College Physics 11th Edition Serway Solutions Manual by Ace Test Banks

College Physics 11th Edition Serway Solutions Manual

richard@qwconsultancy.com

1|Pa ge

Topic 1

Topic 1 Units, Trigonometry, and Vectors

QUICK QUIZZES 1.1 Choice (c). The largest possible magnitude of the resultant occurs when the two vectors are in the same direction. In this case, the magnitude of   the resultant is the sum of the magnitudes of !A and !B : R = A + B = 20

units. The smallest possible magnitude of the resultant occurs when the two vectors are in opposite directions, and the magnitude is the   difference of the magnitudes of !A and !B : R = | A − B | = 4 units.

1.2

Vector

x-component

y-component

 !A

−

 !B

−

  !A + B

−

 1.3 Vector !B . The range of the inverse tangent function includes only the first

and fourth quadrants (i.e., angles in the range −π/2 < θ < π/2). Only

Topic 1

 vector !B has an orientation in this range.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 1.2

Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks.

1.4

1.6

(a)

~0.5 lb ≈ 0.25 kg or ~10−1 kg

(b)

∼4 lb ≈ 0.25 kg or ~100 kg

(c)

~4 000 lb ≈ 2 000 kg or ~103 kg

Let us assume the atoms are solid spheres of diameter 10−10 m. Then, the volume of each atom is of the order of 10−30 m3. (More precisely, volume = 4π r 3 /3 .) Therefore, since 1 cm3 = 10−6 m3, the number of atoms in the 1 cm3 solid is on the order of 10−6 / 10−30 = 1024 atoms. A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10.

1.8

Choice (d). For an angle θ from 0° to 360°, the sine and cosine functions take the values

−1 ≤ sin θ ≤ 1 and − 1 ≤ cosθ ≤ 1 1.10

In the metric system, units differ by powers of ten, so it’s very easy

Topic 1

and accurate to convert from one unit to another. 1.12

Both answers (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be meaningless since quantities can be added or subtracted only if they have the same dimensions.

1.14

The correct answer is (a). The second measurement is more precise but, given the number of reported significant figures, each measurement is consistent with the other.

1.16

The components of a vector will be equal in magnitude if the vector lies at a 45° angle with the two axes along which the components lie.

ANSWERS TO EVEN NUMBERED PROBLEMS L / T2

1.2

(a)

1.4

All three equations are dimensionally incorrect.

1.6

(a)

1.8

1.10 1.12

1.14

(b)

kg ⋅ m/s

(b)

Ft = p

(a)

22.6

(b)

22.7

(a)

3.00 × 108 m/s

(b)

2.997 9 × 108 m/s

(c)

2.997 925 × 108 m/s

(a)

346 m2 ± 13 m3

(b)

66.0 m ± 1.3 m

Topic 1

1.16

(a)

797

(b)

1.1

1.18

132 m2

1.20

3.09 cm/s

1.22

(a)

5.60 × 102 km = 5.60 × 105 m = 5.60 × 107 cm

(b)

0.491 2 km = 4.912 × 104 cm

(c)

6.192 km = 6.192 × 103 m = 6.192 × 105 cm

(d)

2.499 km = 2.499 × 103 m = 2.499 × 105 cm

1.24

10.6 km/L

1.26

9.2 nm/s

1.28

2.9 × 102 m3 = 2.9 × 108 cm3

1.30

2.57 × 106 m3

1.32

~108 steps

1.34

~108 people with colds on any given day

1.36

(a)

1.38

10−14 kg

1.40

2.2 m

1.42

8.1 cm

1.44

Δs = r12 + r22 − 2r1r2 cos(θ1 − θ 2 )

4.2 × 10−18 m3

(b)

~10−1 m3

(c)

17.66

(c)

~1016 cells

Topic 1

1.46

2.33 m

1.48

(a)

1.50

8.60 m

1.52

1.44 × 103 m

1.54

(a) 6.1 units at θ = +113°

(b) 15 units at θ = +23°

1.56

(a) 484 km

(b) 18.1° N of W

1.50 m

(b)

2.60 m

 R ! = 9.5 units at!!57° above!!the!+ x2axis

1.60

(a) 13.4 m

1.62

1.31 km northward, 2.81 km eastward

1.64

(a) 10.0 m

1.66

42.7 yards

1.68

788 mi at 48.0° N of E

1.70

(a) 185 N at 77.8° from the x-axis

(b) 185 N at 258° from the x-axis

1.72

(a) 1.609 km/h

(b) 88 km/h

1.74

(a)

(b) 2.70 × 10−4 m3/s

(b) 19.9 m

(b) 15.7 m

7.14 × 10−2 gal/s

Topic 1

1.76

(a)

A2/A1 = 4

(b) V2/V1 = 8

1.78

(a)

600 yr

(b)

1.80

1 × 106

7 × 104 times

PROBLEM SOLUTIONS 1.1

Substituting dimensions into the given equation T = 2π ℓ/ g , and recognizing that 2π is a dimensionless constant, we have ℓ [T ] = [[ g]]

or T =

L = T2 = T L/T 2

Thus, the dimensions are consistent. 1.2

(a)

From x = Bt2, we find that B = x/t2. Thus, B has units of [B] = [x]/[t2] = L/T2 .

(b) If x = A sin (2π ft), then [A] = [x]/[sin (2π ft)] But the sine of an angle is a dimensionless ratio. Therefore, [A] = [x] = L. 1.3

(a)

The units of volume, area, and height are: [V] = L3, [A] = L2, and [h] = L. We then observe that L3 = L2L or [V] = [A][h]. Thus, the equation V

Topic 1

= Ah is dimensionally correct.

(

)

2 2 (b) Vcylinder = π R h = π R h = Ah , where A = π R 2 .

Vrectangular box = ℓwh = ( ℓw ) h = Ah , where A = ℓw = length × width . 1.4

(a)

2 2 In the equation 12 mv = 12 mv0 + mgh ,

2 ML2 ⎛ L⎞ [mv ] = [mv ] = M ⎜ ⎟ = 2 while ⎡ mgh ⎤ = M ⎛⎜ L ⎞⎟ L = M L . ⎝ T⎠ ⎣ ⎦ T ⎝ T2 ⎠ T

2 0

Thus, the equation is dimensionally incorrect.

(b) In v = v0 + at , [v] = [v0 ] = 2

L ⎛ L⎞ 2 2 2 but [at ] = [a][t ] = ⎜ 2 ⎟ ( T ) = L . ⎝T ⎠ T

Hence, this equation is dimensionally incorrect.

(c)

⎛ L ⎞ ML In the equation ma = v2, we see that [ma] = [m][a] = M ⎜ 2 ⎟ = 2 , ⎝T ⎠ T 2

L2 ⎛ L⎞ while [v ] = ⎜ ⎟ = 2 . Therefore, this equation is also ⎝ T⎠ T 2

dimensionally incorrect. 1.5

From the universal gravitation law, the constant G is G = Fr2/Mm. Its units are then

[ F ] ⎡⎣ r ⎤⎦ = ( kg ⋅ m/s )( m ) = m G = [ ] [ M ][ m] kg ⋅s kg ⋅ kg 2

1.6

(a)

Solving KE = p2/2m for the momentum, p, gives p = 2m ( KE)

Topic 1

where the numeral 2 is a dimensionless constant. Dimensional analysis gives the units of momentum as:

[ p] = [ m][ KE] = M ( M ⋅ L /T ) = M ⋅ L /T = M ( L/T ) 2

Therefore, in the SI system, the units of momentum are kg ⋅ (m/s). (b)

Note that the units of force are kg ⋅ m/s2 or [F] = ML/T2. Then, observe that [F][t] = (ML/T2)T = M(L/T) = [p] From this, it follows that force multiplied by time is proportional to momentum: Ft = p. (See the impulse-momentum theorem in Topic 6, which says that a constant force F multiplied by a duration of time Δt equals the change in momentum, Δp.)

1.7

The rectangular airstrip’s area is its length times its width. The width contains four significant figures while the length contains two. Therefore the answer should be rounded to two significant figures: A = LW = ( 210 m ) ( 32.30 m ) A = 6.8 × 103 m 2 (rounded to two significant figures)

1.8

The number 15 contains no significant figures after the decimal, so the sum should be rounded to no digits after the decimal:

Topic 1

21.4 + 15 + 17.17 + 4.003 = 58 (rounded to no digits past the decimal)

1.9

The area A of the rectangular room is A = LW = (9.72 m)(5.3 m) = 52 m2 .

1.10 (a)

Computing yields

(b)

( 8 ) without rounding the intermediate result 3

( 8 ) = 22.6 to three significant figures. 3

Rounding the intermediate result to three significant figures yields 8 = 2.8284 → 2.83 . Then, we obtain

( 8 ) = ( 2.83) = 22.7 3

to three significant figures. (c)

The answer 22.6 is more reliable because rounding in part (b) was carried out too soon.

1.11 (a)

78.9 ± 0.2 has 3 significant figures with the uncertainty in the tenths position.

(b) 3.788 × 109 has 4 significant figures 4 significant figures (c) 246 × 10−6 has 3 significant figures (d) 0.003 2 = 3.2 × 10−3 has 2 significant figures. The two zeros were originally included only to position the decimal. 1.12

c = 2.997 924 58 × 108 m/s (a) Rounded to 3 significant figures: c = 3.00 × 108 m/s

Topic 1

(b) Rounded to 5 significant figures: c = 2.997 9 × 108 m/s (c) Rounded to 7 significant figures: c = 2.997 925 × 108 m/s 1.13

For a length of L = (2.0 ± 0.2) m and a width of W = (1.5 ± 0.1) m: (a) The maximum value for the area, A = LW, is found using the maximum values of L and W: Amax = (2.2 m)(1.6 m) = 3.52 m2 The minimum area is found using the minimum values of L and W: Amin = (1.8 m)(1.4 m) = 2.52 m2 The area using numbers in the middle of the uncertainty range is Amiddle = LW = (2.0 m)(1.5 m) = 3.0 m2 An uncertainty of about ± 0.5 m2 is needed to cover the range from Amax to Amin so that A = ( 3.0 ± 0.5 ) m . 2

(b) The perimeter of a rectangle is P = 2L + 2W = 2(L + W). As before, the maximum, minimum, and middle values are: Pmax = 2(2.2 m + 1.6 m) = 7.6 m Pmin = 2(1.8 m + 1.4 m) = 6.4 m Pmiddle = 2(2.0 m + 1.5 m) = 7.0 m An uncertainty of ±0.6 is needed to cover the range from Pmax to

Topic 1

Pmin so that P = ( 7.0 ± 0.6 ) m . 1.14

(a)

A = π r2 = π (10.5 m ± 0.2 m)2 = π [(10.5 m)2 ± 2(10.5 m)(0.2 m) + (0.2 m)2]. Recognize that the last term in the brackets is insignificant in comparison to the other two. Thus, we have A = π [110 m2 ± 4.2 m2] = 346 m2 ± 13 m2

(b) C = 2πr = 2π (10.5 m ± 0.2 m) = 66.0 m ± 1.3 m 1.15

The least accurate dimension of the box has two significant figures. Thus, the volume (product of the three dimensions) will contain only two significant figures. V = ℓ ⋅ w ⋅ h = ( 29 cm ) (17.8 cm ) (11.4 cm ) = 5.9 × 10 3 cm 3

1.16

(a) The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal. (b) 0.003 2 × 356.3 = (3.2 × 10−3) × 356.3 = 1.140 16 must be rounded to 1.1 because 3.2 × 10−3 has only two significant figures. (c) 5.620 × π must be rounded to 17.66 because 5.620 has only four significant figures.

1.17

Use the conversion factor 1 cubitus = 0.445 m to find the basketball player’s height:

Topic 1

⎛ 1 cubitus ⎞ 2.00 m = 2.00 m ⎜ = 4.49 cubiti ⎝ 0.445 m ⎟⎠ 1.18

Use the conversion factor 1 m = 3.281 ft to find 2

⎛ 1m ⎞ 1 420 ft 2 = 1 420 ft 2 ⎜ = 132 m 2 ⎝ 3.281 ft ⎟⎠

1.19

⎛ 5 280 ft ⎞ ⎛ 1 fathom ⎞ d = (250 000 mi) ⎜ = 2 × 10 8 fathoms ⎟ ⎜ ⎟ ⎝ 1.000 mi ⎠ ⎝ 6 ft ⎠ The answer is limited to one significant figure because of the accuracy to which the conversion from fathoms to feet is given.

1.20 v=

ℓ 186 furlongs ⎛ 1 fortnight ⎞ ⎛ 1 days ⎞ ⎛ 220 yds ⎞ ⎛ 3 ft ⎞ ⎛ 100 cm ⎞ = t 1 fortnight ⎜⎝ 14 days ⎟⎠ ⎜⎝ 8.64 × 10 4 s ⎟⎠ ⎜⎝ 1 furlongs ⎟⎠ ⎜⎝ 1 yd ⎟⎠ ⎜⎝ 3.281 ft ⎟⎠

giving v = 3.09 cm/s

1.21

⎛ 9 gal ⎞ ⎛ 3.786 L ⎞ ⎛ 10 3 cm 3 ⎞ ⎛ 1 m 3 ⎞ 6.00 firkins = 6.00 firkins ⎜ 3⎟ ⎟⎜ ⎟⎜ ⎟⎜ 6 ⎝ 1 firkin ⎠ ⎝ 1 gal ⎠ ⎝ 1 L ⎠ ⎝ 10 cm ⎠ = 0.204 m 3

1.22

⎛ 1.609 km ⎞ ℓ = (348 mi) ⎜ = 5.60 × 10 2 km = 5.60 × 10 5 m ⎟ ⎝ 1.000 mi ⎠ (a) = 5.60 × 10 7 cm ⎛ 1.609 km ⎞ h = (1 612 ft) ⎜ = 0.491 2 km = 491.2 m ⎝ 5 280 ft ⎟⎠ (b) = 4.912 × 10 4 cm

Topic 1

⎛ 1.609 km ⎞ h = (20 320 ft) ⎜ = 6.192 km = 6.192 × 10 3 m ⎟ 5 280 ft ⎝ ⎠ (c) = 6.192 × 10 5 cm ⎛ 1.609 km ⎞ d = (8 200 ft) ⎜ = 2.499 km = 2.499 × 10 3 m ⎟ ⎝ 5 280 ft ⎠ (d) = 2.499 × 10 5 cm In (a), the answer is limited to three significant figures because of the accuracy of the original data value, 348 miles. In (b), (c), and (d), the answers are limited to four significant figures because of the accuracy to which the kilometers-to-feet conversion factor is given.

1.23

v = 38.0

m ⎛ 1 km ⎞ ⎛ 1 mi ⎞ ⎛ 3 600 s ⎞ = 85.0 mi/h s ⎜⎝ 10 3 m ⎟⎠ ⎜⎝ 1.609 km ⎟⎠ ⎜⎝ 1 h ⎟⎠

Yes, the driver is exceeding the speed limit by 10.0 mi/h.

1.24

efficiency = 25.0

1.25

(a) r =

mi ⎛ 1 km ⎞ ⎛ 1 gal ⎞ = 10.6 km/L gal ⎜⎝ 0.621 mi ⎟⎠ ⎜⎝ 3.786 L ⎟⎠

diameter 5.36 in ⎛ 2.54 cm ⎞ = = 6.81 cm 2 2 ⎜⎝ 1 in ⎟⎠

(b) A = 4π r2 = 4π (6.81 cm)2 = 5.83 × 102 cm2

1.26

4 3 4 3 π r = π ( 6.81 cm ) = 1.32 × 10 3 cm 3 3 3

⎛ 1 in ⎞ ⎛ 1 day ⎞ ⎛ 1 h ⎞ ⎛ 2.54 cm ⎞ ⎛ 10 9 nm ⎞ rate = ⎜ ⎟⎜ ⎟⎜ ⎟ = 9.2 nm / s ⎟⎜ ⎟⎜ 2 ⎝ 32 day ⎠ ⎝ 24 h ⎠ ⎝ 3 600 s ⎠ ⎝ 1.00 in ⎠ ⎝ 10 cm ⎠

Topic 1

This means that the proteins are assembled at a rate of many layers of atoms each second!

1.27

⎛ m ⎞ ⎛ 3 600 s ⎞ ⎛ 1 km ⎞ ⎛ 1 mi ⎞ c = ⎜ 3.00 × 10 8 = 6.71× 10 8 mi/h 3 ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟ s ⎠ ⎝ 1 h ⎠ ⎝ 10 m ⎠ ⎝ 1.609 km ⎠ ⎝

⎛ 2.832 × 10 −2 m 3 ⎞ Volume of house = (50.0 ft)(26 ft)(8.0 ft) ⎜ ⎟⎠ 1 ft 3 ⎝ ⎛ 100 cm ⎞ = 2.9 × 10 m = (2.9 × 10 m ) ⎜ ⎝ 1 cm ⎟⎠

1.28

= 2.9 × 10 8 cm 3

1.29

2 2 ⎛ 1 m ⎞ ⎡⎛ 43 560 ft ⎞ ⎛ 1 m ⎞ ⎤ Volume = 25.0 acre ft ⎜ ⎢⎜ ⎥ ⎟ ⎝ 3.281 ft ⎟⎠ ⎢⎣⎝ 1 acre ⎠ ⎜⎝ 3.281 ft ⎟⎠ ⎥⎦

(

)

= 3.08 × 10 4 m 3

1.30

1 Volume of pyramid = (area of base)(height) 3 1 = [(13.0 acres)(43560 ft 2 / acre)](481 ft) = 9.08 × 10 7 ft 3 3 ⎛ 2.832 × 10 −2 m 3 ⎞ = (9.08 × 10 3 ft 3 ) ⎜ = 2.57 × 10 6 m 3 3 ⎟ 1 ft ⎝ ⎠ 1.31

Volume of cube = L3 = 1 quart (Where L = length of one side of the cube.) ⎛ 1 gallon ⎞ ⎛ 3.786 liter ⎞ ⎛ 1000 cm 3 ⎞ = 947 cm 3 Thus, L = 1 quart ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4 quarts ⎠ ⎝ 1 gallon ⎠ ⎝ 1 liter ⎠ 3

(

)

Topic 1

and L = 3 947 cm 3 = 9.82 cm 1.32

We estimate that the length of a step for an average person is about 18 inches, or roughly 0.5 m. Then, an estimate for the number of steps required to travel a distance equal to the circumference of the Earth would be

or 1.33

Circumference 2π RE 2π (6.38 × 10 6 m) = = ≈ 8 × 10 7 steps Step Length Step Length 0.5 m / step

N ~ 103 steps

We assume an average respiration rate of about 10 breaths/minute and a typical life span of 70 years. Then, an estimate of the number of breaths an average person would take in a lifetime is ⎛ 3.156 × 10 7 s ⎞ ⎛ 1 min ⎞ ⎛ breaths ⎞ n = ⎜ 10 70 yr ⎜ = 4 × 10 8 breaths ⎟ ⎟ ⎜ ⎟ ⎝ min ⎠ 1 yr ⎝ ⎠ ⎝ 60 s ⎠

(

or 1.34

)

n ~ 108 breaths

We assume that the average person catches a cold twice a year and is sick an average of 7 days (or 1 week) each time. Thus, on average, each person is sick for 2 weeks out of each year (52 weeks). The probability that a particular person will be sick at any given time equals the percentage of time that person is sick, or

Topic 1

probability of sickness =

2 weeks 1 = 52 weeks 26

The population of the Earth is approximately 7 billion. The number of people expected to have a cold on any given day is then

⎛ 1⎞ Number sick = ( population)( probability of sickness) = (7 × 10 9 ) ⎜ ⎟ = 3 × 10 8 or 10 8 ⎝ 26 ⎠

1.35

Earth’s current population is about 7.3 billion people. If six people can stand within a 1-m2 area, the total area required to fit all 7.3 billion people is: ⎛ 1 m2 ⎞ Apeople = 7.3 × 10 people ⎜ = 1.2 × 109 m 2 ⎟ ⎝ 6 people ⎠ 9

The required percentage of the habitable part of Earth’s surface is: ⎛ 1.2 × 109 m 2 ⎞ Occupied percent = ⎜ 100% = 2 × 10−3% 12 2⎟ ⎝ 60 × 10 m ⎠

1.36

(a) Vcell =

4 3 4 π r = π (1.0 × 10 −6 m)3 = 4.2 × 10 −18 m 3 3 3

(b) Consider your body to be a cylinder having a radius of about 6 inches (or 0.15 m) and a height of about 1.5 meters. Then, its volume is Vbody = Ah = (π r2)h = π (0.15 m)2 (1.5 m) = 0.11 m3 or ~10−1 m3 (c) The estimate of the number of cells in the body is then

Topic 1

1.37

Vbody 0.11 m 3 = = 2.6 × 1016 or ~ 1016 Vcell 4.2 × 10 −18 m 3

A reasonable guess for the diameter of a tire might be 3 ft, with a circumference (C = 2π r = πD = distance traveled per revolution) of about 9 ft. Thus, the total number of revolutions the tire might make is

1.38

total distance traveled (50 000 mi)(5 280 ft/mi) = = 3 × 10 7 rev, or ~ 10 7 rev distance per revolution 9 ft/rev

An average body mass is about 70 kg and, taking the center of the given range, approximately 2% or 1.4 kg of this mass is made of microorganisms. If there are 100 trillion microorganisms in a human body, the average mass of a microorganism is about

mmicroorganism ≈

1.39

1.4 kg ≈ 1 × 10−14 kg 100 × 10 microorganisms 12

The x coordinate is found as x = r cos θ = (2.5 m)cos 35° = 2.0 m and the y coordinate y = r sin θ = (2.5 m)sin 35° = 1.4 m

1.40

The x distance out to the fly is 2.0 m and the y distance up to the fly is 1.0 m. Thus, we can use the Pythagorean theorem to find the distance from the origin to the fly as

d = x 2 + y 2 = (2.0 m)2 + (1.0 m)2 = 2.2 m 1.41

The distance from the origin to the fly is r in polar coordinates, and this

Topic 1

was found to be 2.2 m in Problem 40. The angle θ is the angle between r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as

tan θ =

y 1.0 m = = 0.50 x 2.0 m

and θ = tan−1 (0.50) = 27°

The polar coordinates are r = 2.2 m and θ = 27° 1.42

The x distance between the two points is |Δx| = |x2 − x1| = |−3.0 cm − 5.0 cm|= 8.0 cm and the y distance between them is |Δy| = |y2 − y1| = | −3.0 cm − 4.0 cm| = 1.0 cm. The distance between them is found from the Pythagorean theorem:

1.43

Δx + Δy = (8.0 cm)2 + (1.0 cm)2 = 65 cm 2 = 8.1 cm 2

Refer to the Figure given in the solution to Problem 1.44 below. The Cartesian coordinates for the two given points are: x1 = r1 cos θ1 = (2.00 m) cos 50.0° = 1.29 m y1 = r1 sin θ1 = (2.00 m) sin 50.0° = 1.53 m x2 = r2 cos θ2 = (5.00 m) cos (−50.0°) = 3.21 m y2 = r2 sin θ2 = (5.00 m) sin (−50.0°) = 3.83 m The distance between the two points is then:

Δs = (Δx)2 + (Δy)2 = (1.29 m − 3.21 m)2 + (1.53 m + 3.83 m)2 = 5.69 m

Topic 1

1.44

Consider the Figure shown below. The Cartesian coordinates for the two points are: x1 =r1 cos θ1

x2 = r2 cos θ2

y1 = r1 sin θ1

y2 = r2 sin θ2

The distance between the two points is the length of the hypotenuse of the shaded triangle and is given by Δs = (Δx)2 + (Δy)2 = (x1 − x2 )2 + (y1 − y2 )2

Δs = (r12 cos 2 θ1 + r22 cos 2 θ 2 − 2r1r2 cosθ1 cosθ 2 ) + (r12 sin 2 θ1 + r22 sin 2 θ 2 − 2r1r2 sin θ1 sin θ 2 ) = r12 (cos 2 θ1 + sin 2 θ1 ) + r22 (cos 2 θ 2 + sin 2 θ 2 ) − 2r1r2 (cosθ1 cosθ 2 + sin θ1 sin θ 2 )

Applying the identities cos2 θ + sin2 θ = 1 and cos θ1 cos θ2 + sin θ1 sin θ2 = cos(θ1 − θ2), this reduces to Δs = r12 + r22 − 2r1r2 (cosθ1 cosθ 2 + sin θ1 sin θ 2 ) =

1.45

(a)

r12 + r22 − 2r1r2 cos(θ1 − θ 2 )

With a = 6.00 m and b being two sides of this right triangle having hypotenuse c = 9.00 m, the Pythagorean theorem gives the

Topic 1

unknown side as b = c 2 − a 2 = (9.00 m)2 − (6.00 m)2 = 6.71 m .

1.46

(b)

tan θ =

a 6.00 m = = 0.894 b 6.71 m

(c)

sin φ =

b 6.71 m = = 0.746 c 9.00 m

From the diagram, cos(75.0°) = d/L Thus, d = L cos(75.0°) = (9.00 m) cos(75.0°) = 2.33 m

1.47

The circumference of the fountain is C = 2πr, so the radius is

C 15.0 m = = 2.39 m 2π 2π

Topic 1

Thus, tan(55.0°) =

h h = which gives r 2.39 m

h = (2.39 m) tan (55.0°) = 3.41 m.

1.48

1.49

1.50

(a) sin θ =

opposite side so, opposite side = (3.00 m) sin (30.0°) = 1.50 m hypotenuse

(b) cosθ =

adjacent side so, adjacent side = (3.00 m) cos (30.0°) = 2.60 m hypotenuse

(a) The side opposite θ = 3.00

(c)

cosθ =

4.00 = 0.800 5.00

(e)

tan φ =

4.00 = 1.33 3.00

(b)

(d)

The side adjacent to φ = 3.00

sin φ =

4.00 = 0.800 5.00

Using the diagram below, the Pythagorean theorem yields

c = (5.00 m)2 + (7.00 m)2 = 8.60 m

Topic 1

1.51

From the diagram given in Problem 1.50 above, it is seen that

tan θ =

1.52

5.00 = 0.714 7.00

and

θ = tan−1 (0.714) = 35.5°

Use the diagram below to equate two expressions for the mountain’s height (units and significant digits have been suppressed for clarity):

y = (x − 1)tan(14.0) = (x)tan(12.0°) Solve for the unknown distance x to find

( x − 1) tan (14.0°) = x tan (12.0°) x ( tan14° − tan12° ) = tan14° x=

tan14° = 6.78 km tan14° − tan12°

Substitute the known value of x into y = (x)tan(12.0°) to find the mountain’s height: y = x tan (12° ) = ( 6.78 km ) tan (12° ) = 1.44 km = 1.44 × 103 m

1.53

Using the sketch below:

Topic 1

w = tan 35.0°, or w = (100 m)tan 35.0° = 70.0 m 100 m 1.54

(a)

 Using graphical methods, place the tail of vector !B at the head of    vector!A . The new vector !A + B has a magnitude of 6.1 units at

113° from the positive x-axis.

(b)

    The vector difference !A − B = A + (−B) is found by placing the   negative of vector !B (a vector of the same magnitude as !B , but  opposite direction) at the head of vector!A . The resultant vector   !A − B has magnitude 15 units at 23° from the positive x-axis.

1.55

   We are given that !R = A + B . When two vectors are added graphically,

the second vector is positioned with its tail at the tip of the first vector. The resultant then runs from the tail of the first vector to the tip of the  second vector. In this case, vector !A will be positioned with its tail at

the origin and its tip at the point (0, 29). The resultant is then drawn, starting at the origin (tail of first vector) and going 14 units in the  negative y-direction to the point (0, −14). The second vector, !B , must   then start from the tip of !A at point (0, 29) and end on the tip of !R at

point (0, −14) as shown in the sketch below. From this, it is seen that © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 1

1.56

(a)

 B is 43 units in!!the!!negative!!y.direction

The distance d from A to C is d = x 2 + y 2 where x = 200 km + (300 ! km)cos 30.0° = 460 km and y = 0 + (300 km)sin 30.0° = 150 km. ∴ d = (460 km)2 + (150 km)2 = 484 km !

(b)

⎛ 150 km ⎞ y φ = tan −1 = tan −1 ⎜ = 18.1° N!of!W x ⎝ 460 km ⎟⎠ !

(c)

Because of the curvature of the Earth, the plane doesn't travel along straight lines. Thus, the answer computed above is only approximately correct.

1.57

(a)

 In your vector diagram, place the tail of vector !B at the tip of    vector!A . The vector sum, !A + B , is then found as shown in the

Topic 1

vector diagram and should be

(b)

  A + B = 5.0 units!!at!!− 53°

     To find the vector difference!A − B = A + (−B) , form the vector !−B   (same magnitude as !B , opposite direction) and add it to vector !A

as shown in the diagram. You should find that

1.58

  A − B = 5.0 units!!at!!+ 53°

    Find the resultant !R = F1 + F2 graphically by placing the tail of !F2 at the  head of !F1 in a scale drawing as shown below. The resultant is the   vector drawn from the tail of !F1 to the head of !F2 to close up the triangle. Measuring the length and orientation of this vector shows

 R that = 9.5 units at 57° above the + x2axis . !

Topic 1

1.59

Your sketch should be drawn to scale, and be similar to that pictured  below. The length of !R and the angle θ can be measured to find,

with use of your scale factor, the magnitude and direction of the resultant displacement. The result should be approximately 421 ft 3° below the horizontal.

1.60

(a) The x-component is x = (24.0 m)cos(56.0°) = 13.4 m. (b) The y-component is y = (24.0 m)sin(56.0°) = 19.9 m.

1.61

(a) The vector’s magnitude is ! A = A = Ax2 + Ay2 =

( −5.00 m ) + ( 9.00 m ) 2

= 10.3 m

(b) The vector is in the second quadrant so its direction is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 1

⎛ Ay ⎞ ⎛ 9.00 m ⎞ θ = tan −1 ⎜ ⎟ + 180° = tan −1 ⎜ + 180° ⎝ −5.00 m ⎟⎠ ⎝ Ax ⎠ = 119°

1.62

The person undergoes a displacement ! A = 3.10 km at 25.0° north of east. Choose a coordinate system with

origin at the starting point, positive x-direction oriented eastward, and positive y-direction oriented northward. Then the components of her displacement are Ax = A cos θ = (3.10 km)cos 25.0° = 2.81 km eastward and Ay = A sin θ = (3.10 km)sin 25.0° = 1.31 km northward

1.63

 The x- and y-components of vector !A are its projections on lines

parallel to the x- and y-axes, respectively, as shown in the sketch. The magnitude of these components can be computed using the sine and cosine functions as shown below:

  Ax = A cos 325° = + A cos 35° = (35.0)cos 35° = 28.7 units ! and

  Ay = A sin 325° = − A sin 35° = −(35.0)sin 35° = −20.1 units !

Topic 1

1.64

(a)

 The skater’s displacement vector, !d , extends in a straight line from her starting point A to the end point B. When she has coasted half way around a circular path as shown in the sketch below, the displacement vector coincides with the diameter of the

 circle and has magnitude d = 2r = 2(5.00 m) = 10.0 m . !

(b)

The actual distance skated, s, is one half the circumference of the

1 circular path of radius r. Thus s = (2π r) = π (5.00 m) = 15.7 m . ! 2 (c)

When the skater skates all the way around the circular path, her end point, B, coincides with the start point, A. Thus, the

 displacement vector has zero length, or d = 0 . ! 1.65

(a)

Her net x (east-west) displacement is −3.00 + 0 + 6.00 = +3.00 blocks, while her net y (north-south) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultant displacement is

Topic 1

R = (∑ x)2 + (∑ y)2 = (3.00)2 + (4.00)2 = 5.00 blocks ! and the angle the resultant makes with the x-axis (eastward direction) is

⎛ ∑y⎞ ⎛ 4.00 ⎞ θ = tan −1 ⎜ = tan −1 ⎜ = tan −1 (1.33) = 53.1° ⎟ ⎟ ⎝ 3.00 ⎠ ⎝ ∑x⎠ ! The resultant displacement is then 5.00 blocks at 53.1° N of E. (b) 1.66

The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks.

  Let !A be the vector corresponding to the 10.0 yd run, !B to the 15.0 yd

 run, and !C to the 50.0 yd pass. Also, we choose a coordinate system with the + y-direction downfield, and the + x-direction toward the sideline to which the player runs. The components of the vectors are then Ax = 0

Ay = −10.0 yds

Bx = 15.0 yds

By = 0

Cx = 0

Cy = +50.0 yds

From these, Rx = ∑x = 15.0 yds, and Ry = ∑y = 40.0 yds, and 1.67

R = Rx2 + Ry2 = (15.0 yds)2 + (40.0 yds)2 = 42.7 yards !

Ax = −25.0

Ay = 40.0

Topic 1

A = Ax2 + Ay2 = (−25.0)2 + (40.0)2 = 47.2 units ! From the triangle, we find that ⎛ Ay ⎞ ⎛ 40.0 ⎞ φ = tan −1 ⎜ = tan −1 ⎜ = 58.0° , so θ = 180° − φ = 122° ⎟ ⎝ 25.0 ⎟⎠ ⎝ Ax ⎠ !  Thus, A = 47.2 units at!!122°!!counterclockwise!!from!!the!!+x8axis !

1.68

Choose the positive x-direction to be eastward and positive y as northward. Then, the components of the resultant displacement from Dallas to Chicago are Rx = ∑x = (730 mi)cos 5.00° − (560 mi)sin 21.0° = 527 mi and Ry = ∑y = (730 mi)sin 5.00° + (560 mi)cos 21.0° = 586 mi

R = Rx2 + Ry2 = (527 mi)2 + (586 mi)2 = 788 mi ! ⎛ ∑y⎞ −1 θ = tan −1 ⎜ ⎟⎠ = tan (1.11) = 48.0° ⎝ ∑ x ! Thus, the displacement from Dallas to Chicago is

Topic 1

1.69

After 3.00 h moving at 41.0 km/h, the hurricane is 123 km at 60.0° N of W from the island. In the next 1.50 h, it travels 37.5 km due north. The components of these two displacements are:

Displacement

x-component (eastward)

y-component (northward)

123 km

−61.5 km

+107 km

37.5 km

+37.5 km

Resultant

−61.5 km

144 km

Therefore, the eye of the hurricane is now R = (−61.5 km)2 + (144 km)2 = 157 km!!from!!the!!island !

1.70

(a)

F1 = 120 N F1x = (120 N)cos 60.0° = 60.0 N F1y = (120 N)sin 60.0° = 104 N

Topic 1

F2 = 80.0 N F2x = −(80 N)cos 75.0° = −20.7 N F2y = (80.0 N)sin 75.0° = 77.3 N

F = (∑ Fx )2 + (∑ Fy )2 = (39.3 N)2 + (181 N)2 = 185 N !R −1 ⎛ 181 N ⎞ = tan −1 (4.61) = 77.8° and θ = tan ⎜ ⎝ 39.3 N ⎟⎠ !

 The resultant force is FR = 185 N at 77.8°!!from!!the!!x0axis ! (b) To have zero net force on the mule, the resultant above must be cancelled by a force equal in magnitude and oppositely directed. Thus, the required force is 185 N at 258° from the x-axis 1.71

   The components of the displacements !a , !b , and !c are

ax = a⋅cos 30.0° = +152 km bx = b⋅cos 110° = −51.3 km cx = c⋅cos 180° = −190 km and ay = a⋅sin 30.0° = +87.5 km by = b⋅sin 110° = +141 km cy = c⋅sin 180° = 0 Thus, Rx = ax + bx + cx = −89.3 km, and Ry = ay + by + cy = 229 km

Topic 1

R = Rx2 + Ry2 = 246 km , and !

(

)

θ = tan −1 |Rx |/Ry = tan −1 (0.390) = 21.3° ! City C is 246 km at 21.3° W of N from the starting point.

1.72

(a) 1

mi ⎛ mi ⎞ ⎛ 1.609 km ⎞ km = ⎜1 ⎟ ⎜ = 1.609 h ⎝ h ⎠ ⎝ 1 mi ⎟⎠ h

(b) vmax = 55

(c)

1.73

mi ⎛ mi ⎞ ⎛ 1.609 km/h ⎞ km = ⎜ 55 ⎟ ⎜ = 88 ⎟ ⎝ ⎠ h h ⎝ 1 mi / h ⎠ h

Δvmax = 65

mi mi ⎛ mi ⎞ ⎛ 1.609 km/h ⎞ km − 55 = ⎜ 10 ⎟ ⎜ = 16 ⎟ ⎝ ⎠ h h h ⎝ 1 mi/h ⎠ h

The term s has dimensions of L, a has dimensions of LT−2, and t has dimensions of T. Therefore, the equation, s = kam tn with k being dimensionless, has dimensions of L = (LT−2)m (T)n

L1T0 = Lm Tn−2m

The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m = 1 Likewise, equating powers of T, we see that n − 2m = 0, or n = 2m = 2

Topic 1

Dimensional analysis cannot determine the value of k, a dimensionless constant. 1.74

(a) The rate of filling in gallons per second is

rate =

30.0 gal ⎛ 1 min ⎞ = 7.14 × 10 −2 gal/s 7.00 min ⎜⎝ 60 s ⎟⎠

1L ⎞ 3 2 3 6 3 ⎛ = 10 3 L . Thus, (b) Note that 1 m = (10 cm) = (10 cm ) ⎜ 3 3⎟ 10 cm ⎝ ⎠

rate = 7.14 × 10 −2

(c)

1.75

gal ⎛ 3.786 L ⎞ ⎛ 1 m 3 ⎞ = 2.70 × 10 −4 m 3 /s s ⎜⎝ 1 gal ⎟⎠ ⎜⎝ 10 3 L ⎟⎠

⎛ 1h ⎞ Vfilled 1.00 m 3 = = 3.70 × 10 3 s ⎜ = 1.03 h −4 3 rate 2.70 × 10 m /s ⎝ 3600 s ⎟⎠

The volume of paint used is given by V = Ah, where A is the area covered and h is the thickness of the layer. Thus,

V 3.79 × 10 −3 m 3 h= = = 1.52 × 10 −4 m = 152 × 10 −6 m = 152 µm 2 A 25.0 m 1.76

(a) For a sphere, A = 4πR2. In this case, the radius of the second sphere is twice that of the first, or R2 = 2R1.

Hence,

A2 4π R22 R22 (2 R1 )2 = = = = 4 A1 4π R12 R12 R12

(b) For a sphere, the volume is V =

4 π R3 3

Topic 1

Thus,

1.77

V2 (4 / 3)π R23 R23 (2 R1 )3 = = = = 8 V1 (4 / 3)π R13 R13 R13

The estimate of the total distance cars are driven each year is d = (cars in use)(distance traveled per car) = (100 × 106 cars)(104 mi/car) = 1 × 1012 mi At a rate of 20 mi/gal, the fuel used per year would be

d 1× 1012 mi V1 = = = 5 × 1010 gal rate1 20 mi/gal If the rate increased to 25 mi/gal, the annual fuel consumption would be

V2 =

d 1× 1012 mi = = 4 × 1010 gal rate2 25 mi/gal

and the fuel savings each year would be savings = V1 − V2 = 5 × 1010 gal − 4 × 1010 gal = 1 × 1010 gal 1.79

(a)

The Sun’s radius is about 6.96 × 105 km and Earth’s radius is about 6.38 × 103 km. The number of Earths that could fit inside the Sun, NES, is approximately equal to the ratio of the Sun’s volume to Earth’s volume:

N ES =

4 VSun π R3 = 43 3Sun ≈ 1 × 106 VEarth 3 π REarth

Topic 1

The Sun’s radius is about 102 times Earth’s radius, so its volume (which scales as radius cubed) is about 106 times Earth’s volume. Approximately a million Earths could fit inside the Sun. (b)

Similarly, the Moon’s radius is about 1.74 × 103 km so that NME, the number of Moons that could fit inside the Earth, is

N ME =

3 VEarth 43 π REarth = 4 3 ≈ 50 VMoon 3 π RMoon

Approximately 50 Moons could fit inside the Earth. 1.80

Assume a sneeze lasts about 3 seconds. One day equals 8.64 × 104 s which can be divided into 8.64/3 × 104 = 2.88 × 104 3-second intervals. If each of Earth’s 7.3 billion people sneezes 3 times per day, then 3(7.3 billion) = 21.9 billion sneezes occur each day. The number of sneezes occurring during any particular 3-second interval is then (rounded to 1 significant figure):

21.9 × 109 sneezes/day ≈ 1 × 106 sneezes/(3-sec interval) 4 2.88 × 10 (3-sec intervals)/day During any given 3-second interval, approximately one million people from around the world will sneeze. 1.81

The volume of the Milky Way galaxy is roughly

⎛ πd2 ⎞ π VG = At = ⎜ t = (10 21 m)2 (1019 m) or VG ~ 10 61 m 3 ⎟ 4 ⎝ 4 ⎠

Topic 1

If, within the Milky Way galaxy, there is typically one neutron star in a spherical volume of radius r = 3 × 1018 m, then the galactic volume per neutron star is

4 4 V1 = π r 3 = π (3 × 1018 m)3 = 1× 10 56 m 3 or V1 ~ 10 56 m 3 3 3 The order of magnitude of the number of neutron stars in the Milky Way is then

VG 10 61 m 3 ~ 56 3 or n  10 5 neutron stars V1 10 m

Topic 2

Topic 2 Motion in One Dimension

QUICK QUIZZES 2.1

(a) 200 yd

(b)

(c)

2.2

(a) False. The car may be slowing down, so that the direction of its acceleration is opposite the direction of its velocity. (b) True. If the velocity is in the direction chosen as negative, a positive acceleration causes a decrease in speed. (c) True. For an accelerating particle to stop at all, the velocity and acceleration must have opposite signs, so that the speed is decreasing. If this is the case, the particle will eventually come to rest. If the acceleration remains constant, however, the particle must begin to move again, opposite to the direction of its original velocity. If the particle comes to rest and then stays at rest, the acceleration has become zero at the moment the motion stops. This is the case for a braking car—the acceleration is negative and goes to zero as the car comes to rest.

Topic 2

2.3

The velocity-vs-time graph (a) has a constant slope, indicating a constant acceleration, which is represented by the acceleration-vs.-time graph (e). Graph (b) represents an object whose speed always increases, and does so at an ever-increasing rate. Thus, the acceleration must be increasing, and the acceleration-vs-time graph that best indicates this behaviour is (d). Graph (c) depicts an object which first has a velocity that increases at a constant rate, which means that the object’s acceleration is constant. The motion then changes to one at constant speed, indicating that the acceleration of the object becomes zero. Thus, the best match to this situation is graph (f).

2.4

Choice (b). According to graph b, there are some instants in time when the object is simultaneously at two different x-coordinates. This is physically impossible.

2.5

(a) The blue graph of Figure 2.14b best shows the puck’s position as a function of time. As seen in Figure 2.14a, the distance the puck has traveled grows at an increasing rate for approximately three time intervals, grows at a steady rate for about four time intervals, and then grows at a diminishing rate for the last two intervals. (b) The red graph of Figure 2.14c best illustrates the speed (distance

Topic 2

traveled per time interval) of the puck as a function of time. It shows the puck gaining speed for approximately three time intervals, moving at constant speed for about four time intervals, then slowing to rest during the last two intervals. (c) The green graph of Figure 2.14d best shows the puck’s acceleration as a function of time. The puck gains velocity (positive acceleration) for approximately three time intervals, moves at constant velocity (zero acceleration) for about four time intervals, and then loses velocity (negative acceleration) for roughly the last two time intervals. 2.6

Choice (e). The acceleration of the ball remains constant while it is in the air. The magnitude of its acceleration is the free-fall acceleration, g = 9.80 m/s2.

2.7

Choice (c). As it travels upward, its speed decreases by 9.80 m/s during each second of its motion. When it reaches the peak of its motion, its speed becomes zero. As the ball moves downward, its speed increases by 9.80 m/s each second.

2.8

Choices (a) and (f). The first jumper will always be moving with a higher velocity than the second. Thus, in a given time interval, the first jumper covers more distance than the second, and the separation distance

Topic 2

between them increases. At any given instant of time, the velocities of the jumpers are definitely different, because one had a head start. In a time interval after this instant, however, each jumper increases his or her velocity by the same amount, because they have the same acceleration. Thus, the difference in velocities stays the same.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2.2

Yes. The particle may stop at some instant, but still have an acceleration, as when a ball thrown straight up reaches its maximum height.

2.4

(a) No. They can be used only when the acceleration is constant. (b) Yes. Zero is a constant.

2.6

(a) In Figure (c), the images are farther apart for each successive time interval. The object is moving toward the right and speeding up. This means that the acceleration is positive in Figure (c). (b) In Figure (a), the first four images show an increasing distance traveled each time interval and therefore a positive acceleration. However, after the fourth image, the spacing is decreasing, showing that the object is now slowing down (or has negative acceleration).

Topic 2

(c) In Figure (b), the images are equally spaced, showing that the object moved the same distance in each time interval. Hence, the velocity is constant in Figure (b). 2.8

(a) At the maximum height, the ball is momentarily at rest (i.e., has zero velocity). The acceleration remains constant, with magnitude equal to the free-fall acceleration g and directed downward. Thus, even though the velocity is momentarily zero, it continues to change, and the ball will begin to gain speed in the downward direction. (b) The acceleration of the ball remains constant in magnitude and direction throughout the ball’s free flight, from the instant it leaves the hand until the instant just before it strikes the ground. The acceleration is directed downward and has a magnitude equal to the freefall acceleration g.

2.10

Once the ball has left the thrower's hand, it is a freely falling body with a constant, nonzero, acceleration of a = −g. Since the acceleration of the ball is not zero at any point on its trajectory, choices (a) through (d) are all false and the correct response is (e).

2.12

The initial velocity of the car is v0 = 0 and the velocity at the time t is v. The constant acceleration is therefore given by

Topic 2

Δv v − v0 v − 0 v = = = Δt t t t

and the average velocity of the car is

( v + v0 ) = ( v + 0 ) = v 2

The distance traveled in time t is Δx = vt = vt/2. In the special case where a = 0 (and hence v = v0 = 0), we see that statements (a), (b), (c), and (d) are all correct. However, in the general case (a ≠ 0, and hence v ≠ 0) only statements (b) and (c) are true. Statement (e) is not true in either case.

ANSWERS TO EVEN NUMBERED PROBLEMS 2.2

(a)

2 × 104 mi

(b)

Δx/2RE = 2.4

2.4

(a)

8.33 yards/s

(b)

2.78 yards/s

2.6

(a)

5.00 m/s

(b)

1.25 m/s

(d)

−3.33 m/s

(e)

(a)

+4.0 m/s

(b)

−0.50 m/s

(d)

(a)

2.3 min

(b)

64 mi

2.8

2.10

(c)

−2.50 m/s

(c)

−1.0 m/s

Topic 2

2.12

(a)

L/t1

(b)

−L/t2

(d)

2L/(t1 + t2)

2.14

(a)

2.16 2.18

(c)

1.3 × 102 s

(b)

13 m

(a)

37.1 m/s

(b)

1.30 × 10−5 m

(a)

Some data points that can be used to plot the graph are as given below:

x (m)

5.75

16.0

35.3

68.0

119

192

t (s)

1.00

2.00

3.00

4.00

5.00

6.00

(b)

41.0 m/s, 41.0 m/s, 41.0 m/s

(c)

17.0 m/s, much smaller than the instantaneous velocity at t = 4.00 s

2.20

(a)

2.00 m/s, 5.0 m/s (b)

2.22

0.391 s

2.24

(i)

(a)

(b)

1.6 m/s2

(c)

0.80 m/s2

(ii)

(a)

(b)

1.6 m/s2

(c)

2.26

263 m

The curves intersect at t = 16.9 s.

Topic 2

2.28

a = 2.74 × 105 m/s2 = (2.79 × 104)g

2.30

(a)

(c)

a = (v 2f − vi2 )/ ( 2 Δx)

(b)

13.5 m (c)

(b)

v 2f = vi2 + 2a(Δx)

(e)

8.00 s

(a)

13.5 m

(d)

22.5 m

(a)

20.0 s

(b)

No, it cannot land safely on the 0.800 km runway.

2.36

(a)

5.51 km

(b)

20.8 m/s, 41.5 m/s, 20.8 m/s, 38.7 m/s

2.38

(a)

107 m

(b)

1.49 m/s2

2.40

(a)

v = a1t1

(b)

Δx =

(c)

Δxtotal =

1 2 1 a1t1 + a1t1t 2 + a2t 22 2 2

2.32

2.34

(d)

1.25 m/s2

13.5 m

1 2 a1t1 2

Topic 2

2.42

8.9 months

2.44

29.1 s

2.46

1.79 s

2.48

(a)

Yes.

(c)

 Δv downward = 2.39 m/s

(d)

! No, Δv upward = 3.71 m/s. The two rocks have the same acceleration,

(b)

vtop = 3.69 m/s

but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval. 2.50

(a)

21.1 m/s

(b)

19.6 m

2.52

(a)

v = | −v0 − gt| = |v0 + gt|

(c)

v = v0 − gt , d =

2.54

(a)

29.4 m/s

(b)

44.1 m

2.56

(a)

−202 m/s2

(b)

198 m

2.58

(a)

4.53 s

(b)

14.1 m/s

2.60

8.4 m/s

2.62

See Solutions Section for Motion Diagrams.

(c)

1.81 m/s, 19.6 m

(b)

1 2 gt 2

Topic 2

2.64

(a)

v = v02 + 2gh

(b)

Δt = 2v0/g

2.66

(a)

2.45 × 10−2 m

(b)

4.67 × 10−2 s

2.68

(a)

3.00 s

(b)

v0,2 = −15.2 m/s

(c)

v1 = −31.4 m/s, v2 = −34.8 m/s

(a)

2.2 s

2.70

(b)

−21 m/s

PROBLEM SOLUTIONS 2.1

We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform speed. The elapsed time is then

Δt =

2.2

Δx 2m = = 2 × 10 −2 s = 0.02 s v 100 m/s

(a) At constant speed, c = 3 × 108 m/s, the distance light travels in 0.1 s is Δx = c(Δt) = (3 × 10 8 m/s)(0.1 s) ⎛ 1 mi ⎞ ⎛ 1 km ⎞ = (3 × 10 7 m ) ⎜ = 2 × 10 4 mi ⎝ 1.609 km ⎟⎠ ⎜⎝ 10 3 m ⎟⎠

(b) Comparing the result of part (a) to the diameter of the Earth, DE, we find

Δx Δx 3.0 × 10 7 m = = ≈ 2.4 DE 2RE 2(6.38 × 10 6 m )

(with RE = Earth's radius)

Topic 2

2.3

Distances traveled between pairs of cities are Δx1 = v1(Δt1) = (80.0 km/h)(0.500 h) = 40.0 km Δx2 = v2(Δt2) = (100.0 km/h)(0.200 h) = 20.0 km Δx3 = v3(Δt3) = (40.0 km/h)(0.750 h) = 30.0 km Thus, the total distance traveled is Δx = (40.0 + 20.0 + 30.0) km = 90.0 km, and the elapsed time is Δt = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h.

(a)

Δx 90.0 km = = 52.9 km/h Δt 1.70 h

(b) Δx = 90.0 km (see above) 2.4

(a) The player runs 100 yards from his own goal line to the opposing team’s goal line. Then he runs an additional 50 yards back to the fiftyyard line, all in 18.0 s. Substitute values into the definition of average speed to find

Average speed =

path length 100 yards + 50 yards = elapsed time 18.0 s

= 8.33 yards/s (b) After returning to the fifty-yard line, the player’s displacement is Δx = xf − xi = 50.0 yards − 0 yards = 50.0 yards. Substitute values into the definition of average velocity to find © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

Δx 50.0 yards = Δt 18.0 s

= 2.78 yards/s 2.5

(a) Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time 2.0 h. Boat B requires 2.0 h to cross the lake at which time the race is over. Boat A wins, being 60 km ahead of B when the race ends. (b) Average velocity is the net displacement of the boat divided by the total elapsed time. The winning boat is back where it started, its displacement thus being zero, yielding an average velocity of zero.

2.6

The average velocity over any time interval is

Δx x f − xi = Δt t f − ti

Δx 10.0 m − 0 = = 5.00 m/s Δt 2.00 s − 0

(b) v =

Δx 5.00 m − 0 = = 1.25 m/s Δt 4.00 s − 0

(a)

Δx 5.00 m − 10.0 m = = −2.50 m/s Δt 4.00 s − 2.00 s

(d) v =

Δx −5.00 m − 5.00 m = = −3.33 m/s Δt 7.00 s − 4.00 s

(c)

Topic 2

(e)

2.7

Δx x2 − x1 0−0 = = = 0 Δt t 2 − t1 8.00s − 0

(a) ⎛ 1h ⎞ Displacement = Δx = (8.50 km/h)(35.0 min) ⎜ + 130 km = 180 km ⎝ 60.0 min ⎟⎠

(b) The total elapsed time is ⎛ 1h ⎞ Δt = (35.0 min + 15.0 min) ⎜ + 2.00 h = 2.83 h ⎝ 60.0 min ⎟⎠

so,

2.8

Δx 180 km = = 63.6 km Δt 2.84 h

The average velocity over any time interval is

(a)

Δx x f − xi = Δt t f − ti

Δx 4.0 m − 0 = = +4.0 m / s Δt 1.0 s − 0

Topic 2

(b) v =

Δx −2.0 m − 0 = = −0.50 m / s Δt 4.0 s − 0

Δx 0 − 4.0 m = = −1.0 m / s Δt 5.0 s − 1.0 s

(c)

(d) v =

2.9

Δx 0−0 = = 0 Δt 5.0 s − 0

The plane starts from rest (v0 = 0) and maintains a constant acceleration of a = +1.3 m/s2. Thus, we find the distance it will travel before reaching the required takeoff speed (v = 75 m/s), from v 2 = v02 + 2a(Δx) , as

Δx =

v 2 − v02 (75 m/s)2 = = 2.2 × 10 3 m = 2.2 km 2 2a 2(1.3 m/s )

Since this distance is less than the length of the runway, the plane takes off safely.

2.10

(a) The time for a car to make the trip is t =

Δx . Thus, the difference in v

the times for the two cars to complete the same 10 mile trip is

Topic 2

Δt = t1 − t 2 =

Δx Δx ⎛ 10 mi 10 mi ⎞ ⎛ 60 min ⎞ − =⎜ − = 2.3 min v1 v2 ⎝ 55 mi/h 70 mi/h ⎟⎠ ⎜⎝ 1 h ⎟⎠

(b) When the faster car has a 15.0 min lead, it is ahead by a distance equal to that traveled by the slower car in a time of 15.0 min. This distance is given by Δx1 = v1(Δt) = (55 mi/h)(15 min). The faster car pulls ahead of the slower car at a rate of vrelative = 70 mi/h − 55 mi/h = 15 mi/h Thus, the time required for it to get distance Δx1 ahead is

Δt =

Δx1 (55 mi/h)(15 min) = = 15 min vrelative 15.0 mi/h

Finally, the distance the faster car has traveled during this time is ⎛ 1h ⎞ Δx2 = v2 (Δt) = (70 mi/h)(55 min) ⎜ = 64 mi ⎝ 60 min ⎟⎠

2.11

(a) From v f + vi + 2a(Δx) , with vi = 0, vf = 72 km/h, and Δx = 45 m, the 2

acceleration of the cheetah is found to be

⎡⎛ km ⎞ ⎛ 10 3 m ⎞ ⎛ 1 h ⎞ ⎤ 72 ⎢ ⎥−0 h ⎟⎠ ⎜⎝ 1 km ⎟⎠ ⎜⎝ 3600 s ⎟⎠ ⎦ v 2f − vi2 ⎣⎜⎝ a= = = 4.4 m/s 2 2(Δx) 2(45 m) (b) The cheetah’s displacement 3.5 s after starting from rest is

Topic 2

1 1 Δx = vi t + at 2 = 0 + (4.4 m/s 2 )(3.5 s) = 27 m 2 2

2.12

v1 =

(Δx)1 +L = = +L /t1 (Δt)1 t1

(b) v2 =

(Δx)2 −L = = −L /t 2 (Δt)2 t2

(a)

(c)

vtotal =

(Δx)total (Δx)1 + (Δx)2 +L − L 0 = = = = 0 (Δt)total t1 + t 2 t1 + t 2 t1 + t 2

(d)

(ave speed)trip =

2.13

+L + −L totaldistance traveled (Δx)1 + (Δx)2 2L = = = (Δt)total t1 + t 2 t1 + t 2 t1 + t 2

(a) The total time for the trip is ttotal = t1 + 22.0 min = t1 + 0.367 h, where t1 is the time spent traveling at v1 = 89.5 km/h. Thus, the distance traveled is Δx = v1t1 = vt total , which gives (89.5 km/h)t1 = (77.8 km/h)(t1 + 0.367 h) = (77.8 km/h)t1 + 28.5 km or, (89.5 km/h − 77.8 km/h)t1 = 28.5 km From which, t1 = 2.44 h for a total time of ttotal = t1 + 0.367 h = 2.81 h (b) The distance traveled during the trip is Δx = v1t1 = vt total , giving Δx = vt total = (77.8 km/h)(2.81 h) = 219 km

Topic 2

2.14

(a) At the end of the race, the tortoise has been moving for time t and the hare for a time t − 2.0 min = t − 120 s. The speed of the tortoise is vi = 0.100 m/s, and the speed of the hare is vh = 20 vt = 2.0 m/s. The tortoise travels distance xt, which is 0.20 m larger than the distance xh traveled by the hare. Hence, xt = xh + 0.20 m which becomes or

vtt = vh(t − 120 s) + 0.20 m

(0.100 m/s)t = (2.0 m/s)(t − 120 s) + 0.20 m

This gives the time of the race as t = 1.30 × 102 s (b) xt = vtt = (0.100 m/s)(1.3 × 102 s) = 13 m 2.15

The maximum allowed time to complete the trip is

t total =

total distance 1600 m ⎛ 1 km/h ⎞ = = 23.0 s required averagespeed 250 km/h ⎜⎝ 0.278 m/s ⎟⎠

The time spent in the first half of the trip is

t1 =

half distance 800 m ⎛ 1 km/h ⎞ = = 12.5 s v1 230 km/h ⎜⎝ 0.278 m/s ⎟⎠

Thus, the maximum time that can be spent on the second half of the trip is

Topic 2

t2 = ttotal − t1 = 23.0 s − 12.5 s = 10.5 s and the required average speed on the second half is

v2 =

2.16

⎛ 1 km/h ⎞ half distance 800 m = = 76.2 m/s ⎜ = 274 km/h t2 10.5 s ⎝ 0.278 m/s ⎟⎠

(a) From the first kinematic equation with v0 = 0, t = 7.00 × 10-7 s, and a = 5.30 × 107 m/s2, the maximum speed is

v = v0 + at = (5.30 × 107 m/s2 )(7.00 × 10−7 s) v = 37.1 m/s (b) Use Δx = v0t + 12 at with v0 = 0 to find the distance travelled during 2

the acceleration:

(

)(

Δx = 12 at 2 = 12 5.30 × 107 m/s2 7.00 × 10−7 s

)

Δx = 1.30 × 10−5 m 2.17

The instantaneous velocity at any time is the slope of the x vs. t graph at that time. We compute this slope by using two points on a straight segment of the curve, one point on each side of the point of interest.

(a)

vt=1.00 s =

10.0 m − 0 = 5.00 m/s 2.00 s − 0

Topic 2

(b) vt=3.00 s =

(c)

vt=4.50 s =

(d) vt=7.50 s =

2.18

(5.00 − 10.0) m = −2.50 m/s (4.00 − 2.00) s (5.00 − 5.00) m = 0 (5.00 − 4.00) s 0 − (−5.00) m = 5.00 m/s (8.00 − 7.00) s

(a) A few typical values are t (s)

x (m)

1.00

5.75

2.00

16.0

3.00

35.3

4.00

68.0

5.00

119

6.00

192

Topic 2

(b) We will use a 0.400 s interval centered at t = 400 s. We find at t = 3.80 s, x = 60.2 m and at t = 4.20 s, x = 76.6 m. Therefore,

Δx 16.4 m = = 41.0 m/s Δt 0.400 s

Using a time interval of 0.200 s, we find the corresponding values to be: at t = 3.90 s, x = 64.0 m and at t = 4.10 s, x = 72.2 m. Thus,

Δx 8.20 m = = 41.0 m/s Δt 0.200 s

For a time interval of 0.100 s, the values are: at t = 3.95 s, x = 66.0 m, and at t = 4.05 s, x = 70.1 m. Therefore,

Δx 4.10 m = = 41.0 m/s Δt 0.100 s

Δx 68.0 m − 0 = = 17.0 m/s Δt 4.00 s − 0

Topic 2

This value is much less than the instantaneous velocity at t = 4.00 s. 2.19

Choose a coordinate axis with the origin at the flagpole and east as the 1 2 positive direction. Then, using x = x0 + v0t at with a = 0 for each runner, 2

the x-coordinate of each runner at time t is xA = −4.0 mi + (6.0 mi/h)t and xB = 3.0 mi + (−5.0 mi/h)t When the runners meet, xA = xB giving −4.0 mi + (60 mi/h)t = 3.0 mi + (−5.0 mi/h)t or

(6.0 mi/h + 5.0 mi/h)t = 3.0 mi + 4.0 mi. This gives the elapsed time

when they meet as t = (7.0 mi)/(11.0 mi/h) = 0.64 h. At this time, xA = xB = −0.18 mi. Thus, they meet 0.18 mi west of the flagpole. 2.20

From the figure below, observe that the motion of this particle can be broken into three distinct time intervals, during each of which the particle has a constant acceleration. These intervals and the associated accelerations are 0 ≤ t ≤ 10.0 s, a = a1 = +2.00 m/s2 10 ≤ t ≤ 15.0 s, a = a2 = 0

Topic 2

and

15.0 ≤ t ≤ 20.0 s, a = a3 = −3.00 m/s2

(a) Applying vf = vi + a(Δt) to each of the three time intervals gives for 0 ≤ t ≤ 10.0 s, v10 + v0 + a1(Δt1) = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s for 10.0 s ≤ t ≤ 15.0 s, v15 + v10 + a2(Δt2) = 20.0 m/s + 0 = 20.0 m/s for 15.0 s ≤ t ≤ 20.0 s, v20 + v15 + a3(Δt3) = 20.0 m/s + (−3.00 m/s2)(5.00 s) = 5.00 m/s 1 2 (b) Applying Δx = vi (Δt) + a(Δt) to each of the time intervals gives 2

for 0 ≤ t ≤ 10.0 s, 1 1 Δx1 = v0 Δt1 + a1 (Δt1 )2 = 0 + (2.00 m/s 2 )(10.0 s)2 = 1.00 × 10 2 m 2 2

for 10.0 s ≤ t ≤ 15.0 s,

Topic 2

1 Δx2 = v10 Δt 2 + a2 (Δt 2 )2 = (20.0 m/s)(5.00 s) + 0 = 1.00 × 10 2 m 2

for 15.0 s ≤ t ≤ 20.0 s,

1 Δx = v15 Δt 3 + a3 (Δt 3 )2 2 1 = (20.0 m/s)(5.00 s) + (−3.00 m/s 2 )(5.00 s)2 = 62.5 m 2 Thus, the total distance traveled in the first 20.0 s is Δxtotal = Δx1 + Δx2 + Δx3 = 100 m + 100 m + 62.5 m = 263 m 2.21

We choose the positive direction to point away from the wall. Then, the initial velocity of the ball is vi = −25.0 m/s and the final velocity is vf = +22.0 m/s. If this change in velocity occurs over a time interval of Δt = 3.50 ms (i.e., the interval during which the ball is in contact with the wall), the average acceleration is

2.22

Δv v f − vi +22.0 m/s − (−25.0 m/s) = = = 1.34 × 10 4 m/s 2 −3 Δt Δt 3.50 × 10 s

From a = Δv/Δt, the required time is

Δt =

2.23

From a =

⎞ ⎛ 0.447 m/s ⎞ Δv ⎛ 60.0 mi/h − 0 ⎞ ⎛ 1g =⎜ = = 0.391 s 2⎟ ⎟ ⎜ ⎠ ⎝ 9.80 m/s ⎠ ⎜⎝ 1 mi/h ⎟⎠ a ⎝ 7g

Δv Δv (60 − 55) mi/h ⎛ 0.447 m/s ⎞ = = 3.7 s , we have Δt = a 0.60 m/s 2 ⎜⎝ 1 mi/h ⎟⎠ Δt

Topic 2

2.24

(i)

(a)

From t = 0 to t = 5.0 s,

v f − vi −8.0 m/s − (−8.0 m/s) = = 0 t f − ti 5.0 s − 0

(b) From t = 5.0 s to t = 15 s,

8.0 m/s − (−8.0 m/s) = 1.6 m/s 2 15 s − 5.0 s

8.0 m/s − (−8.0 m/s) = 0.80 m/s 2 20 s − 0

(ii) At any instant, the instantaneous acceleration equals the slope of the line tangent to the v vs. t graph at that point in time. (a) At t = 2.0 s, the slope of the tangent line to the curve is 0. (b) At t = 10 s, the slope of the tangent line is 1.6 m/s2 . (c) At t = 18 s, the slope of the tangent line is 0.

2.25

(a) a =

Δv 175 mi/h − 0 = = 70.0 mi/h ⋅s Δt 2.5 s

Topic 2

⎛ mi ⎞ ⎛ 1609 m ⎞ ⎛ 1 h ⎞ a = ⎜ 70.0 = = 31.3 m/s 2 h ⋅s ⎟⎠ ⎜⎝ 1 mi ⎟⎠ ⎜⎝ 3600 s ⎟⎠ ⎝

Alternatively,

m⎞ 1g ⎛ a = ⎜ 31.3 2 ⎟ = = 3.19g ⎝ s ⎠ 9.80 m/s 2

1 2 (b) If the acceleration is constant, Δx = v0t + at 2

1⎛ m⎞ Δx = 0 + ⎜ 3.13 2 ⎟ (2.50 s)2 = 97.8 m 2⎝ s ⎠

2.26

⎛ 3.281 ft ⎞ Δx = (97.8 m) ⎜ = 321 ft ⎝ 1 m ⎟⎠

As in the algebraic solution to Example 2.5, we let t represent the time the trooper has been moving.

