.17 T B= er 1.60  10−19 C ( 0.0735m )

(

958

Chapter 27

Magnetism

85. (a) For the rod to be in equilibrium, the gravitational torque and the magnetic torque must be equal and opposite. Since the rod is uniform, the two torques can be considered to act at the same location (the center of mass). Therefore, components of the two forces perpendicular to the rod must be equal and opposite. Since the gravitational force points downward, its perpendicular component will point down and to the right. The magnetic force is perpendicular to the rod and must point towards the left to oppose the perpendicular component of the gravitational force. By the right-hand rule, with a magnetic field pointing out of the page, the current must flow downward from the pivot to produce this force. (b) We set the magnitude of the magnetic force, using Eq. 27–2, equal to the magnitude of the perpendicular component of the gravitational force, F⊥ = mg sin  , and solve for the magnetic field. 2 mg sin  ( 0.150 kg ) 9.80 m/s sin13 I l B = mg sin  → B = = = 0.028T Il (12 A )(1.0m ) (c) The largest magnetic field that could be measured is when  = 90. 2 mg sin 90 ( 0.150 kg ) 9.80 m/s sin 90 = = 0.12T Bmax = Il (12 A )(1.0m )

(

)

959

CHAPTER 28: Sources of Magnetic Field Responses to Questions 1.

Alternating currents will have little effect on the compass needle, due to the rapid change of the direction of the current and of the magnetic field surrounding it. Direct currents will deflect a compass needle. The deflection depends on the magnitude and direction of the current and the distance from the current to the compass. The effect on the compass decreases with increasing distance from the wire.

The magnetic field due to a long straight current is proportional to the current strength. The electric field due to a long straight line of electric charge at rest is proportional to the charge per unit length. Both fields are inversely proportional to the distance from the wire or line of charge. The magnetic field lines form concentric circles around the wire; the electric field lines are directed radially outward if the line of charge is positive and radially inward if the line of charge is negative.

Consider that the two wires are both horizontal, with the lower one carrying a current pointing east and with the upper one carrying a current pointing north. The magnetic field from the upper wire points downward on the east half of the lower wire and points upward on the west half of the lower wire. Using the right-hand rule on the east half of the lower wire, where the magnetic field is downward, the force points to the north. Using the right-hand rule on the west half of the lower wire, where the magnetic field is upward, the force points to the south. Thus, the lower wire experiences a counterclockwise torque about the vertical direction due to the magnetic forces from the upper wire. This torque is attempting to rotate the two wires so their currents are parallel. Likewise, the upper wire would experience a clockwise torque also tending to align the currents in the wires.

Yes. Assume the upper wire is fixed in position. Since the currents in the wires are in the same direction, the wires will attract each other. The lower wire will be held in equilibrium if this force of attraction (upward) is equal in magnitude to the weight of the wire (downward). But the lower wire would not be in stable equilibrium. If the lower wire were to move even lower, the upward magnetic force on it would weaken, and there would be a net downward force on the wire. And if the lower wire were to move upward, the upward magnetic force on it would strengthen, and there would be a net upward force on the wire.

(a) Let I 2 = I1 . 2

0 encl

= 0 ( I 1 + I 2 ) = 2 0 I 1

0 encl

= 0 ( I 1 + I 2 ) = 0

 Bdl =  I (b) Let I = − I .  B  d l =  I 1

Construct a closed path similar to that shown in part (a) of the figure, such that sides ab and cd are perpendicular to the field lines and sides bc and da lie along the field lines. Unlike part (a), the path will not form a rectangle; the sides ab and cd will flare outward so that side bc is longer than side da. Since the field is stronger in the region of da than it is in the region of bc, but da is shorter than bc, the contributions to the integral in Ampère’s law may cancel. Thus, 0 I encl =  B  d l = 0 is possible and the field is consistent with Ampère’s law. The lines could not curve upward instead of downward, because then bc would be shorter than da and it would not be possible for the contributions to sum to zero.

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The equation for the magnetic field strength inside a solenoid is given by B =  0 nI , where n is the number of loops per unit length. If the solenoid’s length is halved and the total number of loops is doubled, then the value of n will increase by a factor of 4. Thus, the magnetic field strength will increase by a factor of 4 also. We assume that the current is held constant.

The Biot-Savart law states that the net field at a point in space is the vector sum of the field contributions due to each infinitesimal current element. As shown in Example 28–11, the magnetic field along the axis of a current loop is parallel to the axis because the perpendicular field contributions cancel. However, for points off the axis, the perpendicular contributions will not cancel. The net field for a point off the axis will be dominated by the current elements closest to it. For example, in Fig. 28–22, the field lines inside the loop but below the axis curve downward, because these points in space are closer to the lower segment of the loop (where the current goes into the page) than they are to the upper segment (where the current comes out of the page).

No. The magnetic field varies in strength for points in the plane of the loop, but always points perpendicularly to the loop. Points near the loop are more influenced by the current in those nearby parts of the loop than by the current from more distant parts of the loop. The diagram shows the field lines closer together near the loop than at the center of the loop, implying that the field is stronger near the elements of the loop. And the field will decrease rapidly as you move away from the loop, outside of the boundary of the loop.

10. The lead-in wires to electrical devices have currents running in opposite directions. The magnetic fields due to these currents are therefore also opposite in direction. If the wires are twisted together, then the distance from a point outside the wires to each of the individual wires is about the same, and the field contributions from the two wires will cancel. If the wires were not twisted and were separate from each other, then a point outside the wires would be a different distance from one of the wires than from the other, and there would be a net field due to the currents in the wires. 11. The Biot-Savart law and Coulomb’s law are both inverse-square in the radius and both contain a proportionality constant. Coulomb’s law describes a central force; the Biot-Savart law involves a cross-product of vectors and so cannot describe a central force. Coulomb’s law can be used for single point charges, but the Biot-Savart law cannot (as discussed in Section 28–7). 12. Determine the magnetic field of the Earth at one of the magnetic poles (North or South), and use Eq. 28–7b to calculate the magnetic moment. In this equation, x will be (approximately) the radius of the Earth. This only uses one data point, however. If you could collect data all along the magnetic axis of the Earth, for many distances from the center of the Earth, then a plot of B vs. r 3 should, by Eq. 28–7b, have a slope of

0 Earth, which could then be solved for the magnetic 2

moment of the Earth. 13. To design a relay, place an iron rod inside of a solenoid, then point one end of the solenoid–rod combination at the piece of iron on a pivot. A spring normally holds the piece of iron away from a switch, making an open circuit where current cannot flow. When the relay is activated with a small current, a relatively strong magnetic field is created inside the solenoid, which aligns most of the magnetic domains in the iron rod and produces an even stronger magnetic field at © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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the end of the solenoid–rod combination. This magnetic field attracts the piece of iron on the springy pivot, which causes it to move toward the switch, connecting it and allowing current to flow through the large current circuit. 14. (a) Eq. 27–12 gives the potential energy of a magnetic dipole in an external magnetic field. The block of iron has many magnetic dipoles. Because the external magnetic field is not uniform, there is a gradient to the potential energy, and as seen in Chapter 8, a potential gradient produces a force. So the block of iron is drawn into the non-uniform magnetic field, and the magnetic potential energy becomes kinetic energy because work is done on it. If the “heavy magnet” produced a uniform magnetic field, there would not be an attractive force, and so the iron block would only rotate, not translate. (b) When the block strikes the magnet, some of the kinetic energy from the block is converted into kinetic energy in the iron atoms in the magnet, randomizing their magnetic moments and decreasing the overall field produced by the magnet. Some of the kinetic energy of the block as a whole is also converted into the kinetic energy of the individual atoms in the block, resulting in an increase in thermal energy. And there is probably some energy that becomes sound energy of sound waves traveling through the air and through the block and magnet. 15. A bar magnet can be considered as a collection of magnetic dipoles, all of which are aligned. If you visualize a magnetic dipole in a uniform magnetic field as a loop of current as in Fig. 27–23, you will note that the opposite sides of the loop will experience equal but opposite forces. This is true for every point on the “loop” that is modeling our bar magnet. The loop may experience a torque, making its magnetic dipole moment align with the uniform external field, but the net force will always be 0. Since the bar magnet is a collection of magnetic dipoles, and each dipole has no net force, the bar magnet will not have a net force on it. Another way to perhaps visualize this is a very lightweight compass needle (representing the paper clip of the question) in the Earth’s magnetic field. The Earth’s field is essentially uniform over small distances (perhaps a few feet or more), and the compass needle is a bar magnet. The compass needle is not attracted towards the north or south directions, but just rotates to line up with the Earth’s magnetic field. This is all very similar to the behavior of an electric dipole in a uniform electric field. 16. A magnet will not attract just any metallic object. For example, while a magnet will attract paper clips and nails, it will not attract coins or pieces of aluminum foil. This is because magnets will only attract other ferromagnetic materials (iron, cobalt, nickel, gadolinium, and some of their oxides and alloys). Iron and its alloys (such as steel) are the only common materials. These ferromagnetic materials contain magnetic domains that can be made to temporarily align when a strong magnet is brought near. The alignment occurs in such a way that the north pole of the domain points toward the south pole of the strong magnet, and vice versa, which creates the attraction. 17. Initially, both the nail and the paper clip have their magnetic domains pointing in random directions (which is why they appear to be unmagnetized). Thus, when you bring them close to each other, they are not attracted to or repelled from each other. Once the nail is in contact with a magnet (let’s say the north pole), many of the nail’s domains will align in such a way that the end of the nail that is touching the magnet becomes a south pole, due to the strong attraction, and the opposite end of the nail then becomes a north pole. Now, when you bring the nail close to the paper clip, there are mainly north poles of nail domains close to the paper clip, which causes some of the domains in the paper clip to align in such a way that the end near the nail becomes a south pole. Since the nail’s domains are only partially aligned, it will not be a strong magnet and thus the alignment of the paper clip’s domains will be even weaker. The attraction of the paper clip to the nail will be weaker than the attraction of the nail to the magnet. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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18. A magnet can attract an iron bar, so by Newton’s third law, an iron bar can attract a magnet. Another consideration is that the iron has domains that can be made slightly aligned by an external magnetic field, as opposed to a substance like plastic or wood. Thus the iron is also a magnet, at least when it is close to the magnet. 19. The magnetic domains in the unmagnetized piece of iron are initially pointing in random directions, as in Fig. 20–42a (which is why it appears to be unmagnetized). When the south pole of a strong external magnet is brought close to the random magnetic domains of the iron, many of the domains will rotate slightly so that their north poles are closer to the external south pole, which causes the unmagnetized iron to be attracted to the magnet. And similarly, when the north pole of a strong external magnet is brought close to the random magnetic domains of the iron, many of the domains will rotate slightly so that their south poles are closer to the external north pole, which causes the unmagnetized iron to now be attracted to the magnet. Thus, either pole of a magnet will attract an unmagnetized piece of iron. 20. Put one end of one rod close to one end of another rod. The ends will either attract or repel. Continue trying all combinations of rods and ends until two ends repel each other. Then the two rods used in that case are the magnets. 21. No, they are not both magnets. If they were both magnets, then they would repel one another when they were placed with like-poles facing each other. However, if one is a magnet and the other isn’t, they will attract each other no matter which ends are placed together. The magnet will cause an alignment of the domains of the non-magnet, causing an attraction. 22. To make the new sketch, take the original picture and replace Original picture all of the arrows (that represent the magnetic moment of each domain region) with an arrow approximately half as long but in the opposite direction as the initial arrows.

Revised picture

23. (a) Yes. Diamagnetism is present in all materials, but in materials that are also paramagnetic or ferromagnetic its effects will not be noticeable because those effects are stronger than the diamagnetic effect. (b) No. Paramagnetic materials are nonferromagnetic materials with a relative permeability greater than 1. (c) No. Ferromagnetic materials are those that can be magnetized by alignment of their domains. 24. (a) The idealized magnetization curve for a paramagnetic substance is a straight line with a positive slope slightly greater than 1. The slope of the idealized line would be 1 +  m , where  m (the magnetic susceptibility) is a small positive number, much less than 1, as given in the paramagnetic column of Table 28–1. The curve passes through the origin; there is no hysteresis. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

963

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The idealized magnetization curve for a diamagnetic substance is a straight line with a positive slope slightly less than 1. The slope of the idealized line would be 1 +  m , where  m (the magnetic susceptibility) is a small negative number, with a magnitude much less than 1, as given in the diamagnetic column of Table 28–1. The curve passes through the origin; there is no hysteresis. The magnetization curve for a ferromagnetic substance is a hysteresis curve (see Fig. 28–30). The magnetic effects for ferromagnetism are several orders of magnitude larger than the effects for paramagnetism or diamagnetism. The vertical scale on the paramagnetic and diamagnetic graphs would be the same as the horizontal scale. We see on the ferromagnetic graphs the vertical scale is several orders of magnitude larger than the horizontal scale. 25. (a) (b) (c) (d)

diamagnetic ferromagnetic, paramagnetic, diamagnetic ferromagnetic, ferromagnetic diamagnetic

Solutions to MisConceptual Questions 1.

(c) Eq. 28–1 states that the magnetic field from a current is directed circularly around the wire, is proportional to the current flowing in the wire, and inversely proportional to the distance from the wire. A constant current produces a magnetic field, so the current does not need to be changing.

(b) The current in the lower wire is opposite in direction to the current in the upper wire.

(c) This question requires a consideration of Newton’s third law. The force that one wire exerts on a second wire must be equal in magnitude, but opposite in direction, to the force that the second exerts on the first.

(e) The magnetic field inside the solenoid does not depend on the diameter of the loops, but it is proportional to the number of turns per unit length. Or, the magnetic field is inversely proportional to the distance between the loops. So doubling that distance means the magnetic field will be halved.

(e) The value of the B is proportional to the number of loops divided by the length of the solenoid. If both the number of loops and the length are doubled, then B doesn’t change.

(a) Imagine that the first wire is fixed, and the current in the first wire is coming out of the page. The magnetic field lines due to the first wire are concentric circles. The second wire is immersed in the magnetic field due to the first wire–see the diagram. The right half of the second wire is in an upward magnetic field, and the left half is in a downward magnetic field. The magnetic force on the right half of the second wire is out of the page, and the force on the left half of the wire will be into the page. Thus, there will be no net force on the second wire. But there will be a torque, tending to align the current in the second wire with the current in the first wire.

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(d) Inside the cavity B = 0. We apply Ampere’s law with the cylindrical geometry to get the following:  B  ds = 0 I encl → B ( 2 R ) = 0 I encl . Since I encl = 0, the magnetic field is 0.

(b) The charge-carrying mosquito is a short current segment. The “mosquito” current is in the same direction as the current in the wire. Thus, there are 2 parallel currents, and as seen in Section 28–2, the two currents will be attracted towards each other. We could also treat the = ( q v )mosquito  B wire to obtain the same result. mosquito as a charge, and use Fon mosquito

(a) This very scenario is discussed in Example 28–6. As seen in that example, the magnetic field is 0 at the center of the wire and increases linearly to the outer edge of the wire.

10. (d) At the center, B due to the upper left current points in the direction of answer (c). B due to the upper right current points in the direction of answer (d). B due to the lower right current points in the direction of answer (b). And B due to the lower left current points in the direction of answer (d). The (b) and (c) vectors add to be 0, so the final magnetic field points in the direction of answer (d). 11. (e) The magnetic field due to a long straight conductor is inversely proportional to the distance from the conductor. At twice the distance, the distance, the magnetic field is reduced by a factor of 2. 12. (c) Electric fields are created by charged objects whether the charges are moving or not. Magnetic fields are created by moving charged objects. Since the electron is charged and moving, it creates both an electric field and a magnetic field. 13. (a, b, e) In each of these regions, an Ampere’s law calculation done using a circular path that is concentric with the conductors would enclose no net current, and so the magnetic field would be 0 in each of those regions. 14. (a, b, c, e) On page 847 the text says that above the Curie temperature, a metal cannot be made into a magnet at all. So answer (a) is true. Also on page 847, the text says that heating a permanent magnet can cause loss of magnetism, thus, answer (b) is true. Figure 28–32 on page 850 illustrates that answer (c) is correct. Page 825 states that dropping a magnet on the floor or striking it with a hammer can cause a magnet to lose some of its magnetism, thus, answer (e) is true. It also states on page 847 that a magnet can be made from an unmagnetized piece of iron by placing it in a strong magnetic field. Thus answer (d) is false.

Solutions to Problems 1.

We assume the jumper cable is a long straight wire, and use Eq. 28–1. 4  10−7 T m A ( 65A ) 0 I = = 2.6  10−4 T Bcable = 2 r 2 ( 0.050 m )

(

)

Compare this to the Earth’s field of 0.5  10−4 T. 2.6  10−4 T Bcable BEarth = = 5.2, so the field of the cable is 5.2 times that of the Earth . 5.0  10−5 T

965

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

We assume that the wire is long and straight, and use Eq. 28–1. −4 2 rBwire 2 ( 0.12 m ) ( 0.50  10 T ) I → I= = = 30 A (2 significant figures) Bwire = 0 0 2 r 4  10 −7 T m A

Since the currents are parallel, the force on each wire will be attractive, toward the other wire. Use Eq. 28–2 to calculate the magnitude of the force. 4  10−7 T m A ( 35 A ) 2  II F2 = 0 1 2 l 2 = ( 35 m ) = 0.21N, attractive 2 d 2 ( 0.040 m )

(

)

Since the force is attractive, the currents must be in the same direction, so the current in the second wire must also be upward. Use Eq. 28–2 to calculate the magnitude of the second current.  II F2 = 0 1 2 l 2 → 2 d 2 F2 d 2 0.090 m = = 15.95 A  16 A upward 7.8  10 −4 N m I2 = −7 0 l 2 I1 4  10 T m A 22 A

(

)

To find the direction, draw a radius line from the wire to the field point. Then at the field point, draw a perpendicular to the radius line, directed so that the perpendicular line would be part of a counter-clockwise circle. The relative magnitude is given by the length of the arrow. The further away a point is from the wire, the weaker the field.

For the experiment to be accurate to  5.0%, the magnetic field due to the current in the cable must be less than or equal to 5.0% of the Earth’s magnetic field. Use Eq. 28–1 to calculate the magnetic field due to the current in the cable. −4 2 r ( 0.050 BEarth ) 2 (1.00 m )( 0.050 ) ( 0.5  10 T ) 0 I  0.050 BEarth → I  = = 12.5 A Bcable = 0 2 r 4  10−7 T m A Thus, the maximum allowable current is 12.5 A  13A . However, we might not want to round up since we are looking for a maximum allowable current. Perhaps we should say 12 A.

Since the magnetic field from a current-carrying wire circles the wire, the individual field at point P from each wire is perpendicular to the radial line from that wire to point P. We define B1 as the field from the top wire, and B 2 as the field from the bottom wire. We use Eq. 28–1 to calculate the magnitude of each individual field. −7 0 I ( 4  10 T m A ) ( 25A ) B1 = = = 8.333  10−5 T 2 r1 2 ( 0.060 m )

0 I ( 4  10 T m A ) ( 25A ) = = 5.000  10−5 T 2 r2 2 ( 0.100 m ) −7

B2 =

We use the law of cosines to determine the angle that the radial line from each wire to point P makes with the vertical. Since the field is perpendicular to the radial line, this is the same angle that the magnetic fields make with the horizontal.

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 ( 0.060 m )2 + ( 0.130 m )2 − ( 0.100 m )2  1 = cos   = 47.7   2 ( 0.060 m )( 0.130 m )   −1

 ( 0.100 m )2 + ( 0.130 m )2 − ( 0.060 m )2   2 = cos   = 26.3   2 ( 0.100 m )( 0.130 m )   Using the magnitudes and angles of each magnetic field we calculate the horizontal and vertical components, add the vectors, and calculate the resultant magnetic field and angle. Bnet x = B1 cos (1 ) − B2 cos 2 = (8.333  10−5 T ) cos 47.7o − ( 5.000  10−5 T ) cos 26.3o = 1.126  10−5 T −1

(

)

(

)

Bnet y = B1 sin (1 ) + B2 sin 1 = 8.333  10−5 T sin 47.7 o + 5.000  10 −5 T sin 26.3o = 8.379  10−5 T

(1.126  10 T ) + (8.379  10 T ) = 8.454  10 T

 = tan −1

Bnet, y

= tan −1

Bnet, y

−5

2 2 B = Bnet, x + Bnet, y =

−5

8.379  10−5 T = 82.3 1.126  10−5 T

B = 8.454  10−5 T @ 82.3  8.5  10−5 T @ 82 or B = (1.1  10−5 T ) ˆi + (8.4  10−5 T ) ˆj 8.

At the location of the compass, the magnetic field caused by the wire will point to the west, and the Earth’s magnetic field points due North. The compass needle will point in the direction of the NET magnetic field. −7  I ( 4  10 T m A ) ( 48 A ) Bwire = 0 = = 4.0  10−5 T 2 r 2 ( 0.24 m )

 = tan −1 9.

−5

BEarth

= tan −1

Bwire

4.5  10 T 4.0  10−5 T

B Earth

 B wire

= 48 N of W

The magnetic field due to the long horizontal wire points straight up at the point in question, and its magnitude is given by Eq. 28–1. The two fields are oriented as shown in the diagram. The net field is the vector sum of the two fields. 4  10−7 T m A ( 24.0 A ) 0 I = = 2.40  10−5 T Bwire = 2 r 2 ( 0.200 m )

(

B net

B wire

)

38

 B net

B Earth

−5

BEarth = 5.0  10 T

(

)

Bnet x = BEarth cos 38o = 5.0  10 −5 T cos38 = 3.94  10 −5 T

(

)

Bnet y = Bwire − BEarth sin 38o = 2.40  10 −5 T − 5.0  10 −5 T sin38 = −6.78  10 −6 T 2 2 Bnet = Bnet + Bnet = x y

 = tan −1

Bnet y Bnet x

= tan −1

( 3.95  10 T ) + ( −6.78  10 T ) = 4.0  10 T −5

−6.78  10 −6 T 3.95  10 −5 T

−6

−5

= 9.7 below the horizontal .

10. (a) If the currents are in the same direction, the magnetic fields at the midpoint between the two currents will oppose each other, and so their magnitudes should be subtracted. 4  10−7 T m A I I Bnet = 0 1 − 0 2 = ( I − 25 A ) = 2.0  10−5 T A ( I − 25 A ) 2 r1 2 r2 2 ( 0.010 m )

(

)

(

)

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) If the currents are in the opposite direction, the magnetic fields at the midpoint between the two currents will reinforce each other, and so their magnitudes should be added. 4  10−7 T m A 0 I 1 0 I 2 + = Bnet = ( I + 25 A ) = 2.0  10−5 T A ( I + 25 A ) 2 r1 2 r2 2 ( 0.010 m )

(

)

(

)

11. The stream of protons constitutes a current, whose magnitude is found by multiplying the proton rate times the charge of a proton. Then use Eq. 28–1 to calculate the magnetic field. 4  10−7 T m A 2.5  109 protons s 1.60  10 −19 C proton 0 I = = 4.4  10−17 T Bstream = 2 r 2 (1.8 m )

(

)(

)

12. Using the right-hand rule we see that if the currents flow in the same direction, the magnetic fields will oppose each other between the wires, and therefore can equal 0 at a given point. Set the sum of the magnetic fields from the two wires equal to 0 at the point 2.2 cm from the first wire and use Eq. 28–1 to solve for the unknown current. I I Bnet = 0 = 0 1 − 0 2 → 2 r1 2 r2

r   6.0cm − 2.2cm  I 2 =  2  I1 =   ( 2.0 A ) = 3.5A, same direction as other current . 2.2cm    r1  13. Use the right-hand rule to determine the direction of the magnetic field from each wire. Remembering that the magnetic field is inversely proportional to the distance from the wire, qualitatively add the magnetic field vectors. The magnetic field at point #2 is 0.

14. The fields created by the two wires will oppose each other, so the net field is the difference of the magnitudes of the two fields. The positive direction for the fields is taken to be into the paper, and so the closer wire creates a field in the positive direction, and the farther wire creates a field in the negative direction. Let d be the separation distance of the wires. 0 I 0 I I 1 1  0 I  1 1  − = 0  − = − Bnet =    1 2 rcloser 2 rfarther 2  rcloser rfarther  2  r − 2 d r + 12 d 

 0 I  d   1 1 2  ( r − 2 d )( r + 2 d ) 

( 4  10 T m A ) ( 35.0 A )  = −7

2

 0.0028 m    ( 0.10 m − 0.0014 m )( 0.10 m + 0.0014 m ) 

= 1.960  10−6 T  2.0  10−6 T Compare this to the Earth’s field of 0.5  10−4 T. 1.960  10−6 T Bnet BEarth = = 0.039 0.5  10−4 T The field of the wires is about 4% that of the Earth. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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15. The center of the third wire is 5.6 mm from the left wire, and 2.8 mm from the right wire. The new wire will attract the wire that is now in the middle, since the currents are in the same direction. The new wire will repel the force on the far left wire, since the currents oppose each other. Use Eq. 28–2 to calculate the force per unit length.  I right I middle Fmiddle = 0 l middle → 2 d middle

0 I right I middle 4  10−7 T m A ( 25.0 A )( 35.0 A ) = = 0.063 N m , attractive . 2 2.8  10−3 m l middle 2 d middle  I right I left Fleft = 0 l left → 2 d left Fmiddle

Fleft l left

0 I right I left 4  10−7 T m A ( 25.0 A )( 35.0 A ) = = 0.031N m , repelling . 2 d left 2 5.6  10−3 m

16. (a) We assume that the power line is long and straight, and use Eq. 28–1. 4  10−7 T m A ( 85 A ) 0 I = = 2.0  10−6 T Bline = 2 r 2 ( 8.5 m )

(

)

The direction at the ground, from the right-hand rule, is south. Compare this to the Earth’s field of 0.5  10−4 T, which points approximately north. Bline BEarth =

2.0  10−6 T

= 0.04 0.5  10−4 T The field of the cable is 4% that of the Earth. (b) We solve for the distance where Bline = BEarth .

4  10−7 T m A ) (85A ) ( 0 I 0 I = BEarth → r = = = 0.34 m Bline = 2 r 2 BEarth 2 ( 0.5  10−4 T )

So about 0.34 m below the wire, the net magnetic field would be 0, assuming the Earth’s field points straight north at this location. 17. The Earth’s magnetic field is present at both locations in the problem, and we assume it is the same at both locations. The field east of a vertical wire must be pointing either due North or due South. The compass shows the direction of the net magnetic field, and it changes from 13o E of N to 55o E of N when taken inside. That is a “southerly” change (rather than a “northerly” change), and so the field due to the wire must be pointing due South. See the diagram. For the angles,  = 13,  +  = 55, and  +  +  = 180, and so  = 42 and  = 125o. Use the

B Earth



 

law of sines to find the magnitude of B wire , and then use Eq. 28–1 to find the magnitude of the current. sin  0 I Bwire BEarth = → Bwire = BEarth = → sin sin sin 2 r I = BEarth

sin  2 sin 0

(

r = 5.0  10−5 T

 B wire

B net

sin 42 2 ( 0.160 m ) = 32.67A  33A ) sin125 4  10 T m A o

−7

Since the field due to the wire is due South, the current in the wire must be down . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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18. The magnetic field at the loop due to the long wire is into the page, and can be calculated by Eq. 28–1. The force on the segment of the loop closest to the wire is towards the wire, since the currents are in the same direction. The force on the segment of the loop farthest from the wire is away from the wire, since the currents are in the opposite direction. Because the magnetic field varies with distance, it is more difficult to calculate the total force on the left and right segments of the loop. Using the right-hand rule, the force on each small piece of the left segment of wire is to the left, and the force on each small piece of the right segment of wire is to the right. If left and right small pieces are chosen that are equidistant from the long wire, the net force on those two small pieces is 0. Thus, the total force on the left and right segments of wire is 0, and so only the parallel segments need to be considered in the calculation. Use Eq. 28–2 to find the net force.  1  II  II  1  Fnet = Fnear − Ffar = 0 1 2 l near − 0 1 2 l far = 0 I1 I 2 l  −  2 d near 2 d far 2  d near d far  =

4  10−7 T m A 2



 = 5.1  10 −6 N, towards wire .   0.030 m 0.080 m  1

( 3.5 A ) ( 0.100 m )  2

−

19. The left wire will cause a field on the x-axis that points in the y-direction, and the right wire will cause a field on the x-axis that points in the negative y-direction. The distance from the left wire to a point on the x-axis is x, and the distance from the right wire is d − x.

B net =

0 I ˆ 0 I ˆj = 0 I  1 − 1  ˆj = 0 I  d − 2 x  ˆj j−     2 x 2 ( d − x ) 2  x d − x  2  x ( d − x ) 

20. The left wire will cause a field on the x-axis that points in the negative y-direction, and the right wire will also cause a field on the x-axis that points in the negative y-direction. The distance from the left wire to a point on the x-axis is x, and the distance from the right wire is d − x. B net = −

  0 ( 2 I ) ˆ 0 I ˆj = − 0 I  2 + 1  ˆj = − 0 I 2d − x ˆj j−     2 x 2 ( d − x ) 2  x d − x  2  x ( d − x ) 

21. The magnetic fields created by the individual currents will be at right angles to each other. The field due to the top wire will be to the right, and the field due to the bottom wire will be out of the page. Since they are at right angles, the net field is the hypotenuse of the two individual fields. 2

 0 I top   0 I bottom  4  10−7 T m A 0 2 2 Bnet =  I I + = + =   top bottom  2 ( 0.100 m )  2 rtop   2 rbottom  2 r

(18.0 A )2 + (12.0 A )2

= 4.33  10−5 T 22. (a) For the net field to be zero, the currents must be in the same direction See the figure, showing that the two fields have opposite directions between the wires.

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(b) We use Eq. 28–1 to calculate the magnetic field of each wire, setting them equal in magnitude. 0 I A 0 I B 0.15I A (15cm )( 20.0 A ) = → I A ( 0.15 − x ) = I B x → x = = = 6.0 cm 2 x 2 ( 0.15 − x ) 50.0 A ( IA + IB ) 23. The net magnetic field is the vector sum of the magnetic fields produced by each current-carrying wire. Since the individual magnetic fields encircle the wire producing it, the field is perpendicular to the radial line from the wire to point P. We let B1 be the field from the left wire, and B 2 designate the field from the right wire. The magnitude of the magnetic field vectors is calculated from Eq. 28–1. −7  I ( 4  10 T m A ) (19.5A ) = 3.25  10−5 T B1 = 0 = 2 r1 2 ( 0.12 m )

0 I ( 4  10 T m A ) (19.5A ) = = 3.00  10−5 T 2 r2 2 ( 0.13m ) −7

B2 =

We use the law of cosines to determine the angle that the radial line from each wire to point P makes with the horizontal. Since the magnetic fields are perpendicular to the radial lines, these angles are the same as the angles the magnetic fields make with the vertical. 2 2 2   −1 ( 0.12 m ) + ( 0.082 m ) − ( 0.13 m ) 1 = cos   = 77.61   2 ( 0.12 m )( 0.082 m )  

 ( 0.13 m )2 + ( 0.082 m )2 − ( 0.12 m )2   2 = cos   = 64.36   2 ( 0.13 m )( 0.082 m )   Using the magnitudes and angles of each magnetic field we calculate the horizontal and vertical components, add the vectors, and calculate the resultant magnetic field and angle. Bnet x = − B1 sin (1 ) − B2 sin  2 = − ( 3.25  10−5 T ) sin 77.61 − ( 3.00  10 −5 T ) sin 64.36 −1

= −5.879  10−5 T Bnet y = B1 cos (1 ) − B2 cos1 = ( 3.25  10−5 T ) cos77.61 − ( 3.00  10−5 T ) cos64.36 = −6.008  10−6 T 2 2 B = Bnet, x + Bnet, y =

 = tan −1

Bnet, y Bnet, y

= tan −1

( −5.879 10 T ) + ( −6.008 10 T ) = 5.9110 T −5

−6

−5

−6.008  10−6 T = 5.83 −5.879  10−5 T

B = 5.91  10−5 T @ 5.83 below the negative x-axis 24. (a) The strip is in the x-z plane in the diagram, and the current is going in the negative z-direction, into the plane of the figure. The net magnetic field at a point y above the center of the strip can be found by dividing the strip into infinitely thin wires of width dx (oriented along the z-axis) and integrating the field contribution from each wire. The cross-section of that wire is shown as a black rectangle in the diagram. Since the point at which we want to calculate the magnetic field is directly above the © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

center of the strip, we see that the vertical contributions to the magnetic field from symmetric points on either side of the center cancel out. Therefore, we only need to integrate the horizontal component of the magnetic field. We use Eq. 28–1 for the magnitude of the I magnetic field, with the current given by dI = dx. l 0 dI 0 dI I y y dB = ; dBx = sin  ; dI = dx ; r = x 2 + y 2 ; sin  = = 2 r l 2 r 2 r x + y2 Bx = 

l 2    0 Iy l 2 dx 0 sin  I dx y dI = 0   =  2 r 2 l − l 2 x 2 + y 2  x 2 + y 2  2 l − l 2 x 2 + y 2 l 2

0 Iy 1 I x  l  = 0 tan −1  tan −1  y −l 2 2 l y l  2y 

(b) In the limit of large y, tan −1 d / 2 y  d / 2 y.

Bx =

 l  I l 0 I I tan −1    0 = 0 l  2 y   l 2 y 2 y

This makes sense, because it is the same as the magnetic field for a long wire. If you are far away from the strip, its width l seems small, and it has the field of a current-carrying wire. 25. We break the current loop into the three branches of the triangle and add the forces from each of the three branches. dF dF The current in the parallel branch flows in the same direction as the long straight wire, so the force is attractive (towards the ds ds y long straight wire) with magnitude given by Eq. 28–2. 60  I I F1 = 0 a 2 d d Consider a small element of the left-hand side of the triangle, labeled as ds on the diagram. The magnetic field due to I current I  is out of the paper, so by using the right-hand rule, the direction of the magnetic force on ds is shown and labeled dF. That magnetic force will have a component parallel to current I , and a component perpendicular to current I . From the symmetry of the problem, the parallel component of force will be exactly cancelled by another small current element on the right-hand side of the triangle. Thus, to find the net force, we need to add up all of the perpendicular components of the magnetic force, on both sides of the triangle.  0 I   dF = I ( BI  ) ds = I   ds ; dFy = dF sin 30 ; dy = ds sin 60 →  2 ( d + y )     0 I I    0 I I   1 2 0 I I  dy dFy =  dy =  = I   dy  sin 30   2 3 ( d + y )  d y d y 2 sin 60 2 +  +   ( ) ( ) 3 2       y = 3a 2

Fy =



0 I I 



0 I I 

  2 3 ( d + y )  dy = 2 3 ln ( d + y )

3a 2 0

0 I I   3a  ln 1 +  2d  2 3 

  This is the vertical force on the left side of the triangle and it is directed perpendicularly away from the long straight wire. The total vertical force is twice this value. To find the net force, add the three vertical forces together, calling the direction towards the long straight wire the positive direction.   I I   I I 3a   0 I I  a 3  3a   ln 1 + ln 1 + Ftotal = 0 a − 2 0   =   − 2 d 2d   2d     2d 3   2 3  y =0

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26. (a) Use the definition of the ampere as 1 coulomb / second. coulomb ampere = → coulomb = ampere second second (b) Use the two equivalent unit definitions for electric field. kilogram meter meter 2 newton volt newton meter kilogram meter 2 second = → volt = = = coulomb meter coulomb ampere second ampere second 3 (c) Use the magnetic force equation, F = I l  B. newton = ampere meter tesla → kilogram meter tesla =

newton ampere meter

kilogram second 2 = ampere meter ampere second 2

(d) Use Ohm’s Law, V = IR. volt = ampere ohm → ohm =

volt ampere

kilogram meter 2 ampere 2 second 3

27. Use Eq. 28–4 for the field inside a solenoid. 0.385  10−3 T ( 0.400 m ) 0 IN Bl → I= = = 0.182 A B= l 0 N 4  10−7 T m A ( 675)

( (

)

28. The field inside a solenoid is given by Eq. 28–4.  IN Bl ( 0.30 T )( 0.32 m ) B= 0 → N= = = 14,150  1.4  104 turns . 0 I l 4  10−7 T m A ( 5.4 A )

(

)

29. (a) We use Eq. 28–1, with r equal to the radius of the wire. −7  I ( 4  10 T m/A ) ( 28A ) B= 0 = = 4.48  10−3 T  4.5mT 2 r 2 (1.25  10−3 m ) (b) We use the results of Example 28–6, for points inside the wire. Note that r = (1.25 − 0.50 ) mm = 0.75mm.

 Ir ( 4  10 T m/A ) ( 28A ) ( 0.75  10 m ) = 2.69  10−3 T  2.7 mT B= 0 2 = 2 −3 2 R 2 (1.25  10 m ) −7

−3

(c) We use Eq. 28–1, with r equal to the distance from the center of the wire. ( 4  10−7 T m/A ) ( 28A ) = 1.49  10−3 T  1.5mT I B= 0 = 2 r 2 (1.25  10−3 m + 2.5  10−3 m )

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Instructor Solutions Manual

30. We use the results of Example 28–9 to find the maximum and minimum fields. −7  NI ( 4  10 T m/A ) ( 785)( 25.0 A ) Bmin = 0 = = 14.5mT 2 rmax 2 ( 0.270 m )

 NI ( 4  10 T m/A ) ( 785)( 25.0 A ) Bmax = 0 = = 15.7 mT 2 rmin 2 ( 0.250 m ) −7

14.5mT  B  15.7 mT

31. (a) The copper wire is being wound about an average diameter that is approximately equal to the outside diameter of the solenoid minus the diameter of the wire, or D − d . See the (not to scale) end-view diagram. The length of each wrapping is  ( D − d ) . We divide the length of the wire L by the length of a single winding to determine the number of loops. The length of the solenoid is the number of loops multiplied by the outer diameter of the wire, d. L 20.0m l =d = ( 2.00  10−3 m ) = 0.554 m −2  (D − d )   2.50  10 m − ( 2.00  10−3 m ) (b) The field inside the solenoid is found using Eq. 28–4. Since the coils are wound closely together, the number of turns per unit length is equal to the reciprocal of the wire diameter. L # turns  ( D − d ) l d 1 n= = = = d l l l −7  I ( 4  10 T m/A ) (15.0 A ) B = 0nI = 0 = = 9.42 mT d 2.00  10−3 m 32. (a) The magnitude of the magnetic field from each wire is found using Eq. 28–1. The direction of the magnetic field is perpendicular to the radial vector from the current to the point of interest. Since the currents are both coming out of the page, the magnetic fields will point counterclockwise from the radial line. The total magnetic field is the vector sum of the individual fields. I I B = B1 + B2 = 0 − sin 1ˆi + cos1ˆj + 0 − sin  2 ˆi + cos 2 ˆj 2 r1 2 r2

(

)

(

)

0 I  sin 1 sin  2  ˆ  cos1 cos 2  ˆ  − +  − i +   j 2  r1 r2   r1 r2  

This equation for the magnetic field shows that the x-component of the magnetic field is symmetric about  = 90, since both terms are related to sin  and sin  is symmetric about  = 90. The equation also shows that the y-component is anti-symmetric about  = 90, since that is the behavior of cos .

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Sources of Magnetic Field

(b) See sketch. (c) The two diagrams are similar in shape, as both form loops around the central axes. However, the magnetic field lines form a vector field, showing the direction, not necessarily the magnitude of the magnetic field. The equipotential lines are from a scalar field showing the points of constant magnitude. The equipotential lines do not have an associated direction.

33. The field due to the solenoid is given by Eq. 28–4. Since the field due to the solenoid is perpendicular to the current in the wire, Eq. 27–2 can be used to find the force on the wire segment. 4  10−7 T m A ( 38 A )( 650 ) I N F = I wire l wire Bsolenoid = I wire l wire 0 solenoid = ( 22 A )( 0.030m ) l solenoid ( 0.15 m )

(

)

= 0.14 N to the south . 34. Because of the cylindrical symmetry, the magnetic fields will be circular. In each case, we can determine the magnetic field using Ampere’s law with concentric loops. The current densities in the wires are given by the total current divided by the cross-sectional area. We take leftward current as positive, which is the “outward” direction in the adjacent figure here. I I0 jinner = 0 2 jouter = − 2  R1  ( R3 − R22 ) (a) Inside the inner wire the enclosed current is determined by the current density of the inner wire. 2  B  d l = 0 I encl = 0 ( jinner R ) B ( 2 R ) = 0

I 0 R 2 I R → B= 0 02 2 2 R1  R1

(b) Between the wires the current enclosed is the current on the inner wire. 0 I 0  B  d l = 0 I encl → B ( 2 R ) = 0 I 0 → B = 2 R (c) Inside the outer wire the current enclosed is the current from the inner wire and a portion of the current from the outer wire. 2 2  B  d l = 0 I encl = 0  I 0 + jouter ( R − R2 ) 2 2   ( R 2 − R22 )  0 I 0 ( R3 − R ) → B= B ( 2 r ) = 0  I 0 − I 0 2 R ( R32 − R22 )  ( R32 − R22 )  

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(d) Outside the outer wire the net current enclosed is 0.  B  d l = 0 Iencl = 0 → B ( 2 R ) = 0 → B = 0 (e) See the adjacent graph.

35. We first find the constants C1 and C2 by integrating the currents over each cylinder and setting the integral equal to the total current. R1 R1 2 3I 0 I 0 =  ( C1R ) 2 RdR = 2 C1  R 2 dR =  R13C1 → C1 = 3 2 R13 0 0 R

3 −3I 0 2 − I 0 = 2 C2  R 2 dR =  ( R33 − R23 ) C2 → C2 = 3 2 ( R33 − R23 ) R2

(a) Inside the inner wire the enclosed current is determined by integrating the current density inside the radius R. R  3I 0  3 2 2 3 l   B  = = d I 0 encl 0  0 ( C1R) 2 RdR = 3 0 C1R = 3 0  2 R13  R B ( 2 R ) = 0

0 I 0 R 2 I 0 R 3 → = B  R13 2 R13

(b) Between the wires the current enclosed is the current on the inner wire. 0 I 0  B  d l = 0 I encl → B ( 2 R ) = 0 I 0 → B = 2 R (c) Inside the outer wire the current enclosed is the current from the inner wire and a portion of the current from the outer wire. R R  B  d l = 0 I encl = 0  I 0 + R2 ( C2 R )2 RdR  = 0  I 0 + R2 ( C2 R )2 RdR   R 3 − R23 )  (  2 3 3   = 0 I 0 1 −  C2 ( R − R2 )  = 0  I 0 − 3  3   ( R3 − R23 ) 

3 3  ( R33 − R23 ) ( R 3 − R23 )  0 I 0 ( R3 − R )  B ( 2 r ) = 0 I 0  3 B − → = 3 3 3 2 R ( R33 − R23 )  ( R3 − R2 ) ( R3 − R2 )  (d) Outside the outer wire the net current enclosed is 0.  B  d l = 0 Iencl = 0 → B ( 2 R ) = 0 → B = 0

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36. Use Eq. 28–7b to write an approximate ratio of the magnetic fields at the surface of the Earth and 13,000 km above the surface, both directly above the North Pole. We assume the “dipole” is located at the Earth’s center. Use the resulting ratio to determine the magnetic field above the surface.

0 

B2 2 x23 x13 = = 3 B1 0  x2 3 2 x1

→ B2 =B1

 6.38  103km  x13 −4  = 1.0 10 T ( )  19.38  103km  = 3.6  10−6T x23  

37. Since the point C is along the line of the two straight segments of the current, these segments do not contribute to the magnetic field at C. We calculate the magnetic field by integrating Eq. 28–5 along the two curved segments. Along each integration the line segment is perpendicular to the radial vector and the radial distance is constant. R R  I 1 d l  rˆ 0 I 0 d l  rˆ 0 I ˆ 1 0 I ˆ 0 ds B= 0  + = k + k ds 4 0 R12 4 R2 R22 4 R12 0 4 R22 R2

R2 I 

0 I ˆ 0 I ˆ 0 I  R2 − R1  ˆ k− k=  k 4 R1 4 R2 4  R1R2 

38. Since the current in the two straight segments flows radially toward and away from the center of the loop, they do not contribute to the magnetic field at the center. We calculate the magnetic field by integrating Eq. 28–5 along the two curved segments. Along each integration segment, the current is perpendicular to the radial vector and the radial distance is constant. By the right-hand rule the magnetic field from the upper portion will point into the page and the magnetic field from the lower portion will point out of the page.  ( R ) ds ds I I  I B = 0 1  2 kˆ + 0 2  2 −kˆ = 0 2 kˆ ( I1 − I 2 ) = 0 kˆ ( 0.25I − 0.75I ) = − 0 kˆ 4 upper R 4 lower R 4 R 4R 8R

( )

39. We assume that the inner loop is sufficiently small that the magnetic field from the larger loop can be considered to be constant across the surface of the smaller loop. The field at the center of the larger loop is illustrated in Example 28–11. Use Eq. 27–10 to calculate the magnetic moment of the small loop, and Eq. 27–11 to calculate the torque. R1 is the radius of the larger loop, and R2 is the radius of the smaller loop I B = 0 ˆi μ = IA = I  R22 ˆj 2 R1

μ2

B1 R1

I   I R2 ˆ τ = μ  B = I  R22 ˆj  0 ˆi = − 0 k 2 R1 2 R1 2

( 4  10 T m/A ) ( 7.0 A ) ( 0.018 m ) kˆ = −9.2  10 kˆ m N =− 2

−7

−8

2 ( 0.25 m )

This torque would cause the inner loop to rotate into the same plane as the outer loop with the currents flowing in the same direction.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

40. (a) The magnetic field at point C can be obtained using the BiotSavart law (Eq. 28–5), integrated over the current. First break the loop into four sections; 1) the upper semi-circle, 2) the lower semi-circle, 3) the right straight segment, and 4) the left straight segment. The two straight segments do not contribute to the magnetic field as the point C is in the same direction that the current is flowing. Therefore, along these segments r̂ and d l are parallel and d l  rˆ = 0. For the upper segment, each infinitesimal line segment is perpendicular to the constant magnitude radial vector, so the magnetic field points downward with constant magnitude.  I d l  rˆ I I B upper =  0 = 0 2 − kˆ ( R1 ) = − 0 kˆ. 2 4 r 4 R1 4 R1 Along the lower segment, each infinitesimal line segment is also perpendicular to the constant radial vector.  I d l  rˆ I I B lower =  0 = 0 2 − kˆ ( R2 ) = − 0 kˆ 2 4 r 4 R2 4 R2 Adding the two contributions yields the total magnetic field.

( )

0 I ˆ

0 I  1

1 ˆ  + k 4 R1 4 R2 4  R1 R2  (b) The magnetic moment is the product of the area and the current. The area is the sum of the two half circles. By the right-hand rule, curling your fingers in the direction of the current, the thumb points into the page, so the magnetic moment is in the − k̂ direction.   R 2  R22  ˆ − I 2 μ = − 1 + ( R1 + R22 ) kˆ I k = 2 2 2   B = B upper + Blower = −

k−

k= −

41. Treat the moving point charge as a small current segment. We can write the product of the charge and velocity as the product of a current and current segment. Inserting these into the Biot-Savart law gives us the magnetic field at point P. d l dq = qv = q d l = Id l dt dt  I d l  rˆ 0 qv  rˆ 0 q v  r B= 0 = = 4 r 2 4 r 2 4 r 3 42. (a) The disk can be broken down into a series of infinitesimal rings. As the charge in each of these rings rotates it produces a current of magnitude dI = ( 2 ) dq, where dq is the surface charge density multiplied by the area of the ring, and  2 is the period of rotation of the disk. We use Eq. 27–10 to calculate the magnetic dipole moment of each current loop and integrate the dipole moments to obtain the total magnetic dipole moment.   Q 3 ˆ  Q 2ˆ 2 rdr dμ = dIA =    r i = 2 r dr i 2 2  R R

(

)

Q 3 ˆ Q R 2 ˆ r dr i = i R2 4 0

μ=

978

Chapter 28

Sources of Magnetic Field

(b) To find the magnetic field a distance x along the axis of the disk, we again consider the disk as a series of concentric currents. We use the results of Example 28–11 to determine the magnetic field from each current loop in the disk, and then integrate to obtain the total magnetic field. 0 r 2 0 r 2 Q dB = rdr 3 dI = 3 2 2 2 2 ( r2 + x2 ) 2 ( r2 + x2 )  R

 r3  Q î R  Q î ( r + 2 x )  Q î  ( R + 2 x ) B= 0 2  dr = 0 2 = 0 2  − 2x 2 R 0 ( r 2 + x 2 ) 2 R 2 R  R 2 + x 2 r2 + x2   2

3 2

(c) When we take the limit x R, our equation reduces to Eq. 28–7b.   Qˆi  R 4   μ   Q ˆi   R 2  R 2 3R 4 B  0 2  2 x  1 + 2  1 − 2 + 4 + ...  − 2 x   0 2  3  = 0 3 2 R   2 x  2 x 8x   2 R  4 x  2 x 43. (a) Choose the y-axis along the wire and the x-axis passing from the center of the wire through the point P. With this definition we calculate the magnetic field at P by integrating Eq. 28–5 over the length of the wire. The origin is at the center of the wire. 1d ˆ ˆ ˆ 0 I d l  rˆ 0 I d l  r 0 I 2 dy j  R i − y j B= = = 4 r 2 4 r 3 4 −1 d ( R 2 + y 2 )3/ 2 2

(

)

1d 2

 IR dy = − 0 kˆ  4 − d ( R 2 + y 2 )3/ 2 1 2

d /2

 IR y = − 0 kˆ 1/ 2 2 2 4 R ( R + y2 )

= − −d /2

0 I d kˆ 2 2 R ( 4 R + d 2 )1/ 2

(b) If we take the limit as d → , this equation reduces to Eq. 28–1.

  I d   0 I B = lim  0 d →  2 R 2 2 1/ 2  ( 4R + d )  2 R  44. (a) The magnetic field at point Q can be obtained by integrating Eq. 28–5 over the length of the wire. In this case, each infinitesimal current segment dl (shown as dx in the diagram) is parallel to the x-axis, as is each radial vector. Since the magnetic field is proportional to the cross-product of the current segment and the radial vector, each segment contributes zero field. Thus the magnetic field at point Q is zero . (b) The magnetic field at point P is found by integrating Eq. 28–5 over the length of the current segment. ˆ ˆ ˆ  I d l  rˆ 0 I d l  r 0 I 0 dx i  − xi + yj 0 I y ˆ 0 dx = = = B= 0 k 3/ 2 2 3   4 r 4 r 4 − l ( x 2 + y 2 ) 4 − l ( x 2 + y 2 )3/ 2

(

)

x =0

Iy x = 0 kˆ 1/ 2 4 y 2 ( x2 + y 2 )

= x =− l

0 I l kˆ 4 y ( y 2 + l 2 )1/ 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

45. We treat the loop as consisting of 5 segments, The first has length d, is located a distance d to the left of point P, and has current flowing toward the right. The second has length d, is located a distance 2d to left of point P, and has current flowing upward. The third has length d, is located a distance d to the left of point P, and has current flowing downward. The fourth has length 2d, is located a distance d below point P, and has current flowing toward the left. Note that the fourth segment is twice as long as the actual fourth current. We therefore add a fifth line segment of length d, located a distance d below point P with current flowing to the right. This fifth current segment cancels the added portion, but allows us to use the results of Problem 44 in solving this problem. Note that the first line points radially toward point P, and therefore by Problem 44(a) does not contribute to the net magnetic field. We add the contributions from the other four segments, with the contribution in the positive z-direction if the current in the segment appears to flows counterclockwise around the point P. The magnitude of the result from problem 44 is (substituting d I d for l) B = 0 , where d is the length of the current segment and y is the distance from 4 y ( y 2 + d 2 )1/ 2 one end of the wire to the point of evaluation. B = B 2 ( using y = 2d , directed into the page )

+ B3 ( using y = d , directed out of the page ) + B 4 ( using y = d , using 2d as the length of the current, directed into the page ) + B5 ( using y = d , directed out of the page ) =−

0 I I I I d d d 2d kˆ + 0 kˆ − 0 kˆ + 0 kˆ 1/ 2 1/ 2 1/ 2 4 ( 2d ) ( 4d 2 + d 2 ) 4 d ( d 2 + d 2 ) 4 d ( d 2 + 4d 2 ) 4 d ( d 2 + d 2 )1/ 2

0 I  5ˆ  2 − k 4 d  2 

46. (a) The angle subtended by one side of a polygon , from the center point P is 2 divided by the number of sides, n. The length of the side L and the distance from the point to the center of the side D, are obtained from trigonometric relations. L = 2 R sin ( / 2 ) = 2 R sin ( / n )

D = R cos ( / 2 ) = R cos ( / n ) The magnetic field contribution from each side can be found using the result of Problem 43, using L in place of d, and using D = R cos ( / n ) in place of R. The result from Problem 43 was B =

d 0 I . 2 2 R ( d + 4 R 2 )1/2 2 R sin ( / n ) 0 I 0 I L = 2 2 2 D ( L2 + 4 D 2 ) 2  R cos ( / n )   2 R sin ( / n )  + 4  R cos ( / n ) 

(

1 2





 )

1 2

0 I tan ( / n ) 2 R

980

Chapter 28

(b)

Sources of Magnetic Field

The contributions from each segment add, so the total magnetic field is n times the field from one side.  In Btotal = 0 tan ( / n ) , into the plane of the figure . 2 R In the limit of large n,  / n becomes very small, so tan( / n)   / n.

0 In  0 I = 2 R n 2R This is the magnetic field at the center of a circle. Btotal =

47. The equation derived in Example 28–11 gives the magnetic field a distance x from a single loop. We expand this single loop to the field of an infinite solenoid by multiplying the field from a single loop by ndx, the density of loops times the infinitesimal thickness, and integrating over all values of x. Use the table in Appendix B–4 to evaluate the integral.  

B= 

−

0 IR 2 ndx

2 ( R2 + x2 ) 2

0 IR 2n  2



−

(R + x ) 2

3 2

0 IR 2n 2

x R2 ( R2 + x2 ) 2

= 0 In

−

48. To find the magnetic field at point (x,y) we break each current segment into two segments and sum fields from each of the eight segments to determine the magnetic field at the center. We use the results of Problem 44(b) to calculate the magnetic field of each segment. We start with the segment along the x-axis whose left end is at the origin, and then do all 8 segments in turn, progressing counterclockwise. (b − x ) x I I B= 0 kˆ + 0 kˆ 1/ 2 2 2 2 4 y ( y + x ) 4 y ( y + (b − x )2 )1/ 2

0 I

4 (b − x ) ( (b − x )2 + y )

2 1/ 2

kˆ +

0 I

(a − y )

4 (b − x ) ( ( a − y ) 2 + (b − x ) 2 )1/ 2

kˆ

x (b − x ) 0 I 0 I kˆ + kˆ 4 ( a − y ) ( ( a − y ) 2 + (b − x ) 2 )1/2 4 ( a − y ) ( ( a − y ) 2 + x 2 )1/2

y (a − y ) 0 I I kˆ + 0 kˆ 1/2 2 2 2 4 x ( ( a − y ) + x ) 4 x ( y + x 2 )1/2

We simplify this equation by factoring out common constants and combining terms with similar roots.

y 2 + (b − x ) 2 ( a − y ) 2 + (b − x ) 2 (a − y ) 2 + x 2  ˆ 0 I  y 2 + x 2 + + +  k 4  xy (b − x ) y (a − y )(b − x ) x (a − y ) 

49. (a) By symmetry we see that on the x-axis the magnetic field can only have an x component. To justify this assertion, imagine that the magnetic field had a component off the axis. If the current loop were rotated by 90 about the x-axis, the loop orientation would be identical to the original loop, but the off-axis magnetic field component would have changed. This is not possible, so the field only has an x component. The contribution to this field is the same for © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

981

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

each loop segment, and so the total magnetic field is equal to 4 times the x component of the magnetic field from one segment. We integrate Eq. 28–5 to find this magnetic field. Note in the figure from the text that the y-axis is parallel to the longer arrow indicated “d,” and the z-axis is parallel to the shorter arrow indicated by “d.” 1 1 d d 2 dy ˆj  12 d kˆ dy 0 I 0 Id ˆi 2 B=4  = 3 / 2  4 ( 1 d )2 + x 2 + y 2  2 − 1 d ( 1 d )2 + x 2 + y 2  3 / 2 − 12 d 2 2    2 

(

)

1 d 2

y 0 Id ˆi 2 2 ( 1 d ) + x 2  ( 1 d )2 + x 2 + y 2 1/ 2  2

 2



= − 12 d

2 2d 2 0 I ˆi

 ( d 2 + 4 x 2 )( d 2 + 2 x 2 )

1/ 2

(b) Let x

d to show that the magnetic field reduces to a dipole field of Eq. 28–7b. 2 2d 2 0 I ˆi d 2 0 I ˆi B = 1/ 2 2 x 3  ( 4 x 2 )( 2 x 2 )

Comparing our magnetic field to Eq. 28–7b we see that it is a dipole field with the magnetic moment μ = d 2 I ˆi . 50. (a) If the iron bar is completely magnetized, all of the dipoles are aligned. The total dipole moment is equal to the number of atoms times the dipole moment of a single atom. The density of iron is found in Table 13–1. N V  = N 1 = A 1 Mm

( 6.022  10 atoms/mole )( 7.80g/cm ) (15cm )(1.2cm )(1.0cm ) 1.8  10 23

55.845g/mole

 

−23

A m2   atom 

= 27.25A m2  27A m 2 (b) We use Eq. 27–11 to find the torque.  =  B sin  = ( 27.25A m2 ) ( 0.80T ) sin 90 = 22 m N 51. The field inside the solenoid is given by Eq. 28–4 with  0 replaced by the permeability of the iron. −7  NI 3000 ( 4  10 T m A ) ( 350 )( 0.35 A ) = = 0.462 T  0.5T B= l (1.0 m ) We assume the factor of 3000 only has one significant figure.

52. The magnetic field of a long, thin torus is the same as the field given by a long solenoid, as in Eq. 28–9. B =  nI = ( 2200 ) ( 4  10−7 T m/A )( 315m −1 ) ( 3.0A ) = 2.6T

982

Chapter 28

Sources of Magnetic Field

53. The magnetic permeability is found from the two fields. B B  B0 = 0 nI ; B =  nI ; = →  = 0 B0 0 B0 Here is a data table with the given values as well as the calculated values of . For the graph, we have not plotted the last three data points so that the structure for low fields is seen. It would appear from the data that the value of  is asymptotically approaching 0 for large fields.

54. The field inside the solenoid is given by Eq. 28–9 with  0 replaced by the permeability of the iron. B=

 NI l

→ =

Bl NI

( 2.2 T )( 0.38 m ) = 3.1  10−5 T m A  250 ( 640)( 42 A )

55. Since the wires all carry the same current and are equidistant from each other, the magnitude of the force per unit length between any two wires is the same and is given by Eq. 28–2. −7 F 0 I 2 ( 4  10 T m/A ) ( 9.50 A ) = = 2 d 2 ( 0.035m ) l

= 5.157  10−4 N m The direction of the force between two wires is along the radial line, and is attractive for currents traveling in the same direction and repulsive for currents traveling in opposite directions. The forces acting on wire M are radially away from the other two wires. By symmetry, the horizontal components of these forces cancel and the net force is the sum of the vertical components. FM = FMP cos30 + FMN cos30 = 2 ( 3.657  10−4 N m ) cos30 = 8.9  10−4 N m at 90

The force on wire N is found by adding the components of the forces from the other two wires. By symmetry we see that this force is directed at an angle of 300. The force on wire P, will have the same magnitude but be directed at 240. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

983

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

FN , x = FNP − FNM cos60 = 5.157  10−4 N m − ( 5.157  10 −4 N m ) cos60 = 2.579  10 −4 N m FN , y = − FNM sin 60 = − ( 5.157  10−4 N m ) sin 60 = −4.466  10 −4 N m FN =

( 2.579  10 N m ) + ( −4.466  10 N m ) = 5.2  10 N m at 300 2

−4

FP = 5.2  10−4 N m at 240 56. The magnetic field at the midpoint between currents M and N is the vector sum of the magnetic fields from each wire, given by Eq. 28–1. Each field points perpendicularly to the line connecting the wire to the midpoint. Let x be to the right, and y be upwards. B net = B M + B N + B P −7 0 I ( 4  10 T m A ) 9.50 A = = 1.0857  10 −4 T BM = BN = 2 rM 2 0.0175 m

BP =

−7 9.50 A 0 I ( 4  10 T m A ) = = 6.268  10 −5 T 2 rP 2 3 ( 0.0175 m )

(

)

(

)

Bx = 2 BM cos 30 + BP cos 60 = 2 1.0857  10−4 T cos 30 + 6.268  10 −5 T cos 60 = 2.194  10−4 T

(

)

(

)

By = −2 BM sin 30 + BP sin 60 = −2 1.0857  10 −4 T sin 30 + 6.268  10 −5 T sin 60 = −5.429  10−5 T 2 2 Bnet = Bnet + Bnet = x y

 net = tan −1

Bnet y Bnet y

= tan −1

( 2.194 10 T ) + ( −5.429 10 T ) = 2.26 10 T −4

−5.429  10−5 T 2.194  10−4 T

−5

−4

= −14

The net field points slightly below the horizontal direction. 57. For the wire to be suspended the net magnetic force must equal the gravitational force. Since the same current flows through the two lower wires, the net magnetic force is the sum of the vertical components of the force from each wire, given by Eq. 28–2. We solve for the unknown current by setting this force equal to the weight of the wire. The mass is found by multiplying the density times the volume of the wire. The density was taken from Table 13–1. I I FM = 2 0 M NP l cos30 =  g ( 14  d 2 l ) 2 r  g 2 rd 2 IM = 4 0 I NP cos30

30o 30o

(8900 kg m )( 9.80 m s ) (0.035m ) (1.00 10 m ) = 198A  2.0 10 A = 4 ( 4  10 T m A ) ( 35.0A ) cos30 3

−3

−7

58. (a) Use Eq. 28–1 to calculate the field due to a long straight wire. 4  10−7 T m A ( 3.0 A ) 0 I A = = 3.333  10−6 T  3.3  10 −6 T BA at B = 2 rA to B 2 ( 0.18 m )

(

)

984

Chapter 28

Sources of Magnetic Field

( 4  10−7 T m A ) ( 6.0 A ) = 6.667  10−6 T  6.7  10−6 T 0 I B = 2 rB to A 2 ( 0.18 m ) (c) The two fields are not equal and opposite. Each individual field is due to a single wire, and has no dependence on the other wire. The magnitude of current in the second wire has nothing to do with the value of the field caused by the first wire. (d) Use Eq. 28–2 to calculate the force due to one wire on another. The forces are attractive since the currents are in the same direction. −7 Fon A due to B Fon B due to A 0 I A I B ( 4  10 T m A ) ( 3.0 A )( 6.0 A ) = = = 2 d A to B 2 lA lB ( 0.18 m ) (b) BB at A =

= 2.0  10−5 N m These two forces per unit length are equal and opposite because they are a Newton’s third law pair of forces. 59. (a) The magnetic field from the long straight wire will be out of the page in the region of the wire loop, and its magnitude is given by Eq. 28–1. By symmetry, the forces from the two horizontal segments are equal and opposite, therefore they do not contribute to the net force. We use Eq. 28–2 to find the force on the two vertical segments of the loop and sum the results to determine the net force. Note that the segment with the current parallel to the straight wire will be attracted to the wire, while the segment with the current flowing in the opposite direction will be repelled from the wire. II II II l1 1  Fnet = F2 + F4 = − 0 1 2 l + 0 1 2 l = 0 1 2  −  2 d 2 2 d1 2  d1 d 2 

( 4 10 T m A ) ( 2.0 A )(10.0 A )( 0.26 m )  −7

1 1  −   0.05 m 0.12 m 

2

= 1.2  10−5 N toward the wire (b) Since the forces on each segment lie in the same plane, the net torque on the loop is zero. 60. The sheet may be treated as an infinite number of parallel wires. The magnetic field at a location y above the wire will be the sum of the magnetic fields produced by each of the wires. If we consider the magnetic field from two wires placed symmetrically on either side of where we are measuring the magnetic field, we see that the vertical magnetic field components cancel each other out. Therefore, the field above the wire must be horizontal and to the left. By symmetry, the field a location y below the wire must have the same magnitude, but point in the opposite direction. We calculate the magnetic field using Ampere’s law with a rectangular loop that extends a distance y above and below the current sheet, as shown in the figure.  B d l =  B d l +  B d l +  B d l = 0 + 2B D = 0 I encl = 0 ( jtD ) sides

top

bottom

→ B = 12 0 jt , to the left above the sheet .

985

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

61. (a) We set the magnetic force, using Eq. 28–2, equal to the weight of the wire and solve for the necessary current. The current must flow in the same direction as the upper current, for the magnetic force to be upward. d2  II l → FM = 0 1 2 l =  g  2 r  4  3 2 2 −3  g 2 rd 2 ( 8900 kg m )( 9.80 m s )  ( 0.050 m ) (1.00  10 m ) I2 = = = 407.8A 40 I1 2 ( 4  10−7 T m A ) ( 42 A ) 2

 410 A, right . (b) The lower wire is in unstable equilibrium, since if it is raised slightly from equilibrium, the magnetic force would be increased, causing the wire to move further from equilibrium. (c) If the wire is suspended above the first wire at the same distance, the same current is needed, but in the opposite direction, as the wire must be repelled from the lower wire to remain in equilibrium. Therefore the current must be 410 A to the left. This is a stable equilibrium for vertical displacement, since if the wire is moved slightly off the equilibrium point the magnetic force will increase or decrease to push the wire back to the equilibrium height. 62. (a) Choose x = 0 at the center of one coil. The center of the other coil will then be at x = R. Since the currents flow in the same direction in both coils, the right-hand rule shows that the magnetic fields from the two coils will point in the same direction along the axis. The magnetic field from a current loop was found in Example 28–11. Adding the two magnetic fields together yields the total field.

B( x ) =

0 NIR 2 2 3/ 2

2  R 2 + x 

0 NIR 2 2 3/ 2

2  R2 + ( x − R )   

(b) Evaluate the derivative of the magnetic field at x = 12 R.

−30 NIR 3 30 NIR 2 x 30 NIR 2 ( x − R ) 30 NIR 3 dB =− − = − − = 0 5/ 2 5/ 2 5/ 2 2 5/ 2 dx 2  R 2 + x 2  4  R 2 + R 2 / 4 4  R 2 + R 2 / 4 2  R2 + ( x − R )    Evaluate the second derivative of the magnetic field at x = 12 R.

d 2B 30 NIR 2 150 NIR 2 x 2 30 NIR 2 150 NIR 2 ( x − R )2 = − + − + 5/ 2 7/2 2 5/ 2 2 7/2 dx 2 2  R 2 + x 2  2  R 2 + x 2  2  R2 + ( x − R )  2  R2 + ( x − R )      =− =

30 NIR 2

150 NIR 4

30 NIR 2

150 NIR 4

2 5R 2 / 4 

8 5R 2 / 4 

2 5R 2 / 4 

8 5R 2 / 4 

+ 5/ 2

− 7/2

+ 5/ 2

7/2

0 NIR 2  3 15  4 5R / 4  2

5/ 2

3 15  4  − + − + = 0  2 85 2 85 

Therefore, at the midpoint

dB d 2B = 0 and = 0. dx dx 2

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Chapter 28

Sources of Magnetic Field

(c) We insert the given data into the magnetic field equation to calculate the field at the midpoint. 0 NIR 2 0 NIR 2 0 NIR 2 + = B ( 12 R ) = 3/ 2 3/ 2 3/ 2 2 2  R 2 + ( 12 R )2  2  R 2 + ( 12 R )  2  R 2 + ( 12 R )       

( 4 10 T m A ) (85)(3.0 A )( 0.10 m ) = 2.3mT to the left. = 2

−7

( 0.10 m )2 + ( 0.05m )2   

3/ 2

63. (a) Use the results of Problem 62(a) to write the magnetic field, and then substitute in the values of R = 0.100 m, N = 280 loops, and I = 2.0 A.   0 NIR 2  1 1  B ( x) = + 2   R 2 + x 2  3/ 2  R 2 + ( x − R )2  3/ 2       

 4  10 T m A ) ( 85 )( 2.0A )( 0.100 m )  ( =

  + 3/ 2 3/ 2   2 2 2 2  R + x   R2 + ( x − R )          1 1  −6 3 = 1.068  10 T m  + 3/ 2 3/ 2  2 2 2 2  0.100 m ) + ( x − 0.100 m )    ( 0.100 m ) + x  (   (b) See the graph. 2

−7

(c) Use the formula to find the % difference. B ( x = 6.0cm ) − B ( x = 5.0cm ) 1.52843mT − 1.52860 mT % diff = (100 ) = (100 ) B ( x = 5.0cm ) 1.52860 mT

= −1.1  10−2 % 64. From Example 28–11, the magnetic field on the axis of a circular loop of wire of radius R carrying 0 IR 2 , where x is the distance along the axis from the center of the loop. current I is B = 3/2 2 R2 + x2

(

)

For the loop described in this problem, we have R = x = REarth .

987

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

0 IR 2

2 R2 + x

)

2 3/ 2

→ I=

= 65.

(

2B R 2 + x 2

0 R 2 ( 2)

3/ 2

)

3/ 2

(

Instructor Solutions Manual

2 2 + REarth 2B REarth

)

3/ 2

0 R

2 Earth

2 ( 2)

3/ 2

BREarth

0

(110 T )( 6.38 10 m ) = 3 10 A −4

4  10 T m A −7

The magnetic field from the wire at the location of the plane is perpendicular to the velocity of the plane since the plane is flying parallel to the wire. We calculate the force on the plane, and thus the acceleration, using Eq. 27–5b, with the magnetic field of the wire given by Eq. 28–1. I F = qv B = qv 0 2 r −3 −7 F qv 0 I (18.0  10 C ) ( 2.8 m/s ) ( 4  10 T m/A ) ( 25A ) a= = = m m 2 r 2 ( 0.175 kg )( 0.078 m )   1g = 1.846  10−5 m/s 2  = 1.9  10−6 g’s 2   9.80 m s 

66. The magnetic field at the center of the square is the vector sum of the magnetic field created by each current. Since the magnitudes of the currents are equal and the distance from each corner to the center is the same, the magnitude of the magnetic field from each wire is the same and is given by Eq. 28–1. The direction of the magnetic field is directed by the right-hand rule and is shown in the diagram. By symmetry, we see that the vertical components of the magnetic field cancel and the horizontal components add.  I  B = B1 + B 2 + B3 + B 4 = −4  0 0  cos 45ˆi  2 r 

   I  2 0 0 ˆi = − 2 0 I 0 ˆi = −4   d  2 2 d  2   2   67. The wire can be broken down into five segments; the two long wires, the left vertical segment, the right vertical segment, and the top horizontal segment. Since the current in the two long wires either flow directly toward or away from point P, they will not contribute to the magnetic field at P. The magnetic field from the top horizontal segment points into the page and is obtained from the solution I1 to Problem 43. I 0 I a Btop = 0 1 = 2 2 a ( a 2 + 4a 2 ) 2 a 5

a I3

a I4

The magnetic fields from the two vertical segments both point into the page with magnitudes obtained from the solution to Problem 44.

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Chapter 28

Sources of Magnetic Field

Bvert =

0 I

4 ( a / 2 ) a 2 + a / 2 2

(

))

1 2

0 I a 5

Summing the magnetic fields from all the segments yields the net field.

B = Btop + 2 Bvert =

0 I I I 5 , into the page . +2 0 = 0 2 a 2 a 5 a 5

68. Use Eq. 28–4 for the field inside a solenoid. −7  IN ( 4  10 T m A ) ( 2.0 A )( 370 ) = = 7.7  10−3 T B= 0 l 0.12 m 69. Since the mass of copper is fixed and the density is fixed, the volume of copper is fixed, and we designate it as VCu = mCu  Cu = l Cu ACu . We call the fixed voltage V0 . The magnetic field in the solenoid is given by Eq. 28–4. For the resistance, we used the resistivity-based definition from Eq. 25–3, with resistivity represented by  R Cu .

0 NI  V N ACu 0V0 N mCu  Cu N V0 N V0 = 0 = 0 = 0 0 = 2 l Cu  R Cu l sol l Cu  R Cu l sol l Cu l sol l sol RCu l sol  R Cu ACu 0V0mCu  Cu N 2  R Cu l sol l Cu

The number of turns of wire is the length of wire divided by the circumference of the solenoid. l Cu l Vm   V m  2 rsol 0V0mCu  Cu 1 N → B = 0 0 Cu Cu = 0 0 Cu Cu = N = Cu 2 2 l sol l Cu l sol l Cu  R Cu  R Cu 2 rsol 2 R Cu l sol rsol l Cu The first factor in the expression for B is made of constants, so we have B 

1 l sol rsol l Cu

. Thus, we

want the wire to be short and thick, so that its length is small. This minimizes the resistance, which increases the current. The radius of the solenoid should be small and the length of the solenoid also small. 70. The magnetic field inside the smaller solenoid will equal the sum of the fields from both solenoids. The field outside the inner solenoid will equal the field produced by the outer solenoid only. We set the sum of the two fields given by Eq. 28–4 equal to − 12 times the field of the outer solenoid and solve for the ratio of the turn density. n 2 0 ( − I ) na + 0 Inb = − 12 ( 0 Inb ) → b = na 3 71. Take the origin of coordinates to be at the center of the semicircle. The magnetic field at the center of the semicircle is the vector sum of the magnetic fields from each of the two long wires and from the semicircle. By the right-hand rule each of these fields point into the page, so we can sum the magnitudes of the fields. The magnetic field for each of the long segments is obtained by integrating Eq. 28–5 over the straight segment.

989

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Bstraight = 

Instructor Solutions Manual

( ˆ ˆ)

0 I d l  Rˆ 0 I d l  R 0 I 0 dx i  − xi − rj 0 Ir ˆ 0 dx = = = − k 3/ 2 2 3   2 2 2 4 R 4 R 4 − ( x + r ) 4 − ( x + r 2 )3 / 2 0

 Ir x = − 0 kˆ 4 r 2 ( x 2 + r 2 )1/ 2

=−

0 I ˆ k 4 r

−

The magnetic field for the curved segment is obtained by integrating Eq. 28–5 over the semicircle.  I  r d l  rˆ 0 I ˆ  r I B curve = 0  = − k  ds = − 0 kˆ 2 2 4 0 r 4 r 0 4r Add all the contributions to find the total magnetic field. I I I B = 2Bstraight + Bcurve = −2 0 kˆ − 0 kˆ = − 0 ( 2 +  ) kˆ 4 r 4r 4 r −7 ( 4 10 T m A ) ( 9.0A ) ( 2 +  ) kˆ = −9.3  10−5 T kˆ =− 4 ( 0.050 m ) 72. (a) Set x = 0 at the midpoint on the axis between the two loops. Since the loops are a distance R apart, the center of one loop will be at x = − 12 R and the center of the other at x = 12 R. The currents in the loops flow in opposite directions, so by the right-hand rule the magnetic fields from the two wires will subtract from each other. The magnitude of each field can be obtained from Example 28–11. 0 NIR 2 0 NIR 2 B( x ) = − 2 3/ 2 2 3/ 2 2  R 2 + ( 12 R − x )  2  R 2 + ( 12 R + x )      3 1 Factoring out 8 R from each of the denominators yields the desired equation.

B( x) =

4 0 NI R  4 + (1 − 2 x / R )    2

3/ 2

−

4 0 NI R  4 + (1 + 2 x / R )    2

3/ 2

−3/ 2 2 −3/ 2   2 x  2   4 0 NI    2 x   = −  4 + 1 +      4 + 1 −   R    R   R        2

4x  2x  (b) For small values of x, we can use the approximation 1    1  . R R  −3/2 −3/2 4  NI   4x  4 x    B( x ) = 0  4 + 1 −  − 4 + 1 +   R   R R    −3/2 −3/2 4 0 NI   4 x   4 x   = − 1 +  1−  5 5 R   5R   5R  

4x   Again we can use the expansion for small deviations 1    5R  4  NI  6 x   6 x   480 NIx B( x ) = 0  1 +  − 1 −  = 5 5 R  5R   5R   25 5 R 2

−3/ 2

1

6x 5R

990

Chapter 28

Sources of Magnetic Field

This magnetic field has the expected linear dependence on x with a coefficient of C = 480 NI / 25R 2 5 .

(

(c)

)

Set C equal to 0.15 T/m and solve for the current. 2 25CR 2 5 25 ( 0.15 T/m )( 0.04 m ) 5 = = 1.5 A I= 480 N 48 ( 4  10−7 T  m/A ) (150 )

73. Example 25–10 estimates the current in a lightning bolt at 100 A. Use Eq. 27–2, with the Earth’s magnetic field. We estimate the flag pole as being 6 m tall. F = I l B = (100 A )( 6 m ) ( 0.5  10−4 T ) = 0.03 N 74. We approximate the magnetic field by using Eq. 28–1. The current is found from Eq. 25–6. −7 6 I  P ( 4  10 T m/A ) ( 45  10 W ) P I= → B= 0 = 0 = = 5.172  10−6 T  5.2  10 −6 T 3 2 r 2 r V 2 (12 m ) V 145 10 V  ( )

Bwire 5.172  10−6 T = = 0.103  10% BEarth 5  10−5 T The power line is almost certainly AC, and so the voltage, power, current, and the magnitude of the magnetic field are most likely rms values. 75. We use the results of Example 28–11 to calculate the magnetic field as a function of position.

N 0 IR 2

2 ( R2 + x2 )

= 3/2

( 250) ( 4 10−7 T m/A ) ( 2.0A )( 0.15m ) 2 ( 0.15m ) + x 2    2

3/2

7.0686  10−6 T m3 ( 0.15m )2 + x 2   

3/2

76. (a) To find the length of wire that will give the coil sufficient resistance to run at maximum power, we write the power equation (Eq. 25–7b) with the resistance given by Eq. 25–3. We divide the length by the circumference of one coil to determine the number of turns. V2 V2 V 2d 2 l Pmax = = → = R  l (d 2 )  Pmax

( 35V ) ( 2.0  10−3 m ) l V 2d 2 N= = = = 31turns  D  D Pmax  ( 2.0 m ) (1.68  10−8  m )(1.5  103 W ) 2

991

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) We use the result of Example 28–11 to determine the magnetic field at the center of the coil, with the current obtained from Eq. 25–6. −7 3  NI 0 N Pmax ( 4  10 T m/A ) ( 31) (1.5  10 W ) = = = 8.3  10−4 T B= 0 2.0 m 35V D D V ( )( ) (c) Increasing the number of turns will proportionately increase the resistance and therefore decrease the current. The net result is no change in the magnetic field. 77. There are three forces on each wire; its weight, the magnetic force of repulsion from the other wire, and the tension in the attached string. See the diagram. The magnetic force is given by Eq. 28–2. The mass of the wire is its density times its volume. The length of the current-carrying wires is l . The net force in both the vertical and horizontal directions is zero. We use d to represent the horizontal distance between the current-carrying wires. mg → FB = FT sin  ; FT cos  = mg → FT = cos  mg sin  = mg tan  =  Al rAl2 l g tan  FB = cos  FB =

 F T

0 I 2  I2 l → 0 l =  Al rAl2 l g tan  → 2 d 2 d 2 d  Al rAl2 g tan 

0

(

) (

2  2 ( 0.50 m ) sin 3o  2700 kg m 3  2.1  10 −4 m

( 4 10 T m A )

) (9.8 m s ) tan 3.0 2

−7

= 7.1A 78. There will be no force on either the top or bottom part of the wire, because the current is either parallel to or opposite to the magnetic field. So the only force is on the left branch. Since the current is perpendicular to the magnetic field, we use Eq. 27–2. The magnetic field can be calculated by Eq. 28–4 for the magnetic field inside a solenoid. By the right-hand rule, the force on the left branch is up out of the paper. We take the 600 turns as having at least 2 significant figures.  ( 4  10−7 T m A ) ( 600 )( 7.0 A )   0 NI solenoid  F = I wire l wire B = I wire l wire    = ( 5.0 A )( 0.125 m )  0.15 m  l solenoid    = 2.2  10−2 N

79. (a) The magnetic field at the center of the loop is the vector sum of the field due to the long straight wire plus the field due to the loop. Using the right-hand rule, both fields point along the z-axis. The field due to the wire is given by Eq. 28–1, and the field due to the loop is given in the Note at the top of page 845 in the textbook.

0 I ˆ 0 I ˆ I k+ k = 0 (1 +  ) kˆ 2 R 2R 2 R (b) We assume that the loop is twisted so that the left side of the original loop comes “out” of the page, and the right side of the original loop goes “into” the page. Then the field due to the loop points along the x-axis, while the field due to the long wire is unchanged. B = B wire + B loop =

992

Chapter 28

Sources of Magnetic Field

B = B wire + B loop =

0 I ˆ 0 I ˆ 0 I k+ i= ( ˆi + kˆ ) 2 R 2R 2 R

0 I 1 +  2 . The angle of inclination is 2 R calculated from the two components. See the diagram. 1  = tan −1 = 17, from the x-axis towards the z -axis .  The magnitude is

80. (a) From Example 28–9, the field inside the windings of the toroid are given by B =

0 NI . 2 r

−7 0 NI ( 4  10 T m A ) ( 320 )( 45 A ) = = 9.505  10−4 T  9.5  10−4 T B= 2 r 2 ( 3.030 m ) (b) Using the right-hand rule for the direction of current due to a current loop, the magnetic field is pointing counterclockwise as seen in Fig. 28–69(a). (c) The resistivity can be found from Eq. 25–3 combined with Ohm’s Law. The length of the wire is the perimeter of one loop, times the number of loops. V l R= = → I A 0.10  10−4 m 2 ) ( V A ( 20.0 V ) = = 5.787  10−8  m  5.8  10−8  m = −2 45 A ( 320 )( 4 ) ( 6.0  10 m ) I l

Looking at Table 25–1, the wire might be made of tungsten, which is a commonly used material for conducting wires.

993

CHAPTER 29: Electromagnetic Induction and Faraday’s Law Responses to Questions 1.

The advantage of using many turns (N = large number) in Faraday’s experiments is that the emf and induced current are proportional to N, which makes it easier to experimentally measure those quantities. Many turns in the primary coil will make a larger magnetic flux, and many turns in the output coil will produce a larger output voltage and current.

Magnetic flux is proportional to the total amount of magnetic field passing through the area of a loop:  B = BA cos  . The magnetic flux depends not only on the field itself, but also on the area and on the angle between the field and the area. Thus, they also have different units (magnetic field = Tesla = T; magnetic flux = T m2 = Wb). Another difference is that magnetic field is a vector, but magnetic flux is a scalar. The flux can be thought of as being proportional to the total number of magnetic field lines passing through an area enclosed by a loop.

(a) A current is induced in the ring when you move the south pole toward the ring. An emf and current are induced in the ring due to the changing magnetic flux. As the magnet gets closer to the ring, more magnetic field lines are going through the ring. Using Lenz’s law and the righthand rule, the direction of the induced current when you bring the south pole toward the ring is clockwise. In this case, the number of magnetic field lines coming through the loop and pointing toward you is increasing (remember, magnetic field lines point toward the south pole of the magnet). The induced current in the loop will oppose this change in flux and will attempt to create magnetic field lines through the loop that point away from you. A clockwise induced current will provide this opposing magnetic field. (b) A current is not induced in the ring when the magnet is held steady within the ring. An emf and current are not induced in the ring since the magnetic flux through the ring is not changing while the magnet is held steady. (c) A current is induced in the ring when you withdraw the magnet. An emf and current are induced in the ring due to the changing magnetic flux. As you pull the magnet out of the ring toward you, fewer magnetic field lines are going through the ring. Using Lenz’s law and the right-hand rule again, the direction of the induced current when you withdraw the south pole from the ring is counterclockwise. In this case, the number of magnetic field lines coming through the loop and pointing toward you is decreasing. The induced current in the loop will oppose this change in flux and will create more magnetic field lines through the loop that point toward you. A counterclockwise induced current will provide this opposing magnetic field.

The current in Fig. 29–12a will flow counterclockwise. If the area of the loop decreases, the flux through the loop (directed out of the page) decreases. By Lenz’s law, the resulting induced current will be counterclockwise to oppose the change. Another way to approach this question is to use the right-hand rule. As the bar moves to the left, the negative electrons in the bar will experience a force down, which results in a counterclockwise current.

(a) There will be no magnetic field lines “piercing” the loop from top to bottom or vice-versa. All of the field lines are parallel to the face of the loop. Thus, there will be no magnetic flux in the loop, and no change of flux in the loop, so there will be no induced current. (b) Now, since the magnet is much thicker than the loop, there will be some field lines that pierce the loop from top to bottom. The flux will increase as the magnet gets closer, so a current will be induced that makes upward-pointing field lines through the loop. The current will be counterclockwise.

994

Chapter 29

Electromagnetic Induction and Faraday’s Law

There will be an attractive force between the two loops. As the right loop moves away, the “inward” flux in it due to the magnetic field of the long wire will be decreasing. Thus by Faraday’s law, current will flow clockwise in the right loop. This induced current in the right loop will create a magnetic field which will be coming up out of the page at the location of the left loop, but will be weaker over time because the right loop is getting farther away. Thus there will be an induced counterclockwise current in the left loop, For the parts of the loops that are the closest to each other, the current is parallel, and so the two loops will attract each other.

With the current I in the wire decreasing, the upward flux through the left loop is decreasing, which induces a counterclockwise current in the left loop. Also, with the current I decreasing, the current in the right hand loop will be even stronger, because current is induced to compensate both for the “moving away” and for the “decreasing I.” Thus, both currents in the loops are now stronger than in Question 6, but in the same direction, and so the force will still be attractive, but larger than in Question 6. There is also now a flux in the right-hand ring due to the flux in the left ring, which also will be decreasing, again causing the current in the right-hand ring to increase, providing even more attractive force.

(a) Yes. As current starts to flow in the front loop, it will create an increasing magnetic field that points away from you and down through the two loops. Because the magnetic flux will be increasing in the second loop, an emf and current will be induced in the second loop. (b) The induced current in the second loop starts to flow as soon as the current in the front loop starts to increase and create a magnetic field (basically, immediately upon the connection of the battery to the front loop). (c) The current in the second loop stops flowing as soon as the current in the front loop becomes steady. Once the battery has increased the current in the front loop from zero to its steady-state value, then the magnetic field it creates is also steady. Since the magnetic flux through the second loop is no longer changing, the induced current goes to zero. (d) The induced current in the second loop is counterclockwise. Since the increasing clockwise current in the front loop is causing an increase in the number of magnetic field lines down through the second loop, Lenz’s law states that the second loop will attempt to oppose this change in flux. To oppose this change, the right-hand rule indicates that a counterclockwise current will be induced in the second loop. (e) Yes. Since both loops carry currents and create magnetic fields while the current in the front loop is increasing from the battery, each current will “feel” the magnetic field caused by the other loop. (f) The force between the two loops will repel each other. The front loop is creating a magnetic field pointed toward the second loop. This changing magnetic field induces a current in the second loop to oppose the increasing magnetic field, and this induced current creates a magnetic field pointing toward the front loop. These two magnetic fields will act like two north poles pointing at each other and repel. We can also explain it by saying that the two currents are in opposite directions, and opposing currents exert a repelling force on each other.

Yes, a current will be induced in the second loop. It will start when the current in the first loop begins to decrease, and will stop when the current falls to zero in the first coil. The current in the second loop will now be clockwise. All of the reasoning is similar to that given for Question 6, except now the current is decreasing instead of increasing.

10. (a) The induced current in RA is to the right as coil B is moved toward coil A. As B approaches A, the magnetic flux through coil A increases (there are now more magnetic field lines in coil A pointing to the left). The induced emf in coil A creates a current to produce a magnetic field © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

that opposes this increase in flux, with the field pointing to the right through the center of the coil. A current through RA to the right will produce this opposing field. (b) The induced current in RA is to the left as coil B is moved away from coil A. As B recedes from A, the magnetic flux through coil A decreases (there are now fewer magnetic field lines in coil A pointing to the left). The induced emf in coil A creates a current to produce a magnetic field that opposes this decrease in flux, pointing to the left through the center of the coil. A current through RA to the left will produce this opposing field. (c) The induced current in RA is to the left as RB in coil B is increased. As RB increases, the current in coil B decreases, which also decreases the magnetic field coil B produces. As the magnetic field from coil B decreases, the magnetic flux through coil A decreases (there are now fewer magnetic field lines in coil A pointing to the left). The induced emf in coil A creates a current to produce a magnetic field pointing to the left through the center of the coil, opposing the decrease in flux. A current through RA to the left will produce this opposing field. 11. One advantage of placing the two insulated wires carrying AC close together is that whatever magnetic fields are created by the changing current moving one way in one wire is approximately cancelled out by the magnetic field created by the current moving in the opposite direction in the second wire. Also, since large loops of wire in a circuit can generate a large self-induced back emf, by placing the two wires close to each other, or even twisting them about each other, the effective area of the current loop is decreased and the induced current is minimized. 12. The net current in the wire and shield combination is 0. Thus to the “outside,” there is no magnetic field created by the combination. If there is an external magnetic field, it will not be influenced by the signal current, and then by Newton’s third law, the signal current will not be influenced by an external magnetic field either. 13. At the moment shown in Fig. 29–15, the armature is rotating clockwise and the magnetic flux through the loop is decreasing. The induced current in the loop will flow “into” the paper at point a and out of the paper at point b. This induced current is (obviously) in a magnetic field, and so there will be a force on that current. That force will be upwards on the “upper right” part of the armature coil, and downwards on the “lower left” part of the coil. This produces a counterclockwise torque, which opposes the motion. 14. Higher voltages, such as 600 V or 1200 V, would be dangerous if they were used in household wires. Having a larger voltage than the typical 120 V would mean that any accidental contact with a “live” wire would send more current through a person’s body. Such a large potential difference between household wires and anything that is grounded (other wires, people, etc.) would more easily cause electrical breakdown of the air and then much more sparking would occur. Basically, this would supply each of the charges in the household wires with much more energy than the lower voltages, which would allow them to arc to other conductors. This would increase the possibility of more short circuits and accidental electrocutions. 15. (a) To determine which leads are paired with which, you could use an ohmmeter, since the two source wires are in no way electrically connected to the two output wires. If the resistance between two wires is very small, then those two wires are a pair. If the resistance between two wires is infinite, then those two wires are not a pair. (b) To determine the ratio of turns on the two coils of a transformer without taking it apart, apply a known AC input source voltage to one pair of leads and carefully measure the output voltage

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across the other two leads. Then, Vsource/Voutput = Ns/No, which provides us with the ratio of turns on the two coils. 16. When 120 V DC is applied to the transformer, there is no induced back emf that would usually occur with 120 V AC. This means that the 120 V DC encounters much less resistance than the 120 V AC, resulting in too much current in the primary coils. This large amount of current could overheat the coils, which are usually wound with many loops of very fine, low-resistance wire, and could melt the insulation and burn out or short out the transformer. 17. When the motor first starts up, there is only a small back emf in the circuit (back emf is proportional to the rotation speed of the motor). This allows a large current to flow to the refrigerator. The power source for the house can be treated as an emf with an internal resistance. This large current to the refrigerator motor from the power source reduces the voltage across the power source because of its internal resistance. Since the power source voltage has decreased, other items (like lights) will have a lower voltage across them, receiving less current, and so may “dim.” As the motor speeds up to its normal operational speed, the back emf increases to its normal level and the current delivered to the motor is now limited to its usual amount. This current is no longer enough to significantly reduce the output voltage of the power source, and so the other devices then get their normal voltage. Thus, the lights flicker just when the refrigerator motor first starts up. A heater, on the other hand, draws a large amount of current (it is a very low-resistance device) at all times. (The heat-producing element is not a motor, and so has very little induction associated with it.) The source is then continually delivering a large current, which continually reduces the output voltage of the power source. In an ideal situation, the source could provide any amount of current to the whole circuit in either situation. In reality, though, the higher currents in the wires causes larger losses of energy along the way to the devices, and so the lights dim. 18. As the aluminum sheet is moved through the magnetic field, eddy currents are created in the sheet. The magnetic force on these induced currents opposes the motion. Thus, it requires some force to pull the sheet out. (See Fig. 29–21.) 19. The speed of the magnet in case (b) will be larger than that in case (a). As the bar magnet falls through the loop it sets up an induced current in the loop, which opposes the change in flux. This current acts like a magnet that is opposing the physical magnet, repelling it, and so reduces its speed. Then, after the midpoint of the magnet passes through the loop, the induced current will reverse its direction and attract the falling magnet, again reducing its speed. 20. Eddy current brakes will work on metallic wheels, such as copper and aluminum. Eddy current brakes do not need to act on ferromagnetic wheels. The external magnetic field of the eddy brake just needs to interact with the “free” conduction electrons in the metal wheels in order to have the braking effect. First, the magnetic field creates eddy currents in the moving metal wheel using the free conduction electrons (the right-hand rule says moving charges in a magnetic field will experience a magnetic force, making them move, and creating an eddy current). This eddy current is also in the braking magnetic field. The right-hand rule says these currents will experience a force opposing the original motion of the piece of metal and the eddy current brake will begin to slow the wheel. Good conductors, such as copper and aluminum, have many free conduction electrons and will allow large eddy currents to be created, which in turn will provide good braking results. 21. The nonferrous materials are not magnetic but they are conducting. As they pass by the permanent magnets, eddy currents will be induced in them. The eddy currents provide a “braking” mechanism which will cause the metallic materials to slide more slowly down the incline than the nonmetallic materials. The nonmetallic materials will reach the bottom with larger speeds. The nonmetallic materials can therefore be separated from the metallic, nonferrous materials by placing bins at © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

different distances from the bottom of the incline. The closest bin will catch the metallic materials, since their projectile velocities off the end of the incline will be small. The bin for the nonmetallic materials should be placed farther away to catch the higher-velocity projectiles. 22. The slots in the metal bar prevent the formation of large eddy currents, which would slow the bar’s fall through the region of magnetic field. The smaller eddy currents then experience a smaller opposing force to the motion of the metal bar. Thus, the slotted bar falls more quickly through the magnetic field. 23. As the metal bar enters (or leaves) the magnetic field during the swinging motion, areas of the metal bar experience a change in magnetic flux. This changing flux induces eddy currents with the “free” conduction electrons in the metal bar. These eddy currents are then acted on by the magnetic field and the resulting force opposes the motion of the swinging metal bar. This opposing force acts on the bar no matter which direction it is swinging through the magnetic field, thus damping the motion during both direction of the swing. 24. As a magnet falls through a metal tube, an increase in the magnetic flux is created in the areas ahead of it in the tube. This flux change induces a current to flow around the tube walls to create an opposing magnetic field in the tube (Lenz’s law). This induced magnetic field pushes against the falling magnet and reduces its acceleration. The speed of the falling magnet increases until the magnetic force on the magnet is the same size as the gravity force. The opposing magnetic field cannot cause the magnet to actually come to a stop, since then the flux would become a constant and the induced current would disappear, as would the opposing magnetic field. Thus, the magnet reaches a state of equilibrium and falls at a constant terminal velocity. The weight of the magnet is balanced by the upward force from the eddy currents. 25. Although in principle you could use a loudspeaker in reverse as a microphone, it would probably not work in actual practice. The membrane of the microphone is very lightweight and sensitive to the sound waves produced by your voice. The cardboard cone of a loudspeaker is much stiffer and would significantly dampen the vibrations so that the frequency of the impinging sound waves would not be translated into an induced emf with the same frequency.

Responses to MisConceptual Questions 1.

(b, d) The right-hand rule shows that a clockwise current will create a flux in the loop that points into the page. Since the initial magnetic flux is into the page, the clockwise current will be induced when the magnetic flux decreases. This happens when the size of the coil decreases, or the magnetic field becomes tilted. Increasing the magnetic field, as in answer (a), will create a counterclockwise current. Moving the coil sideways, as in answer (c), does not change the flux, so no current would be induced.

(c) A common misconception is that a moving loop would experience a change in flux. However, if the loop is moving through a constant field without rotation, the flux through the loop will remain constant and no current will be induced.

(d) A current is induced in the loop when the flux through the loop is changing. As the loop passes through J it enters a region with a magnetic field, so the flux through the loop increases and a counterclockwise current will be induced. When the loop passes K, the flux remains constant as there is no change in field, so no current is induced. As the loop passes L, the magnetic field flux through the loop decreases and a current is again induced, but this time it will be clockwise.

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(c, b) The magnetic field near a long straight wire is inversely proportional to the distance from the wire. For C, each part of the loop remains at a constant distance from the wire, so the magnetic flux through the wire remains constant and no current is induced in the loop. For D, the magnetic field from the long wire points out of the page in the region of the loop. As the loop moves away from the wire the magnetic flux decreases, so a counterclockwise current is induced in the loop to oppose the change in flux.

(c) The induced current creates a magnetic field that opposes the motion of the magnet. The result is that the magnet has less kinetic energy than it would have had if the loop were not present. The change in gravitational potential energy provides the energy for the current.

(c) Since the flux through the loop is increasing, an emf will be produced. However, since plastic is not a conductor, no current will be induced. The emf is produced regardless of whether there is a conducting path for current or not.

(b) A current will be induced in the loop whenever the magnetic flux through the loop is changed. Increasing the current in the wire will increase the magnetic field produced by the wire and therefore the magnetic flux in the loop. Rotating the loop changes the angle between the loop and wire, which will change the flux. Since the magnetic field strength decreases with distance from the wire, moving the loop away from the wire, either with or without rotation, will decrease the flux in the wire. If the loop is moved parallel to the wire, the flux through the loop does not change, so no current is created in the loop.

(b) A generator converts mechanical energy into electric energy. The generator’s magnetic field remains unchanged as the generator operates, so energy is not being pulled from the magnetic field. Resistance in the coils removes electric energy from the system in the form of heat–it does not provide the energy. In order for the generator to work, an external force is necessary to rotate the generator’s axle. This external force does work, which is converted to electric energy.

(c) Increasing the rotation frequency increases the rate at which the flux changes and therefore answer (a) will increase the output voltage. Answers (b), (d), and (e) all increase the maximum flux in the coil. Since the generator’s voltage output is proportional to the rate of change of the flux through the coil, increasing the maximum flux results in a greater voltage output. When the magnetic field is parallel to the generator’s rotation axis, there will be little to no change in the flux passing through the coil. With no flux change, there will be no output voltage.

10. (c) When a steady current flows in the first coil it creates a constant magnetic field, and therefore a constant magnetic flux through the second coil. Since the flux is not changing, a current will not be induced in the second coil. If the current in the first coil changes, the flux through the second will also change, inducing a current in the second coil. 11. (d) A common misconception is that a transformer only changes voltage. However, power is conserved across an ideal transformer, where power is the product of the voltage and current. When a transformer increases the voltage it must also proportionately decrease the current, and vice-versa. 12. (b) Many people may not realize that generators (rotating coils in a magnetic field) are the heart of most electric power plants that produced the alternating current in the wall outlets. 13. (a) In a transformer with no lost flux, the power across the transformer (product of current and voltage) is constant across the transformer. Therefore, if the voltage increases, the current must © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

decrease across the transformer. If the current increases, the voltage must decrease across the transformer. A transformer works due to the induced voltage created by the changing flux. A DC circuit does not have a changing flux, so a transformer does not work with DC current.

Solutions to Problems 1.

The average induced emf is given by Eq. 29–2b. 48 Wb − ( −68 Wb )  B dB e = −N = −N = −2 = −550 V t dt 0.42 s A direction for the emf cannot be specified without more information.

As the coil is pushed into the field, the magnetic flux through the coil increases into the page. To oppose this increase, the flux produced by the induced current must be out of the page, so the induced current is counterclockwise .

The flux changes because the loop rotates. The angle between the field and the normal to the loop changes from 0o to 90o. The average induced emf is given by the “difference” version of Eq. 29–2b.  B

eavg = −

t

=−

AB cos  t

 ( 0.110 m ) 1.5T ( cos 90o − cos 0o ) 2

=−

 ( 0.110 m ) 1.5T ( 0 − 1)

0.35s

=−

0.35s

= 0.16 V

(a) There is an emf induced in the coil since the flux through the coil changes. The current in the coil is the induced emf divided by the resistance of the coil. The resistance of the coil is found from Eq. 25–3. l dB e = NAcoil R= dt Awire

e R

dB dt = NAcoil Awire dB l l dt Awire

NAcoil

(

) (

2 2 36  ( 0.110 m )   1.3  10 −3 m  8.65  10 −3 T s



(1.68  10  m ) 36 ( 2 )( 0.110 m ) −8

)

= 0.1504 A  0.15 A

(b) The rate at which thermal energy is produced in the wire is the power dissipated in the wire. −8 2 1.68  10  m 36 ( 2 )( 0.110 ) 2 2 l P=I R=I = ( 0.1504 A ) = 1.8  10−3 W 2 − 3 Awire  1.3  10 m

(

)

(a) When the plane of the loop is perpendicular to the field lines, the flux is given by the maximum of Eq. 29–1a.  B = BA = B r 2 = ( 0.65T )  ( 0.080 m ) = 1.3  10−2 Wb 2

(b) The angle is  = 55o

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We choose up as the positive direction. The average induced emf is given by the “difference” version of Eq. 29–2a.

e=−

 B

=−

t

AB t

 ( 0.054 m ) ( −0.25T − 0.48T ) 2

=−

0.14 s

= 4.8  10−2 V

As the solenoid is pulled away from the loop, the magnetic flux to the right through the loop decreases. To oppose this decrease, the flux produced by the induced current must be to the right, so the induced current is counterclockwise as viewed from the right end of the solenoid.

(a) As the resistance is increased, the current in the outer loop will decrease. Thus the flux through the inner loop, which is out of the page, will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux inside the inner loop that is out of the page, so the direction of the induced current will be counterclockwise . (b) If the small loop is placed to the left, the flux through the small loop will be into the page and will decrease. To oppose this decrease, the induced current in the inner loop will produce a flux into the page, so the direction of the induced current will be clockwise .

(a) The magnetic flux through the loop is into the paper and decreasing, because the area is decreasing. To oppose this decrease, the induced current in the loop will produce a flux into the paper, so the direction of the induced current will be clockwise . (b) The average induced emf is given by the “difference” version of Eq. 29–2b.

eavg =

 B t

B A t

( 0.65T )  ( 0.100 m )2 − ( 0.040 m )2  0.50s

= 3.431  10−2 V  3.4  10−2 V (c) We find the average induced current from Ohm’s law. e 3.431  10−2 V I= = = 1.4  10−2 A R 2.5  10. As the loop is pulled from the field, the flux through the loop decreases, causing an induced EMF whose magnitude is given by Eq. 29–3, e = Bl v. Because the inward flux is decreasing, the induced flux will be into the page, so the induced current is clockwise, given by I = e R . Because this current in the left-hand side of the loop is in a downward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, given by F = I l B. F = IlB =

e R

lB =

Bl v R

lB =

B2 l 2v R

( 0.650 T ) ( 0.350 m ) ( 3.40 m s ) 2

0.250 

= 0.704 N

11. (a) Because the current is constant, there will be no change in flux, so the induced current will be zero . (b) The decreasing current in the wire will cause a decreasing field into the page through the loop. To oppose this decrease, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(c) The decreasing current in the wire will cause a decreasing field out of the page through the loop. To oppose this decrease, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise . (d) The increasing current in the wire will cause an increasing field out of the page through the loop. To oppose this increase, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise . 12. Use Eq. 29–2a to calculate the emf. Setting the flux equal to the magnetic field multiplied by the area of the loop, A =  r 2, and the emf equal to zero, we can solve for the rate of change in the coil radius. dB d dB dr e=− = − ( B r 2 ) = −  r 2 − 2 Br = 0 dt dt dt dt 0.12 m dr dB r =− = − ( −0.010T/s ) = 0.0015m/s = 1.5 mm/s dt dt 2 B 2 ( 0.400T ) 13. (a) Use Eq. 29–2b to calculate the emf. dB d = ( −75)  9.6t − 0.51t 3  10 −2 T m 2  = −7.2 + 1.1475t 2 V e = −N dt dt

(

)

(

)

 −7.2 + 1.1t 2 V (b) Evaluate at the specific times.

( ) e ( t = 4.0s ) = ( −7.2 + 1.1475 ( 4.0 ) ) V = 11V

e ( t = 1.0s ) = −7.2 + 1.1475 (1.0 ) V = −6.1V 2

14. The energy dissipated in the process is the power dissipated by the resistor, times the elapsed time that the current flows. The average induced emf is given by the “difference” version of Eq. 29–2a. e2  B ; P= ; e=− R t 2

  B  t = A ( B ) =  ( 0.125m )  ( 0.40 T ) = 2.4  10−5 J E = P t = t =   R Rt (135  )( 0.12 s )  t  R 2

15. The induced emf is given by Eq. 29–2a. Since the field is uniform and is perpendicular to the area, the flux is simply the field times the area. dB dA e=− = −B = − ( 0.35T ) −3.50  10−2 m 2 s = 1.2  10 −2 V dt dt Since the area changes at a constant rate, and the area has not shrunk to 0 at t = 2.00 s, the emf is the same for both times.

(

)

16. The induced emf is given by Eq. 29–2a. Since the field is uniform and is perpendicular to the area, the flux is simply the field times the area of a circle. We calculate the initial radius from the initial area. To calculate the radius after one second we add the change in radius to the initial radius. d  r2 A0 dB dr e (t ) = − = −B = −2 Br A0 =  r02 → r0 =  dt dt dt

(

)

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Chapter 29

Electromagnetic Induction and Faraday’s Law

e (0) = −2 ( 0.35 T )

0.285 m 2



( 0.0375 m s ) = −2.5  10−2 V

 0.285 m 2  + ( 0.0375 m s )(1.00 s )  ( 0.0375 m/s ) = −2.8  10 −2 V   

e (1.00 s) = −2 ( 0.35 T ) 

17. The charge that passes a given point is the current times the elapsed time, Q = I t. The current will l e , and be the emf divided by the resistance, I = . The resistance is given by Eq. 25–3, R = Awire R the emf is given by the “difference” version of Eq. 29–2a. Combine these equations to find the charge during the operation. Aloop B Aloop Awire B  B Aloop B l e t = = ; R= ; I= = e = l t t Awire R  l t

Awire Q = I t =

1 2

Aloop Awire B

l

r r 2 loop

2 wire

B

 ( 2 ) rloop

2 B rloop rwire

( 0.132 m )  (1.125  10−3 m ) ( 0.760 T )

2

(

2 1.68  10−8  m

)

= 5.94 C

18. The magnetic field inside the solenoid is given by Eq. 28–4, B = 0 nI . Use Eq. 29–2a to calculate the induced emf. The flux causing the emf is the flux through the small loop. dB dB dI e=− = − A1 solenoid = − A10n = − A10n ( − I 0 sin  t ) = A10n I 0 sin  t dt dt dt 19. (a) If the magnetic field is parallel to the plane of the loop, no magnetic flux passes through the loop at any time. Therefore, the emf and the current in the loop are zero. (b) When the magnetic field is perpendicular to the plane of the loop, we differentiate Eq. 29–1a with respect to time to obtain the emf in the loop. Then we divide the emf by the resistance to calculate the current in the loop.  1 d e 1  dB  I = = − ( t )( A0 +  t ) = −  A0 + 2  t  =− R R  dt  R dt R

=−

( 0.60 T/s ) ( 0.50 m2 ) + 2 ( 0.70 m2 /s ) ( 2.0 s )

= −0.99 A 2.0  Since the magnetic field is pointing down into the page, the downward flux is increasing. The current then flows in a direction to create an upward flux. The resulting current is then 0.99 A in the counterclockwise direction. 20. The sinusoidal varying current in the power line creates a sinusoidal varying magnetic field encircling the power line, given by Eq. 28–1. Using Eq. 29–1b we integrate this field over the area of the rectangle to determine the flux through it. Differentiating the flux as in Eq. 29–2b gives the emf around the rectangle. Finally, by setting the maximum emf equal to 170 V we can solve for the necessary length of the rectangle. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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0 I 0 cos ( 2 f t ) ; 2 r 7.0 m  I 7.0 m dr I Il 0 0  B (t ) =  BdA =  = 0 0 ln (1.4 ) cos ( 2 f t ) cos ( 2 f t ) l dr = 0 0 l cos ( 2 f t )  5.0 m 2 r 5.0 m r 2 2 dB N 0 I 0 d  e = −N ln (1.4 ) l  cos ( 2 ft )  = N 0 I 0 f ln (1.4 ) l sin ( 2 ft ) ; =− dt 2  dt  e0 = N 0 I 0 f ln (1.4 ) l → B (t ) =

l =

e0 170 V = = 27 m −7 N 0 I 0 f ln (1.4 ) 2000 ( 4  10 T m A ) (125A )( 60 Hz ) ln (1.4 )

This is unethical because the current in the rectangle creates a back emf in the initial wire. This results in a power loss to the electric company, just as if the wire had been physically connected to the line. 21. (a)

The magnetic field a distance r from the wire is perpendicular to the wire and given by Eq. 28–1. Integrating this magnetic field over the area of the loop gives the flux through the loop. b+ a  I  Ia  a  0 adr = 0 ln 1 +   B =  BdA =  b 2 r 2  b

(b) Since the loop is being pulled away, v =

db . Differentiate the magnetic flux with respect to dt

time to calculate the emf in the loop.

0 Ia d   a   db 0 Ia 2 v dB d  0 Ia  a   ln 1 +   = − ln 1 +  =−  = e =− dt dt  2 2 db   b   dt 2 b ( b + a )  b  Note that this is the emf at the instant the loop is a distance b from the wire. The value of b is changing with time. (c) Since the magnetic field at the loop points into the page and the flux is decreasing, the induced current will create a downward magnetic field inside the loop. The current then flows clockwise. (d) The power dissipated in the loop as it is pulled away is related to the emf and resistance by Eq. 25–7b. This power is provided by the force pulling the loop away. We calculate this force from the power using Eq. 8–21. As in part (b), the value of b is changing with time.

02 I 2 a 4 v P e2 = = v Rv 4 2 Rb 2 ( b + a )2

22. From Problem 21, the flux through the loop is given by  B =

0 Ia  a  ln  1 +  . The emf is found 2  b

from Eq. 29–2a.  a  a  dI dB d   Ia  a   e =− = −  0 ln  1 +   = − 0 ln  1 +  dt dt  2 2  b  dt  b 

( 4 10 T m A ) ( 0.120 m ) ln 1 + 12.0  (15.0 A )( 2200 rad s ) cos ( 2200 t ) =− −

2

(

 



15.0 

)

= − 4.7  10−4 V cos ( 2200 t )

1004

Chapter 29

Electromagnetic Induction and Faraday’s Law

23. The velocity is found from Eq. 29–3. e 0.12 V e = Bl v → v = = = 0.85 m s Bl ( 0.90 T )( 0.156 m ) 24. Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from Eq. 29–3. e = Bl v = ( 0.650 T )( 0.120 m )( 0.150 m s ) = 1.17  10−2 V

25. (a) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from Eq. 29–3. e = Bl v = ( 0.35 T )( 0.300 m )(1.6 m s ) = 0.168 V  0.17 V (b) Find the induced current from Ohm’s law, using the total resistance. 0.168 V e I= = = 6.340  10 −3 A  6.3 mA R 24.0  + 2.5  (c) The induced current in the rod will be down. Because this current is in an upward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, given by Eq. 27–1.

(

)

F = I l B = 6.340  10−3 A ( 0.300 m )( 0.35T ) = 6.657  10−4 N  6.7  10−4 N 26. The emf is given by Eq. 29–3 as e = Bl v. The resistance of the conductor is given by Eq. 25–3. The length in Eq. 25–3 is the length of resistive material. Since the movable rod starts at the bottom of the U at time t = 0, in a time t it will have moved a distance vt. I=

e R

Bl v Bl v Bl vA = =  L  ( 2v t + l )  ( 2v t + l ) A

27. From the derivation in Example 29–9, we have an expression relating the external force to the magnetic field.

Fexternal =

B2 l 2v R

→ B=

Fexternal R 2

l v

( 0.350 N )( 0.25  ) = 1.297 T  1.3T 2 ( 0.200 m ) (1.30 m s )

28. The rod will descend at its terminal velocity when the magnitudes of the magnetic force (found in Example 29–9) and the gravitational force are equal. We set these two forces equal and solve for the terminal velocity. B 2 l 2 vt = mg → R −3 2 mgR ( 3.6  10 kg )( 9.80 m/s ) ( 0.0013  ) vt = 2 2 = = 0.22 m/s 2 2 B l ( 0.080T ) ( 0.18m ) 29. (a) As the rod moves through the magnetic field an emf will be built up across the rod, but no current can flow. Without the current, there is no force to oppose the motion of the rod, so yes, the rod travels at constant speed. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1005

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) We set the force on the moving rod, obtained in Example 29–9, equal to the mass times the acceleration of the rod. We then write the acceleration as the derivative of the velocity, and by separation of variables we integrate the velocity to obtain an equation for the velocity as a function of time. dv B2 l 2 dv B2 l 2 F = ma = m dt =− v→ =− dt R mR v 2 2

B l t − dv B2 l 2 t B2 l 2 v mR ln ( ) dt t t e = − → = − → = v v 0 v0 v  0 mR mR v0 The magnetic force is proportional to the velocity of the rod and opposes the motion. This results in an exponentially decreasing velocity. v

30. (a)

For a constant current of polarity shown in the figure, the magnetic force will be constant, given by Eq. 27–2. Using Newton’s second law we can integrate the acceleration to calculate the velocity as a function of time. v dv IlB t IlB F =m = I l B →  dv = dt → v (t ) = t  0 0 dt m m (b) For a constant emf, the current will vary with the speed of the rod, as motional emf opposes the motion of the rod. We again use Eq. 27–2 for the force on the rod, with the current given by Ohm’s law, and the induced motional emf given by Eq 29–3. The current produced by the induced emf opposes the current produced by the battery. dv dv dv B2 l 2 lB  e − Bl v  F =m B dt dt → = IlB =  0 → = → = − l  dt R mR e0 − B l v mR v − e0 B l   B l t  −  v − e0 Bl  e0  dv B2 l 2 t B2 l 2 mR ln 1 dt t t e = − → = − → = − v   ( )   0 v − e0 Bl   mR 0 mR Bl   −e0 Bl   2 2

(c)

With constant current, the acceleration is constant and so the velocity does not reach a terminal velocity. However, with constant emf, the increasing motional emf decreases the applied force. This results in a limiting, or terminal velocity of vt = e0 Bl .

31. (a) The magnetic field is perpendicular to the rod, with the magnetic field decreasing with distance from the rod, as in Eq. 28–1. The emf, de, across a short segment, dr, of the rod is given by the differential version of Eq. 29–3. Integrating this emf across the length of the wire gives the total emf. d e = Bvdr → e =  de = 

b+a

0 I  Iv  b + a  v dr = 0 ln   2 r 2  b 

This emf points toward the wire, as positive charges are attracted toward the current. (b) The only change is the direction of the current, so the magnitude of the emf remains the same, but points away from the wire, since positive charges are repelled from the current.

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Chapter 29

Electromagnetic Induction and Faraday’s Law

32. Find the number of turns from Eq. 29–4. The factor multiplying sin  t is the peak output voltage. epeak 24.0 V epeak = NB A → N = = = 63.2 loops 2 B A ( 0.380 T ) ( 2 rad rev ) ( 60.0 rev s )( 0.0515 m ) Assuming that we don’t have fractional loops, we should use 63 loops. 33. From Eq. 29–4, the amplitude of the induced voltage is proportional to the angular speed. Thus, their quotient is a constant. e1 e 2  1850 rpm = → e 2 = e1 2 = (12.4 V ) = 26.2 V 1 2 1 875 rpm 34. From Eq. 29–4, the peak voltage is epeak = NB A. Solve this for the rotation speed.

epeak = NB A →  = f =

epeak NBA

120 V 440 ( 0.550 T )( 0.220 m )

= 10.25 rad s

 10.25 rad s = = 1.6 rev s  98 rpm 2 2 rad rev

35. (a) The rms voltage is found from the peak induced emf. The peak induced emf is calculated from Eq. 29–4. epeak = NB A →

Vrms =

epeak 2

NB A 2

2 280 )( 0.45T )( 2 rad rev )(120 rev s )  ( 0.050m ) ( =

= 527.6 V  530 V (b) To double the output voltage, you must double the rotation frequency to 240 rev/s. 36. (a) The peak current is found from the rms current. I peak = 2 I rms = 2 ( 60.0 A ) = 84.9 A (b) The area can be found from Eq. 29–4. epeak = NB A = A=

2 Vrms NB

2 Vrms → 2 (150 V )

( 950 )( 0.030 T )(85 rev s )( 2 rad rev )

37. We find the peak emf from Eq. 29–4.

(

= 0.01394 m 2  1.4  10 −2 m 2

epeak = NB A = (125)( 0.200T )( 2 rad rev )(120 rev s ) 5.50  10−2 m

) = 57 V 2

38. We assume that the voltage given is an rms value, and that the power is an average power, as in Eq. 25–10a. We calculate the resistance of the wire on the armature using Eq. 25–3. 1.68  10−8  m 85 ( 4 )( 0.060 m ) l = = 1.033  Rwire = 2 A   12 6.5  10−4 m 

(

)

(

I rms =

P Vrms

25.0 W 12.0 V

)

= 2.083A

1007

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Vcircuit = Vbulb + Vwire = Vbulb + IRwire = 12V + ( 2.083A )(1.033  ) = 14.15V

So the generator’s rms emf must be 14.15 volts. We find the frequency of the generator from Eq. 29–4. e 2 fNBA = = 14.15 V → 2 2

f =

(14.15 V ) 2 = 2 NBA

(14.15V ) 2 = 16.01Hz  16 Hz = 16 cycles s 2 2 ( 85)( 0.65T )( 0.060 m )

39. From Eq. 29–4, the induced voltage (back emf) is proportional to the angular speed. Thus, their quotient is a constant. e1 e 2  2500 rpm = → e 2 = e1 2 = ( 72 V ) = 130 V 1 2 1 1400 rpm 40. When the motor is running at full speed, the back emf opposes the applied emf, to give the net across the motor. eapplied − eback = IR → eback = eapplied − IR = 120 V − ( 7.20 A )( 2.65  ) = 100.9 V  1.0  102 V

41. The back emf is proportional to the rotation speed (Eq. 29–4). Thus, if the motor is running at half speed, the back emf is half the original value, or 54 V. Find the new current from writing a loop equation for the motor circuit, from Fig. 29–20. e − eback 120 V − 54 V = = 13 A e − eback − IR = 0 → I = R 5.0  42. The magnitude of the back emf is proportional to both the rotation speed and the magnetic field, e from Eq. 29–4. Thus is constant. B e1 e e B ( 75V ) B1 (1400 rpm ) = 0.54 B = 2 → B2 = 2 1 1 = 1 2 e1 B11 B22 ( 2300 rpm ) (85V ) So reduce the magnetic field to 54% of its original value . 43. (a) The generator voltage rating is the generator emf less the back emf. The ratio of the generator voltage rating to the generator emf is equal to the ratio of the effective resistance to the armature resistance. We solve this ratio for the generator emf, which is the same as the “no load” voltage.  250V 56A  R V I Vnl = e = Vload load = Vload load load = 250V   = 2790V  2.8kV Rnl Rnl  0.40   (b) The generator voltage is proportional to the rotation frequency. From this proportionality we solve for the new generator voltage. V2 2 750 rpm  = → V2 = V1 2 = ( 250 V ) = 156 V  160 V V1 1 1200 rpm 1 44. We find the number of turns in the secondary from Eq. 29–5. 13,500 V VS N S V = → N S = N P S = (172 turns )  19,800 turns 117 V VP N P VP © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1008

Chapter 29

Electromagnetic Induction and Faraday’s Law

45. Use Eqs. 29–5 and 29–6 to relate the voltage and current ratios. 1120 turns VS N S N = → VS = VP S = (120 V ) = 407.3V  410 V 330 turns VP N P NP IS IP

NP NS

→ I P = IS

= (15.0 A )

1120 turns

= 50.9A  51A

330 turns

46. Because N S  N P , this is a step-down transformer. Use Eq. 29–5 to find the voltage ratio, and Eq. 29–6 to find the current ratio. 85 turns VS N S I S N P 680 turns = = = 0.13 = = = 8.0 VP N P 680 turns I P NS 85 turns 47. We find the ratio of the number of turns from Eq. 29–5. N S VS 12000 V = = = 50 N P VP 240 V If the transformer is connected backward, the role of the turns will be reversed: N S VS VS 1 1 = → = → VS = ( 240 V ) = 4.8 V 50 240 V 50 N P VP 48. (a) Use Eqs. 29–5 and 29–6 to relate the voltage and current ratios. 0.35 A VS N S I S N P VS I P I ; = = → = → VS = VP P = (120 V ) = 6.5 V 6.5 A VP N P I P N S VP I S IS (b) Because VS  VP , this is a step-down transformer. 49. (a) The current in the transmission lines can be found from Eq. 25–10a, and then the emf at the end of the lines can be calculated from Kirchhoff’s loop rule. Ptown 75  106 W Ptown = Vrms I rms → I rms = = = 1667A Vrms 45  103 V e − IR − Voutput = 0 →

e = IR + Voutput =

Ptown Vrms

R + Vrms =

75  106 W 45  10 V 3

( 3.0  ) + 45  103 V = 50, 000 V

= 5.0  104 V ( rms ) 2 (b) The power loss in the lines is given by Ploss = I rms R.

Fraction wasted =

Ploss Ptotal

Ploss Ptown + Ploss

2 I rms R 2 Ptown + I rms R

(1667A ) ( 3.0  ) 2 75  106 W + (1667A ) ( 3.0  ) 2

= 0.10 = 10 % ( 2 sig. figs.)

1009

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

50.

Instructor Solutions Manual

(a) If the resistor R is connected between the terminals, then it has a voltage V0 across it and current I0 passing through it. Then by Ohm’s law the equivalent resistance is equal to the resistance of the resistor. V Req = 0 = R I0 (b) We use Eqs. 29–5 and 29–6 to write the voltage drop and current through the resistor in terms of the source voltage and current to calculate the effective resistance. NS V 2 Vs N P 0 V0  N P  R= = → Req = =   R Is N P I I0  NS  0 NS

51. At the power plant, 100 kW of power is delivered at 12,000 V. We first find the initial current from Eq. 25–6. P 100  103 W = 8.333A P = VI → I = = V 12  103 V Then we have a step-up transformer, which steps up the voltage by a factor of 20, so the current is stepped down by a factor of 20. 1 IS N P N = → I S = I P P = ( 8.333 A ) = 0.4167 A 20 I P NS NS The power lost is given by Eq. 25–7a.

(

)

Plost = I 2 R = → 50  103 W = ( 0.4167 A ) 4  10−5  m ( x ) → 2

50  103 W

( 0.4167 A ) ( 4  10  m ) 2

−5

= 7.20  109 m  7  109 m

Note that we are not taking into account any power losses that occur after the step-down transformers. 52. (a) At the power plant, the voltage is changed from 12,000 V to 240,000 V. N S VS 240, 000 V = = = 20 ( 2 significant figures ) N P VP 12, 000 V (b) At the house, the voltage is changed from 7200 V to 240 V. 240 V N S VS = = = 3.3  10 −2 N P VP 7200 V 53. We set the power loss equal to 2% of the total power. Then using Eq. 25–7a we write the power loss in terms of the current (equal to the power divided by the voltage drop) and the resistance. Then, using Eq. 25–3, we calculate the cross-sectional area of each wire and the minimum wire diameter. We assume there are two lines to have a complete circuit, and so we double the length. 2 d2 P l  P   l  2 = → Ploss = 0.020 P = I R =     → A = 2 0.020V 4 V   A 

4 ( 625  106 W )( 2.65  10−8  m ) 2 ( 485  103 m ) 4 P l = = 0.04846 m  4.8cm 2 0.020 V 2 0.020 ( 660  103 V )

The transmission lines must have a diameter greater than or equal to 4.8 cm. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 29

Electromagnetic Induction and Faraday’s Law

54. Without the transformers, we find the delivered current, which is the current in the transmission lines from the delivered power, and the power lost in the transmission lines. P 7.85  105 W Pout = Vout I line → I line = out = = 6542 A Vout 120 V 2 Plost = I line Rline = ( 6542 A ) 2 ( 0.100  ) = 8.5596  106 W 2

Thus there must be 7.85  105 W + 8.5596  106 W = 9.34  106 W  9340 kW of power generated at the start of the process. With the transformers, to deliver the same power at 120 V, the delivered current from the step-down transformer must still be 6542 A. Using the step-down transformer efficiency, we calculate the current in the transmission lines, and the loss in the transmission lines. V I (120 V )( 6542 A ) = 660.8 A Pout = 0.99 Pline → Vout I out = 0.99Vline I line → I line = out out = 0.99Vline ( 0.99 )(1200 V ) end 2 Plost = I line Rline = ( 660.8A ) 2 ( 0.100  ) = 8.733  104 W The power to be delivered is 785 kW. The power that must be delivered to the step-down 785 kW transformer is = 792.9 kW. The power that must be present at the start of the transmission 0.99 must be 792.9 kW + 8.733  104 W = 8.802  105 W to compensate for the transmission line loss. The power that must enter the transmission lines from the 99% efficient step-up transformer is 8.802  105 W = 8.891  105 W. So the power saved is: 0.99 2

9.34  106 W − 8.891 105 W = 8.45  106 W  8.5 MW .

55. We choose a circular path centered at the origin with radius 10 cm. By symmetry the electric field is uniform along this path and is parallel to the path. We then use Eq. 29–8 to calculate the electric field at each point on this path. From the electric field we calculate the force on the charged particle. d r dB 2 dB  E d l = E ( 2 r ) = − dt B = − ( r ) dt → E = − 2 dt r dB 0.10 m F = QE = −Q = − (1.0  10−6 C ) ( −0.10 T/s ) = 5.0 nN 2 dt 2 Since the magnetic field points into the page and is decreasing, Lenz’s law tells us that an induced circular current centered at the origin would flow in the clockwise direction. Therefore, the force on a positive charge along the positive x-axis would be down, or in the − ˆj direction . 56. (a) (b) (c) (d)

The increasing downward magnetic field creates a circular electric field along the electron path. This field applies an electric force to the electron causing it to accelerate. For the electrons to move in a circle, the magnetic force must provide a centripetal acceleration. With the magnetic field pointing downward, the right-hand rule requires the electrons travel in a clockwise direction for the force to point inward. For the electrons to accelerate, the electric field must point in the counterclockwise direction. A current in this field would create an upward magnetic flux. So by Lenz’s law, the downward magnetic field must be increasing. For the electrons to move in a circle and accelerate, the field must be pointing downward and increasing in magnitude. For a sinusoidal wave, the field is downward half of the time and

1011

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

upward the other half. For the half that it is downward, its magnitude is decreasing half of the time and increasing the other half. Therefore, the magnetic field is pointing downward and increasing for only one fourth of every cycle. 57. In Example 29–15 we found the electric field along the electron’s path from Faraday’s law. Multiplying this field by the electron charge gives the force on the electron, and from the force, we calculate the change in tangential velocity. dv F q q r dBavg = =E = dt m m m 2 dt We set the centripetal force on the electron equal to the magnetic force (using Eq. 27–5b) and solve for the velocity. Differentiating the velocity with respect to time (keeping the radius constant) yields a relation for the acceleration in terms of the changing magnetic field. v2 qBr d v q dB0 qvB = m → v= → = r r m dt m dt Equating these two equations for the electron acceleration, we see that the change in magnetic field at the electron must equal 12 of the average change in magnetic field. This relation is satisfied if at all times B0 = 12 Bavg . 58. (a) The electric field is the change in potential across the rod (obtained from Ohm’s law) divided by the length of the rod. V IR E= = l l (b) Again, the electric field is the change in potential across the rod divided by the length of the rod. The electric potential is the supplied potential less the motional emf found using Eq. 29–3 and the results of Problem 30(b). B2 l 2   − t e0 − B l e0 B l 1 − e mR    e − B2 l 2 t V e0 − B l v   = 0 e mR E= = = l l l l 59. The electrical energy is dissipated because there is current flowing in a resistor. The power dissipation by a resistor is given by P = I 2 R, and so the energy dissipated is E = Pt = I 2 Rt. The current is created by the induced emf caused by the changing magnetic field. The average induced emf is given by the “difference” version of Eq. 29–2b. AB AB  B e I = =− =− e=− R Rt t t

E = Pt = I 2 Rt =

A2 ( B ) R 2 ( t )

Rt =

A2 ( B ) R ( t )

( 0.270 m )2  ( 0 − 0.755T )2  = 2

(8.80  )( 0.0400s )

= 8.61  10−3 J 60. Because there are perfect transformers, the power loss is due to resistive heating in the transmission 65 MW lines. Since the town requires 65 MW, the power at the generating plant must be = 65.99 0.985 MW. Thus the power lost in the transmission is 0.99 MW. This can be used to determine the current in the two transmission lines. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 29

Electromagnetic Induction and Faraday’s Law

P = I 2R → I =

0.99  106 W

2 ( 75 km ) 0.10  km

= 256.9 A

To produce 65.99 MW of power at 256.9 A requires the following voltage. P 65.99  106 W V = = = 2.57  105 V  260 kV 256.9 A I 61. The charge on the capacitor can be written in terms of the voltage across the battery and the capacitance using Eq. 24–1. When fully charged, the voltage across the capacitor will equal the emf of the loop, which we calculate using Eq. 29–2b. dB dB Q = CV = C = CA = ( 6.5  10−12 F )(12 m 2 ) (8.0T/s ) = 6.24  10−10 C  0.62 nC dt dt 62. (a) From the efficiency of the transformer, we have PS = 0.85PP . Use this to calculate the current in the primary. 75 W PS PS = 0.85PP = 0.85 I PVP → I P = = = 0.8021A  0.80 A 0.85VP 0.85 (110 V ) (b) The voltage in both the primary and secondary is proportional to the number of turns in the respective coil. The secondary voltage is calculated from the secondary power and resistance since P = V 2 R .

NP NS

VP VS

VP PS RS

110 V

( 75 W )( 2.4  )

= 8.2

63. (a) The voltage drop across the lines is due to the resistance. Vout = Vin − IR = 42000 V − ( 680 A )( 2 )( 0.80  ) = 40912 V  41kV (b) The power input is given by Pin = IVin .

Pin = IVin = ( 680 A )( 42000 V ) = 2.8567  107 W  2.9  107 W (c) The power loss in the lines is due to the current in the resistive wires. Ploss = I 2 R = ( 680 A ) (1.60  ) = 7.40  105 W  7.4  105 W 2

(d) The power output is given by Pout = IVout .

Pout = IVout = ( 680 A )( 40912 V ) = 2.782  107 W  2.8  107 W . This could also be found by subtracting the power lost from the input power. Pout = Pin − Ploss = 2.8567  107 W − 7.40  105 W = 2.783  107 W  2.8  107 W

64. We find the current in the transmission lines from the power transmitted to the user, and then find the power loss in the lines. PT = I LV → I L =

PT V

PT RL  PT   RL = V2 V 

PL = I L2 RL = 

65. (a) Because VS  VP , this is a step-down transformer. (b) Assuming 100% efficiency, the power in both the primary and secondary is 35 W. Find the current in the secondary from the relationship P = IV . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

PS = I SVS → I S =

35 W 12 V

= 2.9 A

35 W

= 0.29 A VP 120 V (d) Find the resistance of the bulb from Ohm’s law. The bulb is in the secondary circuit. 12 V V VS = I S R → R = S = = 4.1  I S 2.9 A

(c)

PP = I PVP → I P =

Instructor Solutions Manual

66. A side view of the rail and bar is shown. From Section 29–3, the emf in FN the bar is produced by the components of B , the length of the bar, and the B v velocity of the bar, which are mutually perpendicular. B and the length of FB the bar are already perpendicular. The component of the velocity of the bar that is perpendicular to B is v cos  , and so the induced emf is given mg  by e = Bl v cos  . This produces a current in the wire, which is found from Ohm’s law. That current is pointing into the page on the diagram. e Bl v cos  I= = R R Because the current is perpendicular to the magnetic field, the force on the wire from the magnetic field can be calculated from Eq. 27–2, and will be horizontal, as shown in the diagram. B l v cos  B 2 l 2 v cos  FB = I l B = lB = R R For the wire to slide down at a steady speed, the net force along the rail must be zero. Write Newton’s second law for forces along the rail, with up the rail being positive. B 2 l 2 v cos 2  = mg sin  → Fnet = FB cos  − mg sin  = 0 → R

( 0.60  )( 0.040 kg ) ( 9.80 m s2 ) sin 6.0o v= 2 2 2 = = 0.57 m s 2 2 B l cos  ( 0.65 T ) ( 0.32 m ) cos 2 6.0o Rmg sin 

67. The induced current in the coil is the induced emf divided by the resistance. The induced emf is found from the changing flux by Eq. 29–2a. The magnetic field of the solenoid, which causes the flux, is given by Eq. 28–4. For the area used in Eq. 29–2a, the cross-sectional area of the solenoid (not the coil) must be used, because all of the magnetic flux is inside the solenoid. dB N I e d I = ind Bsol =0 sol sol eind = N coil = N coil Asol sol R dt dt l sol N coil Asol 0 I=

N sol dI sol l sol

N coil Asol 0 N sol dI sol R

l sol

(160 turns )  ( 0.045 m ) ( 4  10−7 T m A ) ( 230 turns ) 2.0 A = = 4.9  10 −2 A 12  ( 0.01m ) 0.10 s 2

As the current in the solenoid increases, a magnetic field from right to left is created in the solenoid and the loop. The induced current will flow in such a direction as to oppose that field, and so must flow from left to right through the resistor. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 29

Electromagnetic Induction and Faraday’s Law

68. The average induced emf is given by the “difference” version of Eq. 29–2b. Because the coil orientation changes by 180, the change in flux is the opposite of twice the initial flux. The average current is the induced emf divided by the resistance, and the charge that flows in a given time is the current times the elapsed time.

eavg = − N

Q = I t =

 B t eavg R

= − NA

B

( − B ) − ( + B )

= − NA 

t  2 NAB 



t = 

t R

t

  t = 2 NAB → R

2 NAB t QR 2 NA

69. Calculate the current in the ring from the magnitude of the emf (from Eq. 29–2a) divided by the resistance. Setting the current equal to the derivative of the charge, we integrate the charge and flux over the 90 rotation, with the flux given by Eq. 29–1a. This results in the total charge flowing past a given point in the ring. Note that the initial orientation of the ring area relative to the magnetic field is not given. e dQ e dB → dQ = dt → ; I= = e= dt R dt R BA cos ( + 90 ) − cos  1 d 1 1 BA cos( +90) e Q =  dQ =  dt =  dt =  d  =  d = R R dt R R BA cos( ) R

( 0.23T ) ( 0.030 m )

cos ( + 90 ) − cos  = 0.01858C cos ( + 90 ) − cos  0.035  To find the maximum charge, we set the derivative of the charge with respect to the starting angle, , equal to zero to find the extremes. Inserting the maximum angle into our equation, we find the maximum charge passing through the ring. Finally, we divide the maximum charge by the charge of a single electron to obtain the number of electrons passing the point in the ring. dQ = 0.01858C  − sin ( + 90 ) + sin   = 0.01858C  − cos  + sin   = 0 → tan  = 1 → d  = 45 or 225 =

Qmax = 0.01858C cos ( 225 + 90 ) − cos 225 = 0.02628C N max =

0.02628 C Qmax = = 1.6  1017 electrons −19 1.60  10 C/e q

70. The coil should have a diameter roughly equal to that of a standard flashlight D-cell so that it will be simple to hold and use. This gives the coil a radius of about 1.5 cm. As the magnet passes through the coil the field changes direction, so the change in flux for each pass is twice the maximum flux. Let us assume that the magnet is shaken with a frequency of about two shakes per second, so the magnet passes through the coil four times per second. Obtain the number of turns in the coil using Eq. 29–2b. ( 3.0V )( 0.25s ) e e t e t N= = = =  11,000 turns  t  2 B0 A 2 ( 0.050T )  ( 0.015m )2 The answer will vary somewhat based on the approximations used. For example, a flashlight with a smaller diameter, for AA batteries perhaps, would require more turns.