We graph xcar = 24.0 m + (24.0 m/s)t and xtrooper = (1.50 m/s2)t2

Topic 2

The curves intersect at t = 16.9 s

2.27

1 2 Apply Δx = v0 + at to the 2.00-second time interval during which the 2

object moves from xi = 3.00 cm to xf = −5.00 cm. With v0 = 12.0 cm/s, this yields an acceleration of 2 ⎡(x f − xi ) − v0t ⎤⎦ 2[(−5.00 − 3.00) cm − (12.0 cm/s)(2.00 s)] a= ⎣ = t2 (2.00 s)2

or 2.28

a = −16.0 cm/s2

From v 2 = v02 + 2a(Δx) , we have (10.97 × 103 m/s)2 = 0 + 2a(220 m) so that

v 2 − v02 (10.97 × 10 3 m/s)2 − 0 = = 2.74 × 10 5 m/s 2 2(Δx) 2(220 m)

⎛ ⎞ 1g = (2.74 × 10 5 m/s 2 ) ⎜ = 2.79 × 10 4 times g! 2⎟ ⎝ 9.80 m/s ⎠ 2.29

(a)

Δx = v(Δt) = [(v + v0 ) / 2]Δt becomes

⎛ 2.80 m/s + v0 ⎞ 40.0 m= ⎜ ⎟⎠ (8.50 s) ⎝ 2

which yields

(b) a =

v0 =

2 (40.0 m) − 2.80 m/s = 6.61 m/s 8.50 s

v − v0 2.80 m/s − 6.61 m/s = = −0.448 m/s 2 Δt 8.50 s

Topic 2

2.30

(a)

(b) The known quantities are initial velocity, final velocity, and displacement. The kinematics equation that relates these quantities to acceleration is v f = vi + 2a(Δx) 2

(c)

v 2f − vi2 a= 2(Δx)

(d) a =

v 2f − vi2 (30.0 m/s)2 − (20.0 m/s)2 = = 1.25 m/s 2 2(Δx) 2(2.00 × 10 2 m)

(e) Using a = Δv/Δt, we find that

Δt =

2.31

Δv v f − vi 30.0 m/s − 20.0 m/s = = = 8.00 s a a 1.25 m/s 2

(a) With v = 120 km/h, v 2 = v02 + 2a(Δx) yields v 2 − v02 ⎡⎣(120 km/h) − 0 ⎤⎦ ⎛ 0.278 m/s ⎞ a= = = 2.32 m/s 2 ⎜ ⎟ 2(Δx) 2(240 m) ⎝ 1 km/h ⎠ 2

(b) The required time is

Δt =

2.32

v − v0 (120 km/h − 0) ⎛ 0.278 m/s ⎞ = = 14.4 s a 2.32 m/s 2 ⎜⎝ 1 km/h ⎟⎠

(a) From v f = vi + 2a(Δx) , with vi = 6.00 m/s and vf = 12.0 m/s, we find 2

Topic 2

Δx =

v 2f − vi2 (120 m/s)2 − (6.00 m/s)2 = = 13.5 m 2a 2(4.00 m/s 2 )

(b) In this case, the object moves in the same direction for the entire time interval and the total distance traveled is simply the magnitude or absolute value of the displacement. That is, d = |Δx| = 13.5 m (c) Here, vi = −6.00 m/s and vf = 12.0 m/s, and we find

v 2f − vi2 Δx = = 13.5m 2a

[the same as in part (a)]

(d) In this case, the object initially slows down as it travels in the negative x-direction, stops momentarily, and then gains speed as it begins traveling in the positive x-direction. We find the total distance traveled by first finding the displacement during each phase of this motion. While coming to rest (vi = −6.00 m/s, vf = 0), v 2f − vi2 (0)2 − (−6.00 m/s)2 Δx1 = = = −4.50 m 2a 2(4.00 m/s2 )

After reversing direction (vi = 0 m/s, vf = 12.0 m/s), v 2f − vi2 (12.0 m/s)2 − (0)2 Δx2 = = = 18.0 m 2a 2(4.00 m/s 2 ) © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

Note that the net displacement is Δx = Δx1 + Δx2 = −4.50 m + 18.0 m = 13.5 m, as found in part (c) above. However, the total distance traveled in this case is d = |Δx1| + |Δx2| = |−4.50 m| + |18.0 m| = 22.5 m

2.33

v − v0 24.0 m/s 2 − 0 = = 8.14 m / s 2 (a) a = Δt 2.95 s (b) From a = Δv/Δt, the required time is

Δt =

v f − vi 20.0 m/s − 10.0 m/s = = 1.23 s a 8.15 m/s 2

(c) Yes. For uniform acceleration, the change in velocity Δv generated in time Δt is given by Δv = a(Δt). From this, it is seen that doubling the length of the time interval Δt will always double the change in velocity Δv. A more precise way of stating this is: “When acceleration is constant, velocity is a linear function of time.” 2.34

(a) The time required to stop the plane is

v − v0 0 − 100 m/s = = 20.0 s a −5.00 m/s 2

(b) The minimum distance needed to stop is

⎛ v + v0 ⎞ ⎛ 0 + 100 m/s ⎞ Δx = v = ⎜ t=⎜ ⎟ ⎟⎠ (20.0 s) = 1 000 m = 1.00 km ⎝ 2 ⎠ ⎝ 2 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

Thus, the plane requires a minimum runway length of 1.00 km. It cannot land safely on a 0.800 km runway. 2.35

We choose x = 0 and t = 0 at the location of Sue’s car when she first spots the van and applies the brakes. Then, the initial conditions for Sue’s car are x0S = 0 and v0S = 30.0 m/s. Her constant acceleration for aS = −2.00 m/s2. The initial conditions for the van are x0V = 155 m, v0V = 5.00 m/s, and its 1 2 constant acceleration is aV = 0. We then use Δx = x − x0 = v0t + at to write 2

an equation for the x-coordinate of each vehicle for t ≥ 0. This gives Sue’s Car: 1 xS − 0 = (30.0 m/s)t + (−2.00 m/s 2 )t 2 or xS = (30.0 m/s)t − (−1.00 m/s 2 )t 2 2 xV − 155 m = (5.00 m/s)t +

Van:

1 (0)t 2 or xV = 155 m + (5.00 m/s)t 2

In order for a collision to occur, the two vehicles must be at the same location (i.e., xS = xV). Thus, we test for a collision by equating the two equations for the x-coordinates and see if the resulting equation has any real solutions. xS = xV

⇒

(30.0 m/s)t − (1.00 m/s2)t2 = 155 m + (5.00 m/s)t

(1.00 m/s2)t2 − (25.00 m/s) + 155 m = 0

Topic 2

Using the quadratic formula yields −(−25.00 m/s) ± (−25.00 m/s 2 ) − 4(1.00 m/s 2 )(155 m) t= = 13.6s or 11.4 s 2(1.00 m/s 2 )

The solutions are real, not imaginary, so a collision will occur. The smaller of the two solutions is the collision time. (The larger solution tells when the van would pull ahead of the car again if the vehicles could pass harmlessly through each other.) The x-coordinate where the collision occurs is given by xcollision = xS|t= 11.4 s = xV|t= 11.4 s = 155 m + (5.00 m/s)(11.4 s) = 212 m 2.36

The velocity at the end of the first interval is v = v0 + at = 0 + (2.77 m/s2)(15.0 s) = 41.6 m/s This is also the constant velocity during the second interval and the initial velocity for the third interval. Also, note that the duration of the second interval is t2 = (2.05 min)(60.0 s/1 min) = 123 s. 1 2 (a) From Δx = v0t + at , the total displacement is 2

(Δx)total = (Δx)1 + (Δx)2 + (Δx)3 1 ⎡ ⎤ − ⎢ 0 + (2.77 m/s 2 )(15.0 s)2 ⎥ + [(2.77 m/s 2 )(15.0 s)(123 s) + 0] 2 ⎣ ⎦ 1 ⎡ ⎤ + ⎢(2.77 m/s 2 )(15.0 s)(4.39 s) + (−9.47 m/s 2 )(4.39 s)2 ⎥ 2 ⎣ ⎦

Topic 2

(Δx)total = 312 m + 5.11 × 103 m + 91.2 m = 5.51 × 103 m = 5.51 km

(b) v1 =

(Δx)1 312 m = = 20.8 m/s t1 15.0 s

v2 =

(Δx)2 5.11× 10 3 m = = 41.5 m/s t2 123 s

v3 =

(Δx)3 91.2 m = = 20.8 m/s , and the average velocity for the t3 4.39 s

(Δx)total 5.51× 10 3 m = = 38.7 m/s total trip is vtotal = t total (15.0 + 123 + 4.39) s

2.37

1 2 Using the uniformly accelerated motion equation Δx + v0t + at for the 2 1 2 2 full 40 s interval yields Δx = (20 m/s)(40 s) + (−1.0 m/s )(40 s) = 0 , which 2

is obviously wrong. The source of the error is found by computing the time required for the train to come to rest. This time is

v − v0 0 − 20 m/s = = 20 s. a −1.0 m/s 2

Thus, the train is slowing down for the first 20 s and is at rest for the last 20 s of the 40 s interval. The acceleration is not constant during the full 40 s. It is, however, constant during the first 20 s as the train slows to rest. Application of

Topic 2

1 Δx = v0t + at 2 to this interval gives the stopping distance as 2 1 Δx = (20 m/s)(20 s) + (−1.0 m/s 2 )(20 s)2 = 200 m 2

2.38

mi ⎞ ⎛ 0.447 m/s ⎞ ⎛ v0 = 0 and v f = ⎜ 400 ⎟ ⎜ = 17.9 m/s ⎝ h ⎠ ⎝ 1 mi/h ⎟⎠

(a) To find the distance traveled, we use

⎛ v + v0 ⎞ ⎛ 17.9 m/s + 0 ⎞ Δx = vt = ⎜ f t=⎜ ⎟⎠ (12.0 s) = 107 m ⎝ 2 ⎟⎠ ⎝ 2

(b) The constant acceleration is

2.39

v f − v0 17.9 m/s − 0 = = 1.49 m/s 2 t 12.0 s

At the end of the acceleration period, the velocity is v = v0 + ataccel = 0 + (1.5 m/s2)(5.0 s) = 7.5 m/s This is also the initial velocity for the braking period. (a) After braking,

vf = v + atbrake = 7.5 m/s + (−2.0 m/s2)(3.0 s) = 1.5 m/s

(b) The total distance traveled is

⎛ v + v⎞ ⎛ v + v0 ⎞ Δxtotal = (Δx)accel + (Δx)brake = (vt)accel + (vt)brake = ⎜ t accel + ⎜ f t ⎟ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ brake ⎛ 7.5 m/s +0 ⎞ ⎛ 1.5 m/s +7.5 m/s ⎞ Δx total = ⎜ (5.0 s) + ⎜ ⎟ ⎟⎠ (3.0 s) = 32 m ⎝ ⎠ ⎝ 2 2

Topic 2

2.40

For the acceleration period, the parameters for the car are: initial velocity = via = 0, acceleration = aa = a1, elapsed time = (Δt)a = t1, and final velocity = vfa. For the braking period, the parameters are: initial velocity = vib = final velocity of acceleration period = vfa, acceleration = ab = a2, and elapsed time = (Δt)b = t2. (a) To determine the velocity of the car just before the brakes are engaged, we apply vf = vi + a(Δt) to the acceleration period and find vib = vfa = via + aa(Δt)a = 0 + a1t1

vib = a1t1

1 2 (b) We may use Δx = vi (Δt) + a(Δt) to determine the distance traveled 2

during the acceleration period (i.e., before the driver begins to brake). This gives 1 1 (Δx)a = via (Δt)a + aa (Δt)2a = 0 + a1t12 2 2

(Δx)a =

1 2 a1t1 2

Thus, the total displacement of the car during the two intervals combined is

Topic 2

(Δx)total = (Δx)a + (Δx)b =

2.41

1 2 1 a1t1 + a1t1t 2 + a2t 22 2 2

The time the Thunderbird spends slowing down is

Δt1 =

Δx1 2(Δx1 ) 2(250 m) = = = 6.99 s v1 v + v0 0 + 71.5 m/s

The time required to regain speed after the pit stop is

Δt 2 =

Δx2 2(Δx2 ) 2(350 m) = = = 9.79 s. v2 v + v0 71.5 m/s + 0

Thus, the total elapsed time before the Thunderbird is back up to speed is Δt = Δt1 + 5.00 s + Δt2 = 6.99 s + 5.00 s + 9.79 s = 2.18 s During this time, the Mercedes has traveled (at constant speed) a distance ΔxM = v0(Δt) = (71.5 m/s)(21.8 s) = 1 559 m and the Thunderbird has fallen behind a distance d = ΔxM − Δx1 − Δx2 = 1 559 m − 250 m − 350 m = 959 m 2.42

The initial bank account balance is xi = $1.0 × 104 (to two significant figures) and the bank account is empty when xf = 0 with a change of Δx = xf − xi = $(−1.0 × 104). Use Δx = v0t + 12 at 2

Topic 2

with v0 = 0 and a = −2.5 × 102 $/month to find the time, t:

Δx = 12 at 2 → t =

2Δx a

(

)

2.43

2 $(−1.0 × 10 4 )

( −2.5 × 10 $/month ) 2

= 8.9 months

(a) Take t = 0 at the time when the player starts to chase his opponent. At this time, the opponent is distance d = (12 m/s)(30 s) = 36 m in front of the player. At time t > 0, the displacements of the players from their initial positions are 1 1 Δxplayer = (v0 )player t + aplayer t 2 = 0 + (4.0 m/s 2 )t 2 2 2 1 Δxopponent = (v0 )opponent t + aopponent t 2 = (12 m/s)t + 0 2

and

When the players are side-by-side, Δxplayer = Δxopponent + 36 m

[1]

[2] [3]

Substituting Equations [1] and [2] into Equation [3] gives 1 (4.0 m/s 2 )t 2 = (12 m/s)t + 36 m 2

t2 + (−6.0 s)t + (−18 s2) = 0

Applying the quadratic formula to this result gives

−(−6.0 s) ± (−6.0 s)2 − 4(1)(−18 s 2 ) 2(1)

Topic 2

which has solutions of t = −2.2 s and t = +8.2 s. Since the time must be greater than zero, we must choose t = 8.2 s as the proper answer. 1 1 2 2 2 2 (b) Δxplayer =(v0 )player t + aplayer t = 0 + (4.0 m/s )(8.2 s) = 1.3 × 10 m 2 2

2.44

The initial velocity of the train is v0 = 82.4 and the final velocity is v = 16.4 km/h. The time required for the 400 m train to pass the crossing is found from Δx = vt = [(v + v0 )/2]t as

2.45

⎛ 3600 s ⎞ 2(Δx) 2(0.400 km) = = (8.10 × 10 −3 h) ⎜ = 29.1 s v + v0 (82.4 + 16.4) km/h ⎝ 1 h ⎟⎠

(a) From v 2 = v02 + 2a(Δy) with v = 0, we have

v 2 − v02 0 − (25.0 m/s)2 (Δy)max = = = 31.9 m 2a 2(−9.80 m/s 2 ) (b) The time to reach the highest point is

v 2 − v02 0 − 25.0 m/s t up = = = 2.55 s a −9.80 m/s 2 (c) The time required for the ball to fall 31.9 m, starting from rest, is found from 1 Δy = (0)t + at 2 as t = 2

2(Δy) = a

2(−31.9 m) = 2.55 s −9.80 m/s 2

(d) The velocity of the ball when it returns to the original level (2.55 s

Topic 2

after it starts to fall from rest) is v = v0 + at = 0 + (−9.80 m/s2)(2.55 s) = −25.0 m/s 2.46

We take upward as the positive y-direction and y = 0 at the point where the ball is released. Then, v0y = −8.00 m/s, ay = −g = −9.80 m/s2, and Δy = −30.0 m when the ball reaches the ground. From v y = v0 y + 2ay (Δy) , the 2

velocity of the ball just before it hits the ground is

v y = v02 y + 2ay (Δy) = − (8.00 m/s)2 + 2(−9.80 m/s 2 )(−30.0 m) = −25.5 m/s Then, vy = v0y + ayt gives the elapsed time as

2.47

v y − v0 y −25.5 m/s − (−8.00 m/s) = = 1.79 s ay −9.80 m/s 2

(a) The velocity of the object when it was 30.0 m above the ground can be 1 2 determined by applying Δy = v0t + at to the last 1.50 s of the fall. 2

This gives

1⎛ m⎞ −30.0 m = v0 (1.50 s) + ⎜ −9.80 2 ⎟ (1.50 s)2 2⎝ s ⎠

v0 = −12.7 m/s

(b) The displacement the object must have undergone, starting from rest, to achieve this velocity at a point 30.0 m above the ground is given by

v 2 = v02 + 2a(Δy) as

Topic 2

(Δy)1 =

v 2 − v02 (−12.7 m/s)2 − 0 = = −8.23 m 2a 2(−9.80 m/s 2 )

The total distance the object drops during the fall is then |(Δy)total|=|(−8.23 m) + (−30.0 m)| = 38.2 m 2.48

(a) Consider the rock’s entire upward flight, for which v0 = +7.40 m/s, vf = 0, a = −g = −9.80 m/s2, yi = 1.55 m, (taking y = 0 at ground level), and yf = hmax = maximum altitude reached. Then applying v 2f = vi2 + 2a(Δy) to this upward flight gives

0 = (7.40 m/s)2 + 2(−9.80 m/s2)(hmax − 1.55 m) Solving for the maximum altitude of the rock gives

hmax = 1.55 m +

(7.40 m/s)2 = 4.34 m 2(9.80 m/s 2 )

Since hmax > 3.65 m (height of the wall), the rock does reach the top of the wall. (b) To find the velocity of the rock when it reaches the top of the wall, we use v f = vi + 2a(Δy) and solve for vf when yf = 3.65 m (starting with vi 2

= +7.40 m/s at yi = 1.55 m). This yields

v f = vi2 + 2a(y f − yi ) = (7.40 m/s)2 + 2(−9.80 m/s 2 )(3.65 m − 1.55 m) = 3.69 m/s © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

(c) A rock thrown downward at a speed of 7.40 m/s (vi = −7.40 m/s) from the top of the wall undergoes a displacement of (Δy) = yf − yi = 1.55 m − 3.65 m = −2.10 m before reaching the level of the attacker. Its velocity when it reaches the attacker is

v f = − vi2 + 2a(Δy) = − (−7.40 m/s)2 + 2(−9.80 m/s 2 )(−2.10 m) = −9.79 m/s so the change in speed of this rock as it goes between the 2 points located at the top of the wall and the attacker is given by Δ(speed)down = ||vf| − |vi|| = ||−9.79 m/s| − |−7.40 m/s|| = 2.39 m/s. (d) Observe that the change in speed of the ball thrown upward as it went from the attacker to the top of the wall was Δ(speed)up = ||vf| − |vi||=|3.69 m/s − 7.40 m/s| = 3.71 m/s The two rocks do not undergo the same magnitude speed change. The rocks have the same acceleration, but the rock thrown downward has a higher average speed between the two levels, and is accelerated over a smaller time interval. 2.49

The velocity of the child’s head just before impact (after falling a distance of 0.40 m, starting from rest) is given by v 2 = v02 + 2a(Δy) as v I = v02 + 2a(Δy) = − 0 + 2(−9.8 m/s 2 )(−0.40 m) = −2.8 m/s

Topic 2

If, upon impact, the child’s head undergoes an additional displacement Δy = −h before coming to rest, the acceleration during the impact can be found from v 2 = v02 + 2a(Δy) to be a = (0 − v I2 )/2(−h) = v I2 /2h . The duration of the impact is found from v = v0 + at as

t = Δv/a = −v I /(v I2 /2h), or t = −2h/v I . Applying these results to the two cases yields Hardwood Floor (h = 2.0 × 10 −3 m) : a =

and

vI2 (−2.8 m/s)2 = = 2.0 × 10 3 m/s 2 2h 2(2.0 × 10 −3 m)

−2h −2(2.0 × 10 −3 m) = = 1.4 × 10 −3 s = 1.4 ms vI −2.8 m/s

Carpeted Floor (h = 1.0 × 10

and

2.50

−2

m): a =

vI2 (−2.8 m/s)2 = = 3.9 × 10 2 m/s 2 −2 2h 2(1.0 × 10 m)

−2h −2(1.0 × 10 −2 m) = = 7.1× 10 −3 s = 7.1 ms vI −2.8 m/s

(a) After 2.00 s, the velocity of the mailbag is vbag = v0 + at = −1.50 m/s + (−9.80 m/s2)(2.00 s) = −21.1 m/s The negative sign tells us that the bag is moving downward and the magnitude of the velocity gives the speed as 21.1 m/s. (b) The displacement of the mailbag after 2.00 s is

Topic 2

⎛ v + v0 ⎞ ⎡ −21.1 m/s + ( − 1.50 m/s) ⎤ (Δy)bag = ⎜ t=⎢ ⎟ ⎥⎦ (2.00 s) = − 22.6 m ⎝ 2 ⎠ 2 ⎣ During this time, the helicopter, moving downward with constant velocity, undergoes a displacement of 1 (Δy)copter = v0t + at 2 = (−1.5 m/s)(2.00 s) + 0 = −3.00 m 2

The distance separating the package and the helicopter at this time is then d = |(Δy)P − (Δy)h| = |−22.6 m − (−3.00 m)| = |−19.6 m| = 19.6 m (c) Here, (v0)bag = (v0)copter = +1.50 m/s and abag = −9.80 m/s2 while acopter = 0. After 2.00 s, the velocity of the mailbag is

vbag = 1.50

m ⎛ m⎞ m m + ⎜ −9.80 2 ⎟ (2.00 s) = −18.1 and its speed is vbag = 18.1 s ⎝ s ⎠ s s

In this case, the displacement of the helicopter during the 2.00 s interval is Δycopier = (+1.50 m/s)(2.00 s) + 0 = +3.00 m Meanwhile, the mailbag has a displacement of

⎛ v + v0 ⎞ ⎡ −18.1 m/s + 1.50 m/s ⎤ (Δy)bag = ⎜ bag t=⎢ ⎟ ⎥⎦ (2.00 s) = −16.6 m ⎝ 2 ⎠ 2 ⎣

Topic 2

The distance separating the package and the helicopter at this time is then d = |(Δy)p − (Δy)h| = |−16.6 m − (+3.00 m)| = |−19.6 m| = 19.6 m 2.51

(a) From the instant the ball leaves the player’s hand until it is caught, the ball is a freely falling body with an acceleration of a = −g = −9.80 m/s2 = 9.80 m/s2 = 9.80 m/s2 (downward) (b) At its maximum height, the ball comes to rest momentarily and then = 0 . begins to fall back downward. Thus, vmax height 1 2 (c) Consider the relation Δy = v0t + at with a = −g. When the ball is at 2

the thrower’s hand, the displacement is Δy = 0, giving 1 0 = v0t − gt 2 2

This equation has two solutions, t = 0, which corresponds to when the ball was thrown, and t = 2v0/g corresponding to when the ball is caught. Therefore, if the ball is caught at t = 2.00 s, the initial velocity must have been

v0 =

gt (9.80 m/s 2 )(2.00 s) = = 9.80 m/s 2 2

Topic 2

(d) From v 2 = v02 + 2a(Δy) , with v = 0 at the maximum height,

(Δy)max =

2.52

v 2 − v02 0 − (9.80 m/s 2 ) = = 4.90 m 2a 2(−9.80 m/s 2 )

(a) Let t = 0 be the instant the package leaves the helicopter, so the package and the helicopter have a common initial velocity of vi = −v0 (choosing upward as positive). At times t > 0, the velocity of the package (in free-fall with constant acceleration ap = −g) is given by v = vi + at as vp = −v0 − gt = −(v0 + gt) and speed = |vp| = v0 + gt. (b) After an elapsed time t, the downward displacement of the package from its point of release will be

1 1 1 ⎞ ⎛ (Δy)P = vi t + aP t 2 = −v0t − gt 2 = − ⎜ v0t + gt 2 ⎟ ⎝ 2 2 2 ⎠ and the downward displacement of the helicopter (moving with constant velocity, or acceleration ah = 0) from the release point at this time is 1 (Δy)h = vi t + aht 2 = −v0t + 0 = −v0t 2

The distance separating the package and the helicopter at this time is then © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

1 1 ⎛ ⎞ d = (Δy) p − (Δy)h = − ⎜ v0t + gt 2 ⎟ − (−v0t) = gt 2 ⎝ ⎠ 2 2

(c) If the helicopter and package are moving upward at the instant of release, then the common initial velocity is vi = +v0. The accelerations of the helicopter (moving with constant velocity) and the package (a freely falling object) remain unchanged from the previous case (ap = −g and ah = 0). In this case, the package speed at time t > 0 is |vP| = |vi + apt| = |v0 − gt| . At this time, the displacements from the release point of the package and the helicopter are given by 1 1 (Δy) p = vi t + a pt 2 = v0t − gt 2 2 2

and

1 (Δy)h = vi t + aht 2 = v0t + 0 = +v0t 2

The distance separating the package and helicopter at time t is now given by 1 1 d = (Δy) p − (Δy)h = v0t − gt 2 − v0t = gt 2 2 2

2.53

(the same as earlier!)

(a) After its engines stop, the rocket is a freely falling body. It continues

Topic 2

upward, slowing under the influence of gravity until it comes to rest momentarily at its maximum altitude. Then it falls back to Earth, gaining speed as it falls. (b) When it reaches a height of 150 m, the speed of the rocket is v = v02 + 2a(Δy) = (50.0 m/s)2 + 2(2.00 m/s 2 )(150 m) = 55.7 m/s

After the engines stop, the rocket continues moving upward with an initial velocity of v0 = 55.7 m/s and acceleration a = −g = −9.80 m/s2. When the rocket reaches maximum height, v = 0. The displacement of the rocket above the point where the engines stopped (that is, above the 150 m level) is

Δy =

v 2 − v02 0 − (55.7 m/s)2 = = 158 m 2a 2(−9.80 m/s 2 )

The maximum height above ground that the rocket reaches is then given by hmax = 150 m + 158 m = 308 m (c) The total time of the upward motion of the rocket is the sum of two intervals. The first is the time for the rocket to go from v0 = 50.0 m/s at the ground to a velocity of v = 55.7 m/s at an altitude of 150 m. This time is given by

Topic 2

t1 =

(Δy)1 (Δy)1 2(150 m) = = = 2.84 s v1 (v + v0 ) / 2 (55.7 + 50.0) m/s

The second interval is the time to rise 158 m starting with v0 = 55.7 m/s and ending with v = 0. This time is

t2 =

(Δy)2 (Δy)2 2(158 m) = = = 5.67 s v2 (v + v0 ) / 2 0 + 55.7 m/s

The total time of the upward flight is then tap = t1 + t2 = (2.84 + 5.67) s = 8.51 s (d) The time for the rocket to fall 308 m back to the ground, with v0 = 0 1 2 and acceleration a = −g = −9.80 m/s2, is found from Δy = v0t + at as 2

t down =

2(Δy) = a

2(−308 m) = 7.93 s −9.80 m/s 2

so the total time of the flight is tflight = tup + tdown = (8.51 + 7.93) s = 16.4 s. 2.54

(a) For the upward flight of the ball, we have vi = v0, vf = 0, a = −g, and Δt = 3.00 s. Thus, vf = vi + a(Δt) gives the initial velocity as vi = vf − a(Δt) = vf + g(Δt) or v0 = 0 + (9.80 m/s2)(3.00 s) = +29.4 m/s (b) The vertical displacement of the ball during this 3.00-s upward flight

Topic 2

⎛ v + vf ⎞ ⎛ 29.4 m/s + 0 ⎞ Δymax = h = v(Δt) = ⎜ i (Δt) = ⎜ ⎟⎠ (3.00 s) = 44.1 m ⎟ ⎝ ⎝ 2 ⎠ 2 2.55

During the 0.600 s required for the rig to pass completely onto the bridge, the front bumper of the tractor moves a distance equal to the length of the rig at constant velocity of v = 100 km/h. Therefore the length of the rig is ⎡ km ⎛ 0.278 m/s ⎞ ⎤ Lrig = vt = ⎢100 ⎜⎝ 1 km/h ⎟⎠ ⎥ (0.600 s) = 16.7 m h ⎣ ⎦

While some part of the rig is on the bridge, the front bumper moves a distance Δx = Lbridge + Lrig = 100 m + 16.7 m. With a constant velocity of v = 100 km/h, the time for this to occur is

2.56

Lbridge + Lrig 400 m + 16.7 m ⎛ 1 km/h ⎞ = ⎜⎝ 0.278 m/s ⎟⎠ = 15.0 s v 100 km/h

(a) The acceleration experienced as he came to rest is given by v = v0 + at as

v − v0 = t

mi ⎞ ⎛ 0.447 m/s ⎞ ⎛ 0 − ⎜ 632 ⎟ ⎜ ⎝ h ⎠ ⎝ 1 mi/h ⎟⎠ 1.40 s

= −202 m/s 2

(b) The distance traveled while stopping is found from

Topic 2

⎡ ⎛ mi ⎞ ⎛ 0.447 m/s ⎞ ⎤ ⎢ 0 + ⎜⎝ 632 ⎟⎠ ⎜ ⎥ h ⎝ 1 mi/h ⎟⎠ ⎦ ⎛ v + v0 ⎞ ⎣ Δx = vt = ⎜ t= (1.40 s) = 198 m ⎝ 2 ⎟⎠ 2 2.57

(a) The acceleration of the bullet is

v 2 − v02 (300 m/s)2 − (400 m/s)2 = = −3.50 × 10 5 m/s 2 2(Δx) 2(0.100 m)

(b) The time of contact with the board is

2.58

v − v0 (300 − 400) m/s = = 2.86 × 10 −4 s 5 2 a −3.50 × 10 m/s

1 2 (a) From Δx = v0t + at , we have 2 1 100 m = (30.0 m/s)t + (−3.50 m/s 2 )t 2 2

This reduces to 3.50t2 + (−60.0 s)t + (200 s2) = 0, and the quadratic formula gives −(−60.0 s) ± (−60.0 s)2 − 4(3.50)(200 s 2 ) t= 2(3.50)

The desired time is the smaller solution of t = 4.53 s. The larger solution of t = 12.6 s is the time when the boat would pass the buoy moving backwards, assuming it maintained a constant acceleration. (b) The velocity of the boat when it first reaches the buoy is

Topic 2

v = v0 + at = 30.0 m/s + (−350 m/s2)(4.53 s) = 14.1 m/s. 2.59

(a) The keys have acceleration a = −g = −9.80 m/s2 from the release point 1 2 until they are caught 1.50 s later. Thus, Δy = v0t + at gives 2

Δy − at 2 /2 (+4.00 m) − (−9.80 m/s 2 )(1.50 s)2 /2 v0 = = = +10.0 m/s t 1.50 s or

v0 = 10.0 m/s upward

(b) The velocity of the keys just before the catch was v = v0 + at = 10.0 m/s + (−9.80 m/s2)(1.50 s) = −4.70 m/s or 2.60

v = 4.70 m/s downward

(a) While in the air after launching itself from the water, the salmon’s vertical acceleration is ay = −g = −9.80 m/s2. Assume it comes to rest at the top of its vertical leap, a distance Δy = 3.60 m above the bottom of the waterfall. From the time-independent kinematic equation, with the final velocity v = 0, the initial speed v0 is v 2 = v02 − 2g ( Δy )

(

)

v0 = 2g ( Δy ) = 2 9.80 m/s2 ( 3.6 m ) v0 = 8.4 m/s

Topic 2

2.61

(a) From v 2 = v02 + 2a(Δy) , the insect’s velocity after straightening its legs is

v = v02 + 2a(Δy) = 0 + 2(4 000 m/s 2 )(2.0 × 10 −3 m) = 4.0 m/s (b) The time to reach this velocity is

v − v0 4.0 m/s − 0 = = 1.0 × 10 −3 s = 1.0 ms a 4 000 m/s 2

(c) The upward displacement of the insect between when its feet leave the ground and it comes to rest momentarily at maximum altitude is

Δy =

2.62

v 2 − v02 0 − v02 −(4.0 m/s)2 = = = 0.82 m 2a 2(−g) 2(−9.8 m/s 2 )

(a)

For constant speed:

(b)

When speeding up at a constant rate:

Topic 2

(c)

2.63

When slowing down at a constant rate:

The falling ball moves a distance of (15 m − h) before they meet, where h 1 2 is the height above the ground where they meet. Apply Δy = v0t + at , 2

with a = −g, to obtain 1 −(15 m − h) = 0 − gt 2 2

1 h =15 m − gt 2 2

[1]

1 2 Applying Δy = v0t + at to the rising ball gives 2 1 h = (25 m/s)t − gt 2 2

[2]

Topic 2

Combining Equations [1] and [2] gives

(25 m/s)t −

2.64

1 2 1 gt = 15 m − gt 2 2 2

15 m = 0.60 s 25 m/s

(a) When the ball hits the ground, its change in height will be Δy = −h. Solve for the final speed of each ball using the time-independent kinematic equation: v 2 = v02 − 2g ( Δy ) v=

( ±v ) − 2g ( −h) = 2

v02 + 2gh

(b) We’re asked to find an expression for the time difference Δt between the times of flight for the upward- and downward-thrown balls.

For the upward-thrown ball, the path to the ground can be separated into two parts. In the first part the ball rises with initial velocity +v0 and falls back to its original height where its velocity is –v0. In the second part it moves from height h to the ground in exactly the same time it takes the downward-thrown ball to reach the ground. The difference between the two ball’s times of flight is therefore equal to

Topic 2

the time for the first part of the upward-thrown ball’s path. That time is found using the first kinematic equation with v0 = +v0 and v = −v0:

v = v0 − gΔt Δt =

2.65

v − v0 −v0 − v0 2v0 = = −g −g g

From the time-independent kinematic equation with v0 > 0, Δy = 2.00 m, and v = ±1.50 m/s: v 2 = v02 − 2g ( Δy ) v0 = v 2 + 2g ( Δy ) =

2.66

( ±1.50 m/s) + 2g ( 2.00 m ) = 6.44 m/s 2

(a) To find the distance Δx traveled by the blood during the acceleration, apply the time-independent kinematic equation with v0 = 0, v = 1.05 m/s, and a = 22.5 m/s2:

v 2 = v02 + 2aΔx v 2 − v02 (1.05 m/s) − 0 Δx = = = 2.45 × 10−2 m = 2.45 cm 2a 2 22.5 m/s2 2

(

)

(b) Solve for the time t required for the blood to reach its peak speed using the first kinematic equation:

Topic 2

v = v0 + at t=

2.67

v − v0 (1.05 m/s) − 0 = = 4.67 × 10−2 s a 22.5 m/s2

When released from rest (v0 = 0), the bill falls freely with a downward acceleration due to gravity (a = −g = −9.80 m/s2). Thus, the magnitude of its downward displacement during David’s 0.2 s reaction time will be

1 1 Δy = v0t + at 2 = 0 + (−9.80 m/s 2 )(0.2 s)2 = 0.2 m = 20 cm 2 2 This is over twice the distance from the center of the bill to its top edge (≈8 cm), so David will be unsuccessful. 2.68

(a) The velocity with which the first stone hits the water is 2

m⎞ m⎞ m ⎛ ⎛ v1 = − v + 2a(Δy) = − ⎜ −2.00 ⎟ + 2 ⎜ −9.80 2 ⎟ (−5.00 m) = −31.4 ⎝ ⎝ s⎠ s ⎠ s 2 01

The time for this stone to hit the water is

t1 =

v1 − v01 [−31.4 m/s − (−2.00 m/s)] = = 3.00 s a −9.80 m/s 2

(b) Since they hit simultaneously, the second stone, which is released 1.00 s later, will hit the water after an flight time of 2.00 s. Thus,

v02 =

Δy − at 22 /2 −50.0 m − (−9.80 m/s 2 )(2.00 s)2 /2 = = −15.2 m/s t2 2.00 s

Topic 2

(c) From part (a), the final velocity of the first stone is v1 = −31.4 m/s. The final velocity of the second stone is v2 = v02 + at2 = −15.2 m/s + (−9.80 m/s2)(2.00 s) = −34.8 m/s 2.69

When the hare wakes up, the tortoise is a distance L > 0 from the finish line whereas the hare is a distance L + d from the finish line. The hare, running with constant speed v1, reaches the finish line in a time given by thare =

L+d v1

The tortoise, crawling with constant speed v2 < v1, reaches the finish line in a time given by ttortoise =

L v2

The tortoise wins the race if ttortoise < thare, so it follows that L L+d < v2 v1

Rearrange this expression to find a condition on the length L: Lv1 < Lv2 + v2 d → L <

v2 d v1 − v2

The tortoise wins the race if the length to the finish line, L, satisfies that inequality.

Topic 2

2.70

1 2 (a) From Δy = v0t + at with v0 = 0, we have 2

2(Δy) = a

2(−23 m) = 2.2 s −9.80 m/s 2

(b) From the time-independent velocity equation, the final speed is: v 2 = v02 − 2gΔy

(

)

v = 2gΔy = 2 −9.80 m/s2 ( −23 m ) = 21 m/s

Because the man is falling, his final velocity is v = −21 m/s. (c) The time it takes for the sound of the impact to reach the spectator is

tsound =

Δy 23 m = = 6.8 × 10 −2 s vsound 340 m/s

so the total elapsed time is ttotal = 2.2 s + 6.8 × 10–2 s ≈ 2.3 s. 2.71

The time required for the stuntman to fall 3.00 m, starting from rest, is 1 2 found from Δy = v0t + at as 2 1 −3.00 m = 0 + (−9.80 m/s 2 )t 2 2

2(3.00 m) = 0.782 s 9.80 m/s 2

(a) With the horse moving with constant velocity of 10.0 m/s, the horizontal distance is Δx = vhorset = (10.0 m/s)(0.782 s) = 7.82 m © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 2

(b) The required time is t = 0.782 s as calculated above.

Topic 3

Topic 3 Motion in Two Dimensions

QUICK QUIZZES 3.1

Choice (b). If velocity is constant, the acceleration (rate of change in velocity) is zero. An object may have constant speed (magnitude of velocity) but still be accelerating due to a change in direction of the velocity. If an object is following a curved path, it is accelerating because the velocity is changing in direction.

3.2

Choice (a). Any change in the magnitude and/or direction of the velocity is an acceleration. The gas pedal and the brake produce accelerations by altering the magnitude of the velocity. The steering wheel produces accelerations by altering the direction of the velocity.

3.3

(a) Average speed is the path length traveled divided by the time interval. In this case, the path length equals πR, half the circumference of the circular path, with R = 10.0 m. Her average speed is (Average speed) = πR/Δt = π(10.0 m)/15.0 s = 2.09 m/s. (b) Her average velocity has magnitude

Topic 3

! Δr 2R 2 (10.0 m ) ! v av = = = = 1.33 m/s Δt Δt 15.0 s 3.4

Choice (c) is correct. A projectile has constant horizontal velocity. Thus, if the runner throws the ball straight up into the air, the ball maintains the horizontal velocity it had before it was thrown (that is, the velocity of the runner). In the runner’s frame of reference, the ball appears to go straight upward and come straight downward. To a stationary observer, the ball follows a parabolic trajectory, moving with the same horizontal velocity as the runner and staying above the runner’s hand.

3.5

Choice (b) is correct. The velocity is always tangential to the path while the acceleration is always directed vertically downward. Thus, the velocity and acceleration are perpendicular only where the path is horizontal. This only occurs at the peak of the path.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

3.2

(a)

3.4

(a)

(b)

The balls will be closest at the instant the second ball is projected.

(b) The first ball will always be going faster than the second ball. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 3

(c)

There will be a one second time interval between their collisions with the ground.

(d) The two move with the same acceleration in the vertical direction. Thus, changing their horizontal velocities can never make them hit at the same time. 3.6

The equations of projectile motion are only valid for objects moving freely under the influence of gravity. The only acceleration such an object has is the free-fall acceleration, g, directed vertically downward. Of the objects listed, only a and d meet this requirement.

3.8

(a) The passenger sees the ball go into the air and come back in the same way he would if he were at rest on Earth. An observer by the tracks would see the ball follow the path of a projectile. (b) If the train were accelerating, the ball would fall behind the position it would reach in the absence of the acceleration.

3.10

Statement (c) is true. At its highest point, the baseball’s velocity is horizontal and its acceleration is downward.

3.12

Choice (c) is correct. The dolphin is heading due north away from a boat that is moving east at speed v. Relative to a stationary observer, the dolphin is moving both east (at v) and north, so it’s velocity would be

Topic 3

greater than v in a direction north of east.

ANSWERS TO EVEN NUMBERED PROBLEMS 3.2

(a)

3.61 km at 33.7° N of E

(b) 1.44 km/h at 33.7° N of E

(a)

Δx = 0.300 m, Δy = 0.200 m

(b) vav,x = 0.060 0 m/s, vav,y = 0.040 0 m/s (c)

aav,x = 0.020 0 m/s2, aav,y = −0.020 0 m/s2

3.6

(a)

−1.33 m/s2

3.8

2.68 ft

3.10

(a)

76.0°

(b)

−8.00 m/s2

Rmax = 2.13R

3.14

(a)

28.2 m/s

(b)

4.07 s

(c)

v0 and t would increase

(a)

x = 0, y = y0

(b)

v0x = 7.52 m/s, v0y = −2.74 m/s

(c)

x = (7.52 m/s)t, y = y0 − (2.74 m/s)t − (4.90 m/s2)t2

Topic 3

100

(d)

22.6 m

(e)

y0 = 52.3 m

(b)

vCP,y = 0.82 m/s

(f) 1.18 s

3.16

x = 7.22 km, y = 1.69 km

3.18

18.6 m

3.20

(a)

vCP,x = 3.68 m/s

3.22

(a)

57.7 km/h at 60.0° west of vertical

(b)

28.9° km/h vertically downward

3.24

(a)

27.7° east of north

(b)

47.6 min

3.26

(a)

1.26 h

(b)

1.13 h

3.28

(a) 6.80 km

(b)

3.00 km vertically above the impact

point

(c)

66.2°

d tup = v − vs !

(b)

d tdown = v + vs !

3.30

(a)

(c)

1.19 h

See

Solution. (d)

2d tb = v !

(e)

ta > tb as long as vs > 0

3.32

(a)

twoman = L/v1

(b)

L tman = v1 + v2 !

3.34

(a)

hmax = 57.4 m, t = 1.22 s

(b)

Same as for part (a).

3.36

(a) !t = 2h/g

(b)

v = d g/2h !i

Topic 3

101

(c)

v = d 2 g/ ( 2h ) + 2gh

(d)

θ = tan −1 (−2h/d)

3.38

(a)

0.85 m/s

(b)

2.1 m/s

3.40

(a)

0° to the vertical

(b)

8.25 m/s

(c)

straight up and down

(d)

parabolic arc opening downward

(e)

12.6 m/s at 41.0° above the horizontal eastward line

3.42

(a)

14.7 m/s

3.44

10.7 m/s

3.46

⎛ ⎛ v0y ⎞ g ⎞ y = ⎜ − 2 ⎟ x2 + ⎜ x+0 ⎝ 2v0x ⎠ ⎝ v0x ⎟⎠ !

3.48

(a)

3.50

3.95 m/s

3.52

16.7°

3.54

(a)

0.519 m, too narrow for a pedestrian walkway

(b)

0.217 m/s

26.6°

3.56

1.1 m/s

3.58

(a) 14.8 m/s

(b)

29.4 m/s

(b)

t1 bounce/tno bounce = 0.950

(b) 1.28 s

(c)

17.0 m

Topic 3

102

PROBLEM SOLUTIONS 3.1

(a)

After moving half way around the circle, the magnitude of the ! airplane’s displacement Δr is equal to the diameter:

Δr = 2R = 7.00 km = 7.00 × 103 m (b)

The magnitude of the airplane’s average velocity is

vav =

Δr 7.00 × 103 m = Δt 150 s

= 46.7 m/s (c)

The airplane’s average speed equals the path length divided by the elapsed time. After moving half way around the circle, the plane has traveled over a path length equal to half the circle’s circumference. Its average speed is

( 2π R ) = π ( 3.50 × 10 m ) Average speed = 3

1 2

Δt

150 s

= 73.3 m/s

3.2

(a)

The hiker walks 2.00 km north and 3.00 km east. Her displacement has magnitude Δr =

( 3.00 km ) + ( 2.00 km ) = 3.61 km 2

Topic 3

103

−1 ⎛ 2.00 km ⎞ = 33.7° north of east. in the direction θ = tan ⎜ ⎝ 3.00 km ⎟⎠

(b) The hiker’s average velocity is:

! Δr ! v av = Δt vav =

Δr 3.61 km = = 1.44 km/h Δt 2.50 h

at 33.7° north of east (the same direction as her displacement). (c)

The hiker’s average speed equals the path length divided by the elapsed time. The path length is 2.00 km + 3.00 km = 5.00 km so that average speed =

3.3

(a)

5.00 km = 2.00 km/h 2.50 h

Solve for the quadcopter’s x-location at t = 2.00 s using the definition of average velocity:

vav,x =

Δx x f − xi = Δt Δt

→ x f = xi + vav,x Δt

x f = 2.00 m + (1.50 m/s ) ( 2.00 s ) = 5.00 m

(b) Similarly,

Topic 3

104

vav,y =

Δy y f − yi = Δt Δt

→ y f = yi + vav,y Δt

y f = 4.50 m +( −1.00 m/s ) ( 2.00 s ) = 2.50 m

3.4

(a) The components of the ant’s displacement are:

Δx = x f − xi = 0.400 m − 0.100 m = 0.300 m Δy = y f − yi = 0.350 m − 0.150 m = 0.200 m (b) The average velocity has components:

vav,x =

Δx 0.300 m = = 0.060 0 m/s Δt 5.00 s

vav,y =

Δy 0.200 m = = 0.040 0 m/s Δt 5.00 s

aav,x =

aav,y =

3.5

Δv x v x , f − v x ,i 0.100 m/s − 0 = = = 0.020 0 m/s 2 Δt Δt 5.00 s Δv y Δt

v y , f − v y ,i Δt

0 − 0.100 m/s = −0.020 0 m/s 2 5.00 s

The car’s average acceleration has components: aav,x =

aav,y =

Δvx vx , f − vx ,i 0 − 25.0 m/s = = = −4.17 m/s 2 Δt Δt 6.00 s Δv y Δt

v y , f − v y ,i Δt

35.0 m/s − 0 = 5.83 m/s 2 6.00 s

Topic 3

3.6

105

The rabbit’s average acceleration has components: (a) aav,x =

(b) aav ,y =

3.7

(a)

vx , f − vx ,i Δt v y , f − v y ,i Δt

0 − 2.00 m/s = −1.33 m/s 2 1.50 s

−12.0 m/s − 0 = −8.00 m/s 2 1.50 s

With the origin chosen at point O as shown in Figure P3.23, the coordinates of the original position of the stone are x0 = 0 and y0 = +50.0 m.

(b) The components of the initial velocity of the stone are v0x = +18.0 m/s and v0y = 0. (c)

The components of the stone’s velocity during its flight are given as functions of time by vx = v0x + axt = 18.0 m/s + (0)t and

vy = v0y + ayt = 0 + (−g)t or

vx = 18.0 m/s

vy = −(9.80 m/s2)t

(d) The coordinates of the stone during its flight are

1 1 x = x0 + v0xt + axt 2 = 0 + (18.0 m/s)t + (0)t 2 or 2 2 !

x = (18.0 m/s)t

and

1 1 y = y0 + v0yt + ayt 2 = 50.0 m + (0)t + (−g)t 2 or 2 2 !

y = 50.0 m − (4.90 m/s 2 )t 2

Topic 3

106

(e)

1 2 We find the time of fall from Δy = v0yt + ayt with v0y = 0: 2 ! t= !

2(Δy) 2(−50.0 m) = = 3.19 s a −9.80 m/s 2

(f) At impact, vx = v0x = 18.0 m/s, and the vertical component is vy = v0y + ayt = 0 + (−9.80 m/s2)(3.19 s) = −31.3 m/s Thus,

and

v = vx2 + vy2 = (18.0 m/s 2 ) + (−31.3 m/s 2 ) = 36.1 m/s ! ⎛v ⎞ ⎛ −31.3 ⎞ y θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = −60.1° ⎜⎝ v x ⎟⎠ ⎝ 18.0 ⎠

 or v = 36.1 m/s!!at!!60.1°!!below!!the!!horizontal . ! 3.8

⎛ 0.447 m/s ⎞ v0x = (101 mi/h) ⎜ = 45.1 m/s and ⎝ 1 mi/h ⎟⎠ ! ⎛ 1m ⎞ Δx = (60.5 ft) ⎜ = 18.4 m ⎝ 3.281 ft ⎟⎠ ! Δx 18.4 m = = 0.408 s . The time to reach home plate is t = ! v0x 45.1 m/s

In this time interval, the vertical displacement is

1 1 Δy = v0yt + ayt 2 = 0 + (−9.80 m/s 2 )(0.408 s)2 = −0.817 m 2 2 Thus, the ball drops vertically 0.817 m(3.281 ft/1 m) = 2.68 ft. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 3

3.9

107

At the maximum height vy = 0, and the time to reach this height is found from

vy = v0y + ayt

t= !

vy − v0y ay

0 − v0y −g

v0y g

The vertical displacement that has occurred during this time is 2 ⎛ vy + v0y ⎞ ⎛ 0 + v0y ⎞ ⎛ v0y ⎞ v0y (Δy)max = (vy )av t = ⎜ t= = ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ g ⎟⎠ 2g !

If (Δy)max = 3.7 m, we find

v = 2g(Δy)max = 2(9.80 m/s 2 )(3.7 m) = 8.5 m/s ! 0y and if the angle of projection is θ = 45°, the launch speed is

v0y 8.5 m/s v0 = = = 12 m/s sin θ sin 45° !

3.10

(a)

At y = hmax, the vertical component of velocity is zero, so vy = v0y + ayt gives the time to reach the peak as

tP = !

0 − v0y ay

−v0 sin θ v0 sin θ = −g g

Topic 3

108

Then, Δy = (vy)avt = [(vy + v0y)/2]t gives the maximum vertical displacement as 2 2 ⎛ 0 + v0y ⎞ ⎛ v0 sin θ ⎞ ⎛ v0 sin θ ⎞ v0 sin θ hmax = ⎜ t = = ⎜⎝ ⎟⎜ ⎟ P ⎝ 2 ⎟⎠ 2 ⎠⎝ 2 ⎠ 2g !

The total time of flight is ttotal = 2tP = 2v0 sin θ/g, so the horizontal range is ⎛ 2v sin θ ⎞ 2v 2 sin θ cosθ 0 R = v0xttotal = (v0 cosθ ) ⎜ 0 ⎟= g g ⎝ ⎠

If we are to have hmax = R, it is necessary that

v02 sin 2 θ 2v02 sin θ cosθ = . 2g g ! This requirement reduces to

sin θ = 2cosθ or tan θ = 4 ! 2

which gives the required launch angle as θ = tan−1(4) = 76.0°. (b)

For maximum range, the launch angle would be θ = 45°, so

2v02 sin 45°cos 45° v02 Rmax = = g g ! The ratio of Rmax to the range in part (a) (where θ = 76.0°) is

Rmax v02 /g 1 = = or Rmax = 2.13R . 2 ! R [2v sin(76.0°)cos(76.0°)] / g 0.469 0 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 3

109

(c)

Note that in the calculation of the answer to part (a), the acceleration of gravity canceled out. Thus, the result is valid on every planet independent of the local acceleration of gravity.

3.11

(a)

The time required for the ball to travel Δx = 36.0 m horizontally to the goal is found from Δx = v0xt = (v0 cosθ)t as t = Δx/v0 cosθ. At this time, the vertical displacement of the ball is given by

1 Δy = v0yt + ayt 2 . The vertical distance by which the ball clears the bar 2 ! is d = Δy − h, or

⎛ 36.0 m ⎞ 1 ⎛ ⎞ 36.0 m d = v0 sin 53.0° ⎜ + (−9.80 m/s 2 ) ⎜ − 3.05 m ⎟ ⎝ (20.0 m/s)cos 53.0° ⎟⎠ ⎝ v0 cos 53.0° ⎠ 2 2

yielding d = +0.89 m. Thus, the ball clears the crossbars by 0.89 m. (b) The ball reaches the plane of the goal post at t = Δx/v0 cosθ, or 36.0 m t= = 2.99 s (20.0 m/s)cos 53.0° !

At this time, its vertical velocity is given by vy = v0y + ayt as vy = (20.0 m/s)sin 53.0° + (−9.80 m/s2)(2.99 s) = −13.3 m/s. Since vy < 0, the ball has passed the peak of its arc and is descending when it crosses the crossbar.

Topic 3

3.12

110

(a)

When a projectile is launched with speed v0 at angle θ0 above the horizontal, the initial velocity components are v0x = v0 cosθ0 and v0y = v0 sinθ0. Neglecting air resistance, the vertical velocity when the projectile returns to the level from which it was launched (in this case, the ground) will be vy = −v0y. From this information, the total time of flight is found from vy = v0y + ayt to be

ttotal = !

vyf − v0y ay

−v0y − v0y −g

2v0y g

or ttotal =

2v0 sin θ 0 g

Since the horizontal velocity of a projectile with no air resistance is constant, the horizontal distance it will travel in this time (i.e., its range) is given by

⎛ 2v sin θ 0 ⎞ v02 v02 sin(2θ 0 ) R = v0xttotal = (v0 cosθ 0 ) ⎜ 0 = (2 sin θ cos θ ) = 0 0 ⎟⎠ g g g ⎝ ! Thus, if the projectile is to have a range of R = 81.1 m when launched at an angle of θ0 = 45.0°, the required initial speed is Rg (81.1 m)(9.80 m/s 2 ) v0 = = = 28.2 m/s sin(2θ 0 ) sin(90.0°) !

(b) With v0 = 28.2 m/s and θ0 = 45.0°, the total time of flight (as found above) will be

Topic 3

111

ttotal = !

(c)

2v0 sin θ 0 2(28.2 m/s)sin(45.0°) = = 4.07 s g 9.80 m/s 2

Note that at θ0 = 45.0°, sin(2θ0) = 1 and that sin(2θ0) will decrease as θ0 is increased above this optimum launch angle. Thus, if the range is to be kept constant while the launch angle is increased above 45.0°, we see from v0 = Rg/ sin(2θ 0 ) that the required initial velocity will ! increase. Observe that for θ0 < 90°, the function sinθ0 increases as θ0 is increased. Thus, increasing the launch angle above 45.0° while keeping the range constant means that both v0 and sinθ0 will increase. Considering the expression for ttotal given above, we see that the total time of flight will increase.

3.13

We choose our origin at the initial position of the projectile. After 3.0 s, it is at ground level, so the vertical displacement is Δy = −H.

1 2 To find H, we use Δy = v0yt + ayt , which becomes 2 ! 1 −H = [(15 m/s)sin 25°](3.0 s) + (−9.80 m/s 2 )(3.0 s)2 , or H = 25 m. 2 ! 3.14

(a)

With the origin at ground level directly below the window, the original coordinates of the ball are (x, y) = (0, y0).

Topic 3

112

(b) v0x = v0 cosθ0 = (8.00 m/s)cos(−20.0°) = +7.52 m/s v0y = v0 sinθ0 = (8.00 m/s)sin(−20.0°) = −2.74 m/s

(c)

1 1 x = x0 + v0xt + axt 2 = 0 + (7.52 m/s)t + (0)t 2 or x = (7.52 m/s)t 2 2 1 1 y = y0 + v0yt + ayt 2 = y0 + (−2.74 m/s)t + (−9.80 m/s 2 )t 2 2 2 ! or y = y0 − (2.74 m/s)t − (4.90 m/s2)t2 .

(d) Since the ball hits the ground at t = 3.00 s, the x-coordinate at the landing site is xlanding = x|t=3.00s = (7.52 m/s)(3.00 s) = 22.6 m (e)

Since y = 0 when the ball reaches the ground at t = 3.00 s, the result of (c) above gives m⎞ ⎛ m⎞ ⎤ m⎞ m⎞ ⎡ ⎛ ⎛ ⎛ y0 = ⎢ y + ⎜ 2.74 ⎟ t + ⎜ 4.90 2 ⎟ t 2 ⎥ = 0 + ⎜ 2.74 ⎟ (3.00 s) + ⎜ 4.90 2 ⎟ (3.00 s)2 ⎝ ⎝ s⎠ ⎝ s ⎠ ⎦ t=3.00 s s⎠ s ⎠ ⎣ ⎝ !

or (f)

y0 = 52.3 m.

When the ball has a vertical displacement of Δy = −10.0 m it will be moving downward with a velocity given by !vy = v0y + 2ay (Δy) as 2

2 v = − v0y + 2ay (Δy) = − (−2.74 m/s 2 ) + 2(−9.80 m/s 2 )(−10.0 m) = −14.3 m/s ! y

Topic 3

113

The elapsed time at this point is then

t= ! 3.15

vy − v0y ay

−14.3 m/s − (−2.74 m/s) = 1.18 s −9.80 m/s 2

The speed of the car when it reaches the edge of the cliff is

v = v02 + 2a(Δx) = 0 + 2(4.00 m/s 2 )(50.0 m) = 20.0 m/s ! Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as it reaches the water is 2 v y = − v0y + 2ay (Δy)

= − [−(20.0 m/s)sin 24.0°]2 + 2(−9.80 m/s 2 )(−30.0 m) = −25.6 m/s (b) The time of flight is

t= ! (a)

vy − v0y ay

−25.6 m/s − [−(20.0 m/s)sin 24.0°] = 1.78 s −9.80 m/s 2

The horizontal displacement of the car during this time is Δx = v0xt = [(20.0 m/s)cos 24.0°](1.78 s) = 32.5 m

3.16

The initial velocity components of the projectile are v0x = (300 m/s)cos 55.0° = 172 m/s and v0y = (300 m/s)sin 55.0° = 246 m/s while the constant acceleration components are ax = 0 and ay = −g = −9.80

Topic 3

114

m/s2. The coordinates of where the shell strikes the mountain at t = 42.0 s are

1 x = v0xt + axt 2 = (172 m/s)(42.0 s) + 0 = 7.22 × 103 m 2 = 7.22 km and 1 y = v0yt + ayt 2 2 1 = (246 m/s)(42.0 s) + (−9.80 m/s 2 )(42.0 s)2 = 1.69 × 103 m 2 = 1.69 km

3.17

(a)

At the highest point of the trajectory, the projectile is moving horizontally with velocity components of vy = 0 and vx = v0x = v0 cosθ = (60.0 m/s)cos 30.0° = 52.0 m/s

(b) The horizontal displacement is Δx = v0xt = (52.0 m/s)(4.00 s) = 208 m

1 2 and, from Δy = (v0 sin θ )t + ayt the vertical displacement is 2 !

1 Δy = (60.0 m/s)(sin 30.0°)(4.00 s) + (−9.80 m/s 2 )(4.00 s)2 = 41.6 m 2 The straight line distance is d = (Δx)2 + (Δy)2 = (208 m)2 + (41.6 m)2 = 212 m !

Topic 3

3.18

115

The components of the initial velocity are v0x = (40.0 m/s)cos 30.0° = 34.6 m/s and v0y = (40.0 m/s)sin 30.0° = 20.0 m/s

The time for the water to reach the building is Δx 50.0 m t= = = 1.44 s v 34.6 m 0x !

1 2 The height of the water at this time is h = Δy = v0yt + ayt , or 2 ! 1 h = (20.0 m/s)(1.44 s) + (−9.80 m/s 2 )(1.44 s)2 = 18.6 m 2 ! 3.19

(a)

The horizontal displacement of the ball at time t is Δx = vxt = (v0 cos 53.0°)t Thus, if the ball travels 24.0 m horizontally in 2.20 s, the initial speed of the ball must be Δx 24.0 m v0 = = = 18.1 m/s t ⋅cos 53.0° (2.20 s)cos 53.0° !

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116

(b) The vertical displacement of the ball at the instant it passes over the top of the wall (at t = 2.20 s, and 24.0 m horizontally from the launch

1 2 point) is given by Δy = v0yt + ayt , with !v0y = v0 sin 53.0° , as 2 ! 1 Δy = (18.1 m/s)sin 53.0°(2.20 s) + (−9.80 m/s 2 )(2.20 s)2 = 8.09 m 2 ! Thus, the ball clears the top of the wall by: d = 8.09 m − 7.00 m = 1.09 m. (c)

The times when the ball is 6.00 m above ground level are found from

1 Δy = v0yt + ayt 2 , using Δy = 6.00 m, v0y = v0 sin 53.0°, and ay = −g: 2 ! 6.00 m = [(18.1 m/s)sin 53.0°]t − (4.90 m/s2)t2 or

(4.90)t2 − (14.5 s)t + (6.00 s2) = 0

This has solutions of t = 0.497 s and t = 2.46 s. The first solution is when the ball passes the 6.00 m level on the way up and the second solution is the time when it lands on the roof. Note that this is Δt = 2.46 s − 2.20 s = 0.26 s after it crosses over the wall. Thus, the ball © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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lands a horizontal distance beyond the wall given by Δx = vx(Δt) = (v0 cos 53.0°)Δt = (18.1 m/s)cos 53.0°(0.26 s) = 2.8 m 3.20

Identify the cruise ship with the letter C and the patrol boat with the letter P. Identify the stationary reference frame of earth with the letter E. We are

asked to find the components of v CP , the velocity of the cruise ship relative to the patrol boat.

The proper relative velocity equation is v CP = v CE − v PE so that vCP,x = vCE,x − vPE,x and vCP,y = vCE,y − vPE,y Here, the cruise ship is sailing due north at 4.50 m/s so that vCE,x = 0 and vCE,y = 4.50 m/s The patrol boat’s heading is θ = 45.0° north of west at vPE = 5.20 m/s so that

vPE,x = −vPE cosθ =3.68 m/s and vPE,y = vPE sin θ =3.68 m/s

Substitute those values to find (a)

vCP,x = 0 − (−3.68 m/s) = 3.68 m/s and

(b)

vCP,y = (4.50 m/s) − (3.68 m/s) = 0.82 m/s .

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3.21

118

Let east be the positive direction so that the water’s velocity relative to the ground is vWG = +2.50 m/s . When the boat is moving east (with the current), its velocity relative to the water is vBW = +12.0 m/s . When it is moving west (against the current), its velocity relative to the water is

vBW = −12.0 m/s . (a) The boat’s velocity relative to the ground when heading east is

vBG = vBW + vWG = +12.0 m/s + 2.50 m/s = 14.5 m/s Its ground speed is therefore vBG = 14.5 m/s . (b)

The boat’s velocity relative to the ground when heading west is

vBG = vBW + vWG = −12.0 m/s + 2.50 m/s = −9.50 m/s Its ground speed is therefore vBG = 9.50 m/s . 3.22

Choose a reference system with the positive x-axis in the eastward direction and the positive y-axis vertically upward. Then, the velocities of the car and the raindrops relative to Earth are:

 v = 50.0 km/h in!!the! + x0direction , and ! CE  v RE = vrain in! the negative! y-direction. We also know that the velocity of !  !

the rain relative to the car, v RC , is directed downward at θ = 60.0° from the vertical. From Equation 3.16 in the text, these relative velocities are related by © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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   !v RC = v RE − v CE

   !v RE = v CE + v RC

Thus, these relative velocities form a 90°-vector triangle as shown below: We then have  v CE 50.0 km/h tan θ =  = v RE vrain !

and

50.0 km/h vrain = = 28.9 km/h . tan 60.0° !

(a)

 v = ! RC

 2  2 v CE + v RE = (50.0 km/h)2 + (28.9 km/h)2 = 57.7 km/h

and

 v RC = 57.7 km/h at 60.0° west!!of!!vertical

 (b) ! v RE = vrain in the negative y-direction and 3.23

(a)

 v RE = 28.9 km/h vertically!!downward

The jet moves at 3.00 × 102 mi/h due east relative to the air. Choosing a coordinate system with the positive x-direction eastward and the positive y-direction northward, the components of this velocity are

( v ) = 3.00 × 10 mi/h and ( v ) = 0 2

JA x

JA y

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120

(b) The velocity of the air relative to Earth is 1.00 × 102 mi/h at 30.0° north of east. Using the coordinate system adopted in (a) above, the components of this velocity are

(! v AE )x = v AE cosθ = (1.00 × 102 mi/h)cos 30.0° = 86.6 mi/h and (c)

(! v AE )y = v AE sin θ = (1.00 × 102 mi/h)sin 30.0° = 50.0 mi/h

Carefully observe the pattern of the subscripts in Equation 3.11 of the textbook. There, two objects (cars A and B) both move relative to a third object (Earth, E). The velocity of object A relative to object B is given in terms of the velocities of these objects relative to E as

   !v AB = v AE − v BE . In the present case, we have two objects, a jet (J) and the air (A), both moving relative to a third object, Earth (E). Using the same pattern of subscripts as that in Equation 3.11, the velocity of the jet relative to the air is given by

   v JA = v JE − v AE !  (d) From the expression for v JA found in (c) above, the velocity of the jet !    relative to the ground is v JE = v JA + v AE . Its components are then !    (v JE )x = (v JA )x + (v AE )x = 3.00 × 102 mi/h + 86.6 mi/h = 3.87 × 102 mi/h !

   and (v JE )y = (v JA )y + (v AE )y = 0 + 50.0 mi/h = 50.0 mi/h ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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This gives the magnitude and direction of the jet’s motion relative to Earth as

 v JE =

 2  2 v JE x + v JE y = (3.87 × 102 mi/h)2 + (50.0 mi/h)2 = 3.90 × 102 mi/h

and

 ⎛ (v JE )y ⎞ ⎛ 50.0 mi/h ⎞ θ = tan ⎜  = tan −1 ⎜ = 7.36° 2 ⎟ ⎝ 3.87 × 10 mi/h ⎟⎠ ⎝ (v JE )x ⎠ ! −1

 Therefore, v JE = 3.90 × 102 mi/h at 7.36° north!!of!!east . !

3.24

As shown in the figure below, we choose a coordinate system with the yaxis lying at 15.0° east of north (i.e., along the direction of the original  displacement, !d 0 , of the ship from the cutter. During the time t required  for the cutter to intercept the ship, the ship undergoes displacement !dS

(directed at 40.0° east of north and 25.0° east of our y-axis) and the cutter  undergoes displacement !d C (directed at angle θ east of north or angle α

relative to our y-axis).   The magnitude of !d 0 is given to be 20.0 km while the magnitudes of !dS  and !d C will be dS = (26.0 km/h)t and dC = (50.0 km/h)t, respectively.

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(a)

Equating the x-components in the vector triangle formed by the three    displacements !d 0 , !dS , and !d C , we see that

dC sin α = dS sin 25.0° or (50.0 km/h)t sin α = (26.0 km/h)t sin 25.0°

yielding sin α = !

(26.0 km/h)sin 25.0° = 0.220 or α = 12.7° 50.0 km/h

or to intercept the ship, the cutter must steer a course directed at

θ = α + 15.0° = 12.7° + 15.0° = 27.7° east of north (b) To find the time required to intercept the ship, we equate the ycomponents in the vector triangle to find [(50.0 km/h)cos 12.7°]t = 20.0 km + [(26.0 km/h)cos 25.0°]t or

[(48.8 − 23.6) km/h]t = 20.0 km yielding

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123

20.0 km t= = 0.794 h = 47.6 min . (48.8 − 23.6) km/h !

3.25

Choose a reference system with the positive x-axis in the northward direction and the positive y-axis vertically upward. Then, the accelerations of the car and the bolt (in free-fall) relative to Earth are:

 a CE = 2.50 m/s 2 in! the + x0direction , and !  2 a ! BE = 9.80 m/s in! the negative y4direction . Similar to Equation 3.11 in the text for relative velocities, these accelerations are related to the

 !

acceleration of the bolt relative to the car, a BC , by

   a = a − a BC BE CE !

   a = a + a BE CE BC !

Thus, these relative accelerations form a 90°-vector triangle as shown below: (a)



We see that !a BC has components of

  (! a BC )x = −(a CE )x = −2.50 m/s 2 and

  (a ) = −(a BE )y = −9.80 m/s 2 ! BC y

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 a BC = 2.50 m/s 2 southward and 9.80 m/s 2 downward

(b) The acceleration of the bolt relative to Earth is that of a freely falling body, namely 9.80 m/s2 downward. (c)

Observers fixed on Earth see the bolt follow a parabolic path with a vertical axis, the same as any freely falling body having a horizontal initial velocity.

3.26

During the trip of duration t, the displacement of the plane relative to the  ground, !d PG , is to have a magnitude of 750 km and be directed due

north. We choose the positive y-axis to be directed northward and the positive x-axis directed eastward. During the trip, the plane’s

  displacement relative to the air !d PA has magnitude d PA = (630 km/h)t ! and is directed at some angle α relative to the y-axis. The displacement of

  the air relative to the ground, !d AG , has magnitude d AG = (35.0 km/h)t ! and is assumed to be at angle β from the y-axis.

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125

   Since these relative displacements are related by !d PA = d PG − d AG , or    d = d + d PA AG , they form a vector triangle as shown above. Equating the ! PG

  x-components in the vector triangle gives − d PA sin α + d AG sin β = 0 , or ! −(630 km/h)⋅t⋅sinα + (35.0 km/h)⋅t⋅sinβ = 0

and

⎛ 35.0 ⎞ sin α = ⎜ sin β ⎝ 630 ⎟⎠ !

[1]

Equating y-components in the vector triangle gives

   d PA cos α + d AG cos β = d PG , or

[(630 km/h)cosα + (35.0 km/h)cosβ]t = 750 km/h (a)

[2]

The wind blows toward the south (β = 180°) and is a headwind for the plane (α = 0°). Then, Equation [2] gives 750 km/h = 1.26 h . [630 km/h − 35.0 km/h]t = 750 km/h or t = 595 km/h !

(b) The wind blows northward as a tailwind (α = β = 0°, and Equation [2] yields 750 km/h = 1.13 h . [630 km/h + 35.0 km/h]t = 750 km/h or t = ! 665 km/h

(c)

The wind blows due East, so β = 90.0°. Then Equation [1] requires that

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126

⎛ 35.0 ⎞ sin α = ⎜ sin 90.0° = 0.056 ⎝ 630 ⎟⎠ !

and

α = 3.18°

or the plane must fly 3.18° W of N relative to the air to maintain a due north heading relative to the ground. Finally, Equation [2] for this case gives [(630 km/h)cos 3.18° + (35.0 km/h)cos 90.0°]t = 750 km/h or

3.27

(a)

750 km/h t= = 1.19 h . (629 km/h + 0) !

If the salmon (a projectile) is to have vy = 0 when Δy = +1.50 m, the required initial velocity in the vertical direction is given by 2 v 2 = v0y + 2ay Δy as ! y

v0y = + vy2 − 2ay Δy = 0 − 2(−9.80 m/s 2 )(+1.50 m) = 5.42 m/s !

The elapsed time for the upward flight will be

Δt = !

vy − v0y ay

0 − 5.42 m/s = 0.553 s −9.80 m/s 2

If the horizontal displacement at this time is to be Δx = +1.00 m, the required constant horizontal component of the salmon’s velocity must be

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127

Δx 1.00 m v0x = = = 1.81 m/s Δt 0.553 s !

(b) The speed with which the salmon must leave the water is then 2 2 v = v0x + v0y = (1.81 m/s)2 + (5.42 m/s)2 = 5.71 m/s ! 0

Yes, since v0 < 6.26 m/s, the salmon is capable of making this jump. 3.28

(a)

The bomb starts its fall with v0y = 0 and v0x = vplane = 275 m/s. Choosing the origin at the location of the plane when the bomb is released and upward as positive, the y-coordinate of the bomb at ground level is y = −h = −3.00 × 103 m. The time required for the bomb to fall is given

1 2 1 2 by y = y0 + v0yt + ayt as 0 = y0 + 0 + (−g)tfall or tfall = 2y0 /g . ! 2 2 ! ! With ax = 0, the horizontal distance the bomb travels during this time is

d = v0xtfall = v0x

2y0 2(3.00 × 103 m) = (275 m/s) g 9.80 m/s 2

= 6.80 × 103 m = 6.80 km (b) While the bomb is falling, the plane travels in the same horizontal direction with the same constant horizontal speed, vx = v0x = vplane, as the bomb. Thus, the plane remains directly above the bomb as the

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128

bomb falls to the ground. When impact occurs, the plane is directly over the impact point, at an altitude of 3.00 km. (c)

The angle, measured in the forward direction from the vertical, at which the bombsight must have been set is

⎛ 6.80 km ⎞ ⎛ d⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ = 66.2° ⎝ h⎠ ⎝ 3.00 km ⎟⎠ ! 3.29

(a)

Both the student (S) and the water (W) move relative to Earth (E). The velocity of the student relative to the water is given by

     v = v − v v and v , where SW SE WE SE WE are the velocities of the student ! ! relative to Earth and the water relative to Earth, respectively. If we

 !

choose downstream as the positive direction, then v WE = +0.500 m/s ,

 !v SW = −1.20 m/s when the student is going up stream, and  v ! SW = +1.20 m/s when the student moves downstream. The velocity of the student relative to Earth for each leg of the trip is

(! v SE )upstream = v WE + ( v SW )upstream = 0.500 m/s + (−1.20 m/s) = −0.700 m/s and

(v! )

SE downstream

( )

! ! = v WE + v SW

downstream

= 0.500 m/s + (+1.20 m/s) = +1.70 m/s

The distance (measured relative to Earth) for each leg of the trip is d = © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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1.00 km = 1.00 × 103 m. The times required for each of the two legs are d 1.00 × 103 m tupstream =  = = 1.43 × 103 s v SE upstream 0.700 m/s !

and

d 1.00 × 103 m tdownstream =  = = 5.88 × 102 s v SE downstream 0.700 m/s !

so the time for the total trip is ttotal = tupstream + tdownstream = 1.43 × 103 s + 5.88 × 102 s = 2.02 × 103 s . !

 (b) If the water had been still v WE = 0 , the speed of the student relative !

(

)

to Earth would have been the same for each leg of the trip,

   v SE = v SE upstream = v SE downstream = 1.20 m/s . In this case, the time for

each leg and the total time would have been

d 1.00 × 103 m/s tleg = ! = = 8.33 × 102 s, and ttotal = 2tleg = 1.67 × 103 s v SE 1.20 m/s (c)

The time savings going downstream with the current is always less than the extra time required to go the same distance against the current.

3.30

(a)

The speed of the student relative to shore is vup = v − vs while swimming upstream and vdown = v + vs while swimming downstream. The time required to travel distance d upstream is then

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130

d d tup = = vup v − vs ! (b) The time required to swim the same distance d downstream is

d d tdown = = vdown v + vs ! (c)

The total time for the trip is therefore ta = tup + tdown = =

d d d(v + vs ) + d(v − vs ) + = v − vs v + vs (v − vs )(v + vs )

2dv 2d/v = 2 v − vs 1 − vs2 /v 2 2

(d) In still water, vs = 0 and the time for the complete trip is seen to be

tb = ta |vs =0 =

(e)

2d / v 2d = 2 1− 0 / v v

2dv 2d 2dv = 2 and that ta = 2 Note that tb = . Thus, when there is a v − vs2 v v ! !

current (vs > 0), it is always true that ta > tb . 3.31

Choose the positive direction to be the direction of each car’s motion relative to Earth. The velocity of the faster car relative to the slower car is

 !



 !

given by v FS = v FE − v SE , where v FE = +60.0 km/h is the velocity of the

 !

faster car relative to Earth, and v SE = 40.0 km/h is the velocity of the slower car relative to Earth. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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 !

Thus, v FS = +60.0 km/h − 40.0 km/h = +20.0 km/h and the time required for the faster car to move 100 m (0.100 km) closer to the slower car is

3.32

(a)

d vFS

⎛ 3 600 s ⎞ = 5.00 × 10−3 h ⎜ ⎟ = 18.0 s 20.0 km/h ⎝ 1h ⎠ 0.100 km

The time required for the woman, traveling at constant speed v1 relative to the ground, to travel distance L relative to the ground is twoman = L/v1 .

(b) With both the walkway (W) and the man (M) moving relative to Earth (E), we know that the velocity of the man relative to the

 !



moving walkway is v MW = v ME − v WE . His velocity relative to Earth is

 !



then v ME = v MW + v WE . Since all of these velocities are in the same    direction, his speed relative to Earth is v ME = v MW + v WE = v2 + v1 . !

The time required for the man to travel distance L relative to the ground is then

L L tman =  = v ME v1 + v2 ! 3.33

(a)

To find the baseball’s time of flight, consider its motion along the yaxis. The initial y-component of velocity is v0 y = v0 sin θ 0 = ( 3.75 m/s) sin ( −35.0° ) = −2.15 m/s and, because the baseball is

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132

moving down when it hits the ground, its final velocity has vy < 0 so that v y2 = v02y − 2gΔy v y = − ( −2.15 m/s) − 2g ( −2.50 m ) = −7.32 m/s 2

Next, solve for the time of flight, t, using the y-component of the timedependent velocity equation:

v y = v0 y − gt t=

v y − v0 y −g

−7.32 m/s − ( −2.15 ) m/s −9.80 m/s2

= 0.528 s (b)

The baseball’s horizontal velocity component is constant and equal to vx = v0 cosθ0 = ( 3.75 m/s) cos ( −35.0° ) = 3.07 m/s. The horizontal distance, Δx, traveled during time t is therefore

Δx = vxt = ( 3.07 m/s) ( 0.528 s) = 1.62 m 3.34

The vertical displacement from the launch point (top of the building) to 2 the top of the arc may be found from vy2 = v0y + 2ay Δy with vy = 0 at the top !

of the arc. This yields

Δy = !

2 vy2 − v0y

2ay

0 − (12.0 m/s)2 = +7.35 m 2(−9.80 m/s)2

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133

ymax = y0 + Δy = y0 + 7.35 m

(a)

If the origin is chosen at the top of the building, then y0 = 0 and ymax = 7.35 m. Thus, the maximum height above the ground is hmax = 50.0 m + ymax = 50.0 m + 7.35 m = 57.4 m The elapsed time from the point of release to the top of the arc is found from vy = v0y + ayt as

t= !

vy − v0y ay

0 − 12.0 m/s = 1.22 s −9.80 m/s 2

(b) If the origin is chosen at the base of the building (ground level), then y0 = +50.0 m and hmax = ymax, giving hmax = y0 + Δy = 50.0 m + 7.35 m = 57.4 m The calculation for the time required to reach maximum height is exactly the same as that given above. Thus, t = 1.22 s.

Topic 3

3.35

134

!AL = v1t = (90.0 km/h)(2.50 h) = 225 km !BD = AD − AB = ALcos 40.0° − 80.0 km = 92.4 km From the triangle BLD, BL = !

( BD) + ( DL) or 2

BL = (92.4 km)2 + (ALsin 40.0°)2 = 172 km ! Since Car 2 travels this distance in 2.50 h, its constant speed is 172 km v2 = = 68.8 km/h 2.50 h !

3.36

The cup leaves the counter with initial velocity components of (v0x = vi, v0y = 0), and has acceleration components of (ax = 0, ay = −g) while in flight.

(a)

1 2 Applying Δy = v0yt + ayt from when the cup leaves the counter 2 ! (−g) 2 t so the time of the fall is until it reaches the floor gives −h = 0 + 2 !

t = 2h/g .

(b) If the cup travels a horizontal distance d while falling to the floor, Δx

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135

Δx d g . = v0xt gives vi = v0x = t = 2h/g or vi = d 2h ! !

(c)

The components of the cup’s velocity just before hitting the floor are

2h g = − 2gh . vx = v0x = vi = d and vy = v0y + ayt = 0 − g g 2h ! !

Thus, the total speed at this point is v = v + v = ! 2 x

2 y

d2 g + 2gh . 2h

(d) The direction of the cup’s motion just before it hits the floor is

⎛ ⎛ 2h ⎞ ⎞ ⎛ − 2gh ⎞ ⎛ vy ⎞ ⎛ −2h ⎞ −1 −1 −1 θ = tan ⎜ ⎟ = tan ⎜ = tan ⎜ 2 g h ⎜ ⎟ ⎟ = tan −1 ⎜ ⎟ ⎝ d ⎟⎠ ⎝ vx ⎠ ⎜⎝ d ⎝ d g/2h ⎠ ⎝ g ⎠ ⎟⎠ ! −1

3.37

2 2 The pea’s speed when it hits the ceiling is v = vx + v y , where vx and vy are

unknowns to be found.

Just after being launched, v0 = 8.25 m/s and θ0 = 75.0° so that the (constant) horizontal component of its velocity is:

vx = v0 cosθ 0 = 2.14 m/s Use the time-independent kinematic equation for the y-component to find

v y2 at impact: v y2 = v02 sin 2 θ 0 − 2gΔy → v y2 = 24.3 m 2 /s2 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Substitute those values to find the speed when the pea hits the ceiling:

v = vx2 + v y2 = 5.37 m/s 3.38

Taking downstream as the positive direction, the velocity of the water

 !

relative to shore is v WS = +vWS , where vWS is the speed of the flowing water. Also, if vCW is the common speed of the two canoes relative to the water, their velocities relative to the water are

 (! v CW )downstream = +vCW

and

 (v CW )upstream = −vCW !

The velocity of a canoe relative to the water can also be expressed as

   v = v − v CW CS WS . Applying this to the canoe moving downstream gives ! +vCW = +2.9 m/s − vWS

[1]

and for the canoe going upstream −vCW = −1.2 m/s − vWS (a)

[2]

Adding Equations [1] and [2] gives 2vWS = 1.7 m/s,

vWS = 0.85 m/s

(b) Subtracting Equation [2] from [1] yields 2vCW = 4.1 m/s, 3.39

vCW = 2.1 m/s

The distance, s, moved in the first 3.00 seconds is given by

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137

1 1 s = v0t + at 2 = (100 m/s)(3.00 s) + (30.0 m/s 2 )(3.00 s)2 = 435 m 2 2 ! Choosing the origin at the point where the rocket was launched, the coordinates of the rocket at the end of powered flight are x1 = s(cos 53.0°) = 262 m

and

y1 = s(sin 53.0°) = 347 m

The speed of the rocket at the end of powered flight is v1 = v0 + at + 100 m/s + (30.0 m/s2)(3.00 s) = 190 m/s so the initial velocity components for the free-fall phase of the flight are v0x = v1 cos 53.0° = 114 m/s (a)

and

v0y = v1 sin 53.0° = 152 m/s

When the rocket is at maximum altitude, vy = 0. The rise time during the free-fall phase can be found from vy = v0y + ayt as

trise = !

0 − v0y ay

0 − 152 m = 15.5 s −9.80 m/s 2

The vertical displacement occurring during this time is

⎛ vy + v0y ⎞ ⎛ 0 + 152 m/s ⎞ 3 Δy = ⎜ trise = ⎜ ⎟⎠ (15.5 s) = 1.18 × 10 m ⎟ ⎝ ⎝ 2 ⎠ 2 ! The maximum altitude reached is then H = y1 + Δy = 347 m + 1.18 × 103 m = 1.53 × 103 m

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(b) After reaching the top of the arc, the rocket falls 1.53 × 103 m to the ground, starting with zero vertical velocity (v0y = 0). The time for this

1 2 fall is found from Δy = v0yt + ayt as 2 !

tfall = !

2(Δy) 2(−1.53 × 103 m) = = 17.7 s ay −9.80 m/s 2

The total time of flight is t = tpowered + trise + tfall = (3.00 + 15.5 + 17.7) s = 36.2 s (c)

The free-fall phase of the flight lasts for t2 = trise + tfall = (15.5 + 17.7) s = 33.2 s The horizontal displacement occurring during this time is Δx = v0xt2 = (114 m/s)(33.2 s) = 3.78 × 103 m and the full horizontal range is R = x1 + Δx = 262 m + 3.78 × 103 m = 4.04 × 103 m

3.40

(a)

Since the can returns to the same spot on the truck bed that it was thrown from, the can must have zero horizontal velocity relative to the truck. This means that, in the reference frame of the truck, the can was thrown vertically upward or at 0° to the vertical.

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(b) While the can was in the air, the truck moved 16.0 m, at a constant speed of 9.50 m/s, relative to the ground. The time of flight of the can was therefore tflight = (16.0 m)/(9.50 m/s). The time for the can to go from the truck bed to the top of its arc was then

1 1 ⎛ 16.0 m ⎞ tup = tflight = ⎜ = 0.842 s 2 2 ⎝ 9.50 m /s ⎟⎠ ! Thus, vy = v0y + ayt (with vy = 0 at t = tup) gives the initial vertical velocity, and hence the initial speed of the can relative to the truck, as v0y = vy − ayt = 0 − (−9.80 m/s2)(0.842 s) = 8.25 m/s. (c)

Since the boy is stationary in the reference frame of the truck, he sees the can go straight upward and fall straight back downward.

(d) In a frame of reference fixed on the ground, the can has a constant horizontal velocity vx = vtruck = 9.50 m/s as it rises upward and falls back to the truck bed. Thus, the can follows a parabolic path opening downward in this reference frame. (e)

The initial velocity of the can as seen by an observer on the ground is 2 2 v = v0x + v0y = (9.50)2 + (8.25)2 m/s = 12.6 m/s ! 0

directed at θ = tan−1(v0y/v0x) = tan−1(8.25/9.50) = 41.0° above the

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horizontal

 v 0 = 12.6 m/s at!!41.0°!!above!!the!!horizontal!!eastward!!line . !

3.41

(a)

1 2 The time of flight is found from Δy = v0yt + ayt with Δy = 0, as t = 2 ! 2v0y/g.

This gives the range as R = v0xt = !

2v0x v0y g

On Earth this becomes REarth = 2v0xv0y/gEarth and on the Moon, RMoon = 2v0xv0y/gMoon. Dividing RMoon by REarth, we find RMoon = (gEarth/gMoon)REarth. With gMoon = gEarth/6, this gives RMoon = 6REarth = 6(3.0 m) = 18 m. (b) Similarly, RMars = (gEarth/gMars)REarth = 3.0 m/0.38 = 7.9 m. 3.42

(a)

When a projectile returns to the level it was launched from, the time to reach the top of the arc is one half of the total time of flight. Thus, the elapsed time when the first ball reaches maximum height is t = (3.00 s)/2 = 1.50 s. Also, at this time, vy = 0, and vy = v0y + ayt gives 0 = v0y −(9.80 m/s2)(1.50 s)

v0y = 14.7 m/s

(b) In order for the second ball to reach the same vertical height as the

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first, the second must have the same initial vertical velocity as the first. Thus, we find v0 as

v0y 14.7 m /s v0 = = = 29.4 m /s sin 30.0° 0.500 ! 3.43

(a)

The time for the ball to reach the fence is Δx 130 m 159 m t= = = v0 ! v0x v0 cos 35°

At this time, the ball must be Δy = 21 m − 1.0 m = 20 m above its

1 launch position, so Δy = v0yt + ayt 2 gives 2 ! 2 ⎛ 159 m ⎞ 2 ⎛ 159 m ⎞ 20 m = ( v0 sin 35°) ⎜ ⎟ − (4.90 m /s ) ⎜ ⎝ v0 ⎟⎠ ⎝ v0 ⎠ !

(159 m)sin 35° − 20 m = !

(4.90 m /s 2 )(159 m)2 v02

(4.90 m / s 2 )(159 m)2 = 42 m /s From which, v0 = (159 m)sin 35° − 20 m !

(b) From above, t = (159 m)/v0 = (159 m)/(42 m /s) = 3.8 s ! (c)

When the ball reaches the wall (at t = 3.8 s), vx = v0x = (42 m/s)cos 35° = 34 m/s vy = v0y + ayt = (42 m/s)sin 35° − (9.80 m/s2)(3.8 s) = −13 m/s

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v = vx2 + vy2 = (34 m /s 2 ) + (−13 m /s)2 = 36 m /s !

3.44

The horizontal component of the initial velocity is v0x = v0 cos 40° = 0.766 v0 and the time required for the ball to move 10.0 m horizontally is Δx 10.0 m 13.1 m t= = = v 0.766 v v0 0x 0 !

At this time, the vertical displacement of the ball must be Δy = y − y0 = 3.05 m − 2.00 m = 1.05 m

1 2 Thus, Δy = v0yt + ayt becomes 2 ! 1.05 m = ( v0 sin 40.0°) ! or

1.05 m = 8.42 m − !

which yields v0 =

3.45

13.1 m 1 (13.1 m)2 + (−9.80 m/s 2 ) v0 2 v02

841 m 3 /s 2 v02

841 m 3 /s 2 = 10.7 m/s . 8.42 m − 1.05 m

1 2 The time of flight of the ball is given by Δy = v0yt + ayt , with Δy = 0, as 2 !

1 0 = [(20 m/s)sin 30°]t + (−9.80 m/s 2 )t 2 2 yielding a non-zero solution of t = 2.0 s.

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The horizontal distance the football moves in this time is Δx = v0xt = [(20 m/s)cos 30°](2.0 s) = 35 m. (a)

Since Δx > 20 m, the receiver must run away from the quarterback, in the direction the ball was thrown, if he is to catch the ball.

(b) The receiver has a time of 2.0 s to run a distance of d = Δx − 20 m = 15 m, so the required speed is d 15 m v= = = 7.5 m /s t 2.0 s !

3.46

x = v0xt, so the time may be written as t = x/v0x.

1 2 Thus, y = v0yt − gt becomes 2 !

⎛ x ⎞ 1 ⎛ x ⎞ y = v0y ⎜ − g ⎝ v0x ⎟⎠ 2 ⎜⎝ v0x ⎟⎠ !

or !

⎛ ⎛ v0y ⎞ g ⎞ y = ⎜ − 2 ⎟ x2 + ⎜ x+0 ⎝ 2v0x ⎠ ⎝ v0x ⎟⎠

Note that this result is of the general form y = ax2 + bx + c with

⎛ g ⎞ a = ⎜− 2 ⎟ , ⎝ 2v0x ⎠ 3.47

⎛ v0y ⎞ b=⎜ , ⎝ v0x ⎟⎠

c=0

The venom’s initial velocity has components

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144

v0x = v0 cosθ 0 = ( 3.50 m/s) cos ( 50.0° ) = 2.25 m/s v0 y = v0 sin θ 0 = ( 3.50 m/s) sin ( 50.0° ) = 2.68 m/s

To find the venom’s time of flight, consider its motion along the y-axis. Because the venom is moving down when it hits the ground, its final velocity has vy < 0 so that v y2 = v02y − 2gΔy v y = − ( 2.68 m/s) − 2g ( −0.500 m ) = −4.12 m/s 2

Next, solve for the time of flight, t, using the y-component of the timedependent velocity equation:

v y = v0 y − gt t=

v y − v0 y

−g = 0.694 s

−4.12 m/s − 2.68 m/s −9.80 m/s2

The horizontal distance travelled during time t is

Δx = v0xt = ( 2.25 m/s) ( 0.694 s) = 1.56 m 3.48

For the ball thrown at 45.0°, the time of flight is found from

1 Δy = v0yt + ayt 2 2 !

g ⎛ v ⎞ 0 = ⎜ 0 ⎟ t1 − t12 ⎝ 2⎠ 2 !

which has the single non-zero solution of t1 = !

v0 2 . g

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The horizontal range of this ball is 2 ⎛ v ⎞⎛ v 2⎞ v R1 = v0x t1 = ⎜ 0 ⎟ ⎜ 0 ⎟ = 0 ⎝ 2⎠⎝ g ⎠ g !

Now consider the first arc in the motion of the second ball, started at

1 2 angle θ with initial speed v0. Applied to this arc, Δy = v0yt + ayt becomes 2 ! g 2 0 = (v0 sin θ )t21 − t21 2 ! with non-zero solution t21 = !

2v0 sin θ g

Similarly, the time of flight for the second arc (started at angle θ with initial speed v0/2) of this ball’s motion is found to be

t22 = !

2(v0 / 2)sin θ v0 sin θ = g g

The horizontal displacement of the second ball during the first arc of its motion is

⎛ 2v sin θ ⎞ v02 (2 sin θ cosθ ) v02 sin(2θ ) R21 = v0x t21 = (v0 cosθ ) ⎜ 0 = ⎟⎠ = g g g ⎝ ! Similarly, the horizontal displacement during the second arc of this motion is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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R22 = !

(v0 /2)2 sin(2θ ) 1 v02 sin(2θ ) = g 4 g

The total horizontal distance traveled in the two arcs is then

5 v02 sin(2θ ) R2 = R21 + R22 = 4 g ! (a)

Requiring that the two balls cover the same horizontal distance (that is, requiring that R2 = R1) gives

5 v02 sin(2θ ) v02 = 4 g g ! This reduces to sin(2θ) = 4/5, which yields 2θ = 53.1°, so θ = 26.6° is the required projection angle for the second ball. (b) The total time of flight for the second ball is t2 = t21 + t22 = !

2v0 sin θ v0 sin θ 3v0 sin θ + = g g g

Therefore, the ratio of the times of flight for the two balls is t2 (3v0 sin θ )/g 3 = = sin θ t (v 2)/g 2 1 0 !

With θ = 26.6° as found in (a), this becomes t2 3 = sin(26.6°) = 0.950 t 2 1 !

Topic 3

3.49

147

In order to cross the river in minimum time, the velocity of the boat

 !

relative to the water ( v BW ) must be perpendicular to the banks (and

 !

hence perpendicular to the velocity v WS of the water relative to shore).

 !



 !

The velocity of the boat relative to the water is v BW = v BS − v WS , where v BS is the velocity of the boat relative to shore. Note that this vector equation

 !



 !

can be rewritten as v BS = v BW + v WS . Since v BW and v WS are to be perpendicular in this case, the vector diagram for this equation is a right

 !

triangle with v BS as the hypotenuse. Hence, velocity of the boat relative to shore must have magnitude 2 2 v = vBW + vWS = (12 km/h)2 + (5.0 km/h)2 = 13 km/h ! BS

and be directed at

⎛ 12 km/h ⎞ ⎛v ⎞ θ = tan −1 ⎜ BW ⎟ = tan −1 ⎜ = 67° ⎝ vWS ⎠ ⎝ 5.0 km/h ⎟⎠ ! to the direction of the current in the river (which is the same as the line of the riverbank).

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The minimum time to cross the river is

t= !

width!of!river 1.5 km ⎛ 60 min ⎞ = = 7.5 min vBW 12 km/h ⎜⎝ 1 h ⎟⎠

During this time, the boat drifts downstream a distance of

⎛ 1 h ⎞ ⎛ 103 m ⎞ d = vWS t = (5.0 km/h)(7.5 min) ⎜ = 6.3 × 102 m ⎟ ⎜ ⎟ ⎝ 60 min ⎠ ⎝ 1 km ⎠ ! 3.50

The vertical component of the salmon’s velocity as it leaves the water is v0y = +v0 sin θ = +(6.26 m/s)sin 45.0° = +4.43 m/s When the salmon returns to water level at the end of the leap, the vertical component of velocity will be vy = −v0y = −4.43 m/s. The time the salmon is out of the water is given by

t1 = !

vy − v0y ay

−4.43 m/s − 4.43 m/s = 0.904 s −9.80 m/s 2

The horizontal distance traveled during the leap is L = v0xt1 = (v0 cosθ)t1 = (6.26 m/s)cos 45.0°(0.904 s) = 4.00 m To travel this same distance underwater, at speed v = 3.58 m/s, requires a time of L 4.00 m t2 = = = 1.12 s v 3.58 m/s !

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The average horizontal speed for the full porpoising maneuver is then Δx 2L 2(4.00 m) vav = total = = = 3.95 m/s Δt t + t 0.904 s + 1.12 s total 1 2 !

3.51

The initial velocity components for the daredevil are v0x = v0 cos 45° and v0y = v0 sin 45°, or v 25.0 m/s v0x = v0y = 0 = 2 2 !

The time required to travel 50.0 m horizontally is

Δx (50.0 m) 2 t= = =2 2 s 25.0 m/s ! v0x The vertical displacement of the daredevil at this time, and the proper height above the level of the cannon to place the net, is 1 1 ⎛ 25.0 m/s ⎞ Δy = v0yt + ayt 2 = ⎜ (2 2 s) − (9.80 m/s 2 )(2 2 s)2 = 10.8 m ⎟ ⎝ 2 2 2 ⎠ !

3.52

The three objects to consider in this problem are the raindrops, the person, and the Earth. Label these with the letters R, P, and E, respectively. The rain’s velocity relative to the person and the Earth are related by

! ! ! v RP = v RE − v PE

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150

Relative to the Earth, rain is falling straight down at 7.50 m/s and the person is moving horizontally at 2.25 m/s. Rearranging the previous

equation to v RE = v RP + v PE and drawing the vector sum gives

From this figure,

tan θ =

vPE 2.25 m/s = → θ = 16.7° vRE 7.50 m/s

So the rain makes an angle of 16.7° from the vertical, relative to the jogging person. 3.53

(a)

The horizontal range of a projectile launched from the same height at which it lands is

Δx =

v02 sin 2θ g

→ v0 =

gΔx sin 2θ

Substituting Δx = 2.2 m and θ = 45° gives v0 = (b)

4.6 m/s

Solve for the time of flight using Δx = v0cosθ t:

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151

Δx = v0 cos 45°

2.2 m

( 9.80 m/s )( 2.2 m ) cos 45° 2

= 0.67 s

sin 90°

3.54

(a) Consider the falling water to consist of droplets, each following a projectile trajectory. If the origin is chosen at the level of the pool and directly beneath the end of the channel, the parameters for these projectiles are: x0 = 0

y0 = h = 2.35 m

v0x = 0.75 m/s

v0y = 0

ax = 0

ay = −g

The elapsed time when the droplet reaches the pool is found from

1 g y − y0 = v0yt + ayt 2 as 0 − h = 0 − tP2 2 2 ! !

tP = !

2h g

The distance from the wall where the water lands is then

R = xmax = v0xtP = v0x !

2h 2(2.35 m) = (0.75 m/s) = 0.519 m g 9.80 m/s 2

This space is too narrow for a pedestrian walkway . (b) It is desired to build a model in which linear dimensions, such as the height hmodel and horizontal range of the water Rmodel, are one-twelfth

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152

the corresponding dimensions in the actual waterfall. If vmodel is to be the speed of the water flow in the model, then we would have

()

Rmod el = vmod el tp

model

= vmod el

2hmod el g

g R vmod el = Rmod el = actual 2hmod el 12 !

g 2(hactual / 12)

1 ⎛ g ⎞ vactual Ractual = ⎜ 2hactual ⎟⎠ 12 ⎝ 12

and the needed speed of flow in the model is vmod el = !

3.55

(a)

vactual 0.750 m/s = = 0.217 m/s 12 12

Take the origin at the point where the ball is launched. Then x0 = y0 = 0, and the coordinates of the ball at time t later are

!x = v0xt = (v0 cosθ 0 )t

and

1 ⎛ g⎞ y = v0yt + ayt 2 = (v0 sin θ 0 )t − ⎜ ⎟ t 2 ⎝ 2⎠ 2 !

When the ball lands at x = R = 240 m, the y-coordinate is y = 0 and the elapsed time is found from

⎛ g⎞ 0 = (v0 sin θ 0 )t − ⎜ ⎟ t 2 ⎝ 2⎠ ! for which the non-zero solution is

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153

t= !

2v0 sin θ 0 g

Substituting this time into the equation for the x-coordinate gives

⎛ 2v0 sin θ 0 ⎞ ⎛ v02 ⎞ ⎛ v02 ⎞ x = 240 m = (v0 cosθ 0 ) ⎜ ⎟⎠ = ⎜⎝ g ⎟⎠ (2sin θ 0 cosθ 0 ) = ⎜⎝ g ⎟⎠ sin(2θ 0 ) g ⎝ ! Thus, if v0 = 50.0 m/s, we must have

(240 m)g (240 m)(9.80 m/s 2 ) sin(2θ 0 ) = = = +0.941 v02 (50.0 m/s 2 ) ! with solutions of 2θ0 = 70.2° and 2θ0 = 180° − 70.2° = 109.8°. So, the two possible launch angles are θ0 = 35.1° and θ0 = 54.9° (b) At maximum height, vy = 0, and the elapsed time is given by

tpeak = !

(vy )peak − v0y ay

0 − v0 sin θ 0 −g

or tpeak =

v0 sin θ 0 g

The y-coordinate of the ball at this time will be

⎛ g⎞ 2 ymax = (v0 sin θ 0 )tpeak − ⎜ ⎟ tpeak ⎝ 2⎠ ⎛ v sin θ 0 ⎞ ⎛ g ⎞ v02 sin 2 θ 0 v02 sin 2 θ 0 = (v0 sin θ 0 ) ⎜ 0 −⎜ ⎟ = g ⎟⎠ ⎝ 2 ⎠ g2 2g ⎝ The maximum heights corresponding to the two possible launch angles are

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154

(ymax )1 = ! and

3.56

(50.0 m/s)2 sin 2 (35.1°) = 42.2 m 2(9.80 m/s 2 )

(50.0 m/s)2 sin 2 (54.9°) (ymax )2 = = 85.4 m 2(9.80 m/s 2 ) !

The sand grain’s displacement has components given by Δx = ( v0 cosθ 0 ) t and Δy = ( v0 sin θ 0 ) t − 12 gt 2

To hit the prey, the sand grain’s displacement must equal Δx = 0.038 m and Δy = 0.050 m. The two equations above contain two unknowns: the launch speed v0 and the time of flight t. Solve the x-component for time t and substitute into the y-component:

⎛ Δx ⎞ 1 ⎛ Δx ⎞ Δx t= → Δy = ( v0 sin θ 0 ) ⎜ − 2 g⎜ v0 cosθ 0 ⎝ v0 cosθ 0 ⎟⎠ ⎝ v0 cosθ 0 ⎟⎠ ⎛ Δx ⎞ Δy = Δx tan θ 0 − g ⎜ ⎝ v cosθ ⎟⎠

1 2

Solve this expression for the launch speed v0 to find: g ( Δx ) 2 ( Δx tan θ 0 − Δy ) cos2 θ 0 2

v0 =

Substituting Δx = 0.038 m and Δy = 0.050 m gives a launch speed of v0 =

1.1 m/s .

Topic 3

3.57

155

(a)

1 2 Applying Δy = v0yt + ayt to the vertical motion of the first snowball 2 ! gives

1 0 = [(25.0 m/s)sin70.0°]t1 + (−9.80 m/s 2 )t12 2 ! which has the non-zero solution of t1 = !

2(25.0 m/s)sin70.0° = 4.79 s 9.80 m/s 2

as the time of flight for this snowball. The horizontal displacement this snowball achieves is Δx = v0xt1 = [(25.0 m/s)cos 70.0°](4.79 s) = 41.0 m Now consider the second snowball, also given an initial speed of v0 = 25.0 m/s, thrown at angle θ, and in the air for time t2. Applying

1 Δy = v0yt + ayt 2 to its vertical motion yields 2 ! 1 0 = [(25.0 m /s)sin θ ]t2 + − (9.80 m /s 2 )t22 2 ! which has a non-zero solution of t2 = !

2(25.0 m/s)sin θ = (5.10 s)sin θ 9.80 m/s 2

We require the horizontal range of this snowball be the same as that © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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156

of the first ball, namely Δx = v0xt2 = [(25.0 m/s)cosθ][(5.10 s)sinθ] = 41.0 m This yields the equation 41.0 m sin θ cosθ = = 0.321 (25.0 m/s)(5.10 s) !

From the trigonometric identity sin 2θ = 2sinθ cosθ, this result becomes sin2θ = 2(0.321) = 0.642

2θ = 39.9°

and the required angle of projection for the second snowball is

θ = 20.0° above the horizontal. (b) From above, the time of flight for the first snowball is t1 = 4.79 s and that for the second snowball is t2 = (5.10 s)sinθ = (5.10 s)sin 20.0° = 1.74 s Thus, if they are to arrive simultaneously, the time delay between the first and second snowballs should be Δt = t1 − t2 = 4.79 s − 1.74 s = 3.05 s 3.58

(a)

Let x = 0 be at the football’s initial position. Before being caught, the football covers a distance xball equal to its horizontal range given by

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157

xball =

v02 sin 2θ 0 g

When the receiver catches the football, his position, given by xR = xR0 + vRt, equals the football’s position so that

xball = xR v02 sin 2θ 0 = xR0 + vR t g Substitute the football’s time of flight, t = (2v0 sinθ0 )/g, to obtain a quadratic expression for the football’s initial speed: ⎛ 2v sin θ 0 ⎞ v02 sin 2θ 0 = xR0 + vR ⎜ 0 ⎟⎠ g g ⎝

(sin 2θ ) v − ( 2v sin θ ) v − ( x ) g = 0 0

2 0

Substitute known values and suppress the units for clarity to find:

0.766v02 − 4.65v0 − 98.0 = 0 Applying the quadratic equation with a = 0.766, b = −4.65, and c = −98.0 and choosing the positive root gives

−b ± b 2 − 4ac 4.65 + v0 = = 2a

( −4.65) − 4 (0.766)( −98.0) 2 ( 0.766 ) 2

= 14.8 m/s A check of this value verifies that xball = xR to within roundoff error. (b)

The football’s time of flight is

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158

(c)

2v0 sin θ 0 2 (14.8 m/s ) sin ( 25.0° ) = = 1.28 s g 9.80 m/s 2

(

)

As before, taking x = 0 at the football’s (and the quarterback’s) initial location, the receiver makes the catch at

xR = xR0 + vR t = (10.0 m ) + ( 5.50 m/s) (1.28 s) xR = 17.0 m

3.59

(a)

1 2 First, use Δx = v0xt + axt to find the time for the coyote to travel 70.0 2 ! m, starting from rest with constant acceleration ax = 15.0 m/s2:

2Δx 2(70.0 m) = = 3.06 s ax 15.0 m/s 2

t1 =

The minimum constant speed the roadrunner must have to reach the edge in this time is v=

Δx 70.0 m = = 22.9 m/s t1 3.06 s

(b) The initial velocity of the coyote as it goes over the edge of the cliff is horizontal and equal to v0 = v0x = 0 + axt1 = (15.0 m/s2)(3.06 s) = 45.9 m/s

1 2 From Δy = v0yt + ayt , the time for the coyote to drop 100. m, with v0y 2 !

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159

= 0, is

t2 =

2Δy = ay

2(−100. m) = 4.52 s −9.80 m/s 2

The horizontal displacement of the coyote during his fall is

1 1 Δx = v0xt2 + axt22 = (45.9 m/s)(4.52 s) + (15.0 m/s 2 )(4.52 s)2 2 2 = 3.61 × 102 m

Topic 4

160

Topic 4 Newton’s Laws of Motion

QUICK QUIZZES 4.1

Newton’s second law says that the acceleration of an object is directly proportional to the resultant (or net) force acting on. Recognizing this, consider the given statements one at a time. (a) True – If the resultant force on an object is zero (either because no forces are present or the vector sum of the forces present is zero), the object can still move with constant velocity. (b) False – An object that remains at rest has zero acceleration. However, any number of external forces could be acting on it, provided that the vector sum of these forces is zero. (c) True – When a single force acts, the resultant force cannot be zero and the object must accelerate. (d) True – When an object accelerates, a set containing one or more forces with a non-zero resultant must be acting on it. (e) False – Many external forces could be acting on an object with zero

Topic 4

161

acceleration, provided that the vector sum of these forces is zero. (f) False – If the net force is in the positive x-direction, the acceleration will be in the positive x-direction. However, the velocity of an object does not have to be in the same direction as its acceleration (consider the motion of a projectile). 4.2

Choice (b). The newton is a unit of weight, and the quantity (or mass) of gold that weighs 1 newton is m = (1 N)/g. Because the value of g is smaller on the Moon than on the Earth, someone possessing 1 newton of gold on the Moon has more gold than does a person having 1 newton of gold on Earth.

4.3

(a) False – When on an orbiting space station the astronaut is farther from the center of Earth than when on the surface of the Earth. Thus, the astronaut experiences a reduced gravitational force, but that force is not zero. (b) True – On or above the surface of Earth, the acceleration of gravity is inversely proportional to the square of the distance from the center of Earth. Therefore, when this distance is increased by a factor of 3, the acceleration of gravity decreases by a factor of 9.

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162

(c) False – The acceleration of gravity on the surface of a planet of mass M and volume V is g = Fg/m = GM/R2, where R = [3V/4π]1/3. If the mass and volume are doubled (as when two identical planets coalesce), the radius becomes R′ = [3(2V)/4π]1/3 = 21/3R and the surface gravity becomes g' = GM'/R'2 = G(2M)/(21/3R)2 = 22/3g (d) False – Unlike the weight of an object, its mass is independent of the local acceleration of gravity. Thus, one kilogram of gold contains the same amount of gold on the Moon as it does on Earth. 4.4

Choices (c) and (d). Newton’s third law states that the car and truck will experience equal magnitude (but oppositely directed) forces. Newton’s second law states that acceleration is inversely proportional to mass when the force is constant. Thus, the lower mass vehicle (the car) will experience the greater acceleration.

4.5

Choice (b). Friction forces are always parallel to the surfaces in contact, which, in this case, are the wall and the cover of the book. This tells us that the friction force is either upward or downward. Because the tendency of the book is to fall downward due to gravity, the friction force must be in the upward direction.

Topic 4

4.6

163

Choice (b). The static friction force between the bottom surface of the crate and the surface of the truck bed is the net horizontal force on the crate that causes it to accelerate. It is in the same direction as the acceleration, toward the east.

4.7

Choice (b). It is easier to attach the rope and pull. In this case, there is a component of your applied force that is upward. This reduces the normal force between the sled and the snow. In turn, this reduces the friction force between the sled and the snow, making it easier to move. If you push from behind, with a force having a downward component, the normal force is larger, the friction force is larger, and the sled is harder to move.

4.8

Choice (c). In case (i), the scale records the tension in the rope attached to its right end. The section of rope in the man’s hands has zero acceleration, and hence, zero net force acting on it. This means that the tension in the rope pulling to the left on this section must equal the force F the man exerts toward the right on it. The scale reading in this case will then be F. In case (ii), the person on the left can be modeled as simply holding the rope tightly while the person on the right pulls. Thus, the person on the left is doing the same thing that the wall does in case (i). The resulting scale reading is the same whether there is a wall or a person holding the

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164

left side of the scale. 4.9

Choice (c). The tension in the rope has a vertical component that supports part of the total weight of the woman and sled. Thus, the upward normal force exerted by the ground is less than the total weight.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 4.2

If it has a large mass, it will take a large force to alter its motion even when floating in space. Thus, to avoid injuring himself, he should push it gently toward the storage compartment.

4.4

The coefficient of static friction is larger than that of kinetic friction. To start the box moving, you must counterbalance the maximum static friction force. This force exceeds the kinetic friction force that you must counterbalance to maintain constant velocity of the box once it starts moving.

4.6

(a) The barbell always exerts a downward force on the lifter equal in magnitude to the upward force that she exerts on the barbell. Since the lifter is in equilibrium, the magnitude of the upward force exerted on her by the scale (that is, the scale reading) equals the sum of her weight and the downward force exerted by the barbell. As the

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165

barbell goes through the bottom of the cycle and is being lifted upward, the scale reading exceeds the combined weights of the lifter and the barbell. At the top of the motion and as the barbell is allowed to move back downward, the scale reading is less than the combined weights. (b) If the barbell is moving upward, the lifter can declare she has thrown it just by letting go of it for a moment. Thus, the case is included in the previous answer. 4.8

In order for an object to be in equilibrium, the resultant force acting on it must be zero. Thus, it is not possible for an object to be in equilibrium when a single force of non-zero magnitude acts on it.

4.10

The net force acting on the object decreases as the resistive force increases. Eventually, the resistive force becomes equal to the weight of the object, and the net force goes to zero. In this condition, the object stops accelerating, and the velocity stays constant. The rock has reached its terminal velocity.

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166

4.12

 In the free-body diagrams given above, R represents a force due to air   resistance, T is a force due to the thrust of the rocket engine, n is a

  normal force, f is a friction force, and the forces labeled mg are gravitational forces. 4.14

When a tire is rolling, the point on the tire in contact with the ground is momentarily at rest relative to the ground. Thus, static friction exists between the tire and the ground under these conditions. When the brakes lock, the tires begin to skid over the ground and kinetic friction now exists between tires and the ground. Since the kinetic friction force is less than the maximum static friction force (µk < µs), the friction force tending to slow the car is less with the brakes locked than while the tires continue to roll.

4.16

When the crate is held in equilibrium on the incline as shown in the sketch, Newton’s second law requires that ΣFx = ΣFy = 0 . From

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167

! ! ΣFx = + Fg − fs = 0 , the magnitude of the friction force equals the compox

nent of gravitational force acting down the incline, or choice (e) is correct.

! ! Note that fs = fs

max

= µsn only when the crate is on the verge of starting

to slide.

4.18

! An object in equilibrium has zero acceleration (a = 0) , so both the magnitude and direction of the object’s velocity must be constant. Also, Newton’s second law states that the net force acting on an object in equilibrium is zero. The only untrue statement among the given choices is (d), untrue because the value of the velocity’s constant magnitude need not be zero.

4.20

When the truck accelerates forward, its natural tendency is to slip from beneath the crate, leaving the crate behind. However, friction between the crate and the bed of the truck acts in such a manner as to oppose this relative motion between truck and crate. Thus, the friction force acting on the crate will be in the forward horizontal direction and tend to accelerate the

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168

crate forward. The crate will slide only when the coefficient of static friction is inadequate to prevent slipping. The correct response to this question is (c). 4.22

Choice (a). The two forces for an action-reaction pair which, by Newton’s third law, are equal in magnitude (and opposite in direction).

4.24

Choice (c). The answer depends on the motion. A car traveling at con! stant speed in a straight line has a = 0 and, by Newton’s second law, zero

net force acting on it. A car traveling at constant speed along a curved ! path has a ≠ 0 and, again by Newton’s second law, a nonzero net force

acting on it.

ANSWERS TO EVEN NUMBERED PROBLEMS 4.2

25 N

4.4

See Solution.

4.6

7.4 min

4.8

See Solution.

4.10

3.1 × 102 N

4.12

(a)

798 N at 8.79° to the right of the forward direction

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169

0.266 m/s2

(b) 4.14

(a) !t = 2h/g (d)

(b)

ax = F/m

(c)

Δx = Fh/(mg)

(c)

12.5 m/s

a = (F/m)2 + g 2 !

4.16

1.59 m/s2 at 65.2° N of E

4.18

4.90 m/s2

4.20

23.0 N

4.22

(a)

61.1 N

(b)

3.88 m/s2

4.24

(a)

1.4 m/s2

(b)

0.49 m/s2

4.26

(a)

125 N

(b)

0.425

4.28

0.850

4.30

29.7 m/s2

4.32

84.9 N

4.34

amax = 2.1 m/s2

4.36

613 N

4.38

8.71 N

4.40

w2 = 1.7 × 102 N, α = 61°

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170

4.42

(a)

3.43 m/s2

(b)

3.14 m/s2

4.44

(a)

0.368 m/s2

(b)

1.28 m/s2 down the incline

4.46

(a)

1.78 m/s2

(b)

0.367

(d)

2.67 m/s

4.48

(a)

−35.3 N

(b)

−2.94 m/s2

4.50

(a)

98.6 m

(b)

16.4 m

4.52

(a)

ax = −µkg

(b)

Δx = v02 / ( 2 µk g )

4.54

4.45 m/s2

4.56

(a)

T1 = 2m(g + a), T2 = m(g + a)

(b)

upper string breaks first

(a)

49.0 N

(c)

(d)

24.5 N

4.60

(a)

4.43 m/s2

(b)

53.7 N

4.62

(a)

See Solution.

(b)

ΣFx = F, ΣFy = 0

(c)

ΣFx = F − P, ΣFy = 0

(d)

ΣFx = P, ΣFy = 0

(e)

For m1: F − P = m1a; For m2: P = m2a

4.58

(b)

(c)

9.36 N

(c)

10.9 m

98.0 N

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171

(f) a = F/(m1 + m2);

P = [m2/(m1 + m2)]F

(g) P = [m1/(m 1 + m2)]F, larger because m1 > m2 4.64

(a)

a1 = a2 = 6.53 m/s2

(b)

T = 32.7 N

4.66

(a)

36.8 N

(b)

2.44 m/s2

4.68

(a)

See Solution.

(b)

static friction between blocks

(c)

a = F/4m

4.70

(a)

0.125 m/s2

(b)

39.7 N

(c)

0.235

4.72

21.4 N

4.74

(a)

50.0 N

(b)

µs ≥ 0.500

(c)

25.0 N

4.76

(a)

See Solution.

(b)

55.2°

(c)

167 N

4.78

(a)

1.63 m/s2

(b)

T1 = 57.0 N, T2 = 24.5 N

4.80

1.18 × 103 N upward

4.82

(a)

See Solution.

(b)

No. For θ ≤ θc , f ≤ fs,max = µsn.

(c)

See Solution.

(d) !µk = tan θ c′

4.84

(a)

7.25 × 103 N

(b)

4.86

104 N

(c)

1.22 m

4.57 m/s2

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172

0.408 m/s2 upward

4.88

(a)

4.90

5.10 × 103 N

4.92

(a)

7.1 × 102 N

(d)

6.5 × 102 N

(b)

83.3 N

(b)

8.1 × 102 N

(c)

7.1 × 102 N

PROBLEM SOLUTIONS 4.1

4.2

(

)

⎛ 2 000 lbs ⎞ ⎛ 4.448 N ⎞ w = 2 tons ⎜ = 2 × 10 4 N ⎟⎜ ⎟ ⎜⎝ 1 ton ⎟⎠ ⎝ 1 lb ⎠ From v = v0 + at, the acceleration given to the football is

aav = !

v − v0 10 m/s − 0 = = 50 m/s 2 t 0.20 s

Then, from Newton’s second law, we find Fav = maav = (0.50 kg)(50 m/s2) = 25 N 4.3

(a) ΣFx = max = (6.0 kg)(2.0 m/s2) = 12 N ∑ Fx 12 N = = 3.0 m/s 2 (b) ax = m 4.0 kg !

4.4

(a) Action: The hand exerts a force to the right on the spring. Reaction: The spring exerts an equal magnitude force to the left on the hand.

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173

Action: The wall exerts a force to the left on the spring. Reaction: The spring exerts an equal magnitude force to the right on the wall. Action: Earth exerts a downward gravitational force on the spring. Reaction: The spring exerts an equal magnitude gravitational force upward on the Earth. (b) Action: The handle exerts a force upward to the right on the wagon. Reaction: The wagon exerts an equal magnitude force downward to the left on the handle. Action: Earth exerts an upward contact force on the wagon. Reaction: The wagon exerts an equal magnitude downward contact force on the Earth. Action: Earth exerts a downward gravitational force on the wagon. Reaction: The wagon exerts an equal magnitude gravitational force upward on the Earth. (c) Action: The player exerts a force upward to the left on the ball. Reaction: The ball exerts an equal magnitude force downward to the right on the player. Action: Earth exerts a downward gravitational force on the ball. Reaction: The ball exerts an equal magnitude gravitational force upward on the Earth. (d) Action: M exerts a gravitational force to the right on m. Reaction: m exerts an equal magnitude gravitational force to the left on M.

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174

(e) Action: The charge +Q exerts an electrostatic force to the right on the charge −q. Reaction: The charge −q exerts an equal magnitude electrostatic force to the left on the charge +Q. (f) Action: The magnet exerts a force to the right on the iron. Reaction: The iron exerts an equal magnitude force to the left on the magnet. 4.5

The weight of the bag of sugar on Earth is

⎛ 4.448 N ⎞ wE = mgE = (5.00 lbs) ⎜ = 22.2 N ⎝ 1 lb ⎟⎠ ! If gM is the free-fall acceleration on the surface of the Moon, the ratio of the weight of an object on the Moon to its weight when on Earth is wM/wE = mgM/mgE = gM/gE, so wM = wE(gM/gE). Hence, the weight of the bag of sugar on the Moon is ⎛ 1⎞ wM = (22.2 N) ⎜ ⎟ = 3.70 N ⎝ 6⎠ !

On Jupiter, its weight would be ⎛g ⎞ w J = wE ⎜ J ⎟ = (22.2 N)(2.64) = 58.6 N ⎜⎝ g ⎟⎠ E

The mass is the same at all three locations, and is given by

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175

w (5.00 lb)(4.448 N/lb) m= E = = 2.27 kg 2 g 9.80 m/s E !

4.6

∑ F 7.5 × 105 N a= = = 5.0 × 10−2 m/s 2 7 m 1.5 × 10 kg ! and v = v0 + at gives

4.7

v − v0 a

80 km/h − 0 ⎛ 0.278 m/s ⎞ ⎛ 1 min ⎞ ⎜ ⎟⎜ ⎟ = 7.4 min 5.0 × 10−2 m/s 2 ⎝ 1 km/h ⎠ ⎝ 60 s ⎠

The net force has components: Fx = Ax + Bx + Cx + Dx = 40.0 N + 0 − 70.0 N + 0 = −30.0 N and Fy = Ay + By + Cy + Dy = 0 + 50.0 N + 0 − 90.0 N = −40.0 N. (a) The magnitude of the net force is F = Fx2 + Fy2 =

( −30.0 N ) + ( −40.0 N ) = 50.0 N 2

(b) The net force lies in the third quadrant so its direction is

⎛F ⎞ ⎛ −30.0 km ⎞ θ = tan −1 ⎜ x ⎟ + 180° = tan −1 ⎜ + 180° = 217° ⎝ −40.0 km ⎟⎠ ⎝ Fy ⎠ or 36.9° south of west. 4.8

(a) The sphere has a larger mass than the feather. Hence, the sphere experiences a larger gravitational force Fg = mg than does the feather.

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176

(b) The time of fall is less for the sphere than for the feather. This is because air resistance affects the motion of the feather more than that of the sphere. (c) In a vacuum, the time of fall is the same for the sphere and the feather. In the absence of air resistance, both objects have the free-fall acceleration g. (d) In a vacuum, the total force on the sphere is greater than that on the feather. In the absence of air resistance, the total force is just the gravitational force, and the sphere weighs more than the feather. 4.9

2 (a) From vy2 = v0y + 2ay (Δy) , with Δy = (2/3)(1.50 m) = 1.00 m, we find !

ay =

2 v y2 − v0y

2(Δy)

(6.00 m/s)2 − (3.00 m/s)2 2(1.00 m)

= 13.5 m/s

(b) We apply Newton’s second law to the vertical motion of the fish with F being the upward force exerted by the tail fin. This gives ΣFy = may

⇒

F − mg = may

or F = m(ay + g) = (61.0 kg)(13.5 m/s2 + 9.80 m/s2) = 1.42 × 103 N.

4.10

The acceleration of the bullet is given by a =

v 2 − v02 2(Δx)

(320 m/s)2 − 0 2(0.82 m)

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177

⎡ (320 m/s)2 ⎤ −3 2 ∑ F = ma = (5.0 × 10 kg) Then, ⎢ ⎥ = 3.1 × 10 N . 2(0.82 m) ⎣ ⎦ ! 4.11

(a) From the second law, the acceleration of the boat is

∑F m

2 000 N − 1 800 N 1 000 kg

= 0.200 m/s 2

(b) The distance moved is

1 1 Δx = v0t + at 2 = 0 + (0.200 m/s 2 )(10.0 s)2 = 10.0 m 2 2 (c) The final velocity is v = v0 + at = 0 + 2.00 m/s2 (10.0 s) = 2.00 m/s. 4.12

(a) Choose the positive y-axis in the forward direction. We resolve the forces into their components as

Force

x-component

y-component

400 N

200 N

346 N

450 N

−78.1 N

443 N

Resultant

ΣFx = 122 N

ΣFy = 789 N

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178

The magnitude and direction of the resultant force is

F = !R

( ∑ Fx )2 + ( ∑ Fy ) = 798 N 2

⎛ ∑F ⎞ θ = tan −1 ⎜ x ⎟ = 8.79° to right ⎝ ∑ Fy ⎠ !

of y-axis  Thus, F R = 798 N!at!8.79° to!the!right!of!the!forward!direction !

 (b) The acceleration is in the same direction as !F R and has magnitude F 798 N a= R = = 0.266 m/s 2 m 3 000 kg !

4.13

Taking eastward as the positive x-direction, the average horizontal acceleration of the car is

ax =

v x − v0x Δt

25.0 m/s − 0 5.00 s

= +5.00 m/s 2

Thus, the average horizontal force acting on the car during this 5.00-s period is ΣFx = max = (970 kg)(+5.00 m/s2) = +4.85 × 103 N = 4.85 kN eastward.

Topic 4

4.14

179

(a) With the wind force being horizontal, the only vertical force acting on the object is its own weight, mg. This gives the object a downward acceleration of ∑ Fy −mg ay = = = −g m m !

The time required to undergo a vertical displacement Δy = −h, starting with initial vertical velocity v0y = 0, is found from 1 Δy = v0yt + ayt 2 as 2 !

g −h = 0 − t 2 2 !

2h h

(b) The only horizontal force acting on the object is that due to the wind,

∑ Fx F = so ΣFx = F and the horizontal acceleration will be ax = . m m ! (c) With v0x = 0, the horizontal displacement the object undergoes while 1 2 falling a vertical distance h is given by Δx = v0xt + axt as 2 ! 2

1 ⎛ F ⎞ ⎛ 2h ⎞ Fh Δx = 0 + ⎜ ⎟ ⎜ = ⎟ 2 ⎝ m⎠ ⎝ g ⎠ mg ! (d) The total acceleration of this object while it is falling will be

Topic 4

180

a = ax2 + ay2 = !

4.15

( F/m)2 + ( −g ) = ( F/m)2 + g 2 2

Starting with v0y = 0 and falling 30.0 m to the ground, the velocity of the ball just before it hits is 2 v1 = − v0y + 2ay Δy = − 0 + 2 ( −9.80 m/s 2 ) (−30.0 m) = −24.2 m/s

On the rebound, the ball has vy = 0 after a displacement Δy = +20.0 m. Its velocity as it left the ground must have been

v2 = + v y2 − 2ay Δy = + 0 − 2 ( −9.80 m/s 2 ) (20.0 m) = +19.8 m/s Thus, the average acceleration of the ball during the 2.00-ms contact with the ground was

aav =

v2 − v1 +19.8 m/s − ( −24.2 m/s ) = = +2.20 × 10 4 m/s 2 upward −3 Δt 2.00 × 10 s

The average resultant force acting on the ball during this time interval must have been Fnet = maav = (0.500 kg)(2.20 × 104 m/s2) = +1.10 × 104 N or 4.16

!" F net = 1.10 × 10 4 N upward .

Because the two forces are perpendicular to each other, their resultant is F = (180 N)2 + (390 N)2 = 430 N , at !R

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181

⎛ 390 N ⎞ θ = tan −1 ⎜ = 65.2° N!of!E ⎝ 180 N ⎟⎠ ! Thus,

4.17

F 430 N a= R = = 1.59 m/s 2 or a! = 1.59 m/s 2 at 65.2° N of E . m 270 kg !

(a) Solve the x-component of Newton’s second law for the acceleration, a, to find a=

ΣFx 30.0 N = = 3.75 m/s2 m 8.00 kg

(b) Use the time-dependent velocity equation from kinematics to find the block’s speed after 6.00 s: v = v0 + at

(

)

= 0 + 3.75 m/s2 ( 6.00 s) = 22.5 m/s

4.18

A planet’s gravitational acceleration is given by

gp = G

Mp rp2

Construct the ratio with the gravitational acceleration on Earth to find

gp g

Mp /ME

( r /R ) p

(

2 1 = 22 2

g p = 12 g = 12 9.80 m/s2

)

= 4.90 m/s2

Topic 4

4.19

182

(a) This is Case 1 from Section 4.3. Using the free-body diagram to write the sum of forces in the y-direction, Newton’s second law gives

ΣFy = may n − mg = 0

(

n = mg = (15.0 kg ) 9.80 m/s2

)

= 147 N (b) This is Case 4 from Section 4.3. Using the free-body diagram to write the sum of forces in the tilted y’-direction, Newton’s second law gives

ΣFy' = may' n − mgcosθ = 0

(

)

n = mgcosθ = (15.0 kg ) 9.80 m/s2 cos ( 30.0° ) = 127 N

(c) This is Case 3 from Section 4.3 with the block accelerating upward so that ay = +3.00 m/s2. Using the free-body diagram to write the sum of forces in the y-direction, Newton’s second law gives ΣFy = may n − mg = may

(

)

(

n = mg + may = (15.0 kg ) 9.80 m/s2 + (15.0 kg ) 3.00 m/s2

)

= 192 N ! (d) This is Case 2 from Section 4.3 with the force Fapp directed above the

horizontal so that θ = 30.0°. Using the free-body diagram to write the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 4

183

sum of forces in the y-direction, Newton’s second law gives ΣFy' = may' n − mg + Fapp sin θ = 0

(

)

n = mg − Fapp sin θ = (15.0 kg ) 9.80 m/s2 − (125 N ) sin ( 30.0° ) = 84.5 N

4.20

Four forces act on the crate. Use the free-body diagram to add all the xcomponents and apply Newton’s second law:

ΣFx = max Fapp − f k = ma

Solve for the kinetic friction force, fk , and substitute values to find fk = Fapp − ma fk = (95.0 N) − (60.0 kg)( 1.20 m/s2) fk = 23.0 N 4.21

(a) The car is under constant acceleration as it skids to a stop. To find the acceleration, solve the time-dependent velocity kinematic equation for the acceleration a and substitute v = 0, v0 = +30.0 m/s and t = 5.60 s:

Topic 4

184

v = v0 + at a=

v − v0 0 − 30.0 m/s = t 5.60 s

= −5.36 m/s2 (b) Three forces act on the skidding car. Draw a free-body diagram, identifying kinetic friction, fk , as the only horizontal force. Apply the x-component of Newton’s second law to find the magnitude of fk: ΣFx = max f k = ma

(

= ( 875 kg ) −5.36 m/s2

)

= −4690 N f k = 4690 N

−v 2 − ( 30.0 m/s) Δx = 0 = 2a 2 −5.36 m/s2

(

)

= 84.0 m

4.22

(a) The force of kinetic friction is fk = µkn. From Section 4.3, Case 4, the normal force acting on the student is n = mg cosθ. Substitute this result to find

Topic 4

185

f k = µ k n = µ k mgcosθ

(

)

= ( 0.120 ) ( 60.0 kg ) 9.80 m/s2 cos ( 30.0° ) = 61.1 N

(b) Three forces act on the sliding student. Find her acceleration by drawing a free-body diagram with tilted coordinates and applying the x-component of Newton’s second law:

ΣFx = max mgsin θ − f k = ma

Solve for the acceleration a and substitute fk = µkmg cosθ to find

(

)

a = g ⎡⎣sin θ − µ k cosθ ⎤⎦ = 9.80 m/s2 ⎡⎣sin ( 30.0° ) − ( 0.120 ) cos ( 30.0° ) ⎤⎦ = 3.88 m/s2 (c) To find her speed at the bottom of the slide, apply the timeindependent kinematic equation with v0 = 0:

v 2 = v02 + 2aΔx

(

)

v = 2 3.88 m/s2 ( 20.0 m ) = 12.5 m/s 4.23

(a) Since the crate has constant velocity, ax = ay = 0.

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186

Applying Newton’s second law: ΣFx = F cos 20.0° − fk = max = 0 and

fk = (300 N) cos 20.0° = 282 N

ΣFy = n − F sin 20.0° − w = 0

n = (300 N) sin 20.0° + 1000 N = 1.10 × 103 N fk 282 N = 0.256 . The coefficient of friction is then µk = = 3 n 110 × 10 N !

(b) In this case, ΣFy = n + F sin 20.0° − w = 0

n = w − F sin 20.0° − 897 N The friction force now becomes fk = µkn = (0.256)(897 N) = 230. N. Therefore, ΣFx = F cos 20.0° − fk = max = (w/g)ax and the acceleration is (F cos 20.0° − f k )g w [(300. N)cos 20.0° − 230. N)(9.80 m/s 2 ) = = 0.509 m/s 2 1 000 N

4.24

The figure below shows the forces acting on the block. The incline is tilted at θ = 25°, the mass of the block is m = 5.8 kg, while the applied force pulling the block up the incline is F = 32 N.

Topic 4

187

Because ay = 0 for this block, ΣFy = n − mg cosθ = 0 and the normal force is n = mg cosθ. (a) Since the incline is considered frictionless for this part, we take the friction force to be fk = 0 and find

ΣFx = F − mg sinθ = max

giving

F ax = − gsin θ m !

32 N ax = − ( 9.8 m/s 2 ) sin 25° = 1.4 m/s 2 5.8 kg !

(b) If the coefficient of kinetic friction between the block and the incline is µk, then the friction force is fk = µkn = µkmg cosθ, and ΣFx = F − fk − mg sinθ = F − mg (µk cosθ + sinθ) = max

Thus,

F ax = − g ( µk cos θ + sin θ ) m !

and

32 N ax = − ( 9.8 m/s 2 )[(0.10)cos 25° + sin 25° ] = 0.49 m/s 2 5.8 kg !

Topic 4

4.25

188

Two forces act on the astronaut, resulting in an upward acceleration ay = +72.0 m/s2. As in Section 4.3, Case 4, apply the y-component of Newton’s second law to find

ΣFy = may n − mg = may

(

n = mg + may = m g + ay

(

)

= ( 85.0 kg ) 9.80 m/s2 + 72.0 m/s2

)

= 6.95 × 103 N 4.26

(a) To find the static friction force on the crate, apply the x-component of Newton’s second law with ax = 0. From the free-body diagram, two forces have x-components so that

ΣFx = max Fapp − fs = 0 fs = Fapp = 125 N

(b) The minimum possible value of the coefficient of static friction is found using the relation fs ≤ µsn. From Section 4.3, Case 2 with θ = 0, the normal force acting on the crate is n = mg so that fs ≤ µs mg

µs ≥

fs 125 N = mg ( 30.0 kg ) 9.80 m/s 2

(

)

Topic 4

4.27

189

(a) Following Section 4.3, Case 2 with θ = 35.0°, the normal force on the sled is

ΣFy = may n + Fapp sin θ − mg = 0 n = mg − Fapp sin θ

(

)

= ( 236 kg ) 9.80 m/s 2 − (1240 N ) sin ( 35.0° )

= 1.60 × 103 N (b) When Fapp = 1240 N, the static friction force takes its maximum value given by fs,max = µsn. To find the coefficient of static friction, µs, use the free-body diagram with fs = fs,max and apply the x-component of Newton’s second law: ΣFx = max Fapp cosθ − fs,max = 0 fs,max = µsn = Fapp cosθ

µs =

Fapp cosθ

n = 0.635

(1240 N ) cos ( 35.0°) 1.60 × 103 N

(c) With an applied force of Fapp = 6.20 × 102 N, the sled does not move so that ax = 0 and fs < µsn. To find the static friction force, fs, use the freebody diagram and apply the x-component of Newton’s second law:

Topic 4

190

ΣFx = max Fapp cosθ − fs = 0

(

)

fs = Fapp cosθ = 6.20 × 10 2 N cos ( 35.0° ) = 508 N 4.28

Following Section 4.3, Case 3 with θ = 25.0°, the normal force on the block is ΣFy = may n − mgcosθ = 0

(

)

n = mgcosθ = ( 55.0 kg ) 9.80 m/s 2 cos ( 25.0° ) = 488 N

With an applied downhill force of Fapp = 187 N, the block is on the verge of moving so the static friction force takes its maximum value of fs = fs,max =

µsn.

Using the free-body diagram, apply the tilted x-component of Newton’s second law with ax = 0 to find

ΣFx = max Fapp + mgsin θ − fs = 0 Fapp + mgsin θ − µsn = 0

Solving for µs, the coefficient of static friction, gives

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191

µs =

Fapp + mgsin θ n

(

)

187 N + ( 55.0 kg ) 9.80 m/s 2 sin ( 25.0° ) 488 N

= 0.850 4.29

When the block is on the verge of moving, the static friction force has a magnitude fs = (fs)max = µsn. Since equilibrium still exists and the applied force is 75.0 N, we have ΣFx = 75.0 N − fs = 0

(fs)max = 75.0 N

In this case, the normal force is just the weight of the crate, or n = mg. Thus, the coefficient of static friction is

µs =

( f s )max ( f s )max 75.0 N = = = 0.383 n mg (20.0 kg) ( 9.80 m/s 2 )

After motion exists, the friction force is that of kinetic friction, fk = µkn. Since the crate moves with constant velocity when the applied force is 60.0 N, we find that ΣFx = 60.0 N − fk = 0 or fk = 60.0 N. Therefore, the coefficient of kinetic friction is

µk =

4.30

fk f 60.0 N = k = = 0.306 n mg (20.0 kg) ( 9.80 m/s 2 )

Find an expression for the normal force on the quarter using its freebody diagram and the x-component of Newton’s second law:

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192

ΣFx = max n = ma

Because the coin does not fall down the car’s back wall, ay = 0 and the ycomponent of Newton’s second law gives ΣFy = may fs − mg = 0 fs = mg

To find the minimum required acceleration, amin, substitute fs ≤ µsn with n = ma: fs = mg ≤ µsn → µsn ≥ mg

µs ( ma) ≥ mg a≥

g µs

amin =

g 9.80 m/s 2 = µs 0.330

= 29.7 m/s 2

4.31

The crate does not accelerate perpendicular to the incline. Thus,

!∑ F⊥ = ma⊥ = 0 ⇒ n = F + mgcosθ

Topic 4

193

The net force tending to move the crate down the incline is ΣF⎜⎜ = mg sinθ − fs, where fs is the force of static friction between the crate and the incline. If the crate is in equilibrium, then mg sinθ − fs = 0 so fs = Fg sinθ. But, we also know fs ≤ µsn = µs(F + mg cosθ). Therefore, we may write mg sinθ ≤ µs (F + mg cosθ) or

⎛ sin θ ⎞ ⎛ sin 35.0° ⎞ F ≥ mg ⎜ − cos θ ⎟ = (3.00 kg)(9.80 m/s 2 ) ⎜ − cos 35.0°⎟ = 32.1 N ⎝ 0.300 ⎠ ⎝ µs ⎠ 4.32

The strings make the same angle θ on opposite sides of the vertical, so their tensions must be equal. From the free-body diagram, the vertical component of each tension force is Tcosθ. Applying the y-component of Newton’s second law gives

Topic 4

194

ΣFy = may

2T cosθ − mg = 0 T=

(

(15.0 kg) 9.80 m/s mg = 2cosθ 2cos ( 30.0° )

)

= 84.9 N

4.33

If F = 825 N is the upward force exerted on the man by the scales, the upward acceleration of the man (and hence, the acceleration of the elevator) is

∑ Fy 825 N − mman g 825 N 825 N ay = = = −g= − 9.8 m/s 2 = 1.2 m/s 2 m m m 75 kg man man man ! 4.34

With the truck accelerating in a forward direction on a horizontal roadway, the acceleration of the crate is the same as that of the truck as long as the cord does not break. Applying Newton’s second law to the horizontal motion of the block gives ΣFx = max ⇒

Thus

T = ma or a = T/m

T 68 N amax = max = = 2.1 m/s 2 m 32 kg !

Topic 4

4.35

195

(a) Since the burglar is held in equilibrium, the tension in the vertical cable equals the burglar’s weight of 600 N.

Now, consider the junction in the three cables: ΣFy = 0, giving T2 sin 37.0° − 600 N = 0

600 N T2 = = 997 N!in!the!inclined!cable sin 37.0° !

Also, ΣFx = 0, which yields T2 cos 37.0° − T1 = 0 or

T1 = (997 N) cos 37.0° = 796 N in the horizontal cable

(b) If the left end of the originally horizontal cable was attached to a point higher up the wall, the tension in this cable would then have an upward component. This upward component would support part of the weight of the cat burglar, thus decreasing the tension in the cable on the right. 4.36

m = 1.00 kg and mg = 9.80 N

Topic 4

196

⎛ 0.200 m ⎞ α = tan −1 ⎜ = 0.458° ⎝ 25.0 m ⎟⎠ !

Since ay = 0, this requires that ΣFy = T sin α + T sin α − mg = 0, giving 2T sinα = mg 4.37

9.80 N T= = 613 N 2sin α !

(a) When the acceleration is upward, the total upward force T must exceed the total downward force w = mg = (1 500 kg)(9.80 m/s2) = 1.47 × 104 N. (b) When the velocity is constant, the acceleration is zero. The total upward force T and the total downward force w must be equal in magnitude. (c) If the acceleration is directed downward, the total downward force w must exceed the total upward force T. (d) ΣFy = may ⇒ T = mg + may = (1 500 kg)(9.80 m/s2 + 2.50 m/s2) = 1.85 × 104 N . Yes, T > w. (e) ΣFy = may ⇒ T = mg + may = (1 500 kg)(9.80 m/s2 + 0) = 1.47 × 104 N.

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197

Yes, T > w. (f) ΣFy = may ⇒ T = mg + may = (1 500 kg)(9.80 m/s2 − 1.50 m/s2) = 1.25 × 104 N. Yes, T > w. 4.38

Using the reference axis shown in the sketch below, we see that

ΣFx = T cos 14.0° − T cos 14.0° = 0 and ΣFy = –T sin 14.0° − T sin 14.0° = –2T sin 14.0° Thus, the magnitude of the resultant force exerted on the tooth by the wire brace is

R= !

( ∑ Fx )2 + ( ∑ Fy ) = 0 + (−2T sin 14.0°)2 = 2T sin 14.0° 2

or R = 2(18.0 N) sin 14.0° = 8.71 N 4.39

From ΣFx = 0, T1 cos 30.0° − T2 cos 60.0° = 0 or

T2 = (1.73)T1

[1]

Topic 4

198

The tension in the vertical cable is the full weight of the feeder, or Tvertical = 150 N Then ΣFy = 0 becomes T1 sin 30.0° + (1.73 T1) sin 60.0° − 150 N = 0 which gives T1 = 75.1 N in the right side cable. Finally, Equation [1] above gives T2 = 130 N in the left side cable. 4.40

If the hip exerts no force on the leg, the system must be in equilibrium with the three forces shown in the free-body diagram.

Thus ΣFx = 0 becomes w2 cos α = (110 N) cos 40°

[1]

From ΣFy = 0, we find

Topic 4

199

w2 sin α = 220 N − (110 N) sin 40°

[2]

Dividing Equation [2] by Equation [1] yields

⎛ 220 N − (110 N)sin 40° ⎞ α = tan −1 ⎜ ⎟⎠ = 61° (110 N)cos 40° ⎝ ! Then, from either Equation [1] or [2], w2 = 1.7 × 102 N 4.41

(a) First, we consider the glider, jet, and connecting rope to be a single unit having mass mtotal = 276 kg + 1950 kg = 2 226 kg n1 mtotal

F (thrust)

(glider + jet + rope)

mtotal g

For this system, the tension in the rope is an internal force and is not included in an application of Newton’s second law. Applying the second law to the horizontal motion of this combined system gives ΣFx = mtotalax ⇒ F = (2 226 kg)(2.20 m/s2) = 4.90 × 103 N = 4.90 kN. (b) To determine the tension in the rope connecting the glider and the jet, we consider a system consisting of the glider alone. For this

Topic 4

200

system, the rope is an external agent and the tension force it exerts on our system (glider) is included in a second law calculation. ΣFx = mtotalax ⇒ T = (276 kg)(2.20 m/s2) = 607 N. n2

mglider

mglider g

4.42

(a) The free-body diagram of the crate is shown below. Since the crate has no vertical acceleration (ay = 0), we see that ΣFy = may

⇒

n − mg = 0

n = mg

The only horizontal force present is the friction force exerted on the crate by the truck bed. Thus, ΣFx = max

⇒

f = ma

If the crate is not to slip (i.e., the static case is to prevail), it is necessary that the required friction force not exceed the maximum

Topic 4

201

possible static friction force, (fs)max = µsn. From this, we find the maximum allowable acceleration as

amax =

f max ( f y )max µsn µs (mg) = = = m m m m

= µs g = (0.350)(9.80 m/s 2 ) = 3.43 m/s 2 (b) Once slipping has started, the kinetic friction case prevails and f = fk =

µkn. The acceleration of the crate in this case will be f µ n µ (mg) a= = k = k = µk g = (0.320)(9.80 m/s 2 ) = 3.14 m/s 2 m ! m m

4.43

When the load is on the verge of sliding forward on the bed of the slowing truck, the rearward-directed static friction force has its maximum value (fs)max = µsn = µsmloadg.

Since slipping is not yet occurring, this single horizontal force must give the load an acceleration equal to that the truck. Thus, ΣFx = max ⇒ − µs mload g = mload (atruck )max or (atruck)max = −µsg. (a) If slipping is to be avoided, the maximum allowable rearward

Topic 4

202

acceleration of the truck is seen to be (atruck)max = −µsg, and 2 2 !vx = v0x + 2ax ( Δx ) gives the minimum stopping distance as

2 2 0 − v0x v0x (Δx)min = = 2 ( atruck )max 2 µs g !

If v0x = 12 m/s and µs = 0.500, then

(12.0 m/s)2 (Δx)min = = 14.7 m 2 2(0.500)(9.80 m/s ) ! (b) Examining the calculation of part (a) shows that neither mass is necessary. 4.44

 (a) Find the normal force n on the 25.0 kg box: ΣFy = n + (80.0 N) sin 25.0° − 245 N = 0 or

n = 211 N

Now find the friction force, f, as f = µkn = 0.300(211 N) = 63.3 N From the second law, we have ΣFx = ma, or

Topic 4

203

(80.0 N) cos 25.0° − 63.3 N = (25.0 kg)a which yields a = 0.368 m/s2 . (b) When the box is on the incline, ΣFy = n + (80.0 N) sin 25.0° − (245 N) cos 10.0° = 0 giving n = 207 N.

The friction force is f = µkn = 0.300(207 N) = 62.1 N. The net force parallel to the incline is then ΣFx = (80.0 N) cos 25.0° − (245 N) sin 10.0° − 62.1 N = −32.1 N Thus, a =

4.45

ΣFx −32.1 N = = −1.28 m/s 2 or 1.28 m/s2 down the incline. m 25.0 kg

(a) The object will fall so that ma = mg − bv, or a = !

(mg − bv ) , where the m

downward direction is taken as positive.

Topic 4

204

Equilibrium (a = 0) is reached when

mg ( 50 kg ) ( 9.80!m/s ) v = vcriminal = = = 33 m/s b 15!kg/s ! 2

(b) If the initial velocity is less than 33 m/s, then a ≥ 0 and 33 m/s is the largest velocity attained by the object. On the other hand, if the initial velocity is greater than 33 m/s, then a ≤ 0 and 33 m/s is the smallest velocity attained by the object. Note also that if the initial velocity is 33 m/s, then a = 0 and the object continues falling with a constant speed of 33 m/s.

4.46

2(Δx) 2(2.00 m) 1 2 1 2 = 1.78 m/s 2 (a) Δx = v0t + axt = 0 + axt gives ax = 2 = 2 t (1.50 s) 2 2 ! !