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Instructor Solutions Manual

71. (a) Use Ohm’s law to see that the current is 120V 3.0  = 40A ( 2 significant figures ) . (b) Find the back emf from the normal operating current. eapplied − eback = IR → eback = eapplied − IR = 120 V − ( 2.0 A )( 3.0  ) = 114 V (c) Use Eq. 25–7a. 2 P = I 2 R = ( 2.0 A ) ( 3.0  ) = 12 W (d) Use Eq. 25–7a. 2 P = I 2 R = ( 40 A ) ( 3.0  ) = 4800 W 72. (a) Use Eq. 29–2a to calculate the emf induced in the ring, where the flux is the magnetic field multiplied by the area of the ring. Then using Eq. 25–7b, calculate the average power dissipated in the ring as it is moved away. The thermal energy is the average power times the time. B ( 14  d 2 )  B BA e=− =− =− t t t  B ( 14  d 2 )   t  ( B )2  2 d 4  e2    = Q = Pt =   t =    R 16 Rt t  R   2

( 0.60T )  2 ( 0.015m ) 2

16 ( 55  10−6  )( 45  10−3 s )

= 4.542  10−3J  4.5mJ

(b) The temperature change is calculated from the thermal energy using Eq. 19–2. Q 4.542  10−3 J T = = = 2.3  10−3 C mc (15  10−3 kg ) (129 J kg C ) 73. (a) The clockwise current in the left-hand loop produces a magnetic field which is into the page within the loop and out of the page outside the loop. Thus the right-hand loop is in a magnetic field that is directed out of the page. Before the current in the left-hand loop reaches its steady state, there will be an induced current in the right-hand loop that will produce a magnetic field into the page to oppose the increase of the field from the left-hand loop. Thus, the induced current will be clockwise. (b) After a long time, the current in the left-hand loop is constant, so there will be no induced current in the right-hand coil. (c) If the second loop is pulled to the right, the magnetic field out of the page from the left-hand loop through the second loop will decrease. During the motion, there will be an induced current in the right-hand loop that will produce a magnetic field out of the page to oppose the decrease of the field from the left-hand loop. Thus the induced current will be counterclockwise. (d) Since equilibrium has been reached (“a long time”), there is a steady current in the left-hand loop and steady magnetic flux passing through the right-hand loop. There is no emf being induced in the right-hand loop, and so closing a switch in the right-hand loop has no effect. 74. (a)

Since the coils are directly coupled to the wheels, the torque provided by the motor (Eq. 27–9) balances the torque caused by the frictional force. ( 250 N )( 0.29 m ) Fr NIAB = Fr → I = = = 23.31A  23A NAB 320 ( 0.12 m )( 0.15m )( 0.54T )

1016

Chapter 29

Electromagnetic Induction and Faraday’s Law

(b)

To maintain this speed the power loss due to the friction (Eq. 8–21) must equal the net power provided by the coils. The power provided by the coils is the current through the coils multiplied by the back emf. F v ( 250 N )( 35km h )  1000m km  2 P = F v = I eback → eback = =   = 104.27 V  1.0  10 V 23.31A 3600s h I  

(c)

The power dissipated in the coils is the difference between the power produced by the coils and the net power provided to the wheels. Ploss = P − Pnet = I e − I eback = ( 23.31A )(120V − 104.27 V ) = 366.7 W  370W

(d)

We divide the net power by the total power to determine the percent used to drive the car. Pnet I eback 104.27 V = = = 0.8689  87% P Ie 120 V

75. The energy is dissipated by the resistance. The power dissipated by the resistor is given by Eq. 25–7b, and the energy is the integral of the power over time. The induced emf is given by Eq. 29–2a. dB dB NAB0 −t  e 2  N 2 A2 B02  −2t  e ; P = I 2R = e = −N = − NA = = e dt dt R  R 2  

 N 2 A2 B02  −2t   N 2 A2 B02    −2t   ( NAB0 ) E =  Pdt =   e dt =  1 − e −2t  ) ( − e  =   2 2 R 2 R 0   R   2 0  t

18 ( 0.100 m )2 ( 0.50 T )  1 − e −2t ( 0.10 s ) = ( 0.20 J ) 1 − e −20 t = ( ) ( ) 2 ( 2.0  )( 0.10s )

76. The total emf across the rod is the integral of the differential emf across each small segment of the rod. For each differential segment, dr, the differential emf is given by the differential version of Eq. 29–3. The velocity is the angular speed multiplied by the radius. The figure is a top view of the spinning rod. l

d e = Bvd l = B rdr → e =  d e =  B rdr = 12 B l 2 0

77. (a)

(b)

(c)

At startup there is no back emf, so treat the circuit as two parallel resistors, each with the same voltage drop. The current through the battery is the sum of the currents through each resistor. e e 115V 115V I = I0 + I0 = + = + = 41.93A  41.9 A Rfield Rarmature 32.0  3.00  field armature At full speed the back emf decreases the voltage drop across the armature resistor. e e − eback 115V 115V − 105V + = + = 6.927A  6.9 A I = I field + I armature = Rfield Rarmature 32.0  3.00 

1017

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

78. Assume that the electric field does not fringe, but only has a horizontal component between the plates and zero field outside the plates. Apply Faraday’s law (Eq. 29–8) to this situation for a rectangular loop with one horizontal leg inside the plates and the second horizontal leg outside the plates, as shown in the dashed line in the adjacent figure. We integrate around this path in the counterclockwise direction. Since the field only has a horizontal component between the plates, only the horizontal leg will contribute to the electric field integral. Since the field is constant in this region, the integral is the electric field times the length of the leg. l

 E d l =  Ed l = E l 0

For a static electric field, the magnetic flux is unchanging. Therefore

dB = 0. dt

dB → E l = 0, which is not possible. Thus, one of dt the initial assumptions must be false. We conclude that the field must have some fringing at the edges.

Using Faraday’s law, we have  E d l = −

79. The total emf from the center of the disk to the edge of the disk is the integral of the differential emf across each small segment of the radial line passing from the center of the disk to the edge. For each differential segment, dr, the emf is given by the differential version of Eq. 29–3. The velocity is the angular speed multiplied by the radius. Since the disk is rotating in the counterclockwise direction, and the field is out of the page, the emf is increasing with increasing radius. Therefore the rim is at the higher potential. d e = Bvd l = B rdr → R

e =  d e =  B rdr = 12 B R 2 0

80. We set the electric field equal to the negative gradient of the electric potential (Eq. 23–8), with the differential potential given by Eq. 29–3, as in Problem 79. de Brdr E=− rˆ = − rˆ = − Br rˆ dr dr The electric field has magnitude Br and points radially inwards, toward the center of the disk. 81. The emf around the loop is equal to the time derivative of the flux, as in Eq. 29–2a. Since the area of the coil is constant, the time derivative of the flux is equal to the derivative of the magnetic field multiplied by the area of the loop. To calculate the emf in the loop we add the voltage drop across the capacitor to the voltage drop across the resistor. The current in the loop is the derivative of the charge on the capacitor (Eq. 24–1). dQ d ( CV ) d CV0 − t / V0 − t / I= e = e = = CV0 (1 − e − t / )  = dt dt dt R  dB dB dB dB V0 V  e = IR + VC =  0 e − t /  R + V0 (1 − e − t / ) = V0 = =A =  r2 → = dt dt dt dt  r 2 R  Since the charge is building up on the top plate of the capacitor, the induced current is flowing clockwise. By Lenz’s law this produces a downward flux, so the external downward magnetic field must be decreasing. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1018

Chapter 29

82.

Electromagnetic Induction and Faraday’s Law

(a) As the loop falls out of the magnetic field, the flux through the loop decreases with time creating an induced emf in the loop. The current in the loop is equal to the emf divided by the resistance, which can be written in terms of the resistivity using Eq. 25–3. e   d 2 / 4  d  B   d 2  dA  d 2 = = I = = Bl v  B R   4l  dt  16  l  dt 16  l This current induces a force on the three sides of the loop in the magnetic field. The forces on the two vertical sides are equal and opposite and therefore cancel. F = IlB =

d2  d 2 B2 l v Bl vl B = 16  l 16 

By Lenz’s law this force is upward to slow the decrease in flux. (b) Terminal speed will occur when the gravitational force is equal to the magnetic force.  16 m g  d 2 B 2 l vT d2  = → vT = Fg = m  4 l g  4  16  B2  (c) We calculate the terminal velocity using the given magnetic field, the density of copper from Table 13–1, and the resistivity of copper from Table 25–1. 16 ( 8.9  103 kg/m 3 )(1.68  10−8 m )( 9.80 m/s 2 ) vT = = 3.7 cm/s 2 ( 0.80 T ) 83. The magnetic field created by the solenoid is given by Eq. 28–4, using the sinusoidal current given in the problem. The emf developed by the square loop is given by Faraday’s law, Eq. 29–2a. The area of the square loop comes from the diagram shown, using the Pythagorean theorem. 2 l 2 + l 2 = ( 2 R ) → l = 2 R → Asquare = l 2 = 2 R 2 B = 0 nI = o nI 0 cos t ; e = −A

dB = − 0 nI 0 sin  t dt

dB = −2 R 2 ( − 0 nI 0 sin  t ) ; emax = 2 R 2 0 nI 0 dt

84. The emf developed in the loop is given by Faraday’s law, Eq. 29–2a. The instantaneous power dissipated in the loop is given by Eq. 25–7b. To find the energy dissipated, the power is integrated over the time interval. d 2 b e 2 4 2b 2  2 t   2 t  cos  ; P =− = = 2 cos 2  e =−   dt T R T R  T   T  T /2

4 2b 2 4 2b 2 4 2b 2  1 T  4 t   2  2 t  2  2 t  = = cos cos sin  E =  P dt =  dt dt t+      2  2 2 2  8 T R T R 0 T R   T   T   T 0 0 0 T /2

T /2

4 2b 2  T   2b 2 = T 2 R  4  TR

1019

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

85. (a) See the graph, with best fit linear trend line (with the y intercept forced to be 0). (b) The theoretical slope is the induced voltage divided by the velocity. Take the difference between the experimental value found in part (a) and the theoretical value and divide the result by the theoretical value to obtain the percent difference.

 m − mtheory     mexp  0.3532 V s m % diff =  exp − 1100 =  − 1100 100 =     mtheory  BN l   ( 0.126T )( 50 )( 0.0561m )    (c)

= −0.065% Use the theoretical equation to calculate the voltage at each experimental speed. Then calculate the percent difference at each speed. Theoretical Speed Induced induced (m/s) voltage (V) voltage (V) % diff. 0.367 0.128 0.130 – 1.32% 0.379 0.135 0.134 0.78% 0.465 0.164 0.164 – 0.21% 0.623 0.221 0.220 0.37% 0.630 0.222 0.223 – 0.30%

1020

CHAPTER 30: Inductance, Electromagnetic Oscillations, and AC Circuits Responses to Questions 1.

(a) To create the largest amount of mutual inductance with two flat circular coils of wire, you would place them face-to-face and very close to each other. This way, almost all of the magnetic flux from one coil also goes through the other coil. (b) To create the least amount of mutual inductance with two flat circular coils, you would place them with their faces at right angles, as shown, and their centers overlapping. This way, almost none of the magnetic flux from one coil goes through the other coil.

The magnetic field near the end of the first solenoid is less than it is in the center, due to the fringing of the magnetic field lines. Therefore the flux through the second coil would be less than that given by the formula, and the mutual inductance would be lower.

Yes. If two coils have mutual inductance, then they each have the capacity for self-inductance. Any coil that experiences a changing current will have a self-inductance.

The energy density is greater near the center of a solenoid, where the magnetic field is greater.

To create the greatest self-inductance, bend the wire into as many loops as possible. To create the least self-inductance, leave the wire as a straight piece of wire.

(a) No. The time needed for the LR circuit to reach a given fraction of its maximum possible current depends on the time constant, τ = L/R, which is independent of the emf. (b) Yes. The emf determines the maximum value of the current (Imax = V0/R,) and therefore will affect the time it takes to reach a particular value of current.

A circuit with a large inductive time constant is resistant to changes in the current. When a switch is opened, the inductor continues to force current to flow for a small period of time. A large charge can build up on the switch from that current, making a large potential difference across the open switch, and may cause a spark to jump across the switch (similar to a Van de Graaf generator which has been charged).

Although the current is zero at the instant the battery is connected, the rate at which the current is changing is a maximum and therefore the rate of change of flux through the inductor is a maximum. Since, by Faraday’s law, the induced emf depends on the rate of change of flux and not the flux itself, the emf in the inductor is a maximum at this instant.

When the capacitor has discharged completely, energy is stored in the magnetic field of the inductor. The inductor will resist a change in the current, so current will continue to flow and will charge the capacitor again, with the opposite polarity.

10. (a) Yes. (b) Yes. The rms voltages across either an inductor or a capacitor of an LRC circuit can be greater than the rms source voltage because the different voltages are out of phase with each other. At any given instant, the voltage across either the inductor or the capacitor could be negative, for example, thus allowing for a very large positive voltage on the other device. (The rms voltages, however, are always positive by definition.) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

11. The energy comes from the generator. A generator is a device that converts mechanical energy to electrical energy, so ultimately, the energy came from some mechanical source, such as wind or falling water. Some of the energy is dissipated in the resistor and some is stored in the fields of the capacitor and the inductor. The values of L, C, R, and the driver frequency determine the current flow in the circuit, which determines the power supplied by generator. 12. XL = XC at the resonant frequency. If the circuit is predominantly inductive, such that XL > XC, then the frequency is greater than the resonant frequency and the voltage leads the current. If the circuit is predominantly capacitive, such that XC > XL, then the frequency is lower than the resonant frequency and the current leads the voltage. Values of L and C cannot be meaningfully compared, since they are in different units. Describing the circuit as “inductive” or “capacitive” relates to the values of XL and XC, which are both in ohms and which both depend on frequency. 13. Yes. When ω approaches zero, XL approaches zero, and XC becomes infinitely large. This is consistent with what happens in a series LRC circuit connected to a dc power supply. For the dc case, ω is zero and XL will be zero because there is no changing current to cause an induced emf. XC will be infinitely large, because steady direct current cannot flow across a capacitor once it is charged. 14. The impedance in an LRC circuit will be a minimum at resonance, when XL = XC. At resonance, the impedance equals the resistance, so the smallest R possible will give the smallest impedance. 15. Yes. The power output of the generator is P = IV. When either the instantaneous current or the instantaneous voltage in the circuit is negative, and the other variable is positive, the instantaneous power output can be negative. At this time either the inductor or the capacitor is discharging power back to the generator. 16. Yes, the power factor depends on frequency because XL and XC, and therefore the phase angle, depend on frequency. For example, at resonant frequency, XL = XC, the phase angle is 0º, and the power factor is one. The average power dissipated in an LRC circuit also depends on frequency, since it depends on the power factor: Pavg = Irms Vrms cosφ. Maximum power is dissipated at the resonant frequency. The value of the power factor decreases as the frequency gets farther from the resonant frequency. 17. (a) The impedance of a pure resistance is unaffected by the frequency of the source emf. (b) The impedance of a pure capacitance decreases with increasing frequency, since X C = 1 2 fC . (c) The impedance of a pure inductance increases with increasing frequency, since X L = 2 fL. (d) In an LRC circuit near resonance, small changes in the frequency will cause large changes in the impedance. (e) For frequencies far above the resonance frequency, the impedance of the LRC circuit is dominated by the inductive reactance and will increase with increasing frequency. For frequencies far below the resonance frequency, the impedance of the LRC circuit is dominated by the capacitive reactance and will decrease with increasing frequency. 18. In all three cases, the energy dissipated decreases as R approaches zero. Energy oscillates between being stored in the field of the capacitor and being stored in the field of the inductor. (a) The energy stored in the fields (and oscillating between them) is a maximum at resonant frequency and approaches an infinite value as R approaches zero. (b) When the frequency is near resonance, a large amount of energy is stored in the fields but the value is less than the maximum value. (c) Far from resonance, a much lower amount of energy is stored in the fields. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

19. In an LRC circuit, the current in the circuit and the voltage across each element all oscillate. The energy stored in the circuit also oscillates and is alternately stored in the magnetic field of the inductor and the electric field of the capacitor. 20. In an LRC circuit, energy oscillates between being stored in the magnetic field of the inductor and being stored in the electric field of the capacitor. This is analogous to a mass on a spring, with energy alternating between kinetic energy of the mass and spring potential energy as the spring compresses and extends. The energy stored in the magnetic field is analogous to the kinetic energy of the moving mass, and L corresponds to the mass, m, on the spring. The energy stored in the electric field of the capacitor is analogous to the spring potential energy, and C corresponds to the reciprocal of the spring constant, 1/k. 21. The way to make the impedance of an LRC circuit a minimum is to make the resistance very small and make the reactance of the capacitor equal to the reactance of the inductor: X L = X C , or 2 fL =

1 2 fC

→ f =

1 2 LC

Responses to MisConceptual Questions 1.

(a) To create the largest amount of mutual inductance with two flat circular coils of wire, you would place them face-to-face and very close to each other. This way, almost all of the magnetic flux from one coil also goes through the other coil. To create the least amount of mutual inductance with two flat circular coils, you would place them with their faces at right angles, as in answer (c). This way, almost none of the magnetic flux from one coil goes through the other coil.

(d) To create the greatest self-inductance, bend the wire into as many loops as possible. Each loop creates its “own” magnetic field, and so there is more flux if there are more loops. To create the least self-inductance, leave the wire as a straight piece of wire.

(c) A circuit with a large inductive time-constant is resistant to changes in the current. When a switch is opened, the inductor continues to force the current to flow. A large charge can build up on the switch, and the electric field (or emf) of that large charge may be able to ionize a path for current flow across a small air gap, creating a spark.

(c) The period of an oscillating LC circuit is given by T = 2 LC . Thus, we see that doubling either the inductance or the capacitance would increase the period by 2 , but doubling them both would increase the period by 2. And the period doesn’t depend on the maximum charge or current, so the correct answer is (c).

(b) The frequency of an oscillating LRC circuit is given by f =

−

. As R increases 2 LC 4 L2 from 0, the frequency decreases, and so the period increases, up to an infinite period (no oscillation) for R = 2

L C

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(a) The rms current is directly proportional to the source voltage, so answer (a) is correct. As can be seen from a resonance graph, like Fig. 30–22, increasing source frequency can either increase or decrease the rms current, so answer (b) is incorrect. The rms current is inversely proportional to the impedance, and the impedance always increases as the resistance increases, so answer (c) is incorrect. Changing either the inductance or the capacitance will change the resonance frequency, and so can either increase or decrease the current, depending on which way the resonance frequency shifts. Thus answer (a) is the only correct answer.

(e) In an oscillating LC circuit, the charge on the capacitor oscillates, flowing from one plate through the inductor to the other plate. The flowing charge oscillates, so the current (which is the time derivative of the charge passing a point) also oscillates. Since the voltage across the capacitor is proportional the charge on the capacitor, the voltage across the capacitor oscillates. And since the voltage across the inductor is proportion to the rate of change of the current (and the current is changing), the voltage across the inductor also oscillates. See Fig. 30–11.

(d) Yes, always. The instantaneous voltages across the different elements in the circuit will be different, but the current through each element in the series circuit is the same at every instant.

(a) Energy is transformed from being stored in the inductor to being stored in the capacitor, but that total energy is constant on average. The resistor is the only “consumer” of power from the source. This can be seen from the equation right about Eq. 30–30, which is equivalent to it: 2 P = I rms R.

10. (f) Figs. 30–20 and 30–21 show that none of the quantities listed in answers (a) through (d) are in phase with the input voltage in general. 11. (b) In a series circuit (whether ac or dc), the current is the same at every point in the circuit. In an ac circuit the current is out of phase with the voltage across an inductor and the voltage across a capacitor, so (a) is incorrect. In non-resonant ac circuits there is a phase difference between the current and the voltage source, so (c) is not correct. The current and voltage across a resistor are always in phase, so the resistor does not change the phase of the current in a series circuit, so (d) is incorrect. 12. (d) Since all of the voltages are out of phase with each other, the peak voltages of each element will not occur simultaneously, and so the voltages cannot be added up linearly. The relationship is given by V0 = VR20 + (VL 0 − VC 0 ) . 2

13. (c) From the discussion in Section 30–9, the magnitude of the current is the largest at resonance. The current is inversely proportional to the impedance. Thus the impedance is a minimum at resonance. If C = 0 or L = 0, the circuit will not be at resonance.

Solutions to Problems 1.

If we assume the outer solenoid is carrying current I 1 , then the magnetic field inside the outer solenoid is B = 0 n1 I1 . The flux in each turn of the inner solenoid is  21 = B r22 = 0n1 I1 r22 . The mutual inductance is given by Eq. 30–1. N n l  n I  r2 M = 0n1n2 r22 M = 2 21 = 2 0 1 1 2 → l I1 I1

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Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

We find the mutual inductance of the inner loop. If we assume the outer solenoid is carrying current N I 1 , then the magnetic field inside the outer solenoid is B = 0 1 I1 . The magnetic flux through each l loop of the small coil is the magnetic field times the area perpendicular to the field. The mutual inductance is given by Eq. 30–1. NI N 2 0 1 1 A2 sin   N N A sin  NI N l  21 = BA2 sin  = 0 1 1 A2 sin  ; M = 2 21 = = 0 1 2 2 l l I1 I1

(a) The mutual inductance is found in Example 30–1. M =

 N1 N 2 A l

(

)

1850 4  10 −7 T m A ( 265 )(115 )  ( 0.0200 m )

2.44 m

= 3.6488  10−2 H

 3.65  10−2 H (b) The emf induced in the second coil can be found from Eq. 30–3b. dI I ( −12.0A ) = 4.47 V e 2 = − M 1 = − M 1 = −3.6488  10−2 H 0.0980 ms dt t We find the mutual inductance of the system using Eq. 30–1, with the flux equal to the integral of the magnetic field of the wire (Eq. 28–1) over the area of the loop.

(

 w l  12 1 l 2 0 I1 =  wdr = 0 ln  2  l 2  l 1  I1 I1 1 2 r

Use the relationship for the self-inductance of a solenoid, as given in Example 30–3.

L= 6.

)

0 N 2 A l

( 4  10 T m A ) ( 7500 )  (1.45  10 m ) = 7.8  10 H = 2

−7

−2

0.60 m

Use the relationship for the inductance of a solenoid, as given in Example 30–3. 0 N 2 A L= → l N=

0 A

( 0.13 H )( 0.330 m ) = 4963 turns  5.0  103 turns 2 −7 ( 4  10 T m A )  ( 0.021m )

Because the current is increasing, the emf is negative. We find the self-inductance from Eq. 30–5. I t dI 0.0120s e = −L = −L → L = −e = − ( −2.20 V ) = 0.498 H t I dt 0.0250 A − ( −0.0280 A )

(a) The number of turns can be found from the inductance of a solenoid, which is derived in Example 30–3. L=

0 N 2 A l

( 4  10 T m A ) ( 3200 )  ( 0.0125 m ) = 0.02240 H  0.022 H 2

−7

( 0.282 m )

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) Apply the same equation again, solving for the number of turns.

0 N 2 A l

→ N=

0 A

( 0.02240 H )( 0.282 m )  92 turns (1200 ) ( 4  10−7 T m A )  ( 0.0125 m )2

We draw the coil as two elements in series, a pure resistance and a R L pure inductance. There is a voltage drop due to the resistance of + − the coil, given by Ohm’s law, and an induced emf due to the einduced Iincreasing inductance of the coil, given by Eq. 30–5. Since the current is increasing, the inductance will create a potential difference to a b oppose the increasing current, and so there is a drop in the potential due to the inductance. The potential difference across the coil is the sum of the two potential drops. dI Vab = IR + L = ( 3.00 A )( 3.25  ) + ( 0.44 H )( 3.15 A s ) = 11.1V dt

10. We use the result for inductance per unit length from Example 30–5.

2  ( 7510 H m ) 2 − − ( 7510 H m ) 0 r2 ( 4 10 T m A )  −9 = ln  75  10 H m → r1  r2 e = ( 0.0030 m ) e = 0.002062 m l 2 r1 −9

−9

−7

r1  0.0021m

11. The inductance of the solenoid is given by L =

0 N 2 A

0 N 2  d 2

. The (constant) length of the 4 l l wire is given by l wire = N  d sol , and so since d sol 2 = 2.2d sol 1 , we also know that N1 = 2.2 N 2 . The fact

that the wire is tightly wound gives l sol = Nd wire . Find the ratio of the two inductances.

0 N 22 2 N 22 2 l 2wire  2 d sol 2 d sol 2 2.2 N 2 L2 Nd N l 4 l sol 2 l l = = sol22 = 2 sol 2 2 = sol 1 = 1 wire = 1 = = 2.2 2 N1 2 l wire  L1 0 N1 2 N2 l sol 2 N 2 d wire N 2 d sol 1 d sol 1 l sol 1 l sol 1 4 l sol 1 12. (a) When connected in series the voltage drops across each inductor will add, while the currents in each inductor are the same. dI dI dI dI e = e1 + e2 = − L1 − L2 = − ( L1 + L2 ) = − Leq → Leq = L1 + L2 dt dt dt dt (b) When connected in parallel the currents in each inductor add to the equivalent current, while the voltage drop across each inductor is the same as the equivalent voltage drop. dI dI1 dI 2 1 1 1 e e e = + → = + → = + dt dt dt Leq L1 L2 Leq L1 L2 Therefore, inductors in series and parallel add the same as resistors in series and parallel.

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13. We use Eq. 30–4 to calculate the self-inductance, where the flux is the integral of the magnetic field over a crosssection of the toroid. The magnetic field inside the toroid was calculated in Example 28–9.

 N 2 h  r2  N N r2  NI ln    B =  0 hdr = 0 2 I I r1 2 r  r1 

14. The magnetic energy in the field is derived from Eq. 30–7. Energy stored 1 B 2 =2 → u= 0 Volume Energy = 12

0

( Volume ) = 12

0

 r 2 l = 12

( 0.720 T )2

( 4  10 T m A ) −7

 ( 0.0105 m ) ( 0.380 m ) = 27.1J 2

15. (a) We use Eq. 24–6 to calculate the energy density in an electric field and Eq. 30–7 to calculate the energy density in the magnetic field.

uE = 12  0 E 2 = 12 (8.85  10−12 C2 N m2 )(1.0  104 V m ) = 4.425  10−4 J m3 2

 4.4  10−4 J m3

( 2.0T ) B2 uB = = = 1.592  106 J m3  1.6  106 J m3 2 0 2 ( 4  10−7 T m A ) 2

uE 4.425  10−4 J m 3 = = 2.779  10−10  2.8  10−10 uB 1.592  106 J m3 (b) Use Eq. 24–6 to calculate the electric field from the energy density for the magnetic field given in part (a). uE = 12  0 E 2 = uB → E =

2u B

0

2 (1.592  106 J m 3 )

(8.85  10−12 C2 N m2 )

= 6.0  108 V m

Note that this is 60,000 × larger than the “large” electric field used in part (a). 16. We use Eq. 30–7 to calculate the energy density with the magnetic field calculated in Example 28–11. 2 −7 2 B2 1  0 I  0 I 2 ( 4  10 T m A ) (19.0 A ) uB = = = = 7.23  10−4 J m 3   = 2 2 0 2 0  2 R  8R 2 8 ( 0.280 m ) 17. We use Eq. 30–7 to calculate the magnetic energy density, with the magnetic field calculated using Eq. 28–1. 2 2 4  10−7 T m/A ) (18A ) ( 0 I 2 B2 1  0 I  = = = 2.3J m3 uB =   = 2 2 −3 2 0 2 0  2 R  8 2 R 2 8 (1.5  10 m ) To calculate the electric energy density with Eq. 24–6, we must first calculate the electric field at the surface of the wire. The electric field will equal the voltage difference along the wire divided by the length of the wire. We can calculate the voltage drop using Ohm’s law and the resistance from the resistivity and diameter of the wire. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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l V IR I A I = = 2 E= = l l l r 2  (18A ) (1.68  10−8  m )   I  −12 2 2 2  1 1 1  uE = 2  0 E = 2  0  2  = 2 ( 8.85  10 C /N m ) 2 −3   r   (1.5  10 m )  

= 8.1  10−15 J/m3 18. We use Eq. 30–7 to calculate the energy density in the toroid, with the magnetic field calculated in Example 28–9. We integrate the energy density over the volume of the toroid to obtain the total energy stored in the toroid. Since the energy density is a function of radius only, we treat the toroid as cylindrical shells each with differential volume dV = 2 rhdr. 2

uB =

0 N 2 I 2 1  0 NI  B2 = = 2 0 2 0  2 r  8 2 r 2 0 N 2 I 2 0 N 2 I 2 h r dr 0 N 2 I 2 h  r2  2  = rhdr r r = 4 ln  r1  r 8 2 r 2 4

U =  u B dV = 

19. The magnetic field between the cables is given in Example 30–5. Since the magnetic field only depends on radius, we use Eq. 30–7 for the energy density in the differential volume dV = 2 r l dr and integrate over the radius between the two cables. 2

r2 1   I  0 I 2 r2 dr 0 I 2  r2  U 1 0 2 ln   =  uB dV =  = = rdr  r1 2   2 r  4 r1 r 4 l l  0   r1 

20. We create an Amperean loop of radius s < r to calculate the magnetic field within the wire using Eq. 28–3, and following Example 28–6. In the diagram, the inner circle is of radius s, the next circle is of radius s + ds, and and the wire itself is of radius r. Current flows through the wire from left to right. Since the resulting magnetic field only depends on radius, we use Eq. 30–7 for the energy density in the differential volume dV = 2 sl ds and integrate from zero to the radius of the wire.  Is  I  2  B d l = 0 I enc → B ( 2 s ) = 0   r 2  ( s ) → B = 20 r 2 2

r 1   Is  U 1 0 I 2 r 3 0 I 2 0 sds s ds  2 =  uB dV =  = =   2 0 2 4 r 4 0 16 l l 0  2 r 

21. (a) We set I equal to 75% of the maximum value in Eq. 30–9 and solve for the time constant. ( 2.85ms ) = 2.056 ms  2.1ms t I = 0.75 I 0 = I 0 (1 − e − t / ) →  = − =− ln ( 0.25) ln ( 0.25 ) (b) The resistance can be calculated from the time constant using Eq. 30–10. L 31.0mH R= = = 15   2.056 ms

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Inductance, Electromagnetic Oscillations, and AC Circuits

22. The potential difference across the resistor is proportional to the current, and so we set the current in the equation I = I max e − t / equal to 0.020 I0 and solve for the time.

I = 0.020I 0 = I 0e− t / → t = − ln ( 0.020)  3.9 23. (a) We use Eq. 30–6 to determine the energy stored in the inductor, with the current given by Eq. 30–9. 2 LV02 1 − e−t / ) 2 ( 2R (b) Set the energy from part (a) equal to 99.9% of its maximum value and solve for the time. 2 2 LV 2 LV 2 U = 0.999 02 = 02 (1 − e − t / ) → 0.999 = (1 − e − t / ) → t = − ln 1 − 0.999  7.6 2R 2R

U = 12 LI 2 =

(

)

24. (a) At the moment the switch is closed, no current will flow through the inductor. Therefore, the resistors R1 and R2 can be treated as in series. e e = I ( R1 + R2 ) → I1 = I 2 = , I3 = 0 R1 + R2 (b) A long time after the switch is closed, there is no voltage drop across the inductor so resistors R2 and R3 can be treated as parallel resistors in series with R1. I1 = I 2 + I 3 , e = I1R1 + I 2 R2 , I 2 R2 = I 3R3

I R e − I 2 R2 e R3 = I2 + 2 2 → I2 = R1 R3 R2 R3 + R1 R3 + R1 R2 I3 = (c)

I 2 R2 e R2 = R3 R2 R3 + R1R3 + R1R2

I1 = I 2 + I 3 =

e ( R3 + R2 ) R2 R3 + R1R3 + R1R2

Just after the switch is opened the current through the inductor continues with the same magnitude and direction. With the open switch, no current can flow through the branch with the switch. Therefore the current through R2 must be equal to the current through R3, but in the opposite direction. e R2 −e R2 I3 = , I2 = , I1 = 0 R2 R3 + R1 R3 + R1 R2 R2 R3 + R1 R3 + R1 R2

(d) After a long time with no voltage source, the energy in the inductor will dissipate and no current will flow through any of the branches. I1 = I 2 = I 3 = 0 25. (a) The current is the same just before and just after the switch moves from A to B. We use Ohm’s law for a steady-state current to determine I0 before the switch is thrown. We use Eq. 30–5 to determine the emf in the inductor as a function of time. Since the exponential term decreases in time, the maximum emf occurs when t = 0. V dI d  −t  d  − t  R R eL = − L = − L  I 0 e    = − L  I 0 e R/ L  = − L I 0 e −t / = R 0 e −t / → emax = V0 dt dt  dt  L R R    R   45R  (b) emax =  V0 =   (145V ) = 6525V  6500 V R  R 

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Instructor Solutions Manual

26. When the switch is initially closed, the inductor prevents current from flowing, and so the initial current is 0, as shown in Fig. 30–6b. If the current is 0, there is no voltage drop across the resistor (since VR = IR ), and so the entire battery voltage appears across the inductor. Apply Eq. 30–5 to find the initial rate of change of the current.

V0 = VL = L

→

dt dt L The maximum value of the current is reached after a long time, when there is no voltage across the inductor, and so the entire battery voltage appears across the resistor. Apply Ohm’s law. V V0 = I max R → I max = 0 . R dI V0 Find the time to reach the maximum current if the rate of current change remained at = . dt L −1

L  dI  V L I max = I 0 + ( elapsed time ) → elapsed time = ( I max − I 0 )   =  0 − 0  = = dt R  V0 R  dt  dI

27. We use the inductance of a solenoid, as derived in Example 30–3: Lsol =

0 N 2 A

l (a) Both solenoids have the same cross-sectional area and the same length. Because the wire in solenoid 1 is 1.5 times as thick as the wire in solenoid 2, solenoid 2 will have 1.5 times the number of turns as solenoid 1. 0 N 22 A 2 L2 N 22  N 2  L2 l = = 2 = = 1.52 = 2.25 → = 2.25  2 L1 0 N1 A N1  N1  L1 l (b) To find the ratio of the time constants, both the inductance and resistance ratios need to be known. Since solenoid 2 has 1.5 times the number of turns as solenoid 1, the length of wire used to make solenoid 2 is 1.5 times that used to make solenoid 1, or l wire 2 = 1.5l wire 1 , and the

diameter of the wire in solenoid 1 is 1.5 times that in solenoid 2, or d wire 1 = 1.5d wire 2 . Use this to find their relative resistances, and then the ratio of time constants. l wire 1 l wire 1  l wire 1 2 2 2 2  ( d wire 1 2 ) Awire 1 d wire l wire 1  d wire 2   1   1  R1 1 1 = = = = = = →      l wire 2 l wire 2 R2  l wire 2 l wire 2  d wire 1   1.5   1.5  1.53

 ( d wire 2 2 )

Awire 2

R1 R2

1 3

1.5

;

2 d wire 2

 1 L1 R1 L1 R2  1  1 3 = = =1.5 =  1.5 = 1.5 →  2 L2 R2 L2 R1  2.25  2

( )

28. (a) To have maximum current and no charge at the initial time, we set t = 0 in Eqs. 30–13 and 30–15 to solve for the necessary phase factor .    I ( 0 ) = I 0 sin  →  = → I (t ) = I 0 sin   t +  = I 0 cos t 2 2      Q ( 0 ) = Q0 cos   = 0 → Q = Q0 cos   t +  = −Q0 sin (t ) 2 2  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1030

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Inductance, Electromagnetic Oscillations, and AC Circuits

Differentiating the charge with respect to time gives the negative of the current. We use this to write the charge in terms of the known maximum current. dQ I I = −Q0 cos ( t ) = I 0 cos ( t ) → Q0 = − 0 → Q (t ) = − 0 sin ( t ) I =−   dt (b) As in the figure, attach the inductor to a battery and resistor for an extended period so that a steady state current flows through the inductor. Then at time t = 0, flip the switch connecting the inductor in series to the capacitor.

29. (a) We calculate the resonant frequency using Eq. 30–14. 1 1 1 1 f = = = 17,456 Hz  17.5 kHz 2 LC 2 ( 0.175H ) ( 475  10−12 F ) (b) As shown in Eq. 30–15, we set the peak current equal to the maximum charge (from Eq. 24–1) multiplied by the angular frequency. I = Q0 = CV ( 2 f ) = ( 475  10−12 F ) (135V )( 2 )(17,456 Hz )

= 7.033  10−3 A  7.03mA (c)

We use Eq. 30–6 to calculate the maximum energy stored in the inductor. U = 12 LI 2 = 12 ( 0.175H ) ( 7.033  10−3 A ) = 4.33  10 −6 J 2

30. (a) When the energy is equally shared between the capacitor and inductor, the energy stored in the capacitor will be one-half of the initial energy in the capacitor. We use Eq. 24–5 to write the energy in terms of the charge on the capacitor and solve for the charge when the energy is equally shared.

Q 2 1 Q02 2 = →Q = Q0 2C 2 2C 2 (b) We insert the charge into Eq. 30–13 and solve for the time.  2  T   T 2 1 Q0 = Q0 cos t → t = cos −1   =  =  2  2  2  4  8 31. Since the circuit loses 3.5% of its energy per cycle, it is an underdamped oscillation. We use Eq. 24–5 for the energy with the charge as a function of time given by Eq. 30–19. Setting the change in energy equal to a loss of 3.5% and using Eq. 30–18 to determine the period, we solve for the resistance. − RT Q02 e L cos 2 ( 2 ) Q02 cos 2 ( 0 ) − E RT − RT 2 2C C = = e L − 1 = −0.035 → = ln(1 − 0.035) = 0.03563 2 2 Q0 cos ( 0 ) E L 2C

4 L ( 0.03563) 2 R  2  R = →R= 0.03563 =   2 2 2 L     L 1 LC − R 4 L C 16 2 + ( 0.03563)    2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

4 ( 0.085H )( 0.03563)

(1.00 10 F) 16 + ( 0.03563)  −6

Instructor Solutions Manual

= 1.6533   1.7 

32.

As in the derivation of Eq. 30–16, we set the total energy equal to the sum of the magnetic and electric energies, with the charge given by Eq. 30–19. We then solve for the time that the energy is 65% of the initial energy. Q 2 LI 2 Q02 − RL t Q2 − Rt Q2 − Rt + = U = UE +UB = e cos2 ( t +  ) + 0 e L sin 2 ( t +  ) = 0 e L 2C 2 2C 2C 2C 2 2 Q Q −Rt L L L 0.65 0 = 0 e L → t = − ln ( 0.65) = − ln ( 0.65)  0.43 = 0.43 2C 2C R R R

33.

As shown by Eq. 30–18, adding resistance will decrease the oscillation frequency. We use Eq. 30–14 for the pure LC circuit frequency and Eq. 30–18 for the frequency with added resistance to solve for the resistance.

  = (1 − .0025) R=

1 1 R2 − 2 = 0.9975 LC 4 L LC

→

4L 1 − 0.99752 = C

(

)

(

4 ( 0.35H )

1.6  10−9 F

)

→

(1 − 0.9975 ) = 2090   2.1k 2

34. We find the frequency from Eq. 30–23b for the reactance of an inductor. 760  X = 3779 Hz  3800 Hz X L = 2 fL → f = L = 2 L 2 ( 0.0320 H ) 35. The reactance of a capacitor is given by Eq. 30–25b, X C = (a)

XC =

(b)

XC =

1 2 fC 1 2 fC

= =

(

2 ( 60.0 Hz ) 8.4  10−6 F 1

(

)(

)

2 fC

= 316   320 

2 1.00  10 Hz 8.4  10 F 6

−6

)

= 1.9  10−2 

36. We find the reactance from Eq. 30–23b, and the current from Ohm’s law.

(

)

X L = 2 fL = 2 33.3  103 Hz ( 0.0360 H ) = 7532   7530  V = IX L → I =

V XL

350 V 7532 

= 0.04647 A  4.6  10 −2 A

37. We find the reactance from Eq. 30–23a, and the inductance from Eq. 30–23b. Note that we can use either peak values or rms values in the “Ohm’s law” expression, as in Eq. 30–23a. V V = IX L → X L = I X V 240 V X L = 2 fL → L = L = = = 5.9  10 −2 H 2 f 2 fI 2 ( 60.0 Hz ) 10.8 A © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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38. (a) At  = 0, the impedance of the capacitor is infinite. Therefore, the parallel combination of the resistor R and capacitor C behaves as the resistor only, and so is R. Thus, the impedance of the entire circuit is equal to the resistance of the two series resistors. Z = R + R (b) At  = , the impedance of the capacitor is zero. Therefore, the parallel combination of the resistor R and capacitor C is equal to zero. Thus, the impedance of the entire circuit is equal to the resistance of the series resistor only. Z = R 39. If there is no current in the secondary, there will be no induced emf from the mutual inductance. Therefore, we set the ratio of the voltage to current equal to the inductive reactance and solve for the inductance. Vrms Vrms 110 V = X L = 2 fL → L = = = 7.7  10−2 H I rms 2 f I rms 2 ( 60.0 Hz )( 3.8A ) 40. (a) We find the reactance from Eq. 30–25b. 1 1 XC = = = 7368   7400  2 fC 2 ( 720 Hz ) 3.0  10 −8 F

(

)

(b) We find the peak value of the current from Ohm’s law. V 1600 V I peak = 2 I rms = 2 rms = 2 = 0.31A XC 7368  41. (a) Since the resistor and capacitor are in parallel, they will have the same voltage drop across them, with an amplitude called V0 in the equations below. We use Ohm’s law to determine the amplitude of the current through the resistor, and Eqs. 30–25a and 30–25b to determine the amplitude of the current through the capacitor. The amplitude of the total current is the “sum” of the current amplitudes across each element, taking into account the fact that the two currents have a phase difference of 90. V V I R = 0 ; I C = 0 = V0 ( 2 f C ) R XC IC I R2 + I C2

V0 ( 2 f C )

2 f CR

( 2 f CR ) + 1 ( 2 fC ) + (V0 R ) 2 ( 60.0 Hz ) ( 0.35  10−6 F ) ( 530  ) 2

V02

(

)

= 0.06976  7.0%

 2 ( 60.0 Hz ) 0.35  10 F ( 530  )  + 1   (b) We repeat part (a) with a frequency of 60.0 kHz. 2 6.00  104 Hz 0.35  10−6 F ( 530  ) IC = = 0.999898  100% 2 I R2 + I C2  2 6.00  104 Hz 0.35  10−6 F ( 530  ) + 1   We see that the filter “passes” most of the lower frequency current to the resistor, but does not “pass” the higher frequency current. So the capacitor in parallel with the source is a “low pass” filter.

(

−6

)( )(

)

42. The power is only dissipated in the resistor, so we use the power dissipation equation Eq. 25–10b. 2 Pavg = 12 I 02 R = 12 (1.80A ) (1350  ) = 2187 W  2.19 kW © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1033

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

43. The impedance of the circuit is given by Eq. 30–28a without a capacitive reactance. The reactance of the inductor is given by Eq. 30–23b. (a) Z =

R 2 + X L2 =

R 2 + 4 2 f 2 L2 =

(10.0  10  ) + 4 ( 55.0 Hz ) ( 0.0340 H )

R 2 + 4 2 f 2 L2 =

(10.0  10  ) + 4 (5.50  10 Hz ) ( 0.0340 H )

= 1.00  104  (b) Z =

R 2 + X L2 =

= 1.54  104  44. The impedance of the circuit is given by Eq. 30–28a without an inductive reactance. The reactance of the capacitor is given by Eq. 30–25b. We assume that 60 Hz has 2 significant figures. (a) Z =

R 2 + X C2 =

R2 +

1 4 f C 2

( 75  )2 +

(

4 2 ( 60.0 Hz ) 5.8  10−6 F 2

)

= 463 

 460 

(b) Z =

R 2 + X C2 =

R2 +

1 4 f C 2

(

4 2 6.0  106 Hz

)( 2

5.8  10−6 F

)

= 75 

45. We find the impedance from Eq. 30–27. 120 V V Z = rms = = 1700   2000  I rms 70  10 −3 A 46. Use Eq. 30–28c, with no capacitive reactance.

R 2 + X L2

→ R=

( 275  )2 − (145  )2 = 234 

Z 2 − X L2 =

47. The impedance is given by Eq. 30–28a with no capacitive reactance.

R 2 + ( 2 f L )

R 2 + X L2 =

Z f = 2 Z 60 →

R 2 + 4 2 f 2 L2 = 2 R 2 + 4 2 ( 60.0 Hz ) L2 2

→

R 2 + 4 2 f 2 L2 = 4  R 2 + 4 2 ( 60.0 Hz ) L2  = 4 R 2 + 16 2 ( 60.0 Hz ) L2 → 2



3R 2 + 16 2 ( 60.0 Hz ) L2 2

f =

4 L

2 2



3R 2

+ 4 ( 60.0 Hz ) = 4 2 L2 2

3 (1500  )

4 ( 0.37 H ) 2

+ 4 ( 60.0 Hz )

= 1124 Hz  1100 Hz

48. We use the rms voltage across the resistor to determine the rms current through the circuit. Then, using the rms current and the rms voltage across the capacitor in Eq. 30–25 we determine the frequency. VR ,rms I I rms = VC ,rms = rms 2 fC R

1034

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

f =

VR ,rms ( 3.0 V ) I rms = = = 117.9 Hz  120 Hz 2 CVC ,rms 2 CRVC ,rms 2 2.0  10−6 C ( 750  )( 2.7 V )

(

)

Since the voltages across the resistor and capacitor are not in phase, the rms voltage across the power source will not be the sum of their rms voltages. 49. The total impedance is given by Eq. 30–28a.

1   Z = R + ( X L − X C ) = R +  2 fL −  2 fC   2

  1 = ( 8.70  10  ) +  2 (1.00  104 Hz )( 3.20  10 −2 H ) −  2 (1.00  104 Hz )( 6.25  10 −9 F )  

= 8716.5   8.72 k

The phase angle is given by Eq. 30–29a.

 = tan −1

XL − XC R

2 fL − = tan −1

(

2 1.00  104 Hz = tan

−1

1 2 fC

)( 3.20  10 H ) − 2 1.00  10 Hz1 6.25  10 F ( )( ) −2

−9

R −535.9 

= tan −1

= −3.52 8.70  103  The voltage is lagging the current, or the current is leading the voltage. The rms current is given by Eq. 30–27. 625 V V I rms = rms = = 7.17  10 −2 A Z 8716.5  50. (a) The rms current is the rms voltage divided by the impedance. The impedance is given by Eq. 30–28a with no capacitive reactance.

Z = R 2 + X L2 = I rms = =

Vrms Z

R 2 + ( 2 fL ) Vrms

R + 4 f L 2

2 2

120 V

( 965  ) + 4 ( 60.0 Hz )2 ( 0.255 H )2 2

120 V

= 0.1237 A  0.12 A 969.8  (b) The phase angle is given by Eq. 30–29a with no capacitive reactance. 2 ( 60.0 Hz )( 0.255 H ) X 2 fL = tan −1 = 5.69  = tan −1 L = tan −1 R R 965  The current is lagging the source voltage. 2 R = ( 0.1237 A ) ( 965  ) = 14.8 W  15 W . (c) The power dissipated is given by P = I rms 2

1035

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(d) The rms voltage reading is the rms current times the resistance or reactance of the element.

Vrms = I rms R = ( 0.1237 A )( 965  ) = 119.4 V  120 V R

Vrms = I rms X L = I rms 2 fL = ( 0.1237 A ) 2 ( 60.0 Hz )( 0.255 H ) = 11.89 V  12 V L

Note that, because the maximum voltages occur at different times (i.e., the are out of phase), the two individual voltage readings do not add to the applied voltage of 120 V (even if keeping more significant figures). 51. (a) The current is found from the voltage and impedance. The impedance is given by Eq. 30–28a.

R2 + ( X L − X C ) = 2



R 2 +  2 fL −



  2 fC  1

  1 = ( 2.0  ) +  2 ( 60.0 Hz )( 0.035 H ) −  = 113.2  2 ( 60.0 Hz ) ( 21  10−6 F )   2

I rms =

Vrms

45 V

= 0.3975 A  0.40 A 113.2  Z (b) Use Eq. 30–29a to find the phase angle. 1 2 fL − X − XC 2 fC  = tan −1 L = tan −1 R R 2 ( 60.0 Hz )( 0.035 H ) − = tan −1

(

2 ( 60.0 Hz ) 21  10−6 F

2.0 

) = tan −113.1 = −89 −1

2.0 

2 R = ( 0.3975 A ) ( 2.0  ) = 0.32 W . (c) The power dissipated is given by P = I rms 2

52. For the current and voltage to be in phase, the reactances of the capacitor and inductor must be equal. Setting the two reactances equal enables us to solve for the capacitance. 1 1 1 X L = 2 fL = X C = →C = 2 2 = 2 = 5.7  F 2 fC 4 f L 4 ( 420 Hz )2 ( 0.025H ) 53.

The light bulb acts like a resistor in series with the inductor. Using the desired rms voltage across the resistor and the power dissipated by the light bulb we calculate the rms current in the circuit and the resistance. Then using this current and the rms voltage of the circuit we calculate the impedance of the circuit (Eq. 30–27) and the required inductance (Eq. 30–28b). V P 75W 120 V I rms = = = 0.625A R = R ,rms = = 192  VR ,rms 120 V I rms 0.625A

Z= L=

Vrms 2 = R 2 + ( 2 fL ) I rms 1 2 f

 Vrms   240 V  1 2 2   −R =   − (192  ) = 0.88 H 2 ( 60.0 Hz )  0.625A   I rms 

1036

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

54. We multiply the instantaneous current by the instantaneous voltage to calculate the instantaneous power. Then using the trigonometric identity for the summation of sine arguments (inside back cover of text) we can simplify the result. We integrate the power over a full period and divide the result by the period to calculate the average power. P = IV = ( I 0 sin  t )V0 sin ( t +  ) = I 0V0 sin  t ( sin  t cos  + sin  cos  t )

(

= I 0V0 sin 2  t cos  + sin  t cos  t sin  P=

1 T



)

2

 PdT = 2   I V (sin  t cos + sin  t cos  t sin  ) dt T

0 0

2

  I 0V0 cos    sin 2 t dt + I 0V0 sin    sin t cos t dt = 0 0 2 2 2

55.

1    1 2   I 0V0 cos   I 0V0 sin   sin 2  t   = 12 I 0V0 cos  +  0  2  2   2  

Given the resistance, inductance, capacitance, and frequency, we calculate the impedance of the circuit using Eq. 30–28b. X L = 2 f L = 2 ( 660 Hz )( 0.025H ) = 103.67  XC =

1 1 = = 120.57  2 f C 2 ( 660 Hz ) 2.0  10−6 F

(

Z = R2 + ( X L − X C ) = 2

)

(120  ) + (103.67  − 120.57  ) = 121.18  2

(a) From the impedance and the peak voltage we calculate the peak current, using Eq. 30–27. V 340 V I0 = 0 = = 2.806 A  2.8A Z 121.18  (b) We calculate the phase angle of the current from the source voltage using Eq. 30–29a. X − XC 103.67  − 120.57   = tan −1 L = tan −1 = −8.0 R 120  This means that the current is LEADING the source voltage in this circuit. (c) We multiply the peak current times the resistance to obtain the peak voltage across the resistor. The voltage across the resistor is in phase with the current, so the phase angle is the same as in part (b). V0,R = I 0 R = ( 2.806A )(120  ) = 336.7 V  340V ;  = −8.0 (d) We multiply the peak current times the inductive reactance to calculate the peak voltage across the inductor. The voltage in the inductor is 90º ahead of the current. Subtracting the phase difference between the current and source from the 90º between the current and inductor peak voltage gives the phase angle between the source voltage and the inductive peak voltage. V0,L = I 0 X L = ( 2.806 A )(103.67  ) = 290.9 V  290 V

L = 90.0 −  = 90.0 − ( −8.0 )  = 98.0 (e) We multiply the peak current times the capacitive reactance to calculate the peak voltage across the capacitor. Subtracting the phase difference between the current and source from the –90º between the current and capacitor peak voltage gives the phase angle between the source voltage and the capacitor peak voltage. V0,C = I 0 X C = ( 2.806 A )(120.57  ) = 338.3V  340 V

C = −90.0 −  = −90.0 − ( −8.0 ) = −82.0 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1037

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

56. Using Eq. 30–23b we calculate the impedance of the inductor. Then we set the phase shift in Eq. 30–29a equal to 25º and solve for the resistance. We calculate the output voltage by multiplying the current through the circuit, from Eq. 30–27, by the inductive reactance (Eq. 30–23b). X L = 2 fL = 2 ( 225Hz )( 0.055H ) = 77.75  tan  = Voutput V0

XL X 77.75  =→ R = L = = 166.7   170  R tan  tan 25

VR IR R = = = V0 IZ Z

166.7 

(166.7  ) + ( 77.75  ) 2

= 0.91

57. The resonant frequency is found from Eq. 30–32. The resistance does not influence the resonant frequency.

f0 =

1 2

1 LC

1 2

( 26.0  10 H )( 3400  10 F ) −6

−12

= 5.4  105 Hz

58. We calculate the resonant frequency using Eq. 30–32 with the inductance and capacitance given in the example. We use Eq. 30–30 to calculate the power dissipation, with the impedance equal to the resistance. 1 1 f0 = = = 265 Hz 2 LC 2 ( 0.0300 H ) 12.0  10−6 F

(

)

2 ( 90.0V ) = 324 W V  R V P = I rmsVrms cos  =  rms Vrms   = rms = R 25.0   R  R 2

59. (a) We find the capacitance from the resonant frequency, Eq. 30–32.

f0 =

1 2

1 LC

→ C=

1 4 Lf 2

2 0

1 4

( 4.15  10 H )( 38.0  10 Hz ) 3

−3

= 4.23  10−9 F

(b) At resonance the impedance is the resistance, so the current is given by Ohm’s law. Vpeak 136 V I peak = = = 35.8 mA R 3.80  103  60. (a) The peak voltage across the capacitor is the peak current multiplied by the capacitive reactance. We calculate the current in the circuit by dividing the source voltage by the impedance, where at resonance the impedance is equal to the resistance. V0 V 1 V0 1 VC 0 = X C I 0 = = = 0 T0 2 f 0C R 2 ( RC ) f 0 2 (b) We set the amplification equal to 125 and solve for the resistance. 1 1 1 T →R= = = 130  = 0 = 3 2 2 f 0 RC 2 f 0  C 2 5.0  10 Hz (125) 2.0  10−9 F

(

)

(

)

61. (a) We calculate the resonance frequency from the inductance and capacitance using Eq.30–32. 1 1 f0 = = = 21460 Hz  21 kHz 2 LC 2 ( 0.055 H ) 1.0 10−9 F

(

)

1038

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

(b) We use the result of Problem 60 to calculate the voltage across the capacitor. V0 1 2.0V VC 0 = = = 420 V 2 ( RC ) f 0 2 ( 35  ) 1.0  10−9 F ( 21460 Hz )

(

)

(c) We divide the voltage across the capacitor by the voltage source. VC 0 420 V = = 210 V0 2.0 V 62. (a) We write the charge on the capacitor using Eq. 24–1, where the voltage drop across the capacitor is the inductive capacitance multiplied by the circuit current (Eq. 30–25a) and the circuit current is found using the source voltage and circuit impedance (Eqs. 30–27 and 30–28b). CV0 V0 V  Q0 = CVC 0 = CI 0 X C = C  0  X C = = 2 2 Z  C R 2 + ( L − 1  C ) 2 R2 + 2 L − 1 C

(

)

(b) We set the derivative of the charge with respect to the angular frequency equal to zero to calculate the angular frequency at which the charge is a maximum. 2 −1/2 dQ0 d V0 d = = V0  2 R 2 +  2 L − 1 C  2  d d d    2 R 2 +  2 L − 1 C

(

)

( (

) )

)

(

)

2 R + 2   L − 1 C 2 L 2 R 2 + 4 3 L2 − 4 L C 1 1 = − V0 = − V0 =0 3 3 2 2  2 R 2 +  2 L − 1 C 2  2  2 R 2 +  2 L − 1 C 2  2     2

(

)

→ 2 R 2 + 4 3 L2 − 4 L C = 0 → 2 2 L2 = 2 L C − R 2 → 1 R2 − 2 LC 2 L (c) The amplitude in a forced damped harmonic oscillation is given by Eq. 14–23, and is shown here: F0 V0 A= compare to Q0 = 2 2 2  R + ( L − 1  C ) m  2 − 02 + b2 2 m 2

 =

(

)

This is equivalent to the LRC circuit with A  Q0, F0  V0 , m  L, 0 =

k  m

1 and LC

b  R. Specifically, L corresponds to m and C corresponds to 1/k .

63. Since the circuit is in resonance, we use Eq. 30–32 for the resonant frequency to determine the necessary inductance. We set this inductance equal to the solenoid inductance calculated in Example 30–3, with the area of the solenoid equal to the area of a circle of radius r. The number of turns (N) is equal to the length of the wire divided by the circumference of a turn, and the length of the solenoid is equal to the diameter of the wire (d) multiplied by the number of turns. We solve the resulting equation for the number of turns. f0 =

1 2 LC

→ L=

0 N 2 A 1 = = 4 2 f 02C l solenoid

l



0  wire   r 2  2 r  Nd

→

1039

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

 f 02C 0 l 2wire d

( =

 35.0  103 Hz

Instructor Solutions Manual

) ( 2.20  10 F )( 4 10 T m A ) (12.0 m ) 2

−7

1.1  10−3 m

= 139.27loops  140 loops

64. (a) We write the average power using Eq. 30–30, with the current in terms of the impedance (Eq. 30–27) and the power factor in terms of the resistance and impedance (Eq. 30–29b). Finally we write the impedance using Eq. 30–28b.

Vrms R V2 R V02 R Vrms = rms2 = 2 Z Z Z 2  R 2 + ( L − 1  C )    (b) The power dissipation will be a maximum when the inductive reactance is equal to the capacitive reactance, which is the resonant frequency. 1 f res = 2 LC (c) We set the power dissipation equal to ½ of the maximum power dissipation and solve for the angular frequencies. V02 R 1 1  V02 R  P = Pmax → =  2  → ( L − 1 C ) =  R 2 2 2  R 2 + ( L − 1  C )  2  2 R    P = I rmsVrms cos  =

 RC  R 2C 2 + 4 LC 2 LC We require the angular frequencies to be positive and for a sharp peak, R 2C 2 4 LC. The angular width will then be the difference between the two positive frequencies. 2 LC  RC 1 R R   1 R  R  1 = =  →  =  + − −  = 2 LC LC 2 L  LC 2 L   LC 2 L  L → 0 =  2 LC  RC − 1 →  =

65. The power on each side of the transformer must be equal. We replace the currents in the power equation with the number of turns in the two coils using Eq. 29–6. Then we solve for the turn ratio.





Z I Pp = I p2 Z p = Ps = I s2 Z s → p =  s  Z s  I p  →

Np Ns

Zp Zs

N  = p   Ns 

45  103  = 75 8.0 

3V 2 1 2 V1 + V22 + V32 = 0 . Recall that sin ( a + b ) = sin a cos b + cos a sin b. 2R R Also recall that cos ( 2 3) = cos ( 4 3) = −1 2, and sin ( 2 3) = − sin ( 4 3) = 3 2 .

66. We are to show that P =

(

)

(

)

V12 = V02 sin 2  t V22 = V02 sin 2 ( t + 2 3) = V02 sin  t cos ( 2 3) + cos  t sin ( 2 3 )  = V02 sin  t ( −1 2 ) + cos  t 

( 3 2) = V  sin  t − ( 3 4 ) sin  t cos  t + cos  t  2

2 0

1 4

3 4

1040

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

V32 = V02 sin 2 ( t + 4 3) = V02 sin  t cos ( 4 3) + cos  t sin ( 4 3) 

(

)

(

)

= V02 sin  t ( −1 2 ) + cos  t − 3 2  = V02  14 sin 2  t + 3 4 sin  t cos  t + 34 cos 2  t      sin 2  t +  1 sin 2  t − 3 4 sin  t cos  t + 3 cos 2  t   4 4  2 2 2 2 V1 + V2 + V3 = V0   2 2  +  14 sin  t + 3 4 sin  t cos  t + 43 cos  t      

(



)



= V02 sin 2  t +  14 sin 2  t + 43 cos2  t  +  14 sin 2  t + 43 cos2  t 





= 23 sin 2  t + 23 cos2  t = 23 V02

3V02 1 2 1 3 2 2 2 P = V1 + V2 + V3 = V0 = 2R R R 2

(

)

(

)

67. We use Eq. 30–6 to calculate the initial energy stored in the inductor.

U 0 = 12 LI 02 = 12 ( 0.0600 H )( 0.0500A ) = 7.50  10−5 J Set the inductor’s energy equal to 8.0 times the initial energy and solve for the current. We set the current equal to the initial current plus the rate of increase multiplied by time and solve for the time. 2

U = 12 LI 2

→ I=

I = I0 +  t → t =

(

) = 141.4 mA

2U = L

2 8.0  7.50  10−5 J

I − I0

141.4 mA − 50.0 mA = 1.17s  1.2s 78.0 mA/s



0.0600 H

68. When the currents have acquired their steady-state values, the capacitor will be fully charged, and so no current will flow through the capacitor. At this time, the voltage drop across the inductor will be zero, as the current flowing through the inductor is constant. Therefore, the current through R1 is zero, and the resistors R2 and R3 can be treated as in series. V0 12 V I1 = I 3 = = = 2.4 mA ; I 2 = 0 R1 + R3 5.0k 69. (a) The self inductance is written in terms of the magnetic flux in the toroid using Eq. 30–4. We use the magnetic field of a toroid, from Example 28–9. The field is dependent upon the distance from the center of the toroid, but if the diameter of the toroid loops is small compared with the radius of the toroid, as given in the problem, the magnetic field can be can be treated as approximately constant. 2 N  B NAtoroid loop Bcener of toroid loops N  d 4 ( 0 NI 2 r0 ) 0 N 2 d 2 L=  = = 8r0 I I I

(

(b)

)

This is consistent with the inductance of a solenoid with length l = 2 r0 . It should

be consistent, thinking of a toroid as a solenoid that has its two ends joined together. (c) Calculate L from the given data, with r0 equal to half of the diameter. L=

0 N 2 d 2 8r0

( 4  10 T m/A ) (660 ) (0.020 m ) = 83  H = 2

−7

8 ( 0.33m )

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

70.

Instructor Solutions Manual

We use Eq. 30–4 to calculate the self inductance between the two wires. We calculate the flux by integrating the magnetic field from the two wires, using Eq. 28–1, over the rectangular region between the two wires. That rectangular region has a length of “h” and a width of l − 2r (the distance from the outer edge of one wire to the outer edge of the other wire). Dividing the inductance by the length of the wire gives the inductance per unit length. 0 I   h l −r  1  1 l − r  0 I 1  + + L= B = hdr  = 0    dr  I I r  2 r  2 ( l − r  )  2 r  r  ( l − r  ) 



l −r   l −r  L 0  r   0  l − r  ln ( r  ) − ln ( l − r  )  = 0 ln  = − ln  ln    =  r h 2 2   r   l − r    r 

71. The magnetic energy is the energy density (Eq. 30–7) multiplied by the volume of the spherical shell enveloping the earth. The volume of a relatively thin spherical shell, like the first 5 km above the Earth’s surface, is the surface area of the sphere times its thickness.

(

)

0.50  10−4 T B2 2  4 6.38  106 m 2 5.0  103 m  U = uBV = 4 REarth h = −7  2 0 2 4  10 T m A 

(

)

(

)

)(

)

= 2.5  1015 J

72. (a) For underdamped oscillation, the charge on the capacitor is given by Eq. 30–19, with  = 0. Differentiating the current with respect to time gives the current in the circuit. dQ −Rt −Rt R  cos  t +   sin  t  = −Q0e 2 L  Q (t ) = Q0e 2 L cos  t ; I (t ) = dt  2L  The total energy is the sum of the energies stored in the capacitor (Eq. 24–5) and the energy stored in the inductor (Eq. 30–6). Since the oscillation is underdamped (  R / 2L ), the cosine term in the current is much smaller than the sine term and so will be ignored. The frequency of oscillation is approximately equal to the undamped frequency of Eq. 30–14.

(Q e Q LI U =U +U = + = 2

Q02 e

− RL t

(

− 2RL t

cos  t

) + L (Q e ) ( sin t ) 2

− 2RL t

Q02 e

)

− RL t

cos  t +   LC sin  t  2C 2C (b) We differentiate the energy with respect to time to show the average power dissipation. We then set the power loss per cycle equal to the resistance multiplied by the square of the current. For a lightly damped oscillation, the exponential term does not change much in one cycle, while the sine squared term averages to ½. −Rt 2 −Rt RQ02 e L dU d  Q0 e L  = =− 2 LC dt dt  2C    =

−Rt

RQ02 e L 1  1  = −   sin  t  −Q0 e  P = − I R = −Q0 e   2 LC  LC  2  The rate of energy change in the circuit is equal to the power dissipated by the resistor. 2

2 − RL t

(

)

2 − RL t 

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Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

73. The self-inductance of an air-filled solenoid was determined in Example 30–3. We solve this equation for the length of the tube, using the diameter of the wire as the length per turn. L=

l =

0 N 2 A l Ld

=  0 n 2 Al =

 0 r 2

0 Al d2

(1.0H ) ( 0.81  10−3 m )

( 4 10 T m/A )  ( 0.060 m ) −7

= 46.16 m  46 m

The length of the wire is equal to the number of turns (the length of the solenoid divided by the diameter of the wire) multiplied by the circumference of the turn. 46.16 m l L = D =  ( 0.12 m ) = 21,490 m  21km d 0.81  10−3m The resistance is calculated from the resistivity, area, and length of the wire. −8  l (1.68  10  m ) ( 21,490 m ) R= = = 0.70 k = 700  2 A  ( 0.405  10−3 m ) 74. We use Kirchhoff’s loop rule to equate the input voltage to the voltage drops across the inductor and resistor. Note that Vout = IR. We then multiply both sides of the loop rule equation by the integrating Rt

factor e L and integrate the right-hand side of the equation using a u substitution with u = IRe L , Rt du dI R Rt R  dI  Rt considering that I = I ( t ) and Re L + IR e L =  L + IR  e L . = dt dt L L  dt  dI Vin = L + IR → dt Rt L du L L  dI  Rt Vin e L dt =  L + IR e L dt = dt = du = u R dt R R  dt  Rt Rt L L = IRe L = Vout e L R R



For L / R t , e L  1. Setting the exponential term equal to unity on both sides of the equation gives the desired result. L R Vin dt = Vout → Vout = Vin dt → Vout  Vin dt R L We see that when Vin is a positive constant value, the integral (the area under the graph of Vin ) increases linearly to a maximum, and then when Vin switches to a negative constant value, the integral decreases linearly to 0. Then the entire cycle repeats.



75. (a) Since the capacitor and resistor are in series, the impedance of the circuit is given by Eq. 30–28a. Divide the source voltage by the impedance to determine the magnitude of the current in the circuit. Then multiply amplitude of the current by the resistance to determine the amplitude of the voltage drop across the resistor. V Vin R VR = IR = in R = 2 2 Z R + 1 ( 2 fC )

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(130 mV )(550  )

= 40.93mV  41mV 2 2 −6    550 1 2 60.0 Hz 1.6 10 F  +  ( ) )  (  (b) Repeat the calculation with a frequency of 6.0 kHz. (130 mV )(550  ) VR = = 129.9 mV  130 mV 2 2 3 6 − (550  ) + 1 2 6.0  10 Hz 1.6 10 F  Thus, the capacitor allows the higher frequency to pass with its amplitude almost unaffected, but attenuates the amplitude at the lower frequency by about a factor of 3.

(

)

(

)(

)

76. We set the power factor equal to the resistance divided by the impedance (Eq. 30–29b) with the impedance written in terms of the angular frequency (Eq. 30–28b). We rearrange the resulting equation to form a quadratic equation in terms of the angular frequency. We divide the positive angular frequencies by 2 to determine the desired frequencies.

cos  =

R = Z

R R 2 + ( L − 1 C )

(

 1  →  2 LC  C R 2  − 1 − 1 = 0 2  cos  

) (

 2 ( 0.033H ) 55  10−9 F   55  10 −9 F

1  − 1 − 1 = 0 ) (1500  )  0.24  2

(1.815  10 F H )   (3.337 10  F )  − 1 = 0 −9

−4

3.337  10−4 Ω F  3.444  10−4 Ω F 3.337  10 −4 Ω F  3.444  10−4 Ω F = 3.63  10−9 F H 3.63  10−9 F H 1.868  105 rad s,  2.948  103 rad s

=

f =

 1.868  105 rad s 2.948  103 rad s = = 3.0  104 Hz and = 4.7  102 Hz 2 2 2

77. (a) We set V = V0 sin  t and assume the inductive reactance is greater than the capacitive reactance. The current will lag the voltage by an angle . The voltage across the resistor is in phase with the current and the voltage across the inductor is 90º ahead of the current. The voltage across the capacitor is smaller than the voltage in the inductor, and anti-parallel to it. Since V = V0 sin  t, the instantaneous voltage is the projection of the V0 on the y-axis. (b) From the diagram, the current is the projection of the maximum current onto the y-axis, with the current lagging the voltage by the angle . This is the same angle obtained in Eq. 30–29a. The magnitude of the maximum current is the voltage divided by the impedance, Eq. 30–28b. V0 L − 1  C I (t ) = I 0 sin ( t −  ) = sin ( t −  ) ;  = tan −1 2 R R 2 + ( L − 1  C )

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Inductance, Electromagnetic Oscillations, and AC Circuits

78. (a) We integrate the power directly from the current and voltage over one cycle. 2 2     1 T P= IVdt = I 0 sin (t )V0 sin (t + 90 ) dt = I 0 sin (t )V0 cos (t ) dt T 0 2 0 2 0



2

 sin (t )  I 0V0  2  2   2 I 0V0 = = sin    − sin ( 0 )  = 0 2 2 4      2

(b) We apply Eq. 30–30, with  = 90 . P = I rmsVrms cos90 = 0 As expected, the average power is the same for both methods of calculation.

79. Since the current lags the voltage, one of the circuit elements must be an inductor. Since the phase angle is not a right angle, the other element must be a resistor. We use 30–29a to write the resistance in terms of the impedance. Then, using Eq. 30–27 to determine the impedance from the voltage and current and Eq. 30–28b, we solve for the unknown inductance and resistance. 2 fL → R = 2 fL cot  tan = R V 2 2 2 Z = rms = R 2 + ( 2 fL ) = ( 2 fL cot  ) + ( 2 fL ) = 2 fL 1 + cot 2  I rms L=

Vrms

2 f I rms 1 + cot 2 

120 V 2 ( 60 Hz )( 5.6 A ) 1 + cot 2 55

= 46.56 mH  47 mH

R = 2 f L cot  = 2 ( 60 Hz )( 46.56 mH ) cot 55 = 12.29   12 

80. (a) We calculate capacitive reactance using Eq. 30–25b. Then, using the resistance and capacitive reactance we calculate the impedance. Finally, we use Eq. 30–27 to calculate the rms current. 1 1 XC = = = 1474  2 f C 2 ( 60.0 Hz ) 1.80  10−6 F

(

Z = R 2 + X C2 =

)

( 6.40  10  ) + (1474  ) = 6568  3

Vrms 120 V = = 18.27 mA  18mA Z 6568  (b) We calculate the phase angle using Eq. 30–29a. − XC −1474   = tan −1 = tan −1 = −13.0 R 6.40  103  (c) The average power is calculated using Eq. 30–30. P = I rmsVrms cos  = 18.27  10−3 A (120 V ) cos ( −13.0 ) = 2.1W I rms =

(

)

(d) The voltmeter will read the rms voltage across each element. We calculate the rms voltage by multiplying the rms current through the element by the resistance or capacitive reactance. VR = I rms R = (18.27 mA )( 6.40 k ) = 117 V  120 V

VC = I rms X C = (18.27 mA )(1.474 k ) = 26.9 V  27 V Note that since the voltages are out of phase they do not sum to the applied voltage, even if extra significant figures are kept. However, since they are 90º out of phase their squares sum to the square of the input voltage. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

81. We find the resistance using Ohm’s law with the dc voltage and current. We then calculate the impedance from the ac voltage and current, and using Eq. 30–28b. V 45V V 120 V R= = = 20.45   2.0  101  ; Z = rms = = 31.58  I 2.2 A I rms 3.8A

( 31.58  ) − ( 20.45  ) = 63.83mH  64 mH 2 ( 60 Hz ) 2

Z 2 − R2 = 2 f

Z = R + ( 2 fL ) → L = 2

82. (a) From the text of the problem, the Q factor is the ratio of the voltage across the capacitor or inductor to the voltage across the resistor, at resonance. The resonant frequency is given by Eq. 30–32.

VL VR

I res X L I res R

2 f 0 L R

2 =

2

LC R

L =

(b) Find the inductance from the resonant frequency, and the resistance from the Q factor. f0 =

→ 2 LC 1 1 = = 2.533  10 −6 H  2.5  10 −6 H L= 2 2 2 − 2 8 6 4 Cf 0 4 1.0  10 F 1.0  10 Hz

(

83.

1 R

L C

→ R=

1 Q

L C

)(

)

2.533  10 −6 H

550

−8

1.0  10 F

= 2.9  10 −2 

We calculate the period of oscillation as 2 divided by the angular frequency. Then write the total energy of the system at the beginning of each cycle using Eq. 24–5, with the charge given by Eq. 30–19, with cos ( t +  ) = cos   ( t + T ) +   = 1. We take the difference in energies at the beginning and end of a cycle, divided by the initial energy. For small damping, the argument of the resulting exponential term is small and we replace it with the first two terms of the Taylor series expansion. 2 2 T=    −Rt

Q02 e L cos2 ( t +  ) −Rt

U max =

 2  − RL  t + 

U Q02 e L − Q02e  = −Rt U Q02 e L

 

−Rt

Q 2e L = 0 2C

2 R

−  2 R  2 R 2 = 1 − e L  1 −  1 − = = Q  L   L 

84. (a) We write the oscillation frequency in terms of the capacitance using Eq. 30–14, with the parallel plate capacitance given by Eq. 24–2. We then solve the resulting equation for the plate separation distance. 1 1 = → x = 4 2 A 0 f 2 L 2 f = LC L ( 0 A / x ) (b) For small variations we can differentiate x and divide the result by x to determine the fractional change. 2 dx 4 A 0 ( 2 f df ) L 2 df x 2 f dx = 4 2 A 0 ( 2 f df ) L ; = = →  x f x f 4 2 A 0 f 2 L © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1046

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Inductance, Electromagnetic Oscillations, and AC Circuits

(c) Inserting the given data, we can calculate the fractional variation on x. x 2 (1Hz )  = 2  10−6 = 0.0002% x 1MHz 85. Putting an inductor in series with the device will protect it from sudden surges in current. The growth of current in an LR circuit is given is Eq. 30–9. V I= 1 − e −tR L = I max 1 − e −tR L R The maximum current is 33 mA, and the current is to have a value of 7.5 mA after a time of 75 microseconds. Use this data to solve for the inductance. I → I = I max 1 − e −tR L → e −tR L = 1 − I max

(

)

(

L=−

(

)

( 75  10 sec ) (120  ) = 3.491  10 H =− −6

−2

 I  ln  1 −   I max 

 7.5 mA  ln  1 − 33mA  

Put an inductor of value 35 mH in series with the device. 86. (a) We use Eq. 30–28b to calculate the impedance and Eq. 30–29a to calculate the phase angle. X L =  L = ( 688rad s )( 0.0220 H ) = 15.136 

(

)

X C = 1  C = 1 ( 688rad s ) 0.42  10−6 F = 3460.7  Z = R2 + ( X L − X C ) = 2

( 23.2  10  ) + 15.136  − 3460.7  = 23.5k 3

X L − XC 15.136  − 3460.7  = tan −1 = −8.45 R 23.2  103  (b) We use Eq. 30–30 to obtain the average power. We obtain the rms voltage by dividing the maximum voltage by 2. The rms current is the rms voltage divided by the impedance.

 = tan −1

2 ( 0.95V ) cos −8.45 = 19  W Vrms V2 cos  = 0 cos  = ( ) 2Z Z 2 23.5  103  2

P = I rmsVrms cos  =

(

)

(c) The rms current is the peak voltage, divided by 2 , and then divided by the impedance. V 2 0.95V 2 I rms = 0 = = 2.859  10−5 A  29  A 3 Z 23.5  10  The rms voltage across each element is the rms current times the resistance or reactance of the element. VR = I rms R = 2.859  10−5 A 23.2  103  = 0.66V

(

)( ) V = I X = ( 2.859  10 A ) ( 3460.7  ) = 0.099 V V = I X = ( 2.859  10 A ) (15.136  ) = 4.3  10 V C

rms

−5

−4

87. (a) The impedance of the circuit is given by Eq. 30–28b with X L  X C and R = 0. We divide the magnitude of the ac voltage by the impedance to get the magnitude of the ac current in the circuit. Since X L  X C , the voltage will lead the current by  =  2. No dc current will flow through the capacitor since the resistance is basically 0. The time constant for charging the © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1047

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

capacitor would be very small, and so the capacitor charges up almost instantly and blocks the dc current. V V20 2 Z = R 2 + ( L − 1 C ) =  L − 1 C I C = 20 = Z  L − 1 C

I (t ) =

V20 sin ( t −  2 ) L − 1  C

(b) The voltage across the capacitor at any instant is equal to the charge on the capacitor divided by the capacitance. This voltage is the sum of the ac voltage and dc voltage. There is no dc voltage drop across the inductor so the dc voltage drop across the capacitor is equal to the input dc voltage. Q V2 out = Vout,ac = Vout − V1 = − V1 C We treat the emf as a superposition of the ac and dc components. At any instant of time the sum of the voltage across the inductor and capacitor will equal the input voltage. We use Eq. 30–5 to calculate the voltage drop across the inductor. Subtracting the voltage drop across the inductor from the input voltage gives the output voltage. Finally, we subtract off the dc voltage to obtain the ac output voltage.  dI d  V20 V20 L VL = L = L  sin ( t −  2 ) = cos ( t −  2 ) dt dt   L − 1  C  L − 1  C

V20 L sin ( t ) L − 1  C

 V20 L  Vout = Vin − VL = V1 + V20 sin  t −  sin ( t )    L − 1 C     1 C  L = V1 + V20  1 −  sin ( t ) = V1 − V20   sin ( t )  L − 1  C   L − 1  C   1 C   V20  V2 out = Vout,ac = Vout − V1 = −V20   sin ( t ) =  2  sin ( t −  )   LC − 1   L − 1  C 

 V  The amplitude is  2 20  and the phase shift is 180 relative to the input voltage.  LC 1 −   (c) The attenuation of the ac voltage is greatest when the denominator is large. 1  2 LC 1 →  L → X L XC C We divide the output ac voltage by the input ac voltage to obtain the attenuation. V20 V2,out  2 LC − 1 1 1 = = 2  2 V2,in V20  LC − 1  LC (d) The dc output is equal to the dc input, since there is no dc voltage drop across the inductor. V1,out = V1 88. Since no dc current flows through the capacitor, there will be no dc current through the resistor. Therefore the dc voltage passes through the circuit with little attenuation. The ac current in the circuit is found by dividing the input ac voltage by the impedance (Eq. 30–28b) We obtain the output ac voltage by multiplying the ac current by the capacitive reactance. Dividing the result by the input ac voltage gives the attenuation. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1048

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

V2,out = IX C =

V20 X C R

+ X C2

→

V2,out V20

1 R  C +1 2



1 R C

1 1. If there is no R C dc current flowing through the capacitor, then there will be no dc current flowing through the resistor either, and so there is no dc voltage drop from the input voltage. If R is large, then there is significant attenuation of the ac voltage because

89. Since the three elements are connected in parallel, at any given instant in time all three will have the same voltage drop across them. Thus the voltages across each element will be in phase with the source. (a) The current in R is in phase with the voltage source. It’s magnitude is given by Ohm’s law. V I R (t ) = 0 sin  t . R (b) The current through L will lag behind the voltage by /2, with magnitude equal to the voltage source divided by the inductive reactance.

I L (t ) =

 V0  sin   t −  XL 2 

(c) The current through C leads the voltage by /2, with magnitude equal to the voltage source divided by the capacitive reactance.

I C (t ) =

 V0  sin   t +  XC 2 

(d) The total current is the sum of the currents through each element. We use a phasor diagram to add the currents, as was used in Section 30–8 to add the voltages with different phases. The net current is found by subtracting the current through L from the current through C. Then using the Pythagorean theorem to add the current through the resistor. We use the tangent function to find the phase angle between the current and voltage source.

I0 =

I R2 0 +

( I C 0 − I L0 )

V  V 1  V   V  =  0  +  0 − 0  = 0 1 +  RC − R R L   R   XC X L  

 V R  I (t ) = 0 1 +  R C − sin ( t +  )  L  R  V0 V − 0  R R  R  X XL −1  →  = tan −1  − tan  = C  = tan  R C −  V0 L   XC X L  R (e) We divide the magnitude of the voltage source by the magnitude of the current to find the impedance. V V0 R Z= 0 = = 2 2 I0 V R  R    0 1 +  R C − 1 R C  + −  R  L   L    © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1049

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(f)

Instructor Solutions Manual

The power factor is the ratio of the power dissipated in the circuit divided by the product of the rms voltage and current. 2

I R2 ,rms R

Vrms I rms

90.

V R   V0 0 1 +  RC − R  L  

1 R   1 +  RC −  L  

We find the equivalent values for each type of element in series. From the equivalent values we calculate the impedance using Eq. 30–28b. 1 1 1 Req = R1 + R2 Leq = L1 + L2 = + Ceq C1 C2

91.

I R2 R = V0 I 0

 V0  R R  



1   Leq −  = Ceq  

 1 1  − ( R1 + R2 ) +   L1 +  L2 −  C1 C2  

2 Req + 

With the given applied voltage, calculate the rms current through each branch as the rms voltage divided by the impedance in that branch. Vrms Vrms I C ,rms = I L,rms = 2 2 R1 + X C R22 + X L2 Calculate the potential difference between points a and b in two ways. First pass through the capacitor and then through R2. Then pass through R1 and the inductor. V X Vrms R2 Vab = I C X C − I L R2 = rms C − 2 2 R1 + X C R22 + X L2

Vab = − I C R1 + X L I L = −

Vrms R1

Vrms X L

R22 + X L2 Set these voltage differences equal to zero, and rearrange the equations. Vrms X C Vrms R2 − = 0 → X C R22 + X L2 = R2 R12 + X C2 2 2 2 2 R1 + X C R2 + X L −

Vrms R1 R12 + X C2

Vrms X L

R22 + X L2

R12 + X C2

= 0 → R1 R22 + X L2 = X L R12 + X C2

Divide the resulting equations and solve for the product of the resistances. Write the reactances in terms of the capacitance and inductance to show that the result is frequency independent.

X C R22 + X L2 R1 92.

R22 + X L2

R2 R12 + X C2 XL

R12 + X C2

→ R1 R2 = X L X C =

L L → R1 R2 = C C

The peak voltage across either element is the current through the element multiplied by the reactance. We set the voltage across the inductor equal to five times the voltage across the capacitor and solve for the frequency in terms of the resonant frequency, Eq. 30–14. 5.0 I 0 1 5.0 VL = I 0 X L = I 0 2 f L = 5.0VC = 5.0 I 0 X C = → f = = 5.0 f 0  2.2 f 0 2 fC 2 LC

1050

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

93. (a) We use Eq. 24–8 to calculate the capacitance, assuming a parallel plate capacitor. –12 2 2 –4 2 K  o A ( 5.0 ) 8.85  10 C N m 1.0  10 m = = 2.213  10 –12 F  2.2 pF C= –3 d 2.0  10 m (b) We use Eq. 30–25b to calculate the capacitive reactance. 1 1 XC = = = 5.995  106   6.0 M 2 fC 2 (12000 Hz ) 2.2  10 –12 F

(

)(

(

)

(c) Assuming that the resistance in the plasma and in the person is negligible compared with the capacitive reactance, calculate the current by dividing the voltage by the capacitive reactance. V 2500 V Io  o = = 4.17  10 –4 A  0.42 mA X C 5.995  106  This is not a dangerous current level. (d) We replace the frequency with 1.0 MHz and recalculate the current. V I 0  0 = 2 fCV0 = 2 1.0  106 Hz 2.2  10 –12 F ( 2500 V ) = 35mA XC This current level is dangerous.

(

)(

)

94. We calculate the resistance from the power dissipated and the current. Then setting the ratio of the voltage to current equal to the impedance, we solve for the inductance. P 350 W 2 P = I rms R→R= 2 = = 14  I rms ( 5.0 A )2

Vrms 2 = R 2 + ( 2 fL ) → I rms

(Vrms I rms ) − R 2 2

2 f

(120 V 5.0 A ) − (14  ) = 5.171  10−2 H  52 mH 2 ( 60 Hz ) 2

95. We insert the proposed current of I 0 sin (t −  ) into the differential equation and solve for the unknown peak current and phase. d V0 sin t = L  I 0 sin ( t −  ) + RI 0 sin ( t −  ) dt = L I 0 cos ( t −  ) + RI 0 sin ( t −  ) = L I 0 ( cos  t cos  + sin  t sin  ) + RI 0 ( sin  t cos  − cos  t sin  ) = ( L I 0 cos  − RI 0 sin  ) cos  t + ( L I 0 sin  + RI 0 cos  ) sin  t

For the given equation to be a solution for all times, the coefficients of the sine and cosine terms must independently be equal. For the cos  t term:

0 = L I 0 cos  − RI 0 sin  → tan  =

L R

→  = tan −1

L R

For the sin  t term: V0 = L I 0 sin  + RI 0 cos  I0 =

V0 = L sin  + R cos  L

L R + L 2

2 2

R R + L 2

2 2

V0 R +  2 L2 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

96.

We use Kirchhoff’s junction rule to write an equation relating the currents in each branch, and the loop rule to write two equations relating the voltage drops around each loop. We write the voltage drops across the capacitor and inductor in terms of the charge and derivative of the current. I = I R = I L + IC

Instructor Solutions Manual R I

IC C

V = V 0 sin t QC dI L V0 sin  t − I R R − = 0 ; V0 sin  t − I R R − L =0 C dt We combine these equations to eliminate the charge in the capacitor and the current in the inductor to write a single differential equation in terms of the current through the resistor. dI L V0 I R = sin  t − R dt L L 2 2 dI C d QC d d 2IR 2 CV sin t I RC CV sin t RC = =  − = −   − ( 0 ) 0 R dt dt 2 dt 2 dt 2 dI R dI L dI C V0 I R d 2I = + = sin  t − R − CV0 2 sin  t − RC 2R dt dt dt L L dt We set the current in the resistor, I R = I 0 sin ( t +  ) = I 0 ( sin  t cos  + cos  t sin  ) , equal to the current provided by the voltage source and take the necessary derivatives. d V I R I 0 ( sin  t cos  + cos  t sin  ) = 0 sin  t − ( sin  t cos  + cos  t sin  ) 0 − CV0 2 sin  t dt L L 2 d − RCI 0 2 ( sin  t cos + cos  t sin  ) dt V I R I R I 0 cos  t cos  − I 0 sin  t sin  = 0 sin  t − 0 sin  t cos  − 0 cos  t sin  − CV0 2 sin  t L L L 2 + RCI 0 sin  t cos + RCI 0 2 cos  t sin  Setting the coefficients of the time-dependent sine and cosine terms separately equal to zero enables us to solve for the magnitude and phase of the current through the voltage source. We also use Eq. 30–23b and Eq. 30–25b to write the inductance and capacitance in terms of their respective reactances. From the cos( t ) term:

 X L XC  I R I R X L XC I 0 cos  = − 0 sin  + 0 sin  → tan  = →  = tan −1   XL XC R( X L − XC )  R ( X L − X C )  From the sin( t ) term: V  I R V  RI  cos  − 0 + 0 cos  − I 0 sin  = 0 − 0 XL XL XC XC I0 =

( X L − X C )V0  R ( X L − X C ) cos  + X L X C sin  

1052

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

( X L − X C )V0  R( X L − XC ) R ( X − X ) + X L XC L C 2 2  2 − + R X X X X ( L C) ( L C)  ( X L − X C )V0 R2 ( X L − X C ) + ( X L X C ) 2

  2 2  2 R ( X L − X C ) + ( X L X C )  X L XC

We now have the amplitude and phase of the current through the power source and resistor. We insert these values back into the junction and loop equations to determine the current in each element as a function of time. We calculate the impedance of the circuit by dividing the peak voltage by the peak current through the voltage source.

2

V0 = I0

1  +   2C −  R 2 2 2 2 2 + − X X R X X ( L C) ( L C) L ( X L XC ) = R  = R2 + 2 1  2 ( X L − XC ) ( X L − XC )  C −  L  

;

      X L XC  −1  ;  = tan −1   = tan  1    2  R ( X L − X C )   R   C − L     V0 sin ( t +  ) Z dQC d dI IC = = ( CV0 sin  t − I R RC ) = CV0 cos  t − RC R dt dt dt

IR =

V0  R  cos  t − cos (t +  )  XC  Z 

I L = I R − IC = =

V0 V  R  sin ( t +  ) − 0 cos  t − cos ( t +  ) Z XC  Z 

 V V0  R cos ( t +  ) − 0 cos  t sin ( t +  ) + Z  XC  XC

We could of course substitute in for the impedance and reactances, to get final answers in terms of the component values alone. 97. (a) The amplitude of the output voltage is the voltage across the resistor, which is the amplitude of the current through the circuit multiplied by the resistance. We calculate the amplitude of the current by dividing the input voltage amplitude by the impedance. Finally, we divide the output voltage amplitude by the input voltage amplitude to calculate the gain. 2 f C RVin Vin R Vin R = = Vout = IR = 2 2 R 2 + X C2 ( 2 f C R ) + 1  1  R2 +    2 f C 

1053

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Vout = Vin

Instructor Solutions Manual

2 f C R 4 2 f 2C 2 R 2 + 1

(b) As the frequency goes to zero, the gain drops to zero; A ( f = 0 ) =

2 ( 0 ) C R 4 ( 0 ) C R + 1 2

0 = 0. 1

In this instance the capacitor becomes fully charged, so no current flows through the resistor. Therefore the output voltage drops to zero. As the frequency becomes very large, the “+1” in the denominator can be ignored, and so 2 f C R = 1. The capacitive A( f →  ) = 4 2 f 2C 2 R 2 reactance becomes very small, allowing a large current. In this case, most of the voltage drop is across the resistor, and the gain is equal to unity. Although not a part of the problem, here is a graph of the log of the gain as a function of the log of the frequency, with R = 850  and C = 1.0 F. Note that for frequencies greater than about 1000 Hz the gain is ~ 1. For lower frequencies the gain drops off. 98. (a) The amplitude of the output voltage is the voltage across the capacitor, which is the amplitude of the current through the circuit multiplied by the capacitive reactance. We calculate the amplitude of the current by dividing the input voltage amplitude by the impedance. Finally, we divide the output voltage amplitude by the input voltage amplitude to calculate the gain. Vin X C Vin Vin Vout = IX C = = = 2 2 2 2 R + XC ( R X C ) + 1 ( 2 fCR ) + 1

A= (b)

Vout = Vin

1 4 2 f 2C 2 R 2 + 1

As the frequency goes to zero, the gain becomes one: A ( f = 0 ) =

1 4 2 ( 0 ) C 2 R 2 + 1 2

= 1.

In this instance the capacitor becomes fully charged, so no current flows through the resistor. Therefore the output voltage is equal to the input voltage. As the frequency becomes very large, the denominator of the gain becomes very large, and the gain goes to 0. The capacitive reactance becomes very small, allowing a large current. In this case, most of the voltage drop is across the resistor, and the gain goes to zero. Although not a part of the problem, here is a graph of the log of the gain as a function of the log of the frequency, with R = 850  and C = 1.0 F. Note that for frequencies less than about 100 Hz the gain is ~ 1. For higher frequencies the gain drops off. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1054

Chapter 30

Inductance, Electromagnetic Oscillations, and AC Circuits

99. (a) The resonant frequency is given by Eq. 30–32. At resonance, the impedance is equal to the resistance, so the rms voltage of the circuit is equal to the rms voltage across the resistor. 1 1 f = = = 7118 Hz  7.1kHz 2 LC 2 ( 0.0050 H ) 0.10  10−6 F

(

)

(VR )rms = Vrms (b) We set the inductance equal to 90% of the initial inductance and use Eq. 30–28b to calculate the new impedance. Dividing the rms voltage by the impedance gives the rms current. We multiply the rms current by the resistance to determine the voltage drop across the resistor. 1 1 = = 223.6  XC = 2 fC 2 ( 7118 Hz ) 0.10  10−6 F

(

)

X L = 2 fL = 2 ( 7118 Hz )( 0.90 )( 0.0050 H ) = 201.3  Z = R2 + ( X L − X C ) = 2

( 38  ) + ( 201.3 − 223.6  ) = 44.06  2

 38   R Vrms = 0.86Vrms Vrms =  Z  44.06  

(VR )rms = 

(c) The eddy currents in the car will create a flux that opposes the flux through the inductor, thus reducing the self-inductance.

1055

CHAPTER 31: Maxwell’s Equations and Electromagnetic Waves Responses to Questions 1.

The magnetic field will be clockwise in both cases. In the first case, the electric field is away from dE you and is increasing. The direction of the displacement current (proportional to ) is therefore dt away from you and the corresponding magnetic field is clockwise. In the second case, the electric field is directed towards you and is decreasing; the displacement current is still away from you, and the magnetic field is still clockwise.

The displacement current is to the right.

The displacement current is spread out over a larger area than the conduction current in the wires. Thus, the displacement current has a smaller current density (current per unit area) than the current in a wire. If the current sensor has a small cross-sectional area, then it would not intercept as much “current” as it would for a thin wire. The magnetic field the displacement current causes is small if you are near the center of the capacitor, while the magnetic field near a thin wire would be much larger, due to the 1/r dependence. In Example 31–1, the largest magnetic field created by the displacement current is about 1.2 Gauss, which occurs about 5 cm from the center of the capacitor. Alternatively, if you could measure the magnetic field near the wires carrying current to the capacitor, at a distance of about a half-millimeter, the magnetic field would be 100 times stronger. Also note in the example that the time constant is only about 60 picoseconds. Measuring a weak field in such a short time would be very difficult indeed.

One possible reason the term  0

The magnetic field vector will oscillate up and down, perpendicular to both the direction of propagation and to the direction of the electric field vector. For an EM wave, the direction of travel, the electric field, and the magnetic field must all be perpendicular to each other.

No, sound is not an electromagnetic wave. Sound is a longitudinal mechanical (pressure) wave, which requires a medium in which to travel. The medium can be a gas, a liquid, or a solid. EM waves do not need a medium in which to travel.

EM waves can travel through a perfect vacuum. The energy is carried in the oscillating electric and magnetic fields and no medium is required to travel. Sound waves cannot travel through a perfect vacuum. A medium is needed to carry the energy of a mechanical wave such as sound (i.e., to have mechanical energy) and there is no medium in a perfect vacuum.

No. Electromagnetic waves travel at a very large but finite speed. When you flip on a light switch, it takes a very small amount of time for the electrical signal to travel along the wires from the switch to the light bulb.

The wavelengths of radio and TV signals are much longer than visible light. Radio waves are on the order of 3 m–30,000 m. TV waves are on the order of 0.3 m–3 m. Visible waves are on the order of 10–7 m.

dE

can be called a “current” is because it has units of amperes. It dt also creates a magnetic field, and in general we think of magnetic field as being due to moving charge (i.e., current).

1056

Chapter 31

Maxwell’s Equations and Electromagnetic Waves

10. The wavelength of the current is 5000 km; the house is only 200 km away. The phase of the current at the position of the house is 2/25 radians (or about 29o) different from the phase at the source due to the position of the house. 11. It is not necessary to make the lead-in wires to your speakers the exact same length. Since electrical signals in the wires travels at nearly the speed of light, the difference in time between the signals getting to the different speakers will be too small for your ears to detect. 12. Wavelength of 103 km: Sub-radio waves (or very long radio waves; for example, ELF waves for submarine communication fall into this category). Wavelength of 1 km: Radio waves. Wavelength of 1 m: TV signals and microwaves. Wavelength of 1 cm: microwaves and satellite TV signals. Wavelength of 1 mm: microwaves and infrared waves. Wavelength of 1 m: infrared waves. 13. Yes, radio waves can have the same frequencies as sound waves. These 20–20,000 Hz EM waves would have extremely long wavelengths when compared to the sound waves, because of their high speed. A 5000 Hz sound wave has a wavelength of about 70 mm, while a 5000 Hz EM wave has a wavelength of about 60 km. 14. Cordless phones utilize EM waves when sending information back and forth between the handset (the part you hold up to your ear/mouth) and its base (the base is the unit where the phone sits while charging, and is physically connected to the wire phone lines that lead outside to the phone company’s network). These EM waves are usually very weak–you can’t walk very far away from the base before you lose the signal. Cell phones utilize EM waves when sending information back and forth between the phone and the nearest tower in your geographical area (which could be miles away from your location). These EM waves need to be much stronger than cordless phone waves (or the cell phone electronics need to be more sensitive) because of the larger distances involved. Cordless phones and cell phones use different frequency ranges. 15. The receiver antenna should also be vertical for obtaining the best reception. The oscillating carrier electric field is up-and-down, so a vertical antenna would “pick up” that signal better, since the electrons in the metal antenna could oscillate up and down along the entire length of the vertical antenna, creating a stronger signal in the antenna. 16. Diffraction effects (the bending of waves around the edge of an object) are only evident when the size of the wavelength of the wave is on the order of the size of the object (or larger). AM waves have wavelengths that are on the order of 300 m long, while FM waves have wavelengths on the order of 3 m long. Buildings and hills are much larger than FM waves, and so FM waves will not diffract around the buildings and hills. Thus the FM signal will not be received behind the hills or buildings. On the other hand, these objects are smaller than AM waves, and so the AM waves will diffract around them easily. The AM signal can be received behind the objects. 17. Morse code is an amplitude modulated, or AM, signal. The person flashing the light on and off is changing the amplitude of the light (“on” is maximum amplitude and “off” is zero). The carrier wave is visible light, so the frequency of the carrier wave is just the frequency of the visible light, approximately 1014 to 1015 Hz.

1057

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Responses to MisConceptual Questions 1.

(c) Visible light has a wavelength on the order of 10–7 to 10–8 m. This is over one thousand times larger than the size of an atom.

(a, b, c, e, f, g, h) All electromagnetic waves travel at the speed of light. The only listed wave that is not an electromagnetic wave is (d) ultrasonic wave, which is a sound wave and would travel at the speed of sound.

(c) In empty space, X-ray radiation and a radio wave both travel at the speed of light. X-ray radiation has a much smaller wavelength than a radio wave. Since wave speed is equal to the product of the wavelength and the frequency, and both waves travel at the same speed, the X-ray radiation has a higher frequency than a radio wave. The amplitude depends upon the intensity of the wave, so either the X-ray radiation or the radio wave could have the greater amplitude.

(b) Frequency is the measure of the number of oscillations made per second. The rate at which the electrons oscillate will be equal to the oscillation frequency of the radiation emitted.

(d) A common misconception is to think that the radiation decreases linearly with distance, such that the intensity would decrease by a factor of two. However the radiation intensity decreases as the square of the distance, since intensity is power per unit area. Doubling the distance will decrease the radiation intensity to one-fourth the initial intensity.

(a) The direction of the magnetic field is perpendicular to the direction the wave is traveling and perpendicular to the electric field, so only north and south are possible answers. The right-hand rule can be used to determine which direction is correct, by pointing fingers in the direction of the electric field (west), bending them in the direction of the magnetic field (north), resulting in the thumb pointing in the direction of the wave (down).

(c) A misconception that can arise is thinking that the wave intensity is proportional to just one of the field amplitudes. However, the intensity is proportional to the product of the amplitudes, with the field amplitudes proportional to each other. If the intensity doubles, each field (electric and magnetic) must increase by the same factor, 2 .

(a, b, d, e) Maxwell’s equations are used to describe the electric and magnetic fields produced by charges and currents. They are not related to the gravitational field of masses. So answer (c) is incorrect.

(b, c) Electromagnetic waves are produced by accelerating electric charges, which occur in answers (b) and (c). Static charges and constant currents (charges with a constant velocity) do not produce electromagnetic waves, so answer (a) is incorrect. And therefore answer (d) is also incorrect.

10. (d) A common misconception is that every color surface will experience the same radiation pressure. However, the pressure is related to the change in momentum of the light. With a black object, all of the light is absorbed and so the radiation pressure is proportional to the incident momentum. With a colored object, some of the light is reflected, thus increasing the change in momentum of the light and therefore increasing the radiation pressure as compared to the black object. For a white object the maximum amount of light is reflected off the surface, creating the greatest change in momentum of the light, and creating the greatest radiation pressure. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1058

Chapter 31

Maxwell’s Equations and Electromagnetic Waves

11. (c) A common misconception is that the digital broadcast is at a different frequency than the analog broadcast. This is incorrect–the carrier wave frequency is the same. Since the antenna is designed for operation at the broadcast frequency, the antenna will work just as well to receive a digital signal. However, a digital-to-analog “converter box” is needed for older, analog-only TV’s. 12. (a, d) Charges must be moving to make magnetic fields. Thus answer (a) does not make a magnetic field. We also see from Section 31–1 that changing electric fields make magnetic fields, and so and (d) does not make a magnetic field.

Solutions to Problems 1.

The electric field between the plates is given by E =

V d

→

dE dt

 dV d dt



V d

, where d is the distance between the plates.

V m  (120 V s ) = 1.1  105  s  0.0011m 

=

The displacement current is shown in Eq. 31–3 to be I D =  0 ID = 0 A

dE dt

(

)

dE dt

 

= 8.85  10−12 C 2 N  m 2 ( 0.058 m )  2.6  106 2

= 0 A

dE dt

V 

−8

 = 7.7  10 A m s

The current in the wires must also be the displacement current in the capacitor. Use the displacement current to find the rate at which the electric field is changing. ( 2.4 A ) I dE dE V ID = 0 A → = D = = 1.1  1015 dt dt ms  0 A ( 8.85  10−12 C2 N  m 2 ) ( 0.0160 m )2

(a) The current in the wires is the rate at which charge is accumulating on the plates and also is the displacement current in the capacitor. Because the location in question is outside the capacitor, use the expression for the magnetic field of a long wire, with the “wire” location along the line connecting the centers of the two capacitor plates. −7 −3  I    2 I (10 T m A ) 2 ( 38.0  10 A ) = = 7.60  10−8 T B= 0 = 0  2 R  4  R ( 0.100 m ) (b) After the capacitor is fully charged, all currents will be zero, so the magnetic field will be zero .

The electric field between the plates is given by E =

V d

, where d is the distance between the plates.

The displacement current is shown in Eq. 31–3 to be I D =  0

ID = 0 A

dE dt

= 0 A

1 dV d dt

 0 A dV d

= C

dE dt

= 0 A

dE dt

dV dt

1059

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(a) The footnote on page 924 indicates that Kirchhoff’s junction rule is valid at a capacitor plate, and so the conduction current is the same value as the displacement current. Thus the maximum conduction current is 35 A . (b) The charge on the pages is given by Q = CV = C e0 cos  t. The current is the derivative of this. I= e0 =

dQ dt I max

C

= − C e0 sin  t ; I max =  C e0 → =

( 35  10 A )(1.6  10 m ) 2 ( 68.0 Hz ) ( 8.85  10 C N  m )  ( 0.025m ) −6

I max d 2 f  0 A

−3

−12

= 7543V  7500 V

ID = 0 7.

dE

. dt  dE 

→ 

 =  dt  max

( I D ) max 0

35  10−6 A 8.85  10

−12

C Nm 2

= 4.0  106 V m s

(a) We follow the development and geometry given in Example 31–1, using R for the radial distance from a line joining the centers of the two capacitor plates. The electric field between V the plates is given by E = , where d is the distance between the plates. d 2 d  rflux E dE  B d l = 0 0 dt → B ( 2 rpath ) = 0 0 dt The subscripts are used on the radial variable because there might not be electric field flux through the entire area bounded by the amperian path. The electric field between the plates is V V sin ( 2 f t ) , where d is the distance between the plates. given by E = = 0 d d 2 d  rflux E → B ( 2 rpath ) = 0 0 dt

(

)

2 2 d ( E ) 0 0 rflux 0 0 rflux   r 2  f V0 2 f V0 = cos ( 2 f t ) = 0 0 flux cos ( 2 f t ) 2 rpath 2 rpath dt d rpath d

We see that the functional form of the magnetic field is B = B0 ( r ) cos ( 2 f t ) . (b) If r  r0 , then there is electric flux throughout the area bounded by the amperian loop, and so

rpath = rflux = r. B0 ( r  r0 ) =

2 0 0 rflux  f V0

rpath

(

=  0 0

)

 f V0 d

(

 ( 60 Hz )(180 V ) 3.00  108 m s

)( 2

5.0  10 −3 m

)

= 7.5340  10 −11 T m r  7.5  10 −11 T m r If r  r0 , then there is electric flux only out a radial distance of r0 . Thus rpath = r and rflux = r0 .

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B0 ( r  r0 ) =

2 0 0 rflux  f V0

rpath

(

=  0 0

= 6.786  10−14 T m

 f V0 r02 1 d

 ( 60 Hz )(180 V )( 0.030 m ) 1 2

( 3.00 10 m s ) ( 5.0 10 m ) r 8

−3

) 1r  ( 6.8 10 T m ) 1r −14

(c) See the graph of B0 ( r ) vs. r. Note that the magnetic field is continuous at the transition from “inside” to “outside” the capacitor radius. The maximum value of the magnetic field is 2.3  10−12 T. 8.

Use Eq. 31–11 with v = c.

E0 B0 9.

= c → B0 =

E0 c

0.45  10−4 V m 3.00  108 m s

= 1.5  10−13 T

Use Eq. 31–11 with v = c. E0 = c → E0 = B0c = 16.5  10 −9 T 3.00  108 m s = 4.95 V m B0

(

)(

)

10. The frequency of the two fields must be the same: 85.0 kHz . The rms strength of the electric field can be found from Eq. 31–11 with v = c. Erms = cBrms = ( 3.00  108 m s )( 7.75  10−9 T ) = 2.33V m The electric field is perpendicular to both the direction of travel and the magnetic field, so the electric field oscillates along the horizontal north–south line . 11. Use the relationship that d = v t to find the time.

(1.50  10 m ) = 5.00  10 s = 8.33min d = vt → t = = v ( 3.00  10 m s ) d

12. We assume the only significant time is the transit time from the Earth to the Moon, at the speed of light. We use the Earth–Moon distance, but subtract the Earth and Moon radii since the astronaut and the receiver are on the surfaces of the bodies. d Earth − − rEarth − rMoon ( 384106 m − 6.38  106 m − 1.74  106 m ) = 1.25s d Moon d = vt → t = = = c v ( 3.00108 m s ) 13. (a) If we write the argument of the cosine function as kz +  t = k(z + ct), we see that the wave is traveling in the – z-direction, or −kˆ . (b) E and B are perpendicular to each other and to the direction of propagation. At the origin, the electric field is pointing in the positive x-direction. Since E  B must point in the negative z-direction, B must point in the the – y-direction, or −jˆ . The magnitude of the magnetic field is found from Eq. 31–11 as B0 = E0 c = E0k  . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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14. The wave equation to be considered is v 2

Instructor Solutions Manual

2 E 2 E = 2 . x 2 t

(a) Given E ( x, t ) = Ae− ( x −vt ) . 2

2 E = Ae − ( x − vt ) −  2 ( x − vt )  x 2 2 2 2 2 E 2 = Ae − ( x − vt ) ( −2 ) + Ae − ( x − vt ) − 2 ( x − vt )  = −2 Ae − ( x − vt ) 1 − 2 ( x − vt )  2    x 2 2 E 2 ( x − vt )( −v )  = Ae − ( x − vt )  2 v ( x − vt )  = Ae − ( x − vt ) −  t 2 2 2 2 2 E 2 = Ae − ( x − vt ) ( −2 v 2 ) + Ae − ( x − vt )  2 v ( x − vt )  = −2 v 2 Ae − ( x −vt ) 1 − 2 ( x − vt )  2   t 2 2  E  E We see that v 2 2 = 2 , and so the wave equation is satisfied. x t

(b) Given E ( x, t ) = Ae

(

−  x 2 − vt

)

− ( x − v t ) E = Ae ( −2 x ) x − ( x 2 − v t ) −  x 2 − vt −  x 2 − vt 2 E 2 = Ae ( −2 ) + Ae ( ) ( −2 x ) = −2 Ae ( ) 1 − 2 x 2  2 x − ( x 2 − v t ) − ( x 2 − v t ) E 2 E ; = Ave = Av 2 e 2 t t −( x 2 − vt ) −  x 2 − vt ) 2 E 2 E 1 − 2 x 2   Av 2 e ( This does NOT satisfy v 2 2 = 2 , since −2 v 2 Ae in x t general. 2

15. Use Eq. 31–14 to find the wavelength and frequency. 8 c ( 2.998  10 m s ) c f   = → = = = 8.386  10−3 m (a) 9 f ( 35.75  10 Hz ) (b) c =  f

→ f =



( 3.00  10 m s ) = 2.5  10 Hz ( 0.12  10 m ) 8

−9

16. Use Eq. 31–14 to find the wavelength. 8 c ( 3.00  10 m s ) c=f →  = = = 1.13  10−6 m = 1.13 m f ( 2.65  1014 Hz ) This wavelength is in the infrared region of the EM spectrum. 17. (a) Use Eq. 31–14 to find the wavelength. 8 c ( 3.00  10 m s ) c=f →  = = = 3.00  105 m f (1.00  103 Hz ) (b) Again use Eq. 31–14, with the speed of sound in place of the speed of light. ( 341m s ) = 0.341m v v=f →  = = f (1.00  103 Hz ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(c) No, you cannot hear a 1000-Hz EM wave. It takes a pressure wave to excite the auditory system. However, if you applied the 1000-Hz EM wave to a speaker, you could hear the 1000-Hz pressure wave produced by the speaker. 18. The distance is the rate (speed of light) times the time (1 year), with the appropriate units. d = ( 3.00  108 m s ) (1yr ) ( 3.156  107 s yr ) = 9.47  1015 m

19. We convert the units from light years to meters.

d = ( 4.2ly ) ( 3.00  108 m s )( 3.16  107 s yr ) = 4.0  1016 m

20. (a) The radio waves travel at the speed of light, and so d = v t. The distance is found from the radii of the orbits. For Mars when nearest the Earth, the radii should be subtracted. 9 9 d ( 230  10 m − 149.6  10 m ) t = = = 268s  300s  4 min c ( 3.000  108 m s ) (b) For Mars when farthest from Earth, the radii should be added. 9 9 d ( 230  10 m + 149.6  10 m ) t = = = 1265s  1300s  21min c ( 3.000  108 m s ) 21. The length of the pulse is d = ct. Use this to find the number of wavelengths in a pulse. 8 −12 ct ) ( 3.00  10 m s )( 32  10 s ) ( N= = = 9040  9.0  103 wavelengths −9  (1062  10 m ) If the pulse is to be only one wavelength long, then its time duration is the period of the wave, which is the reciprocal of the wavelength. 1062  10−9 m ) ( 1  T= = = = 3.542  10−15 s 8 f c ( 2.998  10 m s ) 22. (a) The general form of a plane wave is given in Eq. 31–7. For this wave, E x = E0 sin ( kz − t ) .

=

2 2 = = 118.55m  120 m k 0.053m −1

 1.6  107 rad s = = 2.546  106 Hz  2.5MHz 2 2 Note that  f = (118.55m ) ( 2.546  106 Hz ) = 3.018 108 m s  c to two significant figures. f =

(b) The magnitude of the magnetic field is given by B0 = E0 c . The wave is traveling in the

k̂ direction, and so the magnetic field must be in the ĵ direction, since the direction of travel is given by the direction of E  B. 275V m E B0 = 0 = = 9.17  10−7 T → c 3.00  108 m s B = ˆj ( 9.17  10−7 T ) sin ( 0.053m −1 ) z − (1.6  107 rad s ) t 

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

23. The eight-sided mirror would have to rotate 1/8 of a revolution for the succeeding mirror to be in position to reflect the light in the proper direction. During this time the light must travel to the opposite mirror and back. 8  18 ( 2 rad ) ( rad ) c ( rad ) ( 3.00  10 m s ) = = = = = 3400 rad s ( 3.2  104 rev min ) 3 t 8x 8 ( 35  10 m ) ( 2x c ) 24. The mirror must rotate at least 1/8 revolution in the time it takes the light beam to travel twice the length of the room. ( 2 )(16 m ) = 1.067  10−7 s ;  = ( 18 revolutions ) = 1.2  106 revolutions s t= ( 3.0  108 m s ) (1.067  10−7 s ) This is a “megahertz” range frequency. I’m not sure that would be feasible! Very high speed “ultracentrifuges” reach about 150,000 rpm or 2500 revolutions/s, which is a factor of 500 smaller than the required angular velocity here. 25. The average energy transferred across unit area per unit time is the average magnitude of the Poynting vector, and is given by Eq. 31–19a.

S = 12  0cE02 = 12 ( 8.85  10−12 C2 N m 2 )( 3.00  108 m s ) ( 0.0235V m ) = 7.33  10 −7 W m 2 2

26. The energy per unit area per unit time is given by the magnitude of the Poynting vector. Let U represent the energy that crosses area A in a time T . cB 2 U → S = rms = 0 At t =

0 U 2 AcBrms

( 4 10 T m A ) ( 415J ) = 3.43  10 s  397 days (1.00  10 m )( 3.00  10 m s )( 22.5  10 T ) −7

−4

−9

27. The energy per unit area per unit time is given by the magnitude of the Poynting vector. Let U represent the energy that crosses area A in a time t. U 2 = → S = c 0 Erms At U 2 = c 0 Erms A t = ( 3.00  108 m s )( 8.85  10−12 C2 N  m 2 ) ( 0.0258V m ) (1.00  10 −4 m 2 ) ( 3600s h ) 2

= 6.36  10−7 J h

28. The intensity is the power per unit area, and also is the time-averaged value of the Poynting vector. The area is the surface area of a sphere, since the wave is spreading spherically. ( 2100 W ) = 6.685W m 2  6.7 W m 2 P S= = A  4 ( 5.0 m )2    2 S = c 0 Erms → Erms =

S = c 0

6.685W m 2 = 50.2 V m  5.0  101 V m 8 2 2 −12 ( 3.00  10 m s )(8.85  10 C N  m )

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29. (a) We find E using Eq. 31–11 with v = c. E = cB = ( 3.00  108 m s )(1.8  10−7 T ) = 54 V m (b) The average power per unit area is given by the Poynting vector, from Eq. 31–19a. ( 54 V m ) (1.8  10−7 T ) E0 B0 I = = = 3.9 W m2 ( 20 ) 2 ( 4  10−7 N s 2 C2 ) 30. From Eq. 31–16b, the instantaneous energy density is u =  0 E 2 . From Eq. 31–17, we see that this instantaneous energy density is also given by S c . The time-averaged value is therefore S c . Multiply that by the volume to get the energy. S 1350 W m 2 U = uV = V = (1.00 m3 ) = 4.50  10−6 J c 3.00  108 m s 31. The power output per unit area is the intensity, and also is the magnitude of the Poynting vector. Use Eq. 31–19a with rms values. P 2 S = = c 0 Erms → A 0.0185W P Erms = = 2 Ac 0  (1.20  10−3 m ) ( 3.00  108 m s )(8.85  10 −12 C 2 N  m 2 ) = 1241.1V m  1240 V m Brms =

1241.1V m Erms = = 4.14  10−6 T 3.00  108 m s c

32. The radiation from the Sun has the same intensity in all directions, so the rate at which it reaches the Earth is the rate at which it passes through a sphere centered at the Sun with a radius equal to the Earth’s orbit radius. The 1350W m 2 is the intensity, or the magnitude of the Poynting vector. 2 P → P = SA = 4 R 2 S = 4 (1.496  1011 m ) (1350 W m 2 ) = 3.80  1026 W S= A 33. (a) The energy emitted in each pulse is the power output of the laser times the time duration of the pulse. W P= → W = Pt = (1.5  1011 W )(1.0  10−9 s ) = 150J t (b) We find the rms electric field from the intensity, which is the power per unit area. That is also the magnitude of the Poynting vector. Use Eq. 31–19a with rms values. P 2 S = = c 0 Erms → A Erms =

(1.5  1011 W ) P = 2 Ac 0  (1.8  10−3 m ) ( 3.00  108 m s )(8.85  10−12 C2 N  m 2 )

= 2.4  109 V m

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

34. (a) Example 31–1 refers back to Example 21–14 and Fig. 21–32. In that figure, and the figure included here, the electric field between the plates is to the right. The magnetic field is shown as a counterclockwise circle. Take any point between the capacitor plates, and find the direction of E  B, which is the direction of S. For instance, at the top of the circle shown in Fig. 31–4, E is toward the viewer, and B is to the left. The cross-product E  B points down, directly to the line connecting the center of the plates. Or take the rightmost point on the circle. E is again toward the viewer, and B is upwards. The cross-product E  B points to the left, again directly to the line connecting the center of the plates. In cylindrical coordinates, E = E kˆ and B = Bφˆ . The cross-product kˆ  φˆ = −rˆ . (b) We evaluate the Poynting vector, and then integrate it over the curved cylindrical surface dE , between the capacitor plates. The magnetic field (from Example 31–1) is B = 12 0 0 r0 dt 1 dE evaluated at r = r0 . E and B are perpendicular to each other, so S = E  B = 12  0 r0 E , dt 0 inward. In calculating  S dA for energy flow into the capacitor volume, note that both S and

dA point inward, and that S is constant over the curved surface of the cylindrical volume. dE dE 2  S dA =  SdA = S  dA = SA = S 2 r0d = 12  0r0 E dt 2 r0d =  0d r0 E dt The amount of energy stored in the capacitor is the energy density times the volume of the capacitor. The energy density is given by Eq. 24–6, u = 12  0 E 2. Take the derivative of the energy stored with respect to time. dU dE =  0 E r02 d U = u ( Volume ) = 12  0 E 2 r02 d → dt dt dU . We see that  S dA = dt 35. (a) The intensity from a point source is inversely proportional to the distance from the source. 2 rStar − I Sun I 1350 W m 2  1ly  Earth = 2 → rStar − = rSun − Sun = (1.496  1011 m )  −23 2  15 I Star rSun − I Star 1  10 W m  9.46  10 m  Earth Earth Earth

= 1.84  108 ly  2  108 ly

(b) Compare this distance to the galactic size. rStar − 1.84  108 ly Earth = = 1840  2000 galactic size 1  105 ly The distance to the star is about 2000 times the size of our galaxy. The Arecibo telescope can detect signals from sources far outside our galaxy.

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36. We assume the light energy is all absorbed, and so use Eq. 31–21a. 5.0 W

−2 S 4 ( 8.0  10 m ) = 2.072  10−7 N m 2  2.1  10 −7 N m 2 . P= = 8 c 3.00 10 m s  ( ) 2

The force is pressure times area. We approximate the area of a fingertip to be 1.0 cm2. F = PA = ( 2.072  10−7 N m 2 )(1.0  10−4 m 2 ) = 2.1  10−11 N

37. (a) Since the cylinder is absorbing the radiation, the pressure (P) is given by Eq. 31–21a, where the intensity is the power ( Pavg ) divided by the cross-sectional area. The force is the pressure multiplied by the cross-sectional area of the cylinder, or the power absorbed divided by the speed of light. I P 1.0 W Pressure = = avg = = 4244 Pa  4  103 Pa c cA ( 3.00  108 m s )  ( 5  10−7 m )2 Pavg

Pavg

1.0 W = 3.3  10−9 N cA c 3.00  108 m s (b) The acceleration of the cylindrical particle will be the force on it (due to radiation pressure) divided by its mass. The mass of the particle is its volume times the density of water. F F 3.3  10−9 N = = 8.49  106 m s2  8  106 m s2 a= = m  H 2O r 3 (1000 kg m 2 )  ( 5  10−7 m )3 F = PA =

38. The intensity from a point source is inversely proportional to the distance from the source. 2 2 rSun − 7.78  1011 m ) ( I Earth Jupiter = 2 = = 27.0 2 I Jupiter rSun − (1.496 1011 m ) Earth

So it would take an area of 27m2 at Jupiter to collect the same radiation as a 1.0-m2 solar panel at the Earth. 39. We convert the horsepower rating to watts, and then use the intensity to relate the wattage to the needed area. 746 W 100 hp  P P 1hp → A= = = 373m 2  400 m 2 I= 2 A I 200 W m That’s a square with a side length of 20 m. It is impractically large. A car isn’t nearly that big. 40. Use Eq. 31–14. Note that the higher frequencies have the shorter wavelengths. (a) For FM radio we have the following. 8 8 c ( 3.00  10 m s ) c ( 3.00  10 m s ) = = = = 3.41m 2.78m = =  to f f (1.08  108 Hz ) (8.8  107 Hz ) (b) For AM radio we have the following. 8 8 c ( 3.00  10 m s ) c ( 3.00  10 m s ) = = = = 561m 180 m = =  to f f (1.7  106 Hz ) (5.35  105 Hz ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1067

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

41. Use Eq. 31–14. 8 c ( 3.00  10 m s ) = 0.16 m = = f (1.9  109 Hz ) 42. The frequencies are 1030 kHz on the AM dial and 103.1 MHz on the FM dial. From Eq. 31–14, c = f  , we see that the lower frequency will have the longer wavelength: the AM station . When we form the ratio of wavelengths, we get 6 2 f1 (103.1 10 Hz ) = = = 100.1  1.00  102 . 3 1 f 2 (1030 10 Hz ) 43. Each wavelength is found by dividing the speed of light by the frequency, as in Eq. 31–14. 8 c ( 3.00  10 m s ) = 5.56 m 2 = = Channel 2: f 2 ( 54.0  106 Hz ) Channel 51:

51 =

( 3.00  108 m s ) = 0.434 m c = f 51 ( 692  106 Hz )

44. The resonant frequency of an LC circuit is given by f =

2



2 . We assume the inductance is LC

constant, and form the ratio of the two frequencies. 2 2 2 LC1  f   550 kHz  f1 C2 = = → C2 =  1  C1 =   (1800 pF ) = 210 pF 2 f2 C1  1610 kHz   f2  LC 2 45. The resonant frequency of an LC circuit is given by Eq. 30–32b. Solve for the capacitance. 1 1 1 → C= 2 2 = = 9.4  10−13 F f = 2 6 4 f F 4 ( 98.3  10 Hz )2 ( 2.8  10−6 H ) 2 LC 46. The resonant frequency of an LC circuit is given by f =

f =

→ L=

 1 = . Solve for the inductance. 2 2 LC

1 4 f 2C 1

2 LC 1 = 4.8  10−9 H L1 = 2 2 = 4 f1 C 4 2 (88  106 Hz )2 ( 680  10−12 F ) L2 =

1 1 = = 3.2  10−9 H 2 4 f 2 C 4 2 (108  106 Hz )2 ( 680  10−12 F ) 2

The range of inductances is 3.2 10−9 H  L  4.8  10−9 H .

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47. (a) The minimum value of C corresponds to the higher frequency, according to Eq. 30–32b. 1 f = → 2 LC 1 1 L= 2 2 = = 1.185  10−6 H  1.2  10−6 H 4 f C 4 2 (15.0  106 Hz )2 ( 95  10−12 F ) (b) The maximum value of C corresponds to the lower frequency. 1 1 1 → C= 2 2 = = 1.1  10−10 F f = 2 2 6 − 6 4 f L  2 LC 4 (14.0  10 Hz ) (1.185  10 H ) 48. The rms electric field strength of the beam can be found from the Poynting vector. P 2 → S = = c 0 Erms A Erms =

P 1.6  104 W = = 1.8V m 2 Ac 0  ( 750 m ) ( 3.00  108 m s )(8.85  10−12 C2 N  m 2 )

49. The electric field is found from the desired voltage and the length of the antenna. Then use that electric field to calculate the magnitude of the Poynting vector. V 1.00  10−3 V Erms = rms = = 7.14  10−4 V m d 1.40 m S = c 0 E

2 rms

2 1.00  10−3 V ) Vrms 8 2 ( −12 2 = c 0 2 = ( 3.00  10 m s )( 8.85  10 C N m ) 2 d (1.40 m )

= 1.35  10−9 W m 2

50. We ignore the time for the sound to travel to the microphone. Find the difference between the time for sound to travel to the balcony and for a radio wave to travel 1800 km.   1.8  106 m   50.0m  d  d t = tradio − tsound =  radio  −  sound  =  −  = − 0.140s, 8  c   vsound   3.00  10 m s   343m s  so the person at the radio hears the voice 0.140 s sooner. 51. The ratio of distance to time must be the speed of light. 9 3m x x −8  10 ns  1 10 s = c → t = = =    = 10ns c 3.0  108 m s t  1s  52. The length is found from the speed of light and the duration of the burst. x = ct = ( 3.00  108 m s )(10−8s ) = 3m 53. (a) The time consists of the time for the radio signal to travel to Earth and the time for the sound to travel from the loudspeaker. We use 343 m/s for the speed of sound. For the distance for the radio signal, we take the Earth–Moon distance from the inside front cover of the book, and subtract away the radius of the Moon and the radius of the Earth. The radio signal is travelling from the surface of the Moon to the surface of the Earth, not from the center of the Moon to the

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Instructor Solutions Manual

center of the Earth. This is the shortest possible time. However, since the orbits are not perfect circles, the Earth and Moon are not perfect spheres, and the sender and receiver might not have been on the line connected the centers, other values are possible.  d  d t = tradio + tsound =  radio  +  sound   c   vsound   3.84  108 m − 6.38  106 m − 1.74  106 m   32 m  = +  = 1.35s 3.00  108 m s    343m s  Note that about 7% of the time is for the sound wave.

Using the radio distance to be just the average Earth–Moon distance gives an answer of   3.84  108 m   32m  d  d t = tradio + tsound =  radio  +  sound  =  +  = 1.37s. 8  c   vsound   3.00  10 m s   343m s  (b) The microphone was inside the helmet. The Moon has essentially no atmosphere, and so there would not be sound waves in free space on the Moon. The helmet is filled with breathable air, and so the microphone has to be inside the helmet so that the sound waves will travel in the air around the astronaut’s head, inside the helmet. 54. The time travel delay is the distance divided by the speed of radio waves (which is the speed of light). 3  106 m d t= = = 0.01s c 3.00  108 m s 55. After the change occurred, we would find out when the change in radiation reached the Earth, traveling at the speed of light. (1.50  1014 m ) = 5.00  102 s = 8.33min d t = = c ( 3.00  108 m s ) 56. (a) The rms value of the associated electric field is found from Eq. 24–6. 2 → Erms = u = 12  0 E 2 =  0 Erms

0

4  10−14 J m3 = 0.0672 V m  0.07 V m 8.85  10−12 C2 N m 2

(b) A comparable value can be found using the magnitude of the Poynting vector. P 2 = → S =  0cErms 4 r 2 1 1 6800 W P = r= −12 2 Erms 4 0c 0.0672 V m 4 (8.85  10 C n m 2 )( 3.00  108 m s )

= 6718m  7 km 57. The light has the same intensity in all directions, so use a spherical geometry centered on the source to find the value of the Poynting vector. Then use Eq. 31–19a to find the magnitude of the electric field, and Eq. 31–11 with v = c to find the magnitude of the magnetic field.

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P0 P = 0 2 = 12 c 0 E02 → A 4 r

E0 =

( 5.0 W ) P0 = = 6.925V m  6.9 V m 2 2 2 r c 0 2 ( 2.50 m ) ( 3.00  108 m s )(8.85  10 −12 C 2 n m 2 )

B0 =

( 6.925V m ) = 2.3  10−8 T E0 = c ( 3.00  108 m s )

58. The radiation from the Sun has the same intensity in all directions, so the rate at which energy passes through a sphere centered at the Sun is P = S 4 R 2 . This rate must be the same at any distance from the Sun. Use this fact to calculate the magnitude of the Poynting vector at Mars, and then use the Poynting vector to calculate the rms magnitude of the electric field at Mars. 2  REarth  2 2 2 S Mars ( 4 RMars S R S S = → =  4 ) Earth ( Earth )  = c 0 Erms → Mars Earth  2 Mars  RMars  S Earth  REarth   = c 0  RMars 

Erms = Mars

1350 W m 2  1  = 469 V m −12 2 8 2  ( 3.00  10 m s )(8.85  10 C n m )  1.52 

59. (a) Since intensity is energy per unit time per unit area, the energy is found by multiplying the intensity by the area of the antenna times the elapsed time. 2  0.33m  −11 U = IAt = (1.0  10−13 W m 2 )    ( 3.0 h )( 3600s h ) = 9.2  10 J  2  (b) The electric field amplitude can be found from the intensity, which is the magnitude of the Poynting vector. The magnitude of the magnetic field is then found from Eq. 31–11 with v = c. I = 12  0cE02 →

E0 =

2I =  0c

2 (1.0 10−13 W m 2 )

(8.8510 C N m )( 3.00 10 m s ) −12

= 8.679  10−6 V m

 8.7  10−6 V m B0 =

E0 8.679  10−6 V m = = 2.9  10−14 T c 3.00 108 m s

60. According to Fig. 31–12, satellite TV waves have a wavelength of ~ 0.03 m. Example 31–4 says that the wavelength is 4 times the antenna’s length. So the antenna might be 0.075 m in length. 61. Use the relationship between average intensity (the magnitude of the Poynting vector) and electric field strength, as given by Eq. 31–19a. Also use the fact that intensity is power per unit area. We assume that the power is spherically symmetric about source. P P → S = 12  0cE0 2 = = A 4 r 2

25  103 W P = = 61,200 m 2 2 0cE0 2 2 ( 8.85  10−12 C2 N m 2 )( 3.00 108 m s ) ( 0.020 V m )

 61km Thus, to receive the transmission one should be within 61 km of the station. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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62. In each case, the required area is the power requirement of the device divided by 20% of the intensity of the sunlight. P 50  10−3 W = 2.5  10−4 m 2  3cm 2 (a) A = = 2 I 200 W m (b)

P 1875W = = 9.375m 2  9 m 2 2 I 200 W m

(c)

P 120 hp ( 746 W hp ) = = 447.6 m 2  400 m 2 200 W m 2 I

( to one sig. fig.)