(b) Considering forces parallel to the incline, Newton’s second law yields ΣFx = (29.4 N) sin 30.0° − fk = (3.00 kg)(1.78 m/s2) or

fk = 9.36 N.

Topic 4

205

Perpendicular to the plane, we have equilibrium, so ΣFy = n − (29.4 N) cos 30.0° = 0

n = 25.5 N

fk 9.36 N = 0.367 . Then, µk = = n 25.5 N !

v = v02 + 2ax (Δx) = 0 + 2(1.78!m/s 2 )(2.00 m) = 2.67!m/s !

4.47

When a person walks in a forward direction on a level floor, the force propelling them is a forward reaction force equal in magnitude to the rearward static friction force, fs, their foot exerts on the floor. The normal force exerted on the person by the level floor is n = mg. If the person is to have maximum acceleration (in order to travel distance d = 3.00 m in minimum time), the static friction force must have its maximum value, (fs)max = µsn = µsmg, so the maximum acceleration possible is

amax =

( f s )max m

µsmg m

= µs g

The minimum time required to travel distance d (starting from rest) is then given by !Δx = v0t + 12 at as 2

Topic 4

206

1 2 3.00 m = 0 + ( µs g)tmin 2

4.48

tmin =

6.00 m

µs g

(a) If µs = 0.500,

6.00 m tmin = = 1.11 s 0.500(9.80 m/s 2 ) !

(b) If µs = 0.800,

6.00 m tmin = = 0.875 s 0.800(9.80 m/s 2 ) !

(a) The magnitude of the kinetic friction force is fk = µkn. From the freebody and the y-component of Newton’s second law with ay = 0, the normal force is ΣFy = may n − mg = 0 n = mg

so that f k = µ k n = µ k mg

(

= ( 0.300 ) (12.0 kg ) 9.80 m/s 2

)

= 35.3 N

The kinetic friction force is directed opposite the motion, so its xcomponent is negative and

( f ) = − f = −35.3 N k x

(b) Find the block’s acceleration using the free-body diagram and the xcomponent of Newton’s second law, substituting fk = µkmg:

Topic 4

207

ΣFx = max − f k = ma − f k − µ k mg = m m = − µ k g = − ( 0.300 ) 9.80 m/s 2

(

)

= −2.94 m/s 2 (c) Use the time-independent kinematic equation to find the block’s displacement as it slows from an initial velocity of v0 = 8.00 m/s to rest at the final velocity v = 0: v 2 = v02 + 2aΔx − ( 8.00 m/s ) −v 2 Δx = 0 = 2a 2 −2.94 m/s 2

(

)

= 10.9 m

4.49

From the free-body diagram of the person, ΣFx = F1 sin(22.0°) − F2 sin(22.0°) = 0, which gives or F1 = F2 = F Then, ΣFy = 2F cos 22.0° + 85.0 lbs − 170 lbs = 0 yields F = 45.8 lb. (a) Now consider the free-body diagram of a crutch tip.

Topic 4

208

ΣFx = fs − (45.8 lb) sin 22.0°, or

fs = 17.2 lb

ΣFy = ntip − (45.8 lb) cos 22.0° = 0, which gives ntip = 42.5 lb. For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so

fs = (fs)max = µsntip and µs =

fs ntip

17.2 lb 42.5 lb

= 0.405

(b) As found above, the compression force in each crutch is F1 = F2 = F = 45.8 lb. 4.50

(a) The force of friction is found as f = µkn = µk(mg). Choose the positive direction of the x-axis in the direction of motion and apply Newton’s second law. We have ΣFx = −f = max

−f ax = = − µk g m !

From !v = v0 + 2a ( Δx ) , with v = 0, v0 = 50.0 km/h = 13.9 m/s, we find 2

!0 = (13.9 m/s) + 2(− µk g)(Δx) 2

(13.9 m/s)2 Δx = [1] 2 µ g k !

(13.9 m/s)2 = 98.6 m With µk = 0.100, this gives Δx = 2 2(0.100)(9.80 m/s ) ! (b) With µk = 0.600, Equation [1] above gives © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 4

209

(13.9 m/s)2 Δx = = 16.4 m 2 2(0.600)(9.80 m/s ) ! 4.51

The forces on the bucket are the tension in the rope and the weight of the bucket, mg = (5.0 kg)(9.80 m/s2) = 49 N. Choose the positive direction upward and use Newton’s second law: ΣFy = may T − 49 N = (5.0 kg)(3.0 m/s2) T = 64 N

4.52

(a) Since the puck is on a horizontal surface, the normal force is vertical. With ay = 0, we see that ΣFy = may

⇒

n − mg = 0

n = mg

Once the puck leaves the stick, the only horizontal force is a friction force in the negative x-direction (to oppose the motion of the puck). The acceleration of the puck is

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210

ΣF − f − µk n − µk (mg) ax = x = k = = = − µk g m m m m !

(b) Then !vx = v0x + 2ax (Δx) gives the horizontal displacement of the puck 2

before coming to rest as

Δx = ! 4.53

2 vx2 − v0x 0 − v02 v02 = = 2ax 2 ( − µk g ) 2 µk g

(a) Assuming frictionless pulleys, the tension is uniform through the entire length of the rope. Thus, the tension at the point where the rope attaches to the leg is the same as that at the 8.00 kg block. Part (a) of the sketch below gives a force diagram of the suspended block.

Recognizing that the block has zero acceleration, Newton’s second law gives ΣFy = T − mg = 0 or

T = mg(8.00 kg)(9.80 m/s2) = 78.4 N.

(b) Part (b) of the sketch above gives a force diagram of the pulley near the foot. Here, F is the magnitude of the force the foot exerts on the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 4

211

pulley. By Newton’s third law, this is the same as the magnitude of the force the pulley exerts on the foot. Applying the second law gives ΣFx = T + T cos 70.0° − F = max = 0 or 4.54

F = T(1 + cos 70.0°) = (78.4 N)(1 + cos 70.0°) = 105 N.

Mass m1 is less than m2 so it rises as m2 descends with a2 = −a1. From the free-body diagrams for masses m1 and m2, Newton’s second law gives

ΣFy = m2 a2

ΣFy = m1 a1

T − m2 g = m2 ( −a1 )

T − m1 g = m1 a1 T = m1 g + m1 a1

T = m2 g − m2 a1

Equate these two expressions for the tension T and solve for the acceleration a1:

m1 g + m1 a1 = m2 g − m2 a1

(m + m ) a = (m − m ) g 2

a1 =

( m − m ) g = 8.00 kg − 3.00 kg 9.80 m/s ) ( m + m ) 8.00 kg + 3.00 kg ( 2

= 4.45 m/s 2 4.55

(a) Consider the System comprised of masses m1 and m2 to be a single unit with the combined mass (m1 + m2). The tension in the horizontal

Topic 4

212

string acts with equal magnitudes and opposite directions on m1 and m2 so that it cancels and only the external force F appears on the System free-body diagram. Apply Newton’s second law to the System to find the acceleration a:

ΣFx = msystem a

F = ( m1 + m2 ) a a=

F 1.20 × 10 2 N = m1 + m2 16.0 kg + 24.0 kg

= 3.00 m/s 2 4.56

(a) Force diagrams of the two blocks are shown below. Note that each block experiences a downward gravitational force Fg = mg.

Also, each has the same upward acceleration as the elevator, ay = +a. Applying Newton’s second law to the lower block: ΣFy = may ⇒ T2 − Fg = may or T2 = mg + ma = m(g + a). Next, applying Newton’s second law to the upper block: © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 4

213

ΣFy = may or

⇒

T1 − T2 − Fg = may

T1 = T2 + Fg + may = (mg + ma) + mg + ma = 2(mg + ma) = 2T2.

(b) Note that T1 = 2T2, so the upper string breaks first as the acceleration of the system increases. (c) When the upper string breaks, both blocks will be in free-fall with a = −g. Then, using the results of part (a), T2 = m(g + a) = m(g − g) = 0 and T1 = 2T2 = 0. 4.57

We choose reference axes that are parallel to and perpendicular to the incline as shown in the force diagrams below. Since both blocks are in equilibrium, ax = ay = 0 for each block.

Then, applying Newton’s second law to each block gives

Topic 4

214

For Block 1 (mass m): ΣFx = max ⇒ or

T1 = T2 + mg sinθ

–T1 + T2 + mg sinθ = 0 [1]

For Block 2 (mass 2m):

ΣFx = max ⇒

–T2 + 2mg sinθ = 0

T2 = 2mg sinθ

[2]

(a) Substituting Equation [2] into Equation [1] gives T1 = 3mg sinθ . (b) From Equation [2] above, we have T2 = 2mg sinθ . 4.58

In each case, the scale is measuring the tension in the cord connecting to it. This tension can be determined by applying Newton’s second law  (with !a = 0 for equilibrium) to the object attached to the end of this cord.

Topic 4

215

(a) ΣFy = 0

⇒

T − mg = 0

or T = mg = (5.00 kg)(9.80 m/s2) = 49.0 N. (b) The solution to part (a) is also the solution to part (b). (c) Neglecting the mass of the pulley and the cord supporting the two objects of mass m, we find ΣFy = 0

⇒

T − mg − mg = 0

or T = 2mg = 2(5.00 kg)(9.80 m/s2) = 98.0 N. (d) In the figure below, n is the normal force exerted on the block by the frictionless incline.

ΣFx = 0

⇒

T − mg sinθ = 0

or T = mg sinθ = (5.00 kg)(9.80 m/s2) sin 30.0° and T = 24.5 N. 4.59

(a) The resultant external force acting on this system, consisting of all three blocks having a total mass of 6.0 kg, is 42 N directed horizontally toward the right. Thus, the acceleration produced is

Topic 4

216

∑F 42 N a= = = 7.0 m/s 2 horizontally!to!the!right m 6.0 kg !

(b) Draw a free body diagram of the 3.0-kg block and apply Newton’s second law to the horizontal forces acting on this block: ΣFx = max ⇒ 42 N − T = (3.0 kg)(7.0 m/s2), and therefore T = 21 N. (c) The force accelerating the 2.0-kg block is the force exerted on it by the 1.0-kg block. Therefore, this force is given by F = ma = (2.0 kg)(7.0  m/s2), or F = 14 N horizontally to the right.

4.60

Let m1 = 10.0 kg, m2 = 5.00 kg, and θ = 40.0°.

(a) Applying the second law to each object gives m1a = m1g − T

[1]

and

[2]

m2a = T − m2g sinθ

Adding these equations yields

m1a + m2 a = m1 g − T + T − m2 gsin θ !

⎛ m − m2 sin θ ⎞ or a = ⎜ 1 g ⎝ m1 + m2 ⎟⎠

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217

⎛ 10.0 kg − ( 5.00 kg ) sin 40.0° ⎞ a=⎜ 9.80 m/s 2 ) = 4.43 m/s 2 . ( ⎟ 15.0 kg ⎠ ! ⎝

(b) Then, Equation [1] yields T = m1 (g − a) = (10.0 kg)[(9.80 − 4.43) m/s2] = 53.7 N.

4.61

Choosing the +x-direction to be horizontal and forward, the +y-direction vertical and upward, the common acceleration of the car and trailer has components of ax = +2.15 m/s2 and ay = 0. (a) The net force on the car is horizontal and given by (ΣFx)car = F − T = mcarax = (1 000 kg)(2.15 m/s2) = 2.15 × 103 N forward. (b) The net force on the trailer is also horizontal and given by (ΣFx)trailer = +T = mtrailerax = (300 kg)(2.15 m/s2) = 645 N forward. (c) Consider the free-body diagrams of the car and trailer. The only horizontal force acting on the trailer is T = 645 N forward, and this is exerted on the trailer by the car. Newton’s third law then states that the force the trailer exerts on the car is 645 N toward the rear. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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(d) The road exerts two forces on the car. These are F and nc shown in the free-body diagram of the car. From part (a), F = T + 2.15 × 103 N = 645 N + 2.15 × 103 N = +2.80 × 103 N. Also, (ΣFy)car = nc − wc = mcaray = 0, so nc = wc = mcarg = 9.80 × 103 N. The resultant force exerted on the car by the road is then

Rcar = F 2 + nc2 =

(

) ( 2

2.80 × 103 N + 9.80 × 103 N

) = 1.02 × 10 N 2

at θ = tan−1(nc/F) = tan−1(3.50) = 74.1° above the horizontal and forward. Newton’s third law then states that the resultant force exerted on the road by the car is 1.02 × 104 N at 74.1° below the horizontal and rearward.

4.62

(a)

(b) Note that the blocks move on a horizontal surface with ay = 0. Thus, the net vertical force acting on each block and on the combined system of both blocks is zero. The net horizontal force acting on the

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combined system consisting of both m1 and m2 is ΣFx = F − P + P = F. (c) Looking at just m1, ΣFy = 0 as explained above, while ΣFx = F − P. (d) Looking at just m2, we again have ΣFy = 0, while ΣFx = +P. (e) For m1:

ΣFx = max

⇒

F − P = m1a.

For m2:

ΣFx = max

⇒

P = m2a.

(f) Substituting the second of the equations found in (e) above into the first gives m1a = F − P = F − m2a or (m1 + m2)a = F and a = F/(m1 + m2). Then substituting this result into the second equation from (e), we have

⎛ F ⎞ P = m2 a = m2 ⎜ ⎝ m1 + m2 ⎟⎠ !

⎛ m2 ⎞ P=⎜ ⎟⎠ F m + m ⎝ 1 2 !

(g) Realize that applying the force to m2 rather than m1 would have the effect of interchanging the roles of m1 and m2. We may easily find the results for that case by simply interchanging the labels m1 and m2 in the results found in (f) above. This gives a = F/(m2 + m1)

⎛ m1 ⎞ F . (the same result as in the first case) and P = ⎜ ⎝ m2 + m1 ⎟⎠ ! We see that the contact force, P, is larger in this case because m1 > m2 . © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 4

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220

(a) First, consider a system consisting of the two blocks combined, with mass m1 + m2. For this system, the only external horizontal force is the tension in cord A pulling to the right. The tension in cord B is a force one part of our system exerts on another part of our system, and is therefore an internal force.

Applying Newton’s second law to this system (including only external forces, as we should) gives ΣFx = ma

⇒

TA = (m1 + m2)a [1]

Now, consider a system consisting of only !m2 . For this system, the tension in cord B is an external force since it is a force exerted on block 2 by block 1 (which is not part of this system). Applying Newton’s second law to this system gives ΣFx = ma

⇒

TB = m2a

[2]

Comparing Equations [1] and [2], and realizing that the acceleration is the same in both cases (see part (b) below), it is clear that Cord A exerts a larger force on block 1 than cord B exerts on block 2.

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(b) Since cord B connecting the two blocks is taut and unstretchable, the two blocks stay a fixed distance apart, and the velocities of the two blocks must be equal at all times. Thus, the rates at which the velocities of the two blocks change in time are equal, or the two blocks must have equal accelerations. (c) Yes. Block 1 exerts a forward force on Cord B, so Newton's third law tells us that Cord B exerts a force of equal magnitude in the backward direction on Block 1. 4.64

Note that if the cord connecting the two blocks has a fixed length, the accelerations of the blocks must have equal magnitudes, even though they differ in directions. Also, observe from the diagrams below, we choose the positive direction for each block to be in its direction of motion.

First consider the block moving along the horizontal. The only force in

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the direction of movement is T. Thus, ΣFx = max ⇒

T = (5.00 kg)a

[1]

Next consider the block which moves vertically. The forces on it are the tension T and its weight, 98.0 N. ΣFy = may ⇒

98.0 N − T = (10.0 kg)a [2]

Equations [1] and [2] can be solved simultaneously to give

(a) 98.0 N − (5.00 kg)a = (10.0 kg)a or

98.0 N a= = 6.53 m/s 2 . 15.0 kg !

(b) Then, Equation [1] yields T = (5.00 kg)(6.53 m/s2) = 32.7 N. 4.65

The acceleration of the system is found from 1 1 Δy = v0yt + at 2 , or 1.00 m = 0 + a(1.20 s)2 2 2 ! !

which gives a = 1.39 m/s2.

Using the force diagram of m2, the second law gives (5.00 kg)(9.80 m/s2) − T(5.00 kg)(1.39 m/s2) or T = 42.1 N.

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Then applying the second law to the horizontal motion of m1, 42.1 N − f = (10.0 kg)(1.39 m/s2) or f = 28.2 N. f 28.2 N = 0.288 . Because n = m1g = 98.0 N, we have µk = = n 98.0 N !

4.66

First, consider the 3.00-kg rising mass. The forces on it are the tension, T, and its weight, 29.4 N. With the upward direction as positive, the second law becomes T − 29.4 N = (3.00 kg)a

[1]

The forces on the falling 5.00-kg mass are its weight and T, and its acceleration has the same magnitude as that of the rising mass. Choosing the positive direction down for this mass, gives 490.0 N − T = (5.00 kg)a

[2]

(a) Solving Equation [2] for a and substituting into [1] gives

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224

⎛ 3.00 kg ⎞ T − 29.4 N = ⎜ (49.0 N − T) ⎝ 5.00 kg ⎟⎠ !

1.60 T = 58.8 N

and the tension is T = 36.8 N. (b) Equation [2] then gives the acceleration as

49.0 N − 36.8 N 5.00 kg

= 2.44 m/s 2

4.67

(a) The static friction force attempting to prevent motion may reach a maximum value of (fs)max = µsn1 = µsm1g = (0.50)(10 kg)(9.80 m/s2) = 49 N. This exceeds the force attempting to move the system, w2 = m2g = 39 N. Hence, the system remains at rest and the acceleration is a = 0. (b) Once motion begins, the friction force retarding the motion is ! fk = µk n1 = µk m1 g = (0.30)(10 kg)(9.80 m/s ) = 29 N 2

This is less than the force trying to move the system, w2 = m2g. Hence, the system gains speed at the rate

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225

F m g − µk m1 g [4.0 kg − 0.30(10 kg)](9.80 m/s 2 ) a = net = 2 = = 0.70 m/s 2 m m + m 4.0 kg + 10 kg total 1 2 !

4.68

(a)

(b)

 The static friction force, f, exerted on the upper block (mass m) by

the lower block causes the upper block to accelerate toward the right. (c) As long as slipping does not occur between the two blocks, both have acceleration a directed toward the right. Apply Newton’s second law to the horizontal motion of the upper block [see the leftmost diagram in part (a)] to find ΣFx = max

⇒

f = ma

Now, we make use of this result as we apply Newton’s second law to the horizontal motion of the lower block (mass 3m): ΣFx = max

⇒

F − f = (3m)a

Thus, F = ma + 3ma = 4ma or 4.69

F = f + 3ma

a = F/4m.

(a) When the block is resting in equilibrium (ay = 0) on a horizontal floor with only two vertical forces acting on it, the upward normal force exerted on the block by the floor must equal the downward

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226

gravitational force. That is, ΣFy = n − mg = 0

n = mg = 15.0 lb.

⇒

(b) When a 10.0-lb object hangs in equilibrium on one end of the rope, the tension in the rope must be T = 10.0 lb. Thus, the 15.0-lb block has now has three vertical forces acting on it: an upward-directed tension force, T = 10.0 lb, a downward gravitational force of magnitude mg = 15.0 lb and the upward normal force, n, exerted by the floor. Since the block is in equilibrium, ay = 0 and we find ΣFy = n + T − mg = 0

n = mg − T = 15.0 lb − 10.0 lb = 5.00 lb.

(c) If the hanging object on the end of the rope is heavier than the 15.0-lb block, the system will not remain in equilibrium. Rather, the hanging object will accelerate downward, lifting the 15.0-lb block off the floor. As soon as the block leaves the floor, the normal force exerted on the block by the floor becomes zero. 4.70

(a) Both objects start from rest and have accelerations of the same magnitude, a. This magnitude can be determined by applying Δy = v0yt + 12 ayt 2 to the motion of m1: ! a= !

2(Δy) 2(1.00 m) = = 0.125 m/s 2 2 2 t (4.00s)

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(b) Consider the free-body diagram of m1 and apply Newton’s second law: ΣFy = may

⇒

T − m1g = m1(+a)

T = m1(g + a) = (4.00 kg)(9.80 m/s2 + 0.125 m/s2) = 39.7 N.

⇒

n − m2g cosθ = 0

n = m2g cosθ

so n = (9.00 kg)(9.80 m/s2) cos 40.0° = 67.6 N. ΣFx = max Then

⇒

m2g sinθ − T − fk = m2(+a)

fk = m2 (g sinθ − a) − T

or fk = (9.00 kg)[(9.80 m/s2) sin 40.0° − 0.125 m/s2] − 39.7 N = 15.9 N. f 15.9 N = 0.235 . The coefficient of kinetic friction is µk = k = n 67.6 N ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 4

4.71

228

(a) Force diagrams of the two blocks are shown below. Note that each block experiences a downward gravitational force

Fg = (3.50 kg)(9.80 m/s2) = 34.3 N. Also, each has the same upward acceleration as the elevator, in this case ay = +1.60 m/s2. Applying Newton’s second law to the lower block: ΣFy = may ⇒ T2 − Fg = may or

T2 = Fg + may = 34.3 N + (3.50 kg)(1.60 m/s2) = 39.9 N.

Next, applying Newton’s second law to the upper block: ΣFy = may

⇒

T1 − T2 − Fg = may.

or T1 = T2 + Fg + may = 39.9 N + 34.3 N + (3.50 kg)(1.60 m/s2) = 79.8 N. (b) As the acceleration of the system increases, we wish to find the value of ay when the upper string reaches its breaking point (i.e., when T1 = 85.0 N). Making use of the general relationships derived in (a) above gives © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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T1 = T2 + Fg + may = (Fg + may) + Fg + may = 2Fg + 2may

4.72

ay = !

T1 − 2Fg 2m

85.0 N − 2(34.3 N) = 2.34 m/s 2 . 2(3.50 kg)

The acceleration of the ball is found from

v 2 − v02 ( 20.0 m/s ) − 0 a= = = 133 m/s 2 2 ( Δy ) 2(1.50 m) ! 2

From the second law, ΣFy = F − w = may, so F = w + may = 1.47 N + (0.150 kg)(133 m/s2) = 21.4 N.

4.73

(a)

(b) Note that the suspended block on the left, m1, is heavier than that on the right, m3. Thus, if the system overcomes friction and moves, the center block will move right to left with each block’s acceleration © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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being in the directions shown above. First, consider the center block, m2, which has no vertical acceleration. Then, ΣFy = n − m2g = 0

n = m2g(1.00 kg)(9.80 m/s2) = 9.80 N.

This means the friction force is: fk = µkn = (0.350)(9.80 N) = 3.43 N. Assuming the cords do not stretch, the speeds of the three blocks must always be equal. Thus, the magnitudes of the blocks’ accelerations must have a common value, a.    a = a2 = a3 = a ! 1

Taking the indicated direction of the acceleration as the positive direction of motion for each block, we apply Newton’s second law to each block as follows: For m1:

m1g − T1 = m1a or T1 = m1(g − a) = (4.00 kg)(g − a)

[1]

For m2:

T1 − T2 − fk = m2a or T1 − T2 = (1.00 kg)a + 3.43 N

[2]

For m3:

T2 − m3g = m3a

[3]

or T2 = m3(g + a) = (2.00 kg)(g + a)

Substituting Equations [1] and [3] into Equation [2], and solving for a yields

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231

(4.00 kg)(g − a) − (2.00 kg)(g + a) = (1.00 kg)a + 3.43 N

(4.00 kg − 2.00 kg)(9.80!!m/s 2 ) − 3.43 N a= = 2.31 m/s 2 4.00 kg + 2.00 kg + 1.00 kg ! so

  a 1 = 2.31 m/s 2 downward , a 2 = 2.31 m/s 2 to!the!left , ! !

 2 and a 2 = 2.31 m/s upward !

(c) Using this result in Equations [1] and [3] gives the tensions in the two cords as T1 = (4.00 kg)(g − a) = (4.00 kg)(9.80 − 2.31) m/s2 = 30.0 N and T2 = (2.00 kg)(g + a) = (2.00 kg)(9.80 + 2.31) m/s2 = 24.2 N. (d) From the final calculation in part (b), observe that if the friction force had a value of zero (rather than 3.53 N), the acceleration of the system would increase in magnitude. Then, observe from Equations [1] and [3] that this would mean T1 would decrease while T2 would increase. 4.74

(a) For the suspended block, ΣFy = T − 50.0 N = 0, so the tension in the rope is T = 50.0 N. Then, considering the horizontal forces on the 100N block, we find ΣFx = T − fs = 0

fs = T = 50.0 N.

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232

(b) If the system is on the verge of slipping, fs = (fs)max = µsn. Therefore, the minimum acceptable coefficient of friction is f 50.0 N µs = s = = 0.500 n 100 N !

(c) If µk = 0.250, then the friction force acting on the 100-N block is fk = µkn = (0.250)(100 N) = 25.0 N Since the system is to move with constant velocity, the net horizontal force on the 100-N block must be zero, or ΣFx = T − fk = T − 25.0 N = 0. The required tension in the rope is T = 25.0 N. Now, considering the forces acting on the suspended block when it moves with constant velocity, ΣFy = T − w = 0, giving the required weight of this block as w = T = 25.0 N. 4.75

(a) The horizontal component of the resultant force exerted on the light by the cables is Rx = ΣFx = (60.0 N) cos 45.0° − (60.0 N) cos 45.0° = 0

The resultant y-component is

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233

Ry = ΣFy = (60.0 N)(sin 45.0° + (60.0 N) sin 45.0° = 84.9 N Hence, the resultant force is 84.9 N vertically upward. (b) The forces on the traffic light are the weight, directed downward, and the 84.9 N vertically upward force exerted by the cables. Since the light is in equilibrium, the resultant of these forces must be zero. Thus, w = 84.9 N downward. 4.76

(a)

 (b) Since the suitcase moves with constant velocity, a = 0 and ax = ay = 0.

Applying Newton’s second law to the horizontal motion of the suitcase gives ΣFx = max = 0 ⇒ (35.0 N) cosθ − 20.0 N = 0

cos θ =

20.0 N = 0.571 35.0 N

and

θ = 55.2°

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234

ΣFy = may = 0 ⇒ n + (35.0 N) sinθ − w = 0 and n = w − (35.0 N) sinθ = 196 N − (35.0 N) sin 55.2° = 167 N. 4.77

On the level surface, the normal force exerted on the sled by the ice equals the total weight, or n = 600 N. Thus, the friction force is fk = µkn = (0.050 0)(600 N) = 30 N. Hence, Newton’s second law yields ΣFx = −fk = max, or

− f k − f k −(30 N)(9.80 m/s 2 ) ax = = = = −0.490 m/s 2 m w/g 600 N The distance the sled travels on the level surface before coming to rest is

Δx =

4.78

2 v x2 − v0x 0 − (7.00 m/s)2 = = 50.0 m 2ax 2(− 0.490 m/s 2 )

Let m1 = 5.00 kg, m2 = 4.00 kg, and m3 = 3.00 kg. Let T1 be the tension in the string between m1 and m2, and T2 be the tension in the string between m2 and m3. Note that the three objects have the same magnitude acceleration:

   a1 = a directed upward, !a 2 = a 3 = a directed downward. We take the positive direction for each object to be the direction of its acceleration. (a) We may apply Newton’s second law to each of the masses.

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235

for m1:

m1a = T1 − m1g

[1]

for m2:

m2a = T2 + m2g − T1

[2]

for m3:

m3a = m3g − T2

[3]

Adding these equations yields (m1 + m2 + m3)a = (−m1 + m2 + m3)g, so

⎛ 2.00!kg ⎞ ⎛ −m1 + m2 + m3 ⎞ a=⎜ g = 9.80!m/s 2 ) = 1.63!m/s 2 ( ⎟ ⎜ ⎟ ⎝ m1 + m2 + m3 ⎠ ⎝ 12.0!kg ⎠ ! (b) From Equation [1], T1 = m1(a + g) = (5.00 kg)(11.4 m/s2) = 57.0 N, and from Equation [3], T2 = m3(g − a) = (3.00 kg)(8.17 m/s2) = 24.5 N. 4.79

(a) The force that accelerates the box is the friction force between the box and truck. (b) We assume the truck is on level ground. Then, the normal force exerted on the box by the truck equals the weight of the box, n = mg. The maximum acceleration the truck can have before the box slides is found by considering the maximum static friction force the truck bed can exert on the box: (fs)max = µsn = µs (mg) Thus, from Newton’s second law,

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236

amax = 4.80

( f s )max m

µs (mg) m

= µs g = (0.300)(9.80 m/s 2 ) = 2.94 m/s 2

Before she enters the water, the diver is in free-fall with an acceleration of 9.80 m/s2 downward. Taking downward as the positive direction, her velocity when she reaches the water is given by

(

)

v = v02 + 2a(Δy) = 0 + 2 9.80 m/s 2 (10.0 m) = 14.0 m/s

This is also her initial velocity for the 2.00 s after hitting the water. Her average acceleration during this 2.00 s interval is

aav = !

v − v0 0 − (14.0 m/s ) = = −7.00 m/s 2 t 2.00 s

Continuing to take downward as the positive direction, the average upward force by the water is found as ΣFy = Fav + mg = maav, or F = m(aav − g) = (70.0!kg) ⎡⎣( − 7.00!m/s 2 ) − 9.80!m/s 2 ⎤⎦ = −1.18 × 103 !N ! av

or Fav = 1.18 × 103 N upward. 4.81

When an object of mass m is on this frictionless incline, the only force acting parallel to the incline is the parallel component of weight, mg sinθ, directed down the incline. The acceleration is then

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a= !

mgsin θ = gsin θ = (9.80 m/s 2 )sin 35.0° = 5.62 m/s 2 (directed down the m

incline). (a) Taking up the incline as positive, the time for the sled projected up the incline to come to rest is given by t= !

v − v0 0 − 5.00 m/s = = 0.890 s a −5.62 m/s 2

The distance the sled travels up the incline in this time is ⎛ v + v0 ⎞ ⎛ 0 + 5.00!m/s ⎞ Δs = vavt = ⎜ t=⎜ ⎟ ⎟⎠ ( 0.890!s ) = 2.23!m ⎝ 2 ⎠ ⎝ 2 !

(b) The time required for the first sled to return to the bottom of the incline is the same as the time needed to go up, that is, t = 0.890 s. In this time, the second sled must travel down the entire 10.0 m length of the incline. The needed initial velocity is found from 1 Δs = v0t + at 2 as 2 !

Δs at −10.0 m ( −5.62 m/s ) (0.890 s) v0 = − = − = −8.74 m/s t 2 0.890 s 2 ! 2

or 8.74 m/s down the incline.

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4.82

238

(a)

(b) No. In general, the static friction force is less than the maximum value of (fs)max = µsn. It is equal to this maximum value only when the coin is on the verge of slipping, or at the critical angle θc. For θ ≤ θc, f ≤ (fs)max = µsn. (c) Recognize that when the y-axis is chosen perpendicular to the incline as shown above, ay = 0 and we find ΣFy = n − mg cosθ = may = 0

n = mg cosθ

Also, when static conditions still prevail, but the coin is on the verge of slipping, we have ax = 0, θ = θc, and f = (fs)max = µsn = µsmg cosθc. Then, Newton’s second law becomes ΣFx = mg sinθc − µsmg cosθc = max = 0 and µs mg cos θ c = mg sin θ c

yielding µs =

sin θ c = tan θ c . cos θ c

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239

(d) Once the coin starts to slide, kinetic conditions prevail and the friction force is f = fk = µkn = µkmg cosθ. At !θ = θ c′ < θ c , the coin slides with constant velocity, and ax = 0 again. Under these conditions, Newton’s second law gives

!ΣFx = mgsin θ c′ − µk mgcos θ c′ = max = 0 ′ ′ and µs mg cos θ c = mg sin θ c yielding µs =

4.83

sin θ c′ = tan θ c′ . ′ cos θ c

(a) Free-body diagrams for the two blocks are given below. The coefficient of kinetic friction for aluminum on steel is µ1 = 0.47 while that for copper on steel is µ2 = 0.36. Since ay = 0 for each block, n1 = w1

Thus,

and n2 = w2 cos 30.0°

f1 = µ1n1 = 0.47(19.6 N) = 9.2 N

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240

and f2 = µ2n2 = 0.36(58.8 N) cos 30.0° = 18 N. For the aluminum block: ΣFx = max giving

⇒

T − f1 = m(+a) or T = f1 + ma

T = 9.2 N + (2.00 kg)a [1]

For the copper block: ΣFx = max or

⇒

(58.8 N) sin 30.0° − T − 18 N = (6.00 kg)a

11 N − T = (6.00 kg)a

[2]

Substituting Equation [1] into Equation [2] gives 11 N − 9.2 N − (2.00 kg)a = (6.00 kg)a

1.8 N a= = 0.23 m/s 2 . ! 8.00 kg

(b) From Equation [1] above, T = 9.2 N + (2.00 kg)(0.23 m/s2) = 9.7 N. 4.84

(a) In the vertical direction, we have ΣFy = (8 000 N) sin 65.0° − w = may = 0 so

w = (8 000 N) sin 65.0° = 7.25 × 103 N

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241

(b) Along the horizontal, Newton’s second law yields

⎛w ⎞ ΣFx = (8 000 N)cos 65.0° = max = ⎜ ax ⎟ ⎝g ⎠ or ax =

4.85

g [(8 000 N)cos 65.0° ] ( 9.80 m/s 2 ) (8 000 N)cos 65.0° = = 4.57 m/s 2 3 w 7.25 × 10 N

Figure 1 is a free-body diagram for the system consisting of both blocks. The friction forces are f1 = µkn1 = µk(m1g)

and f2 = µk(m2g)

FIGURE 1 For this system, the tension in the connecting rope is an internal force and is not included in second law calculations. The second law gives ΣFx = 50.0 N − f1 − f2 = (m1 + m2)a © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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which reduces to a =

50.0 N − µk g m1 + m2

[1]

Figure 2 gives a free-body diagram of m1 alone. For this system, the tension is an external force and must be included in the second law. We find ΣFx = T − f1 = m1a, or T = m1(a + µkg)

[2]

FIGURE 2 (a) If the surface is frictionless, µk = 0. Then, Equation [1] gives a=

50.0 N 50.0 N −0= = 1.67 m/s 2 m1 + m2 30.0 kg

and Equation [2] yields T = (10.0 kg)(1.67 m/s2 + 0) = 16.7 N. (b) If µk = 0.10, Equation [1] gives the acceleration as a=

50.0 N − (0.10)(9.80 m/s 2 ) = 0.69 m/s 2 30.0 kg

while Equation [2] gives the tension as T = (10.0 kg)[0.69 m/s2 + (0.10)(9.80 m/s2)] = 16.7 N.

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4.86

243

Consider the two force diagrams, one of the penguin alone and one of the combined system consisting of penguin plus sled.

The normal force exerted on the penguin by the sled is n1 = w1 = m1g and the normal force exerted on the combined system by the ground is n2 = wtotal = mtotalg = 130 N The penguin is accelerated forward by the static friction force exerted on it by the sled. When the penguin is on the verge of slipping, this acceleration is

amax = !

(f )

1 max

(

)

µsn1 µs m1 g = = µs g = (0.700)(9.80!m/s 2 ) = 6.86!m/s 2 m1 m1

Since the penguin does not slip on the sled, the combined system must have the same acceleration as the penguin. Applying Newton’s second law to this system gives

Topic 4

244

ΣFx = F − f2 = mtotalamax

⎛w ⎞ F = f2 + mtotal amax = µk (wtotal ) + ⎜ total ⎟ amax ⎝ g ⎠ !

⎛ 130 N ⎞ (6.86 m/s 2 ) = 104 N . which yields F = (0.100)(130 N) + ⎜ 2⎟ ⎝ 9.80 m/s ⎠ 4.87

Choose the positive x-axis to be down the incline and the y-axis perpendicular to this as shown in the free-body diagram of the toy.

The acceleration of the toy then has components of ay = 0, and ax =

Δv x +30.0 m/s = = +5.00 m/s 2 Δt 6.00 s

Applying the second law to the toy gives (a) ΣFx = mg sinθ = max

and

⇒

sinθ = max/mg = ax/g,

2 ⎛ ax ⎞ −1 ⎛ 5.00!m/s ⎞ θ = sin ⎜ ⎟ = sin ⎜ = 30.7° ⎝ 9.80!m/s 2 ⎟⎠ ⎝ g⎠ ! −1

(b) ΣFy = T − mg cosθ = may = 0, or T = mg cosθ = (0.100 kg)(9.80 m/s2) cos 30.7° = 0.843 N.

Topic 4

4.88

245

(a) Consider the first free-body diagram in which the child and the chair are treated as a combined system. The weight of this system is wtotal = 480 N, and its mass is mtotal = !

wtotal = 49.0 kg g

Taking upward as positive, the acceleration of this system is found from Newton’s second law as ΣFy = 2T − wtotal = mtotalay

Thus ay = !

2(250 N) − 480 N = +0.408 m/s 2 or 0.408 m/s2 upward. 49.0 kg

(b) The downward force that the child exerts on the chair has the same magnitude as the upward normal force exerted on the child by the chair. This is found from the free-body diagram of the child alone as ΣFy = T + n − wchild = mchilday

so n = mchilday + wchild − T

Topic 4

246

Hence,

4.89

⎛ 320 N ⎞ n=⎜ (0.408 m/s 2 ) + 320 N − 250 N!=! 83.3 N 2⎟ 9.80 m/s ⎝ ⎠ !

The car’s acceleration while coming to a stop is

v 2 − v02 2(Δx)

0 − (35.0 m/s)2 2(1 000 m)

= −0.613 m/s 2

Thus, the magnitude of the total retarding force acting on the car is

⎛ 8 820 N ⎞ ⎛ w⎞ F = m a =⎜ ⎟ a = ⎜ (0.612 5 m/s 2 ) = 5.51 × 102 N 2⎟ ⎝ g⎠ ⎝ 9.80 m/s ⎠ 4.90

 Let R represent the horizontal force of air resistance. Since the helicopter

and bucket move at constant velocity, ax = ay = 0. The second law then gives

ΣFy = T cos 40.0° − mg = 0

Also,

ΣFx = T sin 40.0° − R = 0

mg T= cos 40.0° !

R = T sin 40.0°

Thus,

Topic 4

247

⎛ mg ⎞ R=⎜ sin 40.0° = (620 kg)(9.80 m/s 2 )tan 40.0° = 5.10 × 103 N ⎟ cos 40.0° ⎝ ⎠ 4.91

Since the board is in equilibrium, ΣFx = 0 and we see that the normal forces must have the same magnitudes on both sides of the board. Also, if the minimum normal forces (compression forces) are being applied, the board is on the verge of slipping and the friction force on each side is f = (fs)max = µsn.

The board is also in equilibrium in the vertical direction, so w ΣFy = 2f − w = 0, or f = 2 !

The minimum compression force needed is then f w 95.5 N n= = = = 72.0 N µ 2 µ 2(0.663) s s !

4.92

First, we will compute the needed accelerations:

Topic 4

248

(1) Before it starts to move:

ay = 0

(2) During the first 0.80 s:

ay = !

vy − v0y t

1.2 m/s − 0 = 1.5 m/s 2 0.80 s

(3) While moving at constant velocity: ay = 0

(4) During the last 1.5 s:

ay = !

vy − v0y t

0 − 1.2 m/s = −0.80 m/s 2 1.5 s

The spring scale reads the normal force the scale exerts on the man. Applying Newton’s second law to the vertical motion of the man gives ΣFy = n − mg = may

n = m(g + ay).

(a) When ay = 0, n = (72 kg)(9.80 m/s2 + 0) = 7.1 × 102 N. (b) When ay = 1.5 m/s2, n = (72 kg)(9.80 m/s2 + 1.5 m/s2) = 8.1 × 102 N. (c) When ay = 0, n = (72 kg)(9.80 m/s2 + 0) = 7.1 × 102 N. (d) When ay = −0.80 m/s2, n = (72 kg)(9.80 m/s2 − 0.80 m/s2) = 6.5 × 102 N.

Topic 5

249

Topic 5 Energy

QUICK QUIZZES 5.1

Choice (c). The work done by the force is W = F(Δx)cosθ, where θ is the angle between the direction of the force and the direction of the displacement (positive x-direction). Thus, the work has its largest positive value in (c) where θ = 0°, the work done in (a) is zero since θ = 90°, the work done in (d) is negative since 90° < θ < 180°, and the work done is most negative in (b) where θ = 180°.

5.2

(a) True. The net work equals the change in kinetic energy, which in this case equals zero because the block slides at constant speed. (b) False. The gravitational force of weight on the block does positive work as the block slides down the ramp. The normal force does zero work and the kinetic friction force does negative work. (c) False. The gravitational force of weight on the block does positive work as the block slides down the ramp.

5.3

Choice (d). All three balls have the same speed the moment they hit the

Topic 5

250

ground because all start with the same kinetic energy and undergo the same total change in gravitational potential energy. 5.4

Choice (c). They both start from rest, so the initial kinetic energy is zero for each of them. They have the same mass and start from the same height, so they have the same initial potential energy. Since neither spends energy overcoming friction, all of their original potential energy will be converted into kinetic energy as they move downward. Thus, they will have equal kinetic energies when they reach the ground.

5.5

The elastic potential energy is PEs = 21 kx where x is the displacement 2

from equilibrium. When the spring is compressed, xcompressed = −1.00 × 10−2 m and when it is stretched, xstretched = 1.00 × 10−2 m. In each case, x2 = 1.00 × 10−4 m2 so that: (a)

(

)

(

)

PEs = 21 kx 2 = 21 ( 225 N/m ) 1.00 × 10−4 m 2 = 1.13 × 10−2 J

2 −4 2 −2 (b) PEs = 21 kx = 21 ( 225 N/m ) 1.00 × 10 m = 1.13 × 10 J

5.6

True. PEs = 21 kx where the spring constant, k, and x2 are both positive.

5.7

(a) The elastic potential energy scales with x2. As the stretch is doubled,

the potential energy increases by 22 = 4. (b) As the stretch is tripled, the potential energy increases by 32 = 9. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 5

5.8

251

Choice (c). The decrease in mechanical energy of the system is fk(Δx). This has a smaller value on the tilted surface for two reasons: (1) the force of kinetic friction fk is smaller because the normal force is smaller, and (2) the displacement Δx is smaller because a component of the gravitational force is pulling on the book in the direction opposite to its velocity.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 5.2

Choice (a), True. By the work-energy theorem, the net work equals the change in kinetic energy. A car traveling at constant speed has zero change in kinetic energy and, therefore, zero net work done on it.

5.4

(a) Kinetic energy is always positive. Mass and speed squared are both positive. (b) Gravitational potential energy can be negative when the object is lower than the chosen reference level.

5.6

(a) The ball initially has gravitational potential energy mghi and zero kinetic energy. Because a small amount of energy is always spent overcoming air resistance and friction in the support as the ball swings, it will come to rest (i.e., have zero kinetic energy) on the return swing at a level slightly lower than its initial position.

Topic 5

252

(b) If anyone gives the ball a forward push anywhere along its path, positive work is done on the ball and it will tend to reach a level higher than its initial position before coming to rest again. In this case, the demonstrator may have to duck to avoid injury. 5.8

(a) The chicken does positive work on the ground. (b) No work is done. (c) The crane does positive work on the bucket. (d) The force of gravity does negative work on the bucket. (e) The leg muscles do negative work on the individual.

5.10

The kinetic energy is converted to internal energy within the brake pads of the car, the roadway, and the tires.

5.12

(a) The friction force is directed opposite to the motion, so it does negative work on the plate. (b) The friction force is directed opposite to the motion, so it does negative work on the chair. (c) In this case, friction is responsible for keeping the box at rest on the conveyor belt. The friction force is in the same direction as the motion, so it does positive work on the box.

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253

(d) As always, zero work is done on a stationary object. To do work, a force must act through a displacement. 5.14

Choice (a). The change in gravitational potential energy is ΔPEg = mgΔy. Each stone has the same change in height, Δy, so the stone with twice the mass will have twice the change in gravitational potential energy.

5.16

Since the rollers on the ramp used by David were frictionless, he did not do any work overcoming nonconservative forces as he slid the block up the ramp. Neglecting any change in kinetic energy of the block (either because the speed was constant in the case of sliding the block, or, in the case of lifting the block, the speed at the ground and at the truck bed were both zero), the work done by either Mark or David equals the increase in the gravitational potential energy of the block as it is lifted from the ground to the truck bed. Because they lift identical blocks through the same vertical distance, they do equal amounts of work and the correct choice is (b).

5.18

KEcar =

1 1 1 1 1 1 mcar v 2 = ⎛ mtruck ⎞ v 2 = ⎛ mtruck v 2 ⎞ = KEtruck , so (b) is the ⎠ ⎠ 2 2 2⎝ 2 2⎝ 2

correct answer.

5.20

If the car is to have uniform acceleration, a constant net force F must act

Topic 5

254

on it. Since the instantaneous power delivered to the car is P = Fv, we see that maximum power is required just as the car reaches its maximum speed. The correct answer is (b).

ANSWERS TO EVEN NUMBERED PROBLEMS 5.2

(a)

472 J

(b)

2.76 kN

5.4

(a)

1.6 × 103 J

(b)

(c)

The applied force is smaller, the work done by the shopper is unchanged.

5.6

(a)

900 J

(b)

0.383

5.8

(a)

31.9 J

(b)

(d)

31.9 J

(b)

The speed of the crate will increase.

5.10

160 m/s

5.12

(a)

29.2 N

(c)

The crate would slow down and come to rest.

5.14

(a)

3.17 × 104 J

(b)

7.57 Calories

5.16

(a)

1.2 J

(b)

5.0 m/s

(c)

6.3 J

Topic 5

5.18

255

(a)

47.8 N directed opposite to the motion

(c)

0.22 m/s

5.20

(a)

8.88 × 102 N/m

5.22

(a)

The athlete and the Earth interact via the gravitational force.

(b)

KE = 2.4 × 103 J, PEg = 0 (c)

KE = 0, PEg = 2.4 × 103 J

(d)

ymax = 4.1 m

(e)

v y= 1 y = 6.4 m/s

(b)

1.68 × 105 J

(b)

ΔKE = mgh

(b)

1.38 cm

(b)

–31 J

(c)

2.84 J

2 max

5.24

d = 2(m1 − m2)g/k

5.26

(a)

5.28

≈ 0.5 m

5.30

(a)

Wg = mgh

(c)

1 KE f = mv02 + mgh 2

(d)

No, the previous answers are independent of the initial angle.

0.768 m

5.32

5.1 m

5.34

(a)

1.13 kN/m

5.36

(a)

vB = 5.94 m/s, vC = 7.67 m/s

(b)

52.3 cm (b)

147 J

Topic 5

5.38

5.40

5.42

256

(a)

v f = 2(m1 − m2 )gh/(m1 + m2 )

(c)

2.6 m/s

(a)

No, fk = µk(mg − F sinθ)

(c)

! ! ! Forces that do no work are n, mg and the vertical component of F .

(d)

4.24 N, 47.9 J and −17.0 J

(b)

3.7 m/s

Wfk = −µk(mg − F sinθ)x, WF = Fx cosθ

(a) No, mechanical energy is not conserved when friction forces are present. (b)

77.0 m/s

5.44

(a)

2.29 m/s

5.46

(a) Yes, the only nonconservative force is perpendicular to the motion.

(b)

15.7 J

(b) At the top, KE = 0, PEg = mgh; At launch point, KE = 54 mgh, PEg = 15 mgh ; At pool level, KE = mgh, PEg = 0. (c)

v0 = 8gh/5

(e)

ymax = h(1 − 54 cos 2 θ )

(d)

2 ymax = h − v0x /2g

(f) No, if energy is used overcoming friction, v0, ymax and the final speed

Topic 5

257

are all reduced. 5.48

1.5 m along the incline or 0.17 m vertically

5.50

(a)

5.52

2.9 m/s

5.54

4.5 × 103 N

5.56

2.03 × 108 s (or 6.43 yr)

5.58

(a)

5.60

2.06 × 104 J

(b)

0.920 hp

7.92 hp

(b)

14.9 hp

(a)

7.50 J

(b)

15.0 J

(d)

at x = 5.00 m, v = 2.29 m/s; at x = 15.0 m, v = 4.50 m/s

5.62

90.0 J

5.64

(a)

5.66

0.116 m

5.68

1.4 m/s

5.70

236 s or 3.93 min

5.72

(a)

3.57 m/s

0.225 J

(b)

3.22 kN/m

(b)

0.363 J

(c)

7.50 J

Topic 5

258

5.74

(a) 3.39 × 106 J

(b)

577 trips

5.76

(a)

3.0 × 102 J

(b)

−1.5 × 102 J

(d)

1.5 × 102 J

5.78

0.082 3 W

5.80

(a) 147 J

(b)

1.00 km

5.82

(a) 2.04 × 10−2 m

(b)

2.35 × 103 kg

5.84

0.386 m

5.86

3.93 kJ

(c)

PROBLEM SOLUTIONS 5.1

If the weights are to move at constant velocity, the net force on them must be zero. Thus, the force exerted on the weights is upward, parallel to the displacement, with magnitude 350 N. The work done by this force is W = (F cos θ)s = [(350 N)cos 0°](2.00 m) = 700 J

5.2

(a) We assume the object moved upward with constant speed, so the kinetic energy did not change. Then, the work-energy theorem gives

Topic 5

259

the work done on the object by the lifter as Wnc = ΔKE + ΔPE = 0 + (mgyf − mgyi) = mg(Δy), or Wnc = (281.5 kg)(9.80 m/s2)[(17.1 cm)(1 m/102 cm)] = 472 J (b) If the object moved upward at constant speed, the net force acting on it was zero. Therefore, the magnitude of the upward force applied by the lifter must have been equal to the weight of the object: F = mg = (281.5 kg)(9.80 m/s2) = 2.76 × 103 N = 2.76 kN 5.3

(a) The work done by tension T on the elevator is WT = Td cosθ where T = 1.25 × 104 N is the tension, d = 2.00 m is the vertical displacement, and θ is the angle between the tension and the displacement. Here, both the tension and the displacement are directed vertically upward so that θ = 0° and

(

)

WT = Tdcosθ = 1.25 × 10 4 N ( 2.00 m ) cos 0° = 2.50 × 10 4 J (b) The work done by the force of gravity on the elevator is Wg = (mg)d cosθ where m = 1.00 × 103 kg is the elevator’s mass, d = 2.00 m is the vertical displacement, and θ is the angle between the gravity force and the displacement. The downward gravity force and the upward displacement are in opposite directions so that θ = 180° and © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 5

260

((

)(

Wg = ( mg ) dcosθ = 1.00 × 103 kg 9.80 m/s 2

))( 2.00 m ) cos180°

= −1.96 × 10 4 J

5.4

(a) The 35 N force applied by the shopper makes a 25° angle with the displacement of the cart (horizontal). The work done on the cart by the shopper is then Wshopper = (F cos θ)Δx = (35 N)(cos 25°)(50.0 m) = 1.6 × 103 J (b) Since the speed of the cart is constant, KEf = KEi and Wnet = ΔKE = 0. (c) Since the cart continues to move at constant speed, the net work done on the cart in the second aisle is again zero. With both the net work and the work done by friction unchanged, the work done by the shopper (Wshopper = Wnet − Wfriction) is also unchanged. However, the shopper now pushes horizontally on the cart, making F′ = Wshopper/(Δx ⋅ cos 0°) = Wshopper/Δx smaller than before when the force was F = Wshopper/(Δx ⋅ cos 35°).

5.5

(a) The gravitational force acting on the object is w = mg = (5.00 kg)(9.80 m/s2) = 49.0 N and the work done by this force is wg = −ΔPEg = −mg(yf − yi) = +w(yi − yf)

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261

Wg = w(L sin 30.0°) = (49.0 N)(2.50 m)sin 30.0° = 64.3 J.

(b) The normal force exerted on the block by the incline is n = mg cos 30.0°, so the friction force is fk = µkn = (0.436)(49.0 N)cos 30.0° = 18.5 N This force is directed opposite to the displacement (that is θ = 180°), and the work it does is Wf = (fk cosθ)L = [(18.5 N)cos 180°](2.50 m) = −46.3 J. (c) Since the normal force is perpendicular to the displacement, so the work done by the normal force is Wn = (n cos 90.0°)L = 0. (d) If a shorter ramp is used to increase the angle of inclination while maintaining the same vertical displacement ⎢yf − yi⎢, the work done by gravity will not change, the work done by the friction force will decrease (because the normal force, and hence the friction force, will decrease and also because the ramp length L decreases), and the work done by the normal force remains zero (because the normal © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 5

262

force remains perpendicular to the displacement). 5.6

(a) WF = F(Δx)cosθ = (150 N)(6.00 m)cos 0° = 900 J

(b) Since the crate moves at constant velocity, ax = ay = 0. Thus, ΣFx = 0 ⇒ fk = F = 150 N Also, ΣFy = 0 ⇒ n = mg = (40.0 kg)(9.80 m/s2) = 392 N so

µk =

5.7

fk n

150 N 392 N

= 0.383

(a) The work done on the crate by tension is WT = (Tcosθ)d where T = 175 N, θ = 20.0°, and d = 6.00 m. Substitute values to find

(

)

WT = (T cosθ ) d = (175 N ) cos ( 20.0° ) ( 6.00 m ) = 987 J

(b) Here we’ll solve for the coefficient of kinetic friction in two different ways: first using the work-energy theorem and then using Newton’s

Topic 5

263

second law. Each method requires the magnitude of the normal force on the crate, given in Topic 4 for this case as

(

)

n = mg − T sin θ = ( 40.0 N ) 9.80 m/s2 − (175 N ) sin ( 20.0° ) = 332 N

Four forces act on the crate as it moves: gravity, the normal force, kinetic friction, and tension. According to the work-energy theorem, Wnet = ΔKE. Here, the crate moves at constant speed so that ΔKE = 0 and Wnet = ΔKE = 0 so that Wnet = Wgravity + Wnormal + Wfriction + Wtension = 0

The gravity and normal forces are directed perpendicular to the crate’s displacement so that, for these two forces, θ = 90° and Wgravity = Wnormal = 0. The work-energy theorem then reduces to Wfriction = −Wtension.

The work done by kinetic friction is Wfriction = −fkd = −(µkn)d and the work done by tension is Wtension = (Tcosθ)d. Substitute these results and solve for µk, the coefficient of kinetic friction, to find

Topic 5

264

Wfriction = −Wtension

− ( µ k n ) d = − (T cosθ ) d

T cosθ (175 N ) cos ( 20.0° ) = n 332 N = 0.495

µk =

Alternatively, apply ΣFx = max with ax = 0 to find Tcos(20.0°) − fk = 0 or Tcos(20.0°) = µkn. Solve for µk to find, as above,

µk =

T cosθ (175 N ) cos ( 20.0° ) = n 332 N

= 0.495

5.8

(a) WF = (F cosθ)s = [(16.0 N)cos 25.0°](2.20 m) WF = 31.9 J

(b) Wn = (n cos 90°)s = 0 (c) Wg = (mg cos 90°)s = 0

Topic 5

265

(d) Wnet = WF + Wn + Wg = 31.9 J + 0 + 0 = 31.9 J 5.9

(a) The work-energy theorem, Wnet = KEf − KEi, gives

5 000 J =

(2.50 × 10 kg ) v − 0 , or v = 2.00 m/s 2 1

(b) W = (F cosθ)s = (F cos 0°)(25.0 m) = 5 000 J, so F = 200 N 5.10

Requiring that KEping pong = KEbowling, with KE = 12 mv 2 , we have 1 2

(2.45 × 10 kg ) v = (7.00 g)(3.00 m/s)

giving 5.11

−3

1 2

v = 160 m/s

(a) Use the definition of kinetic energy to find KEi = 12 mvi2 = 12 ( 65.0 kg ) ( 5.20 m/s) = 879 J 2

(b) With the runner’s speed doubled,

(

)

KE f = 12 m ( 2v ) = 4 12 mv 2 = 4 ( KEi )

According to the work-energy theorem, the required net work is then

Wnet = ΔKE = KE f − KEi

= 4 ( KEi ) − KEi

= 3 ( KEi ) = 3 ( 879 J ) = 2.64 × 103 J 5.12

(a) Since the applied force is horizontal, it is in the direction of the

Topic 5

266

displacement, giving θ = 0°. The work done by this force is then

WF = (F0 cosθ )Δx = F0 (cos0°)Δx = F0 (Δx) 0

and

F0 =

Δx

350 J 12.0 m

= 29.2 N

(b) Since the crate originally had zero acceleration, the original applied force was just enough to offset the retarding friction force. Therefore, when the applied force is increased, it has a magnitude greater than the friction force. This gives the crate a resultant force (and hence, an acceleration) in the direction of motion, meaning the speed of the crate will increase with time. (c) If the applied force is made smaller than F0, the magnitude of the friction force will be greater than that of the applied force. This means the crate has a resultant force, and acceleration, in the direction of the friction force (opposite to the direction of motion). The crate will now slow down and come to rest. 5.13

(a) We use the work-energy theorem to find the work.

1 1 1 W = ΔKE = mv 2f − mvi2 = 0 − (70 kg)(4.0 m/s)2 = −5.6 × 102 J 2 2 2 (b) W = (F cosθ)s = (fk cos 180°)s = (−µkn)s = (−µk mg)s, © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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267

(−5.6 × 10 J) = 1.2 m s=− =− µ mg (0.70)(70 kg) ( 9.80 m/s ) 2

5.14

(a) From the work-energy theorem, the required net work equals the change in the cheetah’s kinetic energy. The cheetah starts from rest, so its initial kinetic energy, KEi, equals zero and Wnet = ΔKE = KE f − KEi = KE f = 12 mv 2 = 12 ( 62.0 kg ) ( 32.0 m/s)

= 3.17 × 10 4 J

(b) Converting to Calories gives:

⎛ 1 Calorie ⎞ Wnet = 3.17 × 10 4 J ⎜ = 7.57 Calories ⎝ 4 186 J ⎟⎠ 5.15

(a) As the bullet penetrates the tree trunk, the only force doing work on it is the force of resistance exerted by the trunk. This force is directed opposite to the displacement, so the work done is Wnet = (fav cos180°)Δx = KEf − KEi, and the magnitude of the average resistance force is

fav =

KE f − KEi (Δx)cos180°

0−

(

)

1 7.80 × 10−3 kg (575 m/s)2 2 = 2.34 × 10 4 N −2 − 5.50 × 10 m

(

)

(b) If the friction force is constant, the bullet will have a constant © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 5

268

(

)

acceleration and its average velocity while stopping is v = v f + vi /2 . The time required to stop is then

Δt =

5.16

Δx v

2(Δx) v f + vi

(

2 5.50 × 10−2 m 0 + 575 m/s

) = 1.91 × 10 s −4

1 1 (a) KEA = mv A2 = (0.60 kg)(2.0 m/s)2 = 1.2 J 2 2 2 (b) KEB = 12 mvB , so

vB =

2(KEB ) m

2(7.5 J) 0.60 kg

= 5.0 m/s

5.17

(a) KEi =

1 2

mvi2 =

(6.50 × 10 kg )(12.0 m/s) = 4.68 × 10 J 2 1

(b) Wnet = KEf − KEi = 0 − 4.68 × 109 J = −4.68 × 109 J (c) Wnet = (FRcosθ)Δx, and θ = 180° since the resultant force acting on the ship is a retarding frictional force (note that the normal force the water exerts on the ship simply cancels out the weight of the ship). Thus, if the ship comes to rest after a displacement of Δx = 2.50 km, the resultant force acting on the ship is

Topic 5

269

FR =

5.18

Wnet (Δx)cosθ

−4.68 × 109 J

(2.50 × 10 m)cos180° 3

= 1.87 × 106 N

The crate has zero vertical acceleration, so ΣFy = n − mg = 0, and the normal force is n = mg. Thus, the kinetic friction force is fk = µkn = µkmg.

(a) Note that the two vertical forces acting on the crate cancel, meaning that the net force on the crate is horizontal. Choosing the +x-direction toward the right, we have Fnet = ΣFx = F − fk = F − µkmg or

Fnet = 275 N − (0.358)(92.0 kg)(9.80 m/s2) = −47.8 N Fnet = 47.8 N directed opposite to the motion of the crate

(b) Wnet = Fnet(Δx)cosθ = (47.8 N)(0.65 m)cos180° = −31 J

(c)

(

)

Wnet = KE f − KEi = 12 m v 2f − vi2 , so v 2f = 2Wnet /m + vi2

vf =

2 2(−31 J) + ( 0.850 m/s 2 ) = 0.22 m/s 92.0 kg

Topic 5

5.19

270

(a) PEi = mgyi = (0.20 kg)(9.80 m/s2)(1.3 m) = −2.5 J (b) PEf = mgyf = (0.20 kg)(9.80 m/s2)(−5.0 m) = −9.8 J (c) ΔPE = PEf − PEi = −9.8 J − 2.5 J = −12 J

5.20

(a) The force stretching the spring is the weight of the suspended object. Therefore, the force constant of the spring is

Fg Δx

mg Δx

(2.50 kg)(9.80 m/s 2 ) 2.76 × 10 m −2

= 8.88 × 102 N/m

(b) If a 1.25 kg block replaces the original 2.50 kg suspended object, the force applied to the spring (weight of the suspended object) will be one-half the original stretching force. Since, for a spring obeying Hooke’s law, the elongation is directly proportional to the stretching force, the amount the spring stretches now is

1 1 (Δx)2 = (Δx)1 = (2.76 cm) = 1.38 cm 2 2 (c) The work an external agent must do on the initially unstretched spring to produce an elongation xf is equal to the potential energy stored in the spring at this elongation:

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271

Wdone on = (PEs ) f − (PEs )i = spring

1 2 kx f − 0 2

1 = (8.88 × 102 N/m)(8.00 × 10−2 m)2 = 2.84 J 2 5.21

(a) From the definition of spring potential energy, PEs = 21 kx 2 = 21 ( 875 N/m ) ( 0.070 0 m ) d 2

= 2.14 J

(b) Use conservation of energy to find the block’s speed as it leaves the spring. In this case, the block starts from rest and its spring’s potential energy is converted into kinetic energy so that

( KE + PE + PE ) = ( KE + PE + PS ) g

0 + 0 + kx = mv + 0 + 0 1 2

k x= m

1 2

875 N/m ( 0.070 0 m ) 3.00 kg

= 1.20 m/s

5.22

(a) While the athlete is in the air, the interacting objects are the athlete and the Earth. They interact though the gravitational force that one exerts on the other. (b) If the athlete leaves the trampoline (at the y = 0 level) with an initial speed of vi = 9.0 m/s, her initial kinetic energy is

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272

KEi =

1 mvi2 = (60.0 kg)(9.0 m/s)2 = 2.4 × 103 J 2 2

and her gravitational potential energy is (PEg)i = mgyi = mg(0) = 0 J. (c) When the athlete is at maximum height, she is momentarily at rest and KEf = 0 J. Because the only force acting on the athlete during her flight is the conservative gravitation force, her total energy (kinetic plus potential) remains constant. Thus, the decrease in her kinetic energy as she goes from the launch point (where PEg = 0) to maximum height is matched by an equal size increase in the gravitational potential energy. ΔPEg = −ΔKE ⇒ PEf − PEi = −(KEf − KEi) or PEf = PEi + KEi − KEf and

PEf = 0 + 2.4 × 103 J − 0 = 2.4 × 103 J.

(d) The statement that the athlete’s total energy is conserved is summarized by the equation ΔKE + ΔPE = 0 or KE2 + PE2 = KE1 + PE1. In terms of mass, speed, and height, this becomes 1 2

mv22 + mgy2 = 12 mv12 + mgy1 . Solving for the final height gives 1 1 mgy1 + mv12 − mv22 2 2 y2 = mg

y2 = y1 +

(v12 − v22 ) 2g

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273

The given numeric values for this case are y1 = 0, v1 = 9.0 m/s (at the trampoline level), and v2 = 0(at maximum height). The maximum height attained is then

( v − v ) = 0 + (9.0 m / s) − 0 = 4.1 m y =y + 2 1

2 2

2(9.80 m / s 2 )

(e) Solving the energy conservation equation given in part (d) for the final speed gives

2⎛1 ⎞ v22 = ⎜ mv12 + mgy1 − mgy2 ⎟ ⎝ ⎠ m 2 !

v2 = v12 + 2g(y1 − y2 )

With y1 = 0, v1 = 9.0 m/s, and y2 = ymax/2 = (4.1 m)/2, the speed at half the maximum height is given as

4.1 m ⎞ ⎛ v2 = (9.0 m/s)2 + 2(9.80 m/s 2 ) ⎜ 0 − ⎟⎠ = 6.4 m/s ⎝ 2 ! 5.23

The work the beam does on the pile driver is given by Wnc = (Fcos 180°)Δx = –F(0.120 m) Here, the factor cos 180° is included because the force F exerted on the driver by the beam is directed upward, but the Δx = 12.0 cm = 0.120 m displacement undergone by the driver while in contact with the beam is directed downward.

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From the work-energy theorem, this work can also be expressed as

(

) (

)

(

)

(

1 Wnc = KE f − KEi + PE f − PEi = m v 2f − vi2 + mg y f − yi 2 !

)

Choosing y = 0 at the level where the pile driver first contacts the top of the beam, the driver starts from rest (v1 = 0) at yi = +5.00 m and comes to rest again (vf = 0) at yf = –0.120 m. Therefore, we have

(

)

1 −F(0.120 m) = m 0 − 0 + (2 100 kg)(9.80 m/s 2 )(−0.120 m − 5.00 m) 2 yielding F = 8.78 × 105 N directed upward. 5.24

Note that the system is released from rest, and at the maximum upward displacement of m2, the system is again at rest. Thus, the kinetic energy of the system is zero in both the initial and final states. Since only conservative forces (gravitational forces and a spring force) do work on this system, the total energy is constant. Therefore, the gravitational potential energy given up by m1 as it drops down distance d must equal the sum of the gravitational potential energy gained by m2 as it rises distance d and the elastic potential energy stored in the spring when it is stretched distance d. Mathematically, this is 2 1 m1 g −d = m2 g d + k d 2

1 m1 g = m2 g + kd 2 !

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275

yielding d = ! 5.25

2 ( m1 − m2 ) g . k

While the motorcycle is in the air, only the conservative gravitational force acts on cycle and rider. Thus,

1 1 mv 2f + mgy f = mv i2 + mgyi , which 2 !2

gives

h = y f − yi = ! 5.26

v 2f − vi2 2g

(35.0 m/s)2 − (33.0 m/s)2 = 6.94 m 2(9.80 m/s 2 )

(a) When equilibrium is reached, the total spring force supporting the load equals the weight of the load, or Fs, total = Fs, leaf + Fs, helper = wload. Let kℓ and kh represent the spring constants of the leaf spring and the helper spring, respectively. Then, if xℓ is the distance the leaf spring is compressed, the condition for equilibrium becomes kℓxℓ + kh(xℓ – y0) = wload or

5.00 × 10 N + ( 3.60 × 10 N/m ) (0.500 m) w +k y x = load h 0 = = 0.768 m k + kh 5.25 × 105 N/m + 3.60 × 105 N/m ! 5

(b) The work done compressing the springs equals the total elastic potential energy at equilibrium. Thus, 1 1 2 W = k x2 + kh ( x − 0.500 m ) , or 2 2 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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1 1 5.25 × 105 N/m ) (0.768 m)2 + ( 3.60 × 105 N/m ) (0.268 m)2 ( 2 2

= 1.68 × 105 J

5.27

The total work done by the two bicep muscles as they contract is Wbiceps = 2FavΔx = 2(800 N)(0.075 m) = 1.2 × 102 J The total work done on the body as it is lifted 40 cm during a chin-up is Wchin-up = mgh = (75 kg)(9.80 m/s2)(0.40 m) = 2.9 × 102 J Since Wchin-up > Wbiceps, it is clear that additional muscles must be involved

5.28

Applying Wnc = (KE + PE)f – (KE + PE)i to the jump of the “original” flea gives

Fmd = (0 + mgyf) – (0 + 0)

Fd yf = m mg !

where Fm is the force exerted by the muscle and d is the length of contraction. If we scale the flea by a factor f, the muscle force increases by f 2 and the length of contraction increases by f. The mass, being proportional to the volume which increases by f 3, will also increase by f 3. Putting these factors into our expression for yf gives

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277

( ) !

( f F )( fd) = F d = y ≈ 0.5 m ( f m) g mg 2

y f super = flea

m 3

so the “super flea” cannot jump any higher! This analysis is used to argue that most animals should be able to jump to approximately the same height (~0.5 m). Data on mammals from elephants to shrews tend to support this. 5.29

(a) Taking y = 0, and hence PEg = mgy = 0, at ground level, the initial total mechanical energy of the projectile is

(E ) = KE + PE = 12 mv + mgy total i

2 i

1 = (50.0 kg)(1.20 × 102 m/s)2 + (50.0 kg)(9.80 m/s 2 )(142 m) 2 = 4.30 × 105 J

(b) The work done on the projectile is equal to the change in its total mechanical energy.

(Wnc )rise = ( KE f + PE f ) − ( KEi + PEi ) = 2 m ( v 2f − vi2 ) + mg(y f − yi ) 1

1 = (50.0 kg) ⎡⎣(8.50 m/s)2 − (120 m/s)2 ⎤⎦ + (50.0 kg)(9.80 m/s 2 )(427 m − 142 m) 2 !

= −3.97 × 10 4 J

(c) If, during the descent from the maximum height to the ground, air resistance does one and a half times as much work on the projectile © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 5

278

as it did while the projectile was rising to the top of the arc, the total work done for the entire trip will be

(Wnc )total = (Wnc )rise + (Wnc )descent = (Wnc )rise + 1.50 (Wnc )rise = 2.50 ( −3.97 × 10 4 J ) = −9.93 × 10 4 J

Then, applying the work-energy theorem to the entire flight of the projectile gives

1 1 − (KE + PE)at!lauch = mv 2f + mgy f − mvi2 − mgyi 2 2 hitting!ground

(Wnc )total = (KE + PE)just!before

and the speed of the projectile just before hitting the ground is

vf = = ! 5.30

2 ( Wnc )total m

+ vi2 + 2g(yi − y f )

2(−9.93 × 10 4 J) + (120 m/s)2 + 2(9.80 m/s 2 )(142 m − 0) = 115 m/s 50.0 kg

(a) The work done by the gravitational force equals the decrease in the gravitational potential energy, or Wg = –(PEf – PEi) – PEi – PEf = mg(yi – yf)=mgh (b) The change in kinetic energy is equal to the net work done on the projectile, which, in the absence of air resistance, is just that done by the gravitational force. Thus, Wnet = Wg = ΔKE ⇒ ΔKE = mgh

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279

1 KE f = KEi + mgh = mv02 + mgh 2 !

(d) No. None of the calculations in parts (a), (b), or (c) involve the initial angle. 5.31

(a) The system will consist of the mass, the spring, and the Earth. The parts of this system interact via the spring force, the gravitational force, and the normal force. (b) The points of interest are where the mass is released from rest (at x = 6.00 cm) and the equilibrium point, x = 0. (c) The energy stored in the spring is the elastic potential energy, 1 PEs = kx 2 . 2 ! 1 −2 2 At x = 6.00 cm, PEs = (850 N/m)(6.00 × 10 m) = 1.53 J 2 ! 1 2 and at the equilibrium position (x = 0), PEs = k(0) = 0 2 !

(d) The only force doing work on the mass is the conservative spring force (the normal force and the gravitational force are both perpendicular to the motion). Thus, the total mechanical energy of the mass will be constant. Because we may choose y = 0, and hence

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280

PEg = 0, at the level of the horizontal surface, the energy conservation equation becomes KEf + (PEs)f = KEi + (PEs)i

1 2 1 2 1 2 1 2 mv f + kx f = mvi + kxi 2 2 2 !2

and solving for the final speed gives

v f = vi2 + !

(

k 2 xi − x 2f m

)

If the final position is the equilibrium position (xf = 0) and the object starts from rest (vi = 0) at xi = 6.00 cm, the final speed is 2 850 N/m ⎡ vf = 0 + 6.00 × 10−2 m ) − 0 ⎤ = 3.06 m 2 /s 2 = 1.75 m/s ( ⎣ ⎦ 1.00 kg !

(e) When the object is halfway between the release point and the equilibrium position, we have vi = 0, xi = 6.00 cm and xf = 3.00 cm, giving 2 2 850 N/m ⎡ vf = 0 + 6.00 × 10−2 m ) − ( 3.00 × 10−2 m ) ⎤ = 1.51 m/s ( ⎦ 1.00 kg ⎣ !

This is not half of the speed at equilibrium because the equation for final speed is not a linear function of position. 5.32

Using conservation of mechanical energy, we have

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281

1 2 1 mv f + mgy f = mvi2 + 0 2 !2

5.33

yf = !

vi2 − v 2f 2g

(10 m/s)2 − (1.0 m/s)2 = 5.1 m 2(9.80 m/s 2 )

Since no nonconservative forces do work, we use conservation of mechanical energy, with the zero of potential energy selected at the level of the base of the hill. Then, 1 2 1 mv f + mgy f = mvi2 + mgyi with yf = 0 yields 2 !2

yi = !

v 2f − vi2 2g

(3.00 m/s)2 − 0 = 0.459 m 2(9.80 m/s 2 )

Note that this result is independent of the mass of the child and sled. 5.34

(a) The hanging mass stretches the spring by an amount x = 41.5 cm− 35.0 cm = 6.50 cm, resulting in an upward spring force of magnitude Fs = kx. The mass hangs in equilibrium, acted on by this spring force and its weight so that

ΣFy = may = 0 Fs − mg = 0 Fs = mg kx = mg

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282

(

mg ( 7.50 kg ) 9.80 m/s k= = x 6.50 × 10−2 m

)

= 1.13 × 103 N/m (b) With the hanging mass replaced by a w = 195-N weight, the new equilibrium condition is

ΣFy = may = 0 Fs − w = 0 Fs = w kx = w x=

w 195 N = k 1.13 × 103 N/m

= 0.173 m = 17.3 cm

The total length of the 35.0-cm long spring, stretched by an additional 17.3 cm, is

L = 35.0 cm + 17.3 cm = 52.3 cm 5.35

(a) On a frictionless track, no external forces do work on the system consisting of the block and the spring as the spring is being compressed. Thus, the total mechanical energy of the system is

( )

constant, or KE f + PEg f + ( PEs ) f = KEi + PEg i + ( PEs )i . Because the ! track is horizontal, the gravitational potential energy when the mass © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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283

comes to rest is the same as just before it made contact with the

( ) ( )

spring, or PEg f = PEg i . This gives ! 1 2 1 2 1 2 1 2 mv f + kx f = mvi + kxi 2 2 2 !2

Since vf = 0 (the block comes to rest) and xi = 0 (the spring is initially undistorted),

x f = vi !

m 0.250 kg = (1.50 m/s) = 0.350 cm k 4.60 N/M

(b) If the track was not frictionless, some of the original kinetic energy would be spent overcoming friction between the block and track. This would mean that less energy would be stored as elastic potential energy in the spring when the block came to rest. Therefore, the maximum compression of the spring would be less in this case. 5.36

(a) From conservation of mechanical energy, 1 2 1 mvB + mgyB = mv A2 + +mgy A , or 2 !2

vB = v A2 + 2g(y A − yB ) !

= 0 + 2(9.80 m/s 2 )(1.80 m) = 5.94 m/s

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284

Similarly, vC = v A2 + 2g(y A − yC ) = 0 + 2(9.80 m/s 2 )(5.00 m − 2.00 m) = 7.67 m/s !

(b) 5.37

(W )

A→C

( ) − ( PE ) = mg ( y − y ) = (49.0 N)(3.00 m) = 147 J

= PEg

(a) We choose the zero of potential energy at the level of the bottom of the arc. The initial height of Tarzan above this level is yi = (30.0 m)(1 – cos 37.0°) = 6.04 m

Then, using conservation of mechanical energy, we find 1 2 1 mv f + 0 = mvi2 + mgyi 2 !2

v f = vi2 + 2gyi = 0 + 2(9.80 m/s 2 )(6.04 m) = 10.9 m/s !

(b) In this case, conservation of mechanical energy yields

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285

v f = vi2 + 2gyi = (4.00 m/s)2 + 2(9.80 m/s 2 )(6.04 m) = 11.6 m/s !

5.38

(a) If the string does not stretch, the speeds of the two blocks must be equal at all times until m1 reaches the floor. Also, while m1 is falling, only conservative forces (the gravitational forces) do work on the system of two blocks. Thus, the total mechanical energy is constant, or

KE + KE2, f + (PEg )1, f + (PEg )2, f = KE1,i + KE2,i + (PEg )1,i + (PEg )2,i ! 1, f Choosing y = 0 at floor level, this becomes 1 1 m1v 2f + m2 v 2f + 0 + m2 gh = 0 + 0 + m1 gh + 0 2 !2

vf =

and yields

2(m1 − m2 )gh m1 + m2

(b) Using the provided data values, the answer from part (a) gives 2(6.5 kg − 4.2 kg)(9.80 m/s 2 )(3.2 m) vf = = 3.7 m/s 6.5 kg + 4.2 kg !

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286

vf = !

vf = ! 5.39

(

)

(

)

2g ⎡⎣m1 y1,i − y1, f + m2 y2,i − y2, f ⎤⎦ m1 + m2

2 ( 9.80 m/s 2 ) ⎡⎣( 6.5 kg ) (1.6 m ) + ( 4.2 kg ) ( −1.6 kg ) ⎤⎦ 6.5 kg + 4.2 kg

= 2.6 m/s

(a) Initially, all the energy is stored as elastic potential energy within the spring. When the gun is fired, and as the projectile leaves the gun, most of the energy is in the form of kinetic energy along with a small amount of gravitational potential energy. When the projectile comes to rest momentarily at its maximum height, all of the energy is in the form of gravitational potential energy. (b) Use conservation of mechanical energy from when the projectile is at rest within the gun (vi = 0, yi = 0, and xi = –0.120 m) until it reaches maximum height where vf = 0, yf = ymax = 20.0 m, and xf = 0 (the spring is relaxed after the gun is fired).

(

) (

)

Then, KE + PEg + PEs f = KE + PEg + PEs i becomes ! 1 0 + mgymax + 0 = 0 + 0 + kxi2 2 !

−3 2 2mgymax 2 ( 20.0 × 10 kg ) ( 9.80 m/s ) ( 20.0 m ) k= = = 544 N/m 2 xi2 −0.120 m ) ( !

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287

(c) This time, we use conservation of mechanical energy from when the projectile is at rest within the gun (vi = 0, yi = 0, and xi = –0.120 m) until it reaches the equilibrium position of the spring (yf = +0.120 m and xf = 0). This gives

(

) (

)

KE f = KE + PEg + PEs − PEg + PEs i f !

(

1 2 ⎛ 1 ⎞ mv f = ⎜ 0 + 0 + kxi2 ⎟ − mgy f + 0 ⎝ ⎠ 2 !2

)

⎛ k⎞ v 2f = ⎜ ⎟ xi2 − 2gy f ⎝ m⎠

⎛ 544 N/m ⎞ 2 =⎜ −0.120 m ) − 2(9.80 m/s 2 )(0.120 m) ( −3 ⎟ ⎝ 20.0 × 10 kg ⎠

yielding 5.40

vf = 19.7 m/s

(a) ΣFy = 0 ⇒ n + F sinθ – mg = 0 or

n = mg – F sinθ

The friction force is then fk = µkn = µk(mg – F sinθ)

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288

(b) The work done by the applied force is  W = F Δx cosθ = Fxcosθ ! F

 and the work done by the friction force is !W fk = fk Δx cosφ where φ is  the angle between the direction of !fk !and!Δx . Thus,

W = fk xcos180° = − µk (mg − F sin θ )x ! fk (c) The forces that do no work are those perpendicular to the direction  of the displacement !Δx . These are

! ! ! n, mg, and the vertical component of F .

(d) For part (a): n = mg – F sinθ = (2.00 kg)(9.80 m/s2) – (15.0 N) sin 37.0° = 10.6 N fk = µkn = (0.400)(10.6 N) = 4.24 N For part (b): WF = Fx cosθ = (15.0 N)(4.00 m) cos 37.0° = 47.9 J

W = fk xcosφ = (4.24 N)(4.00 m)cos 180° = −17.0 J ! fk 5.41

(a) When the child slides down a frictionless surface, the only nonconservative force acting on the child is the normal force. At each instant, this force is perpendicular to the motion and, hence, does no work. Thus, conservation of mechanical energy can be used in this

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289

case. (b) The equation for conservation of mechanical energy, (KE + PE)f = (KE + PE)i, for this situation is

1 1 mv 2f + mgy f = mvi2 + mgyi . Notice that 2 !2

the mass of the child cancels out of the equation, so the mass of the child is not a factor in the frictionless case. (c) Observe that solving the energy conservation equation from above

(

)

2 for the final speed gives v f = vi + 2g yi − y f . Since the child starts !

with the same initial speed (vi = 0) and has the same change in altitude in both cases, vf is the same in the two cases. (d) Work done by a neoconservative force must be accounted for when friction is present. This is done by using the work-energy theorem rather than conservation of mechanical energy. (e) From part (b), conservation of mechanical energy gives the final speed as

(

)

v f = vi2 + 2g yi − y f = 0 + 2(9.80 m/s 2 )(12.0 m) = 15.3 m/s ! 5.42

(a) No. The change in the kinetic energy of the plane is equal to the net work done by all forces doing work on it. In this case, there are two

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290

such forces, the thrust due to the engine and a resistive force due to the air. Since the work done by the air resistance force is negative, the net work done (and hence, the change in kinetic energy) is less than the positive work done by the engine thrust. Also, because the thrust from the engine and the air resistance force are nonconservative forces, mechanical energy is not conserved in this case.

( ) ( )

(b) Since the plane is in level flight, PEg f = PEg i and the work-energy ! theorem reduces to Wnc = Wthrust + Wresistance = KEf – KEi, or

( F cos 0°) s + ( f cos180°) s = mv 2f − mvi2 1 2

1 2

This gives

v f = vi2 +

2( F − f ) s m

= (60.0 m/s) + 2

5.43

2 ⎡⎣(7.50 − 4.00) × 10 4 N ⎤⎦ (500. m) 1.50 × 10 4 kg

= 77.0 m/s

Neglecting the mass of the rope and any friction in the pulleys, the tension T1 is uniform throughout the length of the rope and equal to the magnitude of the force F applied to the loose end of the rope. Also, since the load (and hence the moving pulley) have constant velocity, applying Newton’s second law to the three force diagrams below gives

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291

Load:

T2 – mg = 0

moving pulley:

T2 = mg

[1]

T mg 2T1 − T2 = 0 or T1 = 2 = 2 2 !

[2]

fixed pulley: T3 – 2T1 = 0

T3 = 2T1 = mg

[3]

(a) If m = 76.0 kg, Equation [2] gives

F = T1 = !

(76.0 kg)(9.80 m/s 2 ) = 372 N 2

(b) From above, T1 = 372 N. From Equations [1] and [3], T2 = T3 = mg =(76.0 kg)(9.80 m/s2) = 754 N (c) Observe that if the load is raised 1.80 m, this length of rope must be removed from each of the two vertical segments of rope supporting the moving pulley. Thus, the loose end of the rope must be pulled downward a distance d = 3.60 m. The work done by the applied force

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292

is then WF = F ⋅ d ⋅ cosθ = (372 N)(3.60 m)cos 0° = 1.34 × 103 J = 1.34 kJ 5.44

(a) Choose PEg = 0 at the level of the bottom of the arc. The child’s initial vertical displacement from this level is yi = (2.00 m)(1 – cos 30.0°) = 0.268 m In the absence of friction, we use conservation of mechanical energy as

1 (KE + PEg ) f = (KE + PEg )i , or mv 2 + 0 = 0 + mgy , which gives f i ! 2 v f = 2gyi = 2(9.80 m/s 2 )(0.268 m) = 2.29 m/s !

(b) With a nonconservative force present, we use

(

) (

)

⎛1 ⎞ Wnc = KE + PEg − KE + PEg = ⎜ mv 2f + 0⎟ − ( 0 + mgyi ) , or f i ⎝2 ⎠ ! ⎛ v 2f ⎞ Wnc = m ⎜ − gyi ⎟ ⎝ 2 ⎠

⎡ (2.00 m/s)2 ⎤ = (25.0 kg) ⎢ − ( 9.80 m/s 2 ) (0.268 m) ⎥ = 15.7 J 2 ⎣ ⎦

Thus, 15.7 J of energy is spent overcoming friction. 5.45

Choose PEg = 0 at the level of the bottom of the driveway.

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293

(

) (

)

Then Wnc = KE + PEg − KE + PEg becomes f i !

( f cos180°) s = ⎛⎜⎝ 12 mv + 0⎞⎟⎠ − [ 0 + mg ( ssin 20°)]

2 f

Solving for the final speed gives 2 fs v f = (2gs)sin 20° − m !

5.46

v f = 2(9.80 m/s 2 )(5.0 m)sin 20° − !

2 ( 4.0 × 103 N ) (5.0 m) 2.10 × 103 kg

= 3.8 m/s

(a) Yes. Two forces, a conservative gravitational force and a nonconservative normal force, act on the child as she goes down the slide. However, the normal force is perpendicular to the child’s motion at each point on the path and does no work. In the absence of work done by nonconservative forces, mechanical energy is conserved. (b) We choose the level of the pool to be the y = 0 (and hence, PEg = 0) level. Then, when the child is at rest at the top of the slide, PEg = mgh

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and KE = 0. Note that this gives the constant total mechanical energy of the child as Etotal = KE + PEg = mgh. At the launching point (where y = h/5), we have PEg = mgy = mgh/5 and KE = Etotal – PEg = 4mgh/5. At the pool level, PEg = 0 and KE = mgh. (c) At the launching point (i.e., where the child leaves the end of the slide), 1 4mgh KE = mv02 = 2 5 !

meaning that

v0 = !

8gh 5

(d) After the child leaves the slide and becomes a projectile, energy 1 2 conservation gives KE + PEg = mv + mgy = Etotal = mgh where 2 ! v 2 = vx2 + vy2 . Here, vx = v0x is constant, but vy varies with time. At !

maximum height, y = ymax and vy = 0, yielding

1 2 m(v0x + 0) + mgymax = mgh !2

ymax = h −

and

2 v0x 2g

(e) If the child’s launch angle leaving the slide is θ, then v0x = v0 cosθ. Substituting this into the result from part (d) and making use of the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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v02 1 ⎛ 8gh ⎞ 2 2 result from part (c) gives ymax = h − cos θ = h − ⎜⎝ ⎟⎠ cos θ 2g 2g 5 ! or

4 ⎛ ⎞ ymax = h ⎜ 1 − cos 2 θ ⎟ ⎝ ⎠ 5

(f) No. If friction is present, mechanical energy would not be conserved, so her kinetic energy at all points after leaving the top of the waterslide would be reduced when compared with the frictionless case. Consequently, her launch speed, maximum height reached, and final speed would be reduced as well. 5.47

Choose PEg = 0 at the level of the base of the hill and let x represent the distance the skier moves along the horizontal portion before coming to rest. The normal force exerted on the skier by the snow while on the hill is n1 = mg cos 10.5° and, while on the horizontal portion, n2 = mg. Consider the entire trip, starting from rest at the top of the hill until the skier comes to rest on the horizontal portion. The work done by friction forces is Wnc = ⎡⎣( fk )1 cos 180° ⎤⎦ (200 m) ⎡⎣( fk )2 cos180° ⎤⎦ x !

= − µk ( mgcos 10.5° ) (200 m) − µk ( mg ) x

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(

) (

)

Applying Wnc = KE + PEg − KE + PEg to this complete trip gives f i ! –µk(mg cos 10.5°)(200 m) – µk(mg)x = [0 + 0] – [0 + mg(200 m) sin 10.5°]

⎛ sin10.5° ⎞ x=⎜ − cos10.5°⎟ (200 m) . If µk = 0.075 0, then x = 289 m ⎝ µk ⎠ !

5.48

The normal force exerted on the sled by the track is n = mg cosθ and the friction force is fk = µkn = µkmg cosθ. If s is the distance measured along the incline that the sled travels,

(

) (

)

applying Wnc = KE + PEg − KE + PEg to the entire trip gives f i !

⎛1 ⎞ ⎡⎣ µk ( mgcos θ ) cos 180° ⎤⎦ s = [0 + mg s(sin θ )] − ⎜ mv12 + 0⎟ ⎝ ⎠ 2 !

5.49

(

vi2

2g sin θ + µk cos θ

(4.0 m/s)2

) 2(9.80 m/s )(sin 20° + 0.20cos 20°)

= 1.5 m

(

)

) (

(a) Consider the entire trip and apply Wnc = KE + PEg − KE + PEg to f i ! obtain

( f cos 180°) d + ( f cos 180°) d = ⎛⎜⎝ 12 mv + 0⎞⎟⎠ − ( 0 + mgy )

2 f

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⎛ fd +fd ⎞ v f = 2 ⎜ gyi − 1 1 2 2 ⎟ m ⎝ ⎠ ⎛ (50.0 N)(8 000 m) + (3 600 N)(200 m) ⎞ = 2 ⎜ 9.80 m/s 2 (1 000 m) − ⎟ 80.0 kg ⎝ ⎠

(

)

which yields vf = 24.5 m/s. (b) Yes, this is too fast for safety.

(

) (

)

(c) Again, apply Wnc = KE + PEg f − KE + PEg i , now with d2 considered ! to be a variable, d1 = 1 000 m – d2, and vf = 5.00 m/s. This gives

⎛

⎞

( f cos180°)(1 000 m − d ) + ( f cos180°) d = ⎜⎝ 12 mv + 0⎟⎠ − (0 + mgy ) 1

(

2 f

)

1 which reduces to − 1 000 m f1 + f1d2 − f 2 d2 = mv 2f − mgyi . Therefore, 2 1 (mg)yi − (1 000 m) f1 − mv 2f 2 d2 = f 2 − f1

1 (784 N)(1 000 m) − (1 000 m)(50.0 N) − (80.0 kg)(5.00 m/s)2 2 = = 206 m 3 600 N − 50.0 N

(d) In reality, the air drag will depend on the skydiver’s speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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she touches down, whenever she moves near terminal speed. 5.50

(a) Wnc = ΔKE + ΔPE, but ΔKE = 0 because the speed is constant. The skier rises a vertical distance of Δy = (60.0 m) sin 30.0° = 30.0 m. Thus, Wnc = (70.0 kg)(9.80 m/s2)(30.0 m) = 2.06 × 104 J (b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus, the required power input is W mg(Δy) (70.0 kg) ( 9.80 m/s ) (30.0 m) P = nc = = Δt Δt 30.0 s 2

⎛ 1 hp ⎞ = (686 W) ⎜ = 0.920 hp ⎝ 746 W ⎟⎠

5.51

The mountain climber’s average power is P = W /Δt where W is the work done climbing the hill and Δt is the elapsed time in seconds. Using the conversion factor 1 min = 60 s, the elapsed time is Δt = 45.0 min = 2.70 × 103 s.

The work W done in climbing the hill is found using the work-energy theorem. Assuming the climber starts from rest at yi = 0 and stops at the top of the hill where yf = 325 m,

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W + ( KE + PE)i = ( KE + PE) f

(

W + ( 0 + 0 ) = 0 + mgy f

)

(

)

W = mgy f = ( 70.0 kg ) 9.80 m/s2 ( 325 m ) = 2.23 × 105 J

The average power is then

W 2.23 × 105 J = Δt 2.70 × 103 s

= 82.6 W 5.52

Let ΔN be the number of steps taken in time Δt. We determine the number of steps per unit time by Power = !

work!done (work!per!step!per!unit!mass)(mass)(# steps) = Δt Δt

⎛ J/step ⎞ ΔN ⎛ ΔN ⎞ 70 W = ⎜ 0.60 (60 kg) ⎜ = 1.9 steps/s , giving ⎟ ⎟ ⎝ Δt ⎠ kg ⎠ Δt ⎝ ! !

The running speed is then vav =

Δx ⎛ ΔN ⎞ =⎜ ⎟ (distance traveled per step) Δt ⎝ Δt ⎠

step ⎞ ⎛ m ⎞ ⎛ = ⎜ 1.9 1.5 = 2.9 m/s ⎟ ⎜ ⎝ s ⎠⎝ step ⎟⎠

5.53

Assuming a level track, PEf = PEi, and the work done on the train is

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Wnc = (KE + PE) f − (KE + PE)i !

(

)

1 1 2 = m v 2f − vi2 = (0.875 kg) ⎡( 0.620 m/s ) − 0 ⎤ = 0.168 J ⎣ ⎦ 2 2

The power delivered by the motor is then W 0.168 J P = nc = = 8.00 W −3 Δt 21.0 × 10 s !

5.54

When the car moves at constant speed on a level roadway, the power used to overcome the total frictional force equals the power input from the engine, or Poutput = ftotalv = Pinput. This gives

f total =

5.55

Pinput

175 hp ⎛ 746 W ⎞ = 4.5 × 103 N 29 m/s ⎜⎝ 1 hp ⎟⎠

(a) There are 8.64 × 104 s in a day. The heart consumes 13.0 J each second so its daily energy requirement is

⎛ 13.0 J ⎞ ⎛ 8.64 × 10 4 s ⎞ E=⎜ 1 day ) = 1.12 × 106 J ( ⎟ ⎜ ⎟ ⎝ 1 s ⎠ ⎝ 1 day ⎠

Using the conversion factor 1 Calorie = 4 186 J, the daily energy requirement in Calories is

⎛ 1 Calorie ⎞ E = 1.12 × 106 J ⎜ = 268 Calories ⎝ 4186 J ⎟⎠

(

)

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301

(b) Metabolizing 1 kg of fat can release about 9 000 Calories. Rounded to one significant figure, the mass of fat required to provide one day’s worth of Calories for the heart is

E 268 Calories = 9 000 Calories/kg 9 000 Calories/kg

= 2.98 × 10−2 kg 5.56

Neglecting any variation of gravity with altitude, the work required to lift a 3.20 × 107 kg load at constant speed to an altitude of Δy = 1.75 km is W = ΔPEg = mg(Δy) = (3.20 × 107 kg)(9.80 m/s2)(1.75 × 103 m) = 5.49 × 1011 J The time required to do this work using a P = 2.70 kW = 2.70 × 103 J/s pump is

Δt =

W 5.49 × 1011 J = = 2.03 × 108 s P 2.70 × 103 J/s

⎛ ⎞ 1 yr = ( 2.03 × 108 s ) ⎜ = 6.43 yr 7 ⎟ ⎝ 3.156 × 10 s ⎠ 5.57

(a) The acceleration of the car is a= !

v − v0 18.0 m/s − 0 = = 1.50 m/s 2 t 12.0 s

Thus, the constant forward force due to the engine is found from ΣF = Fengine – Fair = ma as

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302

Fengine = Fair + ma = 400 N + (1.50 × 103 kg)(1.50 m/s2) = 2.65 × 103 N. The average velocity of the car during this interval is vav = (v + v0)/2 = 9.00 m/s, so the average power input from the engine during this time is

Pav = Fengin e vav = ( 2.65 × 103 N ) (9.00 m/s) ⎛ 1 hp ⎞ = ( 2.39 × 10 4 W ) ⎜ = 32.0 hp ⎝ 746 W ⎟⎠ (b) At t = 12.0 s, the instantaneous velocity of the car is v = 18.0 m/s and the instantaneous power input from the engine is

P = Fengine v = ( 2.65 × 103 N ) (18.0 m/s) ⎛ 1 hp ⎞ = ( 4.77 × 10 4 W ) ⎜ = 63.9 hp ⎝ 746 W ⎟⎠ 5.58

(a) The acceleration of the elevator during the first 3.00 s is a= !

v − v0 1.75 m/s − 0 = = 0.583 m/s 2 t 3.00 s

so Fnet = Fmotor – mg = ma gives the force exerted by the motor as Fmotor = m(a + g) = (650 kg)[(0.583 + 9.80) m/s2] = 6.75 × 103 N The average velocity during this interval is vav = (v + v0)/2 = 0.875 m/s so the average power input from the motor during this time is

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303

⎛ 1 hp ⎞ Pav = Fmotor vav = ( 6.75 × 103 N ) (0.875 m/s) ⎜ = 7.92 hp ⎝ 746 W ⎟⎠ ! (b) When the elevator moves upward with a constant speed of v = 1.75 m/s, the upward force exerted by the motor is Fmotor = mg and the instantaneous power input from the motor is

⎛ 1 hp ⎞ P = (mg)v = ( 650 kg ) (9.80 m/s 2 )(1.75 m/s) = ⎜ = 14.9 hp ⎝ 746 W ⎟⎠ ! 5.59

The work done on the particle by the force F as the particle moves from x = xi to x = xf is the area under the curve from xi to xf.

(a) For x = 0 to x = 8.00 m, 1 W = area of triangle ABC = AC × altitude ! 2 1 W0→8 = (8.00 m)(6.00 N) = 24.0 J 2 !

(b) For x = 8.00 m to x = 10.0 m,

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1 W0→8 = area of triangle CDE = CE × altitude ! 2 1 (2.00 m)(−3.00 N) = −3.00 J !2

The work done by a force equals the area under the Force versus Displacement curve.

(a) For the region 0 ≤ x ≤ 5.00 m, W0 to 5 = !

(3.00 N)(5.00 m) = 7.50 J 2

(b) For the region 5.00 m ≤ x ≤ 10.0 m, W5 to 10 = (3.00 N)(5.00 m) = 15.0 J

x=0

(3.00 N)(5.00 m) = 7.50 J 2

= W0 to x f = area under F vs. x curve from x = 0 m to x = xf

, or

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305

1 2 1 2 mv f = mv0 + W0!to!x f 2 !2

giving

⎛ 2⎞ v f = v02 + ⎜ ⎟ W0!to!x f ⎝ m⎠ !

For xf = 5.00 m:

⎛ 2 ⎞ ⎛ 2⎞ v f = v02 + ⎜ ⎟ W0!to! 5 = (0.500 m/s)2 + ⎜ (7.50 J) = 2.29 m/s ⎝ m⎠ ⎝ 3.00 kg ⎟⎠ ! For xf = 15.0 m

⎛ 2⎞ ⎛ 2⎞ v f = v02 + ⎜ ⎟ W0!to!15 = v02 + ⎜ ⎟ ( W0!to! 5 + W5!to!10 + W10!to!15 ) ⎝ m⎠ ⎝ m⎠ ! or ⎛ 2 ⎞ v f = (0.500 m/s)2 + ⎜ ⎟ (7.50 J + 1.50 J + 7.50 J) = 4.50 m/s ⎝ 3.00 kg ⎠

5.61

(a) Fx = (8x – 16) N see the graph below.

(b) The net work done is the total area under the graph from x= 0 to x = 3.00 m. This consists of two triangular shapes, one below the axis (negative area) and one above the axis (positive). The net work is

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306

then

(2.00 m)(−16.0 N) (1.00 m)(8.00 N) + 2 2 = −12.0 J

Wnet = ! 5.62

At the top of the arc, vy = 0 and vx = v0x = v0 cos 30.0° 2 + 0 = ( v0 cos 30.0° ) , and Therefore v 2 = vx2 + vy2 = v0x ! 2

1 1 KE = mv 2 = (0.150 kg) ⎡⎣(40.0 m/s)2 cos 2 (30.0°) ⎤⎦ = 90.0 J 2 2 !

5.63

(a) If y = 0 at point B, then yA = (35.0 m) sin 50.0° = 26.8 m and yB = 0. Thus, PEA = mgyA = (1.50 × 103 kg)(9.80 m/s2)(26.8 m) = 3.94 × 105 J PEB = mgyB = 0 and ΔPEA→B = PEB – PEA = 0 – 3.94 × 105 J = 3.94 × 105 J (b) If y = 0 at point C, then yA = (50.0 m) sin 50.0° = 38.3 m and yB = (15.0 m) sin 50.0° = 11.5 m. In this case, PEA = mgyA = (1.50 × 103 kg)(9.80 m/s2)(38.3 m) = 5.63 × 105 J PEB = mgyB = (1.50 × 103 kg)(9.80 m/s2)(11.5 m) = 1.69 × 105 J and ΔPEA→B = PEB – PEA = 1.69 × 105 J – 5.63 × 105 J = –3.94 × 105 J

5.64

(a) Taking y = 0 at the initial level of the upper end of the spring, and

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307

applying conservation of energy from the instant the ball is released from rest to the instant just before it contacts the spring gives KEf = KEi + (PEgi – PEgf) + (PEsi – PEsf)

and

1 2 mv f = 0 + (mgh − 0) + (0 − 0) !2 v f = 2gh = 2(9.80 m/s 2 )(0.650 m) = 3.57 m/s !

(b) Since the spring is very light, we neglect any energy loss during the collision between the ball and spring, and apply conservation of energy from the instant the ball is released from rest at y = +h to the instant the ball comes to rest momentarily at y = –d. This yields KEf = KEi + (PEgi – PEgf) + (PEsi – PEsf)

1 ⎛ ⎞ 0 = 0 + [mgh − mg(−d)] + ⎜ 0 − kd 2 ⎟ ⎝ ⎠ 2 !

Thus, the force constant of the spring is

or 5.65

2mg(h + d) d

(

)

2(1.80 kg) 9.80 m/s 2 (0.650 m + 0.090 0 m) (0.090 0 m)2

k = 3.22 × 103 N/m = 3.22 kN/m

(a) The equivalent spring constant of the bow is given by F = kx as

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308

Ff 230 N k= = = 575 N/m x f 0.400 m ! (b) The work done pulling the bow is equal to the elastic potential energy stored in the bow in its final configuration, or 1 1 W = kx 2f = (575 N/m) + (0.400 m)2 = 46.0 J 2 2 !

5.66

Choose PEg = 0 at the level where the block comes to rest against the spring. Then, in the absence of work done by nonconservative forces, the conservation of mechanical energy gives

( KE + PE + PE ) + ( KE + PE + PE )

! or

1 0 + 0 + kx 2f = 0 + mg!L! sin θ + 0 2 !

Thus, xf = !

5.67

2mg!L! sin θ 2(12.0 kg)(9.80 m/s 2 )(3.00 m)sin 35.0° = = 0.116 m k 3.00 × 10 4 N/m

(a) From v 2 = v02 + 2ay (Δy) , we find the speed just before touching the ! ground as v = 0 + 2(9.80 m/s 2 )(1.0 m) = 4.4 m/s !

(b) Choose PEg = 0 at the level where the feet come to rest. Then

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309

(

) (

)

Wnc = KE + PEg − KE + PEg becomes f i !

( Fav cos180°) s = (0 + 0) − ⎛⎜⎝ 2 mvi2 + mg s⎞⎟⎠ 1

! or

mvi2 (75 kg)(4.4 m/s)2 Fav = + mg = + (75 kg)(9.80m/s 2 ) = 1.5 × 105 N 2s 2 ( 5.0 × 10−3 m ) ! 5.68

From the work-energy theorem,

(

) (

)

Wnc = KE + PEg + PEs − KE + PEg + PEs f i ! we have

( f cos180°) s = ⎛⎜⎝ 12 mv + 0 + 0⎞⎟⎠ − ⎛⎜⎝ 0 + 0 + 12 kx ⎞⎟⎠

2 f

2 i

or kxi2 − 2 fk s (8.0 N/m)(5.0 × 10−2 m)2 − 2(0.032 N)(0.15 m) vf = = = 1.4 m/s m 5.3 × 10−3 kg !

5.69

(a) The two masses will pass when both are at yf = 2.00 m above the table. From conservation of energy,

( KE + PE + PE ) = ( KE + PE + PE )

1 m1 + m2 ) v 2f + ( m1 + m2 ) gy f + 0 = 0 + m1 gy1i + 0 , or ( !2

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310

vf = = !

2m1 gy1i − 2gy f m1 + m2 2(5.00 kg)(9.80 m/s 2 )(4.00 m) − 2(9.80 m/s 2 )(2.00 m) 8.00 kg

This yields the passing speed as vf = 3.13 m/s. (b) When m1 = 5.00 kg reaches the table, m2 = 3.00 kg is y2f = 4.00 m above

(

) (

)

the table. Thus, KE + PEg + PEs f = KE + PEg + PEs i becomes !

(

1 2g m1y1i − m2 y2 f m1 + m2 ) v 2f + m2 gy2 f + 0 = 0 + m1 gy1i + 0 , or v = ( f !2 m1 + m2 !

)

Thus,

2(9.80 m/s 2 )[(5.00 kg)(4.00 m) − (3.00 kg)(4.00 m)] vf = = 4.43 m/s 8.00 kg ! (c) When the 5.00-kg object hits the table, the string goes slack and the 3.00-kg object becomes a projectile launched straight upward with initial speed !v0y = 4.43 m/s . At the top of its arc, 2 v = 0!and!vy2 = v0y + 2ay (Δy) gives ! y

Δy = ! 5.70

2 vy2 − v0y

2ay

υy =

0 − (4.43 m/s)2 = 1.00 m 2(−9.80 m/s 2 )

As the piano is lifted at constant speed up to the apartment, the total work that must be done on it is

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311

Wnc = ΔKE + ΔPEg = 0 + mg(yf – yi) = (3.50 × 103 N)(25.0 m) = 8.75 × 104 J The three workmen (using a pulley system with an efficiency of 0.750) do work on the piano at a rate of

⎛ ⎞ Pnet = 0.750 ⎜ 3Psingle ⎟ = 0.750[3(165 W)] = 371 W = 371 J/s ⎝ worker ⎠ ! so the time required to do the necessary work on the piano is

Δt =

Wnc 8.75 × 10 4 J = = 236 s Pnet 371 J/s

⎛ 1 min ⎞ = (236 s) ⎜ = 3.93 min ⎝ 60 s ⎟⎠ 5.71

Since the bowl is smooth (that is, frictionless), mechanical energy is conserved or (KE + PE)f = (KE + PE)i. Also, if we choose y = 0 (and hence, PEg = 0) at the lowest point in the bowl, then yA = +R, yB = 0, and yC = 2R/3

(a)

(PE ) = mgy = mgR g

(PE ) = (0.200 kg)(9.80 m/s )(0.300 m) = 0.588 J 2

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312

(b) KEB = KEA + PEA – PEB = 0 + mgyA – mgyB = 0.588 J – 0 = 0.588 J

2KEB

2(0.588 J)

(c)

KEB =

(d)

m) ⎥ = 0.392 J (PE ) = mgy = (0.200 kg)(9.80 m/s ) ⎢⎢⎣ 2(0.300 3 ⎥⎦

mvB2 ⇒ vB =

0.200 kg

= 2.42 m/s

⎡

⎤

(e) KEC = KEB + PEB – PEC = 0.588 J + 0 – 0.392 J = 0.196 J

5.72

1 1 (a) KEB = mvB2 = (0.200 kg)(1.50 m/s)2 = 0.225 J 2 2

(b) The change in the altitude of the particle as it goes from A to B is yA – yB = R, where R = 0.300 m is the radius of the bowl. Therefore, the work-energy theorem gives Wnc = (KEB – KEA) + (PEB – PEA) = KEB – 0 + mg(yB – yA) = KEB + mg(–R) or

Wnc = 0.225 J + (0.200 kg)(9.80 m/s2)(–0.300 m) = –0.363 J

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313

The loss of mechanical energy as a result of friction is then 0.363 J. (c) No. Because the normal force, and hence the friction force, varies with the position of the particle on its path, it is not possible to use the result from part (b) to determine the coefficient of friction without using calculus. 5.73

(a) The person walking uses Ew = (220 kcal)(4 186 J/1 kcal) = 9.21 × 105 J of energy while going 3.00 miles. The quantity of gasoline which could furnish this much energy is

9.21 × 105 J V1 = = 7.08 × 10−3 gal 8 1.30 × 10 J/gal ! This means that the walker’s fuel economy in equivalent miles per gallon is fuel economy =

3.00 mi = 424 mi/gal 7.08 × 10−3 gal

(b) In 1 hour, the bicyclist travels 10.0 miles and uses

⎛ 4 186 J ⎞ 6 EB = (4 000 kcal) ⎜ ⎟ = 1.67 × 10 J ⎝ 1 kcal ⎠ which is equal to the energy available in

1.67 × 106 J V2 = = 1.29 × 10−2 gal 8 1.30 × 10 J/gal ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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314

of gasoline. Thus, the equivalent fuel economy for the bicyclist is 10.0 mi = 775 mi/gal −2 1.29 × 10 gal !

5.74

(a) Using the numbers provided, the chemical energy available from 0.450 kg of fat is

⎛ 3.77 × 107 J ⎞ Echem = ( 0.450 kg ) ⎜ = 1.70 × 107 J ⎝ 1.00 kg ⎟⎠

The available mechanical energy is 20% of that number so that

⎛ 3.77 × 107 J ⎞ Emech = ( 0.200 ) ( 0.450 kg ) ⎜ = 3.39 × 106 J ⎟ ⎝ 1.00 kg ⎠ (b) Ignoring any energy used in coming down the stairs, and assuming the student starts and stops from rest, the mechanical energy required for each trip up the stairs equals the change in the student’s gravitational potential energy:

(

)

Etrip = mgΔy = ( 50.0 kg ) 9.80 m/s2 (12.0 m ) = 5.88 × 103 J

For the student to lose 0.450 kg of fat, she must make a number of trips, N, so that Emech = N(Etrip). Solve for N to find

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315

Emech 3.39 × 106 J = Etrip 5.88 × 103 J

= 577 trips 5.75

(a) Use conservation of mechanical energy, (KE + PEg)f = (KE + PEg)i , from the start to the end of the track to find the speed of the skier as she leaves the track. This gives

(

1 2 mv + mgy f = 0 + mgyi , or !2

)

v = 2g yi − y f = 2(9.80 m/s 2 )(40.0 m) = 28.0 m/s ! (b) At the top of the parabolic arc the skier follows after leaving the track, vy = 0 and vx = (28.0 m/s) cos 45.0° = 19.8 m/s. Thus, v = vx2 + vy2 = 19.8 m/s . Applying conservation of mechanical ! top

energy from the end of the track to the top of the arc gives 1 1 m(19.8 m/s)2 + mgymax = m(28.0 m/s)2 + mg(10.0 m) , or 2 !2

ymax = 10.0 m+ !

(28.0 m/s)2 − (19.8 m/s)2 = 30.0 m 2(9.80 m/s 2 )

1 2 (c) Using Δy = v0yt + ayt for the flight from the end of the track to the 2 !

ground gives 1 −10.0 m = [(28.0 m/s)sin 45.0° ]t + (−9.80 m/s 2 )t 2 2 !

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316

The positive solution of this equation gives the total time of flight as t = 4.49 s. During this time, the skier has a horizontal displacement of Δx = v0xt = [(28.0 m/s) cos 45.0°](4.49 s) = 88.9 m 5.76

First, determine the magnitude of the applied force by considering a freebody diagram of the block. Since the block moves with constant velocity, ΣFx = ΣFy = 0.

From ΣFx = 0, we see that n = F cos 30°. Thus, fk = µkn = µkF cos 30°, and ΣFy = 0 becomes F sin 30° = mg + µkF cos 30°, or

mg (5.0 kg)(9.80 m/s 2 ) F= = = 2.0 × 102 N sin 30° − µk cos 30° sin 30° − (0.30)cos 30° ! (a) The applied force makes a 60° angle with the displacement up the wall. Therefore, WF = (F cos 60°)s = [(2.0 × 102 N) cos 60°](3.0 m) = 3.0 × 102 J (b) Wg = (mg cos 180°)s = (49 N)(–1.0)(3.0 m) = –1.5 × 102 J

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317

We choose PEg = 0 at the level where the spring is relaxed (x = 0), or at the level of position B. (a) At position A, KE = 0 and the total energy of the system is given by

(

E = 0 + PEg + PEs

) = mg x + 12 kx

2 1

or 1 E = (25.0 kg)(9.80 m/s 2 )(−0.100 m) + ( 2.50 × 10 4 N/m ) (−0.100 m)2 = 101 J 2 !

(b) In position C, KE = 0 and the spring is uncompressed, so PEs = 0. Hence,

(

E = 0 + PEg + 0

) = mg x C

E 101 J x2 = = = 0.412 m 2 mg (25.0 kg)(9.80 m/s ) !

1 (c) At Position B, PEg = PEs = 0 and E = (KE + 0 + 0)B = mvB2 2

Therefore, vB =

2E m

2(101 J) 25.0 kg

= 2.84 m/s

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318

(d) Where the velocity (and hence the kinetic energy) is a maximum, the acceleration (slope of the velocity versus time graph) is zero. Thus, ΣFy = –kx –mg = 0 and we find

mg k

245 kg 4

2.50 × 10 N/m

= −9.80 × 10−3 m = −9.80 mm

1 1 (e) From the total energy, E = KE + PEg + PEs = mv 2 + mgx + kx 2 , we 2 2 find

v= !

2E k − 2gx − x 2 m m

Where the speed, and hence kinetic energy, is a maximum (that is, at x = –9.80 mm), this gives 2(101 J) (2.50 × 10 4 N/m) 2 −3 vmax = − 2(9.80 m/s )(−9.80 × 10 m) − (−9.80 × 10−3 m)2 25.0 kg 25.0 kg !

or 5.78

vmax = 2.86 m/s

The average delivered power is P = W /Δt where W is the work done per stroke and Δt is the elapsed time.

The hummingbird’s weight has magnitude w = mg and a downward force of this average magnitude is exerted over the length of each stroke. Using © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 5

319

W = (Fcosθ)Δx with F = mg, θ = 0°, and d = 0.035 0 m, the bird does W = mgd Joules of work each stroke. Each stroke takes, on average, 1/(80.0 s) so that the average power is

(

)(

)

3.00 × 10−3 kg 9.80 m/s 2 ( 0.035 0 m ) W P= = Δt (1/80.0 s ) = 0.082 3 W 5.79

Choose PEg = 0 at the level of the river. Then yi = 36.0 m, yf = 4.00, the jumper falls 32.0 m, and the cord stretches 7.00 m. Between the balloon and the level where the diver stops momentarily, (KE + PEg + PEs)f = (KE + PEg + PEs)i , gives 1 0 + (700 N)(4.00 m) + k(7.00 m)2 = 0 + (700 N)(36.0 m) + 0 2 !

or 5.80

k = 914 N/m.

(a) A projectile launched at speed v0 and angle θ0 has a constant xcomponent of velocity given by v0x = v0 cosθ0. Here, the golf ball travels a horizontal distance of 4.00 × 103 m in 70.0 s so that v0x = (4.00 × 103 m)/ 70.0 s = 57.1 m/s and

v0 =

v0x 57.1 m/s = cosθ 0 cos ( 45.0° )

= 80.8 m/s

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320

With a mass of m = 0.045 0 kg, its initial kinetic energy is then KEi = 12 mv02 = 12 ( 0.045 0 kg ) ( 80.0 m/s)

= 147 J

(b) The maximum altitude can be found in a variety of ways. Taking yi = 0 at the launch point and yf at the maximum altitude where v = v0x (because vy = 0 at the maximum altitude), conservation of energy gives

( KE + PE) = ( KE + PE) i

1 2

mv + 0 = 12 mv + mgmoon ymax 2 0

2 0x

2 v02 − v0x ( 80.8 m/s ) − ( 57.1 m/s ) ymax = = 2gmoon 2 1.63 m/s 2 2

(

)

= 1.00 × 103 m = 1.00 km 5.81

(a) From the work-energy theorem, Wnet = KEf –KEi. Since the package moves with constant velocity, KEf = KEi giving Wnet = 0. Note that the above result can also be obtained by the following reasoning: Since the object has zero acceleration, the net (or resultant) force acting on it must be zero. The net work done is Wnet = Fnetd = 0. (b) The work done by the conservative gravitational force is

Topic 5

321

Wgravity = –ΔPEg = –mg(yf – yi) = –mg(d sinθ) or

Wgravity = –(50 kg)(9.80 m/s2)(340 m)sin 7.0° = –2.0 × 104 J

(d) Since the package moves up the incline at constant speed, the net force parallel to the incline is zero. Thus, ΣF|| = 0 ⇒ fs – mg sinθ = 0 or fs = mg sinθ. The work done by the friction force in moving the package distance d up the incline is Wfriction = fkd = (mg sin θ)d = [(50 kg)(9.80 m/s2)sin 7.0°](340 m) = 2.0 × 104 J 5.82

(a) Two forces act on the man standing in equilibrium on the scale: the downward force of his weight and the upward spring force of magnitude Fs = kx where x = 0.650 mm = 6.50 × 10−4 m. Apply the ycomponent of Newton’s second law with ay = 0 to find the spring constant, k:

Topic 5

322

ΣFy = may kx − mg = 0 k=

(

mg ( 75.0 kg ) 9.80 m/s = x 6.50 × 10−4 m

)

= 1.13 × 106 N/m Use conservation of energy to find the spring’s maximum compression. Taking y = 0 at the spring’s equilibrium point, the person’s initial height is yi = 0.300 m where KEi = PEs,i = 0. When the spring is compressed its maximum distance xmax, the person is below the spring’s equilibrium point with gravitational potential energy PEg,f = −mgxmax and kinetic energy KEf = 0. Substitute these results to find

( KE + PE + PE ) = ( KE + PE + PE ) g

0 + mgyi + 0 = 0 − mgxmax + kx 1 2

1 2

2 max

2 kxmax − mgxmax − mgyi = 0

Solve for xmax, the spring’s maximum compression, using the quadratic formula. Substitute known values, suppressing units for clarity, to find 1 2

(1.13 × 10 ) x 6

2 max

− ( 75.0 ) ( 9.80 ) xmax − ( 75.0 ) ( 9.80 ) ( 0.300 ) = 0

( 5.65 × 10 ) x 5

2 max

− ( 735 ) xmax − 221 = 0

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323

Taking the positive root of this equation, the spring’s maximum compression is

xmax =

735 +

( −735) − 4 ( 5.65 × 10 )( −221) 2

(

2 5.65 × 105

)

= 2.04 × 10−2 m

(b) When a person in equilibrium stands on the scale, the spring force and weight have the same magnitude and Fs = kx = mg. In general, Fs = Rg where R is the scale’s reading in kilograms.

At maximum compression, the spring force exerted by the scale on the person is Fs = kxmax = 2.31 × 104 N. Solve for the scale reading, R, to find R=

Fs,max kxmax = g g

(1.13 × 10 N/m)( 2.04 × 10 m) = −2

9.80 m/s2

= 2.35 × 103 kg

5.83

(a) While the car moves at constant speed, the tension in the cable is F = mg sinθ, and the power input is P = Fv = mgv sinθ or

Topic 5

324

P = (950 kg)(9.80 m/s2)(2.20 m/s)sin 30.0° = 1.02 × 104 W = 10.2 kW (b) While the car is accelerating, the tension in the cable is

Δv ⎞ ⎛ Fa = mgsin θ + ma = m ⎜ gsin θ + ⎟ ⎝ Δt ⎠ ⎡ 2.20 m/s − 0 ⎤ = (950 kg) ⎢(9.80 m/s 2 )sin 30.0° + = 4.83 × 103 N ⎥ 12.0 s ⎦ ⎣ Maximum power input occurs the last instant of the acceleration phase. Thus, Pmax = Favmax = (4.83 × 103 N)(2.20 m/s) = 10.6 kW (c) The work done by the motor in moving the car up the frictionless track is

( )

1 Wnc = (KE + PE) f − (KE + PE)i = KE f + PEg − 0 = mv 2f + mg(Lsin θ ) f 2 !

⎡1 ⎤ Wnc = (950 kg) ⎢ (2.20 m/s)2 + ( 9.80 m/s 2 ) (1 250 m)sin 30.0° ⎥ ⎣2 ⎦ = 5.82 × 106 J 5.84

Because no nonconservative forces do work on the toy as it swings upward, mechanical energy is conserved and KEi + PEi = KEf + PEf

Topic 5

325

Taking the toy’s initial height as yi = 0, its kinetic energy equals zero at the maximum height of yf. Substitute these results and solve for yf to find the maximum height: 1 2

mvi2 + 0 = 0 + mgy f vi2 ( 2.75 m/s ) yf = = 2g 2 9.80 m/s 2

(

)

= 0.386 m

5.85

Observe that when m3 moves downward distance d, m1 must move upward distance d and m2 must slide distance d to the right across the horizontal tabletop. Also, each block must always have the same speed as each of the other blocks. Therefore, if the system starts from rest, and f is the friction force the table exerts on m2, the work energy theorem (Wnc = ΔKE + ΔPEg) gives

− fd =

(2 m + m + m )(v − v ) + ⎡⎣m g(+d) + m g(0) + m g(−d)⎤⎦ 1

2 f

2 i

With vi = 0, m1 = 5.00 kg, m2 = 10.0 kg, m3 = 15.0 kg, f = 30.0 N, and d = 4.00 m, this yields a final speed of

vf = =

2d ⎡⎣( m3 − m1 ) g − f ⎤⎦ m1 + m2 + m3

2(4.00 m) ⎡⎣(10.0 kg)(9.8 m/s 2 ) − (30.0 N) ⎤⎦ 30.0 kg

= 4.26 m/s

Topic 5

5.86

326

The normal force the incline exerts on block A is nA = (mAg) cos 37.0°, and the friction force is fk = µknA = µkmAg cos 37.0°. The vertical distance block A rises is ΔyA = (20.0 m) sin 37.0° = 12.0 m, while the vertical displacement of block B is ΔyB = –20.0 m. We find the common final speed of the two blocks by use of

(

) (

)

Wnc = KE + PEg − KE + PEg = ΔKE + ΔPEg f i ! This gives

⎡1 ⎤ − ( µk mA gcos 37.0° ) s = ⎢ ( mA + mB ) v 2f − 0 ⎥ + ⎡⎣ mA g ( Δy A ) + mB g ( ΔyB ) ⎤⎦ ⎣2 ⎦ or

( )

( ) (

)

2g ⎡ −mB ΔyB − mA Δy A − µk mA cos 37° s ⎤ ⎣ ⎦ v = mA + mB 2 f

2(9.80 m/s 2 ) ⎡⎣ −(100. kg)(−20.0 m) − (50.0 kg)(12.0 m) − 0.250(50.0 kg)(20.0 m)cos 37.0° ⎤⎦ 150. kg

= 157 m / s

The change in the kinetic energy of block A is then 1 1 ΔKEA = mA v 2f − 0 = (50.0 kg)(157 m 2 /s 2 ) = 3.93 × 103 J = 3.93 kJ 2 2

Topic 6

327

Topic 6 Momentum and Collisions

QUICK QUIZZES 6.1

Choice (b). The relation between the kinetic energy of an object and the magnitude of the momentum of that object is KE = p2/2m. Thus, when two objects having masses m1 < m2 have equal kinetic energies, we may write p12 p2 = 2 !2m1 2m2

6.2

p1 m1 = < 1 and p1 < p2 p2 m2

Choice (c). Because the momentum of the system (boy + raft) remains constant with zero magnitude, the raft moves towards the shore as the boy walks away from the shore.

6.3

Choice (c). The total momentum of the car-truck system is conserved. Hence, any change in momentum of the truck must be counterbalanced by an equal magnitude change of opposite sign in the momentum of the car.

6.4

Choice (a). The total momentum of the two-object system is zero before collision. To conserve momentum, the momentum of the combined object must be zero after the collision. Thus, the combined

Topic 6

328

object must be at rest after the collision. 6.5

(a)

Perfectly inelastic. Any collision in which the two objects stick together afterwards is perfectly inelastic.

(b)

Inelastic. Both the Frisbee and the skater lose speed (and hence, kinetic energy) in this collision. Thus, the total kinetic energy of the system is not conserved.

(c)

Inelastic. The kinetic energy of the Frisbee is conserved. However, the skater loses speed (and hence, kinetic energy) in this collision. Thus, the total kinetic energy of the system is not conserved.

6.6

Choice (a). If all of the initial kinetic energy is transformed, then nothing is moving after the collision. Consequently, the final momentum of the system is necessarily zero. Because momentum of the system is conserved, the initial momentum of the system must be zero, meaning that the two objects must have had equal magnitude momenta in opposite directions before the collision.

6.7

Choice (c). In a one-dimensional elastic collision, the relative speed between the two objects is unchanged during the collision. Before the collision, the two object are coming together at relative speed v; after they collision they are separating at relative speed v. In this case, the stationary bowling ball remains at rest after the collision so the pingpong ball’s speed does not change (although its velocity changes sign).

Topic 6

329

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 6.2

(a)

No. One of the objects was in motion before collision, so the system consisting of the two particles had a nonzero momentum before impact. Since momentum is always conserved in collisions, the system must have nonzero momentum after impact, meaning that at least one of the particles must be in motion.

(b)

Yes. It is possible for one of the particles to be at rest after collision, provided the other particle leaves the collision with a momentum equal to the total momentum of the two-particle system before impact.

6.4

Choice (c). The puck’s momenta are in opposite directions and their kinetic energies are equal.

6.6

Since the total momentum of the skater-Frisbee system is conserved, the momentum transferred to the skater equals the magnitude of the change in the Frisbee’s momentum. This is greatest when the skater throws the Frisbee back after catching it.

6.8

(a)

No. In most collisions, there is some loss of kinetic energy (meaning the collision is not elastic), but the objects fail to stick together as they would in a completely inelastic collision (a collision in which the maximum possible kinetic energy loss,

Topic 6

330

consistent with conservation of momentum, occurs). (b)

A greater portion of the incident kinetic energy is transformed to other forms of energy in a head-on collision than in a glancing collision. Thus, the expectation of damage to passengers is greatest in head-on collisions.

6.10

Choice (a). Each cart receives the same impulse, so each has the same change in momentum.

6.12

The passenger must undergo a certain momentum change in the collision. This means that an impulse, I = Fav ⋅ Δt = Δp, must be imparted the passenger by the steering wheel, the window, an air bag, or something. By increasing the time Δt during which this momentum change occurs, the resulting force on the passenger can be decreased.

6.14

Choice (b). Its speed decreases as its mass increases. No external horizontal forces act on the box-rainwater system, so its horizontal momentum cannot change as the box moves along the surface. Because the product mvx must be constant, and because the mass of the box (m) is increasing as it slowly fills with water, the horizontal speed of the box must decrease.

6.16

No. The change in kinetic energy of an object is equal to the net work done on it. This net work is the product of the net force acting on the object and the displacement in the direction of the force. Thus, a small

Topic 6

331

magnitude force acting through a large distance may do more work (and hence produce a greater change in kinetic energy) than a large force acting through a small distance.

6.18

Since the same net force acts on the two particles through equal displacements, they have equal amounts of work done on them. From the work-energy theorem, we see that the two particles, both having started with zero kinetic energy, will have equal final kinetic energies. Choice (c) is the correct response.

ANSWERS TO EVEN NUMBERED PROBLEMS 6.2

1.7 kN

6.4

(a) 10.4 kg ⋅ m/s in the direction of the ball’s final velocity (b) 173 N

6.6

See Solution.

6.8

(a) 13.5 N ⋅ s

6.10

(a) Fav = 6.4 × 103 N = 1.4 × 103 lb

(b)

9.00 kN

(b) It is unlikely that the man has sufficient arm strength to guarantee the safety of the child during a collision. The violent forces during the collision would tear the child from his arms. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 6

332

(a) 5.40 N ⋅ s in the direction of the final velocity

6.14

(a) 117 N ⋅ s upward

(b)

637 N upward

(b)

6.00 m/s

(b)

−27.0 J

(c)

4.00 m/s

(a) 12.0 N ⋅ s

6.18

260 N in the −x-direction or perpendicular to the wall

6.20

(a) 13 N ⋅ s

(b)

2.6 kN toward the pitcher

(a) 0.490 m/s

(b)

2.01 × 10−2 m/s

6.24

⎛ m ⎞! ! G (a) v PI = − ⎜ ⎟ vGP ⎝ mG + mP ⎠

(b)

⎛ m ⎞! ! P vGI = ⎜ ⎟ vGP ⎝ mG + mP ⎠

6.26

0.34 m/s toward the left

6.28

v1 = −1.7 m/s (to the left); v2 = 1.1 m/s (to the right)

6.30

vx = 3.33 m/s and vy = 3.11 m/s

6.32

1.67 m/s

6.34

No. The average force exerted is 3.75 × 103 N, which is less than 4 500 N.

Topic 6

333

6.36

(a) vf = (v1 + 2v2)/3

6.38

(a) 90.0°

(b)

KEi − KEf = (M/3)(v1 − v2)2

(b) vcue ball = 3.46 m/s, vtarget ball = 2.00 m/s 6.40

vOf = 3.99 m/s, vGf = 3.01 m/s

6.42

143 m/s

6.44

(a) 20.9 m/s eastward (b) 1.50 × 104 J, which becomes internal energy

6.46

(a) See Solution. (b) x-direction: m1v1 cos 0° + m2v2 cos 105° + m3v3 cosθ = 0 y-direction: m1v1 sin 0° + m2v2 sin 105° + m3v3 sinθ = 0 (c) p1x = 576 kg ⋅ m/s, p2x = −241 kg ⋅ m/s (d) p1y = 0, p2y = 898 m/s (e) x-direction: 576 kg ⋅ m/s − 241 kg ⋅ m/s + (112 kg) v3 cosθ = 0 y-direction: 0 + 898 kg ⋅ m/s + (112 kg) v3 sinθ = 0 (f) v3 cosθ = −2.99 m/s, v3 sinθ = −8.02 m/s, v3 = 8.56 m/s (g) θ = 250° (h) Because the third fragment must have a momentum equal in magnitude and opposite direction to the resultant of the other two

Topic 6

334

fragments momenta, all three pieces must travel in the same plane. 6.48

(a) v1f = 0, v2f = 1.50 m/s

(b)

v1f = −1.00 m/s, v2f = 1.50 m/s

v = 0.655 m/s;! vBf = 0.537 m/s ! Af

6.52

No, his initial speed was 41.5 mi/h.

6.54

263 kg/s

6.56

2.57 × 103 m/s

6.58

(a) 13.0 m/s2

(b) 1.54 × 104 kg

(d) 2.14 × 103 m/s 6.60

40.5 g

6.62

0.556 m

6.64

v = 4M g /m !

6.66

0.961 m

6.68

 7 !v = 1.3 × 10 m/s at 220° counterclockwise from the +x-axis

6.70

(a) v1i = +9.90 m/s, v2i = −9.90 m/s (b) v1f = −16.5 m/s, v2f = +3.30 m/s (c) h1f = 13.9 m, h2f = 0.556 m

Topic 6

335

6.72

(a) !vm = v0 2 ; v3m = v0 2/3

6.74

(a) vf = (v1 + 2v2)/3

6.76

(a) See Solution.

(b)

θ3m = 35.3°

(b) !ΔKE = −m ( v1 − 2v1v2 + v2 ) /3 = −m(v1 − v2)2/3 2

(b) From Newton’s third law, the two horizontal forces are equal in magnitude and opposite in direction.

   (c) !Δp A = −2Mv/3, ΔpB = +2Mv/3, ΔpC = 0 (d) Kinetic energy is not conserved in this inelastic collision. (b)

2.9 × 103 m/s2

(b)

2.54 m

6.78

(a) 33 m/s

6.80

6.15 m/s

6.82

(a) 7.06 m/s

6.84

(a) Conservation of mechanical energy of the bullet-block-Earth system from just after impact until maximum height is reached may be used to relate the speed of the block and bullet just after collision to the maximum height. Then, conservation of momentum from just before to just after impact can be used to relate the initial speed of the bullet to the speed of the block and bullet just after collision. (b) !vi = [( M + m) /m] 2gh

Topic 6

336

PROBLEM SOLUTIONS 6.1

6.2

Use p = mv (a)

p = (1.67 × 10−27 kg)(5.00 × 106 m/s) = 8.35 × 10−21 kg ⋅ m/s

(b)

p = (1.50 × 10−2 kg)(3.00 × 102 m/s) = 4.50 kg ⋅ m/s

(c)

p = (75.0 kg)(10.0 m/s) = 750 kg ⋅ m/s

(d)

p = (5.98 × 1024 kg)(2.98 × 104 m/s) = 1.78 × 1029 kg ⋅ m/s

From the impulse-momentum theorem, Fav(Δt) = Δp = mvf − mvi, we find the average force to be

( F =

m v f − vi

6.3

(a)

( Δt)

) = (55 × 10 kg)(2.0 × 10 ft/s − 0) ⎛ 1 m/s ⎞ = 1.7 kN 2

−3

0.002 0 s − 0

⎜ ⎟ ⎝ 3.281 ft/s ⎠

If Pball = Pbullet, then −3 3 mbullet v bullet ( 3.00 × 10 kg ) (1.50 × 10 m/s ) v ball = = = 31.0!m/s m 0.145!kg ball !

(b)

The kinetic energy of the bullet is

3.00 × 10−3 !kg ) (1.50 × 103 !m/s ) ( 1 2 KEbullet = mbullet v bullet = = 3.38 × 103 !J 2 2 ! 2

while that of the baseball is 1 (0.145 kg)(31.0 m/s)2 2 KEball = mball v ball = = 69.7 J 2 2 !

The bullet has the larger kinetic energy by a factor of 48.5. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 6

6.9

337

(a)

We choose the positive direction to be the direction of the final velocity of the ball. I = Δp = m(vf − vi) = (0.280 kg)[+22.0 m/s − (−15.0 m/s)] or I = +10.4 kg ⋅ m/s = 10.4 kg ⋅ m/s in the direction of the final velocity.

(b)

The average force the player exerts on the ball is I 10.4 kg ⋅ m/s Fav = = = 173 N Δt 0.060 0s !

By Newton’s third law, the ball exerts a force of equal magnitude back on the player’s fist. 6.5

(a)

If Δm is the mass of rain hitting the roof in time Δt, the impulse imparted to the rain by the roof is

    Irain = Fav Δt = (Δm)v f − (Δm)v i rain !

( )

or (taking upward as positive)  Fav

( )

 0 − (Δm)v i = = (0.035!kg/s)[ 0 − (−12!m/s)] = +0.42!N rain Δt

From Newton’s third law, the average force the rain exerts on the roof is



(F ) !

av roof

(b)

 = − Fav

( )

rain

= −0.42 N = 0.42 N downward

Hailstones striking the roof would rebound upward, and hence

Topic 6

338

experience a greater change in momentum than that experienced by an equal mass of liquid water which strikes the roof without   rebounding. Thus, the impulse-momentum theorem, !F = Δp/Δt ,

tells us that the hail will experience a greater average force than that experienced by an equal mass of water striking the roof. Newton’s third law then tells us that the hailstones will exert the greater force on the roof

6.6

6.7

6.8

mv 2 m2 v 2 (mv)2 p2 KE = = = = 2 2m 2m 2m ! (a)

KE 12 mv 2 v = = p mv 2 !

(b)

p 25.0 kg ⋅ m/s m= = = 1.14 kg v 22.0 m/s !

(a)

The impulse delivered by a force is equal to the area under the

2(KE) 2(275 J) = = 22.0 m/s p 25.0 kg ⋅ m/s

Force versus Time curve. From the figure below, this is seen to be a triangular area having a base of 1.50 ms = 1.50 × 10−3 s and altitude of 18 000 N. Thus,

(b)

(1.50 × 10 s)(18 000 N) = 13.5 N ⋅s 2 1

−3

1 13.5 N ⋅s Fav = = = 9.00 × 103 N = 9.00 kN −3 Δt 1.50 × 10 s !

Topic 6

6.9

339

! ! ! (a) The soccer ball’s momentum change is Δp = p f − pi . Here, the

! soccer ball is initially at rest with pi = 0 so that pi,x = pi,y = 0.

The x- and y-component of the momentum change are

Δpx = p f ,x − pi ,x

= mvx − 0 = ( 0.425 kg ) ( 22.5 m/s) cos ( 35.0° ) = 7.83 kg ⋅ m/s

Δpy = p f ,y − pi ,y = mv y − 0

= ( 0.425 kg ) ( 22.5 m/s) sin ( 35.0° ) = 5.49 kg ⋅ m/s

(b) From the impulse-momentum theorem, the magnitude of the ! average force is Fav = Δp /Δt where Δt is the time interval the

player’s foot is in contact with the ball. Substitute known values to find

! Δp Fav = Δt

( Δp ) + ( Δp ) 2

Δt

(7.83 kg ⋅ m/s) + ( 5.49 kg ⋅ m/s) = 2

5.00 × 10−2 m/s

Topic 6

6.10

340

(a)

Fav =

I , where I is the impulse the man must deliver to the Δt

child: |I| = mchild|vf − v0|

Fav =

mchild v f − v0 Δt

(12.0 kg ) 0 − 120 mi/h ⎛ 0.447 m/s ⎞ 0.10 s

⎜⎝ ⎟ 1 mi/h ⎠

= 6.4 × 103 N

! (b)

⎛ 0.224 8 lb ⎞ Fav = ( 6.4 × 103 N ) ⎜ = 1.4 × 103 lb ⎟ ⎝ 1N ⎠

It is unlikely that the man has sufficient arm strength to guarantee the safety of the child during a collision. The violent forces during the collision would tear the child from his arms.

(c)

The laws are soundly based on physical principles: always wear a seat belt when in a car.

6.11

The velocity of the ball just before impact is found from 2 v 2 = v0y + 2ay Δy as ! y

2 v = − v0y + 2ay Δy = − 0 + 2 ( −9.80 m/s 2 ) ( −1.25 m ) = −4.95 m/s !1

and the rebound velocity with which it leaves the floor is v = + v 2f − 2ay Δy = − 0 − 2 ( −9.80 m/s 2 ) (+0.960 m) = +4.34 m/s ! 2

The impulse given the ball by the floor is then

Topic 6

341

! ! ! ! ! I = FΔt = Δ(mv) = m(v 2 − v1 )

= (0.150 kg)[ +4.34 m/s − (−4.95 m/s)] = +1.39 N ⋅ s = 1.39 N ⋅s upward

6.12

Take the direction of the ball’s final velocity (toward the net) to be the +x-direction. (a)

I = Δp = m(vf − vi) = (0.060 0 kg)[40.0 m/s − (−50.0 m/s)] giving I = +5.40 kg ⋅ m/s = 5.40 N ⋅ s in direction of final velocity.

(b)

6.13

(a)

(

)

1 Wact = ΔKE = m v 2f − vi2 2 2 2⎤ ⎡ 0.060 0 kg ⎢ 40.0 m/s − 50.0 m/s ⎥ ⎣ ⎦ = −27.0 J = 2

(

)(

) (

)

Taking forward as the positive direction, I = m(Δv) = (70.0 kg)(5.20 m/s − 0) = +364 kg ⋅ m/s = 364 N ⋅ s forward

6.14

(b)

1 +364!kg ⋅ m/s Fav = = = 438!kg ⋅ m/s 2 = 438!N!forward Δt 0.832!s !

(a)

Choose upward as the positive direction: I = m(vf − vi) = (65.0 kg)(+1.80 m/s − 0) = +117 kg ⋅ m/s = 117 kg ⋅ m/s upward

(b)

Before the jump, the player is in equilibrium, so ΣFy = may = 0 ⇒ F1 = mg

Topic 6

342

F1 = (65.0 kg)(9.80 m/s2) = +637 N = 637 N upward

or (c)

From the impulse-momentum theorem, the net force the player experiences during the jump is 1 +117 kg ⋅ m/s Fnet = = = +260 N Δt 0.450 s !

But Fnet = F2 − mg = F2 − F1, where F2 is the upward force the floor exerts on the player during the jump and F1 is the force exerted by the floor before the jump. Thus, F2 = Fnet + F1 = +260 N + 637 N = +897 N = 897 N upward

6.15

(a)

The impulse equals the area under the F versus t graph. This area is the sum of the area of the rectangle plus the area of the triangle. Thus, 1 I = (2.0 N)(3.0 s) + (2.0 N)(2.0 s) = 8.0 N ⋅s 2 !

(b)

I = Fav(Δt) = Δp = m(vf − vi) 8.0 N ⋅ s = (1.5 kg)vf − 0, giving vf = 5.3 m/s

Topic 6

343

(c)

(

)

I I = Fav (Δt) = Δp = m v f − vi ,!so!v f = vi + m ! 8.0 N ⋅s v f = −2.0 m/s!+ = 3.3 m/s 1.5 kg !

6.16

(a)

Impulse = area under curve = (two triangular areas of altitude 4.00 N and base 2.00 s) + (one rectangular area of width 1.00 s and height of 4.00 N). Thus, ⎡ (4.00 N)(2.00 s) ⎤ I = 2⎢ ⎥⎦ + (4.00 N)(1.00 s) = 12.0 N ⋅s 2 ⎣ !

(b)

I = Fav(Δt) = Δp = m(vf − vi), so vf = vi + I/m 12.0 N ⋅s vf = 0 + = 6.00 m/s 2.00 kg !

6.17

(c)

I 12.0 N ⋅s v f = vi + = −2.00 m/s + = 4.00 m/s m 2.00 kg !

(a)

The impulse is the area under the curve between 0 and 3.0 s. This is:

(b)

I = (4.0 N)(3.0 s) = 12 N ⋅ s

The area under the curve between 0 and 5.0 s is I = (4.0 N)(3.0 s) + (−2.0 N)(2.0 s) = 8.0 N ⋅ s For parts (c) and (d), we use I = Fav(Δt) = Δp = m(vf − vi), giving vf = vi + I/m

(c)

At 3.0 s:

I 12 N ⋅s v f = vi + = 0 + = 8.0 m/s m 1.50 kg !

Topic 6

344

(d)

6.18

At 5.0 s:

  Δp Fav = so Δt !

I 8.0 N ⋅s v f = vi + = 0 + = 5.3 m/s m 1.50 kg !

Δp

( Fav )x = Δtx

and

Δp

( Fav )y = Δty

( ) ( )

⎡m v − v ⎤ y y ⎥ m[ vcos !60.0° − vcos !60.0° ] f i⎦ = ( Fav )y = ⎢⎣ Δt Δt !

⎡m ( vx ) − ( vx ) ⎤ m[( −vsin !60.0° ) − ( +vsin !60.0° )] f i⎦ = ( Fav )x = ⎣ Δt Δt ! = !