(d) Calculator: A typical calculator is about 17 cm × 8 cm, which is about 140cm 2 . So yes, the solar panel can be mounted directly on the calculator. Hair dryer: A house of floor area 1000ft 2 would have on the order of 100 m 2 of roof area. So yes, a solar panel on the roof should be able to power the hair dryer. Car: For the car, a square panel of side length about 20 m would be required. So no, this panel could not be mounted on a car and used for real-time power. 63. The light has the same intensity in all directions. Use a spherical geometry centered at the source with the definition of the Poynting vector. S=

 1  P0 P = 0 2 = 12 c 0 E02 = 12 c  2  E02 → A 4 r  c 0 

1 2

 1  P c  2  E02 = 0 2 4 r  c 0 

→

E0 =

0cP0 2 r 2

c ( 3.0 10 m s ) = = 300 m. f (106 Hz ) 8

64. The wavelength of the AM radio signal is given by Eq. 31–14,  = (a)

1 2

 = 12 ( 300m ) = 150 m

(b)

1 4

 = 14 ( 300m ) = 75 m

So there must be another way to pick up this signal–we don’t see antennas this long for AM radio. dU = 3.0 W. dt dU dt . From That power is related to the average magnitude of the Poynting vector by S = A 2S Eq. 31–21b, that causes a pressure on the suit of P = , and the force due to that pressure is c Flaser = PA. Combine these relationships to calculate the force. 2 ( 3.0 W ) 2S 2 dU = = 2.0  10−8 N Flaser = PA = A= 8 c c dt 3.00  10 m s (b) Use Newton’s law of universal gravitation, Eq. 6–1, to estimate the gravitational force. We take the 20 m distance as having 2 significant figures. (1.03  105 kg ) (120 kg ) m m Fgrav = G shuttle 2 astronaut = ( 6.67  10−11 N m 2 kg 2 ) 2 r ( 30 m )

65. (a) The light is delivering electromagnetic energy to an area A of the suit at a rate of

= 9.160  10−7 N  9.2  10 −7 N (c) The gravity force is larger, by a factor of approximately 50. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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66. We take the units of  0 times the units of electric flux, divided by the units of time.

 C2   N 2  m  0  E  =  N m2   C  = C = Amperes s s t  67. For a lightning flash 2.0 km away, it will take the sound ( 3.0s km )( 2.0 km ) = 6.0s to arrive, using the given “rule of thumb.” But the light was also “delayed” by this amount of time: ( 2.0  103 m ) = 6.67  10−6 s. So instead of the sound arriving 6.0 seconds after the light flash, it 3.0  108 m s

actually arrived ( 6.0 − 6.67  10−6 ) s after the light flash. Calculate the % error due to the “mistake.” 6.67  10−6 s  100 = 1.1  10−4 % . 6.0s

68. The power output of the antenna would be the intensity at a given distance from the antenna, times the area of a sphere surrounding the antenna. The intensity is the magnitude of the Poynting vector. S = 12 c 0 E02

P = 4 r  S = 2 r c 0 E02 = 2 ( 0.75m ) ( 3.00  108 m s )(8.85  10−12 C2 N m 2 )( 3  106 V m ) 2

= 8.45  1010 W  8  1010 W This is many orders of magnitude higher than the power output of commercial radio stations, which are no higher than the 10’s of kilowatts. 69. We calculate the speed of light in water according to the relationship given. 1 1 1 1 1 c= vwater = = = ( 3.00  108 m s ) = 2.25  108 m s K  0 0 K  0 0 K 1.77

vwater = c

1 c K = 1 = 0.752 = 75.2% c K

70. A standing wave has a node every half-wavelength, including the endpoints. For this wave, the nodes would occur at the spacing calculated here. c 1 3.00  108 m s 1 1 = =2 = 0.0612 m  2 2 f 2.45  109 Hz Thus there would be nodes at the following distances from a wall: 0, 6.12 cm, 12.2 cm, 18.4 cm, 24.5 cm, 30.6 cm, and 36.7 cm (approximately the other wall). So there are 5 nodes, not counting the ones at (or near) the walls. 71. (a) Assume that the wire is of length l and cross-sectional area A. There must be a voltage across the ends of the wire to make the current flow (V = IR ) , and there must be an electric field associated with that voltage ( E = V l ) . Use these relationships with the definition of displacement current, Eq. 31–3.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

ID = 0

Instructor Solutions Manual

d ( EA) d (V l ) dE dE A dV A d ( IR ) = 0 = 0 A = 0 A = 0 =  0  l dt l dt dt dt dt dt

1 dI dI R =  0 R dt dt (b) Calculate the displacement current found in part (a). dI  1.0 A  = ( 8.85  10−12 C2 N m 2 )(1.68  10−8  m )  ID =  0  −3  dt  1.0  10 s  =  0

= 1.4868  10−16 A  1.5  10−16 A (c) From Example 28–6, Ampere’s law gives the magnetic field created by a cylinder of current as I B = 0 at a distance of r from the axis of the cylindrical wire. This is true whether the current 2 r is displacement current or steady current. −7 2 2 −16 0 I D ( 4  10 N s C )(1.486  10 A ) BD = = = 2.97  10−20 T  3.0  10 −20 T −3 2 r 2 (1.0  10 m )

0 I D 1.486  10−16 A BD I = 2 r = D = = 1.486  10−16  1.5  10−16  I 1.0 A Bsteady I steady 0 steady 2 r 72. The direction of the wave velocity is the direction of the cross product E  B. “South” crossed into “west” gives the direction downward. The electric field is found from the Poynting vector, Eq. 31–19a, and then the magnetic field is found from Eq. 31–11 with v = c. S = 12 c 0 E0 2 → 2 ( 520 W m 2 )

E0 =

2S = c 0

B0 =

( 625.9 V m ) = 2.1  10−6 T E0 = c ( 3.00  108 m s )

( 3.00  10 m s )(8.85  10 C n m ) 8

−12

= 625.9 V m  630 V m

73. (a) To show that E and B are perpendicular, calculate their dot product. E B =  E0 sin ( kx −  t ) ˆj + E0 cos ( kx −  t ) kˆ   B0 cos ( kx −  t ) ˆj − B0 sin ( kx −  t ) kˆ  = E0 sin ( kx −  t ) B0 cos ( kx −  t ) − E0 cos ( kx −  t ) B0 sin ( kx −  t ) = 0

Since E B  0, E and B are perpendicular to each other at all times. (b) The wave moves in the direction of the Poynting vector, which is given by S =

0

E  B.

ˆi ˆj kˆ 1 1 S = EB = 0 E0 sin ( kx −  t ) E0 cos ( kx −  t ) 0 0 0 B0 cos ( kx −  t ) − B0 sin ( kx −  t ) =

1 ˆ 1 i  − E0 B0 sin 2 ( kx −  t ) − E0 B0 cos2 ( kx −  t ) + ˆj ( 0 ) + kˆ ( 0 ) = − E0 B0ˆi

0

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We see that the Poynting vector points in the negative x-direction, and so the wave moves in the negative x-direction, which is perpendicular to both E and B. (c) We find the magnitude of the electric field vector and the magnetic field vector.

(

E = E =  E0 sin ( kx −  t ) +  E0 cos ( kx −  t ) 2

1/2

=  E02 sin 2 ( kx −  t ) + E02 cos2 ( kx −  t )

(

= E0

B = B =  B0 cos ( kx −  t ) +  B0 sin ( kx −  t ) 2

1/2

=  B02 cos2 ( kx −  t ) + B02 sin 2 ( kx −  t )

)

2 1/2

)

2 1/2

= B0

(d) At x = 0 and t = 0, E = E0kˆ and B = B0ˆj. See the figure. The x-axis is coming out of the page toward the reader. As time increases, the component of the electric field in the z-direction electric field begins to get smaller and the component in the negative y-direction begins to get larger. At the same time, the component of the magnetic field in the y-direction begins to get smaller, and the component in the z-direction begins to get larger. The net effect is that both vectors rotate counterclockwise. 74. (a) The radiation pressure when there is total reflection is given in Eq. 31–21b. 2 2 I avg 2 (1350 W m ) = = 9.00  10−6 N m2 P= c 3.00  108 m s (b) The force on the sail is the pressure times the area, and then use Newton’s second law to calculate the acceleration. F ma PA P 9.00  10−6 N m 2 → a= = = = 9  10−3 m s 2 P= = −3 2 A A m m A 1  10 kg m This answer is not an accurate description of the full situation. There is also a gravitational attraction due to the Sun’s mass, and it might be big enough to change the acceleration calculated here. (c) The required force is the total mass (the payload plus the sail), and is provided by the radiation pressure times the area of the sail.

(

)

m payload + ( 0.001kg m2 ) A a F ma ma ( m payload + msail ) a → A= = = → P= = A A P P P (100 kg ) (1 10−3 m s 2 ) m payload a A= =  P − ( 0.001kg m 2 ) a  ( 9.00  10−6 N m 2 ) − ( 0.001kg m 2 )(1  10−3 m s 2 )     

= 1.25  104 m 2  1  104 m 2 Again, the Sun’s gravitational attraction should also be figured into this before we claim to have a full understanding of the situation. 75. (a) The radio waves have the same intensity in all directions. The power crossing a given area is the intensity times the area. The intensity is the total power through the area of a sphere centered at the source. P 25  103 W P = IA = 0 A = 1.0 m 2 ) = 1.989  10−3 W  2.0 mW 2 ( 3 Atotal 4 (1.0  10 m ) (b) We find the rms value of the electric field from the intensity, which is the magnitude of the Poynting vector. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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2 S = c 0 Erms =

Erms =

P0 4 r 2

Instructor Solutions Manual

→

P0 25  103 W = 2 4 r 2 c 0 4 (1.0  103 m ) ( 3.00  108 m s )(8.85  10 −12 C 2 N m 2 )

= 0.8656 V m  0.87 V m

(c)

The voltage over the length of the antenna is the electric field times the length of the antenna. Vrms = Ermsd = ( 0.8656V m )(1.0m ) = 0.87 V

(d)

We calculate the electric field at the new distance, and then calculate the voltage. We also assume that 50 km has 2 significant figures. Erms =

P0 25  103 W = 2 4 r 2 c 0 4 ( 5.0  104 m ) ( 3.00  108 m s )(8.85  10−12 C2 N m 2 )

= 1.731  10−2 V m ; Vrms = Erms d = (1.731  10−2 V m ) (1.0 m ) = 1.7  10−2 V

76. From the hint, we use Eq. 29–4, which says e = e0 sin  t = NBA sin  t. The intensity is given, and this can be used to find the magnitude of the magnetic field. 0 S E B cB 2 S = rms rms = rms → Brms = ; e = e0 sin  t = NBA sin  t → 0 0 c erms = NA Brms = NA

0 S c

= ( 280 )  ( 0.011m ) 2 ( 810  103 Hz ) 2

( 4  10 N s C )(1.0  10 W m ) −7

−4

3.00  108 m s

= 3.5  10−4 V

77. We find the average intensity at the 9.0-km distance from Eq. 31–19a. Then the power can be found from the intensity, assuming the signal is spherically symmetric, using the surface area of a sphere of radius 9.0 km. 2 I = 12  0cE02 = 12 (8.85  10−12 C2 N m2 )( 3  108 m s ) ( 0.12 V m ) = 1.9116  10−5 W m2

P = IA = (1.9116  10−5 W m2 ) 4 ( 9.0  103 m ) = 1.946  104 W  19 kW 2

 A B   A B 78. (a) Use the sin A  sin B = 2sin   cos   from page A–4 in Appendix A.  2   2  E y = E0 sin ( kx −  t ) + sin ( kx +  t )

 ( kx −  t ) + ( kx +  t )   ( kx −  t ) − ( kx +  t )  = 2 E0 sin   cos   = 2 E0 sin ( kx ) cos ( − t ) 2 2     = 2 E0 sin ( kx ) cos ( t )

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Bz = B0 sin ( kx −  t ) − sin ( kx +  t )  ( kx −  t ) − ( kx +  t )   ( kx −  t ) + ( kx +  t )  = 2 B0 sin   cos   = 2 B0 sin ( − t ) cos ( kx ) 2 2     = −2 B0 cos ( kx ) sin ( t ) (b) The Poynting vector is given by S =

0

E  B. Use the fact that 2sin  cos  = sin 2 .

ˆi ˆj kˆ 0 2 E0 sin ( kx ) cos ( t ) 0 S = EB = 0 0 0 0 −2 B0 cos ( kx ) sin ( t ) 1

1 ˆ 1 i  −4 E0 B0 sin ( kx ) cos ( kx ) sin ( t ) cos ( t )  = − ˆi E0 B0 sin ( 2kx ) sin ( 2 t )

0

This is 0 for all times at positions where sin ( 2kx ) = 0. sin ( 2kx ) = 0 → 2kx = n →

n , n = 0, 1, 2, 2k

79. (a) We note that − x −  t + 2 xt = − ( x −  t ) and so E y = E0e 2

2 2

−( x −  t )

= E0e

   − 2  x − t    

. Since

the wave is of the form f ( x − v t ) , with v =   , the wave is moving in the +x-direction. (b) The speed of the wave is v =   = c, and so  =  c . (c) The electric field is in the y-direction, and the wave is moving in the x-direction. Since E  B must be in the direction of motion, the magnetic field must be in the z-direction. The magnitudes are related by B = E c . Bz =

E 0 − ( x −  t ) 2 e c

1077

CHAPTER 33: Lenses and Optical Instruments Responses to Questions 1.

To make a sharp image of an object that is very far away, the film of a camera must be placed at the focal point of the lens. Objects that are very far away have light rays coming into the camera that are basically parallel, and these rays will be bent and focused to create an image at the focal point of the lens, which is where the film should be in order to record a focused image.

The lens moves farther away from the film. When the photographer moves closer to his subject, the object distance decreases. The focal length of the lens does not change, so the image distance must 1 1 1 = + . To get the image and the film at the same place, they need to increase, by Eq. 33–2, f do di move the lens away from the film, increasing the image distance.

A real image is only formed by a converging lens, and only if the object distance is greater than the focal length. The image is on the opposite side of the lens as the object, and will always be inverted as shown in the first diagram. If the object were moved back, the three rays drawn would never intersect above the axis of symmetry to make an upright image, but only intersect below the axis of symmetry, only making an inverted image. A virtual image is formed by a converging lens if the object distance is less than the focal length. In the second diagram, the outgoing rays will never intersect since ray 2 is parallel to the symmetry axis and rays 1 and 3 are sloping downward. Also, ray 3 will always slope more than ray 1, so none of those rays ever meet. Thus they can only form a virtual image, and it has to be upright, as seen in the diagram. A virtual image is formed by a diverging lens, no matter where a real object is placed. Since the rays leaving the lens diverge, they cannot form a real image. And because the ray that hits the lens at point B is sloping upwards, it will always (virtually) intersect the other ray to make an upright image.

Yes, to say that light rays are “reversible” is consistent with the thin lens equation. Since 1 1 1 + = , if you interchange the image and object positions, the equation is unchanged. do di f

(a) As a thin converging lens is moved closer to a nearby object, the position of the real image changes. As the object distance decreases the thin lens equation says that the image distance increases, in order for the calculation for the focal length to not change. Thus, the position of the image moves farther away from the lens, until the object is at the focal point. At that condition, the image is at infinity.

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(b) As a thin converging lens is moved closer to a nearby object, the size of the real image changes. As the object distance decreases the thin lens equation says that the image distance increases, which means that the magnification ( m = − d i d o ) gets larger. Thus, the size of the image increases until the object reaches the focal point, in which case a real image will no longer be formed. 6.

A virtual image created by a lens can serve as a virtual object for a second lens. If the previous lens creates an image behind the position of the second lens, relative to the incoming light, that image is a virtual object for the second lens.

Assuming that the lens remains fixed and the screen is moved, the dog’s nose will have the greater magnification. The object distance for the nose is less than the object distance for the tail, because the dog is facing the mirror. The image distance for the nose will therefore be greater than the image distance for the tail. Magnification is the ratio of the image distance to the object distance, and so will be greater for the nose.

If the cat’s nose is closer to the lens than the focal point and the tail is farther from the lens than the focal point, the image of the nose will be virtual and the image of the tail will be real. The virtual image of the front part of the cat will be spread out from the image of the nose to infinity on the same side of the lens as the cat. The real image of the back part of the cat will be spread out from the image of the tail to infinity on the opposite side of the lens.

A double convex lens causes light rays to converge because the light bends towards the normal as it enters the lens and away from the normal as it exits the lens. The result, due to the curvature of the sides of the lens, is that the light bends towards the principal axis at both surfaces. The more strongly the sides of the lens are curved (which means the thicker the lens), the greater the refraction, and the shorter the focal length. This can also be 1 1  1 seen from the lensmaker’s equation: = ( n − 1)  +  . To f  R1 R2  make a lens that is thick in the middle, the radii of the two convex sides should be small. If R1 and R2 decrease in the lensmaker’s equation, the focal length decreases.

10. If the object of the second lens (the image from the first lens) is exactly at the focal point, then a virtual image will be formed at infinity and could be viewed with a relaxed eye, looking “into” the two-lens combination. 11. (a) For the converging lens to be “stronger” means that it must have a shorter focal length then the absolute value of the focal length of the diverging lens. If the diverging lens is “stronger,” then instead of the focal point simply being moved back from its original location in Fig. 33–14, the rays would never focus. The diverging lens would diverge the entering rays so much that again, a virtual image would be formed, and no direct measure would be possible. (b) The focal length of a diverging lens cannot be measured directly because the diverging lens cannot form a real image. Therefore you can’t measure the image distance by making the image focus on a screen. Some indirect measure (as described in this problem) must be used.

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12. The image point of an unsymmetrical lens does not change if you turn the lens around. The 1 1  1 lensmaker’s equation, = ( n − 1)  +  , is symmetrical (even if the lens is not). Turning the f  R1 R2  lens around interchanges R1 and R2 , which does not change the value of the focal length or the location of the image point. 13. Because there is less refraction when light passes between water and glass than when light passes between air and glass, the light will not be bent as much when the lens is in water, and so its focal length will be longer in the water. The correct answer is (a). 14. (a) Yes, the focal length of a lens depends on the surrounding fluid. The relative values of the index of refraction of the fluid and the index of refraction of the lens will determine the refraction of light as it passes from the fluid through the lens and back into the fluid. The amount of refraction of light determines the focal length of the lens, so the focal length will change if the lens is immersed in a fluid. (b) No, the focal length of a spherical mirror does not depend on the surrounding fluid. The image formation of the spherical mirror is determined by reflection, not refraction, and is independent of the medium in which the mirror is immersed. 15. No. If a lens with n = 1.25 is a converging lens in air, it will become a diverging lens when placed in water, with n = 1.33. The figure on the left shows that as parallel light rays enter the lens when it is in air, at the first surface the ray is bent toward the normal and at the second surface the ray is bent away from the normal, which is a net converging situation. The figure on the right shows that as parallel light rays enter the lens when it is in water, at the first surface the ray is bent away from the normal and at the second surface the ray is bent toward the normal, which is a net diverging situation. 16. The lens material is air and the medium in which the lens is placed is water. Air has a lower index of refraction than water, so the light rays will bend away from the normal when entering the lens and towards the normal when leaving the lens. (a) A converging lens can be made by a shape that is thinner in the middle than it is at the edges. (b) A diverging lens will be thicker in the middle than it is at the edges. 17. No, a nearsighted person will not be able to see clearly if they wear their corrective lenses underwater. A nearsighted person has a far point that is closer than infinity, and they wear corrective lenses to bring the image of a far away object to their far point so they can see it clearly. See the pair of diagrams. The object is at infinity. In air, the image is at the (relatively close) far point. If the person’s eyes and glasses are underwater, and since the index of refraction of glass is closer to that of water than to that © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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of air, the glasses will not bend the light as much as they did in the air. Therefore, the image of the faraway object will now be at a position that is beyond the person’s far point. The image will now be out of focus. 18. If the person moves the glasses slightly away from their eyes, then the image distance becomes smaller in absolute value, or larger in actual value. This larger image distance means that the object distance will be smaller, meaning the person can hold the reading material closer to their face, which might make it easier to hold. Using the values from Example 33–2, if we move the glasses 3.0 cm from the eye instead of 2.0 cm (using the same focal length for the glasses), then the new near point is 22.5 cm instead of 30 cm. The paragraph on contact lenses shows the same idea. With the contact lenses right at the eye, the near point was even further out, at 41 cm. 19. This person is nearsighted. Diverging lenses are used to correct nearsightedness and converging lenses are used to correct farsightedness. If the person’s face appears narrower through the glasses, then the image of the face produced by the lenses is smaller than the face, virtual, and upright. Thus, the lenses must be diverging, and therefore the person is nearsighted. 20. To see far objects clearly, you want your eye muscles relaxed, which makes your lens relatively flat (large focal length). When you f-stop your eye down by closing your eyelids partially (squinting), you are only using the middle of your lens, where it is the most flat. This creates a smaller circle of confusion for the lens, which helps you see distant objects more clearly. 21. The images formed on our retinas are real, and so are inverted. The implication of this is that our brains must flip this inverted image for us so that we can perceive objects as upright. 22. Both reading glasses and magnifiers are converging lenses. A magnifier, generally a short focal length lens, is typically used by adjusting the distance between the lens and the object so that the object is exactly at or just inside the focal point. An object exactly at the focal point results in an image that is at infinity and can be viewed with a relaxed eye. If the lens is adjusted so that it makes a virtual image at the eye’s near point, the magnification is slightly greater. The lenses in reading glasses typically are a fixed distance from the eye. They are used when the near point is too far away to be convenient–such as when reading a book. These lenses make a virtual, magnified, upright image at the near point of the eye. It allows the eye to focus on an object that is inside the near point. The focal length of the lens needed for reading glasses will depend on the individual eye. For both reading glasses and magnifiers, the lenses allow the eye to focus on an object that is located closer than the near point. 23. The glasses that near-sighted people wear are designed to help with the viewing of distant objects, by moving the focus point further back in the eye. Near-sighted eyes in general have good vision for close objects whose focus point already lies on the retina. They can actually see close objects better without the corrective lenses. The near point of a near-sighted person can be very close to their eye, and so they bring small objects close to their eyes in order to see them most clearly. 24. As people get older, their eyes can no longer accommodate as well. It becomes harder for the muscles to change the shape of the lens, since the lens becomes less flexible with age. In general, people first lose the ability to see far objects, and so they need corrective diverging lenses to move their “far point” back towards infinity. Then, as people get older, their near point increases and becomes greater than the ideal value of 25 cm. They may still need the diverging portion of their corrective lenses (kept as the upper part of the corrective lens) so they can have a far point at infinity, but now also need a converging lens to move the near point back towards 25 cm for seeing close objects conveniently. Thus, as people get older, their far point is too close and their near point is too far. Bifocals (with two different focal lengths) can correct both of these problems. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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25. Stopping down a lens to a larger f-number means that the lens opening is smaller and only light rays coming through the central part of the lens are accepted. These rays are more nearly parallel than if a smaller f-number was used. These rays form smaller circles of confusion, which means a greater range of object distances will be more sharply focused. 26. (a) Light from a real image actually passes through the image location. Light from a virtual image only appears to pass through the image location. A real image can be displayed on a screen. To see a virtual image you must look through the mirror or lens that is forming the image, in order to see the image. (b) No, your eyes cannot tell the difference. In each case the light entering your eye appears to come from the position of the image. (c) On a ray diagram, the ray traces pass through the real image location. You must extrapolate the rays backward through the mirror/lens to find the location of the virtual image. (d) You could place a screen at the location of the image. If it is a real image, it will appear on the screen. If it is a virtual image, it will not appear on the screen. (e) Yes, if the photograph is taken through the lens or mirror, the image would appear in the photograph. (f) No. The light does not actually pass through the location of the virtual image, so no image would be captured on the sensor. (g) Light appears to come from the position of the virtual image, and that light can be captured by the camera. However, the light does not pass through the location of the virtual image, so the film placed at that position would not capture the image. 27. The optical zoom is preferable because it uses all of the available pixels to maximize the resolution. When a digital zoom is used to take a picture, the size of the pixels is enlarged on the final print and so there is a loss of sharpness. 28. The curved surface should face the object. If the flat surface faces the object and the rays come in parallel to the optical axis, then no bending will occur at the first surface and all the bending will occur at the second surface. Bending at the two surfaces will clearly not be equal in this case. The bending at the two surfaces may be equal (or at least less different) if the curved surface faces the object. If the parallel rays from the distant object come in above or below the optical axis with the flat side towards the object, then the first bending is actually away from the axis. In this case also, bending at both surfaces can be equal (or at least less different) if the curved side of the lens faces the object. 29. For both converging and diverging lenses, the focal point for violet light is closer to the lens than the focal point for red light. The index of refraction for violet light is slightly greater than for red light for glass, so the violet light bends more, resulting in a smaller magnitude focal length. 30. i) ii)

A reflecting telescope can be built with a much larger diameter, as the mirror can be made of many small mirrors connected together. The lens, being one single piece of glass, is more difficult to manufacture. A lens must be supported around the outer edge and therefore, due to the large weight of the lens, its size is limited. A mirror can be supported everywhere behind the mirror and is much lighter. Therefore, the mirrors can be much larger in diameter.

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iii) A lens experiences spherical aberrations, while the shape of a mirror can be adjusted to eliminate spherical aberrations. iv) A lens experiences chromatic aberrations since different colors of light refract in the lens at slightly different angles. The image created by a mirror does not depend upon an index of refraction, since the light does not travel through a different material when reflecting from a telescope mirror. 31. Spherical aberrations can be present in a simple lens. To correct this in a simple lens, usually many lenses are used together in combination to minimize the bending at each of the surfaces. Your eye minimizes spherical aberrations by also bending the light at many different interfaces as it makes its way through the different parts of the eye (cornea, aqueous humor, lens, vitreous humor, etc.), each with their own n. Also, the cornea is less curved at the edges than it is at the center and the lens is less dense at the edges than at the center. Both of these reduce spherical aberrations in the eye since they cause the rays at the outer edges to be bent less strongly. Off-axis astigmatism is not a problem in the eye because objects are clearly seen only at the fovea, which is on the lens axis. Curvature of field occurs when the focal plane is not flat. Our curved retina helps with this distortion, whereas a flat piece of film in a camera, for example, wouldn’t be able to fix this. Distortion is a result of the variation of the magnification at different distances from the optical axis. This is most common in wide-angle lenses, where it must be corrected for. This is compensated for in the human eye because it is a very small lens and our retina is curved. Chromatic aberrations are mostly compensated for in the human eye because the lens absorbs shorter wavelengths, and the retina is not very sensitive to most blue and violet wavelengths where most chromatic aberrations occur.

Responses to MisConceptual Questions 1.

(c) The answer is defined in Fig. 33–3. It can be calculated from the lensmaker’s equation, but that equation does not directly involve the diameter or thickness of the lens, so answers (a) and (b) are incorrect. And there are various locations where real images are formed by a converging lens, so answer (d) is also incorrect.

(e) Virtual images are formed by plane mirrors, convex mirrors, concave mirrors (when the object is within the focal distance), diverging lenses, and converging lenses (when the object is within the focal distance). Therefore, virtual images can be formed with plane and curved mirrors and lenses.

(c) Virtual images are seen and can be photographed–for instance, you can take a picture of what you see when you look into a plane mirror. But they cannot be projected on a screen–there are no rays actually intersecting at the virtual image point, so they don’t focus on a screen.

(d) The fact that a lens has two focal points of the same size, one on each side, is discussed on the bottom of page 981 and also in the opening paragraphs of Section 33–4. One way to think about it is that the lens causes the same “bending” of parallel light, no matter which side the light comes from.

(c) A single diverging lens cannot create a real image, so answers (b), (d), and (e) must be incorrect. From Fig. 33–12, an object placed inside the focal point of a converging lens makes a virtual image, so answer (a) is also incorrect. The object must be beyond the focal length of a single converging lens to create a real image.

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(c) A common misconception is that each half of the lens produces half of the image. Actually, light from each part of the image passes through each part of the lens. When half of the lens is covered then the image is less intense, but the full image is still present on the screen.

(d) A common misconception would be to equate power with magnification. However, power is the term used for the reciprocal of the focal length.

(e) It might seem reasonable that a lens must be curved on both sides to produce an image. However, images can be produced by plane mirrors, curved mirrors, and lenses curved on one or both sides.

(d) An object at infinity would produce an image at the focal point. Objects closer than infinity produce images that are farther from the lens than the focal point.

10. (d) The f-stop value determines the amount of light entering the camera and therefore affects how dark the image is. It does not affect the focus. Since the focal length of the lens is fixed, when the object distance increases, the image distance must decrease. Therefore the lens must be moved closer to the sensor or film. 11. (e) A converging lens can form images as described in parts (a), (b), (c), and (d), but none of them are exclusive. Depending on the object position, any of those four results might be obtained–a magnified image, a reduced image, an upright image, or an inverted image. 12. (a) If the distance from the lens to the retina is shorter than normal, the lens must contract to focus distant objects. It must contract even more to view closer objects. Since it is already partially contracted to view far objects, it will not be able to sufficiently contract to view nearby objects. 13. (b) A nearsighted person cannot see objects clearly beyond their far point. Therefore glasses create images of distant objects at the observer’s far point. 14. (a) A nearsighted person’s retina is farther from the front of the eye than usual. As such, a diverging lens is needed to make the image focus further from the front of the eye. 15. (a) Because people perceive objects as upright they may think that the image is upright. However, when a converging lens creates a real image, the image is inverted. 16. (b) Statement (a) is false. Contact lenses are placed on the eye, while glasses are typically placed about 2.0 cm in front of the eyes. This extra distance is taken into account when determining the power of the prescription. Example 33–12 and the following paragraphs illustrate this. Statement (b) is true–see the bottom of page 998. Statement (c) is false–nearsighted people can see nearby objects clearly (see page 998). Statement (d) is false. Astigmatism is corrected by using a non-spherical lens to correct the asymmetric shape of the cornea. See page 999.

Solutions to Problems 1.

(a) From a ray diagram drawn to scale, the object distance is about 630 mm.

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(b) We find the object distance from Eq. 33–2. 1 1 1 + = → do di f do =

( 235mm )( 373mm ) = 635mm fd i = di − f 373mm − 235mm

(a) To form a real image from parallel rays requires a converging lens . (b) We find the power of the lens from Eqs. 33–1 and 33–2. We treat the object distance as infinite. 1 1 1 1 1 + = =P= + = 6.06 D do di f  0.165m

(a) The power of the lens is given by Eq. 33.1. 1 1 P= = = 3.77 D f 0.265m This lens is converging. (b) We find the focal length of the lens from Eq. 33.1. 1 1 1 P= → f = =− = −0.160 m f D 6.25D This lens is diverging.

To form a real image from a real object requires a converging lens. We find the focal length of the lens from Eq. 33–2. ( 2.15m )( 0.483m ) = 0.394 m 1 1 1 dd + = → f = o i = 2.15m + 0.483m do di f do + di Because di  0, the image is real.

(a) We find the image distance from Eq. 33–2. (10.0 m )( 0.105m ) = 0.106 m = 106 mm d f 1 1 1 + = → di = o = do di f do − f 10.0 m − 0.105m (b) Use the same general calculation. ( 3.0 m )( 0.105m ) = 0.109 m = 109 mm d f di = o = do − f 3.0 m − 0.105m (c) Use the same general calculation. (1.0 m )( 0.105m ) = 0.117 m = 117 mm d f di = o = do − f 1.0 m − 0.105m (d) We find the smallest object distance from the maximum image distance. di − f (132 mm )(105mm ) = 513mm = 0.513m 1 1 1 max + = → do = = do di f di f 132 mm − 105mm min min

max

(a) We locate the image using Eq. 33–2. (15cm )( 28cm ) = −32.3cm  −32cm 1 1 1 d f + = → di = o = 15cm − 28cm do di f do − f The negative sign means the image is 32 cm, on the same side of the lens as the stamp (virtual) .

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(b) We find the magnification from Eq. 33–3. ( − 32.3cm ) = 2.15  +2.2 d m=− i =− do (15cm ) 7.

(a) A qualitatively-correct ray diagram is shown. The image should be upright for reading. The image will be virtual, upright, and magnified. (b) To form a virtual, upright, magnified image requires a converging lens. (c) We find the image distance, then the focal length, and then the power of the lens. The object distance is given. d m = − i → d i = −md o do P=

1 1 1 d i + d o −md o + d o m − 1 3.0 − 1 = + = = = = = +7.4 D f do di d od i d o ( −md o ) md o ( 3.0 )( 0.090 m )

For reference, the focal length is 13.5 cm. 8.

From Eq. 33–3, hi = ho when di = d o . So find do from Eq. 33–2.

1 1 1 1 1 + = + = do di do do f 9.

→ d o = 2 f = 50cm

First, find the original image distance from Eqs. 33–1 and 33–2. (1.35m ) do 1 1 1 + = = P → di = = = 0.1378m do di f Pd o − 1 ( 8.0 D )(1.35m ) − 1 (a) With d o = 0.45m, find the new image distance.

( 0.45m ) do 1 1 1 + = = P → di = = = 0.1731m do di f Pd o − 1 ( 8.0 D )( 0.45m ) − 1

Thus, the image has moved 0.1731m − 0.1378m = 0.0353m  0.04 m away from the lens. (b) With d o = 2.25m, find the new image distance.

( 2.25m ) do 1 1 1 + = = P → di = = = 0.1324 m do di f Pd o − 1 ( 8.0 D )( 2.25m ) − 1

The image has moved 0.1378m − 0.1324 m = 0.0054 m  0.01m toward the lens. 10. (a) If the image is real, the focal length must be positive, the image distance must be positive, and the magnification is negative. Thus d i = −md o = − ( −2.50 ) d o = 2.50d o . Use Eq. 33–2. 1 1 1 1 1  3.50   3.50  + = + = → do =   f =  ( 75.0 mm ) = 105mm d o d i d o 2.50d o f 2.50    2.50  (b) If the image is magnified, the lens must have a positive focal length, because negative lenses always form reduced images. Since the image is virtual, the magnification is positive. Thus, d i = −2.50d o . Again use Eq. 33–2. 1 1 1 1 1 + = − = d o d i d o 2.5d o f

 1.50   1.50  → do =   f =  ( 75.0 mm ) = 45.0 mm  2.50   2.50 

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11. Since a concave (negative focal length) lens always forms reduced images of a real object, the object in these cases must be virtual (formed by another piece of optics) and have negative object distances in order to form enlarged images. Using Eqs. 33–2 and 33–3, we get this relationship. − di − di 1 1 1 m 1 1 → do = + = → + = → d i = (1 − m ) f m= ; − di di f do m do di f (a) If the image is real, d i  0. With f  0, we must have m  1 from the above relationship. Thus, m = + 2.50. d i = 1 − ( + 2.50 )  ( − 75.0 mm ) = 112.5mm do =

− d i − (112.5mm ) = = −45.0 mm m ( + 2.50 )

(b) If the image is virtual, di  0. With f  0, we see that m  1. Thus, m = − 2.50. d i = 1 − ( − 2.50 ) ( − 75.0 mm ) = −262.5mm do = −

( − 262.5mm ) = −105mm di =− m ( − 2.50 )

12. (a) Use Eqs. 33–2 and 33–3. (1.30 m )( 0.135m ) = 0.1506 m  151mm d f 1 1 1 + = → di = o = do di f do − f 1.30 m − 0.135m

hi d d 0.1506 m = − i → hi = − i ho = − ( 3.10cm ) = −0.359cm ho do do 1.30 m The image is behind the lens a distance of 151 mm, is real, and is inverted. (b) Again use Eqs. 33–2 and 33–3. (1.30 m )( −0.135m ) = −0.1223m  −122 mm 1 1 1 d f + = → di = o = do di f d o − f 1.30 m − ( −0.135m ) m=

( −0.1223m ) 3.10cm = 0.292cm hi d d = − i → hi = − i ho = − ( ) 1.30 m ho do do The image is in front of the lens a distance of 122 mm, is virtual, and is upright. m=

13. The sum of the object and image distances must be the distance between object and screen, which we label as dT . We solve this relationship for the image distance, and use that expression in Eq. 33–2 in order to find the object distance. 1 1 1 1 1 do + di = dT → di = dT − do ; + = + = → d o2 − d T d o + fd T = 0 → do di do ( dT − do ) f d  d T2 − 4 fd T ( 86.0cm )  (86.0cm ) − 4 (16.0cm )(86.0cm ) do = T = = 21.3cm, 64.7cm 2 2 Note that to have real values for d o , we must in general have d T2 − 4 fd T  0 → d T  4 f . 2

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14. For a real image both the object distance and image distances are positive, and so the magnification is negative. Use Eqs. 33–2 and 33–3 to find the object and image distances. Since they are on opposite sides of the lens, the distance between them is their sum. − di m= → d i = − md o = 2.95d o do

 3.95   3.95  → do =   f =  ( 78cm ) = 104.4cm  2.95   2.95  d i = 2.95d o = 2.95 (104.4cm ) = 308.1cm 1 1 1 1 1 + = + = d o d i d o 2.95d o f

d o + d i = 104.4cm + 308.1cm = 412.5cm  410cm 15. (a) Use Eq. 33–2 to write an expression for the image distance in terms of the object distance and focal length. We then use Eq. 33–3 to write an expression for the magnification. 1 1 1 d f d f + = → di = o ; m=− i =− do di f do − f do do − f These expressions show that when do  f , the image distance is positive, producing a real image, and the magnification is negative, which gives an inverted image. (b) From the above equations, when do  f , the image distance is negative, producing a virtual image, and the magnification is positive, which gives an upright image. (c) We set −d o = f and calculate the limiting image distance and magnification. (− f ) f = f d f 1 = di = m=− i =− −f − f 2 −f − f 2 do We also take the limit of large negative object distance. ( − ) f = f d f di = m=− i =− =0 do − − f − − f From these limiting cases, we see that when −d o  f , the image is real and upright with 1 1 2 f  d i  f and 0  m  2 . (d) We take the limiting condition d o → 0 , and determine the resulting image distance and magnification. ( 0) f = 0 d f di = m=− i =− =1 do 0− f 0− f From this limit and that found in part (c), we see that when 0  −do  f , the image is real and upright, with 0  d i  12 f and 12  m  1. 16. Find the object distance from Eq. 33–2. ( 0.105m )( 22.50 m ) = 0.1055m  0.105m 1 1 1 1 1 1 fd i + = → + = → do = = 22.50 m − 0.105m do di f do di f di − f Find the size of the image from Eq. 33–3. h d d 22.50 m m = i = − i → hi = i ho = ( 24 mm ) = 5118mm  5.1m ho do do 0.1055m

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17. (a) With the definitions as given in the problem, x = d o − f → d o = x + f and x = d i − f d i = x + f . Use Eq. 33–2. ( x + f ) + ( x + f ) = 1 → 1 1 1 1 1 + = + = → d o d i x + f x + f f ( x + f )( x + f ) f

( 2 f + x + x) f = ( x + f )( x + f ) → 2 f 2 + xf + xf = xx + xf + fx + f 2 →

→

f 2 = x x

(b) Use Eq. 33–2. ( 58.0cm )( 36.0cm ) = 94.9cm d f 1 1 1 + = → di = o = do di f do − f 58.0cm − 36.0cm (c) Use the Newtonian form.

( 36.0cm ) f2 → x = = = 58.9cm x ( 58.0cm − 36.0cm ) 2

xx = f

d i = x + f = 58.9cm + 36.0cm = 94.9cm 18. (a) Use Eq. 33–2 with d o + d i = d T → d i = d T − d o .

1 1 1 1 1 + = + = do di do ( dT − do ) f

→ d o2 − d T d o + fd T = 0 → d o =

d T  d T2 − 4 fd T 2

There are only real solutions for d o if d T2 − 4 fd T  0 → d T  4 f . If that condition is met,

)

(

then there will be two locations for the lens, at distances d o = 12 d T  d T2 − 4 fd T from the object, that will form sharp images on the screen. (b) If d T  4 f , then Eq. 33–2 cannot be solved for real values of d o or d i .

(

)

(

)

d o2 = 12 d T − d T2 − 4 fd T .

(

) (

)

d = d o1 − d o2 = 12 d T + d T2 − 4 fd T − 12 d T − d T2 − 4 fd T =

Find the ratio of image sizes using Eq. 33–3. d − ho i2 hi2 d o2 d i2 d o1 d T − d o2 d o1 = = = hi1 − h d i1 d o2 d i1 d o2 d T − d o1 o d o1

(

)

(

)

d T2 − 4 fd T

2  d − 1 d − d 2 − 4 fd   1 d + d 2 − 4 fd   T T T T T T T 2 2 d T + d T2 − 4 fd T       = =   1 d − d 2 − 4 fd   d − 1 d + d 2 − 4 fd   d − d 2 − 4 fd  T T T  T T T T   2 T   T 2 T  

(

)

(

)

19. (a) Find the image formed by the first lens, using Eq. 33–2. ( 35.0cm )( 28.0cm ) = 140cm d f 1 1 1 + = → d i1 = o1 1 = d o1 d i1 f1 d o1 − f1 ( 35.0cm ) − ( 28.0cm ) Note that the value of 140 cm has just 2 significant figures. This image is the object for the second lens. Because it is beyond the second lens, it has a negative object distance. d o2 = 16.5cm − 140cm = −123.5cm Find the image formed by the second lens, again using Eq. 33–2. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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1 1 1 + = d o2 d i2 f 2

→ d i2 =

Instructor Solutions Manual

( −123.5cm )( 28.0cm ) = 22.8cm d o2 f 2 = d o2 − f 2 ( −123.5cm ) − ( 28.0cm )

Thus the final image is real, 22.8 cm beyond the second lens . (b) The total magnification is the product of the magnifications for the two lenses.  d  d  d d m = m1m2 =  − i1  − i2  = i1 i2  d o1  d o2  d o1d o2 =

( +140cm )( + 22.8cm ) = − 0.738  − 0.74  inverted ( ) ( + 35.0cm )( −123.5cm )

20. The first lens is the converging lens. An object at infinity will form an image at the focal point of the converging lens, by Eq. 33–2. 1 1 1 1 1 + = = + → d i1 = f1 = 20.0cm d o1 d i1 f1  d i1 This image is the object for the second lens. Since this image is behind the second lens, the object distance for the second lens is negative, and so d o2 = −6.0cm. Again use Eq. 33–2. ( −6.0cm )( −36.5cm ) = 7.2cm d f 1 1 1 + = → d i2 = o2 2 = d o2 d i2 f 2 d o2 − f 2 ( −6.0cm ) − ( −36.5cm ) Thus the final image is real, 7.2 cm beyond the second lens . 21. From the ray diagram, the image from the first lens is a virtual image at the focal point of the first lens. This is a real object for the second lens. Since the light is parallel after leaving the second lens, the object for the second lens must be at its focal point. Let the separation of the lenses be l. Note that the focal length of the diverging lens is negative. f1 + l = f 2 →

f1  0

f2  0

f1 f2

f1 = f 2 − l = 38.0cm − 24.0cm = 14.0cm → f1 = − 14.0cm 22. (a) Find the image formed by the converging lens, using Eq. 33–2. ( 29cm )(18cm ) = 47.45cm d f 1 1 1 + = → d i1 = o1 1 = d o1 d i1 f1 d o1 − f1 ( 29cm ) − (18cm ) This image is the object for the second lens. The image is to the right of the second lens, and so is virtual. Use that image to find the final image. 1 1 1 d o2 = 12cm − 47.45cm = − 35.45cm ; + = → d o2 d i2 f 2 d i2 =

( −35.45cm )( −14cm ) = −23.14cm d o2 f 2 = d o2 − f 2 ( −35.45cm ) − ( −14cm )

So the final image is 23 cm to the left of the diverging lens, or 23 cm to the left of the diverging lens, or 11 cm to the left of the converging lens .

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(b) The initial image is unchanged. With the change in the distance between the lenses, the image distance for the second lens has changed. 1 1 1 d o2 = 38cm − 47.45cm = − 9.45cm ; + = → d o2 d i2 f 2 d i2 =

( −9.45cm )( −14cm ) = 29.1cm  30cm 1 sig. fig. d o2 f 2 = ( ) d o2 − f 2 ( −9.45cm ) − ( −14cm )

Now the final image is 30 cm to the right of the diverging lens . 23. (a) The first lens is the converging lens. Find the image formed by the first lens. ( 60.0cm )( 20.0cm ) = 30.0cm d f 1 1 1 + = → d i1 = o1 1 = d o1 d i1 f1 d o1 − f1 ( 60.0cm ) − ( 20.0cm ) This image is the object for the second lens. Since this image is behind the second lens, the object distance for the second lens is negative, and so d o2 = 25.0cm − 30.0cm = −5.0cm. Use Eq. 33–2. ( −5.0cm )( −10.0cm ) = 10cm d f 1 1 1 + = → d i2 = o2 2 = d o2 d i2 f 2 d o2 − f 2 ( −5.0cm ) − ( −10.0cm ) Thus the final image is real, 10 cm beyond the second lens . The distance has two significant figures. (b) The total magnification is the product of the magnifications for the two lenses.  d  d  d d ( 30.0cm )(10.0cm ) = −1.0  m = m1m2 =  − i1  − i2  = i1 i2 =  d o1  d o2  d o1d o2 ( 60.0cm )( −5.0cm ) (c) See the diagram here.

24. We find the focal length of the combination by finding the image distance for an object very far away. For the converging lens, we have the following from Eq. 33–2. 1 1 1 1 1 + = = + → d i1 = f C d o1 d i1 f C  d i1 The first image is the object for the second lens. Since the first image is real, the second object distance is negative. We also assume that the lenses are thin, and so d o2 = − d i1 = − f C . For the second diverging lens, we have the following from Eq. 33–2. 1 1 1 1 1 + = =− + d o2 d i2 f D f C d i2 Since the original object was at infinity, the second image must be at the focal point of the combination, and so d i2 = f T . 1 1 1 1 1 1 1 =− + =− + = − fD f C d i2 fC fT fT fC © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

25. (a) We see that the image is real and upright. We estimate that it is 30 cm beyond the second lens, and that the final image height is half the original object height. (b) Find the image formed by the first lens, using Eq. 33–2. 1 1 1 + = → d o1 d i1 f1 d i1 =

( 36cm )(13cm ) = 20.35cm d o1 f1 = d o1 − f1 ( 36cm ) − (13cm )

This image is the object for the second lens. Because it is between the lenses, it has a positive object distance. d o2 = 56cm − 20.35cm = 35.65cm Find the image formed by the second lens, again using Eq. 33–2. ( 35.65cm )(16cm ) = 29.25cm d f 1 1 1 + = → d i2 = o2 2 = d o2 d i2 f 2 d o2 − f 2 ( 35.65cm ) − (16cm ) Thus the final image is real, 29 cm beyond the second lens . The total magnification is the product of the magnifications for the two lenses.  d  d  ( 20.35cm )( 29.25cm ) m = m1m2 =  − i1  − i2  = = 0.46  ( 36cm )( 35.65cm )  d o1  d o2  26. Use Eq. 33–4, the lensmaker’s equation.  1 1 1  = ( n − 1)  +  → f  R1 R2 

f =

 ( −33.4cm )( −28.8cm )  1  R1R2  1   = −24.55cm  −25cm  = ( n − 1)  R1 + R2  (1.63 − 1)  ( −33.4cm ) + ( −28.8cm ) 

27. Find the index from Eq. 33–4, the lensmaker’s equation. 1  1  1 1 1  1 RR  = ( n − 1)  +  → n = 1 +  1 2  = 1 +   ( 2 ( 31.4cm ) ) = 1.597 f f  R1 + R2   26.3cm   R1 R2  28. (a) Find the focal length from Eq. 33–4, the lensmaker’s equation.  1 1 1  = ( n − 1)  +  → f  R1 R2 

f =

(b)

 ( −22.0cm )( +18.5cm )  1  R1 R2  1   = 223.6cm  = ( n − 1)  R1 + R2  (1.52 − 1)  ( −22.0cm ) + ( +18.5cm ) 

( 96.0cm )( 223.6 m ) = −168.2cm  −170cm 1 1 1 d f + = → di = o = 96.0cm − 223.6cm do di f do − f Use Eq. 33–3 to find the magnification. −168.2cm d = 1.752  +1.8 m=− i =− do 96.0cm The image is virtual, in front of the lens, and upright.

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29. Find the radius from the lensmaker’s equation, Eq. 33–4.  1  1 1 1  1  = ( n − 1)  +  → P = ( n − 1)  +  → f  R1 R2   R1 R2  R2 =

( n − 1) R1 = (1.56 − 1)( 0.300 m ) = 0.34 m PR1 − ( n − 1) ( 3.50 D )( 0.300 m ) − (1.56 − 1)

30. Find the radius from the Eq. 33–4, the lensmaker’s equation, with R1 = R2 = R .

 1 1 1  2 = ( n − 1)  +  = ( n − 1)   → f R  R1 R2  R = 2 f ( n − 1) = 2 ( 20.5cm )(1.52 − 1) = 21.32cm  21cm 31. Follow the derivation given in Section 33–4, but set the index of refraction of the lens equal to n2, and the index of the medium equal to n1 . Use the same diagram, make the small angle approximation ( sin   tan    ), and the thin lens approximation. Start with Snell’s law and the small angle approximation. n1 sin 1 = n2 sin  2 → n11 = n2 2

n1 sin  4 = n2 sin 3 → n1 4 = n23

1  sin 1 =

h1 h ;   sin  = 2 ; R1 R3

h2 f From the diagram,  = 1 −  2,  = 3 −  , and  4 =  +  . Combine all of these equations. n n n  = 3 −  = 1  4 − (1 −  2 ) = 1  + 1  − 1 +  2 → n2 n2 n2

  tan  =

h2 n1 h2 n1 h2 h1 n1 h1 = + − + R2 n2 R2 n2 f R1 n2 R1

Since the lens is thin, we have h1  h2, and so all of the h factors cancel. And then multiply by

n2 . n1

1 n1 1 n1 1 1 n1 1 = + − + → R2 n2 R2 n2 f R1 n2 R1 n  1 1 n2 1 1 1 n2 1 1 1 n = + − + →  2 − 1 = −  2 − 1 → n1 R2 R2 f n1 R1 R1  n1  R2 f  n1  R1  1 1  n2 1  =  − 1 +  f  n1  R1 R2 

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32. From the definition of the f-stop, we have f -stop =

Instructor Solutions Manual

f f , and so D = . D f -stop

65mm 65mm = 46 mm ; D22 = = 3.0 mm 1.4 22 Thus, the range of diameters is 3.0mm  D  46mm . D1.4 =

33. As discussed in Example 33–8, to maintain an exposure, the product of the area of the lens opening times the exposure time must be the same. The area is inversely proportional to the f-stop value 2 squared. Thus, exposure time divided by ( f -stop ) must be constant. t1

( f -stop1 )

( f -stop2 )

( f -stop2 ) = 1 s 112 = 7.72  10−3 s  1 s → t2 = t1 ( 500 ) 2 125 2 5.6 ( f -stop1 ) 2

34. We find the f-number from f -stop = f D .

f -stop =

(17cm ) = f f = D ( 6.0cm ) 2.8

35. We use similar triangles, created from the distances between the centers of the two objects (H) and their ray traces to the hole ( L1 ) , and the distance between the centers of the two images (h) and the distance of the screen to the hole ( L2 ) to determine h. Note that the diagram is not to scale, since L1 should be 50  the size of H.

H h L 7.0 cm = → h = H 2 = ( 2.0 cm ) = 0.14 cm = 1.4 mm L1 L2 L1 100 cm Now consider similar triangles from the two rays from just one of the two sources. The base of one triangle is equal to the diameter of the hole (d), and the base of the second triangle equal to the D d diameter of the image circle (D). The heights for these two triangles are the distance from object to hole (L1) and the distance from object to image (L1 + L2). d D L + L2 100cm + 7.0cm = →D=d 1 = (1.0mm ) = 1.07 mm L1 L1 + L2 L1 100cm Since the separation distance of the two images (1.4 mm) is greater than their diameters, the two circles do not overlap. 36. We calculate the effective f-number for the pinhole camera by dividing the focal length by the diameter of the pinhole. The focal length is equal to the image distance. Setting the exposures equal for both cameras, where the exposure is proportional to the product of the exposure time and the area of the lens opening (which is inversely proportional to the square of the f-stop number), we solve for the exposure time. f ( 70mm ) f f -stop 2 = = = . D (1.0mm ) 70

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exposure1 = exposure 2 → A1t1 = A2t2 →



t1 f 2

4 ( f -stop1 )2 37.



t2 f 2

4 ( f -stop2 )2

 4

D12t1 = 2

 4

D22t2 → 2

 f -stop2  1  70  1 → t2 = t1   =   = 0.16 s  6 s 250 s  11   f -stop1 

Consider an object located a distance d o from a converging lens of focal length f , and its real image is formed at distance d i. If the distance d o is much greater than the focal 20 do 20 length, the lens equation tells us that the focal length and image distance are equal. 1 1 1 fd o fd + = → di =  o = f do di f do − f do Thus, in a camera, the recording medium of spatial extent x is placed a distance equal to f behind the lens to form a focused image of a distant object. Assume the distant object subtends an angle of 40 at the position of the lens, so that the half-angle subtended is 20, as shown in the figure. We then use the tangent of this angle to determine the relationship between the focal length and half the image height. 1 x x tan 20o = 2 → f = 2 tan 20o f (a) For a 35-mm camera, we set x = 36 mm to calculate the focal length. 36 mm f = = 49 mm 2 tan 20o (b) For a digital camera, we set x = 1.0 cm = 10 mm. 10 mm f = = 14 mm 2 tan 20o

38. The image distance is found from Eq. 33–3, and then the focal length from Eq. 33–2. The image is inverted. ( −24 mm ) = 38.8mm h d h m = i = − i → d i = − d o i = − ( 55m ) ho do ho ( 34 m )

( 55m )( 0.0388m ) = 0.0388m = 39 mm 1 1 1 dd + = → f = o i = 55m − 0.0388m do di f do + di The object is essentially at infinity, so the image distance is equal to the focal length. 39. (a) Because the Sun is very far away, the image will be at the focal point, or di = f . We find the magnitude of the size of the image using Eq. 33–3, with the image distance equal to 28 mm. 6 hi − d i hod i (1.4  10 km ) ( 28mm ) = → hi = = = 0.26 mm ho do do 1.5  108 km (b) We repeat the same calculation with a 50-mm image distance. 1.4  106 km ) ( 50 mm ) ( hi = = 0.47 mm . 1.5  108 km (c) Again, with a 135-mm image distance. 1.4  106 km ) (135mm ) ( hi = = 1.3 mm . 1.5  108 km © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

(d) The equations show that image height is directly proportional to focal length. Therefore, the relative magnifications will be the ratio of focal lengths. 28mm 135mm = 0.56  for the 28 mm lens ; = 2.7  for the 135 mm lens. 50mm 50mm 40. We calculate the maximum and minimum image distances from Eq. 33–2, using the given focal length and maximum and minimum object distances. Subtracting these two distances gives the distance over which the lens must move relative to the plane of the sensor or film. fd o,min (155mm )(1.30 m ) = 0.176 m = 176 mm 1 1 1 = + → d i,max = = f do di d o,min − f 1300 mm − 155mm

d i,min =

fd o,max (155mm )(  ) = 155mm =  − 155mm d o,max − f

d = d i,max − d i,min = 176 mm − 155mm = 21mm 41. We calculate the range object distances from Eq. 33–2 using the given focal length and maximum and minimum image distances. fdi,max ( 200.0mm )( 208.2mm ) = 5078mm  5.1m 1 1 1 = + → do,min = = 208.2mm − 200.0mm f d o di di,max − f

d o,max =

fd i,min ( 200.0mm )( 200.0mm ) =  = 200.0mm − 200.0mm d i,min − f

Thus, the range of object distances is 5.1m  d o   . 42. When an object is very far away, the image will be at the focal point. We set the image distance in Eq. 33–3 equal to the focal length to show that the magnification is proportional to the focal length. d f  1 m = − i = − =  −  f = ( constant ) f → m  f do do  do  43. The length of the eyeball is the image distance for a far object, i.e., the focal length of the lens. We find the f-number from f -stop = f D .

f -stop =

( 20mm ) = 2.5 or f . f = 2.5 D (8.0mm )

44. Find the far point of the eye by finding the image distance FROM THE LENS for an object at infinity, using Eq. 33–2. 1 1 1 1 1 1 + = → + = → d i1 = f1 = −23.0cm d o1 d i1 f1  d i1 f1 Since the image is 23.0 in front of the lens, the image is 24.8 cm in front of the eye. The contact lens must put the image of an object at infinity at this same location. Use Eq. 33–2 for the contact lens with an image distance of –24.8 cm and an object distance of infinity. 1 1 1 1 1 1 + = → + = → f1 = d i2 = −24.8cm d o2 d i2 f 2  d i2 f1

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45. The screen placed 55 cm from the eye, or 53.2 cm from the lens, is to produce a virtual image 105 cm from the eye, or 103.2 cm from the lens. Find the power of the lens from Eqs. 33–1 and 33–2. 1 1 1 1 1 P= = + = + = 0.91D f d o d i 0.532 m −1.032 m 46. The actual near point of the person is 55 cm. With the lens, an object placed at the normal near point, 25 cm, or 23 cm from the lens, is to produce a virtual image 55 cm from the eye, or 53 cm from the lens. We find the power of the lens from Eqs. 33–1 and 33–2. 1 1 1 1 1 P= = + = + = 2.5D f d o d i 0.23m −0.53m 47. (a) Since the lens power is negative, the lens is diverging, so it produces images closer than the object. Thus the person is nearsighted . (b) We find the far point by finding the image distance for an object at infinity. Since the lens is 2.0 cm in front of the eye, the far point is 2.0 cm farther than the absolute value of the image distance. 1 1 1 1 1 1 + = =P → + = − 3.50 D → di = − = −0.286 m = −28.6cm do di f  di 3.50 D FP = −28.6cm + 2.0cm = 30.6cm from eye.

48. (a) The lens should put the image of an object at infinity at the person’s far point of 88 cm. Note that the image is still in front of the eye, so the image distance is negative. Use Eqs. 33–1 and 33–2. 1 1 1 1 1 P= = + = + = −1.136 D  −1.1D f d o d i  ( −0.88m ) (b) To find the near point with the lens in place, we find the object distance to form an image 25 cm in front of the eye. ( −0.25m ) di 1 1 1 + = = P → do = = = 0.349 m  35cm do di f Pd i − 1 ( −1.136 D )( −0.25m ) − 1 49. The image of an object at infinity is to be formed 13 cm in front of the eye. So for glasses, the image distance is to be di = −11 cm, and for contact lenses, the image distance is to be di = −13 cm. 1 1 1 1 1 1 1 + = = + → f = di → P = = do di f  di f di

Pglasses =

1 1 = −9.1D ; Pcontacts = = −7.7 D −0.11m −0.13m

50. (a) We find the focal length of the lens for an object at infinity and the image on the retina. The image distance is thus 2.0 cm. Use Eq. 33–2. 1 1 1 1 1 1 + = → + = → f = 2.0cm do di f  2.0cm f (b) We find the focal length of the lens for an object distance of 45 cm and an image distance of 2.0 cm. Again use Eq. 33.2. ( 45cm )( 2.0cm ) = 1.9cm dd 1 1 1 + = → f = o i = do di f d o + d i ( 45cm ) + ( 2.0cm ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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51. Find the object distance for the contact lens to form an image at the eye’s near point, using Eqs. 33–1 and 33–2. 1 1 1 di −0.106 m + = = P → do = = = 0.184 m = 18.4cm do di f Pd i − 1 ( −4.00 D )( −0.106 m ) − 1 Likewise, find the object distance for the contact lens to form an image at the eye’s far point. di −0.180 m do = = = 0.643m = 64.3cm Pd i − 1 ( −4.00 D )( −0.180 m ) − 1 52. The diagram shows rays from each of the two points as they pass directly through the cornea and onto the lens. These two rays, the distance between the objects (l) and the distance between the images ( 4  m ) create similar triangles. We set the ratio of the bases and altitudes of these two triangles equal to solve for l. l 4 m 4 m = → l = 25 cm = 50  m 25 cm 2.0 cm 2.0 cm

4 m

25cm

2.0cm

53. The 2.0 cm of a normal eye is the image distance for an object at infinity; thus it is the focal length of the lens of the eye. To find the length of the nearsighted eye, find the image distance (distance from lens to retina) for an object at the far point of the eye. Use Eq. 33–2. ( 2.0cm )(17 cm ) = 2.27cm fd o 1 1 1 + = → di = = do di f do − f 15cm

 1  1 1 , which gives di = 2.27cm.   + =  17cm   di  2.0cm Thus, the difference is 2.27cm − 2.0cm = 0.27cm  0.3cm. 54. (a) We find the focal length with the image at the near point from Eq. 33–6b. 25cm N N M = 1+ → f = = = 16.7cm  0.17 m f M − 1 2.5 − 1 (b) If the eye is relaxed, the image is at infinity, and so use Eq. 33–6a. N N 25cm M= → f = = = 10cm = 0.10m 2.5 f M 55. Find the magnification from Eq. 33–6a. N ( 25cm ) M= = = 1.6  f (16cm ) 56. Maximum magnification is obtained with the image at the near point (which is negative). We find the object distance from Eq. 33–2, and the magnification from Eq. 33–6b. ( −25cm )(8.80cm ) = 6.509cm  6.5cm d f 1 1 1 + = → do = i = d o di f di − f ( −25cm ) − ( 8.80cm )

M =1+

N 25cm =1+ = 3.8  f 8.80cm

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57. (a) We use Eq. 33–6b to calculate the angular magnification. ( 25cm ) = 3.717  3.7  N M =1 + =1 + f ( 9.20cm ) (b)

The image is at the near point, so the image distance is –25 cm. Use Eq. 33–2 to find the object distance, and then use that to find the image width. ( 9.20cm )( −25cm ) = 6.73cm 1 1 1 fd i + = → do = = do di f di − f −25cm − 9.20cm

hi = −ho

( −25cm ) = 12.6 mm  13mm di = − ( 3.40 mm ) do ( 6.73cm )

Alternatively, it can also be seen from Fig. 33–33 in the text that, for small angles, M = hi ho . h M = i → hi = Mho = 3.72 ( 3.40 mm ) = 12.6mm  13mm ho (c) We use Eq. 33–2 to calculate the object distance, with the image distance at –25.0 cm. ( 9.20cm )( −25cm ) = 6.73cm  6.7cm fd i 1 1 1 + = → do = = −25cm − 9.20cm do di f di − f 58. (a) We find the image distance using Eq. 33–2. ( 9.8cm )(8.3cm ) = −54cm 1 1 1 fd o + = → di = = 8.3cm − 9.8cm do di f do − f (b) The angular magnification is found using Eq. 33–5, with the angles given as defined in Fig. 33–33. Note that the image is neither at infinity nor at the near point, so this is an approximation.   ( h d ) N 25 cm M= = o o = = = 3.0   ( ho N ) d o 8.3 cm 59. First, find the focal length of the magnifying glass from Eq. 33–6a, for a relaxed eye (focused at infinity). N N 25.0cm M= → f = = = 8.33cm 3.0 f M (a) Again use Eq. 33–6a for a different near point. ( 57 cm ) = 6.8  N M1 = 1 = f (8.33cm ) (b) Again use Eq. 33–6a for a different near point. (17cm ) = 2.0  N M2 = 2 = f (8.33cm ) Without the lens, the closest an object can be placed is the near point. A farther near point means a smaller angle subtended by the object without the lens, and thus greater magnification. 60. The focal length is 12 cm. First, find the object distance for an image at infinity. Then, find the object distance for an image 25 cm in front of the eye. 1 1 1 1 1 1 + = → + = → d o = f = 12cm Initial: do di f do  f Final:

1 1 1 + = do di f

→ do =

( −25cm )(12cm ) = 8.1cm di f = d i − f ( −25cm ) − (12cm )

The lens was moved 12cm − 8.1cm = 3.9cm  4 cm toward the fine print. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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61. The magnification of the telescope is given by Eq. 33–7. (86cm ) = − 31  f M =− o =− fe ( 2.8cm ) For both object and image far away, the separation of the lenses is the sum of the focal lengths. f o + f e = 86cm + 2.8cm = 88.8cm  89cm 62. We find the focal length of the eyepiece from the magnification by Eq. 33–7. f f 78cm M = − o → fe = − o = − = 2.23cm  2.2cm fe M −35  For both object and image far away, the separation of the lenses is the sum of the focal lengths. f o + f e = 78cm + 2.23cm = 80.23cm  0.80m 63. We find the focal length of the objective from Eq. 33–7. M = − f o f e → f o = − M f e = − ( −8.0 )( 3.0cm ) = 24cm 64. For a distant object and a relaxed eye (which means the image is at infinity), the separation of the eyepiece and objective lenses is the sum of their focal lengths. Use Eq. 33–7 to find the magnification. f f 36.0cm l = fo + fe ; M = − o = − o = − = +16  fe l − fo 33.8cm − 36.0cm 65. For a distant object and a relaxed eye (which means the image is at infinity), the separation of the eyepiece and objective lenses is the sum of their focal lengths. Use Eq. 33–7 to find the magnification. f f 75.5cm l = fo + fe ; M = − o = − o = − = −30  (2 sig. figs.) fe l − fo 78.0cm − 75.5cm 66. The focal length of the mirror is found from Eq. 33–7. The radius of curvature is twice the focal length. f M = − o → f o = − M f e = − ( −140 )( 0.031m ) = 4.34 m  4.3m ; r = 2 f = 8.7 m fe 67. The focal length of the objective is just half the radius of curvature. Use Eq. 33–7 for the magnification. 1 1 r ( 6.1m ) = −108.9   −110  f M =− o =−2 =− 2 fe fe 0.028m 68. The relaxed eye means that the image is at infinity, and so the distance between the two lenses is 1.15 m. Use that relationship with Eq. 33–7 to solve for the focal lengths. Note that the magnification for an astronomical telescope is negative. We assume that the magnification has 3 significant figures. −120 (1.15m ) f f Ml l = fo + fe ; M = − o = − o → fo = = = 1.14 m l − fo fe M −1 −120 − 1

l = f o + f e → 1.15m = 120 f e + f e = 121 f e → f e =

1.15m = 0.950cm 121

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69. We use Eq. 33–6a and the magnification of the eyepiece to calculate the focal length of the eyepiece. We set the sum of the focal lengths equal to the length of the telescope to calculate the focal length of the objective. Then, using both focal lengths in Eq. 33–7, we calculate the maximum magnification. N 25cm = = 5cm ; l = f e + f o → f o = l − f e = 50cm − 5cm = 45cm fe = M 5 f 45cm = −9  M =− o =− fe 5cm 70. Since the star is very far away, the image of the star from the objective mirror (radius of curvature Ro ) will be at the focal length of the objective, which is equal to one-half its radius of curvature (Eq. 32–1). We subtract this distance from the separation distance to determine the object distance for the secondary mirror (radius of curvature Rs ). Then, using Eq. 33–2, we calculate the final image distance, which is where the sensor should be placed. R 3.25m = 1.625m ; d o2 = l − d i1 = 0.90 m − 1.625m = −0.725m d i1 = f o = o = 2 2 ( −1.75m )( −0.725m ) = 4.229 m  4.2 m Rsd o2 1 1 1 2 + = = → di = = d o2 d i2 f e Rs 2d o2 − Rs 2 ( −0.725m ) − ( −1.75m ) 71. We assume a prism binocular so the magnification is positive, but simplify the diagram by ignoring the prisms. We find the focal length of the eyepiece using Eq. 33–7, with the design magnification. f 26cm fe = o = = 4.0cm 6.5 M Using Eq. 33–2 and the objective focal length, we calculate the intermediate image distance. With the final image at infinity (relaxed eye), the secondary object distance is equal to the focal length of the eyepiece. We calculate the angular magnification using Eq. 33–5, with the angles shown in the diagram. ( 26cm )( 400cm ) = 27.81cm f d 1 1 1 + = → di1 = o o = 400cm − 26cm d o1 d i1 f o d o1 − f o   h f e di1 27.81cm M= = = = = 6.95  7.0  4.0cm  h di1 f e 72. The magnification of the microscope is given by Eq. 33–10b. ( 25cm )(19.5cm ) = 500  2 sig. figs. Nl M= = ( ) f o f e ( 0.65cm )(1.50cm ) 73. We find the focal length of the eyepiece from the magnification of the microscope, using the approximate results of Eq. 33–10b. We already know that f o l . ( 25cm )(17.5cm ) = 1.736cm  1.7cm Nl Nl M → fe = = fo fe Mf o ( 630 )( 0.40cm ) Note that this also satisfies the assumption that f e

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74. (a) The total magnification is found from Eq. 33–10a. M = mo M e = ( 62.0)(13.0) = 806  (b) With the final image at infinity, we find the focal length of the eyepiece using Eq. 33–9. N N 25.0cm Me = → fe = = = 1.923cm  1.92cm fe Me 13.0 Since the image from the objective is at the focal point of the eyepiece, set the image distance from the objective as the distance between the lenses minus the focal length of the eyepiece. Using the image distance and magnification in Eq. 33–3, we calculate the initial object distance. Then, using the image and object distance in Eq. 33–2, we calculate the objective focal length. d i = l − f e = 20.0cm − 1.92cm = 18.08cm d d 18.08cm mo = i → d o = i = = 0.2916cm 62.0 do mo

( 0.2916cm )(18.08cm ) = 0.287 cm dd 1 1 1 = + → fo = o i = 0.2916cm + 18.08cm fo do di do + di

(c) We found the object distance, in part (b), d o = 0.292cm . 75. (a) The total magnification is the product of the magnification of each lens, with the magnification of the eyepiece increased by one, as in Eq. 33–6b. M = mo ( M e + 1) = ( 62.0)(13.0 + 1.0) = 868  (b) We find the focal length of the eyepiece using Eq. 33–6b. N N 25cm = 1.9cm ( M e + 1) = +1 → fe = = fe M e 13.0 Since the image from the eyepiece is at the near point, we use Eq. 33–2 to calculate the location of the object. This object distance is the location of the image from the objective. Subtracting this object distance from the distance between the lenses gives us the image distance from the objective. Using the image distance and magnification in Eq. 33–3, we calculate the initial object distance. Then, using the image and object distance in Eq. 33–2, we calculate the objective focal length. (1.92cm )( −25cm ) fd 1 1 1 = + → d o2 = e i2 = = 1.78cm −25cm − 1.92cm f e d o2 di2 di2 − f e

di1 = l − d o2 = 20.0cm − 1.78cm = 18.22cm mo =

di do

→ do =

1 1 1 = + f o do di

di 18.22cm = = 0.2934cm mo 62.0

→ fo =

( 0.2934cm )(18.22cm ) = 0.289cm d o di = d o + di 0.2934cm + 18.22cm

(c) We found the object distance, in part (b), d o = 0.293cm . 76. (a) Since the final image is at infinity (relaxed eye) the image from the objective is at the focal point of the eyepiece. We subtract this distance from the distance between the lenses to calculate the objective image distance. Then using Eq. 33–2, we calculate the object distance. d i1 = l − f e = 16.8cm − 1.8cm = 15.0cm 1 1 1 = + f o d o1 d i1

→ d o1 =

( 0.80cm )(15.0cm ) = 0.85cm f o d i1 = 15.0cm − 0.80cm d i1 − f o

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(b) With the final image at infinity, the magnification of the eyepiece is given by Eq. 33–10a. N  l − f e  ( 25.0cm )  16.8cm − 1.8cm  M=  =   = 247  250  0.85cm f e  d o  (1.8cm )   77. (a) We find the image distance from the objective using Eq. 33–2. For the final image to be at infinity (viewed with a relaxed eye), the objective image distance must be at the focal distance of the eyepiece. We calculate the distance between the lenses as the sum of the objective image distance and the eyepiece focal length. ( 0.740cm )( 0.790cm ) = 11.69cm 1 1 1 fd + = → d i1 = o o1 = 0.790cm − 0.740cm d o1 d i1 f o d o1 − f o

l = d i1 + f e = 11.69cm + 2.80cm = 14.49cm  14cm (b) We use Eq. 33–10a to calculate the total magnification. N  l − f e  ( 25cm )  14.49cm − 2.80cm  M=  =   = 132  130  0.790cm f e  d o  ( 2.80cm )   78. For each objective lens we set the image distance equal to the sum of the focal length and 160 mm. Then, using Eq. 33–2, we write a relation for the object distance in terms of the focal length. Using this relation in Eq. 33–3 we write an equation for the magnification in terms of the objective focal length. The total magnification is the product of the magnification of the objective and focal length. f ( f + 160 mm ) 1 1 1 1 1 1 1 1 1 + = → = − → = − → do = o o do di fo do fo di d o f o f o + 160 mm 160 mm mo =

di f o + 160 mm 160 mm = = d o  f o ( f o + 160 mm )  fo   160 mm  

Since the objective magnification is inversely proportional to the focal length, the objective with the smallest focal length ( f o = 3.9 mm ) combined with the largest eyepiece magnification ( M e = 15  ) yields the largest overall magnification. The objective with the largest focal length ( f o = 32 mm ) coupled with the smallest eyepiece magnification ( M e = 5  ) yields the smallest overall magnification. 160 mm 160 mm M largest = (15 ) = 615.4  620  ; M smallest = ( 5 ) = 25  3.9 mm 32 mm 79. (a) For this microscope both the objective and eyepiece have focal lengths of 12 cm. Since the final image is at infinity (relaxed eye) the image from the objective must be at the focal length of the eyepiece. The objective image distance must therefore be equal to the distance between the lenses less the focal length of the objective. We calculate the object distance by inserting the objective focal length and image distance into Eq. 33–2. d i1 = l − f e = 55cm − 12cm = 43cm

(12cm )( 43cm ) = 16.65cm  17cm 1 1 1 fd = + → d o = o i1 = 43cm − 12cm f o d o d i1 d i1 − f o (b) We calculate the magnification using Eq. 33–10a. N  l − f e  ( 25cm )  55cm − 12cm  M=  =   = 5.38  5.4  f e  d o  (12cm )  16.65cm 

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(c) We calculate the magnification using Eq. 33–10b, and divide the result by the answer to part (b) to determine the percent difference. ( 25cm )( 55cm ) = 9.55  ; M approx − M = 9.55 − 5.38 = 0.775  78% Nl = M approx  f e f o (12cm )( 12cm ) M 5.38 80. We use Eq. 33–4 to find the focal length for each color, and then Eq. 33–2 to find the image distance. For the plano-convex lens, R1  0 and R2 = .

 1   1    1 1 1  = ( nred − 1)  +  = (1.5106 − 1)   +    → f red = 29.965cm f red  R1 R2   15.3cm       1   1    1 1  = ( nyellow − 1)  +  = (1.5226 − 1)   +    → f orange = 29.277cm f yellow  R1 R2   15.3cm      We find the image distances from Eq. 33–2. ( 62.0cm )( 29.965cm ) = 57.994cm  58.0cm 1 1 1 d f + = → d i = o red = do di f red d o − f red ( 62.0cm ) − ( 29.965cm ) red 1

red

1 1 + do di

yellow

1 f yellow

→ di

yellow

d o f yellow d o − f yellow

( 62.0cm )( 29.277cm ) = 55.471cm  55.5cm ( 62.0cm ) − ( 29.277cm )

The images are 2.5 cm apart, an example of chromatic aberration. 81. From Problem 24 we have a relationship between the individual focal lengths and the focal length of the combination. ( 25cm )( −28cm ) = 233cm f f 1 1 1 1 1 1 =− + → = + → fT = C D = fD fC fT fT f D fC f C + f D ( 25cm ) + ( −28cm ) (a) The combination is converging, since the focal length is positive. Also, the converging lens is “stronger” than the diverging lens since it has a smaller absolute focal length (or higher absolute power). (b) From above, f T  230cm . 82. We use Eq. 33–2 with the final image distance and focal length of the converging lens to determine the location of the object for the second lens. Subtracting this distance from the separation distance between the lenses gives us the image distance from the first lens. Inserting this image distance and object distance into Eq. 33–2, we calculate the focal length of the diverging lens. (17.0cm )(12.0cm ) = 40.8cm 1 1 1 d f + = → d o2 = i2 2 = 17.0cm − 12.0cm d o2 d i2 f 2 d i2 − f 2

d i1 = l − d o2 = 30.0cm − 40.8cm = −10.8cm 1 1 1 + = d o1 d i1 f1

→ f1 =

( −10.8cm )( 25.0cm ) = −19.0cm d i1d o1 = d i1 + d o1 −10.8cm + 25.0cm

83. The relationship between two lenses in contact was found in Problem 24. We use this resulting equation to solve for the combination focal length. ( −18.0cm )(13.0cm ) = 46.8cm f f 1 1 1 = + → fT = D C = −18.0cm + 13.0cm fT f D fC f D + fC Since the focal length is positive, the combination is a converging lens. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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84. (a) The eye must be acting as a convex mirror–otherwise you couldn’t see the image. No light would be coming “out” from the eye. (b) The eye is being viewed right side up. Convex mirrors make upright images. Also notice the eyelashes. (c) The eye is convex-shaped. If it were acting as a convex lens, then the object would have to be inside the eye, sending light from inside the eye to outside the eye. That is not possible. So it is acting like a convex mirror, making a virtual, reduced object. This is similar to Fig. 32–19b, where the “mirror” in that figure is the eyeball, and the “eye” in that figure is the camera taking the photo. 85. The left lens is a converging lens, acting as a magnifying glass, forming a magnified virtual image. The object (the face) is closer to the lens than the focal point. The right lens is a diverging lens, forming a reduced virtual image. For the left lens: O is the person’s face. The observer (the camera taking the picture) is to the right of the lens in this ray diagram. For the right lens: O is the person’s face. The observer (the camera taking the picture) is to the right of the lens in the ray diagram.

86. (a) For the image to be between the object and the lens, the lens must be diverging (see Fig. 33–8). Find the focal length from Eq. 33–2. The image distance is negative. ( 37.5cm )( −8.60cm ) = −11.2cm dd 1 1 1 + = → f = o i = do di f do + di 37.5cm − 8.60cm The image is in front of the lens, so it is virtual. (b) For the image to be farther from the lens than the object, but on the same side of the lens as the object, the lens must be converging (see Fig. 33–12). The image distance again is negative. ( 37.5cm )( −44.5cm ) = 238cm  240cm dd f = o i = do + di 37.5cm − 44.5cm The image is in front of the lens, so again it is virtual.

1 1 = = 0.20 m = 20cm. Approach this problem in steps– P 5.0m −1 find the first image formed by the lens, let that image be the object for the mirror, and then use the image from the mirror as the object for the “reversed” light that goes back through the lens. Use Eq. 33–2 for each image formation.

87. The lens has a focal length of f =

Calculate the location of the first image, using d o1 = 30.0cm and f1 = 20.0 cm. ( 30.0cm )( 20.0cm ) = 60.0cm d f 1 1 1 + = → d i1 = o1 1 = d o1 d i1 f1 d o1 − f1 30.0cm − 20.0cm This image is 5.0 cm from the mirror. Calculate the location of the second image, using d o2 = 5.0cm and f 2 = 25.0 cm .

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( 5.0cm )( 25cm ) = −6.25cm 2 sig. figs. d f 1 1 1 + = → d i2 = o2 2 = ( ) d o2 d i2 f 2 d o2 − f 2 5.0cm − 25cm This is a virtual image, 6.25 cm behind the mirror, or 71.25 cm from the lens. The diverging light rays from the mirror will be moving from the mirror towards the lens, and so this is a real object for the lens. So calculate the location of the third image, using do3 = 71.25cm and f3 = 20.0 cm. 1 1 1 + = → d o3 d i3 f 3 ( 71.25cm )( 20.0cm ) = 27.8cm  28cm d o3 f 3 = d o3 − f 3 71.25cm − 20.0cm This would be on the same side of the lens as the original object, and so this final image is 2 cm from the original object, towards the lens. d i3 =

Here is a ray diagram, showing some of the rays used to locate the various images. It is not exactly to scale. 88. (a) When two lenses are placed in contact, the negative of the image of the first lens is the object distance of the second. Using Eq. 33–2, we solve for the image distance of the first lens. Inserting the negative of this image distance into the lens equation for the second lens we obtain a relationship between the initial object distance and final image distance. Again using the lens equation with this relationship, we obtain the focal length of the lens combination. 1 1 1 1 1 1 1 = + → = − =− f1 d o1 d i1 d i1 f1 d o1 d o2

1 1 1 1 1 1  1 1 1 1 1 = + = − − + =  + = f 2 d o2 d i2 d o2  f1 d o1  f 2 f1 d o2 d o1 f T 1 1 1 = + f T f1 f 2

→

fT =

f1 f 2 f1 + f 2

(b) Setting the power equal to the inverse of the focal length gives the relationship between powers of adjacent lenses. 1 1 1 = +  PT = P1 + P2 f T f1 f 2 89. Solve this by working through the lenses “backwards.” Use the image distances and focal lengths to calculate the object distances. Let the right lens be lens # 2, and the left lens be lens # 1. Since the final image from the lens # 2 is halfway between the lenses, set the image distance of the second lens equal to the negative of half the distance between the lenses. Using Eq. 33–2, solve for the object distance lens # 2. By subtracting this object distance from the distance between the two lenses, find the image distance for lens # 1. Then using Eq. 33–2 again, solve for the initial object distance. d i2 = − 12 l = − 12 ( 30.0cm ) = −15.0cm 1 1 1 + = d o2 d i2 f 2

→ d o2 =

( −15.0cm )( 20.0cm ) = 8.57cm d i2 f 2 = −15.0cm − 20.0cm d i2 − f 2

d i1 = l − d o2 = 30.0 cm − 8.57 cm = 21.4 cm 1 1 1 + = d o1 d i1 f1

→ d o1 =

( 21.4cm )(15.0cm ) = 50.0cm d i1 f1 = d i1 − f1 21.4cm − 15.0cm

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90. Since the distance to the Sun is much larger than the telescope’s focal length, the image distance is essentially equal to the focal length. Rays from the top and bottom edges of the Sun pass through the lens unrefracted. These rays form similar triangles along with the object and image heights. Calculate the focal length of the telescope by setting the ratio of height to base for each triangle equal. 1.5  108 km di f do d = = → f = hi o = (15mm ) = 1607 mm  1.6 m 1.4  106 km hi hi ho ho 91. We use Eq. 33–3 to write the image distance in terms of the object distance, image height, and object height. Then using Eq. 33–2 we solve for the object distance, which is the distance between the photographer and the subject. h d 1 h 1 m= i =− i → =− o ho do di hi d o

1 1 1 1  ho 1   ho  1 = + = +−  = 1 −  f d o d i d o  hi d o   hi  d o

→

 h   1750mm  do = 1 − o  f = 1 −  ( 220mm ) = 45,250mm  45m 8.55mm  h −  i   92. The exposure is proportional to the intensity of light, the area of the shutter, and the time. The area of the shutter is proportional to the square of the diameter or inversely proportional to the square of the f-stop. Setting the two proportionalities equal, with constant time, we solve for the change in intensity. I1t

( f -stop1 )

= 2

I 2t

( f -stop2 )

→

I 2  f -stop2   16  =  =  = 8.2 I1  f -stop1   5.6 

93. Since the microscope is adjusted for a good image with a relaxed eye, the image must be at infinity. Thus the object distance for the camera should be set to infinity. 94. The maximum magnification is achieved with the image at the near point, using Eq. 33–6b. (15.0cm ) = 2.8  N M1 = 1 + 1 = 1 + f (8.2cm ) For an adult we set the near point equal to 25 cm. ( 25cm ) = 4.0  N M2 =1 + 2 =1 + f (8.2cm ) The person with the normal eye (adult) sees more detail. 95. The actual far point of the person is 175cm. With the lens, an object far away would produce a virtual image 175 cm from the eye, or 173 cm from the lens. We calculate the power of the upper part of the bifocals using Eq. 33–2 with the power equal to the inverse of the focal length in meters.  1  1   1  1  1 P1 = =   +  =  +   = −0.578D (upper part) f1  d o1   d i1      −1.73m 

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The actual near point of the person is 45cm. With the lens, an object placed at the normal near point, 25 cm, or 23 cm from the lens, would produce a virtual image 45cm from the eye, or 43cm from the lens. We again calculate the power using Eq. 33–2. 1  1   1   1   1  P2 = =   +  = +  = +2.0 D (lower part) f 2  do2   di2   0.23 m   −0.43 m  96. We use Eq. 33–6b to calculate the necessary focal length for a magnifying glass held at the near point ( N = 25 cm ) to have a magnification of M = 3.5. 25 cm N N M = +1 → f = = = 10 cm f M − 1 3.5 − 1 In the text, the lensmaker’s equation (Eq. 33–4) is derived assuming the lens is composed of material with index of refraction n and is surrounded by air, whose index of refraction is na = 1. We now modify this derivation, with the lens composed of air with index of refraction na = 1 surrounded by water, whose index of refraction is nw = 1.33. In the proof of the lensmaker’s equation, Snell’s law at small angles is first applied at both surfaces of the lens. n nw sin 1 = na sin  2 → nw1  na 2 → 1  a  2 nw na sin  3 = nw sin  4 → na3  nw 4 →  4 

na 3 nw

na . The rest of nw the derivation is the same, so we can rewrite the lensmaker’s equation with this single modification. We assume the radii are equal, insert the necessary focal length, and solve for the radius of curvature.  1  1 1   n  2  1  na 1 n = − 1  +  =  a − 1  +  =  a − 1   f  nw   R1 R2   nw   R R   nw R These equations are the same as those following Fig. 33–16, but with n replaced by

n   1  − 1  = −4.96cm  −5.0cm R = 2 f  a − 1 = 2 (10cm )  n 1.33    w  The lens is therefore a concave lens with radii of curvature –5.0 cm.

97. (a) The magnification of the telescope is given by Eq. 33–7. The focal lengths are expressed in terms of their powers. ( 4.5D ) = − 2.25   − 2.3  f P M =− o =− e =− fe Po ( 2.0 D ) (b) To get a magnification greater than 1, for the eyepiece we use the lens with the smaller focal length, or greater power: 4.5 D . 98. We calculate the man’s near point ( d i ) using Eq. 33–2, with the initial object at 0.32 m with a 2.5 D lens. To give him a normal near point, we set the final object distance as 0.25 m and calculate the power necessary to have the image at his actual near point. 1 1 1 1 0.32 m d o1 P1 = + → = P1 − → di = = = −1.6 m d i d o1 di d o1 Pd ( 2.5D )( 0.32 m ) − 1 1 o1 − 1 P2 =

1 1 1 1 + = + = +3.4 D d i d o2 −1.6 m 0.25 m

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99. (a) We solve Eq. 33–2 for the image distance. Then taking the time derivative of the image distance gives the image velocity. If the velocity of the object is taken to be positive, then the d object distance is decreasing, and so vo = − ( d o ) . dt fd o 1 1 1 = + → di = f di d o do − f

vi = =

f ( d o − f ) − fd o fd o d d  fd  f −vo ) = ( di ) =  o  = ( −vo ) − ( −vo ) 2 ( 2 dt dt  d o − f  d o − f ( do − f ) ( do − f ) f 2 vo

( do − f )

(b) The velocity of the image is positive, which means the image is moving the same direction as the object. But since the image is on the opposite side of the lens as the object, the image must be moving away from the lens. (c) We set the image and object velocities equal and solve for the image distance. f 2 vo 2 vi = vo → = vo → ( d o − f ) = f 2 → d o − f = f → d o = 2 f 2 ( do − f ) 100. (a) The length of the telescope is the sum of the focal lengths. The magnification is the ratio of the focal lengths (Eq. 33–7). For a magnification greater than one, the lens with the smaller focal length should be the eyepiece. Therefore the 4.0 cm lens should be the eyepiece. l = f o + f e = 41cm + 4.0cm = 45cm M =−

( 41cm ) = −10  fo =− fe ( 4.0cm )

(b) We use Eq. 33–10b to solve for the length, l, of the microscope. − Mf e f o − ( 25)( 4.0cm )( 41cm ) Nl  l = = = 164cm = 1.6 m M =− fe f o N 25cm This is far too long to be practical. Shorter focal length lenses would be more practical. 101. (a) For a make-up mirror the object O that is to be magnified is also the observer. For a concave mirror, the ray diagram would be as shown here (Fig. 32–17), and the observer can see the magnified virtual image.

For a converging lens, one possibility is that the magnified image would be virtual and behind the observer, as shown in the adjacent diagram (Fig. 33–12, with O as the observer and the object, looking towards the lens). The observer could not see that image, however, so it is not useful.

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The final possibility is a converging lens making a real, inverted, and somewhat distant image as shown in the last image (Fig. 33–6c, with O as the observer and the object, looking towards the lens). For the observer (also the object) to see the real image, a screen would need to be placed at the image location. The resulting image would be inverted (making it very inconvenient) and somewhat distant from the observer. Also, for this diagram to be accurate, the object should be closer to the focal point, and the image further away, probably making it even more inconvenient. (b) Both objects will have the same lateral magnification. If the object distance is 12 f , then the mirror equation and the thin lens equation give the same value for the image distance. −1

−1

1 1  1 1  = + → di =  −  =  −  =−f 1 f do di f 2f   f do 

The lateral magnification equation for mirrors and lenses is the same, which gives the same result for both. d −f m=− i =− =2 1 do f 2 For both the mirror and the lens the image will be virtual with a lateral magnification of +2. 102. The magnification for a relaxed eye is given by Eq. 33–6a. M = N f = NP = ( 0.25m )( + 4.0D ) = 1.0  103. (a) The focal length of the lens is the inverse of the power. 1 1 f = = = 0.27027 m  27.0cm P 3.70 D (b) The lens produces a virtual image at his near point. We set the object distance at 23 cm from the glass (25 cm from the eyes) and solve for the image distance. We add the two centimeters between the glass and eyes to determine the uncorrected near point. 1 1 1 + P= = f do di

−1

  1  1  → d i =  P −  =  3.70 D −  = −1.54 m 0.23m  do   

N = d i + 0.02 m = 1.54 m + 0.02 m = 1.56 m  1.6 m

−1

 1 1   → d o =  P −  =  3.70 D −  = 0.1243m di  −0.23 m    Pam’s near point with the glasses is 12 cm from the glasses or 14 cm from her eyes. 1 1 1 P= = + f do di

104. As shown in the image, the parallel rays will pass through a single point located at the focal distance from the lens. The ray passing through the edge of the lens (a distance D/2 from the principal axis) makes an angle  with the principal axis. We set the tangent of this angle equal to the ratio of the opposite side (D/2) to the adjacent side (f) and solve for the focal length. D2 D 5.0 cm →f = = = 41 cm tan  = f 2 tan  2 tan 3.5 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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105. We use Eq. 33–4 with Eq. 33–1 to find the desired radius.  1 1 1  = P = ( n − 1)  +  → f  R1 R2 

 ( −3.5m −1 )  P 1 1  R2 =  −  = −  = −7.82  10−2 m  −7.8cm n 1 R 1.62 1 0.140 m − − ( ) ( )  1   −1

−1

106. (a) In this case, f = + 40mm, do = 100mm, and ho = 10mm . Use Eqs. 33–2 and 33–3 to calculate the image distance and image height. 1 1 1 = + → f do di −1

−1

1 1  1 1  di =  −  =  −   67 mm  40 mm 100 mm   f di   66.67 mm  di ho = −  10 mm  − 6.7mm do  100 mm  See the ray diagram (not to scale) illustrating this case. (b) In this case, f = − 40mm, do = 100mm, and ho = 10mm . Use the same equations as above. 1 1 1 = + → f do di hi = −

−1

1 1  1 1  di =  −  =  −   − 29 mm  − 40 mm 100 mm   f di   −29 mm  di ho = −  10 mm  2.9 mm do  100 mm  See the ray diagram (not to scale) illustrating this case. hi = −

107. The magnifications in the two figures appear to have the same magnitude with the magnification positive in (a) and negative in (b). In (a) the object distance is 5 cm and in (b) the object distance is 15 cm. Using the magnification equation we write the image distance in (b) in terms of the image and object distances in (a). 15cm d d d d mb = − ma = − ib = ia → d ib = − ob ia = − d ia = −3d ia 5cm d ob d oa d oa Then using the thin lens equation for each image, calculate the image distance in figure (a). 1 1 1 1 1 1 1 1 1 = + = + → + = + f d oa d ia d ob d ib d oa d ia d ob −3d ia d ia =

4 1

−

1 

−1

 =

4

−

1 

−1

   = −10 cm 3  d ob d oa  3  15cm 5cm  Now the image distance in figure (b) and the focal length can be determined. d ib = −3d ia = −3 ( −10 cm ) = 30 cm 1 f

1 d oa

1 d ia

−1

 1 1  1   1 =  → f = + +   = 10 cm  5cm −10 cm   d oa d ia 

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The focal length is 10 cm. In (a) the object is only half the focal length from the lens and therefore creates a virtual image 10 cm behind the lens, with a magnification of +2. In (b) the object is 1.5 times the focal length and therefore creates are real image 30 cm in front of the lens, with a magnification of –2. 108. Use the 150 cm lens as the objective and the 1.5 cm lens as the eyepiece. These two lenses produce the magnification of 100  according to Eq. 33–7. The third lens is placed between the objective and the eyepiece to invert the image. The objective lens creates an image of a distance object at its focal point. The image becomes the object of the intermediate lens. The intermediate lens is placed so that this object is twice the lens’ focal length away. The lens then creates a real, inverted image located at twice its focal length with no magnification. This image becomes the object of the eyepiece, which is placed at is focal distance away from the image. The image of the eyepiece is virtual and infinitely far away.

Calculate the length of the telescope from the figure. l = f o + 4 f 3 + f e = 150 cm + 4 (10 cm ) + 1.5cm = 191.5cm  190 cm Note that this is very long–almost 2 meters! 109. We use Eq. 33–2, and then look at the limit as the object distance goes to 0. Since the object is inside the focal point, we expect the image to be virtual, and so on the same side of the lens as the object. d f 1 1 1 1 1 1 do − f = + → = − = → di = o f do di di f d o do f do − f do f = −do −f We see that the image is at the same location as the object: virtual, on the same side of the lens as the −d f d object, with a magnification of m = − i = − o = +1. do do do

f → di =

110. The two students chose different signs for the magnification, i.e., one upright and one inverted. The focal length of the lens is f = 12cm. We relate the object and image distances from the magnification. d d m = − i →  3 = − i → d i = 3d o do do Use this result in the mirror equation. 1 1 1 1 1 1 2f 4f + = → + = → do = , = 8.0cm,16cm do di f d o ( 3d o ) f 3 3 So the object distances are +8.0 cm (produces virtual image) , and +16 cm (produces real image).

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111. Example 33–9 assumes that the sensor has 12 MP. The 6 MP sensor will have half as many total pixels, so the number of pixels in each direction would be decreased by the square root of two, 3000 pixels 4000 pixels making the pixel dimensions to be = 2121pixels. We = 2828 pixels by 2 2 calculate the new resolution. 2828 pixels 2121pixels = 88 pixels mm or = 88 pixels mm 32 mm 24 mm For the enlargement to 8  10 inches, as in Example 33–10, the size would be magnified by a factor 88 pixels mm of 8, which reduces the resolution by a factor of 8, giving = 11pixels mm , which is 8 right at the limit of creating a sharp image. 112. (a) We use Eqs. 33–2 and 33–3. d d 1 1 1 m 1 m = − i → do = − i ; + = =− + do m do di f di di This is a straight line with slope = −

→ m=−

di +1 f

1 and y-intercept = 1. f

(b) A plot of m vs. d i is shown here. 1 1 f =− =− slope −0.0726cm −1

= 13.8cm  14cm The y-intercept is 1.028. Yes, it is close to the expected value of 1.

113. (a) We use Eq. 33–2 to calculate the image distance and then use the object and image distances in Eq. 33–3 to calculate the magnification. We finally make the approximation that the object distance is much larger than the focal length. 1 1 1 + = d o d i f1 m1 = −

−1

1 1  fd → di =  −  = 1 o d o − f1  f1 d o 

di 1  fd  f1 f =−  1 o =−  − 1 do d o  d o − f1  d o − f1 do

This real image, located near the focal distance from lens 1, becomes the object for the second lens. We subtract the focal length from the separation distance to determine the object distance for lens 2. Using Eq. 33–2, we calculate the second image distance and Eq. 33–3 to calculate the second magnification. Multiplying the two magnifications gives the total magnification. fd 1 1 1 + = → di2 = 2 o2 d o2 di2 f 2 d o2 − f 2

m2 = −

( − 12 f1 ) di2 f2 1  f 2 d o2  =− =− 3 =2  =− d o2 d o2  d o2 − f 2  do2 − f 2 ( 4 f1 − f1 ) − ( − 12 f1 )

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 f  2f m1m2 =  − 1  ( 2 ) = − 1 do  do  (b) If the object is at infinity, the image from the first lens will form a focal length behind that lens. Subtracting this distance from the separation distance gives the object distance for the second lens. We use Eq. 33–2 to calculate the image distance from the second lens. Adding this distance to the separation distance between the lenses gives the distance the image is from the first lens. (− 1 f )( 3 f − f ) 1 1 1 fd + = → d i2 = 2 o2 = 3 2 1 4 1 1 1 = 12 f1 d o2 d i2 f 2 d o2 − f 2 ( 4 f1 − f1 ) − ( − 2 f1 )

d = l + d i2 = 43 f1 + 12 f1 = 45 f1 (c) Set the magnification equal to the total magnification found in part (a) and solve for the focal length. 250 mm 2f 250 mm m=− = − 1  f1 = = 125mm do do 2 Use the results of part (b) to determine the distance between the lens and film. Subtract this distance from 250 mm to determine how much closer the lens can be to the film in the two lens system. d = 45 f1 = 45 (125mm ) = 156mm ; d = 250mm − 156mm = 94 mm

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CHAPTER 34: The Wave Nature of Light: Interference and Polarization Responses to Questions 1.

Huygens’ principle applies both to sound waves and water waves. Huygens’ principle applies to all waves that form a wave crest. Sound, water, and light waves all can be represented in this way. The maximum or crest of a water wave is its highest point above the equilibrium level of the water, so a wave front can be formed by connecting all of the local maximum points of a wave.

Light from the Sun can be focused by a converging lens onto a piece of paper and burn a hole in the paper. This provides evidence that light is energy. You can also feel the heat from the Sun beating down on you on a hot summer day. When you move into the shade you may still feel hot, but you don’t feel the Sun’s energy directly, and the air is cooler in the shade.

A ray shows the direction of propagation of a wave front. If this information is enough for the situation under discussion, then light can be discussed as rays. This is true in particular for geometric optics. Sometimes, however, the wave nature of light is essential to the discussion. For instance, the double-slit interference pattern depends on the interference of the waves, and could not be explained by examining light as only rays.

The bending of waves around corners or obstacles is called diffraction. Diffraction is most prominent when the size of the obstacle is on the order of the size of the wavelength. Sound waves have much longer wavelengths than do light waves. As a result, the diffraction of sound waves around a corner of a building or through a doorway is noticeable and we can hear the sound in the “shadow region,” but the diffraction of light waves around a corner is not noticeable because of the very short wavelength of the light.

The wavelength of light cannot be determined from reflection measurements alone, because the law of reflection is the same for all wavelengths. However, thin-film interference, which involves interference of the rays reflecting from the front and back surfaces of the film, can be used to determine wavelength. Refraction can also be used to determine wavelength because the index of refraction for a given medium is different for different wavelengths.

For destructive interference, the path lengths must differ by an odd number of half wavelengths, such as λ/2, 3λ/2, 5λ/2, 7λ/2, etc. In general, the path lengths must differ by λ(m + ½), where m is an integer with m  0. Under these conditions, the wave crests from one ray match up with the wave troughs from the other ray and cancellation occurs (destructive interference).

As red light is switched to blue light, the wavelength of the light is decreased. Thus, d sin  = m says that  is decreased for a constant m and d. This means that the bright fringes on the screen are more closely packed together with blue light than with red light.

The wavelength of light in a medium such as water is decreased when compared to the wavelength in air. Thus, d sin  = m says that  is decreased for a particular m and d. This means that the bright fringes on the screen are more closely packed together in water than in air.

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Figure 34–5: • The light is traveling from left to right. • The wave crests are located a regular intervals along the rays of light. A line connecting the wave crests from one ray to the next would run vertically up and down the page. • The wave crests are not represented in this figure. • The advantage of this figure is to compare the particle theory prediction for what appears on the screen with the actual image. Figure 34–6: • The light is traveling from left to right. • The wave crests are in the plane of the paper, perpendicular to the direction of motion. • The wave crests are represented by the vertical lines to the left of the slits and barrier, and by the curved lines to the right of the slits and barrier. • An advantage to this figure is that it shows the two slits as point sources due to diffraction, and showing the overlap (implying interference) of the waves from the two slit sources when they reach the viewing screen. Figure 34–7: • The light is traveling from left to right. • The wave crests are in the plane of the paper, perpendicular to the direction of motion. • The wave crests are the maxima of each sinusoidal representation of the two individual rays that are shown in part (a), (b), and (c). They are not shown in part (d). • An advantage is that this figure shows two rays in phase [(a) and (b)] or out of phase (c) when they reach the screen, illustrating constructive and destructive interference, respectively.

10. The two experiments are the same in principle. Each requires coherent sources and is simplest to study with a single frequency source. Each produces a pattern of alternating high and low intensity. Sound waves have much longer wavelengths than light waves, so the appropriate source separation for the sound experiment would be larger. Also, sound waves are mechanical waves which require a medium through which to travel, so the sound experiment could not be done in a vacuum while the light experiment could. 11. The red light and the blue light coming from the two different slits will have different wavelengths (and different frequencies) and will not have a constant phase relationship. In order for a double-slit pattern to be produced, the light coming from the slits must be coherent. No distinct double-slit © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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interference pattern will appear. However, each slit will individually produce a “single-slit diffraction” pattern, as will be discussed in Chapter 35. 12. The reason you do not get an interference pattern from the two headlights of a distant car is that they are not coherent light sources. The phase relationship between the two headlights is not constant– they have randomly changing phases relative to each other. Thus, you cannot produce zones of destructive and constructive interference where the crests and troughs match up or where the crests and crests match up. Also, the headlights are far enough apart that even if they were coherent, the interference pattern would be so tightly packed that it would not be observable with the unaided eye. 13. For a very thin film, there are only a select few distinct visible wavelengths that meet the constructive interference criteria, because the film is only a few wavelengths thick. But for a thick film, there might be many different wavelengths that meet the constructive interference criteria for many different “m” values. Accordingly, many different colors will constructively interfere, and the reflected light will be white (containing all visible wavelengths). See Problem 36 for a detailed analysis of this condition. 14. Bright colored rings will occur when the path difference between the two interfering rays is λ/2, 3λ/2, 5λ/2, and so forth. A given ring, therefore, has a path difference that is exactly one wavelength longer than the path difference of its neighboring ring to the inside and one wavelength shorter than the path difference of its neighboring ring to the outside. Newton’s rings are created by the thin film of air between a glass lens and the flat glass surface on which it is placed. Because the glass of the lens is curved, the thickness of this air film does not increase linearly. The farther a point is from the center, the smaller the horizontal distance that corresponds to an increase in vertical thickness of one wavelength. The horizontal distance between two neighboring rings therefore decreases with increasing distance from the center. An “air wedge,” as in Fig. 34–20, has equally spaced interference patterns because as the observation point is moved farther from the contact point of the flat piece of glass on top, the path differences change linearly. 15. These lenses probably are designed to eliminate reflected wavelengths at both the red and the blue ends of the spectrum. The thickness of the coating is designed to cause destructive interference for reflected red and blue light. The reflected light then appears yellow-green. 16. The index of refraction of the oil must be less than the index of refraction of the water. If the oil film appears bright at the edge, then the interference between the light reflected from the top of the oil film and from the bottom of the oil film at that point must be constructive. The light reflecting from the top surface (the air/oil interface) undergoes a 180° phase shift since the index of refraction of the oil is greater than that of air. The thickness of the oil film at the edge is negligible, so for there to be constructive interference, the light reflecting from the bottom of the oil film (the oil/water interface) must also undergo a 180° phase shift. This will occur only if the index of refraction of the oil is between those of air and water: 1.00 < noil < 1.33. 17. Polarization demonstrates the transverse wave nature of light, and cannot be explained if light is considered as a longitudinal wave or as classical particles. 18. Polarized sunglasses completely block horizontally polarized glare at certain reflected angles, and also block unpolarized light by 50%. Regular tinted sunglasses reduce the intensity of all incoming light, but don’t preferentially reduce the “glare” from smooth reflective surfaces.

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19. Take the sunglasses outside and look up at the sky through them. Rotate the sunglasses (about an axis perpendicular to the lens) through at least 180. If the sky seems to lighten and darken as you rotate the sunglasses, then they are polarizing. You could also look at a liquid crystal display or reflections from a tile floor (with a lot of “glare”) while rotating the glasses, and again look for the light to be lighter or darker depending on the rotation angle. Finally, you could put one pair of glasses on top of the other as in Fig. 34–41 and rotate them relative to each other. If the intensity of light that you see through the glasses changes as you rotate them, then the glasses are polarizing. 20. If Earth had no atmosphere, the “color” of the sky would be black (and dotted with stars and planets) at all times. This is the condition of the sky that the astronauts found on the Moon, which has no atmosphere. If there were no air molecules to scatter the light from the Sun, the only light we would see would be from the stars, the planets, the Moon, and direct sunlight. The rest of the sky would be black. 21. If the atmosphere were 50 times more dense, sunlight (after passing through the atmosphere) would be much redder than it is now. As the atmosphere increased in density, more and more of the blue light would be scattered away in all directions, making the light that reaches the ground very red. Think of the color of a deep red sunset, but this might be the color even when the sun was at high elevations.

Responses to MisConceptual Questions 1.

(c) Diffraction is a basic wave property. Huygens’ principle tells us that as a wave front encounters an obstacle, each point along the opening will act as a tiny source of waves that will spread out in all forward directions.

(a) The width of the fringes is proportional to the wavelength and inversely proportional to the slit spacing. Therefore, since red light has the longer wavelength, red light with small slit spacing will have the largest fringe width.

(a) The location of the fringes is proportional to the wavelength (at least for small angles), and so as the wavelength increases, the pattern spreads out.

(c) The location of the fringes is inversely proportional to the slit separation (at least for small angles), and so as the slits are moved farther apart, the interference pattern shrinks together.

(b) Whether there is air or water on the “far” side of the glass, there will be no phase change due to reflection at that surface (see Fig. 34–19b). This is because the index of refraction of the glass is greater than that of either air or water.

(c) This is analogous to the situation described on page 1027, with Fig. 34–17. In that discussion, constructive interference occurs when ABC = 2toil = n , 2n , 3n , or toil = 12 n , n , 32 n , Destructive interference occurs when ABC = 2toil =  ,  , n , or toil =  ,  , n , So if the thickness of the oil is increased by 50% from the destructive case, we would have . This doesn’t produce either constructive or toil = 32  14 n , 43 n , 54 n ,  = 83 n , 89 n , 158 n , 1 2

3 n 2

5 n 2

1 4

3 n 4

5 n 4

. .

destructive interference. So answer (c) is the correct answer. 7.

(a) In both cases there is a half-cycle phase change at the first reflection and there is no phase change at the second reflection. Thus there is no difference in the pattern that is seen.

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(a, d) We assume “thin film in air” means that there is air on BOTH sides of the film. Thus there will be a half-cycle phase change at the initial reflection, but not at the second reflection. The  path difference is twice the thickness of the film, or 2t. If 2t = m = mn , destructive n interference will occur. The minimum non-zero thickness for this is t = 12 n , and so answer (d) is correct and answer (b) is incorrect. If 2t = ( m + 12 ) n, constructive interference will occur.

The minimum thickness for this is t = 14 n , and so answer (a) is correct. The next thickness for constructive interference is t = 34 n , and so answer (c) is incorrect. 9.

(c) The detector is exactly half-way between the two sources, and yet there is destructive interference. Thus, there must be a half-cycle phase difference between the waves when they leave their respective sources. To bring the two waves into phase, there needs to be a half-cycle difference in their distances. To accomplish this, move the detector ¼  towards one of the sources. Then the DIFFERENCE in distance from the two sources to the detector is ½  Thus, the total phase difference is a ½ cycle from the distance difference, and a ½ cycle from the phase difference at the sources, resulting in constructive interference.

10. (b) In the diagram, one full cycle is the distance from one vertical dashed line to the SECOND vertical dashed line from it. The top wave has its peaks at the vertical dashed lines, while the bottom wave has been shifted ¼ of a cycle to the right. Thus, the phase difference is ¼ of a cycle, or 90°. The correct answer is (b). 11. (b, c) We assume “thin film in air” means that there is air on BOTH sides of the film. Thus, there will be a half-cycle phase change at the initial reflection, but not at the second reflection. For destructive interference to occur, twice the thickness (2t) must be an integer number of  m wavelengths in the film. So 2t = m , or t = . Answer (b) corresponds to m = 1, and n 2 n answer (c) corresponds to m = 2. 12. (b) If two successive polarizers are perpendicular to each other light cannot get through. In case 1 the first two polarizers are perpendicular and in case 2 the last two polarizers are perpendicular. In case 2 no successive polarizers are perpendicular, so some light can pass through.

Solutions to Problems 1.

Consider a wave front traveling at an angle 1 relative to a surface. At time t = 0, the wave front touches the surface at point A, as shown in the figure. After a time t, the wave vt vt front, moving at speed v, has moved forward such that the contact position has moved to point B. The distance between the two contact points is calculated using simple geometry: vt AB = . sin 1 By Huygens’ principle, at each point the wave front touches the surface, it creates a new wavelet. These wavelets expand out in all directions at speed v. The line passing through the surface of each of these wavelets is the reflected wave front. Using the radius of the wavelet created at t = 0, the center of the wavelet created at time t, and the distance between the two contact points (AB) we create a right

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triangle. Dividing the radius of the wavelet centered at AB (vt) by the distance between the contact points gives the sine of the angle between the contact surface and the reflected wave,  2 . vt vt = = sin 1 →  2 = 1 sin  2 = v t AB sin 1 Since these two angles are equal, their complementary angles (the incident and reflected angles) are also equal. 2.

For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. Apply this to the fifth order. −5 d sin  (1.8  10 m ) sin12 = = 7.5  10−7 m d sin  = m →  = 5 m

For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. Apply this to the third order. 3 ( 610  10−9 m ) m = = 4.3  10−6 m d sin  = m → d = sin  sin 25

For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . Adjacent fringes will have m = 1.  ml x d sin  = m → d = m → x = l d  ( m + 1) l  ( m + 1) l  ml  l  m1l → x = x2 − x1 = − = x1 = ; x2 = d d d d d −5 d x ( 4.8  10 m ) ( 0.085m ) = = = 6.277  10−7 m  6.3  10−7 m l 6.50 m

f = 5.



3.00  108 m s = 4.8  1014 Hz 6.277  10−7 m

We have the same setup as in Example 34–4, so the two slits are 0.50 mm apart, and the screen is 2.5 m away. We use Eq. 34–2a with the small angle approximation as described in that example, and solve for the wavelength. −4 −3 x dx ( 5.0  10 m )( 2.9  10 m ) = = 5.8  10−7 m d sin  = d = m →  = l ml (1)( 2.5m ) From Fig. 32–26, that wavelength is yellow.

The slit spacing and the distance from the slits to the screen is the same in both cases. The distance between bright fringes can be taken as the position of the first bright fringe (m = 1) relative to the central fringe. We indicate the lab laser with subscript 1, and the laser pointer with subscript 2. For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l .

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l l x  ml ; x1 = 1 ; x2 = 2 = m → x = d d d l  5.04 mm d 2 = x2 = 1 x2 = ( 632.8nm ) = 531.55nm  532 nm 6.00 mm x1 l d sin  = m → d

→

For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . Second order means m = 2. x  ml  ml  ml d sin  = m → d = m → x = ; x1 = 1 ; x2 = 2 → d d d l −9 2 − 1 ) ml ( 720 − 660 )  10 m  ( 2 )(1.0 m ) ( x = x2 − x1 = = = 1.967  10−4 m  0.2 mm −4 d 6.1 10 m  ( ) This justifies using the small angle approximation, since x

8. The wavelength is 4.5 cm, and the “slit” separation is 7.5 cm. For destructive interference, the path difference is as follows. ( m + 12 )  = ( m + 12 )( 4.5cm ) = m + 1 0.60 , m = 0, 1, 2, 3, ... d sin  = ( m + 12 )  → sin  = . ) ( 2 )( d ( 7.5cm ) The angles for the first three regions of complete destructive interference are calculated. 0 = sin −1 ( 0 + 12 ) ( 0.60 ) = sin −1 0.30 = 17.46  17

1 = sin −1 (1 + 12 ) ( 0.60 ) = sin −1 0.90 = 64.16  64  2 = sin −1 ( 2 + 12 ) ( 0.60 ) = sin −1 1.50 = impossible There are only two regions of destructive interference, at 17 and 64 Note that the 3.0-m distance was not used in the solution. 9.

−9 x  ml ( 680  10 m ) ( 3)( 2.2 m ) = m → d = = = 1.2  10−4 m l 38  10−3 m x

10. For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . For adjacent fringes, m = 1.  ml x → d sin  = m → d = m → x = d l 633  10−9 m ) ( 4.2 m ) ( l x = m = (1) = 0.0391m  3.9cm d ( 6.8  10−5 m ) 11. For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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−9  ml ( 633  10 m ) (1)( 4.6 m ) x = = 1.2  10−5 m d sin  = m → d = m → d = l x ( 0.25m )

12. Pearson acknowledges that this problem cannot be solved using the referenced image. This problem and related image will be updated during the first reprint. We apologize for any inconvenience. 13. The 180° phase shift produced by the glass is equivalent to a path length of 12  . For constructive interference on the screen, the total path difference is a multiple of the wavelength: 1  + d sin max = m , m = 0, 1, 2, → d sin max = ( m − 12 )  , m =1, 2, 2 We could express the result as d sin max = ( m + 12 )  , m = 0, 1, 2,

For destructive interference on the screen, the total path difference is 1  + d sin min = ( m + 12 )  , m = 0, 1, 2, → d sin min = m , m = 0, 1, 2, 2 Thus the pattern is just the reverse of the usual double-slit pattern. There will be a dark central line instead of a bright central maximum. Every place there was a bright fringe will now have a dark line, and vice versa. Fig. 34–10 is reproduced here, and then immediately to the right is the pattern that would appear from the situation described by this problem. Notice that maxima have changed to minima, and vice-versa.

14. To change the center point from constructive interference to destructive interference, the phase shift produced by the introduction of the plastic must be equivalent to half a wavelength. The wavelength of the light is shorter in the plastic than in the air, so the number of wavelengths in the plastic must be ½ wavelength greater than the number of wavelengths in the same thickness of air. The number of wavelengths in the distance equal to the thickness of the plate is the thickness of the plate divided by the appropriate wavelength. N plastic − N air = t=



plastic

2 ( nplastic − 1)

−



t nplastic



−



plastic

− 1) = 12 →

640 nm = 530 nm 2 (1.60 − 1)

15. For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . For adjacent fringes, m = 1.

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 ml x = m → x = → d l (544  10−9 m ) (5.0 m ) = 1.9  10−3 m l x = m = (1) d (1.4  10−3 m )

d sin  = m → d

16. An expression is derived for the slit separation from the data for the 500-nm light (subscript 1). That expression is then used to find the location of the maxima for the 650-nm light (subscript 2). For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . We assume that both wavelengths have 2 significant figures.  ml 1m1l  ml x d sin  = m → d = m → d = = → x= → x x1 d l

x2 =

( 650 nm )( 2 ) = 10.4 mm  0.010 m 2 m2 l m = x1 2 2 = (12 mm ) 1m1l 1m1 ( 500 nm )( 3) x1

17. The presence of the water changes the wavelength according to Eq. 34–1, and so we must change  to n =  n . For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . Adjacent fringes will have m = 1.

 ( m + 1) l x  ml  m1l ; x1 = ; x2 = = mn → x = n → d d d l −9  ( m + 1) l n ml n l  l ( 474  10 m ) ( 0.600 m ) x = x2 − x1 = n − = = = = 3.56  10−3 m d d d nd (1.33) ( 6.00  10−5 m )

d sin  = mn → d

18. We equate the expression from Eq. 34–2a for the second-order blue light to Eq. 34–2b, since the slit separation and angle must be the same for the two conditions to be met at the same location. d sin  = mb = ( 2 )( 480 nm ) = 960 nm ; d sin  = ( m + 12 )  , m = 0, 1, 2,

( m + 12 )  = 960 nm

m = 0 →  = 1920 nm ; m = 1 →  = 640 nm m = 2 →  = 384 nm

The only one visible is 640 nm . 384 nm is near the low-wavelength limit for visible light. 19. The intensity is proportional to the square of the amplitude. Let the amplitude at the center due to one slit be E0 . The amplitude at the center with both slits uncovered is 2 E0 . 2

I1 slit  E0  1 =  = I 2 slits  2 E0  4 Thus the amplitude due to a single slit is one-fourth the amplitude when both slits are open. 20. (a) The phase difference is given in Eq. 34–4. We are given the path length difference, d sin  .  d sin  1.25 = →  = 2 = 2.50 2   Note that this is equivalent to  = 0.50 . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) The intensity is given by Eq. 34–6.   I = I 0 cos2   = I 0 cos2 (1.25 ) = 0.500 I 0 2 21. The intensity of the pattern is given by Eq. 34–6. We find the angle where the intensity is half its maximum value.   d sin   1   d sin 1/ 2  1 2   d sin 1/ 2  1 → I = I 0 cos2   = 2 I 0 → cos   = 2 → cos  =   2         d sin 1/ 2  1  = cos−1 = → sin 1/ 2 =  4d 2 4  1 and so sin    . This is the angle from the central maximum to the If  d , then sin  = 4d location of half intensity. The angular displacement from the half-intensity position on one side of the central maximum to the half-intensity position on the other side would be twice this.  = 2 = 2

 4d

 2d

22. The intensity as a function of angle from the central maximum is given by Eq. 34–6. 1   d sin   1   d sin   2   d sin   1 I = I 0 cos2   = 2 I 0 → cos   = 2 → cos  = 2          1  d sin     1    d sin   cos  → = cos−1   = 45  n ( 90 ) =  n → =  4 2  2 2    

2d sin  = ( 12  n ) 

To only consider   0, we consider just the plus sign. 2d sin  = ( n + 12 )  , n = 0, 1, 2,

23. A doubling of the intensity means that the electric field amplitude has increased by a factor of 2. We set the amplitude of the electric field of one slit equal to E0 and of the other equal to 2E0 . We use Eq. 34–3 to write each of the electric fields, where the phase difference,  , is given by Eq. 34–4. Summing these two electric fields gives the total electric field. E = E0 sin  t + 2 E0 sin ( t +  ) = E0 sin  t + 2 E0 sin  t cos  + 2 E0 cos  t sin 

(

)

= E0 1 + 2 cos  sin  t + 2 E0 cos  t sin 

We square the total electric field intensity and integrate over the period to determine the average intensity. T T 2 1 1 E2 =  E2 dt =   E0 1 + 2 cos  sin  t + 2 E0 cos  t sin   dt  T 0 T 0

(

)

(

)

2 E02   + 1 2 cos sin 2  t + 2cos2  t sin 2  + 2 2 1 + 2 cos  sin  sin  t cos  t  dt    T 0

2 E02  E2 1 + 2 cos  + 2sin 2   = 0 3 + 2 2 cos    2 2 

(

)

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Since the intensity is proportional to this average square of the electric field, and the intensity is maximum when  = 0, we obtain the relative intensity by dividing the square of the electric field by the maximum square of the electric field. I E2 3 + 2 2 cos  2 d sin  = 2 = , with  = I 0 E = 0  3+ 2 2

24. (a) If the sources have equal intensities, their electric fields will have the same magnitudes. We show a phasor diagram with each of the electric fields shifted by an angle  . As shown in the sketch, the three electric fields and their sum form a symmetric trapezoid. Since E20 and E 0 are parallel, and E20 is rotated from E10 and E30 by the angle  , the magnitude of E 0 is the sum of the components of E10 , E20 , and E30 that are parallel to E20 .

E30

 E 0

E 20



E10 E 0 = E10 cos  + E20 + E30 cos  = E10 (1 + 2cos  ) We set the intensity proportional to the square of the electric field magnitude and divide by the maximum intensity (at  = 0 ) to determine the relative intensity.

 E10 (1 + 2cos  ) (1 + 2cos  ) ,  = 2 d sin  I E2 = 2 0 =  = 2  I 0 E =0  E10 (1 + 2cos0 ) 9   (b) The intensity will be at its maximum when cos  = 1. In this case the three phasors are all in line. 2 m cos  max = 1 →  max = 2m = d sin  max → sin  max = , m = 0, 1, 2,  d The intensity will be a minimum when 1 + 2cos  = 0. In this case the three phasors add to 0 and form an equilateral triangle as shown in the diagram at the right, for the case of k = 1, where k is defined below. 1 + 2cos  min = 0 → 2

2 1   + 2m = 2 ( m + 3 ) , m = 0, 1, 2, 2  3  + 2m = 2 ( m + 3 )

 min = cos −1 ( − 12 ) =  43

This can be written as one expression with two parameters. 2 d sin  min , k = 1, 2; m = 0, 1, 2,  min = 2 ( m + 13 k ) =   sin  min = ( m + 13 k ) = , k = 1, 2; m = 0, 1, 2, d

→

25. From Example 34–7, we see that the thickness is related to the bright color wavelength by t =  4n .

t =  4n →  = 4nt = 4 (1.35)(120nm ) = 648nm  650nm

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26. Example 34–6 tells us that there is a dark band at the end where the plates touch and a bright band at the end with the wire. Between the 25 dark lines there are 24 intervals. When we add the halfinterval at the wire end, we have 24.5 intervals over the length of the plates. 26.3cm = 1.07cm 24.5intervals 27. (a) An incident wave that reflects from the outer surface of the bubble has a phase change of  1 =  . An incident wave

1 = 

2 = ( 2t film ) 2 + 0 that reflects from the inner surface of the bubble has a phase change due to the additional path length, so  2t  2 =   2 . For destructive interference with a  film  minimum non-zero thickness of bubble, the net phase change must be  .   2t    480 nm 1 net = 2 − 1 =  = = 180 nm  2  −  =  → t = 2 film = n 2 2 1.33  ( ) film     (b) For the next two larger thicknesses, the net phase change would be 3 and 5 .   2t    480 nm net = 2 − 1 =  = 360 nm  2  −  = 3 → t = film = = n (1.33)  film  

  2t    3 480 nm = 540 nm  2  −  = 5 → t = film = = 2 n 1.33  ( ) film     (c) If the thickness were much less than one wavelength, then there would be very little phase change introduced by additional path length, and so the two reflected waves would have a phase difference of about  1 =  . This would produce destructive interference, which would look

net = 2 − 1 = 

black. 28. When illuminated from above, the light ray reflected from the air–oil interface undergoes a phase shift of  1 =  . A ray reflected at the oil–water interface undergoes no phase shift due to reflection,

 2t  but has a phase change due to the additional path length of 2 =   2 . For constructive  oil  interference to occur, the net phase change must be a multiple of 2 .   2t    1 1 1 1 net = 2 − 1 =   2  −  = m ( 2 ) → t = 2 ( m + 2 ) oil = 2 ( m + 2 ) no  oil   For  = 653nm, the possible thicknesses are as follows. 653nm t653 = 12 ( m + 12 ) = 109 nm, 327 nm, 544 nm, 762 nm, 1.50 For  = 392 nm, the possible thicknesses are as follows. 392 nm t392 = 12 ( m + 12 ) = 65.3nm, 196 nm, 327 nm, 457 nm, 1.50 The minimum thickness of the oil slick must be  327 nm .

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29. (a) When illuminated from above at A, a light ray reflected from the air–oil interface undergoes a phase shift of  1 =  . A ray reflected at the oil–water interface undergoes no phase shift. If the oil thickness at A is negligible compared to the wavelength of the light, then there is no significant shift in phase due to a path distance traveled by a ray in the oil. Thus, the light reflected from the two surfaces will destructively interfere for all visible wavelengths, and the oil will appear black when viewed from above. (b) From the discussion in part (a), the ray reflected from the air–oil interface undergoes a phase shift of  1 =  . A ray that reflects from the oil–water interface has no phase change due to

 2t  reflection, but has a phase change due to the additional path length of 2 =   2 . For  oil  constructive interference, the net phase change must be a multiple of 2 .   2t    1 1 1 1 net = 2 − 1 =   2  −  = m ( 2 ) → t = 2 ( m + 2 ) oil = 2 ( m + 2 ) no  oil   From the diagram, we see that point B is the second thickness that yields constructive interference for 580 nm, and so we use m = 1. (The first location that yields constructive interference would be for m = 0.) 580 nm  t = 12 ( m + 12 ) = 12 (1 + 12 ) = 290 nm no 1.50 30. An incident wave that reflects from the top surface of the coating has a phase change of  1 =  . An incident wave that reflects from the glass ( n = 1.56 ) at the bottom surface of the coating has a phase change due to both the additional path length and a phase change of  on  2t  reflection, so 2 =   2 +  . For constructive  film  interference with a minimum non-zero thickness of coating, the net phase change must be 2 .  2t      1 1 net = 2 − 1 =   2 +   −  = 2 → t = 2 film = 2  . n  film film       The lens reflects the most for  = 570nm. The minimum non-zero thickness occurs for m = 1. ( 570 nm ) = 228nm  230 nm  = tmin = 2nfilm 2 (1.25) Since the middle of the spectrum is being selectively reflected, the transmitted light will be stronger in the red and blue portions of the visible spectrum. 31. An incident wave that reflects from the convex surface of the lens has no phase change, so  1 = 0. An incident wave that reflects from the glass underneath the lens has a phase change due to both the additional path length and a phase change of  2t   on reflection, so 2 =   2 +  . For destructive  interference (dark rings), the net phase change must be an odd-integer multiple of  , so © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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net = 2 − 1 = ( 2m + 1)  , m = 0, 1, 2, . Because m = 0 corresponds to the dark center, m represents the number of the ring.  2t   net = 2 − 1 =   2 +   − 0 = ( 2m + 1)  , m = 0, 1, 2, →     t = 12 mair = 12 ( 35 )( 560 nm ) = 9800 nm = 9.8 m The thickness of the lens at its center is the thickness of the air at the edge of the lens. 32. An incident wave that reflects from the second surface of the upper piece of glass has no phase change, so  1 = 0. An incident wave that reflects from the first surface of the second piece of glass has a phase change due to both the additional path length and a  2t  phase change of  on reflection, so 2 =   2 +  .  For destructive interference (dark lines), the net phase change must be an odd-integer multiple of  , so net = 2 − 1 = ( 2m + 1)  , m = 0, 1, 2, . Because m = 0 corresponds to the left edge of the diagram, the 24th dark line corresponds to m = 23. The 24th dark line also has a gap thickness of d.   2t   net = 2 − 1 =   2 +   − 0 = ( 2m + 1)  → t = 12 m →     d = 12 ( 23)( 670 nm ) = 7705nm  7.7  m

33. (a) With respect to the incident wave, the wave that reflects from the air at the top surface of the air layer has a phase change of 1 = 0. With respect to the incident wave, the wave that reflects from the glass at the bottom surface of the air layer has a phase change due to both the additional  2t  path length and reflection, so 2 =   2 +  . For  constructive interference, the net phase change must be an even non-zero integer multiple of  .  2t   net = 2 − 1 =   2 +   − 0 = 2m → t = 12 ( m − 12 )  , m = 1, 2,     m The minimum thickness is with = 1. tmin = 12 ( 480nm ) (1 − 12 ) = 120nm (b) For destructive interference, the net phase change must be an odd-integer multiple of  .  2t   net = 2 − 1 =   2 +   − 0 = ( 2m + 1)  → t = 12 m , = 0, 1, 2,     The minimum non-zero thickness is tmin = 12 ( 480nm )(1) = 240nm .

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34. With respect to the incident wave, the wave that reflects from the top surface of the alcohol has a phase change of 1 =  . With respect to the incident wave, the wave that reflects from the glass at the bottom surface of the alcohol has a phase change due to both the additional path length and a phase change of  on reflection, so  2t  2 =   2 +  . For constructive interference, the  film  net phase change must be an even non-zero integer multiple of  .  2t    2 +   −  = m1 2 → t = 12  1film m1 = 12 1 m1 , m1 = 1, 2, 3, ... . net = 2 − 1 =     n  film  1film   For destructive interference, the net phase change must be an odd-integer multiple  .   2t   2 1 net = 2 − 1 =  ( 2m2 + 1) , m2 = 0, 1, 2, ... .  2 +   −  = ( 2m2 + 1) → t = 4 nfilm  2film   Set the two expressions for the thickness equal to each other. 1  2m2 + 1 1 ( 615nm ) 5 1 m1 = 14 2 ( 2m2 + 1) → = = = 1.25 = 2 nfilm nfilm 2m1 4 2 ( 492 nm ) Thus, we see that m1 = m2 = 2, and the minimum thickness of the film is t = 12

2 492 nm   615nm  1 m1 = 12  ( 2m2 + 1) = 14   ( 2 ) = 452 nm or t = 4  ( 5 ) = 452 nm . nfilm nfilm  1.36   1.36  1

35. Since the wedge is now filled with water, we adjust the wavelength in the water, as instructed in the problem statement. Since the index of refraction of the water is less than that of the glass, there are no new phase shifts introduced by reflections, and so we can use the relationships developed in that example. We “count” the number of wavelengths at the position of the wire. 2 ( 7.35  10−6 m ) 2t = = 32.6 wavelengths n 6.00  10−7 m 1.33 Thus there will be 33 dark bands across the plates, including the one at the point of contact. 36. (a) Assume the indices of refraction for air, water, and glass are 1.00, 1.33, and 1.50, respectively. When illuminated from above, a ray reflected from the air– water interface undergoes a phase shift of 1 =  , and a ray reflected at the water–glass interface also undergoes a phase shift of  . Thus, the two rays are unshifted in phase relative to each other due to reflection. For constructive interference, the path difference 2t must equal an integer number of wavelengths in water. 2n t  2t = mwater = m , m = 0, 1, 2, →  = water nwater m (b) The above relation can be solved for the m-value associated with the reflected color. If this m-value is an integer the wavelength undergoes constructive interference upon reflection. 2n t 2n t  = water → m = water  m © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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For a thickness t = 200 m = 2  105 nm the m-values for the two wavelengths are calculated.

m700 nm = m400 nm =

2nwater t



2nwater t

700 nm

= 760

2 (1.33) ( 2  105 nm )

= 1330 400 nm Since both wavelengths yield integers for m, they are both reflected. (c) All m-values between m = 760 and m = 1330 will also produce reflected visible colors. There are 1330 – (760 – 1) = 571 such values. For instance, a value of m = 924 would be associated 5 2nwater t 2 (1.33) ( 2  10 nm ) = = 576 nm. with  = m 923 (d) This mix of such a large number of wavelengths from throughout the visible spectrum will give the thick layer a white or grey appearance.



2 (1.33) ( 2  105 nm )

37. With respect to the incident wave, the wave that reflects from point B in the first diagram will not undergo a phase change, and so B = 0. With respect to the incident wave, the wave that reflects from point C in the first diagram has a phase change due to both the additional path length in air, and a phase 2y change of  on reflection, and so we say that D = ( 2 ) +  , where y is  the thickness of the air gap from B to C (or C to D). For dark rings, the net phase difference of the waves that recombine as they leave the glass moving upwards must be an odd-integer multiple of  . 2y net = D − B = ( 2 ) +  = ( 2m + 1)  →



ydark = m , m = 0, 1, 2, Because m = 0 corresponds to the dark center, m represents the number of the dark ring. 1 2

Let the air gap of y be located a horizontal distance r from the center of the lens, as seen in the second diagram. Consider the dashed right triangle in the second diagram. 2 R2 = r2 + ( R − y ) →

R 2 = r 2 + R 2 − 2 Ry + y 2 → r 2 = 2 Ry − y 2 If we assume that y

R, then r 2  2 Ry.

2 r 2 = 2Ry → rdark = 2Rydark = 2R ( 12 m ) → rdark = m R , m = 0, 1, 2,

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38. From Problem 37, we have r = m R = ( m R )

1/ 2

assume m  1 → m = 1

r = ( m R )

1/ 2

;

m. Since m

. To find the distance between adjacent rings, we

m, r 

dr m. dm

dr 1 −1/ 2 = 2 ( m R )  R dm 1/ 2

  2 R2  dr −1/ 2 r  m =  12 ( m R )  R  (1) =     dm  4m R 

R 4m

39. The radius of the mth ring in terms of the wavelength of light and the radius of curvature is derived in Problem 37 as r = m R . Using this equation, with the wavelength of light in the liquid given by Eq. 34–1, we divide the two radii and solve for the index of refraction. 2

 rair   2.92 cm 2 rair m R = = n → n=  =   = 1.32 r rliquid m ( / n ) R  liquid   2.54 cm  40. We use the equation derived in Problem 37, where r is the radius of the lens (1.7 cm) to solve for the radius of curvature. The center is dark, as in Fig. 34–18. The 1st bright ring is between the dark center and the 1st dark ring. Since the outer edge is the 41st bright ring, which would be halfway between the 40th and 41st dark rings, we set m = 40.5.