−2mvsin 60.0° −2(3.00 kg)(10.0 m/s)sin 60.0° = = −260 N Δt 0.200s

 Thus, !Fav = 260 N in the negative x-direction or perpendicular to the wall

6.19

(a)

Δx 2 ( Δx ) 2(1.20!m) Δt = = = = 9.60 × 10−2 !s vav v f + vi 0 + 25.0!m/s !

(b)

Δp m(Δv) (1400 kg)(25.0 m/s) Fav = = = = 365 × 105 N −2 Δt Δt 9.60 × 10 s !

6.20

Δv Δt

⎛ ⎞ 1g 2 2 = 260 m/s = 260 m/s ⎜ ⎟ = 26.5 g 9.60 × 10−2 s ⎝ 9.80 m/s 2 ⎠ 25.0 m/s

(

)

We shall take toward the pitcher as the positive direction. Then, the

Topic 6

345

velocity of the ball just before contact with the bat is vi = −42 m/s, and its velocity just as it leaves the bat is vf = +48 m/s (a)

I = Δp = m(Δv) = (0.14 kg)[48 m/s − (−42 m/s)] = (0.14 kg)(90 m/s) yielding I = 13 kg ⋅ m/s = 13 N ⋅ s.

(b)

Also, I = Fav ⋅ Δt, so

Fav =

(c)

I Δt

+13 N ⋅s 0.005 0 s

= +2.6 × 103 N = 2.6 kN toward the pitcher

The magnitude of the impulsive force is much larger than the weight, with Fav = 2.6 kN and the weight being w = mg = (0.14 kg)(9.80 m/s2) = 1.4 N.

6.21

Requiring that total momentum be conserved gives

(mclubvclub + mball vball ) f = (mclubvclub + mball vball )i

(200. g)(40.0 m/s) + (46.0 g)vball = (200. g)(55.0 m/s) + 0

and vball = 65.2 m/s 6.22

(a)

The mass of the rifle is m=

w 30.0 N ⎛ 30.0 ⎞ = =⎜ ⎟ kg 2 g 9.80 m/s ⎝ 9.80 ⎠

We choose the direction of the bullet’s motion to be negative. Then, conservation of momentum gives

Topic 6

346

(mrifle vrifle + mbullet vbullet ) f = (mrifle vrifile + mbullet vbullet )i

! or

[(30.0/9.80) kg]vrifle + (5.00 × 10−3 kg)( −300. m/s) = 0 + 0

and vrifle =

(b)

9.80 ( 5.00 × 10−3 kg ) (300. m/s) 30.0 kg

= 0.490 m/s

The mass of the man plus rifle is

730 N 9.80 m/s 2

= 74.5 kg

We use the same approach as in (a), to find

⎛ 5.00 × 10−3 kg ⎞ v=⎜ (300. m/s) = 2.01 × 10−2 m/s ⎟ ⎝ 74.5 kg ⎠ 6.23

The velocity of the girl relative to the ice, vGI, is vGI = vGP + vPI, where vGP = velocity of girl relative to plank, and vPI = velocity of plank relative to ice Since we are given that vGP = 1.50 m/s, this becomes vGI = 1.50 m/s + vPI (a)

[1]

Conservation of momentum gives mGvGI + mPvPI = 0, or vPI = − (mG/mP)vGI

[2]

Then, Equation [1] becomes ⎛m ⎞ vGI = 1.50 m/s − ⎜ G ⎟ vGI ⎝ mP ⎠ !

⎛ m ⎞ or ⎜ 1 + G ⎟ vGI = 1.50 m/s mP ⎠ ⎝

Topic 6

347

vGI = ! (b)

1.50 m/s = 1.15 m/s ⎛ 45.0 kg ⎞ 1+ ⎜ ⎝ 150 kg ⎟⎠

Then, using Equation [2] above,

⎛ 45.0 kg ⎞ vPI = − ⎜ (1.15 m/s) = −0.345 m/s ⎝ 150 kg ⎟⎠ ! or 6.24

vPI = 0.345 m/s directed opposite to the girl's motion

Originally, with both girl and plank at rest, the total momentum of the girl-plank system is zero. With negligible friction between the plank and the ice, the total momentum of the girl-plank system is conserved. (a)

The velocity of the girl relative to the ice is given by

! ! ! ! v GI = v GP + v PI where vGP is the velocity of the girl relative to the

! plank and v PI is the velocity of the plank relative to the ice. Conservation of momentum of the girl-plank system then gives ! ! ! ! ! 0 = mG vGI + mP v PI = mG ( vGP + v PI ) + mP v PI

(b)

! ! 0 = mG v GP + ( mG + mP ) v PI

and

⎛ mG ⎞ ! ! v PI = − ⎜ v GP ⎝ mG + mP ⎟⎠

The velocity of the girl relative to the ice is

⎛ mG ⎞ ! ⎛ m + mP − mG ⎞ ! ! ! ! vGI = v PI + v PI = ⎜ vGP = ⎜ G vGP ⎟ ⎝ mG + mP ⎠ ⎝ mG + mP ⎟⎠

Topic 6

348

6.25

⎛ mP ⎞ ! ! v GI = ⎜ v GP ⎝ mG + mP ⎟⎠

After the 1.50-kg squid has taken in the 0.100 kg of water, the squidwater system is stationary and has no momentum. After ejecting the water, conservation of momentum gives pi = p f 0 = ms vs + mw vw vs = −

mw 0.100 kg vw = − ( 3.25 m/s) ms 1.50 kg vs = −0.217 m/s

so the squid achieves a speed of 0.217 m/s . 6.26

The boat and fisherman move as a single unit having mass mBF = mB + mF = 125 kg + 75 kg = 200 kg Before the package is thrown, all parts of the system, (boat + fisherman) and package, are at rest, so the total initial momentum is zero. Neglecting water resistance, the final momentum of the system must also be zero, or mBFvBF + mpvp = 0

giving

⎛ 15 kg ⎞ ⎛ mp ⎞ vBF = − ⎜ vp = − ⎜ (+4.5 m/s) ⎟ ⎝ mBF ⎠ ⎝ 200 kg ⎟⎠ !

Topic 6

349

and 6.27

vBF = −0.34 m/s

0.34 m/s toward the left

Consider the thrower first, with velocity after the throw of vthrower. Applying conservation of momentum yields (65.0 kg)vthrower + (0.045 0 kg)(30.0 m/s) = (65.0 kg + 0.045 0 kg)(2.50 m/s) or

vthrower = 2.48 m/s.

Now, consider the (catcher + ball), with velocity of vcatcher after the catch. From momentum conservation, (60.0 kg + 0.045 0 kg)vcatcher = (0.045 0 kg)(30.0 m.s) + (60.0 kg)(0) or 6.28

vcatcher = 2.25 × 10−2 m/s.

Note: We consider the spring to have negligible mass and ignore any energy or momentum it may possess after being released. Also, we take toward the right as the positive direction.

  Conservation of momentum, p f = pi , gives ! m1v1 + m2v2 = 0

⎛m ⎞ v2 = − ⎜ 1 ⎟ v1 ⎝ m2 ⎠ !

[1]

Since the surface is frictionless, conservation of energy, KE1f + KE2f + PEs,f = KE1i + KE2i + PEs,i with KE1i = KE2i = PEs,f = 0 gives © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 6

350

1 1 1 m1v12 + m2 v22 = kd 2 2 2 !2

[2]

where d = 9.8 cm is the initial compression of the spring. Substituting Equation [1] into Equation [2] gives: 1 1 ⎛ m2 ⎞ 1 m1v12 + m2 ⎜ 12 v12 ⎟ = kd 2 2 ⎝ m2 ⎠ 2 !2

⎛ m ⎞ or m1 ⎜ 1 + 1 ⎟ v12 = kd 2 m2 ⎠ ⎝

Choosing the negative sign (since m1will recoil to the left), this result yields

k 280 N/m v1 = −d = − ( 9.8 × 10−2 m ) m1 (1 + m1 /m2 ) (0.56 kg)(1 + 0.56/0.88) ! and v1 = −1.7 m/s

1.7 m/s to the left

Then, Equation [1] gives the velocity of the second object as ⎛ 0.56 ⎞ v2 = − ⎜ ( −1.7 m/s ) = +1.1 m/s ⎝ 0.88 ⎟⎠ !

6.29

(a)

1.1 m/s to the right.

Using a subscript a for the astronaut and t for the tank, conservation of momentum gives mavaf + mtvtf = mavai + mtvti. Since both astronaut and tank were initially at rest, this becomes

mavaf + mtvtf = 0 + 0

⎛m ⎞ vaf = − ⎜ t ⎟ vtf ⎝ ma ⎠ !

The mass of the astronaut alone (after the oxygen tank has been discarded) is ma = 75.0 kg. Taking toward the spacecraft as the

Topic 6

351

positive direction, the velocity imparted to the astronaut is

⎛ 12.0 kg ⎞ vaf = − ⎜ (−8.00 m/s) = +1.28 m/s ⎝ 75.0 kg ⎟⎠ ! and the distance she will move in 2.00 min is d = vaf t = (1.28 m/s)(120 s) = 154 m (b)

By Newton’s third law, when the astronaut exerts a force on the tank, the tank exerts a force back on the astronaut. This reaction force accelerates the astronaut towards the spacecraft.

6.30

! The momentum vector p is conserved so pi,x = pf,x and pi,y = pf,y. While the skaters are at rest, pi,x = 0 and pi,y = 0. Conservation of momentum gives pi ,x = p f ,x 0 = mB vB + mC vx vx = −

mB vB (75.0 kg)( −4.00 m/s ) =− mC 90.0 kg

= 3.33 m/s

and pi ,y = p f ,y 0 = mA vA + mC v y

vy = −

mA vA ( 80.0 kg)( −3.50 m/s ) =− mC 90.0 kg

= 3.11 m/s

Topic 6

6.31

352

(a)

(b)

The collision is best described as perfectly inelastic, because the skaters remain in contact after the collision.

(c)

m1v1 + m2v2 = (m1 + m2) vf

(d)

vf =

m1v1 + m2 v2 m1 + m2

vf = !

(70.0 kg)(8.00 m/s)!+!(50.0 kg)(4.00 m/s) = 6.33 m/s 70.0 kg + 50.0 kg

(e)

6.32

Consider a system consisting of arrow and target from the instant just before impact until the instant after the arrow emerges from the target. No external horizontal forces act on the system, so total horizontal momentum must be conserved, or

(mava + mt vt ) f = (mava + mt vt )i

Thus,

( va ) f = = !

6.33

ma ( va )i + mi ( vi )t − mt ( vt ) f ma

( 22.5 g )( +35.0 m / s ) + ( 300 g )( −2.50 m / s ) − 0 = 1.67 m / s 22.5 g

When Gayle jumps on the sled, conservation of momentum gives (50.0 kg + 5.00 kg)v2 = (50.0 kg)(4.00 m/s) + 0

Topic 6

353

or the speed of Gayle and the sled as they start down the hill is v2 = 3.64 m/s. After Gayle and the sled glide down 5.00 m, conservation of mechanical energy (taking y = 0 at the level of the top of the hill) gives

(

)

(

)

(

)

1 1 55.0 kg v32 + 55.0 kg ( 9.80 m/s 2 ) (−5.00 m) = 55.0 kg (3.64 m/s)2 + 0 2 !2

so Gayle’s speed just before the brother hops on is !v3 = 111 m/s . After her brother jumps on, conservation of momentum yields

(

)

(55.0 kg + 30.0 kg) v4 = (55.0 kg) 111 m/s + 0 ! and the speed of Gayle, brother, and sled just after her brother hops on is v4 = 6.82 m/s. After all slide an additional 10.0 m down (to a level 15.0 m below the level of the hilltop), conservation of mechanical energy from just after her brother hops on to the end gives the final speed as

( 85.0 kg ) v + ( 85.0 kg )(9.80 m/s )(−15.0 m) 2 1 = ( 85.0 kg ) ( 6.82 m/s ) + ( 85.0 kg ) ( 9.80 m/s ) ( −5.00 m ) 2 1

or 6.34

v5 = 15.6 m/s

For each skater, the impulse-momentum theorem gives

Topic 6

354

Fav = !

Δp m Δv (75.0 kg ) ( 5.00 m/s ) = = = 3.75 × 103 N Δt Δt 0.100 s

Since Fav < 4 500 N, there are no broken bones. 6.35

(a)

If M is the mass of a single car, conservation of momentum gives (3M)vf = M(3.00 m/s) + (2M)(1.20 m/s) or

(b)

vf = 1.80 m/s

The kinetic energy lost is KElost = KEi − KEf, or 1 1 1 2 2 2 KElost = M ( 3.00 m / s ) + ( 2M ) (1.20 m / s ) − ( 3M ) (1.80 m / s ) 2 2 2 !

With M = 2.00 × 104 kg, this yields KElost = 2.16 × 104 J 6.36

(a)

From conservation of momentum, (3M)vf = Mv1 + (2M)v2

(b)

1 v f = (v1 + 2v2 ) 3

The kinetic energy before is 1 1 M KEi = Mv12 + (2M)v22 = ( v12 + 2v22 ) 2 2 2 !

Topic 6

355

and after collision 1 3M ⎡ (v1 + 2v2 )2 ⎤ M 2 2 KE f = (3M)v f = = ( v1 + 4v1v2 + 4v22 ) ⎢ ⎥ 2 2 ⎣ 9 ⎦ 6 ! M 2 2M 2M 2 KE f = v1 + v1v2 + v2 6 3 3 !

The kinetic energy lost is 2⎞ 2 ⎛ 1 1⎞ ⎛ KEi − KE f = ⎜ − ⎟ Mv12 + ⎜ 1 − ⎟ Mv22 − Mv1v2 ⎝ ⎠ ⎝ ⎠ 2 6 3 3 !

6.37

(a)

KEi − KE f =

M 2 M 2 v1 + v22 − 2v1v2 ) = v1 − v2 ) ( ( 3 3

Because momentum is conserved even in a perfectly inelastic collision such as this, the ratio is pf/pi = 1.

(b)

p f = pi !

⇒

(m1 + m2 ) v f = m1v1i + m2 (0)

1 1 1 KEi = m1v1i2 + m2 (0) = m1v1i1 2 2 2 !

vf =

KE f =

1 m1 + m2 ) v 2f ( 2

(m + m ) v = (m + m ) m v = m = KE mv m v (m + m ) m + m

KE f i

6.38

and

m1v1i m1 + m2

2 l li

2 f

2 l

2 li

Choose the positive x-axis in the direction of the initial velocity of the cue ball. Let vci be the initial speed of the cue ball, vcf be the final speed of the cue ball, vTf be the final speed of the target, and θ be the angle the target’s final velocity makes with the x-axis.

Topic 6

356

Conservation of momentum in the x-direction, recognizing that all billiard balls have the same mass, gives mvTf cosθ + mvcf cos 30.0° = 0 + mvci or vTf cosθ = vci − vcf cos 30.0°

[1]

The conservation equation for momentum in the y-direction is mvTf sinθ + mvcf sin 30.0° = 0 + 0

vTf sinθ = −vcf sin 30.0°

[2]

Since this is an elastic collision, kinetic energy is conserved, giving 1 1 1 mvTf2 + mvcf2 = mvci2 2 2 !2

(b)

or vTf2 = vci2 − vcf2

[3]

To solve, square Equations [1] and [2] and add the results to obtain

(

)

(

vTf2 cos 2 θ + sin 2 θ = vci2 − 2vci vcf cos 30.0° + vcf2 cos 2 30.0° + sin 2 30.0° or

)

v 2 = vci2 − 2vci vcf cos 30.0° + vcf2 ! Tf

Now, substitute this result into Equation [3] to get

vci2 − 2vci vcf cos 30.0° + vcf2 = vci2 − vcf2

(

)

or 2vcf vcf − vci cos 30.0° = 0

Since vcf ≠ 0, it is necessary that

v = vci cos 30.0° = (4.00 m/s)cos 30.0° = 3.46 m/s ! cf

Topic 6

357

2 2 Then, Equation [3] yields vTf = vci − vcf , or !

v = ! Tf (a)

( 4.00 m/s ) − ( 3.46 m/s ) = 2.00 m/s 2

With the results found above, Equation [2] gives

⎛ vcf ⎞ ⎛ 3.46 m/s ⎞ sin θ = − ⎜ ⎟ sin 30.0° = − ⎜ sin 30.0° = −0.866 ⎝ 2.00 m/s ⎟⎠ ⎝ vTf ⎠ or θ = −60.0°. Thus, the angle between the velocity vectors after collision is

φ = 30.0° − (−60.0°) = 90.0°. 6.39

The leftmost part of the sketch below depicts the situation from when the actor starts from rest until just before he makes contact with his costar.

Using conservation of energy over this period gives (KE + PE)1 = (KE + PE)i

Topic 6

358

1 m1v12 + 0 = 0 + mgR !2

so his speed just before impact is v = 2gR = 2 ( 9.80 m/s 2 ) (3.75 m) = 8.57 m/s !1

Now, employing conservation of momentum from just before to just after impact gives

!(m1 + m2 )v0 = m1v1 + m2 (0) or m1v1 (80.0 kg)(8.57 m/s) v0 = = = 5.08 m/s m1 + m2 80.0 kg + 55.0 kg !

Finally, using conservation of energy from just after impact to the end yields

( KE + PE ) f = ( KE + PE )0

and

0 + ( m1 + m2 ) gh =

1 m1 + m2 ) v02 ( 2

v2 (5.08 m/s)2 h= 0 = = 1.32 m 2g 2 ( 9.08 m/s 2 ) !

6.40

Consider the sketches above, which show the situation just before and just after collision.

Topic 6

359

Conserving momentum in y-direction:

pyf = pyi ⇒ mvOf sin 37.0° − mvGf sin 53.0° = 0

⎛ sin 37.0° ⎞ vGf = ⎜ ⎟ vOf = 0.754vOf ⎝ sin 53.0° ⎠

Now, conserving momentum in the x-direction:

pxf = pxi ⇒ mvOf cos 37.0° + mvGf cos 53.0° = mvOi + 0 or

vOf cos 37.0° + (0.754vOf)cos 53.0° = vOi

and vOf =

vOi 5.00 m/s = = 3.99 m/s cos 37.0° + ( 0.754 ) cos 53.0° cos 37.0° + ( 0.754 ) cos 53.0°

Then,

vGf = 0.754vOf = 0.754 (3.99 m/s) = 3.01 m/s

Now, we can verify that this collision was indeed an elastic collision:

KEi =

1 2

2 mvOi =

(5.00 m/s) = m(12.5 m /s ) 2

and

KE f =

so 6.41

1 m m 2 2 mvOf + mvOf = (3.99 m/s)2 + (3.01 m/s)2 = m(12.5 m 2 /s 2 ) 2 2 2 2

KEf = KEi, which is the criterion for an elastic collision.

Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V =

Topic 6

360

the initial velocity of the ball-bullet combination. Then, using conservation of momentum from just before to just after collision gives ⎛ m ⎞ (M + m)V = mv + 0 or V = ⎜ ⎟⎠ v ⎝ M = m !

Now, we use conservation of mechanical energy from just after the collision until the ball reaches maximum height to find 2

1 V2 1 ⎛ m ⎞ 2 0 + ( M + m) ghmax = ( M + m) V 2 + 0 or hmax = = ⎜⎝ ⎟⎠ v 2 2g 2g M + m !

With the data values provided, this becomes 2

⎛ ⎞ 1 0.030 kg 2 hmax = 200 m/s ) = 57 m ( 2 ⎜ ⎟ 2 ( 9.80 m/s ) ⎝ 0.15 kg + 0.030 kg ⎠ ! 6.42

First, we will find the horizontal speed, v0x, of the block and embedded bullet just after impact. After this instant, the block-bullet combination is a projectile, and we find the time to reach the floor by use of Δy = v0yt + 12 ayt 2 , which becomes ! 1 −1.00m = 0 + ( −9.80 m/s 2 ) t 2 , giving 2 !

t = 0.452 s

Thus, Δx 2.00 m v0x = = = 4.42 m/s t 0.452 s !

Now use conservation of momentum for the collision, with vb = speed © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 6

361

of incoming bullet: (8.00 × 10−3 kg)vb + 0 = [(250 + 8) × 10−3 kg](4.42 m/s) so 6.43

vb = 143 m/s

(about 320 mph)

First, we use conservation of mechanical energy to find the speed of the block and embedded bullet just after impact:

(KE + PE ) = (KE + PE ) becomes 12 (m + M )V + 0 = 0 + 12 kx 2

s f

s i

and yields

(150 N/m)(0.800 m) = 29.3 m/s V= = m+ M (0.012 0 + 0.100) kg 2

kx 2

Now, employ conservation of momentum to find the speed of the bullet just before impact: mv + M(0) = (m + M)V, or

⎛ 0.112 kg ⎞ ⎛ m+ M⎞ v=⎜ V = ⎜ ⎟ (29.3 m/s) = 273 m/s ⎟ ⎝ m ⎠ ⎝ 0.012 0 kg ⎠ 6.44

(a)

Conservation of momentum gives mTvfT + mcvfc = mTviT + mcvic, or

v fT =

(

mT viT + mc vic − v fc mT

)

9 000 kg ) ( 20.0 m/s ) + (1 200 kg ) ⎡( 25.0 − 18.0 ) m/s ⎤ ( ⎣ ⎦ = 9 000 kg

vfT = 20.9 m/s east

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362

(b) 1 1 ⎡1 ⎤ ⎡1 ⎤ KElost = KEi − KE f = ⎢ mc vic2 + mT viT2 ⎥ − ⎢ mc v 2fc + mT v 2fT ⎥ 2 2 2 2 ⎣ ⎦ ⎣ ⎦ 1⎡ = ⎣ mc vic2 − v 2fc + mT viT2 − v 2fT ⎤⎦ 2 1 2 2 2 2 = (1 200 kg ) ⎡⎣( 25.0 ) − (18.0 ) ⎤⎦ ( m 2 /s 2 ) + ( 9 000 kg ) ⎡⎣( 20.0 ) − ( 20.9 ) ( m 2 /s 2 ) ⎤⎦ 2

(

)

(

)

= 1.50 × 10 4 J, which becomes internal energy

6.45

(a)

The speed v of both balls just before the basketball reaches the 2 2 ground may be found from !vy = v0y + 2ay Δy as

v = v02 ,+2a,Δy = 0 + 2 ( −g ) ( −h ) = 2gh = 2 ( 9.80 m/s 2 ) (1.20 m ) = 4.85 m/s (b)

Immediately after the basketball rebounds from the floor, it and the tennis ball meet in an elastic collision. The velocities of the two balls just before collision are: for the tennis ball: vti = −v for the basketball: vbi = +v We determine the velocity of the tennis ball immediately after this elastic collision as follows: Momentum conservation gives mtvtf + mbvbf = mtvti + mbvbi or mtvtf + mbvbf = (mb − mt)v

[1]

From the criteria for a perfectly elastic collision: © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 6

363

vti − vbi = −(vtf − vbf) or vbf = vtf + vti − vbi = vtf − 2v

[2]

Substituting Equation [2] into [1] gives mtvtf + mb(vtf − 2v) = (mb − mt)v or the upward speed of the tennis ball immediately after the collision is

⎛ 3m − mt ⎞ ⎛ 3mb − mt ⎞ vtf = ⎜ b v = ⎜⎝ m + m ⎟⎠ 2gh ⎝ mt + mb ⎟⎠ t b ! The vertical displacement of the tennis ball during its rebound following the collision is given by !vy = v0y + 2ay Δy as 2

Δy =

2 v y2 − v0y

2ay

0 − vtf2

( )

2 −g

( )

⎛ 3m − m ⎞ 1 ⎛ 3mb − mt ⎞ b t ⎜ ⎟ 2g h = ⎜ ⎟h 2g ⎝ mt + mb ⎠ ⎝ mt + mb ⎠ 2

6.46

⎡ 3(590 g) − (57.0 g) ⎤ Δy = ⎢ ⎥ (1.20 m) = 8.41 m ⎢⎣ 57.0 g + 590 g ⎥⎦

(a)

(b)

x-direction: Σpxf = Σpxi ⇒ m1v1 cos 0° + m2v2 cos 105° + m3v3 cosθ = 0

Topic 6

364

y-direction: Σpyf = Σpyi ⇒ m1v1 sin 0° + m2v2 sin 105° + m3v3 sinθ = 0 (c)

p1x = m1v1 cos 0° = (48.0 kg)(12.0 m/s)(1) = 576 kg ⋅ m/s p2x = m2v2 cos 105° = (62.0 kg)(15.0 m/s)(−0.259) = −241 kg ⋅ m/s

(d)

p1y = m1v1 sin 0° = (48.0 kg)(12.0 m/s)(0) = 0 p2y = m2v2 sin 105° = (62.0 kg)(15.0 m/s)(+0.966) = 898 kg ⋅ m/s

(e)

(f)

x-direction:

576 kg ⋅ m/s − 241 kg ⋅ m/s + (112 kg)v3 cosθ = 0

y-direction:

0 + 898 kg ⋅ m/s + (112 kg)v3 sinθ = 0

x-direction: v3 cosθ = !

−576 kg ⋅ m/s + 241 kg ⋅ m/s 112 kg

v3 cos θ = −2.99 m/s

y-direction: v3 sin θ = !

−898 kg ⋅ m/s 112 kg

or v3 sinθ = −8.02 m/s

Then, squaring and adding these results, recognizing that cos2θ + sin2θ = 1, gives

v 2 cos 2 θ + sin 2 θ ) = ( −2.99 m/s ) + ( −8.02 m/s ) and ! 3( 2

v3 = 73.3 m 2 /s2 = 8.56 m/s !

Topic 6

365

(g)

v3 sin θ v3 cos θ

= tan θ =

−8.02 m/s −2.99 m/s

= 2.68

θ = tan−1(2.68) + 180° = 250°

Note that the factor of 180° was included in the last calculation because it was recognized that both the sine and cosine of angle q were negative. This meant that θ had to be a third quadrant angle. Use of the inverse tangent function alone yields only the principle angles (−90° ≤ θ ≤ +90°) that have the given value for the tangent function. (h)

Because the third fragment must have a momentum equal in magnitude and opposite direction to the resultant of the other two fragments momenta, all three pieces must travel in the same plane.

6.47

Conservation of momentum gives (25.0 g)v1f + (10.0 g)v2f = (25.0 g)(20.0 cm/s) + (10.0 g)(15.0 cm/s) or

2.50v1f + v2f = 65.0 cm/s

[1]

For head-on, elastic collisions, we know that v1i − v2i = −(v1f − v2f). Thus, 20.0 cm/s − 15.0 cm/s = −v1f + v2f or v2f = v1f + 5.00 cm/s

[2]

Topic 6

366

Substituting Equation [2] into [1] yields 3.50v1f = 60.0 cm/s or v1f = 17.1 cm/s. Equation [2] then gives v2f = 17.1 cm/s + 5.00 cm/s = 22.1 cm/s. 6.48

First, consider conservation of momentum and write m1v1i + m2v2i = m1v1f + m2v2f Since m1 = m2, this becomes v1i + v2i = v1f + v2f

[1]

For an elastic head-on collision, we also have v1i − v2i = −(v1f − v2f), which may be written as v1i − v2i = −(v1f + v2f)

[2]

Adding Equations [1] and [2] yields v2f = v1i

[3]

Subtracting Equation [2] from [1] gives v1f = v2i

[4]

Equations [3] and [4] show us that, under the conditions of equal mass objects striking one another in a head-on, elastic collision, the two objects simply exchange velocities. Thus, we may write the results of the various collisions as

Topic 6

6.49

367

(a)

v1f = 0, v2f = 1.50 m/s

(b)

v1f = −1.00 m/s, v2f = 1.50 m/s

(c)

v1f = 1.00 m/s, v2f = 1.50 m/s

(a)

Over the short time interval of the collision, external forces have no time to impart significant impulse to the players. The two players move together after the tackle, so the collision is completely inelastic.

(b)

pxf = Σpxi ⇒ (m1 + m2)vf cosθ = m1v1i + 0

m1v1 f ( 90.0 kg )( 5.00 m/s ) = or v f cosθ = (m1 + m2 ) 90.0 kg + 95.0 kg ! and

vf cosθ = 2.43 m/s pyf = Σpyi ⇒ (m1 + m2)vf sinθ = 0 + m2v2i

giving

m2 v2 f (95.0 kg)(3.00 m/s) v f sin θ = = (m1 + m2 ) 90.0 kg + 95.0 kg !

and

Topic 6

368

Therefore, v 2f ( sin 2 θ + cos 2 θ ) = v 2f = (1.54 m/s ) + ( 2.43 m/s ) , and ! 2

v f = 8.28 m 2 / s 2 = 2.88 m/s Also,

v f sin θ 1.54 m/s tan θ = = = 0.634 and θ = tan −1 (0.634) = 32.4° v f cos θ 2.43 m/s ! Thus,

(c)

 v f = 2.88 m/s!at!32.4° north!of!east

1 1 1 KElost = KEi − KE f = m1v1i2 + m2 v2i2 − ( m1 + m2 ) v 2f 2 2 2 !

1 1 = ⎡⎣(90.0 kg)(5.00 m/s)2 + (95.0 kg)(3.00 m/s)2 ⎤⎦ − (185 kg)(2.88 m/s)2 = 785 J 2 ! 2

(d)

The lost kinetic energy is transformed into other forms of energy, such as thermal energy and sound.

6.50

Consider conservation of momentum in the first event (twin A tossing the pack), taking the direction of the velocity given the backpack as positive. This yields mAvAf + mpackvPack = (mA + mpack)(0) = 0 or

vAf = !

−mpack vpack mA

⎛ 12.0 kg ⎞ = −⎜ (+3.00 m/s) = −0.655 m/s and ⎝ 55.0 kg ⎟⎠

|vAf| = 0.655 m/s

Topic 6

369

Conservation of momentum when twin B catches and holds onto the backpack yields (mB + mpack)vBf = mB(0) + mpackvpack

6.51

mpack vpack (12.0 kg)(+3.00 m/s) vBf = = = 0.537 m/s mB + mpack 55.0 kg + 12.0 kg !

Choose the +x-axis to be eastward and the +y-axis northward. If vi is the initial northward speed of the 3 000-kg car, conservation of momentum in the y-direction gives 0 + (3 000 kg)vi = (3 000 kg + 2 000 kg)[(5.22 m/s)sin 40.0°] or

vi = 5.59 m/s, north

Observe that knowledge of the initial speed of the 2 000-kg car was unnecessary for this solution. 6.52

We use conservation of momentum for both eastward and northward components. For the eastward direction: M(13.0 m/s) = 2MVf cos 55.0° For the northward direction: Mv2i = 2MVf sin 55.0° Divide the northward equation by the eastward equation to find

M v2i M (13.0 m/s)

2MV f sin 55.0° 2MV f cos 55.0°

v2i = (13.0 m/s) tan 55.0°

Topic 6

370

yielding

⎡ ⎛ 2.237 mi/h ⎞ ⎤ v2i = ⎢(13.0 m/s ) ⎜ ⎟⎠ ⎥ tan 55.0° = 41.5 mi/ h . 1 m/s ⎝ ⎣ ⎦ !

Thus, the driver of the northbound car was untruthful. 6.53

Choose the x-axis to be along the original line of motion. (a)

From conservation of momentum in the x-direction, m(5.00 m/s) + 0 = m(4.33 m/s)cos 30.0° + mv2f cosθ or

v2f cosθ = 1.25 m/s

[1]

Conservation of momentum in the y-direction gives 0 = m(4.33 m/s)sin 30.0° + mv2f sinθ or

v2f sinθ = −2.16 m/s

[2]

−2.16 = −1.73 and Dividing Equation [2] by [1] gives tan θ = 1.25 !

θ = −60.0° Then, either Equation [1] or [2] gives v2f = 2.50 m/s, so the final

 velocity of the second ball is v 2 f = 2.50 m/s at − 60.0° ! (b)

1 1 2 KEi = mv1i2 + 0 = m ( 5.00 m/s ) = m (12.5 m 2 /s 2 ) 2 2 !

1 1 KE f = mvlf2 + mv2f2 2 2 1 1 2 2 = m ( 4.33 m/s ) + ( 2.50 m/s ) = m (12.5 m 2 s 2 ) 2 2 !

Topic 6

371

Since KEf = KEi, this is an elastic collision 6.54

A rocket engine’s instantaneous thrust T is given by

T = ve

ΔM Δt

Solve for the change in mass per unit time and substitute to find

ΔM T 8.01 × 105 N = = Δt ve 3.05 × 103 m/s = 263 kg/s 6.55

A rocket engine’s instantaneous thrust T is given by

T = ve

ΔM Δt

Substitute values to find to find

T = ve

ΔM = 3.04 × 10 4 m/s −3.03 × 10−6 kg/s Δt

(

)(

)

= 9.21 × 10−2 N 6.56

A rocket engine’s instantaneous thrust T is given by

T = ve

ΔM Δt

Solve for the exhaust speed and substitute to find ve =

T 6.77 × 106 N = ΔM 2.63 × 103 kg/s Δt

= 2.57 × 103 m/s

6.57

An acceleration of 5g equals 49.0 m/s2. By Newton’s second law, this

Topic 6

372

acceleration requires a net force, assumed equal to the thrust T, of magnitude

(

)(

)

Fnet = T = ma = 3.00 × 10 4 kg 49.0 m/s2 = 1.47 × 106 N A rocket engine’s instantaneous thrust T is given by

T = ve

ΔM Δt

Solve for the burn rate (change in mass per unit time) and substitute to find

ΔM T 1.47 × 106 N = = Δt ve 2.50 × 103 m/s = 588 kg/s 6.58

(a) At t = 0, the spaceship’s mass is m0 = 2.50 × 104 kg and its instantaneous thrust T is

T = ve

ΔM = 4.25 × 103 m/s ( −76.7 kg/s) = 3.26 × 105 N Δt

(

)

From Newton’s second law, its acceleration at t = 0 is

a0 =

T 3.26 × 105 N = m0 2.50 × 10 4 kg

= 13.0 m/s2 (b) After steadily burning fuel for time t = 125 s, the spaceship’s mass will be

Topic 6

373

m f = m0 −

ΔM t Δt

= 2.50 × 10 4 kg − ( −76.7 kg/s) (125 s) = 1.54 × 10 4 kg

T 3.26 × 105 N af = = m f 1.54 × 10 4 E = 21.2 m/s2 (d) As the spaceship burns fuel with constant thrust, its acceleration steadily increases from 13.0 m/s2 to 21.2 m/s2 so that its average acceleration is

(13.0 m/s ) + ( 21.2 m/s ) = 17.1 m/s a = 2

Apply the definition of average acceleration, with vi = 0, to find the spaceship’s final speed:

aav =

Δv v f − vi v f = = Δt t t

(

)

v f = aav t = 17.1 m/s2 (125 s) = 2.14 × 103 m/s

6.59

The spaceship’s speed increase is given by ⎛M ⎞ v f − vi = ve ln ⎜ i ⎟ ⎝ Mf ⎠

Topic 6

374

Solve for the mass ratio and substitute values to find ⎛ M ⎞ v f − vi 1.20 × 103 m/s ln ⎜ i ⎟ = = = 0.480 ve 2.50 × 103 m/s ⎝ Mf ⎠ Mi /M f = exp ( 0.480 ) Mi /M f = 1.62

6.60

The recoil speed of the subject plus pallet after a heartbeat is

Δc 6.00 × 10−5 m V= = = 3.75 × 10−4 m/s Δt 0.160 s ! From conservation of momentum, mv − MV = 0 + 0, so the mass of blood leaving the heart is

⎛ 3.75 × 10−4 m/s ⎞ ⎛V⎞ m = M ⎜ ⎟ = (54.0 kg) ⎜ = 4.05 × 10−2 kg = 40.5 g ⎟ ⎝ v⎠ ⎝ 0.500 m/s ⎠ ! 6.61

Choose the positive direction to be the direction of the truck’s initial velocity. Apply conservation of momentum to find the velocity of the combined vehicles after collision: (4 000 kg + 800 kg)vf = (4 000 kg)(+8.00 m/s) + (800 kg)(−8.00 m/s) which yields vf = +5.33 m/s. Use the impulse-momentum theorem, I = Fav(Δt) = Δp = m(vf − vi), to find the magnitude of the average force exerted on each driver during

Topic 6

375

the collision. Truck Driver:

Fav = !

m v f − vi Δt

truck

(80.0 kg ) 5.33 m/s − 8.00 m/s = 1.78 × 10 N 3

0.120 s

Car Driver:

Fav = ! 6.62

m v f − vi Δt

car

(80.0 kg ) 5.33 m/s − 8.00 m/s = 8.89 × 10 N 3

0.120 s

First, we use conservation of mechanical energy to find the speed of m1 at B just before collision.

This gives

1 m1v12 + 0 = 0 + m1 ghi , !2

v12 = 2ghi = 2 ( 9.80 m/s 2 ) ( 5.00 m ) = 9.90 m/s !

Next, we apply conservation of momentum and knowledge of elastic collisions to find the velocity of m1 at B just after collision. From conservation of momentum, with the second object initially at rest, we have

m1v1f + m2v2f = m1v1i + 0,

(

m v2 f = 1 v1i − v1 f m2 !

)

[1]

For head-on elastic collisions, v1i − v2i = −(v1f − v2f). Since v2i = 0 in this case, this becomes v2f = v1f + v1i, and, combining this with Equation [1]

Topic 6

376

above, we obtain

(

m v1 f + v1i = 1 v1i − v1 f m2 !

)

(m1 + m2)v1f = (m1 − m2)v1i

⎛ m − m2 ⎞ ⎛ 5.00 − 10.0 ⎞ v1 f = ⎜ 1 v1i = ⎜ (9.90 m/s) = −3.30 m/s ⎟ ⎝ 5.00 + 10.0 ⎟⎠ ⎝ m1 + m2 ⎠ !

Finally, use conservation of mechanical energy for m1 after the collision to find the maximum rebound height. This gives (KE + PEg)f = 1 2 (KE + PEg)I or 0 + m1 ghmax = m1v1 f + 0 and 2 !

v12 f

hmax = 2g ! 6.63

( −3.30 m/s ) = 0.556 m = 2

2 ( 9.80 m/s 2 )

Note that the initial velocity of the target particle is zero (that is, v2i = 0). From conservation of momentum m1v1f + m2v2f = m1v1i + 0

[1]

For head-on elastic collisions, v1i = v2i = −(v1f − v2f), and with v2i = 0, this gives v2f = v1i + v1f

[2]

Substituting Equation [2] into [1] yields m1v1f + m2(v1i + v1f) = m1v1i

Topic 6

377

(m1 + m2)v1f = (m1 − m2)v1i

⎛ m − m2 ⎞ v1 f = ⎜ 1 v1i ⎝ m1 + m2 ⎟⎠ !

and

[3]

Now, we substitute Equation [3] into [2] to obtain ⎛ m − m2 ⎞ v2 f = v1i + ⎜ 1 ⎟⎠ v1i m + m ⎝ 1 2 !

⎛ 2m1 ⎞ v2 f = ⎜ v1i ⎝ m1 + m2 ⎟⎠

[4]

Equations [3] and [4] can now be used to answer both parts (a) and (b). (a)

If m1 = 2.0 g, m2 = 1.0 g, and v1i = 8.0 m/s, then 8 32 v1 f = m/s !and!v2 f = m/s 3 3 !

(b)

If m1 = 2.0 g, m2 = 10 g, and v1i = 8.0 m/s, we find 16 8 v1 f = − m/s !and!v2 f = m/s 3 3 !

(c)

The final kinetic energy of the 2.0 g particle in each case is: 2

⎛8 ⎞ Case (a): KE1 f = m1v12 f = 2.0 × 10−3 kg ⎜ m/s⎟ = 7.1 × 10−3 J 2 2 ⎝3 ⎠ 1

(

)

1 1 ⎛ 16 ⎞ Case (b): KE1 f = m1v12 f = ( 2.0 × 10−3 kg ) ⎜ − m/s⎟ = 2.8 × 10−2 J ⎝ ⎠ 2 2 3 !

Since the incident kinetic energy is the same in cases (a) and (b), we observe that the incident particle loses more kinetic energy in case (a)

Topic 6

6.64

378

If the pendulum bob barely swings through a complete circle, it arrives at the top of the arc (having risen a vertical distance of 2 ℓ ) with essentially zero velocity. From conservation of mechanical energy, we find the minimum velocity of the bob at the bottom of the arc as (KE + PEg)bottom = (KE + 2 PEg)top, or !12 MV = 0 + Mg(2) . This gives !V = 2 g as the needed

velocity of the bob just after the collision. Conserving momentum through the collision then gives the minimum initial velocity of the bullet as

(

)

4M ⎛ v⎞ m ⎜ ⎟ + M 2 g = mv + 0 or v = g ⎝ 2⎠ m !

6.65

(a)

The total momentum of the system (girl plus gloves) is zero before the gloves are thrown. Neglecting friction between the girl and the ice, the total momentum is also zero after the gloves are thrown, giving

  (M − m)v girl + mv gloves = 0 ! (b)

and

 ⎛ m ⎞ v girl = − ⎜ v ⎝ M − m ⎟⎠ gloves

As she throws the gloves, she exerts a force on them. As described by Newton’s third law, the gloves exert a force of equal magnitude in the opposite direction on the girl. This force causes

 her to accelerate from rest to reach the velocity !v girl . © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 6

6.66

379

(

Use conservation of mechanical energy, KE + PEg

) = (KE + PE ) , to g

find the speed of the blue bead at point B just before it collides with the green bead. This gives

1 2

mv1i2 + 0 = 0 + mgyA , or

(

)

v1i = 2 g yA = 2 9.80 m/s 2 (1.50 m) = 5.42 m/s

Conservation of momentum during the collision gives (0.400 kg)v1f + (0.600 kg)v2f = (0.400 kg)(5.42 m/s) + 0 or

v1f + 1.50v2f = 5.42 m/s

[1]

For a head-on elastic collision, we have vii − v2i = −(v1f − v2f), and with v2i = 0, this becomes v1f = v2f − v1i

v1f = v2f − 5.42 m/s

[2]

Substitute Equation [2] into [1] to find the speed of the green bead just after collision as v2f − 5.42 m/s + 1.50v2f = 5.42 m/s v2 f = !

2(5.42 m/s) = 4.34 m/s 2.50

Now, we use conservation of energy for the green bead after collision to find the maximum height it will achieve. This gives

v22 f ( 4.34 m/s ) = 0.961 m 1 0 + m2 g ymax = m2 v22 f + 0 or ymax = = 2 2g 2 ( 9.80 m/s 2 ) ! 2

Topic 6

6.67

380

We shall choose southward as the positive direction. The mass of the man is w 730 N m= = = 74.5 kg 2 g 980 m/s !

Then, from conservation of momentum, we find (mmanvman + mbookvbook)f = (mmanvman + mbookvbook)i or (74.5 kg)vman + (1.2 kg)(−5.0 m/s) = 0 + 0

and vman = 8.1 × 10−2 m/s

Therefore, the time required to travel the 5.0 m to shore is Δx 5.0 m t= = = 62 s . −2 v 8.1 × 10 m/s man !

6.68

The mass of the third fragment must be m3 = mnucleas − m1 − m2 = (17 − 5.0 − 8.4) × 10−27 kg = 3.6 × 10−27 kg Conserving momentum in both the x- and y-directions gives y-direction: m1v1y + m2v2y + m3v3y = 0 or

v3y = −

m1v1y + m2 v2y

x-direction:

(5.0 × 10 kg )(6.0 × 10 m/s) + 0 = − 30 × 10 m/s =− 6

−27

3.6 × 10

−27

3.6

m1v1x + m2v2x + m3v3x = 0

Topic 6

381

0 + ( 8.4 × 10−27 kg ) ( 4.0 × 106 m/s ) m1v1x + m2 v2x 34 v3x = − =− =− × 106 m/s −27 m3 3.6 × 10 kg 3.6 !

and 2 2 v = v3x + v3y = !3

( − ( 34 3.6) × 10 m/s ) + ( − ( 30 / 3.6) × 10 ) = 1.3 × 10 m/s 6

⎛v ⎞ ⎛ 30 ⎞ 3y Also, θ = tan ⎜ ⎟ + 180° = tan −1 ⎜ ⎟ + 180° = 2.2 × 102 degrees = 220° ⎜⎝ v3x ⎟⎠ ⎝ 34 ⎠ −1

! 7 Therefore, v 3 = 1.3 × 10 m/s at 220° counterclockwise from the +x-axis Note that the factor of 180° was included in the calculation for θ because it was recognized that both v3x and v3y were negative. This meant that θ had to be a third quadrant angle. Use of the inverse tangent function alone yields only the principal angles (−90° ≤ θ ≤ +90°) that have the given value for the tangent function. 6.69

The sketch below gives before and after views of the collision between these two objects. Since the collision is elastic, both kinetic energy and momentum must be conserved.

Topic 6

382

Conservation of Momentum: m1v1f + m2v2f = m1v1i + m2v2i m1(0) + m2v = m1v0 + m2(−v0)

⎛m ⎞ v = ⎜ 1 − 1⎟ v0 ⎠ ! ⎝ m2

[1]

Since this is a perfectly elastic collision, v1i − v2i = −(v1f − v2f), and with the given velocities this becomes v0 − (−v0) = −(0 − v)

(a)

v = 2v0

[2]

⎛m ⎞ 1 − 1⎟ v0 Substituting Equation [2] into [1] gives 2 v0 = ⎜ ⎝ m2 ⎠

6.70

m1/m2 = 3

(b)

From Equation [2] above, we have

v/v0 = 2

(a)

Let v1i and v2i be the velocities of m1 and m2 just before the collision. Then, using conservation of mechanical energy,

( KE + PE ) = ( KE + PE ) or

gives

1 2

mvi2 + 0 = 0 + mgh0 ,

v = −v2i = 2 g h0 = 2 ( 9.80 m/s 2 ) (5.00 m) = 9.90 m/s ! 1i

v1i = +9.90 m/s and v2i = −9.90 m/s (b)

From conservation of momentum:

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(2.00 g)v1f + (4.00 g)v2f = (2.00 g)(9.90 m/s) + (4.00 g)(−9.90 m/s) or

v1f + (2.00)v2f = −9.90 m/s

[1]

For a perfectly elastic, head-on collision, v1i − v2i = −(v1f − v2f), giving +9.90 m/s − (−9.90 m/s) = −v1f + v2f or

v2f = v1f + 19.8 m/s

[2]

Substituting Equation [2] into [1] gives v1f + (2.00)(v1f + 19.8 m/s) = −9.90 m/s or

v1 f = !

−9.90 m/s − 39.6 m/s = −16.5 m/s 3.00

Then, Equation [2] yields v2f = −16.5 m/s + 19.8 m/s) = +3.30 m/s. (c)

Applying conservation of energy to each block after the collision gives v 2f 1 1 m(0)2 + mghmax = mv 2f + mg(0) or hmax = 2 2g !2

Thus,

and

6.71

(a)

v12 f

( −16.5 m/s ) = 13.9 m =

v22 f

( 3.30 m/s ) = 0.556 m =

h1 f = 2g !

h2 f = 2g !

2 ( 9.80 m/s 2 ) 2

2 ( 9.80 m/s 2 )

Use conservation of mechanical energy to find the speed of m1 just

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before collision. Taking y = 0 at the tabletop level, this gives !2 m1v1i + mg(0) = 2 m1 (0) + m1 gh1 , or 1

v = 2 g h1 = 2 ( 9.80 m/s 2 ) (2.50 m) = 7.00 m/s ! 1i

Apply conservation of momentum from just before to just after the collision: (0.500 kg)v1f + (1.00 kg)v2f = (0.500 kg)(7.00 m/s) + 0 or

v1f + 2v2f = 7.00 m/s

[1]

For a perfectly elastic head-on collision, v1i − v2i = −(v1f − v2f). With

!v2i = 0 , this becomes v2f = v1f + v1i

v2f = v1f + 7.00 m/s

[2]

Substituting Equation [2] into [1] yields v1f + 2(v1f + 7.00 m/s) = 7.00 m/s

v1 f = !

and

−7.00 m/s = −2.33 m/s 3

Then, from Equation [2], v2f = −2.33 m/s + 7.00 m/s = 4.67 m/s (b)

Apply conservation of mechanical energy to m1 after the collision to find the rebound height of this object.

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1 1 m1 (0) + m1 gh1′ = m1v12 f + mg(0) 2 !2

v12 f

h1′ = 2g ! (c)

( −2.33 m/s ) = 0.277 m = 2

2 ( 9.80 m/s 2 )

2 From !Δy = v0yt + 12 ayt , with v0y = 0, the time for m2 to reach the

floor after it flies horizontally off the table is

2(Δy) ay

(

2 −2.00 m −9.80 m/s

) = 0.639 s

During this time it travels a horizontal distance Δx = v0xt = (4.67 m/s)(0.639 s) = 2.98 m (d)

After the 0.500 kg mass comes back down the incline, it flies off the table with a horizontal velocity of 2.33 m/s. The time of the flight to the floor is 0.639 s as found above and the horizontal distance traveled is Δx = v0xt = (2.33 m/s)(0.639 s) = 1.49 m

6.72

We label the two objects such that object 1 has mass m while object 2 has mass 3m. Conservation of the x-component of momentum gives (3m)v2x + 0 = −mv0 + (3m)v0

2 v2x = v0 3 !

[1]

Likewise, conservation of the y-component of momentum gives

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−mv1y + (3m)v2y = 0

and

v1y = 3v2y

[2]

Since the collision is elastic, (KE)f = (KE)i,

(

)

1 2 1 1 1 2 2 mv1y + ( 3m) v2x + v2y = mv02 + (3m)v02 2 2 2 !2

which reduces to

(

)

2 2 v 2 + 3 v2x + v2y = 4v02 ! 1y

[3]

Substituting Equations [1] and [2] into [3] yields ⎛4 2 2 ⎞ 2 9v2y + 3 ⎜ v02 + v2y ⎟⎠ = 4v0 ⎝ 9 !

(a)

v2y = v0

2 3

From Equation [2], the particle of mass m has final speed

v = 3v2y = v0 2 , and the particle of mass 3m ! 1y moves at

4 2 2 2 2 2 2 v2 = v2x + v2y = v0 + v0 = v0 9 9 3 !

6.73

(b)

⎛ v 2 3⎞ ⎛ v2y ⎞ ⎛ 1 ⎞ θ = tan −1 ⎜ = tan −1 ⎜ 0 = tan −1 ⎜ = 35.3° ⎟ ⎟ ⎝ 2 ⎟⎠ 2v0 3 ⎠ ⎝ v2x ⎠ ⎝ !

(a)

The momentum of the system is initially zero and remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have m2v2 + m1v1 = 0, or

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⎛m ⎞ ⎛ 0.500 ⎞ 2.00 v2 = − ⎜ 1 ⎟ v f = − ⎜ m/s ⎟ 4.00 m/s = − 3.00 ⎝ 3.00 ⎠ ⎝ m2 ⎠

(

meaning (b)

)

! v 2 = 0.667 m/s to the left

Using conservation of energy as the block slides down the wedge, we have

( KE + PE ) = ( KE + PE ) ! g

1 1 or 0 + m1 gh = m1v12 + m2 v22 + 0 2 2

Thus, h= = !

6.74

(a)

1 ⎡ 2 ⎛ m2 ⎞ 2 ⎤ ⎢ v1 + ⎜ ⎟ v2 ⎥ 2g ⎣ ⎝ m1 ⎠ ⎦ 2 ⎡ 1 ⎛ 3.00 ⎞ ⎛ 2.00 ⎞ ⎤ 2 (4.00 m/s) + − m/s ⎜⎝ ⎟⎜ ⎟⎠ ⎥ = 0.952 m ⎢ 19.6 m/s 2 ⎣ 0.500 ⎠ ⎝ 3.00 ⎦

Momentum is conserved, even in a perfectly inelastic collision. Thus, pf = pi or

(m + 2m)vf = mv1 + 2mv2

vf =

giving !

(b)

1 ( v1 + 2v2 ) 3

The kinetic energy before impact is 1 1 m KEi = mv12 + (2m)v22 = ( v12 + 2v22 ) 2 2 2 !

After impact, the total kinetic energy is 1 3 ⎡1 4 4 ⎤ ⎤ m ⎡1 KE f = (3m)v 2f = m ⎢ ( v12 + 4v1v2 + 4v22 ) ⎥ = ⎢ v12 + v1v2 + v22 ⎥ 2 2 ⎣9 3 3 ⎦ ⎦ 2 ⎣3 !

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388

The change that has occurred in the kinetic energy is then

ΔKE = KE f − KEt = =

6.75

(a)

m⎡ 2 2 4 2 2⎤ − v + v v − v2 1 1 2 2 ⎢⎣ 3 3 3 ⎥⎦

ΔKE = −

m ⎡⎛ 1 ⎞ 2 4 ⎛ 4 ⎞ 2⎤ ⎜⎝ − 1⎟⎠ v1 + v1v2 + ⎜⎝ − 2⎟⎠ v2 ⎥ ⎢ 2⎣ 3 3 3 ⎦

m 2 m 2 v1 − 2v1v2 + v22 ) = − ( v1 − v2 ) ( 3 3

Use conservation of the horizontal component of momentum from just before to just after the cannon is fired.

( ∑ p ) = ( ∑ p ) gives

x f

x i

mshell(vshell cos 45.0°) + mcannonvrecoil = 0,

⎛ m ⎞ vrecoil = − ⎜ shell ⎟ vshell cos 45.0° ⎝ mcannon ⎠ ⎛ 200 kg ⎞ = −⎜ (125 m/s)cos 45.0° = −3.54 m/s ⎝ 5000 kg ⎟⎠ (b)

Use conservation of mechanical energy for the cannon-spring system from right after the cannon is fired to the instant when the

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389

cannon comes to rest.

( KE + PE + PE ) = ( KE + PE + PE )

1 2 1 2 0 + 0 + kxmax = mcannon vrecoil +0+0 2 2 ! 2 mcannon vrecoil xmax = = k !

( 5 000 kg )( −3.54 m s ) = 1.77 m 2

2.00 × 10 4 N m

(c)

|Fmax| = kxmax = (2.00 × 104 N/m)(1.77 m) = 3.54 × 104 N

(d)

No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it from recoiling vertically. Momentum is not conserved in the vertical direction. The spring does not have time to stretch during the cannon firing. Thus, no external horizontal force is exerted on the system (cannon plus shell) from just before to just after firing. Momentum is conserved in the horizontal direction during this interval.

6.76

(a)

(b)

 From Newton’s third law, the force !FBA exerted by B on A is at each instant equal in magnitude and opposite in direction to the

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390

(c)

There are no horizontal external forces acting on System C, which

  consists of both blocks. The forces !FBA and !FAB are internal forces exerted on one part of System C by another part of System C.

! ! ΔpC ∑ Fexternal = =0 ⇒ Δt

Thus,

! ΔpC = 0

This gives

(! p ) = ( p ) = ( p ) + ( p ) or C

C i

A i

(M + 2M)V = M(+v) + 0

B i

so the velocity of the combined blocks after collision is

V = +v/3

The change in momentum of A is then    ⎛v ⎞ Δp A = ( p A ) f − ( p A )i = MV − Mv = M ⎜ − v ⎟ = −2Mv/3 ⎝3 ⎠ !

and the change in momentum for B is    ⎛ +v ⎞ ΔpB = ( pB ) f − ( pB )i = 2MV − 0 = 2M ⎜ ⎟ = +2Mv/3 ⎝ 3 ⎠ !

(d) 2

⎛ v⎞ ⎡1 ⎤ 1 1 ΔKE = KEC − ⎡ KEA + KEB ⎤ = 3M ⎜ ⎟ − ⎢ Mv 2 + 0 ⎥ = − Mv 2 ⎣ ⎦l 2 f 3 ⎝ 3 ⎠ ⎢⎣ 2 ⎥⎦

( ) ( ) ( )

( )

Thus, kinetic energy is not conserved in this inelastic collision 6.77

(a)

The owner’s claim should be denied. Immediately prior to impact, the total momentum of the two-car system had a

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391

northward component and an eastward component. Thus, after impact, the wreckage moved in a northeasterly direction and could not possibly have damaged the owner’s property on the southeast corner. (b)

Choose east as the positive x-direction and north as the positive ydirection. From conservation of momentum: (px)after = (px)before ⇒ (m1 + m2)vx = m1(v1i)x + m2(v2i)x or

vx =

( ) + m (v ) = (1 300 kg )(30.0 km h ) + 0 = 16.3 km h

m1 v1i

2i x

m1 + m2

1 300 kg + 1 100 kg

(py)after = (py)before ⇒ (m1 + m2)vy = m1(v1i)y + m2(v2i)y or

vy =

( ) + m (v ) = 0 + (1 100 kg )(20.0 km h ) = 9.17 km h

m1 v1i

m1 + m2

2i y

1 300 kg + 1 100 kg

Thus, the velocity of the wreckage immediately after impact is

⎛ vy ⎞ v = vx2 + vy2 = 18.7 km/h and θ = tan −1 ⎜ ⎟ = tan −1 ( 0.563 ) = 29.4° ⎝ vx ⎠ ! or 6.78

(a)

 !v = 18.7 km/h at 29.4° north of east, consistent with part (a)

Ignore any change in velocity due to the force of gravity during the brief collision time, and let vbf denote the velocity of the ball

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just after impact while vpf is that of the player. We use the conservation of momentum to obtain

(0.45 kg ) v + (60 kg ) v = (0.45 kg )(−25 m s) + (60 kg )( 4.0 m s) bf

vpf = 3.8 m/s − (7.5 × 10−3)vbf

[1]

Also, for an elastic collision ⇒ vbf − vpf = −(vbi − vpi) = −(−25 m/s − 4.0 m/s) or

vbf = 29 m/s + vpf

[2]

Substituting Equation [1] into [2] yields vbf = !

(b)

The average acceleration of the ball during the collision is

aav =

6.79

(a)

29 m/s + 3.8 m/s = 33 m/s 1 + 7.5 × 10−3

vbf − vbi Δt

(

33 m s − −25 m s 20 × 10 s −3

) = 2.9 × 10 m s 3

We choose east (the direction of the girl’s velocity) to be the positive direction. Since momentum is conserved in the event and both skaters were initially at rest, it is necessary that mbvb + mgvg = 0

giving

vb = −(mg/mb)vg

Thus, the boy will recoil toward the west with speed

(

)

v = mg mb vx . ! b

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393

(b)

The girl’s kinetic energy is

For the boy:

(

KEg = 12 mg v g2

2 1 1 ⎛ mg ⎞ KEb = mbvb2 = mb ⎜ 2 v g2 ⎟ 2 2 ⎝ mb ⎠ !

)

KEb = mg2 /2mb v g2 !

Therefore,

KEg !

(c)

KEb

1 2

mg v g2

2 g

(m /2m ) v

2 g

mb > 1,!since!mb > mg mg

Mechanical energy is gained because work is done by the skater's muscles as they push each other apart. The origin of this work is chemical energy within their bodies.

6.80

The woman starts from rest (v0y = 0) and drops freely with ay = −g for 2.00 m before the impact with the toboggan. Then, !v2i = v0y + 2ay ( Δy ) 2

gives her speed just before impact as 2 v = v0y + 2ay ( Δy ) = 0 + 2 ( −9.80 m s 2 ) ( −2.00 m ) = 6.26 m s ! 2i

The sketches below show the situation just before and just after the woman’s impact with the toboggan. Since no external forces impart any significant impulse directed parallel to the incline (+x-direction) to the system consisting of man, woman, and toboggan during the very brief duration of the impact, we will consider the total momentum

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parallel to the incline to be conserved. That is,

(m1 + m2 ) v f = m1v1i + m2 ( v2i )x = m1v1i + m2v2i sin 30.0°

or the speed of the system immediately after impact is

vf = =

m1v1i + m2 v2i sin 30.0° m1 + m2

( 90.0 kg )(8.00 m s ) + ( 55.0 kg )( 6.26 m s ) sin 30.0° 90.0 kg + 55.0 kg

= 6.15 m s 6.81

First consider the motion of the block and embedded bullet from immediately after impact until the block comes to rest after sliding distance d across the horizontal table. During this time, a kinetic friction force fk = µkn = µk(M + m)g, directed opposite to the motion, acts on the block. The net work done on the block and embedded bullet during this time is 1 Wnet = ( fk cos 180° ) d = KE f − KEi = 0 − (M + m)V 2 2 !

so the speed, V, of the block and embedded bullet immediately after

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impact is

(

)

2 µk M + m gd −2 f1d V= = = 2 µk gd −(M + m) M + m ! Now, make use of conservation of momentum from just before to just after impact to obtain p = pxf ! xi

⇒ mv0 = (M + m)V = (M + m) 2 µk gd

and the initial velocity of the bullet was

! 6.82

(a)

⎛ M + m⎞ v0 = ⎜ 2 µk gd ⎝ m ⎟⎠

Apply conservation of momentum in the vertical direction to the squid-water system from the instant before to the instant after the water is ejected. This gives msvs + mwvw = (ms + mw)(0) or

⎛ 0.300 kg ⎞ ⎛m ⎞ vs = − ⎜ w ⎟ vw = − ⎜ ( −20 m s ) = 7.06 m s ⎝ ms ⎠ ⎝ 0.850 kg ⎟⎠ (b)

Apply conservation of mechanical energy to the squid from the instant after the water is ejected until the squid reaches maximum height to find

1 0 + ms gy f = msvs2 + mgyi 2

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396

(7.06 m/s ) = 2.54 m vs2 Δy = y f − yi = = 2g 2 ( 9.80 m/s 2 ) 2

6.83

(a)

The situations just before and just after the collision are shown above. Conserving momentum in both the x- and y-directions gives

( p ) = ( p ) ⇒ m v sin 53° − m v sin φ = 0

1 1f

2 2f

or m2v2f sinφ = m1v1f sin 53°

(p ) = (p )

! or

x f

x i

[1]

⇒ m1v1 f cos 53° + m2 v2 f cos φ = m1v1i + 0

m2v2f cosφ = m1v1i − m1v1f cos 53°

[2]

Dividing Equation [1] by [2] yields

v1 f sin 53° (1.0 m s ) sin 53° tan φ = = = 0.57 v1i − v1 f cos 53° ( 2.0 m s ) − (1.0 m s ) cos 53° ! or

φ = 30°

Equation [1] then gives

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397

v2 f = ! (b)

m1v1 f sin 53° m2 sin φ

( 0.20 kg )(1.0 m s ) sin 53° = 1.1 m s ( 0.30 kg ) sin 30°

The fraction of the incident kinetic energy lost in this collision is

KE f ΔKE KEi − KEt = = 1− KEi KEi KEi 1 2 1 2 0.20 kg ) (1.0 m/s ) + ( 0.30 kg ) (1.1 m/s ) ( 2 = 1− 2 1 2 0.20 kg ) ( 2.0 m/s ) ( 2 = 0.30 or 30% 6.84

(a)

Conservation of mechanical energy of the bullet-block-Earth system from just after impact until maximum height is reached may be used to relate the speed of the block and bullet just after collision to the maximum height. Then, conservation of momentum from just before to just after impact can be used to relate the initial speed of the bullet to the speed of the block and bullet just after collision.

(b)

Conservation of energy from just after impact until the block and embedded bullet come to rest momentarily at height h gives KEi + PEg,i = KEf + PEg,f

1 (M + m)V 2 + 0 = 0 + (M + m)gh !2

and the speed of the block + bullet just after collision is !V = 2gh . Now, using conservation of momentum during the collision gives

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mvi + 0 = (M + m)V

⎛ M + m⎞ vi = ⎜ V ⎝ m ⎟⎠ !

The initial speed of the bullet is then

! 6.85

(a)

⎛ M + m⎞ vi = ⎜ 2gh ⎝ m ⎟⎠

(b)

Conservation of energy from just after impact until the block and embedded bullet come to rest momentarily at height h = 22.0 cm gives KEi + PEg,i = KEf + PEg,f

1 (M + m)V 2 + 0 = 0 + (M + m)gh 2 !

and the speed of the block + bullet just after collision is V = 2gh . ! Now, using conservation of momentum during the collision gives mvi + 0 = (M + m)V

⎛ M + m⎞ ⎛ M + m⎞ vi = ⎜ V =⎜ 2gh ⎟ ⎝ m ⎠ ⎝ m ⎟⎠ ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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The initial speed of the bullet is then

⎛ 1.25 kg + 5.00 × 10−3 kg ⎞ vi = ⎜ 2 ( 9.80 m/s 2 ) ( 0.220 m ) = 521 m/s −3 ⎟ 5.00 × 10 kg ⎝ ⎠ !  and the initial velocity of the bullet is v i = 521 m/s upward . !

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400

Topic 7 Rotational Motion and the Law of Gravity

QUICK QUIZZES 7.1

Choice (c). For a rotation of more than 180°, the angular displacement must be larger than π = 3.14 rad. The angular displacements in the three choices are (a) 6 rad − 3 rad = 3 rad, (b) 1 rad − (−1) rad = 2 rad, (c) 5 rad − 1 rad = 4 rad.

7.2

Choice (b). Because all angular displacements occurred in the same time interval, the displacement with the lowest value will be associated with the lowest average angular speed.

7.3

Choice (b). From

α= !

ω 2 − ω 02 ω 2 − 0 ω 2 = = 2Δθ 2Δθ 2Δθ

it is seen that the case with the smallest angular displacement involves the highest angular acceleration. 7.4

Choice (b). All points in a rotating rigid body have the same angular speed.

7.5

Choice (a). Andrea and Chuck have the same angular speed ω, but Andrea moves in a circle with twice the radius of the circle followed

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401

by Chuck. Thus, from vt = rω, it is seen that Andrea’s tangential speed is twice Chuck’s. 7.6

Choice (e). Since the tangential speed is constant, the tangential acceleration is zero.

Choice (a). The centripetal acceleration, !ac = vt /r , is inversely 2

proportional to the radius when the tangential speed is constant. 3.

Choice (b). The angular speed, ω = vt/r, is inversely proportional to the radius when the tangential speed is constant.

7.7

Choice (c). Both the velocity and acceleration are changing in direction, so neither of these vector quantities is constant.

7.8

Choices (b) and (c). According to Newton’s law of universal gravitation, the force between the ball and the Earth depends on the product of their masses, so both forces, that of the ball on the Earth, and that of the Earth on the ball, are equal in magnitude. This follows also, of course, from Newton’s third law. The ball has large motion compared to the Earth because, according to Newton’s second law, the force gives a much greater acceleration to the small mass of the ball.

7.9

Choice (e). From F = GMm/r2, the gravitational force is inversely proportional to the square of the radius of the orbit.

7.10

Choice (d). The semimajor axis of the asteroid’s orbit is 4 times the size

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402

of Earth’s orbit. Thus, Kepler’s third law (T2/r3 = constant) indicates that its orbital period is 8 times that of Earth.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 7.2

(a) No, she would not be weightless inside the space station. Her weight is w = mg and g is not zero inside the space station. ( g = GMP /r 2 goes to zero only as r → ∞ .) However, she will feel weightless because her surroundings are also in free-fall with the same acceleration of magnitude g. (b) The value of g falls off as the distance squared, so at twice the distance the value of g will be reduced by a factor of 4. Therefore, worbit/wsurface = ¼.

7.4

(a) Gravity is the only appreciable external force acting on a planet, so it is the force providing the centripetal acceleration. (b) In the absence of friction, a moving car travels in a straight line. Friction between the tires and the road is required for it to navigate an unbanked, circular turn. (c) Two forces act on the rock: tension and gravity. These two forces, added together, are the net force that accelerates the rock. (d) The normal force between the sock and the clothes dryer is

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403

directed toward the center of the sock’s circular path and is the only horizontal force. The normal force is therefore the net radial force responsible for the centripetal acceleration. 7.6

Consider one end of a string connected to a spring scale and the other end connected to an object, of true weight w. The tension T in the string will be measured by the scale and construed as the apparent weight. We have w − T = mac. This gives T = w − mac. Thus, the apparent weight is less than the actual weight by the term mac. At the poles the centripetal acceleration is zero, and T = w. However, at the equator the term containing the centripetal acceleration is non-zero, and the apparent weight is less than the true weight.

7.8

(a) If the acceleration is constant in magnitude and perpendicular to the velocity, the object is moving in a circular path at constant speed. (b) If the acceleration is parallel to the velocity, the object moves in a straight line, and is either speeding up (v and a in same direction) or slowing down (v and a in opposite directions).

7.10

(a) Net centripetal force is proportional to the mass, so for equal centripetal acceleration (v2/r), the ratio of their net centripetal forces will equal the ratio of their masses. Therefore, Ftruck/Fcar = 2.

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(b) If the net centripetal forces are equal, 2 2 vtruck vcar 2m = m ( ) r ( ) r so that vtruck = v/ 2

7.12

(a) Velocity is north at A, west at B, and south at C. (b) The acceleration is west at A, nonexistent at B, east at C, to be radially inward.

ANSWERS TO EVEN NUMBERED PROBLEMS 7.2

(a)

5.00 rad/s

(b)

4.00 rad/s2

7.4

(a)

0.209 rad/s2

(b)

Yes.

7.6

−226 rad/s2

7.8

(a) The tire is slowing down, so the second drop was released with a smaller initial velocity. (b)

−0.322 rad/s2

7.10

51 revolutions

7.12

(a)

5.75 rad/s

(c)

76° counterclockwise from the +x-axis

7.14

(a)

5.25 s

7.16

4.9 × 10−2 rad/s

(b)

vt = 1.29 m/s, at = 0.563 m/s2

27.6 rad

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405

7.18

No. The required maximum tension in the vine is 1.38 × 103 N.

7.20

(a)

4.43 rad/s

(b) 19.9 m/s

7.22

(a)

4.81 m/s

(b)

7.24

1.5 × 102 rev/s

7.26

(a)

a normal force exerted by the “floor” of the cabin

(b)

∑ F = n = mvt2 /r ! c

(c)

(e)

0.700 rad/s

(f) 8.98 s

700 N

294 N

(d)

7.00 m/s

(g) slower; 5.74 m/s

7.28

See solution for parts (a)-(c)

7.30

(a)

the gravitational force and a contact force exerted by the pail

(b)

the contact force (the gravitational force alone would produce projectile motion)

(c)

3.13 m/s.

(d) No, the motion would be identical to projectile motion. 7.32

(a)

25 kN

(b)

12 m/s

7.34

(a)

3.62 × 106 N

(b)

−2.46 × 1013 J

7.36

(a)

1.83 × 109 kg/m3

(b)

3.26 × 106 m/s2

(c)

7.38

2.59 × 108 m from the center of the Earth

7.40

The two masses are 2.00 kg and 3.00 kg.

−2.08 × 1013 J

Topic 7

406

7.42

(a)

1.23 × 106 m

(b)

6.89 m/s2

7.44

(a)

5.59 × 103 m/s

(b)

3.98 h

7.46

8.91 × 107 m

7.48

1.63 × 104 rad/s

7.50

(a)

56.5 rad/s

(b)

22.4 rad/s

(c)

−7.62 × 10−3 rad/s2

(d)

1.77 × 105 rad

(e)

5.81 km

7.52

(a)

3.13 rad/s

(b)

7.96 rev

7.54

(a)

8.42 N

(b)

64.8°

7.56

(a)

3.34 × 10−2 rad/s

(b)

159 rev

7.58.