( 0.017 m ) r2 = = 12.303m  12 m m ( 40.5) ( 580  10−9 m ) 2

r = m R → R =

We calculate the focal length of the lens using Eq. 33–4 (the lensmaker’s equation) with the index of refraction of Lucite taken from Table 32–1.  1  1 1  1 1 = ( n − 1)  +  = (1.50 − 1)  +  = 0.0406 m −1 → f  12.303m    R1 R2 

f =

1 = 24.63  25m 0.0406 m −1

41. According to page 1030, at each surface approximately 4% of the light is reflected. So passing each surface of a lens results in only 96% transmission either into the lens (at the front surface) or on to the next piece of optics (at the second surface). So the light exiting a lens would experience 2 of these reductions, and so only pass 96% of 96% of the light incident on the lens. This is a factor of 2 ( 0.96 ) for each lens. To reduce to 50% would involve that factor for each lens. Let “n” be the number of lenses. ln ( 0.50 ) 2 n 2n = 8.49 0.50 = ( 0.96 )  = ( 0.96 ) → ln ( 0.50 ) = 2n ln ( 0.96 ) → n =   2ln ( 0.96 ) This it would take 9 lenses for the light to be reduced to 50% or less. 42. We assume n1  n2  n3 and that most of the incident light is transmitted. If the amplitude of an incident ray is taken to be E0 , then the amplitude of a reflected ray is rE0 , with r  1. The light reflected from the top surface of the film therefore has an amplitude of rE0 and is phase shifted by 1 =  from the incident wave, due to the higher index of

11 == n1 n2

22 = + +  = (22t /l film 2)2+ ))2 film 2 = ((2 l film

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refraction. The light transmitted at that top surface has an amplitude of (1 − r ) E0 . That light is then reflected off the bottom surface of the film, and we assume that it has the same reflection coefficient. Thus the amplitude of that second reflected ray is r (1 − r ) E0 = ( r − r 2 ) E0  rE0 , the same amplitude as the first reflected ray. Due to traveling through the film and reflecting from the glass, the second ray has a phase shift of 2 =  + 2 ( 2 l / film ) =  + 4 l n2  , where l is the thickness of the film. Summing the two reflected rays gives the net reflected wave. E = rE0 cos (t +  ) + rE0 cos (t +  + 4 l / n ) = rE0 (1 + cos 4 l n2  ) cos (t +  ) − sin ( 4 l n2  ) sin (t +  ) 

As with the double-slit experiment, we set the intensity proportional to the square of the wave amplitude and integrate over one period to calculate the average intensity. 2 1 T 1 T I   E 2 dt =   rE0 (1 + cos 4 l n2  ) cos (t +  ) − sin ( 4 l n2  ) sin (t +  )   dt T 0 T 0 2 2 2 2 r 2 E02 T (1 + cos 4 l n2  ) cos (t +  ) + sin ( 4 l n2  ) sin (t +  ) =   dt T 0  −2 (1 + cos 4 l n2  ) sin ( 4 l n2  ) cos (t +  ) sin (t +  )    2 2 r E0  2 = (1 + cos 4 l n2  ) + sin 2 ( 4 l n2  ) = r 2 E02 (1 + cos 4 l n2  )  2 The reflected intensity without the film is proportional to the square of the intensity of the single reflected electric field. 2 1 T 2 1 T r 2 E02 T r 2 E02 2   cos cos I 0   Eno dt rE t dt t dt     = + = + =   ( ) ( ) film 0   2 T 0 T 0  T 0  Dividing the intensity with the film to that without the film gives the factor by which the intensity is reduced. 2 2 I r E0 (1 + cos 4 l n2  ) = = 2 (1 + cos 4 l n2  ) 1 2 2 I0 2 r E0 To determine the thickness of the film, the phase difference between the two reflected waves with  = 550nm must be an odd-integer multiple of  so that there is destructive interference. The minimum thickness will be for m = 0.   550 nm = 2 − 1 =  + 4 l n2   −  = ( 2m + 1)  → l = n = 4 4n 4n It is interesting to see that the same result is obtained if we set the reflected intensity equal to zero for a wavelength of 550 nm. I 550 nm = 2 (1 + cos 4 l n2  ) = 0 → cos 4 l n2  = −1 → 4 l n2  =  → l = I0 4n Finally, we insert the two given wavelengths (430 nm and 670 nm) into the intensity equation to determine the reduction in intensities.  I 550 nm/4n  550 nm   = 2  1 + cos 4 For  = 430 nm,  = 0.721  72%  = 2  1 + cos 4 I0 430 nm / n  4 ( 430 nm )    For  = 670 nm,

 I 550 nm  = 2  1 + cos 4  = 0.308  31% I0 4 670 nm ( )  

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43. From the discussion in Section 34–6, we see that the path length change is twice the distance that the mirror moves. One fringe shift corresponds to a change in path length of  , and so corresponds to a mirror motion of 12  . Let N be the number of fringe shifts produced by a mirror movement of x. We assume it is an exact number. x → x = 12 N  = 12 ( 730 ) ( 589  10−9 m ) = 2.15  10−4 m N=1 2



44. From the discussion in Section 34–6, we see that the path length change is twice the distance that the mirror moves. One fringe shift corresponds to a change in path length of  , and so corresponds to a mirror motion of 12  . Let N be the number of fringe shifts produced by a mirror movement of x. We assume it is exact. −4 2x 2 (1.25  10 m ) x N=1 → = = = 6.79  10−7 m = 679 nm N 368 2 45. From the discussion in Section 34–6, we see that the path length change is twice the distance that the mirror moves. One fringe shift corresponds to a change in path length of  , and so corresponds to a mirror motion of 12  . Let N be the number of fringe shifts produced by a mirror movement of x. The thickness of the foil is the distance that the mirror moves during the 272 fringe shifts. x → x = 12 N  = 12 ( 272 ) ( 611  10−9 m ) = 8.31  10−5 m N=1 2



46. One fringe shift corresponds to an effective change in path length of  . The actual distance has not changed, but the number of wavelengths in the depth of the cavity has. If the cavity has a length d, d the number of wavelengths in vacuum is , and the (greater) number with the gas present is  ngas d d . Because the light passes through the cavity twice, the number of fringe shifts is twice = gas  the difference in the number of wavelengths in the two media. The number of fringe shifts is taken to be exact. (163) ( 632.8  10−9 m ) n d d  d N N = 2  gas −  = 2 ( ngas − 1) → ngas = +1 = + 1 = 1.004465   2d 2 (1.155  10−2 m )   47. There are two interference patterns formed, one by each of the two wavelengths. The fringe patterns overlap but do not interfere with each other. Accordingly, when the bright fringes of one pattern occurs at the same locations as the dark fringes of the other patterns, there will be no fringes seen, since there will be no dark bands to distinguish one fringe from the adjacent fringes. To shift from one “no fringes” occurrence to the next, the mirror motion must produce an integer number of fringe shifts for each wavelength, and the number of shifts for the shorter wavelength must be one more than the number for the longer wavelength. From the discussion in Section 34–6, we see that the path length change is twice the distance that the mirror moves. One fringe shift corresponds to a change in path length of  , and so corresponds to a mirror motion of 12  . Let N be the number of fringe shifts produced by a mirror movement of x.

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N1 = 2 x =

x

1

; N2 = 2

x

2

; N 2 = N1 + 1 → 2

x

2

x

1

Instructor Solutions Manual

+1 →

( 589.6 nm )( 589.0 nm ) = 2.89  105 nm  2.9  10−4 m 12 = 2 ( 1 − 2 ) 2 ( 0.6 nm )

48. Use Eq. 34–8. Since the initial light is unpolarized, the intensity after the first polarizer will be half the initial intensity. Let the initial intensity be I 0 . I1 = 12 I 0 ; I 2 = I1 cos2  = 12 I 0 cos2  →

I 2 cos2 75 = = 0.033 , or 3.3% 2 I0

49. Use Eq. 34–9a. First consider the light passing from air to glass. n 1.56 tan  p = glass = →  p = tan −1 1.56 = 57.3 nair 1.00 Then consider the light passing from glass to air. n 1.00  1.00  →  p = tan −1  tan  p = air =  = 32.7 nglass 1.56  1.56  We notice that the two angles are compliments of each other, because of the reciprocal of the tangents. 50. First assume the light is traveling from water to diamond. Use Eq. 34–9a. n 2.42 = 1.82 →  p = tan −1 1.82 = 61.2 tan  p = diamond = nwater 1.33 Then assume the light is traveling from diamond to water. Again use Eq. 34–9a. n 1.33 = 0.550 →  p = tan −1 0.550 = 28.8 tan  p = water = ndiamond 2.42 The diamond will not sparkle as much because there will be less refraction when it is immersed in water, and thus less dispersion. 51. For the first transmission, the angle between the light and the polarizer is 22.0. For the second transmission, the angle between the light and the polarizer is 44.0. Use Eq. 34–8 twice. I1 = I 0 cos2 22.0 ; I 2 = I1 cos2 44.0 = I 0 ( cos2 22.0 )( cos2 44.0 ) = 0.445I 0 Thus the transmitted intensity is 44.5% of the incoming intensity. 52. Let the initial intensity of the unpolarized light be I 0 . The intensity after passing through the first Polaroid will be I1 = 12 I 0 . Then use Eq. 34–8.

I 2 = I1 cos2  = 12 I 0 cos2  →  = cos −1 (a)  = cos−1

2I2 2 = cos−1 = 35.3 I0 3

(b)  = cos−1

2I2 2 = cos−1 = 63.4 I0 10

2I2 I0

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53. The critical angle exists when light passes from a material with a higher index of refraction ( n1 ) into a material with a lower index of refraction ( n2 ) . Use Eq. 32–7. n2 = sin  C = sin 48 n1 To find the Brewster angle, use Eq. 34–9a. If light is passing from high index to low index, we have the following. n2 = tan  p = sin 48 →  p = tan −1 ( sin 48 ) = 37 n1 If light is passing from low index to high index, we have the following. 1 n1  1  = tan  p = →  p = tan −1   = 53 n2 sin 48  sin 55 

54. First case: the light is coming from water to air. Use Eq. 34–9a. n n 1.00 = 36.9 tan  p = air →  p = tan −1 air = tan −1 nwater nwater 1.33 Second case: for total internal reflection, the light must also be coming from water into air. Use Eq. 32–7. n n 1.00 = 48.8 sin C = air →  p = sin −1 air = sin −1 nwater nwater 1.33 Third case: the light is coming from air to water. Use Eq. 34–9a. n n tan  p = water →  p = tan −1 water = tan −1 1.33 = 53.1 nair nair Note that the two Brewster angles add to give 90.0. 55. If I 0 is the intensity passed by the first Polaroid, the intensity passed by the second will be I 0 when the two axes are parallel. To calculate a reduction to half intensity, we use Eq. 34–81. I = I 0 cos2  = 12 I 0 → cos2  = 12 →  = 45 56. The polarizing angle  p is found using Eq. 34–9a, tan  p =

n2 . For an oil–diamond interface, n1

2.42 , which gives  p = 59.4. The material does NOT appear to be diamond . 1.43 Alternatively, we could solve for the index of refraction of the suspected diamond. n tan  p = 2 → n2 = n1 tan  p = 1.43tan ( 56 ) = 2.12 n1 Since the index of refraction is significantly different from that of diamond, it again appears that the material is NOT diamond. tan  p =

57. For the first transmission, since the incoming light is unpolarized, the transmission is the same as if the angle is 45; so I1 = 12 I 0 . The second polarizer is at an angle of 30 relative to the first one, so I 2 = I1 cos2 30. The third polarizer is at an angle of 60 relative to the second one, so I 3 = I 2 cos2 60. Combine these to find the percent that is transmitted out through the third polarizer.

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I 3 = I 2 cos2 60 = I1 cos2 30 cos2 60 = 12 I 0 cos2 30 cos2 60 → I3 1 3 = 2 cos2 30 cos2 60 = 12 ( 43 ) ( 14 ) = = 0.09375  9.4% I1 32

58. When plane-polarized light passes through a sheet oriented at an angle  , the intensity decreases according to Eq. 34–8, I = I 0 cos2  . For the first sheet, with unpolarized light incident, we can treat

 = 45, cos2  = 12 . Then sheets 2 through 6 will each reduce the intensity by a factor of cos2 38. Thus, we have the following.

I = I 0 ( cos2 45 )( cos2 38 ) = 0.046 I 0 5

The transmitted intensity is 4.6% of the incident intensity. 59. We assume vertically polarized light of intensity I 0 is incident on the polarizer A. The angle between the polarization direction and the middle polarizer is  . After the light passes the middle polarizer, the angle between that light and polarizer B will be 90 −  . Apply Eq. 34–8. I1 = I 0 cos2  ; I = I1 cos2 ( 90 −  ) = I 0 cos2  cos 2 ( 90 −  ) = I 0 cos 2  sin 2 

We can also use the trigonometric identity sin  cos = 12 sin 2 to write the final intensity as I = I 0 cos2  sin 2  = 14 I 0 sin 2 2 .

dI d 1 I sin 2 2 ) = 14 I 0 ( 2sin 2 )( cos 2 ) 2 = I 0 sin 2 cos 2 = 12 I 0 sin 4 = ( 4 0 d d 1 I sin 4 = 0 → 4 =  , 360 →  = 0, 45, 90 2 0 Substituting the three angles back into the intensity equation, we see that the angles 0 and 90 both give minimum intensity. The angle 45 gives the maximum intensity of 14 I 0 . 60. We set the intensity of the beam as the sum of the maximum and minimum intensities. Using Eq. 34–8, we determine the intensity of the beam after it has passed through the polarizer. Since Imin is polarized perpendicular to Imax and the polarizer is rotated at an angle  from the polarization of Imax, the polarizer is oriented at an angle of ( 90 −  ) from Imin.

I 0 = I max + I min I = I 0 cos 2  = I max cos 2  + I min cos 2 ( 90 −  ) = I max cos 2  + I min sin 2  We solve the percent polarization equation for Imin and insert the result into our intensity equation. I −I 1− p p = max min → I min = I max I max + I min 1+ p  (1 + p ) cos 2  + (1 − p ) sin 2   1− p  I = I max cos 2  +  I max  sin 2  = I max   1+ p 1+ p     ( cos 2  + sin 2  ) + p ( cos 2  − sin 2  )  1 + p cos 2   = I max  = I max   1+ p    1+ p 

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61. We assume the luminous flux is uniform, and so is the same in all directions.

Fl = El A = El 4 r 2 = (105 lm m 2 ) 4 (1.496  1011 m ) = 2.81  1028 lm  3  1028 lm 2

Il =

2.81  1028 lm Fl = = 2.24  1027 cd  2  1027 cd 4 sr 4 sr

62. (a) The wattage of the bulb is the electric power input to the bulb. F 1600lm luminous efficiency = l = = 16lm W P 100 W (b) The illuminance is the luminous flux incident on a surface, divided by the area of the surface. Let N represent the number of lamps, each contributing an identical amount of luminous flux. We include the factor of ½ since only half of their flux reaches the floor. N  12 ( luminous efficiency ) P  F → El = l =  A A 2 ( 250lm m 2 ) ( 22 m )( 30 m ) 2E l A = = 137.5lamps  140lamps N= ( luminous efficiency ) P ( 60lm W )( 40 W ) 63. If the original intensity is I 0 , the first polarizer will reduce the intensity to I1 = 12 I 0 . Each subsequent polarizer oriented at an angle  to the preceding one will reduce the intensity by cos 2  , as given by Eq. 34–8. For n polarizers (including the first one), I n = 12 I 0 ( cos2  )

n −1

. We set the final

intensity equal to 16 I 0 , and solve for n with  = 10 . I n = 16 I 0 = 12 I 0 ( cos2 10 ) n = 1+

ln (1 3)

ln ( cos2 10 )

n −1

→

n −1 1 1 = ( cos2 10 ) → ln   = ( n − 1) ln ( cos 2 10 ) → 3  3

= 36.88

Thus 37 polarizers are needed for the intensity to drop below 1 6 of its original value.

v ( 3.00  10 m s ) 64. The wavelength of the signal is  = = = 4.00 m. f ( 75 106 Hz ) 8

(a) There is a phase difference between the direct and reflected signals h from both the path difference,   2 , and the reflection, .  The total phase difference is the sum of the two. (122 m ) 2 +  = 62 = 31 2 h  =   2 +  = ( ) ( 4.00 m )   Since the phase difference is an integer multiple of 2 , the interference is constructive . (b) When the plane is 22 m closer to the receiver, the phase difference is as follows.  (122 m − 22 m )   (h − y )  51 = 2 +  =   2 +  = 51 = ( 2 )  2     ( 4.00 m )  Since the phase difference is an odd-half-integer multiple of 2 , the interference is destructive .

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65. The phase difference caused by the path difference back and forth through the coating must correspond to half a wavelength in order to produce destructive interference. We are instructed to “ignore any change in wavelength in the coating,” so we do not consider the index of refraction. 2t = 12  → t = 14  = 14 ( 2cm ) = 0.5cm 66. With respect to the incident wave, the wave that reflects from the top surface of the coating has a phase change of 1 =  . With respect to the incident wave, the wave that

reflects from the glass ( n  1.5 ) at the bottom surface of the coating has a phase change due to both the additional path  2t  length and reflection, so 2 =   2 +  . For destructive  film  interference, the net phase change must be an odd-integer multiple of  .   2t   net = 2 − 1 =   2 +   −  = ( 2m + 1)  →  film   t = 14 ( 2m + 1) film = 14 ( 2m + 1)

 nfilm

, m = 0, 1, 2, ...

The minimum thickness has m = 0, and so tmin = 14

 nfilm

( 450 nm ) = 79.23nm  79 nm . (1.42 ) ( 720 nm ) = 126.8nm  130 nm . tmin = 14 (1.42 )

(a) For the blue light: tmin = 14 (b) For the red light:

67. (a) For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . For adjacent fringes, m = 1.  ml l x → x = m → d sin  = m → d = m → x = d d l −7  l m ( 5.0  10 m ) ( 5.0 m )(1) = = 1.25  10−4 m  1.3  10−4 m d= −2 x ( 2.0  10 m ) (b) For minima, we use Eq. 34–2b. The fourth minimum corresponds to m = 3, and the fifth minimum corresponds to m = 4 (see Fig. 34–9b). The slit separation, screen distance, and location on the screen are the same for the two wavelengths. x d sin  = ( m + 12 )  → d = ( m + 12 )  → ( mA + 12 ) A = ( mB + 12 ) B → l mA + 12 ) ( 3.5 B = A = ( 5.0  10−7 m ) = 3.9  10−7 m 1 m 4.5 + ( B 2)

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68. Because the measurements are made far from the antennas, we can use the analysis for the doubleslit. Use Eq. 34–2a for constructive interference, and Eq. 34–2b for destructive interference. The 8 v ( 3.00  10 m s ) wavelength of the signal is  = = = 3.243m. f (92.5 106 Hz ) For constructive interference, the path difference is a multiple of the wavelength: m d sin  = m , m = 0, 1, 2, 3, ... ; →  = sin −1 d 1 3.243m 2 )( 3.243m ) ( )( ) ( 1 = sin −1 = 21 ;  2 = sin −1 = 46 ; 9.0 m 9.0 m max max

3

= sin −1

( 3)( 3.243m ) = impossible

9.0 m For destructive interference, the path difference is an odd multiple of half a wavelength: ( m + 12 )  d sin  = ( m + 12 )  , m = 0, 1, 2, 3, ... ; →  = sin −1 d 3 1 ( )( 3.243m ) = 10 ;  = sin −1 ( 2 ) ( 3.243m ) = 33 ; 0 = sin −1 2 1 9.0 m 9.0 m min min max

( 5 ) ( 3.243m ) = 64 ; 

 2 = sin −1 2

3 min

= sin −1

( 72 ) ( 3.243m ) = impossible

9.0 m 9.0 m min These angles are applicable both above and below the midline (assuming no reflection from the ground), and both to the left and the right of the antennas. 69. For constructive interference, the path difference is a multiple of the wavelength, as given by Eq. 34–2a. The location on the screen is given by x = l tan  , as seen in Fig. 34–7(c). For small angles, we have sin   tan   x l . Second order means m = 2. x  ml  ml  ml → d sin  = m → d = m → x = ; x1 = 1 ; x2 = 2 l d d d

x = x1 − x2 =

1ml d

−

2 ml d

→

6.6  10−4 m )(1.23  10−3 m ) ( d x −9 2 = 1 − = 690  10 m − = 4.581  10−7 m  460 nm ml 2 (1.75m ) 70. With respect to the incident wave, the wave that reflects at the top surface of the film has a phase change of 1 =  . With respect to the incident wave, the wave that reflects from the bottom surface of the film has a phase change due to the additional path length and no phase change due to reflection, so  2t  2 =   2 + 0. For constructive interference, the net phase change must be an integer multiple  film  of 2 .  2t   1 1 1 1 , m = 0, 1, 2, ... . net = 2 − 1 =   2 −  = 2 m → t = 2 ( m + 2 ) film = 2 ( m + 2 ) n  film  film  Evaluate the thickness for the two wavelengths. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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t = 12 ( m1 + 12 )

1 nfilm

= 12 ( m2 + 12 )

2 nfilm

Instructor Solutions Manual

→

( m2 + 12 ) = ( m1 + 23 ) = 1 = 688.0 nm = 1.40 = 7 → 5 m + 3 = 7 m + 1 → m = 2 ( 1 2) ( 1 2) 1 5 ( m1 + 12 ) ( m1 + 12 ) 2 491.4 nm Thus m2 = 3 and m1 = 2. Evaluate the thickness with either value and the corresponding wavelength. 688.0nm 491.4 nm   t = 12 ( m1 + 12 ) 1 = 12 ( 25 ) = 531nm ; t = 12 ( m2 + 12 ) 2 = 12 ( 72 ) = 531nm nfilm 1.62 nfilm 1.62 71. The text says that LED bulbs produce from 160 to 180 lumens / watt. We use 170 lumens / watt. 1W 800lumens  = 4.7Ws  5 W 170lumens 72. We use the Brewster angle, Eq. 34–9b, for light coming from air to water. tan p = n → p = tan −1 n = tan −1 1.33 = 53.1 This is the angle from the normal, as seen in Fig. 34–35, so the angle above the horizontal is the complement of the Brewster angle, 90.0 − 53.1 = 36.9 . 73. At point A the green laser light is at a maximum. The next maximum (B) occurs when the path difference between point and each of the two slits is equal to one full wavelength, or 530 nm. 74. From the discussion in Section 34–6, we see that the path length change is twice the distance that the mirror moves. The phase shift is 2 for every wavelength of path length change. The intensity as a function of phase shift is given by Eq. 34–6.   path change 2 x 4 x  2 x  ; I = I 0 cos2 = I 0 cos2  = = → =     2 2    75. To maximize reflection, the three exiting rays shown in the figure should be in phase. We first compare rays 2 and 3. Ray 2 reflects from n2  n1 , and so has a phase shift of 2 =  . Ray 3 will have a phase change due to the additional path length in material 2, and a phase shift of  because of reflecting from n  n2 . Thus

 2d 2   2 +  . For constructive interference the net phase  2  change for rays 2 and 3 must be a non-zero integer multiple of 2 .  2 d   2−3 = 3 − 2 =  2  2 +   −  = 2m → d 2 = 12 m2 , m = 1, 2, 3  2  

3 = 

The minimum thickness is for m = 1, and so d 2 = 12 m2 =

 2n2

Now consider rays 1 and 2. The exact same analysis applies, because the same relationship exists between the indices of refraction: n1  n and n2  n1. Thus d1 =

 2n1

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76. With respect to the incident wave, the wave that reflects from the top surface of the material has a phase change of 1 =  . If we assume that the film has an index less than that of the glass, the wave that reflects from the glass has a phase change due to the additional path-length and a phase change on reflection, for a total phase change of  2t   2t  2 =   2 +  =   2 +    n  film 

For destructive interference, the net phase change must be an odd-integer multiple of  .  2t   1 1 net = 2 − 1 =   2 +  −  = ( 2m − 1) → n = 2   ( m − 2 ) , m = 1, 2, ... t  n The minimum index of refraction is found from m = 1.  655nm  1 nmin = 12   (1 − 2 ) = 1.31  125nm  This index is smaller than the typical index of glass (about 1.5), so choose a material with an index of refraction of n = 1.31. If we now assume that the film has an index greater than glass, the wave that reflects from the glass has a phase change due to the additional path-length and no phase change on reflection, for a total  2t  phase change of 2 =   2 + 0.  n Repeat the calculation from above for this new phase change.  2t  m , m = 1, 2, ... net = 2 − 1 =   2 −  = ( 2m − 1)  → n = 2t  n



 655nm  = 12   = 2.62 2t  125nm  That is a very high index–higher than diamond. It probably is not realistic to look for this material. nmin =

77. We consider a figure similar to Fig. 34–12, but with the incoming rays at an angle of  i to the normal. Ray s2 will travel an extra distance l 1 = d sin i before reaching the slits, and an extra distance l 2 = d sin  after leaving the slits. There will be a phase difference between the waves due to the path difference l 1 + l 2 . When this total path difference is a multiple of the wavelength, constructive interference will occur. l 1 + l 2 = d sin  i + d sin  = m → m − sin  i , m = 0, 1, 2, sin  = d

i 

i

d s2

l 1

l 2



Since the rays leave the slits at all angles in the forward direction, we could have drawn the leaving rays with a downward tilt instead of an upward tilt. This would make the ray s2 traveling a longer distance from the slits to the screen. In this case the path difference would be l 2 − l 1 , and would result in the following expression.

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m + sin i , m = 0, 1, 2, d m l 1 − l 2 = d sin i − d sin  = m → sin  = − + sin i , m = 0, 1, 2, d l 2 − l 1 = d sin  − d sin i = m → sin  =

We combine the statements as follows. m sin  =  sin i , m = 0, 1, 2, d

Because of an arbitrary choice of taking l 2 − l 1 , we could also have formulated the problem so that the result would be expressed as sin  = sin i 

m , m = 0, 1, 2, d

78. The signals will be out of phase when the path difference equals an odd number of half-wavelengths. Let the 175-m distance be represented by d.

y 2 + d 2 − y = ( m + 12 )  , m = 0, 1, 2, 3, ... →

y 2 + d 2 = y + ( m + 12 )  →

d 2 − ( m + 12 )  2 y + d = y + 2 y (m + )  + (m + )  → y = 2 ( m + 12 )  We evaluate this for the first five values of m. The wavelength is c 3.00  108 m s = 44.12 m. = = f 6.8  106 Hz 2

d 2 − ( m + 12 )  2 (175m ) − ( m + 12 ) ( 44.12 m ) = = 683m, 198m, 84 m, 22 m, −22 m 2 ( m + 12 )  2 ( m + 12 )( 44.12 m ) 2

1 2 2

1 2

There are only four points on the positive y-axis where the signals are out of phase. They are at y = 22 m, 84 m, 200 m ( 2 sig. fig.) , and 680 m . 79. As explained in Example 34–6 the 12 - cycle phase change at the lower surface means that destructive interference occurs when the thickness t is such that 2t = m , m = 0, 1, 2, ... . Set m = 1 to find the smallest nonzero value of t. t = 12  = 12 ( 650nm ) = 325nm  330nm As also explained in Example 34–6, constructive interference will occur when 2t = ( m + 12 )  , m = 0, 1, 2, ... . We set m = 0 to find the smallest value of t:

t = 14  = 14 ( 650nm ) = 162.5nm  160nm 80. If we consider the two rays shown in the diagram, we see that the first ray passes through with no reflection, while the second ray has reflected twice. If nfilm  nglass , the first reflection from the glass produces a phase shift equivalent to 12 film , while the second reflection from the air produces no shift. When we compare the two rays at the film-glass surface, we see that the second ray has a total shift in phase, due to its longer path length (2t) and reflection ( 12 film ) . We set this path difference equal to an integer number of wavelengths for maximum intensity and equal to a half-integer number of wavelengths for minimum intensity. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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max: 2t + 12 film = mfilm , m = 1, 2, 3, ... →

t= 2

( m − 12 )  , m = 1, 2, 3, ... nfilm

min: 2t + 12 film = ( m + 12 ) film , m = 0, 1, 2, 3, ... →

1 m t= 2 , m = 0, 1, 2, 3, ... nfilm

At t = 0 , or in the limit t  nfilm , the transmitted beam will be at a minimum. Each time the thickness increases by a quarter wavelength the intensity switches between a maximum and a minimum. If nfilm  nglass , the first reflection from the glass produces no shift, while the second reflection from the air also produces no shift. When we compare the two rays at the film-glass surface, we see that the second ray has a total shift due solely to the difference in path lengths, 2t. For maxima, we have 1 m max: 2t = mfilm , m = 0, 1, 2, 3, ... → t = 2 , m = 0, 1, 2, 3, ... nfilm min: 2t = ( m − 12 ) film , m = 1, 2, 3, ... →

t= 2

( m − 12 )  , m = 1, 2, 3, ... nfilm

At t = 0 , or in the limit t  nfilm , the transmitted beam will be at a maximum. Each time the thickness increases by a quarter wavelength the intensity switches between a maximum and a minimum. In Section 34–5, it states that a single coating of film can reduce the reflection from about 4% to 1% of the incident light. The transmission, then would have a minimum value of about 96% of the incident light to a maximum of 99%. The minimum would not appear much different than the maxima. 81. Since the two sources are 180 out of phase, destructive interference will occur when the path length difference between the two sources and the receiver is 0, or an integer number of wavelengths. Since the antennae are separated by a distance of d =  / 2, the path length difference can never be greater than  / 2, so the only points of destructive interference occur when the receiver is equidistant from each antenna, that is, at destructive = 0 and 180 . Constructive interference occurs when the path difference is a half integer wavelength. Again, since the separation distance between the two antennas is d =  / 2, the maximum path length difference is  / 2, which occurs along the line through the antennae, therefore the constructive interference only occurs at constructive = 90 and 270 . As expected, these angles are reversed from those in phase, found in Example 34–5c. 82. The reflected wave appears to be coming from the virtual image, so this corresponds to a double-slit, with the separation being d = 2S. The reflection from the mirror produces a  phase shift, however, so the maxima and minima are interchanged, as described in Problem 13.   sin  max = ( m + 12 ) , m = 0, 1, 2, ; sin  min = m , m = 0, 1, 2, 2S 2S

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Instructor Solutions Manual

83. With respect to the incident wave, the wave that reflects from the top surface of the film has a phase change of 1 =  . With respect to the incident wave, the wave that reflects from the glass ( n = 1.52 ) at the bottom surface of the film has a phase change due to both the additional path length and reflection, so  2t  2 =   2 +  . For constructive interference, the net  film  phase change must be an even non-zero integer multiple of  .   2t    1 1 , m = 1, 2, 3, ... . net = 2 − 1 =   2 +   −  = m2 → t = 2 mfilm = 2 m nfilm  film   The minimum non-zero thickness occurs for m = 1. 643nm  = = 240 nm = 2.40  102 nm tmin = 2nfilm 2 (1.34 ) 84. The slit separation is d = 7.0  10−5 m, the wavelength of the light is  = 4.88  10−7 m, and the m distance to the screen is l = 4.0 m. To be very accurate, we will NOT assume that sin  = is d x d   equal to tan  = . Thus x = l tan  = l tan  sin −1  . Also, x0 = 0 (the location of the central l m   maximum).  4.88  10−7 m    −2 (a) x = x1 − x0 = l tan  sin −1  − 0 = ( 4.0m ) tan  sin −1  = 2.789  10 m −5 d 7.0  10 m   

 2.8  10−2 m 101    −1 101  (b) x = x101 − x100 = l tan  sin −1  − l tan  sin  d  d       101( 4.88  10−7 m )  100 ( 4.88  10 −7 m )   −1     = 7.677  10−2 m tan sin = ( 4.0 m )  tan  sin −1 −    7.0  10−5 m 7.0  10−5 m       = 7.7  10−2 m 85. (a) Let the initial unpolarized intensity be I 0 . The intensity of the polarized light after passing the first polarizer is I1 = 12 I 0 . Apply Eq. 34–8 to find the final intensity. I 2 = I1 cos2  = I1 cos2 90 = 0 . (b) Now the third polarizer is inserted. The angle between the first and second polarizers is 58, so the angle between the second and third polarizers is 32. It is still true that I1 = 12 I 0 .

I 2 = I1 cos2 58 = 12 I 0 cos2 58 ; I 3 = I 2 cos2 32 = 12 I 0 cos 2 58 cos 2 32 = 0.101I 0 → I3 = 0.10 I1 (c) The two crossed polarizers, which are now numbers 2 and 3, will still not allow any light to pass I through them if they are consecutive to each other. Thus 3 = 0 . I1 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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86. (a) We apply Eq. 34–8 through the successive polarizers. The initial light is unpolarized. Each polarizer is then rotated 30.0 from the previous one. I1 = 12 I 0 ; I 2 = I1 cos 2  2 = 12 I 0 cos 2  2 ; I 3 = I 2 cos 2 3 = 12 I 0 cos 2  2 cos 2 3 ;

I 4 = I 3 cos 2  4 = 12 I 0 cos 2  2 cos 2 3 cos 2  4 = 12 I 0 cos 2 30.0 cos 2 30.0 cos 2 30.0 = 0.211 I 0 (b) If the second polarizer is removed, then the angle between polarizers # 1 and # 3 is now 60.0 I1 = 12 I 0 ; I 3 = I1 cos2 3 = 12 I 0 cos2 3 ;

I 4 = I 3 cos2 4 = 12 I 0 cos2 3 cos2  4 = 12 I 0 cos2 60.0 cos2 30.0 = 0.0938 I 0 The same value would result by removing the third polarizer, because then the angle between polarizers # 2 and # 4 would be 60 The intensity can be decreased by removing either the second or third polarizer. (c) If both the second and third polarizers are removed, there are still two polarizers with their axes perpendicular, so no light will be transmitted. 87. For both configurations, we have d sin  = m. The angles and the orders are to be the same. The slit separations and wavelengths will be different. Use the fact that frequency and wavelength are related by v = f . The speed of sound in room-temperature air is given in Chapter 16 as 343 m/s. We use subscripts of “L” or light and “S” for sound. sin  L S d sin  = m → = = = → m d d L dS vS  343m s   4.6  1014 Hz  f v f S dS = d L = d L S = d L S L = (1.0  10−4 m )   = 200.7 m  8 vL vL fS L  3.00  10 m s  262 Hz  fL  2.0  102 m

88. Let I 0 be the initial intensity. Use Eq. 34–8 for both transmissions of the light. I1 = I 0 cos2 1 ; I 2 = I1 cos2  2 = I 0 cos2 1 cos2  2 = 0.25 I 0 →  0.25   0.25  −1  = cos   = 29  cos  2   cos55 

1 = cos −1 

89. The reduction being investigated is that which occurs when the polarized light passes through the second Polaroid. Let I1 be the intensity of the light that emerges from the first Polaroid, and I 2 be the intensity of the light after it emerges from the second Polaroid. Use Eq. 34–8. (a) I 2 = I1 cos2  = 0.25I1 →  = cos −1 0.25 = 60 (b)

I 2 = I1 cos2  = 0.10 I1 →  = cos −1 0.10 = 72

(c)

I 2 = I1 cos2  = 0.010 I1 →  = cos −1 0.010 = 84

90. The path difference to a point on the x-axis from the two sources is d = d 2 − d1 = x 2 + d 2 − x. For the two signals to be out of phase, this path difference must be an odd number of half-wavelengths, so d = ( m + 12 )  , m = 0, 1, 2, . Also, the maximum path difference is d = 3 . Thus the path difference must be 12  , 23  , or 25  for the signals to be out of phase (m = 0, 1, or 2). We solve for x for the three path differences. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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d = x 2 + d 2 − x = ( m + 12 )  →

Instructor Solutions Manual

x 2 + d 2 = x + ( m + 12 )  →

x 2 + d 2 = x 2 + 2 x ( m + 12 )  + ( m + 12 )  2 → 2

d 2 − ( m + 12 )  2 9 2 − ( m + 12 )  2 9 − ( m + 12 ) = =  x= 2 ( m + 12 )  2 ( m + 12 )  2 ( m + 12 ) 2

9 − ( 0 + 12 ) 9 − (1 + 12 )    = 2.25 = = = 8.75 ; 1: m = 0: x = m x 2 ( 0 + 12 ) 2 (1 + 12 ) 2

9 − ( 2 + 12 )  = 0.55 m = 2: x = 2 ( 2 + 12 ) 2

91. In order for the two reflected halves of the beam to be 180 out of phase with each other, the minimum path difference (2t) should be 12  in the plastic. Notice that there is no net phase difference between the two halves of the beam due to reflection, because both halves reflect from the same material.   780 nm → t= = = 126 nm 2t = 12 n 4n 4 (1.55) 92. We have N polarizers providing a rotation of 90. Thus, each polarizer must rotate the light by an angle of  N = ( 90 N ) . As the light passes through each polarizer, the intensity will be reduced by a factor of cos2  N . Let the original intensity be I 0 . I1 = I 0 cos2  N ; I 2 = I1 cos2  N = I 0 cos4  N ; I 3 = I 2 cos2  N = I 0 cos6  N I N = I 0 ( cos N )

= 0.90 I 0 → cos ( 90 N )

We evaluate cos ( 90 N )

= 0.90

for various values of N. A table for a few

values of N is shown here. We see that N = 24 satisfies the criteria, and so  N = ( 90 24 N )  = ( 90 24 N )  = 3.75. So we need to put 24 polarizers in the path of the original polarized light, each rotated 3.75 from the previous one . 93. About 4% of light is reflected off each surface. Therefore of 100% of the light incident on the front surface only 96% of that light reaches reach the back surface. 4% of this value or .04(0.96) = 3.84% reflects off the back surface, so after one lens only 100% – 4% – 3.94% = 92.2% of the light is transmitted. Thus for each lens only 92% of the light incident on the lens would transmit through the lens. After multiple lenses the transmission can become small. If 12 there are 6 lenses (12 surfaces) then only ( 0.96 ) = 0.61 or 61% is transmitted. To minimize this degradation, it is suggested to coat the lens with a thin film that will cause destructive interference for the reflected light at the center of the visible spectrum. Since the light cannot reflect, a greater percentage of light is transmitted. For the lens in air, this reduces the reflected light from 4% to 1%. For six lenses (12 surfaces) this increases the transmission percentage from 61% to 89%, a significant improvement. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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94. Light traveling from a region 12 from the vertical would have to travel a slightly longer distance to reach the far antenna. Using trigonometry we calculate this distance, as was done in Young’s double-slit experiment. Dividing this additional distance by the speed of light gives us the necessary time shift. d l sin  ( 55 m ) sin12 t = = = = 3.81 10−8 s = 38.1 ns 8 3.00  10 m/s c c

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CHAPTER 35: Diffraction Responses to Questions 1.

Radio waves have a much longer wavelength than visible light and will diffract around normal-sized objects (like hills). The wavelengths of visible light are very small and will not diffract around normal-sized objects. Thus diffraction allows the radio waves to be “picked up” even if the broadcasting tower is not visible on a line of sight.

You see a pattern of dark and bright lines parallel to your fingertips, in the narrow opening between your fingers, due to diffraction.

Light from all points of an extended source produces diffraction patterns, and these many different diffraction patterns overlap each other so that no distinct pattern can be easily seen. When using white light, the diffraction patterns of the different wavelengths will overlap because the locations of the fringes depend on wavelength. Monochromatic light will produce a more distinct diffraction pattern.

(a) When you increase the slit width in a single-slit diffraction experiment, the spacing of the m fringes decreases. The equation for the location of the minima, sin  = , indicates that  is D decreased for a particular m and  when the width D increases. This means that the bright spots on the screen are more closely packed together for a wider slit. (b) When you increase the wavelength of light used in a single-slit diffraction experiment, the m spacing of the fringes increases. The equation for the location of the minima, sin  = , D indicates that  is increased for a particular m and D when the wavelength increases. This means that the bright spots on the screen are spread further apart for a longer wavelength.

(a) A slit width of 50 nm would produce a central maximum so spread out that it would cover the entire width of the screen. No minimum (and therefore no diffraction pattern) would be seen, because the first minimum would have to satisfy sin  =



 10, which is not possible. The D different wavelengths will all overlap, so the light on the screen will be white. The light on the screen will also be dim compared to the source, because it is spread out over a large area, reducing the intensity. (b) For the 50,000-nm slit, the central maximum will be very narrow, about one degree in width for the blue end of the spectrum and about one and a half degrees for the red. The diffraction pattern will not be distinct, because most of the intensity will be in the small central maximum and the fringes for the different wavelengths of white light will not coincide. 6.

 

(a) If the apparatus is immersed in water, the wavelength of the light will decrease    =

  and n

the diffraction pattern will become more compact. (b) If the apparatus is placed in a vacuum, the wavelength of the light will increase very slightly, and the diffraction pattern will also spread out very slightly.

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 sin  2  The intensity pattern is actually a function of the form I = I 0   (see Eqs. 35–7 and 35–8),   2  2 where  = D sin  . The maxima of the intensity function do not coincide exactly with the  maxima of sin  2 , which is why the first maximum do not occur exactly at sin  = 32  D. Mathematical details are given in the solution to Problem 14.

Similarities: double-slit interference and single-slit diffraction are both caused by the interference of light, as superposition of coherent waves of different paths. Both have a regular pattern of light and dark fringes (alternating intensity maxima and minima). They both have a maximum at the center point. The sine of the angle of diffraction in each case is proportional to the wavelength of light and inversely proportional to the slit separation (for double-slit interference) or the slit width (for singleslit diffraction). Differences: in double-slit interference the central maximum has the same angular width as each of the other maxima. In single-slit diffraction the central maximum is much wider than any of the other maxima. Double-slit interference has simple equations for the location of both the maxima and minima, and the maxima are located exactly between the minima. Single-slit diffraction only has a simple equation for the location of the minima. The maxima are not located exactly at the midpoint between each minimum. Maxima for the double slit interference pattern would be equally bright (if single-slit diffraction could be ignored), but higher-order maximum for single-slit diffraction get progressively dimmer.

No. D represents the slit width and d the distance between the centers of the slits. It is possible for the distance between the slit centers to be greater than the width of the slits; it is not possible for the distance between the slit centers to be less than the width of the slits.

10. (a) Increasing the wavelength, λ, will spread out the diffraction pattern, since the locations of the minima are given by sin θ = mλ/D. The interference pattern will also spread out; the interference maxima are given by sin θ = mλ/d. The number of interference fringes in the central diffraction maximum will not change. (b) Increasing the slit separation, d, will decrease the spacing between the interference fringes without changing the diffraction, so more interference maxima will fit in the central maximum of the diffraction envelope. (c) Increasing the slit width, D, will decrease the angular width of the diffraction central maximum without changing the interference fringes, so fewer bright fringes will fit in the central maximum. 11. (a) Immediately below Eq. 35–10 the text says that the resolution limit due to diffraction also applies to spherical mirrors. Mirrors are used to make enlarged images to see more detail, but there will be a resolution limit due to diffraction of the light. (b) Plane mirrors are not used for magnification, but the human eye (or a camera lens, for example) would limit the resolution that could be achieved by a plane mirror. As stated in Section 35–6, “It is not possible to resolve detail of objects smaller than the wavelength of the radiation being used.” So any images that are formed will not be able to show details that are smaller than the wavelength of light being used.

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12. Yes. Diffraction effects will occur for both real and virtual images. Even virtual images are created due to the directions of real light rays, such as passing through lenses. In particular, it is not possible to resolve details of objects that are smaller than the wavelength of the radiation being used. So any image will have diffractive effects. 13. A large mirror has better resolution (as evidenced by Eq. 35–10, with a large D) and gathers more light than a small mirror, thus revealing fainter objects. Large mirrors can be relatively thin and thus be lighter than large lenses, while the large lenses would be very thick. The mass of the large lens can cause the lens to sag under its own weight. Large mirrors can be supported at many points across the full back side of the mirror, while a large lens must be supported just at its edges. A mirror does not cause dispersion of light like a lens does, which means the images from a mirror do not have chromatic aberration. 14. No. The resolving power of a lens is on the order of the wavelength of the light being used, so it is not possible to resolve details smaller than the wavelength of the light. Atoms have diameters of about 10−8 cm and the wavelength of visible light is on the order of 10−5 cm. 15. Violet light would give the best resolution in a microscope, because according to the Rayleigh criterion, a smaller wavelength gives better resolution. Violet light has the shortest wavelength of visible light. 16. Yes. (See the opening paragraph of Section 35–8.) The analysis for a diffraction grating of many slits is essentially the same as for Young’s double-slit interference. However, the bright maxima of a multiple-slit grating are much sharper and narrower than those in a double-slit pattern. 17. The answer depends on the slit spacing of the grating being used. If the spacing is small enough, only the first order will appear so there will not be any overlap. For wider slit spacing there can be overlap. If there is overlap, it will be the higher orders of the shorter wavelength light overlapping with lower orders of the longer wavelength light. See, for instance, Example 35–9, which shows the overlap of the third order blue light with the second order red light. 18. The interference pattern created by the diffraction grating with 104 slits/cm has bright maxima that are more sharply defined and narrower than the interference pattern created by the two slits 10−4 cm apart. The spacing of the bright maxima would be the same in both patterns, but for the grating, each maximum would be essentially the same brightness, while for the 2-slit pattern, slit-width effects would make the maxima for m > 1 much less bright than the central maximum. 19. The Rayleigh criterion gives us the resolution limit for two objects a distance D apart when using 1.22 light of wavelength :  = , which can be interpreted as the smaller the angle, the better the D 1.22 ( 450 nm ) 1.22 ( 700 nm ) resolution. Looking at the two wavelengths given,  blue = and  red = , D D 1.22 ( 450 nm )  450 D we have blue = = = 0.64. This could be expressed as saying that the resolution  red 1.22 ( 700 nm ) 700

D with blue light is

1 0.64

= 1.56 times the resolution with red light.

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20. Radio telescopes are large primarily because the wavelength of radio waves is large–on the order of meters to hundreds of meters. To get better resolution with a large wavelength, the Rayleigh criterion says that a large aperture is needed. Making such large optical telescopes is not done for several reasons: i) The theoretical resolution obtained by such a large aperture would not be achievable because of atmospheric turbulence that does not affect radiotelescopes; ii) the much smaller wavelength of light means that the mirror has to be more perfectly shaped than the radiotelescope, and that is very difficult to do with huge mirrors. Some large-aperture optical telescopes (such as the “Very Large Telescope” in Chile) are now being made with multiple small mirrors being combined to provide some of the capabilities of a single large mirror. 21. (a) The advantage of having many slits in a diffraction grating is that this makes the bright maxima in the interference pattern more sharply defined, brighter, and narrower. It also allows more light through the grating (as opposed to a simple double-slit apparatus), and so the interference pattern is inherently brighter for a given source intensity. (b) The advantage of having closely-spaced slits in a diffraction grating is that this spreads out the bright maxima in the interference pattern and makes them easier to resolve. 22. (a) Violet light will be at the top of the rainbow created by the diffraction grating. Principal m maxima for a diffraction grating are at positions given by sin  = . Violet light has a shorter d wavelength than red light and so will appear at a smaller angle away from the direction of the horizontal incident beam. (b) Red light will appear at the top of the rainbow created by the prism. The index of refraction for violet light in a given medium is slightly greater than for red light in the same medium, and so the violet light will bend more and will appear farther from the direction of the horizontal incident beam.

Responses to MisConceptual Questions 1.

(d) In Eq. 35–2a the angles of destructive interference are proportional to the wavelength and inversely proportional to the slit width. Therefore if the slit width and wavelength are changed by the same factor, the diffraction pattern will not change.

(b) A common misconception is that the center of the shadow will be the darkest. However, since the center of the disk is equidistant from all points on the edge of the disk, the diffracting light constructively interferes at the center making a bright point. This bright point is independent of the wavelength of the light.

(b) It might be common to think that the lines are due to the inability of the eye to focus on the point. However, the lines are due to diffraction of the light as it passes through the very small opening between your fingers.

(a) The cause of single-slit diffraction can be difficult to understand. The slit can be divided into a number of individual regions. If the path difference between each pair of regions and a point on the screen is a half wavelength, then destructive interference occurs at that point.

(d) This is the “classic” single-slit diffraction pattern, as illustrated in Fig. 35–5, or in the adjacent figure. In the figure, it is more of an “oval” than a “rectangle.”

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Instructor Solutions Manual

(b) The width of the fringe is inversely proportion to the width of the slit, as given by Eq. 35–1. So as the slit gets narrower, the fringe gets wider.

(c) It can be difficult to recognize the different scales of length involved with the diffraction of sound vs. the diffraction of light. The sound waves of a person’s voice have wavelengths about the same size as a doorway and therefore diffract easily around the corner. The wavelength of light is much smaller, approximately 10 −7 m, and so light does not diffract around the corner.

(b) The spreading out of the light is due to diffraction, with diffraction being significant when the slit size is about the same size as the wavelength of light. For visible light the wavelength is about 5  10 −7 m, which is on the order of the slit width. Therefore the light will spread out wider than the slit. The height opening, which is 4 orders of magnitude larger than the wavelength, will have almost no discernible diffraction, and so the light will remain about the same height.

(d) The width of the fringe is inversely proportional to the width of the slit, as given by Eq. 35–1. So as the slit gets wider, the fringes get narrower, and the pattern spacing will be compressed.

10. (a) Red light has a longer wavelength than blue light and therefore will not have a better resolving power, since resolving power is inversely proportional to the wavelength. The resolving power is related to the angular separation of the objects. Two objects separated by the same distance would have a larger angular separation when they are closer to the observer, and therefore a better resolution when they are closer to the observer. Objects that are farther apart from each other are easier to resolve. Therefore, the only correct answer is that the larger the lens diameter the better resolving power. 11. (d) The limit on magnification using visible light is not due to the lenses, but due to the nature of light. Diffraction limits resolution to the wavelength of the light used. For visible light the wavelength is about 500 nm. Crystal structures are much smaller (1–10 nm) in size and require X-rays to resolve their structure. 12. (a) It is illustrated in Example 35–6 that the resolution of the eye is dependent on the diameter of the pupil. 13. (a) The resolving power is inversely proportional to the wavelength of the light. Of the four types of light listed, ultraviolet light has the smallest wavelength and therefore the greatest resolving power.

Solutions to Problems 1.

We use Eq. 35–1 to calculate the angular distance from the middle of the central peak to the first minimum. The width of the central peak is twice this angular distance.  680  10−9 m    sin 1 = → 1 = sin −1   = sin −1   = 1.012 −3 D D  0.0385  10 m   = 21 = 2 (1.012 ) = 2.024  2.0

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The angle from the central maximum to the first dark fringe is equal to half the width of the central maximum. Using this angle and Eq. 35–1, we calculate the wavelength used. 1 = 12  = 12 ( 29.0 ) = 14.5

sin 1 = 3.

 D

→  = D sin 1 = ( 2.60  10−3 mm ) sin (14.5 ) = 6.509  10 −4 mm = 651nm

Because the width of the pattern is much smaller than the distance to the screen, the angles from the diffraction pattern for this first order will be small. Thus, we may make the approximation that sin  = tan  . We find the angle to the first minimum from the distances, using half the width of the full first-order pattern. Then we use Eq. 35–2 to find the slit width. ( 9.60cm ) = 0.01684 = sin  tan 1min = 12 1min ( 285cm ) D sin  = m → D =

(1)( 415nm ) = 2.464  104 nm  2.46  10−5 m m = sin  0.01684

3 . We then find the distance on the screen from the 2D central maximum by multiplying the distance to the screen by the tangent of the angle.  3 ( 580  10−9 m )   3  −1  max = sin −1  = sin   = 14.39  −6  2D   2 ( 3.50  10 m ) 

The first bright region occurs at sin  

x = l tan 1 = (10.0 m ) tan (14.39 ) = 2.566 m  2.6 m 5.

(a) We use Eq. 35–2, using m = 1,2,3,… to calculate the possible diffraction minima, when the wavelength is 0.50 cm.  m  D sin  m = m →  m = sin −1    D 

 1  0.50cm   = 16.1  1.8cm 

 2 = sin −1 

 3  0.50cm   = 56.4  1.8cm 

 4 = sin −1 

1 = sin −1 

3 = sin −1 

 2  0.50cm   = 33.7  1.8cm 

 4  0.50cm   → no solution  1.9cm 

There are three diffraction minima: 16, 34, and 56 . (b) We repeat the process from part (a) using a wavelength of 1.0 cm.  1  1.0cm   2  1.0cm  1 = sin −1  2 = sin −1   = 33.7  = no real solution  1.8cm   1.8cm  The only diffraction minima is at 34 . (c) We repeat the process from part (a) using a wavelength of 3.0 cm.  1  3.0cm  1 = sin −1   = no real solution  1.8cm  There are no diffraction minima.

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The angle from the central maximum to the first bright maximum is half the angle between the first bright maxima on either side of the central maximum. The angle to the first maximum is about halfway between the angles to the first and second minima. We use Eq. 35–2b, setting m = 3 2, to calculate the slit width, D. 1 = 12  = 12 ( 32 ) = 16

D sin  m = m → D = 7.

Instructor Solutions Manual

( 3 2 )( 633 nm ) = 3445nm  3.4  m m = sin 1 sin16

We use the distance to the screen and half the width of the diffraction maximum to calculate the angular distance to the first minimum. Then, using this angle and Eq. 35–1, we calculate the slit width. Then, using the slit width and the new wavelength we calculate the angle to the first minimum and the width of the diffraction maximum. We assume that the angles are small enough that the sine and tangent are interchangeable. ( 1 y ) ( 1 y ) ( 1  0.06 m ) = 0.6875 tan 1 = 2 1 → 1 = tan −1 2 1 = tan −1 2 2.50 m l l 580 nm   sin 1 = 1 → D = 1 = = 48,337 nm D sin 1 sin 0.6875

sin  2 =

2 D

 460 nm    →  2 = sin −1  2  = sin −1   = 0.5453 D    48,337 nm 

y2 = 2 l tan  2 = 2 ( 2.50 m ) tan ( 0.5453 ) = 0.04758m  4.8cm 8.

(a) There will be no diffraction minima if the angle for the first minimum is greater than 90. We set the angle in Eq. 35–1 equal to 90 and solve for the slit width.   sin  = → D = =  D sin 90 (b) For no visible light to exhibit a diffraction minimum, the slit width must be equal to the shortest visible wavelength, which is about 400 nm. D = min = 400 nm

We set the angle to the first minimum equal to half of the separation angle between the dark bands. We insert this angle into Eq. 35–1 to solve for the slit width.  = 12  = 12 ( 51.0 ) = 25.5

sin  =

 D

→ D=

 sin 

440 nm = 1022 nm  1.0  10−6 m sin 25.5

10. We find the angle to the first minimum using Eq. 35–1. The distance on the screen from the central maximum is found using the distance to the screen and the tangent of the angle. The width of the central maximum is twice the distance from the central maximum to the first minimum.  450  10−9 m    sin 1 = → 1 = sin −1   = sin −1   = 0.2578 −3 D D  0.10  10 m  y1 = l tan 1 = ( 5.0 m ) tan 0.2578 = 0.02250 m

y = 2 y1 = 2 ( 0.02250 m ) = 0.0450 m = 4.5cm

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11. (a) For vertical diffraction we use the height of the slit (1.5 m) as the slit width in Eq. 35–1 to calculate the angle between the central maximum to the first minimum. The angular separation of the first minima is equal to twice this angle.   680  10−9 m → 1 = sin −1 = sin −1 = 26.96 sin 1 = D D 1.5  10−6 m  = 21 = 2 ( 26.96 )  54 (b) To find the horizontal diffraction we use the width of the slit (3.0 m) in Eq. 35–1.   680  10−9 m → 1 = sin −1 = sin −1 = 13.10 sin 1 = D D 3.0  10−6 m  = 21 = 2 (13.10 )  26 12. (a) If we consider the slit made up of N wavelets each of amplitude E0 , the total amplitude at the central maximum where they are all in phase, is NE0 . Doubling the size of the slit doubles the number of wavelets and thus the total amplitude of the electric field. Because the intensity is proportional to the square of the electric field amplitude, the intensity at the central maximum is increased by a factor of 4. 2 I  E 2 = ( 2 E0 ) = 4 E02  4 I 0 (b) From Eq. 35–1 we see that, for small angles, the width of the central maximum is inversely proportional to the slit width. Therefore, doubling the slit width will cut the area of the central peak in half. Since the intensity is spread over only half the area, where the intensity is four times the initial intensity, the average intensity (or energy) over the central maximum has doubled. This is true for all fringes, so when the slit width is doubled, allowing twice the energy to pass through the slit, the average energy within each slit will also double, in accord with the conservation of energy. 13. We use Eq. 35–8 to calculate the intensity, where the angle  is found from the displacement from the central maximum (18 cm) and the distance to the screen. y  18cm  tan  = →  = tan –1   = 35.75 l  25cm 

= I

2



D sin  = 2

(1.0  10 m) sin 35.75 = 4.895 rad (750  10 m) 2

−6

−9

 sin  2   sin ( 4.895 rad 2 )   =  = 0.06831  0.068   2   4.895 rad 2 

=

I0 So the light intensity at 18 cm is about 6.8% of the maximum intensity.

14. (a) The secondary maxima do not occur precisely where sin (  / 2 ) is a maximum, that is at

 / 2 = (m + 12 ) where m = 1, 2,3,..., because the diffraction intensity (Eq. 35–7) is proportional  sin (  2 )  to   . Near the maximum of the sine function, the denominator of the intensity   2  function causes the intensity to decrease more rapidly than the sine function causes it to increase. This results in the intensity reaching a maximum slightly before the sine function reaches its maximum. 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) We set the derivative of Eq. 35–7 with respect to  equal to zero to determine the intensity extrema.  sin (  2 )   sin (  2 )   cos (  2 ) sin (  2 )  dI d 0= = Io  −  = 2Io     2 2  d d   2    2  2

When the first term in brackets is zero, the intensity is a minimum, so the intensity is a maximum when the second term in brackets is zero. cos (  2 ) sin (  2 ) − →  2 = tan (  2 ) 0=  2 2 (c) The first and secondary maxima are found where these two curves intersect, or 1 = 8.987 and

 2 = 15.451. We calculate the percent difference between these and the maxima of the sine curve, 1 = 3 and  2 = 5 . 

 1 

 2

1 − 1 8.987 − 3 = = −0.0464 = −4.64% 3 1

15.451 − 5 = −0.0164 = −1.64% 5

15. If the central diffraction peak contains eleven fringes, there will be the central interference fringe and then five interference fringe maxima (m = 1, 2, 3, 4, 5) on each side of the central peak. Since the first diffraction minimum occurs at an interference minimum, we use Eq. 34–2b with m = 5 to find the angle of the m = 5 interference minimum and set that angle equal to the first diffraction minimum, given by Eq. 35–1, to solve for the ratio of the slit separation to slit width. 5.5  5.5 → d = 5.5D d sin  = ( m + 12 )  → sin  = ; sin  = = d D d 16. (a) If the central diffraction peak is to contain fifteen fringes, there will be seven fringes on each side of the central peak. Thus, the m = 7 minimum of the double-slit pattern must coincide with the first minimum of the diffraction pattern. We use Eq. 34–2b with m = 7 to find the angle of that interference minimum and set that angle equal to the first diffraction minimum, given by Eq. 35–1, to solve for the ratio of the slit separation to slit width. ( 7 + 12 )  = 7.5 d sin  = ( m + 12 )  → sin  = d d  7.5 sin  = = → d = 7.5D D d Therefore, for the first diffraction minimum to be at the m = 7, the eighth interference minimum, the separation of slits should be 7.5 times the slit width.

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(b) We still see 15 fringes, so the first diffraction minimum is to occur at the m = 8 side interference maximum. We use Eq. 34–2a with m = 8 to find the angle of the interference maximum and set that angle equal to the first diffraction minimum, given by Eq. 35–1, to solve for the ratio of the slit separation to slit width. 8 8  8 = → d = 8D d sin  = m → sin  = ; sin  = = d d D d Therefore, for the first diffraction minimum to be at the m = 8 interference maximum, the separation of slits should be 8 times the slit width. 17. In this problem we will be using integers for both double-slit extrema and diffraction minima. Let us use m for double-slit extrema, and m  for diffraction minima. In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, there is the m = 0 central fringe, along with fringes up to m = 6 on each side of  = 0. Then, at angle  , the m = 6 interference minimum coincides with the first diffraction minima, m = 1. We set this angle in Eqs. 34–2b and 35–2 equal to solve for the relationship between the slit width and separation. 6.5   6.5 → d = 6.5D d sin  = ( m + 12 )  → sin  = ; sin  = m = (1) = d D D d Now we use these equations again to find the m value at the second diffraction minimum, using Eq. 35–2 with m = 2. m d 6.5D  m sin  2 = =2 = → m=2 =2 = 13 D D d D D Thus, the seven fringes corresponding to m = 7 to m = 13 will occur within the first and second diffraction minima. 18. (a) The angle to each of the maxima of the double slit are given by Eq. 34–2a. The distance of a fringe on the screen from the center of the pattern is equal to the distance between the slit and screen multiplied by the tangent of the angle. For small angles, we can set the tangent equal to the sine of the angle. The slit spacing is found by subtracting the distance between two adjacent fringes. m m ym = l tan  m  l sin  m = l sin  m = d d ( m + 1)  − l m = l  = (1.0 m ) ( 520  10−9 m ) = 0.017 m y = ym +1 − ym = l d d d 0.030  10−3m (b) We use Eq. 35–1 to determine the angle between the center and the first minimum. Then, by multiplying the distance to the screen by the tangent of the angle, we find the distance from the center to the first minima. The distance between the two first-order diffraction minima is twice the distance from the center to one of the minima.   520  10−9 m → 1 = sin −1 = sin −1 = 2.981 sin 1 = D D 0.010  10−3m y1 = l tan 1 = (1.0 m ) tan 2.981 = 0.05207 m y = 2 y1 = 2 ( 0.05207 m ) = 0.104 m  0.10 m

19. We set d = D in Eqs. 34–4 and 35–6 to show  =  . Replacing  with  in Eq. 35–9 and using the double angle formula, we show that Eq. 35–9 reduces to Eq. 35–7, with   = 2  . Finally, using Eq. 35–6 again, we show that   = 2 implies that the new slit width D  is simply double the initial slit width. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

=

2



d sin  =

2



Instructor Solutions Manual

D sin  = 

1 sin 2 (  2 ) cos 2 (  2 )  sin  2  4 sin   2 (  2 )  2  cos 2 I = I 0  I I = = ( ) 0  0 2 2 (  2) (  2)   2  2

= I0

 =

sin 2 

2

= I0

sin 2 (   2 )

(   2)

, where   = 2 .

2

 2  D sin  = 2  D sin   → D = 2 D    

20. Using Eq. 34–2a we determine the angle at which the third-order interference maximum occurs. Then we use Eq. 35–9 to determine the ratio of the intensity of the third-order maximum, where  is given by Eq. 35–6 and  is given by Eq. 34–4.  m   −1  3 d sin  = m   = sin −1   = sin   = 4.301  40.0 d      40.0  5 ( ) sin 4.301 = 1.885 rad  2 = D sin  = ( ) 2 2   ( 40.0 )  2 d = sin  = sin ( 4.301 ) = 9.424 rad 2 2   sin (  2 )        sin (1.885 rad )  2 I = Io   cos    = I o   cos ( 9.424 )  = 0.255 I o   2    2   1.885 rad  2

21. We use Eq. 34–2a to determine the order of the double-slit maximum that corresponds to the same angle as the first order single slit minimum, from Eq. 35–1. Since this double-slit maximum is darkened, inside the central diffraction peak, there will be the zeroth-order fringe, and on either side of the central peak a number of maximum equal to one less than the double-slit order. Therefore, there will be 2(m – 1)+1, or 2m – 1 fringes. d sin  d    d d d sin  = m → m = ; N = 2m − 1 = 2 − 1 =  =  D D D (a) We first set the slit separation equal to twice the slit width, d = 2.00 D. 2.00 D N =2 −1 = 3 D (b) In part (a), the ratio of slits is an integer value. This corresponds to the single-slit minimum overlapping the double-slit maximum. Now that d = 4.50 D, the single slit minimum overlaps a double-slit minimum. Thus, the last order maximum, m = 4, is not darkened and N = 2m + 1. N = 2m + 1 = 2(4) + 1 = 9 (c) In this case the ratio of the slit separation to slit width is not an integer or a half-integer value. The first order single-slit minimum falls between the seventh-order maximum and the seventhorder minimum. Thus, the seventh-order maximum will partially be seen as a fringe. N = 2m + 1 = 2(7) + 1 = 15 (d) Set d = 16.0 D and proceed as in part (a). 16.00 D N =2 − 1 = 31 D

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22. (a) If D  , the central maximum of the diffraction pattern will be very wide. Thus we need consider only the interference between slits. We 2 d sin  as construct a phasor diagram for the interference, with  =  the phase difference between adjacent slits. The magnitude of the electric fields of the slits will have the same magnitude, E10 = E20 = E30 = E0 . From the symmetry of the phasor diagram we see that  =  . Adding the three electric field vectors yields the net electric field. E 0 = E10 cos  + E20 + E30 cos  = E0 (1 + 2cos  )

E30

 E 0

E 20



E10

The central peak intensity occurs when  = 0. We set the intensity proportional to the square of the electric field and calculate the ratio of the intensities. 2 (1 + 2cos  ) I E20 E0 (1 + 2cos  ) = 2 = 2 = 2 9 I 0 E00 E0 (1 + 2cos0 ) 2

(b) We find the locations of the maxima and minima by setting the first derivative of the intensity equal to zero. 2I dI d I0 2 = (1 + 2cos  ) = 0 (1 + 2cos  )( −2sin  ) = 0 d d 9 9 This equation is satisfied when either of the terms in parentheses is equal to zero. When 1 + 2cos  = 0, the intensity equals zero and is a minimum. 2 4 8 10 1 + 2cos  = 0 →  = cos −1 ( − 12 ) = , , , ,... 3 3 3 3 Maxima occur for sin  = 0, which also says cos  = 1. sin  = 0 →  = sin −1 0 = 0,  , 2 ,3 ,... When cos  = 1, the intensity is a principal maximum. When cos  = −1, the intensity is a secondary maximum. 2 2 1 + 2cos  ) 1 + 2cos 0 ) ( ( = I0 = I0 I (0) = I 0 9 9 I ( ) = I 0

(1 + 2cos  ) 9

(1 + 2 ( −1) ) = I =I 2

I (2 ) = I 0

(1 + 2cos 2 ) = I (1 + 2 ) = I 2

9 9 Thus, we see that, since cos  alternates between +1 and –1, there is only a single secondary maximum between each principal maximum.

23. The angular resolution is given by Eq. 35–10. 560  10−9 m   180  3600"  = 2.69  10−7 rad   = 1.22 = 1.22   = 0.055" D 254  10−2 m   rad  1  This is also equal to (1.54  10−5 )  . 24. The resolving power is given by Eq. 35–11. ( 550 nm )(8mm ) = 1074 nm  1000 nm = 1 m f RP = 1.22 = 1.22 D ( 5mm ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

25. The angular resolution is given by Eq. 35–10. The distance between the stars ( l ) is the angular resolution ( ) times the distance to the stars from the Earth ( r ) .

 = 1.22

 D

; l = r = 1.22

r = 1.22 D

 9.46  1015 m  550  10−9 m ) (  1ly   = 1.7  1011 m ( 0.66 m )

(18 ly ) 

26. We find the angle  subtended by the planet by dividing the orbital radius by the distance of the star to the earth. Then using Eq. 35–10, we calculate the minimum diameter aperture needed to resolve this angle. r 1.22 → = = d D −9 15 1.22 d 1.22 ( 550  10 m ) ( 4ly ) ( 9.461  10 m/ly ) = = 0.17 m  20cm D= r (1AU ) (1.496  1011 m/AU ) 27. We find the angular half-width of the flashlight beam using Eq. 35–10 with D = 5 cm and  = 550 nm. We set the diameter of the beam equal to twice the radius, where the radius of the beam is equal to the angular half-width multiplied by the distance traveled by the beam, 3.84  108 m. –9 1.22 1.22 ( 550  10 m ) = = 1.3  10 –5 rad = D 0.050 m d = 2 ( r ) = 2 ( 3.84  108 m )(1.3  10 –5 rad ) = 1.0  104 m

28. To find the focal length of the eyepiece we use Eq. 33–7, where the objective focal length is 2.0 m, ′ is the ratio of the minimum resolved distance and 25 cm, and  is the ratio of the object on the moon and the distance to the moon. We ignore the inversion of the image. ( d l ) = 2.0 m ( 6.5 km 384,000 km ) = 0.0846m  8.5cm fo    = → fe = fo = fo o ( )  fe  (d N ) ( 0.10 mm 250 mm ) We use Eq. 35–10 to determine the resolution limit. 560  10−9 m  = 6.2  10−6 rad  = 1.22 = 1.22 D 0.11 m This corresponds to a minimum resolution distance, r = ( 384,000km ) ( 6.2  10−6 rad ) = 2.4km , which is smaller than the 6.5 km object we wish to observe. 29. We set the resolving power as the focal length of the lens multiplied by the angular resolution, as in Eq. 35–11. The resolution is the inverse of the resolving power. −1 1 25 mm D 1.22 f  = = = = 730 lines/mm  1.22 f 1.22 ( 560  10−6 mm ) ( 50.0 mm ) RP ( f / 2)  D 

1 3.0 mm = = 88 lines/mm RP ( f /16) 1.22 ( 560  10−6 mm ) ( 50.0 mm )

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30. The resolving power of a microscope is given by Eq. 35–11. We use a wavelength of 550 nm as was done in Exercise D. RP = 1.22

f D

= 1.22

( 550 nm ) (8.25  10−3 m ) 6.0  10−3 m

= 922 nm  920 nm

31. (a) Assuming diffraction-limited vision, we use the Rayleigh criterion and Eq. 35–10. The angle is also given by the separation of the ants (s), divided by the distance to the ants (l).

4  10−3 m )( 2  10−2 m ) ( s Ds = = 119m  100m  = 1.22 = → l = D l 1.22 1.22 ( 550 10−9 m )



(b) This does not agree with everyday experience–we can’t see two ants, separated by 2 cm, over the length of a football field. There must be other issues, such as aberrations of the eye, which limit our eye’s resolving ability to something less than the theoretical limit. 32. Assuming diffraction-limited vision, we use the Rayleigh criterion and Eq. 35–10. The angle is also given by the separation of the headlights, divided by the distance to the headlights (l).

7.5  10−3 m ) (1.5m ) ( s Ds = = 16770m  1.7  104 m or 17 km  = 1.22 = → l = −9 D l 1.22 1.22 ( 550  10 m )



33. We use Eq. 35–13 to calculate the angle for the second-order maximum.  2 ( 480  10−9 m )  −1  m  −1  = 4.4 d sin  = m →  = sin   = sin  1.25  10−5 m   d    34. We use Eq. 35–13 to calculate the wavelengths from the given angles. The slit separation, d, is the inverse of the number of lines per cm, N. We assume that 12,000 is good to 3 significant figures. sin  d sin  = m →  = Nm sin 28.8 sin 36.7 1 = 2 = = 4.01  10−5 cm = 401 nm = 4.98  10−5 cm = 498 nm 12,000 /cm 12,000 /cm sin 38.6 sin 47.9 3 = 4 = = 5.201  10−5 cm = 520 nm = 6.18  10 −5 cm = 618 nm 12,000 /cm 12,000 /cm 35. We use Eq. 35–13 to find the wavelength, where the number of lines, N, is the inverse of the slit separation, or d = 1/N. d sin  sin  sin 35.0 d sin  = m →  = = = = 5.031  10−5 cm  5.0  10−7 m m mN 3 ( 3800 /cm ) 36. Because the angle increases with wavelength, we use the largest wavelength to have a complete order. We set the maximum angle at 90° to determine the largest integer m in Eq. 35–13. sin  sin 90 d sin  = m → m = = = 2.27 −9  N ( 700  10 m ) ( 6300 /cm )(100 cm/m ) Thus, two full spectral orders (from blue to red) can be seen on each side of the central maximum, and a portion of the third order. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

We can also evaluate for the shortest wavelength. sin  sin 90 d sin  = m → m = = = 3.97 −9  N ( 400  10 m ) ( 6300 cm )(100 cm/m ) Thus, the violet part (and some of the other shorter wavelengths) of the third order will be visible, and perhaps the extreme violet part of the fourth order will be visible. The longest wavelength that can be seen for the third order is given by using an angle of 90. sin  sin 90 d sin  = m →  = = = 5.3  10−7 m, which is green. mN 3 ( 6300 /cm )(100 cm/m ) 37. We find the slit separation from Eq. 35–13. Then set the number of lines per centimeter equal to the inverse of the slit separation, N = 1/d. 1 sin  sin18.0 d sin  = m → N = = = = 1565slits cm  1600slits cm d m 3 ( 650  10−7 cm ) 38. We find the first-order angles for the maximum and minimum wavelengths using Eq. 35–13, where the slit separation distance is the inverse of the number of lines per centimeter. Then we set the distance from the central maximum of the maximum and minimum wavelength equal to the distance to the screen multiplied by the tangent of the first-order angle. The width of the spectrum is the difference in these distances.  m  −1 d sin  = m →  = sin −1   = sin ( m N )  d 

1 = sin −1 ( 410  10−7 cm ) ( 7800 lines/cm ) = 18.65

 2 = sin −1 ( 750  10−7 cm ) ( 7800 lines/cm ) = 35.80 y = y2 − y1 = l ( tan  2 − tan 1 ) = ( 2.50 m )( tan 35.80 − tan18.65 ) = 0.96 m 39. We find the second-order angles for the maximum and minimum wavelengths using Eq. 35–13, where the slit separation distance is the inverse of the number of lines per centimeter. Subtracting these two angles gives the angular width.  m  −1 d sin  = m →  = sin −1   = sin ( m N )  d  1 = sin −1  2 ( 4.5  10−7 m )( 6.5  105 /m ) = 35.80  2 = sin −1  2 ( 7.0  10−7 m )( 6.5  105 /m ) = 65.50

 =  2 − 1 = 65.50 − 35.80 = 29.7  30 Note that the answer has 2 significant figures. 40. The m = 1 brightness maximum for the wavelength of 1200 nm occurs at angle  . At this same angle m = 2, m = 3, etc. brightness maxima may exist for other wavelengths. To find these wavelengths, use Eq. 35–13, keeping d sin  constant, and solve for the wavelengths of higher order. m  d sin  = m11 = mm  m = 1 1 = 1 m m 1200 nm 1200 nm 1200 nm 2 = 3 = = 600 nm = 400 nm 4 = = 300 nm 2 3 4 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Higher-order maxima will have shorter wavelengths. Therefore, in the range 360 nm to 2000 nm, the only wavelengths that have maxima at the angle  are 600 nm and 400 nm besides the 1200 nm. 41. Because the angle increases with wavelength, we compare the maximum angle for the second order with the minimum angle for the third order, using Eq. 35–13, by calculating the ratio of the sines for each angle. Since this ratio is greater than one, the maximum angle for the second order is larger than the minimum angle for the first order and the spectra overlap. sin  2 22 / d 22 2 ( 700 nm )  m  d sin  = m → sin  =  = = = = 1.2  ; sin  3 33 / d 33 3 ( 400 nm )  d  To determine which wavelengths overlap, we set this ratio of sines equal to one and solve for the second-order wavelength that overlaps with the shortest wavelength of the third order. We then repeat this process to find the wavelength of the third order that overlaps with the longest wavelength of the second order. sin  2 2 / d 22 2 2 =1= 2 = → 3 = 2,max = ( 700 nm ) = 467 nm sin 3 33 / d 33 3 3

3 3 → 2 = 3,min = ( 400 nm ) = 600 nm 2 2 Therefore, the wavelengths 600 nm–700 nm of the second order overlap with the wavelengths 400 nm–467 nm of the third order . Note that these wavelengths are independent of the slit spacing. 42. We set the diffraction angles as one half the difference between the angles on opposite sides of the center. Then we solve Eq. 35–13 for the wavelength, with d equal to the inverse of the number of lines per centimeter.  −  2638 − ( −2618 ) 1 = r l = = 2628 = 26 + 28 / 60 = 26.467 2 2 sin  sin 26.467 1 = d sin  = = = 4.767  10−5 cm = 477 nm 9350 lines/cm N 4102 − ( −4027 )  − 2 = 2r 2l = = 4044.5 = 40 + 44.5 / 60 = 40.742 2 2 sin 40.742 2 = = 6.980  10−5 cm = 698nm 9350 lines/cm 43. If the spectrometer were immersed in water, the wavelengths calculated in Problem 42 would be wavelengths in water. To change those wavelengths into wavelengths in air, we must multiply by the index of refraction. 1 = ( 4.618  10−5 cm ) (1.33) = 614 nm ; 2 = ( 6.763 10−5 cm ) (1.33) = 899 nm air

air

Note that the second wavelength is not in the visible range. 44. We solve Eq. 35–13 for the slit separation width, d, using the given information. Then setting m = 3, we solve for the angle of the third-order maximum. m m 1( 589 nm ) sin  = → d= = = 2204 nm = 2.20  m d sin  sin15.5  m  −1  3  589 nm   3 = sin −1   = 53.3  = sin   d   2204 nm 

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45. Since the same diffraction grating is being used for both wavelengths of light, the slit separation will be the same. We solve Eq. 35–13 for the slit separation for both wavelengths and set the two equations equal. The resulting equation is then solved for the unknown wavelength. m m m sin  2 2 sin 21.2 d sin  = m  d = 1 1 = 2 2  2 = 1 1 = ( 632.8nm ) = 572 nm m2 sin 1 sin 1 sin  2 1 sin 53.2 46. Because the angle of the diffracted light increases with wavelength, to have a full order we use the largest wavelength, 700 nm. See Fig. 35–20 for an illustration. The maximum angle of diffraction is 90°. Use that angle with the largest wavelength to find the minimum slit separation. The reciprocal of the slit separation gives the number of slits per cm. 2 ( 7.00  10−7 m ) m d sin  = m → d = = = 1.40  10−6 m = 1.40  10 −4 cm sin  sin 90 1 1 = = 7140slits cm d 1.40  10−4 cm We have assumed that the original wavelength is known to 3 significant figures. 47. We find the angle for each “boundary” color from Eq. 35–13, and then use the fact that the y displacement on the screen is given by tan  = , where y is the displacement on the screen from the L central maximum, and L is the distance from the grating to the screen. m 1  1m   −1 m  5 sin  = ; d=  3  = (1 6.4  10 ) m ; y = L tan  = L tan sin d d  640lines mm  10 mm   m  mviolet    l 1 = L tan sin −1 red  − L tan sin −1 d d    

   −1 (1) ( 400  10 −9 m )   (1) ( 700  10−9 m )  −1 = ( 0.32 m )  tan sin  − tan sin  (1 6.4  105 ) m   (1 6.4  105 ) m     = 0.0756 m  8cm m  mviolet    l 2 = L tan sin −1 red  − L tan sin −1 d  d       −1 ( 2 ) ( 400  10−9 m )   ( 2 ) ( 700  10−9 m )  −1 = ( 0.32 m )  tan sin  − tan sin  (1 6.4  105 ) m   (1 6.4  105 ) m     = 0.45495m  45cm The second-order rainbow is dispersed over a larger distance. More calculations would show that the third-order violet starts at about 38 cm from the central maximum, which would overlap with part of the second order and make the second order harder to see. 48. (a) Missing orders occur when the angle to the interference maxima (Eq. 34–2a) is equal to the angle of a diffraction minimum (Eq. 35–2). We set d = 2 D and show that the even interference orders are missing. m m  m1 d 2 D sin  = 1 = 2 → = = = 2 → m1 = 2m2 d D m2 D D © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Since m2 = 1,2,3,4, ..., all even orders of m1 correspond to the diffraction minima and will be missing from the interference pattern. (b) Setting the angle of interference maxima equal to the angle of diffraction minimum with the orders equal to integers, we determine the relationship between the slit size and separation that will produce missing orders. m m  d m1 → = sin  = 1 = 2 d D D m2 (c) When d = D, all interference maxima will overlap with diffraction minima so that no fringes will exist. This is expected because if the slit width and separation distance are the same, the slits will merge into one single opening. 49. (a) Diffraction maxima occur at angles for which the incident light constructive interferes. That is, when the path length difference between two rays is equal to an integer number of wavelengths. Since the light is incident at an angle  relative to the grating, each succeeding higher ray, as shown in the diagram, travels a distance l 1 = d sin  farther to reach the grating. After passing through the grating the higher rays travel a distance to the screen that is again longer by l 2 = d sin  . By setting the total path length difference equal to an integer number of wavelengths, we are able to determine the location of the bright fringes. l = l 1 + l 1 = d ( sin  + sin  ) =  m , m = 0,1, 2,.... (b) The  allows for the incident angle and the diffracted angle to have positive and negative values. (c) We insert the given data, with m = 1, to solve for the angles . m  550  10−9 m   −1   = sin −1  − sin   = −   sin sin15    d  0.01m 4500lines   

= −0.65 and − ( 3.0  101 )  50. Using Eq. 35–13, we calculate the maximum order possible for this diffraction grating by setting the angle equal to 90. Then we set the resolving power equal to the product of the number of grating lines and the order, where the resolving power is the wavelength divided by the minimum separation in wavelengths (Eq. 35–19), and solve for the separation. m d sin  ( 0.01m 6500lines ) sin 90 sin  = → m= = = 2.47  2 624  10−9 m  d   624 nm = Nm   = = = 0.015nm  Nm ( 6500 lines/cm )( 3.26cm )( 2 ) The resolution is best for the second order, since it is more spread out than the first order. Also the value of  is smaller for m = 2 than m = 1. 51. (a) The resolving power is given by Eq. 35–19.

R = Nm → R1 = (15,000 )(1) = 1.5  104 ; R 2 = (15,000 )( 2 ) = 3.0  104

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(b) The wavelength resolution is also given by Eq. 35–19.

  = Nm →  =  Nm

410 nm 410 nm = 2.7  10−2 nm = 27 pm ; 1 = = 1.37  10−2 nm  14 pm (15,000 )(1) ( 30,000 )(1) The minimum wavelength resolution is for second order, and is 14 pm. 1 =

52. (a) We use Eq. 35–13, with the angle equal to 90 to determine the maximum order. m d sin  (1050 nm ) sin 90 sin  = → m= = = 1.81 d  580 nm Since the order must be an integer number there will only be one principal maximum on either side of the central maximum. Counting the central maximum and the two other principal maxima there will be a total of three principal maxima . (b) We use Eq. 35–17 to calculate the peak width, where the full peak width is double the half-peak width and the angle to the peak is given by Eq. 35–13. 0 = 0

 0 = 2

 Nd cos 0

2 ( 580 nm ) 2 = = 6.4  10−5 rad = 0.0037 −2 l cos 0 (1.80  10 m ) cos0

 m  −1  1  580 nm   = 33.5  = sin   d   1050 nm 

 1 = sin −1   1 =

2 ( 580 nm ) 2 = = 7.7  10−5 rad = 0.0044 −2 l cos 1 (1.80  10 m ) cos ( 33.5 )

53. We use Eq. 35–20, with m = 1. m = 2d sin  →  = sin −1

(1)( 0.138nm ) = 12.7 m = sin −1 2d 2 ( 0.315nm )

54. We use Eq. 35–20 for X-ray diffraction. (a) Apply Eq. 35–20 to both orders of diffraction. m  m1 sin 1 2  m = 2d sin  → = → 2 = sin −1  2 sin 1  = sin −1  sin 23.6  = 53.2 1 m2 sin 2 m    1  (b) Use the first-order data.

m = 2d sin  →  =

2d sin  2 ( 0.24 nm ) sin 23.6 = = 0.19 nm 1 m

55. For each diffraction peak, we can measure the angle and count the order. Consider Eq. 35–20. m = 2d sin  →  = 2d sin 1 ; 2 = 2d sin 2 ; 3 = 2d sin 3  From each equation, all we can find is the ratio = 2sin  = sin 2 = 23 sin 3 . No, we cannot d separately determine the wavelength or the spacing.

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56. (a) With parallel rays, the X-ray images are just shadows. Thus, the image will be the same size as the object, and so the magnification is 1 . (b) The rays from the point source will not refract, so we use similar triangles to compare the image size to the object size for the front of the body. ( d + d2 ) = (15cm + 25cm ) = 2.7 . h m1 = i = 1 ho1 d1 (15cm ) For the back of the body, the image and object have the same size, so the magnification is 1 . 57. The lines act like a reflection grating, similar to a CD or DVD. We assume that we see the first diffractive order, so m = 1. Use Eq. 35–13. m (1)( 520nm ) = = 630nm d sin  = m → d = sin  sin56 58. We assume the sound is diffracted when it passes through the doorway, and find the angles of the minima from Eq. 35–2. mv mv v = ; D sin  = m = →  = sin −1 , m = 1, 2, 3, ... f f Df

m = 1:  = sin −1

(1)( 340 m s ) = 28 mv = sin −1 Df ( 0.88m )(820 Hz )

m = 2 :  = sin −1

( 2 )( 340 m s ) = 70 mv = sin −1 Df ( 0.88m )(820 Hz )

m = 3:  = sin −1

( 3)( 340 m s ) = sin −1 1.41 = impossible mv = sin −1 Df ( 0.88m )(820 Hz )

Thus, the whistle would not be heard clearly at angles of 28° and 70° on either side of the normal . Note that both angles have 2 significant figures. 59. We find the angles for the first order from Eq. 35–13. (1) ( 4.4  10−7 m ) m 1 = sin −1 = sin −1 = 21.15 0.01 m 8200 d

 2 = sin −1

(1) ( 6.8  10−7 m )

= 33.89 0.01 m 8200 The distances from the central white line on the screen are found using the tangent of the angle and the distance to the screen. y1 = l tan 1 = ( 2.5m ) tan 21.15 = 0.967 m y2 = l tan  2 = ( 2.5m ) tan 33.89 = 1.679 m Subtracting these two distances gives the linear separation of the two lines. y2 − y1 = 1.679 m − 0.967 m = 0.712 m  0.7 m

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60. We find the angles for the two first-order peaks from the distance to the screen and the distances along the screen to the maxima from the central peak. ( 3.32cm ) = 3.065 y y tan 1 = 1 → 1 = tan −1 1 = tan −1 l l ( 62.0cm ) tan  2 =

y2 l

→  2 = tan −1

( 3.71cm ) = 3.424 y2 = tan −1 l ( 62.0cm )

Inserting the wavelength of yellow sodium light and the first-order angle into Eq. 35–13, we calculate the slit separation on the grating. Then, using the slit separation and the second angle, we calculate the wavelength of the second source. Finally, we take the inverse of the line separation to determine the number of slits per centimeter on the grating. m1 1( 589 nm ) = = 11,016 nm d sin 1 = m1 → d = sin 1 sin 3.065

d sin 1 = (11,016 nm ) sin 3.424 = 658nm m 1 1 line = = 908 lines/cm d 11,016  10−7 cm

2 =

61. We find the angles for the first order from Eq. 35–13, with m = 1. The slit spacing is the inverse of the slits/cm of the grating. 1 1m 1 m m ; d sin  = m →  = sin −1  = → d= 5 8500lines cm 100cm 8.5  10 d

 = sin −1

1 d

− sin −1

2 d

= sin −1

656  10−9 m 410  10−9 m − sin −1 = 13.49  13 1 1     m m     5 5  8.5  10   8.5  10 

62. (a) This is very similar to Example 35–6. We use the same notation as in that Example, and solve for the distance l. 6.0  10−3 m ) ( 2.0 m ) ( Ds 1.22 s = l = l → l = = = 1.8  104 m = 18km  11miles D 1.22 1.22 ( 560  10−9 m ) (b) We use the same data for the eye and the wavelength. −9  180   3600  1.22 1.22 ( 560  10 m ) = = = 1.139  10−4 rad    = 23 −3 D ( 6.0  10 m )   rad   1  Our answer is less than the real resolution, because of atmospheric effects and aberrations in the eye. 63. We first find the angular half-width for the first order, using Eq. 35–1, sin  =



. Since this angle D is small, we may use the approximation that sin   tan  . The width from the central maximum to the first minimum is given by x = l tan  . That width is then doubled to find the width of the beam, from the first diffraction minimum on one side to the first diffraction minimum on the other side. y = l tan  = l sin 

y = 2 y = 2 l sin  = 2 L

 D

2 ( 3.8  108 m )( 633  10−9 m ) 0.010 m

= 4.8  104 m = 48km

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64. The distance between lines on the diffraction grating is found by solving Eq. 35–13 for d, the grating spacing. The number of lines per meter is the reciprocal of d. m 1 sin  sin18.5 d= → = = = 5.01  105 lines m −7 sin  d m 1 6.328 10 m  () 65. This is very similar to Example 35–6. We use the same notation as in that Example, and solve for the separation s. We use 450 nm for the wavelength of the blue light. 1.22 ( 450  10−9 m ) 1.22 s = l = l = (18m ) = 9.9  10−4 m D 1.0  10−2 m 66. We find the angles for the first order from Eq. 35–13, d sin  = m . We are considering first-order, so m = 1, and the slit separation (in meters) is the reciprocal of the lines per meter value. sin  H = 6.56  10−7 m →  H = sin −1 0.2165 = 12.5 H = d sin  H = 3.30  105 slits m

Ne = d sin  Ne =

sin  Ne = 6.50  10−7 m →  Ne = sin −1 0.2145 = 12.4 3.30  105 slits m

Ar = d sin  Ar =

sin  Ar = 6.97  10−7 m →  Ar = sin −1 0.2300 = 13.3 3.30  105 slits m

67. From Eq. 35–10 we calculate the minimum resolvable separation angle. We then multiply this angle by the distance between Mars and Earth to obtain the minimum distance between two objects that can be resolved by a person on Mars. −9 1.22 1.22 ( 550  10 m ) = = 1.34  10−4 rad = D 0.005 m

l = s = (8  1010 m )(1.34  10−4 rad ) = 1.07  107 m

Since the minimum resolvable distance is much less than the Earth–Moon distance of 4  108 m, a person standing on Mars could resolve the Earth and Moon as two separate objects without a telescope. s 68. The required angular resolution from the geometry is  = , where s is the separation of the two l objects to be resolved, and l is the distance to the objects being viewed. This must equal the resolution from the Rayleigh criterion, Eq. 35–10. −9 3 1.22  l 1.22 ( 550  10 m )(130  10 m ) s 1.22  = = → D= = = 4.36 m  4 m l D s ( 0.02 m )

69. The maximum angle is 90°. The slit separation is the reciprocal of the line spacing.  1  cm  sin 90  d sin   6800  = = 2.32 d sin  = m → m = 633  10−7 cm  Thus, the second order is the highest order that can be seen.

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70. (a) For the refraction at the first surface, apply Snell’s law (twice) to find the exit angles. We estimated the indices of refraction from Fig. 32–28 by magnifying that figure to get more accurate readings. Subscript “1” is the 420-nm light (n = 1.587), and subscript “2” is the 680-nm light (n = 1.567). Subscript “a” is the incident angle on the left side, subscript “b” is the angle after the first refraction, subscript “c” is the incident angle on the right side, and subscript “d” is the final refracted angle. For the entrance face: nair sin  a = n sin b →

(1.00 ) sin 45 = (1.587 ) sin b1 → b1 = 26.459 (1.00 ) sin 45 = (1.567 ) sin b 2 → b 2 = 26.824

We find the angle of incidence at the second surface, leaving the prism, from the second diagram. ( 90 − b ) + ( 90 − c ) + A = 180

c1 = A − b1 = 60 − 26.459 = 33.541 c 2 = A − b 2 = 60 − 26.824 = 33.176 Use Snell’s law again for the refraction at the second surface. n sin  c = nair sin  d

(1.587 ) sin 33.541 = (1.00 ) sin d 1 → d 1 = 1 = 61.267 (1.567 ) sin 33.176 = (1.00 ) sin d 2 → d 2 = 2 = 59.035

 = 61.267 − 59.035 = 2.232  2.2 The values may vary due to reading the graph in Fig. 32–28. (b) We use Eq. 35–13 to determine the angles for each wavelength. The slit separation distance is the inverse of the number of slits per centimeter, and m = 1. 1 m   sin  = →  = sin −1   = 1724 nm d= 5800 slit/cm d d   420 nm   = 14.1  1724 nm 

 680 nm   = 23.2  1724 nm 

1 = sin −1 

 2 = sin −1 

 = 23.2 − 14.1 = 9.1 (c) One obvious advantage for using the diffraction grating is that much larger separation angles are possible: 9.1° vs. 2.2° in this example. Also, with a diffraction grating it is possible to observe second-order diffractions, which give a second, even larger separation angle that can be measured. 71. Assume the pupil is 3.0 mm in diameter and we use light of wavelength 550 nm. The minimum angular separation that we could observe is given in Eq. 35–10. −9 1.22 1.22 550  10 m  180   60 arc min  = 0.77 arc min = = 2.24  10−4 rad  =   −3 D 3.0  10 m 1   rad    This is less than the angular separation of Alcor and Mizar, so they are resolvable by the naked eye.

(

)

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72. (a) We calculate the wavelength of the mother’s sound by dividing the speed of sound by the frequency of her voice. We use Eq. 34–2b to determine the double-slit interference minima with d = 3.0 m.  = v f = ( 340 m s ) ( 400 Hz ) = 0.85 m 1    ( m + 12 )   −1 –1 ( m + 2 )( 0.85 m ) 1 = sin   = sin 0.2833 ( m + 2 )  , m = 0,1,2,...  d 3.0 m ( )    

 = sin –1 

= 8.1, 25, 45, and 83 We use Eq. 35–2 to determine the angles for destructive interference from single-slit diffraction effects, with D = 1.0 m.  m ( 0.85 m )   m  −1  = sin –1   = sin –1   = sin  0.85m  , m = 1,2,... 1.0 m ( )  D   

 = 58 (b) We use the depth and length of the room to determine the angle the sound would need to travel to reach the son.  8.0 m   = tan −1   = 58  5.0 m  This angle is close to the single-slit diffraction minimum, so the son has a good explanation for not hearing her. In reality, there would be many reflections from the other walls in the room, which still could carry sound to the boy’s location. 73. Because the angle increases with wavelength, to miss a complete order we use the smallest visible wavelength, 400 nm. The maximum angle is 90°. With these parameters we use Eq. 35–13 to find the slit separation, d. The inverse of the slit separation gives the number of lines per unit length. 2 ( 400 nm ) m = = 800 nm d sin  = m → d = sin  sin 90 1 1 = = 12,500lines/cm d 800  10−7 cm 74. For the minimum aperture the angle subtended at the lens by the smallest feature is the angular resolution, given by Eq. 35–10. We let l represent the spatial separation, and r represent the altitude of the camera above the ground. −9 1.22 l 1.22 r 1.22 ( 580  10 m ) ( 25000 m ) = = → D= = = 0.3538m  0.4 m l D r ( 0.05m ) 75. We find the spacing from Eq. 35–20.

( 2 ) ( 9.73  10−11 m ) m = = 2.62  10−10 m m = 2d sin  → d = 2sin  2sin 21.8

76. The angles for Bragg scattering are found from Eq. 35–20, for m = 1 and m = 2. If the distance from the crystal to the screen is l, the radius of the diffraction ring is given by r = l tan 2.

X-ray

Crystal 2

l Screen © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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  m   2d sin  = m ; r = l tan 2 = l tan  2sin −1    2d      m   r1 = l tan  2sin −1    2d   

  (1) ( 0.10  10−9 m )     = 0.074 m = ( 0.15m ) tan  2sin −1   2 ( 0.22  10−9 m )        ( 2 ) ( 0.10  10−9 m )     m   −1     = 0.21m = 0.15m tan 2sin r2 = l tan  2sin −1  )  (  2 ( 0.22  10−9 m )    2d      

77. From Eq. 35–10 we calculate the minimum resolvable separation angle. We then multiply this angle by the distance between the Earth and Moon to obtain the minimum distance between two objects on the Moon that the Hubble can resolve. −9 1.22 1.22 ( 550  10 m ) = = = 2.796  10−7 rad D 2.4 m

l = s = ( 3.84  108 m )( 2.796  10−7 rad ) = 110 m 78. The distance x is twice the distance to the first minima. We can write x in terms of the slit width D  using Eq. 35–2, with m = 1. The ratio is small, so we may approximate sin   tan    . sin  =

 D

  ; x = 2 y = 2l tan  = 2l  = 2l

 D

When the plate is heated up the slit width increases due to thermal expansion. Eq. 17–1b is used to determine the new slit width, with the coefficient of thermal expansion, , given in Table 17–1. Each slit width is used to determine a value for x. Subtracting the two values for x gives the change x. We use the binomial expansion to approximate and simplify the evaluation.    2l     2l   −1 1  x = x − x0 = 2 l  = − 1 =  − 2l   (1 + T ) − 1   D0 (1 + T )    D D 1 T D +   )  0 ( 0  0  

(

)

2 ( 2.0 m ) ( 620 10−9 m ) 2l  2l   25  10−6 ( C )−1  ( 55C ) 1 − T − 1) = − T = − ( −6  D0 D0 22 10 m  ( ) 

= −1.55  10−4 m  −1.6 10−4 m 79. We use the Rayleigh criterion, for an angular separation of 2o. D ( 0.5 m )  o  rad  1.22 → = = = 1.43  10 −2 m  1cm = 2  o  D 1.22 1.22  180 

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CHAPTER 36: The Special Theory of Relativity Responses to Questions 1.

No. Since the windowless car in an exceptionally smooth train moving at a constant velocity is an inertial reference frame and the basic laws of physics are the same in all inertial reference frames, there is no way for you to tell if you are moving or not. The first postulate of the special theory of relativity can be phrased as “no experiment can tell you if an inertial reference frame is at rest or moving uniformly at constant velocity.”

The fact that you instinctively think you are moving is consistent with the relativity principle applied to mechanics. Even though you are at rest relative to the ground, when the car next to you creeps forward, you are moving backward relative to that car.

Since the railroad car is traveling with a constant velocity, the ball will land back in his hand. Both the ball and the car are already moving forward (relative to the ground), so when the ball is thrown straight up into the air with respect to the car, it will continue to move forward at the same rate as the car and fall back down to land in his hand.

Whether you say the Earth goes around the Sun or the Sun goes around the Earth depends on your reference frame. It is valid to say either one, depending on which frame you choose. The laws of physics, though, won’t be the same in each of these reference frames, since the Earth is accelerating as it goes around the Sun. The Sun is nearly an inertial reference frame, but the Earth is not.

The starlight would pass at c, regardless of your spaceship’s speed. This is consistent with the second postulate of relativity, which states that the speed of light through empty space is independent of the speed of the source or the observer.

The clocks are not at fault and they are functioning properly. Time itself is actually measured to pass more slowly in moving reference frames when compared to a rest frame. Any measurement of time (heartbeats or decay rates, for instance) would be measured as slower than normal when viewed by an observer outside the moving reference frame.

Time actually passes more slowly in the moving reference frame, including aging and other life processes. It is not just that it seems this way–time has actually been measured to pass more slowly in the moving reference frame, as predicted by special relativity.

This situation is an example of the “twin paradox” applied to parent–child instead of to twins. This situation would be possible if the woman was traveling at high enough speeds during her trip. Time would have passed more slowly for her and she would have aged less than her son, who stayed on Earth. (Note that the situations of the woman and son are not symmetric; she must undergo acceleration during her journey.)

You would not notice a change in your own heartbeat, mass, height, or waistline. No matter how fast you are moving relative to Earth, you are at rest in your own reference frame. Thus, you would not notice any changes in your own characteristics. To observers on Earth, you are moving away at 0.6c, which gives  = 1.25. If we assume that you are standing up, so that your body is perpendicular to the direction of motion, then to the observers on Earth, it would appear that your heartbeat has slowed by a factor of 1/1.25 = 0.80 and that your waistline has decreased by a factor of 0.80 (due to time dilation and length contraction). Your height would be unchanged (since there is no relative motion between you and Earth in that direction). Also note the comments in Section 36–9 of the text on “Rest Mass

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and Relativistic Mass” for comments about mass change and relativity. Your actual mass has not changed. 10. Yes, they do occur. However, at a speed of only 90 km/hr, v c is extremely small, and therefore γ is very close to one, so the effects would not be noticeable. 11. Length contraction and time dilation would not occur. If the speed of light were infinite, v c would be 0 for all finite values of v, and therefore γ would always be 1, resulting in t = t0 and l = l 0 . 12. Both the length contraction and time dilation formulas include the term 1 − v 2 c 2 . If c were not the limiting speed in the universe, then it would be possible to have a situation with v  c. However, this would result in a negative number under the square root, which gives an imaginary number as a result, indicating that c must be the limiting speed. Also, assuming the relativistic formulas were still correct, as v gets very close to c, an outside observer should be able to show that l = l 0 1 − v 2 c 2 is getting smaller and smaller and that the limit as v → c is l → 0. This would show that c is a limiting speed, since nothing can get smaller than having a length of 0. A similar to analysis for time dilation should show that t = is getting longer and longer and that the 1 − v 2 c2 limit as v → c is t → . This would show that c is a limiting speed, since the slowest that time can pass is that it comes to a stop.

13. If the speed of light was 25 m/s, then we would see relativistic effects all the time, something like the Chapter opening figure or Figure 36–16 with Question 21. Everything moving relative to us would be length contracted and time dilation would have to be taken into account for many events. There would be no “absolute time” on which we would all agree, so it would be more difficult, for instance, to plan to meet friends for lunch at a certain time. Many “twin paradox” kind of events would occur, and the momentum of moving objects would become very large, making it very difficult to change their motion. One of the most unusual changes for today’s modern inhabitants of Earth would be that nothing would be able to move faster than 25 m/s, which is only about 56 mi/h. 14. No. The relativistic momentum of the electron is given by p =  mv =

. At low speeds 1 − v 2 c2 (compared to c) this reduces to the classical momentum, p = mv. As v approaches c, γ approaches infinity so there is no upper limit to the electron’s momentum.

15. No. To accelerate a particle with nonzero rest mass up to the speed of light would require an infinite amount of kinetic energy, according to Eq. 36–10a, and so is not possible. 16. No, E = mc2 does not conflict with the conservation of energy, it actually completes it. Since this equation shows us that mass and energy are interconvertible, it says it is now necessary to include mass as a form of energy in the analysis of energy conservation in physical processes. 17. Every observer will measure the speed of a beam of light to be c. Check it with Eq. 36–7d. “Away” from the Earth is taken as the positive direction, so “towards” the Earth is the negative direction.

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( − c ) + 0.70c = − c. v + u = vu  1 + ( −1)( 0.70 ) 1+ 2 c The beam’s speed (magnitude of velocity), relative to Earth, is c. u=

18. Yes. One way to describe the energy stored in the compressed spring is to say it is a mass increase (although it would be so small that it could not be measured). This “mass” will convert back to energy when the spring is uncompressed. 19. Matter and energy are interconvertible (matter can be converted into energy and energy can be converted into matter). Thus we should say, “Energy can neither be created nor destroyed.” 20. No, our intuitive notion that velocities simply add is not completely wrong. Our intuition is based on our everyday experiences, and at these everyday speeds our intuition is correct regarding how velocities add. Our intuition does break down, though, at very high speeds, where we have to take into account relativistic effects. Relativity does not contradict classical mechanics, but it is a more general theory whereas classical mechanics is a limiting case. 21. (a) From the girlfriend’s frame of reference, she and her Vespa are at rest while the observer and the streetscape are moving to the left at 70 km/h. As a result the observer and the streetscape will be narrower (in the horizontal direction), and she and her Vespa appear at their original width. The observer and streetscape will appear unchanged in the vertical direction.

Responses to MisConceptual Questions 1.

(e) Answer (e) is one of the postulates of special relativity: Light propagates through empty space with a definite speed c independent of the speed of the source or observer. The other answers contradict this postulate.

(c, d) Page 1078 says: “Rotating or otherwise accelerating frames of reference are noninertial frames,” and “A reference frame that moves with constant velocity with respect to an inertial frame is itself also an inertial frame.” So answers (a) and (b) describe inertial frames (answer a with a relative velocity of 0), and answers (c) and (d) describe noninertial frames.

(a) Proper length is the length measured by a person at rest with the object measured. The ship’s captain is at rest with the ship, so that measurement is the proper length.

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(c) A common misconception is that the distance between the objects is important when measuring relativistic effects. The important parameter is the relative velocity. The ship’s captain is at rest with the flashlight and therefore the captain measures the proper time between flashes. The space-dock personnel measure the dilated time, which is always longer than the proper time. Since the dilated time is 1.00 s, the proper time must be shorter, which is answer (c).

(b) The ship’s crew will have aged less. While the ship is moving at constant speed, both the crew and the space-dock personnel record that the other’s clock is running slow. However, the ship must accelerate (and therefore change frames of reference) when it turns around. The spacedock personnel do not accelerate and therefore they are correct in their measurement that they have aged more than the ship’s crew.

(d) A common misconception is that the relativistic formulas are only valid for speeds close to the speed of light. The relativistic formulas are always valid. However, for speeds much smaller than the speed of light, time dilations, length contractions, and changes in momentum are insignificant.

(d) Although different values are measured for energy and momentum in different frames, the values of the particle’s momentum and energy are conserved in each of the respective reference frames. The laws of conservation of energy and momentum pertain to events such as collisions and energy exchanges. If a collision or energy exchange is observed from a single inertial reference frame, the momentum and/or energy will be conserved in that reference frame, meaning the initial and final energy and/or momentum will be the same before and after the event. Observers from different frames will measure different values for the energy or momentum according to the equations of special relativity (Eqs. 36–7 for speeds, for example). But the observers in a different frame will still find that the final energy or momentum in their frame is equal to the initial energy or momentum in that frame, confirming that energy and momentum are conserved in that different frame. In other words, conservation of energy and conservation of momentum apply WITHIN a frame of reference, not between different frames of reference.

(a, d, e) The speed of light in a vacuum is the same in all reference frames, so answer (a) is NOT “relative”–its value does not change from one reference frame to another. The time interval does change (i.e., time dilation), so answer (b) is relative. The length of an object in its direction of motion does change (i.e., length contraction), so answer (c) is relative. The length of an object perpendicular to its direction of motion does not change from one reference frame to another, and so answer (d) is NOT relative. The laws of physics have the same form in all inertial reference frames–that’s the first postulate of special relativity–and so answer (e) is NOT relative. Since time intervals are relative, so is simultaneity, so answer (f) is relative.

(c) It might be assumed that the speed of the ship should be added to the speed of light to obtain a relative speed of 1.5c. However, the second postulate of relativity is that the speed of light is 1.0c, as measured by any observer.

10. (f)

A common misconception is that the person on the ship would observe the change in their measurement of time, or that the person on the spacecraft would see the Earth clocks running fast. Actually, the person on the spacecraft would observe their time running normally. Since the spacecraft and Earth are moving relative to each other, observers on the spacecraft and observers on the Earth would both see the other’s clocks running slow.

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11. (d, e) Due to relativistic effects, the observers will not necessarily agree on the time an event occurs, the distance between events, the time interval between events, or the simultaneity of two events. However, the postulates of relativity state that both observes agree on the validity of the laws of physics and on the speed of light. 12. (d) It is common to think that one frame of reference is preferable to another. Due to relativistic effects, observers in different frames of reference may make different time measurements. However, each measurement is correct in that frame of reference. 13. (d) A common misconception is that there is a stationary frame of reference and that motion relative to this frame can be measured. Special relativity demonstrates that a stationary frame does not exist, but all motion is relative. If the spaceship has no means to observe the outside (and make a reference to another object), then there is no way to measure your velocity. 14. (b) To an observer on Earth, the pendulum would appear to be moving slowly. Therefore the period would be longer than 2.0 seconds. 15. (a) A common misconception is to add the velocities together to obtain a speed of 1.5c. However, due to relativistic effects, the relative speed between two objects must be less than the speed of light. Since the objects are moving away from each other the speed must be greater than the speed of each object, so it must be greater than 0.75c. The only option that is greater than 0.75c and less than 1.0c is 0.96c.

Solutions to Problems 1.

We find the lifetime at rest from Eq. 36–1a. 2

 2.70  108 m s  −6 t0 = t 1 − v c = ( 5.16  10 s ) 1 −   = 2.25  10 s 8 3.00 10 m s    2

−6

You measure the contracted length. Find the rest length from Eq. 36–3a. l 112 m l0 = = = 169 m 2 2 2 1− v c 1 − ( 0.750 )

The measured distance is the contracted length. Use Eq. 36–3a. 2

 2.90  108 m s  l = l 0 1 − v c = (115 ly ) 1 −   = 29.44 ly  29ly 8  3.00  10 m s  2

The speed is determined from the time dilation relationship, Eq. 36–1a.

t0 = t 1 − v 2 c 2 → 2

 2.60  10−8 s   t  v = c 1−  0  = c 1−  = 0.841c = 2.52  108 m s −8   t   4.80  10 s 

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The speed is determined from the length contraction relationship, Eq. 36–3a. 2

l = l 0 1− v c 2

Instructor Solutions Manual

 l   32ly  8 → v = c 1−   = c 1−   = 0.757c  0.76c = 2.3  10 m s 49ly l    0

The speed is determined from the length contraction relationship, Eq. 36–3a. 2

l = l 0 1− v c 2

 l  2 → v = c 1 −   = c 1 − ( 0.900 ) = 0.436 c = 1.31  108 m s l0 

(a) We use Eq. 36–3a for length contraction with the contracted length 99.0% of the rest length. 2

 l  2 l = l 0 1 − v c → v = c 1 −   = c 1 − ( 0.990 ) = 0.141c l0  (b) We use Eq. 36–1a for time dilation with the time as measured from a relative moving frame 1.00% greater than the rest time. 2

 t   1  t0 = t 1 − v 2 c 2 → v = c 1 −  0  = c 1 −   = 0.140 c  1.0100   t  We see that a speed of 0.14 c results in about a 1% relativistic effect. 8.

The speed is determined from the length contraction relationship, Eq. 36–3a. Then the time is found from the speed and the contracted distance.

l = l 0 1 − v 2 c2 →

 l  v = c 1−   l0 

; t=

l = v

l 2

25ly 2

( 25 y ) c = 26.9 y  27 y c ( 0.930 )

 l   25ly  c 1−  c 1−     68ly  l0  That is the time according to you, the traveler. The time according to an observer on Earth would be as follows. ( 68 yr ) c = 73yr l t= 0 = v c ( 0.930 ) 9.

(a) The measured length is the contracted length. We find the rest length from Eq. 36–3a. l 4.80 m l0 = = = 6.47 m 2 2 2 1− v c 1 − ( 0.670 ) Distances perpendicular to the motion do not change, so the rest height is 1.35m . (b) The time in the spacecraft is the proper time, found from Eq. 36–1a. t0 = t 1 − v 2 c 2 = ( 20.0s ) 1 − ( 0.670 ) = 14.8s 2

(c) To your friend, you moved at the same relative speed: 0.670 c . (d) She would measure the same time dilation: 14.8 s .

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10. (a) To an observer on Earth, 23.5 ly is the rest length, so the time will be the distance divided by the speed. ( 23.5ly ) = 24.74 yr  24.7 yr l tEarth = 0 = v 0.950c (b) We find the proper time (the time that passes on the spacecraft) from Eq. 36–1a. t0 = t 1 − v 2 c 2 = ( 22.74 yr ) 1 − ( 0.950 ) = 7.725 yr  7.73 yr 2

l = l 0 1 − v 2 c 2 = ( 23.5ly ) 1 − ( 0.950 ) = 7.338ly  7.34ly 2

(d) To the spacecraft observer, the speed of the spacecraft is v =

( 7.338ly ) = 0.95c l = , t 7.725 yr

as expected. 11. (a) In the Earth frame, the clock on the Enterprise will run slower. Use Eq. 36–1a. t0 = t 1 − v 2 c 2 = ( 5.0 yr ) 1 − ( 0.80 ) = 3.0 yr 2

(b) Now we assume the 5.0 years is the time as measured on the Enterprise. Again use Eq. 36–1a. ( 5.0 yr ) t0 t0 = t 1 − v 2 c 2 → t = = = 8.3yr 2 1 − v 2 c2 1 − ( 0.80 ) 12. The dimension along the direction of motion is contracted, and the other two dimensions are unchanged. Use Eq. 36–3a to find the contracted length. l = l 0 1 − v 2 c2 ; V = l ( l 0 ) = ( l 0 ) 2

1 − v 2 c 2 = ( 2.4 m )

1 − ( 0.80 ) = 8.29 m3  8.3m3 2

Note that the original volume was ( 2.4m ) = 13.8m3 . 3

13. The change in length is determined from the length contraction relationship, Eq. 36–3a. The speed is very small compared to the speed of light, so we use the binomial expansion.

l = l 0 1 − v 2 c2 → 1/ 2

 v2  l = 1 − v 2 c 2 = 1 − 2  l0  c 

 11.2  103 m s  v2 −10  1 − 2 = 1 − 12   = 1 − 6.97  10 8 3.00 10 m s c    1 2

So the percent decrease is ( 6.97  10−8 ) % .

14. We find the speed of the particle in the lab frame, and use that to find the rest frame lifetime and distance. x 1.00 m v = lab = = 2.6316  108 m s = 0.8772 c tlab 3.80  10−9 s (a) Find the rest frame lifetime from Eq. 36–1a.

t0 = tlab 1 − v 2 c 2 = ( 3.80  10−9 s ) 1 − ( 0.8772 ) = 1.824  10−9 s  1.82 10−9 s 2

(b) In its rest frame, the particle will travel the distance given by its speed and the rest lifetime. x0 = vt0 = ( 2.6316  108 m s )(1.824 10−9 s ) = 0.480m

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15. In the Earth frame, the average lifetime of the pion will be dilated according to Eq. 36–1a. The speed of the pion will be the distance moved in the Earth frame divided by the dilated time. d d v= = 1 − v 2 c2 → t t0 v=c

= 0.9633 c 2  ( 3.00  108 m s )( 2.6  10−8 s )   1+    28m ) (   Note that the significant figure addition rule give 4 sig. figs. for the value under the radical sign.  ct0  1+    d 

16. The length of all three sides of the triangle will change due to length contraction, but the vertical dimension will not change. We designate the rest length of the equal sides as l , and so the rest length of the base is 1.5 l . When moving, we designate the length of each side as s. Note that the altitude of the triangle, designated as h, does not change. See the diagram. Use the Pythagorean theorem to determine h2 in each triangle, and equate those two expressions to find the relationship between l and s. Then determine the speed from the length contraction formula, Eq. 36–3.

h2 = l 2 − ( 0.75l )

(1.5 l )

=1− 2

v2 c2

7 2 3 2 s2 7 l = s → 2 = 16 4 12 l 2 2 7 7 20 20 v s → 2 =1− =1− =1− = → v= c = 0.86c 2 2 27 27 27 c 12 (1.5 ) (1.5 l )

l 2 − ( 0.75l ) = s 2 − ( 0.50s ) 2

h2 = s 2 − ( 0.50s )

→ (1 − 0.752 ) l 2 = (1 − 0.502 ) s 2 →

17. Since the number of particles passing per second is reduced from N to N / 2, a time T0 must have elapsed in the particles’ rest frame. The time T elapsed in the lab frame will be greater, according to Eq. 36–1a. The particles moved a distance of 2cT0 in the lab frame during that time. T0 2cT0 x ; v= = T0 = T 1 − v 2 c 2 → T = → v = 54 c = 0.894 c 2 2 T   1− v c T0    1 − v 2 c2    18. We take the positive direction as the direction of motion of the Enterprise. Consider the alien vessel as reference frame S, and the Earth as reference frame S. The velocity of the Earth relative to the alien vessel is v = −0.70 c. The velocity of the Enterprise relative to the Earth is u = 0.90 c. Solve for the velocity of the Enterprise relative to the alien vessel, u, using Eq. 36–7d. ( v + u) ( 0.90c − 0.70c ) u= = = 0.54c  v u   1 + ( − 0.70 )( 0.90 )  1 + 2  c   We could also have made the Enterprise as reference frame S, with v = −0.90 c, and the velocity of the alien vessel relative to the Earth as u = 0.70 c. The same answer would result. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Choosing the two spacecraft as the two reference frames would also work. Let the alien vessel be reference frame S, and the Enterprise be reference frame S. Then we have the velocity of the Earth relative to the alien vessel as u = −0.70 c, and the velocity of the Earth relative to the Enterprise as u = −0.90 c. We solve for v, the velocity of the Enterprise relative to the alien vessel.

( u + v )  vu   1 + 2  c  

→ v=

( −0.70c ) − ( −0.90c ) u − u = = 0.54c   u u   ( −0.90c )( −0.70c )  − 1 1−    c 2    c2 

19. The Galilean transformation is given in Eq. 36–4. (a) ( x, y, z ) = ( x + vt , y , z  ) = ( 25m + ( 30 m s )( 3.5s ) , 20 m,0 ) = (130 m, 20 m,0 ) (b)

( x, y, z ) = ( x + vt , y, z  ) = ( 25m + ( 30 m s )(10.0s ) , 20 m,0 ) = (325m, 20 m,0 )

20. (a) The person’s coordinates in S are found using Eq. 36–6, with x = 25 m , y = 20 m , z  = 0, and

t  = 3.5  10−6 s. We set v = 1.60 108 m/s. 25m + (1.60  108 m/s )( 3.5  10−6 s ) x + v t  = = 690 m x= 2 2 1 − v 2 c2 1 − (1.60  108 m/s ) ( 3.00  108 m/s ) y = y  = 20 m ; z = z  = 0

(b) We repeat part (a) using the time t  = 10.0  10−6 s. x=

x + v t  1 − v 2 c2

25m + (1.60  108 m/s )(10.0  10−6 s )

1 − (1.60  108 m/s )

( 3.00 108 m/s )

= 1920 m

y = y  = 20 m ; z = z  = 0

21. We determine the components of her velocity in the S frame using Eq. 36–7, where ux = uy = 1.30 108 m/s and v = 1.60 108 m/s. Then using trigonometry we combine the components to determine the magnitude and direction. u x + v 1.30  108 m/s + 1.60  108 m/s = = 2.356  108 m/s ux = 2 2 8 8 8 1 + vu x / c 1 + (1.60  10 m/s )(1.30  10 m/s ) / ( 3.00  10 m/s ) 8 8 8 u y 1 − v 2 c 2 (1.30  10 m/s ) 1 − (1.60  10 m/s ) ( 3.00  10 m/s ) = = 8.932  107 m/s uy = 2 2 8 8 8  1 + vux / c 1 + (1.60  10 m/s )(1.30  10 m/s ) / ( 3.00  10 m/s ) 2

u = u x2 + u 2y =

 = tan −1

uy ux

( 2.356  10 m/s ) + (8.932  10 m/s ) = 2.52 10 m/s

= tan −1

8.932  107 m/s = 20.8 2.356  108 m/s

22. (a) Take the positive direction to be the direction of motion of spaceship 1. Consider spaceship 2 as reference frame S, and the Earth reference frame S. The velocity of the Earth relative to spaceship 2 is v = 0.50 c. The velocity of spaceship 1 relative to the Earth is u = 0.50 c. Solve for the velocity of spaceship 1 relative to spaceship 2, u, using Eq. 36–7d. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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( 0.50 c + 0.50 c ) = = 0.800 c   vu  1 + ( 0.50 )( 0.50 )  1 +   c2   (b) Now consider spaceship 1 as reference frame S. The velocity of the Earth relative to spaceship 1 is v = −0.60 c. The velocity of spaceship 2 relative to the Earth is u = −0.60 c. Solve for the velocity of spaceship 2 relative to spaceship 1, u, using Eq. 36–7d. ( v + u) ( −0.50 c − 0.50 c ) u= = = −0.800 c  vu   1 + ( − 0.50 )( −0.50 )  1 + 2  c   As expected, the two relative velocities are the opposite of each other. u=

( v + u)

23. (a) We take the positive direction in the direction of the first spaceship. We choose reference frame S as the Earth, and reference frame S as the first spaceship. So v = 0.66 c. The speed of the second spaceship relative to the first spaceship is u = 0.87 c. We use Eq. 36–7d to solve for the speed of the second spaceship relative to the Earth, u. ( v + u) ( 0.66c + 0.87c ) u= = = 0.972 c  vu   1 + ( 0.66 )( 0.87 )  1 + 2  c   (b) The only difference now is that u  = − 0.87 c. ( v + u) ( 0.66c − 0.87c ) u= = = −0.49 c  vu   1 + ( 0.66 )( − 0.87 )  1 + 2  c   The problem asks for the speed, which would be 0.49 c . 24. (a) The Galilean transformation is given in Eq. 36–4. x = x + vt  = x + vt = 100m + ( 0.88 ) ( 3.00  108 m s )(1.00 10−6 s ) = 364m (b) The Lorentz transformation is given in Eq. 36–6. Note that we are given t, the clock reading in frame S. v x  t v x  t =   t + 2  → t = − 2  c c  

   t v x   v  ct vx   x =  ( x + v t  ) =   x + v  − 2   =   x +  −  c c    c    1 (100 m ) + ( 0.88 ) 1 − 0.882 ( 3.00  108 m s )(1.00  10−6 s ) − 0.88 (100 m )  =  2  1 − 0.88 

)

(

= 311m

25. Choose frame S as the frame at rest with the spaceship. In this frame the module has speed u = uy = 0.85 c. Frame S is the frame that is stationary with respect to the Earth. The spaceship, and therefore frame S moves in the x-direction with speed 0.76c in this frame, or v = 0.76 c. We use Eqs. 36–7a and 36–7b to determine the components of the module velocity in frame S. Then using trigonometry we combine the components to determine the speed and direction of travel. u y 1 − v 2 c 2 0.85 c 1 − 0.762 u x + v 0 + 0.76 c = = 0.76 c ; u y = = = 0.552 c ux = 1 + vu x / c 2 1+ 0 1 + vu x / c 2 1+ 0 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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u = ux2 + u 2y =

( 0.76 c ) + ( 0.552 c ) = 0.94 c ;  = tan −1 y = tan −1 2

0.552 c = 36 0.76 c

26. We assume that the given speed of 0.85c is relative to the planet that you are approaching. We take the positive direction in the direction that you are traveling. Consider your spaceship as reference frame S, and the planet as reference frame S. The velocity of the planet relative to you is v = −0.85 c. The velocity of the probe relative to the planet is ux = 0.95 c. Solve for the velocity of the probe relative to your spaceship, u x , using Eq. 36–7a. ( ux + v ) ( 0.95 c − 0.85 c ) ux = = = 0.5195 c  0.52 c   vu x  1 + ( − 0.85)( 0.95)  1 +  c 2   27. (a) In frame S the horizontal component of the stick length will be contracted, while the vertical component remains the same. We use the trigonometric relations to determine the x- and y-components of the length of the stick. Then using Eq. 36–3a we determine the contracted length of the x-component. Finally, we use the Pythagorean theorem to determine stick length in frame S. l x = l 0 cos  ; l y = l 0 sin  = l y ; l x = l x 1 − v 2 c 2 = l 0 cos  1 − v 2 c 2 l = l x 2 + l y 2 = l 0 2 cos 2  (1 − v 2 c 2 ) + l 0 2 sin 2  = l 0 1 − ( v cos  c )

(b) We calculate the angle from the length components in the moving frame.    tan   l l 0 sin   = tan −1   = tan −1 (  tan  )   = tan −1 y = tan −1   l cos 1 − v 2 c 2   1 − v 2 c2  l x  0    28. We set up the two frames such that in frame S, the first event takes place at the origin and the second event takes place 180 meters from the origin, so xA = 0 and xB = 180 m. We set the time when event A occurred equal to zero, so tA = 0 and tB = 0.80 s. We then set the location of the two events in frame S equal, and using Eq. 36–6 we solve for the velocity. x − xB 0 − 180 m = = 2.25  108 m/s = 0.75c xA = xB →  ( xA − vt A ) =  ( xB − vtB ) ; v = A tA − tB 0 − 0.80s 29. (a) We choose the train as frame S and the Earth as frame S. Since the guns fire simultaneously in S, we set these times equal to zero, that is tA = tB = 0. To simplify the problem we also set the location of gunman A equal to zero in frame S when the guns were fired, xA = 0. This places gunman B at xB = 55.0m. Use Eq. 36–6 to determine the time that each gunman fired his weapon in frame S. v x  v0    tA =   tA + 2A  =   0 + 2  = 0 c  c     35m/s )( 55.0 m )  ( vxB  1   = 2.14  10−14 s 0+ tB =   tB + 2  = 2 2  8  c  8  3.00  10 m/s  1 − 35.0 m/s 3.00  10 m/s 

(

)

(

)

Therefore, in Frame S, A fired first. (b) As found in part (a), the difference in time is 2.14  10−14 s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1219

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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(c) In the Earth frame of reference, since A fired first, B was struck first. In the train frame, A is moving away from the bullet fired toward him, and B is moving toward the bullet fired toward him. Thus B will be struck first in this frame as well. 30. The velocity components of the particle in the S frame are ux = u cos and u y = u sin  . We find the components of the particle in the S frame from the velocity transformations given in Eqs. 36–7a and 36–7b. Those transformations are for the S frame moving with speed v relative to the S frame. We can find the transformations from the S frame to the S frame by simply changing v to –v and primed to unprimed variables. u y 1 − v 2 c 2 uy 1 − v 2 c2 ux + v ) ux − v ) ( (   → ux = → uy = ux = ; u = (1 + vux c2 ) (1 − vux c2 ) y (1 + vux c2 ) (1 − vux c 2 ) uy 1 − v 2 c2

tan   =

u y (1 − vux c 2 ) = u y 1 − v 2 c 2 = u sin  1 − v 2 c 2 = sin  1 − v 2 c 2 = u x ( ux − v ) ( ux − v ) ( u cos − v ) ( cos − v u )

(1 − vu c ) 2

31. We set frame S as the frame moving with the observer. Frame S is the frame in which the two light bulbs are at rest. Frame S is moving with velocity v with respect to frame S. We solve Eq. 36–6 for the time t  in terms of t, x, and v. Using the resulting equation we determine the time in frame S that each bulb is turned on, given that in frame S the bulbs are turned on simultaneously at tA = tB = 0. Taking the difference in these times gives the time interval as measured by the observing moving with velocity v. x x =  ( x + v t  ) → x = − v t 



 v 2  vx t  vx  v x   v x   vx  t =   t  + 2  = t  + 2  − vt    =  t  1 − 2  + 2 = + 2 → t  =   t − 2   c c   c    c    c  c

vx  vx  v0  vl  vl     t A =   t A − 2A  =   0 − 2  = 0 ; t B =   t B − 2B  =   0 − 2  = − 2 c  c  c  c  c     vl c2 According to the observer, bulb B turned on first. t  = t B − t A = −

32. From the boy’s frame of reference, the pole remains at rest with respect to him. As such, the pole will always remain 13.0 m long. As the boy runs toward the barn, relativity requires that the (relatively moving) barn contract in size, making the barn even shorter than its rest length of 10.0 m. Thus it is impossible, in the boy’s frame of reference, for the barn to be longer than the pole. So according to the boy, the pole will never completely fit within the barn. In the frame of reference at rest with respect to the barn, it is possible for the pole to be shorter than the barn. We use Eq. 36–3a to calculate the speed that the boy would have to run for the contracted length of the pole, l, to equal the length of the barn. l = l 0 1 − v 2 c 2 → v = c 1 − l 2 l 02 = c 1 − (10.0 m )

(13.0 m ) = 0.6390 c 2

If persons standing at the front and back door of the barn were to close both doors exactly when the pole was completely inside the barn, we would have two simultaneous events in the barn’s rest frame © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 36

The Special Theory of Relativity

S with the pole completely inside the barn. Let us set the time for these two events as tA = tB = 0. In frame S these two events occur at the front and far side of the barn, or at xA = 0 and xB = 10.0m. Using Eq. 36–6, we calculate the times at which the barn doors close in the boy’s frame of reference. vx  v0   t A =   t A − 2A  =   0 − 2  = 0 c  c   

0.6390 (10.0 m )   vx  1  −8 t B =   t B − 2B  = 0 −  = −2.769  10 s 8 2 3.00 10 m/s  c   1 − 0.6390   Therefore, in the boy’s frame of reference the far door of the barn closed 27.7 ns before the front door. If we multiply the speed of the boy by this time difference, we calculate the distance the boy traveled between the closing of the two doors. x = vt = 0.6390 ( 3.00  108 m/s )( 2.769  10−8 s ) = 5.31 m. We use Eq. 36–3a to determine the length of the barn in the boy’s frame of reference.

l = l 0 1 − v 2 c 2 = (10.0 m ) 1 − 0.63902 = 7.69 m Subtracting the distance traveled between closing the doors from the length of the pole, we find the length of the barn in the boy’s frame of reference. l 0,pole − x = 13.0 m − 5.31 m = 7.69 m = l barn Therefore, in the boy’s frame of reference, when the front of the pole reached the far door it was closed. Then 27.7 ns later, when the back of the pole reached the front door, that door was closed. In the boy’s frame of reference these two events are not simultaneous. 33. The momentum of the proton is given by Eq. 36–8. (1.67 10−27 kg ) ( 0.72 ) (3.00 108 m s ) = 5.2 10−19 kg  m s mv p =  mv = = 1 − v 2 c2 1 − 0.722 34. (a) We compare the classical momentum to the relativistic momentum (Eq. 36–8). p classical mv 2 = = 1 − v 2 c 2 = 1 − ( 0.10 ) = 0.995 prelativistic   mv    1 − v 2 c2    The classical momentum is about −0.5% in error. The negative sign means the classical momentum is smaller than the relativistic momentum (b) We again compare the two momenta. p classical mv 2 = = 1 − v 2 c 2 = 1 − ( 0.50 ) = 0.866 mv prelativistic 1 − v 2 c2

The classical momentum is −13% in error. 35. The momentum at the higher speed is to be twice the initial momentum. Designate the initial state with a subscript “0,” and the final state with a subscript “f.” Use Eq. 36–8 for relativistic momentum.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

  mvf   vf2    2 2  2 2   1 − vf c  pf  1 − vf c  = =2 →  =4 → p0     v02 mv0     1 − v02 c 2   1 − v 2 c2   0   2  ( 0.32 c )2   v02   vf   0.456  2 = = 4 4   = 0.456c 2 → vf2 =     c → vf = 0.56 c 2 2 2  2 2  c c 1 1 − − v v  1.456  1 − ( 0.32 )  f 0    

36. The two momenta, as measured in the frame in which the particle was initially at rest, will be equal to each other in magnitude. The lighter particle is designated with a subscript “1,” and the heavier particle with a subscript “2.” Use Eq. 36–8 for relativistic momentum. m1v1 m2 v2 = → p1 = p2 → 2 2 1 − v1 c 1 − v22 c 2 2 2 2  m2   6.68  10−27 kg   ( 0.53 c )  v22 = =  = 6.25c 2 →    2 −27 2 2 2 2 (1 − v1 c )  m1  (1 − v2 c )  1.67 10 kg  1 − ( 0.53) 

v12

7.25 v12 = 6.25c 2 → v1 =

6.25 c = 0.93 c 7.25

37. We find the proton’s momenta using Eq. 36–8. mp v1 mp ( 0.45 c ) mp v2 mp ( 0.80 c ) p0.45 = = = 0.5039mp c ; p0.80 = = = 1.3333mp c 2 2 2 2 v1 v2 1 − 0.45 1 − 0.80 ( ) ( ) 1− 2 1− 2 c c mp v3 mp ( 0.98 c ) p0.98 = = = 4.9247 mp c 2 v32 1 0.98 − ( ) 1− 2 c  1.3333mp c − 0.5039mp c   p − p1  (a)  2 100 = 164.6  160% 100 =  0.5039mp c  p1     4.9247mp c − 1.3333mp c   p − p1  (b)  2 100 = 269.4  270% 100 =  1.3333mp c  p1    38. The rest energy of the electron is given by Eq. 36–12.

E = mc 2 = ( 9.11  10−31 kg )( 3.00  108 m s ) = 8.199  10−14 J  8.20  10−14 J 2

(8.199 10

(1.60 10

−13

−14

J MeV )

= 0.5124 MeV  0.512 MeV

This does not exactly agree with the “endpaper” value because of significant figures. If more significant digits were used for the given values, a value of 0.511 MeV would be obtained. 39. We find the loss in mass from Eq. 36–12. −13 E ( 200 MeV ) (1.60  10 J MeV ) m= 2 = = 3.56  10−28 kg  4  10−28 kg 2 8 c ( 3.00 10 m s ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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40. We find the mass conversion from Eq. 36–12. (11020 J ) = 1111kg  1000 kg E m= 2 = 2 c ( 3.00 108 m s ) 41. Each photon has momentum 0.75 MeV/c. Since the photon is massless, Eq. 36–14 says each photon has an energy of 0.75 MeV. Assuming the photons have opposite initial directions, then the total momentum is 0, and so the product mass will not be moving. Thus, all of the photon energy can be converted into the rest mass energy of the particle, and so the heaviest particle would have a mass of 1.50 MeV c 2. Convert this to kg using values from the front of the book for the atomic mass unit.

 1.6605  10−27 kg  1.50MeV c 2  = 2.67  10−30 kg 2  931.49MeV c   42. (a) The work is the change in kinetic energy. Use Eq. 36–10b. The initial kinetic energy is 0.   1 − 1 ( 938 MeV ) = 8.454  103 MeV W = K = K final = ( − 1) mc 2 =  2  1 − 0.995 

= 8.45GeV (b) The momentum of the proton is given by Eq. 36–8. 1 p =  mv = (938MeV c2 ) ( 0.995 c ) = 9.3447  103 MeV c = 9.34GeV c 2 1 − 0.995 43. Use Eq. 36–10b to calculate the kinetic energy of the proton. Note that the classical answer would be that a doubling of speed would lead to a four-fold increase in kinetic energy. Subscript “1” represents the lower speed ( v = 13 c ) and subscript “2” represents the higher speed ( v = 23 c ) .

( (

) )

2 −1/2

2 − 1 ( 5 9 )−1/2 − 1 0.3416 1 − ( 23 ) K 2 ( 2 − 1) mc 2 = = = = = 5.6 K = ( − 1) mc → −1/2 2 −1/2 K1 ( 1 − 1) mc 2 − 1 ( 8 9 ) − 1 0.0607 1− ( 1 ) 3

44. Since the electron was accelerated by a potential difference of 28 kV, its potential energy decreased by 28 keV, and so its kinetic energy increased from 0 to 28 keV, or 0.028 MeV. Use Eq. 36–10b to find the speed from the kinetic energy.   1 − 1 mc 2 → K = (  − 1) mc 2 =   1 − v 2 c2   

v = c 1−

1  K   2 + 1  mc 

= c 1−

1  0.028MeV  + 1   0.511MeV 

= 0.32 c

45. The work is the change in kinetic energy. Use Eq. 36–10b. The initial kinetic energy is 0. W1 = (  0.90 − 1) mc 2 ; W2 = K 0.99 c − K 0.90 c = (  0.99 − 1) mc 2 − (  0.90 − 1) mc 2

−

2  −  0.90 W2 (  0.99 − 1) mc − (  0.90 − 1) mc 1 − 0.902 = 3.7 = = 0.99 = 1 − 0.99 2 1  0.90 − 1 W1 ( 0.90 − 1) mc −1 1 − 0.902 2

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46. We find the energy equivalent of the mass from Eq. 36–12.

E = mc 2 = (1.0  10−3 kg )( 3.00  108 m s ) = 9.0  1013 J 2

We assume that this energy is used to increase the gravitational potential energy. E 9.0  1013 J E = mgh → m = = = 9.2  109 kg 3 2 hg (1.0  10 m )( 9.80 m s ) 47. The total energy of the proton is the kinetic energy plus the mass energy. Use Eq. 36–14 to find the momentum. E = K + mc 2 ;

( pc ) = E 2 − ( mc 2 ) = ( K + mc 2 ) − ( mc 2 ) = K 2 + 2 K ( mc 2 ) 2

pc = K 2 + 2 K ( mc 2 ) = K 1 + 2

938.3MeV mc 2 = ( 950 MeV ) 1 + 2 = 1638 MeV 950 MeV K

p = 1638 MeV c  1.6GeV c 48. The kinetic energy is given by Eq. 36–10b.

K = (  − 1) mc 2 = mc 2 →  = 2 =

→ v=

1 − v 2 c2 We see that the mass of the particle does not affect the result.

3 c = 0.866 c 4

49. We use Eq. 36–10b to find the speed from the kinetic energy.   1 − 1 mc 2 → K = (  − 1) mc 2 =   1 − v 2 c2   

v = c 1−

1  K   2 + 1  mc 

= c 1−

1  1.45MeV  + 1   0.511MeV 

= 0.9655 c

50. We let M represent the rest mass of the new particle. The initial energy is due to both incoming particles, and the final energy is the rest energy of the new particle. Use Eq. 36–11b for the initial energies. 2m E = 2 (  mc 2 ) = Mc 2 → M = 2 m =  2m 1 − v 2 c2 We assumed that energy is conserved, and so there was no loss of energy in the collision. The final kinetic energy is 0, so all of the kinetic energy was lost.

  1 K lost = Kinitial = 2 (  − 1) mc 2 =  − 1 2mc 2  1 − v 2 c2   

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51. We use Eqs. 36–11a and 36–14 in order to find the mass.

E 2 = p 2 c 2 + m 2 c 4 = ( K + mc 2 ) = K 2 + 2 Kmc 2 + m 2 c 4 → 2

p 2 c 2 − K 2 (121MeV c ) c − ( 45MeV ) = = 140 MeV c 2  2.5  10−28 kg m= 2 Kc 2 2 ( 45MeV ) c 2 2

The particle is most likely a probably a  0 meson. 52. (a) Since the kinetic energy is half the total energy, and the total energy is the kinetic energy plus the rest energy, the kinetic energy must be equal to the rest energy. We also use Eq. 36–10b. K = 12 E = 12 ( K + mc 2 ) → K = mc 2

K = (  − 1) mc 2 = mc 2 →  = 2 =

→ v = 34 c = 0.866 c 1 − v 2 c2 (b) In this case, the kinetic energy is half the rest energy. 1 → v = 95 c = 0.745 c K = (  − 1) mc 2 = 12 mc 2 →  = 23 = 1 − v 2 c2 53. We use Eq. 36–10b for the kinetic energy and Eq. 36–8 for the momentum. v 7.85  107 m s = = 0.2617 c 3.00  108 m s     1 1 − 1 mc 2 =  − 1 ( 938.3MeV ) K = (  − 1) mc 2 =   1 − 0.2617 2  1 − v 2 c2  ( )     = 33.88Mev  34 MeV

p =  mv =

mv 1 − v 2 c2

2 1 mc ( v c ) 1 ( 938.3MeV )( 0.2617 ) = = 254MeV c 2 c 1 − v 2 c2 c 1 − ( 0.2617 )

Evaluate with the classical expressions. 2 2 2 2v 1 1 K c = 2 mv = 2 mc   = 12 ( 938.3MeV )( 0.2617 ) = 32.1MeV c

1 v pc = mv = mc 2   = ( 938.3MeV c )( 0.2617 ) = 246 MeV c c c Calculate the percent error. 32.1MeV − 33.9 MeV K −K errorK = c  100 =  100 = −5% 33.9 MeV K 246 MeV c − 254 MeV c p −p errorp = c  100 =  100 = −3% p 254 MeV c 54. (a) The kinetic energy is found from Eq. 36–10b.     2 1 1 K = (  − 1) mc  =  − 1 mc 2 =  − 1 (1.7  104 kg )( 3.00  108 m s ) 2  1 − v 2 c2   1 − 0.22   

= 3.843  1019 J  3.8  1019 J © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

(b) Use the classical expression and compare the two results.

K = 12 mv  = 12 (1.7  104 kg ) ( 0.22 ) ( 3.00  108 m s )  = 3.703  1019 J 2

( 3.703 10 J ) − ( 3.843 10 J )  100 = −3.6% % error = ( 3.843  10 J ) 19

The classical value is 3.6% too low. 55. By conservation of energy, the rest energy of the americium nucleus is equal to the rest energies of the other particles plus the kinetic energy of the alpha particle. mAm c 2 = ( mNp + m ) c 2 + K →

mNp = mAm − m −

 5.5MeV  1u K = 241.05682 u − 4.00260 u − = 237.0483u  2 2 2  c c  931.49 MeV c 

56. (a) For a particle of non-zero mass, we derive the following relationship between kinetic energy and momentum.

E = K + mc 2 ; ( pc ) = E 2 − ( mc 2 ) = ( K + mc 2 ) − ( mc 2 ) = K 2 + 2 K ( mc 2 ) 2

−2mc 2  4 ( mc 2 ) + 4 ( pc ) 2

K + 2 K ( mc ) − ( pc ) = 0 → K = 2

2 For the kinetic energy to be positive, we take the positive root. −2mc 2 + 4 ( mc 2 ) + 4 ( pc ) 2

= − mc 2 +

( mc ) + ( pc ) 2 2

2 This is the relationship that is graphed in the first graph below. But notice these 2 “extreme” cases. First, if the momentum is large, we have the following relationship.

K = −mc 2 +

( mc ) + ( pc )  pc − mc 2 2

Thus, there should be a linear relationship between kinetic energy and momentum for large values of momentum, with a negative y-intercept on the graph. If the momentum is small, we use the binomial expansion to derive the classical relationship.

K = −mc 2 +

2 pc  ( mc2 ) + ( pc )2 = −mc2 + mc2 1 +  mc 2  

2 2   pc   p  −mc 2 + mc 2  1 + 12  2   =   mc   2m  Thus, we expect a quadratic relationship for small values of momentum. The adjacent graph verifies these approximations.

(b) For a particle of zero mass, the relationship is simply K = pc. See the second graph.

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57. All of the energy, both rest energy and kinetic energy, becomes electromagnetic energy. We use Eq. 36–11. Both masses are the same.   1 1 + Etotal = E1 + E2 =  1mc 2 +  2mc 2 = ( 1 +  2 ) mc 2 =   (105.7 MeV ) 2 2 1 − 0.55   1 − 0.38

= 240.8 MeV  240 MeV 58. We use Eqs. 36–11 and 36–13.

E = K + mc 2 ; ( pc ) = E 2 − ( mc 2 ) = ( K + mc 2 ) − ( mc 2 ) = K 2 + 2 K ( mc 2 ) → 2

K 2 + 2 K ( mc 2 ) c

59. Use Eq. 36–10b for kinetic energy, and Eq. 36–12 for rest energy. K = (  − 1) mEnterprise c 2 = mconverted c 2 →     1 1 mconverted =  − 1 mEnterprise =  − 1 ( 4  109 kg ) = 2.015  107 kg  2  107 kg 2 2 2  1− v c   1 − 0.10   

60. (a) We assume the mass of the particle is m, and we are given that the velocity only has an x-component, u x . We write the momentum in each frame using Eq. 36–8, and we use the velocity transformation given in Eq. 36–7. Note that there are three relevant velocities: u x , the velocity in reference frame S; ux , the velocity in reference frame S; and v, the velocity of one frame relative to the other frame. There is no velocity in the y or z-directions, in either frame. 1 We reserve the symbol  for , and also use Eq. 36–11b for energy. 1 − v 2 c2 px = ux = px =

mu x 1 − u x2 c 2 u x + v 1 + v u x c 2

; p y = 0 ; pz = 0 → u x =

mu x

ux − v 1 + vu x c 2 1 + vu x c 2   ; u u 0 ; u u = = = =0 y y z z 1 − vu x c 2 1 − v 2 c2 1 − v 2 c2

; py = 0 ( since u y = 0 ) ; pz = 0 ( since u z = 0 )

1 − u x2 c 2

Substitute the expression for ux into the expression for px . px =

mu x 1 − u x c 2

( ux − v )

(1 − vu c ) 2

1 ( ux − v ) 1− 2 c (1 − v u c 2 )2 2

( ux − v )

(1 − vu c ) 2

( ux − v ) 2

2 2

( ux − v )

(1 − vu c ) 2

1 1 (1 − vux c 2 )

(1 − vu c ) − c (1 − vu c ) (1 − vu c ) 2 2

(1 − vu c ) − ( c ) 2 2

ux − v 2

m (ux − v )

(1 − vu c ) − ( c ) 2 2

ux − v

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

m ( ux − v )

px =

1− 2

2 x 2

m (ux − v )

vux  vux  u 2u v v +  2  − + 2x − 2 2 c c c  c  c mu x

Instructor Solutions Manual

2 v2  vu  u 1 +  2x  − 2x − 2 c  c  c mv

−

(1 − u c ) (1 − u c ) = = (1 − v c )(1 − u c ) (1 − v c ) m ( ux − v ) 2

2 x

mu x

2 x

−

v 2 (1 − u 2 c 2 ) c

2 x

mc 2

(1 − u c ) (1 − v c ) 2 x

mc 2

v 2 (1 − u 2 c 2 ) c

px −

(1 − v c ) 2

px − vE c 2

(1 − v c ) 2

It is obvious from the first few equations of the problem that py = p y ( = 0 ) and pz = pz ( = 0 ) .

E =

mc 2

1 − u x2 c 2

1 ( ux − v ) 1− 2 c (1 − v u c 2 )2 2

mc 2

( ux − v )

(1 − vu c ) − c (1 − vu c ) (1 − vu c ) 2 2

2 2

mc 2

mv u x

−

( mc − mvu ) = (1 − u c ) (1 − u c ) = = u − v) (1 − v c )(1 − u c ) (1 − v c ) (1 − vu c ) − ( mc 2 (1 − vu x c 2 ) 2 2

2 x

E − px v

(1 − v c ) 2

(b) We summarize these results, and write the Lorentz transformation from Eq. 36–6, but solve in terms of the primed variables. That can be easily done by interchanging primed and unprimed quantities, and changing v to −v. p − vE c 2 E − px v px = x ; py = p y ; py = p y ; E  = 2 2 (1 − v c ) (1 − v 2 c2 ) x =

x − vt

(1 − v c ) 2

; y = y ; z = z ; t  =

t − vx c 2

(1 − v c ) 2

These transformations are identical if we exchange p x with x, p y with y, p z with z, and E c 2 with t (or E c with ct). 61. The galaxy is moving away from the Earth, and so we use Eq. 36–16b. f 0 − f = 0.1015 f 0 → f = 0.8985 f 0

f = f0

c−v → c+v

1 − ( f f 0 )2  2  c =  1 − 0.8985  c = 0.1066 c v=  2  1+ ( f f 0 )2   1 + 0.8985   

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62. For source and observer moving towards each other, use Eq. 36–15b. 1+ v c 1 + 0.65 c+v f = f0 = f0 = ( 95.0MHz ) = 206MHz  210MHz 1− v c 1 − 0.65 c−v 63. We use Eq. 36–16a, and assume that v

 = 0

1+ v c c+v = 0 = 0 1− v c c−v

= 0 (1 + v c ) (1 − v 2 c 2 )

−1/ 2

(1 + v c )(1 + v c ) =  1 + v c 1 ) 0( − + v v 1 1 c c ( )( ) (1 − v 2 c2 )  − 0  v = = 0 0 c

 0 (1 + v c ) = 0 + 0 v c →

64. (a) We apply Eq. 36–15b to determine the received/reflected frequency f. Then we apply this same equation a second time using the frequency f as the source frequency to determine the Dopplershifted frequency f . We subtract the initial frequency from this Doppler-shifted frequency to obtain the beat frequency. The beat frequency will be much smaller than the emitted frequency when the speed is much smaller than the speed of light. We then set c − v  c and solve for v. c+v c+v c+v c+v c+v f = f0 f= f = f0 = f0   c−v c−v c−v c−v c−v  cf 2v 2v c+v c−v  f beat = f  − f 0 = f 0   f0 → v  beat  − f0   = f0 c−v c 2 f0 c−v  c−v 

( 3.00  10 m/s ) ( 6670 Hz ) = 27.8m/s 2 ( 36.0  10 Hz ) 8

v

(b) We find the change in velocity and solve for the resulting change in beat frequency. Setting the change in the velocity equal to 1 km/h we solve for the change in beat frequency. cf cf beat 2 f v → f beat = 0 v = beat → v = c 2 f0 2 f0 f beat =

2 ( 36.0  109 Hz ) (1km/h )  1m/s  = 70 Hz ( 3.00 108 m/s )  3.600 km/h 

65. (a) We consider the difference between Doppler-shifted frequencies for atoms moving directly towards the observer and atoms moving directly away. Use Eqs. 36–15b and 36–16b.  2v c   c+v  2v  c+v c−v c−v   f = f 0 − f0 = f 0  − = f0   = f 0  2  2  1 − v 2 c2  c−v c+v c+v   c −v   c−v   We take the speed to be the rms speed of thermal motion, given by Eq. 18–5. We also assume that the thermal energy is much less than the rest energy, and so 3kT mc 2 . v = vrms =

3kT m

→

3kT v = c mc 2

f 3kT  3kT  =2 1 −  f0 mc 2  mc 2 

→

−1/ 2

 2

3kT mc 2

(b) We evaluate for a gas of H atoms (not H 2 molecules) at 650 K. Use Appendix G to find the mass.

f 3kT =2 =2 f0 mc 2

3 (1.38  10−23 J K ) ( 650 K )

(1.008u ) (1.66 10

−27

kg u )( 3.00  10 m s ) 8

= 2.7  10−5

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66. We choose the North Pole location as a stationary frame of reference, so the clock at the North Pole is at rest, while the clock on the equator travels the circumference of the Earth each day. We divide the circumference of the Earth by the length of the day to determine the speed of the equatorial clock. We set the dilated time equal to 1.5 years and solve for the change in rest times for the two clocks. 6 2 R 2 ( 6.38  10 m ) v= = = 464 m/s T ( 24 hr )( 3600s/hr )

t0,eq

t =

 v2  → t0,eq = t 1 − v 2 / c 2  t 1 + 2   2c 

1 − v 2 / c2 t0,pole t = → t0,pole = t 1− 0

 v2  t0,eq − t0,pole = t 1 + 2  − t  2c 

7 v 2 (1.5 yr )( 464 m/s ) ( 3.156  10 s/yr ) = t 2 = = 57  s 2 2c 2 ( 3.00  108 m/s ) 2

67. We treat the Earth as the stationary frame, and the airplane as the moving frame. The elapsed time in the airplane will be dilated to the observers on the Earth. Use Eq. 36–1a. 2 rEarth 2 rEarth tEarth = ; tplane = tEarth 1 − v 2 c 2 = 1 − v 2 c2 v v 2 rEarth 2 rEarth   1 v 2    rEarth v t = tEarth − tplane = 1 − 1 − v 2 c2  1 − 1 −  = v v   2 c2  c2

)

(



 1m s     3.6 km h  

 ( 6.38  106 m ) 1500 km h  =



( 3.00 10 m s ) 8

= 9.3  10−8 s

68. (a) To travelers on the spacecraft, the distance to the star is contracted, according to Eq. 36–3a. This contracted distance is to be traveled in 5.5 years. Use that time with the contracted distance to find the speed of the spacecraft. xspacecraft xEarth 1 − v 2 c 2 = → v= tspacecraft tspacecraft

v=c

1 2

= 0.6159 c  0.62 c

 ctspacecraft   5.5ly  1+  1+     4.3ly   xEarth  (b) Find the elapsed time according to observers on Earth, using Eq. 36–1a. tspaceship 5.5 y tEarth = = = 6.981y  7.0 y 2 2 1− v c 1 − 0.61592 Note that this agrees with the time found from distance and speed. x 4.3ly = 7.0 yr tEarth = Earth = v 0.6159 c © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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The Special Theory of Relativity

69. The increase in kinetic energy comes from the decrease in potential energy:   1 − 1 mc 2 → K = (  − 1) mc 2 =  2 2    1− v c  1/ 2

1/ 2

    1  v = c 1 − 2   K     2 + 1  mc    

      1   = c 1 − 2      7.20  10−14 J   + 1   2 −31 8     9.11  10 kg )( 3.00  10 m s )    (

= 0.8465c

70. We assume that some kind of a light signal is being transmitted from the astronaut to Earth, with a frequency of the heartbeat. That frequency will then be Doppler shifted, according to Eq. 36–16b. We express the frequencies in beats per minute. f 02 − f 2 ) 602 − 522 ) ( ( c−v → v=c 2 =c = 0.14 c f = f0 c+v ( f + f02 ) ( 602 + 522 ) 71. (a) The velocity components of the light in the S frame are ux = 0 and uy = c. We transform those velocities to the S frame according to Eq. 36–7. ux =

u y 1 − v 2 c 2 c 1 − v 2 c 2 u x + v 0+v v = = = = = c 1 − v 2 c2 ; u y 2 2   1 + vu x c 1+ 0 1 + vu x c 1+ 0

 = tan −1

uy ux

= tan −1

c 1 − v 2 c2 c2 = tan −1 −1 v v2

(b) u = ux2 + u y2 = v 2 + c 2 (1 − v 2 c 2 ) = v 2 + c 2 − v 2 = c (c) In a Galilean transformation, we would have the following. u x = u x + v = v ; u y = u y = c ; u = v 2 + c 2 (  c ) ;  = tan −1

c v

72. We take the positive direction as the direction of motion of rocket A. Consider rocket A as reference frame S, and the Earth as reference frame S. The velocity of the Earth relative to rocket A is v = −0.65 c. The velocity of rocket B relative to the Earth is ux = 0.90 c. Solve for the velocity of rocket B relative to rocket A, u x , using Eq. 36–7a. ( u x + v ) ( 0.90c − 0.65c ) ux = = = 0.60c  vu x  1 + ( − 0.65 )( 0.90 )  1 + c 2    Note that a Galilean analysis would have resulted in ux = 0.25c. 73. (a) We find the speed from Eq. 36–10a.   1 − 1 mc 2 = 14,000mc 2 → K = (  − 1) mc 2 =   1 − v 2 c2   

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

c 1   1  v = c 1−  c−   2  14,001   14,001 

Instructor Solutions Manual

→

c  1  ( 3.00  10 m s )  1  c−v =  =  14,001  = 0.77 m s 2  14,001  2   (b) The tube will be contracted in the rest frame of the electron, according to Eq. 36–3a. 8

  1 2  l 0 = l 1 − v c = ( 3.0  10 m ) 1 − 1 −    = 0.21m   14,001   2

74. The electrostatic force provides the radial acceleration. We solve that relationship for the speed of the electron. 1 e 2 melectron v 2 Felectrostatic = Fcentripetal → = → r 4 0 r 2

4 0 melectron

(8.99 10 N  m C ) (1.60 10 C ) = 2.18 10 m s = 0.0073 c = r 9.11  10 kg ( 0.53 10 m ) 9

−19

−31

−10

Because this is much less than 0.1c, the electron is not relativistic . 75. The minimum energy required would be the energy to produce the pair with no kinetic energy, so the total energy is their rest energy. They both have the same mass. Use Eq. 36–12.

E = 2mc 2 = 2 ( 0.511MeV ) = 1.022 MeV (1.637  10−13 J )

76. The wattage times the time is the energy required. We use Eq. 36–12 to calculate the mass. 7 Pt (15W ) ( 3.16  10 s )  1000g  2 −6 E = Pt = mc → m = 2 =  = 5.3  10 g 2  8 c ( 3.00 10 m s )  1kg  77. Use Eqs. 36–14, 36–8, and 36–11b.

E 2 = p 2 c 2 + m2 c 4 → E = ( p 2c 2 + m2c 4 )

1/ 2

→

−1/ 2 dE 1 2 2 pc  mvc = 2 ( p c + m2 c 4 ) 2 pc 2 = = = v  mc 2 dp E 2

78. The kinetic energy available comes from the decrease in rest energy. K = mn c 2 − ( mpc 2 + mec 2 + mv c 2 ) = 939.57 MeV − ( 938.27 MeV + 0.511MeV + 0 ) = 0.79 MeV 79. (a) We find the rate of mass loss from Eq. 36–13. E = mc 2 → E = ( m ) c 2 →

m 1  E  4  1026 J s = 2 = = 4.44  109 kg s  4  109 kg s  t c  t  ( 3.00  108 m s )2 (b) Find the time from the mass of the Sun and the rate determined in part (a). (5.98  1024 kg ) m t = Earth = = 4.27  107 y  4  107 y 9 7 m t ( 4.44  10 kg s )( 3.156  10 s y ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(c) We find the time for the Sun to lose all of its mass at this same rate. (1.99  1030 kg ) mSun t = = = 1.42  1013 y  1  1013 y m t ( 4.44  109 kg s )( 3.156  107 s y ) 80. Use Eq. 36–8 for the momentum to find the mass. mv p =  mv = → 1 − v 2 c2 2

 2.24  108 m s  3.07 10 kg m s 1  − ( )   8 p 1 − v 2 c2  3.00  10 m s  m= = = 9.12  10−31 kg 2.24  108 m s v This particle has the mass of an electron, and a negative charge, so it must be an electron. −22

81. The total binding energy is the energy required to provide the increase in rest energy. E = ( 2mp+e + 2mn ) − mHe  c 2

 931.5MeV c 2  =  2 (1.00783u ) + 2 (1.00867 u ) − 4.00260 u  c 2   = 28.32 MeV u   82. The momentum is given by Eq. 36–8, and the energy is given by Eqs. 36–11b and 36–14.

p =  mv =

 mc 2 v c

Ev pc 2 pc 2 → v = = = c2 E m2 c 4 + p 2 c 2

pc m c + p2 2 2

83. (a) The magnitudes of the momenta are equal. We use Eq. 36–8. 2 mv 1 mc ( v c ) 1 ( 938.3MeV )( 0.945 ) = = = 2711MeV c p =  mv = 1 − v 2 c2 c 1 − v 2 c2 c 1 − 0.9452

   1.602  10−10 J GeV  1c  2.71GeV c = ( 2.711GeV c )    8 1GeV  3.00  10 m s   = 1.45  10−18 kg m s (b) Because the protons are moving in opposite directions, the vector sum of the momenta is 0 . (c) In the reference frame of one proton, the laboratory is moving at 0.945c. The other proton is moving at + 0.945c relative to the laboratory. We find the speed of one proton relative to the other, and then find the momentum of the moving proton in the rest frame of the other proton by using that relative velocity. ( v + ux ) = 0.945 c + ( 0.945 c ) = 2 ( 0.945 c )  0.9984 c ux = v u x  1 + ( 0.945 )( 0.945 )  1 + ( 0.945 )2  + 1  c 2   2 1 mc ( u x c ) 1 ( 938.3MeV )( 0.9984 ) mux = = = 16567 MeV c p =  mu x = 1 − ux2 c 2 c 1 − u x2 c 2 c 1 − 0.99842

   1.602  10−10 J GeV  1c  16.6GeV c = (16.567GeV c )    8 1GeV  3.00  10 m s   = 8.85  10−18 kg m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

84. The kinetic energy is given by Eq. 36–10b.

    2 1 1 K = (  − 1) mc 2 =  − 1 mc 2 =  − 1 (14,500 kg ) ( 3.00  108 m s )  1 − 0.80 2  1 − v 2 c2  ( )     = 8.7  1020 J The spaceship’s kinetic energy is almost 9 times as great as the annual U.S. energy use. 85. The pi meson decays at rest, and so the momentum of the muon and the neutrino must each have the same magnitude (and opposite directions). The neutrino has no rest mass, and the total energy must be conserved. We combine these relationships using Eq. 36–14. Ev = ( pv 2 c 2 + mv 2 c 4 )

1/ 2

= pv c ; p  = p v = p

E = E + Ev → m c 2 = ( p 2c 2 + m 2c 4 )

+ pv c = ( p 2 c 2 + m  2 c 4 )

1/ 2

m c 2 − pc = ( p 2 c 2 + m 2 c 4 )

1/ 2

→

1/ 2

+ pc →

( m c − pc ) = ( p c + m c ) 2



2 2



2 4

Solve for the momentum. m 2 c 4 − 2m c 2 pc + p 2 c 2 = p 2 c 2 + m 2 c 4 → pc =

m 2 c 2 − m 2 c 2

2m Write the kinetic energy of the muon using Eqs. 36–11a and 36–14. K  = E − m c 2 ; E = E − Ev = m c 2 − pc →

K  = ( m c 2 − pc ) − m c 2 = m c 2 − m c 2 − =

2m ( m c 2 − m c 2 ) 2m

−

(m c − m c ) 

2 2



2 2

2m

(m c − m c ) 



2 2

2m

( 2 m − 2 m m − m + m ) c = ( m − 2m m + m ) c = ( m − m ) c = 









2m





2m







2m

86. We find the loss in mass from Eq. 36–13. E 484  103 J m = 2 = = 5.38  10−12 kg 2 8 c ( 3.00  10 m s ) Two moles of water has a mass of 36.0  10−3 kg. Find the percentage of mass lost.

5.38  10−12 kg = 1.49  10−10 = 1.49  10−8 % −3 36.0  10 kg 87. (a) We use Eq. 36–16a. To get a longer wavelength than usual means that the object is moving away from the Earth. 1.0852 − 1) ( c+v c = 0.0814 c  = 0 = 1.0850 → v = c−v (1.0852 + 1) (b) We assume that the quasar is moving and the Earth is stationary. Then we use Eq. 16–9b. f0 c c  1  → =  f =  →  = 0 (1 + v c ) = 1.0850 → v = 0.085 c 1+ v c  0  1 + v c  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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88. (a) We set the kinetic energy of the spacecraft equal to the rest energy of an unknown mass, m. Use Eqs. 36–10b and 36–12. K = (  − 1) mship c 2 = mc 2 →     1 1 m = (  − 1) mship =  − 1 mship =  − 1 (1.6  105 kg ) = 6.4  10 4 kg 2  1 − v 2 c2   1 − 0.70    This is 40% of the spacecraft’s entire mass. (b) From the Earth’s point of view, the distance is 32 ly and the speed is 0.70c. That data is used to calculate the time from the Earth frame, and then Eq. 36–1a is used to calculate the time in the spaceship frame. d ( 32 yr ) c t = = = 45.7 yr ; t0 = t 1 − v 2 c 2 = ( 45.7 yr ) 1 − 0.702 = 32.6 yr  33yr 0.70c v

89. We consider the motion from the reference frame of the spaceship. The passengers will see the trip distance contracted, as given by Eq. 36–3a. They will measure their speed to be that contracted distance divided by the year of travel time (as measured on the ship). Use that speed to find the work done (the kinetic energy of the ship). v=

l 1 − v 2 c2 l = 0 t0 t0

 W = K = (  − 1) mc 2 =   

v 1 1 = = = 0.9887 c 2  2  c       1.0ly  1 +  ct0    1 +     6.6ly   l0           1 − 1 mc 2  1 − v 2 c2 

→

  2 1 = − 1 ( 4.2  104 kg )( 3.00  108 m s ) = 2.1  1022 J  1 − 0.9887 2    Note that, according to Problem 84, this is about 200  the annual energy consumption of the U.S. 90. For the classical expression, use K = 12 mv 2 . For v = 0.6c, the value is K = 12 mv 2 = 12 (1.0 kg )( 0.6 ) ( 3.00  108 m s ) = 1.6  1016 J. 2

  1 For the relativistic expression, use Eq. 36–10a, K = mc 2  − 1 . Evaluate for v = 0.6c.  1 − v 2 c2       2 1 1 − 1 = (1.0kg ) ( 3.00  108 m s )  − 1 = 2.3  1016 J K = mc 2  2 2 2  1− v c   1 − 0.6    The requested plot is shown here.

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Instructor Solutions Manual

91. In O2’s frame of reference, O2 observes that both trains are the same length. But O2 is observing the contracted length of O1’s train. Thus, the proper length of O1’s train is longer than O2’s train. In O1’s frame of reference, O2 is moving and therefore O2’s train is length contracted, making it even shorter. When the end of O2’s train passes the end of O1’s train, the first lightning bolt strikes and both observers record the strike as occurring at the end of their train. A short while later O1 observes that O2’s train moves down the track until the front ends of the trains are at the same location. When this occurs, the second lightning bolt strikes and both observers note that it occurs at the front of their train. Since to observer O2 the trains are the same length, the strikes occurred at the same time. To observer O1, the trains are not the same length and therefore train O2 had to move between strikes so that the ends were aligned during each strike. 92. The change in length is the difference between the initial length and the contracted length. The binomial expansion is used to simplify the square root expression.

(

l = l 0 1 − v 2 c2 = l 0 1 − v 2 c2

(

)  l (1 − v c ) 1/2

1 2

)

l = l 0 − l  l 0 1 − 1 − 12 v 2 c 2  = 12 l 0 v 2 c 2

 m/s  = ( 500 m )(100 km/hr )    3.6 km/hr  2

1 2

( 3.00 10 m/s ) = 2.14 10 m 2

−12

This length contraction is less than the size of an atom, which is on the order of 10−9 m. 93. If the ship travels 0.90c at 35 above the horizontal, we can consider the motion as having two components: horizontal motion of 0.90c cos 35 = 0.737c and vertical motion of 0.90c sin 35 = 0.516c. The length of the painting will be contracted due to the horizontal motion, and the height will be contracted due to the vertical motion.

l = l 0 1 − v x2 c 2 = 1.50 m 1 − 0.737 2 = 1.01m  1.0 m h = h0 1 − v y2 c 2 = 1.00 m 1 − 0.5162 = 0.857 m  0.86 m 94. Using non-relativistic mechanics, we would find that the muon could only travel a maximum distance of ( 3.00  108 m s )( 2.20  10−6 s ) = 660m before decaying. But from an Earth-based reference frame, the muon’s “clock” would run slowly, so the time to use is the dilated time. The speed in the Earth’s reference frame is then the distance of 30 km divided by the dilated time.

l 0

t

l 0

t0

l 0

(1 − v c ) 2

t0

l 0 v = = 2 2 c c 2 ( t0 ) + ( l 0 )

→

( 3.0  10 m ) ( 3.00  10 m s ) ( 2.20 10 s ) + ( 3.0  10 m ) 4

−6

= 0.9997581 → v = 0.99976 c © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1236

Chapter 36

The Special Theory of Relativity

Alternatively, we could find the speed for the muon to travel a contracted length during it’s “normal” l l  l 0  1 = = 0 , lifetime. The contracted length is l = l 0  , so the speed would be v =  t0    t0 t0 which is the same as the earlier expression. The kinetic energy is found from Eq. 36–10a.  K = (  − 1) mc =    2

   2  − 1 mc =  (1 − v 2 c 2 )    1

(1 − ( 0.9997581) ) 2

  − 1 (105.7 MeV )  

= 4700 MeV

95. (a) Earth observers see the ship as contracted, as given by Eq. 36–3a. l = l 0 1 − v 2 c 2 = ( 35m ) 1 − ( 0.75 ) = 23.15m  23m 2

(b) Earth observers see the duration of the lunch as lengthened in time, as given by Eq. 36–1a. t0 25min t = = = 37.8min  38min 2 2 2 1 −v c 1 − ( 0.75 ) 96. Use the intensity of sunlight reaching the Earth times the area of a sphere with the radius of the Earth’s orbit (1.5  1011 m ) to calculate the power (rate of energy) leaving the sun. Then use Eq. 36–12 to relate the energy radiated per second to the mass loss per second. 2  1300 J s   11 26  P = IA =    4 (1.5  10 m )  = 3.676  10 J s 2  1m 

mass energy second 3.676  1026 J s E → = = = 4.084  109 kg s  4.1 109 kg s 2 8 second c2 c2 ( 3.0 10 m s )

97. The “gamma” factor for the motion of the meter stick is calculated from its contracted length using Eq. 36–3, and then its kinetic energy is calculated from Eq. 36–10. 2  1.000m  l  = 0 ; K = (  − 1) mc 2 =  − 1 ( 0.30kg ) ( 3.0  108 m s ) = 2.9  1016 J l  0.480m  98. (a) The relative speed can be calculated in either frame, and will be the same value in both frames. The time as measured on the Earth will be longer than the time measured on the spaceship, as given by Eq. 36–1a. tspaceship tspaceship x = → v = Earth ; tEarth = 2 2 2 tEarth 1− v c  xEarth  1−    ctEarth  2

2 x 2 ( tEarth ) −  Earth  = ( tspaceship ) →  c  2

2  x  tEarth =  Earth  + ( tspaceship ) =  c 

( 6.00 y ) + ( 3.25 y ) = 6.82 y 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The distance as measured by the spaceship will be contracted. xspaceship tspaceship x 3.25 y v = Earth = → xspaceship = xEarth = ( 6.00ly ) = 2.86ly tEarth tspaceship tEarth 6.82 y 99. (a) The kinetic energy is 5.00 times as great as the rest energy. Use Eq. 36–10b. 1 1 = 6.00 → v = c 1 − = 0.986c K = (  − 1) mc 2 = 5.00mc 2 → 2 2 2 1− v c ( 6.00 ) (b) The kinetic energy is 999 times as great as the rest energy. We use the binomial expansion. 1 K = (  − 1) mc 2 = 999mc 2 → = 1000 → 1 − v 2 c2 v = c 1−

 1  1 2  −7 c  1 − 2    = c (1 − 5.0  10 ) 2 1000    (1000 )  1

100. (a) To observers on the ship, the period is non-relativistic. Use Eq. 14–7b. T = 2

( 2.42 kg ) = 1.0652s  1.07s (84.2 N m )

m = 2 k

(b) The oscillating mass is a clock. According to observers on Earth, clocks on the spacecraft run slow. (1.0652s ) T TEarth = = = 2.44s 2 2 2 1− v c 1 − ( 0.900 ) 101. We use the Lorentz transformations relating two reference frames moving with respect to each other along the x-axis to derive the result. v x  vx    x =  ( x + vt  ) → x =  ( x + vt  ) ; t =   t  + 2  → t =   t  + 2  c  c    2 2   2  v x  v x   2 2   x t v −  +  ( ct ) − ( x ) = c  t  + 2   −  ( x + vt  ) =  2   ct  + ( )   c  c        2  2 v x  v x  2 2 2  =  c ( t  ) + 2ct  +  − ( x ) − 2xvt  − ( vt  )  c  c    2

  2  v  2  1 2 2 2   c t x 1 = −  + −  v ( ) ( )   ( )     (1 − v 2 c 2 )   c    =



1 2 2 c 2 (1 − v 2 c 2 ) ( t  ) − (1 − v 2 c 2 ) ( x ) 2 2 1− v c

(1 − v c ) ct  − x = ct  − x = ( ) ( ) ( ) ( ) (1 − v c )  2



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Chapter 36

The Special Theory of Relativity

102. We assume that the left edge of the glass is even with point A when the flash of light is emitted. There is no loss of generality with that assumption. We do the calculations in the frame of reference in which points A and B are at rest, and the glass is then moving to the right with speed v. If the glass is not moving (the “ v = 0” case), we would have this “no motion” result. distance in glass distance in vacuum d l −d tv =0 = tglass + tvacuum = + = + speed in glass speed in vacuum c vglass

d l − d nd l − d nd + l − d l + ( n − 1) d + = + = = cn c c c c c If the index of refraction is n = 1, then the glass will have no effect on the light, and the time would simply be the distance divided by the speed of light. distance in glass distance in vacuum d l − d d + l − d l tn =1 = tglass + tvacuum = + = + = = speed in glass speed in vacuum c c c c This is the same result we obtain from the “ v = 0” if n = 1 . =

Now, let us consider the problem from a relativistic point of view. The speed of light in the glass will be the relativistic sum of the speed of light in stationary glass, c n , and the speed of the glass, v, by Eq. 36–7a. We define  to simplify further expressions.   vn   vn   c c +v +v 1 +  1 +    c c c c    n n  =   vlight = = = = v n cv v  v  n  in moving  1 1 + + glass 1 +  1 +    nc 2 nc  nc   nc   The contracted width of the glass, from the Earth frame of reference, is given by Eq. 36–3a. d d moving = d 1 − v 2 c 2 =



glass

Again, we assume the light enters the block when the left edge of the block is at point A, and write simple equations for the displacement of the leading edge of the light, and the leading edge of the block. Set them equal and solve for the time when the light exits the right edge of the block. c d xlight = vlight t =  t ; xright = + v t ;  n in glass edge c d d n xlight = xright →  tglass = + vtglass → tglass =   (  c − nv ) n edge Where is the front edge of the block when the light emerges? Use tglass =

n with either  (  c − nv )

expression–for the leading edge of the light, or the leading edge of the block. cd n  cd xlight = vlight tglass =  = n  (  c − nv )  (  c − nv ) in glass

xright = edge



+ vtglass =



d (  c − nv ) + vdn n  cd = =  (  c − nv )  (  c − nv )  (  c − nv ) d

 cd , will be traveled at speed c by the light. We express  (  c − nv ) that time, and then find the total time. The part of the path that is left, l −

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

l− tvacuum =

 cd  (  c − nv ) c

ttotal = tglass + tvacuum = =

Instructor Solutions Manual

l−

 (  c − nv )

l ( n − 1) d + c c

 cd  (  c − nv ) c

l d n− + c  (  c − nv )

c−v c+v

We check this for the appropriate limiting cases. l ( n − 1) d c − v l ( n − 1) d c − c l = + = Case 1: ttotal = + c c c+v c c c+c c v =c This result was expected, because the speed of the light would always be c. l + ( n − 1) d l ( n − 1) d c − v l ( n − 1) d = + Case 2: ttotal = + (1) = c c c+v c c c v =0 This result was obtained earlier in the solution. l ( n − 1) d c − v l = Case 3: ttotal = + c c c+v c n =1 This result was expected, because then there is no speed change in the glass. 103. The kinetic energy of 998 GeV is used to find the speed of the protons. Since the energy is over 1000 times the rest mass, we expect the speed to be very close to c. Use Eq. 36–10b.   1 − 1 mc 2 → K = (  − 1) mc 2 =   1 − v 2 c2   

v = c 1−

1  K   2 + 1  mc 

= c 1−

1  998GeV  + 1   0.938GeV 

= c ( to 7 sig. figs.)

 998GeV   K  + 1 (1.673  10−27 kg )( 3.00  108 m s ) + 1 mc   2 0.938GeV   mv  mv  mc  =  = = 3.3T B= rqv rq rq (1.0 103 m )(1.60 10−19 C ) 2

1240

CHAPTER 37: Early Quantum Theory and Models of the Atom Responses to Questions 1.

Bluish stars are the hottest, whitish-yellow stars are hot, and reddish stars are the coolest. This follows from Wien’s law, which says that stars with the shortest wavelength peak in their spectrum have the highest temperatures. Blue is the shortest visible wavelength and red is the longest visible wavelength, so blue is the hottest, and red is the coolest.

The difficulty with seeing objects in the dark is that although all objects emit radiation, only a small portion of the electromagnetic spectrum can be detected by our eyes. Usually objects are so cool that they only give off very long wavelengths of light (infrared), which our eyes are unable to detect.

The lightbulb will not produce light as white as the Sun, since the peak of the lightbulb’s emitted light is in the infrared. The lightbulb will appear more yellowish than sunlight. The Sun has a spectrum that peaks in the visible range.

According to the wave theory, light of any frequency can cause electrons to be ejected as long as the light is intense enough. A higher intensity corresponds to a greater electric field magnitude and more energy. Therefore, there should be no frequency below which the photoelectric effect does not occur. According to the particle theory, however, each photon carries an amount of energy which depends upon its frequency. Increasing the intensity of the light increases the number of photons but does not increase the energy of the individual photons. The cutoff frequency is that frequency at which the energy of the photon equals the work function. If the frequency of the incoming light is below the cutoff, then the electrons will not be ejected because no individual photon has enough energy to eject an electron.

If the threshold wavelength increases for the second metal, then the second metal has a smaller work function than the first metal. Longer wavelength corresponds to lower energy. It will take less energy for the electron to escape the surface of the second metal.

Cesium will give a higher maximum kinetic energy for the ejected electrons. Since the incident photons bring in a given amount of energy, and in cesium less of this energy goes to releasing the electron from the material (the work function), it will give off electrons with a higher kinetic energy.

Individual photons of ultraviolet light are more energetic than photons of visible light and will deliver more energy to the skin, causing burns. UV photons also can penetrate farther into the skin and, once at the deeper level, can deposit a large amount of energy that can cause damage to cells.

No, fewer electrons are emitted, but each one is emitted with higher kinetic energy, when the 400-nm light strikes the metal surface. The intensity (energy per unit time) of both light beams is the same, but the 400-nm photons each have more energy than the 450-nm photons. Thus there are fewer photons hitting the surface per unit time. This means that fewer electrons will be ejected per unit time from the surface with the 400-nm light. The maximum kinetic energy of the electrons leaving the metal surface will be greater, though, since the incoming photons have shorter wavelengths and more energy per photon, and it still takes the same amount of energy (the work function) to remove each electron. This “extra” energy goes into higher kinetic energy of the ejected electrons.

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(a) No. The energy of a beam of photons depends not only on the energy of each individual photon but also on the total number of photons in the beam. It is possible that there could be many more photons in the IR beam than in the UV beam. In this instance, even though each UV photon has more energy than each IR photon, the IR beam could have more total energy than the UV beam. (b) Yes. A photon’s energy depends on its frequency: E = hf. Since IR light has a lower frequency than UV light, a single IR photon will always have less energy than a single UV photon.

10. Yes, an X-ray photon that scatters from an electron does have its wavelength changed. The photon gives some of its energy to the electron during the collision and the electron recoils slightly. Thus, the photon has less energy and its wavelength is longer after the collision, since the energy and wavelength are inversely proportional to each other ( E = hf = hc  ). 11. In the photoelectric effect, the photons (typically visible frequencies) have only a few eV of energy, whereas in the Compton effect, the photons (typically X-ray frequencies) have more than 1000 times greater energy and a correspondingly smaller wavelength. In the photoelectric effect, the incident photons eject electrons completely out of the target material while the photons are completely absorbed (no scattered photons). In the Compton effect, the photons are not absorbed, but are scattered from the electrons. 12. According to both the wave theory and the particle theory, the intensity of a point source of light decreases as the inverse square of the distance from the source. In the wave theory, the intensity of the waves obeys the inverse square law. In the particle theory, the surface area of a sphere increases with the square of the radius, and therefore the density of particles decreases with distance, obeying the inverse square law. The variation of intensity with distance cannot be used to help distinguish between the two theories. 13. The proton will have the shorter wavelength, since it has a larger mass than the electron and therefore a larger momentum (  = h p ). 14. Light demonstrates characteristics of both waves and particles. Diffraction and interference are wave characteristics, and are demonstrated, for example, in Young’s double-slit experiment. The photoelectric effect and Compton scattering are examples of experiments in which light demonstrates particle characteristics. We can’t say that light IS a wave or a particle, but it has properties of each. 15. We say that electrons have wave properties since we see them act like waves when they are diffracted, or exhibit 2-slit interference. We say that electrons have particle properties since we see them act like particles when they are bent by magnetic fields or accelerated and fired into materials where they scatter other electrons. 16. Both a photon and an electron have properties of waves and properties of particles. They can both be associated with a wavelength and they can both undergo scattering. An electron has a negative charge and a rest mass, obeys the Pauli exclusion principle, and travels at less than the speed of light. A photon is not charged, has no rest mass, does not obey the Pauli exclusion principle, and travels at the speed of light. Property Photon Electron 9.11  10 −31 kg Mass None

−1.60  10−19 C Charge None 3  108 m s  3  108 m s Speed There are other properties, such as spin, that will be discussed in later chapters. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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17. In Rutherford’s planetary model of the atom, the Coulomb force (electrostatic force) keeps the electrons from flying off into space. Since the protons in the center are positively charged, the negatively charged electrons are attracted to the center by the Coulomb force and orbit around the center just like the planets orbiting a sun in a solar system due to the attractive gravitational force. 18. To tell if there is oxygen near the surface of the Sun, you need to collect light coming from the Sun and spread it out using a diffraction grating or prism so you can see the spectrum of wavelengths. If there is oxygen near the surface, there will be dark (absorption) lines at the wavelengths corresponding to electron transitions in oxygen. 19. At room temperature, nearly all the atoms in hydrogen gas will be in the ground state. When light passes through the gas, photons are absorbed, causing electrons to make transitions to higher states and creating absorption lines. These lines correspond to the Lyman series since that is the series of transitions involving the ground state or n = 1 level. Since there are virtually no atoms in higher energy states, photons corresponding to transitions from n > 2 to higher states will not be absorbed. 20. The closeness of the spacing between energy levels near the top of Fig. 37–26 indicates that the energy differences between these levels are small. Small energy differences correspond to small wavelength differences, leading to the closely spaced spectral lines in Fig. 37–21. 21. (a) The Bohr model successfully explains why atoms emit line spectra; it predicts the wavelengths of emitted light for hydrogen; it explains absorption spectra; it ensures the stability of atoms (by decree); and it predicts the ionization energy of hydrogen. (b) The Bohr model did not give a reason for orbit quantization; it was not successful for multielectron atoms; it could not explain why some emission lines are brighter than others; and it could not explain the “fine structure” of some very closely-spaced spectral lines. 22. It is possible for the de Broglie wavelength (  = h p ) of a particle to be larger than the dimension of the particle. If the particle has a very small mass and a slow speed (like a low-energy electron or proton) then the wavelength may be larger than the dimension of the particle. It is also possible for the de Broglie wavelength of a particle to be smaller than the dimension of the particle if it has a large momentum and a moderate speed (like a baseball). There is no direct connection between the size of a particle and the size of the de Broglie wavelength of a particle. For example, you could also make the wavelength of a proton much smaller than the size of the proton by making it go very fast. 23. The lines in the spectrum of hydrogen correspond to all the possible transitions that the electron can make. The Balmer lines, for example, correspond to an electron moving from all higher energy levels to the n = 2 level. Although an individual hydrogen atom only contains one electron, a sample of hydrogen gas contains many atoms and all the different atoms could be undergoing different transitions, leading to many lines being visible simultaneously. 24. Both particles will have the same kinetic energy. But since the proton has a larger mass, it will have

(

)

a larger momentum than the electron K = p 2 2m → p = 2mK . Wavelength is inversely proportional to momentum, so the proton will have the shorter wavelength, and the electron will have the longer wavelength. 25. The Balmer series spectral lines are in the visible light range and could be seen by early experimenters without special detection equipment. It was only later that the UV (Lyman) and IR (Paschen) regions were explored thoroughly, using detectors other than human sight. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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26. When a photon is emitted by a hydrogen atom as the electron makes a transition from one energy state to a lower one, not only does the photon carry away energy and momentum, but to conserve momentum, the atom must also take away some momentum. If the atom carries away some momentum, then it must also carry away some of the available energy, which means that the photon takes away less energy than Eq. 37–9 predicts. 27. (a) (b) (c) (d) (e)

continuous line, emission continuous line, absorption continuous with absorption lines (like the Sun)

28. No, the two spectra will not contain identical lines. At room temperature, virtually all the atoms in a sample of hydrogen gas will be in the ground state. Thus, the absorption spectrum will contain primarily just the Lyman lines, as photons corresponding to transitions from the n = 1 level to higher levels are absorbed. Hydrogen at very high temperatures will have atoms in excited states. The electrons in the higher energy levels will fall to all lower energy levels, not just the n = 1 level. Therefore, emission lines corresponding to transitions to levels higher than n = 1 will be present as well as the Lyman lines. In general, you would expect to see only Lyman lines in the absorption spectrum of room-temperature hydrogen, but you would find Lyman, Balmer, Paschen, and other lines in the emission spectrum of high-temperature hydrogen. 29. On average, the electrons of helium are closer to the nucleus than the electrons of hydrogen. The nucleus of helium contains two protons (positive charges), and so attracts each electron more strongly than the single proton in the nucleus of hydrogen. (There is some shielding of the nuclear charge by the “other” electron, but each electron still feels an average attractive force of more than one proton’s worth of charge.) 30. (a) (b) (c) (d)

particle wave particle wave

Responses to MisConceptual Questions 1.

(b) A common misconception is that the maximum wavelength increases as the temperature increases. However, the temperature and maximum wavelength are inversely proportional. As the temperature increases the intensity increases and the wavelength decreases.

(b) The blue light has a shorter wavelength and therefore more energy per photon. Since both beams have the same intensity (energy per unit time per unit area), the red light will have more photons.

(d) A common misconception is that violet light has more energy than red light. However, the energy is the product of the energy per photon and the number of photons. A single photon of violet light has more energy than a single photon of red light, but if a beam of red light has more photons than the beam of violet light, the red light could have more energy.

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(d) The energy of the photon E is equal to the sum of the work function of the metal and the kinetic energy of the released photons. If the energy of the photon is more than double the work function, then cutting the photon energy in half will still allow electrons to be emitted. However, if the work function is more than half of the photon energy, then no electrons would be emitted if the photon energy were cut in half.

(c) An increase in intensity of the incoming light means that more photons are incident, so more electrons will be ejected, but since the photon energy is unchanged, the maximum energy of the ejected electrons will also be unchanged.

(a) Since no electrons are being ejected, the incoming photons do not have enough energy to overcome the work function. Higher energy photons are needed, which means a higher frequency of photon, which is a shorter wavelength photon.

(b) The momentum of a photon is inversely proportional to its wavelength. Therefore doubling the momentum would cut the wavelength in half.

(d) A common misconception is to think that only light behaves as both a particle and as a wave. De Broglie postulated, and experiments have confirmed that in addition to light, electrons and protons (and many other particles) have both wave and particle properties that can be observed or measured.

(d) This is the central idea of the “wave-particle duality” discussed in Section 37–6.

10. (d) A thrown baseball has a momentum on the order of 1 kg m s. The wavelength of the baseball is Planck’s constant divided by the momentum. Due to the very small size of Planck’s constant, the wavelength of the baseball would be much smaller than the size of a nucleus. 11. (d) This chapter demonstrates the wave–particle duality of light and matter. Both electrons and photons have momentum that is related to their wavelength by p = h  . Young’s double-slit experiment demonstrated diffraction with light and later experiments have demonstrated electron diffraction. Therefore, all three statements are correct. 12. (a, c, d) Alpha particles are positive. To produce large angle scattering, the nucleus of the atom would have to repel the alpha particle, thus the nucleus must be positive. If the nuclear charge was spread out over a large area there would be more small angle scattering, but insufficient force to create large angle scattering. With a small nuclei, large scattering forces are possible. Since most of the alpha particles pass through the foil undeflected, most of the atom must be empty space. The scattering does not require quantized charge, but would be possible with a continuous charge distribution. Therefore answer (b) is incorrect. 13. (d) The current model of the atom is the quantum mechanical model. The plum-pudding model was rejected as it did not explain Rutherford scattering. The Rutherford atom was rejected as it did not explain the spectral lines emitted from atoms. The Bohr atom did not explain fine structure. Each of these phenomena are explained by the quantum mechanical model. 14. (d) A photon is emitted when the electron transitions from a higher state to a lower state. Photons are not emitted when an electron transitions from 2 → 5 or from 5 → 8. The other two transitions are to states that are three states below the original state. The energy levels change more rapidly for lower values of n, so the 5 → 2 transition will have a higher energy, and therefore a shorter wavelength than the 8 → 5 transition. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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15. (b) The lowest energy cannot be zero, if zero energy has been defined when the electron and proton are infinitely far away. As the electron and proton approach each other, their potential energy decreases. The energy levels are quantized and therefore cannot be any value. As shown after Eq. 37–14b, the lowest energy level of the hydrogen atom is –13.6 eV.

Solutions to Problems In several problems, the value of hc is needed. We often use the result of Problem 24, hc = 1240eV nm. 1.

We use Wien’s law, Eq. 37–1. ( 2.90 10−3 m K ) = ( 2.90 10−3 m K ) = 1.06  10−5 m = 10.6 m (a) P = T ( 273K ) This wavelength is in the far infrared . 2.90  10−3 m K ) ( 2.90  10−3 m K ) ( (b) P = = = 8.29  10−7 m = 829 nm T ( 3500 K ) This wavelength is in the infrared . 2.90  10−3 m K ) ( 2.90  10−3 m K ) ( (c) P = = = 6.90  10−4 m = 0.69 mm T ( 4.2 K ) This wavelength is in the microwave region. ( 2.90  10−3 m K ) = ( 2.90 10−3 m K ) = 1.06  10−3 m = 1.06 mm (d) P = T ( 2.725K ) This wavelength is in the microwave region.

Use Wien’s law, Eq. 37–1, to find the temperature for a peak wavelength of 490 nm. ( 2.90  10−3 m K ) = ( 2.90 10−3 m K ) = 5900 K T= P ( 490  10−9 m )

Because the energy is quantized according to Eq. 37–2, the difference in energy between adjacent levels is simply E = hf . E = hf = ( 6.63  10−34 J s )(8.1  1013Hz ) = 5.37  10 −20 J  5.4  10 −20 J 1eV   5.37  10−20 J  = 0.34eV −19   1.60  10 J 

Use Wein’s law, Eq. 37–1. ( 2.90  10−3 m K ) = ( 2.90 10−3 m K ) = 1.81  105 K (a) T = P (16.0  10−9 m )

( 2.90  10 m K ) = ( 2.90  10 m K ) = 1.318  10 m  1.3  m (b)  = −3

−3

−6

( 2200 K )

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The potential energy is “quantized” in units of mgh. (a) U1 = mgh = ( 68.0kg ) ( 9.80m s2 ) ( 0.200m ) = 133.28J  133J (b) U 2 = mg ( 2h ) = 2U1 = 2 (133.28J ) = 267J (c) U 3 = mg ( 3h ) = 3U1 = 3 (133.28J ) = 4.00  102 J (d) U n = mg ( nh ) = nU1 = n (133.28J ) = (133n ) J (e)

E = U 2 − U 6 = ( 2 − 6 )(133.28J ) = −533J

Use Eq. 37–1 with a temperature of 98.6 F = 37.0 C = ( 273 + 37 ) K = 310 K ( 3 sig. figs.)

( 2.90  10 m K ) = ( 2.90 10 m K ) = 9.35  10 m = 9.35  m  = −3

−3

−6

( 310 K )

(a) Wien’s displacement law says that PT = constant. We must find the wavelength at which I (  , T ) is a maximum for a given temperature. This can be found by setting I  = 0.

 −5  I   2 hc 2 −5  2    = = hc 2        e hc /  kT − 1    e hc /  kT − 1   hc /  kT  hc   − 1)( −5 −6 ) −  −5e hc /  kT  − 2   (e  kT    = 2 hc 2  2   ( ehc/kT − 1)   hc  2 hc 2   hc   = − 5   = 0 → 5 = e hc /  kT  5 − 5 + e hc /  kT   → 2  hc /  kT 6 kT    kT     (e − 1)  e x (5 − x ) = 5 ; x =

hc P kT

This transcendental equation will have some solution x = constant, and so

hc = constant, and P kT

so PT = constant . The constant could be evaluated from solving the transcendental equation. (b) To find the value of the constant, we solve e x ( 5 − x ) = 5, or 5 − x = 5e− x . This can be done graphically, by graphing both y = 5 − x and y = 5e − x on the same set of axes and finding the intersection point. Or, the quantity 5 − x − 5e− x could be calculated, and find for what value of x that expression is 0. The answer is x = 4.966. We use this value to solve for h. hc = 4.966 → P kT

h = 4.966

PTk

( 2.90  10 m K )(1.38 10 J K ) = 6.62  10 J s = 4.966 −3

−23

−34

c 3.00  108 m s (c) We integrate Planck’s radiation formula over all wavelengths.    2 hc 2  −5  hc hc hc   = I , T d 0 ( ) 0  ehc / kT − 1  d  ; let  kT = x ;  = xkT ; d  = − x2 kT dx

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−5   2  hc   hc 2     2 −5 4 4  3  2 hc   xkT    hc  2 k T  x       = = − = I T d d dx , ( )      0 0  ehc / kT − 1    e x − 1   x 2 kT  h3c 2 0  e x − 1  dx     



 2 k 4   x3   4 4    dx  T  T h3c 2  0  e x − 1  

Thus the total radiated power per unit area is proportional to T 4 . Everything else in the expression is constant with respect to temperature. 8.

We use Eq. 37–3 to find the energy of the photons.

We use Eq. 37–3 along with the fact that f = c  for light. The longest wavelength will have the lowest energy. −34 8 hc ( 6.626  10 J s )( 3.00  10 m / s ) 1eV   E1 = hf1 = = = 4.85  10−19 J  = 3.03eV −19  −9  1 1.60 10 J   ( 410  10 m )

E = hf = ( 6.626  10−34 J s )(88.5  106 Hz ) = 5.86  10 −26 J

E2 = hf 2 =

2

( 6.626  10 J s )( 3.00 10 m / s ) = 2.65  10 J  1eV  = 1.66eV    1.60  10 J  ( 750  10 m ) −34

−19

−9

Thus, the range of energies is 2.7 10−19 J  E  4.9 10−19 J or 1.7eV  E  3.0eV . 10. At the minimum frequency, the kinetic energy of the ejected electrons is 0. Use Eq. 37–4a. 5.2  10−19 J W K = h f min − W0 = 0 → f min = 0 = = 7.8  1014 Hz h 6.626  10−34 J s 11. The longest wavelength corresponds to the minimum frequency. That occurs when the kinetic energy of the ejected electrons is 0. Use Eq. 37–4a. We check our result with a result from Problem 24. c W K = hf min − W0 = 0 → f min = = 0 → h max 8 −34 ch ( 3.00  10 m s )( 6.626  10 J s ) max = = = 3.50  10−7 m = 350 nm or W0 ( 3.55eV ) (1.60  10−19 J eV )

max =

hc 1240eV nm = = 349 nm 3.55eV W0

12. We use Eq. 37–3 with the fact that f = c  for light. We utilize a result from Problem 24. c hc 1240eV nm = = = = 2.952  10−3 nm  3.0  10−3 nm 3 f E 420  10 eV Significant diffraction occurs when the opening is on the order of the wavelength. Thus, there would be negligible diffraction through the doorway.

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13. We use Eq. 37–3 with the fact that f = c  for light. Emin = hf min

max =

−19 Emin ( 0.1eV ) (1.60  10 J eV ) → f min = = = 2.41  1013 Hz  2  1013 Hz −34 h  6.63 10 J s ( )

( 3.00  10 m s ) = 1.24  10 m  1  10 m = ( 2.41  10 Hz ) 8

−5

f min

−5

14. The energy of the photon will equal the kinetic energy of the baseball. We use Eq. 37–3. −34 8 c 2hc 2 ( 6.626  10 J s )( 3.00  10 m s ) 2 1 K = hf → 2 mv = h → = = = 3.76  10−27 m 2  mv 2 ( 0.145kg )( 27.0 m s ) 15. We divide the minimum energy by the photon energy at 550 nm to find the number of photons. 10−18 J )( 550  10−9 m ) ( Emin Emin  E = nhf = Emin → n = = = = 2.77  3 photons hf hc ( 6.63  10−34 J s )( 3.00 108 m s ) 16. The photon of visible light with the maximum energy has the least wavelength. We use 400 nm as the lowest wavelength of visible light, and calculate the energy for that wavelength. ( 6.63  10−34 J s )( 3.00  108 m/s ) = 3.11eV hc hf max = = min (1.60  10−19 J/eV )( 400  10−9 m ) Electrons will not be emitted if this energy is less than the work function. The metals with work functions greater than 3.11 eV are copper and iron . 17. (a) At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so the work function is equal to the energy of the photon. We use a result from Problem 24. hc 1240eV nm W0 = hf − K max = hf = = = 2.296eV  2.3eV  540 nm (b) The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy. We use Eq. 37–4b to calculate the maximum kinetic energy, along with a result from Problem 24. hc 1240eV nm K max = hf − W0 = − W0 = − 2.296eV = 0.287eV ( only 1 significant figure )  480 nm

V0 =

K max 0.287eV = = 0.287 V  0.3V e e

18. The photon of visible light with the maximum energy has the minimum wavelength. We use Eq. 37–4b to calculate the maximum kinetic energy, with a result from Problem 24. hc 1240eV nm K max = hf − W0 = − W0 = − 2.48eV = 0.54eV  410 nm 19. We use Eq. 37–4b to calculate the work function, along with a result from Problem 24. hc 1240eV nm W0 = hf − K max = − K max = − 1.55eV = 2.80eV  285nm

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20. Electrons emitted from photons at the threshold wavelength have no kinetic energy. We use Eq. 37–4b with the threshold wavelength to determine the work function. hc hc 1240eV nm W0 = − K max = = = 3.758eV  max 330 nm (a) We now use Eq. 36–4b with the work function determined above to calculate the kinetic energy of the photoelectrons emitted by 290 nm light. hc 1240eV nm K max = − W0 = − 3.758eV = 0.5179eV  0.5eV  290 nm The subtraction rule for significant figures is used here. (b) Because the wavelength is greater than the threshold wavelength, the photon energy is less than the work function, so there will be no ejected electrons . 21. The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy of the photoelectrons. We use Eq. 37–4b to calculate the work function where the maximum kinetic energy is the product of the stopping voltage and electron charge. We also use a result from Problem 24. hc hc 1240eV nm W0 = − K max = − eV0 = − (1.74 V ) e = 3.651eV  3.7eV   230 nm The subtraction rule for significant figures is used here. 22. The energy required for the chemical reaction is provided by the photon. We use Eq. 37–3 for the energy of the photon, where f = c / . We also use a result from Problem 24. hc 1240eV nm E = hf = = = 1.851eV  1.9eV  670 nm Each reaction takes place in a molecule, so we use the appropriate conversions to convert eV/molecule to kcal/mol. −19 23  1.851eV   1.60  10 J  6.02  10 molecules  kcal  E =    = 42.59 kcal mol  eV mol  molecule      4186 J 

 43kcal/mol 23. (a) Since f = c  , the photon energy given by Eq. 37–3 can be written in terms of the wavelength as E = hc  . This shows that the photon with the largest wavelength has the smallest energy. The 700-nm photon then delivers the minimum energy that will excite the retina. –34 8 hc ( 6.63  10 J s )( 3.00  10 m s )  1 eV  E= = = 1.776eV  1.8eV  –19  –9  1.60 10 J   700 10 m   ( ) (b) The eye cannot see light with wavelengths less than 410 nm. Obviously, these wavelength photons have more energy than the minimum required to initiate vision, so they must not arrive at the retina. So we speculate that photons with a wavelength less than 410 nm are absorbed by the eye before the photon reaches the retina. The threshold photon energy is that of a 410-nm photon. –34 8 hc ( 6.63  10 J s )( 3.00  10 m s )  1 eV  E= = = 3.032eV  3.0eV  –19  –9   1.60  10 J  ( 410  10 m ) We assumed both wavelengths were measured to the nearest 10 nm.

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24. (a) Since f = c  , the energy of each emitted photon is E = hc  . We insert the values for h and c and convert the resulting units to eV nm. –34 8 –19 eV nm hc ( 6.626  10 J s )( 2.998  10 m s ) (1eV 1.602  10 J ) = = 1239.997 E= –9    ( in nm ) (10 m 1nm )



1.240  103 eV nm  ( in nm )

(b) Insert 610 nm into the above equation. 1240 eV nm E= = 2.03eV  2.0eV 610 nm 25. (a) Since f = c  , the photon energy is E = hc  and the largest wavelength has the smallest energy. In order to eject electrons for all possible incident visible light, the metal’s work function must be less than or equal to the energy of a 750-nm photon. Thus the maximum value for the metal’s work function Wo is found by setting the work function equal to the energy of the 750-nm photon. –34 8 hc ( 6.63  10 J s )( 3.00  10 m s )  1 eV  Wo = = = 1.7 eV  –19  –9   1.60  10 J  ( 750  10 m ) (b) If the photomultiplier is to function only for incident wavelengths less than 410 nm, then we set the work function equal to the energy of the 410-nm photon. –34 8 hc ( 6.63  10 J s )( 3.00  10 m s )  1 eV  Wo = = = 3.0 eV  –19  –9   1.60  10 J  ( 410  10 m ) 26. Since f = c  , the energy of each emitted photon is E = hc  . We multiply the energy of each photon by n = 1.3  106 s to determine the average power output of each atom. At a distance of

r = 25 cm, the light sensor measures an intensity of I = 1.6 nW 1.0 cm 2. Since light energy emitted from atoms radiates equally in all directions, the intensity varies with distance as a spherical wave. Thus, from Section 15–3 in the text, the average power emitted is P = 4 r 2 I . Dividing the total average power by the power from each atom gives the number of trapped atoms. 2 4 ( 25cm ) (1.6  10−9 W/cm 2 ) P 4 r 2 I N= = = Patom nhc  (1.3  106 /s )( 6.63  10−34 J s )( 3.00  108 m/s ) / ( 780  10−9 m ) = 3.8  107 atoms

27. We set the kinetic energy in Eq. 37–4b equal to the stopping voltage, eV0 , and write the frequency of the incident light in terms of the wavelength, f = c  . We differentiate the resulting equation and solve for the fractional change in wavelength, and we take the absolute value of the final expression. hc hc d e dV0  e  eV0 = − W0 → e dV0 = − 2 d  → =−  = V0     hc hc





(1.60  10 C )(550 10 m ) ( 0.01V ) = 0.004 ( 6.63  10 J s )( 3.00 10 m s ) −19

−34

−9 8

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28. We use Eq. 37–6b. h  = (1 − cos  ) → me c

 

 = cos−1  1 −

 ( 9.11  10−31 kg )( 3.00  108 m s )(1.2  10 −13 m )  me c  −1 1 −  = 18 = cos  −34   h   6.63 10 J s ( )  

29. The Compton wavelength for a particle of mass m is h mc . (a) (b)

6.63  10−34 J s ) ( h = = 2.43  10−12 m −31 8 me c ( 9.11  10 kg )( 3.00  10 m s )

( 6.63  10−34 J s ) h = = 1.32  10−15 m mp c (1.67  10−27 kg )( 3.00  108 m s )

(c) The energy of the photon is given by Eq. 37–3. hc hc Ephoton = hf = = = mc 2 = rest energy  ( h mc ) 30. We find the Compton wavelength shift for a photon scattered from an electron, using Eq. 37–6b. The Compton wavelength of a free electron is given in the text right after Eq. 37–6b, and was calculated in Problem 29 above. The angles are taken as exact, and so do not limit the number of significant figures in the answers.  h  −3  −  =   (1 − cos ) = C (1 − cos ) = ( 2.43  10 nm ) (1 − cos  )  me c  (a)  a −  = ( 2.43  10−3 nm ) (1 − cos60 ) = 1.22  10−3 nm

(b)  b −  = ( 2.43  10−3 nm ) (1 − cos90 ) = 2.43  10−3 nm (c)

 c −  = ( 2.43  10−3 nm ) (1 − cos180 ) = 4.86  10−3 nm

31. (a) In the Compton effect, the maximum change in the photon’s wavelength occurs when the scattering angle  = 180. We use Eq. 37–6b to determine the maximum change in wavelength. Dividing the maximum change by the initial wavelength gives the maximum fractional change. h  = (1 – cos ) → me c

6.63  10 –34 J s ) (1 − cos180 ) ( h = = 8.8  10−6 (1 – cos ) =  me c ( 9.1110–31 kg )(3.00 108 m s )(550 10−9 m )



(b) We replace the initial wavelength with  = 0.10 nm. 



( 6.63  10–34 J s ) (1 − cos180 ) h = 0.049 (1 – cos ) = me c ( 9.11  10–31 kg )( 3.00  108 m s )( 0.10  10−9 m )

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32. We find the change in wavelength for each scattering event using Eq. 37–6b, with a scattering angle of  = 0.50. To calculate the total change in wavelength, we subtract the initial wavelength, obtained from the initial energy, from the final wavelength. We divide the change in wavelength by the wavelength change from each event to determine the number of scattering events. 6.63  10 –34 J s ) (1 − cos0.5 ) ( h o 1 – cos0.5 ) =  = = 9.24  10 –17 m = 9.24  10−8 nm ( –31 8 me c (9.1110 kg )(3.00 10 m s ) –34 8 hc ( 6.63  10 J s )( 3.00 10 m s ) 0 = = = 1.24  10 –12 m = 0.00124nm. 6 –19 E0 (1.0  10 eV )(1.60  10 J eV )

 − 0 ( 555 nm ) – ( 0.00124 nm ) = = 6  109 events –8  9.24  10 nm

33. (a) We use conservation of momentum to set the initial momentum of the photon equal to the sum of the final momentum of the photon and electron, where the momentum of the photon is given by Eq. 37–5 and the momentum of the electron is written in terms of the total energy (Eq. 36–14). We multiply this equation by the speed of light to simplify. h hc h  hc  + 0 = −   + pe → = −   + E 2 − E02         Using conservation of energy, we set the initial energy of the photon and rest energy of the electron equal to the sum of the final energy of the photon and the total energy of the electron.  hc   hc    + E0 =   + E      By summing these two equations, we eliminate the final wavelength of the photon. We then solve the resulting equation for the kinetic energy of the electron, which is the total energy less the rest energy. 2

  hc    hc  2   + E0 = E 2 − E0 2 +E →  2   + E0 − E  = E 2 − E0 2       2

  hc     hc   2 2 2  2    + E0  − 2 E  2    + E 0  + E = E − E 0         2

  hc   2  2    + E0  + E0    → E=   hc   2  2   + E0      2

  hc     hc   2  hc  2   2    + E0  + E0 2  2    + E0  E0       =   −  K = E − E0 =    hc     hc     hc   2  2   + E0  2  2   + E0   2    + E0                2

 1240eV nm  2   0.140 nm  = = 297eV   1240eV nm   5 2   + 5.11  10 eV    0.140 nm  

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(b) We solve the energy equation for the final wavelength.  hc   hc    + E0 =   + E     

 =

−1

 1 297eV  1 K  = −  = − = 0.145nm hc 0.140 nm 1240eV nm   hc        + E0 − E   hc

34. First we use conservation of energy, where the energy of the photon is written in terms of the wavelength, to relate the initial and final energies. Solve this equation for the electron’s final energy.  hc   hc   hc   hc  Einitial + Einitial = Efinal + Efinal →   + mc 2 =   + Ee  Ee =   −   + mc 2 photon electron photon electron            Next, we define the x-direction as the direction of the initial motion of the photon. We write equations for the conservation of momentum in the horizontal and vertical directions, where  is the angle the photon makes with the initial direction of the photon and  is the angle the electron makes. h h h px : = pe cos  + cos  p y : 0 = pe sin  + sin     To eliminate the variable  we solve the momentum equations for the electron’s momentum components, square the resulting equations, and add the two equations together using the identity cos 2  + sin 2  = 1.

h h   −  cos     

= ( pe cos )

( pe cos ) + ( pe sin  ) =  2

2  h    sin   = ( pe sin  )  



−

h   h  cos   +  sin        

2 2  h  2h h   h   h  2h h cos  +  cos   +  sin   =   − cos  +   p =  −                    2 We now apply the relativistic equation relating energy and momentum, E = p 2 c 2 + m 2 c 4 (Eq. 36–14), to write the electron momentum in terms of the electron energy. Then using the electron energy obtained from the conservation of energy equation (the first line of equations in this solution), we eliminate the electron energy and solve for the change in wavelength. E 2 − m2 c 4 Ee2 = pe2 c 2 + m2c 4 → pe2 = e 2 → c 2 e

2 Ee2 − m 2 c 4  h   h    h  2h h cos  − + = =   −   + mc  − m 2 c 2     2 c                 2

2 h  h   1 1  2h =   +   + m 2 c 2 + 2hmc  −  − − m2c 2              2 2 2h     1 1  2h  cos  = 2hmc  −  − multiply by −   2h           

− h cos  = mc (   −  ) − h →   −  =

h (1 − cos  ) mc

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35. The photon energy must be equal to the kinetic energy of the products plus the mass energy of the products. The mass of the positron is equal to the mass of the electron. Ephoton = K products + mproducts c 2 →

K products = Ephoton − mproducts c 2 = Ephoton − 2melectron c 2 = 2.85MeV − 2 ( 0.511MeV ) = 1.83MeV 36. The photon with the longest wavelength has the minimum energy in order to create the masses with no additional kinetic energy. Use Eq. 37–5. 6.63  10−34 J s ) ( hc hc h = = = = 6.62  10−16 m max = Emin 2mc 2 2mc 2 (1.67  10−27 kg )( 3.00  108 m s ) This must take place in the presence of some other object in order for momentum to be conserved. 37. The minimum energy necessary is equal to the rest energy of the two muons. Emin = 2mc 2 = 2 ( 207 )( 0.511MeV ) = 212 MeV The wavelength is given by Eq. 37–5. −34 8 hc ( 6.63  10 J s )( 3.00  10 m s ) = = 5.86  10−15 m = E (1.60  10−19 J eV )( 212  106 eV ) 38. Since v  0.001c, the total energy of the particles is essentially equal to their rest energy. Both particles have the same rest energy of 0.511 MeV. Since the total momentum is 0, each photon must have half the available energy and equal momenta. E Ephoton = melectron c 2 = 0.511MeV ; pphoton = photon = 0.511MeV c c 39. The energy of the photon is equal to the total energy of the two particles produced. Both particles have the same kinetic energy and the same mass. Ephoton = 2 ( K + mc 2 ) = 2 ( 0.345MeV + 0.511MeV ) = 1.712 MeV The wavelength is found from Eq. 37–5. 6.63  10−34 J s )( 3.00  108 m s ) ( hc = = 7.26  10−13 m = −19 6 E (1.60  10 J eV )(1.712  10 eV ) Or, using Problem 24,  =

hc 1240eV nm = = 7.24  10−4 nm = 7.24  10 −13 m 6 E 1.712  10 eV

40. We find the wavelength from Eq. 37–7. ( 6.63 10−34 J s ) = 3.2  10−32 m h h = = = p mv ( 0.21kg )( 0.10 m s ) 41. The neutron is not relativistic, so we can use p = mv. We also use Eq. 37–7.

6.63  10−34 J s ) ( h h = = 2.5  10−12 m = = p mv (1.67  10−27 kg )(1.6  105 m s )

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42. We use the relativistic expression for momentum, Eq. 36–8. mv mv h p= = = → 1 − v 2 c2 1 − v 2 c2 

6.63  10−34 J s ) 1 − ( 0.95 ) ( h 1 − v 2 c2 = = 8  10−13 m = 8 −31 mv 9.11 10 kg 0.95 3.00 10 m s   ( )( )( ) 2

The subtraction rule for significant figures is used here. 43. Since the particles are not relativistic, we may use K = p 2 2m . We then form the ratio of the kinetic energies, using Eq. 37–7. h2 2 2 p h K e 2me 2 mp 1.67  10−27 kg = = = = = 1840 K= ; h2 Kp me 9.11  10−31 kg 2m 2m 2 2mp 2 44. For diffraction, the wavelength must be on the order of the opening. Find the speed from Eq. 37–7. 6.63  10−34 J s ) ( h h h → v= = = 3.9  10−38 m s = = p mv m (1400 kg )(12 m ) This is on the order of 10−39 times ordinary highway speeds. 45. We assume the neutron is not relativistic. If the resulting velocity is small, our assumption will be valid. We use Eq. 37–7. 6.63  10−34 J s ) ( h h h → v= = = 1323m s  1000m s = = p mv m (1.67  10−27 kg )( 0.3  10−9 m ) This is not relativistic, so our assumption was valid. 46. (a) We find the momentum from Eq. 37–7. h 6.63  10−34 J s p= = = 9.5  10−25 kg m s 7.0  10−10 m  (b) We assume the speed is non-relativistic. 6.63  10−34 J s h h h = = → v= = = 1.0  106 m s p mv m ( 9.11 10−31 kg )( 7.0  10−10 m ) Since v c = 3.47  10−3 , our assumption is valid. (c) We calculate the kinetic energy classically.

K = 12 mv 2 = 12 ( mc 2 ) ( v c ) = 12 ( 0.511MeV ) ( 3.47  10−3 ) = 3.08  10−6 MeV = 3.08eV 2

This is the energy gained by an electron if accelerated through a potential difference of 3.1 V. 47. Because all of the energies to be considered are much less than the rest energy of an electron, we can use non-relativistic relationships. We use Eq. 37–7 to calculate the wavelength. p2 h h → p = 2mK ;  = = K= 2m p 2mK −34 h 6.63  10 J s = = 2.7  10−10 m  3  10−10 m (a)  = −31 −19 2mK 2 ( 9.11  10 kg ) ( 20eV ) (1.60  10 J eV ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b)  = (c)

=

h 6.63  10−34 J s = = 8.7  10−11 m  9  10−11 m − 31 − 19 2mK 2 ( 9.11  10 kg ) ( 200eV ) (1.60  10 J eV ) h = 2mK

6.63  10−34 J s 2 ( 9.11  10

−31

kg )( 2.0  10 eV )(1.60  10 3

−19

J eV )

= 2.7  10−11 m

48. Since the particles are not relativistic, we may use K = p 2 2m . We then form the ratio of the wavelengths, using Eq. 37–7. h 2mp K p me h h = = 1 = = ; h p mp e 2mK 2me K Thus the proton has the shorter wavelength, since me  mp . 49. We assume the electrons are non-relativistic, and then check the result in light of that assumption. The electron’s wavelength is found from Eq. 34–2a. The speed can then be found from Eq. 37–7. d sin  h h d sin  = morder  →  = ; = = → morder p me v

6.63  10−34 J s ) ( 2 ) ( hmorder v= = = 690 m s me d sin  ( 9.11  10−31 kg )( 3.0  10−6 m ) ( sin 45 )

This is far from being relativistic, so our original assumption was fine. 50. We relate the kinetic energy to the momentum with a classical relationship, since the electrons are non-relativistic. We also use Eq. 37–7. We then assume that the kinetic energy was acquired by electrostatic potential energy. p2 h2 K= = = eV → 2m 2m 2 6.63  10−34 J s ) ( h2 V= = = 13.85V  14 V 2me 2 2 ( 9.11  10−31 kg )(1.60  10−19 C )( 0.33  10−9 m )2 2

51. The kinetic energy is 6450 eV. That is small enough compared to the rest energy of the electron for the electron to be non-relativistic. We use Eq. 37–7. (1.240  103 eV nm ) h h hc = = = = p ( 2mK )1/2 ( 2mc 2 K )1/2  2 ( 0.511  106 eV ) ( 6450eV ) 1/2  

= 1.527  10−2 nm = 15.3pm 52. The energy of a level is En = −

(13.6 eV ) .

n2 (a) The transition from n = 1 to n' = 3 is an absorption , because the final state , n' = 3, has a higher energy. The photon energy is the difference between the energies of the two states.  1   1   hf = En  − En = − (13.6 eV )  2  −  2   = 12.1 eV  3   1  

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(b) The transition from n = 6 to n' = 2 is an emission , because the initial state , n' = 2, has a higher energy. The photon energy is the difference between the energies of the two states.  1   1   hf = − ( En  − En ) = (13.6 eV )   2  −  2   = 3.0 eV  2   6   (c) The transition from n = 4 to n' = 5 is an absorption , because the final state , n' = 5, has a higher energy. The photon energy is the difference between the energies of the two states.  1   1   hf = En  − En = − (13.6 eV )  2  −  2   = 0.31 eV  5   4   The photon for the transition from n = 1 to n' = 3 has the largest energy. 53. To ionize the atom means removing the electron, or raising it to zero energy. ( −13.6eV ) = (13.6eV ) = 1.51eV Eionization = 0 − En = 0 − n2 32 54. We use the equation that appears above Eq. 37–15 in the text. (a) The second Balmer line is the transition from n = 4 to n = 2. hc 1240eV nm = = = 490 nm ( E4 − E2 )  − 0.85eV − ( −3.4eV ) (b) The third Lyman line is the transition from n = 4 to n = 1. hc 1240eV nm = = = 97.3nm ( E4 − E1 )  − 0.85 eV − ( −13.6 eV ) 55. Doubly ionized lithium is similar to hydrogen, except that there are three positive charges (Z = 3) in the nucleus. Use Eq. 37–14b with Z = 3. Z 2 (13.6eV ) 32 (13.6eV ) (122eV ) En = − = − =− 2 2 n n n2  (122eV )  Eionization = 0 − E1 = 0 −  −  = 122eV 2 (1)   56. We evaluate the Rydberg constant using Eqs. 37–8 and 37–15. We use hydrogen so Z = 1.  1 1 1  Z 2e4m  1 1  = R − = −    →  ( n)2 ( n )2  8 02 h 3c  ( n)2 ( n )2       2 4 Z em R= 2 3 8 0 h c

(1) (1.602176634  10−19 C ) ( 9.1093837015  10−31 kg ) 4

8 (8.8541878128  10−12 C2 N m 2 ) ( 6.62607015  10 −34 J s ) ( 2.99792458  108 m s )

= 1.0973731568  107

C4 kg C4 J 3s3 m s N 2 m4

 1.0974  107 m −1

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57. The longest wavelength corresponds to the minimum energy, which is the ionization energy. hc 1240eV nm = = = 91.2 nm Eion 13.6eV 58. Singly ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus. Use Eq. 37–14b. Z 2 (13.6 eV ) 22 (13.6 eV ) (54.4 eV ) En = − = − =− 2 2 n n n2 We find the energy of the photon from the n = 6 to n = 2 transition in singly-ionized helium.  1   1   E = E6 − E2 = − ( 54.4 eV )  2  −  2   = 12.1 eV  6   2   Because this is the energy difference for the n = 1 to n = 3 transition in hydrogen, the photon can be absorbed by a hydrogen atom which will jump from n = 1 to n = 3 . 59. The energy of the photon is the sum of the ionization energy of 13.6 eV and the kinetic energy of 17.5 eV. The wavelength is found from Eq. 37–3. hc hc 1240eV nm = Etotal →  = = = 39.9 nm hf =  Etotal (13.6 + 17.5) eV 60. A collision is elastic if the kinetic energy before the collision is equal to the kinetic energy after the collision. If the hydrogen atom is in the ground state, then the smallest amount of energy it can absorb is the difference in the n = 1 and n = 2 levels. So as long as the kinetic energy of the incoming electron is less than that difference, the collision must be elastic.  13.6eV  K  E2 − E1 =  −  − ( −13.6eV ) = 10.2eV 4   61. Singly-ionized helium is like hydrogen, except that there are two positive charges (Z = 2) in the nucleus. Use Eq. 37–14b. Z 2 (13.6eV ) 22 (13.6eV ) ( 54.4eV ) En = − = − =− 2 2 n n n2 E1 = −54.5eV, E2 = −13.6eV, E3 = −6.0eV, E4 = −3.4eV

62. The potential energy for the ground state is given by the charge of the electron times the electric potential caused by the proton.

9.00  109 N m 2 C2 )(1.60  10−19 C ) (1eV 1.60  10−19 J ) ( e U = ( −e )Vproton = ( −e ) =− 4 0 r1 ( 0.529  10−10 m ) 2

= − 27.2eV The kinetic energy is the total energy minus the potential energy. K = E1 − U = −13.6eV − ( −27.2eV ) = + 13.6eV © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

63. The value of n is found from rn = n 2 r1 , and then find the energy from Eq. 37–14b. r rn = n r1 → n = n = r1 2

E=−

1 2

( 0.10  10 m ) = 972 −3

0.529  10−10 m

(13.6 eV ) = − (13.6 eV ) = − (13.6 eV ) = −1.4  10−5 eV n2

9722

13752

64. The velocity is found from Eq. 37–10 evaluated for n = 1. nh → mvrn = 2 6.63  10−34 J s ) ( h = = 2.190  106 m s = 7.30  10−3 c v= 2 r1me 2 ( 0.529  10−10 m )( 9.11  10−31 kg ) Since v

c, we say yes, non-relativistic formulas are justified. The relativistic factor is as follows. 2

 2.190  106 m s  −5 1 − v 2 c 2  1 − 12 ( v 2 c 2 ) = 1 − 12   = 1 − 2.66  10  0.99997 8  3.00 10 m s   We see that

1 − v 2 c 2 is essentially 1. The answer is yes, non-relativistic formulas are justified.

65. Hydrogen atoms start in the n = 1 orbit (“ground state”). Using Eqs. 37–9 and 37–14b, we determine the state to which the atom is excited when it absorbs a photon of 12.75 eV via collision with an electron. Then, using Eq. 37–15, we calculate all possible wavelengths that can be emitted as the electron cascades back to the ground state. 13.6 eV E = EU − EL → EU = − = EL + E → n2 −13.6 eV −13.6 eV = =4 n= −13.6 eV + 12.75 eV EL + E Starting with the electron in the n = 4 orbit, the following transitions are possible: n = 4 to n = 3 ; n = 4 to n = 2 ; n = 4 to n = 1 ; n = 3 to n = 2 ; n = 3 to n = 1 ; n = 2 to n = 1. 1 1 1  = (1.097  107 m –1 )  2 – 2  = 5.333  105 m –1   = 1875 nm  3 4  1 1   1 = (1.097  107 m –1 )  2 – 2  = 2.057  106 m –1   = 486.2 nm 4   2 1 1 1  = (1.097  107 m –1 )  2 – 2  = 1.028  107 m –1   = 97.23 nm  1 4  1  1 1 = (1.097  107 m –1 )  2 – 2  = 1.524  106 m –1   = 656.3 nm  2 3  1 1 1 = (1.097  107 m –1 )  2 – 2  = 9.751 106 m –1   = 102.6 nm  1 3  1 1 1  = (1.097  107 m –1 )  2 – 2  = 8.228  106 m –1   = 121.5 nm  1 2 

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66. When we compare the gravitational and electric forces we see that we can use the same expression for the Bohr orbits, Eqs. 37–11 and 37–14a, if we replace Ze 2 4 0 with Gme m p . r1 =

h 2 0 h 2 4 0 =  me Ze 2 4 2 me Ze 2

→

6.626  10−34 J s ) ( h2 = r1 = 2 4 Gm 2 e m p 4 2 ( 6.67  10−11 N m 2 / kg 2 )( 9.11  10 −31 kg )2 (1.67  10 −27 kg ) 2

= 1.20  1029 m 2

 Ze 2  2 2 me Z 2 e 4 me = − E1 = −   2 8 02 h 2  4 0  h

→ E1 = −

2 2 ( 6.67  10−11 N m 2 kg 2 ) ( 9.11  10 −31 kg ) (1.67  10 −27 kg ) 2

=−

2 2G 2 me 3m p 2 3

( 6.626  10 J s ) −34

= −4.22  10−97 J

67. We know that the radii of the orbits are given by rn = n 2 r1. Find the difference in radius for adjacent orbits. 2 r = rn − rn −1 = n 2 r1 − ( n − 1) r1 = n 2 r1 − ( n 2 − 2n + 1) r1 = ( 2n − 1) r1 r 2r 1, we have r  2nr1 = 2n n2 = n . n n In the classical limit, the separation of radii (and energies) should be very small. We see that letting n →  accomplishes this. If we substitute the expression for r1 from Eq. 37–11, we have this.

If n

2nh 2 0  me2 We see that r  h 2 , and so letting h → 0 is equivalent to considering n → . r  2nr1 =

68. (a) Calculate the energy from the light bulb that enters the eye by calculating the intensity of the light at a distance of l = 1.0m, by dividing the power in the visible spectrum by the surface area of a sphere of radius l = 1.0m . Multiply the intensity of the light by the area of the pupil (diameter = D) to determine the energy entering the eye per second. Finally, divide this energy by the energy of a photon (Eq. 37–3) to calculate the number of photons entering the eye per second (n). I= n=

P 4 l 2

Pe = I ( D 2 / 4 ) =

P D   16  l 

2 ( 4.5W ) ( 550  10−9 m )  4.0  10−3 m  Pe P  D  = =     hc /  16hc  l  16 ( 6.626  10−34 J s )( 3.00  108 m s )  1.0 m 

= 1.2  1013 photons/sec

(b) Repeat the above calculation for a distance of l = 1.0 km instead of 1.0 m. I=

P 4 l 2

Pe = I ( D 2 / 4 ) =

P D   16  l 

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

2 ( 4.5 W ) ( 550 10−9 m )  4.0  10−3 m  P P  D  n= e = =     hc /  16hc  l  16 ( 6.626  10−34 J s )( 3.00  108 m s )  1.0  103 m 

= 1.2  107 photons/sec 69. To produce a photoelectron, the hydrogen atom must be ionized, so the minimum energy of the photon is 13.6 eV. We find the minimum frequency of the photon from Eq. 37–3. −19 E Emin (13.6eV ) (1.60  10 J eV ) → f min = = = 3.28  1015 Hz E = hf → f = −34 h h  6.63 10 J s ( ) 70. From Section 35–11, the spacing between planes, d, for the first-order peaks is given by Eq. 35–20,  = 2d sin . The wavelength of the electrons can be found from their kinetic energy. The electrons are not relativistic at the energy given. p2 h2 h = → = = 2d sin  → K= 2m 2m 2 2mK d=

( 6.63  10 J s ) = 8.1  10 m 2 ( sin 43 ) 2 ( 9.11  10 kg ) (125eV ) (1.60  10 J/eV ) −34

h 2sin  2mK

−11

−31

−19

71. The power rating is the amount of energy produced per second. If this is divided by the energy per photon, then the result is the number of photons produced per second. (830 W ) (12.2  10−2 m ) hc P P Ephoton = hf = = = = 5.1  1026 photons s ;  Ephoton hc ( 6.63  10−34 J s )( 3.00  108 m s ) 72. The intensity is the amount of energy per second per unit area reaching the Earth. If that intensity is divided by the energy per photon, the result will be the photons per second per unit area reaching the Earth. We use Eq. 37–3. hc Ephoton = hf =



I photons =

I sunlight Ephoton

I sunlight  hc

(1350 W m )(550 10 m ) = 3.7  10 photons s m ( 6.63  10 J s )( 3.00 10 m/s ) 2

−9

−34

73. First find the area of a sphere whose radius is the Earth–Sun distance. A = 4 r 2 = 4 (150  109 m ) = 2.83  1023 m 2 2

Multiply this by the given intensity to find the Sun’s total power output. ( 2.83 1023 m2 )(1350W m2 ) = 3.82 1026 W Multiplying by the number of seconds in a year gives the annual energy output. E = ( 3.82  1026 W ) ( 3600s h )( 24 h d )( 365.25d yr ) = 1.20 1034 J yr Finally we divide by the energy of one photon to find the number of photons per year. 34 −9 E E  (1.20  10 J yr )( 550  10 m ) = = = 3.3  1052 photons yr −34 8 hf hc ( 6.63  10 J s )( 3.00  10 m s )

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74. We find the peak wavelength from Wien’s law, Eq. 37–1. ( 2.90 10−3 m K ) = ( 2.90 10−3 m K ) = 1.1  10−3 m = 1.1mm P = T ( 2.7 K ) 75. If we ignore the recoil motion, at the closest approach the kinetic energy of both particles is zero. The potential energy of the two charges must equal the initial kinetic energy of the  particle:

K = U = rmin =

( Z e ) ( Z Age )

4 0

rmin

( Z e ) ( Z Age )

4 0

KE

→

( 9.00  10 N m C ) ( 2 )( 79 ) (1.60 10 C ) = 4.74  10 m = ( 4.8 MeV ) (1.60  10 J MeV ) 9

−19

−14

−13

The distance to the “surface” of the gold nucleus is then 4.74 10−14 m − 7.0 10−15 m = 4.0 10−14 m . 76. The electrostatic potential energy is given by Eq. 23–5. The kinetic energy is given by the total energy, Eq. 37–14a, minus the potential energy. The Bohr radius is given by Eq. 37–11. 1 Ze 2 1 Ze 2 mZe2 Z 2e4m =− = − U = − eV = − 4 0 rn 4 0 n 2 h 2 0 4n 2 h 2 02 K = E −U = −

Z 2e 4 m  Z 2 e 4 m  Z 2 e 4 m −− = 8 02 h 2 n 2  4n 2 h 2 02  8n 2 h 2 02

Z 2e4m U Z 2 e 4 m 8n 2 h 2 2 4 n 2 h 2 2 = 2 4 0 = 2 2 2 2 4 0 = 2 ; Z em K 4n h  0 Z e m 8n 2 h 2 02

77. We calculate the ratio of the forces.  Gme m p  2 2 −11 −31 −27 Fgravitational  r 2  Gme m p ( 6.67  10 N m kg )( 9.11  10 kg )(1.67  10 kg ) = = = 2 Felectric ke 2  ke 2  9.00  109 N m 2 C2 )(1.60  10−19 C ) (  r2   

= 4.40  10−40 Yes , the gravitational force may be safely ignored. 78. (a) The electron has a charge e, so the potential difference produces a kinetic energy of eV. The shortest wavelength photon is produced when all the kinetic energy is lost and a photon is emitted. hc hc hf max = = eV → 0 = eV 0 (b) 0 =

hc 1240eV nm = = 0.040 nm eV 31  103 eV

79. The wavelength is found from Eq. 35–13. The velocity of electrons with the same wavelength (and thus the same diffraction pattern) is found from their momentum, assuming they are not relativistic. We use Eq. 37–7 to relate the wavelength and momentum.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

d sin  = n →  = v=

d sin  h h = = → n p mv

Instructor Solutions Manual

d sin    −3  = n = ( 0.010  10 m ) ( sin 3.5 ) = 610.5 nm   

( 6.63 10−34 J s ) (1) hn = = 1192 m s  1200 m s md sin  ( 9.11  10−31 kg )( 0.010  10−3 m ) ( sin 3.5 )

80. (a) See the adjacent figure. (b) Absorption of a 5.1 eV photon represents a transition from the ground state to the state 5.1 eV above that, the third excited state. Possible photon emission energies are found by considering all the possible downward transitions that might occur as the electron makes its way back to the ground state. −6.4eV − ( −6.8eV ) = 0.4eV

−6.4eV − ( −9.0eV ) = 2.6eV

−6.4eV − ( −11.5eV ) = 5.1eV −6.8eV − ( −9.0eV ) = 2.2eV −6.8eV − ( −11.5eV ) = 4.7eV

−9.0eV − ( −11.5eV ) = 2.5eV 81. (a) We use Eq. 37–4b to calculate the maximum kinetic energy of the electron and set this equal to the product of the stopping voltage and the electron charge. hf − W0 hc /  − W0 = K max = hf − W0 = eV0 → V0 = e e − 1240eV nm 408nm 2.28eV ( )( ) = 0.7592 V  0.76 V V0 = e (b) We calculate the speed from the non-relativistic kinetic energy equation and the maximum kinetic energy found in part (a). 2 K max = 12 mvmax →

2 ( 0.7592eV ) (1.60  10 −19 J eV ) 2 K max = = 5.164  105 m s  5.2  105 m/s m 9.11  10−31 kg This is only about 0.002c, so we are justified in using the classical definition of kinetic energy. (c) We use Eq. 37–7 to calculate the de Broglie wavelength. 6.63  10−34 J s h h = = = = 1.409  10−9 m = 1.4nm p mv ( 9.11 10−31 kg )( 5.164  105 m s ) vmax =

82. (a) Use Bohr’s analysis of the hydrogen atom, replacing the proton mass with Earth’s mass, the ke2 electron mass with the Moon’s mass, and the electrostatic force Fe = 2 with the gravitational r GmE mM . To account for the change in force, replace ke2 with GmE mM . With force, Fg = 2 r these replacements, write expressions similar to Eqs. 37–11 and 37–14a for the Bohr radius and energy. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Early Quantum Theory and Models of the Atom

rn =

h 2n 2 4 2 mke 2

→

( 6.626  10−34 J s ) h 2n 2 rn = 2 n2 = 4 GmM2 mE 4 2 ( 6.67  10−11 N m 2 / kg 2 )( 7.35  1022 kg )2 ( 5.98  1024 kg ) 2

= n 2 ( 5.16  10−129 m ) En = −

2 2 e 4 mk 2 n 2h 2

→

2 ( 6.67  10 2 2G 2 mE2 mM3 En = − =− 2 2 nh 2

−11

N m 2 / kg 2 ) ( 5.98  1024 kg ) ( 7.35  1022 kg ) 2

n 2 ( 6.626  10−34 J s )

2.84  10165 J n2 (b) Insert the known masses and Earth–Moon distance into the Bohr radius equation to determine the Bohr state. = −

4 2GmM2 mE rn h2 4 2 ( 6.67  10−11 Nm 2 / kg 2 )( 7.35  1022 kg ) ( 5.98  1024 kg )( 3.84  108 m ) 2

( 6.626  10

−34

Js )

= 2.73  1068 Since n 1068 , a value of n = 1 is negligible compared to n. Hence the quantization of energy and radius is not apparent .

83. We use Eqs. 36–14, 36–11a, and 37–7 to derive the expression.

p 2 c 2 + m 2 c 4 = E 2 ; E = K + mc 2 → p 2c 2 + m 2c 4 = ( K + mc 2 ) = K 2 + 2mc 2 K + m 2c 4 → 2

K 2 + 2mc 2 K = p 2 c 2 =

h 2c 2



→ 2 =

h 2c 2 → ( K 2 + 2mc 2 K )

=

hc K 2 + 2mc 2 K

84. The theoretical resolution limit is the wavelength of the electron. We find the wavelength from the momentum, and find the momentum from the kinetic energy and rest energy. We use the result from Problem 83. The kinetic energy of the electron is 78 keV. hc 1240eV nm = = 2 2 2 K + 2mc K ( 78  103 eV ) + 2 ( 0.511  106 eV )( 78  103 eV ) = 4.233  10−3 nm  4.2  10−3 nm

85. As light leaves the flashlight it gains momentum. This change in momentum is given by Eq. 31–20a. Dividing the change in momentum by the elapsed time gives the force the flashlight must apply to the light to produce this momentum. This is equal to the reaction force that light applies to the flashlight. p U P 2.8W = = = = 9.3  10−9 N t ct c 3.00  108 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1265

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

86. (a) We write the Planck time as tP = G h  c , and the units of t P must be T  . 



 ML2   L   + 2  +  M  − T −2 −  −     = L  T   T 

 L3  → T  =  2  MT 

  

tP = G h c

5 +

There are no mass units in T  , and so  =  , and T  =  L  −3 + 5

length units in T  , and so  = −5 and T  = T 

T −3 − . There are no

2

= T 

. Thus,  = 1 =  and 2

 = − 25 . Gh c5

tP = G1/ 2 h1/ 2 c −5 / 2 =

( 6.67 10 N m kg ) ( 6.63 10 −11

Gh = c5

(b) tP =

( 3.00 10 m s ) 8

−34

J s)

= 1.35  10−43 s

P = G h c

 L3  →  L =    MT 2 





 ML2   L   + 2  +  M  − T −2 − −     =  L  T   T  5 +

There are no mass units in  L  , and so  =  , and  L  =  L  5 − 3

units in  L  , and so  = −3 and  L  =  L 

. Thus  = 1 =  and  = − 3 . 2

Gh c3

tP = G1/ 2 h1/ 2 c −3 / 2 =

Gh (d) P = = c3

2

= L

T −3 − . There are no time

( 6.67 10 N m kg ) ( 6.63 10 −11

(3.00 10 m s ) 8

−34

J s)

= 4.05  10−35 m

87. For standing matter waves, there are nodes at the two walls. For the ground state (first harmonic), the wavelength is twice the distance between the walls, or l = 12  (see Fig. 15–26b). We use Eq. 37–7 to find the velocity and then the kinetic energy. 2

h p2 h2 1  h  ; K= l =  →  = 2l ; p = = =   = 2m 2m  2 l  8ml 2  2l For the second harmonic, the distance between the walls is a full wavelength, and so l =  . h

1 2

l = → p=



h p2 h2 1 h ; K= = =   2m 2m  l  2ml 2 l

88. Find the energy of the photon from Eq. 37–3 and Eq. 37–7. 1eV c   = 0.662eV E = hf = h = pc = ( 3.53  10−28 kg m s )( 3.00  108 m s )  −19    1.60  10 J 

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Looking at Fig. 37–26, the Lyman series photons have 13.6eV  E  10.2eV. The Balmer series photons have 3.4eV  E  1.9eV . The Paschen series photons have 1.5eV  E  0.65eV. It appears that this photon belongs to the Paschen series, ejected from energy level 4. 89. (a) We solve Eq. 34–2a for the slit separation with the wavelength given in terms of the electron momentum from Eq. 37–5. The electron momentum is written in terms of the kinetic energy. m = d sin  →

 sin 

h p sin 

2mc 2 eV sin 

1240 eV nm

(

)

sin10 2 0.511  106 eV (18eV )

= 1.665  10 −9 m  1.7 nm (b) No. The slit separation distance is only about 10 atomic diameters. That would probably be impossible to create. 90. The impulse on the wall is due to the change in momentum of the photons. Each photon is absorbed, and so its entire momentum is transferred to the wall. nh Fon wall t = pwall = −pphotons = − ( 0 − npphoton ) = npphoton = →



n F  ( 7.5  10 N )( 633  10 m ) = = = 7.2  1018 photons s −34 h t 6.63 10 J s  ( ) −9

−9

91. The kinetic energy of the hydrogen gas would have to be the difference between the n = 1 and n = 2 states of the hydrogen atom, 10.2 eV. Use Eq. 18–4. −19 2 K 2 (10.2eV ) (1.60  10 J eV ) = = 7.88  104 K K = 23 kT → T = 3k 3 (1.38  10−23 J K ) 92. (a) The minimum energy necessary to initiate the chemical process on the retina is the energy of a single photon of red light. Red light has a wavelength of approximately 750 nm. 1eV c  3.00  108 m/s    = 1.7 eV Emin = hf = h = 6.626  10−34 J s    −9 −19    750  10 m   1.60  10 J  (b) The maximum energy is the energy of a photon of violet light of wavelength 400 nm. 1eV  3.00  108 m/s    = 3.1eV Emax = 6.626  10−34 J s    −9 −19   400  10 m   1.60  10 J 

(

)

(

)

93. (a) Here are standing wave diagrams for the first three modes of vibration.

(b) From the diagram we see that the wavelengths are given by n = The momentum is pn =



2L , n = 1, 2, 3, … . n

p2 nh n2 h2 , and so the kinetic energy is K n = n = . 2L 2m 8mL2

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(c) Because the potential energy is zero inside the box, the total energy is the kinetic energy. The ground state energy has n = 1.

(1) ( 6.63  10−34 J s ) 1eV n 2h 2   E1 = = = 150 eV 2  2 −19  31 10 − − 8mL 8 ( 9.11  10 kg )( 0.50  10 m )  1.60  10 J  2

(d) Do the same calculation for the baseball, and then find the speed from the kinetic energy. −34 n 2h 2 (1) ( 6.63  10 J s ) = = 1.297  10−66 J  1.3  10−66 J E1 = 8mL2 8 ( 0.14 kg )( 0.55m )2 2

2 (1.297  10−66 J ) 2K = = 4.3  10−33 m s E1 = mv → v = m ( 0.14 kg ) 2

1 2

(e) Find the width of the box from E1 = E1 =

n2 h2 . 8mL2

n 2h 2 → 8mL2

(1) ( 6.63  10−34 J s ) nh = = 1.447  10−10 m = 0.14 nm L= − − 31 19 8mE1 8 ( 9.11  10 kg ) (18eV ) (1.60  10 J eV ) L=

(1)(1240eV nm ) nh nhc = = = 1.446  10−1 nm = 0.14 nm 2 6 8mE1 8mc E1 8 ( 0.511  10 eV ) (18eV )

94. The photon will be moving at the speed of light, c, regardless of its wavelength. The speed of the electron can be found from Eq. 37–7. We assume the electron is non-relativistic. We also use a result from Problem 24. h h h hc 2 e = = → ve = = → pe me ve me e me c 2 e ve

= hc

1 1 = (1240eV nm ) = 4.3  10−6 6 mc 2 e ( 0.511 10 eV ) ( 560 nm )

95. The work function is the amount of energy needed to ionize a hydrogen atom, which is the ionization enegy of 13.6eV . 96. (a) Apply conservation of momentum before and after the emission of the photon to determine the recoil speed of the atom, where the momentum of the photon is given by Eq. 37–7. The initial momentum is 0. ( 6.63 10–34 J s ) h h 0 = − mv → v = = = 6.0  10 –3 m s –27 –9 m 85 (1.66  10 kg )( 780  10 m )  (b) We solve Eq. 18–5 for the lowest achievable temperature, where the recoil speed is the rms speed of the rubidium gas.

3kT v= m

–27 –3 mv 2 85 (1.66  10 kg )( 6.0  10 m s ) → T= = = 1.2  10 –7 K = 0.12  K –23 3k 3 (1.38  10 J K ) 2

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97. Each time the rubidium atom absorbs a photon its momentum decreases by the momentum of the photon. Dividing the initial momentum of the rubidium atom by the momentum of the photon, Eq. 37–7, gives the number of collisions necessary to stop the atom. Multiplying the number of collisions by the absorption time, 25 ns per absorption, provides the time to completely stop the atom. −27 −9 mv mv ( 85u ) (1.66  10 kg/u ) ( 260 m s ) ( 780  10 m ) = = = 43,160 n= h h 6.63  10−34 J s T = 43,160 ( 25  10−9 s ) = 1.1  10−3 s

98. The decrease in mass occurs because a photon has been emitted. We calculate the fractional change. Since we are told to find the amount of decrease, we use the opposite of the change. The mass of an H atom is its atomic weight (1.008 u) times the value of “u” in MeV. 1   1  −E  ( 13.6 eV )  2  −  2   2  −m  c  −E  1   3   = 1.29  10−8 = = = 2 m0 m0 m0c 1.008u ( 931.5  106 eV u ) 99. The intensity is the amount of energy hitting the surface area per second. That is found from the number of photons per second hitting the area, and the energy per photon, from Eq. 37–3. 1.0  1012 photons ) hc (1.0  1012 ) ( 6.626  10−34 J s )( 3.00  108 m s ) ( I = = = 3.815  10−7 W m 2 2 2 −9  1m 1s 1m 1s 521  10 m ( )( ) ( )( ) ( )

 3.8  10−7 W m 2 The magnitude of the electric field is found from Eq. 31–19a. I = 12  0 cE02 → E0 =

2I =  0c

2 ( 3.815  10−7 W m 2 )

(8.85  10

−12

C N m )( 3.00  10 m s ) 2

= 1.7  10−2 V m

1269

CHAPTER 38: Quantum Mechanics Responses to Questions 1.

(a) A matter wave ψ does not need a medium, but a wave on a string does. The square of the wave function for a matter wave ψ describes the probability of finding a particle within a certain spatial range, whereas the expressions equation for a wave on a string describes the displacement of a piece of string from its equilibrium position. (b) Neither an EM nor a matter wave needs a medium in which to exist. The expression for the EM wave describes the way in which the amplitudes of the electric and magnetic fields change as the wave passes a point in space. An EM wave represents a vector field and can be polarized. A matter wave is a scalar and cannot be polarized.

According to Bohr’s theory, each electron in an atom travels in a circular orbit and has a precise position and momentum at any point in time. This view is inconsistent with the postulates of quantum mechanics and the uncertainty principle, which does not allow both the position and momentum to be known precisely. According to quantum mechanics, the “orbitals” of electrons do not have precise radii but describe the probability of finding an electron in a given spatial range.

As a particle becomes more massive, the uncertainty of its momentum ( p = mv ) becomes larger, and so the Heisenberg uncertainty principle predicts a reduction in the uncertainty of its position (due to x  p ). Thus, with a small uncertainty in its position, we can do a better job of predicting its future position. Because of the very small value of , the uncertainties in both a baseball’s momentum and position can be very small compared to typical macroscopic positions and momenta without being close to the limit imposed by the uncertainty principle. For instance, the uncertainty in the baseball’s position −10

could be 10−10 m, and the uncertainty in the baseball’s momentum 10 kg m s , and still easily satisfy the uncertainty principle. Yet we never try to measure the position or momentum of a baseball to that level of precision. For an electron, however, typical values of uncertainty in position and momentum can be close to the uncertainty principle limit, so the relative uncertainty can be much higher for that small object. 5.

No. According to the uncertainty principle, if the needle were balanced, then the position of the center of mass would be known exactly, and there would have to be some uncertainty in its momentum. The center of mass of the needle could not have a zero momentum, and therefore would fall over. If the initial momentum of the center of mass of the needle were exactly zero, then there would be uncertainty in its position, and the needle could not be perfectly balanced (with the center of mass exactly over the tip).

Yes, some of the air escapes the tire (and goes into the tire gauge) in the act of measuring the pressure. It is impossible to avoid this escape. The act of measuring the air pressure in a tire therefore actually changes the pressure, although not by much since very little air escapes compared to the total amount of air in the tire. This is similar to the uncertainty principle, in which one of the two factors limiting the precision of measurement is the interaction between the object being observed, or measured, and the observing instrument. By making the measurement, you have affected the state of the system.

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Yes, this is consistent with the uncertainty principle for energy ( E t  ) . The ground-state energy can be precisely determined because the electrons remain in that state for a very long time. Thus, as t →  , E → 0 . However, the electrons do not remain in the excited states for very long, thus the t in this case is relatively small, making the E , the energy width, relatively large.

If Planck’s constant were much larger than it is, then the consequences of the uncertainty principle would be noticeable with macroscopic objects. For instance, attempts to determine a baseball’s speed would mean that you could not find its position very accurately. Using a radar gun to find the speed of a pitcher’s fastball could significantly change the actual course of the ball.

According to Newtonian mechanics, all objects have an exact position and momentum at a point in time. This information can be used to predict the future motion of an object, and so Newtonian mechanics is said to be “deterministic”. According to quantum mechanics, there is unavoidable uncertainty in the position and momentum of all objects. It is impossible to exactly determine both position and momentum at the same time, which introduces uncertainty into the prediction of the future motion of the object. Newtonian mechanics is still relevant for macroscopic objects because there even small macroscopic uncertainties means that the consequences of the uncertainty principle are not noticed. Thus we don’t notice quantum effects in macroscopic situations like the motion of a car. This is an example of the correspondence principle.

10. If you knew the position precisely (meaning a very, very small uncertainty), then you would know essentially nothing about the momentum (because it would have a very large uncertainty). 11. No. Some of the energy of the soup would be used to heat up the thermometer, so the temperature registered on the thermometer (the equilibrium temperature of the soup and thermometer) would be slightly less than the original temperature of the soup. This is a situation where making a measurement affects the physical situation and causes uncertainty in the measurement. 12. No. However, the greater the precision of the measurement of position, the greater the uncertainty in the measurement of the object’s momentum will be. 13. A particle in a box is confined to a region of space. Since the uncertainty in position is limited by the box, there must be some uncertainty in the particle’s momentum, and the momentum cannot be zero. The zero-point energy reflects the uncertainty in momentum. 14. Yes, the probability of finding the particle at these points is zero. It is possible for the particle to pass by these points. Since the particle is acting like a wave, these points correspond to the nodes in a standing wave pattern in the box. 15. For large values of n, the probability density varies rapidly between zero and the maximum value. It can be averaged easily to the classical result (a uniform probability density for all points in the well) as n becomes large.

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16. As n increases, the energy of the corresponding state increases, and the separation between states increases, but the ratio ΔE/E is approximately 2/n, which approaches zero.

h 2 ( n + 1) h2 n2 2 − 2 2 n + 1) − n 2 2n + 1 ( E h2 n2 8 8 l l m m ; E = En +1 − En ; = = = 2 En = h2 n2 8ml 2 E n2 n 2 8ml 2

E 2n + 1 2 = lim 2  lim = 0 n → E n → n n → n

lim

17. As the potential decreases, the wave function extends into the forbidden region as an exponential decay function. When the potential drops below the particle energy, the wave function outside the well changes from an exponential decay function to an oscillating function with a longer wavelength than the function within the well. When the potential is zero, the wavelengths of the wave function will be the same everywhere. The ground state energy of the particle in a well becomes the energy of the free particle.

Responses to MisConceptual Questions 1.

(a) In considering electrons, if only the particle property of electrons is considered, then the electron is viewed as a point particle passing through the slit. The electron also has wave properties and therefore undergoes diffraction as it passes through the slit.

(c) The uncertainty principle allows for a precise measurement of the position, provided that there is a large uncertainty in the momentum. It also allows for a precise measurement of the momentum, provided that the simultaneous measurement of the position has a large uncertainty. It does not allow for the precise measurement of position and momentum simultaneously.

(c) The Heisenberg uncertainty principle deals with our inability to know the position and momentum of a particle precisely. The more precisely you can determine the position, the less precisely you can determine the momentum, and vice versa.

(b) If you knew the position precisely, then you would know nothing about the momentum.

(e) According to the uncertainty principle, it is impossible to measure a system without disturbing, or affecting, the system. The more precisely you measure the position, the greater the disturbance in the momentum, and hence the greater uncertainty in the momentum. The more precisely you measure the momentum, the greater the disturbance in the position, and hence the greater the uncertainty in the position. When measuring position and momentum, there is a minimum value for the product of the uncertainties in momentum and position. Similar statements could be made for energy and time (instead of position and momentum).

(a) The hydrogen atom will have a greater probability of tunneling through the barrier because it has a smaller mass and therefore a larger transmission coefficient. (See Eqs. 38-17a and 3817b.)

(b) The de Broglie wavelength of the baseball is given by  = h p . For a moving baseball, the momentum is on the order of 1kg m s , and so the wavelength is on the order of 10−34 m. This is far too small to be detected.

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(a) Energy and momentum are conserved in both Newtonian and quantum mechanics, so answers (b) and (d) are not true. And quantum mechanics can be used for macroscopic objects – it gives the same results as Newtonian mechanics. Answer (a) highlights a main difference between the two formulations of mechanics.

(d) The “waving” of , or it’s variation in space and time, is a variation in the probability of finding a particle (like an electron).

10. (b), (d) The electron cannot be at rest. The ground state energy is greater than 0, and so the electron is always moving, and thus answer (a) is false. The energy values are quantized (take on specific discrete values), and so answer (c) is also false. Answer (b) is true (for the same reason that answer (c) is false). And, the electron has a wave function that describes the relative probability of the electron being at a certain position. 11. (b) Classically, the answer would be (a), because the particle could not escape from the potential well. But, due to quantum mechanical “tunneling”, the particle may be found occasionally outside of the well (answer (b)). However, it cannot completely escape from the well. 12. (c) For a free particle, the wave function of the particle is a simple sinusoidal function. Because there are no boundaries on the particle, the energy is not quantized. The energy of a free particle can have any value. 13. (c) The normalization condition calculates the probability of finding the particle somewhere in all of space. That probability is “1” since the particle must be found somewhere.

Solutions to Problems 1.

We find the wavelength of the neutron from Eq. 37–7, using the classical relationship between momentum and kinetic energy for the low-energy neutrons. The peaks of the interference pattern are given by Eq. 34–2a and Fig. 34–10. For small angles, we have sin   tan  . h h = = ; d sin  = n , n = 1, 2, ... ; x = l tan  p 2mK n x n l = → x= sin  = tan  → , n = 1, 2, ... → l d d 6.63  10−34 J s ) (1.0 m ) ( hl l x = = = d d 2mK ( 3.5  10−4 m ) 2 (1.675  10−27 kg ) ( 0.025eV ) (1.60  10−19 J eV ) = 5.17  10−7 m  5.2  10−7 m

The ratio 2.

x is very small, so our use of sin   tan  is justified. l

We find the wavelength of a pellet from Eq. 37–7. The half-angle for the central circle of the diffraction pattern is given in Section 35–5 as sin  =

1.22 , where D is the diameter of the D

opening. Assuming the angle is small, the diameter of the spread of the bullet beam is

d = 2l tan = 2l sin .

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=

h h 1.22 1.22h = ; d = 2l tan  = 2l sin  = 2l = 2l → p mv D Dmv

−3 −3 Dmvd ( 3.0  10 m )( 2.0  10 kg ) (150 m s )( 0.010 m ) = = 5.6  1027 m −34 2.44h 2.44 ( 6.63  10 J s )

This is almost 1012 light years. The small angle approximation is definitely valid. 3.

We find the wavelength of the electrons from their kinetic energy, using nonrelativistic mechanics. Classical formulas are justified because the kinetic energy of 45 keV = 0.045 MeV is much less than the rest energy of 0.511 MeV. Eq. 37–7 is used to find the wavelength, and Eq. 34–2a is used for the two-slit interference, with a small-angle approximation. h h ; d sin  = m , m = 0,1, 2, ... ; x = l tan  = = p 2me K

m x m l = → x= , m = 0,1, 2, ... → d d l ( 6.63 10−34 J s ) ( 0.450 m ) hl l x = = = d d 2me K (1.6  10−6 m ) 2 ( 9.11  10−31 kg )( 45  103 eV )(1.60  10−19 J eV )

sin  = tan  →

= 1.63  10−6 m  1.6  m 4.

The uncertainty in the velocity is given. Use Eq. 38–1 to find the uncertainty in the position. 1.055  10−34 J s x  = = = 3.9  10−11 m −27 p mv 1.67  10 kg (1600m s )

(

)

The uncertainty in position is given. Use Eq. 38–1 to find the uncertainty in the momentum. 1.055  10−34 J s p = mv  → v  = = 5515m s  5.5  103 m s −31 −8 mx 9.11 10 kg 2.1  10 m x

(

)(

)

The minimum uncertainty in the energy is found from Eq. 38–2. 1.055  10−34 J s 1eV   E  = = 1.055  10−26 J  = 6.59  10−8 eV  10−7 eV −19  −8 t 1  10 s  1.60  10 J 

(

)

We find the uncertainty in the muon’s energy from Eq. 38–2, and then find the uncertainty in the mass. E 

t

; E = ( m ) c 2 →

(1.055  10 J s ) =  4.7955  10  c ( 2.20  10 s )  −34

m 

)

c t 2

−6

−29

J  1eV  = 3.00  10−10 eV c 2 2  −19  c  1.60  10 J 

We find the uncertainty in the energy of the free neutron from Eq. 38–2, and then the mass uncertainty from Eq. 36–13. We assume that the lifetime of the neutron is good to 2 significant figures. The current experimental lifetime of the neutron is between 881 and 882 seconds.

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E 

; E = ( m ) c 2 → m 

t

c t 2

(1.055  10

−34

J s)

( 3.00 10 m s ) (880s ) 2

= 1.3  10−54 kg

The uncertainty in the position is found from the uncertainty in the velocity and Eq. 38–1. The uncertainty in velocity is found by multiplying the velocity by its precision. 1.055  10−34 J s xelectron  = = = 1.287  10−3 m  1.3  10−3 m p mv 9.11  10−31 kg (120 m s ) 7.5  10−4

(

xbaseball 

p

mv

) (

)

(1.055 10

−34

)

J s)

( 0.14 kg )(120 m s ) ( 7.5  10 ) −4

= 8.373  10−33 m  8.4  10−33 m

( 0.14 kg ) = 1.5  1029 xelectron melectron v mbaseball = = = xbaseball melectron ( 9.11  10−31 kg ) mbaseball v The uncertainty for the electron is greater by a factor of 1.5  1029. 10. The uncertainty in the energy is found from the lifetime and the uncertainty principle. 1.055  10−34 J s  1eV  E = = = 5.49  10−11 eV  −19  −6 1.60 10 J t  12  10 s  

(

)

E 5.49  10−11 eV = = 1.056  1017  1.1  10−17 = 1.1  10−15 % 6 E 5.2  10 eV 11. Use Eq. 38–1 (the uncertainty principle) and Eq. 37–7 (the de Broglie wavelength). 1.055  10−34 J s p  = = 4.058  10−27 kg m s  4.1 10−27 kg m s x 2.6  10−8 m

(

)

( 6.63 10 J s ) = 1.634  10−7 m  1.6  10−7 m h = p 4.058  10−27 kg m s −34

=

12. We use the radius as the uncertainty in position for the neutron. We find the uncertainty in the momentum from Eq. 38–1. If we assume that the lowest value for the momentum is the least uncertainty, we estimate the lowest possible kinetic energy (non-relativistic) by the following calculation. 2

  2 −34   1.055 10 kg m s  ( ) p ) ( x   1MeV  E= =  = = 2.314  10−12 J  2 −13  27 15 − − 2m 2m  1.60  10 J  2 (1.67  10 kg )(1.2  10 m ) 2

= 14.46 MeV  14 MeV 13. We assume the electron is non-relativistic. The momentum is calculated from the kinetic energy, and the position uncertainty from the momentum uncertainty, Eq. 38–1. Since the kinetic energy is known to 1.00%, we have K K = 1.00  10−2.

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p = 2mK ;

2mK dp  1  = 2m  =  2K dK 2 K 

→ p 

Instructor Solutions Manual

2mK K K = 12 2mK 2K K

(1.055  10 J s ) 2 ( 9.11  10 kg ) ( 4.75keV ) (1.60  10 −34

x 

p

K K

2mK

1 2

−31

1 2

−16

J keV )(1.00  10 −2 )

= 5.67  10−10 m

14. Use the radius as the uncertainty in position for the electron. We find the uncertainty in the momentum from Eq. 38–1, and then find the energy associated with that momentum from Eq. 36–14. 1.055  10−34 J s p  = = 1.055  10−19 kg m s. −15 x 1.0  10 m

(

)

If we assume that the lowest value for the momentum is the least uncertainty, we can estimate the lowest possible energy. E = ( p 2 c 2 + m0 2 c 4 )

1/ 2

= ( p ) c 2 + m02 c 4    2

1/ 2

2 2 2 4 = (1.055  10−19 kg m s ) ( 3.00  108 m s ) + ( 9.11  10 −31 kg ) ( 3.00  108 m s )   

 1MeV  = 3.175  10−11 J   200 MeV −13   1.60  10 J 

15. (a) The minimum uncertainty in the energy is found from Eq. 38–2. 1.055  10−34 J s 1eV   E  = = 1.055  10−26 J  = 6.59  10−8 eV  10−7 eV −19  −8 t  1.60 10 J 1  10 s  

(

)

(b) The transition energy can be found from Eq. 37–14b. Z = 1 for hydrogen.  Z2 12   12  En = − (13.6eV ) 2 → E2 − E1 =  − (13.6eV ) 2  −  − (13.6eV ) 2  = 10.2eV n 2   1   E 6.59  10−8 eV = = 6.46  10−9  10−8 E2 − E1 10.2eV (c) The wavelength is given by Eq. 37–3. hc E = hf = →



−34 8 hc ( 6.63  10 J s )( 3.00  10 m s ) = = = 1.22  10−7 m = 122 nm  100 nm E  1.60  10−19 J  (10.2eV )   eV   Take the derivative of the above relationship to find  . E hc hc hc = → d  = − 2 dE →   − 2 E = − → E E E E E    = (122 nm ) ( 6.46  10−9 ) = 7.88  10−7 nm  10−6 nm E

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16. Assume the electron has an initial x momentum px , and thus a wavelength of  = h px . The maxima of the double-slit interference pattern occur at locations satisfying Eq. 34–2a, d sin = m , m = 0,1,2, . If the angles are small, then we replace sin  by  , and so the maxima are given by  = m  d . The angular separation of the maxima is then  =  d , and the angular separation between a maximum and the adjacent minimum is  =  2d . The separation of a maximum and the adjacent minimum on the screen is then yscreen =  l 2d , where l is the distance from the slits to the detection screen. This means that many electrons hit the screen at a maximum position, and very few electrons hit the screen a distance  l 2d to either side of that maximum position. If the particular slit that an electron passes through is known, then y for the electrons at the h location of the slits is d 2. The uncertainty principle says p y  = 1 = . We assume that yslits 2 d  d p y for the electron must be at least that big. Because of this uncertainty in y momentum, the

h yscreen p y l = → yscreen = l  d = . electron has an uncertainty in its location on the screen, as h l px d



Since this is about the same size as the separation between the maxima and minima, the interference pattern will be “destroyed.” The electrons will not be grouped near the maxima locations. They will instead be “spread out” on the screen, and no interference pattern will be visible. 17. We are given that 1 ( x, t ) and  2 ( x , t ) are solutions to the Schrödinger equation. Substitute the function A1 ( x, t ) + B 2 ( x, t ) into the Schrödinger equation. 2 2  ( A1 )  ( B 2 ) 2 A B U x A B  +  +  +  = − − + U ( x ) A1 + U ( x ) B 2 ( )     1 2 1 2 2 2 2m x 2m x 2m x 2 2 2  2 1 22 B = −A − + AU ( x ) 1 + BU ( x )  2 2m x 2 2m x 2 2 2      2 1 22 = A − + U ( x ) 1  + B  − + U ( x ) 2  2 2  2m x   2m x 

−

  1    2  = A i  A1 + B 2   + Bi =i t  t  t  

2  A1 + B 2  + U ( x )  A1 + B 2  = i  A1 + B 2  , the combination 2 2m x t A1 ( x, t ) + B 2 ( x, t ) is also a solution to the time-dependent Schrödinger equation. So, since −

18. (a) Substitute  ( x, t ) = Ae into both sides of the time-dependent Schrödinger equation, Eq. 38–7, and compare the functional form of the results. 2 2  2k 2  i kx −t ) 2  2 Aei ( kx −t ) i ( kx −t ) U U Ae − +  = − + = + U 0  Ae ( 0 0  2 2 2m x 2m x  2m  i ( kx −t )

i kx −t )

 Ae ( =i t t

=  Aei ( kx −t )

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Both sides of the equation give a result of ( constant ) Ae valid solution if the constants are equal.

i ( kx −t )

Instructor Solutions Manual

, and so t  ( x, t ) = Ae (

i kx −t )

is a

Now repeat the process for  ( x, t ) = A cos ( kx −  t ) . −

2  A cos ( kx −  t ) 2 +  = − + U 0 A cos ( kx −  t ) U 0 2 x 2 2m x 2m  2k 2  = + U 0  A cos ( kx −  t )  2m  2

A cos ( kx −  t )  =i = i  A sin ( kx −  t ) t t Because cos ( kx −  t )  sin ( kx −  t ) for arbitrary values of x and t,  ( x, t ) = A cos ( kx −  t ) is NOT a valid solution. i

Now repeat the process for  ( x, t ) = A sin ( kx −  t ) . 2  A sin ( kx −  t ) 2 − +  = − + U 0 A sin ( kx −  t ) U 0 2 x 2 2m x 2m  2k 2  = + U 0  A sin ( kx −  t )  2m  2

A sin ( kx −  t )  =i = −i  A cos ( kx −  t ) t t Because cos ( kx −  t )  sin ( kx −  t ) for arbitrary values of x and t,  ( x, t ) = A sin ( kx −  t ) is NOT a valid solution. (b) Conservation of energy gives the following result. 2 2 p2 h hk k + U0 ; p = = = k → = + U0 E = K +U =  2 2m 2m We equate the two results from the valid solution. 2 2  2k 2  i ( kx −t ) k i ( kx −t ) + =  → + U0 =  U Ae Ae 0  2m  2m  i

The expressions are the same. 19. The wave function is given in the form  ( x ) = A sin kx.

2 2 = = 2.618  10−10 m  2.6  10 −10 m 10 −1 k 2.4  10 m h 6.63  10−34 J s p= = = 2.532  10−24 kg m s  2.5  10−24 kg m s  2.618  10−10 m

(a)  = (b) (c)

p 2.532  10−24 kg m s = = 2.779  106 m s  2.8  106 m s −31 m 9.11  10 kg

Note that the electron is not relativistic – its speed is less than 1% of the speed of light. −24 1 p 2 ( 2.532  10 kg m s ) K= = = 21.99eV  22eV −31 2m 2 ( 9.11  10 kg ) (1.60  10−19 J eV ) 2

(d)

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20. The general expression for the wave function of a free particle is given by Eq. 38–3a. The particles are not relativistic. −31 5 2 2 p mv 9.11 10 kg 2.7 10 m s = = = = 2.3  109 m −1 (a) k = −34 h  1.055  10 J s

(

)(

)

 = A sin ( 2.3  109 m −1 ) x  + B cos ( 2.3  109 m −1 ) x 

(b) k =

2



−27 5 2 p mv (1.67  10 kg )( 2.7  10 m s ) = = = 4.3  1012 m −1 −34 h (1.055  10 J s )

 = A sin ( 4.3  1012 m −1 ) x  + B cos ( 4.3  1012 m −1 ) x 

21. This is similar to the analysis performed in Chapter 16 Section 16–6 for beats. Referring to Fig. 16– 17, we see the distance from one node to the next can be considered a wave packet. We add the two wave functions, employ the trigonometric identity for the sine of a sum of two angles, and then find the distance between the nodes. The wave numbers are related to the wavelengths by k =

1  2 , it is also true that k1  k2 and so 12 ( k1 + k2 )  kavg . We define k = k1 − k2 .

2 . Since 

 =  1 + 2 = A sin k1 x + A sin k2 x = A ( sin k1 x + sin k2 x ) = 2 A sin  12 ( k1 + k2 ) x  cos  12 ( k1 − k2 ) x  = 2 A sin kavg x cos  12 ( k ) x 

The sum function will take on a value of 0 if 12 ( k ) x = ( n + 12 )  , n = 0,1,2 . The distance between these nodal locations is found as follows. 2 ( n + 12 )  ( 2n + 1)  ( 2 ( n + 1) + 1) − ( 2n + 1) = 2 = x= , n = 0,1,2 . → x = k k k k k Now use the de Broglie relationship between the wavelength and momentum. h p 2 2 p = = k → k = ; x = = → xp = h  k p 22. The minimum speed corresponds to the lowest energy state. The energy is given by Eq. 38–13. 6.63  10−34 J s h2 h 2 Emin = = 12 mvmin → vmin = = = 1.5  106 m s 2 −31 −9 8ml 2ml 2 ( 9.11  10 kg )( 0.25  10 m ) Note that this is less than 1% of the speed of light, so the classical formula for kinetic energy is justified. 23. We assume the particle is not relativistic. The energy levels are given by Eq. 38–13, and the wave functions are given by Eq. 38–14. 2 2 h 2n 2 p2 hn n  n x  ; n = sin  En = = → pn = = kn = →  → 2 8ml 2m 2l l l n  l 

n =

2l h = , which is the de Broglie wavelength n pn

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

24. We assume the particle is not relativistic. The energy levels provide the kinetic energy of the particles in the box. h2 p12 h2 h h 2 E1 = = → p = → p1 = → p  2 p1 = 1 2 2 8ml 2m 4l 2l l h xp  l = h l This is consistent with the uncertainty principle. 25. The energy levels for a particle in an infinite potential well are given by Eq. 38–13. The wave 2 functions are given by Eq. 38–14 with A = . l 6.63  10−34 J s ) ( h2 E1 = = = 6.544  10−2 eV  0.065eV 2 2 − 31 − 9 − 19 8m l 8 ( 9.11  10 kg )( 2.4  10 m ) (1.60  10 J eV ) 2

E2 = 22 E1 = 4 ( 6.544  10 −2 eV ) = 0.26eV E3 = 32 E1 = 9 ( 6.544  10 −2 eV ) = 0.59eV E4 = 42 E1 = 16 ( 6.544  10 −2 eV ) = 1.0eV

n =

   2 2  n  sin  sin  x  ; 1 = x  = ( 0.91nm −1/ 2 ) sin (1.3nm −1 ) x  l 2.4 nm  l   2.4 nm 

 2 = ( 0.91nm −1/ 2 ) sin ( 2.6 nm −1 ) x  ;  3 = ( 0.91nm −1/ 2 ) sin ( 3.9 nm −1 ) x  ;  4 = ( 0.91nm −1/ 2 ) sin ( 5.2 nm −1 ) x  26. The wave functions for an infinite square well are given by Eq. 38–14. 2  n   n  x  ;  n = A2 sin 2  x  n = A sin  l    l  2

(a) The maxima occur at locations where  n = A2 . n  n  x =1 → x = ( m + 12 )  , m = 0,1,2, sin 2  l  l   2m + 1  xmax =   l , m = 0,1,2,  2n 

n −1 →

n −1

The values of m are limited because x  l . (b)

The minima occur at locations where  n = 0.

n  n  sin 2  x = 0 → x = m , m = 0,1,2, n → l  l 

xmin =

m l , m = 0,1,2, n n

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Quantum Mechanics

27. The energy levels for a particle in a rigid box are given by Eq. 38–13. We substitute the appropriate mass in each part of the problem. (a) For an electron, we have the following:

6.63  10−34 J s ) ( h2 n2 E= = = 1163MeV 8ml 2 8 ( 9.11  10−31 kg )(1.8  10−14 m )2 (1.60  10−13 J MeV ) 2

 1200 MeV (b) For a neutron, we have the following:

6.63  10−34 J s ) ( h2 n2 E= = = 0.63MeV 8ml 2 8 (1.675  10−27 kg )(1.8  10 −14 m )2 (1.60 10 −13 J MeV ) 2

6.63  10−34 J s ) ( h2 n2 = = 0.63MeV E= 8ml 2 8 (1.673  10−27 kg )(1.8  10 −14 m )2 (1.60  10 −13 J MeV ) 2

28. (a) We use the ground state energy and Eq. 38–13 to find the width of the well. h2 E1 = → 8m l 2 ( 6.63 10−34 J s ) h l = = = 2.1  10−10 m = 0.21nm 31 19 − − 8mE1 8 ( 9.11  10 kg ) ( 8.4eV ) (1.60  10 J eV ) (b) The longest wavelength photon will be the photon with the lowest frequency, and thus the lowest energy. The difference between energy levels increases with high states, so the lowest energy transition is from n = 2 to n = 1. The energy levels are given by Eq. 38–13. h2 En = n 2 = n 2 E1 8ml 2 6.63  10−34 J s 3.00  108 m s c hc hc hc = = = = =  E E2 − E1 4 E1 − E1 3 ( 8.4eV ) 1.60  10−19 J eV

(

)(

)

= 4.9  10−8 m = 49 nm 29. The energy levels for a particle in a rigid box are given by Eq. 38–13. Use that equation, evaluated for n = 4 and n = 1, to calculate the width of the box. We also use Eq. 37–3. hc h2 E = hf = = E4 − E1 = ( 42 − 12 ) → 8m l 2  15 ( 6.63  10−34 J s )( 370  10−9 m ) 15h l = = = 1.3  10−9 m = 1.3nm 8mc 8 ( 9.11  10−31 kg )( 3.00  108 m s )

30. The longest wavelength photon will be the photon with the lowest frequency, and thus the lowest energy. The difference between energy levels increases with high states, so the lowest energy transition is from n = 2 to n = 1. The energy levels are given by Eq. 38–13.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

E = hv = l =



= E2 − E1 =

Instructor Solutions Manual

h2 ( 22 − 12 ) → 8ml 2

3 ( 6.63  10−34 J s )( 580  10−9 m ) 3h = = 7.3  10−10 m = 0.73nm 8mc 8 ( 9.11  10−31 kg )( 3.00  108 m s )

31. The energy released is calculated by Eq. 38–13, with n = 2 for the initial state and n = 1 for the final state. We switch the order of the initial and final states in order to get a positive answer, since the problem asks for the “energy released”.