(a)

See Solution.

(b)

See Solution.

(c)

1.47 × 103 N

(c)

1.67 N

ntop = 546 N ; nbottom = 826 N

7.60

12.8 N

7.62

2.06 × 103 rev/min

7.64

(a)

(b)

1.3 kN

7.66

(a)

1.62 × 103 N

(b)

196 N

(c)

2.1 kN

Topic 7

407

(c) With the right speed, the needed centripetal force at the top of the loop can be made exactly equal to the gravitational force. At this speed, the normal force exerted on the pilot by the seat (his apparent weight) will be zero, and the pilot will have the sensation of weightlessness. (d)

177 m/s

7.68

(a)

static friction

7.70

(a) a downward gravitation force and a tension force that is directed

(b)

6.55 s

toward the center of the circular path

7.72

(b)

See Solution.

(c)

6.05 N

(d)

7.82 m/s

(a)

4.1 m/s

(b)

80°

(c)

1.7 m

(d) No. Flying slower increases the minimum radius of achievable turns. 7.74

0.75 m

PROBLEM SOLUTIONS 7.1

Use the conversion factors π rad = 180° and 1 rev = 2π rad to find:

⎛ π rad ⎞ = 0.820 rad (a) 47.0° = 47.0° ⎜ ⎝ 180° ⎟⎠

Topic 7

408

⎛ 1 rev ⎞ = 1.91 rev (b) 12.0 rad=12.0 rad ⎜ ⎝ 2π rad ⎟⎠

7.2

rev rev ⎛ 1min ⎞ ⎛ 2π rad ⎞ = 75.0 = 7.85 rad/s min min ⎜⎝ 60s ⎟⎠ ⎜⎝ 1 rev ⎟⎠

(a) The change in the tire’s angular velocity is Δω = ωf − ωi . Substitute

ωf = +2.50 rad/s and ωi = −2.50 rad/s to find Δω = +2.50 rad/s − (−2.50 rad/s) = +5.00 rad/s . (b) The average angular acceleration is

α av =

7.3

(a) θ =

s r

Δω +5.00 rad/s = = 4.00 rad/s2 Δt 1.25 s

60 000 mi ⎛ 5 280 ft ⎞ 8 ⎜ ⎟ = 3.2 × 10 rad 1.0 ft ⎝ 1 mi ⎠

(b) The car travels a distance equal to the circumference of the tire for every revolution the tire makes if there is no slipping of the tire on the roadway. Thus, the number of revolutions made during the warranty period is S 60 000 miles ⎛ 5 280 ft ⎞ n= = = 5.0 × 107 rev ⎜ ⎟ 2π r 2π (1.0 ft ) ⎝ 1 mile ⎠ !

7.4

Δω 1.00 rev/s − 0 ⎛ rev ⎞ ⎛ 2π rad ⎞ = = ⎜ 3.33 × 10−2 2 ⎟ ⎜ = 0.209 rad/s 2 (a) α = ⎟ Δt 30.0 s s ⎠ ⎝ 1 rev ⎠ ⎝ !

(b) Yes. When an object starts from rest, its angular speed is related to the angular acceleration and time by the equation ω = α (Δt). Thus,

Topic 7

409

the angular speed is directly proportional to both the angular acceleration and the time interval. If the time interval is held constant, doubling the angular acceleration will double the angular speed attained during the interval.

( 2.51 × 10 rev/min − 0) ⎛ 2π rad ⎞ ⎛ 1 min ⎞ = 821 rad/s 4

7.5

(a) α = !

⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60.0 s ⎟⎠

3.20 s

⎛ ω +ω ⎞ 0 θ = ωt = ⎜ t ⎟t ⎝ 2 ⎠ ⎡ 2.51 × 10 4 rev/min 2π rad 1 rev 1 min 60.0 s + 0 ⎤ (b) ⎥ 3.20 s =⎢ ⎢ ⎥ 2 ⎣ ⎦

(

)(

) (

)

= 4.21 × 103 rad

7.6

rev ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ω t = 3 600 = 377 rad/s min ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60.0 s ⎟⎠ ! ⎛ 2π rad ⎞ Δθ = 50.0 rev ⎜ = 314 rad ⎝ 1 rev ⎟⎠ !

( ) = −226 rad/s Thus, α = = 2Δθ 2 ( 314 rad ) w 2 − wi2

7.7

0 − 377 rad/s

(a) The bicycle’s linear acceleration is at = rα. Substitute r = 38.0 cm = 0.380 m and α = 1.60 rad/s2 to find

(

)

at = rα = ( 0.380 m ) 1.60 rad/s2 = 0.608 m/s2 (b) The linear speed is vt = rω. Solve for ω to find

Topic 7

410

ω=

vt 11.0 m/s = = 28.9 rad/s r 0.380 m

ω 2 − ω i2 ( 28.9 rad/s) − 0 Δθ = = = 261 rad 2α 2 1.60 rad/s2 2

(

)

(d) The bicycle will have traveled a distance d equal to the arc length s traced out by a point on the rim: d = s = r Δθ :

d = r Δθ = ( 0.380 m ) ( 261 rad) = 99.2 m 7.8

(a) The maximum height h depends on the drop’s vertical speed at the instant it leaves the tire and becomes a projectile. The vertical speed at this instant is the same as the tangential speed, vt = rω, of points on the tire. Since the second drop rose to a lesser height, the tangential speed decreased during the intervening rotation of the tire. (b) From !v = v0 + 2ay ( Δy ) , with v0 = vt, ay = −g, and v = 0 when Δy = h, 2

the relation between the tangential speed of the tire and the maximum height h is found to be

0 = vt2 + 2 ( −g ) h !

vt = 2gh

Thus, the angular speed of the tire when the first drop left was

Topic 7

411

ω1 = !

( vt )1 = 2gh1 , and when the second drop left, the angular r

speed was ω 2 = !

( vt )2 = r

2gh2 2 2 . From !ω = ω 0 + 2α ( Δθ ) , with Δθ = r

2π rad, the angular acceleration is found to be

α=

ω 22 − ω 12

( )

2 Δθ

2gh2 r 2 − 2gh1 r 2

( )

2 Δθ

h −h ) ( ( ) g

r 2 Δθ

( 9.80 m/s ) 2

α= ( 0.510 m − 0.540 m ) = −0.322 rad/s2 2 ( 0.381 m ) ( 2π rad ) !

7.9

Main Rotor:

rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎛ v = rw = ( 3.80 m ) ⎜ 450 = 179 m/s ⎟ ⎝ min ⎠ ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠ !

⎞ m⎞⎛ v ⎛ v = ⎜ 179 ⎟ ⎜ sound ⎟ = 0.522 vsound ⎝ s ⎠ ⎝ 343 m/s ⎠ !

Tail Rotor:

⎛ rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ v = rω = 0.510 m ⎜ 4 138 ⎟⎜ ⎟ = 221 m/s ⎟⎜ min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠ ⎝

(

)

⎞ m⎞⎛ v ⎛ v = ⎜ 221 ⎟ ⎜ sound ⎟ = 0.644 vsound ⎝ s ⎠ ⎝ 343 m/s ⎠ !

7.10

We will break the motion into two stages: (1) an acceleration period and (2) a deceleration period. The angular displacement during the acceleration period is

Topic 7

412

(

)(

) ( )

⎡ 5.0 rev/s 2π rad 1 rev + 0 ⎤ ⎛ ω +ω ⎞ f i ⎥ 8.0 s = 1.3 × 102 rad θ 1 = ω avt ⎜ ⎟t= ⎢ ⎢ ⎥ ⎜⎝ 2 ⎟⎠ 2 ⎣ ⎦

and while decelerating,

⎡ 0 + ( 5.0 rev/s ) ( 2π rad 1 rev ) ⎤ ⎛ ω f + ωi ⎞ 2 θ2 = ⎜ t=⎢ ⎥ (12 s ) = 1.9 × 10 rad ⎟ ⎝ 2 ⎠ 2 ⎣ ⎦ ! The total displacement is ⎛ 1 rev ⎞ θ = θ 1 + θ 2 = ⎡⎣(1.3 + 1.9 ) × 102 rad ⎤⎦ ⎜ = 51 rev ⎝ 2π rad ⎟⎠ !

7.11

(a) The linear distance the car travels in coming to rest is given by v 2 = v02 + 2a ( Δx ) as ! f

Δx = !

v 2f − v02 2a

0 − ( 29.0 m/s ) = = 240 m 2 ( −1.75 m/s 2 ) 2

Since the car does not skid, the linear displacement of the car and the angular displacement of the tires are related by Δx = r(Δθ). Thus, the angular displacement of the tires is

Δθ =

Δx r

⎛ 1 rev ⎞ = 727 rad ⎜ ⎟ = 116 rev 0.330 m ⎝ 2π rad ⎠ 240 m

(

)

(b) When the car has traveled 120 m (one half of the total distance), the linear speed of the car is v = v02 + 2a ( Δx ) = !

( 29.0 m/s ) + 2 ( −1.75 m/s )(120 m ) = 20.5 m/s 2

Topic 7

413

and the angular speed of the tires is

v 20.5 m/s ω= = = 62.1 rad/s r 0.330 m ! 7.12

(a) The angular speed is ω = ω0 + αt = 0 + (2.50 rad/s2)(2.30 s) = 5.75 rad/s. (b) Since the disk has a diameter of 45.0 cm, its radius is r = (0.450 m)/2 = 0.225 m. Thus, vt = rω = (0.225 m)(5.75 rad/s) = 1.29 m/s and

at = rα = (0.225 m)(2.50 rad/s2) = 0.563 m/s2

Δθ = θ f − θ 0 =

ω 2f − ω 02 2α

(

2 2.50 rad/s 2

⎛ 360° ⎞ = 6.61 rad ⎜ ⎟ = 379° ⎝ 2π rad ⎠

(

)

5.75 rad/s − 0

)

and the final angular position of the radius line through point P is

θf = θ0 + Δθ = 57.3° + 379° = 436° or it is at 76° counterclockwise from the +x-axis after turning 19° beyond one full revolution. 7.13

From Δθ = ωavt = [(ωf + ωi)/2]t, we find the initial angular speed to be

Topic 7

414

ωi =

2Δθ t

=ωf =

(

)(

2 37.0 rev 2π rad 1 rev 3.00 s

) − 98.0 rad/s = 57.0 rad/s

The angular acceleration is then

α=

7.14

ω f − ωi t

98.0 rad/s − 57.0 rad/s 3.00 s

= 13.7 rad/s 2

(a) The initial angular speed is rev ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ω 0 = 1.00 × 102 = 10.5 rad/s min ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60.0 s ⎟⎠ !

The time to stop (i.e., reach a speed of ω = 0) with α = −2.00 rad/s2 is

ω − ω0 α

0 − 10.5 rad/s −2.00 rad/s 2

= 5.25 s

⎛ ω + ω 0 ⎞ ⎛ 0 + 10.5 rad/s ⎞ (b) Δθ = ω avt = ⎜ ⎟⎠ t = ⎜⎝ ⎟⎠ ( 5.25 s ) = 27.6 rad ⎝ 2 2 ! 7.15

(a) The car travels 235 m at constant speed in an elapsed time of 36.0 s. Its constant speed is therefore

Δs 235 m v= = = 6.53 m/s Δt 36.0 s ! (b) The angular displacement of the car during the 36.0 s time interval is one-fourth of a full circle or π/2 radians. Thus, the radius of the circular path is

Topic 7

415

Δs 235 m 470 r= = = m Δ θ π /2 rad π ! During the 36.0 s interval, the car has zero tangential acceleration, but does have a centripetal acceleration of constant magnitude

(6.53 m / s) = π (6.53 m / s) = 0.285 m / s a = = r 470 m ( 470 / π ) m v2

This acceleration is always directed toward the center of the circle. Therefore, when the car is at point B, the vector expression for the  2 car’s acceleration is a c = 0.285 m/s !at!35.0°!north!of!west . !

7.16

⎛ 1 609 m ⎞ 3 The radius of the cylinder is r = 2.5 mi ⎜ ⎟ = 4.0 × 10 m . Thus, ⎝ 1 mi ⎠ from ac = rω2, the required angular velocity is

ω= ! 7.17

ac = r

9.80 m/s 2 = 4.9 × 10−2 rad/s 3 4.0 × 10 m

(a) The tangential acceleration of the bug as the disk speeds up is

⎛ ω −ω ⎞ f 0 at = rα = r ⎜ ⎟ ⎜⎝ Δt ⎟⎠ ⎛ 78.0 rev min − 0 ⎞ ⎛ 1 m ⎞ ⎛ 1 min ⎞ ⎛ 2π rad ⎞ = 5.00 in ⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎜⎝ ⎟⎠ ⎜⎝ 39.37 in ⎟⎠ ⎜⎝ 60 s ⎟⎠ ⎜⎝ 1 rev ⎟⎠ 3.00 s

(

)

= 0.346 m/s 2 (b) The final tangential speed of the bug is

Topic 7

416

(

)

⎛ rev ⎞ ⎛ 1 m ⎞ ⎛ 1 min ⎞ 2π rad vt = rω f = 5.00 in ⎜ 78.0 ⎟⎜ ⎟ ⎟⎜ ⎝ min ⎠ ⎜⎝ 39.37 in ⎟⎠ ⎜⎝ 60 s ⎟⎠ 1 rev = 1.04 m/s (c) Since the bug has constant angular acceleration, and hence constant tangential acceleration (at = rα), the tangential acceleration at t = 1.00 is at = 0.346 m/s2 as above. (d) At t = 1.00 s, the tangential velocity of the bug is vt = v0 + att = 0(0.346 m/s2)(1.00 s) = 0.346 m/s and the radial or centripetal acceleration is

( 0.346 m/s ) v2 ac = t = = 0.943 m/s 2 r ( 5.00 in ) (1 m 39.37 in ) ! 2

2 2 2 (e) The total acceleration is a = ac + at = 1.00 m/s , and the angle this !

 acceleration makes with the direction of !a c is ⎛a ⎞ ⎛ 0.346 ⎞ θ = tan −1 ⎜ t ⎟ = tan −1 ⎜ = 20.1° ⎝ 0.943 ⎟⎠ ⎝ ac ⎠ !

7.18

In order for the archeologist to make it safely across the river, the vine must be capable of giving him a net acceleration of !ac = vmax /r upward 2

as he passes through the lowest point on the swing with a speed of vmax = 8.00 m/s. Thus, with T being the tension in the vine, the net force acting on the archeologist at the lowest point is ΣFy = T − mg = mac,

Topic 7

417

giving the required minimum tensile strength of the vine as

(

)

2⎞ ⎛ 2 ⎛ ⎞ 8.00 m/s vmax m ⎜ ⎟ T = mg + mac = m ⎜ g + ⎟ = 85.0 kg ⎜ 9.80 2 + r ⎠ 10.0 m ⎟⎟ s ⎝ ⎜⎝ ⎠

(

)

= 1.38 × 103 N Since he chose a vine with a breaking strength of 1 000 N, he does not make it across. 7.19

(a) The tension in the string must counteract the radial component of the object’s weight, and also supply the needed centripetal acceleration.

mv 2 ∑ Fc = T − mgcos θ = mac = r ! or

T = m ( v 2 r + gcos θ ) ⎡ ( 8.00 m/s )2 ⎤ = ( 0.500 kg ) ⎢ + ( 9.80 m/s 2 ) cos 20.0° ⎥ ⎢⎣ 2.00 m ⎥⎦

= 20.6 N

(b) The net tangential force acting on the object is Ft = mg sin θ, so the tangential acceleration has magnitude F at = t = gsin θ = ( 9.80 m/s 2 ) sin 20.0° = 3.35 m/s 2 m ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 7

418

and is directed downward, tangential to the circular path. The radial component of the acceleration is

v 2 ( 8.00 m/s ) ac = = = 32.0 m/s 2 !toward!the!center!of!the!path r 2.00 m ! 2

( 3.35 m/s ) + ( 32.0 m/s ) 2 2

atotal = 32.2 m/s2

⎛a ⎞ ⎛ 3.35 ⎞ θ = tan −1 ⎜ t ⎟ = tan −1 ⎜ = 5.98° ⎝ 32.0 ⎟⎠ ⎝ ac ⎠ !

Thus, !

2 2

 a total = 32.2 m/s 2 !at!5.98°!to!the!cord!and!pointing!below the!center!of!the!circular!path

(d) No change in answers if the object is swinging toward the equilibrium point instead of away from it. (e) If the object is swinging toward the equilibrium position, it is gaining speed, whereas it is losing speed if it is swinging away from the equilibrium position. In both cases, when the cord is 20.0° from the vertical, the tangential, centripetal, and total accelerations have the magnitudes and directions calculated in parts (a) through © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 7

419

(c).

7.20

(a) Use the expression ac =

ω=

ac = r

v2 = rω 2 to solve for ω : r 9g = 4.43 rad/s 4.50 m

(b) The linear speed is

vt = rω = ( 4.50 m ) ( 4.43 rad/s) = 19.9 m/s 7.21

(a) From ΣFr = mac, we have

⎛ vt2 ⎞ ( 55.0 kg ) ( 4.00 m/s ) T = m⎜ ⎟ = = 1.10 × 103 N = 1.10 kN ⎝ r⎠ 0.800 m ! 2

(b) The tension is larger than her weight by a factor of T mg

7.22

1.10 × 103 N

(55.0 kg )(9.80 m/s ) 2

= 2.04 times

(a) If T is the tension in each of the two support chains, the net force acting on the child at the lowest point on the circular path is ⎛ v2 ⎞ ∑ Fy = 2T − mg = mac = m ⎜ ⎟ ⎝ r⎠ !

so the speed at this point is

⎛ 2T ⎞ v = r⎜ − g⎟ = ⎝ m ⎠ !

⎛ 2 ( 350 N ) ⎞ − 9.80 m/s ⎟ = 4.81 m/s ( 3.00 m ) ⎜ 40.0 kg 2

⎝

⎠

(b) The upward force the seat exerts on the child at this lowest point is

Topic 7

420

Fseat = 2T = 2(350 N) = 700 N 7.23

Friction between the tires and the roadway is capable of giving the truck a maximum centripetal acceleration of

vt2,max ( 32.0 m/s ) ac ,max = = = 6.83 m/s 2 r 150 m ! 2

If the radius of the curve changes to 75.0 m, the maximum safe speed will be v = r ac ,max = ! t ,max

7.24

(75.0 m )( 6.83 m/s ) = 22.6 m/s 2

v2 Since Fc = m t = mr ω 2 , the needed angular velocity is r ! Fc ω= = mr !

4.0 × 10−11 N ( 3.0 × 10−16 kg )( 0.150 m )

⎛ 1 rev ⎞ = ( 9.4 × 102 rad s ) ⎜ = 1.5 × 102 rev/s ⎟ ⎝ 2π rad ⎠ !

7.25

(a) ac = rω2 = (2.00 m)(3.00 rad/s)2 = 18.0 m/s2 (b) Fc = mac = (50.0 kg)(18.0 m/s2) = 900 N (c) We know the centripetal acceleration is produced by the force of friction. Therefore, the needed static friction force is fs = 900 N. Also, the normal force is n = mg = 490 N. Thus, the minimum coefficient of friction required is

Topic 7

421

µs = !

(f )

s max

900 N = 1.84 490 N

Such a large coefficient of friction is unrealistic, and she will not be able to stay on the merry-go-round. 7.26

(a) The only force acting on the astronaut is the normal force exerted on him by the “floor” of the cabin.

2 (b) ΣF = n= mac or n = mvt /r !

(d) From the equation in Part (b),

vt = !

nr = m

( 294 N)(10.0 m ) = 7.00 m/s 60.0 kg

(e) Since vt = rω, we have

v 7.00 m/s ω= t= = 0.700 rad/s r 10.0 m ! (f) The period of rotation is

Topic 7

422

2π 2π T= = = 8.98 s ω 0.700 rad/s ! (g) Upon standing, the astronaut’s head is moving slower than his feet because his head is closer to the axis of rotation. When standing, the radius of the circular path followed by the head is rhead = 10.0 m − 1.80 m = 8.20 m, and the tangential speed of the head is (vt)head = rhead ω = (8.20 m)(0.700 rad/s) = 5.74 m/s 7.27

(a) Since the 1.0-kg mass is in equilibrium, the tension in the string is T = mg = (1.0 kg)(1.0 kg)(9.8 m/s2) = 9.8 N (b) The tension in the string must produce the centripetal acceleration of the puck. Hence, Fc = T = 9.8 N. 2 (c) From !Fc = mpuck ( vt R ) , we find

R Fc vt = = mpuck ! 7.28

(1.0 m )( 9.8 N) = 6.3 m/s 0.25 kg

(a) Taking y = 0 at the bottom of the frictionless halfpipe of radius R, a snowboarder starting from rest will have a speed at the bottom equal to

Wnc + KEi + PEi = KE f + PE f 0 + 0 + mgR = 12 mv 2 + 0 v = 2gR

Topic 7

423

(b) The snowboarder’s centripetal acceleration has magnitude ac = v2/r. At the bottom of the halfpipe, v = 2gR so that

v 2 2gR ac = = → ac = 2g R R (c) At the bottom of the halfpipe, two forces act on the snowboarder: w = mg is directed down (in the positive radial direction) and the normal force is directed up (in the negative radial direction). Applying Newton’s second law in the radial direction gives: mv 2 r mv 2 mg − N = − R mv 2 N = mg + R ΣFr = −

Substitute v = 2gR to find

N = mg +

7.29

m ( 2gR ) = 3mg R

(a) The force of static friction acting toward the road’s center of curvature must supply the briefcase’s required centripetal acceleration. The condition that it be able to meet this need is that F = mvt2 r ≤ ( fs )max = µsmg , or µs ≥ vt2 rg . When the tangential ! !c

speed becomes large enough that µs = vt2 rg , the briefcase will begin to slide.

Topic 7

424 2 (b) As discussed above, the briefcase starts to slide when !µs = vt rg . If

this occurs at the speed, vt = 15.0 m/s, the coefficient of static friction must be

(15.0 m/s ) µ = = 0.370 62.0 m ) ( 9.80 m/s ) ( ! 2

7.30

(a) The external forces acting on the water are the gravitational force and the contact force exerted on the water by the pail. (b) The contact force exerted by the pail is the most important in causing the water to move in a circle. If the gravitational force acted alone, the water would follow the parabolic path of a projectile. (c) When the pail is inverted at the top of the circular path, it cannot hold the water up to prevent it from falling out. If the water is not to spill, the pail must be moving fast enough that the required centripetal force is at least as large as the gravitational force. That is, we must have

v2 r

≥ mg or v ≥ rg =

(1.00 m)(9.80 m s ) = 3.13 m/s 2

(d) If the pail were to suddenly disappear when is it at the top of the circle and moving at 3.13 m/s, the water would follow the parabolic arc of a projectile launched with initial velocity © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 7

425

components of v0x = 3.13 m/s, v0y = 0. 7.31

(a) The centripetal acceleration is 2

⎡⎛ rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎤ 2 ac = rω = ( 9.00 m ) ⎢⎜ 4.00 ⎥ = 1.58 m/s ⎟ ⎜ ⎟ ⎜ ⎟ min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠ ⎦ ⎣⎝ ! 2

(b) At the bottom of the circular path, we take upward as positive and apply Newton’s second law. This yields ΣFy = n − mg = m(+ac), or n = m(g + ac) = (40.0 kg)[(9.80 + 1.58) m/s2] = 4.55 N upward (c) At the top of the path, we again take upward as positive and apply Newton’s second law to find ΣFy = n − mg = m(−ac), or n = m(g − ac) = (40.0 kg)[(9.80 − 1.58) m/s2] = 329 N upward (d) At a point halfway up, the seat exerts an upward vertical component equal to the child’s weight (392 N) and a component toward the center having magnitude Fc = mac = (40.0 kg)(1.58 m/s2) = 63.2 N. The total force exerted by the seat is

F = !R

( 392 N) + ( 63.2) = 397 N directed inward and at 2

⎛ 392 N ⎞ θ = tan −1 ⎜ = 80.8° above!the!horizontal ⎝ 63.2 N ⎟⎠ !

7.32

(a) At A, taking upward as positive, Newton’s second law gives ΣFy = n − mg = m(+ac). Thus,

Topic 7

426

⎛ vt2 ⎞ n = mg + mac = m ⎜ g + ⎟ r⎠ ⎝ 2⎤ ⎡ 20.0 m s ⎥ ⎢ = 500 kg ⎢ 9.80 m s 2 + ⎥ = 25 kN 10 m ⎢ ⎥ ⎣ ⎦

(

)

(b) At B, still taking upward as positive, Newton’s second law yields 2 ΣFy = n − mg = m(−ac) or !mg = n + mac = n + mvt r. If the car is on the 2 verge of leaving the track, then n = 0 and !mg = mvt r , giving

(15 m )( 9.80 m s ) = 12 m/s

v = rg = !t

7.33

(a) Substitute values into Newton’s law of universal gravitation:

F=G

(

m1m2 r2

= 6.673 × 10

(7.50 × 10 kg )( 2.70 × 10 kg ) N ⋅ m /kg ) ( 2.80 × 10 m ) 24

−11

= 1.72 × 10 20 N (b) From Newton’s second law (with gravity as the only acting force), the moon’s acceleration is

amoon =

F mmoon

mplanet r2

= 6.38 × 10−3 m/s2 (c) From Newton’s second law (with gravity as the only acting force), the planet’s acceleration is

Topic 7

427

aplanet =

F mplanet

mmoon r2

= 2.30 × 10−5 m/s2 7.34

(a) Substitute values into Newton’s law of universal gravitation:

F=G

MEm r2

5.98 × 10 kg ) ( 4.19 × 10 kg ) )( (6.79 × 10 m ) 24

(

= 6.67 × 10−11 N ⋅ m 2 /kg 2

= 3.62 × 106 N (b) The gravitational potential energy is PE = −G

MEm r

( 5.98 × 10 kg )( 4.19 × 10 kg ) N ⋅ m /kg ) 24

(

= − 6.67 × 10

−11

6.79 × 106 m

= −2.46 × 1013 J

MEm r2

(

= 6.67 × 10

( 5.98 × 10 kg )( 80.0 kg ) N ⋅ m /kg ) (6.79 × 10 m ) 24

−11

= 692 N

7.35

The forces exerted on the 2.0-kg by the other bodies are Fx and Fy as

Topic 7

428

shown in the diagram at the below.

The magnitudes of these forces are

(6.67 × 10 N ⋅ m /kg )(2.0 kg )( 4.0 kg ) F = ( 4.0 m) 2

−11

= 3.3 × 10−11 N

and

( 6.67 × 10 F = y

−11

N ⋅ m 2 /kg 2 ) ( 2.0 kg ) ( 3.0 kg )

( 2.0 m )

= 1.0 × 10−10 N

2 2 −10 The resultant force exerted on the 2.0-kg is F = Fx + Fy = 1.1 × 10 N !

⎛F ⎞ y −1 θ = tan ⎜ ⎟ = tan −1 3.0 = 72° above the + x-axis . directed at ⎜⎝ Fx ⎟⎠

( )

7.36

(a) The density of the white dwarf would be

ρ=

M MSun MSun 3MSun = = = 3 V VEarth 4π RE / 3 4π RE3

Using data from Table 7.3,

ρ= !

3 (1.991 × 1030 kg )

4π ( 6.38 × 10 m ) 6

= 1.83 × 109 kg/m 3

(b) Fg = mg = GMm/r2, so the acceleration of gravity on the surface of © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 7

429

the white dwarf would be

GMS RE2

(6.67 × 10 N ⋅ m kg )(1.991 × 10 kg ) = 3.26 × 10 m s = (6.38 × 10 m) 2

−11

(c) The general expression for the gravitational potential energy of an object of mass m at distance r from the center of a spherical mass M is PE = −GMm/r. Thus, the potential energy of a 1.00 kg mass on the surface of the white dwarf would be

PE = − =−

GMSun (1.00 kg) RE

(6.67 × 10

−11

N ⋅ m 2 kg 2 ) ( 1.991 × 1030 kg ) (1.00 kg) 6.38 × 106 m

= −2.08 × 1013 J 7.37

(a) At the midpoint between the two masses, the forces exerted by the 200-kg and 500-kg masses are oppositely directed, so from F = GMm/r2 and r1 = r2 = r, we have

∑F = !

GM1m GM2m Gm − = 2 ( M1 − M2 ) r12 r22 r

( 6.67 × 10 ∑F = !

−11

N ⋅ m 2 kg 2 ) ( 500 kg ) ( 500 kg − 200 kg ) (0.200 m)2

= 2.5 × 10−5 N toward the 500-kg (b) At a point between the two masses and distance d from the 500-kg © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 7

430

mass, the net force will be zero when

(

)(

G 50.0 kg 200 kg

(0.400 m − d)

) = G (50.0 kg )(500 kg ) or d = 0.245 m d2

Note that the above equation yields a second solution d = 1.09 m. At that point, the two gravitational forces do have equal magnitudes, but are in the same direction and cannot add to zero. 7.38

The equilibrium position lies between the Earth and the Sun on the line connecting their centers. At this point, the gravitational forces exerted on the object by the Earth and Sun have equal magnitudes and opposite directions. Let this point be located distance r from the center of the Earth. Then, its distance from the Sun is (1.496 × 1011 m − r), and we may determine the value of r by requiring that

GmE m GmSm = 2 r (1.496 × 1011 m − r)2 where mE and mS are the masses of the Earth and Sun, respectively. This reduces to

(1.496 × 10 m − r ) = m = 577 11

or 1.496 × 1011 m = 577r, which yields r = 2.59 × 108 m from center of the Earth.

Topic 7

7.39

431

(a) If air resistance is ignored, the only force acting on the projectile during its flight is the gravitational force. Since the gravitational force is a conservative force, the total energy of the projectile remains constant. At r = RE, the projectile has speed v = vesc 3 = 2GME /RE / 3 , and its total energy is

E = KE + PEg =

1 2 ⎛ GME m ⎞ 1 ⎛ 1 2 GME ⎞ GME m mv + ⎜ − = m ⋅ − 2 RE ⎟⎠ 2 ⎜⎝ 9 RE ⎟⎠ RE ⎝

8 GMEm E=− ⋅ . 9 RE !

When the projectile reaches maximum height at r = rmax, and is momentarily at rest, the kinetic energy is zero and we have

GMEm 8 GMEm E = KE + PEg = 0 − =− ⋅ rmax 9 RE ! or

9 9 rmax = RE = ( 6.38 × 106 m ) = 7.18 × 106 m . 8 8 !

(b) The altitude of the projectile when at r = rmax is

9 R 6.38 × 106 m h = rmax − RE = RE − RE = E = = 7.98 × 105 m 8 8 8 ! 7.40

We know that m1 + m2 = 5.00 kg, or m2 = 5.00 kg − m1.

2 ⎛ Gm1m2 −8 −11 N ⋅ m ⎞ m1 (5.00 kg − m1 ) ⇒ 1.00 × 10 N = 6.67 × 10 ⎜⎝ r2 kg 2 ⎟⎠ (0.200 m)2

Topic 7

432

1.00 × 10 N ) ( 0.200 m ) = 6.00 kg (5.00 kg ) m − m = ( 2

−8

2 1

6.67 × 10

−11

N ⋅ m kg

Thus, !m1 − ( 5.00 kg ) m1 + 6.00 kg = 0 , or (m1 − 3.00 kg)(m1 − 2.00 kg) = 0. 2

This yields m1 = 3.00 kg, so m2 = 2.00 kg. The answer m1 = 2.00 kg and m2 = 3.00 kg is physically equivalent. 7.41

(a) The radius of the satellite’s orbit is r = h + RE = 2.80 × 106 m + 6.38 × 106 m = 9.18 × 106 m Then, modifying Equation 7.23 (Kepler’s third law) for orbital motion about the Earth rather than the Sun, we have

4π ( 9.18 × 10 m ) ⎛ 4π 2 ⎞ 3 T =⎜ r = −11 ⎟ 6.67 × 10 N ⋅ m 2 kg 2 ) ( 5.98 × 1024 kg ) ⎝ GME ⎠ ( ! 2

yielding

T2 = 7.66 × 107 s2 and ⎛ 1h ⎞ T = 8.75 × 103 s ⎜ = 2.43 h ⎝ 3600 s ⎟⎠ !

(b) The constant tangential speed of the satellite is

vt =

circumference of orbit period

2π r T

(

2π 9.18 × 106 m 5

8.75 × 10 s

)

vt = 6.59 × 103 m/s = 6.59 km/s.

Topic 7

433

v 2 ( 6.59 × 10 km/s ) a = ac = t = = 4.73 m/s 2 !toward!center!of!Earth 6 r 9.18 × 10 m ! 2

7.42

(a) The satellite’s period is T = 110 min(60.0s/1.00 min) = 6.60 × 103 s. Using the form of Kepler’s third law (Equation 7.23) suitable for objects orbiting the Earth, we have ⎛ 4π 2 ⎞ 3 T2 = ⎜ ⎟⎠ r GM ⎝ E !

⎛ T 2GME ⎞ 3 r=⎜ ⎝ 4π 2 ⎟⎠

Thus, the radius of the orbit must be

(

)(

)

2 ⎛ ⎞5 6.60 × 103 s 6.67 × 10−11 N ⋅ m 2 kg 2 5.98 × 1024 kg ⎟ r=⎜ ⎜ ⎟ 4π ⎜⎝ ⎟⎠

= 7.61 × 106 m

so that the altitude of the satellite is h = r − RE = 7.61 × 106 m − 6.38 × 106 m = 1.23 ×106 m (b) The gravitational force is

⎛ GM ⎞ Fg = ⎜ 2 E ⎟ m = mg ⎝ r ⎠ !

GME r2

Thus, at the altitude of the satellite, the acceleration due to gravity is

g= !

( 6.67 × 10

−11

N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg )

(7.61 × 10 m ) 6

= 6.89 m/s 2

Topic 7

7.43

434

From Kepler’s third law (Equation 7.23), written in the form suitable for bodies orbiting Mars, we have T2 = (4π2/GMMars)r3, so the mass of Mars, computed from the given data, must be ⎛ 4π 2 ⎞ 2 MMars = ⎜ r 2⎟ ⎝ GT ⎠ ⎛ ⎞ 2 4 π ⎟ 9.4 × 106 m 3 =⎜ 2⎟ ⎜ ⎜⎝ 6.67 × 10−11 N ⋅ m 2 kg 2 2.8 × 10 4 s ⎟⎠

(

)(

)

(

)

= 6.3 × 1023 kg

7.44

(a) The satellite moves in an orbit of radius r = 2RE and the gravitational force supplies the required centripetal acceleration. Hence, m(vt / 2RE ) = GmE m/(2RE ) , or 2

2 24 ⎛ GmE kg) −11 N ⋅ m ⎞ (5.98 × 10 vt = = ⎜ 6.67 × 10 = 5.59 × 103 m/s 2 ⎟ 6 ⎝ 2RE kg ⎠ 2(6.38 × 10 m)

(b) The period of the satellite’s motion is

(

)

2π ⎡ 2 6.38 × 106 m ⎤ ⎛ 1h ⎞ ⎣ ⎦ T= = ⎜ ⎟ = 3.98 h 3 vt 5.59 × 10 m/s ⎝ 3 600 s ⎠ 2π r

⎛ N ⋅ m 2 ⎞ (5.98 × 1024 kg)(600 kg) F = ⎜ 6.67 × 10−11 = 1.47 × 103 N 2 ⎝ kg 2 ⎟⎠ ⎡⎣ 2(6.38 × 106 m) ⎤⎦

Topic 7

7.45

435

(a) Use Kepler’s third law to find the semimajor axis, a: 1

T 2 = KS a 3

⎛ T2 ⎞ 3 → a = ⎜ ⎟ where, for distances measured in AU and ⎝K ⎠ S

time measured in Earth years, KS = 1 yr2/AU3. Substituting values gives 1

⎛ ( 76.3 yr )2 ⎞ 3 a=⎜ → 2 3⎟ ⎝ 1 yr /AU ⎠

a = 18.0 AU

(b) The comet orbits in an ellipse with a major axis, 2a, equal to the sum of the farthest and closest distances from the Sun (aphelion ra and perihelion rp, respectively): 2a = ra + rp. Solve for the aphelion to find

ra = 2a − rp = 2 (18.0 AU ) − 0.610 AU = 35.4 AU 7.46

A synchronous satellite will have an orbital period equal to Jupiter’s rotation period, so the satellite can have the red spot in sight at all times. Thus, the desired orbital period is ⎛ 3 600 s ⎞ T = 9.84 h ⎜ = 3.54 × 10 4 s ⎟ ⎝ 1h ⎠ !

Kepler’s third law gives the period of a satellite in orbit around Jupiter as T2 = (4π2/GMJupiter)r3. The required radius of the circular orbit is therefore

Topic 7

436

⎛ GMJupiterT 2 ⎞ r=⎜ ⎟⎠ 4π 2 ⎝

1/3

⎡ (6.67 × 10−11 N ⋅ m 2 / kg 2 )(1.90 × 1027 kg)(3.54 × 10 4 s)2 ⎤ =⎢ ⎥ 4π 2 ⎣ ⎦

1/3

r = 1.59 × 108 m

The altitude of the satellite above Jupiter’s surface should be h = r − RJupiter = 1.59 × 108 m − 6.99 × 107 m = 8.91 × 107 m 7.47

From Kepler’s third law, the mass of Jupiter can be expressed in terms of one of its satellite’s orbital radius and period as MJupiter = (4π2/GT2)r3. (a) For Io, r = 4.22 × 108 m and T = 1.77 days [(8.64 × 104 s)/1 day] = 1.53 × 105 s, giving

MJupiter =

4π 2 (4.22 × 108 m)3 = 1.90 × 1027 kg −11 2 2 5 2 (6.67 × 10 N ⋅ m / kg )(1.53 × 10 s)

(b) For Ganymede, r = 1.07 × 109 m and T = 7.16 days [(8.64 × 104 s)/1 day] = 6.19 × 105 s, giving

MJupiter =

4π 2 (1.07 × 109 m)3 = 1.89 × 1027 kg −11 2 2 5 2 (6.67 × 10 N ⋅ m / kg )(6.19 × 10 s)

(c) Yes. The results of parts (a) and (b) are consistent. They predict the same mass within the limits of uncertainty of the data used to compute these results. 7.48

The gravitational force on a small parcel of material at the star’s

Topic 7

437

equator supplies the centripetal acceleration, or

Gms m/ Rs2 = m ( vt2 / Rs ) = m(Rsω 2 ) . Hence, ω = Gms / Rs3 and 6.67 × 10 N ⋅ m /kg ) ⎡ 2 (1.99 × 10 kg ) ⎤ ( ⎣ ⎦ = 1.63 × 10 rad/s ω= (10.0 × 10 m) −11

7.49

vt ( 98.0 mi/h ) ⎛ 0.447 m/s ⎞ ⎛ 1 rev ⎞ = 9.40 rev/s (a) ω = = r 0.742 m ⎜⎝ 1 mi/h ⎟⎠ ⎜⎝ 2π rad ⎟⎠ !

(b) α =

ω 2 − ω i2 2 Δθ

(

) 2 (1 rev )

9.40 rev/s − 0

= 44.2 rev/s 2

⎡ ⎛ 0.447 m/s ⎞ ⎤ 98.0 mi h ( ) ⎢ ⎜⎝ 1 mi/h ⎟⎠ ⎥ vt2 ⎣ ⎦ = 2.59 × 103 m/s 2 ac = r 0.742 m !

⎡ rev ⎛ 2π rad ⎞ ⎤ 2 ac = rα = ( 0.742 m ) ⎢ 44.2 2 ⎜ ⎥ = 206 m/s ⎟ s ⎝ 1 rev ⎠ ⎦ ⎣ ! (c) In the radial direction at the release point, the hand supports the weight of the ball and also supplies the centripetal acceleration. Thus, Fr = mg + mar = m(g + ar) or Fr = (0.198 kg)(9.80 m/s2 + 2.59 × 103 m/s2) = 515 N In the tangential direction, the hand supplies only the tangential acceleration, so Ft = mat = (0.198 kg)(206 m/s2) = 40.8 N

Topic 7

7.50

438

vt 1.30 m/s = 56.5 rad/s (a) ω i = = −2 r 2.30 × 10 m i ! vt 1.30 m/s = 22.4 rad/s (b) ω f = = rf 5.80 × 10−2 m ! (c) The duration of the recording is Δt = (74 min)(60 s/min) + 33 s = 4 473 s

Thus, α av =

(d) Δθ =

ω f − ωi

ω 2f − ω i2 2α

Δt

(22.4 − 56.5) rad/s = −7.62 × 10 rad/s −3

4 473 s

(22.4 rad s) − (56.5 rad s) = 1.77 × 10 rad = 2 ( −7.62 × 10 rad s ) 2

−3

(e) The track moves past the lens at a constant speed of vt = 1.30 m/s for 4 473 seconds. Therefore, the length of the spiral track is Δs = vt(Δt) = (1.30 m/s)(4 473 s) = 5.81 × 103 m = 5.81 km 7.51

The angular velocity of the ball is ω = 0.500 rev/s = π rad/s. (a) vt = rω = (0.800 m)(π rad/s) = 2.51 m/s

(b) ac =

vt2 2 rω = (0.800 m)(π rad/s)2 = 7.90 m/s 2 r

(c) We imagine that the weight of the ball is supported by a frictionless platform. Then, the rope tension need only produce the centripetal acceleration. The force required to produce the needed

Topic 7

439 2 centripetal acceleration is !F = m ( vt r ) . Thus, if the maximum force

the rope can exert is 100 N, the maximum tangential speed of the ball is

( vt )max = 7.52

rFmax (0.800 m)(100 N) = = 4.00 m/s m 5.00 kg

(a) The angular speed is related to the linear speed by vt = rω . Solve for ω to find

ω=

vt 6.25 cm/s = → ω = 3.13 rad/s r 2.00 cm

(b) When the dungball has rolled 1.00 m, the arclength s = 1.00 m so that

θ=

s 1.00 m = = 50.0 rad r 2.00 × 10−2 m

⎛ 1 rev ⎞ 50.0 rad = 50.0 rad ⎜ = 7.96 rev ⎝ 2π rad ⎟⎠

7.53

The radius of the satellite’s orbit is r = RE + h = 6.38 × 106 m + (1.50 × 102 mi)(1 609 m/1 mi) = 6.62 × 106 m (a) The required centripetal acceleration is produced by the gravitational force, so m(vt / r) = GME m/ r , which gives 2

vt = GME / r .

Topic 7

440

(

)

24 2 ⎞ 5.98 × 10 kg ⎛ −11 N ⋅ m vt = ⎜ 6.67 × 10 = 7.76 × 103 m/s 2 ⎟ 6 kg ⎠ 6.62 × 10 m ⎝

(b) The time for one complete revolution is

7.54

2π r vt

2π (6.62 × 106 m) 3

7.76 × 10 m/s

= 5.36 × 103 s = 89.3 min

(a) At the lowest point on the path, the net upward force (i.e., the force directed toward the center of the path and supplying the 2 centripetal acceleration) is !∑ Fup = T − mg = m ( vt r ) , so the tension

in the cable is 2 ⎛ ⎞ ⎛ vt2 ⎞ 2 ( 3.00 m/s ) T = m ⎜ g + ⎟ = ( 0.400 kg ) ⎜ 9.80 m/s + ⎟ = 8.42 N ⎝ r⎠ 0.800 m ⎠ ⎝ !

(b) Using conservation of mechanical energy, (KE + PEg)f = (KE + PEg)i, as the bob goes from the lowest to the highest point on the path, gives

0 + mg ⎡⎣ L ( 1 − cosθ max ) ⎤⎦ =

1 2 v2 mvi + 0, or cosθ max = 1 − i 2 2gL

⎛ ⎛ ⎞ v2 ⎞ (3.00 m/s)2 θ max = cos −1 ⎜ 1 − i ⎟ = cos −1 ⎜ 1 − = 64.8° 2 ⎝ 2gL ⎠ 2(9.80 m/s )(0.800 m) ⎟⎠ ⎝ (c) At the highest point on the path, the bob is at rest and the net radial force is

⎛ v2 ⎞ ΣFr = T − mg cosθ max = m ⎜ t ⎟ = 0 ⎝ r ⎠

Topic 7

441

Therefore, T = mg cosθ max = (0.400 kg)(9.80 m/s 2 )cos(64.8°) = 1.67 N

7.55

(a) When the car is at the top of the arc, the normal force is upward and the weight downward. The net force directed downward, toward the center of the circular path and hence supplying the 2 centripetal acceleration, is !∑ Fdown = mg − n = m ( vt r ) . Thus, the 2 normal force is n = m ( g − vt r ) . !

2 (b) If r = 30.0 m and n → 0, then !g − vt r → 0 . For this to be true, the

speed of the car must be vt = rg = (30.0 m)(9.80 m/s 2 ) = 17.1 m/s

7.56

(a) Solve vt = rω for the angular speed to find

ω min =

vt 1.67 × 10−7 m/s = → ω min = 3.34 × 10−2 rad/s r 5.00 × 10−6 m

(b) When the cell has rolled 5.00 × 10−3 m, the arclength s = 5.00 × 10−3 m so that

θ=

s 5.00 × 10−3 m = = 1.00 × 103 rad −6 r 5.00 × 10 m

⎛ 1 rev ⎞ 1.00 × 103 rad = 1.00 × 103 rad ⎜ = 159 rev ⎝ 2π rad ⎟⎠

7.57

The speed the person has due to the rotation of the Earth is vt = rω,

Topic 7

442

where r is the distance from the rotation axis and ω is the angular velocity of rotation. The person’s apparent weight, Fg, apparent, equals the magnitude of the upward normal force exerted on him by the scales. The true weight, Fg, true = mg, is directed downward. The net downward force produces the needed centripetal acceleration, or

⎛ v2 ⎞ ΣF down = −n + Fg, true = Fg, apparent + Fg, true = m ⎜ t ⎟ mrω 2 ⎝ r ⎠ (a) At the equator, r = RE, so Fg, true = Fg, apparent + mREω

> Fg, apparent .

(b) At the equator, it is given that ac = REω2 = 0.034 0 m/s2, so the apparent weight is

Fg, apparent = Fg, true − mREω 2 = (75.0 kg) ⎡⎣( 9.80 − 0.034 0 ) m/s 2 ⎤⎦ = 732 N At either pole, r = 0 (the person is on the rotation axis) and

Fg, apparent = Fg, true = mg = (75.0 kg)(9.80 m/s 2 ) = 735 N 7.58

(a) When the passenger is at the top, the radial forces producing the centripetal acceleration are the upward force of the seat and the downward force of gravity. The downward force must exceed the upward force to yield a net force toward the center of the circular path.

Topic 7

443

(b) At the lowest point on the path, the radial forces contributing to the centripetal acceleration are again the upward force of the seat and the downward force of gravity. However, the upward force must now exceed the downward force to yield a net force directed toward the center of the circular path.

(c) The seat must exert the greatest force on the passenger at the lowest point on the circular path. (d) At the top of the loop, ΣFr = Fg − n = mv2/r, or

2 ⎛ 4.00 m s ) ⎞ ⎛ ( v2 v2 ⎞ 2 n = Fg − m = m ⎜ g − ⎟ = (70.0 kg ) ⎜ 9.80 m s − ⎟ = 546 N r ⎝ r⎠ 8.00 m ⎠ ⎝ !

At the bottom of the loop, ΣFr = n − Fg = mv2/r, or

Topic 7

444 2 ⎛ 4.00 m s ) ⎞ ⎛ ( v2 v2 ⎞ 2 n = Fg + m = m ⎜ g + ⎟ = (70.0 kg ) ⎜ 9.80 m s + ⎟ = 826 N r ⎝ r⎠ 8.00 m ⎠ ⎝ !

7.59

Define the following symbols: MM = mass of the Moon, ME = mass of the Earth, RM = radius of the Moon, RE = radius of the Earth, and r = radius of the Moon’s orbit around the Earth. We interpret “lunar escape speed” to be the escape speed from the surface of a stationary Moon alone in the universe. Then,

vlaunch = 2vescape = 2

2GM M 8GM M 2 , or vlaunch = RM RM

Applying conservation of mechanical energy from launch to impact

( )

gives 2 mvimpact + PEg f = 2 mvlaunch + PEg i , or ! 2

( ) ( )

2 2 vimpact = vlaunch + ⎡ PEg − PEg ⎤ i f ⎥ ⎦ m ⎢⎣ ! The needed potential energies are

(PE ) = − GMR m − GMr m and (PE ) = − GMR m − GMr m M

Using these potential energies and the expression for !vlaunch from 2

above, the equation for the impact speed reduces to

(

)

⎛3M ME − M M ⎞ ME M ⎜ ⎟ vimpact = 2G + − RE r ⎜⎝ RM ⎟⎠

Topic 7

445

−11 With numeric values of G = 6.67 × 10

N ⋅ m2 kg

, ME = 5.98 × 1024 kg,

MM = 7.36 × 1022 kg, RM = 1.74 × 106 m, RE = 6.38 × 106 m, and r = 3.84 × 108 m, we find vimpact = 1.18 × 104 m/s = 11.8 km/s. 7.60

Since the airplane flies in a horizontal circle, its vertical acceleration is zero, and ΣFy = may ⇒ F cos θ − T sin θ − mg = 0

mg F= + T tan θ cos θ !

[1]

Also, the components of the lift force and the tension in the wire directed toward the center of the circular path must supply the required centripetal acceleration. Hence, ⎛ vt2 ⎞ ∑ Fradial = mac = m ⎜ ⎟ ⎝ r⎠ !

mvt2 F sin θ + T sin θ = r

[2]

Substituting Equation [1] into [2] yields

Topic 7

446

2 t

(mg tan θ + T sin θ tan θ ) + T cos θ = mvr

or the tension in the wire is m ( vt2 r ) − mg tan θ

T= sin θ tan θ + cos θ !

where

r = cos θ

If θ = 20.0°, m = 0.750 kg, v1 = 35.0 m/s and ! = 60.0 m , this gives

(

) (

)

(

)

2 ⎡ ⎡ 60.0 m cos 20.0° ⎤ − 9.80 m/s 2 tan 20.0° ⎤ 35.0 m/s ⎢ ⎥ ⎣ ⎦ T = 0.750 kg ⎢ ⎥ sin 20.0°tan 20.0° + cos 20.0° ⎢ ⎥ ⎣ ⎦

(

or 7.61

)

T = 12.8 N.

The item of clothing will fall away from the rotating drum when the component of its weight directed toward the center of the drum exceeds the needed centripetal force.

That is, when mg! sin θ ≥ m ( vt2 r ) = mrω 2 . Thus, if the clothing is to lose ! contact with the drum at θ = 68.0°, we must have mrω = mg sin 68.0°. The required angular velocity, expressed in revolutions per second, is

Topic 7

447

therefore

ω= ! 7.62

( 9.80 m/s ) sin 68.0° = 0.835 rev/s

⎛ 1 rev ⎞ gsin 68.0° rad/s = ⎜ r ⎝ 2π rad ⎟⎠

0.330 m

The centripetal acceleration of a particle at distance r from the axis is 2 2 !ac = vt r = rω . If we are to have ac = 100 g, then it is necessary that

rω 2 = 100g or ω = 100 g/r ! The required rotation rate increases as r decreases. In order to maintain the required acceleration for all particles in the casting, we use the minimum value of r and find

ω=

100g rmin

= 216

7.63

(

100 9.80 m/s 2 2.10 × 10 m

)

−2

rad ⎛ 1 rev ⎞ ⎛ 60.0 s ⎞ 3 rev ⎜ ⎟⎜ ⎟ = 2.06 × 10 s ⎝ 2π rad ⎠ ⎝ 1 min ⎠ min

Choosing PEg = 0 at the top of the hill, the speed of the skier after dropping distance h is found using conservation of mechanical energy as

1 2 mvt − mgh = 0 + 0, or vt2 = 2gh . 2

Topic 7

448

The net force directed toward the center of the circular path, and providing the centripetal acceleration, is

ΣFr = mg cosθ − n = m ( vt2 / R ) Solving for the normal force, after making the substitutions vt = 2gh 2

and cos θ = (R − h)/R = 1 − (h/R), gives n = mg(1 − h/R) − m(2gh/R) = mg(1 − 3h/R) The skier leaves the hill when n → 0. This occurs when 3h 1− = 0 or h = R/3. ! R

7.64

When the rope makes angle θ with the vertical, the net force directed toward the center of the circular path is ΣFr = T − mg cos θ as shown in the sketch below. This force supplies the needed centripetal acceleration, so ⎛ v2 ⎞ ⎛ v2 ⎞ T − mgcos θ = m ⎜ t ⎟ , or T = m ⎜ gcos θ + t ⎟ ⎝ r⎠ ⎝ r⎠ !

Using conservation of mechanical energy, (KE +PEg)f = (KE +PEg)i, with

Topic 7

449

KE = 0 at θ = 90° and PEg = 0 at the bottom of the arc, the speed when the rope is at angle θ from the vertical is given by 1 2 mvt + mg ( r − r cos θ ) = 0 + mgr , or !vt2 = 2gr cos θ . The expression for 2 !

the tension in the rope at angle θ then reduces to T = 3mg cos θ. (a) At the beginning of the motion, θ = 90° and T = 0. (b) At 1.5 m from the bottom of the arc, cos θ = 2.5 m/r = 2.5 m/4.0 m = 0.63 and the tension is T = 3(70 kg)(9.8 m/s2)(0.63) = 1.3 ×103 N = 1.3 kN (c) At the bottom of the arc, θ = 0° and cos θ = 1.0, so the tension is T = 3(70 kg)(9.8 m/s2)(1.0) = 2.1 × 103 N = 2.1 kN 7.65

The sketch below shows the car as it passes the highest point on the bump.

Taking upward as positive, we have

∑ F = may ⇒ n − mg = m ( −v 2 r ) ! y or

n = m(g − v2/r).

Topic 7

450

(a) If v = 8.94 m/s, the normal force exerted by the road is

(

)

2⎤ ⎡ 8.94 m/s m ⎢ ⎥ n = 1 800 kg ⎢ 9.80 2 − = 1.06 × 10 4 N = 10.6 kN ⎥ 20.4 m ⎥ s ⎢ ⎣ ⎦

(

)

(b) When the car is on the verge of losing contact with the road, n = 0. This gives g = v2/r and the speed must be v = rg = !

7.66

( 20.4 m )( 9.80 m/s ) = 14.1 m/s 2

(a) Consider the sketch below. At the bottom of the loop, the net force toward the center (i.e., the centripetal force) is

mv 2 Fc = = n − Fg R ! so the pilot’s apparent weight (normal force) is

(

)

Fg g v 2 ⎛ mv 2 v2 ⎞ n = Fg + = Fg + = Fg ⎜ 1 + R R gR ⎟⎠ ⎝ ! or

(

)

2 ⎛ ⎞ 2.00 × 102 m/s ⎜ ⎟ n = 712 N 1 + 2 3 2 ⎟ ⎜ 9.80 m/s 3.20 × 10 m/s ⎟ ⎜⎝ ⎠

(

) (

)(

)

= 1.62 × 103 N

Topic 7

451

(b) At the top of the loop, the centripetal force is Fc = mv2/R = n + Fg, so the apparent weight is

mv 2 R

( F g) v − F = F ⎛ v − 1⎞ −F = 2

(

⎜ ⎝ gR

)

⎟ ⎠

2 ⎛ ⎞ 2.00 × 102 m/s = 712 N ⎜ − 1⎟ = 196 N ⎜ 9.80 m/s 2 3.20 × 103 m/s 2 ⎟ ⎜⎝ ⎟⎠

(

)(

)

mg v= = Rg = m/R !

( 3.20 × 10 m )( 9.80 m/s ) = 177 m/s 3

Topic 7

7.67

452

(a) The desired path is an elliptical trajectory with the Sun at one of the foci, the departure planet at the perihelion, and the target planet at the aphelion. The perihelion distance rD is the radius of the departure planet’s orbit, while the aphelion distance rT is the radius of the target planet’s orbit. The semimajor axis of the desired trajectory is then a = (rD + rT)/2.

If Earth is the departure planet, rD = 1.496 × 1011 m = 1.00 AU. With Mars as the target planet, ⎛ ⎞ 1 AU rT = 2.28 × 1011 m ⎜ = 1.52 AU 11 ⎝ 1.496 × 10 m ⎟⎠ !

Thus, the semimajor axis of the minimum energy trajectory is r + r 1.00 AU + 1.52 AU a= D T = = 1.26 AU 2 2 !

Kepler’s third law, T2 = a3, then gives the time for a full trip around this path as T = a3 = !

(1.26 AU ) = 1.41 y 3

so the time for a one-way trip from Earth to Mars is Δt = T/2 = 1.41

Topic 7

453

y/2 = 0.71 y. (b) This trip cannot be taken at just any time. The departure must be timed so that the spacecraft arrives at the aphelion when the target planet is located there. 7.68

(a) The static friction force provides the centripetal acceleration required to keep the coin stationary relative to the turntable. The coin will move relative to the turntable if the required static friction force exceeds its maximum value of fs,max = µs N . With N = mg, the coin will begin to move if fs,max = µs mg < mv 2 /r = mrω 2 . So the condition for relative motion is: rω 2 > µs g

(b) Substitute ω = ωi + αt (with ωi = 0) into the result from (a) to find

r (α t ) > µs g → t > 2

µs g = rα 2

( 0.350)( 9.80 m/s ) ( 0.150 m )( 0.730 rad/s ) 2

t > 6.55 s 7.69

From Figure (a) below, observe that the angle the strings make with the vertical is θ = cos

−1

⎛ 1.50 m ⎞ = 41.4° . ⎝ 2.00 m ⎠

Topic 7

454

Also, the radius of the circular path is r = !

( 2.00 m ) − (1.50) = 1.32 m . 2

Figure (b) gives a force diagram of the object with the +y-axis vertical and the +x-axis directed toward the center of the circular path. (a) Since the object has zero vertical acceleration, Newton’s second law gives

∑ Fy = T1 cos θ − T2 cos θ − mg = 0 or T1 − T2 =

mg cos θ

[1]

In the horizontal direction, the object has the centripetal acceleration ac = v2/r directed in the +x-direction (toward the center of the circular path). Thus,

∑ Fx = T1 sin θ + T2 sin θ =

mv 2 r

or T1 + T2 =

mv 2 r sin θ

[2]

⎛ g v2 ⎞ + Adding Equations [1] and [2] gives 2T1 = m ⎜ , so the ⎝ cos θ r sin θ ⎟⎠ !

tension in the upper string is

Topic 7

455

( 4.00 kg ) ⎡⎢ 9.80 m s + ( 6.00 m/s ) ⎤⎥ = 109 N T = 2 ⎢⎣ cos 41.4° (1.32 m ) sin 41.4° ⎥⎦ 2 2

(b) To compute the tension T2 in the lower string, subtract Equation [1] above from Equation [2] to obtain !2T2 = m ( v /r sin θ − g/ cos θ ) . 2

Thus,

( 4.00 kg ) ⎡⎢ ( 6.00 m/s ) − 9.80 m/s ⎤⎥ = 56.4 N T = 2 ⎢⎣ (1.32 m ) sin 41.4° cos 41.4° ⎥⎦ 2 2

! 7.70

(a) At each point on the vertical circular path, two forces are acting on the ball. These are: (1) the downward gravitational force with constant magnitude Fg = mg, and (2) the tension force in the string, always directed toward the center of the path. (b) The sketch below shows the forces acting on the ball when it is at the bottom of the circular path, and when it is at the highest point on the path. Note that the gravitational force has the same magnitude and direction at each point on the circular path. The tension force varies in magnitude at different points and is always directed toward the center of the path.

Topic 7

456

T = mv 2 r − Fg = mv 2 r − mg = m ( v 2 r − g ) ⎡ ( 5.20 m/s )2 ⎤ = ( 0.275 kg ) ⎢ − 9.80 m s 2 ⎥ = 6.05 N ⎢⎣ 0.850 m ⎥⎦ ! (d) At the bottom of the circle, Fc = T − Fg = T − mg = mv2/r, and solving for the speed gives r ⎛T ⎞ v 2 = (T − mg) = r ⎜ − g ⎟ ⎝ ⎠ m m !

⎛T ⎞ v = r ⎜ − g⎟ ⎝m ⎠

and

If the string is at the breaking point at the bottom of the circle, then T = 22.5 N, and the speed of the object at this point must be

v= ! 7.71

⎛

⎞

22.5 N − 9.80 m/s ⎟ = 7.82 m/s ( 0.850 m ) ⎜ 0.275 kg ⎝

⎠

The angular speed of the luggage is ω = 2π/T, where T is the time for one complete rotation of the carousel. The resultant force acting on the luggage must be directed toward the center of the horizontal circular path (that is, in the +x-direction) and supply the needed centripetal

Topic 7

457

force Fc = mac = mrω2 = 4π2mr/T2.

Thus,!∑ Fx = max ⇒ fs cos θ − nsin θ = 4π mr/T 2

[1]

and !∑ Fy = may ⇒ fs sin θ + ncos θ − mg = 0 or

n= !

mg − fs sin θ cos θ

[2]

Substituting Equation [2] into Equation [1] gives

⎛ ⎛ sin 2 θ ⎞ 4π 2 r ⎞ f s ⎜ cos θ + = m gtan θ + ⎟ ⎜ ⎟ cos θ ⎠ T2 ⎠ ⎝ ⎝ or, multiplying by cos θ and using the identity cos2 θ + sin2 θ = 1, we have ⎛ 4π 2 r cos θ ⎞ fs = m ⎜ gsin θ + ⎟⎠ ⎝ T2 !

[3]

(a) With r = 7.46 m and T = 38.0 s, Equation [3] gives the required friction force as

⎡ 4π 2 (7.46 m) cos 20.0° ⎤ f s = (30.0 kg) ⎢( 9.80 m/s 2 ) sin 20.0° + ⎥ (38.0 s)2 ⎣ ⎦ = 106 N (b) If the bag is on the verge of slipping when r = 7.94 m and T = © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 7

458

34.0 s, then fs = (fs)max = µsn and (using Equations [2] and [3]) the coefficient of static friction is

µs = =

fs n

f s cos θ mg − f s sin θ

(

cos θ

(mg f ) − sin θ s

cos θ

)

⎡ g gsin θ + 4π 2 r cos θ T 2 ⎤ − sin θ ⎣ ⎦

µs = !

cos 20.0° ⎡ ⎤ 9.80 m/s 2 − sin 20.0° ⎢ 2 2 2⎥ ⎢⎣ ( 9.80 m/s ) sin 20.0° + 4π (7.94 m ) cos 20.0° (34.0 s) ⎥⎦

giving 7.72

µs = 0.396

The maximum lift force is (FL)max = Cv2, where C = 0.018 N ⋅ s2/m2 and v is the flying speed. For the bat to stay aloft, the vertical component of the lift force must equal the weight, or FL cos θ = mg, where θ is the banking angle. The horizontal component of this force supplies the centripetal acceleration needed to make a turn, or FL sin θ = mv2/r where r is the radius of the turn. (a) To stay aloft while flying at minimum speed, the bat must have θ = 0 (to give cos θ = (cos θ)max = 1) and also use the maximum lift force possible at that speed. That is, we need (FL)max(cos θ)max = mg,

2 !Cvmin (1) = mg

Topic 7

459

Thus, we see that minimum flying speed is

mg vmin = = C !

( 0.031 kg )( 9.8 m s ) = 4.1 m/s 2

0.018 N ⋅s 2 m 2

(b) To maintain horizontal flight while banking at the maximum possible angle, we must have (FL)max cos θmax = mg, or Cv2 cos θmax = mg. For v = 10 m/s, this yields

cosθ max =

mg (0.031 kg)(9.8 m/s 2 ) = = 0.17 or θ max = 80° Cv 2 (0.18 N ⋅s 2 / m 2 )(10 m/s)2

(c) The horizontal component of the lift force supplies the centripetal acceleration in a turn, FL sin θ = mv2/r. Thus, the minimum radius turn possible is given by

mv 2 m v2 m rmin = = = 2 ( FL )max (sin θ )max C v sin θ max C sin θ max ! where we have recognized that sin θ has its maximum value at the largest allowable value of θ. For a flying speed of v = 10 m/s, the maximum allowable bank angle is θmax = 80° as found in part (b). The minimum radius turn possible at this flying speed is then

0.031 kg rmin = = 1.7 m 0.018 N ⋅s 2 /m 2 ) sin 80.0° ( ! (d) No. Flying slower actually increases the minimum radius of the achievable turns. As found in part (c), rmin = m/C sin θmax. To see

Topic 7

460

how this depends on the flying speed, recall that the vertical component of the lift force must equal the weight, or FL cos θ = mg. At the maximum allowable bank angle, cos θ will be a minimum. This occurs when FL = (FL)max = Cv2. Thus, cos θmax = mg/Cv2 and ⎛ mg ⎞ sin θ max = 1 − cos θ max = 1 − ⎜ 2 ⎟ ⎝ Cv ⎠ !

This gives the minimum radius turn possible at flying speed v as

rmin =

m mg C 1− ⎛ 2 ⎞ ⎝ Cv ⎠

Decreasing the flying speed v will decrease the denominator of this expression, yielding a larger value for the minimum radius of achievable turns. 7.73

The normal force exerted on the person by the cylindrical wall must provide the centripetal acceleration, so n = m(rω2). If the minimum acceptable coefficient of friction is present, the person is on the verge of slipping and the maximum static friction force equals the person’s weight, or (fs)max = (µs)min n = mg. mg g 9.80 m s 2 = 2= Thus, ( µs )min = 2 = 0.131 . n rω 3.00 m ) ( 5.00 rad s ) ( !

7.74

If the block will just make it through the top of the loop, the force

Topic 7

461

required to produce the centripetal acceleration at point C must equal

(

)

2 the block’s weight, or m vC R = mg . This gives vC = Rg as the

required speed of the block at point C. We apply the work-energy theorem in the form Wnc = (KE + PEg + PEs)f − (KE + PEg + PEs)i from when the block is first released until it reaches point C to obtain

( )

1 1 fk AB cos 180° = mvC2 + mg ( 2R ) + 0 − 0 − 0 − kd 2 2 2 !

The friction force between points A and B is fk = uk(mg), and for 2 minimum initial compression of the spring, !vC = Rg as found above.

Thus, the work-energy equation reduces to

dmin =

( )

( ) = mg(2 µ AB + 5R)

2 µk mg AB + mRg + 2 mg 2R k

0.50 kg ) ( 9.8 m s ) ⎡ 2 ( 0.30 ) ( 2.5 m ) + 5 (1.5 m ) ⎤ ( ⎣ ⎦ = 0.75 m d = 2

min

7.75

78.4 N m

The horizontal component of the tension in the cord is the only force directed toward the center of the circular path, so it must supply the centripetal acceleration. Thus, ⎛ vt2 ⎞ ⎛ v2 ⎞ T sin θ = m ⎜ t ⎟ = m ⎜ ⎝ r⎠ ⎝ Lsin θ ⎟⎠ !

Topic 7

462

mvt2 T sin 2 θ = L !

[1]

Also, the vertical component of the tension must support the weight of the ball, or T cos θ = mg

[2]

(a) Dividing Equation [1] by [2] gives sin 2 θ vt2 = cos θ Lg !

vt = sin θ

Lg cos θ

[3]

With L = 1.5 m/s and θ = 30°,

(1.5 m )( 9.8 m/s ) = 2.1 m/s 2

vt = sin 30° !

cos 30°

(b) From Equation [3], with sin2 θ = 1 − cos2 θ, we find 1 − cos 2 θ vt2 = cos θ Lg !

⎛ vt2 ⎞ cos θ + ⎜ ⎟ cos θ − 1 = 0 ⎝ Lg ⎠ 2

Solving this quadratic equation for cos θ gives 2

⎛ v2 ⎞ ⎛ v2 ⎞ cos θ = − ⎜ t ⎟ ± ⎜ t ⎟ + 1 ⎝ 2 Lg ⎠ ⎝ 2 Lg ⎠ !

If L = 1.5 m and vt = 4.0 m/s, this yields solutions cos θ = −1.7

Topic 7

463

(which is impossible), and cos θ = +0.59 (which is possible). Thus,

θ = cos−1(0.59) = 54°. (c) From Equation [2], when T = 9.8 N and the cord is about to break, the angle is

⎛ 0.50 kg ) ( 9.8 m/s 2 ) ⎞ ⎛ mg ⎞ −1 ( θ = cos −1 ⎜ = cos ⎜ ⎟ = 60° ⎝ T ⎟⎠ 9.8 N ⎝ ⎠ ! Then Equation [3] gives

Lg vt = sin θ = sin 60° cos θ !

(1.5 m )( 9.8 m s ) = 4.7 m/s 2

cos 60°

Topic 8

464

Topic 8 Rotational Equilibrium and Rotational Dynamics

QUICK QUIZZES 8.1

Choice (d). A larger torque is needed to turn the screw. Increasing the radius of the screwdriver handle provides a greater lever arm and hence an increased torque.

8.2

Choice (b). Since the object has a constant net torque acting on it, it will experience a constant angular acceleration. Thus, the angular velocity will change at a constant rate.

8.3

Choice (b). The hollow cylinder has the larger moment of inertia, so it will be given the smaller angular acceleration and take longer to stop.

8.4

Choice (a). The hollow sphere has the larger moment of inertia, so it will have the higher rotational kinetic energy, KEr = 12 Iω 2 . !

8.5

Choice (c). Apply conservation of angular momentum to the system (the two disks) before and after the second disk is added to get the result: I1ω1 = (I1 + I2)ω.

8.6

Choice (a). Earth already bulges slightly at the Equator, and is slightly flat at the poles. If more mass moved towards the Equator, it would

Topic 8

465

essentially move the mass to a greater distance from the axis of rotation, and increase the moment of inertia. Because conservation of angular momentum requires that Izωz = constant, an increase in the moment of inertia would decrease the angular velocity, and slow down the spinning of Earth. Thus, the length of each day would increase.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 8.2

(a) A point mass has moment of inertia I = mR2. When RB = RA, the ratio mB /mA = I B /I A = 4 . (b) The moment of inertia increases linearly with mass m and as the square of the distance R. For IA to equal IB with one quarter the mass, the distance must be rA = 2R so that mBR 2 = mA rA2 = ( mB /4 ) ( 2R ) . 2

8.4

(a) The lever arm of a particular force is found with respect to some reference point or axis. Thus, an origin must be chosen to compute the torque of a force.   (b) If the object is in translational equilibrium Fnet = ∑ Fi = 0 , the net !

(

)

torque acting on the system is independent of the origin or axis considered. However, if the resultant force acting on the object is not zero, the net torque has different values for different axis of © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

466

rotation. 8.6

The magnitude of torque is proportional to the distance and the magnitude of the force. (a) If the distance is doubled and the force is doubled, the torque increases by a factor of 2 × 2 = 4 . (b) If the distance is doubled and the force is unchanged, the torque increases by a factor of 2 × 1 = 2 . (c) If the distance is unchanged and the force is halved, the torque changes by a factor of 1 × 12 = 12 . (d) If the distance is halved and the force is halved, the torque changes by a factor of 12 × 12 = 14 .