3 ( 6.63  10−34 J s ) h2 E = E2 − E1 = ( 2 − 1 ) = 8ml 2 8 (1.67  10−27 kg )(1.2  10−14 m )2 (1.60  10−13 J MeV ) 2

= 4.3MeV 32. (a) The ground state energy is given by Eq. 38–13 with n = 1. An oxygen molecule has a mass of 32 amu. 6.63  10−34 J s ) (1) ( h2 n2 E1 = = 8ml 2 n =1 8 ( 32 u ) (1.66  10−27 kg u )( 3.0  10−3 m )2 (1.60  10−19 J eV ) 2

= 7.183  10−19 eV  7.2  10−19 eV (b) We equate the thermal energy expression to Eq. 38–13 in order to find the quantum number. 1 2

kT =

h2 n2 8m l 2

→

n = 2 kTm

( 3.0 10−3 m ) l = 2 (1.38  10−23 J K ) ( 300 K )( 32 u ) (1.66  10 −27 kg u ) h ( 6.63 10−34 J s )

= 1.342  108  1  108 (c) Use Eq. 38–13 with a large-n approximation. h2  h2 h2 2 2 E = En +1 − En = + − = +  = 2nE1 n 1 n 2 n 1 2 n ( ) ( )  8m l 2 8m l 2  8m l 2 = 2 (1.342  108 )( 7.183  10−19 eV ) = 1.9  10−10 eV

33. Because the wave function is normalized, the probability is found as in Example 38–8. Change the l  l n x  n   variable to  = dx  or dx = d  . The lower limit is  or x =  , and then d = l  l n  n   x1 = 0.60nm − 0.15nm = 0.45nm, and the upper limit is x2 = 0.60nm + 0.15nm = 0.75nm. The limits x 0.75nm 0.45nm on the variable  are 1 = n 1 = n = 0.375n and  2 = n = 0.625n . 1.20 nm 1.20 nm l x2

P =   n dx = x1

n x l

0.625 n

2 2 2 2 n x 2 l 2 2 = dx sin sin 2  d =    n sin  d l x1 l l n n x1 l n 0.375

2 1 0.625 n  − 14 sin 2 0.375 n  2 n

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(a) For the n = 1 state, we have the following:

2 1 2 0.625  − 14 sin 2 0.375 =  12 ( 0.25 ) − 14 ( sin1.25 − sin 0.75 )  = 0.47508  0.48  2  

Pn =1 =

(b) For the n = 5 state, we have the following:

Pn =5 =

2 1 2 3.125  − 14 sin 2 1.875 =  12 (1.25 ) − 14 ( sin 6.25 − sin 3.75 )  = 0.20498  0.20  2 5 5

2 1 1 1 12.5  − 14 sin 2 7.5 =  2 ( 5 ) − 14 ( sin 25 − sin15 )  = 0.25  2 20 10

Pn = 20 =

(d) The classical prediction would be that the particle has an equal probability of being at any 0.75nm − 0.45nm location, so the probability of being in the given range is P = = 0.25 . We 1.20 nm see that the probabilities approach the classical value for large n. 34. Consider Fig. 38–9, partially copied here. To consider the problem with the boundaries shifted, we would not expect any kind of physics to change. So, we expect the same wave functions in terms of their actual shape, and we expect the same energies if all that is done is to change the labeling of the walls to x = − 12 l and x = 12 l . The mathematical descriptions of the wave functions would change because of the change of coordinates. All we should have to do is shift the origin of coordinates to the right by 12 l . Thus, we might expect the following wave functions and energies. 2  n  sin  x → n = l  l 

n =

2 2  n   n  sin  ( x + 12 l )  = sin  x + 12 n  l l  l   l 

n = 1:  1 =

2   sin  x + 12   = l l 

n = 2: 2 =

2 2 4h 2  2   2  x +  = − x  ; E2 = sin  sin  8m l 2 l l  l   l 

n = 3:  3 =

2 2 2  3   3   3  x + 32   = x + 12  +   = − x + 12   sin  sin  sin  l l l l l l      

= −

n = 4: 4 =

h2 2   cos  x  ; E1 = 8m l 2 l l 

2 9h 2  3  cos  x  ; E3 = 8m l 2 l  l 

2  4  sin  x + 2  = l  l 

2  4  16h 2 sin   ; E4 = 8m l 2 l  l 

For any higher orders, we simply add another 2 of phase to the arguments of the above functions. They can be summarized as follows. n2 h2 ( n −1) / 2  2  n x  cos  n odd:  = ( −1)  , En = 2 



 l



8ml

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

n even:  = ( −1) 

n/ 2

Instructor Solutions Manual

2 2  2 sin  n x  , En = n h  l 8ml 2  l 

Of course, this is not a “solution” in the sense that we have not derived these solutions from the Schrödinger equation. We now show a solution that arises from solving the Schrödinger equation. We follow the development as given in Section 38–8. As suggested, let  ( x ) = A sin ( kx +  ) . For a region where U ( x ) = 0, k =

2mE 2

(Eq. 38–11a). The

boundary conditions are  ( − 12 l ) = A sin ( − 12 k l +  ) = 0 and  ( 12 l ) = A sin ( 12 k l +  ) = 0. To guarantee the boundary conditions, we must have the following:

A sin ( − 12 k l +  ) = 0 → − 12 k l +  = m ; A sin ( 12 k l +  ) = 0 →

1 2

k l +  = n

Both n and m are integers. Add these two results, and subtract the two results, to get two new expressions. 1 1 2 k l +  = n →  = 12 ( n + m )  ; k = ( n − m )  − 12 k l +  = m l So again, we have an energy quantization, with E =

k2 = 2m

 2 (n − m)

2ml 2

h2 (n − m ) . Note that 8ml 2 2

m = n is not allowed, because this leads to k = 0,  = n , and  ( x )  0.

Next, we normalize the wave functions. We use an indefinite integral from Appendix B–4. 1   ( x ) = A sin ( kx +  ) = A sin  ( n − m )  x + 12 ( n + m )   l  

1 l 2

1



  dx =  A sin  l ( n − m ) x + ( n + m ) dx = 1 2

−

1 2

− 12 l

let  =

l 1 d ( n − m ) x + 12 ( n + m ) → dx = l ( n − m )

x = 12 l →  = n ; x = − 12 l →  = m 

1 =   dx = −

( n − m )  = A2 l A2 l A2 l A2 l n 2 1 1 = −     sin sin 2 d ( ) m 2 2 ( n − m ) m ( n − m ) 2 4 ( n − m ) n

2 2 → A= l l This is the same as in Section 38–8. Finally, let us examine a few allowed cases. A2 =

n = 1, m = 0 : k =

 2   ,  = 12  →  1,0 = sin  x + 12   = l l l 

h2 2   cos  x  ; E1,0 = 8m l 2 l l 

2 2  2 2  2  4h 2  ,  =  →  2,0 = sin  n = 2, m = 0 : k = x +   = − sin  x  ; E2,0 = 8m l 2 l l l  l   l  n = 3, m = 0 : k = =−

3 2 2  3   3  ,  = 32  →  3,0 = sin  sin  x + 23   = x + 12  +   l l l  l   l 

2 2 9h 2  3   3  sin  cos  x + 12   = − x  ; E3,0 = 8ml 2 l l  l   l 

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4 2  4 2  4  16h 2  n = 4, m = 0 : k = x + 2  = ,  = 2 →  4,0 = sin  sin   ; E4,0 = 8m l 2 l l l  l   l  These are the same results as those obtained in the less formal method. Other combinations of m and n would give essentially these same results for the lowest four energies and the associated wave functions. For example, consider n = 4, m = 1. n = 4, m = 1: k = =

3 2 2  3   3  x + 25   = x + 12   ,  = 25  →  4,1 = sin  sin  l l l l l    

2 9h 2  4  E cos  ; = 4,1  8ml 2 l  l 

We see that  4,1 = − 3,0 and that both states have the same energy. Since the only difference in the wave functions is the algebraic sign, any physical measurement predictions, which depend on the absolute square of the wave function, would be the same. 35. We choose the zero of potential energy to be at the bottom of the well. Thus in free space, outside the well, the potential is U 0 = 48eV. Thus, the total energy of the electron is E = K + U 0 = 180eV + 48eV = 228 eV. We assume both energies are to the nearest 1 eV. (a) In free space (regions I and III in the diagram), the kinetic energy of the particle is 180 eV. Use that to find the momentum and then the wavelength. p2 h → p = 2mK = → K= 2m  ( 6.63  10−34 J s ) h = = = 9.15  10−11 m 1/ 2 2mK  2 ( 9.11  10−31 kg ) (180eV ) (1.60  10−19 J eV )   (b) Over the well (region II in the diagram), the kinetic energy is 228 eV.

=

h 2mK

( 6.63 10 J s )  2 ( 9.11  10 kg ) ( 228eV ) (1.60 10 J eV )    −34

−31

−19

1/2

= 8.13  10−11 m

(c) The diagram is qualitatively the same as Fig. 38–14, reproduced here. Notice that the wavelength is longer when the particle is not over the well, and shorter when the particle is over the well.

36. We pattern our answer after Fig. 38–13.

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37. (a) We assume U that the lowest three states are bound in the well, so that E  U 0 . See the diagrams for the proposed wave functions. Note that, in the well, the wave functions are U similar to those for the 0 infinite well. Outside the well, for x  l , the wave functions E3 are drawn with an exponential decay, similar to the right side of Fig. 38–13. E2 (b) In the region x  0,  = 0 . E1 x of a In the well, with 0  x  l , the wave function is similar to that 0 L free particle or a particle in an infinite potential well, since U = 0. I II III 2mE Thus  = A sin k x + B cos k x , where k = .

2m (U 0 − E )

In the region x  l ,  = De − Gx , where G =

3

2 1 x=0

x=l

38. We will consider the “left” wall of the square well, using Fig. 38–12, and assume that our answer is applicable at either wall due to the symmetry of the potential well. As in Section 38–9, let  = CeGx for x  0, with G given in Eq. 38–16. Since the wave function must be continuous,  ( x = 0 ) = C. The energy of the electron is to be its ground state energy, approximated by Eq. 38–13 for the infinite well. If that energy is much less than the depth of the well, our approximation will be reasonable. We want to find the distance x for which  ( x ) = 0.010 ( 0 ) .

( 6.63 10−34 J s ) h2 = = 14.73eV E= 8ml 2 8 ( 9.11  10−31 kg )( 0.16  10−9 m )2 (1.60  10−19 J eV ) 2

 ( x ) CeGx ln ( 0.010 ) ln ( 0.010 ) = = eGx = 0.010 → x = =  ( 0) C G 2m (U 0 − E )

(1.055  10 J s ) ln ( 0.010 ) 2 ( 9.11  10 kg ) (1800eV − 14.73eV ) (1.60  10 −34

−31

−19

J eV )

= −2.1  10−11 m

The wave function will be 1.0% of its value at the walls at a distance of 2.1 10−11 m = 0.021nm from the walls. 39. We use Eqs. 38–17a and 38–17b. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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T = e −2Gl

→ G=−

2m (U 0 − E ) ln T ln T =− ; G= 2 2l 2l

→

2 (1.055  10−34 J s )  − ln ( 0.00080 )    1  ln T  − = − E = U0 − 14eV     −19 −31 −9 2m  2 l  2 ( 9.11  10 kg )  2 ( 0.85  10 m )   1.60  10 J eV  2

= 14eV − 0.672eV = 13.328eV  13eV

40. We use Eqs. 38–17a and 38–17b to solve for the particle’s energy.

T = e −2Gl → G = −

2m (U 0 − E ) ln T ln T =− ; G= 2 2l 2l

→

2 1.055  10−34 J s )  ( ln ( 0.010 )    1  ln T     − = − − E = U0 − 18eV    −19 −31 −9 2m  2 l  2 ( 9.11  10 kg )  2 ( 0.48  10 m )   1.60  10 J eV  2

= 18eV − 0.879eV = 17.12eV  17 eV 41. We use Eqs. 38–17a and 38–17b to solve for the transmission coefficient, which can be interpreted in terms of probability. The energy difference U 0 − E = 25.0MeV − 7.5MeV = 17.5MeV. For the mass of the helium nucleus, we take the mass of 2 protons and 2 neutrons, ignoring the (small) binding energy. Proton:

2m (U 0 − E )

2 (1.67  10−27 kg ) (17.5MeV ) (1.60  10 −13 J MeV )

(1.055  10 J s ) =e 2G l = 2 ( 9.166  10 m )( 3.6  10 m ) = 6.600 ; T 14

−34

−1

−15

−2 G l

proton

= 9.166  1014 m −1

= e −6.600 = 1.4  10 −3

Helium: Mass = 2mproton + 2mneutron = 2 (1.673  10−27 kg ) + 2 (1.675  10−27 kg ) = 6.70  10−27 kg.

2m (U 0 − E )

2 ( 6.70  10−27 kg ) (17.5MeV ) (1.60  10 −13 J MeV )

(1.055  10 J s ) 2G l = 2 (1.836  10 m )( 3.6  10 m ) = 13.219 ; T = e 15

−34

−1

−15

−2 G l

= 1.836  1015 m −1

= e −13.219 = 1.8  10 −6

It is much less likely that the heavier particle will tunnel through the barrier. 42. (a) The probability of the electron passing through the barrier is given by Eqs. 38–17a and 38–17b. The energy difference is U 0 − E = 9.2eV − 7.2eV = 2.0eV . T =e 2l

−2 l

2 m (U 0 − E )

2m (U 0 − E )

= 2 ( 0.25  10−9 m )

2 ( 9.11  10−31 kg ) ( 2.0eV ) (1.60  10−19 J eV )

(1.055  10

−34

J s)

= 3.619

T = e −3.619 = 2.681  10−2  2.7% (b) The probability of reflecting is the probability of NOT tunneling, and so it is 97.3% .

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43. The transmitted current is caused by protons that tunnel through the barrier. Since current is directly proportional to the number of charges moving, the transmitted current is the incident current times the transmission coefficient. We use Eqs. 38–17a and 38–17b. T =e 2l

−2 l

2 m (U 0 − E )

2m (U 0 − E )

= 2 ( 2.8  10−13 m )

2 (1.67  10−27 kg ) (1.0 MeV ) (1.60  10−13 J eV )

(1.055  10

−34

J s)

= 122.71

T = e −122.71 → I = I 0T = (1.5mA ) ( e −122.71 ) = 7.7  10−54 mA

This is so incredibly small that it could be considered 0 – no protons will get through the barrier. Note that using more digits on the constants will produce a slightly different answer. 44. The transmission coefficient is given by Eqs. 38–17a and 38–17b. (a) The barrier height is now 1.03 ( 70 eV ) = 72.1eV, so the energy difference is now 22.1 eV/

2m (U 0 − E )

2 ( 9.11  10−31 kg ) ( 22.1eV ) (1.60  10−19 J eV )

= 2 ( 0.10  10 m ) −9

(1.055  10

−34

J s)

= 4.812 2 m (U − E )

T 8.132  10−3 = = 0.81 → 19% decrease 0.010 T0 (b) The barrier width is now 1.03 ( 0.10 nm, ) = 0.103nm, and the energy difference is again 20 eV. T =e

−2 l

= e −4.812 = 8.132  10−3 ;

2m (U 0 − E )

= 2 ( 0.103  10 m ) −9

2 ( 9.11  10−31 kg ) ( 20eV ) (1.60  10−19 J eV )

(1.055  10

−34

J s)

= 4.715 T =e

−2 l

2 m (U 0 − E )

= e −4.715 = 8.960  10−3

T 8.960  10−3 = = 89.6 → 10% decrease ( 2 sig fig ) T0 0.010 45. We assume that the wave function inside the barrier is given by a decaying exponential, so  ( x ) = Ae − Gx .

 (x = l )

 ( x = 0)

( Ae ) = e = − Gl

−2 G l

46. (a) We assume that the transmission coefficient for a single barrier is small; Eq. 38–17a applies, and so T = e−2Gl . With two separate barriers, there can be multiple reflections at each barrier,

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assuming the possibility that the electron may pass through the second barrier on the first time it encounters that barrier, or the second time, or the third, etc. (assuming it doesn’t pass through the first barrier having reflected from the second barrier initially). Note that the separation distance between the barriers does not enter the calculation. Consider the diagram. We take the probability of the electron encountering the first barrier as “1”. Then the 1 (100% ) T T2 probability that the electron passes through the first barrier is T, and thus the probability that the electron encounters the second TR barrier is also T. Then, the probability that the electron passes 2 T 2 R2 TR 2 through the second barrier on this encounter is T , and the probability that the electron reflects from the second barrier is TR3 TR. If the electron has reflected from the second barrier, then the probability that it reflects upon reaching the first barrier again is (TR ) R = TR 2 . The probability that this “twice reflected” electron will reflect again (from the second barrier) is (TR2 ) R = TR3 , and the probability that this “twice reflected”

TR 4

T 2 R4

TR5 TR 6

T 2 R6

electron will pass through the second barrier is (TR 2 ) T = T 2 R 2 . So, the probability of the electron passing through the second barrier on either the first or second 2 2 2 encounter is T + T R . From the diagram, we can see that with each pair of reflections, the 2 “additional” probability of passing through is R times the previous probability of passing through. 2 Thus, the total probability of passing through the two separated barriers is T times an infinite 2 geometric sum, with a first term of 1 and a common ratio of R . Note that R < 1, T < 1, and R = 1 – T.  n  1 − R  T2 2  1− 0  P = T 2 + T 2 R2 + T 2 R4 + T 2 R6 + = T 2  ( R2 ) = T 2  T = =  2  2   1 − R  2T (1 − 12 T ) n=0  1− R  T 2 −T Note that if T << 1 (a small transmission possibility), then the probability of passing through both barriers is 12 T , half the probability of going through just the first barrier. =

(b)

Now there is only one transmission, with twice the thickness, so we use Eq. 38–17a again.

Tdouble = e−2G( 2 l ) = e−4Gl = e−2Gl e−2Gl = T 2 If T is small, 12 T  T 2 , and the transmission probability is greater through two thinner barriers than one thick barrier. 47. Please note – because the final answer is very sensitive to the actual numbers used, we have used more significant figures in the constants than we typically do in these solutions. (a) We assume that the alpha particle is at the outer edge of the nucleus. The potential energy is electrostatic potential energy and is found from Eq. 23–10. 2  2 )( 90 ) (1.602  10 −19 C )  ( 9 2 2 ( 8.988  10 N  m C )  −15 1 Q1Q2   8 10 m  U surface = =  4  rsurface of   1MeV     nucleus    1.602  10−13 J   = 32.40 MeV  30 MeV © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) The kinetic energy of the free alpha particle is also its total energy. Since the free alpha has 4 MeV, by conservation of energy the alpha particle had 4 MeV of potential energy at the exit from the barrier. See the diagram, a copy of Fig. 38–17, modified to show U = 0 inside the barrier, and stated in part (c). r of 1 Q1Q2 1 Q1Q2 surface nucleus = U exit = 4  rexit from 4  rsurface of rexit from barrier

nucleus

rsurface of = U surface nucleus rexit from

rexit from barrier

barrier

→

barrier

U 32.40 MeV rexit from = surface rsurface of = ( 8fm ) = 64.80fm U exit nucleus 4 MeV barrier r = rexit from − rsurface of = 64.80fm − 8fm = 56.80fm  57 fm barrier

nucleus

width of rbarrier = 0.33 ( 56.80fm ) = 18.74fm. The barrier exists at both “edges” of the nucleus, if we imagine the alpha particle as moving along a diameter inside the nucleus, from wall to wall. See the diagram (not to scale). We calculate the speed of the alpha particle and use that to find the frequency of collision with the barrier.

“out”

“in” 18.7 fm

“out” 16 fm

18.7 fm

E = K = 12 mv 2 → v=

2 ( 4 MeV ) (1.602  10−13 J MeV ) 2E = = 1.3839  107 m s  1.4  107 m s −27 m 4 (1.673  10 kg )

Note that the speed of the alpha is less than 5% of the speed of light, so we can treat the alpha without using relativistic concepts. The time between collisions is the diameter of the nucleus (16 fm) divided by the speed of the alpha particles. The frequency of collision is the reciprocal of the time between collisions. v 1.3839 107 m s = 8.6494  1020 collisions s  8.6 1020 collisions s f = = −15 d 16  10 m If we multiply this collision frequency times the probability of tunneling, T, then we will have an estimate of “effective” collisions/s, or in other words, the decays/s. The reciprocal of this effective frequency is an estimate of the time the alpha spends inside the nucleus – the life of the uranium nucleus. G=

2m (U 0 − E )

2 ( 4 ) (1.673  10−27 kg ) ( 28.40 MeV ) (1.602  10−13 J MeV )

(1.05510

−34

J s)

= 2.3390  1015 m −1  2.34  1015 m −1

T = e−2Gl = e

(

)(

−2 2.3391015 m−1 18.7410−15 m

)

= 8.4579  10−39

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Lifetime =

1 1 = 20 fT ( 8.6494  10 collisions s )(8.4579  10−39 )

1yr   = 1.3669  1017 s  = 4.33  109 yr  4.3  109 yr 7   3.156  10 s  While the agreement with experiment is quite good for this model, the calculated answer is highly dependent on the width of the barrier used. For instance, if we used a factor of 0.32 instead of 0.33, the lifetime calculation gives about 3.0  108 yr , and if a factor of 0.34 is used, the calculation gives about 6.2  1010 yr . The textbook points this out in Example 38–11. 48. We find the lifetime of the particle from Eq. 38–2.

(1.055  10 J s ) t  = = 2.6  10 s. E ( 2.5GeV ) (1.60  10 J GeV ) −34

−25

−10

49. We find the wavelength of the protons from their kinetic energy, and then use the two-slit interference formulas from Chapter 34, with a small angle approximation. If the protons were accelerated by a 650-volt potential difference, then they will have 650 eV of kinetic energy, and thus be non-relativistic.

=

h h = ; d sin  = m , m = 1, 2, ... ; y = l tan  p 2m0 K

m y m l = → y= , m = 1, 2, ... → l d d ( 6.63 10−34 J s ) (15m ) l hl y = = = d d 2m0 K ( 8.0  10−4 m ) 2 (1.67  10−27 kg ) ( 650eV ) (1.60 10 −19 J eV )

sin  = tan  →

= 2.1  10−8 m = 21nm 50. We assume that the particles are not relativistic. Conservation of energy is used to find the speed of each particle. That speed can then be used to find the momentum and finally the de Broglie wavelength. We let the magnitude of the accelerating potential difference be V. h h h h 2eV U initial = K final → eV = 12 mv 2 → v = ; = = = = = x m p mv 2eV 2meV m m 2 h h xp = → p = 2 x h 2 h mproton pproton xproton 2melectron eV x 1.67  10−27 kg = 42.8 = = electron = = = 2 h h melectron 9.11  10−31 kg pelectron xproton xelectron 2mproton eV 51. The energy levels for a particle in an infinite potential well are given by Eq. 38–13. The wave 2 functions are given by Eq. 38–14. with A = . l

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( 6.63  10−34 J s ) h2 E1 = = = 32.90 MeV  33MeV 8ml 2 8 (1.67  10−27 kg )( 2.5  10−15 m )2 (1.60  10 −13 J MeV ) 2

(a)

E2 = 22 E1 = 4 ( 32.90 MeV ) = 130 MeV ; E3 = 32 E1 = 9 ( 32.90 MeV ) = 300 MeV ( 2 sig. fig. ) E4 = 42 E1 = 16 ( 32.90 MeV ) = 530 MeV

(b)  n =

(c)

2  n  sin  x l  l 

1 =

 2   sin  x  = ( 2.8  107 m −1/ 2 ) sin (1.3  1015 m −1 ) x  −15 −15 2.5  10 m  2.5  10 m 

2 =

2 2   sin  x  = ( 2.8  107 m −1/ 2 ) sin ( 2.5  1015 m −1 ) x  −15 −15 2.5  10 m  2.5  10 m 

3 =

2 3   sin  x  = ( 2.8  107 m −1/ 2 ) sin ( 3.8  1015 m −1 ) x  −15 −15 2.5  10 m  2.5  10 m 

4 =

2 4   sin  x  = ( 2.8  107 m −1/ 2 ) sin ( 5.0  1015 m −1 ) x  −15 −15 2.5  10 m  2.5  10 m 

E = E2 − E1 = 4 E1 − E1 = 3E1 = 3( 32.90 MeV ) = 98.7 MeV  99 MeV

E =

−34 8 hc hc ( 6.63  10 J s )( 3.00  10 m s ) → = = = 1.3  10−14 m = 13fm −13 E ( 98.7 MeV ) (1.60  10 J MeV ) 

This is in the gamma-ray region of the EM spectrum, as seen in Fig. 31–12. 52. The uncertainty in the energy is found from the lifetime and the uncertainty principle. h hc E = = ; E = hv =  t 2t h  E 2t 400  10−9 m = = = = 2.12  10−8  2  10−8 hc E 2 ct 2 3.00  108 m s 10  10−9 s



(

)(

)

hc hc hc E E  → dE = − 2 d  → E  − 2  = −  → =−      E  = 2  10−8 The wavelength uncertainty is the absolute value of this expression, and thus  E=

53. We assume the alpha particle is in the ground state. The energy is given by Eq. 38–13.

6.63  10−34 J s ) ( h2 E1 = = = 0.3343MeV 8ml 2 8 ( 4 ) (1.67  10−27 kg )(1.24  10−14 m )2 (1.60  10−13 J MeV ) 2

 0.334 MeV The speed can be found from the kinetic energy. The alpha is non-relativistic. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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E1 =

( 6.63 10−34 J s ) h2 h 2 1 m = → = = = 4.00  106 m s v v 2 2 −27 −14 8ml 2ml 2 ( 4 ) (1.67  10 kg )(1.24 10 m )

The result justifies the assumption that the alpha is non-relativistic. 54. We use Eq. 37–10, Bohr’s quantum condition, with n = 1 for the ground state. h → mv r1 = → mv = p = = p mv rn = n 2 r1 xp 

→ x 

= = r1 p r1 The uncertainty in position is comparable to the Bohr radius.

55. The time-independent Schrödinger equation with U = 0 is −

d 2 = E . 2m dx 2 2

2 2 2 2 2 d 2 d2 k ikx ikx 2 Ae k Ae = − = − − = = ( ) ( ) 2 2 2m dx 2m dx 2m 2m We see that the function solves the Schrödinger equation.

−

56. The wave functions for the particle in the infinite well are  n = 38–8. A table of integrals was consulted to find  x 2 ( sin 2 ax ) dx. x =  x n 2

 2mE       = E 2m

2  n  as derived in Section sin  x , l  l 

l  2 2 2 2  n   n   sin  dx =  x  x   dx =  x sin  x  dx l 0  l   l  0  l l

Note that  x 2 ( sin 2 ax ) dx =

1  x3  x2 x cos 2ax . − −  sin 2ax − 6  4a 8a 3  4a 2 l

    2n   x cos  x  3  2  l 2 2x x 1  n   l   sin  2n x  − − x 2 =  x 2 sin 2  x  dx =  −   3 2  n   l 0 l l l 6      n   n    4 8 4  l        l l        0 −2 = l 2  13 − 12 ( n )   

See the adjacent graph.

0.34 0.33 0.32 0.31 0.30 0.29 0.28 0

n © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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57. The de Broglie wavelength is determined by Eq. 37–3.

=

h h h 6.63  10−34 J s = → v= = = 2.07  10−35 m s  2.1  10 −35 m s p mv m ( 64.0 kg )( 0.50 m )

Since  is comparable to the width of a typical doorway, yes , you would notice diffraction effects. However, assuming that walking through the doorway requires travel through a distance of 0.2 m, the time t required is huge.

vt = d → t =

0.20 m d = = 9.7  1034 s  3  1026 yr −35 v 2.07  10 s

This is many orders of magnitude larger than the approximate age of the universe (  1 1010 yr ) . 58. (a) See the diagram. (b) We use the solution  ( x ) = Ae − Bx in the Schrödinger equation. 2 2 d = −2 ABxe − Bx  = Ae − Bx ; dx 2 2 2 d = −2 ABe − Bx − 2 ABx −2 Bxe − Bx 2 dx

(

= 2 ( 2 Bx 2 − 1) ABe − Bx

U (x )

) x

2 2 2 2 d 2 2 ( 2 Bx 2 − 1) ABe − Bx  + 12 Cx 2 Ae − Bx = EAe − Bx → +  = − U 2   2m dx 2m 2  B  1 2 2 B2  2 E C − + −   2 x =0 m   m  

−

This is a solution if

2 2 B2 B . Solve these two equations for E in terms of C, = E and 12 C = m m

and let   C m . 2 2 2 2 B2 mC B mC 1 → B= = =2 ; E= 2 m m m 2 m mC m = B= C m= 2 2 2 1 2

C m = 12 

59. From energy conservation, the speed of a ball after falling a height H, or the speed needed to rise to a height H, is v = 2 gH . We say that the starting height is H 0 , and so the speed just before the ball hits the ground before the first bounce is v0 = 2 gH 0 . After that bounce, the ball rebounds to H1 = 0.65H 0 , and so the speed right after the first bounce, and right before the second bounce, is v1 = 2 gH1 = 2 g ( 0.65 ) H 0 . Repeated application of this idea gives the maximum height after n

bounces as H n = ( 0.65) H 0 , and the maximum speed after n bounces as vn = 2 g ( 0.65 )n H 0 . The uncertainty principle will come into play in the problem when the maximum speed after a bounce is of the same order as the uncertainty in the speed. We take the maximum height as the uncertainty in the position. n

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mv y = p y ; yp y 

m 2 g ( 0.65 ) H 0 

→ mv y 

y

→ mvn 

→ 2m 2 g ( 0.65 ) H 0 

( 0.65) H 0 n

→

( 0.65)

H 02

→

2   1.055  10−34 J s ) (   ln   ln  2 3   2 ( 3.0  10−6 kg )2 ( 9.80 m s 2 ) ( 2.0 m )3  2 2m gH 0  3n  = 105 =  ( 0.65)  2 3 → n   2m gH 0 3ln ( 0.65 ) 3ln ( 0.65 ) After about 105 bounces, the uncertainty principle will be important to consider. 2

60. The difference in measured energies is found from Eq. 37–3. hc hc E = E2 − E1 = − 2 1 1 1   −21 = ( 6.63  10−34 J s )( 3.00  108 m s )  −  = 1.670  10 J −9 −9 487 10 m 489 10 m     The lifetime of the excited state is determined from Eq. 38–2.

1.055  10−34 J s = 6.317  10−14 s  6.32  10−14 s E 1.670  10−21 J This solution assumes that the entire spread in energy is the uncertainty. E t 

→ t 

61. Find the uncertainty in the position from Eq. 38–1.

x =

mv

1.055  10−34 J s = 3.2  10−37 m 1500 kg 0.22 m s ( )( )

62. We model the electrons as being restricted from leaving the surface of the sodium by an energy barrier, similar to Fig. 38–15a. The difference between the barrier’s height and the energy of the electrons is the work function, and so U 0 − E = W0 = 2.28eV. But quantum mechanically, some electrons will “tunnel” through that barrier without ever being given the work function energy, and thus get outside the barrier, as shown in Fig. 38–15b. This is the tunneling current as indicated in Fig. 38–18. The distance from the sodium surface to the tip of the microscope is the width of the barrier, l . We calculate the transmission probability as a function of barrier width by Eqs. 38–17a and 38–17b. The barrier is then increased to l + l , which will lower the transmission probability. 2G l = 2l

2m (U 0 − E )

= 2 ( 0.02  10−9 m )

2 ( 9.11  10−31 kg ) ( 2.28eV ) (1.60  10−19 J eV )

(1.055  10 J s ) −34

= 0.3091 T e ( ) = = e −2Gl = e −0.3091 = 0.734 −2 G l T0 e The tunneling current is caused by electrons that tunnel through the barrier. Since current is directly proportional to the number of electrons making it through the barrier, any change in the transmission probability is reflected as a proportional change in current. So we see that the change in the transmission probability, which will be reflected as a change in current, is a decrease of 27%. Note that this change is only a fraction of the size of an atom. T0 = e −2Gl ; T = e −2G ( l +l ) ;

−2 G l +l

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Instructor Solutions Manual

2   sin  x  . Let l l  1 1 1 1 xcenter = 2 l , so the region of interest extends from xmin = 2 l − 2 x to xmax = 2 l + 12 x . We are to

63. The ground state wave function for the particle in the infinite well is  =

find the largest value of x so that the approximate probability of  ( x = 12 l ) x = 2

2x (from l

 2    sin  x   dx , the exact probability. We Example 38–7) is no more than 10% different than   l  l  xmin  have calculated the integral result two ways. xmax

First, we calculated the value of the integral using numeric integration (as described in Appendix C), first finding the number of steps needed between xmin and xmax that gives a stable value. Then we compared the integral to the approximation. (Note that the integral could be evaluated exactly.) To aid in the evaluation of the integral, we make the substitution that u = x l . Then the integral becomes as follows: I =

 x  umax = 12 1−   l 



 x  umin = 12 1−   l 

 2sin 2 ( u )  du . In doing the numeric integrations, we found that

for any value of x up to l , breaking the numeric integration into 50 steps gave the same answer to 3 significant digits as breaking it into 100 steps. So we did all numeric integrations with 50 steps. We then numerically calculated the integral for values of x l , starting at 0.01, and increasing by steps of 0.01, until we found a 10% difference between the approximation and the numeric integration. This happens at x l = 0.34, and so the approximation is good within 10% up to

x = 0.34l = 0.34 ( 0.10nm ) = 0.034 nm . This is much broader than we might have guessed initially, indicating that the wave function is varying rather slowly over the central region of the potential well. Secondly, we calculated the integral exactly. Here is that calculation.

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xmax = 12 l + 12 x

1 l + 1 x

2 2  2 2 2     sin sin  x  dx x dx =      l 1 l − 1 x l  l  l  xmax = 12 l − 12 x  2 2

1   2 1  Note that  sin ( ax ) dx = 2 x − sin ( 2ax )  4a   2   2 12 ( l + x )    l 1 1 sin x l +  −  ( )    − 1 l +x 2 2 ) 2( 4 l  l      2 1 l  2 x   sin  I = 2 x − =   4 a  l   1 ( l −x )  2  l   2 12 ( l − x )    l 2 1 1    l  2 2 ( l − x ) − 4 sin  l         ( l + x )    1   ( l − x )    2   l l sin  sin  =   14 ( l + x ) −   −  4 ( l − x ) −   4 4 l   l l         1   x 1  x     x    x 1 =  1 + − sin  1 + − sin  1 −     − 1 − l      2  l l   l    x 1   x   x    = + sin  1 −  − sin  1 +    l  l    2     l The following table compares I to I approx =

2x 2x for various values of . We see that the 10% l l

discrepancy shows up at about the same place as with the numeric integration.

1.0

Probability

64. (a) See the graph. (b) From the graph (and the spreadsheet used to create the graph), these results are found.

0.8 Transmission

0.6

Reflection

0.4 0.2 0.0 0

E /U 0

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Instructor Solutions Manual

T = 10% at E U 0 = 0.146 T = 20% at E U 0 = 0.294 T = 50% at E U 0 = 0.787 T = 80% at E U 0 = 1.56 65. Note that

has SI units of J s . And J s =

( E )( t ) has “obvious” SI units of J s . ( x )( px ) has SI units of m ( Lz ) has SI units of

kg m2 kg m 2 s = . s2 s

kg m kg m 2 = . s s

kg m2 kg m2 , and (  ) is unitless, so ( Lz )(  ) has SI units of . s s

66. The duration of shortest pulse that can be produced is found from the uncertainty principle for energy and time, converted to relate wavelength to time. The wavelength is found from Eq. 37–3, and then the derivative of that relationship is used to relate  to t . We also take an absolute value to eliminate the negative sign from the differentiation. hc dE hc hc hc E = hf = → = − 2 → E = − 2  → ( E )( t ) = 2 (  ) ( t )  → d    

( 704.1  10 m ) t  = = 2.63  10 s  3  10 s 2 c  2 ( 3.00  10 m s )( 0.1  10 m ) −9

2

−12

−9

The length of the pulse is the distance that it would travel, at the speed of light, times its time duration. x = ct = ( 3.00  108 m s )( 2.63  10 −12 s ) = 7.89  10 −4 m = 7.89  105 nm  8  105 nm

67. The minimum speed corresponds to the lowest energy state (the ground state). The energy is given by Eq. 38–13. 6.63  10−34 J s h2 h 2 1 v v Emin = m = → = = = 2.9  10−31 m s min min 8ml 2 2 2ml 2 (15.0  10−3 kg )( 7.5  10−2 m ) We can check this result to see if it satisfies the uncertainty principle, Eq. 38–1. 1.055  10−34 J s = = 9.4  10−32 ( x )( px ) = ( x )( mvx )  → ( vx )  m ( x ) (15.0  10−3 kg )( 7.5  10−2 m ) The speed has to be at least as big as its uncertainty, so the uncertainty principle is satisfied. 68. The uncertainty in the electron position is x = r1 . The minimum uncertainty in the velocity, v , can be found by solving Eq. 38–1, where the uncertainty in the momentum is the product of the electron mass and the uncertainty in the velocity. xpmin = r1 ( mv ) 

v 

mr1

(

1.055  10−34 J s

)(

9.11  10−31 kg 0.529  10 −10 m

)

= 2.19  106 m/s

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

69. (a) To check that the wave function is normalized, we calculate   ( x ) dx. 2

−



  ( x)

−



1 x dx =  e b b −

x2 − 2 b



2 dx = 2  xe b 0

x2 − 2 b



x 2  b2 − 2  dx = 2  − e b  = − ( 0 − 1) = 1 b  2  2

We see that the function is normalized, independent of the value of b. 2 (b) The most probable position is that for which  ( x ) is maximized. That point can be found by solving

d  ( x) dx

= 0 for x. Since we are only considering x  0, we need not use the absolute

value signs in the function. 2  x − x2  b d e   2 2 2 x2 d  ( x)  b  1 − bx2 2x2 x  2 x  − b2 = = 2 e + 2  − 2 e = 0 → 1= 2 → dx dx b b  b  b b 1.0 nm = = 0.71nm x= 2 2 This value for x maximizes the function because the function must be positive, and the function is 0 at x = 0 and x = . Thus, this single local extreme point must be a maximum. (c) To find the probability, we integrate the probability density function between the given limits. 0.25nm

0.25nm

  ( x ) dx =  2

x x  − 2  x − b2 1 b = − e dx e   2 b2   2

0.25nm

(

= − 12 e −( 0.25)

) − ( − ) = 0.030 1 2

70. (a) We assume the pencil is a uniform rod, and that it makes an angle of  with the vertical. If the bottom point is fixed, then the torque due to its weight about the bottom point will cause an angular acceleration. See the diagram. l 2 2 d  1 1  = I → mg ( 2 l ) sin  = 3 ml  dt 2 mg From the equation, if the pencil is exactly upright, so that  = 0, then the angular acceleration will be exactly 0, and the pencil will remain stationary. But, according to the uncertainty principle (as expressed near the end of Section 38–4), the angle cannot be known with 0 uncertainty. Let the z axis be coming out of the page. Lz  

→  

Lz

0

Thus, the pencil cannot have exactly  = 0, and so there will be a torque and hence rotation. (b) For the initial part of the motion, the angle will be very small, and so the differential equation d 2 3g  = 2 . The solutions to this differential equation are of the form can be expressed as 2l dt  = Ae kt + Be − kt , where k =

3 ( 9.80 m s2 ) 3g = = 9.037s −1 . Since the angle will be 2l 2 ( 0.18m )

increasing in time, we ignore the second term, which decreases in time. Thus   Aekt , with

 ( t = 0 ) = 0  A. The angular velocity of the pencil is approximated as  =

d = kAe kt , and dt

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the initial angular velocity is 0 = kA. We take the initial position and the initial angular velocity as their smallest possible values, which are their uncertainties – the magnitude of a quantity must be at least as big as its uncertainties. Apply the uncertainty principle in angular form. Lz   → ( I  )  = 13 ml 2 ( kA )( A ) = 13 ml 2 kA2  → 3 A = ml 2 k

3 (1.055  10−34 J s )

( 5.0 10−3 kg ) ( 0.18m ) ( 9.037s−1 ) 2

= 4.650  10−16 rad = 0

0 = kA = ( 9.037s −1 )( 4.650  10−16 rad ) = 4.202  10−15 rad s With this initial position and initial angular velocity, we can then perform a numeric integration to find the time when the angle is  =  2 rad. For a step size of 0.01 s, the time of fall is about 4.05 s. For a step size of 0.001 s, the time of fall is about 3.97 s. This is only about a 2% change in the final result, so the time is quite stable around 4 s. Even changing the starting angle to a value 100 times bigger than that above (so 0 = 3.930  10−14 rad ) still gives a time of fall of 3.46 s. So, within a factor of 2, we estimate the time of fall as 4 seconds. Note that if the solution of the approximate differential equation is used,   Aekt , we get the following time of fall. 1   1  2   = 3.96s. tmax = ln  max  = ln  −1 −16  k  A  9.037s  4.650  10 

1300

CHAPTER 39: Quantum Mechanics of Atoms Responses to Questions 1.

(a) The quantum-mechanical theory retained the aspect that electrons only exist in discrete states of definite energy levels. It also retained the feature that an electron absorbs or emits a photon when it changes from one energy state to another. (b) In the Bohr model, the electrons moved in circular orbits around the nucleus. In the QM model, the electrons occupy orbitals about the nucleus and do not have exact, well-defined orbits.

The quantity  is maximum at r = 0 because of its dependence on the factor e − r r . In the ground

state, the electron is expected to be found near the nucleus. The radial probability density 4 r  gives the probability of finding the electron in a thin spherical shell located at r. Since r = 0 is at the center of the nucleus, the radial probability density is zero at the center. 2

The quantum-mechanical model predicts that the electron spends more time near the nucleus. In the Bohr model, the electron in the ground state is in a fixed orbit of definite radius. The electron cannot come any closer to the nucleus than that distance. In the quantum-mechanical model, the most probable location of the electron is the Bohr radius, but it can also be found closer to the nucleus (and farther away). See Fig. 39–1 for a visual representation of this question.

As the number of electrons goes up, the number of protons in the nucleus increases, which increases the attraction of the electrons to the center of the atom. Even though the outer electrons are partially screened from the increased nuclear charge by the inner electrons, they are all pulled closer to the more positive nucleus. Also, more states are available in the upper shells to accommodate many more electrons at approximately the same radius.

Because the nuclei of hydrogen and helium are different, the energy levels of the atoms are different. The presence of the second electron in helium will also affect its energy levels. If the energy levels are different, then the energy difference between the levels will be different and the spectra will be different.

The two levels have different orbital quantum numbers. The orbital quantum number of the upper level is ℓ = 2. This results in five different possible values of 𝑚ℓ (–2, –1, 0, 1, and 2), so the energy level is split into five separate levels in the presence of a magnetic field. The lower level shown has an orbital quantum number of ℓ = 1, so only three different values of 𝑚ℓ (–1, 0, and 1) are possible. Therefore, the energy level is split into only three separate levels.

The Zeeman effect is the splitting of an energy level in the presence of a magnetic field. In the reference frame of the electron, the nucleus orbits the electron. The “internal” Zeeman effect, as seen in sodium, is caused by the magnetic field produced by the “orbiting” nucleus.

Configuration (a) is not allowed. Only six electrons are allowed in the 2p state. Configurations (b) and (c) are allowed for atoms in an excited state.

The complete electron configuration for a uranium atom is as follows. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 4f 14 5s 2 5p 6 5d 10 5f 3 6s 2 6p 6 6d 1 7s 2

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10. For noble gases, the electrons fill all available states in the outermost shell. Helium has two electrons, which fill the 1s shell. For the other noble gases, there are six electrons in the p subshells, filling each of those shells. For each of these gases, the next electron would have to be added to an s subshell at much higher energy. Since the noble gases have a filled shell, they do not react chemically with other elements. Lithium, which has three electrons and is the first alkali metal, has its third electron in the 2s shell. This shell has significantly higher energy than the 1s shell. All of the alkali metals have one electron in their outer s subshell and therefore react well with halogens. Halogens all have five electrons in their outer p subshell. They react well with atoms that have a donor electron in their outer shell, like the alkali metals. 11. (a) (b) (c) (d)

Group II. Outer electrons are 3s2: magnesium. Group VII. Outer electrons are 2s2 2p5: fluorine. Group VIII. Outer electrons are 3s2 3p6: argon. Group I. Outer electron is 4s1: potassium.

12. The periodicity of the periodic table depends on the number and arrangement of the electrons in the atom. The Pauli exclusion principle requires that each electron have distinct values for the principal quantum number, angular orbital quantum number, magnetic quantum number, and spin quantum number. As electrons are added to the atom, the electrons fill the lowest available energy states. Atoms with the same number of electrons in the outer shells have similar properties, which accounts for the periodicity of the periodic table. The quantization of the angular momentum determines the number of angular momentum states allowed in each shell. These are represented by the letters s, p, d, f, etc. The magnitude of the angular momentum also limits the number of magnetic quantum states available for each orbital angular state. Finally, the electron spin allows for two electrons (one spin up and one spin down) to exist in each magnetic quantum state. Thus, the Pauli exclusion principle, quantization of angular momentum, direction of angular momentum, and spin all play a role in determining the periodicity of the periodic table. 13. If there were no electron spin, then according to the Pauli exclusion principle, s-subshells would be filled with one electron, p-subshells with three electrons, and d-subshells with five electrons. The first 20 elements of the periodic table would look like the following: H 1 1s1 He 2 2s1 C 6 3s1 Ne 10 4s1 K 19 5s1

Na 11 3d1 Ca 20 4d1

Mg 12 3d2

Al 13 3d3

Si 14 3d4

P 15 3d5

Li 3 2p1 N 7 3p1 S 16 4p1

Be 4 2p2 O 8 3p2 Cl 17 4p2

B 5 2p3 F 9 3p3 Ar 18 4p3

14. Neon is a “closed shell” atom—a noble gas. All of its shells are completely full and spherically symmetric, so its electrons are not readily ionized. Sodium is a Group I atom (an alkali metal) that has one lone electron in the outermost shell, and this electron spends a considerable amount of time far away and shielded from the nucleus. The result is that sodium can be ionized much easier than neon, as demonstrated by their two ionization energies. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 39

Quantum Mechanics of Atoms

15. Both chlorine and iodine are in the same column of the periodic table (group VII). Thus, both of their outer electron configurations are p5 (5 electrons in their outer p shell), which makes them both halogens. Halogens are one electron short of a filled shell, so both of their outer electron orbit shapes are similar and can readily accept an additional electron from another atom. This makes their chemical reaction properties extremely similar. 16. Both sodium and potassium are in the same column of the periodic table (group I). Thus, both of their outer electron configurations are s1 (1 electron in their outer s shell), which makes them both alkali metals. Alkali metals have one lone electron in their outermost shell, so both of their outer electron orbit shapes are similar, and they can readily share this electron with another atom (especially since this outermost electron spends a considerable amount of time very far away from the nucleus). This makes their chemical reaction properties extremely similar. 17. Noble gases are nonreactive because they have completely filled shells or subshells. In that configuration, other electrons are not attracted nor are electrons readily lost. Thus, they do not react (share or exchange electrons) with other elements. Alkali metals all have a single outer s electron. This outer electron is easily removed and can spend much of its time around another atom, forming a bond. Thus, alkali metals are highly reactive. 18. Rare earth elements have similar chemical properties because the electrons in the filled 6s or 7ssubshells serve as the valence electrons for all these elements. They all have partially filled inner fsubshells, which are very close together in energy. The different numbers of electrons in the fsubshells have little effect on the chemical properties of these elements. 19. When we use the Bohr theory to calculate the X-ray line wavelengths, we estimate the nuclear charge seen by the transitioning electron as Z – 1, assuming that the second electron in the ground state is partially shielding the nuclear charge. This is only an estimate, so we do not expect the calculated wavelengths to agree exactly with the measured values. 20. In helium and other complex atoms, electrons interact with other electrons in addition to their interactions with the nucleus. The Bohr theory only works well for atoms that have a single outer electron in an s state. X-ray emissions generally involve transitions to the 1s or 2s states. In these cases, the Bohr theory can be modified to correct for screening from a second electron by using the factor Z – 1 for the nuclear charge and can yield good estimates of the transition energies. Transitions involving outer electrons in more complex atoms will be affected by additional complex screening effects and cannot be adequately described by the Bohr theory. 21. The continuous portion of the X-ray spectrum is due to the “bremsstrahlung” radiation. An incoming electron gives up energy in the collision and emits light. Electrons can give up all or part of their kinetic energy. The maximum amount of energy an electron can give up is its total amount of kinetic energy. In the photon description of light, the maximum electron kinetic energy will correspond to the energy of the shortest wavelength (highest energy) photons that can be produced in the collisions. The result is the existence of a “cut-off” wavelength in the X-ray spectrum. An increase in the number of electrons will not change the cut-off wavelength. According to wave theory, an increase in the number of electrons could result in the production of shorter wavelength photons, which is not observed experimentally. 22. To figure out which lines in an X-ray spectrum correspond to which transitions, you would use the Bohr model to estimate the differences in the sizes of the energy level jumps that the falling electrons will make. Then, you just need to match the n = 2 → n = 1 energy to the K line, the n = 3 → n = 1 energy to the K line, the n = 3 → n = 2 jump energy to the L line, and so on. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

23. The characteristic X-ray spectra occur when inner electrons are knocked out of their shells. X-rays are high-energy photons emitted when other electrons fall to replace the knocked-out electrons. Because the shells involved are close to the nucleus, Z will have a direct influence on the energies. The visible spectral lines due to transitions between upper levels have energies less influenced by Z because the inner electrons shield the outer electrons from the nuclear charge. 24. The difference in energy between n = 1 and n = 2 levels is much bigger than that between any other combination of energy levels. Thus, electron transitions between the n = 1 and n = 2 levels produce photons of higher energy and frequency, which means these photons have shorter wavelengths (since frequency and wavelength are inversely proportional to each other). 25. The electron has a negative charge. 26. Consider a silver atom in its ground state for which the entire magnetic moment is due to the spin of only one of its electrons. In a uniform magnetic field, the dipole will experience a torque that would tend to align it with the field. In a non-uniform field, each pole of the dipole will experience a force of different magnitude. Consequently, the dipole will experience a net force that varies with the spatial orientation of the dipole. The Stern-Gerlach experiment provided the first evidence of space quantization since it clearly indicated that there are two opposite spin orientations for the outermost electron in the silver atom. 27. Spontaneous emission occurs when an electron is in an excited state of an atom and it spontaneously (with no external stimulus) drops back down to a lower energy level. To do this, it emits a photon to carry away the excess energy. Stimulated emission occurs when an electron is in an excited state of an atom, but a photon strikes the atom and causes or stimulates the electron to make its transition to a lower energy level sooner than it would have done so spontaneously. The stimulating photon has to have the same energy as the difference in energy levels of the transition. The result is two photons of the same frequency—the original one and the one due to the emission. 28. No, the intensity of light from a laser beam does not drop off as the inverse square of the distance. Laser light is much closer to being a plane wave rather than a spherically symmetric wave, and so its intensity is nearly constant along the entire beam. It does spread slightly over long distances, so it is not a perfect plane wave. 29. Different: Laser light is coherent (all of the photons are in the same phase), and ordinary light is not coherent (the photons have random phases relative to each other). Laser light is nearly a perfect plane wave, while ordinary light is spherically symmetric (which means that the intensity of laser light remains nearly constant as it moves away from the source, while the intensity of ordinary light drops off as 1/r2). Laser light is always monochromatic, while ordinary light can be monochromatic, but it usually is not. Similar: Both travel at c, and both display the wave–particle duality of photons. Both are created when electrons fall to lower energy levels and emit photons. Both are electromagnetic waves. 30. Since laser light is a plane wave, its intensity remains approximately constant with distance. The light produced by a street lamp spreads out with an intensity that decreases as 1/r2. Thus, at a sufficient distance, the laser light will be more intense than the light from a street lamp.

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Responses to MisConceptual Questions 1.

(b) A common error is to add up the coefficients (principal quantum numbers), which would give an answer of 15. However, the exponents represent the number of electrons in each subshell. In this case, there are 19 electrons.

(b) The orbital quantum numbers are represented by the subshell letters: s = 0 and p =1. Since only s and p subshells are present, only the orbital quantum numbers 0 and 1 are in the electron configuration.

(c) A common misconception is that all of the electrons have their lowest quantum numbers when in the ground state. However, due to the Pauli exclusion principle, no two electrons can have the same quantum numbers. The ground state of an atom is when each electron occupies the lowest energy state available to it without violating the exclusion principle.

(e) The Pauli exclusion principle applies to all electrons within the same atom. No two electrons in an atom can have the same set of four quantum numbers. Electrons from different atoms can have the same set of quantum numbers (for example, two separate atoms in the same container, or two atoms from different parts of the periodic table).

(a) In an atom, multiple electrons will be in the same shell and can therefore have the same value of n. Electrons in different shells can have the same value of 𝑚ℓ . Electrons in different atoms can also have the same set of four quantum numbers. The Pauli exclusion principle does not allow electrons within the same atom to have the same identical set of four quantum numbers.

(e) The Heisenberg uncertainty principle prohibits the exact measurement of position and velocity at the same time; the product of their uncertainties will always be greater than a constant value.

(d) Laser light is monochromatic (all photons have nearly identical frequencies), is coherent (all photons have the same phase), moves as a beam, and is created by a population inversion. A laser beam usually is intense, but it does not have to be brighter than other sources of light.

(d) As stated in Table 39–3, the letters s, p, d, f, g, … designate the orbital quantum number, ℓ. They do not summarize all of the quantum numbers for electrons, and they do not designate the principal quantum number. Also, they apply to electrons in any element, not just hydrogen.

(c) The “forbidden transitions” are discussed on page 1207. On that page, it is mentioned that such transitions can occur but with low probability as compared to the “allowed transitions”.

10. (c) From Eq. 39–2, we see that En =

E1 . Consider the energy difference between two adjacent n2

states as a function of n.

 n2 − ( n + 1)2   2n + 1  E1  = − E1  2  E = En +1 = En = − 2 = E1  2 2 2  n ( n + 1)   n ( n + 1)2  ( n + 1) n     E1

 2n + 1   decreases since the denominator grows faster than the As n increases, the value of  2  n ( n + 1)2    numerator, and so E decreases.

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Instructor Solutions Manual

Solutions to Problems 1.

The value of ℓ ranges from 0 to n − 1. Thus, for n = 6, ℓ = 0,1,2,3,4,5 .

The value of ml can range from − to + . Thus, for ℓ = 3, 𝑚ℓ = −3, −2, −1,0,1,2,3 . The possible values of ms are − 12 , + 12 .

The value of 𝑚ℓ can range from −ℓ to +ℓ, so we have 𝑚ℓ = −4, −3, −2, −1,0,1,2,3,4 . The value of ℓ can range from 0 to n − 1. Thus, we have n  5 . There are two values of ms : ms = − 12 , + 12 .

The value of 𝑚ℓ can range from −ℓ to +ℓ, so ℓ ≥ 5 . The value of ℓ can range from 0 to n − 1. Thus, 𝑛 ≥ ℓ + 1(min𝑖𝑚𝑢𝑚6) . There are two values of ms : ms = − 12 , + 12 .

The value of ℓ ranges from 0 to n − 1. Thus for n = 4, ℓ = 0,1,2,3. For each ℓ, the value of 𝑚ℓ can range from −ℓ to +ℓ, or 2ℓ + 1 values. For each 𝑚ℓ , there are 2 values of ms . Thus, the number of states for each ℓ is 2(2ℓ + 1). The number of states is therefore given by the following. N = 2 ( 0 + 1) + 2 ( 2 + 1) + 2 ( 4 + 1) + 2 ( 6 + 1) = 32states . We start with ℓ = 0, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ , 𝑚𝑠 ). (4, 0, 0, − 12 ), (4, 0, 0, + 12 ), (4, 1, –1, − 12 ), (4, 1, –1, + 12 ), (4, 1, 0, − 12 ), (4, 1, 0, + 12 ), (4, 1, 1, − 12 ), (4, 1, 1, + 12 ), (4, 2, – 2, − 12 ), (4, 2, – 2, + 12 ), (4, 2, –1, − 12 ), (4, 2, –1, + 12 ), (4, 2, 0, − 12 ), (4, 2, 0, + 12 ), (4, 2, 1, − 12 ), (4, 2, 1, + 12 ), (4, 2, 2, − 12 ), (4, 2, 2, + 12 ), (4, 3, – 3, − 12 ), (4, 3, – 3, + 12 ), (4, 3, – 2, − 12 ), (4, 3, – 2, + 12 ), (4, 3, −1, − 12 ), (4, 3, – 1, + 12 ), (4, 3, 0, − 12 ), (4, 3, 0, + 12 ), (4, 3, 1, − 12 ), (4, 3, 1, + 12 ), (4, 3, 2, − 12 ), (4, 3, 2, + 12 ), (4, 3, 3, − 12 ), (4, 3, 3, + 12 ).

The magnitude of the angular momentum depends only on l , which is 4. 𝐿 = √ℓ(ℓ + 1)ℏ = √20ℏ = √20(1.055 × 10−34 𝐽•𝑠) = 4.718 × 10−34 𝐽•𝑠

(a) The principal quantum number is n = 8 . (b) The energy of the state is found from Eq. 39–2. (13.6eV ) = − (13.6eV ) = − 0.213eV E7 = − n2 82 (c) The “g” subshell has ℓ = 4 . The magnitude of the angular momentum depends on l only: 𝐿 = ℏ√ℓ(ℓ + 1) = √20ℏ = √20(1.055 × 10−34 𝐽•𝑠) = 4.718 × 10−34 𝐽•𝑠

(d) For each ℓ, the value of 𝑚ℓ can range from −ℓ to +ℓ: 𝑚ℓ = −4, −3, −2, −1,0,1,2,3,4 . Photon emission means a jump to a lower state, which has a lower principal quantum number, so for the final state, n = 1, 2, 3, or 4. For a d subshell, ℓ = 2, and because 𝛥ℓ = ±1, the new value of ℓ must be 1 or 3. (a) ℓ = 1 corresponds to a p subshell, and ℓ = 3 corresponds to an f subshell. Keeping in mind that

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0 ≤ ℓ ≤ 𝑛 − 1, we find the following possible destination states: 2 p,3 p,4 p,4 f . Using the notation requested in the problem, the destination states are ( 2,1) , ( 3,1) , ( 4,1) , ( 4,3 ) . (b) In a hydrogen atom, ℓ has no appreciable effect on energy (ignoring fine structure), and so for energy purposes, there are four possible destination states, corresponding to n = 2, 3, and 4. Thus, there are three different photon wavelengths corresponding to three possible changes in energy. 9.

(a) For each ℓ, the value of 𝑚ℓ can range from −ℓ to +ℓ, or 2ℓ + 1 values. For each of these, there are two values of ms. Thus, the total number of states in a subshell is 𝑁 = 2(2ℓ + 1) .

= 0, 1, 2, 3, 4, 5, and 6, N = 2, 6, 10, 14, 18, 22, and 26 , respectively.

(b) For

10. For a given n, 0 ≤ ℓ ≤ 𝑛 − 1. Since for each ℓ the number of possible states is 2(2ℓ + 1), as shown in Problem 9, the number of possible states for a given n is as follows. 𝑛−1 𝑛−1 ∑𝑛−1 ℓ=0 2(2ℓ + 1) = 4 ∑ℓ=0 ℓ + ∑ℓ=0 2 = 4 (

𝑛(𝑛−1) 2

) + 2𝑛 = 2𝑛2

The summation formulas can be found (and proven) in a pre-calculus mathematics textbook. 11. We use Eq. 39–3 to find l and Eq. 39–4 to find ml . 𝐿2

(6.84×10−34 𝐽•𝑠)

𝐿 = √ℓ(ℓ + 1)ℏ → ℓ(ℓ + 1) = 2 = (1.055×10−34 ℏ

ℓ2 + ℓ − 42 = 0 → ℓ =

𝐽•𝑠)2

= 42

−1 ± √1 − 4(1)(−42) −1 ± 13 = = 6, −7 → ℓ = 6 2 2 𝐿𝑧 2.11 × 10−34 𝐽•𝑠 = = 2 ℏ 1.055 × 10−34 𝐽•𝑠

𝐿𝑧 = 𝑚ℓ ℏ → 𝑚ℓ = Since ℓ = 6, we must have n  7 .

12. The state 𝑛 = 2, ℓ = 0 must have 𝑚ℓ = 0, and so the wave function is  200 =

 r − 2 −  e 2 r0 . 3  r0  32 r0  1

To obtain numerical values, we use r0 = 0.0529 nm = 5.29  10−11 m . 4r

 4r  − 0 1 2 − 0  e 2 r0 = − e−2 = −7.02  1013 m −1.5 (a) ( 200 )r = 4 r =  0 3 3 r 32 r0  8 r0 0  1

(

(c)

Pr = 4 r 2  200

 200

(

)

r = 4 r0

4r0  −2 2 r00 1  1 −4 e = 4.92  1027 m −3 2 − =  e 3  3 r0  32 r0  8 r0

(b)

)

r = 4 r0

1 −4  8 −4 2 = 4 ( 4r0 )  e  = e = 2.77  109 m −1 3  8 r0  r0

13. The ground state wave function is  100 =

r

3 0

e − r / r0 . To obtain numerical values, we use

−7

r0 = 0.0529 nm = 5.29  10 m . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(a)

( 100 )r =1.5r =

(b)

( )

(c)

Pr = 4 r 2  100

r

3 0

100

r =1.5 r0

(

Instructor Solutions Manual

e−1.5 = 3.27  1014 m −1.5

1 −3 e = 1.07  1029 m −3 3  r0 2

)

 9 2 1 = 4 (1.5r0 )  3 e−3  = e−3 = 8.47  109 m −1 r =1.5 r0   r0  r0 2

14. To show that the ground-state wave function is normalized, we integrate  100 over all space. Use substitution of variables and an integral from Appendix B–5. 2r



1 − r0 e 4 r 2 dr ; 3  r0 0

  100 dV =  2

all space

let x =

2r → r = 12 r0 x , dr = 12 r0 dx r0

Note that if r = 0, x = 0 and if r = , x = .

  100

all space



1 − 4 dV =  3 e r0 4 r 2 dr = 3  e − x ( 14 r02 x 2 ) ( 12 r0dx ) = 12  e − x x 2dx = 12 ( 2!) = 1  r0  r0 0 0 0

And so, we see that the ground-state wave function is normalized. 15. The factor is found from the ratio of the radial probability densities for  100 . Use Eq. 39–7.

 r02 − 2rr0   r 2 − 2rr  4 e 0  4 2 e 0   r0   r02  Pr ( r = r0 )   r = r0   r =r0 e −2 e2 = = = =  1.85 4e −4 4 Pr ( r = 2r0 )  r 2 − 2rr   ( 2r )2 − 2( 2r r0 )  4 2 e 0   4 02 e 0   r0   r0   r =2 r0   r =2 r0 16. (a) To find the probability, integrate the radial probability distribution for the ground state. We follow Example 39–4 and use the last integral in Appendix B–4. 2r r0 r 2 − r0 r → r = 12 r0 x , dr = 12 r0 dx P =  4 3 e dr ; let x = 2 r r 0 0 0

( 1 r x) − r2 − P =  4 3 e r0 dr =  4 2 03 e r0 r0 0 0 2r

2( 12 r0 x ) r0

r dx = 12  x 2 e − x dx = 12  −e − x ( x 2 + 2 x + 2 ) 

1 2 0

2 0

= 1 − 5e−2  0.32 = 32% (b) We follow the same process here. 2 r0

r2 − r P =  4 3 e r0 dr ; let x = 2 r0 r0 r0 4

→

P = 12  x 2 e− x dx = 12  −e − x ( x 2 + 2 x + 2 )  = 5e−2 − 13e −4  0.44 = 44% 2

17. (a) To find the probability for the electron to be within a sphere of radius r, we must integrate the radial probability density for the ground state from 0 to r. The density is given in Eq. 39–7. Since r r0 , we approximate e−2 r r0  1. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Quantum Mechanics of Atoms −15 2r 1.1fm 4 r 3 4 (1.1  10 m ) r 2 − r0 r2 P =  4 3 e dr   4 3 dr = = = 1.2  10−14 3 3 10 − 3 r0 3 ( 0.529  10 m ) r0 r0 0 0 3

1.1fm

(b) The Bohr radius, r0 , is inversely proportional to the mass of the particle. So now, the Bohr radius is smaller by a factor of 207. 2r 1.1fm (1.1  10−15 m ) = 1.1  10−7 4 r3 4 r2 − r2 P =  4 3 e r0 dr   4 3 dr = = 3 r03 3 ( 0.529  10−10 m 207 )3 r0 r0 0 0 3

1.1fm

18. To show that  200 is normalized, we integrate  200 over all space. Use substitution of variables and an integral from Appendix B–5. r  1  r  − r0 r 2 → r = r0 x , dr = r0dx  200 dV =  2 −  e 4 r 2 dr ; let x = 3   32 r0  r0  r0 all 0 space

Note that if r = 0, x = 0 and if r = , x = . 2



 1  4 r  − r0 2 2  2 4 2 − x ) e − x r02 x 2 r0dx e r dr − =  3  3 (   32 32 r r r 0  0  0 0 0

  200 dV =  2

all space



1 1 1 1 1 2 ( 2 − x ) e − x x 2dx =  ( 4 x 2 − 4 x 3 + x 4 ) e − x dx =  x 2e − x dx −  x 3e − x dx +  x 4e − x dx  80 80 20 20 80

1 1 1 ( 2!) − ( 3!) + ( 4!) = 1 − 3 + 3 = 1 2 2 8 And so, we see that  200 is normalized. =

19. We follow the directions as given in the problem. We use the first integral listed in Appendix B–5. 



r =  r  100 4 r 2 dr =  r 2



r3 r = 4 3 e r 0 0

r −2 r0



dr = 4 0



1 −2 r0 r 3 −2 r0 r 2 4 4 e r dr e dr ; let x = 2  = → r = 12 r0 x , dr = 12 r0 dx 3 3  r r0  r0 0 0

( 12 r0 x )

r03



e − x 12 r0 dx = 14 r0  x 3e − x dx = 14 r0 ( 3!) = 23 r0 0

20. The probability is found by integrating the radial probability density over the range of radii given. We use integrals from Appendix B–4. 2r 1.01 r0 ( 0.99r0 ) = 1.98 r2 − r → r = 12 r0 x , dr = 12 r0 dx ; r = 0.99r0 → x = 2 P =  4 3 e r0 dr ; let x = 2 r0 r0 r0 0.99 r0

( 1 r x) r2 − P =  4 3 e r0 dr =  4 2 03 e − x 12 r0 dx = 12  x 2 e − x dx r0 r0 x =1.98 x =1.98 0.99 r0 1.01 r0

x = 2.02

= 12  −e − x ( x 2 + 2 x + 2 ) 

2.02 1.98

x = 2.02

= e −1.98 ( 4.9402 ) − e −2.02 ( 5.0602 ) = 0.0108  1.1%

Because the range of radii is small and the radial probability density is relatively constant over that range (see Fig. 39–7), we could have approximated the probability as follows.

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1.01r0

Instructor Solutions Manual

r2 − r2 − 0 P =  4 3 e r0 dr  P ( r = r0 ) r = 4 03 e r0 ( 0.02r0 ) = 0.08e −2 = 0.0108  1.1% r0 r0 0.99 r0 21. We follow the directions as given in the problem. The three wave functions are given in Eq. 39–9. We explicitly show the expressions involving the complex conjugate. 2 2 2 Pr = 4 r 2  13  210 + 13  211 + 13  21−1    r   z2 −   ( x + iy )( x − iy ) − rr   ( x + iy )( x − iy ) − rr   r0 1 = 4 r 2  13  + e e 0  + 13  e 0   3 5 5 5       32 64 64 r r r      0 0 0     

r  z2 − ( x 2 + y 2 ) e− rr0 + 1 ( x 2 + y 2 ) e− rr0  = r 2 z 2 + x 2 + y 2 e− rr0 r0 1 = 4 r 2  13 + e ) 3 3 5 ( 5 64 r05 64 r05  32 r0  24 r0 r

r 4 − r0 = e 24 r05 r

22. From Problem 21, we know that Pr = by setting

r 4 − r0 e for the 2p state. We find the most probable distance 24r05

dPr = 0 and solving for r. This is very similar to Example 39–3. dr r

−

r 4 − r0 dPr r 3e r0 4r 3 − r0 1 r 4 − r0 = − = − =0 → Pr = e e e r r ; 4 ( ) 0 dr 24r05 r0 24r05 r06 24 24r05

r = 4r0

23. To find the probability for the electron to be within a sphere of radius r, we must integrate the radial probability density for the ground state from 0 to r. The density is given in Eq. 39–7. rsphere 2r r r 2 − r0 r P =  4 3 e dr ; let x  = 2 ; let 2 sphere = x → r0 r0 r0 0 x





P = 12  x 2 e − xdx  = 12  −e − x ( x 2 + 2 x  + 2 ) = 12  −e − x ( x 2 + 2 x + 2 ) + 2 = 1 − e − x ( 12 x 2 + x + 1) x

We solve this equation numerically for values of x that give P = 0.50, 0.90, and 0.99. (a) The equation for P = 0.50 is solved by x = 2.674, and so rsphere = 12 ( 2.674 ) r0  1.3 r0 . (b) The equation for P = 0.90 is solved by x = 5.322, and so rsphere = 12 ( 5.322 ) r0  2.7 r0 . (c) The equation for P = 0.99 is solved by x = 8.406, and so rsphere = 12 (8.406 ) r0  4.2 r0 . 24. The probability is found by integrating the radial probability density over the range of radii given. The radial probability density is given after Eq. 39–8. 2

5.00 r0

r2  r  − r0 r − P=  2   e dr ; let x = 3 r0  r0 8r0  4.00 r0 5.00

1 8

→

5.00

 x ( 2 − x ) e dx =  ( x − 4 x + 4 x ) e dx 2

4.00

−x

1 8

−x

4.00

There are some difficult integrals to evaluate. We use integration by parts. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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 x e dx : u = x ; d v = e ; du = 2 xdx; v = −e  x e dx = − x e +  2 xe dx = − x e + 2 ( −e ) ( x + 1) = −e ( x + 2 x + 2 )  x e dx : u = x ; d v = e ; du = 3x dx; v = −e  x e dx = − x e + 3 x e dx = −e ( x + 3x + 6 x + 6 )  x e dx : u = x ; d v = e ; du = 4 x dx; v = −e  x e dx = − x e + 4 x e dx = −e ( x + 4 x + 12 x + 24 x + 24 ) 2 −x

−x

2 −x

3 −x

2 −x

4 −x

3 −x

2 −x

4 −x

−x

2 −x

−x

3 −x

−x

4 −x

−x

3 −x

−x

We substitute the integrals above into the expression for the probability. We are not showing the algebra. P = − 18 e − x ( x 4 + 4 x 2 + 8 x + 8)

5.00 4.00

= − 18 ( 773e −5 − 360e −4 ) = 0.173 = 17.3%

25. The wave function is given in Eq. 39–5a. Note that r = ( x 2 + y 2 + z 2 ) . We will need the derivative 1/ 2

relationship derived in the first line below. r

− − x 1 1 1 x r 1 2   r e r0 ; e r0 ;  100 = = 2 ( x + y2 + z2 ) 2x = = =− 3 3 r r0  r0 r x x r x  r0 r r r   1 1  − − 1 −  2   1 1  1 1 r0 x r0 r0 x r       e e e = − = − + − r   r0  r03 r  r  r0  r03 r  x x 2 x  r0  r03       r

−   1 1  −1 1 1  1  2 1 + 2  e r0 1 − x 2  = + 2   = − 1 − x  3 rr0  r0 rr0   rr0 r    rr0 r   

Similarly, we would have

1  1  r y  2 r z 2 1 ; = = − = ; and 1 − y  + 2   ; 2 rr0  y r y z r  rr0 r  

1  1   2 2 1 = − + 2   . Substitute into the time-independent Schrödinger equation. 1 − z  2 rr0  z  rr0 r   E = −

  2  2  2  1 e2   2 + 2 + 2 − 2m  x x x  4 0 r 2

2  1  1   1    1   1   1    e2 2 1 = − + 2   + 1 − y 2  + 2   + 1 − z 2  + 2   −    1 − x    − rr0     2m   rr0 r     rr0 r     rr0 r    4 0 r

 2 1  1   e2  2 2 2  1 =  − + + + − 3 x y z )  rr r 2   4 r   ( 0   2m rr0   0     2 1   2 1  1   e2  r2  e2  2 1 =  − + − = − −  3 r 2        2  rr0  4 0 r   2m rr0   rr0 r   4 0 r   2m rr0  2  2 e2  =  − −  2  mrr0 2mr0 4 0 r  Since the factor in square brackets must be a constant, the terms with the r dependence must cancel. 2 e2 4 0 2  0h 2 − = 0 → r0 = = mrr0 4 0 r me 2  me 2

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Note from Eq. 37–11 that this expression for r0 is the same as the Bohr radius. Since those two terms cancel, we are left with the following.

E = −

 → E=− 2

2mr0

=− 2

2mr0

  h2  2m  0 2    me 

= − 2

me 4 8h 2 02

26. (a) The probability is found by integrating the radial probability density over the range of radii given. The radial probability density is given after Eq. 39–8. 2

r2  r − r P =  3  2 −  e r0 dr ; let x = 8 r r r 0  0  0 0 1

→

P = 18  x 2 ( 2 − x ) e − x dx = 18  ( x 4 − 4 x 3 + 4 x 2 ) e − x dx 2

The following integrals are derived in the solution to Problem 24. 2 −x −x 2 3 −x −x 3 2  x e dx = −e ( x + 2 x + 2 ) ;  x e dx = −e ( x + 3x + 6 x + 6 )

 x e dx = −e ( x + 4 x + 12 x + 24 x + 24 ) 4 −x

−x

P = 18  ( x 4 − 4 x 3 + 4 x 2 ) e − x dx 0





= 18 − e − x ( x 4 + 4 x 3 + 12 x 2 + 24 x + 24 ) − 4 ( x 3 + 3x 2 + 6 x + 6 ) + 4 ( x 2 + 2 x + 2 ) 

1 0

= 18  e − x ( x 4 + 4 x 2 + 8 x + 8 )  = 18 (8 − 21e −1 ) = 0.0343  3.4% 0

r 4 − r0 (b) From Problem 21, we know that the radial probability density for this state is Pr = e . We 24r05 proceed as in part (a). r r0 r 4 − r0 r P= → e dr ; let x = 5 24r0 r0 0 1

P = 241  x 4 e − x dx = 241  −e − x ( x 4 + 4 x 3 + 12 x 2 + 24 x + 24 ) = 241 ( 24 − 65e −1 ) = 3.66  10−3 1

 0.37%

27. (a) The radial probability distribution is given by Eq. 39–6. Use the wave function given. 2

Pr = 4 r  300 2

1  2r 2r 2  − 3r0 4r 2  2r 2r 2  − 3r0 = 4 r − + = − + 1 1 e     e 27 r03  3r0 27r02  27r03  3r0 27r02  2

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Quantum Mechanics of Atoms

(b) See the graph. (c) The most probable distance is the radius for which the radial probability distribution has a global maximum. We dP find that location by setting r = 0 and dr solving for r. We see from the graph that the global maximum is approximately at r = 13r0 .

2.0

-1

Pr (nm )

1.5 1.0 0.5 0.0 0

r/r0

8r  2r 2 r 2  − 3r0 8r 2  2r 2 r 2  2 4 r  − 3r0 dPr 1 1− e + = − + + − + e 3  2  3  2  27 r0  3r0 27 r0  3r0 27 r02  dr 27r0  3r0 27r0  2

4r 2  2 r 2r 2   2  − 3r0 1 + − +   − e 27 r03  3r0 27 r02   3r0  2r

8r  2r 2r 2   5r 12r 2 2r 3  − 3r0 1 1 − + − + − =0   e 27 r03  3r0 27 r02   3r0 27 r02 81r03  The above system has 6 non-infinite solutions. One solution is r = 0, which leads to Pr = 0, which is not a maximum for the radial distribution. The second-order polynomial,  2r 2r 2  , is a factor of the radial probability distribution, and so its zeros also give + 1 − 2   3r0 27r0  =

locations where Pr = 0. Thus, the maxima must be found from the roots of the third-order 2 3 2 polynomial. A spreadsheet was used to find the roots of 1 − 53 x + 12 27 x − 81 x = 0. Those roots

are x = 0.74, 4.19, and 13.07. Therefore, the most probable distance is r = 13.1 r0 . 28. The number of electrons in the subshell is determined by the value of ℓ. For each ℓ, the value of 𝑚ℓ can range from −ℓ to +ℓ which is 2ℓ + 1 values. For each 𝑚ℓ value, there are two values of ms . Thus, the total number of states for a given ℓ is 𝑁 = 2(2ℓ + 1). 𝑁 = 2(2ℓ + 1) = 2[2(4) + 1] = 18 electrons 29. For carbon, Z = 6. We start with the n = 1 shell, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ , 𝑚𝑠 ).

(1,0,0, − 12 ) , (1,0,0, + 12 ) , ( 2,0,0, − 12 ) , ( 2,0,0, + 12 ) , (2,1, −1, − 12 ) , (2,1, −1, + 12 )

Note that without additional information, there are other possibilities for the last two electrons. The value of 𝑚ℓ for the last two electrons could be –1 (as shown), 0, or 1. For example, those last two electrons might be ( 2,1,0, − 12 ) and ( 2,1,1, + 12 ) . 30. (a) For oxygen, Z = 8. We start with the n = 1 shell, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ , 𝑚𝑠 ).

(1,0,0, − 12 ) , (1,0,0, + 12 ) , ( 2,0,0, − 12 ) , ( 2,0,0, + 12 ) , ( 2,1, −1, − 12 ) , ( 2,1, −1, + 12 ) , ( 2,1,0, − 12 ) , ( 2,1,0, + 12 ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Note that without additional information, there are two other possibilities that could substitute for any of the last four electrons. The third quantum number could be changed to 1. (b) For aluminum, Z = 13. We start with the n = 1 shell, and list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ , 𝑚𝑠 ).

(1,0,0, − 12 ) , (1,0,0, + 12 ) , ( 2,0,0, − 12 ) , ( 2,0,0, + 12 ) , ( 2,1, −1, − 12 ) , ( 2,1, −1, + 12 ) , ( 2,1,0, − 12 ) , ( 2,1,0, + 12 ) , ( 2,1,1, − 12 ) , ( 2,1,1, + 12 ) , ( 3,0,0, − 12 ) , ( 3,0,0, + 12 ) , ( 3,1, −1, − 12 ) Note that without additional information, there are other possibilities for the last electron. The third quantum number could be –2, –1, 0, 1, or 2, and the fourth quantum number could be either +1/2 or –1/2. 31. Since the electron is in its lowest energy state, we must have the lowest possible value of n. Since 𝑚ℓ = 3, the smallest possible value of ℓ is ℓ = 3 , and the smallest possible value of n is n = 4 . 32. The third electron in lithium is in the 2s subshell, which is outside the more tightly bound filled 1s shell. This makes it appear as if there is a “nucleus” with a net charge of +1e. Use the energy of the hydrogen atom, Eq. 39–2. (13.6eV ) = − (13.6eV ) = − 3.4eV E2 = − n2 22 We predict the binding energy to be 3.4 eV. Our assumption of complete shielding of the nucleus by the 2s electrons is evidently not correct. The partial shielding means the net charge of the “nucleus” is higher than +1e, and so it holds the outer electron more tightly, requiring more energy to remove it. 33. The configurations can be written with the subshells in order by energy (4s before 3d, for example), or with the subshells in principal number order (3d before 4s, for example, as in Table 39–4). Both are shown here. (a) Cobalt has Z = 27. 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 7 4s 2 or 1s 2 2s 2 2 p 6 3s 2 3 p 6 4s 2 3d 7

(b) Gold has Z = 79 . 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 4 f 14 5s 2 5 p 6 5d 10 6s1 or 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 6 5s 2 4d 10 5 p 6 4 f 14 6 s1 5d 10

1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4 p 6 4d 10 4 f 145s 2 5 p 6 5d 10 6s 2 6 p 6 5 f 3 6d 1 7 s 2 or 1s 2 2s 2 2 p 6 3s 2 3 p 6 4s 2 3d 10 4 p 6 5s 2 4d 10 5 p 6 4 f 146s 2 5d 10 6 p 6 7 s 2 5 f 36d 1 34. Limiting the number of electron shells to six would mean that the periodic table stops with radon (Rn) since the next element, francium (Fr), begins filling the seventh shell. Including all elements up through radon means 86 elements. 35. We use Eq. 37–13, which says that the radius of a Bohr orbit is inversely proportional to the atomic number. We also use Eq. 37–14b, which says that the energy of a Bohr orbit is proportional to the square of the atomic number. The energy to remove the electron is the opposite of the total energy. A simple approximation is used in which the electron is attracted by the full effect of all 92 protons in the nucleus. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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1 n2 0.529  10−10 m ) = ( 0.529  10−10 m ) = 5.75  10−13 m ( 92 Z 2 922 Z En = (13.6eV ) 2 = (13.6eV ) 2 = 1.15  105 eV = 115keV 1 n

rn =

36. The energy levels of the infinite square well are given in Eq. 38–13. Each energy level can have a maximum of two electrons since the only quantum numbers are n and ms. Thus, the lowest energy level will have two electrons in the n = 1 state, two electrons in the n = 2 state, and 1 electron in the n = 3 state. 𝐸 = 2𝐸1 + 2𝐸2 + 𝐸3 = [2(1)2 + 2(22 ) + 1(3)2 ]

ℎ2 2

8𝑚ℓ

= 19

ℎ2 8𝑚ℓ2

37. In a filled subshell, there are an even number of electrons. All of the possible quantum number combinations for electrons in that subshell represent an electron that is present. Thus for every m value, both values of ms are filled, representing a spin “up” state and a spin “down” state. The total angular momentum of that pair is zero, and since all of the electrons are paired, the total angular momentum is zero. 38. The shortest wavelength X-ray has the most energy, which is the maximum kinetic energy of the electron in the tube: ( 6.63 10−34 J s )(3.00 108 m/s ) = 5.405 104 eV = 54keV hc E= =  (1.60  10−19 J/eV )( 0.023 10−9 m ) Thus, the operating voltage of the tube is 54 kV . 39. The maximum energy is emitted when the shortest wavelength is emitted. That appears to be about 0.025 nm. The minimum energy is emitted when the longest wavelength is emitted. That appears to be about 0.2 nm. Use Eq. 37–3. −34 8 (1eV ) hc ( 6.63  10 J s )( 3.00  10 m s ) Emax = hf max = = = 49725eV −9 min ( 0.025  10 m ) (1.60 10−19 J )

 5.0  104 eV Emin = hf min =

max

( 6.63 10 J s )( 3.00 10 m s ) (1eV ) = 6215eV  6  10 eV = (1.60 10 J ) ( 0.2  10 m ) −34

−9

−19

40. With the shielding provided by the remaining n = 1 electron, we use the energies of the hydrogen atom with Z replaced by Z − 1. The energy of the photon is found, and then the wavelength. 2  1   1  hf = E = − (13.6eV )( 26 − 1)  2  −  2   = 6.40  103 eV.  2   1   −34 6.63  10 J s )( 3.00  108 m/s ) ( hc = = 1.94  10−10 m = 0.194 nm = E (1.60  10−19 J/eV )( 6.40  103 eV )

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

41. The energy of the photon with the shortest wavelength must equal the maximum kinetic energy of an electron. We assume V is in volts. hc E = hf 0 = = eV →

0

hc ( 6.63  10 = eV

0 =

−34

J s )( 3.00  108 m s )(109 nm m )

(1.60  10 C ) (V V ) −19

1243nm 1240 nm  V V

42. We follow the procedure of Example 39–6, of using the Bohr formula, Eq. 37–15, with Z replaced by Z – 1. 1  e4m  1  1 2 1 2 1 =  2 3  ( Z − 1)  2 − 2  = (1.097  107 m −1 ) ( 27 − 1)  2 − 2  = 5.562  109 m −1 →   8 0 h c   n n  1 2 

=

1 = 1.798  10−10 m −1 9 5.562  10 m

43. The wavelength of the K line is calculated for molybdenum in Example 39–6. We use that same procedure. Note that the wavelength is inversely proportional to ( Z − 1) . 2

 ( Z Fe − 1) unknown Fe 194 pm  = → Z unknown = ( Z Fe − 1) + 1 = ( 26 − 1)  + 1 = 24 2 229 pm  Fe unknown ( Z unknown − 1)  The unknown material has Z = 24, and so it is chromium. 2

44. We assume “shielding” is provided by the 1s electron that is already at that level. Thus, the effective charge “seen” by the transitioning electron is 42 – 1 = 41. We use Eqs. 37–9 and 37–14b. 1  2 1 hf = E = (13.6eV )( Z − 1)  2 − 2   n  n

=

hc = E

( 6.63  10 J s )( 3.00  10 m/s ) 1 1 (13.6eV ) ( 41 )  −  (1.60  10 J eV ) 1 3  −34

(13.6eV )( Z − 1)  2

1 1  − 2 2  n n 

−19

= 6.12  10−11 m = 0.0612 nm We do not expect perfect agreement because there is some partial shielding provided by the n = 2 shell, which was ignored when we replaced Z by Z − 1. That would make the effective atomic number a little smaller, which would lead to a larger wavelength. The amount of shielding could be estimated by using the actual wavelength and solving for the effective atomic number. 45. Momentum and energy will be conserved in any inertial reference frame. Consider the frame of reference that is moving with the same velocity as the electron’s initial velocity. In that frame of reference, the initial momentum of the electron is 0, and its initial total energy is mc 2 . Let the emitted photon have frequency f, and let the direction of motion of that photon be considered the h hf positive direction. The momentum of the photon is then p = = , and so the momentum of the  c hf electron must be pe = − . The final energy of the photon is E = hf = p c, and the final energy of c

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the electron is, from Eq. 36–14, Eelectron = final

pe2 c 2 + m 2 c 4 . We write the conservation conditions, and

then solve for the frequency of the emitted photon. hf hf + pe → pe = − Momentum: 0 = c c Energy:

mc 2 = hf + pe2c 2 + m2 c 4 2

 hf  mc 2 − hf =   c 2 + m 2 c 2 = h 2 f 2 + m 2 c 4  c  m 2 c 4 − 2mc 2 hf + h 2 f 2 = h 2 f 2 + m 2 c 4 → 2mc 2 hf = 0 → f = 0 Since the photon must have f = 0, no photon can be emitted and still satisfy the conservation laws. Another way to consider this situation is that if an electron at rest emits a photon, the energy of the electron must decrease for energy to be conserved. But the energy of a stationary electron cannot decrease unless its mass were to change. Then it would no longer be an electron. Thus, we conclude that a third object (with mass) must be present in order for both energy and momentum to be conserved. 46. The Bohr magneton is given by Eq. 39–12. 1.602  10−19 C )(1.054  10−34 J s ) ( e = = 9.268  10−24 J T  9.27  10−24 J T B = −31 2m 2 ( 9.109  10 kg ) 47. We use Eq. 39–14 for the magnetic moment since the question concerns spin angular momentum. The energy difference is the difference in the potential energies of the two spin states. (9.27 10−24 J T ) 3.5T − 1 − 1 = 4.110−4 eV spin down U = (  z B )spin up = − g B Bms = − ( 2.0023) ( )( 2 2 ) (1.60 10−19 J eV ) 48. (a) The energy difference is the difference in the potential energies of the two spin states. Use Eq. 39–14 for the magnetic moment. 9.27  10−24 J T ) ( spin down U = (  z B )spin up = − g B Bms = − ( 2.0023) ( 2.0T ) ( − 12 − 12 ) (1.60 10−19 J eV )

= 2.320  10−4 eV  2.3  10 −4 eV (b) Calculate the wavelength associated with this energy change. c U = E = h →



=

( 6.63  10−34 J s )(3.00  108 m/s ) = 5.358  10−3 m  5.4 mm hc = U ( 2.320  10−4 eV )(1.60  10−19 J eV )

(c) The answer would be no different for hydrogen. The splitting for both atoms is due to an sstate electron: 1s for hydrogen and 5s for silver. See the discussion on page 1216 concerning the Stern-Gerlach experiment.

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Instructor Solutions Manual

49. (a) Refer to Fig. 39–14 and the equation following it. A constant magnetic field gradient will produce a constant force on the silver atoms. Atoms with the valence electron in one of the spin states will experience an upward force, and atoms with the valence electron in the opposite spin state will experience a downward force. That constant force will produce a constant acceleration, leading to the deflection from the original direction of the atoms as they leave the oven. We assume the initial direction of the atoms is the x direction, and the magnetic field gradient is in the z direction. If undeflected, the atoms would hit the screen at z = 0. dB dB 2 2 2 z z − g B ms ) z ( F  x  dz  x  = 1 dz  x  1 z = 12 at 2 = 12 =       2 2 mAg  v  mAg  v  mAg  v  One beam is deflected up, and the other down. Their separation is the difference in the two deflections due to the two spin states. dB 2  − g B ( − 12 − 12 )  z dz  x  z = zms =− 1 − zms = 1 = 12   2 2 mAg  v 

( 2.0023) ( 9.27  10−24 J T ) (1800T m )  0.040 m 2 −4 =   = 2.453  10 m  0.25mm −27 (107.87 u ) (1.66  10 kg u )  780 m s  1 2

(b) The separation is seen in the above equation to be proportional to the g-factor. So, to find the new deflection, divide the answer to part (a) by the original g-factor. 2.453  10−4 m z g =1 = = 1.225  10−4 m  0.12mm 2.0023 1

7 9

2 2

50. For the 5g state, ℓ = 4 and s = 12 . Thus, the possible values of j are 𝑗 = ℓ ± 𝑠 = 4 ± =

, .

Let j = . Then we have the following. 7 2

m j = − 72 , − 25 , − 23 , − 12 , 12 , 23 , 25 , 72

7 2

( 72 + 1) =

3 7 2

J z = m j = − 72 , − 25 , − 23 , − 12 , 12 , 23 , 25 , 72 Let j = 92 . Then we have the following.

m j = − 92 , − 72 , − 25 , − 23 , − 12 , 12 , 23 , 52 , 72 , 92

9 2

( 92 + 1) =

3 11 2

J z = m j = − 92 , − 72 , − 25 , − 23 , − 12 , 12 , 23 , 25 , 27 , 29 51. (a) For the 4p state, ℓ = 1. Since s = 12 , the possible values for j are 𝑗 = ℓ + 𝑠 = 𝑗 =ℓ−𝑠 =

and

7 2

and

5 . 2

1 . 2

(b) For the 4f state, ℓ = 3. Since s = 12 , the possible values for j are 𝑗 = ℓ + 𝑠 = 𝑗 =ℓ−𝑠 =

5 2

and

3 . 2

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Quantum Mechanics of Atoms

(d) The values of J are found from Eq. 39–15.

4p: J =

j ( j + 1) =

15 2

and

3 2

4f : J =

j ( j + 1) =

63 2

and

35 2

3d : J =

j ( j + 1) =

35 2

and

15 2

52. (a) Gallium has Z = 31. We list the quantum numbers in the order (𝑛, ℓ, 𝑚ℓ , 𝑚𝑠 ).

(1,0,0, − 12 ) , (1,0,0, + 12 ) , ( 2,0,0, − 12 ) , ( 2,0,0, + 12 ) , ( 2,1, −1, − 12 ) , ( 2,1, −1, + 12 ) , ( 2,1,0, − 12 ) , ( 2,1,0, + 12 ) , ( 2,1,1, − 12 ) , ( 2,1,1, + 12 ) , ( 3,0,0, − 12 ) , (3,0,0, + 12 ) , ( 3,1, −1, − 12 ) , ( 3,1, −1, + 12 ) , ( 3,1,0, − 12 ) , ( 3,1,0, + 12 ) , ( 3,1,1, − 12 ) , (3,1,1, + 12 ) , ( 3, 2, −2, − 12 ) , ( 3, 2, −2, + 12 ) , ( 3, 2, −1, − 12 ) , ( 3, 2, −1, + 12 ) , ( 3, 2,0, − 12 ) , ( 3, 2,0, + 12 ) , ( 3, 2,1, − 12 ) , ( 3, 2,1, + 12 ) , ( 3, 2, 2, − 12 ) , (3, 2, −2, + 12 ) , ( 4,0,0, − 12 ) , ( 4,0,0, + 12 ) , ( 4,1,0, + 12 )

The last electron listed could have other quantum numbers for 𝑚ℓ and ms . (b) The 1s, 2s, 2p, 3s, 3p, 3d, and 4s subshells are filled. (c) For a 4p state, ℓ = 1. Since s = 12 , the possible values for j are 𝑗 = ℓ + 𝑠 =

3 2

and

1 . 2 (d) The 4p electron is the only electron not in a filled subshell. The angular momentum of a filled subshell is zero, so the total angular momentum of the atom is the angular momentum of the 4p electron. (e) When the beam passes through the magnetic field gradient, the deflecting force will be proportional to mj . If j = 12 , the values of mj are  12 , and there will be two lines. If j = 23 , the values of mj are  12 ,  23 , and there will be four lines. The number of lines indicates the value of 𝑗 =ℓ−𝑠 =

j. 53. (a) The additional term for the spin–orbit interaction is given in the text as U spin = −μ Bn =  z Bn . orbit

The separation of the energy levels due to the two different electron spins is twice this. spin down U spin = (  z Bn )spin up = − g B Bn ms → orbit

U spin

( 5 10 eV )(1.60 10 J eV ) = 0.431T  0.4T B = = − g  m ( 2.0023) ( 9.27  10 J T ) ( − − ) −5

−19

orbit

−24

1 2

(b) If we consider the nucleus to be a loop of current with radius r, then the magnetic field due to

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

the nucleus at the center of the loop (the location of the electron) is given in Example 28–11 as I Bn = 0 . Model the current as the charge of the nucleus moving in a circle, with a period as 2r given by circular motion. e ( me r ) v q e e ev = = = = I= t T 2 r v 2 r 2 r ( me r ) Note that classically, me r v = Le , the angular momentum of the electron, and so 𝑚𝑒 𝑟𝑣 = √ℓ(ℓ + 1)ℏ with ℓ = 2. Thus, we have the following: 𝐵𝑛 =

𝜇0 𝐼 2𝑟

𝜇0 𝑒√ℓ(ℓ+1)ℏ 2𝑟 2𝜋𝑚𝑒 𝑟 2

𝑒ℏ 𝜇0 √ℓ(ℓ+1) 𝜇 𝜇 √ℓ(ℓ+1) = 𝐵 0 3 2𝜋𝑟 3 2𝜋𝑟

2𝑚𝑒

From Fig. 39–9(b), we see that the most probable radius for the 𝑛 = 2, ℓ = 1 state is approximately r = 4 r0 . We can now calculate the approximate magnetic field. 𝑇• 𝑚

𝐽

𝐵𝑛 =

(9.27×10−24 𝑇)(4𝜋×10−7 𝐴 )√6 𝜇𝐵 𝜇0 √ℓ(ℓ+1) = = 0.479𝑇 ≈ 2𝜋(4𝑟0 )3 2𝜋[4(0.529×10−10 𝑚)]3

0.5𝑇

0.479T − 0.431T  100  11% different, and so they are consistent. 2 ( 0.479T + 0.431T )

The two values are about 1

54. The energy of a pulse is the power of the pulse times the duration in time. E = P t = ( 0.68W ) 23  10−3 s = 0.01564J  0.016J

(

)

The number of photons in a pulse is the energy of a pulse divided by the energy of a photon, as given in Eq. 37–3. ( 0.01564J ) ( 640 10−9 m ) E E N= = = = 5.0  1016 photons hf hc ( 6.63  10−34 J s )( 3.00 108 m/s ) 55. The angular half-width of the beam can be found in r 1.22 , where d Section 35–5 and is given by 1/ 2 =  d is the diameter of the diffracting circle. The angular width of the beam is twice this. The linear diameter of the beam is then the angular width times the distance from the source of the light to the observation point, D = r . See the diagram.

2.44 ( 658  10 m ) 2.44 = ( 380  103 m ) = 170 m d 3.6  10−3 m 2.44 ( 658  10−9 m ) 2.44 6 = ( 384  10 m ) = 1.7  105 m (b) D = r = r d 3.6  10−3 m −9

(a) D = r = r

56. Intensity equals power per unit area. The area of the light from the laser is assumed to be in a circular area, while the area intercepted by the light from a light bulb is the surface area of a sphere. P P 0.50  10−3 W (a) I = = 2 = = 70.74 W m 2  71W m 2 2 −3 S r  (1.5  10 m ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) I =

P P 2.2 W = = = 4.377  10−2 W m 2  0.044 W m 2 2 2 S 4 r 4 ( 2.0 m )

The laser beam is more intense by a factor of

70.74 W m 2 = 1616  1600 . 4.377  10−2 W m 2

57. Transition from the E3  state to the E2  state releases photons with energy 1.96 eV, as shown in Fig. 39–21. The wavelength is determined from the energy. −34 8 hc ( 6.63  10 J s )( 3.00  10 m/s ) = = 6.34  10−7 m = 634 nm = −19 E 1.60 10 J/eV 1.96eV  ) ( )( 58. We use Eq. 39–16b. N2 =e N0 N1 =e N0

 E −E  − 2 0   kT 

(

)

 ( 2.2 eV ) 1.6010−19 J eV   −  1.3810−23 J K ( 300 K )   

(

)

 (1.8 eV ) 1.6010−19 J eV   −  1.3810−23 J K ( 300 K )   

(

)

= e −85.0 = 1.2  10−37

= e −69.6 = 6.1  10−31

59. The relative numbers of atoms in two energy states at a given temperature is given by Eq. 39–16b. N From Fig. 39–20, the energy difference between the two states is 2.2 eV. We set 2 = 0.5. N0

N2 =e N0

 E −E  − 2 0   kT 

  E −E   ( 2.2eV ) (1.60  10−19 J eV )  0  = − → T = − 2  = 3.7  104 K −23 N  k ln 2   (1.38  10 J K ) ln 0.5    N 0  

60. Consider Eq. 39–16b, with En  En . To have a population inversion means that N n  N n .  E −E 

− n n   Nn N En − E n   E − En   = e  kT   1 → ln n = −  n 0 0 → N n N n kT  kT  Since En  En , to satisfy the last condition, we must have T < 0, a negative temperature.

This negative “temperature” is not a contradiction. The Boltzmann distribution assumes that a system is in thermal equilibrium, and the inverted system is not in thermal equilibrium. The inversion cannot be maintained without adding energy to the system. If left to itself, the excited states will decay, and the inversion will not be maintained. 61. An h subshell has ℓ = 5. For a given ℓ value, ml ranges from −ℓ to +ℓ, taking on 2ℓ + 1 different values. For each 𝑚ℓ , there are 2 values of ms . Thus, the number of states for a given ℓ value is 2(2ℓ + 1). Thus, there are 2(2ℓ + 1) = 2(11) = 22 possible electron states. 62. (a) Z = 24 is chromium. 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 5 4s1

(b) Z = 35 is bromine. 1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4 p 5 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

63. (a) Boron has Z = 5, so the outermost electron has n = 2. We use the Bohr result with an effective Z. We might naively expect to get Z eff = 1, indicating that the other four electrons shield the outer electron from the nucleus, or Z eff = 3, indicating that only the inner two electrons accomplish the shielding.

(13.6eV )( Z eff ) E =−

(13.6eV )( Z eff ) − 8.26eV = −

→ → Z eff = 1.56 22 n2 This indicates that the other two electrons in the n = 2 shell partially shield the electron that is to be removed. (b) We find the average radius from the expression below. 2 −10 n 2 r1 2 ( 0.529  10 m ) r= = = 1.36  10−10 m Z eff (1.56 ) 2

64. The value of ℓ can range from 0 to n − 1. Thus, for n = 6, we have 0 ≤ ℓ ≤ 5. The magnitude of L is given by Eq. 39–3, 𝐿 = √ℓ(ℓ + 1)ℏ.

Lmin = 0 ; Lmax = 30 65. (a) We treat the Earth as a particle in rotation about the Sun. The angular momentum of a particle is given in Eq. 37–10 as 𝐿 = 𝑚𝑣𝑟𝑛 , where r is the orbit radius. We equate this to the quantum mechanical expression in Eq. 39–3. We anticipate that the quantum number will be very large, so we approximate √ℓ(ℓ + 1) as l . 𝐿 = 𝑀𝐸𝑎𝑟𝑡ℎ 𝑣𝑟Sun− = 𝑀 Earth

2𝜋𝑟 𝑇

𝑟 = ℏ[ℓ(ℓ + 1)]2 = ℏℓ →

24 11 M Earth 2 r 2 ( 5.98  10 kg ) 2 (1.496  10 m ) = = 2.5255  1074  2.53  1074 l = 7 −34 T (1.055  10 J s )( 3.156  10 s) 2

(b) There are 2ℓ + 1values of 𝑚ℓ for a value of ℓ, so the number of orientations is as follows. 𝑁 = 2ℓ + 1 = 2(2.5255 × 1074 ) + 1 = 5.051 × 1074 ≈ 5.05 × 1074 . 66. Eq. 37–15 gives the Bohr-theory result for the wavelength of a spectral line. For the Mosley plot, the wavelengths are for the K line, which has n = 2 and n  = 1. We assume that the shielding of the other n = 1 electron present reduces the effective atomic number to Z – 1. We use the value of the Rydberg constant from Section 37–11. 1  Z 2e4m   1 1  1  e4m  2  1 1  1  3e 4 m  2 =  2 3  2 − 2  → =  2 3 Z  −  → = Z →   8 0 h c   n n    8 0 h c   1 4    32 02 h 3c   3e 4 m  2 =  ( Z − 1) →   32 02 h 3c  1

1/ 2

 3e 4 m  a= 2 3   32 0 h c 

= ( 43 R )

1/ 2

 3e 4 m  = a ( Z − b) , a =  2 3    32 0 h c  1

=  43 (1.0974  107 m −1 )

1/ 2

, b =1

= 2868.9 m −1/ 2

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67. The magnitude of the angular momentum is given by Eq. 39–3, and Lz is given by Eq. 39–4. The cosine of the angle between L and the z axis is found from L and Lz . 𝐿

𝑚ℓ

𝐿

√ℓ(ℓ+1)

𝐿 = √ℓ(ℓ + 1)ℏ ; 𝐿𝑧 = 𝑚ℓ ℏ ; 𝜃 = 𝑐𝑜𝑠 −1 𝑧 = 𝑐𝑜𝑠 −1

(a) For ℓ = 1, 𝑚ℓ = −1, 0, 1. 1 0 1,1 = cos−1 = 45 ; 1,0 = cos −1 = 90 ; 2 2 −1 1,−1 = cos−1 = 135 x 2 (b) For ℓ = 2, 𝑚ℓ = −2, −1, 0, 1, 2. 2 1 0 2,2 = cos−1 = 35.3 ;  2,1 = cos −1 = 65.9 ; 2,0 = cos −1 = 90 6 6 2 −1 −2 2,−1 = cos−1 = 114.1 ; 2,2 = cos −1 = 144.7 6 6 (c) For ℓ = 3, 𝑚ℓ = −3, −2, −1, 0, 1, 2, 3. 3 2 1 = 30 ; 3,2 = cos−1 = 54.7 ; 3,1 = cos−1 = 73.2 3,3 = cos−1 12 12 12 −1 −2 0 = 90 ; 3,−1 = cos−1 = 106.8 ; 3,−2 = cos−1 = 125.3 ; 3,0 = cos−1 12 12 12 −3 = 150 3,−3 = cos−1 12 (d) We see from the previous parts that the smallest angle occurs for 𝑚ℓ = 𝑙. 100 100,100 = cos−1 = 5.71 (100)(101)

10 ,10 = cos −1 6

106

= 0.0573

(10 )(10 + 1) 6

This is consistent with the correspondence principle, which would say that the angle between L and the z axis could be any value classically, which is represented by letting ℓ → ∞ (which also means n →  ). 68. (a) Since Lz = 0,  → , and so  is unknown . We can say nothing about the value of . (b) Since  is completely unknown, we have no knowledge of the component of the angular momentum perpendicular to the z-axis. Thus Lx and L y are unknown . (c) The square of the total angular momentum is given by L = L2x + L2y + L2z . Use this with the quantization conditions for L and Lz given in Eqs. 39–3 and 39–4. 𝐿2 = 𝐿2𝑥 + 𝐿2𝑦 + 𝐿2𝑧 → ℓ(ℓ + 1)ℏ2 = 𝐿2𝑥 + 𝐿2𝑦 + 𝑚ℓ2 ℏ2 → 𝐿2𝑥 + 𝐿2𝑦 = [ℓ(ℓ + 1) − 𝑚ℓ2 ]ℏ2 1/2

√𝐿2𝑥 + 𝐿2𝑦 = [ℓ(ℓ + 1) − 𝑚ℓ2 ]

ℏ

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

69. This is very similar to Example 39–3. The radial probability distribution for the n = 3, ℓ = 0 wave function is derived in Problem 27(a), and then we need to find the position at which that distribution has a maximum. We see from the figure in the solution to Problem 27(b) that there are three local maxima in the probability distribution function, and the global maximum is at approximately 13r0 . The wave function is given in Problem 27. r  1 2r 2r 2  − 3r0 − + e 1  300 =  2  27 r03  3r0 27r0 

2.0

-1

Pr (nm )

1.5 1.0 0.5 0.0 0

Pr = 4 r  300 2

Instructor Solutions Manual

r/r0

1  2r 2r 2  − 3r0 4r 2  2r 2r 2  − 3r0 = 4 r + = + e 1− 1−  e 3  2  3  27 r0  3r0 27r0  27r0  3r0 27r02  2

We see that Pr is a sixth-order polynomial, multiplied by a decaying exponential. Finding the local dP extrema analytically, by solving r = 0 , is at best very difficult (or perhaps even impossible). So, dr instead we simply use the Excel graph from Problem 27(b), calculating with a small step size near the global maximum. Doing that, we find that the maximum occurs at 13.074r0 .

13.074r0 = 13.074 ( 0.0529nm ) = 0.692nm 70. The “location” of the beam is uncertain in the transverse (y) direction by an amount equal to the aperture opening, D. This gives a value for the uncertainty in the transverse momentum. The momentum must be at least as big as its uncertainty, so we obtain a value for the transverse momentum. p y y 

→ p y 

= → py  y D D The momentum in the forward direction is related to the h wavelength of the light by p x = . See the diagram to

 relate the momentum to the angle. p D    = ; “spread” = 2 =  y   px h  2 D D D



71. (a) The mean value can be found as described in Problem 19. We use the first definite integral given in Appendix B–5, with n = 1 and a = 1. r r    1 1 1 −2 r0 4 r −2 r0 r 2 1 2 2 → e 4 r dr = 2  e dr ; let x = 2   =   100 4 r dr =  3 r  r0 r0 0 r0 r0 r 0 r 0 

1 1 1 xe − x dx = (1) =  r0 0 r0 r0

1 e2 U =− 4 0 r

→ U =−



 4  r

2 100

0 0

4 r 2 dr = −

1 e2 4 0 r0

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Chapter 39

Quantum Mechanics of Atoms

(b) For the ground state of hydrogen, Eq. 37–14a gives the energy, and Eq. 37–11 gives the Bohr radius. Substitute those expressions into E = U + K . 1 e2 e4m E=− 2 2 ; U =− 8 0 h 4 0 r0 2 4  h 0  e m   me 2  1 e2  e4m  e2 e4m e2  K = E −U = − 2 2 −  − − = − = 8 0 h  4 0 r0  4 0 r0 8 02 h 2 4 0 r0 8 02 h 2 r0

e2 4 0 r0

−

e2 8 0 r0

72. In the Bohr model, LBohr = n

e2 8 0 r0

h = 2 . In quantum mechanics, 𝐿QM = √ℓ(ℓ + 1)ℏ. For n = 2, ℓ = 0 2

or ℓ = 1, so that LQM = 0 ( 0 + 1) 73. (a) (b) (c) (d) (e)

= − 12 U

= 0 or LQM = 1(1 + 1)

2 .

The 4 p → 3 p transition is forbidden, because 𝛥ℓ = 0 ≠ ±1. The 3 p → 1s transition is allowed, because 𝛥ℓ = −1. The 4d → 3d transition is forbidden, because 𝛥ℓ = 0 ≠ ±1. The 4d → 3s transition is forbidden, because 𝛥ℓ = −2 ≠ ±1. The 4s → 2 p transition is allowed, because 𝛥ℓ = +1.

74. The binding energy is given by the opposite of Eq. 37–14b. Z 2 13.6eV − En = (13.6eV ) 2 = = 6.72  10−3 eV 2 n 45 The radius is given by Eq. 37–13. n2 rn = ( 0.529  10−10 m ) = 452 ( 0.529  10−10 m ) = 1.07  10−7 m Z The effective cross-sectional area is as follows.

 =  r 2 =  (1.07  10−7 m ) = 3.60  10−14 m 2 2

75. The K line is from the n = 2 to n = 1 transition. We use the energies of the hydrogen atom with Z replaced by Z − 1, as in Example 39–6, using Eq. 37–15.  2 2 e 4 mk 2  1  2 1 =  ( Z − 1)  2 − 2  → 3 n    hc   n 1

−1

1  2 2 e 4 mk 2   1 1  Z = 1+    2 − 2 3 n    h c   n

−1

= 1+

( 0.154  10 m ) −9

(1.097  10 m ) 7

−1 −1

1 1  2 − 2 2  1

−1

= 1 + 28.09  29 The element is copper.

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Instructor Solutions Manual

1   1  Z 2  2 − 2  , and so we seek Z and nX so that for every fourth value of nX , n n    there is an nH such that the Lyman formula equals the formula for the unknown element.

76. From Eq. 37–15,



 1 1 1 1  Z2 Z2 − 2 = Z2  2 − 2  = 2 − 2 . 1 nH  nX nX  nX nX Consider atoms (with Z > 1) for which Z 2 n2X = 1 → nX = Z , and for which nX2 = Z 2 nH2 → nX = ZnH for every fourth line. This is met when Z = 4 and nx = 4. The resulting spectral lines for nx = 4, 8, 12, ... match the hydrogen Lyman series lines for n = 1,2,3, ... . The element is beryllium. 77. (a) We use Eq. 39–16b, with the “note” as explained in the problem, multiplying the initial expression times 82 .

(

N2 8 = 2e N1

 E −E  − 2 1   kT 

= 82 e

)

  −13.6 eV −13.6 eV   −19 −  1.6010 J eV   4 1   −  −23   1.3810 J K ( 300 K )    

(

)

= 4e −394.2

Many calculators will not directly evaluate e −394.2 , so we do the following. x = e −394.2 ; log x = −394.2log e = −394.2 ( 0.4343) = −171.2 ; x = 10−171.2 = 10−0.210−171 = 0.631  10−171 N2 = 4e −394.2 = 4 ( 0.631  10−171 ) = 2.52  10−171  3  10−171 N1 There are 18 states with n = 3, so we multiply by 182 .

(

 E −E  − 3 1   kT 

)

  −13.6 eV −13.6 eV   −19 −  1.6010 J eV   9 1   −    1.3810−23 J K ( 300 K )    

(

)

N 3 18 = 2e = 182 e = 9e −467.2 = 1.13  10−202  1  10−202 N1 (b) We repeat the evaluations for the higher temperature.   −13.6 eV −13.6 eV   −19 − (1.6010 J eV )   4 1   −     E −E  (1.3810−23 J K )(6000 K )  N 2 8 − 2kT 1  8   = 2e = 2e = 4e −19.71 = 1.10  10−8  1  10−8 N1

( )

)

  −13.6 eV −13.6 eV   −19 −  1.6010 J eV   9 1   −    1.3810−23 J K ( 6000 K )    

( N 3 18 = 2e = 182 e = 9e −23.36 = 6.44  10−10  6  10−10 N1 (c) Since the fraction of atoms in each excited state is very small, we assume that N 1 is the number  E −E  − 3 1   kT 

of hydrogen atoms given. 1.0 g of H atoms contains 6.02  1023 atoms.

N 2 = N1 (1.10  10−8 ) = ( 6.02  1023 )(1.10  10−8 ) = 6.62  1015  7  1015 N 3 = N1 ( 6.44  10−10 ) = ( 6.02  1023 )( 6.44  10−10 ) = 3.88  1014  4  1014

(d) We assume the lifetime of an excited state atom is 10−8 s. Each atom would emit one photon as its electron goes to the ground state. The number of photons emitted per second can be estimated by the number of atoms, divided by the lifetime. 4  1014 N N 2 7  1015 22 4 10 photons s ;  = =  7  1023 photons s n3 = 3 = n 2 10−8 s 10−8 s   78. (a) The additional energy due to the presence of a magnetic field is derived in Section 39–7, as © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Quantum Mechanics of Atoms

𝑈 = 𝜇𝐵 𝑚ℓ 𝐵𝑧 . We use this to calculate the energy spacing between adjacent 𝑚ℓ values. 𝑈 = 𝜇𝐵 𝑚ℓ 𝐵𝑧 → 𝐽 (9.27 × 10−24 ) 𝑇 (2.6𝑇) = 1.506 × 10−4 eV ≈ 1.5 × 10−4 eV 𝑈 = 𝜇𝐵 𝐵𝑧 𝛥𝑚ℓ = 𝐽 −19 (1.60 × 10 ) eV (b) As seen in Fig. 39–4, the 𝑛 = 3, ℓ = 2 level will split into 5 levels, and the 𝑛 = 2, ℓ = 1 level will split into n = 3, = 2 3 levels. With no restrictions, there would be 15 different transitions possible. All transitions would have the same n = 1. Thus, there are only three unique wavelengths possible: the one corresponding to a transition with 𝛥𝑚ℓ = +1 (a slightly larger energy change than in the B = 0 case); the one corresponding to a transition with 𝛥𝑚ℓ = 0 (the same n = 2, = 1 energy change as in the B = 0 case); and the one corresponding to a transition with 𝛥𝑚ℓ = −1 (a ml = slightly smaller energy change than in the B = 0 case). See the diagram showing 9 possible transitions grouped into 3 actual energy changes. The value along the right side is the change in energy level due to the magnetic field interaction. (c) Eq. 37–15 gives the wavelength for hydrogen, considering only a change in principal quantum number. The energies for those transitions are on the order of eV. The energy change due to the magnetic field interaction is much smaller than that, so we can use an approximation, knowing E from part (a). We obtain E from Eq. 37–14b. −1

−1   1 1  − 2   = 1.0974  107 m −1 ( 14 − 19 )  = 6.56096  10 −7 m  656.10 nm 2   2 3 

n =3→ 2 =  R  =h

c E

→  = −

 = −

hc E  1 1 E = − ; En =3→ 2 = 13.6eV  2 − 2  = 1.889eV 2 E E 2 3 

 1.506  10−4 eV  E = − ( 656.10 nm )   = 0.052 nm E  1.889eV 

m =+1 = n =3→ 2 +  = 656.10 nm − 0.052nm = 656.05nm l

m =+1 = n =3→2 + 0 = 656.10 nm l

m =−1 = n =3→2 +  = 656.10 nm + 0.052nm = 656.15nm l

79. The uncertainty in the electron position is x = r1 . The minimum uncertainty in the velocity, v , can be found by solving Eq. 38–1, where the uncertainty in the momentum is the product of the electron mass and the uncertainty in the velocity. xpmin = r1 ( mv ) 

v 

mr1

1.055  10−34 J s

( 9.1110 kg )( 0.529 10 m ) −31

−10

= 2.19  106 m/s

80. (a) An electron in a 4s state has ℓ = 0, and so due to the selection rule of 𝛥ℓ = ±1, it must transition © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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to an ℓ = 1 state (a p state). Thus, it could transition to a 2p or a 3p state (there is no 1p state). 4s → 3 p , 4s → 2 p (b) To get to the ground state, it must get to a 1s state. Since again 𝛥ℓ = ±1, it must transition from either 3p or 2p to the 1s (ground) state. Thus, it must make 2 transitions to get to the ground state. 4 s → 3 p →1s , 4 s → 2 p → 1s 81. The wave functions are given in Eq. 39–9. To show that the desired quantity is spherically symmetric, there must not be any dependence on x, y, or z except through x 2 + y 2 + z 2 = r 2 . z

 210 =

−

32 r05

x + iy

e 2 r0 ;  211 =

−

64 r05

e 2 r0 ;  21−1 =

 210

64 r05

−

e 2 r0

− ( x + iy )( x − iy ) e− r0 = ( x + y ) e− r0 z2 2 = e r0 ;  211 = 5 32 r0 64 r05 64 r05 r

x − iy

 21−1 =

( x − iy )( x + iy ) − rr

 210 +  211 +  21−1

64 r05

(x + y ) e = 2

64 r05

−

r r0

− ( x + y ) e− r0 + ( x + y ) e− r0 z2 e r0 + = 5 32 r0 64 r05 64 r05 2

− r2 r0 e , which is spherically symmetric 32 r05

82. From Fig. 39–20, we see that the energy difference from E1 to E0 is 1.8 eV. Use Eq. 37–3 to calculate the wavelength. −34 8 hc hc ( 6.63  10 J s )( 3.00  10 m/s ) E = hf = → = = = 6.906  10−7 m  690nm −19 E  (1.8eV ) (1.60  10 J/eV ) The textbook quotes an actual wavelength of 694.3 nm. 83. The classical angular momentum of a particle moving in a circular path is given in Section 39–7 as 𝐿 = 𝑚𝑣𝑟. We want the magnetic moment to be equal to 1 Bohr magneton. And, note from Eq. 37– 11 that the radius of an electron in orbit around a hydrogen atom is given by rn = n 2 r1 . 𝜇=

1 𝑒 2𝑚

𝐿=

1 𝑒 2𝑚

𝑚𝑣𝑟 → 𝑣 =

2𝜇𝐵 𝑒𝑟

𝐽 𝑇

2(9.27×10−24 )

= (1.60×10−19

𝐶)4(5.29×10−11 𝑚)

= 5.48 × 105

𝑚 𝑠

1328

CHAPTER 40: Molecules and Solids Responses to Questions 1.

(a) The bond in an N 2 molecule is expected to be covalent. (b) The bond in the HCl molecule is expected to be ionic. (c) The bond between Fe atoms in a solid is expected to be metallic.

A neutral calcium atom has 20 electrons. Its outermost electrons are in the 4s2 state. The inner 18 electrons form a spherically symmetric distribution and partially shield the outer two electrons from the nuclear charge. A neutral chlorine atom has 17 electrons; it lacks just one electron to have its outer shell filled. A CaCl2 molecule could be formed when the outer two electrons of the calcium atom are “shared” with two chlorine atoms. These electrons will be attracted by both the Ca and the Cl nuclei and will spend part of their time between the Ca and Cl nuclei. The nuclei will be attracted to this negatively charged area, forming a covalent bond. As is the case with other asymmetric covalent bonds, this bond will have a partial ionic character as well. The two electrons will partly orbit the Ca nucleus and partly orbit each of the two Cl nuclei. Since each Cl nucleus will now have an extra electron part of the time, it will have a net negative charge. The Ca nucleus will “lose” two electrons for part of the time, giving it a net positive charge.

No, neither the H2 nor the O2 molecule has a permanent dipole moment. The outer electrons are shared equally between the two atoms in each molecule, so there are no polar ends that are more positively or negatively charged. The H2O molecule does have a permanent dipole moment. The electrons associated with the hydrogen atoms are pulled toward the oxygen atom, leaving each hydrogen with a small net positive charge and the oxygen with a small net negative charge. Because of the shape of the H2O molecule (see Fig. 40–6), one end of the molecule will be positive and the other end will be negative, resulting in a permanent dipole moment.

The molecule H3 has three electrons. According to the Pauli exclusion principle, no two of these electrons can be in the same quantum state. Two of them will be 1s2 electrons and will form a “closed shell” and a spherically symmetric distribution, and the third one will be outside this distribution and unpaired. This third electron will be partially shielded from the nucleus and will thus be easily “lost,” resulting in an unstable molecule. The ion H3+ only has two electrons. These 1s2 electrons will form a closed shell and a spherically symmetric distribution, resulting in a stable configuration.

Yes, H 2+ should be stable. The two positive nuclei share the one negative electron. The electron spends most of its time between the two positive nuclei (basically holding them together).

The two diagrams are copied here for easy reference. The H2 is on the left, and the ATP on the right.

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Note that both diagrams show a “dip” in potential energy indicating a stable bond is formed at bond length r0, but because the “dip” in the ATP diagram is above the U = 0 line, stored potential energy can be released from that bond when it is broken. That energy is used in other chemical processes in living cells. The ATP diagram also has an “activation energy,” which means that an energy input (perhaps from kinetic energy of the components) is necessary to make the bond. The H 2 molecule does not have an activation energy. Thus, the two hydrogen atoms do not need an initial energy input (from kinetic energy, for example) for the atoms to bond. If the separation of the two H atoms is greater than r0, the atoms can attract each other and bond. 7.

The carbon atom (Z = 6) usually forms four bonds because carbon requires four additional electrons to form a closed 2p shell, and each hydrogen-like atom contributes one electron.

The four categories of molecular energy are: translational kinetic energy, electrostatic potential energy, rotational kinetic energy, and vibrational kinetic energy.

Metallic bond theory states that the free electrons in metallic elements can vibrate at any frequency, so when light of a range of frequencies falls on a metal, the electrons can vibrate in response and reemit light of those same frequencies. Hence, the reflected light will largely consist of the same frequencies as the incident light, and so it will not have a distinct color but will be “shiny” instead. Nonmetallic materials only absorb and emit light at specific frequencies, and so they have distinct colors.

10. The conduction electrons are not strongly bound to particular nuclei, so a metal can be viewed as a collection of positive ions and a negative electron “gas.” (The positive ions are just the metal atoms without their outermost electrons since these “free” electrons make up the gas.) The electrostatic attraction between the freely roaming electrons and the positive ions keeps the electrons from leaving the metal. 11. As temperature increases, the thermal motion of ions in a metal lattice will increase. More electrons collide with the ions, increasing the resistivity of the metal. When the temperature of a semiconductor increases, more electrons are able to move into the conduction band, making more charge carriers available and thus decreasing the resistivity. The thermal motion also increases the resistance in semiconductors, but the increase in the number of charge carriers is a larger effect. 12. For an ideal pn junction diode connected in reverse bias, the holes and electrons that would normally be near the junction are pulled apart by the reverse voltage, preventing current flow across the junction. The resistance is essentially infinite. A real diode does allow a small amount of reverse current to flow if the voltage is high enough, so the resistance in this case is very high but not infinite. A pn junction diode connected in forward bias has a low resistance (the holes and electrons are close together at the junction) and current flows easily. 13. The last valence electron of a sodium atom is shielded from most of the sodium nuclear charge and experiences a net nuclear charge of +1e. The outer shell of a chlorine atom is the 3p shell, which contains five electrons. Due to shielding effects, the 3p electrons of chlorine experience a net nuclear charge of +5e. In NaCl, the last valence electron of sodium is strongly bound to a chlorine nucleus. This strong ionic bonding produces a large energy gap between the valence band and the conduction band in NaCl, characteristic of a good insulator. 14. The general shape of Fig. 40–28 is the same for most metals. The scale of the graph (especially the x-axis scale and the Fermi energy) is peculiar to copper and will change from metal to metal. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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15. The main difference between n-type and p-type semiconductors is the type of atom used for the doping impurity. When a semiconductor such as Si or Ge (each atom of which has four electrons to share) is doped with an element that has five electrons to share (such as As or P), then it is an n-type semiconductor since an extra electron has been inserted into the lattice. When a semiconductor is doped with an element that has three electrons to share (such as Ga or In), then it is a p-type semiconductor since an extra hole (the lack of an electron) has been inserted into the lattice. 16. The phosphorus atoms will be donor atoms. Phosphorus has five valence electrons. It will form four covalent bonds with the silicon atoms around it and will have one “extra” electron that is weakly bound to the atom and can be easily excited up to the conduction band. This process results in extra electrons in the conduction band. Silicon doped with phosphorus is therefore an n-type semiconductor. 17. When the top branch of the input circuit is at the high voltage (current is flowing in this direction for half the cycle), then the bottom branch of the output is at the high voltage. The current follows the path through the bridge in the diagram on the left. When the bottom branch of the input circuit is at the high voltage (current is flowing in this direction during the other half of the cycle), then the bottom branch of the output is still at high voltage. The current follows the path through the bridge in the diagram on the right.

18. The energy comes from the power supplied by the collector/emitter voltage source ℰC. The input signal to the base just regulates how much current, and therefore power, can be drawn from the collector’s voltage source. 19. In the circuit shown in Fig. 40–49, the base-emitter is forward biased (the current will easily flow from the base to the emitter), and the base-collector is reverse biased (the current will not flow easily from the base to the collector). 20. No. Ohmic devices (those that obey Ohm’s law) have a constant resistance and therefore a linear relationship between voltage and current. The voltage–current relationship for diodes is not linear (see Fig. 40–38). The resistance of a diode operated in reverse bias is very large. The same diode operated in a forward-bias mode has a much smaller resistance. Since a transistor can be thought of as made up of diodes, it is also non-ohmic. 21. A diode cannot be used to amplify a signal. A diode does let current flow through it in one direction easily (forward biased), and it does not let current flow through it in the other direction (reversed bias), but there is no way to connect a source of power to use it to amplify a signal (which is how a transistor amplifies a signal). Combinations of diodes with additional power sources, as in a transistor, are able to amplify a signal. 22. The base current (between the base and the emitter) controls the collector current (between the collector and the emitter). If there is no base current, then no collector current flows. Thus, © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

controlling the relatively small base current allows the transistor to act as a switch, turning the larger collector current on and off.

Responses to MisConceptual Questions 1.

(c) In the H 2 molecule, the two electrons orbit both atoms. In order for those two electrons to not violate the Pauli exclusion principle, they must have different quantum numbers. Each atom initially had one electron and the molecule has two, so no electrons are lost. When the atoms are separated by one bond length, the energy is a minimum (not a maximum), and the molecule has less total energy than the two atoms separately. This decrease in energy is the binding energy. Thus, energy has to be added to the molecule in order to separate it into individual atoms.

(b) Covalent bonding is the sharing of atoms between molecules. When one atom has excess electrons in its outer shell and another atom lacks electrons in the outer shell, the atom with excess transfers the electrons to the other atom, making a positive ion and a negative ion, and thus creating an ionic “bond”—an attraction between the two ions.

(a, d) Because the ADP molecule has a positive activation energy, as the phosphate group approaches the ADP molecule, it is first repelled and then attracted. The phosphate group must initially have kinetic energy to overcome the repulsion. Some of this kinetic energy is stored as potential energy in the ATP molecule. The binding energy is negative as some of the initial kinetic energy is stored as positive potential energy. When the ATP molecule is broken apart, this energy is released and is available to instigate other reactions.

(d) For the DNA to replicate properly, the bond holding the two strands together must be a very weak bond. Ionic and covalent bonds are strong bonds. The Van der Waals bond is a weak bond that holds the DNA together.

(a) As stated in Section 40–5, the outer electrons in a metallic solid “roam” rather freely among all the metal atoms, acting like an electron “gas.”

(a) A common misconception is that metals have free electrons because they have more electrons than protons. Actually, metals are electrically neutral in that they have the same number of electrons as protons. In metals, the outer electrons are not tightly bound to a single atom but can move between atoms in the metal lattice. Since the electrons can move easily between atoms, metals make good conductors of electricity.

(b) As discussed in Example 40–10, the speed of the electrons in a metal is determined by the Fermi energy. That speed is significantly lower than the speed of light and is much higher than the speed determined from the absolute temperature. And in an earlier chapter it was shown the drift speed of electrons in current is much smaller than the speed as determined by the Fermi energy.

(a) In an insulator, the electrons fill the upper valence bands and are tightly bound to their atoms. A large energy gap exists between the filled valence band and the next higher (conduction) band,

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so a large energy would need to be added to the atom to move an electron to the conduction band. 10. (d) A hole is a positive region in the semiconductor that is formed when an electron is missing from the periodic molecular structure. 11. (a) To be used for doping silicon, the element should have one more or one less electron in its outer shell than silicon. Silicon has two electrons in its outer p shell. Boron and gallium each have one electron in their outer p shell, one fewer than silicon. Phosphorus and arsenic each have three electrons in their outer p shells, one more than silicon. Germanium has two electrons in its outer p shell, the same as silicon, and so would not be a good choice as a doping impurity. 12. (a) A diode allows current to flow through it in one direction, but not in the other (until the breakdown voltage is reached). The diode is not linear, so the current flowing is not proportional to the voltage applied. Thus, answer (b) is not correct. A diode is essentially a junction between n and p type semiconductors, and so it is labeled as a pn junction. Thus, answer (c) is not correct. And, from Fig. 40–38, the current in a forward-biased diode is highly dependent on the applied voltage, and so answer (d) is incorrect.

Solutions to Problems Note:

A factor that appears in the analysis of electron energies is 2 e2 = ( 9.00  109 N m 2 C2 )(1.60  10−19 C ) = 2.30  10−28 J m. 4 0

We calculate the binding energy as the opposite of the electrostatic potential energy. We use Eq. 23– 10 for the potential energy. e2  1 Q1Q2 1 2.30  10−28 J m  =− = − Binding energy = −U = −   4 0 r 4 0  0.28  10−9 m  0.28  10−9 m

= 8.214  10−19 J  8.2  10−19 J 1eV   = 8.214  10−19 J  = 5.134eV  5.1eV −19   1.60 10 J   2.

From Problem 1, the “point electron” binding energy is 5.134 eV. With the repulsion of the electron clouds included, the actual binding energy is 4.43 eV. Use these values to calculate the contribution of the electron clouds. 5.134eV − 4.43eV = 0.704eV  0.70eV

We follow the procedure outlined in the statement of the problem. HN: 12 ( d H 2 + d N 2 ) = 12 ( 74 pm + 145pm ) = 110 pm

( d + d ) = (154 pm + 145pm ) = 150 pm NO: ( d + d ) = (145pm + 121pm ) = 133pm

CN:

1 2

We convert the units from kcal/mole to eV/molecule.

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kcal 4186J 1eV 1mole eV    = 4.339  10−2 23 −19 mole 1kcal 1.602  10 J 6.022  10 molecules molecule Now convert 4.43 eV per molecule into kcal per mole. eV 1kcal mol 4.43  = 102 kcal mol molecule 4.339  10−2 eV molecule 1

We calculate the binding energy as the difference between the energy −0.66 e of two isolated hydrogen atoms and the energy of the bonded p p combination of particles. We estimate the energy of the bonded combination as the negative potential energy of the two electron– d proton combinations, plus the positive potential energy of the proton– proton combination. We approximate the electrons as a single object with a charge of 0.33 of the normal charge of two electrons, since the electrons only spend that fraction of time between the nuclei. A simple picture illustrating our bonded model is shown. When the electrons are midway between the protons, each electron will have a potential energy U ep due to the two protons.

U ep =

− ( 2 )( 0.33) ke2

( 12 d )

=−

( 4 )( 0.33) ( 2.30  10−28 J  m )

( 0.074  10 m )(1.60  10 −9

−19

J eV )

= − 25.6eV

The protons themselves have this potential energy: 2.30  10−28 J  m ke2 U pp = + =+ = +19.4eV. r 0.074  10−9 m 1.60  10−19 J eV

(

)(

)

When the bond breaks, each hydrogen atom will be in the ground state with an energy E1 = −13.6eV. Thus the binding energy is as follows.

Binding energy = 2 E1 − ( 2 PEep + PE pp ) = 2 ( −13.6eV ) − 2 ( −25.6eV ) + 19.4eV  = 4.6eV

This is close to the actual value of 4.5 eV quoted in the text. 6.

According to the problem statement, 5.39 eV of energy is required to make an Li + ion from neutral Li, and 3.41 eV of energy is released when an F atom becomes an F− ion. That means that a net energy input of 5.39 eV – 3.41 eV = 1.98 eV is needed to form the ions. We calculate the negative potential energy of the attraction between the two ions. ( 2.30  10−28 J  m ) 1 e2 =− = −9.21eV U =− 4 0 r ( 0.156  10−9 m )(1.60  10−19 J eV ) The binding energy should therefore be 9.21 eV – 1.98 eV = 7.23 eV. But, the actual binding energy is only 5.95 eV. Thus, the energy associated with the repulsion of the electron clouds is 7.23 eV – 5.95 eV = 1.28 eV.

(a) The neutral He atom has two electrons in the ground state, 𝑛 = 1, ℓ = 0, 𝑚ℓ = 0. Thus, the two electrons have opposite spins, ms =  12 . If we try to form a covalent bond, we see that an electron from one of the atoms will have the same quantum numbers as one of the electrons on the other atom. From the exclusion principle, this is not allowed, so the electrons cannot be shared. (b) We consider the He 2 + molecular ion to be formed from a neutral He atom and an He+ ion. It will have three electrons. If the electron on the ion has a certain spin value, it can have the opposite spin as one of the electrons on the neutral atom. Thus, those two electrons can be in the

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same spatial region (because their quantum numbers are not identical), and so a bond can be formed. 2

The MKS units of

are

( J s)

( kg m ) ( kg m s ) m 2

J2 J2 = = J. The final unit is joules, which ( N m) J

is an energy unit. 9.

The reduced mass is given in Section 40–5 as  =

m1m2 . We calculate in atomic mass units. m1 + m2

(a) KCl:

=

( 39.10 u )( 35.45u ) = 18.59 u m1m2 = m1 + m2 ( 39.10 u ) + ( 35.45u )

(b) O2:

=

(16.00 u )(16.00 u ) = 8.00 u m1m2 = m1 + m2 (16.00 u ) + (16.00 u )

=

(1.008u )( 35.45u ) = 0.9801u m1m2 = m1 + m2 (1.008u ) + ( 35.45u )

r 10. The moment of inertia of H 2 about its CM is I = 2mH   2 2 m r = H . The bond length is given in the text as 74 pm. 2

2 2 (1.008) (1.66  10−27 kg )( 7.4 10−11 m )  r  mH r I = 2mH   = = = 4.6  10−48 kg m 2 2 2 2   2

11. (a) Use the quantization of angular momentum to determine the angular velocity. The moment of 2 mr 2 r . inertia of the dumbbell about its CM is I = 2m   = 2 2 𝐼𝜔 = √ℓ(ℓ + 1)ℏ =

𝜔=

𝑚𝑟 2 2

𝜔 →

2√ℓ(ℓ + 1)ℏ 2√1(2)(1.055 × 10−34 𝐽 · 𝑠) rad = = 3.3 × 10−23 2 −6 −3 2 ( )( ) 𝑚𝑟 1.0 × 10 kg 3.0 × 10 𝑚 𝑠

(b) We must convert the angular velocity to rad/s for the calculation. We anticipate that ℓ will be quite large, so we approximate ℓ(ℓ + 1) ≈ ℓ2 . 𝐼𝜔 = √ℓ(ℓ + 1)ℏ ≈ ℓℏ = ℓ=

𝑚𝑟 2 2

𝜔 →

(1.0 × 10−6 kg)(3.0 × 10−3 𝑚)2 𝑚𝑟 2 rev 2𝜋rad 1min (50 ) ≈ 2.2 × 1023 𝜔= × × −34 2ℏ 2(1.055 × 10 𝐽 · 𝑠) min rev 60𝑠

Our assumption about ℓ is justified.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition mO

12. (a) The moment of inertia of O 2 about its CM is given by 2

Instructor Solutions Manual



m r r I = 2mO   = O . 2 2

(1.055  10 J s ) = = 2I m r (16 ) (1.66  10 kg )( 0.121  10 m ) (1.60  10 2

−34

−27

−9

−19

J eV )

= 1.789  10−4 eV

 1.79  10−4 eV (b) From Fig. 40–17, we see that the energy involved in the ℓ = 3 to ℓ = 2 transition is 3 2 I . 2 3 2 = 6 = 6 (1.789  10−4 eV ) = 1.0734  10−3 eV  1.07  10−3 eV 2I I ( 6.63  10−34 J s )( 3.00 108 m s ) = 1.16  10−3 m c hc E = h → = =  E (1.0734  10−3 eV )(1.60  10−19 J eV )

E =

13. Use the rotational energy and the moment of inertia of N 2 about its CM to find the bond length. See the adjacent diagram.

I = 2mN ( 12 r ) = 12 mN r 2 ; Erot = 2

Erot mN

( 2.48  10 eV )(1.60 10

−19



→

mN r 2

(1.055  10

−4

−34

J s)

J eV ) (14.01u ) (1.66  10

−27

kg u )

= 1.10  10−10 m

14. The energies involved in the transitions are given in Fig. 40–17. We find the rotational inertial from Eq. 40–4. The basic amount of rotational energy is 2 I . 2 (1.055  10−34 J s ) 2 2 = = = = I  m1m2 2  ( 12 mH r 2 ) mH r 2 (1.008u ) (1.66  10−27 kg u )( 0.074  10 −9 m )2 r    m1 + m2  = 2.429  10−21 J (a) For ℓ = 1 to ℓ = 0: 2   1 −2 E = = 2.429  10−21 J   = 1.5  10 eV −19 1.60 10 J eV I    2

8 −34 hc ( 6.63  10 J s )( 3.00  10 m s ) = = 8.2  10−5 m = −21 2.429  10 J E (b) For ℓ = 2 to ℓ = 1: 2   1 −2 E = 2 = 2 ( 2.429  10−21 J )   = 3.0  10 eV −19 I 1.60 10 J eV   

=

8 −34 hc ( 6.63  10 J s )( 3.00  10 m s ) = = 4.1  10−5 m E 2 ( 2.429  10−21 J )

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Molecules and Solids

  1 −2 = 3 ( 2.429  10−21 J )   = 4.6  10 eV −19 I  1.60  10 J eV  2

E = 3

8 −34 hc ( 6.63  10 J s )( 3.00  10 m s ) = = = 2.7  10−5 m E 3 ( 2.429  10−21 J )

15. Use the value of the rotational inertia as calculated in Example 40–5. We also use Eq. 40–3. 𝛥𝐸 =

ℏ2

ℓ=

𝐼

(6.63×10−34 𝐽·𝑠)

4𝜋2 (1.46×10−46 kg·𝑚2 )

(5) = 3.813 × 10−22 𝐽

−34 8𝑚 𝑐 ℎ𝑐 ℎ𝑐 (6.63 × 10 𝐽 · 𝑠) (3.00 × 10 𝑠 ) 𝜆= = = = = 5.22 × 10−4 𝑚 (3.813 × 10−22 𝐽) 𝑓 ℎ𝑓 𝛥𝐸

16. We first find the energies of the transitions represented by the wavelengths. 6.63  10−34 J s 3.00  108 m/s hc E1 = = = 5.38  10−5 eV −19 −3  1.60  10 J/eV 23.1 10 m

( )( ) ( )( ) hc ( 6.63  10 J s )( 3.00  10 m/s ) E = = = 10.72  10 eV  (1.60  10 J/eV )(11.6  10 m ) hc ( 6.63  10 J s )( 3.00  10 m/s ) E = = = 16.12  10 eV  (1.60  10 J/eV )( 7.71  10 m ) −34

−5

−19

−34

−3

−5

−19

−3

E2 10.72 E3 16.12 =  2 and =  3, from the energy levels indicated in Fig. 40–17, and E1 E1 5.38 5.38 from the selection rule that 𝛥ℓ = ±1, we see that these three transitions must represent the ℓ = 1 to ℓ = 0 transition, the ℓ = 2 to ℓ = 1 transition, and the ℓ = 3 to ℓ = 2 transition. Thus, E1 = 2 I . We use that relationship along with Eq. 40–4 to find the bond length.

Since

E1 =

r2

E1

→

(1.055  10

−34

J s)

 ( 22.990 u )( 35.453u )  −27 −5 −19   (1.66  10 kg u )( 5.38  10 eV )(1.60  10 J eV )  ( 22.990 u + 35.453u ) 

= 2.36  10−10 m

17. The longest wavelength emitted will be due to the smallest energy change. From Fig. 40–17, the smallest rotational energy change is E = 2 I . We find the rotational inertia from Eq. 40–4.

E =

→ I  hc 4 2 c  m1m2  2 = 2I=  r h  m1 + m2 

4 2 ( 3.00  108 m s )  ( 6.941u )(1.008u )  2 −27 −9 =   (1.66  10 kg u )( 0.16  10 m ) −34 ( 6.63 10 J s )  6.941u + 1.008u  = 6.7  10−4 m

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Instructor Solutions Manual

18. The energy change for transitions between combined rotational and vibrational states is given above Eq. 40–8a. E = Evib + Erot = hf + Erot If E = hf is to be in the spectrum, then Erot = 0. But, the selection rules state that 𝛥ℓ = ±1 for a transition. It is not possible to have 𝛥ℓ = 0 for a transition. The only way to have Erot = 0 is for 𝛥ℓ = 0, which is forbidden. Thus, E = hf is not possible. Here is a mathematical statement as well. 𝛥𝐸rot = 𝐸(ℓ+𝛥ℓ) − 𝐸ℓ = (ℓ + 𝛥ℓ)(ℓ + 𝛥ℓ + 1)

ℏ2 2𝐼

− ℓ(ℓ + 1)

ℏ2 2𝐼

= 𝛥ℓ(2ℓ + 𝛥ℓ + 1)

ℏ2 2𝐼

For Erot = 0, mathematically we must have𝛥ℓ = 0, which is forbidden. 19. (a) The reduced mass is defined in Eq. 40–4. (12.01u )(16.00 u ) = 6.86 u m m = C O = mC + mO (12.01u ) + (16.00 u ) (b) We find the effective spring constant from Eq. 40–5. 1 k f = → 2 

k = 4 f 2  = 4 2 ( 6.42  1013 Hz ) ( 6.86 u ) (1.66  10−27 kg u ) = 1850 N m 2

The spring constant for H 2 is estimated in Example 40–6 as 550 N/m. kCO 850 N m = = 3.4 kH2 550 N m 20. The effective spring constant can be found from Eq. 40–5 using the vibrational frequency and the reduced mass. 1 k mm → k = 4 f 2  = 4 2 f 2 1 2 f = 2  m1 + m2 = 4 2 (1.7  1013 Hz )

( 6.941u )( 79.904 u ) 1.66  10−27 kg u = 120 N m ( ) ( 6.941u + 79.904 u )

21. (a) The curve for the function 2 U ( r ) = 12 k ( r − r0 ) − 4.5eV is shown in Fig. 40– 18 as a dotted line. Measuring on the graph in the textbook with a ruler gives the distance from the origin to the 0.1 nm mark as 38 mm. The measured distance from the origin to the largest r-intercept of the parabola is 45 mm. Taking a ratio gives the distance from the origin to the largest r-intercept as 0.118 nm. We fit a parabolic curve to data.

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Chapter 40

Molecules and Solids

U = 12 k ( r − r0 ) − 4.5eV ; U r = r0 = 12 k ( 0 ) − 4.5eV = −4.5eV ( check ) 2

U r = 0.118nm = 12 k ( 0.118nm − r0 ) − 4.5eV = 0 → 2

2 ( 4.5eV )

eV  1.60  10−19 J  109 nm  k= = 4649    = 743.8 N m 2 nm 2  eV ( 0.118nm − 0.074 nm )  1m   740 N m (b) The frequency of vibration is given by Eq. 14–7a using the reduced mass. Use this relationship to find the wavelength. 1 k c = → f = 2  

 = 2 c

 k

= 2 ( 3.00  10 m s ) 8

0.5 (1.00794 u ) (1.66  10−27 kg u ) 743.8 N m

22. Consider the system in equilibrium to find the center of mass. See the first diagram. The dashed line represents the location of the center of mass.

ℓ = ℓ1 + ℓ2 ; 𝑚1 ℓ1 = 𝑚2 ℓ2 = 𝑚2 (ℓ − ℓ1 ) → 𝑚2 𝑚1 ℓ ; ℓ2 = ℓ 𝑚1 + 𝑚2 𝑚1 + 𝑚2 Now let the spring be stretched to the left and right, but let the center of mass be unmoved. ℓ1 =

= 2.0  10−6 m

m2 1

𝑥 = 𝑥1 + 𝑥2 ; 𝑚1 (ℓ1 + 𝑥1 ) = 𝑚2 (ℓ2 + 𝑥2 ) → 𝑚1 ℓ1 + 𝑚1 𝑥1 = 𝑚2 ℓ2 + 𝑚2 𝑥2 → 𝑚1 𝑥1 = 𝑚2 𝑥2 This is the second relationship requested in the problem. Now use the differential relationships. d 2x d 2x d 2 x1 k d 2 x2 k m1 21 = −kx ; m2 22 = −kx → x ; x = − =− dt dt dt 2 m1 dt 2 m2 d 2 ( x1 + x2 )  1 d 2 x1 d 2 x2 m1 + m2 k d 2x 1   kx kx x + = − + → = − = − → = − kx    dt 2 dt 2 dt 2 m1m2 dt 2  m1 m2 

This last equation is the differential equation for simple harmonic motion, as in Eq. 14–3, with m 1 k , which is the same as Eq. 40–5. replaced by . The frequency is given by Eq. 14–7a, f = 2  23. The ionic cohesive energy is given right after Eq. 40–9 and derived in the solution to Problem 28. The text states that the Madelung constant is 1.75 for NaCl. ( 2.30  10−28 J m )  e2  1 U0 = − 1 1.75 − = − ( ) (1 − 18 ) = −7.9eV   −9 −19 4  r0  m  0.28  10 m 1.60  10 J eV ) ( )(

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

24. Each ion is at the corner of a cube of side length s, the distance between ions. From Fig.40–24, a cube of side length s would have 4 NaCl pairs. But, each ion is part of 8 cubes that share a common corner. Thus, any one “cube” has only the equivalent of one-half of an NaCl molecule. Use this to find the density, which is mass per unit volume. 1 (m ) →  = 2 NaCl s3 1/3

 1m  s =  2 NaCl    

1/3

 ( 1 molecule cube ) ( 58.44g mole )  1mole  = 2   23  6.022  10 molecule    ( 2.165g cm3 )

= 2.819  10−8 m Note that Problem 26 quotes this value as 2.4  10−8 m . 25. Each ion is at the corner of a cube of side length s, the distance between ions. From Fig. 40–24, a cube of side length s would have 4 KCl pairs. But, each ion is part of 8 cubes that share a common corner. So, any one “cube” has only the equivalent of one-half of a KCl molecule. Use this to find the density, which is mass per unit volume. From the periodic table, the molecule weight of KCl is 74.55. 1 (m )  = 2 3KCl → s 1/3

 1m  s =  2 KCl    

1/3

 ( 1 molecule cube )( 74.55g mole )  1mole  = 2   23  6.02  10 molecule   (1.99 g cm3 ) 

= 3.15  10−8 m

26. According to Section 40–6, the NaCl crystal is face-centered cubic. It is illustrated in Fig. 40–24. We consider four of the labeled ions from Fig. 40–24. See the adjacent diagram. The distance from an Na ion to a Cl ion is labeled as d, and the distance from an Na ion to the nearest neighbor Na ion is called D. D = d 2 = ( 0.24 nm ) 2 = 0.34 nm 27. See the diagram. Select a charge in the middle of the + – + chain. There will be two charges of opposite sign a distance r away, two charges of the same sign a distance 2r away, etc. Calculate the potential energy of the chosen charge. 1  −2e 2  1  +2e 2  1  −2e 2  1  +2e 2  U= + + +        + 4 0  r  4 0  2r  4 0  3r  4 0  4 r 

=−

2e 2 (1 − 12 + 13 − 14 + 4 0 r

–

)

From Appendix A, ln (1 + x ) = x − 12 x + 13 x − 14 x + 2

ln (1 + 1) = 1 − 12 + 13 − 14 +

–

= ln 2 → U = −

. Evaluate this expansion at x = 1. 2

2e (1 − 12 + 13 − 14 + 4 0 r

)=−

2e 2 ln 2 4 0 r

1340

–

Chapter 40

Molecules and Solids

From Section 40–6, the potential energy is given as U = −

 e2 . Equate the two expressions for the 4 0 r

potential energy to evaluate the Madelung constant. 2e 2  e2 U =− ln 2 →  = 2ln 2 =− 4 0 r 4 0 r 28. (a) Start with Eq. 40–9 and find the equilibrium distance, which minimizes the potential energy. Call that equilibrium distance r0 .

U =−

  e2  e2 B  e2 dU B  B m + m ; = − = − m m +1 = 0 → m +1  2 2 dr r = r  4 0 r r  r = r 4 0 r0 r0 4 0 r r 0

 er 4 0 m

2 m −1 0

U 0 = U ( r = r0 ) = −

 e  e B + =− + 4 0 r0 r0m 4 0 r0 2

 e2 r0m −1 4 0 m r0m

=−

 e2  1  1 −  4 0 r0  m 

(b) For NaI, we evaluate U 0 with m = 10,  = 1.75, and r0 = 0.33nm.

U0 = −

2.30  10−28 J m  e2  1  (1 − 1 ) = −6.861eV  1 −  = − (1.75) 4 0 r0  m  ( 0.33  10−9 m )(1.60 10−19 J eV ) 10

 −6.9eV (c) For MgO, we evaluate U 0 with m = 10,  = 1.75, and r0 = 0.21nm.

U0 = −

2.30  10−28 J m  e2  1  − = − 1 1.75 ( ) (1 − 101 ) = −10.78eV   −9 −19 4 0 r0  m    0.21 10 m 1.60 10 J eV ( )( )

 −11eV (d) Calculate the % difference using m = 8 instead of m = 10.  e2  e2 − U0 − U0 (1 − 18 ) − − (1 − 101 ) 1 − 1 − 1 − 1 ( 8 ) ( 10 ) = 101 − 18 = −0.0278 4 0 r0 4 0 r0 m =8 m =10 = = 2  e U0 (1 − 101 ) (1 − 101 ) − (1 − 101 ) m =10 4 0 r0

 −2.8% 29. We follow Example 40–9. The density of states (number of states per unit volume in an infinitesimal energy range) is given by Eq. 40–10. Because we are using a small energy range, we estimate the calculation with a difference expression. We let N represent the number of states and V represent the volume under consideration. 8 2 m 3 / 2 1/ 2 N  g ( E )V E = E V E h3

8 2 ( 9.11  10−31 kg )

( 6.63  10 J s ) −34

3/ 2

( 7.025eV ) (1.0  10−6 m3 ) ( 0.05eV ) (1.60  10−19 J eV )

3/ 2

= 9.0  1020 states © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

30. We follow Example 40–9. The density of occupied states (number of states per unit volume in an infinitesimal energy range) is given by Eq. 40–10. Because we are using a small energy range, we estimate the calculation with a difference expression. We let N represent the number of states and V represent the volume under consideration. 8 2 m3/ 2 1/ 2 N  g ( E )V E = E V E h3 E = 12 ( EF + 0.975EF ) = 0.9875EF ; E = EF − 0.975EF = 0.025EF = 0.137 eV N=

8 2 ( 9.11  10−31 kg )

( 6.63  10 J s ) −34

3/ 2

0.9875 ( 5.48eV ) (1.0  10−6 m3 ) ( 0.137 eV ) (1.60  10−19 J eV )

3/ 2

= 2.2  1021 states

31. The density of molecules in an ideal gas can be found from the ideal gas law. P 1.013  105 Pa N PV = NkT →   = = = 2.576  1025 m −3 1.38  10−23 J K ( 285K )  V gas kT

(

)

We assume that each copper atom contributes one free electron and use the density of copper as given in Table 13–1. 23 3 N  1e   6.02  10 Cu atoms  8.9  10 kg  28 −3 =   = 8.431  10 m     3 −3 m  V e's  Cu atom   63.546  10 kg   Compare the two densities. N    V gas 2.576  1025 m −3 = = 3.1  10−4 8.431  1028 m −3 N    V e's 32. We use Eq. 40–14 for the occupancy probability, and solve for the energy. The Fermi energy is 7.0 eV. (a) Evaluate for T = 295 K. 1 f ( E ) = ( E − EF ) kT → e +1

 1  (1.38  10−23 J K ) ( 295K ) ln  1 − 1 + 7.0eV E = kT ln  − 1 + EF =    0.150  (1.60  10−19 J eV )  f (E)  = 7.04eV

(b) Evaluate for T = 950 K. 1 f ( E ) = ( E − EF ) kT → e +1

1.38  10−23 J K )  1  ( 1  E = kT ln  − 1 + EF = − 1 + 7.0eV ( 950 K ) ln  −19  0.150  (1.60  10 J eV )  f (E)  = 7.14eV

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Chapter 40

Molecules and Solids

33. We use Eq. 40–14 for the occupancy probability, and solve for the energy. The Fermi energy is 7.0 eV. (a) Evaluate for T = 295 K. 1 f ( E ) = ( E − EF ) kT → e +1  1  (1.38  10−23 J K ) ( 295K ) ln  1 − 1 + 7.0eV E = kT ln  − 1 + EF =    0.850  (1.60  10−19 J eV )  f (E)  = 6.96eV

(b) Evaluate for T = 750 K. 1 f ( E ) = ( E − EF ) kT → e +1

 1  (1.38  10−23 J K ) ( 750 K ) ln  1 − 1 + 7.0eV E = kT ln  − 1 + EF =    0.850  (1.60  10−19 J eV )  f (E)  = 6.89eV

34. The occupancy probability is given by Eq. 40–14. The Fermi level for copper is 7.0 eV. 1 1 1 f = ( E − EF ) / kT = (1.015 EF − EF ) / kT =  = 0.0159 0.015( 7.0eV )(1.6010−19 J eV ) (1.3810−23 J K )( 295 K ) e +1 e + 1 e    +1

 1.6% 35. We follow Example 40–10. We need the number of conduction electrons per unit volume of sodium.  6.02  1023 atoms mol  N 28 −3 = ( 970kg m3 )   (1free electrons atom ) = 2.540  10 m −3 V 22.99 10 kg mol    The Fermi energy is given by Eq. 40–12.

h2  3 N  EF =   8m   V 

2/3

( 6.63 10 J s )  3 2.540  10 m  = ( ) 8 ( 9.11  10 kg )   −34

2/3

−3

−31

  1   = 3.159eV −19  1.60  10 J eV 

 3.2eV The Fermi speed is the speed of electrons with the Fermi energy.

2 ( 3.159eV ) (1.60  10−19 J eV ) 2 EF vF = = = 1.1  106 m s −31 m  9.11 10 kg ( )

36. We follow Example 40–10. (a) Because each zinc atom contributes two free electrons, the density of free electrons is twice the density of atoms.  6.02  1023 atoms mol  N 29 −3 = ( 7100 kg m 3 )   ( 2free electrons atom ) = 1.307  10 m −3 V 65.409  10 kg mol  

 1.3  1029 m −3 (b) The Fermi energy is given by Eq. 40–12.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

h2  3 N  EF =   8m   V 

2/3

Instructor Solutions Manual

( 6.63  10 J s )  3 1.307  10 m  = ( ) 8 ( 9.11  10 kg )   2

−34

2/3

−3

−31

  1   −19  1.60  10 J eV 

= 9.414eV  9.4eV (c) The Fermi speed is the speed of electrons with the Fermi energy.

vF =

2 ( 9.414eV ) (1.60  10−19 J eV ) 2 EF = = 1.8  106 m s −31 m ( 9.1110 kg )

37. (a) Find the density of free electrons from Eq. 40–12. 2/3 h2  3 N  EF = →   8m   V  N   8mEF  =  2  3 h  V

3/ 2

−31 −19   8 ( 9.11  10 kg ) (11.63eV ) (1.60  10 J eV ) 

( 6.63  10 J s )

3 

−34

3/ 2

 

= 1.7945  1029 m −3  1.79  1029 m −3 (b) Let n represent the valence number, so there are n free electrons per atom.  6.02  1023 atoms mol  N = ( 270 kg m 3 )   ( n free electrons atom ) → −3 V  27.0  10 kg mol 

 27.0  10−3 kg mol   1 n= 1.7945  1029 m −3 ) = 2.981  3  23 3 (  6.02 10 atoms mol 270 kg m    This agrees nicely with aluminum’s position in the periodic table, and its electron configuration of 1s22s22p63s23p1. The level 3 electrons are the valence electrons. 38. We calculate the given expression with T = 0 so that the maximum energy is EF. The value of f (E) at T = 0 is given below Eq. 40–14. EF EF EF EF  8 2 m 3 / 2 1/ 2  1 E E dE ( ) En E dE Eg E f E dE E 3 / 2dE ( ) ( ) ( )   0  h 3 0 0 0   = EF = EF = E0F E = EF  8 2 m 3 / 2 1/ 2  1/ 2 0 n0 ( E ) dE 0 g ( E ) f ( E ) dE 0  h3 E  (1) dE 0 E dE   EF5 / 2 3 = 5 EF 3/ 2 3 EF

= 25

39. We start with Eq. 38–13 for the energy level as a function of n. If we solve for n, we have the number of levels with energies between 0 and E. Taking the differential of that expression will give the number of levels with energies between E and dE. Finally, we multiply by 2 since there can be 2 electrons (with opposite spins) in each energy level. 𝐸 = 𝑛2

ℎ2 8𝑚ℓ2

→ 𝑛=√

8𝑚ℓ2 ℎ2

8𝑚ℓ2 1 𝑑𝐸

√𝐸 → 𝑑𝑛 = √ ℎ2 √

𝑔ℓ = 2

𝑑𝑛 =2 𝑑𝐸

2 √𝐸

8𝑚ℓ2 1 𝑑𝐸 ℎ2 2 √𝐸 𝑑𝐸

→

= √

8𝑚ℓ2 ℎ2 𝐸

1344

Chapter 40

Molecules and Solids

40. We first find the density of neutrons, and then use Eq. 40–12.   N  1 neutron  1  2.5 (1.99  1030 kg )   = = 4.103  1044 m −3   3 −27   4 V  1.675  10 kg   12,000 m )   3 (

h2  3 N  EF =   8m   V 

2/3

( 6.63  10 J s )  3 4.103  10 m  = ( ) 8 (1.675  10 kg )   2

−34

−27

−3

2/3

  1   −13  1.60  10 J MeV 

= 109.8 MeV  110 MeV 41. We use Eq. 40–14, with E = EF . 1 1 1 1 f = ( E − EF ) / kT = 0 = = e +1 e +1 1+1 2 The result is independent of the value of T. 42. (a) We use Eq. 40–14 with the data as given in the problem. −19 E − EF ( 0.12eV ) (1.60  10 J eV ) = = 4.74848 kT k (1.38  10−23 J K ) ( 293K ) 1

1 = 8.590  10−3  8.6  10−3 + 1 116.409 e +1 e This is reasonable. Very few states this far above the Fermi energy are occupied at this relatively low temperature. (b) Use a similar calculation to part (a). −19 E − EF ( −0.12eV ) (1.60  10 J eV ) = = −4.74848 kT k (1.38  10−23 J K ) ( 293K ) f =

( E − EF ) / kT

4.74848

1 = 0.991409  0.99 + 1 1.008665 e +1 e (c) Since the probability that the state is occupied is 0.991409, the probability that the state is f =

( E − EF ) / kT

−4.74848

unoccupied is 1 − 0.991409 = 8.591 10−3  8.6 10−3 . This is the same as part (a). 43. (a) Eq. 38–13 gives the energy levels as a function of n, the number of levels. Since there are 2 electrons in every energy level, n = N / 2. The Fermi energy will be the highest energy level occupied with N electrons. 𝐸𝑛 = 𝑛2

ℎ2

8𝑚ℓ2

= ( 𝑁)

ℎ2 8𝑚ℓ2

ℎ2 𝑁 2 32𝑚ℓ2

→ 𝐸𝐹 =

ℎ2 𝑁 2 32𝑚ℓ

(b) The smallest amount of energy that this metal can absorb is the spacing between energy levels. 𝛥𝐸 = 𝐸𝑛+1 − 𝐸𝑛 =

ℎ2 8𝑚ℓ2

[(𝑛 + 1)2 − 𝑛2 ] =

ℎ2 8𝑚ℓ2

(2𝑛 + 1) =

ℎ2 8𝑚ℓ2

(𝑁 + 1)

ℎ2 (𝑁+1) 8𝑚ℓ2 ℎ2 𝑁2 32𝑚ℓ2

4 𝑁

For large N, this is a very small change in energy. Thus, a very small change in energy will allow an electron to change energy levels, and so the metal conducts very easily. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1345

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

44. We consider the cube to be a three-dimensional infinite well, with a width of ℓ in each dimension. We apply the boundary conditions as in Section 38–8 separately to each dimension. Each dimension gives a quantum number which we label as n1 , n2 , and n3 . We then have a contribution to the energy of the bound particle from each quantum number, as in Eq. 38–13. 𝐸 = 𝐸1 + 𝐸2 + 𝐸3 = 𝑛12

ℎ2 ℎ2 ℎ2 ℎ2 2 2 (𝑛2 + 𝑛22 + 𝑛32 ), 𝑛1 , 𝑛2 , 𝑛3 = 1,2,3, ⋯ + 𝑛 + 𝑛 = 2 3 8𝑚ℓ2 8𝑚ℓ2 8𝑚ℓ2 8𝑚ℓ2 1

Specifying the three quantum numbers gives a state and the corresponding energy. Choosing axes as specified in the problem, the equation of a sphere of radius R in that coordinate system is R 2 = n12 + n22 + n32 . Each state “contained” in that sphere could be indicated by a cube of side length ℓ, and each state can have two electrons (two spin states). The “volume” of that sphere is 18 of a full sphere. From that we calculate the number of states in one octant, and then g(E). 1 4

𝜋 8𝑚ℓ2

𝜋

𝑁 = 2 ( ) 𝜋𝑅3 = (𝑛12 + 𝑛22 + 𝑛32 )3/2 = ( 8 3

2 3/2

𝑔 (𝐸 ) =

1 𝑑𝑁 1 𝜋 8𝑚ℓ = 3 ( 2 ) 𝑉 𝑑𝐸 ℓ 3 ℎ

ℎ

3/2

𝐸)

3 1/2 𝜋 8𝑚 3/2 1/2 8√2𝜋𝑚3/2 1/2 𝐸 = ( 2) 𝐸 = 𝐸 2 2 ℎ ℎ3

45. The photon with the minimum frequency for conduction must have an energy equal to the energy gap. −34 8 hc ( 6.63  10 J s )( 3.00  10 m/s ) E g = hf = = = 2.0eV  (1.60  10−19 J/eV )( 620  10−9 m ) 46. The photon with the longest wavelength or minimum frequency for conduction must have an energy equal to the energy gap. −34 8 c hc hc ( 6.63  10 J s )( 3.00  10 m/s ) = = = 1.11  10−6 m = 1.11  m = = f hf E g (1.60  10−19 J/eV ) (1.12eV ) 47. The energy of the photon must be greater than or equal to the energy gap. The lowest energy corresponds to the longest wavelength that will excite an electron. −34 8 c hc hc ( 6.63  10 J s )( 3.00  10 m/s ) = = = 1.7  10−6 m = 1.7  m = = f hf E g (1.60  10−19 J/eV ) ( 0.72eV ) Thus, the wavelength range is   1.7  m. 48. The minimum energy provided to an electron must be equal to the energy gap. Divide the total available energy by the energy gap to estimate the maximum number of electrons that can be made to jump. 3 hf ( 710  10 eV ) N= = = 9.9  105 Eg ( 0.72eV ) 49. (a) In the 2s shell of an atom, ℓ = 0, so there are two states: ms =  12 . When N atoms form bands, each atom provides two states, so the total number of states in the band is 2N. (b) In the 2p shell of an atom, ℓ = 1, so there are three states from the 𝑚ℓ values: m = 0,  1; each © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1346

Chapter 40

Molecules and Solids

of which has two states from the ms values: ms =  12 , for a total of six states. When N atoms form bands, each atom provides six states, so the total number of states in the band is 6N. (c) In the 3p shell of an atom, ℓ = 1, so there are three states from the ml values: 𝑚ℓ = 0, ±1; each of which has two states from the ms values: ms =  12 , for a total of six states. When N atoms form bands, each atom provides six states, so the total number of states in the band is 6N. (d) In general, for a value of ℓ, there are 2ℓ + 1 states from the 𝑚ℓ values: 𝑚ℓ = 0, ±1, . . . , ±𝑙. For each of these there are two states from the ms values: ms =  12 , for a total of 2(2ℓ + 1) states. When N atoms form bands, each atom provides 2(2ℓ + 1) states, so the total number of states in the band is 2𝑁 (2ℓ + 1). 50. Calculate the number of conduction electrons in a mole of pure silicon. Also calculate the additional conduction electrons provided by the doping, and then take the ratio of those two numbers of conduction electrons.  ( 28.09  10−3 kg mol )  16 N Si =   (10 electrons m3 ) = 1.206  1011 electrons mole 3 ( 2330 kg m )  

( 6.02  10 atoms ) = 4.631 10 added conduction electrons per mole. = 23

N doping

N doping NSi

1.3  106

( 4.63110 ) = 3.84 10  4 10 (1.206 10 ) 17

51. The wavelength is found from the energy gap. 6.63  10−34 J s 3.00  108 m/s c hc hc = = 6.91  10−7 m  690nm = = = f hf Eg 1.60  10−19 J/eV (1.8eV )

(

)(

(

)

52. The photon will have an energy equal to the energy gap: 6.63  10−34 J s 3.00 108 m/s hc Eg = hf = = = 2.0eV  1.60  10−19 J/eV 630 10−9 m

( (

)( )(

) )

53. From the current–voltage characteristic graph in Fig. 40–38, we see that a current of 16 mA means a voltage of about 0.69 V across the diode. The battery voltage is the sum of the voltages across the diode and the resistor. Vbattery = Vdiode + VR = 0.69V + ( 0.016A )( 860  ) = 14.45V  14V 54. (a) For a half-wave rectifier without a capacitor, the current is zero for half the time. We approximate the average current as half of the full rms current. (120V ) = 1.4mA V I av = 12 rms = 12 R ( 42k ) (b) For a full-wave rectifier without a capacitor, the current is positive all the time. We approximate the average current as equal to the full rms current. (120V ) = 2.9mA V I av = rms = R ( 42k ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1347

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

55. The battery voltage is the sum of the voltages across the diode and the resistor. Vbattery = Vdiode + VR ; Vdiode 2.0 V + 120  120  This is the equation for a straight line which passes through the points (0 V, 16.7 mA) and (0.8 V, 10 mA). The line has a y-intercept of 16.7 mA and a slope of 8.33 mA/V. If we assume the operating voltage of the diode is about 0.7 V, then the current is given from the equation above. 0.7 V 2.0 V I =− + = 1.08  10−2 A  11mA 120  120  There is some approximation involved in the answer. 2.0 V = Vdiode + I (120Ω ) → I = −

56. The band gap is the energy corresponding to the emitted wavelength. 6.63  10−34 J s 3.00 108 m s hc E= = = 1.13eV  1.1eV  1.1 10−6 m 1.60 10−19 J eV

( (

)(

) )

57. We have copied the graph for V > 0 and rotated it so that it shows V as a function of I. This is the first diagram below. The resistance is the slope of that first graph. The slope, and thus the resistance, is very high for low currents and decreases for larger currents, approaching 0. As an approximate value, we see that the voltage changes from about 0.55 V to 0.65 V as the current goes from 0 to 10 mA. That makes the resistance about 10 ohms when the current is about 5 mA. The second diagram is a sketch of the resistance. 0.8 V ( volts ) 0.6

0.4 0.2

I ( mA ) 10

58. There will be a current in the resistor while the ac voltage varies from 0.6 V to 9.0 V rms. Because the 0.6 V is small, the voltage across the resistor will be almost sinusoidal, so the rms voltage across the resistor will be close to 9.0V − 0.6V = 8.4 V. (a) For a half-wave rectifier without a capacitor, the current is zero for half the time. We ignore the short time it takes for the voltage to increase from 0 to 0.6 V, and so current is flowing in the resistor for about half the time. We approximate the average current as half of the full rms current. V 8.4 V I av = 12 rms = 12 = 28mA R 0.150 kΩ (b) For a full-wave rectifier without a capacitor, the current is positive all the time. We ignore the short times it takes for the voltage to increase from 0 to 0.6 V, and so current is flowing in the resistor all the time. We approximate the average current as the full rms current. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1348

Chapter 40

Molecules and Solids

I av =

Vrms 8.4 V = = 56 mA R 0.150 kΩ

(

)(

)

59. (a) The time constant for the circuit is  1 = RC1 = 24  103  35  10−6 F = 0.84s. As seen in Fig. 40–40(c), there are two peaks per cycle. The period of the rectified voltage is 1 T = 120 s = 0.0083s. Because  1 T , the voltage across the capacitor will be essentially constant during a cycle, so the average voltage is the same as the peak voltage. The average current is basically constant. Vavg Vpeak 2 (120 V ) 2Vrms I avg = = = = = 7.07 mA  7.1mA R R R ( 24 103  ) (b) With a different capacitor, the time constant for the circuit changes.  2 = RC2 = 24  103  0.10  10−6 F = 0.0024s

(

)(

)

Now the period of the rectified voltage is about 3.5 time constants, and so the voltage will decrease to about 3% e −3 of the peak value during each half-cycle. We approximate the

( )

voltage as dropping linearly from its peak value to 0 over each half-cycle, and so take the average voltage as half the peak voltage. Vavg 1 2Vrms 1 2 (120 V ) I avg = =2 =2 = 3.54 mA  3.5mA R R ( 24 103  ) 60. For a pnp transistor, both the collector and the base voltages are negative, and holes move from the emitter to the collector. The diagram for a pnp amplifier looks just like Fig. 40–49, with the polarity of eB and eC reversed, I B and I C flowing in opposite directions, and the emitter arrow pointing toward the base.

61. By Ohm’s law, the ac output (collector) current times the output resistor will be the ac output voltage. V V 0.40V Vout = iC RC → RC = out = out = = 4211  4200  iC  I iB 95 (1.0  10−6 A ) 62. (a) The voltage gain is the collector ac voltage divided by the base ac voltage.  7.8k  V i R R V = C = C C =  I C = 65   = 133.4  130 VB iB RB RB  3.8k  (b) The power amplification is the output power divided by the input power. iV  P = C C =  I V = ( 65)(133.4 ) = 8672  8700 iBVB 63. By Ohm’s law, the ac output (collector) current times the output resistor will be the ac output voltage. V 75 ( 0.080V ) V = 2.4  10−4 A = 0.24 mA Vout = iC R → iC = out = V input = 3 25  10  R R © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1349

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

64. By Ohm’s law, the ac output (collector) current times the output resistor will be the ac output voltage. Vout = iC RC =  I iB RC = ( 85) 2.0  10−6 A ( 3900  ) = 0.663V  0.66V

(

)

65. The arrow at the emitter terminal, E, indicates the direction of current I E . The current into the transistor must equal to the current out of the transistor. I B + IC = I E 66. (a) The potential energy for the point charges is found from Eq. 23–10. ( −1.60 10−19 C )(1.60 10−19 C ) 1 e2 = ( 9.0  109 N m 2 C2 ) U= 4 0 r ( 0.27 10−9 m )(1.60 10−19 J eV ) = − 5.33eV  − 5.3eV

(b) Because the potential energy of the ions is negative, 5.33 eV is released when the ions are brought together. The other energies quoted involve the transfer of the electron from the K atom to the F atom. 3.41 eV is released, and 4.34 eV is absorbed in the individual electron transfer processes. Thus, the total binding energy is as follows. Binding energy = 5.33eV + 3.41eV − 4.34eV = 4.4eV 67. We find the temperature from the given relationship. −19 2 K 2 ( 4.0eV ) (1.60  10 J eV ) (a) K = 23 kT → T = = = 3.1  104 K 3k 3 (1.38  10−23 J K ) (b) K = 23 kT → T =

−19 2 K 2 ( 0.12 eV ) (1.60  10 J eV ) = = 930 K 3k 3 (1.38  10−23 J K )

68. For an electron confined in one dimension, we find the uncertainty in the momentum from Eq. 38–1, p 

x

. The momentum of the electron must be at least as big as the uncertainty in the momentum,

so we approximate p 

. Finally, we calculate the kinetic energy by K =

x between the two kinetic energies based on the two position uncertainties. 2 p2 K= = 2m 2m ( x )2

K = K in atoms − K molecule =

  1 1 −   2 2 2m  ( x ) ( x )molecule  in atoms  2

1.055  10 J s )  ( 1  = 2 ( 9.11  10 kg )  ( 0.053  10 m )  −34

−31

p2 . Find the difference 2m

−9

   1  −   2 2 −19 −9 ( 0.074  10 m )molecule   1.60  10 J eV  in atoms 1

= 6.62eV There are two electrons, and each one has this kinetic energy difference, so the total kinetic energy difference is 2 ( 6.62eV ) = 13.2eV  13eV . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1350

Chapter 40

Molecules and Solids

69. The diagram here is similar to Fig. 40–9 and Fig. 40–11. The activation energy is the energy needed to get the (initially) stable U system over the barrier in the potential energy. The activation energy is 1.4 eV for this molecule. The dissociation energy is the energy that is released when the bond is broken. The dissociation energy is 1.6 eV for this molecule. 70. Vibrational states have a constant energy difference of Evib = 0.54eV, as found in Example 40–7. ℏ2 ℓ

Rotational states have a varying energy difference, depending on the ℓ value, of 𝛥𝐸rot = , where l 𝐼 represents the upper energy state, as given in Eq. 40–3.

(1.055 10 J s ) E = = = r I ( 0.5)(1.008u ) (1.66 10 kg u )( 0.074 10 m ) (1.60 10 2

−34

rot

2 0

−27

−9

−19

J eV )

= ( 0.0152eV ) Each gap, as represented in Fig. 40–17, is larger. We add those gaps until we reach 0.54eV. ℓ = 1 → ℓ = 0: 𝛥𝐸rot = 0.0152eV

∑ 𝛥𝐸rot = 0.0152eV

ℓ = 2 → ℓ = 1: 𝛥𝐸rot = 2(0.0152eV) ∑ 𝛥𝐸rot = 3(0.0152eV) = 0.0456eV ℓ = 3 → ℓ = 2: 𝛥𝐸rot = 3(0.0152eV) ∑ 𝛥𝐸rot = 6(0.0152eV) = 0.0912eV ℓ = 4 → ℓ = 3: 𝛥𝐸rot = 4(0.0152eV) ∑ 𝛥𝐸rot = 10(0.0152eV) = 0.152eV ℓ = 5 → ℓ = 4: 𝛥𝐸rot = 5(0.0152eV) ∑ 𝛥𝐸rot = 15(0.0152eV) = 0.228eV ℓ = 6 → ℓ = 5: 𝛥𝐸rot = 6(0.0152eV) ∑ 𝛥𝐸rot = 21(0.0152eV) = 0.3192eV ℓ = 7 → ℓ = 8: 𝛥𝐸rot = 7(0.0152eV) ∑ 𝛥𝐸rot = 28(0.0152eV) = 0.4256eV ℓ = 8 → ℓ = 7: 𝛥𝐸rot = 8(0.0152eV) ∑ 𝛥𝐸rot = 36(0.0152eV) = 0.5472eV So, we see that rotational states from ℓ = 0 to ℓ = 7 can be “between” vibrational states, or a total of 8 rotational states.

2 71. The kinetic energy of the baton is 12 I  , and the quantum number can be found from Eq. 40–2. Let the length of the baton be d. We assume the quantum number will be very large. The rotational inertia about the center of mass is the sum of the inertias for a uniform rod 121 mbar d 2 and two point

(

masses 2mend ( 12 d )

)

1351

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

1 2

2 2 l+ ) 2 ( I =  2

I

Instructor Solutions Manual

→

2 f 2 =  2mend ( 12 d ) + 121 mbar d 2  =  

2 (1.6s −1 ) 2 2 1   = 2 ( 0.35kg )( 0.16 m ) + 12 ( 0.26 kg )( 0.32 m ) = 1.92  1033   (1.055  10−34 J s )

The spacing between rotational energy levels is given by Eq. 40–3. We compare that value to the rotational kinetic energy. ℓℏ2 𝛥𝐸 2 2 = 2𝐼 2 = = = 1.0 × 10−33 𝐸 2ℓ (1.92 × 1033 ) ℓ ℏ 2𝐼 This is such a small difference that it would not be detectable, so no, we do not need to consider quantum effects. 72. (a) The equilibrium position is the location where the potential energy is a minimum. We find that location by setting the derivative of the potential energy equal to 0. 2 dU − a r −r − a r −r − a r −r − a r −r U = U 0 1 − e ( 0 )  ; = 2U 0 1 − e ( 0 )  ae ( 0 ) = 0 → 1 − e ( 0 ) = 0 dr − a ( r − r0 ) e = 1 → a ( r − r0 ) = 0 → r = r0

(

)

The dissociation energy is the energy difference between the two states of equilibrium separation and infinite separation.

U = U r = − U r =r0 = U 0 1 − e

− a ( − r0 )

 − U 0 1 − e − a ( r0 −r0 )  = U 0 1 − 02 − U 0 1 − 12 = U 0   

(b) See the included graph.

U (eV)

6 4 2 0 0.0

0.5

1.0

1.5

2.0

r/r 0

2.5

3.0

3.5

4.0

73. The Boltzmann factor indicates that the population of a state decreases as the energy of the state increases, according to N n  e − En / kT . The rotation energy of states increases with higher ℓ values, according to Eq. 40–2. Thus states with higher values of ℓ have higher energies, and so there are fewer molecules in those states. Since the higher states are less likely to be populated, they are less likely to absorb a photon. As an example, the probability of absorption between ℓ = 1 and ℓ = 2 is more likely than that between ℓ = 2 and ℓ = 3, and so the peak representing the transition between ℓ = 1 and ℓ = 2 is higher than that between ℓ = 2 and ℓ = 3. The molecule is not rigid, and so the distance between the two ions is not constant. The moment of inertia depends on the bond length, and the energy levels depend on the moment of inertia. Thus, the energy levels are not exactly equally spaced.

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74. From Fig. 40–17, a rotational absorption spectrum would show peaks at energies of 3 2 I , etc. Adjacent peaks are separated by an energy of energy to determine the rotational inertia.

E =

I, 2 2 I,

I . We use the photon frequency at that

( 6.63 10 J s ) = 1.8 10 kg m → I= = = = E hf 4 f 4 ( 9.2  10 Hz ) 2

−34

−47

75. (a) The reduced mass is defined in Eq. 40–4. (1.008u )( 35.453u ) = 0.9801u m m  = H Cl = mH + mCl (1.008u ) + ( 35.453u ) (b) We find the effective spring constant from Eq. 40–5. 1 k f = → 2 

k = 4 f 2  = 4 2 (8.66  1013 Hz ) ( 0.9801u ) (1.6605  10−27 kg u ) = 482 N m 2

The spring constant for H 2 is estimated in Example 40–6 as 550 N/m. kHCl 482 N m = = 0.88 kH2 550 N m 76. An O 2 molecule can be treated as two point masses, 16 u each, and each a distance of 6.05  10−11 m from the molecule’s center of mass. 2 I =  mr 2 = 2 (16 u ) (1.66  10−27 kg u ) ( 6.05  10−11 m ) = 1.94  10−46 kg m 2 77. From the diagram of the cubic lattice, we see that an atom inside the cube is bonded to the six nearest neighbors. Because each bond is shared by two atoms, the number of bonds per atom is 3 (as long as the sample is large enough that most atoms are in the interior, and not on the boundary surface). We find the heat of fusion from the energy required to break the bonds:  number of bonds  number of atoms  Lfusion =    Ebond atom mol    = ( 3) ( 6.02  1023 atoms mol )( 3.6  10 −3 eV )(1.60  10 −19 J eV ) = 1040 J mol  1.0  103 J mol

78. (a) We calculate the Fermi temperature for a Fermi energy of 7.0 eV, the Fermi energy of copper (as calculated in Example 40–10). −19 EF ( 7.0eV ) 1.60  10 J/eV = = 8.1  104 K TF = −23 k 1.38  10 J K

(

)

(b) We are given that T TF , and we assume that e E / kT  1. 1 1 1 1 =  E kTF   E   E  = e − E / kT f ( E ) = ( E − EF ) / kT +1 e      −  e kT kT  + 1 e kT  + 1 e kT  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

This is not useful for conductors like copper because the Fermi temperature is higher than the melting point, and we would no longer have a solid conductor. 79. From Eq. 25–13, the number of charge carriers per unit volume in a current is given by n =

I evdrift A

where vdrift is the drift velocity of the charge carriers, and A is the cross-sectional area through which e the carriers move. From Eq. 27–14, the drift velocity is given by vdrift = H , where eH is the HallBd effect voltage and d is the width of the strip carrying the current (see Fig. 27–32). The distance d is the shorter dimension on the “top” of Fig. 40–54. We combine these equations to find the density of charge carriers. We define the thickness of the current-carrying strip by t = A d .

0.28  10−3 A ) (1.3T ) ( IBd IB n= = = = = 1.264  1020 electrons m3 −19 −3 evdrift A eeH A eeH t (1.60  10 C ) ( 0.018V ) (1.0 10 m ) I

The actual density of atoms per unit volume in the silicon is found from the density and the atomic weight. We let that density be represented by N. That density is used to find the number of charge carriers per atom.    6.02  1023 atoms  1mole 28 3 N = ( 2330 kg m3 )    = 4.994  10 atoms m  −3 1mole  28.0855  10 kg  

n 1.264  1020 electrons m3 = = 2.5  10−9 electrons atom N 4.994  1028 atoms m3 80. To use silicon to filter the wavelengths, wavelengths below the IR should cause the electron to be raised to the conduction band, so the photon is absorbed in the silicon. We find the shortest wavelength that will cause the electron to jump. 6.63  10−34 J s 3.00 108 m/s c hc hc = = 1.11 10−6 m = 1.11  m = =  f hf Eg 1.60  10−19 J/eV (1.12eV )

(

)(

(

)

Because this is in the IR region of the spectrum, the shorter wavelengths of visible light will excite the electron, and the photon would be absorbed. Thus, silicon can be used as a window. 81. The longest wavelength will be for the photon with the minimum energy, which corresponds to the gap energy. −34 8 hc hc ( 6.63  10 J s )( 3.00  10 m/s ) Eg = → max = = = 3.5  10−7 m −19 Eg max ( 3.6eV ) (1.60  10 J/eV ) So, the photon must have   3.5  10−7 m . 82. We use Eq. 40–11 with the limits given in order to determine the number of states. E2

N = V  g ( E ) dE = E1

8 2 m 3 / 2 2 1/ 2 8 2 m3 / 2 2 3 / 2 V E dE V ( E2 − E13 / 2 ) = 3 E h3 h 3 1

16 2 ( 9.11  10−31 kg ) 3 ( 6.63  10−34 J s )

3/ 2

( 0.1m ) ( 6.2eV ) 3

3/ 2

− ( 4.0eV )

3/ 2

 (1.60  10−19 J eV ) 

3/ 2

= 3.365  1025  3  1025 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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83. The photon with the longest wavelength has the minimum energy, which should be equal to the gap energy. −34 8 hc ( 6.63  10 J s )( 3.00  10 m/s ) Eg = = = 1.130eV  1.1eV  (1.60  10−19 J/eV )(1100  10−9 m ) If the energy gap is any larger than this, some solar photons will not have enough energy to cause an electron to jump levels. So, they will not be absorbed, making the solar cell less efficient. 84. The energy gap is related to photon wavelength by E g = hf = hc  . Use this for both colors of LED.

( 6.63  10 J s )( 3.00  10 m/s ) = 2.37 eV (1.60  10 J/eV )(525  10 m ) ( 6.63  10 J s )( 3.00  10 m/s ) = 2.67 eV E = (1.60  10 J/eV )( 465  10 m ) −34

Green:

Eg =

−19

−9

−34

Blue:

−19

−9

85. The photon with the maximum wavelength for conduction has an energy equal to the energy gap. −34 8 hc ( 6.63  10 J s )( 3.00  10 m/s ) Eg = hf = = = 5.50eV  (1.60  10−19 J/eV )( 226  10−9 m ) 86. (a) For the glass to be transparent to the photon, the photon’s energy must be < 1.12 eV, so the wavelength of the photon must be longer than the wavelength corresponding to 1.12 eV. hc → Eband gap =

min

min =

hc Eband gap

( 6.63  10 J s )( 3.00  10 m s ) = 1.11  10 m →   1.11  10 m (1.12eV ) (1.60  10 J eV ) −34

−6

−19

The minimum wavelength for transparency is in the infrared region of the spectrum. Since IR has longer wavelengths than visible light, the silicon would not be transparent for visible light. The silicon would be opaque, as in Example 40–14. (b) The minimum possible band gap energy for light to be transparent would mean that the band gap energy would have to be larger than the most energetic visible photon. The most energetic photon corresponds to the shortest wavelength, which is 400 nm in this problem. We treat the wavelength as being accurate to 2 significant figures. ( 6.63  10−34 J s )( 3.00 108 m s ) = 3.1078eV  3.1eV hc = Eband gap  Emin = min ( 400  10−9 m )(1.60  10−19 J eV ) 87. We assume that the value of 120 mA has 3 significant figures. (a) The current through the load resistor must be maintained at a constant value. Voutput (125V ) = 5.952mA I load = = Rload ( 21.0k ) At the minimum supply voltage, there will be no current through the diode, so the current through R is also 5.952 mA. The supply voltage is equal to the voltage across R plus the output voltage. VR = I load R = ( 5.952mA )( 2.80k ) = 16.67 V

Vsupply = VR + Voutput = 16.67 V + 125V = 142V min

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Instructor Solutions Manual

At the maximum supply voltage, the current through the diode will be 120 mA, and so the current through R is I load + I diode = 5.952mA + 120mA = 125.952mA .

VR = I load R = (125.952 mA )( 2.80 k ) = 353V Vsupply = VR + Voutput = 353V + 125V = 478V max

(b) The supply voltage is fixed at 175 V, and the output voltage is still to be 125 V. The voltage across R is then fixed at 175V − 125V = 50V. We calculate the current through R.

( 50V ) = 17.9mA VR = R ( 2.80k ) If there is no current through the diode, this 17.9-mA current will be in the load resistor. Note that the value of 17.9 mA only has 2 significant figures. V 125V Rload = load = = 6.98k  7.0 k I load 17.9 mA IR =

If Rload is less than this, there will be a greater current through R, meaning a greater voltage drop across R and a smaller voltage across the load. Thus, regulation would be lost, so 7.0k is the minimum load resistance for regulation. If Rload is greater than 7.0k, the current through Rload will have to decrease in order for the voltage to be regulated, which means current must flow through the diode. The current through the diode is 17.9 mA when Rload is infinite, which is less than the diode maximum of 120 mA. Thus, the range for load resistance is 7.0k  Rload  . 88. The voltage as graphed in Fig. 40–40c decays exponentially according to Eq. 26–9b. As suggested in the problem, we use a linear approximation for the decay using an expansion from Appendix A–3. From Fig. 40–40c, we see that the decay lasts for approximately one-half of a cycle before it increases back to the peak value. 1 1 Vmin t t 2 ( 60 s ) = e−t /  1 − = 1 − = 1− = 0.97 Vmin = Vpeak e − t /  → 3  Vpeak RC ( 7.8  10  )( 36  10−6 F ) The voltage will decrease 3% from its maximum, or 1.5% above and below its average. 89. (a) The total potential energy is due to the electron–electron interaction, the proton–proton interaction, and 4 electron–proton interactions.  e2  1 e2 1 e2 1 U = U e-e + U p-p + 4U p-e = + + 4 −  2 2 4 0 d 4 0 r0  4 0  ( 1 d ) + ( 1 r ) 2

2 0

 8 e2  1 1  + −  2 2  4 0  d r0 r d + 0  

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(b) U has a minimum at d  0.043nm . U  0 for the

200 150

approximate range 0.011nm  d  0.51nm. U (eV)

(c) To find the point of greatest stability, set the derivative of U with respect to d (indicated by U  ) equal to 0 and solve for d.

100 50 0 -50 -100 0

0.2

0.4

0.6

0.8

d (nm)

 8 e2  1 1  + −  → 4 0  d r0 r02 + d 2  

    8d 1  e 2  1  1  8 ( 2d )  e2  − 2 −−  = − =0 → 3/ 2 3/ 2 4 0  d d2   2  ( r02 + d 2 )  4 0  ( r02 + d 2 )     3 / 2 8d 1 = 2 → 8d 3 = ( r02 + d 2 ) → 2d = ( r02 + d 2 ) → 2 2 3/ 2 d (r + d )

U =

4d 2 = r02 + d 2 → d =

0.074 r0 = = 0.0427 nm 3 3

1357

CHAPTER 41: Nuclear Physics and Radioactivity Responses to Questions 1.

Different isotopes of a given element have the same number of protons and electrons. Because they have the same number of electrons, they have almost identical chemical properties. Each isotope has a different number of neutrons from other isotopes of the same element. Accordingly, they have different atomic masses and mass numbers. Since the number of neutrons is different, they may have different nuclear properties, such as whether they are radioactive or not.

With 88 nucleons and 50 neutrons, there must be 38 protons. The number of protons is the atomic number, and so the element is strontium. The nuclear symbol is 88 Sr . 38

Identify the element based on the atomic number. (a) Uranium (Z = 92) (b) Nitrogen (Z = 7) (c) Hydrogen (Z = 1) (d) Strontium (Z = 38) (e) Berkelium (Z = 97)

The number of protons is the same as the atomic number, and the number of neutrons is the mass number minus the number of protons. (a) Uranium: 92 protons, 232 – 9 = 140 neutrons (b) Nitrogen: 7 protons, 18 – 7 = 11 neutrons (c) Hydrogen: 1 proton, 1 – 1 = 0 neutrons (d) Strontium: 38 protons, 86 – 38 = 48 neutrons (e) Berkelium: 97 protons, 252 – 97 = 155 neutrons

The atomic mass of an element as shown in the periodic table is the average atomic mass of all 35 37 naturally occurring isotopes. For example, chlorine occurs as roughly 75% 17 Cl and 25% 17 Cl , and so its atomic mass is about 35.5 (= 0.75  35 + 0.25  37). Other smaller effects would include the fact that the masses of the nucleons are not exactly 1 atomic mass unit, and that some small fraction of the mass energy of the total set of nucleons is in the form of binding energy.

The nucleus of an atom consists of protons (which carry a positive electric charge) and neutrons (which are electrically neutral). The electric force between protons is repulsive and much larger than the force of gravity. If the electric and gravitational forces are the only two forces present in the nucleus, the nucleus would be unstable as the electric force would push the protons away from each other. Nuclei are stable, and therefore, another force must be present in the nucleus to overcome the electric force. This force is the strong nuclear force.

The strong force and the electromagnetic (EM) force are two of the four fundamental forces in nature. They are both involved in holding atoms together: the strong force binds quarks into nucleons and binds nucleons together in the nucleus; the EM force is responsible for binding negatively charged electrons to positively charged nuclei and for binding atoms into molecules. The strong force is the strongest fundamental force; the EM force is about 100 times weaker at distances on the order of 10−17 m. The strong force operates at short range and is negligible for distances greater than about the size of the nucleus. The EM force is a long-range force that decreases as the inverse square of the distance between the two interacting charged particles. The EM force operates

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Nuclear Physics and Radioactivity

only between charged particles. The strong force is always attractive; the EM force can be attractive or repulsive. Both these forces have mediating field particles associated with them – the gluon for the strong force and the photon for the EM force. 8.

Quoting from Section 41–3, “… radioactivity was found in every case to be unaffected by the strongest physical and chemical treatments, including strong heating or cooling and the action of strong chemicals.” Chemical reactions are a result of electron interactions, not nuclear processes. The absence of effects caused by chemical reactions is evidence that the radioactivity is not due to electron interactions. Another piece of evidence is the fact that the -particle emitted in many radioactive decays is much heavier than an electron and has a different charge than the electron, so it can’t be an electron. Therefore, it must be from the nucleus. Finally, the energies of the electrons or photons emitted from radioactivity are much higher than those corresponding to electron orbital transitions. All of these observations support radioactivity being a nuclear process.

The resulting nuclide for gamma decay is the same isotope in a lower energy state: 64 64 29 Cu* →29 Cu + γ . 64

The resulting nuclide for beta-minus decay is an isotope of zinc, 30 Zn : 64 29

64 Cu →30 Zn + e− + .

The resulting nuclide for beta-plus decay is an isotope of nickel, 28 Ni : 64 29

10.

238 92

+ Cu →64 28 Ni + e +  .

U decays by alpha emission into 23490Th, which has 144 neutrons.

11. The alpha particle is a very stable nucleus. It has less energy when bound together than when split apart into separate nucleons. In most cases, more energy is required to emit separate nucleons than an alpha particle. It is usually true that the emission of a single nucleon is energetically not possible. See Example 41–5. 12. Alpha (α) particles are helium nuclei. Each α particle consists of 2 protons and 2 neutrons, and therefore, it has a charge of +2e and an atomic mass value of 4 u. They are the most massive of the three. Beta (β) particles are electrons  − or positrons  + . Electrons have a charge of –e and

( )

positrons have a charge of +e. In terms of mass, beta particles are much lighter than protons or neutrons, by a factor of about 2000, so are lighter than alpha particles by a factor of about 8000. Their emission is always accompanied by either an anti-neutrino (in  − decay) or a neutrino (in  + decay). Gamma (γ) particles are photons. They have no rest mass and no charge. 13. (a) Magnesium is formed: 11 Na → 12 Mg + e + . 24

−

(b) Neon is formed: 11 Na → 10 Ne + e + . 22

206

14. (a) Sulfur is formed: 15 Pb → 16 S + e + . 32

−

(b) Chlorine is formed: 16 S → 17 Cl + e + . 35

−

207

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Sc + e − + v

Scandium-45 is the missing nucleus.

Cu + 

Copper-58 is the missing nucleus.

15. (a)

45 20

Ca →

(b)

58 29

Cu* →

(c)

46 24

Cr → 4623V + e + + v

(d)

234 94

Pu →

(e)

239 93

Np → 239 Pu + e − + v 94

45 21

58 29

230 92

U +

Instructor Solutions Manual

The positron and the neutrino are the missing particles. Uranium-230 is the missing nucleus. The electron and the anti-neutrino are the missing particles.

16. The two extra electrons held by the newly formed thorium will be very loosely held, as the number of protons in the nucleus will have been reduced from 92 to 90, reducing the nuclear charge. It will be easy for these extra two electrons to escape from the thorium atom through a variety of mechanisms. They are in essence “free” electrons. They do not gain kinetic energy from the decay. They might get captured by the alpha nucleus, for example. 17. When a nucleus undergoes either  − or  + decay it becomes a different element, since it has either converted a neutron to a proton or a proton to a neutron. Thus, its atomic number (Z) has changed. The energy levels of the electrons are dependent on Z, and so all of those energy levels change to become the energy levels of the new element. Photons (with energies on the order of a few eV) are likely to be emitted from the atom as electrons change energies to occupy the new levels. 18. Alpha particles from an alpha-emitting nuclide are part of a two-body decay. The energy carried off by the decay fragments is determined by the principles of conservation of energy and of momentum. With only two decay fragments, these two constraints require the alpha particles to be monoenergetic. Beta particles from a beta-emitting nucleus are part of a three-body decay. Again, the energy carried off by all of the decay fragments is determined by the principles of conservation of energy and of momentum. However, with three decay fragments, the energy distribution between the fragments is not determined by these two constraints. The beta particles will therefore have a range of energies. 19. In electron capture, the nucleus will effectively have a proton change to a neutron. This isotope will then lie to the left and above the original isotope. Since the process would only occur if it made the nucleus more stable, it must lie BELOW the line of stability in Fig. 41–2. 20. Neither hydrogen nor deuterium can emit an  particle. Hydrogen has only one nucleon (a proton) in its nucleus, and deuterium has only two nucleons (one proton and one neutron) in its nucleus. Neither one has the necessary four nucleons (two protons and two neutrons) to emit an  particle. 21. Many artificially produced radioactive isotopes are rare in nature because they have decayed away over time. If the half-lives of these isotopes are relatively short in comparison with the age of Earth (which is typical for these isotopes), there won’t be any significant amount of these isotopes left to be found in nature. Also, many of these isotopes have a very high energy of formation, which is generally not available under natural circumstances. 22. After two months, the sample will not have completely decayed. After one month, half of the sample will remain, and after two months, one-fourth of the sample will remain. Each month, half of the remaining atoms decay. 23. For Z > 92, the short range of the attractive strong nuclear force means that no number of neutrons is able to overcome the electrostatic repulsion of the large concentration of protons. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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24. There are a total of 4 protons and 3 neutrons in the reactants. The -particle has 2 protons and 2 neutrons, and so 2 protons and 1 neutron are in the other product particle. It must be 23 He . 6 3

Li + 11 p → 24 + 23 He 14

25. The technique of 6 C could not be used to measure the age of stone walls and tablets. Carbon-14 dating is only useful for measuring the age of objects that were living at some earlier time. Stone walls and tablets were never alive. 26. The decay series of Fig. 41–12 begins with a nucleus that has many more neutrons than protons and lies far above the line of stability in Fig. 41–2. In a β+ decay, a proton is converted to a neutron, which would take the nuclei in this decay series farther from the line of stability and is not energetically preferred. 27. There are four alpha particles and four β– particles (electrons) emitted, no matter which decay path is 222 206 chosen. The nucleon number drops by 16 as 86 Rn decays into 82 Pb , indicating the presence of four alpha decays. The proton number only drops by four, from Z = 86 to Z = 82, but four alpha decays would result in a decrease of eight protons. Four β– decays will convert four neutrons into protons, making the decrease in the number of protons only four, as required. (See Fig. 41–12.) 28. (i)

Since the momentum before the decay was 0, the total momentum after the decay will also be 0. Since there are only 2 decay products, they must move in opposite directions with equal magnitude of momentum. Thus, (c) is the correct choice: both the same. (ii) Since both products have the same momentum, the one with the smallest mass will have the greater velocity. Thus, (b) is the correct choice: the alpha particle. (iii) We assume that the products are moving slowly enough that classical mechanics can be used. In 𝑝2

that case, k = . Since both particles have the same momentum, the one with the smallest 2𝑚 mass will have the greater kinetic energy. Thus, (b) is the correct choice: the alpha particle. 29. Fig. 41–7 shows the potential energy curve for an alpha particle and daughter nucleus for the case of radioactive nuclei. The alpha particle tunnels through the barrier from point A to point B in the figure. In the case of stable nuclei, the probability of this happening must be essentially zero. The maximum height of the Coulomb potential energy curve must be larger and/or the Q-value of the reaction must be smaller so that the probability of tunneling is extremely low. 30. If the Earth had been bombarded with additional radiation several thousand years ago, there would have been a larger abundance of carbon-14 in the atmosphere at that time. Organisms that died in that time period would have had a greater percentage of carbon-14 in them than organisms that die today. Since we assumed the starting carbon-14 was the same, we would have previously underestimated the age of the organisms. With the new discovery, we would re-evaluate the organisms as older than previously calculated. 31. (a) An einsteinium nucleus has 99 protons and a fermium nucleus as 100 protons. If the fermium undergoes either electron capture or + decay, a proton would in effect be converted into a neutron. The nucleus would now have 99 protons and be an einsteinium nucleus. (b) If the einsteinium undergoes  − decay, a neutron would be converted into a proton. The nucleus would now have 100 protons and be a fermium nucleus. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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32. In  decay, an electron is ejected from the nucleus of the atom, and a neutron is converted into a proton. The atomic number of the nucleus increases by one, and the element now has different chemical properties. In internal conversion, an orbital electron is ejected from the atom. This does not change the atomic number of the nucleus, nor its chemical properties. Also, in  decay, both a neutrino and an electron will be emitted from the nucleus. Because there are three decay products (the neutrino, the  particle, and the nucleus), the momentum of the  particle can have a range of values. In internal conversion, since there are only two decay products (the electron and the nucleus), the electron will have a unique momentum and, therefore, a unique energy.

MisConceptual Questions 1.

(a) A common misconception is that the elements of the period table are distinguished by the number of electrons in the atom. This misconception arises because the number of electrons in a neutral atom is the same as the number of protons in its nucleus. Elements can be ionized by adding or removing electrons, but this does not change what type of element it is. When an element undergoes a nuclear reaction that changes the number of protons in the nucleus, the element does transform into a different element.

(b) The role of energy in binding nuclei together is often misunderstood. As protons and neutrons are added to the nucleus, they release some mass energy, which usually appears as radiation or kinetic energy. This lack of energy is what binds the nucleus together. To break the nucleus apart, the energy must be added back in. As a result, a nucleus will have less energy than the protons and neutrons that constitute the nucleus.

(c) Regarding (a), large nuclei (with many more neutrons than protons) are typically unstable, so increasing the number of nuclei does not necessarily make the nucleus more stable. Regarding (b), nuclei such as 148 O have more protons than neutrons, yet 148 O is not as stable as 168 O . Therefore, having more protons than neutrons does not necessarily make a nucleus more stable. The large Coulomb repulsion between the protons helps lead to instability. Regarding (d), there are no electrons in the nucleus. Electrons do not affect the stability of nuclei. Regarding (e), large unstable nuclei, such as 238 94 Pu, have a much larger total binding energy than small stable nuclei such as 42 He, so the total binding energy is not a measure of stability. The correct answer is (c). Stable nuclei typically have large binding energies per nucleon, which means that each nucleon is more tightly bound to the others.

(e) The Coulomb repulsive force does act inside the nucleus pushing the protons apart. Another larger attractive force is thus necessary to keep the nucleus together. The force of gravity is far too small to hold the nucleus together. Neutrons are not negatively charged. It is the attractive strong nuclear force that overcomes the Coulomb force to hold the nucleus together.

(b) The exponential nature of radioactive decay is a concept that can be misunderstood. It is sometimes thought that the decay is linear, such that the time for a substance to completely decay is twice the time for half of the substance to decay. Radioactive decay is not linear but exponential. That is, during each half-life, 1/2 of the remaining substance decays. If half the original decays in the first half-life, then 1/2 remains. During the second half-life, 1/2 of what is left decays, which would be 1/4 of the initial substance. In each subsequent half-life, half of the remaining decays, so it takes many half-lives for a substance to effectively decay away. The decay constant is inversely related to the half-life.

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(e) The half-life is the time it takes for half of the substance to decay away. The half-life is a constant determined by the composition of the substance and not on the quantity of the initial substance. As the substance decays, the number of nuclei decreases, the activity (number of decays per second) decreases, but the half-life remains constant.

(d) A common misconception is that it would take twice the half-life, or 20 years, for the substance to completely decay. This is incorrect because radioactive decay is an exponential process. That is, during each half-life, 1/2 of the remaining substance decays. After the first 10 years, 1/2 remains. After the second ten-year period, 1/4 remains. After each succeeding half-life, another half of the remaining substance decays, leaving 1/8, then 1/16, then 1/32, and so forth. The decays stop when none of the substance remains. But after a time, there will be too few nuclei to reliably use statistics. Thus, the time cannot be exactly determined.

(c) A common misconception is that after the second half-life, none of the substance remains. However, during each half-life, 1/2 of the remaining substance decays. After one half-life, 1/2 remains. After two half-lives, 1/4 remains. After three half-lives, 1/8 remains.

(a) The decay constant is proportional to the probability of a particle decaying and is inversely proportional to the half-life. Therefore, the substance with the shorter half-life (Sr) has the larger decay constant and the larger probability of decaying. The activity is proportional to the amount of the substance (number of atoms) and the decay constant, so the activity of Tc will be smaller than the activity of Sr.

10. (d) The element with the largest decay constant will have the shortest half-life. Converting each of the choices to decays per second yields: (a) 100/s, (b) 1.610–7/s, (c) 2.510–9/s, (d) 1.2104/s. Answer (d) has the largest decay rate, so it will have the smallest half-life. 11. (b) The half-life of radium remains constant and is not affected by temperature. Radium that existed several billion years ago will have essentially completely decayed. Small isotopes, such as carbon-14, can be created from cosmic rays, but radon is a heavy element. Lightning is not energetic enough to affect the nucleus of the atoms. Heavy elements such as plutonium and uranium can decay into radium and thus are the source of present-day radium. 12. (a) The nature of mass and energy in nuclear physics is often misunderstood. When the neutron and proton are close together, they bind together by releasing mass energy that is equivalent to the binding energy. This energy comes from a reduction in their mass. Therefore, when the neutron and proton are far from each other, their net mass is greater than their net mass when they are bound together.

Solutions to Problems 1.

Convert the units from MeV c 2 to atomic mass units.

  1u m = ( 775MeV c 2 )  = 0.832u 2   931.49MeV c 

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The  particle is a helium nucleus and has A = 4. Use Eq. 41–1. 1

r = (1.2  10−15 m ) A 3 = (1.2  10−15 m ) ( 4 ) 3 = 1.9  10−15 m = 1.9fm

The radii of the two nuclei can be calculated with Eq. 41–1. Take the ratio of the two radii. 1/3 −15 1/3 r214 (1.2  10 m ) ( 214 )  214  = =  = 1.0128 r206 (1.2  10−15 m ) ( 206 )1/3  206  206 So, the radius of 214 82 Pb is 1.28%  1% larger than the radius of 82 Pb.

Use Eq. 41–1 for both parts.

(a) r = (1.2  10−15 m ) A1/3 = (1.2  10−15 m ) (120 )

1/3

= 5.9  10 −15 m = 5.9fm 3

−15 r    3.7  10 m  = (b) r = (1.2  10−15 m ) A1/ 3 → A =   = 29.3  29   −15 −15  1.2  10 m   1.2  10 m 

To find the rest mass of an  particle, we subtract the rest mass of the two electrons from the rest mass of a helium atom: m = mHe − 2me

= ( 4.002603u ) ( 931.5MeV uc 2 ) − 2 ( 0.511MeV c 2 ) = 3727 MeV c 2 This is less than the sum of the masses of two protons and two neutrons because of the binding energy. 6.

Each particle would exert a force on the other through the Coulomb electrostatic force. The distance between the particles is twice the radius of one of the particles. The Coulomb force is given by Eq. 21–2. F=

q q

(8.988  10 N  m C ) ( 2 ) (1.60 10 9

−19

C ) 

= 63.41N  63 N 2 ( 2 ) ( 41/3 )(1.2  10−15 m )    The acceleration is found from Newton’s second law. We use the mass of a “bare” alpha calculated in Problem 5. 63.41N F F = ma → a = = = 9.5  1027 m s 2 −27 m  1.6605  10 kg  3727 MeV c 2  2   931.49 MeV c 

4 0 ( 2r )2

First, we calculate the density of nuclear matter. The mass of a nucleus with mass number A is approximately (A u) and its radius is r = (1.2  10−15 m ) A1/ 3 . Calculate the density.

=

m A (1.6605  10 = 3 4 V 3r

−27

kg u )

A (1.6605  10−27 kg u ) 4 3

 (1.2  10

−15

m) A 3

= 2.294  1017 kg m3

We see that this is independent of A. The value has 2 significant figures.

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(a) We set the density of the Earth equal to the density of nuclear matter. M  Earth =  nuclear = 4 Earth → 3 matter 3  REarth 1/3

1/3     M Earth  5.98  1024 kg    REarth =  4  =  4  ( 2.294  1017 kg m3 )  = 183.9 m  180 m 3  3  nuclear    matter   (b) Set the density of Earth equal to the density of uranium, and solve for the radius of the uranium. Then compare that to the actual radius of uranium, using Eq. 41–1, with A = 238. M m M Earth mU = →  = 4 Earth 3 = 4 U 3 → REarth 3 rU 3 3  REarth 3  rU

 ( 238u ) (1.6605  10−27 kg u )   = ( 6.38  10 m )    ( 5.98 1024 kg ) rU 2.58  10−10 m = = 3.5  104 rU actual (1.2  10−15 ) ( 238 )1/3

1/3

 m  rU = REarth  U   M Earth 

Use Eq. 41–1 to find the value for A. We use uranium-238 since it is the most common isotope. 1.2  10−15 m ) A1/3 ( runknown 3 = = 0.5 → A = 238 ( 0.5) = 29.75  30 1/3 −15 rU (1.2  10 m ) ( 238) From Appendix G, a stable nucleus with A  30 is

= 2.58  10−10 m

31 15

The basic principle to use is that of conservation of energy. We assume that the centers of the two particles are located a distance from each other equal to the sum of their radii. That distance is used to calculate the initial electrical potential energy. Then, we also assume that since the Rf nucleus is much heavier than the alpha, the alpha has all of the final kinetic energy when the particles are far apart from each other (and so have no potential energy). q qRf 1 = K + 0 → Ki + U i = K f + U f → 0 + 4 0 ( r + rRf )

K = ( 8.988  10 N  m C ) 9

1/3

( 2 )(104 ) (1.60  10−19 C ) + 263

1/3

)(1.2 10 m )(1.60 10 −15

−19

J eV )

= 3.118  107 eV

 31MeV 10. (a) The hydrogen atom is made of a proton and an electron. Use values from Appendix G. mp 1.007276 u = = 0.9994553  99.95% mH 1.007825u (b) Compare the volume of the nucleus to the volume of the atom. The nuclear radius is given by Eq. 41–1. For the atomic radius, we use the Bohr radius given in Eq. 37–12. 3 −15 3  rnucleus   (1.2  10 m )  Vnucleus 43  rnucleus = 4 3 = =   = 1.2  10−14  −10  Vatom r r 0.53 10 m  )  atom 3  atom   ( 3

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11. Electron mass is negligible compared to nucleon mass, and one nucleon weighs about 1.01 atomic mass unit. Therefore, in a 1.0-kg object, (1.0 kg) ( 6.022  1026 u kg ) N= = 5.96  1026  6  1026 nucleons 1.01u nucleon No, it does not matter what the element is because the mass of one nucleon is essentially the same for all elements. 12. The initial kinetic energy of the alpha must be equal to the electrical potential energy when the alpha just touches the thorium nucleus. Assume the two nuclei are at rest when they “touch”. The distance between the two particles is the sum of their radii. q qTh 1 → K i + U i = K f + U f → K + 0 = 0 + 4 0 ( r + rTh )

K = ( 8.988  10 N  m C ) 9

( 2 )( 90 ) (1.60  10−19 C )

( 41/3 + 2321/3 )(1.2 10−15 m )(1.60 10−19 J eV )

= 2.790  107 eV

 28MeV 13. From Fig. 41–1, we see that the average binding energy per nucleon at A = 63 is about 8.7 MeV. Multiply this by the number of nucleons in the nucleus. ( 63)(8.7 MeV ) = 548.1MeV  550 MeV 14. (a) From Fig. 41–1, we see that the average binding energy per nucleon at A = 238 is 7.5MeV. Multiply this by the 238 nucleons. ( 238)( 7.5MeV ) = 1785MeV  1800MeV (b) From Fig. 41–1, we see that the average binding energy per nucleon at A = 84 is 8.7 MeV. Multiply this by the 84 nucleons. (84)(8.7 MeV ) = 730.8MeV  730MeV 15.

18 8

O consists of 8 protons and 10 neutrons. We find the binding energy from the masses of the components and the mass of the nucleus, from Appendix G. Note that the electron masses from the hydrogens cancel out the electron masses for the oxygen. Binding energy = 8m 11 H +10m 01 n − m 188 O  c 2

( )

= 8 (1.007825u ) +10 (1.008665u ) − (17.999160 u )  c 2  931.5MeV c 2  = 0.150090   = 139.81MeV u   Binding energy per nucleon = (139.81MeV ) 18 = 7.767 MeV

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16. Deuterium consists of 1 proton, 1 neutron, and 1 electron. Ordinary hydrogen consists of 1 proton and 1 electron. Use the atomic masses from Appendix G. The electron masses cancel. Binding energy =  m ( 11 H ) + m ( 01 n ) − m ( 21 H ) c 2

 931.5MeV c 2  = (1.007825u ) + (1.008665u ) − ( 2.014102 u ) c 2   u   = 2.224 MeV 17. We find the binding energy of the last neutron from the masses of the isotopes. 31  2 Binding energy =  m ( 15 P ) + m ( 10 n ) − m ( 32 15 P )  c

= ( 30.973762 u ) + (1.008665u ) − ( 31.973908u )  c 2 (931.5MeV / c 2 ) = 7.935MeV

18. (a)

7 4

Be consists of 4 protons and 3 neutrons. We find the binding energy from the masses, using hydrogen atoms in place of protons so that we account for the mass of the electrons. Binding energy =  4m 11 H + 3m 10 n − m 74 Be  c 2

( )

(

)

=  4 (1.007825u ) + 3 (1.008665u ) − ( 7.016929 u )  c 2 ( 931.5MeV / c 2 ) = 37.60 MeV

Binding energy 37.60 MeV = = 5.372 MeV nucleon nucleon 7 nucleons

(b)

197 79

Au consists of 79 protons and 118 neutrons. We find the binding energy as in part (a).

 2 Binding energy = 79m ( 11 H ) + 118m ( 10 n ) − m ( 197 79 Au )  c

=  79 (1.007825u ) + 118 (1.008665u ) − (196.966570 u )  c 2 ( 931.5MeV / c 2 ) = 1559.4 MeV  1559 MeV

Binding energy 1559.4 MeV = 7.916 MeV nucleon = nucleon 197 nucleons

19.

31 15

P consists of 15 protons and 16 neutrons. We find the binding energy from the masses: 31 Binding energy = 15 m ( 11 H ) + 16m ( 01 n ) − m ( 15 P )  c 2

= 15 (1.007825u ) + 16 (1.008665u ) − ( 30.973762 u )  c 2 ( 931.5MeV uc 2 )

= 262.9 MeV Binding energy 262.9 MeV = = 8.481MeV nucleon nucleon 31 We do a similar calculation for 32 15 P, consisting of 15 protons and 17 neutrons.

 2 Binding energy = 15 m ( 11 H ) + 17 m ( 01 n ) − m ( 32 15 P )  c

= 15 (1.007825u ) + 17 (1.008665u ) − ( 31.973908u )  c 2 (931.5MeV uc 2 ) = 270.85MeV

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Binding energy 270.85MeV = = 8.464 MeV nucleon nucleon 32 31 31 By this measure, the nucleons in 15 P are more tightly bound than those in 32 15 P. Thus, we expect 15 P 32 31 to be more stable than 32 15 P. Appendix G bears that out – 15 P is radioactive, while 15 P is stable. 20. We find the required energy by calculating the difference in the masses. (a) Removal of a proton creates an isotope of nitrogen. To balance electrons, the proton is included as a hydrogen atom: 168 O → 11 H + 157 N. Energy needed =  m ( 157 N ) + m ( 11 H ) − m ( 168 O )  c 2

= (15.000109 u ) + (1.007825u ) − (15.994915u )  ( 931.5MeV uc 2 ) = 12.13MeV

(b) Removal of a neutron creates another isotope of oxygen: 168 O → 01 n + 158 O. Energy needed =  m ( 158 O ) + m ( 01 n ) − m ( 168 O )  c 2

= (15.003066 u ) + (1.008665u ) − (15.994915u )  ( 931.5MeV uc 2 ) = 15.66 MeV

The nucleons are held by the attractive strong nuclear force. It takes less energy to remove the proton because there is also the repulsive electric force from the other protons “helping” to remove the proton. 21. (a) We find the binding energy from the masses. Binding Energy =  2m ( 42 He ) − m ( 48 Be ) c 2

=  2 ( 4.002603u ) − ( 8.005305u )  c 2 ( 931.5MeV uc 2 ) = − 0.092 MeV

Because the binding energy is negative, the nucleus is unstable. It will be in a lower energy state as two alphas instead of a beryllium. (b) We find the binding energy from the masses. Binding Energy = 3m ( 42 He ) − m ( 126 C ) c 2

= 3 ( 4.002603u ) − (12.000000 u ) c 2 ( 931.5MeV uc 2 ) = + 7.3MeV

Because the binding energy is positive, the nucleus is stable. 22. The wavelength is determined from the energy change between the states. −34 8 c hc ( 6.63  10 J s )( 3.00  10 m s ) E = hf = h → = = = 2.6  10−12 m E ( 0.48 MeV ) (1.60  10−13 J MeV )  23. The nuclear decay is 31 H → 23 He + −01 e + v . When we add one electron to both sides in order to use atomic masses, we see that the mass of the emitted  particle is included in the atomic mass of 23 He. The energy released is the difference in the masses. Energy released =  m ( 31 H ) − m ( 23 He ) c 2

= ( 3.016049 u ) − ( 3.016029 u ) c 2 ( 931.5MeV uc 2 ) = 0.019 MeV

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24. For the decay 116 C → 105 B + 11 p, we find the difference of the final and initial final masses. To balance the electrons (so that we can use atomic masses), we assume the proton is an atom of 11 H.

m = m ( 105 B ) + m ( 11 H ) − m ( 116 C )

= (10.012937 u ) + (1.007825u ) − (11.011433u ) = 0.009329 u Since the final masses are more than the original mass, energy would not be conserved. 25. The decay is 01 n → 11 p + −01 e + v . The electron mass is accounted for if we use the atomic mass of 1 1

H as the combination of the proton and electron. If we ignore the recoil of the proton and the neutrino, and any possible mass of the neutrino, we get the maximum kinetic energy. K max =  m 01 n − m 11 H  c 2 = (1.008665u ) − (1.007825u ) c 2 931.5MeV uc 2

( )

(

)

= 0.782 MeV 26. For each decay, we find the difference of the initial and the final masses. If the final mass is more than the initial mass, the decay is not possible. 1 233 (a) m = m ( 232 92 U ) + m ( 0 n ) − m ( 92 U ) = 232.037155u + 1.008665u − 233.039634u = 0.006816u Because an increase in mass is required, the decay is not possible. (b) m = m ( 137 N ) + m ( 01 n ) − m ( 147 N ) = 13.005739 u + 1.008665u − 14.003074 u = 0.011330 u Because an increase in mass is required, the decay is not possible. (c)

1 40 m = m ( 39 19 K ) + m ( 0 n ) − m ( 19 K ) = 38.963706u + 1.008665u − 39.963998u = 0.008373u

Because an increase in mass is required, the decay is not possible. 24 27. (a) From Appendix G, 11 Na is a  − emitter .

(b) The nuclear decay reaction is

24 11

24 Na → 12 Mg +  − + v . We add 11 electrons to both sides in

order to use atomic masses. Then, the mass of the beta is accounted for in the mass of the magnesium. The maximum kinetic energy of the  − corresponds to the neutrino having no kinetic energy (a limiting case) and no mass. We also ignore the recoil of the magnesium. 24  2 K  − =  m ( 24 11 Na ) − m ( 12 Mg )  c

= ( 23.990963u ) − ( 23.985042 u )  c 2 ( 931.5MeV uc 2 ) = 5.515MeV

28. (a) We find the final nucleus by balancing the mass and charge numbers. Z ( X ) = Z ( U ) − Z ( He ) = 92 − 2 = 90

A ( X ) = A ( U ) − A ( He ) = 238 − 4 = 234 Thus, the final nucleus is

234 90

Th .

(b) If we ignore the recoil of the thorium, the kinetic energy of the  particle is equal to the Qvalue of the reaction. The electrons are balanced. 234 4  2 → K = Q =  m ( 238 92 U ) − m ( 90Th ) − m ( 2 He )  c

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4 m ( 23490Th ) = m ( 238 92 U ) − m ( 2 He ) −

Instructor Solutions Manual

K c2

  4.20 MeV  1u =  238.050787 u − 4.002603u −  2 2  c  931.5MeV c    = 234.043675u This answer assumes that the 4.20 MeV value does not limit the sig. fig. of the answer. 60 − 29. The nuclear reaction is 60 27 Co → 28 Ni +  +  . We add 27 electrons to both sides of the equation so that we can use atomic masses. The kinetic energy of the  − will be maximum if the (essentially) massless neutrino has no kinetic energy. We also ignore the recoil of the nickel. 60  2 Co ) − m ( 60 K  =  m ( 27 28 Ni )  c  931.5MeV c 2  = ( 59.933816 u ) − ( 59.930785u )  c 2   = 2.823MeV u  

30. We add three electron masses to each side of the nuclear reaction 74 Be + −01 e → 73 Li + v. Then for the mass of the product side, we may use the atomic mass of 73 Li. For the reactant side, including the three electron masses and the mass of the emitted electron, we may use the atomic mass of 74 Be. The energy released is the Q-value. Q =  m ( 74 Be ) − m ( 73 Li ) c 2  931.5MeV c 2  = ( 7.016929 u ) − ( 7.016003u ) c 2   = 0.863MeV u   214 4 31. For alpha decay, we have 218 84 Po → 82 Pb + 2 He. We find the Q value, which is the energy released.

214 4  2 Q =  m ( 218 84 Po ) − m ( 82 Pb ) − m ( 2 He )  c

 931.5MeV c 2  =  218.008972 u − 213.999804 u − 4.002603u  c 2   u   = 6.115MeV 218 0 For beta decay, the nuclear reaction is 218 84 Po → 85 At + −1 e + v . We add 84 electrons to both sides of the equations so that we can use the atomic masses. We assume the neutrino is massless, and find the Q value. 218  2 Q =  m ( 218 84 Po ) − m ( 85 At )  c  931.5MeV c 2  =  218.008972 u − 218.008694 u  c 2   = 0.259 MeV u  

32. (a) We find the final nucleus by balancing the mass and charge numbers. Z ( X ) = Z ( P ) − Z ( e ) = 15 − ( − 1) = 16

A ( X ) = A ( P ) − A ( e ) = 32 − 0 = 32 32 S. Thus, the final nucleus is 16

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(b) If we ignore the recoil of the sulfur and the energy of the neutrino, the maximum kinetic energy 32 − of the electron is the Q-value of the reaction. The reaction is 32 15 P → 16 S +  + v . We add 15 electrons to each side of the reaction, and then we may use atomic masses. The mass of the emitted beta is accounted for in the mass of the sulfur. 32  2 → K = Q =  m ( 32 15 P ) − m ( 16 S )  c  1.71MeV 1u K  32 m ( 32 = 31.97207 u 16 S ) = m ( 15 P ) − 2 = ( 31.973908u ) − 2 2  931.5MeV c  c c  33. We find the energy from the wavelength. 6.63  10−34 J s )( 3.00  108 m s ) ( hc E= = = 11.8MeV  (1.05  10−13 m )(1.602  10−19 J eV ) This has to be a  ray from the nucleus rather than a photon from the atom. Electron transitions do not involve this much energy. Electron transitions involve energies on the order of a few eV. 34. The nuclear decay reaction is

23 10

− Ne → 23 11 Na +  + v . We add 10 electrons to both sides in order to

use atomic masses for the calculations. Then the mass of the beta is accounted for in the mass of the sodium. The maximum kinetic energy of the  − corresponds to the neutrino having no kinetic energy (a limiting case) and no mass, and it is the Q-value of the reaction. We also ignore the recoil of the sodium. 23  2 ( 22.9945u ) − ( 22.9898u )  c 2 ( 931.5MeV uc 2 ) K max = Q =  m ( 10 Ne ) − m ( 23 11 Na )  c = 

= 4.4 MeV If the neutrino were to have all of the kinetic energy, the minimum kinetic energy of the electron is 0. The sum of the kinetic energy of the electron and the energy of the neutrino must be the Q-value, and so the neutrino energies are 0 and 4.4 MeV, respectively. 35. A  + emission can be modeled as a proton changing into a neutron and a positron. The nuclear reaction here is 137 N → 136 C + 10 + + v . If we add 7 electrons to each side, the nitrogen atomic mass can be used, and the carbon atomic mass can be used, but there will the mass of an electron and the mass of a positron as products that must be included. Thus, the atomic reaction is 13 13 0 + 0 −1 + v . See Problem 36 for a general  + discussion. 7 N → 6 C + 1  + −1 e The kinetic energy of the  + particle will be maximum if the (almost massless) neutrino and the electron have no kinetic energy. We also ignore the recoil of the carbon. K =  m ( 137 N ) − m ( 136 C ) − m ( −01 e ) − m ( 10 + )  c 2 =  m ( 137 N ) − m ( 136 C ) − 2m ( −01 e )  c 2

= (13.005739 u ) − (13.003355 ) − 2 ( 0.00054858u )  c 2 (931.5MeV uc 2 ) = 1.199 MeV

If the  + and electron have no kinetic energy, the maximum kinetic energy of the neutrino is also

1.199MeV . The minimum energy of each is 0 when the other has the maximum. 36. For the positron-emission process, ZA N → Z −A1 N + e + + v. We must add Z electrons to the nuclear mass of N to be able to use the atomic mass, and so we must also add Z electrons to the reactant side. On the reactant side, we use Z − 1 electrons to be able to use the atomic mass of N. Thus, we © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

have 1 “extra” electron mass and the -particle mass, which means that we must include 2 electron masses on the right-hand side. We find the Q-value given this constraint. Q =  M P − ( M D + 2me ) c 2 = ( M P − M D − 2me ) c 2 . 37. We assume that the energies are low enough that we may use classical kinematics. In particular, we 234 4 will use p = 2mK . The decay is 238 92 U → 90Th + 2 He. If the uranium nucleus is at rest when it decays, the magnitude of the momentum of the two daughter particles must be the same.  4u  2m K p 2 m p 2 p = pTh ; K Th = Th =  =   =  K =   ( 4.20MeV ) = 0.0718MeV 2mTh 2mTh 2mTh mTh  234u  The Q-value is the total kinetic energy produced. Q = K + KTh = 4.20MeV + 0.0718MeV = 4.27 MeV 38. Both energy and momentum are conserved. Therefore, the momenta of the product particles are equal in magnitude. We assume that the energies involved are low enough that we may use classical kinematics; in particular, p = 2mK . 2 m  2m K p2 pRn 4.002603 =  =   =    K = K 2mRn 2mRn 2mRn 222.017576  mRn  The sum of the kinetic energies of the product particles must be equal to the Q-value for the reaction. 4.002603 222 4  2 K Rn + K =  m ( 226 K + K → 88 Ra ) − m ( 86 Rn ) − m ( 2 He )  c = 222.017576 222 4  m ( 226  2 88 Ra ) − m ( 86 Rn ) − m ( 2 He )  c  K = 4.002603   1 +   222.017576 

p = pRn ; K Rn =

( 226.025409 u ) − ( 222.017576 u ) − ( 4.002603u )  c 2 = 931.5MeV uc 2 ) = 4.79 MeV ( 4.002603   1 +  222.017576   The rest energy (mc2) of an alpha particle is about 4000 MeV, so our assumption of classical kinematics is valid. 39. (a) The decay constant can be found from the half-life, using Eq. 41–8. ln 2 ln 2 = = = 1.5  10−10 yr −1 = 4.9  10−18 s −1 T1/ 2 4.5  109 yr (b) The half-life can be found from the decay constant, using Eq. 41–8.  1h  ln 2 ln 2 = = 18240s  T1/2 =   5.1h −5 −1  3.8  10 s  3600s  40. We find the half-life from Eq. 41–7d and Eq. 41–8. ln 2 t T1/2

ln 2 ln 2 t=− ( 4.2 h ) = 1.4 h 140 R ln ln R0 1120 We can see this also from the fact that the rate dropped by a factor of 8, which takes 3 half-lives. R = R0 e

− t

= R0 e

−

→ T1/ 2 = −

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41. We use Eq. 41–6 to find the fraction remaining. N = N0e

− t

→

 ( ln 2 )( 2.2 yr )(12mo yr )   9 mo 

− N = e − t = e  N0

= 0.1309  0.13 = 13%

42. The activity at a given time is given by Eq. 41–7b. The half-life is found in Appendix G. ln 2 ln 2 dN = N = 5.6  1020 nuclei ) = 2.1 109 decays s N= ( 7 dt T1/ 2 ( 5730 yr ) ( 3.16 10 s yr ) 43. Every half-life, the number of nuclei left in the sample is multiplied by one-half. N n 5 = ( 12 ) = ( 12 ) = 0.03125 = 1 32 = 3.125% N0 44. We need the decay constant and the initial number of nuclei. The half-life is found in Appendix G and is about 8 days. Use Eq. 41–8 to find the decay constant. ln 2 ln 2 = = 9.99668  10−7 s−1 = T1/2 ( 8.0252days )( 24 h day )( 3600s h )

 ( 846  10−6 g )   ( 6.022  1023 atoms mol ) = 3.8918 1018 nuclei. N0 =  130.906126g mol )   ( (a) We use Eq. 41–7b to evaluate the initial activity. Activity =  N =  N 0 e − t = ( 9.99668  10−7 s −1 )( 3.8918  1018 ) e −0 = 3.8905  1012 decays s  3.89  1012 decays s

(b) We evaluate Eq. 41–7d at t = 1.50h. This is much less than one half life, so we expect that most of the initial nuclei remain, and the activity is essentially unchanged.

R = R0 e − t = ( 3.8905  1012 decays s ) e

(

)

 3600s  − 9.9966810−7 s −1 (1.50 h )   h 

= 3.8696  1012 decays s

 3.87  1012 decays s (c) We evaluate Eq. 41–7d at t = 3.0 months. We use a time of 1/4 year for the 3.0 months – other approximations (like 1 month = 30 days) for that time period will give a slightly different final answer. In any event, this time period is more than 10 half-lives, so we expect the activity to be significantly lower than the initial value.

R = R0 e − t = ( 3.8905  1012 decays s ) e

(

)

(

− 9.9966810−7 s −1 ( 0.25yr ) 3.156107 s yr

)

= 1.4607  109 decays s

 1.46  109 decays s 45. The activity of a sample is given by Eq. 41–7a. There are two different decay constants involved. Note that Appendix G gives half-lives, not activities.   − t − t − t  I N I =  Co N Co →  I N 0 e I =  Co N 0 e Co → I = e( I Co ) →

 Co

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 ( 5.2712 y )( 365.25d y )    ln  ln  TCo TI   ln (  I  Co ) (8.0252d ) 1/2 1/2     = = = 63.715d t= ln 2 ln 2  I −  Co   1 1 − − ln 2   TI TCo 8.0252d ) ( 5.2712 y )( 365.25d y )  (  1/2 1/2 46. We find the number of nuclei from the activity of the sample and the half-life. The half-life is found in Appendix G. dN ln 2 N → = N = dt T1/ 2

T1/ 2 dN ( 4.468  10 yr )( 3.156  10 s yr ) = ( 370decays s ) = 7.5  1019 nuclei ln 2 dt ln 2 9

47. Each  emission decreases the mass number by 4 and the atomic number by 2. The mass number changes from 235 to 207, which is a change of 28. Thus, 7  particles must be emitted. With the 7  emissions, the atomic number would have changed from 92 to 78. Each  − emission increases the atomic number by 1. Thus, to have a final atomic number of 82, 4  − particles must be emitted. 48. We will use the decay constant frequently, so we calculate it here. ln 2 ln 2 = = = 0.022432s −1 T1/ 2 30.9s (a) We find the initial number of nuclei from an estimate of the atomic mass. ( 9.6 10−6 g ) 6.022 1023 atoms mol = 4.662 1016  4.7 1016 nuclei. N0 = ( ) (124g mol ) (b) Evaluate Eq. 41–6 at t = 2.6min.

N = N0e− t = ( 4.662 1016 ) e

(

)

− 0.022432s−1 ( 2.6min )( 60s min )

= 1.409 1015  1.4 1015 nuclei

 N = ( 0.022432s −1 )(1.409  1015 ) = 3.161  1013  3.2 1013 decays s

(d) We find the time from Eq. 41–7a, relative to the time when we had the initial 9.6 g of 124 55 Cs .

 N =  N 0 e − t →   1decay/s  N    ln ln  16 −1  0.022432s 4.662 10 decay/s   ( )( ) N  0   = 1542s  26 min t=−  =−  −1 0.022432s  This is about 50 half-lives.

49. We find the mass from the initial decay rate and Eq. 41–7b. dN 6.022  1023 nuclei mole =  N0 =  m → dt 0 ( atomic weight ) g mole

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dN 1 ( atomic weight ) dN T1/ 2 ( atomic weight ) = dt 0  ( 6.022  1023 ) dt 0 ln 2 ( 6.022  1023 )

= (1.6  105 s −1 )

(1.248 10 yr )( 3.156  10 s yr ) ( 39.963998g ) = 0.6034g  0.60g ( ln 2 ) ( 6.022 10 ) 9

50. The number of nuclei is found from the mass and the atomic weight. The activity is then found from number of nuclei and the half-life, using Eq. 41–7b.  ( 7.8 10−6 g )  dN ln 2 ln 2   ( 6.022  1023 atoms mol ) N= = N = dt T1/ 2 1.23  106 s  ( 31.973908g mol )   

= 8.279  1010 decays s  8.3  1010 decays s 51. (a) The decay constant is found from Eq. 41–8. ln 2 ln 2 = = = 1.381 10−13 s −1  1.38  10−13 s −1 5 T1/2 (1.59  10 yr )( 3.156  107 s yr ) (b) The activity is the decay constant times the number of nuclei.

 60s    1min 

 N = (1.381 10−13 s −1 )( 3.50  1018 ) = 4.8335  105 decays s   2.90  107 decays min 52. We use Eq. 41–7d, with R = 16 R0 . −

ln 2

R = R0 e − t = R0 e T1/2

→ T1/ 2 = −

ln 2 ln 2 t = − 1 ( 9.8min ) = 3.8min R ln 6 ln R0

53. We assume the initial sample of pure carbon is naturally occurring carbon, and so use the atomic weight of naturally occurring carbon to find the number of atoms in the sample. The activity is found from Eq. 41–7a, and the half-life is found in Appendix G.  ( 315g )  6.022  1023 atoms mol = 1.579 1025 atoms N = ) (  (12.0109g mol ) 

 1.3  N14 =  12  (1.579  1025 ) = 2.053  1013 nuclei of 146 C  10  

 ln 2  ( 2.053  1013 ) = 78.69decays s  79decays s 7  ( 5730 yr ) ( 3.156  10 s yr ) 

N = 

54. We find the mass from the activity. Note that N A is used to represent Avogadro’s number. The half-life is found in Appendix G. ln 2 mN A R = N = → T1/ 2 A

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RT1/ 2 A ( 475decays s ) ( 4.468  10 yr )( 3.156  10 s yr ) ( 238.050787 g mole ) = N A ln 2 ( 6.022 1023 nuclei mole ) ln 2 9

Instructor Solutions Manual

= 3.8199  10−2 g  3.82  10−2 g 87 55. We assume that we currently have N nuclei of 87 37 Rb and 0.0260 N nuclei of 38 Sr. The sum of those

two would be the initial number of 87 37 Rb nuclei, N 0 . Thus, N 0 = 1.0260 N . Use Eq. 41–6 and Eq. 87 41–8 to find the time for N 0 nuclei of 87 37 Rb to decay to N nuclei of 37 Rb , which is the estimated age of the fossils. N = N 0 e − t →

 T1/ 2  N  T1/ 2  1  N  4.97  1010 yr  1  N ln  ln  ln  t = − ln  =− =− =−    N0  ln 2  N 0  ln 2  1.0260 N 0  ln 2  1.026  = 1.84  109 yr

56. We are not including the neutrinos that will be emitted during beta decay. First sequence:

Second sequence:

232 90

4 228 228 − 228 228 − Th → 228 ; 88 Ra + 2 ; 88 Ra → 89 Ac +  ; 89 Ac → 90 Th + 

228 90

224 220 4 4 Th → 224 88 Ra + 2 ; 88 Ra → 86 Rn + 2

235 92

231 231 − 227 4 4 ; 231 U → 231 91 Pa → 89 Ac + 2 ; 90 Th + 2 ; 90 Th → 91 Pa + 

227 89

− 223 4 Ac → 227 ; 227 90 Th +  90Th → 88 Ra + 2

57. Because the fraction of atoms that are 146 C is so small, we use the atomic weight of 126 C to find the number of carbon atoms in 67 g. We then use the ratio to find the number of 146 C atoms present when the club was made. Finally, we use the activity as given in Eq. 41–7c to find the age of the club. The half-life is found in Appendix G.  ( 67 g )  23 24 12 N 12 C =   ( 6.022  10 atoms mol ) = 3.362  10 atoms of 6 C 6  (12g mol ) 

N 14 C = (1.3  10−12 )( 3.362  1024 ) = 4.371  1012 nuclei of 146 C 6

( N ) 14 6C

today

(

=  N 14 C 6

)e 0

− t

→

N ) N ) ( ( T t = − ln =− ln  (  N ) ln 2  ln 2 N  1

14 6C

=−

today

1/ 2

14 6C

today

 T1/ 2 6 0 ( 7.0decays s )

5730 yr = 7214 yr  7200 yr ln ln 2   ln 2 12  ( 4.37110 nuclei ) 7  ( 5730 yr ) ( 3.156  10 s yr )  0

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58. The decay rate is given by Eq. 41–4b,

N = − N . We assume equal numbers of nuclei decaying by t

 emission for the two isotopes.  N  T1/2   (1.6 10−4 s ) = 8.6  10−7  t 218 = −218 N 218 = 218 = 214 = −214 N 214 214 T1/2 ( 3.1min )( 60s min )  N    218  t  214

59. The activity is given by Eq. 41–7a. The original activity is  N 0 , so the activity 351.0 hours later is 0.945 N 0 . ln 2 t → 0.945 N 0 =  N 0 e − t → ln 0.945 = −t = − T1/ 2

T1/ 2 = −

 1d  ln 2 ( 35.0 h ) = 428.85h   = 17.9d ln 0.945  24 h 

60. The activity is given by Eq. 41–7a. (a) Use Eq. 41–7d. T 1 R 53d  25decays s  R = − 1/2 ln =− R = R0 e− t → t = − ln  ln  = 194.94d  190d  R0 ln 2 R0 ln 2  320decays s  (b) We find the mass from the activity. Note that N A is used to represent Avogadro’s number, and A is the atomic weight. ln 2 m0 N A → R0 =  N 0 = T1/ 2 A

R0T1/ 2 A ( 320decays s )( 53d )( 86, 400s d )( 7.016929g mole ) = = 2.5  10−14 g 23 N A ln 2  6.022 10 nuclei mole ln 2 ( )

61. The number of radioactive nuclei decreases exponentially, and every radioactive nucleus that decays becomes a daughter nucleus. N = N 0 e − t

N D = N 0 − N = N 0 (1 − e − t ) 62. The activity is given by Eq. 41–7d, with R = 0.01050R0 . R ln ( 4.00 h ) ln 2 R0 ln 2 t ln 2 = → T1/ 2 = − =− = 0.6085h = 36.5min R = R0 e − t →  = − R ln 0.01050 t T1/ 2 ln R0 From Appendix G, we see that the isotope is

211 82

Pb .

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63. Because the carbon is being replenished in living trees, we assume that the amount of 146 C is constant until the wood is cut, and then it decays. We use Eq. 41–6. The half-life is given in Appendix G. The N ratio is 0.062. N0

ln N = N 0 e − t →  = −

N N 0 ln 2 = → t T1/ 2

( 5730 yr ) ln 0.062 = 22,986 yr  23,000 yr T N =− t = − 1/ 2 ln ln 2 N 0 ln 2 64. (a)

The mass number is found from the radius, using Eq. 41–1. r = (1.2  10−15 m ) A1/3 → 3

   7500 m  r 56 56 A=  =  = 2.441  10  2.4  10 −15 −15  1.2  10 m   1.2  10 m  (b) The mass of the neutron star is the mass number times the atomic mass unit conversion in kg. m = ( 2.441  1056 u )(1.6605  10−27 kg u ) = 4.053  10 29 kg  4.1  10 29 kg

Note that this is about 20% of the mass of the Sun. (c) The acceleration of gravity on the surface of the neutron star is found from Eq. 6–4 applied to the neutron star. 2 2 29 −11 Gm ( 6.67  10 N m kg )( 4.053  10 kg ) g= 2 = = 4.806  1011 m s 2  4.8  1011 m s 2 2 r ( 7500 m ) 65. Because the tritium in water is being replenished, we assume that the amount is constant until the N wine is made, and then it decays. We use Eq. 41–6. The value of is 0.10. N0 N ln (12.3yr ) ln 0.10 N 0 ln 2 T N N = N 0 e − t →  = − = → t = − 1/ 2 ln =− = 41yr t T1/ 2 ln 2 N 0 ln 2 66. (a) We assume a mass of 70 kg of water, and find the number of protons, given that there are 10 protons in a water molecule.  ( 70  103 g water )   molecules water  10 protons  N protons =    6.02  1023   mol water  water molecule   (18g water mol water )   = 2.34  1028 protons We assume that the time is much less than the half-life so that the rate of decay is constant.  ln 2  N = N =  N → t  T1/2   1033 yr  N  T1/2  1 proton 4 =   = 6.165  10 yr  60,000 yr   28 N  ln 2  2.34  10 protons  ln 2  This is about 880 times longer than a life expectancy of 70 years. t =

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(b) Instead of waiting 61,650 consecutive years for one person to experience a proton decay, we could interpret that number as there being 880 people, each living 70 years, to make that 61,650 years (since 880  70 = 61,600). We would then expect one person out of every 880 to experience a proton decay during their lifetime. Divide 7 billion by 880 to find out how many people on Earth would experience proton decay during their lifetime. 7  109 = 7.95  106  8 million people 880 67. We assume that all of the kinetic energy of the alpha particle becomes electrostatic potential energy at the distance of closest approach. Note that the distance found is the distance from the center of the alpha to the center of the gold nucleus. The alpha particle has a charge of +2e, and the gold nucleus has a charge of +79e. 1 q qAu K i + U i = K f + U f → K + 0 = 0 + → 4 0 r

( 2 )( 79 ) (1.60  10−19 C ) 1 q qAu 9 2 2 r= = ( 8.988  10 N  m C ) = 2.951  10−14 m −13 4 0 K ( 7.7MeV ) (1.60 10 J MeV ) 2

 3.0  10−14 m

We use Eq. 41–1 to compare to the size of the gold nucleus. rapproach 2.951 10−14 m = = 4.2 rAu 1971/3 (1.2  10−15 m ) So the distance of approach is about 4.2  the radius of the gold nucleus. 68. We find the number of half-lives from the change in activity. dN ln 0.0150 n dt = ( 12 ) = 0.0150 → n = = 6.06 half-lives dN ln 12 dt 0 It takes 6.06 half-lives for a sample to drop to 1.50% of its original activity. 69. We can find the mass of 40 19 K from its activity. Note that N A is Avogadro’s number, N 40 is the 40 39 number of particles of 40 19 K , and A40 is the atomic mass of 19 K . Similar notation is used for 19 K . ln 2 m40 R40 =  N 40 = NA → T1/ 2 A40

R40T1/ 2 A40 ( 48decays s ) (1.248  10 yr )( 3.156  10 s yr ) ( 39.963998g mole ) = N A ln 2 ( 6.022 1023 nuclei mole ) ln 2 9

m40 =

= 1.8101  10−4 g  1.8  10−4 g 40 We find the number of 39 19 K atoms from the number of 19 K atoms and the abundances given in

Appendix G. That is then used to find the mass of 39 19 K. R R T R40 =  N 40 → N 40 = 40 = 40 1/ 2 ;  ln 2 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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0.932581 N 40 0.000117 A  0.932581 R40T1/ 2  A39  0.932581 A m39 = N 39 39 =  N 40  39 =   N A  0.000117  N A  0.000117 ln 2  N A

N 40 = 0.000117 N K ; N 39 = 0.932581N K =

 0.932581  ( 48decays s ) (1.248  10 yr )( 3.156  10 s yr ) ( 38.963706g mole ) =  ln 2  0.000117  ( 6.022 1023 nuclei mole ) 9

= 1.407 g  1.4g

70. We see from the periodic table that Sr is in the same column as calcium. If strontium is ingested, the body may treat it chemically as if it were calcium, which means it might be stored by the body in bones. We use Eq. 41–6 to find the time to reach a 1% level. N ln N 0 ln 2 = → N = N 0 e − t →  = − t T1/ 2

( 29 yr ) ln 0.01 = 192.67 yr  200 yr T N =− t = − 1/ 2 ln ln 2 N 0 ln 2 The decay reactions are as follows. The strontium reaction is beta decay, as given in Appendix G. We assume the daughter also undergoes beta decay. 90 38

Sr → 9039Y + −01 e + v ;

0 Y →90 40 Zr + −1 e + v

90 39

71. We find the mass from the initial decay rate and Eq. 41–7b. dN 6.022  1023 nuclei mole =  N0 =  m → dt 0 ( atomic weight ) g mole m=

dN 1 ( atomic weight ) dN T1/ 2 ( atomic weight ) = dt 0  ( 6.022  1023 ) dt 0 ln 2 ( 6.022  10 23 )

= ( 3.94  104 s −1 )

( 87.37 d )(86, 400s d ) ( 34.969032g ) = 2.49  10−11 g ( ln 2 ) ( 6.022  1023 )

72. (a) We find the daughter nucleus by balancing the mass and charge numbers: 191 Z ( X ) = Z ( Os ) − Z ( e − ) = 76 − ( − 1) = 77 76 Os

A ( X ) = A ( Os ) − A ( e − ) = 191 − 0 = 191

The daughter nucleus is 191 77 Ir . (b) See the included diagram. (c) Because there is only one  energy, the  decay must be to the higher excited state.

 – (0.14 MeV)  (0.042 MeV)  (0.129 MeV)

191 77 Ir* 191 77 Ir* 191 77 Ir

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73. The activity is the decay constant times the number of nuclei, as given by Eq. 41–7a. (a) We calculate the activity for 32 15 P.

R = N =

  ln 2 ln 2 1.0g 23 N=   ( 6.022  10 nuclei mol ) T1/ 2 (14.3d )(86, 400s d )  31.973908g mol 

= 1.057  1016 decays s  1.1  1016 decays s (b) We calculate the activity for 23290Th. ln 2 R= N T1/ 2 =

  ln 2 1.0g 23   ( 6.022  10 nuclei mol ) 7 232.038054g mol (1.40  10 yr )( 3.156 10 s yr )   10

= 4.1  103 decays s

74. From Fig. 41–1, the average binding energy per nucleon at A = 63 is ~8.6 MeV. We use the mass average atomic weight as the average number of nucleons for the two stable isotopes of copper. That gives a binding energy of ( 63.546 )(8.6MeV ) = 546.5MeV  550MeV . The number of copper atoms in a penny is found from the atomic weight. ( 3.0g ) N= ( 6.022 1023 atoms mol ) = 2.843 1022 atoms ( 63.546g mol ) Thus the total energy needed is the product of the number of atoms times the binding energy.

( 2.843 10 atoms ) ( 546.5MeV atom ) (1.60 10 22

−13

J MeV ) = 2.5  1012 J

75. (a)  ( 42 He ) = m ( 42 He ) − A ( 42 He ) = 4.002603u − 4 = 0.002603u = ( 0.002603u ) ( 931.5MeV uc 2 ) = 2.425Mev c 2

(b)  ( 126 C ) = m ( 126 C ) − A ( 126 C ) = 12.000000u − 12 = 0 for both units. (c)

86 86  ( 86 38 Sr ) = m ( 38 Sr ) − A ( 38 Sr ) = 85.909261u − 86 = − 0.090739 u

= ( − 0.090739 u ) ( 931.5MeV uc 2 ) = − 84.52 MeV c 2

235 235 (d)  ( 235 92 U ) = m ( 92 U ) − A ( 92 U ) = 235.043928u − 235 = 0.043928u

= ( 0.043928u ) ( 931.5MeV uc 2 ) = 40.92 MeV c 2

(e) From Appendix G we see that   0 for 0  Z  7 and Z  86;

  0 for 8  Z  85.

  0 for 0  A  15 and A  214;   0 for 16  A  214.

76. The reaction is 11 H + 01 n → 21 H. If we assume the initial kinetic energies are small, then the energy of the gamma is the Q-value of the reaction. Q =  m ( 11 H ) + m ( 01 n ) − m ( 21 H ) c 2

= (1.007825u ) + (1.008665u ) − ( 2.014102 u )  c 2 ( 931.5MeV uc 2 ) = 2.224 MeV

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Instructor Solutions Manual

77. The mass of carbon 60,000 years ago was 1.0 kg. Find the total number of carbon atoms at that time, and then find the number of 146 C atoms at that time. The text says that the fraction if C-14 atoms is

1.3  10−12 . Use that with the half-life to find the present activity, using Eq. 41–7d and Eq. 41–8. (1.0 103 g )( 6.022 1023 atoms mol ) = 5.014 1025 C atoms N= 12.0109g mol N14 = ( 5.014  1025 )(1.3  10−12 ) = 6.518  1013 nuclei of 146 C = N 0 ln 2

− t − ln 2 ln 2 ln 2 13 R = N = N= N 0 e T1/2 = 6.518 10 e  ( ) T1/ 2 T1/ 2 ( 5730 yr ) ( 3.156 107 s yr )

ln 2( 60,000yr )

( 5730 yr )

= 0.1759decays s  0.2decays s Since the time only has 1 significant figure, we only keep 1 significant figure in the answer. 78. The energy to remove the neutron would be the difference in the masses of the 42 He and the combination of 23 He + n. It is also the opposite of the Q-value for the reaction. The number of electrons doesn’t change, so atomic masses can be used for the helium isotopes.  931.5MeV c 2  −QHe = ( mHe-3 + mn − mHe-4 ) c 2 = ( 3.016029 u + 1.008665u − 4.002603u ) c 2   u  

= 20.58 MeV Repeat the calculation for the carbon isotopes.

 931.5MeV c 2  −QC = ( mC-12 + mn − mC-13 ) c 2 = (12.000000 u + 1.008665u − 13.003355u ) c 2   u   = 4.946 MeV The helium value is 20.58MeV 4.946MeV = 4.161 greater than the carbon value. 79. (a) Take the mass of the Earth and divide it by the mass of a nucleon to find the number of nucleons. Then, use Eq. 41–1 to find the radius. 1/3

 5.98  1024 kg  r = (1.2  10 m ) A = (1.2  10 m )   = 183.6 m  180 m −27 1.67  10 kg  (b) Follow the same process as above, just using the Sun’s mass. −15

1/3

−15

1/3

 1.99  1030 kg  r = (1.2  10−15 m ) A1/3 = (1.2  10−15 m )   −27  1.67  10 kg 

= 12700 m  1.3  104 m

80. (a) The usual fraction of 146 C is 1.3  10−12. Because the fraction of atoms that are 146 C is so small, we use the atomic weight of 126 C to find the number of carbon atoms in 84 g. We use Eq. 41–6 to find the time.  84g  23 24 N12 =   ( 6.022  10 atoms mol ) = 4.215  10 atoms 12g mol   N14 = ( 4.215  1024 atoms )(1.3  10−12 ) = 5.4795  1012 atoms

( 5730 yr ) T N 1 N 1 N = N 0 e − t → t = − ln ln = − 1/ 2 ln =− = 2.4  105 yr ln 2 N 0 ln 2 5.4795  1012  N0 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) We do a similar calculation for an initial mass of 340 grams.  340g  23 −12 13 N14 =   ( 6.022  10 atoms mol )(1.3  10 ) = 2.218  10 atoms  12g mol 

( 5730 yr ) ln T 1 N 1 N N = N 0 e − t → t = − ln = − 1/ 2 ln =− = 2.5  105 yr 13  N0 ln 2 N 0 ln 2 2.218  10

This shows that, for times on the order of 105 yr, the sample amount has fairly little effect on the age determined. Thus, times of this magnitude are not accurately measured by carbon dating. 81. (a) This reaction would turn the protons and electrons in atoms into neutrons. This would eliminate chemical reactions, and thus eliminate life as we know it. (b) We assume that there is no kinetic energy brought into the reaction, and solve for the increase of mass necessary to make the reaction energetically possible. For calculating energies, we write the reaction as 11 H → 10 n + v, and we assume the neutrino has no mass or kinetic energy.

Q =  m ( 11 H ) − m ( 01 n )  c 2 = (1.007825u ) − (1.008665u ) c 2 ( 931.5MeV uc 2 )

= − 0.782 MeV This is the amount that the proton would have to increase in order to make this energetically possible. We find the percentage change.  ( 0.782 MeV c 2 )   m  = 100  (100 ) = 0.083% )   ( 2  m   ( 938.27 MeV c )  82. We assume the particles are not relativistic, so that p = 2mK . The radius is given in Example 27–7 mv . Set the radii of the two particles equal. Note that the charge of the alpha particle is twice as r = qB that of the electron (in absolute value). We also use the “bare” alpha particle mass, subtracting the two electrons from the helium atomic mass. m v mv r = r →   = → m v = 2m v → p = 2 p eB 2eB 4 p2 p2 4m 4 ( 0.000549 u ) K 2m 2m = 2 = 2 = = = 5.48  10−4 p p 4.002603u − 2 ( 0.000549 u ) K m

2m

83. Natural samarium has an atomic mass of 150.36 grams per mole. We find the number of nuclei in the natural sample, and then take 15% of that to find the number of 147 62 Sm nuclei. We first find the number of 147 62 Sm nuclei from the mass and proportion information.

N 147 Sm = ( 0.15 ) N natural =

( 0.15)(1.00g ) ( 6.022  1023 nuclei / mol )

150.36g mol The activity level is used to calculate the half-life. 62

= 6.008  1020 nuclei of 147 62 Sm

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Activity = R =  N = T1/ 2 =

Instructor Solutions Manual

ln 2 N → T1/ 2

ln 2 ln 2 1yr   N= 6.008  1020 ) = 3.44069  1018 s  = 1.1  1011 yr ( 7  R 120decays s 3.156 10 s   

84. Since amounts are not specified, we will assume that “today” there is 0.720 g of 235 92 U and

100.000 − 0.720 = 99.280g of 238 92 U. We use Eq. 41–6. (a) Relate the amounts today to the amounts 1.0 109 years ago. t

N = N 0 e − t → N 0 = Net = Ne T1/2

ln 2

(1.010 yr ) ln 2 ( 7.0410 yr ) 9

( N 0 )235 = ( N 235 ) e T

ln 2

1/2

= ( 0.720g ) e

= 1.927 g

(1.010 ) ln 2 ( 4.46810 ) 9

( N0 )238 = ( N 238 ) e T

ln 2

1/2

= ( 99.280g ) e

= 115.94g

1.927  100% = 1.63% 1.927 + 115.94

The percentage of 235 92 U was

(b) Relate the amounts today to the amounts 100  106 years from now.

(10010 yr ) ln 2 ( 7.0410 yr ) 6

N = N 0 e − t →

−

( N 235 ) = ( N 0 )235 e T

ln 2

1/2

= ( 0.720g ) e

(10010 yr ) ln 2 ( 4.46810 yr )

−

= 0.6525g

−

( N 238 ) = ( N0 )238 e T

1/2

ln 2

= ( 99.280g ) e

The percentage of 235 92 U will be

−

= 97.752g

0.6525  100% = 0.663% 0.6525 + 97.752

85. (a) The decay constant is calculated from the half-life using Eq. 41–8. ln 2 ln 2 = = = 3.83  10−12 s −1 7 T1/2 ( 5730 yr ) ( 3.156  10 s yr ) (b) In a living organism, the abundance of 146 C atoms is 1.3  10−12 per carbon atom. Multiply this abundance by Avogadro’s number and divide by the molar mass of carbon-12 to find the number of carbon-14 atoms per gram of carbon. 14 14 6.022  10 23 atoms 126 C mole  C atoms  N A  −12 6 C atoms  N =  1.3  10 −12 126 = 1.3  10    12 C atoms  M  C atoms  12.000000 g 126 C mole  6 6

(

)

= 6.524 atoms 146 C g 126 C  6.5  1010 atoms 146 C g 126 C (c) The activity in natural carbon for a living organism is the product of the decay constant and the 14

number of 6 C atoms per gram of 6 C . Use Eq. 41–7d and Eq. 41–4b.

(

R =  N = 3.83  10−12 s −1

)( 6.524 10 atoms C g C ) = 0.2499 decays s g C 10

14 6

12 6

 0.25 decays s g 126 C

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(d) Take the above result as the initial decay rate (while Otzi was alive), and use Eq. 41–7d to find the time elapsed since he died. R = R0 e − t → t=−

R 1 1yr  0.121   11  = 6000 yr ln   = 1.894  10 s  =− −12 −1 7    R 3.83 10 s 0.2499 3.156 10 s      0

ln 

 Otzi lived approximately 6000 years ago.

86. (a) We use the definition of the mean life given in the problem. We use a definite integral formula from Appendix B-5. 

 tN ( t ) dt

 = 0



− t  tN 0e dt

= 0



 te dt

= 0

− t

 N ( t ) dt  N e dt  e dt − t

− t



1 − e− t

(b) We evaluate at time t =  =



2 

2 1 = = 1 



 − N ( t =  ) N 0e −   e = = = e −1 = 0.368 N ( t = 0 ) N 0e −0

87. The mass number changes only with  decay, and changes by − 4. If the mass number is 4n, the new number is 4n − 4 = 4 ( n − 1) = 4n. There is a similar result for each family, as shown here. 4n → 4n − 4 = 4 ( n − 1) = 4n 4n + 1 → 4n − 4 + 1 = 4 ( n − 1) + 1 = 4n + 1 4n + 2 → 4n − 4 + 2 = 4 ( n − 1) + 2 = 4n + 2 4n + 3 → 4n − 4 + 3 = 4 ( n − 1) + 3 = 4n + 3

Thus, the daughter nuclides are always in the same family. 88. (a) If the initial nucleus is at rest when it decays, momentum conservation says that the magnitude of the momentum of the alpha particle will be equal to the magnitude of the momentum of the daughter particle. We use that to calculate the (non-relativistic) kinetic energy of the daughter particle. The mass of each particle is essentially equal to its atomic mass number, in atomic mass units. Note that classically, p = 2mK .

p = pD ; K D =

2m K p2 m A pD2 4 K =  =   =  K =  K = mD AD AD 2mD 2mD 2mD

4 K AD

K KD 1 = = = 1 A K + K D  1 +   D  4 4 AD  K + A K   4 K + K  D   (b) We specifically consider the decay of 226 88 Ra. The daughter has AD = 222. KD 1 1 = = = 0.017699  1.8% 1 1 K + K D 1 + 4 AD 1 + 4 ( 222 ) © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Thus, the alpha particle carries away 1 − 0.018 = 0.982 = 98.2% . 89. We determine the number of 40 19 K nuclei in the sample, and then use the half-life to determine the activity. ( 400  10−3 g )( 6.022 1023 atoms mol ) = 7.208 1017 N 40 K = ( 0.000117 ) N naturally = ( 0.000117 ) 19 39.0983g mol occurring K R = N =

ln 2 ln 2 N= ( 7.208  1017 )  3.1561yr107 s  = 12.51decay s  13decay s T1/ 2 1.265  109 yr

232 4 90. (a) The reaction is 236 92 U → 90Th + 2 He. If we assume the uranium nucleus is initially at rest, the magnitude of the momenta of the two products must be the same. The kinetic energy available to the products is the Q-value of the reaction. We use the non-relativistic relationship that p 2 = 2mK .

pHe = pTh ; K Th =

2 p2 2mHe K He mHe pTh K He = He = = 2mTh 2mTh 2mTh mTh

m  Q = K Th + K He =  He + 1 K He →  mTh   mTh   mTh  2 K He =  Q =    mU − mTh − mHe  c m + m m + m Th  Th   He  He   232.038054 u 2 =   236.045566 u − 232.038054 u − 4.002603u  c 4.002603u + 232.038054 u   = 0.004826u ( 931.5MeV uc 2 ) = 4.495MeV

(b) We use Eq. 41–1 to estimate the radii.

rHe = (1.2  10−15 m ) ( 4 )

1/3

= 1.905  10−15 m  1.9  10−15 m

rTh = (1.2  10−15 m ) ( 232 )

1/3

= 7.374  10−15 m  7.4  10−15 m

(c) The maximum height of the Coulomb barrier will correspond to the alpha particle and the thorium nucleus being separated by the sum of their radii. We use Eq. 23–10. qHe qTh 1 qHe qTh 1 = U= 4 0 rA 4 0 ( rHe + rTh )

= ( 8.988  10 N  m C ) 9

1/3

( 2 )( 90 ) (1.60  10−19 C )

+ 2321/3 )(1.2  10−15 m )(1.60  10 −13 J MeV )

= 27.898MeV  28MeV (d) At position “A”, the product particles are separated by the sum of their radii, about 9.3 fm. At position “B”, the alpha particle will have a potential energy equal to its final kinetic energy, 4.495 MeV. Use Eq. 23–10 to solve for the separation distance at position “B”. 1 qHe qTh UB = → 4 0 RB

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( 2 )( 90 ) (1.60  10−19 C ) 1 qHe qTh 9 2 2 RB = = ( 8.988  10 N  m C ) 4 0 U B ( 4.495MeV ) (1.60  10−13 J MeV ) 2

= 57.57  10−15 m RB − RA = 57.6fm − 9.3fm = 48.3fm

Note that this is a center-to-center distance. 91. We take the momentum of the nucleon to be equal to the uncertainty in the momentum of the nucleon, as given by the uncertainty principle. The uncertainty in position is estimated as the radius of the nucleus. With that momentum, we calculate the kinetic energy using a classical formula. Iron has about 56 nucleons (depending on the isotope).

px 

→ p  p 

x

(

)

2 (1.055 10−34 J s ) p2 K= = = 2m 2mr 2 2 (1.67  10−27 kg ) ( 561/3 )(1.2  10−15 m )  2 (1.60  10−13 J MeV )   2

= 0.988MeV  1MeV 92. (a) From the figure, the initial force on the detected particle is down. Using the right-hand rule, the force on a positive particle would be upward. Thus, the particle must be negative, and the decay is  − decay. (b) The magnetic force is producing circular motion. Set the expression for the magnetic force equal to the expression for centripetal force, and solve for the velocity. −19 −3 mv 2 qBr (1.60  10 C ) ( 0.012T ) ( 4.7  10 m ) qvB = → v= = = 9.9  106 m s 9.11  10−31 kg r m 93. If the original nucleus is of mass 256 u, the daughter nucleus after the alpha decay will have a mass of 252 u. Moreover, the alpha has a mass of 4 u. Use conservation of momentum – the momenta of the daughter nucleus and the alpha particle must be equal and opposite since there are only two products. The energies are small enough that we may use non-relativistic relationships. m 4 p = p X → 2m K = 2mX K X → K X =  K = ( 4.5MeV ) = 0.071MeV 252 mX 94. We use Eq. 41–7a to relate the activity to the half-life. We estimate the atomic weight at 152 grams/mole since this isotope is not given in Appendix G. ln 2 dN = R = N = N → dt T1/ 2

T1/ 2 =

 6.022  1023 nuclei   ln 2 ln 2 1yr  7  = 1.3  1021 yr 1.5 10 g N= ( )   7   1decay s 152g 3.156 10 s R   

95. We calculate the initial number of nuclei from the initial mass and the atomic mass. 1 nucleus N 0 = ( 2.20  10−9 kg ) = 1.0187  1017 nuclei  1.02  1017 nuclei −27 (13.005739u ) (1.6605  10 kg u ) See the adjacent graph. From the graph, the half-life is approximately 600 seconds. Note that the © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1387

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half-life as given in Appendix G is T1 2 = 9.965min

Instructor Solutions Manual

60s = 597.9s . 1min

96. The emitted photon and the recoiling nucleus have the same magnitude of momentum. We find the recoil energy from the momentum. We assume the energy is small enough that we can use classical relationships. E = pK = 2mK K K → p = c

KK =

E2 2mK c

(1.46 MeV )

 931.5MeV c  2 2 ( 39.963998u )  c u   2

= 2.86  10−5 MeV = 28.6eV

97. Section 41–1 states that the neutron, proton, and electron are all spin 12 particles. Consider the reaction n → p + e − + v . If the proton and neutron spins are aligned (both are 12 , for example), the electron and anti-neutrino spins must cancel. Since the electron is spin 12 , the anti-neutrino must be spin − 12 , which means the magnitude of the neutrino spin must be 12 in this case. The other possibility is if the proton and neutron spins are opposite of each other. Consider the case of the neutron having spin 12 and the proton having spin − 12 . If the electron has spin 12 , the spins of the electron and proton cancel, and the neutrino must have spin 12 for angular momentum to be conserved. If the electron has spin − 12 , the spin of the neutrino must be 32 for angular momentum to be conserved. A similar argument could be made for positron emission, with p → n + e + + v.

1388

CHAPTER 41: Nuclear Physics and Radioactivity Responses to Questions 1.

With 88 nucleons and 50 neutrons, there must be 38 protons. The number of protons is the atomic number, and so the element is strontium. The nuclear symbol is 88 Sr . 38

Identify the element based on the atomic number. (a) Uranium (Z = 92) (b) Nitrogen (Z = 7) (c) Hydrogen (Z = 1) (d) Strontium (Z = 38) (e) Berkelium (Z = 97)

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Instructor Solutions Manual

The resulting nuclide for gamma decay is the same isotope in a lower energy state: 64 64 29 Cu* →29 Cu + γ . 64

The resulting nuclide for beta-minus decay is an isotope of zinc, 30 Zn : 64 29

64 Cu →30 Zn + e− + .

The resulting nuclide for beta-plus decay is an isotope of nickel, 28 Ni : 64 29

10.

238 92

+ Cu →64 28 Ni + e +  .

U decays by alpha emission into 23490Th, which has 144 neutrons.

( )

−

(b) Neon is formed: 11 Na → 10 Ne + e + . 22

206

14. (a) Sulfur is formed: 15 Pb → 16 S + e + . 32

−

(b) Chlorine is formed: 16 S → 17 Cl + e + . 35

−

207

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Chapter 41

Nuclear Physics and Radioactivity

Sc + e − + v

Scandium-45 is the missing nucleus.

Cu + 

Copper-58 is the missing nucleus.

15. (a)

45 20

Ca →

(b)

58 29

Cu* →

(c)

46 24

Cr → 4623V + e + + v

(d)

234 94

Pu →

(e)

239 93

Np → 239 Pu + e − + v 94

45 21

58 29

230 92

U +

The positron and the neutrino are the missing particles. Uranium-230 is the missing nucleus. The electron and the anti-neutrino are the missing particles.

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Instructor Solutions Manual

Li + 11 p → 24 + 23 He 14

1392

Chapter 41

Nuclear Physics and Radioactivity

MisConceptual Questions 1.

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Solutions to Problems 1.

Convert the units from MeV c 2 to atomic mass units.

  1u m = ( 775MeV c 2 )  = 0.832u 2   931.49MeV c 

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The  particle is a helium nucleus and has A = 4. Use Eq. 41–1. 1

r = (1.2  10−15 m ) A 3 = (1.2  10−15 m ) ( 4 ) 3 = 1.9  10−15 m = 1.9fm

Use Eq. 41–1 for both parts.

(a) r = (1.2  10−15 m ) A1/3 = (1.2  10−15 m ) (120 )

1/3

= 5.9  10 −15 m = 5.9fm 3

−15 r    3.7  10 m  = (b) r = (1.2  10−15 m ) A1/ 3 → A =   = 29.3  29   −15 −15  1.2  10 m   1.2  10 m 

To find the rest mass of an  particle, we subtract the rest mass of the two electrons from the rest mass of a helium atom: m = mHe − 2me

= ( 4.002603u ) ( 931.5MeV uc 2 ) − 2 ( 0.511MeV c 2 ) = 3727 MeV c 2 This is less than the sum of the masses of two protons and two neutrons because of the binding energy. 6.

q q

(8.988  10 N  m C ) ( 2 ) (1.60 10 9

−19

C ) 

4 0 ( 2r )2

First, we calculate the density of nuclear matter. The mass of a nucleus with mass number A is approximately (A u) and its radius is r = (1.2  10−15 m ) A1/ 3 . Calculate the density.

=

m A (1.6605  10 = 3 4 V 3r

−27

kg u )

A (1.6605  10−27 kg u ) 4 3

 (1.2  10

−15

m) A 3

= 2.294  1017 kg m3

We see that this is independent of A. The value has 2 significant figures.

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(a) We set the density of the Earth equal to the density of nuclear matter. M  Earth =  nuclear = 4 Earth → 3 matter 3  REarth 1/3

 ( 238u ) (1.6605  10−27 kg u )   = ( 6.38  10 m )    ( 5.98 1024 kg ) rU 2.58  10−10 m = = 3.5  104 rU actual (1.2  10−15 ) ( 238 )1/3

1/3

 m  rU = REarth  U   M Earth 

= 2.58  10−10 m

31 15

K = ( 8.988  10 N  m C ) 9

1/3

( 2 )(104 ) (1.60  10−19 C ) + 263

1/3

)(1.2 10 m )(1.60 10 −15

−19

J eV )

= 3.118  107 eV

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K = ( 8.988  10 N  m C ) 9

( 2 )( 90 ) (1.60  10−19 C )

( 41/3 + 2321/3 )(1.2 10−15 m )(1.60 10−19 J eV )

= 2.790  107 eV

18 8

( )

= 8 (1.007825u ) +10 (1.008665u ) − (17.999160 u )  c 2  931.5MeV c 2  = 0.150090   = 139.81MeV u   Binding energy per nucleon = (139.81MeV ) 18 = 7.767 MeV

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= ( 30.973762 u ) + (1.008665u ) − ( 31.973908u )  c 2 (931.5MeV / c 2 ) = 7.935MeV

18. (a)

7 4

( )

(

)

=  4 (1.007825u ) + 3 (1.008665u ) − ( 7.016929 u )  c 2 ( 931.5MeV / c 2 ) = 37.60 MeV

Binding energy 37.60 MeV = = 5.372 MeV nucleon nucleon 7 nucleons

(b)

197 79

Au consists of 79 protons and 118 neutrons. We find the binding energy as in part (a).

 2 Binding energy = 79m ( 11 H ) + 118m ( 10 n ) − m ( 197 79 Au )  c

=  79 (1.007825u ) + 118 (1.008665u ) − (196.966570 u )  c 2 ( 931.5MeV / c 2 ) = 1559.4 MeV  1559 MeV

Binding energy 1559.4 MeV = 7.916 MeV nucleon = nucleon 197 nucleons

19.

31 15

P consists of 15 protons and 16 neutrons. We find the binding energy from the masses: 31 Binding energy = 15 m ( 11 H ) + 16m ( 01 n ) − m ( 15 P )  c 2

= 15 (1.007825u ) + 16 (1.008665u ) − ( 30.973762 u )  c 2 ( 931.5MeV uc 2 )

= 262.9 MeV Binding energy 262.9 MeV = = 8.481MeV nucleon nucleon 31 We do a similar calculation for 32 15 P, consisting of 15 protons and 17 neutrons.

 2 Binding energy = 15 m ( 11 H ) + 17 m ( 01 n ) − m ( 32 15 P )  c

= 15 (1.007825u ) + 17 (1.008665u ) − ( 31.973908u )  c 2 (931.5MeV uc 2 ) = 270.85MeV

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= (15.000109 u ) + (1.007825u ) − (15.994915u )  ( 931.5MeV uc 2 ) = 12.13MeV

(b) Removal of a neutron creates another isotope of oxygen: 168 O → 01 n + 158 O. Energy needed =  m ( 158 O ) + m ( 01 n ) − m ( 168 O )  c 2

= (15.003066 u ) + (1.008665u ) − (15.994915u )  ( 931.5MeV uc 2 ) = 15.66 MeV

=  2 ( 4.002603u ) − ( 8.005305u )  c 2 ( 931.5MeV uc 2 ) = − 0.092 MeV

= 3 ( 4.002603u ) − (12.000000 u ) c 2 ( 931.5MeV uc 2 ) = + 7.3MeV

= ( 3.016049 u ) − ( 3.016029 u ) c 2 ( 931.5MeV uc 2 ) = 0.019 MeV

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m = m ( 105 B ) + m ( 11 H ) − m ( 116 C )

( )

(

)

1 40 m = m ( 39 19 K ) + m ( 0 n ) − m ( 19 K ) = 38.963706u + 1.008665u − 39.963998u = 0.008373u

Because an increase in mass is required, the decay is not possible. 24 27. (a) From Appendix G, 11 Na is a  − emitter .

(b) The nuclear decay reaction is

24 11

24 Na → 12 Mg +  − + v . We add 11 electrons to both sides in

= ( 23.990963u ) − ( 23.985042 u )  c 2 ( 931.5MeV uc 2 ) = 5.515MeV

28. (a) We find the final nucleus by balancing the mass and charge numbers. Z ( X ) = Z ( U ) − Z ( He ) = 92 − 2 = 90

A ( X ) = A ( U ) − A ( He ) = 238 − 4 = 234 Thus, the final nucleus is

234 90

Th .

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Nuclear Physics and Radioactivity

4 m ( 23490Th ) = m ( 238 92 U ) − m ( 2 He ) −

K c2

214 4  2 Q =  m ( 218 84 Po ) − m ( 82 Pb ) − m ( 2 He )  c

32. (a) We find the final nucleus by balancing the mass and charge numbers. Z ( X ) = Z ( P ) − Z ( e ) = 15 − ( − 1) = 16

A ( X ) = A ( P ) − A ( e ) = 32 − 0 = 32 32 S. Thus, the final nucleus is 16

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23 10

− Ne → 23 11 Na +  + v . We add 10 electrons to both sides in order to

= (13.005739 u ) − (13.003355 ) − 2 ( 0.00054858u )  c 2 (931.5MeV uc 2 ) = 1.199 MeV

If the  + and electron have no kinetic energy, the maximum kinetic energy of the neutrino is also

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Nuclear Physics and Radioactivity

p = pRn ; K Rn =

ln 2 ln 2 t=− ( 4.2 h ) = 1.4 h 140 R ln ln R0 1120 We can see this also from the fact that the rate dropped by a factor of 8, which takes 3 half-lives. R = R0 e

− t

= R0 e

−

→ T1/ 2 = −

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41. We use Eq. 41–6 to find the fraction remaining. N = N0e

− t

→

 ( ln 2 )( 2.2 yr )(12mo yr )   9 mo 

− N = e − t = e  N0

= 0.1309  0.13 = 13%

(b) We evaluate Eq. 41–7d at t = 1.50h. This is much less than one half life, so we expect that most of the initial nuclei remain, and the activity is essentially unchanged.

R = R0 e − t = ( 3.8905  1012 decays s ) e

(

)

 3600s  − 9.9966810−7 s −1 (1.50 h )   h 

= 3.8696  1012 decays s

R = R0 e − t = ( 3.8905  1012 decays s ) e

(

)

(

− 9.9966810−7 s −1 ( 0.25yr ) 3.156107 s yr

)

= 1.4607  109 decays s

 Co

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Nuclear Physics and Radioactivity

T1/ 2 dN ( 4.468  10 yr )( 3.156  10 s yr ) = ( 370decays s ) = 7.5  1019 nuclei ln 2 dt ln 2 9

N = N0e− t = ( 4.662 1016 ) e

(

)

− 0.022432s−1 ( 2.6min )( 60s min )

= 1.409 1015  1.4 1015 nuclei

 N = ( 0.022432s −1 )(1.409  1015 ) = 3.161  1013  3.2 1013 decays s

(d) We find the time from Eq. 41–7a, relative to the time when we had the initial 9.6 g of 124 55 Cs .

49. We find the mass from the initial decay rate and Eq. 41–7b. dN 6.022  1023 nuclei mole =  N0 =  m → dt 0 ( atomic weight ) g mole

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dN 1 ( atomic weight ) dN T1/ 2 ( atomic weight ) = dt 0  ( 6.022  1023 ) dt 0 ln 2 ( 6.022  1023 )

= (1.6  105 s −1 )

(1.248 10 yr )( 3.156  10 s yr ) ( 39.963998g ) = 0.6034g  0.60g ( ln 2 ) ( 6.022 10 ) 9

 60s    1min 

 N = (1.381 10−13 s −1 )( 3.50  1018 ) = 4.8335  105 decays s   2.90  107 decays min 52. We use Eq. 41–7d, with R = 16 R0 . −

ln 2

R = R0 e − t = R0 e T1/2

→ T1/ 2 = −

ln 2 ln 2 t = − 1 ( 9.8min ) = 3.8min R ln 6 ln R0

 1.3  N14 =  12  (1.579  1025 ) = 2.053  1013 nuclei of 146 C  10  

 ln 2  ( 2.053  1013 ) = 78.69decays s  79decays s 7  ( 5730 yr ) ( 3.156  10 s yr ) 

N = 

54. We find the mass from the activity. Note that N A is used to represent Avogadro’s number. The half-life is found in Appendix G. ln 2 mN A R = N = → T1/ 2 A

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RT1/ 2 A ( 475decays s ) ( 4.468  10 yr )( 3.156  10 s yr ) ( 238.050787 g mole ) = N A ln 2 ( 6.022 1023 nuclei mole ) ln 2 9

= 3.8199  10−2 g  3.82  10−2 g 87 55. We assume that we currently have N nuclei of 87 37 Rb and 0.0260 N nuclei of 38 Sr. The sum of those

56. We are not including the neutrinos that will be emitted during beta decay. First sequence:

Second sequence:

232 90

4 228 228 − 228 228 − Th → 228 ; 88 Ra + 2 ; 88 Ra → 89 Ac +  ; 89 Ac → 90 Th + 

228 90

224 220 4 4 Th → 224 88 Ra + 2 ; 88 Ra → 86 Rn + 2

235 92

231 231 − 227 4 4 ; 231 U → 231 91 Pa → 89 Ac + 2 ; 90 Th + 2 ; 90 Th → 91 Pa + 

227 89

− 223 4 Ac → 227 ; 227 90 Th +  90Th → 88 Ra + 2

N 14 C = (1.3  10−12 )( 3.362  1024 ) = 4.371  1012 nuclei of 146 C 6

( N ) 14 6C

today

(

=  N 14 C 6

)e 0

− t

→

N ) N ) ( ( T t = − ln =− ln  (  N ) ln 2  ln 2 N  1

14 6C

=−

today

1/ 2

14 6C

today

 T1/ 2 6 0 ( 7.0decays s )

5730 yr = 7214 yr  7200 yr ln ln 2   ln 2 12  ( 4.37110 nuclei ) 7  ( 5730 yr ) ( 3.156  10 s yr )  0

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58. The decay rate is given by Eq. 41–4b,

Instructor Solutions Manual

N = − N . We assume equal numbers of nuclei decaying by t

T1/ 2 = −

 1d  ln 2 ( 35.0 h ) = 428.85h   = 17.9d ln 0.945  24 h 

R0T1/ 2 A ( 320decays s )( 53d )( 86, 400s d )( 7.016929g mole ) = = 2.5  10−14 g 23 N A ln 2  6.022 10 nuclei mole ln 2 ( )

61. The number of radioactive nuclei decreases exponentially, and every radioactive nucleus that decays becomes a daughter nucleus. N = N 0 e − t

211 82

Pb .

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Nuclear Physics and Radioactivity

ln N = N 0 e − t →  = −

N N 0 ln 2 = → t T1/ 2

( 5730 yr ) ln 0.062 = 22,986 yr  23,000 yr T N =− t = − 1/ 2 ln ln 2 N 0 ln 2 64. (a)

The mass number is found from the radius, using Eq. 41–1. r = (1.2  10−15 m ) A1/3 → 3

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( 2 )( 79 ) (1.60  10−19 C ) 1 q qAu 9 2 2 r= = ( 8.988  10 N  m C ) = 2.951  10−14 m −13 4 0 K ( 7.7MeV ) (1.60 10 J MeV ) 2

 3.0  10−14 m

R40T1/ 2 A40 ( 48decays s ) (1.248  10 yr )( 3.156  10 s yr ) ( 39.963998g mole ) = N A ln 2 ( 6.022 1023 nuclei mole ) ln 2 9

m40 =

= 1.8101  10−4 g  1.8  10−4 g 40 We find the number of 39 19 K atoms from the number of 19 K atoms and the abundances given in

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Chapter 41

Nuclear Physics and Radioactivity

0.932581 N 40 0.000117 A  0.932581 R40T1/ 2  A39  0.932581 A m39 = N 39 39 =  N 40  39 =   N A  0.000117  N A  0.000117 ln 2  N A

N 40 = 0.000117 N K ; N 39 = 0.932581N K =

 0.932581  ( 48decays s ) (1.248  10 yr )( 3.156  10 s yr ) ( 38.963706g mole ) =  ln 2  0.000117  ( 6.022 1023 nuclei mole ) 9

= 1.407 g  1.4g

Sr → 9039Y + −01 e + v ;

0 Y →90 40 Zr + −1 e + v

90 39

71. We find the mass from the initial decay rate and Eq. 41–7b. dN 6.022  1023 nuclei mole =  N0 =  m → dt 0 ( atomic weight ) g mole m=

dN 1 ( atomic weight ) dN T1/ 2 ( atomic weight ) = dt 0  ( 6.022  1023 ) dt 0 ln 2 ( 6.022  10 23 )

= ( 3.94  104 s −1 )

( 87.37 d )(86, 400s d ) ( 34.969032g ) = 2.49  10−11 g ( ln 2 ) ( 6.022  1023 )

72. (a) We find the daughter nucleus by balancing the mass and charge numbers: 191 Z ( X ) = Z ( Os ) − Z ( e − ) = 76 − ( − 1) = 77 76 Os

A ( X ) = A ( Os ) − A ( e − ) = 191 − 0 = 191

The daughter nucleus is 191 77 Ir . (b) See the included diagram. (c) Because there is only one  energy, the  decay must be to the higher excited state.

 – (0.14 MeV)  (0.042 MeV)  (0.129 MeV)

191 77 Ir* 191 77 Ir* 191 77 Ir

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73. The activity is the decay constant times the number of nuclei, as given by Eq. 41–7a. (a) We calculate the activity for 32 15 P.

R = N =

  ln 2 ln 2 1.0g 23 N=   ( 6.022  10 nuclei mol ) T1/ 2 (14.3d )(86, 400s d )  31.973908g mol 

= 1.057  1016 decays s  1.1  1016 decays s (b) We calculate the activity for 23290Th. ln 2 R= N T1/ 2 =

  ln 2 1.0g 23   ( 6.022  10 nuclei mol ) 7 232.038054g mol (1.40  10 yr )( 3.156 10 s yr )   10

= 4.1  103 decays s

( 2.843 10 atoms ) ( 546.5MeV atom ) (1.60 10 22

−13

J MeV ) = 2.5  1012 J

75. (a)  ( 42 He ) = m ( 42 He ) − A ( 42 He ) = 4.002603u − 4 = 0.002603u = ( 0.002603u ) ( 931.5MeV uc 2 ) = 2.425Mev c 2

(b)  ( 126 C ) = m ( 126 C ) − A ( 126 C ) = 12.000000u − 12 = 0 for both units. (c)

86 86  ( 86 38 Sr ) = m ( 38 Sr ) − A ( 38 Sr ) = 85.909261u − 86 = − 0.090739 u

= ( − 0.090739 u ) ( 931.5MeV uc 2 ) = − 84.52 MeV c 2

235 235 (d)  ( 235 92 U ) = m ( 92 U ) − A ( 92 U ) = 235.043928u − 235 = 0.043928u

= ( 0.043928u ) ( 931.5MeV uc 2 ) = 40.92 MeV c 2

(e) From Appendix G we see that   0 for 0  Z  7 and Z  86;

  0 for 8  Z  85.

  0 for 0  A  15 and A  214;   0 for 16  A  214.

= (1.007825u ) + (1.008665u ) − ( 2.014102 u )  c 2 ( 931.5MeV uc 2 ) = 2.224 MeV

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− t − ln 2 ln 2 ln 2 13 R = N = N= N 0 e T1/2 = 6.518 10 e  ( ) T1/ 2 T1/ 2 ( 5730 yr ) ( 3.156 107 s yr )

ln 2( 60,000yr )

( 5730 yr )

= 20.58 MeV Repeat the calculation for the carbon isotopes.

 5.98  1024 kg  r = (1.2  10 m ) A = (1.2  10 m )   = 183.6 m  180 m −27 1.67  10 kg  (b) Follow the same process as above, just using the Sun’s mass. −15

1/3

−15

1/3

 1.99  1030 kg  r = (1.2  10−15 m ) A1/3 = (1.2  10−15 m )   −27  1.67  10 kg 

= 12700 m  1.3  104 m

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(b) We do a similar calculation for an initial mass of 340 grams.  340g  23 −12 13 N14 =   ( 6.022  10 atoms mol )(1.3  10 ) = 2.218  10 atoms  12g mol 

( 5730 yr ) ln T 1 N 1 N N = N 0 e − t → t = − ln = − 1/ 2 ln =− = 2.5  105 yr 13  N0 ln 2 N 0 ln 2 2.218  10

Q =  m ( 11 H ) − m ( 01 n )  c 2 = (1.007825u ) − (1.008665u ) c 2 ( 931.5MeV uc 2 )

2m

N 147 Sm = ( 0.15 ) N natural =

( 0.15)(1.00g ) ( 6.022  1023 nuclei / mol )

150.36g mol The activity level is used to calculate the half-life. 62

= 6.008  1020 nuclei of 147 62 Sm

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Nuclear Physics and Radioactivity

Activity = R =  N = T1/ 2 =

ln 2 N → T1/ 2

ln 2 ln 2 1yr   N= 6.008  1020 ) = 3.44069  1018 s  = 1.1  1011 yr ( 7  R 120decays s 3.156 10 s   

84. Since amounts are not specified, we will assume that “today” there is 0.720 g of 235 92 U and

100.000 − 0.720 = 99.280g of 238 92 U. We use Eq. 41–6. (a) Relate the amounts today to the amounts 1.0 109 years ago. t

N = N 0 e − t → N 0 = Net = Ne T1/2

ln 2

(1.010 yr ) ln 2 ( 7.0410 yr ) 9

( N 0 )235 = ( N 235 ) e T

ln 2

1/2

= ( 0.720g ) e

= 1.927 g

(1.010 ) ln 2 ( 4.46810 ) 9

( N0 )238 = ( N 238 ) e T

ln 2

1/2

= ( 99.280g ) e

= 115.94g

1.927  100% = 1.63% 1.927 + 115.94

The percentage of 235 92 U was

(b) Relate the amounts today to the amounts 100  106 years from now.

(10010 yr ) ln 2 ( 7.0410 yr ) 6

N = N 0 e − t →

−

( N 235 ) = ( N 0 )235 e T

ln 2

1/2

= ( 0.720g ) e

(10010 yr ) ln 2 ( 4.46810 yr )

−

= 0.6525g

−

( N 238 ) = ( N0 )238 e T

1/2

ln 2

= ( 99.280g ) e

The percentage of 235 92 U will be

−

= 97.752g

0.6525  100% = 0.663% 0.6525 + 97.752

(

)

= 6.524 atoms 146 C g 126 C  6.5  1010 atoms 146 C g 126 C (c) The activity in natural carbon for a living organism is the product of the decay constant and the 14

number of 6 C atoms per gram of 6 C . Use Eq. 41–7d and Eq. 41–4b.

(

R =  N = 3.83  10−12 s −1

)( 6.524 10 atoms C g C ) = 0.2499 decays s g C 10

14 6

12 6

 0.25 decays s g 126 C

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Instructor Solutions Manual

(d) Take the above result as the initial decay rate (while Otzi was alive), and use Eq. 41–7d to find the time elapsed since he died. R = R0 e − t → t=−

R 1 1yr  0.121   11  = 6000 yr ln   = 1.894  10 s  =− −12 −1 7    R 3.83 10 s 0.2499 3.156 10 s      0

ln 

 Otzi lived approximately 6000 years ago.

86. (a) We use the definition of the mean life given in the problem. We use a definite integral formula from Appendix B-5. 

 tN ( t ) dt

 = 0



− t  tN 0e dt

= 0



 te dt

= 0

− t

 N ( t ) dt  N e dt  e dt − t

− t



1 − e− t

(b) We evaluate at time t =  =



2 

2 1 = = 1 



 − N ( t =  ) N 0e −   e = = = e −1 = 0.368 N ( t = 0 ) N 0e −0

p = pD ; K D =

2m K p2 m A pD2 4 K =  =   =  K =  K = mD AD AD 2mD 2mD 2mD

4 K AD

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ln 2 ln 2 N= ( 7.208  1017 )  3.1561yr107 s  = 12.51decay s  13decay s T1/ 2 1.265  109 yr

pHe = pTh ; K Th =

2 p2 2mHe K He mHe pTh K He = He = = 2mTh 2mTh 2mTh mTh

(b) We use Eq. 41–1 to estimate the radii.

rHe = (1.2  10−15 m ) ( 4 )

1/3

= 1.905  10−15 m  1.9  10−15 m

rTh = (1.2  10−15 m ) ( 232 )

1/3

= 7.374  10−15 m  7.4  10−15 m

= ( 8.988  10 N  m C ) 9

1/3

( 2 )( 90 ) (1.60  10−19 C )

+ 2321/3 )(1.2  10−15 m )(1.60  10 −13 J MeV )

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Instructor Solutions Manual

( 2 )( 90 ) (1.60  10−19 C ) 1 qHe qTh 9 2 2 RB = = ( 8.988  10 N  m C ) 4 0 U B ( 4.495MeV ) (1.60  10−13 J MeV ) 2

= 57.57  10−15 m RB − RA = 57.6fm − 9.3fm = 48.3fm

px 

→ p  p 

x

(

)

2 (1.055 10−34 J s ) p2 K= = = 2m 2mr 2 2 (1.67  10−27 kg ) ( 561/3 )(1.2  10−15 m )  2 (1.60  10−13 J MeV )   2

T1/ 2 =

 6.022  1023 nuclei   ln 2 ln 2 1yr  7  = 1.3  1021 yr 1.5 10 g N= ( )   7   1decay s 152g 3.156 10 s R   

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Chapter 41

Nuclear Physics and Radioactivity

half-life as given in Appendix G is T1 2 = 9.965min

60s = 597.9s . 1min

KK =

E2 2mK c

(1.46 MeV )

 931.5MeV c  2 2 ( 39.963998u )  c u   2

= 2.86  10−5 MeV = 28.6eV

1419

CHAPTER 43: Elementary Particles Responses to Questions 1.

A reaction between two nucleons that would produce a  − is: p + n → p + p +  − .

No, the decay is still impossible. In the rest frame of the proton, this decay is energetically impossible, because the proton’s mass is less than the mass of the products. Since it is impossible in the rest frame, it is impossible in every other frame as well. In a frame in which the proton is moving very fast, the decay products must be moving very fast as well to conserve momentum. With this constraint, there will still not be enough energy to make the decay energetically possible. In order for these products to be produced, the proton would have to interact with some other particle.

Antiatoms would be made up of negatively charged antiprotons and neutral antineutrons in the nucleus with positively charged positrons surrounding the nucleus. If antimatter and matter came into contact, the particle–antiparticle pairs would annihilate, converting their mass into energetic photons.

The photon signals the electromagnetic interaction.

(a) Yes. If a neutrino is produced during a decay, the weak interaction is responsible. (b) No. For example, a weak interaction decay could produce a Z0 instead of a neutrino.

The neutron decay process also produces an electron and an antineutrino; the antineutrino will only be present in a weak interaction.

An electron takes part in the electromagnetic, weak, and gravitational interactions. A neutrino takes part in the weak and gravitational interactions. A proton takes part in the strong, electromagnetic, weak, and gravitational interactions.

The chart here shows charge and baryon conservation checks for many of the decays in Table 43–2. Particle W

Z0 Higgs

Decay W+ → e + + ve (others are similar) Z0 → e + + e − (others are similar) H0 → b + b

Charge conservation +1 = +1 + 0

Baryon conservation 0=0+0

0 = +1 + ( −1)

0=0+0

0 = ( − 13 ) + 13

H0 → W+ + W− H 0 → Z0 + Z0

0 = +1 + ( −1) 0=0+0

0 = 13 + ( − 13 )

(others are similar)

0=0+0 0=0+0

muon

 − → e − + ve + v

−1 = ( −1) + 0 + 0

0=0+0+0

tau

 − →  − + v + v

−1 = ( −1) + 0 + 0

0=0+0+0

 + →  + + v

+1 = +1 + 0

0=0+0

0 → +

0=0+0

(other is similar) pion

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Chapter 43

Elementary Particles

kaon

K + →  + + v

+1 = +1 + 0

0=0+0

K+ →  + +  0 KS0 →  + +  −

+1 = +1 + 0

0=0+0

0 = +1 + ( −1)

0=0+0

K → +

0=0+0

0 = +1 + ( −1) + 0

0=0+0+0

K →  +  + v

0 = +1 + ( −1) + 0

0=0+0+0

K → + +

0 = +1 + ( −1) + 0

0=0+0+0

K → + +

0=0+0+0

0=0+0

0 S

K →  + e + ve +

0 L

 → +  → + +

0=0+0+0

0=0+0 0=0+0+0

 → + +

0 = +1 + ( −1) + 0

0=0+0+0

0 →  + + −

0 = +1 + ( −1)

0=0+0

0 →  0 + 0

0=0+0

 → +

+1 = +1 + 0

0=0+0

−

neutron

n → p + e − + ve

0 = +1 + ( −1) + 0

+1 = + 1 + 0 + 0

lambda

0 → p +  −

0 = +1 + ( −1)

+1 = + 1 + 0

sigma

 → n + + → p +  0

0=0+0 +1 = +1 + 0

+1 = + 1 + 0 +1 = + 1 + 0

+ → n +  + 0 →  0 + 

+1 = 0 + 1 0=0+0

+1 = + 1 + 0 +1 = + 1 + 0

− → n +  −

−1 = 0 + ( −1)

+1 = + 1 + 0

0 →  0 +  0 − → 0 +  −

0=0+0

+1 = + 1 + 0

−1 = 0 + ( −1)

+1 = + 1 + 0

− → 0 +  −

−1 = 0 + ( −1)

+1 = + 1 + 0

− →  0 + K −

−1 = 0 + ( −1)

+1 = + 1 + 0

 →  +

−1 = ( −1) + 0

+1 = + 1 + 0

xi omega

−

rho

−

0 L

eta

−

Since decays via the electromagnetic interaction are indicated by the production of photons, the decays in Table 43–2 that occur via the electromagnetic interaction are those of the Higgs boson, the  0 , the  0 , and the  0 .

10. All of the decays listed in Table 43–2 with a neutrino or antineutrino as a decay product occur via 0 the weak interaction. These include the W, muon, tau, pion, kaon, K L , and neutron. In addition, the Z particle and the Higgs boson both decay via the weak interaction. In each case, we could also list the corresponding antiparticle.

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Instructor Solutions Manual

11. Since the  baryon has B = 1, it is made of three quarks. Since the spin of the  baryon is 3/2, none of these quarks can be antiquarks. Thus, since the charges of quarks are either +2/3 or –1/3, the only charges that can be created with this combination are Q = –1 (= –1/3 – 1/3 – 1/3), Q = 0 (= +2/3 – 1/3 – 1/3), Q = +1 (= +2/3 + 2/3 – 1/3), and Q = +2 (= +2/3 + 2/3 + 2/3). There is no way to combine three quarks for a total charge of Q = –2. 12. Based on lifetimes shown in Table 43–4, the particle decays that occur via the electromagnetic interaction are J  ( 3097 ) and  ( 9460 ) , which have shorter lifetimes than the other mesons. 13. All of the particles in Table 43–4, except for J  ( 3097 ) ,  ( 9460 ) , and the sigma particles decay via the weak interaction, based on their lifetimes. 14. Baryons are formed from three quarks or antiquarks, each of spin 12 or − 12 , respectively. Any combination of quarks and antiquarks will yield a spin magnitude of either 12 or 32 . Mesons are formed from two quarks or antiquarks. Any combination of two quarks or antiquarks will yield a spin magnitude of either 0 or 1. 15. If a neutrinolet was massless, it would not interact via the gravitation force; if it had no electrical charge, it would not interact via the electromagnetic force; if it had no color charge, it would not interact via the strong force; and if it does not interact via the weak force, it would not interact with matter at all. It would be very difficult to say that it exists at all. However, a similar argument could be made for photons. Photons have no color, no mass, and no charge, but they do exist. 16. (a) No. Leptons are fundamental particles with no known internal structure. Baryons are made up of three quarks. (b) Yes. All baryons are hadrons. (c) No. A meson is a quark–antiquark pair. (d) No. Hadrons are made up of quarks and leptons are fundamental particles. 17. No. A particle made up of two quarks would have a particular color. Three quarks or a quark– antiquark pair are necessary for the particle to be white or colorless. A combination of two quarks and two antiquarks is possible, as the resulting particle could be white or colorless. The neutral pion can in some ways be considered a 4-quark combination. 18. In the nucleus, the strong interaction with the other nucleons does not allow the neutron to decay. When a neutron is free, the weak interaction is the dominant force and can cause the neutron to decay. −

19. No, the reaction e + p → n + ve is not possible. The electron lepton number is not conserved: The reactants have L e = 1 + 0 = 1 , but the products have L e = 0 − 1 = −1 . Thus, this reaction is not possible. If the product were an electron neutrino (instead of an antineutrino), the reaction would be possible. +

−

20. The reaction  → p + e + ve proceeds via the weak force. We know this is the case since an electron antineutrino is emitted, which only happens in reactions governed by the weak interaction. 0

21. (a) The two major classes of fundamental particles are quarks and leptons. (b) Quarks: up, down, strange, charm, bottom, and top. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Elementary Particles

Leptons: electron, muon, tau, electron neutrino, muon neutrino, and tau neutrino. (c) Gravity, electromagnetic, weak nuclear, and strong nuclear. (d) Gravity is carried by the graviton; the electromagnetic force is carried by the photon; the weak nuclear force is carried by the W+, W–, and Z0 bosons; the strong nuclear force is carried by the gluon. The gravitational force is much weaker than the other three forces. 22. (a) Hadrons interact via the strong nuclear force (as well as the other three fundamental forces) and are made up of quarks. (b) Baryons are hadrons, are made of three quarks, and have a baryon number of either +1 or –1. (c) Mesons are hadrons, are made of one quark and one anti-quark, and have a baryon number of 0.

Responses to MisConceptual Questions 1.

(c) Quarks are NOT leptons, nor are they composed of leptons, so answer (a) is incorrect. They also are not components of electrons, but electrons are part of “ordinary matter”. Thus, answer (b) is also incorrect. Table 43–5 lists quarks as fundamental particles, so they have no internal structure. If they had internal structure, they would not be fundamental. Thus, answer (c) is correct.

(a) Atoms interact with other atoms to form molecules by means of the electromagnetic force, which binds positive and negative (nuclei and electrons) charges together. The gravitational force is too weak to affect atomic structure, and the strong and weak forces are “short range” and only affect nuclei.

(a, d) From Table 43–5, we see that electrons and quarks are considered to be fundamental. Protons, neutrons, pions, and mesons are all made of quarks and so are not fundamental.

(a) A common misconception is that all six quarks make up most of the known matter. However, charmed, strange, top, and bottom quarks are unstable. Protons and neutrons are made of up and down quarks, so they make up most of the ordinary matter.

(b, e) Quarks combine together to form mesons (such as the  meson) and baryons (such as the proton and the neutron). The electron and Higgs boson are fundamental particles and therefore cannot be made of quarks.

(c) All of the fundamental forces act on a variety of objects, including our bodies. Although the strong and weak forces are very short range, the electromagnetic force is a long-range force, just like gravity. One reason we notice the “weak” gravity force more than the electromagnetic force is that most objects are electrically neutral and so do not have significant net electromagnetic forces on them. It is true that the gravitational force between people and other objects of similar size is too small for us to notice, but due to the huge mass of the Earth, we are always aware of the influence of the Earth’s gravitational force on us.

(d) A tau lepton has a tau lepton number of one. When it decays, the lepton number must be conserved, so it cannot decay into only hadrons, at least a tau neutrino would have to be one of the byproducts.

(b, e, f, g) Atoms are not fundamental, because they are made up of protons, neutrons, and electrons. Protons and neutrons are not fundamental as they are made from quarks. The fundamental

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

particles are leptons (including electrons), quarks, and gauge bosons (including the photon and the Higgs boson) 9.

(a) Unlike the electron, the positron has positive charge and a negative lepton number. Unlike the charge and lepton number, the mass of the positron has the same sign (and magnitude) as the mass of the electron.

10. (e) A common misconception is that the strong force is a result of just the exchange of  mesons between the protons and neutrons. This is correct on the scale of the nucleons. However, when the quark composition of the protons, neutrons, and  mesons are considered at the elementary particle scale, it is seen that the transfer is due to the exchange of gluons. Therefore, both answers can be considered correct at different scales. Students who answer (d) should be given credit for their answer as well. 11. (c) Pions are not fundamental particles and are made up of quark and anti-quark pairs. Leptons and bosons (including photons) are fundamental particles, but are not a constituent of protons and neutrons. Protons and neutrons are comprised of up and down quarks. 12. (d) Quarks, gluons, neutrons, and the Higgs boson interact through the strong force. Electrons and muons are charged particles and interact through the electromagnetic force. Neutrinos only interact through the weak force. 13. (a) Strangeness is conserved in strong interactions but not in weak interactions. The other quantities (energy, charge, and momentum) are listed in the Summary section as being conserved in all nuclear and particle reactions. 14. (c) Neutrinos, electrons, and positrons (antielectrons) are all leptons (and thus fundamental particles). Mesons are made of quark–anti-quark pairs and so are not leptons.

Solutions to Problems 1.

The total energy is given by Eq. 36–11a. E = m0 c 2 + K = 0.938 GeV + 5.25 GeV = 6.19 GeV

Because the energy of the electrons is much greater than their rest mass, we have K = E = pc. Combine that with Eq. 43–1 for the de Broglie wavelength. 6.63  10−34 J s 3.00 108 m s h hc hc E = pc ; p = → E= → = = = 5.2  10−17 m 9 −19 E   24  10 eV 1.60 10 J eV

( (

) )

Use Eq. 43–2 to calculate the frequency. The alpha particle has a charge of +2e and a mass of 4 times the proton mass. 2 1.60  10−19 C (1.7 T ) qB f = = = 1.3  107 Hz = 13MHz 2 m 2  4 1.67  10−27 kg 

(

)(

(

)

The time for one revolution is the period of revolution, which is the circumference of the orbit divided by the speed of the protons. Since the protons have very high energy, their speed is essentially the speed of light.

1424

Chapter 43

Elementary Particles

T= 5.

2 r v

27  103 m 3.0  10 m s 8

= 9.0  10−5 s

The frequency is related to the magnetic field in Eq. 43–2. −27 7 qB 2 mf 2 1.67  10 kg 3.5  10 Hz f = → B= = = 2.3T q 2 m 1.60  10−19 C

(

)(

(a) The maximum kinetic energy is K =

q2 B2 R2

)

= 12 mv 2 . Compared to Example 43–2, the charge

2m has been doubled and the mass has been multiplied by 4. These two effects cancel each other in the equation, and so the maximum kinetic energy is unchanged. The kinetic energy from that example was 8.653 MeV. K = 8.7 MeV 2K

(

2 ( 8.653 MeV ) 1.60  10−13 J MeV

(

−27

4 1.67  10 kg

(b) The maximum kinetic energy is K =

)

q2 B2 R2

) = 2.036 10 m s  2.0 10 m s 7

= 12 mv 2 . Compared to Example 43–2, the charge

2m is unchanged, and the mass has been multiplied by 2. Thus, the kinetic energy will be half of what it was in Example 43–2 (8.653 MeV). K = 4.3MeV 2K

(

2  12 ( 8.653 MeV ) 1.60  10−13 J MeV

(

−27

2 1.66  10 kg

)

) = 2.036 10 m s  2.0 10 m s 7

The alpha and the deuteron have the same charge to mass ratio and so move at the same speed. qB (c) The frequency is given by f = . Since the charge to mass ratio of both the alpha and the 2 m deuteron is half that of the proton, the frequency for the alpha and the deuteron will both be half the frequency found in Example 43–2 for the proton. f = 13 MHz 7.

From Eq. 41–1, the diameter of a nucleon is about dnucleon = 2.4 10 m. The 25-MeV alpha particles −15

and protons are not relativistic, so their momentum is given by p = mv = 2mK . The wavelength is given by Eq. 43–1,  =

 =

p =

h 2 m K

h 2 mp K

h p

2mK

6.63  10−34 J s

(

)(

2 ( 4 ) 1.66  10 kg 25  10 eV 1.6  10 −27

−19

J eV

6.63  10−34 J s

(

−27

)(

2 1.67  10 kg 25  10 eV 1.6  10 6

−19

J eV

)

= 2.88  10−15 m

= 5.75  10−15 m

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

We see that   d nucleon and p  2d nucleon . Thus, the alpha particle will be better for picking out details in the nucleus. 8.

Because the energy of the protons is much greater than their rest mass, we have K = E = pc. Combine that with Eq. 43–1 for the de Broglie wavelength. That is the minimum size that protons of that energy could resolve. 6.63  10−34 J s 3.0 108 m s h hc hc E = pc ; p = → E= → = = = 1.9  10−19 m 12 −19 E   6.5  10 eV 1.60 10 J eV

( (

)(

) )

If the speed of the protons is c, the time for one revolution is found from uniform circular motion. The number of revolutions is the total time divided by the time for one revolution. The energy per revolution is the total energy gained divided by the number of revolutions. 2 r 2 r 2 r t ct v= → T= = n= = v T c T 2 r

Energy revolution =

E n

( E ) 2 r ct

(1.0  10 eV − 150  10 eV ) 2 (1.0  10 m ) ( 3.00  10 m s ) ( 20s ) 12

= 8.9  105 eV rev  0.9 MeV rev 10. (a) At an energy of 6.5 TeV, the protons are moving at practically the speed of light. From uniform circular motion, we find the time for the protons to complete one revolution of the ring. Then the total charge that passes any point in the ring during that time is the charge of the entire group of stored protons. The current is then the total charge divided by the period. 2 R 2 R 2 R v= → T= = T v c

( 2  10 protons )(1.60  10 C proton )(3.0  10 m s ) = 0.355A  0.4 A I= = = T 2 R 2 ( 4.3  10 m ) Ne

−19

Nec

(b) The 6.5 TeV is equal to the kinetic energy of the proton beam. We assume the car would not be moving relativistically. K beam = K car → K beam = 12 mv 2 →

2 K beam m

(

)(

2 6.5  1012 eV proton 2  1014 protons 1.60  10−19 J eV 1600 kg

) = 509.9 m s

 500 m s Our assumption that the car is not relativistic is confirmed. This is about 1200 mi/hr. 11. These protons will be moving at essentially the speed of light for the entire time of acceleration. The number of revolutions is the total gain in energy divided by the energy gain per revolution. Then the distance is the number of revolutions times the circumference of the ring, and the time is the distance of travel divided by the speed of the protons.

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Elementary Particles

( 6.5 10 eV − 450 10 eV ) = 7.56 10 rev 12

E E rev

8.0  10 eV rev 6

(

) (

)

d = N ( 2 R ) = 7.56  105 2 4.3  103 m = 2.043  1010 m  2.0  1010 m t=

d c

2.043  10 m 10

3.00  108 m s

= 68s

12. Because the energy of the protons is much greater than their rest mass, we have K = E = pc. A relationship for the magnetic field is given right before Eq. 43–2. We assume the LHC is like a cyclotron with a constant magnetic field around the entire ring. qBr E v= → mv = qBr → p = qBr → = qBr → m c 6.5  1012 eV 1.60 10−19 J eV E = = 5.1T B= qrc 1.60  10−19 C 4.25  103 m 3.00 108 m s

(

)(

)

)(

)

13. (a) The magnetic field is found from the maximum kinetic energy as derived in Example 43–2.

K= B=

q2 B 2 R 2

2mK

→ B=

(

→

)(

2 ( 2.014 ) 1.66  10−27 kg 14  106 eV 1.60  10−19 J eV

(1.60 10 C ) (1.0 m ) −19

)

= 0.7649 T  0.76 T

(b) The cyclotron frequency is given by Eq. 43–2. 1.60  10−19 C ( 0.7649 T ) qB f = = = 5.826  106 Hz  5.8 MHz −27 2 m 2 ( 2.014 ) 1.66  10 kg

(

)

(c) The deuteron will be accelerated twice per revolution, and so will gain energy equal to twice its charge times the voltage on each revolution. 14  106 eV number of revolutions = n = 1.60  10 −19 J eV 2 1.60  10 −19 C 22  103 V

(

)(

)

(

)

= 318 revolutions  320 revolutions (d) The time is the number of revolutions divided by the frequency (which is revolutions per second). 318 revolutions n t = = = 5.458  10−5 s  5.5  10 −5 s = 55 s 6 f 5.826  10 rev s (e) If we use an average radius of half the radius of the cyclotron, the distance traveled is the average circumference times the number of revolutions. distance = 12 2 rn =  (1.0 m )( 318 ) = 999 m  1.0 km

14. Start with an expression from Section 43–1, relating the momentum and radius of curvature for a particle in a magnetic field, with q replaced by e. eBr v= → mv = eBr → p = eBr m © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1427

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

In the relativistic limit, p = E c and so

E c

Instructor Solutions Manual

= eBr . To put the energy in electron volts, divide the

energy by the charge of the object.

E c

= eBr →

E e

= B rc

15. The energy released is the difference in the mass energy between the products and the reactant. E = m c 2 − mn c 2 − m c 2 = 1115.7 MeV − 939.6 MeV − 135.0 MeV = 41.1MeV 0

16. The energy released is the difference in the mass energy between the products and the reactant. E = m c 2 − m c 2 − m c 2 = 139.6 MeV − 105.7 MeV − 0 = 33.9 MeV +



17. Use Eq. 43–3 to estimate the range of the force based on the mass of the mediating particle. 6.63  10−34 J s 3.00 108 m s hc hc mc 2  → d =  3.98  10−16 m 2 6 −19 2 d 2 mc 2 497.6 10 eV 1.60 10 J eV

( (

)( )(

)

18. The energy required is the mass energy of the two particles. E = 2mn c 2 = 2 ( 939.6 MeV ) = 1879.2 MeV 19. The reaction is multi-step, and can be written as shown here:  0 →  0 + The energy released is the initial rest energy minus the → p +  − final rest energy of the proton and pion, using Table 43–2. E = m0 − mp − m − c 2 = 1192.6 Mev − 938.3MeV − 139.6 MeV = 114.7 MeV

(

)

20. Because the two protons are heading towards each other with the same speed, the total momentum of the system is 0. The minimum kinetic energy for the collision would result in all three particles at rest, and so the minimum kinetic energy of the collision must be equal to the mass energy of the  0 . Each proton will have half of that kinetic energy. From Table 43–2, the mass of the  0 is 135.0 MeV c 2 . 2 K proton = m c 2 = 135.0 MeV → K proton = 67.5 MeV 0

21. Because the two particles have the same mass and they are traveling towards each other with the same speed, the total momentum of the system is 0. The minimum kinetic energy for the collision would result in all four particles at rest, and so the minimum kinetic energy of the collision must be equal to the mass energy of the K + K − pair. Each initial particle will have half of that kinetic energy. From Table 43–2, the mass of both the K + and the K − is 493.7 MeV c 2 .

2 ( K p or p ) = 2mK c 2 → K p or p = mK c 2 = 493.7 MeV 22. The energy of the two photons (assumed to be equal so that momentum is conserved) must be the combined rest mass energy of the proton and antiproton. 6.63  10−34 J s 3.00  108 m s c hc → = = = 1.32  10−15 m 2m0c 2 = 2hf = 2h 2 6 −19  938.3  10 eV 1.60  10 J eV m0c

(

)( )(

)

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Chapter 43

Elementary Particles

23. By assuming that the kinetic energy is approximately 0, the total energy released is the rest mass energy of the annihilating pair of particles. (a) Etotal = 2m0 c 2 = 2 ( 0.511MeV ) = 1.022 MeV (b) Etotal = 2m0 c 2 = 2 ( 938.3 MeV ) = 1876.6 MeV 24. The total energy is the sum of the kinetic energy and the mass energy. The wavelength is found from the relativistic momentum. E = K + mc 2 = 12  109 eV + 938  106 eV = 1.294  1010 eV  13GeV

=

h p

(

E 2 − mc 2

)

(

E 2 − mc 2

)

( 6.63 10 J s )( 3.00 10 m s ) 1 = = 9.6  10 m  1.60 10 J eV ( ) (1.294 10 eV ) − ( 938 10 eV ) −34

−17

−19

Charge conservation is violated, since 0  0 − 1 Strangeness is violated, since −1  0 + 0 Energy conservation is violated, since

25. (a)  0 → n +  − (b)  0 → p + K −

1115.7 MeV c 2  938.3 MeV c 2 + 493.7 MeV c 2 = 1432.0 MeV c 2

(c)

Baryon number conservation is violated, since 1  0 + 0 Strangeness is violated, since −1  0 + 0 Spin is violated, since 12  0 + 0

0 →  + +  −

26. The total momentum of the electron and positron is 0, so the total momentum of the two photons must be 0. Thus, each photon has the same momentum, so each photon also has the same energy. The total energy of the photons must be the total energy of the electron / positron pair. c Ee /e pair = Ephotons → 2 m0 c 2 + K = 2hf = 2h → +

=

(

−

hc m0 c 2 + K

)

 ( 6.63 10 J s )( 3.00 108 m s ) −34

(

)(

0.511  106 eV + 390  103 eV 1.60  10−19 J eV

)

= 1.380  10−12 m  1.4  10−12 m

27. (a) For the reaction  − + p → n +  0 , the conservation laws are as follows. Charge: −1 + 1 = 0 + 0 Charge is conserved. Baryon number: 0 + 1 = 1 + 0 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 Lepton number is conserved. Strangeness: 0 + 0 = 0 + 0 Strangeness is conserved. The reaction is possible.

(b) For the reaction  + + p → n +  0 , the conservation laws are as follows. Charge: 1 + 1  0 + 0 Charge is NOT conserved. The reaction is forbidden because charge is not conserved. (c) For the reaction  + + p → p + e + , the conservation laws are as follows. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Instructor Solutions Manual

Charge: 1 + 1 = 1 + 1 Charge is conserved. Baryon number: 0 + 1 = 1 + 0 Baryon number is conserved. Lepton number: 0 + 0  0 + 1 Lepton number is NOT conserved. The reaction is forbidden because lepton number is not conserved. +

(d) For the reaction p → e + ve , the conservation laws are as follows. Charge: 1 = 1 + 0 Charge is conserved. Baryon number: 1  0 + 0 Baryon number NOT conserved. Mass energy is fine because mp  me + mv . The reaction is forbidden because baryon number is not conserved. (e) For the reaction  + → e + +   , the conservation laws are as follows. Charge: 1 = 1 + 0 Charge is conserved. Baryon number: 0 = 0 + 0 Baryon number is conserved. Electron lepton number: 0  −1 + 0 Lepton number is NOT conserved. Mass energy is fine because m  me + mv . The reaction is forbidden because lepton number is not conserved. (f)

For the reaction p → n + e + ve , the conservation laws are as follows. Mass energy: 938.3 MeV c 2  939.6 MeV c 2 + 0.511MeV c 2 Mass energy is NOT conserved. The reaction is forbidden because energy is not conserved. This reaction will not happen because charge is not conserved ( 2  0 ) ,

28. p + p → p + p :

and baryon number is not conserved ( 2  0 ) . This reaction will not happen because charge is not conserved ( 2  1) ,

p+p → p+p+ p :

and baryon number is not conserved ( 2  1) .

p+p → p+p+p+p: +

This reaction is possible. All conservation laws are satisfied. This reaction will not happen. Baryon number is not conserved ( 2  0 )

p+p → p+e +e + p:

, and lepton number is not conserved ( 0  −2 ) . 29. Since the  − decays from rest, the momentum before the decay is zero. Thus, the momentum after the decay is also zero, so the momenta of the  0 and  − are equal in magnitude. Use energy and momentum conservation along with the relativistic relationship between energy and momentum. m c 2 = E + E → E = m c 2 − E −

−

( p c) = ( p c) → E − m c = E − m c = (m c − E ) − m c → E − m c = ( m c − 2E m c + E ) − m c → p = p 0



−

E = 0



→







−



m2 c 4 + m2 c 4 − m2 c 4 −

−

2m c 2 −

−



−



−



−



−

(1321.7 MeV ) + (1115.7 MeV ) − (139.6 MeV ) = = 1124.4 MeV 2 (1321.7 MeV ) 2

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Elementary Particles

E = m c 2 − E = 1321.7 MeV − 1124.4 MeV = 197.3MeV −

−

K  = E − m c 2 = 1124.4 MeV − 1115.7 MeV = 8.7 MeV 0

K = E − m c 2 = 197.3MeV − 139.6 MeV = 57.7 MeV −

−

30. Since the pion decays from rest, the momentum before the decay is zero. Thus, the momentum after the decay is also zero, so the magnitudes of the momenta of the positron and the neutrino are equal. We also treat the neutrino as massless. Use energy and momentum conservation along with the relativistic relationship between energy and momentum. m c 2 = Ee + E ; pe = p → pe2 c 2 = p2 c 2 → Ee2 − me2 c 4 = E2 +

(

Ee2 − me2 c 4 = m c 2 − Ee +

me2 c 2 +

Ee = 12 m c 2 + +

2m

)

) = m c − 2E m c + E 2

) (

+

→ K e + me c 2 = 12 m c 2 + +

→ 2 Ee m c 2 = m2 c 4 + me2 c 4

2 e

me2 c 2 +

2m

→

( 0.511MeV c ) ( 0.511MeV ) K = m c −m c + = (139.6 MeV ) − 0.511MeV + 2m 2 (139.6 MeV c ) e

1 2

+

me2 c 2 +

1 2

+

= 69.3 MeV Here is an alternate solution using the momentum.

(

)

Q =  m − me + mv  c 2 = 139.6 MeV − 0.511MeV = 139.1MeV +

( p c) + (m c ) − m c ; K = E = p c ; p = p = p 2

K e = Ee − me c 2 = +

( pc )2 + ( 0.511MeV )2 − ( 0.511MeV ) + pc

Q = K e + K v → 139.1MeV = +

( pc ) + ( 0.511MeV ) 2

139.6 − pc =

→ (139.6 ) − 2 (139.6 ) pc + p 2 c 2 = ( pc ) + ( 0.511MeV )

(139.6 MeV ) − ( 0.511MeV ) = pc = 69.8 MeV 2 (139.6 MeV ) 2

Ev = K v = 69.8 MeV ; Q − K v = K e = 139.1MeV − 69.8 MeV = 69.3 MeV +

31. (a) The Q-value is the mass energy of the reactants minus the mass energy of the products. Q = m c 2 − mp c 2 + m c 2 = 1115.7 MeV − ( 938.3MeV + 139.6 MeV ) = 37.8 MeV 0

(

)

−

(b) Energy conservation for the decay gives the following. m c 2 = Ep + E → E = m c 2 − Ep −

−

Momentum conservation says that the magnitudes of the momenta of the two products are 2 2 2 2 4 equal. Then convert that relationship to energy using E = p c + m0 c with energy conservation.

( p c) = ( p c) → E − m c = E − m c = (m c − E ) − m c pp = p 2 p

−

2 4 p

→

−



−

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Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

)

Instructor Solutions Manual

Ep2 − mp2 c 4 = m2 c 4 − 2 Ep m c 2 + Ep2 − m2 c 4 → Ep =

m2 c 4 + mp2 c 4 − m2 c 4 −

2m c 2 0

−

(1115.7 MeV ) + ( 938.3MeV ) − (139.6 MeV ) = = 943.7 MeV 2 (1115.7 MeV ) 2

E = m c − Ep = 1115.7 MeV − 943.7 MeV = 172.0 MeV 2

−

K p = Ep − mp c 2 = 943.7 MeV − 938.3MeV = 5.4 MeV K = E − m c 2 = 172.0 MeV − 139.6 MeV = 32.4 MeV −

−

32. The two neutrinos must move together, in the opposite direction of the electron, in order for the electron to have the maximum kinetic energy. Thus, the total momentum of the neutrinos will be equal in magnitude to the momentum of the electron. Since a neutrino is (essentially) massless, we have Ev = pv c. We assume that the muon is at rest when it decays. Use conservation of energy and momentum along with their relativistic relationship. pe = p + p −



(

)

m c = Ee + E + E = Ee + p c + p c = Ee + p + p c = Ee + pe c → 2

−



−

−

−

m c − 2m c Ee + Ee = Ee − me c → Ee = 2



−

(m c − E ) = ( p c) = E − m c →

m c 2 − Ee = pe c → −

−



−

m2 c 4 + me2 c 4 −

−

2m c

= K e + me c 2 → −

−

Ke = −

m2 c 4 + me2 c 4 −

−

(105.7 MeV ) + ( 0.511MeV ) − me c = − ( 0.511MeV ) = 52.3 MeV 2 (105.7 MeV ) 2

−

2m c 2 −

33. A  + could NOT be produced by p + p → p + n +  + . The pion has a mass energy of 139.6 MeV, so the extra 130 MeV of energy could not create it. The Q-value for the reaction is as follows.

(

)

Q = 2mp c 2 − mp c 2 + mn c 2 + m c 2 = mp c 2 − mn c 2 − m c 2 +

= 938.3MeV − 939.6 MeV − 139.6 MeV = −140.9 MeV More than 140.9 MeV of kinetic energy is needed to create the pion. The minimum initial kinetic energy would produce the particles all moving together at the same speed, having the same total momentum as the incoming proton. We consider the products to be one mass M = mp + mn + m +

since they all move together with the same velocity. We use energy and momentum conservation, 2 2 2 2 4 along with their relativistic relationship, E = p c + m0 c . Ep + mp c 2 = EM

; pp = pM → ( pp c ) = ( pM c )

(

Ep2 − mp2 c 4 = Ep + mp c 2 Ep =

M 2 c 4 − 2mp2 c 4 2mp c 2

→ Ep2 − mp2 c 4 = EM2 − M 2c 4 →

) − M c = E + 2E m c + m c − M c → 2

2 4

2 p

2 4 p

2 4

= K p + mp c 2 →

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Chapter 43

Elementary Particles

Kp =

M 2 c 4 − 2mp2 c 4 2 mp c 2

M 2c4

− mp c 2 =

2 mp c 2

− 2 mp c 2

( 938.3 MeV + 939.6 MeV + 139.6 MeV ) 2 = − 2 ( 938.3 MeV ) = 292.4 MeV 2 ( 938.3 MeV ) 34. We use the uncertainty principle to estimate the uncertainty in rest energy. 6.63  10−34 J s h E  = = 9420eV  0.009 MeV 2t 2 7  10−20 s 1.60  10−16 J eV

(

)

)(

)

35. We estimate the lifetime from the energy width and the uncertainty principle. 6.63  10−34 J s h h E  → t  = = 2  10−21 s −19 2t 2E  1.60  10 J  2 300  103 eV    1eV 

(

)

36. Apply the uncertainty principle, which says that E  E 

h 2t

→ t 

2t −34 6.63  10 J s

. = 7.5  10 −21 s

(

 10 J  )  1.601eV 

−1 = − 13 + ( − 32 )

Spin:

0 = 12 − 12

Strangeness:

0 = 0+0 −1 = − 1 + 0

2E

−19

2 88  103 eV 





−

37. (a) For B = b u , we have Charge: Baryon number: Charm: Topness:

0 = 13 − 13 0 = 0+0 0 = 0+0

Bottomness:

(b) Because B+ is the antiparticle of B− , B+ = b u . The B0 must have an anti-bottom quark but must be neutral. Therefore, B0 = b d . Because B0 is the antiparticle to B0 , we must have

B0 = b d . 38. We find the energy width from the lifetime in Table 43–2 and the uncertainty principle. 6.63  10−34 J s h (a) 0 : t = 5.1 10−19 s ; E  = = 1293eV  1300 eV 2t 2 ( 5.1  10 −19 s )(1.60  10 −19 J eV ) (b)  + : t = 4.4  10−24 s ; E 

h 2t

(

6.63  10 −34 J s

)(

2 4.4  10 s 1.60  10 J eV −24

−19

)

= 1.499 108 eV  150 MeV

1433

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Spin:

( 0 ) = ( +1) + ( −1) ( +1) = ( +1) + ( 0 ) ( 0) = ( 0) + ( 0) ( −2 )  ( −1) + ( 0 ) ( 12 ) = ( 12 ) + ( 0 )

Energy:

1314.9 Mev c  1189.4 Mev c 2 + 139.6 Mev c 2 = 1329 Mev c 2

39. (a) Charge: Baryon number: Lepton number: Strangeness:

Charge is conserved. Baryon number is conserved. Lepton number is conserved. Strangeness is NOT conserved. Spin is conserved.

Energy is NOT conserved.

The decay is not possible. Neither strangeness or energy is conserved. (b) Charge: Baryon number: Lepton number: Strangeness: Spin:

( −1) = ( 0 ) + ( −1) + ( 0 ) ( +1) = ( +1) + ( 0 ) + ( 0 ) ( 0 )  ( 0 ) + ( 0 ) + (1) ( −3)  ( −1) + ( 0 ) + ( 0 ) ( 23 )  ( 12 ) + ( 0) + ( 12 )

Charge is conserved. Baryon number is conserved. Lepton number is NOT conserved. Strangeness is NOT conserved. Spin is NOT conserved.

Energy: 1672.5 Mev c  1192.6 Mev c + 139.6 Mev c 2 + 0 Mev c 2 = 1332.2 Mev c 2 Energy is conserved. The decay is not possible. Neither lepton number, strangeness, or spin is conserved. 2

(0) = (0) + (0) + (0) (0) = (0) + (0) + (0) (0) = (0) + (0) + (0) ( −1) = ( −1) + ( 0 ) + ( 0 ) ( 12 ) = ( 12 ) + (1) + ( −1)

Charge is conserved. Baryon number is conserved. Lepton number is conserved. Strangeness is conserved.

Spin is conserved, if the gammas have opposite spins. 2 2 2 2 2 Energy: 1192.6 Mev c  1115.7 Mev c + 0 Mev c + 0 Mev c = 1115.7 Mev c Energy is conserved. The decay is possible. 40. The Q-value is the mass energy of the reactants minus the mass energy of the products. 0 For the first reaction, p + p → p + p +  :

(

)

Q = 2mp c 2 − 2mp c 2 + m c 2 = − m c 2 = −135.0 MeV 0

For the second reaction, p + p → p + n +  : +

(

)

Q = 2mp c 2 − mp c 2 + mn c 2 + m c 2 = mp c 2 − mn c 2 − m c 2 a +

= 938.3 MeV − 939.6 MeV − 139.6 MeV= −140.9 MeV

41. The expected lifetime of the virtual W particle is found from the Heisenberg uncertainty principle, as given in Eq. 38–2 and Example 43–11. 6.63  10 −34 J s 1eV E t  → t  =  = 8.20  10 −27 s  8  10 −27 s −19 9 E 2 ( 80.379  10 eV ) 1.60  10 J © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1434

Chapter 43

Elementary Particles

42. (a) The  0 has a strangeness of –2, so it must contain two strange quarks. In order to make a neutral particle, the third quark must be an up quark. So  0 = u ss . (b) The  − has a strangeness of –2, so it must contain two strange quarks. In order to make a particle with a total charge of –1, the third quark must be a down quark. So  − = d s s . 43. (a) The neutron has a baryon number of 1, so there must be three quarks. The charge must be 0, as must be the strangeness, the charm, the bottomness, and the topness. Thus, n = u d d . (b) The antineutron is the antiparticle of the neutron, and so n = u d d . (c) The  0 has a strangeness of −1, so it must contain an “s” quark. It is a baryon, so it must contain three quarks. And, it must have charge, charm, bottomness, and topness equal to 0. Thus,  0 = u d s . (d) The  0 has a strangeness of +1, so it must contain an s quark. It is a baryon, so it must contain three quarks. And, it must have charge, charm, bottomness, and topness equal to 0. Thus, 0 = u d s . 44. (a) The combination u u d has charge = +1, baryon number = +1, and strangeness, charm, bottomness, and topness all equal to 0. Thus, u u d = p . (b) The combination u u s has charge = –1, baryon number = –1, strangeness = +1, and charm, bottomness, and topness all equal to 0. Thus, u u s =  − . (c) The combination u s has charge = –1, baryon number = 0, strangeness = –1, and charm, bottomness, and topness all equal to 0. Thus, u s = K − . (d) The combination d u has charge = –1, baryon number = 0, and strangeness, charm, bottomness, and topness all equal to 0. Thus, d u =  − (e) The combination c s has charge = –1, baryon number = 0, strangeness = –1, charm = –1, and bottomness and topness of 0. Thus, c s = D −S 45. To form the D 0 meson, we must have a total charge of 0, a baryon number of 0, a strangeness of 0, and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we must have a “c” quark, with a charge of + 23 e . To have a neutral meson, there must be another quark with a charge of − 23 e . To have a baryon number of 0, that second quark must be an antiquark. The only candidate with those properties is an anti-up quark. Thus, D 0 = c u . +

46. To form the DS meson, we must have a total charge of +1, a baryon number of 0, a strangeness of +1, and a charm of +1. We assume that there is no topness or bottomness. To get the charm, we must have a “c” quark, with a charge of + 23 e . To have a total charge of +1, there must be another quark © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1435

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

with a charge of + 13 e . To have a baryon number of 0, that second quark must be an antiquark. To have a strangeness of +1, the other quark must be an anti-strange. Thus, DS+ = c s . 47. Here is a Feynman diagram for the reaction  − + p →  0 + n. There are other possibilities, since the  0 also can be represented as a d d combination or as a mixture of d d and u u combinations.

48. Since leptons are involved, the reaction n + v → p +  − is a weak interaction. Since there is a charge change in the lepton, a W boson must be involved in the interaction. If we consider the neutron as having emitted the − boson, it is a W , which interacts with the neutrino. If we consider the +

neutrino as having emitted the boson, it is a W , which interacts with the neutron. 49. To find the length in the lab, we need to know the speed of the particle which is moving relativistically. Start with Eq. 36–10a.





K = m0 c 2 

 1− v c 

tlab =

t0 1− v c 2

− 1 → v = c 1 −

 

2.90  10 −13 s 1 − ( 0.7522 )

(

 K   m c 2 + 1  0 

= c 1−

 920 MeV   1777 MeV + 1  

= 0.7522c

= 4.401  10 −13 s

)(

)

xlab = vtlab = ( 0.7522 ) 3.00  108 m s 4.401 10 −13 s = 9.93  10 −5 m

50. (a) For the reaction  + p → K + p +  , the conservation laws are as follows. Charge: −1 + 1  0 + 1 + 0 Charge is NOT conserved. 1 1 Spin: 0 + 2 = 0 + 2 + 0 Spin is conserved. −

Baryon number: 0 + 1 = 1 + 0 + 0 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 + 0 Lepton number is conserved. Strangeness: 0 + 0  1 + 0 + 0 Strangeness is NOT conserved. The reaction is not possible because neither charge nor strangeness is conserved.

Note that the reactants would need significant kinetic energy to be able to “create” the K . − 0 0 (b) For the reaction K + p →  +  , the conservation laws are as follows. Charge: −1 + 1 = 0 + 0 Charge is conserved. 1 1 0 + = + 0 Spin: Spin is conserved. 2 2 0

1436

Chapter 43

Elementary Particles

Baryon number: 0 + 1 = 1 + 0 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 Lepton number is conserved. Strangeness: −1 + 0 = −1 + 0 Strangeness is conserved. The reaction is possible via the strong interaction. (c) For the reaction K + n →  +  +  , the conservation laws are as follows. Charge: 1 + 0 = 1 + 0 + 0 Charge is conserved. 1 1 Spin: 0 + 2 = − 2 + 0 + 1 Spin is conserved. +

Baryon number: 0 + 1 = 1 + 0 + 0 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 + 0 Lepton number is conserved. Strangeness: 1 + 0  −1 + 0 + 0 Strangeness is NOT conserved. The reaction is not possible via the strong interaction because strangeness is not conserved. It is possible via the weak interaction. (d) For the reaction K + →  0 +  0 +  + , the conservation laws are as follows. Charge: 1 = 0 + 0 + 1 Charge is conserved. Spin: 0 = 0 + 0 + 0 Spin is conserved. Baryon number: 0 = 0 + 0 + 0 Baryon number is conserved. 0 = 0 + 0 + 0 Lepton number: Lepton number is conserved. Strangeness: 1  0 + 0 + 0 Strangeness is NOT conserved. Mass: 493.7 MeV/c2 > 135.0 + 135.0 + 139.6 = 409.6 MeV/c2 The reaction is possible energetically. The reaction is not possible via the strong interaction because strangeness is not conserved. It is possible via the weak interaction. (e) For the reaction 

→ e+ + e , the conservation laws are as follows. Charge: 1 = 1 + 0 Charge is conserved. 1 1 Spin: 0 = − 2 + 2 Spin is conserved. Baryon number: 0 = 0 + 0 Baryon number is conserved. Lepton number: 0 = −1 + 1 Lepton number is conserved. Strangeness: 0 + 0 = 0 + 0 + 0 Strangeness is conserved. Mass: 139.6 MeV/c2 > 0.511 MeV/c2 The reaction is possible energetically. The reaction is possible via the weak interaction. +

51. (a) For the reaction  − + p → K + +  − , the conservation laws are as follows. Charge: −1 + 1 = 1 − 1 Charge is conserved. 1 1 Spin: 0 + 2 = 0 + 2 Spin is conserved. Baryon number: 0 + 1 = 0 + 1 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 Lepton number is conserved. Strangeness: 0 + 0 = 1 − 1 Strangeness is conserved. The reaction is possible via the strong interaction. (b) For the reaction  + + p → K + +  + , the conservation laws are as follows. Charge: 1 + 1 = 1 + 1 Charge is conserved. 1 1 Spin: 0 + 2 = 0 + 2 Spin is conserved. Baryon number: 0 + 1 = 0 + 1 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 Lepton number is conserved. Strangeness: 0 + 0 = 1 − 1 Strangeness is conserved. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1437

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The reaction is possible via the strong interaction. (c) For the reaction  − + p →  0 + K 0 +  0 , the conservation laws are as follows. Charge: −1 + 1 = 0 + 0 + 0 Charge is conserved. 1 1 Spin: 0 + 2 = 2 + 0 + 0 Spin is conserved. Baryon number: 0 + 1 = 1 + 0 + 0 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 + 0 Lepton number is conserved. Strangeness: 0 + 0 = −1 + 1 + 0 Strangeness is conserved. The reaction is possible via the strong interaction. (d) For the reaction  + + p →  0 +  0 , the conservation laws are as follows. Charge: 1 + 1  0 + 0 Charge is NOT conserved. 1 1 Spin: 0 + 2 = 2 + 0 Spin is conserved. Baryon number: 0 + 1 = 1 + 0 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 0 Lepton number is conserved. Strangeness: 0 + 0  −1 + 0 Strangeness is NOT conserved. The reaction is not possible because neither charge nor strangeness is conserved.

(e) For the reaction  + p → p + e + ve , the conservation laws are as follows. Charge: −1 + 1 = 1 − 1 + 0 Charge is conserved. 1 1 1 1 Spin: 0 + 2 = 2 + 2 − 2 Spin is conserved. −

−

Baryon number: 0 + 1 = 1 + 0 + 0 Baryon number is conserved. Lepton number: 0 + 0 = 0 + 1 − 1 Lepton number is conserved. Strangeness: 0 + 0 = 0 + 0 + 0 Strangeness is conserved. The reaction is possible via the weak interaction. Note that we did not check mass conservation because in a collision, there is always some kinetic energy brought into the reaction. Thus, the products can be heavier than the reactants. 52. The  is the anti-particle of the  , so the reaction is  − →  − + v . The conservation rules are −

as follows. Charge: −1 = −1 + 0 Baryon number: 0 = 0 + 0 Lepton number: 0 = 1 − 1 Strangeness: 0 = 0 + 0 Spin: 0 = 12 − 12

Charge is conserved. Baryon number is conserved. Lepton number is conserved. Strangeness is conserved. Spin is conserved

53. Use Eq. 43–3 to estimate the mass of the particle based on the given distance. 6.63  10−34 J s 3.0 108 m s  hc 1  2 mc  = = 1.98  1011 eV  200 GeV   −19 −18 2 d 2 10 m  1.60 10 J eV 

(

)(

)



This value is of the same order of magnitude as the mass of the W , which is about 80 GeV. 54. The fundamental fermions are the quarks and electrons. In a water molecule, there are 2 hydrogen atoms consisting of 1 electron and 1 proton each, and 1 oxygen atom consisting of 8 electrons, 8 protons, and 8 neutrons. Thus, there are 18 nucleons consisting of 3 quarks each and 10 electrons. The total number of fermions is thus 18  3 + 10 = 64 fundamental fermions. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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55. As mentioned in Example 43–9, the  0 can be considered as either u u or d d. There are various models to describe this reaction. Four are shown here.

Another model that shows the  0 as a combination of both u u and d d is also shown.

56. (a) To conserve charge, the missing particle must be neutral. To conserve baryon number, the missing particle must be a meson. To conserve strangeness, charm, topness, and bottomness, the missing particle must be made of up and down quarks and antiquarks only. With all this information, the missing particle is  0 . (b) This is a weak interaction since one product is a lepton. To conserve charge, the missing particle must be neutral. To conserve the muon lepton number, the missing particle must be an antiparticle in the muon family. With this information, the missing particle is v . 57. We assume that the interaction happens essentially at rest, so that there is no initial kinetic energy or 0 momentum. Thus, the momentum of the neutron and the momentum of the  will have the same magnitude. From energy conservation, we find the total energy of the  . m c 2 + mp c 2 = E + mn c 2 + K n → 0

−

(

)

E = m c + mp c − mn c 2 + K n = 139.6 MeV + 938.3MeV − ( 939.6 MeV + 0.60 MeV ) 2

−

= 137.7 MeV From momentum conservation, we can find the mass energy of the  . We utilize Eq. 36–14 to relate momentum and energy. 0

(

pn = p → ( pn c ) = p c 2

) → E −m c = E −m c → m c = E −E +m c → 2

2 n

2 4 n

0

2 n

2 4 n

m c 2 = E2 − En2 + mn2 c 4 = (137.7 MeV ) − ( 939.6 MeV + 0.60 MeV ) + ( 939.6 MeV )  0



1/ 2



= 133.5 MeV → m = 133.5 MeV c 2 0

The value from Table 43–2 is 135.0 MeV.

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58. (a) First, we use the uncertainty principle, Eq. 38–1. The energy is so high that we assume E = pc,

E

and so p =

xp 

c h

. E

→ x

2



→

2

( 6.63  10 J s )( 3.00  10 m s ) (1GeV 10 eV ) = 2  10 GeV E  = 2x 2 (10 m ) (1.60  10 J eV ) −34

−31

−19

Next, we use de Broglie’s wavelength formula. We take the de Broglie wavelength as the unification distance. h h = = → p Ec

( 6.63  10 J s )( 3.00  10 m s ) (1GeV 10 eV ) = 1  10 GeV E= =  (10 m ) (1.60  10 J eV ) −34

−31

−19

Both energies are reasonably close to 1016 GeV . This energy is the amount that could be violated in conservation of energy if the universe were the size of the unification distance. (b) From Eq. 18–4, we have E = 32 kT .

E = 23 kT → T =

2E 3k

(

)(

2 1025 eV 1.6  10−19 J eV

(

3 1.38  10

−23

J K

)

) = 7.7  10 K  10 K 28

59. The Q-value is the mass energy of the reactants minus the mass energy of the products. Q = m c 2 + mp c 2 − ( m c 2 + mK c 2 ) = 139.6 MeV + 938.3 MeV − (1115.7 MeV + 497.6 MeV ) −

= −535.4 MeV

We consider the products to be one mass M = m + mK = 1613.3MeV c 2 since they both have the 0

same velocity. Energy conservation gives the following: E + mp c 2 = EM . Momentum conservation −

says that the incoming momentum is equal to the outgoing momentum. Then convert that 2 2 2 2 4 relationship to energy using the relativistic relationship that E = p c + m c .

( p c) = ( p c) → E − m c = E − M c → E − m c = ( E + m c ) − M c = E + 2E m c + m c − M c → 2

p = pM → −

−

E = −

K = −



−

−

−

M 2 c 4 − m2 c 4 − mp2 c 4 −

2 M

−

2 4

2 4 p

2 4

= K + m c 2 → −

−

M 2 c 4 − m2 c 4 − mp2 c 4 2 mp c

2 4

−

−

− m c 2 −

2 mp c 2

(1613.3MeV ) − (139.6 MeV ) − ( 938.3 MeV ) = − (139.6 MeV ) = 767.8 MeV 2 ( 938.3 MeV ) 2

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60. Since there is no initial momentum, the final momentum must add to zero. Thus, each of the pions must have the same magnitude of momentum and therefore the same kinetic energy. Use energy conservation to find the kinetic energy of each pion. 2mp c 2 = 2 K + 2m c 2 → K = mpc 2 − m c 2 = 938.3 MeV − 139.6 MeV = 798.7 MeV 61. The Q-value is the energy of the reactants minus the energy of the products. We assume that one of the initial protons is at rest, and that all four final particles have the same velocity and therefore the same kinetic energy since they all have the same mass. We consider the products to be one mass M = 4mp since they all have the same speed.

Q = 2mp c 2 − 4mp c 2 = 2mp c 2 − Mc 2 = −2mp c 2 Energy conservation gives the following, where K th is the threshold energy.

( K + m c ) + m c = E = K + Mc 2

Momentum conservation says that the incoming momentum is equal to the outgoing momentum. 2 2 2 2 4 Then convert that relationship to energy using the relativistic relationship that E = p c + m0 c .

( K + m c ) − m c = ( K + Mc ) − M c → K + 2 K m c + m c − m c = K + 4 K m c + 4m c − ( 4m ) c → pp = pM → ( pp c ) = ( p M c ) 2

2 th

2 4 p

→

2 4 p

2 th

2 4 p

2 4

2 K th mp c 2 = 4 K th mp c 2 + 4mp2 c 4 − 16mp2 c 4 → 2 K th mp c 2 = 12mp2 c 4 → K th = 6mp c 2 = 3 Q 62. We use 0 to represent the actual wavelength, and  to define the approximate wavelength. The approximation is to ignore the rest mass in the expression for the total energy, E = K + mc . We also use Eqs. 36–11a, 36–14, and 43–1. 2 2 2  mc 2  p 2 c 2 = E 2 − mc 2 = K + mc 2 − mc 2 = K 2  1 + 2  K   2

(

0 =

hc K

h p

1/ 2

 2mc 2  K 1 +  K   hc



K =

1/ 2



2mc 



K 

K 1 + hc

) (

= 1.02

=

) (



; =

hc K

→ K=

)

 0 ;  = 1.020 →

2mc 2

(1.02 ) − 1 2

(

2 9.38  108 eV 0.0404

) = 4.644 10 eV  4.6 10 eV 10

( 6.63 10 J s )( 3.00 10 m s ) = 2.677 10 m  2.7 10 m ( 4.644 10 eV )(1.60 10 J eV ) −34

−19

−17

 1 − v 2 c2 



−2

−17

−2

 4.644  1010 eV    K = c 1 − + 1   9.38  108 eV + 1  = 0.9998c 2  mc   

− 1  mc 2 → v = c 1 − 

 

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63. A relationship between total energy and speed is given by Eq. 36–11b. mc 2 E= → 1 − v 2 c2 2

 mc 2   9.38  108 eV   9.38  108 eV  1 = 1−  = 1 −  1 − = 1 − 1.2  10−8 = 0.999999988    2  12 12 c  E   6.0  10 eV   6.0  10 eV 

This is about 3.6 m/s slower than the speed of light. 64. According to Section 43–1, the energy involved in the LHC’s collisions will reach 13 TeV. Use that with the analysis used in Example 43–1. From Eq. 41–1, nuclear sizes are on the order of 1.2  10−15 m . 6.63  10 −34 J s 3.00  108 m s hc = = = 9.56  10 −20 m 12 −19 E 13  10 eV 1.60  10 J eV

( (

 rnucleus

)(

9.56  10 −20 m −15

1.2  10 m

)(

) )

= 8.0  10 −5 

1 12, 500

65. (a) The nucleus undergoes 2 beta-decays. Thus, in the nucleus, 2 neutrons change into protons. The daughter nucleus must have 2 more protons than the parent but the same number of nucleons. Thus, the daughter nucleus is

96 42

−

Mo . The reaction would be 40 Zr → 42 Mo + 2 −1 e . 96

(b) Since the reaction is neutrinoless, lepton conservation would be violated in this decay. Checking isotope masses at www.nist.gov/pml/data/comp.cfm shows that the reaction is 96 96 energetically possible. 40 Zr has a mass of 95.908271 amu and 42 Mo has a mass of 95.904676 amu. (c) If 9640 Zr would undergo 2 beta decays simultaneously and emit 2 electron antineutrinos, it could 96

decay to 42 Mo without violating any conservation laws. 66. We write equations for both conservation of energy and conservation of momentum. The magnitudes of the momenta of the products are equal. We also use Eqs. 36–11a and 36–14. p1 = p2 ; E0 = mc 2 = E1 + E2 →

( mc − E ) = E = p c + m c = p c + m c = E − m c + m c → 2

2 2

2 2 2

2 4 2

2 2 1

2 4 2

2 1

m 2 c 4 − 2mc 2 E1 + E12 = E12 − m12 c 4 + m22 c 4 → E1 = K1 = E1 − m1c 2 =

m 2 c 4 + m12 c 4 − m22 c 4 2mc 2

− m1c 2 =

2 4 1

2 4 2

m 2 c 4 + m12 c 4 − m22 c 4 2mc 2

m 2 c 4 + m12 c 4 − m22 c 4 − 2mc 2 m1c 2 2mc 2

( mc − m c ) − m c = 2

2 4 2

2mc 2

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67. The conservation laws that must hold are the conservation of momentum, angular momentum (spin), mass-energy, charge, baryon number, and lepton number. Momentum can be conserved by the two decay products having equal but opposite momentum in the rest frame of the parent particle. There are no measurements given to enable us to check conservation of momentum, so we assume that it holds. The other conservation laws are evaluated for each decay. Spin can of course be positive or negative, so our “check” means seeing if there is a way for spins to be equal, using either positive or negative values for the spin. For the mass-energy to be conserved the mass of the initial particle must be greater than the mass of the resulting particles, with the remaining mass becoming kinetic energy 2 of the particles. Those values are in GeV c , but units are omitted below. +

For the reaction t → W + b : 1 2

= 1 + ( − 12 )

Spin: Mass-energy: Charge:

173  80.4 + 4.18

Baryon number:

1 1 3 = 0+ 3

Baryon number is conserved.

0=0+0 − For the reaction t → W + b :

Lepton number is conserved.

2 3

Spin is conserved. Mass-energy is conserved. Charge is conserved.

= 1 + ( − 13 )

Lepton number:

1 2

Spin: Mass-energy: Charge:

Spin is conserved. Mass-energy is conserved. Charge is conserved.

173  80.4 + 4.18

Baryon number: Lepton number: For the reaction W

= 1 + ( − 12 )

Spin: Mass-energy: Charge:

( − 23 ) = ( −1) + 13 − 13 = 0 + ( − 13 )

Baryon number is conserved.

0=0+0 → u + d:

Lepton number is conserved.

1 = 12 + 12

Spin is conserved. Mass-energy is conserved. Charge is conserved.

80.4  0.0023 + 0.0048

1 = 23 + 13

0 = 13 + ( − 13 ) 0=0+0 Lepton number: − For the reaction W →  − + v : Baryon number:

Spin: Mass-energy: Charge: Baryon number: Lepton number:

Baryon number is conserved. Lepton number is conserved.

1 = 12 + 12

Spin is conserved. Mass-energy is conserved. Charge is conserved.

80.4  0.1057 + 0

−1 = ( −1) + 0 0=0+0 0 = 1 + ( −1)

Baryon number is conserved. Lepton number is conserved.

68. (a) Each tau will carry off half of the released kinetic energy. The kinetic energy is the difference in mass-energy of the two tau particles and the original Higgs boson. Q = mH − 2m c 2 = 125  103 MeV − 2 (1777 MeV ) = 121.4  103 MeV = 121.4 GeV

(

)

K = 12 Q = 60.7 GeV (b) The total charge must be zero, so one will have a positive charge and the other will have a negative charge. One of them must be an anti-tau particle. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(c) No. The mass of the two Z bosons is greater than the mass of the initial Higgs boson. Thus, the decay would violate the conservation of energy. 69. (a) We work in the rest frame of the isolated electron, so that it is initially at rest. Energy conservation gives the following. me c 2 = Ke + me c 2 + E → Ke = − E → Ke = E = 0 Since the photon has no energy, it does not exist and so has not been emitted. Here is an alternate explanation: Consider the electron in its rest frame. If it were to emit a single photon of energy E , the photon would have a momentum p = E c . For momentum to be conserved, the electron would move away in the opposite direction as the photon with the same magnitude of momentum. Since the electron is moving (relative to the initial frame of reference), the electron will now have kinetic energy in that frame of reference, with a total energy given by Eq. 36–14, E =

( m c ) + ( p c ) = ( m c ) + E . With the added kinetic 2



energy, the electron now has more energy than it had at rest, and the total energy of the photon and electron is greater than the initial rest energy of the electron. This is a violation of the conservation of energy, so an electron cannot emit a single photon. (b) For the photon exchange in Fig. 43–8, the photon exists for such a short time that the Heisenberg uncertainty principle allows energy conservation to be violated during the exchange. So if  t is the duration of the interaction, and E is the amount by which energy is not conserved during the interaction; as long as E t  , the process can happen.

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CHAPTER 44: Astrophysics and Cosmology Responses to Questions 1.

Long ago, without telescopes, it was difficult to see the individual stars in the Milky Way. The stars in this region of the sky were so numerous and so close together and so tiny that they all blended together to form a cloudy or milky stripe across the night sky. Now using more powerful telescopes, we can see the individual stars that make up the Milky Way galaxy.

If a star generates more energy in its interior than it radiates away, its temperature will increase, indicating a higher average kinetic energy of the particles. Consequently, there will be greater outward pressure opposing the gravitational force directed inward. To regain equilibrium, the star will expand. If a star generates less energy than it radiates away, then its temperature will decrease, reflecting a lower average kinetic energy of the particles. There will be a smaller outward pressure opposing the gravitational force directed inward, and, in order to regain equilibrium, the star will contract. In both situations a new equilibrium will be reached, which means the star will not produce either a runaway heating and explosion, or a runaway total collapse.

Red giants are extremely large stars with relatively cool surface temperatures, resulting in their reddish colors. These stars are very luminous because they are so large. When the Sun becomes a red giant, for instance, its radius will be on the order of the distance from the Earth to the Sun. A red giant has run out of hydrogen in its inner core and is fusing hydrogen to helium in a shell surrounding the core. Red giants have left their main sequence positions on the H–R diagram and moved up (more luminous) and to the right (cooler).

Although the H-R diagram only directly relates the surface temperature of a star to its absolute luminosity (and thus doesn’t directly reveal anything about the core), the H-R diagram does provide clues regarding what is happening at the core of a star. Using the current model of stellar evolution and the H-R diagram, we can infer that the stars on the main sequence are fusing hydrogen nuclei to helium nuclei at the core and that stars in the red giant region are fusing helium and beryllium to make heavier nuclei such as carbon and that this red giant process will continue until fusion can no longer occur and the star will collapse.

The initial mass of a star determines its final destiny. If, after the red giant stage of a star’s life, its mass is less than 1.4 solar masses, then the star cools as it shrinks and it becomes a white dwarf. If its mass is between 1.4 and 2-3 solar masses, then the star will condense down to a neutron star, which will eventually explode as a supernova and become a white dwarf. If its mass is greater than 2-3 solar masses, then the star will collapse even more than the neutron star and form a black hole.

When measuring parallaxes from the Moon, there are two cases: (1) If you did the measurements two weeks apart (one at full moon and one at new moon), you would need to assume that the Earth did not move around the Sun very far, and then the d shown in Fig. 44–11 would be the Earth-Moon distance instead of the Sun-Earth distance. (2) If you did the measurements six months apart and at full moon, then the d shown in Fig. 44–11 would be the Sun-Earth distance plus the Earth-Moon distance instead of just the Sun-Earth distance. From Mars, then the d shown in Fig. 44–11 would be the Sun-Mars distance instead of the Sun-Earth distance. You would also need to know the length of a Mars “year” so you could take your two measurements at the correct times.

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Measure the period of the changing luminosity of a Cepheid variable star. Use the known relationship between period and luminosity to find its absolute luminosity. Compare its absolute luminosity to its apparent brightness (the observed brightness) to determine the distance to the galaxy in which it is located.

A geodesic is the shortest distance between two points. For instance, on a flat plane the shortest distance between two points is a straight line, and on the surface of a sphere the shortest distance is an arc of a great circle. According to general relativity, space–time is curved. Determining the nature of a geodesic, for instance, by observing the motion of a body or light near a large mass, will help determine the nature of the curvature of space–time near that large mass.

If the redshift of spectral lines of galaxies were discovered to be due to something other than expansion of the universe, then the Big Bang theory and the idea that the universe is expanding would be called into question. However, the evidence of the cosmic background microwave radiation would conflict with this view, unless it too was determined to result from some cause other than expansion.

10. No. In an expanding universe, all galaxies are moving away from all other galaxies on a large scale. (On a small scale, neighboring galaxies may be gravitationally bound to each other.) Therefore, the view from any galaxy would be the same. Our observations do not indicate that we are at the center. (See Fig. 44–22.) Here is an analogy, frim Fig. 44–21. If we were sitting on the surface of a balloon and more air was put into the balloon causing it to expand, every other point on the balloon moves away from you. The points close to you are farther away because of the expansion of the rubber and the points on the other side of the balloon are farther away from you because the radius of the balloon is now larger. 11. If you were located in a galaxy near the boundary of our observable universe, galaxies in the direction of the Milky Way would be receding from you. The outer “edges” of the observable universe are expanding at a faster rate than the points more “interior”. Accordingly, due to a relative velocity argument, the slower galaxies in the direction of the Milky Way would look like they are receding from your faster galaxy near the outer boundary. Also see Fig. 44–22. 12. An explosion on Earth blows pieces out into the space around it, but the Big Bang was the start of the expansion of space itself. In an explosion on Earth, the pieces that are blown outward will slow down due to air resistance, the farther away they are the slower they will be moving, and then they will eventually come to rest. But with the Big Bang, the farther away galaxies are from each other the faster they are moving away from each other. In an explosion on Earth, the pieces with the higher initial speeds end up farther away from the explosion before coming to rest, but the Big Bang appears to be relatively uniform where the farthest galaxies are moving fastest and the nearest galaxies are moving the slowest. An explosion on Earth would correspond to a closed universe, since the pieces would eventually stop, but we would not see a “big crunch” due to gravity as we would with an actual closed universe. 13. To “see” a black hole in space we need indirect evidence. If a large visible star or galaxy was rotating quickly around a non-visible gravitational companion, the non-visible companion could be a massive black hole. Also, as matter begins to accelerate toward a black hole, it will emit characteristic X-rays, which we could detect on Earth. Another way we could “see” a black hole is if it caused gravitational lensing of objects behind it. Then we would see stars and galaxies in the “wrong” place as their light is bent as it passes past the black hole on its way to Earth.

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Astrophysics and Cosmology

14. Both the formation of the Earth and the time during which people have lived on Earth are on the far right edge of Fig. 44–29, in the era of dark energy. 15. Atoms were unable to exist until hundreds of thousands of years after the Big Bang because the temperature of the universe was still too high. At those very high temperatures, the free electrons and nuclei were moving so fast and had so much kinetic energy, and they had so many high energy photons colliding with them, that they could never combine together to form stable atoms. Once the universe cooled below 3000 K, this coupling could take place, and atoms were formed. 16. (a) Type Ia supernovae are expected to all be of nearly the same luminosity. Thus they are a type of “standard candle” for measuring very large distances. (b) The distance to a supernova can be determined by comparing the apparent brightness to the intrinsic luminosity, and then using Eq. 44–1 to find the distance. 17. If the average mass density of the universe is above the critical density, then the universe will eventually stop its expansion and contract, collapsing on itself and ending finally in a “big crunch.” This scenario corresponds to a closed universe, or one with positive curvature. 18. (a) Gravity between galaxies should be pulling the galaxies back together, slowing the expansion of the universe. (b) Astronomers could measure the redshift of light from distant supernovae and deduce the recession velocities of the galaxies in which they lie. By obtaining data from a large number of supernovae, they could establish a history of the recessional velocity of the universe, and perhaps tell whether the expansion of the universe is slowing down. 19. Many methods are available. • For nearby stars (from about 100 ly to as much as 1000 ly away) we can use parallax. In this method we measure the angular distance that a star moves relative to the background of stars as the Earth travels around the Sun. Half of the angular displacement is then equal to the ratio of Earth-Sun distance and the distance between the Earth and that star. See Fig. 44–11. • Apparent brightness of the brightest stars in galaxies, combined with the inverse square law, can be used to estimate distances to galaxies, assuming they have the same intrinsic luminosity. • The H-R diagram can be used for distant stars. Determine the surface temperature using its blackbody radiation spectrum and Wien’s law, and then estimate its luminosity from the H-R diagram. Using its apparent brightness with Eq. 44–1 will give its distance. • Variable stars, like Cepheid variables, can be used by relating the period to its luminosity. The luminosity and apparent brightness can be used to find the distance. • The largest distances are measured by measuring the apparent brightness of Type Ia supernovae. All supernovae are thought to have nearly the same luminosity, so the apparent brightness can be used to find the distance. • The redshift in the spectral lines of very distant galaxies can be used to estimate distances that are further than 107 to 108 ly. The Cepheid variable method gives the most accurate distances for far-away stars. For the most distant galaxies, redshifts give the best measurements. 20. (a) All distant objects in the universe are moving away from each other, as indicated by the galactic redshift, indicating that the universe is expanding. If the universe has always expanded, it must have started as a point. The 25% abundance of He supports the Standard Big Bang Model. The Big Bang Theory predicted the presence of background radiation, which has since been observed. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(b) The curvature of the universe determines whether the universe will continue expanding forever (open) or eventually collapse back in on itself (closed). (c) Dark energy increases the total energy of the universe increasing the probability that it is an open universe. 21. About 380,000 years after the Big Bang, the initially very hot temperature of the universe would have cooled down to about 3000 K. At that temperature, electrons could orbit bare nuclei and remain there, without being ejected by collisions. Thus stable atoms were able to form. This allowed photons to travel unimpeded through the universe, and so the universe became transparent. The radiation at this time period would have been blackbody radiation at a temperature of about 3000 K. As the universe continued to expand and cool, the wavelengths of this radiation would have increased according to Wien’s law. The peak wavelength now is much larger and corresponds to the blackbody radiation at a temperature of 2.7 K.

Responses to MisConceptual Questions 1.

(c) At the end of the hydrogen-fusing part of the star’s life, it will move toward the upper right of the diagram as it becomes a red giant. Low mass stars, not large enough to end as neutron stars, then move to the lower left as they become white dwarf stars. The position of the star on the main sequence is a result of the size (mass) of the star. This does not change significantly during its main sequence lifetime and therefore the star does not change positions along the main sequence.

(b) Parallax requires that the star moves relative to the background of distant stars. This only happens when the distance to the star is relatively small, less than about 100 light years.

(b) If the universe were expanding in every direction, then any location in the universe would observe that all other points in the universe are moving away from it with the speed increasing with distance. This is observed when galaxy speeds are measured, implying that indeed the universe is expanding. The observation does not say anything about whether the expansion will continue forever or eventually stop.

4. (c)

The Sun is primarily made of hydrogen, not heavy radioactive isotopes. As gravity initially compressed the Sun, its core heated sufficiently for the hydrogen atoms in the core to fuse to create helium. This is now the primary source of energy in the Sun. The Sun is too hot for molecules, such as water, to be formed by the oxidation of hydrogen. When fusion began in the core, the energy release balanced the force of gravity to stop the gravitational collapse. Since the Sun is no longer contracting, the gravitational potential energy is no longer decreasing and is not a significant source of energy.

(b, c, d) Parallax is only useful to find distances to stars that are less than 100 light years distant, much less than inter-galactic distances. The H-R diagram with luminosity and temperature of the star can tell us the absolute brightness of the distant star. The relationship between the relative brightness and absolute brightness tells us the distance to the star. Certain supernova explosions have a standard luminosity. A comparison of the apparent brightness and absolute brightness of the supernova can give the distance to the supernova. Finally, since the universe is expanding and the velocity of objects increases with distance, a measure of the redshift in light from distant stars can be used to find the speed of the stars and therefore the distance.

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(b) Due to the speed of light, light takes a finite time to reach Earth from the stars, with light from more distant objects requiring more time to reach Earth. Therefore when an object is observed that is a million light years away, the observers sees the object as it was a million years ago. If the object is 100 million light years away, the observer sees it as it was 100 million years ago. The farther the observed object, the earlier the time that it is being observed.

(d) At the time of the Big Bang, all of space was compacted to a small region. The expansion occurred throughout all of space. And none of the specific items mentioned, like the Earth, the Milky Way Galaxy, or the Andromeda Galaxy even existed at the time of the Big Bang.

(d) In the first few seconds of the universe matter did not exist, only energy; the first atoms were not formed until many thousands of years later, and these atoms were primarily hydrogen. Stars on the main sequence and novae convert hydrogen into helium. They do not create the heavier elements. These heavy elements are typically created in the huge energy bursts of supernovae, or in neutron star mergers, as mentioned on page 1353.

(c) The rotational period of a star in a galaxy is related to the mass in the galaxy. As astronomers observed the rotational periods, they were found to be much greater than could be accounted for by the visible matter (stars and dust clouds). Dust clouds are not dark matter, as they can reflect light and are therefore visible. The acceleration of the expansion of the universe is the foundation for the theory of dark energy, which is different from dark matter.

10. (a) If the Big Bang were a single large explosion that gave the universe an initial large kinetic energy and since currently the only large-scale force in the universe is the attractive force of gravity, we expect that the kinetic energy of the universe would be decreasing as that energy becomes gravitational potential energy. The expansion, however, is accelerating so there must be another energy source fueling the acceleration. This source has been dubbed “dark energy.”

Solutions to Problems Note – since the textbook says that the Hubble parameter is ( 23  0.4 ) km s Mly , we will treat the Hubble parameter as having 3 significant figures. 1.

Convert the angle to seconds of arc, reciprocate to find the distance in parsecs, and then convert to light years. o  3600   = 3.5  10−4  o  = 1.26  1 

(

d ( pc ) =

)

 3.26 ly  1 1 = = 0.794 pc   = 2.6 ly   1.26  1pc   3.26 ly   9.461 1015 m  16  = 2.4  10 m  ly  1pc  

0.794 pc  2.

Use the angle to calculate the distance in parsecs, and then convert to light years.  3.26 ly  1 1 = = 4.348 pc → 4.348 pc  d ( pc ) =  = 14.17 ly  14 ly   0.23  1pc 

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The parallax angle is smaller for the further star. Since tan  = d D , as the distance D to the star increases, the tangent decreases, so the angle decreases. And since for small angles, tan    , we have that   d D . Thus if the distance D is doubled, the angle  will be smaller by a factor of 2 .

The apparent brightness of an object is inversely proportional to the square of the observer’s distance from the object, given by Eq. 44–1. To find the relative brightness at one location as compared to another, take a ratio of the apparent brightness at each location. L 2

2 2 d   1  4 d Jupiter d Earth = = 2 =  Earth  =   = 0.037 or only about 4%. L bEarth d Jupiter  d Jupiter   5.2  2 4 d Earth

bJupiter

The density is the mass divided by the volume. M M 1.99  1030 kg = = = = 2  10−3 kg m3 3 3 4 10 4 V r  6  10 m 3 3

(

)

(a) The apparent brightness is the solar constant, 1.3  10 W m . 3

(b) Use Eq. 44-1 to find the intrinsic luminosity. 2 L → L = 4 d 2b = 4 1.496  1011 m2 1.3  103 W m2 = 3.7  1026 W b= 2 4 d

(

)(

)

The angular width is the inverse tangent of the diameter of our Galaxy divided by the distance to the nearest galaxy. According to Fig. 44–2, our Galaxy is about 100,000 ly in diameter. Galaxy diameter 1.0  105 ly Galaxy = tan −1 = tan −1 = 0.042 rad  2.4o Distance to nearest galaxy 2.4  106 ly

Moon = tan −1

Moon diameter Distance to Moon

= tan −1

3.48  106 m 3.84  108 m

= 9.1  10−3 rad  0.52o

The Galaxy width is about 4.5 times the Moon width. 8.

The text says that there are about 4  1011 stars in the galaxy, and about 1011 galaxies, so that means about 4  1022 stars in the observable universe.

The density is the mass divided by the volume. M M 1.99  1030 kg = 4 Sun3 = = 1.83  109 kg m3 = 3 6 4 V R   6.38  10 m Earth 3 3

(

)

Since the volumes are the same, the ratio of the densities is the same as the ratio of the masses. 1.99  1030 kg  M = = = 3.33  105 times larger 24  Earth M Earth 5.98  10 kg

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10. The density of the neutron star is its mass divided by its volume. Use the proton to calculate the density of nuclear matter. The radius of the proton is taken from Eq. 41–1. 30 M 1.5 1.99  10 kg  neutron = = = 5.354  1017 kg m3  5.4  1017 kg m 3 3 3 4 V star  11  10 m 3

(

 neutron star

 white

(

5.354  1017 kg m3 1.83  109 kg m 3

)

 neutron star

= 2.9  10

 nuclear

dwarf

matter

5.354  1017 kg m 3 1.673  10−27 kg 4 3

 (1.2  10−15 m )

= 2.3

11. The reciprocal of the distance in parsecs is the angle in seconds of arc. 1 1 (a)   = = = 0.02083  0.021 d ( pc ) 48 pc

 1o

o o  = ( 5.786  10−6 )  ( 5.8  10−6 )   3600 

(b) 0.02083 

12. Convert the light years to parsecs, and then take the reciprocal of the number of parsecs to find the parallax angle in seconds of arc.  1pc  1 = 0.044 = 74 ly   = 22.70 pc  23 pc 22.70 pc  3.26 ly  13. Find the distance in light years. That value is also the time for light to reach us.  3.26 ly  77 pc   = 251ly  250 ly → It takes light 250 years to reach us.  1pc  14. The Q-value is the mass energy of the reactants minus the mass energy of the products. The masses are found in Appendix G. 4 He + 42 He → 48 Be 2

(

)

Q = 2mHe c 2 − mBe c 2 =  2 ( 4.002603 u ) − 8.005305 u  c 2 931.5 Mev c 2 = −0.092 MeV 4 2

He + Be → 8 4

12 6

(

Q = mBe c + mHe c − mC c 2 =  4.002603 u + 8.005305 u − 12.000000  c 2 931.5 Mev c 2 2

)

= 7.366 MeV

15. The angular width is the inverse tangent of the diameter of the Moon divided by the distance to the Sun. o Moon diameter 3.48  106 m = tan −1 = 2.33  10−5 rad  1.33  10−3  4.79  = tan −1 11 Distance to Sun 1.496  10 m

(

)

16. A: The temperature increases, the luminosity stays the same, and the size decreases. B: The temperature stays the same, the luminosity decreases, and the size decreases. C: The temperature decreases, the luminosity increases, and the size increases.

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17. Wien’s law (Eq. 37-1) says that the PT =  , where  is a constant, and so P1T1 = P2T2 . The Stefan–Boltzmann equation (Eq. 19–18) says that the power output of a star is given by P =  AT , where  is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the same as the luminosity L in this chapter. The luminosity L is related to the apparent brightness b by Eq. 44-1. It is given that b1 = b2 , r1 = r2 , P1 = 750 nm, and P 2 = 450 nm. 4

P1T1 = P2T2 →

b1 = b2 →

4 d12

P1 P2

 A2T24 4 r22T24 T24  T2  → = = = = = =  4 d12 L1 P1  AT 4 r12T14 T14  T1  1 1 d 22

L2 4 d 22

→

 T      750  =  2  =  P1  =   = 2.8 d1  T1   P2   450 

The star with the peak at 450 nm is 2.8 times further away than the star with the peak at 750 nm. 18. Wien’s law (Eq. 37-1) says that the PT =  , where  is a constant, and so P1T1 = P2T2 . The Stefan–Boltzmann equation (Eq. 19–18) says that the power output of a star is given by P =  AT , where  is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the same as the luminosity L in this chapter. The luminosity L is related to the apparent brightness b by Eq. 44–1. It is given that b1 b2 = 0.091 , d1 = d 2 , P1 = 470 nm, and P 2 = 720 nm. 4

P1T1 = P2T2 → 1=

d 22

0.091L2

= 2

T1 =

P1 L1 L = 0.091 2 2 ; b1 = 0.091 b2 → 2 P2 4 d1 4 d 2

0.091P2 P1

0.091 A2T24

4 AT 1 1

( 0.091) 4 r22T24 4 r12T14

→

T 4 r2 = 0.091 24 22 T1 r1

→

T     470 nm  = 0.091  2  = 0.091  P1  = 0.091   = 0.1285 r2  720 nm   T1   P2  r1

The ratio of the diameters is the same as the ratio of radii, so

= 0.13 .

19. The Schwarzschild radius is 2GM c . REarth =

2GM Earth c2

( =

)( 5.98 10 kg ) = 8.86 10 m  8.9 mm ( 3.00 10 m s )

2 6.67  10−11 N m 2 kg 2

in Example 44–1.

2GM c2

( =

2GM c2

−3

20. The Schwarzschild radius is given y R =

. An approximate mass for our Galaxy is calculated

)( 2  10 kg ) = 3  10 m ( 3.00  10 m s )

2 6.67  10−11 N m 2 kg 2 8

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21. The Schwarzschild radius is given by R = 2GM c , so M = 2

Rc 2 2G

. The radius is given in Eq. 37–

12.

( 5.29  10 m )( 3.00  10 m s ) = 3.57  10 kg M = = 2G 2 ( 6.67  10 N m kg ) −11

Rc 2

−16

−11

22. The limiting value for the angles in a triangle on a sphere is 540o . Imagine drawing an equilateral triangle near the north pole, enclosing the north pole. If that triangle were small, the surface would be approximately flat, and each angle in the triangle would be 60o . Then imagine “stretching” each side of that triangle down towards the equator, while keeping sure that the north pole stayed inside the triangle. The angle at each vertex of the triangle would expand, with a limiting value of 180o . The three 180o angles in the triangle would sum to 540o . 23. (a) For the vertices of the triangle we choose the North pole and two points on a latitude line on nearly opposite sides of the Earth, as shown on the diagram. Let the angle at the North pole be 179°. (b) It is not possible to draw a triangle on a sphere with the sum of the angles less than 180°. To draw a triangle like that, a hyperbolic surface like a saddle (similar to Fig. 44–18) would be needed.

180° 179o 90° 90o

90°o 90

24. To just escape from an object, the kinetic energy of the body at the surface of the body must be equal to the magnitude of the gravitational potential energy at the surface. Use Eq. 8–19. 2GM

vesc =

RSchwarzchild

2GM 2GM c 2

= c

25. We find the time for the light to cross the elevator, and then find how far the elevator moves during that time due to its acceleration.

t =

x c

; y = g ( t ) = 2

1 2

g ( x ) 2c

( 9.80 m s ) ( 3.2 m ) = 5.6  10 m = 2 ( 3.00  10 m s ) 2

−16

Note that this is smaller than the size of a proton. 26. Use Eq. 44–4, Hubble’s law. v 2250 km s v = H 0d → d = = = 97.8 Mly = 9.78  107 ly H 0 23 km s Mly 27. Use Eq. 44–4, Hubble’s law. v = H0d → d =

v H0

( 0.022 ) ( 3.00  108 m s ) 2.3  10 m s Mly 4

= 287.0 Mly  2.9  10 2 Mly = 2.9  108 ly

28. (a) Use Eq. 44–6 to solve for the speed of the galaxy.

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obs − rest v  455 nm − 434 nm   → v = c  = 0.04839c  0.048 c rest c 434 nm  

(b) Use Hubble’s law, Eq. 44–4, to solve for the distance. 0.04839 3.00  108 m s v v = H0d → d = = = 631Mly  6.3  108 ly H0 23000 m s Mly

(

)

29. We find the velocity from Hubble’s law, Eq. 44–4, and the observed wavelength from the Doppler shift, Eq. 44–3a. 2.3  104 m s Mly ( 6.0 Mly ) v H0d (a) = = = 4.6  10−4 8 c c 3.00  10 m s

(

1+ v c

 = 0 (b)

v c

1− v c

H0d c

 = 0

)

= ( 656.3 nm )

1 + 4.6  10−4

= 656.6 nm →  =  − 0 = 0.3 nm

1 − 4.6  10−4

( 2.3 10 m s Mly ) ( 60 Mly ) = 4.6 10 4

−3

3.00  10 m s 8

1+ v c 1− v c

= ( 656.3 nm )

1 + 4.6  10−3 1 − 4.6  10−3

= 659.3 nm →  =  − 0 = 3.0 nm

30. Use Eqs. 44–3a and 44–4 to solve for the distance to the galaxy. 2 2 obs − rest 1+ v c obs = rest → v=c 2 2 1− v c obs + rest

( (

) )

(  −  ) = ( 3.00 10 m s ) ( 419.7 nm ) − ( 393.4 nm )  d= = H H (  +  ) ( 2.30  10 m s Mly ) ( 419.7 nm ) + ( 393.4 nm )    v

2 obs

2 rest

2 obs

2 rest

= 842.9 Mly  8.43  108 ly

If instead we use Eq. 44–3b, we get the following.  v  26.3 nm = → v= c= c = 6.685  10 −2 c rest c rest 393.4 nm d=

v H0

(

= 6.685  10 −2

( 3.00 10 m s ) 8

)H (

= 6.685  10 −2

) 2.3 10 m s Mly  8.72 10 ly ( ) 8

31. Use Eqs. 44–3a and 44–5a to solve for the speed of the galaxy.

z= v c

obs − rest obs 1+ v c = −1 = −1 → rest rest 1− v c

( z + 1) − 1 1.0682 − 1 = = 0.06569 → v = 0.066 c 2 ( z + 1) + 1 1.0682 + 1 2

The approximation of Eq. 44–6 gives v = zc = 0.068 c .

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32. Use Eqs. 44–3a and 44–5a to solve for the redshift parameter.

obs − rest obs 1+ v c 1 + 0.068 = −1 = −1 = − 1 = 0.07048  0.070 rest rest 1− v c 1 − 0.068 Or, we use the approximation given in Eq. 44–6. v z  = 0.068 c z=

33. We cannot use the approximation of Eq. 44–6 for this circumstance. Instead, we use Eq. 44–3a combined with Eq, 44–5c, and then use Eq. 44–4 for the distance. 1+ v c 1+ v c 2 2 = 2.15 → = ( 2.15 ) → 1 + v c = ( 2.15 ) (1 − v c ) → z +1 = 1− v c 1− v c

3.6225  ( 2.15 )2 + 1 v c = ( 2.15 )2 − 1 → 5.6225 v c = 3.6225 → v =   c = 0.6443c    5.6225  d=

v H0

( 0.6443) ( 3.00  108 m s ) 23  103 m s Mly

= 8404 Mly  8.40  109 ly

34. Use Eqs. 44–3a and 44–5a to solve for the speed of the galaxy.  = 625 nm − 434 nm = 191nm 2

    191nm  + 1 − 1  + 1 − 1  434 nm   v  0 1+ v c 1.0739   = −1 → = = = = 0.3494 → z= 2 2 rest 1− v c 3.0739 c     191nm  + 1 + 1 + 1 + 1   434 nm     0 

(

)

v  0.349c or v = ( 0.3494 ) 3.00  108 m s = 1.05  108 m s

Use Hubble’s law, Eq. 44–4, to solve for the distance. v v c 0.3494 v = H0d → d = = = = 4.557  103 Mly  4.56  109 ly H 0 H 0 c ( 23000 m s Mly )

3.00  108 m s 35. For small relative wavelength shifts, we may use Eq. 44–6 to find the speed. We use Eq. 44–4 to find the distance. v   ; v = H 0d → = → v=c c rest rest

 3.00  108 m s  0.10 cm  = = d=   = 62 Mly H 0 H 0 rest  23, 000 m s Mly   21cm  v

c 

We also use the more exact relationship of Eq. 44–3a.

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 obs   21.1cm    −1   −1 21cm  obs v  rest  1+ v c  = → = = = 4.75  10 −2 2 2 rest c   1− v c  21.1cm  obs   + 1  21cm  + 1    rest  v = H0d → d =

v H0

v c H0 c

4.75  10−2 = 61.96 Mly  62 Mly ( 23000 m s Mly ) 3.00  108 m s

36. Eq. 44–3a states obs = rest

obs = rest

1+ v c 1− v c

. 1/ 2

1+ v c

 v  v = rest  1 +   1 −  1− v c  c  c

−1/ 2

v  v v    rest  1 + 12  1 − ( − 12 )  = rest  1 + 12  c  c c  

v v  v   v   v obs  rest  1 + 2  12   = rest  1 +  = rest + rest →  − rest =  = rest → = rest c c c  c   c  37. Wien’s law is given in Eq. 37–1.

PT = 2.90  10−3 m K → P =

2.90  10−3 m K T

2.90  10−3 m K 2.7 K

= 1.1 10−3 m

38. (a) We approximate the temperature–kinetic energy relationship by kT = K as given on page 1368. 13  1012 eV 1.60  10−19 J eV K kT = K → T = = = 1.5  1017 K 1.38  10−23 J K k

(

)(

)

(b) From Fig. 44–29, this might correspond to a time around 10−14 s . Note that this is just a very rough estimate due to the qualitative nature of Fig. 44–29. (c) According to Fig. 44–29, this is in the “Electroweak era” . 39. We use Wien’s law, Eq. 37–1. From Fig. 44–29, the temperature is about 1010 K. 2.90  10−3 m K 2.90  10 −3 m K =  3  10−13 m PT = 2.90  10−3 m K → P = T 1010 K From Fig. 31–12, that wavelength is in the gamma ray region of the EM spectrum. 40. We use the proton as typical nuclear matter.  −26 kg   1nucleon  3  10  = 6 nucleons m 3  −27 m 1.67 10 kg     41. If the scale factor of the universe is inversely proportional to the temperature, then the scale times the temperature should be constant. If we call the current scale factor “1”, and knowing the current temperature to be about 3 K, then the product of the scale factor and the temperature is about 3. Use Fig. 44–29 to estimate the temperature at various times. For purposes of illustration, we assume the 10 universe has a current size of about 10 ly . There will be some variation in the answer due to reading the figure. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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(a) At t = 10 yr , we estimate the temperature to be is about 1000 K. Thus the scale factor is found as follows. 3 3 = = 3  10 −3 → ( Scale )( Temperature ) = 3 → Scale = Temperature 1000 6

(

)(

)

Size  3  10−3 1010 ly = 3  107 ly

(b) At t = 1 s , the temperature is about 1010 K . 3 3 = 10 = 3  10 −10 Scale = Temperature 10

(

)(

)

→ Size  3  10 −10 1010 ly = 3ly

(

)(

)

Size  3  10−12 1010 ly = 3  10 −2 ly  3  1014 m −33

(d) At t = 10 s , the temperature is about 1027 K . 3 3 Scale = = 27 = 3  10−27 Temperature 10

(

)(

)

Size  3  10−27 1010 ly = 3  10−17 ly  0.3 m 42. We approximate the temperature–energy relationship by kT = E = mc 2 as suggested on page 1368. mc 2 2 . kT = mc → T = k 500 MeV c 2 c 2 1.60  10 −13 J MeV mc 2 (a) T = = = 6  1012 K −23 1.38  10 J K k

(

) (

)

From Fig. 44–29, this corresponds to a time of  10−5 s . (b) T =

mc 2 k

( 9500 MeV c ) c (1.60  10 J MeV ) = 1  10 K = 2

−13

1.38  10 −23 J K

From Fig. 44–29, this corresponds to a time of  10−11 s . (c) T =

mc 2 k

(100 MeV c ) c (1.60  10 J MeV ) = 1  10 K = 2

−13

1.38  10

−23

J K

From Fig. 44–29, this corresponds to a time of  10−6 s . There will be some variation in the answers due to reading the figure. 43. (a) According to the text, near Fig. 44–33, baryons (protons and neutrons) make up about 5% of the total mass-energy in the universe, in the form of stars and galaxies. M stars & 1011 galaxies 1011 stars galaxy 2.0  1030 kg star galaxies  baryon = stars & = 4 3 = 3 4 R galaxies   14  109 ly 9.46  1015 m ly  3 3

(

)( (

)(

)

= 2.055  10−27 kg m 3  2.1  10 −27 kg m 3

This is less than the critical density, suggesting a negative curvature. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1457

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Instructor Solutions Manual

(b) Again, according to the text, dark matter is about 27/5 = 5.4 times more plentiful than normal matter.

dark = 5.4baryon = 5.4 ( 2.055 10−27 kg m3 )  1.1 10−26 kg m3

This is somewhat larger than the critical density, suggesting a positive curvature. 44. Each black hole is twice the Schwarzschild radius 2GM    RS = 2  from the other, and is moving in a circular orbit, c  

2 RS

with the orbit radius = RS . The gravitational attraction of one star on the other will produce uniform circular motion about their center of mass, the center of their orbits. M v2 M2 M 2  RS  GM GM c2 2 G v =G → = = = =  2 2  RS ( 2 RS ) ( 2 RS )  M  4 RS 4  2GM  8 2  c  c

 0.35 c =

3.00  108 m s 8

 1.1  108 m s

45. The angular momentum is the product of the rotational inertia and the angular velocity. ( I  )initial = ( I  )final → 2

 2 MR 2  I   Rinitial   6  106 m  1rev month final = initial  initial  = initial  52 initial  = = ( )   initial    2 3  8  10 m   I final   Rfinal   5 MRfinal  = 5.625  105 rev month = 5.625  105

rev month



1month 30 d



1d 24 h



1h 3600 s

= 0.217 rev s

 0.2 rev s 46. The rotational kinetic energy is given by 12 I  . The final angular velocity, from Problem 45, is 2

5.625  105 rev month .  R   = 1 = 2 =  final final  2 2 2 K initial 2 I initialinitial 5 MRinitialinitial  Rinitialinitial  K final

1 2

2 I finalfinal

2 5

2 2 final MRfinal

 ( 8  103 m )( 5.625  105 rev month )  5 5 =  = 5.625  10  6  10 6   ( 6 10 m ) (1rev month )   2

47. The apparent luminosity is given by Eq. 44–1. Use that relationship to derive an expression for the absolute luminosity, and equate that for two stars.

1458

Chapter 44

Astrophysics and Cosmology

L 4 d

2 2 bdistant = 4 dSun bSun → → L = 4 d 2b ; Ldistant = LSun → 4 d distant

star

bSun

d distant = dSun

bdistant

star

(

= 1.5  1011 m

star

 ) 101  9.4611ly  = 5 ly  10 m  −11

star

48. The power output is the energy loss divided by the elapsed time. 2 2 2 K K initial ( fraction lost ) 12 I  ( fraction lost ) 12 25 MR  ( fraction lost ) = = = P= t t t t =

(

)(

30 3 1 (1.5) 1.99  10 kg 8.0  10 m

) ( 2 rad s ) (1  10 ) = 1.74610 W  1.7  10 W 2

−9

(1d )( 24 h d )( 3600s h )

49. Use Newton’s law of universal gravitation. F =G

m1m2 r2

( 4  10 kg ) = ( 6.67  10 N m kg ) = 2.98  10 N ( 2  10 ly )( 9.46  10 m ly ) 2

−11

 3  1028 N

50. We use the Sun’s mass and given density to calculate the size of the Sun. M M = → = 3 4 V  rSun 3 1/ 3

 3M  rSun =    4  rSun d Earth-Sun

 3 (1.99  1030 kg )  = −26 3   4 (10 kg m ) 

3.62  1018 m 1.50  10 m 11

 2  107 ;

1/ 3

1ly   = 382 ly  400 ly  15  9.46  10 m 

= 3.62  1018 m  rSun

d galaxy

382 ly 100,000 ly

 4  10−3

51. The temperature of each star can be found from Wien’s law, Eq. 37–1. The peak wavelength is used as a subscript to designate each star’s properties. PT = 2.90  10−3 m K → T660 =

2.90  10−3 m K

= 4394 K

T480 =

2.90  10 −3 m K

660  10−9 m 480  10−9 m The luminosity of each star can be found from the H–R diagram.

L660  7  1025 W

= 6042 K

L480  5  1026 W

The Stefan–Boltzmann equation says that the power output of a star is given by P =  AT , where  is a constant, and A is the radiating area. The P in the Stefan–Boltzmann equation is the same as the luminosity L given in Eq. 44–1. Form the ratio of the two luminosities. 4

L480 L660

4 2 4  A480T480 T480 4 r480 = 4 2 4  A660T660 4 r660 T660

→

r480 r660

2 L480 T660 2 L660 T480

5  1026 W ( 4394 K )

7  1025 W ( 6042 K )

= 1.413

The diameters are in the same ratio as the radii. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1459

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

d 480 d 660

Instructor Solutions Manual

= 1.413  1.4

The luminosities are fairly subjective, since they are read from the H–R diagram. Different answers may arise from different readings of the H–R diagram.

52. (a) First calculate the number of parsecs. Then use the fact that the number of parsecs is the reciprocal of the angular resolution in seconds of arc.

1  1pc  =30.67 pc = →     3.26 ly 

100 ly 

 1   1   1  = 9.06  10−6   9  10−6   = ) ( )     (  30.67   60   60  (b) We use the Rayleigh criterion, Eq. 35–10, which relates the angular resolution to the diameter of the optical element. We choose a wavelength of 550 nm, in the middle of the visible range. 1.22 550  10−9 m 1.22 1.22 → D= = = 4.24 m  4 m = D   9.06  10−6  ( rad 180 )

(

)

The largest optical telescopes with single mirrors are about 10 m in diameter. 53. (a) To find the energy released in the reaction, we calculate the Q-value for this reaction. From Eq. 42–2a, the Q-value is the mass energy of the reactants minus the mass energy of the products. The masses are found in Appendix G.

(

)

Q = 2mC c 2 − mMg c 2 =  2 (12.000000 u ) − 23.985042 u  c 2 931.5 Mev c 2 = 13.93 MeV

(b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching. The distance between the two nuclei will be twice the nuclear radius, from Eq. 41–1. Each nucleus will have half the total kinetic energy. 2 1 qnucleus 1/ 3 1/ 3 r = 1.2  10−15 m ( A ) = 1.2  10 −15 m (12 ) U= 4 0 2 r

(

K = 12 U = 12

)

(

2 qnucleus

4 0

)

( 6 )2 (1.60  10−19 C )  1MeV  = 4.711MeV = ( 8.988  10 N n C ) 1/ 3  −13  2 (1.2  10 −15 m ) (12 )  1.60  10 J  2

1 2

 4.7 MeV

K k

( 4.711MeV ) (1.60  10−13 J MeV ) 1.38  10−23 J K

= 5.5  1010 K

54. (a) Find the Q-value for this reaction. From Eq. 42–2a, the Q-value is the mass energy of the reactants minus the mass energy of the products.

1460

Chapter 44

Astrophysics and Cosmology

16 8

28 O + 168 O → 14 Si + 42 He

(

Q = 2mO c 2 − mSi c 2 − mHe c 2 =  2 (15.994915 u ) − 27.976927 u − 4.002603 c 2 931.5 Mev c 2

)

= 9.594 MeV

(b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching. The distance between the two nuclei will be twice the nuclear radius, from Eq. 41–1. Each nucleus will have half the total kinetic energy. 2 1 qnucleus 1/3 1/3 −15 −15 r = 1.2  10 m ( A ) = 1.2  10 m (16 ) U= 4 0 2r

(

)

K nucleus = 12 U = 12

(

2 qnucleus

4 0

)

( 8 ) (1.60  10−19 C ) 1eV 1 = 2 ( 8.988  10 N n C )  = 7.609 MeV 1/3 −15 1.60  10−19 J 2 (1.2  10 m ) (16 ) 2

 7.6 MeV (c) We approximate the temperature–kinetic energy relationship by kT = K as given on page 1368.  1.60  10−19 J  7.609  106 eV   1eV K   = 8.8  1010 K kT = K → T = = −23 1.38  10 J K k

(

)

55. We must find a combination of c, G, and

that has the dimensions of time. The dimensions of c are

L  L   T  , the dimensions of G are  MT 2  , and the dimensions of   3





 ML2  are  .  T 



3 2  + 3  + 2  L   L   ML  tP = c G → T  =    =  L  M  −  T − − 2  −  2    T   MT   T   + 3 + 2 = 0 ;  −  = 0 ; −  − 2  −  = 1 →  + 5 = 0 ;  = −1 − 3 →







−5 = −1 − 3 →  = 12 ;  = 12 ;  = − 52

( 6.67 10 N m kg ) 21 ( 6.63 10 J s ) = = 5.38  10 s c 3.00  10 m s ( ) −11

tP = c

−5/ 2

1/ 2

−34

−44

56. The radius of the universe is estimated as the speed of light (c) times the age of the universe (T). This radius is used to find the volume of the universe, and the critical density times the volume gives the mass.

r = cT ; V = 43  r 3 = 43  ( cT ) ; 3

  3.156  107 s   8 9 m = V  =  ( cT )  =  ( 3.00  10 m s )(13.8  10 yr )  10 −26 kg m 3 ) (   1yr    4 3

4 3

= 9.34  1052 kg  1053 kg © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1461

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

57. (a) Assume that the nucleons make up only 2% of the critical mass density.

(

)

nucleon mass density = 0.02 10 −26 kg m 3 nucleon number density =

(

0.02 10 −26 kg m 3 1.67  10

−27

)

kg nucleon

= 0.12 nucleon m 3

neutrino number density = 109 ( nucleon number density ) = 1.2  108 neutrino m 3

(

0.98 10 −26 kg m 3

) = 8.17 10

−35

1.2  108 neutrino m 3

neutrino



9.315  108 eV c 2 1.66  10 −27 kg

= 46 eV c 2

(b) Assume that the nucleons make up only 5% of the critical mass density. nucleon mass density = 0.05 10−26 kg m3

(

nucleon number density =

)

(

0.05 10−26 kg m 3 1.67  10

−27

)

= 0.30 nucleon m3

kg nucleon

neutrino number density = 109 ( nucleon number density ) = 3.0  108 neutrino m3

(

0.95 10−26 kg m3

) = 3.17  10

3.0  10 neutrino m 8

−35

neutrino



9.315  108 eV c 2 1.66  10

−27

= 18 eV c 2

58. (a) From page 1358, a white dwarf with a mass equal to that of the Sun has a radius about the size of the Earth’s radius, 6380 km . From page 1358, a neutron star with a mass equal to 1.5 solar masses has a radius of about 20 km . For the black hole, we use the Schwarzschild radius formula. R=

2GM c

(

) ( ( 3.00  10 m s )

2 6.67  10−11 N m 2 kg 2 3 1.99  1030 kg 8

) = 8849m  8.85 km

(b) The ratio is 6380 : 20 : 8.85 = 721: 2.26 : 1  700 : 2 : 1 . 59. Each helium atom requires 2 protons and 2 neutrons and has a total mass of 4u. Each hydrogen atom requires one proton and has a mass of 1u. If there are 7 times more protons than neutrons, then for every two neutrons there are 14 protons. Two protons combine with the two neutrons to produce a helium atom. The other 12 protons produce 12 hydrogen atoms. Therefore there is 12 times as much hydrogen as helium, by number of atoms. Each helium atom has a mass 4 times that of the hydrogen atom, so the total mass of the hydrogen is12 u, and the mass of the helium is 4 u. The mass ratio of hydrogen:helium is thus 12:4, or 3:1. 60. Because Venus has a more negative apparent magnitude, Venus is brighter. We write the logarithmic relationship as follows, letting m represent the magnitude and b the brightness. m = k log b ; m2 − m1 = k ( log b2 − log b1 ) = k log ( b2 b1 ) →

m2 − m1

log ( b2 b1 )

+5 log ( 0.01)

m2 − m1 = k log ( b2 b1 ) →

= −2.5 b2 b1

= 10

m2 − m1 k

→

bVenus bSirius

= 10

mVenus − mSirius

−4.4 +1.4

−2.5

= 10 −2.5 = 16

1462

Chapter 44

Astrophysics and Cosmology

61. We assume that gravity causes a centripetal force on the gas. Solve for the speed of the rotating gas, and then use Eq. 44–6 to estimate a z value for the gas that is moving away from Earth, near the receding outer edge of the cloud. mgas mblack 2 mgas vgas hole Fgravity = Fcentripetal → G = → r2 r mblack 2  109 1.99  1030 kg hole 2 2 −11 = 6.67  10 N m kg = 6.42  105 m s vgas = G r  9.46  1015 m  ( 68 ly )   1ly  

(

z

v c

)

6.42  105 m s 3.00  10 m s 8

(

)(

)

= 2.14  10−3  2  10−3

62. We treat the energy of the photon as a “rest mass,” and so mphoton " = " Ephoton c . To just escape from 2

a spherical mass M of radius R, the energy of the photon must be equal to the magnitude of the gravitational potential energy at the surface. 2 GMmphoton GMmphoton GM Ephoton c GM Ephoton = → R= = = 2 R Ephoton Ephoton c

(

63. (a) The photon enters from the left, grazing the Sun, and moving off in a new direction. The deflection is assumed to be very small. Consider a small part of the motion in which the photon moves a horizontal distance dx = cdt while located at (x,y) relative to the center of the Sun. Note that y  R and

)

E = mc c

x 

y=R GMm F= 2 r



r = x + y . If the photon has energy E, it has 2

a “mass” of m = E c , and a momentum of p = E c = mc. To find the change of momentum in the y-direction, we use the impulse produced by the y-component of the gravitational force. GMm dx GMm R dx GMmR dx =− 2 =− dp y = Fy dt = − 2 cos  3/2 r c r r c c x2 + R2 2

(

)

To find the total change in the y-momentum, we integrate over all x (the entire path of the photon). We use an integral from Appendix B-4.  

p y =  − −

GMmR

(x + R ) 2

2 3/ 2

=−

GMmR c

(

R2 x2 + R

)

=−

2 1/ 2

2GMm cR

=−

2GMp c2 R

−

The total magnitude of deflection is p y p .

1463

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

2GMp

p y

2 2GM = c R = 2 p p c R (b) We use data for the Sun.

 =



N m2 

 (1.99  10 kg ) kg  2GM  180   3600    = 2 =    = 0.87 2 8 8 c R rad 1       ( 3.00 10 m s ) ( 6.96 10 m ) 2  6.67  10 −11

64. Eq. 44–3b states states obs = rest



rest

obs − rest v  v c rest

1+ v c 1− v c

c  , and Eq. 44-6 also states that the ratio is z. Eq. 44–3a

. It is given that v = 0.1c . Calculate v = 0.1c obs using both Eq. 44–3b

and Eq. 44–3a, and compare via percent difference.  − rest = 0.1 → obs = 44 − 3b : obs

rest

44 − 3a : obs = rest % diff =

1+ v c 1− v c

1.1

= rest

1.10554rest − 1.1 rest 1.10554rest

0.9

= 1.10554rest

 100 = 5.01%  5%

65. If there are N nucleons, we assume that there are approximately 2 N neutrons and 2 N protons. Thus, 1

for the star to be neutral, there would also be 2 N electrons. (a) From Eq. 40–12 and 40–13, we find that if all electron levels are filled up to the Fermi energy 1

EF , the average electron energy is 53 EF . Ee = N e ( EF ) = ( N ) 3 5

1 2

3 5

h2  3 Ne 



2/3



8me   V 

31

 h 3 N  =  N   5  2  8me   2V 

2/3

(b) The Fermi energy for nucleons would be a similar expression, but the mass would be the mass of a nucleon instead of the mass of the electron. Nucleons are about 2000 times heavier than electrons, so the Fermi energy for the nucleons would be on the order of 1/1000 the Fermi energy for the electrons. We will ignore that small correction.

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Chapter 44

Astrophysics and Cosmology

To calculate the potential energy of the star, think about the mass in terms of shells. We assume the star is spherical and has a uniform mass density. Consider the inner portion of the star with radius r < R and mass m, surrounded by a shell of thickness dr and mass dM. See the diagram. From Gauss’s law applied to gravity, the gravitational effects of the inner portion of the star on the shell are the same as if all of its mass were at the geometric center. Likewise, the spherically-symmetric outer portion of the star has no gravitational effect on the shell. Thus the gravitational energy of the inner portion–shell) combination is given by a form of Eq. 8–17, dU = −G

R r

m dM . The density of the star r

M . We use that density to calculate the masses, and then integrate over the 3 3R

is given by  = 4

full radius of the star to find the total gravitational energy of the star. M r3 3 4  = m =  43  r 3 = 4 r M 3  R3 R3 3

(

)

(

)

dM =  4 r 2 dr = 4 3

dU = −G R

mdM r

 3GM 2

U = −

 R3 M

= −G



)

(

)

4 r 2 dr =

r 3 3Mr 2 R3

r 4  dr = −

dr =−

r 3GM 2

3Mr 2

3GM 2 R6

( )

4  r dr = −

r 4 dr

3GM 2 R 5

= −

3 GM 2

5 R R6 R6 0 R6 5  (c) The total energy is the sum of the two terms calculated above. The mass of the star is primarily due to the nucleons, and so M = Nmnucleon . 0



3  1  h2  3 N  Etotal = Ee + U =  N    5  2  8me   2V  = Let a =

2/3

−

  3 M    3 80me  mnucleon   2mnucleon 43  R  3h 2 

92/33h 2 M 5/3

320 me m equilibrium radius. 4/3

5/3 nucleon

3 GM 2 R

2/3

−

3 GM 2 5

92/33h 2 M 5/3 5/3 R 320 4/3 me mnucleon

− 2

3 GM 2 5

3 a b and b = GM 2 , so Etotal = 2 − . We set dEtotal dR = 0 to find the 5 R R 2

92/33h 2 M 5/3

5/3 320 4/3 me mnucleon 92/3 h 2 = 5/3 3 32 4/3GM 1/3 me mnucleon dR R R b GM 2 5 We evaluate the equilibrium radius using the Sun’s mass.

dEtotal

=−

+ 3

= 0 → Req = 2

1465

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Req =

Instructor Solutions Manual

92/3 h 2 5/3 32 4/3GM 1/3 me mnucleon

(

92/3 6.63  10−34 J s

= 32

)

2 1/3 5/3  30 −11 N m  −31 −27  6.67  10 kg 2  ( 2.0  10 kg ) ( 9.11  10 kg )(1.67  10 kg )  

4/3

= 7.178  106 m  7.2  103 km The text states on page 1358 says that the radius of a white dwarf with a mass equal to that of the Sun would be about the size of the Earth. The Earth has a radius of about 6400 km, so the calculation gives a value in agreement with the text.

66. There are N neutrons. The mass of the star is due only to neutrons, and so M = Nmn . From Eqs. 40– 12 and 40–13, we find that if all energy levels are filled up to the Fermi energy EF , the average energy is 53 EF . We follow the same procedure as in Problem 65. The expression for the gravitational energy does not change.

En = N n ( EF ) = ( N ) 3 5

Etotal = En + U = Let a =

3 (18 )

2/3

160

3 (18 )

h2  3 N 



8/3 n

2/3



8mn   V 

2/3

h 2 M 5/3

160 4/3 mn8/3 R

h 2 M 5/3

4/3

3 5

− 2

 M  3h 2  3 M  =  4 3    mn  40mn   3  R mn 

2/3

3 (18 )

2/3

h 2 M 5/3

160 4/3 mn8/3 R 2

3 GM 2 5

3 a b and b = GM 2 , so Etotal = 2 − . We set dEtotal dR = 0 to find the 5 R R

equilibrium radius. 2

3 (18 )

2/3

h 2 M 5/3

160 4/3 mn8/3 (18 ) h = − 3 + 2 = 0 → Req = = = 3 dR R R b 16 4/3GM 1/3 mn8/3 GM 2 5 We evaluate the equilibrium radius for a mass of 1.5 solar masses. dEtotal

Req = =

(18 )

2/3

16 4/3GM 1/3 mn8/3

(

182/3 6.63  10−34 J s



N m2 



16 4/3  6.67  10−11

)

 1.5 ( 2.0  10 kg )   30

1/3

(1.67  10 kg ) −27

8/3

= 1.086  104 m  11km

67. (a) Because the speed of the galaxy is small compared to the speed of light, we can use Eq. 44–6 instead of Eq. 44–3a. The velocity of the galaxy is negative.  −130  103 m s  v  = rest   = 656 nm   = −0.284 nm 8 c  3.00  10 m s  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 44

Astrophysics and Cosmology

(b) This is a blue shift because the wavelength has decreased, and because the galaxy is approaching. (c) The time is equal to the distance between the galaxies divided by their relative speed. This is assuming that they continue to move at the same present speed. (They would actually accelerate due to the gravitational force between them). 2.5  106 ly  9.46  1015 m   1yr d  = 5.8  109 yr t= =    3 7  ly v 130  10 m s    3.156  10 s 

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