8.8

8.10

(a)

No , it is not possible for the object to have zero acceleration. A single nonzero force F will produce an acceleration of magnitude F/m.

(b) Yes , it is possible for the object to have zero angular acceleration.

Topic 8

467

The angular acceleration has magnitude α = τ /I where τ is the magnitude of torque from F and I is the moment of inertia. The torque produced by a single force may be zero. (c)

No , it is not possible for the object to be in mechanical equilibrium. In this case, a ≠ 0 and an object in mechanical equilibrium has both a = 0 and α = 0.

8.12

(a) Consider two people, at the ends of a long table, pushing with equal magnitude forces directed in opposite directions perpendicular to the length of the table. The net force will be zero, yet the net torque is not zero. (b) Consider a falling body. The net force acting on it is its weight, yet the net torque about the center of gravity is zero.

8.14

As the cat falls, angular momentum must be conserved. Thus, if the upper half of the body twists in one direction, something must get an equal angular momentum in the opposite direction. Rotating the lower half of the body in the opposite direction satisfies the law of conservation of angular momentum.

8.16

Since the axle of the turntable is frictionless, no external agent exerts a torque about this vertical axis of the mouse-turntable system. This means that the total angular momentum of the mouse-turntable

Topic 8

468

system will remain constant at its initial value of zero. Thus, as the mouse starts walking around the axis (and developing an angular momentum, Lmouse = Imωm, in the direction of its angular velocity), the turntable must start to turn in the opposite direction so it will possess an angular momentum, Ltable = Itωt, such that Ltotal = Lmouse + Ltable = Imωm + Iiωi = 0. Thus, the angular velocity of the table will be ωt = −(ImIi)ωm The negative sign means that if the mouse is walking around the axis in a clockwise direction, the turntable will be rotating in the opposite direction, or counterclockwise. The correct choice for this question is (d).

ANSWERS TO EVEN NUMBERED PROBLEMS 8.2

34.6 N ⋅ m

8.4

3.55 N · m clockwise

8.6

0.642 N · m counterclockwise

8.8

(−1.5 m, −1.5 m)

8.10

xcm = 0.394 m, ycm = 0.200 m

8.12

xcg = 0, ycg = −9.64 × 10−2 m

8.14

21.0 xiles

Topic 8

469

8.16

12.0 m

8.18

(a)

8.20

139 g

8.22

(a) (d)

226 N

(b)

117 N upward

See Solution.

(b)

at x = 0

(c)

n2 = 1.42 × 103 N

(e)

5.64 m

(f) yes

n1 = 0

8.24

T = 1.68 × 103 N, R = 2.33 × 103 N, θ = 21.0°

8.26

567 N, 333 N

8.28

(a)

See Solution.

(b)

T = 343 N, 172 N to the right, 683 N upward

(c)

xmax = 5.14 m

(a)

See Solution.

(b)

To simplify equations by eliminating 2 unknowns.

(c)

Στ

(e)

ΣFx = Fx − T = 0, ΣFy = Fy − mg = 0

8.30

⎛L ⎞ = 0 + 0 − mg ⎜ cosθ ⎟ + T(Lsin θ ) = 0 hinge ⎝2 ⎠

(d)

T = 136 N

(f) Fx = 136 N, Fy = 157 N (g) Yes, this would reduce tension in the cable and stress on the hinge.

Topic 8

8.32

470

ΣFx = Fx − Rx = 0, ΣFy = Fy + Ry − Fg = 0, Στ O = +Fy (ℓcosθ ) − Fx (ℓsin θ ) − Fg (ℓ/2)cosθ = 0

8.34

(a)

T = 1.47 kN

(b)

1.33 kN to the right, 2.58 kN upward

8.36

xmin = 2.8 m

8.38

τx = 149 N ⋅ m, τy = 66.0 N ⋅ m, τO = 215 N ⋅ m

8.40

2 2 (a) !I = MR + 12 mr

(c)

τT > 0, α > 0, a < 0

(d)

α = − a/r

(f) rT = Iα

(h)

a = − 2.72 m/s2

(b)

(e)

T − (2M + m)g = (2M + m)a

(g)

−(2M + m)g a= 2 ! 2M + M(R/r) + 3m/2

(i)

T = 35.4 N

(b)

1.40 kN

(j)

t = 0.857 s

8.42

µk = 0.30

8.44

(a)

872 N

8.46

(a)

To provide the net clockwise torque that accelerates the pulley.

(b)

2.88 m/s2

(c)

T1 = 127 N, T2 = 138 N

8.48

(a)

17.6 J

(b)

8.79 J

8.50

(a)

Ihoop = 0.254 kg ⋅ m2, Icyl = 0.127 kg ⋅ m2, Isphere = 0.102 kg ⋅ m2,

Topic 8

471

Ishell = 0.169 kg ⋅ m2 (b)

solid sphere; solid cylinder; thin spherical shell; and hoop

(c)

hoop; thin spherical shell; solid cylinder; and solid sphere

8.52

36 rad/s

8.54

(a)

8.56

30.2 rev/s

8.58

10.9 rad/s

8.60

(a)

1.37 × 108 J

(b)

8.00 h

2.72 kg ⋅ m2/s

(b)

1.36 kg ⋅ m2/s

(c)

1.09 kg ⋅ m2/s

(d)

1.81 kg ⋅ m2/s

8.62

(a)

3.00 rad/s

(b)

2.00 rad/s

8.64

(a)

Yes, the bullet effectively moves in a circle around the hinge just before impact.

(b)

No, the bullet undergoes an inelastic collision with the door.

(c)

0.749 rad/s

(d)

8.66

0.91 km/s

8.68

(a)

0.360 rad/s, counterclockwise

8.70

(a)

1.9 rad/s

(b)

KEi = 2.5 J, KEf = 6.3 J

KEf = 1.68 J << KEi = 2.50 × 103 J

(b)

99.9 J

Topic 8

472

8.72

12.3 m/s2

8.74

(a)

0.433 kg ⋅ m2/s

8.76

(a)

780 N

(b)

 !R = 716 N at 70.4° above the horizontal to the right

8.78

(a)

46.8 N

(b)

0.234 kg ⋅ m2

8.80

(a)

3.75 × 103 kg ⋅ m2/s

(b)

1.88 kJ

(c)

3.75 × 103 kg⋅m2/s

(d)

10.0 m/s

(e)

7.50 kJ

(f) 5.62 kJ

(b)

1.73 kg ⋅ m2/s

(c)

40.0 rad/s

8.82

Tleft = 1.01 kN, Tright = 1.59 kN

8.84

(a)

E = (m1 − m2)gL

(b)

2 2 1 !E = 2 ( m1 + m2 ) L ω + ( m1 − m2 ) gLsin θ

(c)

Equate the results of parts (a) and (b) and solve for ω.

(d)

τnet|vertical = 0; τnet|rotated = (m1 − m2)gL cos θ ; since ΔL/Δt = τnet, the angular momentum changes at a non-uniform rate.

(e)

α = (m1 − m2)g cos θ/(m1 + m2)L; Yes, α → 0 as θ → 90° and α is a maximum at θ = 0°.

( )

3gL /2

vlower = 3gL = 2vcg ! end

8.86

(a)

vcg =

8.88

(a)

A frictionless wall cannot exert a component of force parallel to its

(b)

Topic 8

473

surface. (b)

L sin θ

(c)

1 2

Lcosθ

(d)

2.5 m

8.90

Answers are given in the problem statement.

8.92

(a)

T = 10.2 N

(b)

Rx = 6.56 N, Ry = 7.84 N

9.94

(a)

See Solution.

(b)

218 N

(c)

72.4 N

(d)

2.41 m

(e) The analysis would need to include a horizontal friction force at the lower end of the ladder. The coefficient of static friction between the ladder and the floor would have to be known.

PROBLEMS SOLUTIONS 8.1

The angle between the position and force vectors is θ = 90° so that sin θ = 1 and τ = rFsinθ = rF. (a) With the force applied at the center of the door, r = (1.00 m)/2 and F = 50.0 N so that

τ = rF =

(1.00 m ) ( 50.0 N ) = 25.0 N ⋅ m 2

(b) With the force applied at the edge of the door farthest from the hinges, r = (1.00 m) and F = 50.0 N so that © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

474

τ = rF = (1.00 m ) ( 50.0 N ) = 50.0 N ⋅ m . 8.2

The angle between the position vector and the force is θ = 60.0° so that

τ = rF sin θ = ( 0.500 m ) ( 80.0 N ) sin 60.0° = 34.6 N ⋅ m 8.3

Resolve the 100-N force into components parallel to and perpendicular to the rod, as F|| = F cos(20.0° + 37.0°) = F cos 57.0°

and F⊥ = F sin (20.0° + 37.0°) = F sin 57.0° The lever arm of F⊥ about the indicated pivot is 2.00 m, while that of F|| is zero. The torque due to the 100-N force may be computed as the sum of the torques of its components, giving

τ = F(0) − F⊥(2.00 m) = 0 − [(100 N) sin 57.0°](2.00 m) = −168 N ⋅ m or 8.4

τ = 168 N ⋅ m clockwise

Note that each of the forces is perpendicular to the radius line of the wheel at the point where the force is tangent to the wheel. Thus, when considering torques about the center of the wheel, the radius line is

Topic 8

475

the lever arm of the force. Taking counterclockwise torques as positive, ΣτO = + (12.0 N)a − (10.0 N)b − (9.00 N)b = + (12.0 N)(0.100 m) − (19.0 N)(0.250 m) = −3.55 N ⋅ m or 3.55 N ⋅ m clockwise

8.5

First resolve all of the forces shown in Figure P8.5 into components parallel to and perpendicular to the beam as shown in the sketch below.

(a) τO = +[(25 N) cos 30°](2.0 m) − [(10 N) sin 20°](4.0 m) = +30 N ⋅ m or

τO = 30 N ⋅ m counterclockwise

(b) τC = +[(30 N)sin 45°](2.0 m) − [(10 N)sin 20°](2.0 m) = +36 N ⋅ m

Topic 8

476

or 8.6

τC = 36 N ⋅ m counterclockwise

The lever arm is d = (1.20 × 10−2 m) cos 48.0° = 8.03 × 10−3 m, and the torque is

τ = Fd = (80.0 N)(8.03 × 10−3 m) = 0.642 N ⋅ m counterclockwise 8.7

(a)

τ = Fg ⋅(lever arm) = (mg)⋅[ℓsin θ ] = (3.0 kg)(9.8 m/s2) ⋅[(2.0 m) sin 5.0°] = 5.1 N ⋅ m

(b)

The magnitude of the torque is proportional to sin θ, where θ is the angle between the direction of the force and the line from the pivot to the point where the force acts. Note from the sketch that this is the same as the angle the pendulum string makes with the vertical. Since sin θ increases as θ increases, the torque also increases with the angle.

8.8

Requiring that xcg =

Σmi xi Σmi

= 0 gives

Topic 8

477

( 5.0 kg )( 0) + ( 3.0 kg )( 0) + ( 4.0 kg )( 3.0 m ) + (8.0 kg ) x = 0 ( 5.0 + 3.0 + 4.0 + 8.0) kg

or 8.0x + 12 m = 0 which yields x = −15 m Also, requiring that ycg = Σmiyi/Σmi = 0 gives

( 5.0 kg )( 0) + ( 3.0 kg )( 4.0 m ) + ( 4.0 kg )( 0) + (8.0 kg ) y ( 5.0 + 3.0 + 4.0 + 8.0) kg

or 8.0y + 12 m = 0 yielding y = −15 m. Thus, the 8.0-kg object should be placed at coordinates (−1.5 m, −1.5 m). 8.9

All objects in this problem are uniform so their centers of mass are located at their geometric centers. Take x = 0 at the left end of the plank. For the plank, xplank,cm = 0.500 m and mplank = 5.00 kg. For the left bowling ball, xA,cm = 0.300 m and mA = 3.75 kg. For the right bowling ball, xB,cm = 0.850 m and mB = 5.50 kg. The x-component of the center of mass is then:

xcm =

Σmi xi mplank xplank,cm + mA xA ,cm + mB xB,cm = Σmi mplank + mA + mB

( 5.00 kg)(0.500 m ) + ( 3.75 kg)(0.300 m ) + ( 5.50 kg)(0.850 m ) ( 5.00 kg) + ( 3.75 kg) + ( 5.50 kg)

= 0.582 m

Topic 8

8.10

478

All boxes in this problem are uniform so their centers of mass are located at their geometric centers. Take x = 0 and y = 0 at the bottom left corner of box A. For each box, ybox,cm = (0.400 m)/2 = 0.200 m so that ycm = 0.200 m .

For box A, mA = 0.800 kg, xA,cm = 0.125 m. For box B, mB = 1.00 kg, xB,cm = 0.250 m + (0.350 m)/2 = 0.425 m. For box C, mC = 0.600 kg, xC,cm = 0.250 m + 0.350 m + (0.200 m)/2 = 0.700 m. The x-component of the center of mass is then:

xcm =

Σmi xi mA xA + mB xB + mC xC = Σmi mA + mB + mC

(0.800 kg)(0.125 m ) + (1.00 kg)(0.425 m ) + (0.600 kg)(0.700 m ) (0.800 kg) + (1.00 kg) + (0.600 kg)

= 0.394 m

8.11

Consider the remaining plywood to consist of two parts: A1 is a 4.00-ft by 4.00-ft section with center of gravity located at (2.00 ft, 2.00 ft), while A2 is a 2.00-ft by 4.00-ft section with center of gravity at (6.00 ft, 1.00 ft).

Since the plywood is uniform, its mass per area σ is constant and the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

479

mass of a section having area A is m = σA. The center of gravity of the remaining plywood has coordinates given by

xcg =

∑m x = σ A x + σ A x σA +σA ∑m i

1 1

2 2

(16.0 ft )(2.00 ft ) + (8.00 ft )(6.00 ft ) = 3.33 ft = (16.0 ft ) + (8.00 ft ) 2

and ycg =

∑m y = σ A y + σ A y σA +σA ∑m i

1 1

2 2 2

(16.0 ft )(2.00 ft ) + (8.00 ft )(1.00 ft ) = 1.67 ft = (16.0 ft ) + (8.00 ft ) 2

8.12

Treating the boomerang legs as uniform thin rods, the center of gravity of each leg is located at its geometric center. By symmetry, the xcomponent of the center of gravity lies at x = 0 so that xcg = 0 . Also by symmetry, the y-component of the center of gravity lies at half the boomerang’s height below y = 0 so that ycg = −((0.300 m)/2) sin 40.0°)

= −9.64 × 10−2 m . 8.13

For the block-ramp system, there is no horizontal external force so that ax = 0 for the system. Therefore, starting from rest at xcm = 0, the center of mass remains at xcm = 0. Apply the definition of center of mass to find:

Topic 8

480

xcm = 0 =

=−

mxblock + Mxramp m+ M

m 1.50 kg xblock = − ( −0.300 m ) M 4.50 kg

= 0.100 m

8.14

There is no external force acting on the mothership-cruiser system so ! that a = 0 and the center of mass, initially at rest, remains at rest. From

the problem figure: xcm =

Σmi xi Σmi

(185 xons)(0 m ) + ( 20.0 xons)( 215 xiles) (185 xons) + ( 20.0 xons)

= 21.0 xiles

After the tractor beam has pulled the two spacecraft to the same final location xf, we have xcm =

(185 xons + 20.0 xons) x (185 xons + 20.0 xons)

x f = xcm = 21.0 xiles

8.15

Take y = 0 to be at ground level. The frog-stick center of mass moves as a particle in free-fall, dropped from rest with an initial height of 1.85 m. When the stick hits the ground at t = 0.450 s, the center of mass is located at

Topic 8

481

ycm = y0 + v0 yt − 12 gt 2 ycm = 1.85 m − 12 g ( 0.450 s)

ycm = 0.858 m From the definition of center of mass, ycm = =

Σmi yi Σmi mfrog yfrog + mstick ystick mfrog + mstick

Solve for the frog’s height when ystick = 0 to find

yfrog =

mfrog + mstick mfrog

⎛ 7.25 g + 4.50 g ⎞ ycm = ⎜ ⎟⎠ ( 0.858 m ) 7.25 g ⎝

= 1.39 m 8.16

First, apply kinematics to the falling rider to find the maximum height reached by the bicycle-rider system. The rider’s initial height is ymax and he begins his fall with vy0 = 0 so that:

y = y0 + v0 yt − 12 gt 2 0 = ymax + 0 − 12 g ( 0.550 s)

ymax = 1.48 m Next, use kinematics to find the bicycle-rider’s launch speed off the ramp. Use the fact that the time to maximum height equals the time to fall back to the ground:

ymax = y0 + v0 sin θ 0t − 12 gt 2

1.48 m = 0 + ( v0 sin 60.0° ) ( 0.550 s) − 12 g ( 0.550 s)

v0 = 6.22 m/s

Topic 8

482

Conserve the x-component of the bicycle-rider system’s momentum to find the bicycle’s x-component of velocity after the rider pushes the bicycle away: pix = p fx

bike

+ mrider ) v0 cos60.0° = mbike vbike,f + mrider ( 0 m/s)

vbike,f =

(12 kg + 72 kg)( 6.22 m/s) cos 60.0° = 21.8 m/s 12.0 kg

After t = 0.550 s has passed, the bicycle has traveled a distance x = vbike,f t away from the rider: x = ( 21.8 m/s) ( 0.550 s) = 12.0 m

8.17

Requiring that Στ = 0, using the shoulder joint at point O as a pivot, gives Στ = (Ft sin 12.0°)(0.080 m) − (41.5 N)(0.290 m) = 0, or Ft = 724 N Then ΣFy = 0 ⇒ −Fsy + (724 N) sin 12.0° − 41.5 N = 0 yielding Fsy = 109 N. ΣFx = 0 gives Fsx − (724 N) cos 12.0° = 0, or Fsx = 708 N

Topic 8

483

Therefore, 8.18

F = Fsx2 + Fsy2 = !s

(708 N) + (109 N) = 716 N 2

Since the beam is uniform, its center of gravity is at its geometric center.

Requiring that Στ = 0 about an axis through point O and perpendicular to the page gives

!F(0) + mg( / 2) − T( − d) = 0 (a)

The tension in the rope must then be

T= !

(b)

mg ( /2 ) ( 343 N ) ( 2.50 m ) = = 226 N −d 5.00 m − 1.20 m

The force the column exerts is found from ΣFy = 0 ⇒ F + T − mg = 0 or

8.19

F = mg − T = 343 N − 226 N = 117 N upward

Require that Στ = 0 about an axis through the elbow and perpendicular to the page. This gives

Στ = + [(2.00 kg)(9.80 m/s2)](25.0 cm + 8.00 cm) − (FB cos 75.0°)(8.00 cm) = 0

Topic 8

484

8.20

(19.6 N)( 33.0 cm ) = 312 N FB = (8.00 cm ) cos 75.0° !

Since the bare meter stick balances at the 49.7 cm mark when placed on the fulcrum, the center of gravity of the meter stick is located 49.7 cm from the zero end. Thus, the entire weight of the meter stick may be considered to be concentrated at this point. The force diagram of the stick when it is balanced with the 50.0-g mass attached at the 10.0 cm mark is as given below.

Requiring that the sum of the torques about point O be zero yields + ⎡( 50.0 g ) g ⎤⎦ ( 39.2 cm − 10.0 cm ) − Μ g ( 49.7 cm − 39.2 cm ) = 0 ! ⎣

8.21

⎛ 39.2 cm − 10.0 cm ⎞ M = ( 50.0 g ) ⎜ = 139 g ⎝ 49.7 cm − 39.2 cm ⎟⎠ !

We neglect the weight of the board and assume that the woman’s feet are directly above the point of support by the rightmost scale. Then, the force diagram for the situation is as shown below. From ΣFy = 0, we have Fg1 + Fg2 − w = 0, or w = 380 N + 320 N = 700 N.

Topic 8

485

Choose an axis perpendicular to the page and passing through point P. Then Στ = 0 gives w ⋅ x − Fg1(2.00 m) = 0, or

8.22

(

Fg1 2.00 m w

) = (380 N)(2.00 m) = 1.09 m 700 N

(a)

Mg = (90.0 kg)(9.80 m/s2) = 882 N (b)

mg = (55.0 kg)(9.80 m/s2) = 539 N

The woman is at x = 0 when n1 is greatest. With this location of the woman, she exerts her maximum possible counterclockwise torque about the center of the beam. Thus, n1 must be exerting its maximum clockwise torque about the center to hold the beam in rotational equilibrium.

(c)

n1 = 0 As the woman walks to the right along the beam, she will eventually reach a point where the beam will start to rotate clockwise about the rightmost pivot. At this point, the beam is starting to lift up off of the leftmost pivot and the normal force

Topic 8

486

exerted by that pivot will have diminished to zero. (d)

When the beam is about to tip, n1 = 0, and ΣFy = 0 gives 0 + n2 − Mg − mg = 0 n2 = Mg + mg = 882 N + 539 N = 1.42 × 103 N

or (e)

Requiring that ∑ τ rightmost = 0 when the beam is about to tip (n1 = pivot

0) gives − (4.00 m − x)mg − (4.00 m − 3.00 m)Mg = 0 or

(f)

(mg) x = (1.00 m)Mg + (4.00 m) mg

and

M x = (1.00 m ) + 4.00 m m !

Thus,

( 90.0 kg ) + 4.00 m = 5.64 m x = (1.00 m ) ( 55.0 kg ) !

When n1 = 0 and n2 = 1.42 × 103 N, requiring that ∑ τ left = 0 gives end

0 − (539 N)x − (882 N)(3.00 m) + (1.42 × 103 N)(4.00 m) = 0

−3.03 × 103 N ⋅ m x= = 5.62 N −539 N !

which, within limits of rounding errors, is the same as the answer to part (e). 8.23

The force diagram for the spine is shown below.

Topic 8

487

(a)

When the spine is in rotational equilibrium, the sum of the torques about the left end (point O) must be zero. Thus,

⎛ 2L ⎞ ⎛ L⎞ +Ty ⎜ ⎟ − ( 350 N ) ⎜ ⎟ − ( 200 N ) ( L) = 0 ⎝ 2⎠ ! ⎝ 3⎠ yielding Ty = T sin 12.0° = 563 N The tension in the back muscle is then 563 N T= = 2.71 × 103 N = 2.71 kN . sin12.0° !

(b)

The spine is also in translational equilibrium, so ΣFx = 0 ⇒ Rx − Tx = 0 and the compression force in the spine is Rx = Tx = T cos 12.0° = (2.71 kN) cos 12.0° = 2.65 kN

8.24

 In the force diagram of the foot given below, note that the force !R (exerted on the foot by the tibia) has been replaced by its horizontal and vertical components.

Topic 8

488

Employing both conditions of equilibrium (using point O as the pivot point) gives the following three equations: ΣFx = 0 ⇒ R sin 15.0° − T sin θ = 0

T sin θ

[1]

sin 15.0°

ΣFy = 0 ⇒ 700 N − Rcos 15.0° + Tcos θ = 0

[2]

ΣτO = 0 ⇒ −(700 N) [(18.0 cm) cos θ] + T(25.0 cm − 18.0 cm) = 0 or

T = (1 800 N) cos θ

[3]

Substituting Equation [3] into Equation [1] gives

⎛ 1 800 N ⎞ R=⎜ ⎟ sin θ cos θ ⎝ sin 15.0∞ ⎠

[4]

Substituting Equations [3] and [4] into Equation [2] yields

⎛ 1 800 N ⎞ 2 ⎜ ⎟ sin θ cos θ − 1 800 Ν cos θ = 700 N ⎝ tan 15.0° ⎠

(

)

which reduces to: sin θ cos θ = (tan 15.0°)cos2θ + 0.104 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

489

Squaring this result and using the identity sin2θ = 1 − cos2θ gives [tan2 (15.0)° + 1]cos4 θ + [(2tan 15.0°)(0.104) − 1]cos2 θ + (0.104)2 = 0 In this last result, let u = cos2 θ and evaluate the constants to obtain the quadratic equation: (1.07)u2 − (0.944)u + (0.010 8) = 0 The quadratic formula yields the solutions u = 0.871 and u = 0.011 6. Thus θ = cos !

−1

( 0.871 ) = 21.0° or θ = cos ( 0.011 6 ) = 83.8° . We −1

ignore the second solution since it is physically impossible for the human foot to stand with the sole inclined at 83.8° to the floor. We are the left with θ = 21.0° Equation [3] then yields:

T = (1 800 N)cos 21.0° = 1.68 × 103 N

and Equation [1] gives:

(1.68 × 10 N) sin 21.0° = 2.33 × 10 N R= 3

! 8.25

sin 15.0°

Consider the torques about an axis perpendicular to the page through the left end of the rod.

Topic 8

490

Στ = 0 ⇒ T = !

(100 N)( 3.00 m ) + ( 500 N)( 4.00 m ) ( 6.00 m ) cos 30.0°

T = 443 N ΣFx = 0 ⇒ Rx = T sin 30.0° = (443 N) sin 30.0° Rx = 222 N toward the right ΣFy = 0 ⇒ Ry = Tcos 30.0° − 100 N − 500 N = 0 Ry = 600 N − (443 N) cos 30.0° = 216 N upward 8.26

Consider the torques about an axis perpendicular to the page through the left end of the scaffold.

Στ = 0 ⇒ T1(0) − (700 N)(1.00 m) − (200 N)(1.50 m) + T2(3.00 m) = 0 From which,

T2 = 333 N

Then, from ΣFy = 0, we have T1 + T2 − 700 N − 200 N = 0 or T1 = 900 N − T2 = 900 N − 333 N = 567 N 8.27

Consider the torques about an axis perpendicular to the page and

Topic 8

491

through the left end of the plank. Στ = 0 gives − (700 N)(0.500 m) − (294 N)(1.00 m) + (T1 sin 40.0°)(2.00 m) = 0 or

T1 = 501 N

Then, ΣFx = 0 gives −T3 + T1cos 40.0° = 0, or

T3 = (501 N)cos 40.0° = 384 N

From ΣFy = 0, T2 − 994 N + T1sin 40.0° = 0,

8.28

T2 = 994 N − (501 N) sin 40.0° = 672 N

(a)

See the diagram below:

(b)

If x = 1.00 m, then Στ|left end = 0 ⇒ − (700 N)(1.00 m) − (200 N)(3.00 m) − (80.0 N)(6.00 m) + (T sin 60.0°)(6.00 m) = 0 giving T = 343 N

Topic 8

492

Then, ΣFx = 0 ⇒ H − T cos 60.0° = 0,

H = (343 N) cos 60.0° =

172 N to the right and

ΣFy = 0 ⇒ V − 980 N + (343 N) sin 60.0° = 0, or V = 683 N

upward (c)

When the wire is on the verge of breaking, T = 900 N and Στ|left end = − (700 N)xmax − (200 N)(3.00 m) − (80.0 N)(6.00 m) + [(900 N)sin 60.0°](6.00 m) = 0 which gives xmax = 5.14 m

8.29

(a) Considering a pivot at the lower end of the beam, we get

⎛ℓ ⎞ Στ lower = 0 ⇒ +Fsℓsin θ − mg ⎜ cos θ ⎟ = 0 ⎝2 ⎠ end and the spring force is

Fs = !

mg ( cos θ /2 ) mg = sin θ 2tan θ

From Hooke’s law, Fs = kd, the distance the spring is stretched is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

493

then F d= s k !

mg 2k tan θ

(b) From the first condition of equilibrium,

ΣFx = 0 ⇒ Rx − Fs = 0

Rx = Fs =

or !

and

8.30

ΣFy = 0 ⇒ Ry − mg = 0

mg 2tan θ

Ry = mg

(a)

(b)

The point of intersection of two unknown forces is always a good choice as the pivot point in a torque calculation. Doing this eliminates these two unknowns from the calculation (since they have zero lever arms about the chosen pivot) and makes it easier to solve the resulting equilbrium equation

=0 ⇒

⎛L ⎞ 0 + 0 − mg ⎜ cos θ ⎟ + T Lsin θ = 0 ⎝2 ⎠

(

)

(c)

Στ

(d)

Solving the above result for the tension in the cable gives

hinge

Topic 8

494

T= !

(mg/2) Lcos θ = Lsin θ

mg 2tan θ

(16.0 kg )( 9.80 m/s ) = 136 N T= 2

2tan 30.0°

(e)

ΣFx = 0 ⇒ Fx − T = 0

and

ΣFy = 0 ⇒ Fy − mg = 0

(f)

Solving the above results for the components of the hinge force gives Fx = T = 136 N

(g)

and

Fy = mg = (16.0 kg)(9.80 m/s2) = 157 N

Attaching the cable higher up would allow the cable to bear some of the weight, thereby reducing the stress on the hinge. It would also reduce the tension in the cable.

8.31

When the refrigerator is on the verge of tipping (i.e., rotating counterclockwise about point O in the sketch below), the center of gravity of the refrigerator will be directly above point O and the line of action of the gravitational force will pass through point O. When this is true, increasing angle θ by any amount will cause the line of action of Fg to pass to the left of point O, producing a counterclockwise torque with no force available to produce a counterbalancing clockwise torque about this point.

Topic 8

495

At this critical value of angle θ, we have

w/2 tan θ = h/2 ! 8.32

θ = tan−1 (w/h)

The object is in both translational and rotational equilibrium. Thus, we may write ΣFx = 0 ⇒ Fx − Rx = 0 ΣFy = 0 ⇒ Fy + Ry − Fg = 0

and

8.33

⎛ℓ ⎞ Στ O = 0 ⇒ Fy ℓcosθ − Fx ℓsin θ − Fg ⎜ cosθ ⎟ = 0 ⎝2 ⎠

(

) (

)

Consider the torques about an axis perpendicular to the page and

 through the point where the force !T acts on the jawbone.

Στ = 0 ⇒ (50.0 N)(7.50 cm) − R(3.50 cm) = 0 which yields R = 107 N

Topic 8

496

Then, ΣFy = 0 ⇒ −(50.0 N) + T − 107 N = 0 , or T = 157 N 8.34

(a)

Observe that the cable is perpendicular to the boom. Then, using Στ = 0 for an axis perpendicular to the page and through the lower end of the boom gives

⎛L ⎞ ⎛3 ⎞ − (1.20 kN ) ⎜ cos 65.0°⎟ + T ⎜ L⎟ − ( 2.00 kN ) ( Lcos 65.0° ) = 0 ⎝2 ⎠ ⎝4 ⎠ ! or

(b)

T = 1.47 kN

From ΣFx = 0′ H = Tcos 25.0° = 1.33 kN to the right and ΣFy = 0 gives V = 3.20 kN − T sin 25.0° = 2.58 kN upward

8.35

Choose a reference frame with the x-axis parallel to the tibia and the yaxis perpendicular to it. Then, resolve all forces into their x- and ycomponents, as shown. Note that θ = 40.0° and wy = (30.0 N)sin 40.0° = 19.3 N

Topic 8

497

Fy = (12.5 N)sin 40.0° = 8.03 N and

Ty = Tsin 25.0°

Using Στ = 0 for an axis perpendicular to the page and through the upper end of the tibia gives

(T sin 25.0°) − (19.3 N ) − (8.03 N ) d = 0 , d 5

! or

8.36

d 2

T = 209 N

When x = xmin, the rod is on the verge of slipping, so f = (fs)max = µsn = 0.50 n From ΣFx = 0, n − T cos 37° = 0, or n = 0.80T. Thus, f = 0.50(0.80 T) = 0.40T. From ΣFy = 0, f + Tsin 37° − 2w = 0, or 0.40T + 0.60T − 2w = 0, giving T = 2w. Using Στ = 0 for an axis perpendicular to the page and through the left

Topic 8

498

end of the beam gives −w ⋅ xmin − w (2.0 m) + [(2w) sin 37°](4.0 m) = 0, which reduces to xmin = 2.8 m.

8.37

2 The moment of inertia for rotations about an axis is !I = Σmi ri , where ri

is the distance mass mi is from that axis. (a) For rotation about the x-axis, Ix = (3.00 kg)(3.00 m2) + (2.00 kg)(3.00 m)2 + (2.00 kg)(3.00 m)2 + (4.00 kg)(3.00 m)2 = 99.0 kg ⋅ m2 (b) When rotating about the y-axis, Iy = (3.00 kg)(2.00 m)2 + (2.00 kg)(2.00 m)2 + (2.00 kg)(2.00 m)2 + (4.00 kg)(2.00 m)2 = 44.0 kg ⋅ m2 (c) For rotations about an axis perpendicular to the page through point O, the distance ri for each mass is r= !i

( 2.00 m ) + ( 3.00 m ) = 13.0 m . Thus, 2

IO = [(3.00 + 2.00 + 2.00 + 4.00)kg](13.0 m2) = 143 kg ⋅ m2

Topic 8

8.38

499

The required torque in each case is τ = Iα. Thus,

τx = Ixα = (99.0 kg ⋅ m2)(1.50 rad/s2) = 149 N ⋅ m τy = Iyα = (44.0 kg ⋅ m2)(1.50 rad/s2) = 66.0 N ⋅ m and

8.39

τO = IOα = (143 kg ⋅ m2)(1.50 rad/s2) = 215 N ⋅ m τ net

rF sin 90°

(0.330 m)(250 N) = 87.8 kg ⋅ m

(a)

τ net = Iα ⇒ I =

(b)

For a solid cylinder, I = Mr2/2, so

0.940 rad/s

2I 2 ( 87.8 kg ⋅ m ) M= 2 = = 1.61 × 103 kg 2 r 0.330 m ( ) ! 2

8.40

(c)

ω = ω0 + αt = 0 + (0.940 rad/s2)(5.00 s) = 4.70 rad/s

(a)

I = 2Idisk + Icylinder = 2(MR2/2) + mr2/2

(b)

τg = 0 Since the line of action of the gravitational force passes

I = MR2 + mr2/2

through the rotation axis, it has zero lever arm about this axis and zero torque. (c)

The torque due to the tension force is positive. Imagine gripping the cylinder with your right hand so your fingers on the front side of the cylinder point upward in the direction of the tension force. The thumb of your right hand then points toward the left   (positive direction) along the rotation axis. Because!τ = Iα , the

Topic 8

500

torque and angular acceleration have the same direction. Thus, a positive torque produces a positive angular acceleration. When released, the center of mass of the yoyo drops downward, in the negative direction. The translational acceleration is negative. (d)

Since, with the chosen sign convention, the translational acceleration is negative when the angular acceleration is positive, we must include a negative sign in the proportionality between these two quantities. Thus, we write: a = −rα or α = −a/r

(e)

Translation:

ΣFy = mtotala ⇒ T − (2M + m)g = (2M + m)a

[1]

(f)

Rotational:

Στ = Iα ⇒ rTsin 90° = Iα or rT = Iα

[2]

(g)

Substitute the results of (d) and (a) into Equation [2] to obtain

a 2 ⎛α⎞ ⎛ −a/r ⎞ T = I⎜ ⎟ = I⎜ = − ( MR2 + mr 2 2 ) 2 !or!T = − ⎡⎣ M ( R/r ) + m/2 ⎤⎦ a ⎟ ⎝ ⎠ ⎝ ⎠ r r r ! [3] Substituting Equation [3] into [1] yields − [M(R/r)2 + m/2]a − (2M + m)g = (2M + m)a

(h)

− ( 2M + m) g 2 2M + M ( R/r ) + 3m/2 − ⎡⎣ 2 ( 2.00 kg ) + 1.00 kg ⎤⎦ ( 9.80 m/s 2 )

a= = −2.72 m/s 2 2 2 ( 2.00 kg ) + ( 2.00 kg ) (10.0 / 4.00 ) + 3 (1.00 kg ) 2 !

Topic 8

501

(i)

From Equation [1], T = (2M + m)(g + a) = (5.00 kg)(9.80 m/s2 − 2.72 m/s2) = 35.4 N

(j) Δy = ( 0 ) t + at 2 2 ⇒ t = ! 8.41

2 ( Δy ) = a

2 ( −1.00 m ) = 0.857 s −2.72 m/s 2

(a) The fan’s moment of inertia about an axis through its center is I = Idisk + 4I rod

(

2 = 12 Mdisk Rdisk + 4 13 MrodL2rod

)

(

= 12 ( 2.50 kg ) ( 0.200 m ) + 4 13 ( 0.850 kg ) ( 0.750 m ) 2

)

= 0.687 kg ⋅ m 2

(b) Apply rotational kinematics with ω0 = 0 to find the angular acceleration: Δθ = ω 0t + 12 α t 2 → α =

α=

⎛ 2π rad ⎞ 2 (18.5 rev ) ⎜ ⎝ 1 rev ⎟⎠

(15.0 s)

2Δθ t2

= 1.03 rad/s2

Use the rotational analog of Newton’s second law to find the torque provided by the fan’s motor, τmotor. Take the frictional torque to be negative:

Στ = Iα

τ friction + τ motor = Iα τ motor = Iα − τ friction

(

)(

)

= 0.687 kg ⋅ m 2 1.03 rad/s2 − ( −0.115 N ⋅ m ) = 0.823 N ⋅ m

Topic 8

8.42

502

The angular acceleration is α = (ωf − ωi)/Δt = −(ωi/Δt) since ωf = 0. The torque is τ = Iα = −(Iωi/Δt). But the torque is also τ = −fr, so the magnitude of the required friction force is

12 kg ⋅ m 2 ) ( 50 rev min ) ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ( Iω i f= = ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠ = 21 N r ( Δt ) ( 0.50 m )( 6.0 s )

Therefore, the coefficient of friction is f 21 N µk = = = 0.30 n 70 N !

8.43

8.44

(a)

τ = F ⋅ r sin θ = (0.800 N)(30.0 m)sin 90.0° = 24.0 N ⋅ m

(b)

α=

(c)

at = rα = (30.0 m)(0.035 6 rad/s2) = 1.07 m/s2

24.0 N ⋅ m τ τ = = = 0.035 6 rad/s 2 2 2 I mr 0.750 kg 30.0 m

(

)(

)

I = MR2 = (1.80 kg)(0.320 m2) = 0.184 kg⋅m2

τnet = τapplied−τresistive = Iα, or F ⋅ r − f ⋅ R = Iα yielding

F = (Iα + f ⋅ R)/r

( 0.184 kg ⋅ m )( 4.50 rad/s ) + (120 N)( 0.320 m ) = 872 N F= 2

(a)

4.50 × 10−2 m

( 0.184 kg ⋅ m )( 4.50 rad/s ) + (120 N)( 0.320 m ) = 1.40 kN (b) F = 2

! 8.45

2.80 × 10−2 m

1 1 I = MR2 = (150 kg ) (1.50 m 2 ) = 169 kg ⋅ m 2 2 ! 2

Topic 8

503

and

α= !

ω f − ωi Δt

( 0.500 rev/s − 0) ⎛ 2π rad ⎞ = π rad/s ⎜⎝ 1 rev ⎟⎠

2.00 s

Thus, τ = F ⋅ r = Iα gives

Iα F= = r ! 8.46

(a)

(169 kg ⋅ m ) ⎛⎜⎝ π2 rad/s ⎞⎟⎠ 2

1.50 m

= 177 N

It is necessary that the tensions T1 and T2 be different in order to provide a net torque about the axis of the pulley and produce an angular acceleration of the pulley. Since intuition tells us that the system will accelerate in the directions shown in the below when m2 > m1, it is necessary that T2 > T1.

(b)

We adopt a sign convention for each object with the positive direction being the indicated direction of the acceleration of that object in the diagrams below. Then, apply Newton’s second law to each object: For m1: ΣFy = m1a ⇒ T1 − m1g = m1a

T1 = m1(g + a) [1]

Topic 8

504

For m2: ΣFy = m2a ⇒ m2g − T2 = m2a

T2 = m2(g − a) [2]

For M: Στ = Iα ⇒ rT2 − rT2 = Iα

T2 − T1 = Iα/r [3]

Substitute Equations [1] and [2], along with the relations I = Mr2/2 and α = a/r, into Equation [3] to obtain

(

)

(

)

m2 g − a − m1 g + a =

Mr 2 ⎛ a ⎞ Ma ⎜ ⎟= 2r ⎝ r ⎠ 2

⎛ M⎞ m + m + ⎜ 1 ⎟ a = m2 − m1 g 2 2⎠ ⎝

(

)

and

( 20.0 kg − 10.0 kg )( 9.80 m/s ) = 2.88 m/s a= = m + m + M/2 20.0 kg + 10.0 kg + ( 8.00 kg ) 2 ! (m2 − m1 ) g 1

8.47

T1 = (10.0 kg)(9.80 m/s2 + 2.88 m/s2) = 127 N

From Equation [2]:

T2 = (20.0 kg)(9.80 m/s2 − 2.88 m/s2) = 138 N

(a) The rotational analog of Newton’s second law is Στ = Iα , where

I = 13 ML2 is the rod’s moment of inertia about an axis through the pin. Summing torques on the rod about the pin gives a single nonzero torque due to the rod’s weight acting at its center of gravity: Στ = Iα

(

)

MgL 3g = 13 ML2 α → α = = 14.7 rad/s2 2 2L

Topic 8

505

(b) The rod is uniform so its center of mass is at its geometric center. The tangential acceleration at that point is at,cm = rcmα where rcm = L/2 so that at ,cm =

(

Lα (1.00 m ) 14.7 rad/s = 2 2

)

= 7.35 m/s2

(

at ,end = Lα = (1.00 m ) 14.7 rad/s2

)

= 14.7 m/s2 8.48

(a) The disk’s translational kinetic energy is

KEt = 12 mv 2 = 12 ( 2.50 kg ) ( 3.75 m/s)

= 17.6 J (b) The disk’s moment of inertia about its center is Idisk = 12 MR 2 so that it has a rotational kinetic energy of

KEr = Iω = 1 2

1 2

(

1 2

⎛ v⎞ MR ⎜ ⎟ = 12 ( KEt ) ⎝ R⎠ 2

)

= 8.79 J 8.49

The moment of inertia of the cylinder is

1 1 ⎛ w⎞ 1 ⎛ 800 N ⎞ I = MR2 = ⎜ ⎟ R2 = ⎜ 1.50 m 2 ) = 91.8 kg ⋅ m 2 2⎟( 2 2⎝ g⎠ 2 ⎝ 9.80 m s ⎠ !

Topic 8

506

The angular acceleration is given by

τ F ⋅ R ( 50.0 N ) (1.50 m ) α= = = = 0.817 rad/s 2 2 I I 91.8 kg ⋅ m ! At t = 3.00 s, the angular velocity is

ω = ωi + αt = 0 + (0.817 rad/s2)(3.00 s) = 2.45 rad/s and the kinetic energy is 1 1 2 KEr = Iω 2 = ( 91.8 kg ⋅ m 2 ) ( 2.45 rad/s ) = 276 J 2 2 !

8.50

(a) Hoop: I = MR2 = (4.80 kg)(0.230 m)2 = 0.254 kg ⋅ m2 1 1 2 2 2 Solid Cylinder: I = MR = ( 4.80 kg ) ( 0.230 m ) = 0.127 kg ⋅ m 2 2 ! 2 2 2 2 2 Solid Sphere: I = MR = ( 4.80 kg ) ( 0.230 m ) = 0.102 kg ⋅ m 5 ! 5

Thin Spherical Shell: 2 2 2 I = MR2 = ( 4.80 kg ) ( 0.230 m ) = 0.169 kg ⋅ m 2 3 ! 3

(b) When different objects of mass M and radius R roll without slipping (⇒ a = Rα) down a ramp, the one with the largest translational acceleration a will have the highest translational speed at the bottom. To determine the translational acceleration for the various objects, consider the force diagram below: ΣFx = Ma ⇒ Mg sin θ − f = Ma

[1]

Topic 8

507

τ = Iα ⇒ fR = I(a/R) or f = Ia/R2

[2]

Substitute Equation [2] into [1] to obtain

Mg sin θ − Ia⁄R2 = Ma

Mgsin θ a= 2 ! M + I/R

Since M, R, g, and θ are the same for all of the objects, we see that the translational acceleration (and hence, the translational speed) increases as the moment of inertia decreases. Thus, the proper rankings from highest to lowest by translational speed will be: solid sphere; solid cylinder; thin spherical shell; and hoop (c)

When an object rolls down the ramp without slipping, the friction force does no work and mechanical energy is conserved. Then, the total kinetic energy gained equals the gravitational potential 1 2 energy given up: KEr + KEt = −ΔPEg = Mgh and KEr = Mgh − Mv 2 !

, where h is the vertical drop of the ramp and v is the translational speed at the bottom. Since M, g, and h are the same for all of the objects, the rotational kinetic energy decreases as the translational speed increases. Using this fact, along with the result of part (b), © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

508

we rank the object’s final rotational kinetic energies, from highest to lowest, as: hoop; thin spherical shell; solid cylinder; and solid sphere 8.51

(a)

Treating the particles on the ends of the rod as point masses, the total moment of inertia of the rotating system is 2 !I = I rod + I1 + I 2 = mrod  12 + m1 ( /2 ) + m2 ( /2 ) . If the mass of the 2

rod can be ignored, this reduces to I = 0 + ( m1 + m2 ) ( /2 ) , and the ! 2

rotational kinetic energy is

(

)(

) (

)

2⎤ 2 1 1⎡ KEr = Iω 2 = ⎢ 3.00 kg + 4.00 kg 0.500 m ⎥ 2.50 rad/s 2 2⎣ ⎦

= 5.47 J (b)

If the rod has mass mrod = 2.00 kg, the rotational kinetic energy is

1 KEr = Iω 2 2 2 2⎤ 2 1⎡ 1 = ⎢ 2.00 kg 1.00 m + 3.00 kg + 4.00 kg 0.500 m ⎥ 2.50 rad/s 2 ⎢⎣ 12 ⎥⎦

(

)(

) (

)(

) (

)

= 5.99 J

8.52

Using conservation of mechanical energy, (KEt + KEr + PEg)f = (KEt + KEr + PEg)i

1 1 Mvt2 + Iω 2 + 0 = 0 + 0 + Mg(Lsin θ ) 2 !2

2 2 Since I = MR for a solid sphere and vt = Rω when rolling without ! 5 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

509

slipping, this becomes 1 1 MR2ω 2 + MR2ω 2 = Mg(Lsin θ ) 5 !2

and reduces to

ω= ! 8.53

(a)

10 ( 9.8 m s 2 ) ( 6.0 m ) sin 37° 10gLsin θ = = 36 rad/s 2 7R2 7 ( 0.20 m )

Assuming the disk rolls without slipping, the angular speed of the disk is ω = v/R where v is the translational speed of the center of the disk. Also, if the disk does not slip, the friction force between disk and ramp does no work and total mechanical energy is conserved. Hence, (KEt + KEr + PEg)f = (KEt + KEr + PEg)i or 1 1 Mv 2 + Iω 2 + 0 = 0 + 0 + mghi 2 !2

Since I = MR2/2, and hi = L sin θ = (4.50 m) sin 15.0°, we have

1 1 ⎛ M R2 ⎞ ⎛ v 2 ⎞ Mv 2 + ⎜ = mgLsin θ 2 2 ⎝ 2 ⎟⎠ ⎜⎝ R2 ⎟⎠ ! and v= !

4gLsin θ = 3

4 ( 9.80 m s 2 ) ( 4.50 m ) sin 15.0° 3

= 3.90 m/s

(b) The angular speed of the disk at the bottom is

Topic 8

510

v 3.90 m/s ω= = = 15.6 rad/s R 0.250 m !

8.54

(a) Convert 5.00 × 103 rev/min to rad/s:

5.00 × 103

rev rev ⎛ 1 min ⎞ ⎛ 2π rad ⎞ = 5.00 × 103 = 524 rad/s min min ⎜⎝ 60 sec ⎟⎠ ⎜⎝ 1 rev ⎟⎠

The flywheel’s kinetic energy is:

(

)

KEr = 12 Iω 2 = 12 12 MR 2 ω 2

(

)

= 14 5.00 × 10 2 kg ( 2.00 m ) ( 524 rad/s) 2

= 1.37 × 108 J

(b) Use the conversion 1 hp = 476 W to find the average rate at which rotational kinetic energy is removed from the flywheel:

⎛ 476 W ⎞ P = (10.0 hp ) ⎜ = 4.76 × 103 W ⎟ ⎝ 1 hp ⎠ From the definition of average power, the time before the flywheel would have to be brought back up to speed is

KEr Δt

→ Δt =

KEr 1.37 × 108 J = 4.76 × 103 W P

Δt = 2.88 × 10 4 s Use the conversion 1 h = 3600 s to find the length of time in hours:

⎛ 1h ⎞ Δt = 2.88 × 10 4 s ⎜ = 8.00 h ⎝ 3600 s ⎟⎠

(

8.55

)

1 2 Using Wnet = KE f − KEi = Iω f − 0 , we have 2 !

Topic 8

511

ωf = ! 8.56

2 ( 5.57 N ) ( 0.800 m ) 2Wnet 2F ⋅ s = = = 149 rad/s I I 4.00 × 10−4 kg ⋅ m 2

The work done on the grindstone is Wnet = F ⋅ s = F ⋅ (rθ) = (F ⋅ r)θ = τ ⋅ θ 1 1 Thus, Wnet = ΔKE becomes τ ⋅ θ = Iω 2f − Iω i2 , or 2 2 !

⎛

⎞

⎝

⎠

rad 1 ( 25.0 N ⋅ m )(15.0 rev ) ⎜ 21πrev ⎟ = 2 ( 0.130 kg ⋅ m )ω − 0

2 f

This yields

rad ⎞ ⎛ 1 rev ⎞ ⎛ ω f = ⎜ 190 = 30.2 rev/s ⎟ ⎝ s ⎠ ⎜⎝ 2π rad ⎟⎠ ! 8.57

(a)

1 1 2 KEt = mvt2 = (10.0 kg ) (10.0 m/s ) = 500 J 2 2 !

(b)

⎞ ⎛ vt2 ⎞ 1⎛ 1 2 KEr = Iω = ⎜ m R ⎟ ⎜ ⎟ 2 2⎝ 2 ⎠ ⎝ R2 ⎠ 1

1 1 2 = mvt2 = (10.0 kg ) (10.0 m/s ) = 250 J 4 ! 4

KEtotal = KEt + KEr = 750 J

As the bucket drops, it loses gravitational potential energy. The spool gains rotational kinetic energy and the bucket gains translational kinetic energy. Since the string does not slip on the spool, v = rω where r is the radius of the spool. The moment of inertia of the spool is 1 I = Mr 2 , where M is the mass of the spool. Conservation of energy ! 2

Topic 8

512

gives (KEt + KEr + PEg)f = (KEt + KEr + PEg)i 1 2 1 2 mv + Iω + mgy f = 0 + 0 + mgyi 2 !2

(

1 1⎛ 1 2 ⎞ m ( rω ) + ⎜ Mr 2 ⎟ ω 2 = mg yi − y f ⎝ ⎠ 2 2 !2

)

This gives

ω= !

8.59

(

2mg yi − y f

1 ⎞ 2 ⎛ ⎜⎝ m + M ⎟⎠ r 2

2 ( 3.00 kg ) ( 9.80 m/s 2 ) ( 4.00 m ) = 10.9 rad/s ⎡ 3.00 kg + 1 5.00 kg ⎤ 0.600 m 2 ( )⎥ ( ) ⎢⎣ 2 ⎦

(a) The ball’s moment of inertia is I = 52 MR 2 . Taking the ground to be at y = 0, apply conservation of mechanical energy to the ball as it rolls from point A to point B:

Topic 8

513

KEtA + KErA + mgyA = KEtB + KErB + mgyB 0 + 0 + Mg(1.50 m) = 12 MvB2 + 12 Iω B2 + Mg ( 0.500 m ) ⎛ vB2 ⎞ 1 1 2 (1.00 m ) Mg = 2 Mv + 2 5 MR ⎜ R ⎟ ⎝ ⎠ 2 B

(1.00 m ) g =

7 10

(

)

vB2 → vB = 3.74 m/s

(b) Use ω = v/r to find

ωB =

vB 3.74 m/s = → ω B = 37.4 rad/s R 0.100 m

(c) No torque acts on the ball as it rises from B to C, so the change in its angular velocity is zero and ω C = ω B = 37.4 rad/s . (d) Apply conservation of energy between points A and C, using ωB =ωC = vB/R : KEtA + KErA + mgyA = KEtC + KErC + mgyC 0 + 0 + Mg(1.50 m) = 0 + 12 Iω C2 + Mghmax

(1.50 m ) Mg = ( 1 2

2 5

⎛v ⎞ MR ⎜ B ⎟ + Mghmax ⎝ R⎠ 2

)

(1.50 m ) g = 15 vB2 + ghmax → hmax = 1.50 m −

vB2 5g

hmax = 1.21 m

8.60

(a) L = Iω = (MR2)ω = (2.40 kg)(0.180 m)2 (35.0 rad/s) = 2.72 kg ⋅ m2/s

Topic 8

514

⎛1 ⎞ L = Iω = ⎜ MR 2 ⎟ ω ⎝2 ⎠ (b) 2 1 = 2.40 kg 0.180 m 35.0 rad/s = 1.36 kg ⋅ m 2 /s 2

(

(c)

)

(2.40 kg )(0.180 m) (35.0 rad/s) = 1.09 kg ⋅ m /s 5 2

⎛2 ⎞ L = Iω = ⎜ MR 2 ⎟ ω ⎝3 ⎠ =

8.61

)(

⎛2 ⎞ L = Iω = ⎜ MR 2 ⎟ ω ⎝5 ⎠ =

(d)

)(

(2.40 kg )(0.180 m) (35.0 rad/s) = 1.81 kg ⋅ m /s 3 2

(a) For a single torque τ acting on the hoop, the change in the hoop’s angular momentum is

(

)

ΔL = τ Δt = 1.25 × 10−2 N ⋅ m ( 2.00 s) = 2.50 × 10−2 kg ⋅ m 2 /s (b) Use ΔL = I Δω with I = MR2 (for a hoop) to find ΔL 2.50 × 10−2 kg ⋅ m 2 /s Δω = = MR 2 ( 0.250 kg ) ( 0.100 m )2 = 10.0 rad/s

8.62

(a) Because no net external torque acts on the colliding, two-disk system, angular momentum is conserved so that

Topic 8

515

Li = L f

Iω 1 = ( I + I )ω 2

ω 1 6.00 rad/s = 2 2 = 3.00 rad/s

ω2 =

(b) As in part (a), angular momentum is conserved for the colliding, three-disk system so that

Li = L f

Iω 1 = ( 3I )ω 3

ω 1 6.00 rad/s = 3 3 = 2.00 rad/s

ω3 =

8.63

(a) The rotational speed of Earth is

ωE = !

⎞ 2π rad ⎛ 1d = 7.27 × 10−5 rad/s 4 ⎟ ⎜ 1 d ⎝ 8.64 × 10 s ⎠

⎛2 ⎞ Lspin = I sphereω E = ⎜ ME RE2 ⎟ ω E ⎝5 ⎠ 2⎤ ⎡2 = ⎢ 5.98 × 1024 kg 6.38 × 106 m ⎥ 7.27 × 10−5 rad/s ⎢⎣ 5 ⎥⎦

(

)(

) (

)

= 7.08 × 1033 J ⋅s (b)

For Earth’s orbital motion, ωorbit = 2π rad/y and

L =I ω = M R2 ω ! orbit point orbit ( E orbit ) orbit . Using data from Table 7.3, we find

Topic 8

516

(

)(

Lorbit = 5.98 × 1024 kg 1.496 × 1011 m

⎛

⎞

) (2π rad/y )⎜⎝ 3.1561×y10 s ⎟⎠ 2

= 2.66 × 10 40 J ⋅s

8.64

(a)

Yes, the bullet has angular momentum about an axis through the hinges of the door before the collision. Consider the sketch below, showing the bullet the instant before it hits the door.

The physical situation is identical to that of a point mass mB moving in a circular path of radius r with tangential speed vt = vi. For that situation the angular momentum is

⎛v ⎞ Li = Iiω i = ( mBr 2 ) = ⎜ i ⎟ = mBrvi ⎝ r⎠ ! and this is also the angular momentum of the bullet about the axis through the hinge at the instant just before impact. (b)

No, mechanical energy is not conserved in the collision. The bullet embeds itself in the door with the two moving as a unit after impact. This is a perfectly inelastic collision in which a significant amount of mechanical energy is converted to other

Topic 8

517

forms, notably thermal energy. (c)

Apply conservation of angular momentum with Li = mBrvi as discussed in part (a). After impact,

⎛1 ⎞ L f = I f ω f = ( Idoor + I bullet ) = ω f = ⎜ Mdoor L2 + mBr 2 ⎟ ω f where L = 1.00 ⎝ ⎠ 3 ! m = the width of the door and r = L − 10.0 cm = 0.900 m. Then,

L f = Li ⇒

ωf =

(

mBrvi

)

1 Mdoor L2 + mBr 2 3

(0.005 00 kg )(0.900 m)(1.00 × 10 m/s) = 1 (18.0 kg )(1.00 m) + (0.005 kg )(0.900 m) 3 3

yielding ωf = 0.749 rad/s. (d)

The kinetic energy of the door-bullet system immediately after impact is

1 KE f = I f ω 2f 2 2 2 ⎤ 1 ⎡1 = ⎢ 18.0 kg 1.00 m + 0.005 00 kg 0.900 m 2 ⎥ 0.749 rad/s 2 ⎢⎣ 3 ⎥⎦

(

)(

) (

)(

KEf = 1.68 J

The kinetic energy (of the bullet) just before impact was

)

Topic 8

518

)(

(

)

2 1 1 KEi = mBvi2 = 0.005 00 kg 1.00 × 103 m/s 2 2 3 = 2.50 × 10 J ≈ 1490 ⋅ KE f

8.65

Each mass moves in a circular path of radius r = 0.500 m/s about the center of the connecting rod. Their angular speed is v 5.00 m/s ω= = = 10.0 m/s r 0.500 m/s !

Neglecting the moment of inertia of the light connecting rod, the angular momentum of this rotating system is L = Iω = [m1r2 + m2r2]ω = (4.00 kg + 3.00 kg)(0.500 m2)(10.0 rad/s) = 17.5 J ⋅ s 8.66

Using conservation of angular momentum, Laphelion = Lperihelion. Thus,

(mr )ω = (mr )ω . Since ω = v /r at both aphelion and perihelion, this

2 a

2 p

( )

is equivalent to ( mra ) va ra = mrp vp rp , giving ! 2

⎛ 0.59 AU ⎞ ⎛ rp ⎞ va = ⎜ ⎟ v p = ⎜ ( 54 km/s ) = 0.91 km/s ⎝ ra ⎠ ⎝ 35 AU ⎟⎠ ! 8.67

(a) The stool is free to rotate about the vertical axis and no external torque acts as the student turns the wheel over. Angular momentum around the vertical axis is therefore conserved. Taking the positive direction to be vertically up, conservation of angular momentum gives:

Topic 8

519

Li = L f I wheelω wheel = −I wheelω wheel + Istudentω student

(

)

2 2I wheelω wheel 2 0.150 kg ⋅ m (17.5 rad/s) ω student = = Istudent 3.00 kg ⋅ m 2

= 1.75 rad/s (b) The student’s angular velocity is positive, so his angular momentum is directed up . 8.68

(a) The table turns counterclockwise, opposite to the way the woman walks. Its angular momentum cancels that of the woman so the total angular momentum maintains a constant value of Ltotal = Lwoman + Ltable = 0. Since the final angular momentum is Ltotal = Iwωw + Itωt = 0, we have

⎛I ⎞ ⎛ m r2 ⎞ ⎛ v ⎞ ⎛ m r⎞ ω t = − ⎜ w ⎟ ω w = − ⎜ w ⎟ ⎜ w ⎟ = − ⎜ w ⎟ vw ⎝ It ⎠ ⎝ It ⎠ ⎝ r ⎠ ⎝ It ⎠ ! or (taking counterclockwise as positive), ⎡ ( 60.0 kg ) ( 2.00 m ) ⎤ ωt = − ⎢ ⎥ ( −1.50 m/s ) = +0.360 rad/s 500 kg ⋅ m 2 ⎣ ⎦ !

Hence,

ωtable = 0.360 rad/s counterclockwise

1 1 (b) Wnet = ΔKE = KE f − 0 = mvw2 + I tω t2 2 2

Topic 8

520

1 1 2 Wnet = ( 60.0 kg ) (1.50 m/s 2 ) + ( 500 kg ⋅ m 2 ) ( 0.360 rad/s ) = 99.9 J 2 2 !

8.69

The moment of inertia of the cylinder before the putty arrives is 1 1 2 Ii = MR2 = (10.0 kg ) (1.00 m ) = 5.00 kg ⋅ m 2 2 2 !

After the putty sticks to the cylinder, the moment of inertia is If = Ii + mr2 = 5.00 kg ⋅ m2 + (0.250 kg)(0.900 m)2 = 5.20 kg ⋅ m2 Conservation of angular momentum gives Ifωf = Iiωi,

8.70

⎛I ⎞ ⎛ 5.00 kg ⋅ m 2 ⎞ ω f = ⎜ i ⎟ ωi = ⎜ 7.00 rad/s ) = 6.73 rad/s 2 ( If ⎠ ⎝ 5.20 kg ⋅ m ⎟⎠ ⎝ !

The total moment of inertia of the system is

( )

I total = I masses + I student = 2 mr 2 + 3.0 kg ⋅ m 2 plus stool

Initially, r = 1.0 m, and Ii = 2[(3.0 kg)(1.0 m)2] + 3.0 kg ⋅ m2 = 9.0 kg ⋅ m2 Afterward, r = 0.30 m so If = 2[(3.0 kg)(0.30 m)2] + 3.0 kg ⋅ m2 = 3.5 kg ⋅ m2 (a)

From conservation of angular momentum, Ifωf = Iiωi, or

⎛I ⎞ ⎛ 9.0 kg ⋅ m 2 ⎞ ω f = ⎜ i ⎟ ωi = ⎜ 0.75 rad/s ) = 1.9 rad/s 2 ( If ⎠ ⎝ 3.5 kg ⋅ m ⎟⎠ ⎝ ! (b)

1 1 2 KEi = Iiω i2 = ( 9.0 kg ⋅ m 2 ) ( 0.75 rad/s ) = 2.5 J 2 2 !

Topic 8

521

1 1 2 KE f = I f ω 2f = ( 3.5 kg ⋅ m 2 ) (1.9 rad/s ) = 6.3 J 2 2 !

8.71

The initial angular velocity of the puck is (v ) 0.800 m/s rad ωi = t i = = 2.00 ri 0.400 m s !

Since the tension in the string does not exert a torque about the axis of revolution, the angular momentum of the puck is conserved, or Ifωf = Iiωi. Thus, 2

⎛ Ii ⎞ ⎛ mri2 ⎞ ⎛r⎞ ⎛ 0.400 m ⎞ ω f = ⎜ ⎟ ωi = ⎜ 2 ⎟ ωi = ⎜ i ⎟ ωi = ⎜ ( 2.00 rad/s ) = 5.12 rad/s If ⎠ mrf ⎠ rf ⎠ ⎝ 0.250 m ⎟⎠ ⎝ ⎝ ⎝ ! 2

The net work done on the puck is 1 1 Wnet = KE f − KEi = I f ω 2f − I iω i2 2 2 1 m = ⎡⎢ mrf2 ω 2f − mri2 ω i2 ⎤⎥ = ⎡ rf2ω 2f − ri2ω i2 ⎤ ⎦ ⎦ 2⎣ 2⎣

( ) ( )

or Wnet = !

( 0.120 kg ) ⎡( 0.250 m ) 5.12 rad/s − ( 0.400 m ) ( 2.00 rad/s ) ⎤ 2

⎣

(

)

⎦

This yields Wnet = 5.99 × 10−2 J. 8.72

With all crew members on the rim of the station, the apparent acceleration experienced is the centripetal acceleration, ac = rω2 = g. Thus, the initial angular velocity of the station is ω i = g/r . !

Topic 8

522

The initial moment of inertia of the rotating system is Ii = Icrew + Istation = 150 mr2 + Istation After most of the crew move to the rotation axis, leaving only the managers on the rim, the moment of inertia is If = Imanagers + Istation = 50 mr2 + Istation Thus, conservation of angular momentum (Ifωf = Iiωi) gives the angular velocity during the union meeting as ⎛ 150 ( 65.0 kg ) (100 m )2 + 5.00 × 108 kg ⋅ m 2 ⎞ ⎛ Ii ⎞ ω f = ⎜ ⎟ ωi = ⎜ ⎟ 50 ( 65.0 kg ) (100 m 2 ) + 5.00 × 108 kg ⋅ m 2 ⎠ ⎝ If ⎠ ⎝ !

g g = 1.12 r r

The centripetal acceleration experienced by the managers still on the rim is 2 g 2 ac = rω 2f = r (1.12 ) = (1.12 ) ( 9.80 m/s 2 ) = 12.3 m/s 2 r !

8.73

(a) From conservation of angular momentum, Ifωf = Iiω i,

⎛I ⎞ ⎛ I ⎞ ω f = ⎜ i ⎟ ωi = ⎜ 1 ⎟ ω0 ⎜⎝ I f ⎟⎠ ⎝ I1 + I 2 ⎠ 2

⎛ I ⎞ KE f = I f ω 2f = I1 + I 2 ⎜ 1 ⎟ ω 02 2 2 ⎝ I1 + I 2 ⎠ (b) ⎛ I ⎞ ⎡1 ⎤ ⎛ I ⎞ = ⎜ 1 ⎟ ⎢ I1ω 02 ⎥ = ⎜ 1 ⎟ KEi ⎥⎦ ⎝ I1 + I 2 ⎠ ⎝ I1 + I 2 ⎠ ⎢⎣ 2 1

(

)

Topic 8

523

KE f !

KEi

I1 <1 I1 + I 2

Since this is less than 1.0, kinetic energy was lost. 8.74

(a) When rotating about axis a − a':

I a−a$ = Istick + I particle = !

1 2 ⎛ m M⎞ mL2 + M ( L − L/2 ) = ⎜ + ⎟ L2 ⎝ 12 4 ⎠ 12

Thus, La−a' = I c−a'ω

(

) (

)(

⎡ 0.100 kg 0.400 kg ⎤ 2 ⎢ ⎥ 1.00 m 4.00 rad/s = + ⎢ ⎥ 12 4 ⎣ ⎦

)(

)

= 0.433 kg ⋅ m 2 /s

(b) When rotating about axis b − b′:

1 2 ⎛m ⎞ Ib−b$ = Istick + I particle = mL2 + M ( L − 0 ) = ⎜ + M ⎟ L2 ⎝ ⎠ 3 3 ! and Lb−b' = I b−b'ω

(

)

⎡ 0.100 kg ⎤ 2 ⎢ = + 0.400 kg ⎥ 1.00 m 4.00 rad/s = 1.73 kg ⋅ m 2 /s ⎢ ⎥ 3 ⎣ ⎦

(

)(

)

Topic 8

8.75

524

rev ⎛ 2π rad ⎞ ⎛ 1 min ⎞ = 2.6 rad/s (a) ω = 25 min ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠ ! (b) Treat each blade as a long, thin rod rotating about an axis perpendicular to its length and passing through its end. Then, Iblade = mL2/3 and ⎛ mL2 ⎞ 2 I = 3I blade = 3 ⎜ = ( 420 kg ) ( 35 m ) = 5.1 × 105 kg ⋅ m 2 ⎟ ⎝ 3 ⎠ !

1 2 1 2 5 2 6 (c) KEr = Iω = ( 5.1 × 10 kg ⋅ m ) ( 2.6 rad/s ) = 1.7 × 10 J 2 2 !

8.76

 (a) In the sketch below, the force !P is the force the nail exerts on the claws of the hammer. It is equal in magnitude and oppositely directed to the force the claws exert on the nail. Choose an axis perpendicular to the page and passing through the indicated

 pivot. Then, with θ = 30.0° the lever arm of the force !P is 5.00 cm 5.00 cm = = = 5.77 cm cos θ cos 30.0° !

and Στ = 0 gives + P(5.77 cm) − (150 N)(30.0 cm) = 0

P= !

(150 N)( 30.0 cm ) = 780 N 5.77 cm

Topic 8

525

(b) ΣFy = 0 ⇒ n − P cos 30.0° = 0, giving n = P cos 30.0° = (780 N)cos 30.0° = 675 N ΣFx = 0 ⇒ f + F − P sin 30.0° = 0, or f = P sin 30.0° − F = (780 N)sin 30.0° − 150 N = 240 N The resultant force exerted on the hammer at the pivot is R= !

f 2 + n2 =

( 240 N) + ( 675 N) = 716 N 2

at θ = tan−1 (n/f ) = tan−1(675 N/240 N) = 70.4°, or  R = 716 N at 70.4° above!the!horizontal!to!the!right !

8.77

(a) Since no horizontal force acts on the child-boat system, the center of gravity of this system will remain stationary, or

xcg =

mchild xchild + mboat x boat mchild + mboat

= constant

The masses do not change, so this result becomes mchild xchild +

Topic 8

526

mboat xboat = constant. Thus, as the child walks to the right, the boat will move to the left. (b) Measuring distances from the stationary pier, with away from the pier being positive, the child is initially at (xchild)i = 3.00 m and the center of gravity of the boat is at (xboat)i = 5.00 m. At the end, the child is at the right end of the boat, so (xchild)f = (xboat)f + 2.00 m. Since the center of gravity of the system does not move, we have (mchildxchild + mboatxboat)f = (mchildxchild + mboatxboat)i, or mchild(xchild)f + mboat[(xchild)f − 2.00 m] = mchild(3.00 m)mboat(5.00 m) and

mchild ( 3.00 m ) + mboat ( 5.00 m + 2.00 m ) mchild + mboat

( xchild ) f =

( 40.0 kg )( 3.00 m )(70.0 kg )( 5.00 m + 2.00 m ) = 5.55 m

( xchild ) f =

40.0 kg + 70.0 kg

(c) When the child arrives at the right end of the boat, the greatest distance from the pier that he can reach is xmax = (xchild)f + 1.00 m = 5.55 m + 1.00 m = 6.55 m. This leaves him 0.45 m short of reaching the turtle. 8.78

(a) Consider the force diagram of the block given below.

Topic 8

527

If the x-axis is directed down the incline, ΣFx = max gives Mg sin θ − T − Mat, or T = M(g sin θ − at) T = (12.0 kg)[(9.80 m/s2)sin 37.0° − 2.00 m/s2] = 46.8 N (b) Now, consider the force diagram of the pulley.

Choose an axis perpendicular to the page and passing through the center of the pulley,

⎛ at ⎞ Στ = Iα gives T ⋅ r = I ⎜ ⎟ , or ⎝ r⎠ ! T ⋅ r 2 ( 46.8 N ) ( 0.100 m ) I= = = 0.234 kg ⋅ m 2 2 a 2.00 m/s t ! 2

2 ⎛ at ⎞ ⎛ 2.00 m/s ⎞ ( 2.00 s ) = 40.0 rad/s (c) ω = ω i + α t = 0 + ⎜⎝ ⎟⎠ t = ⎜ r ⎝ 0.100 m ⎟⎠ !

8.79

If the ladder is on the verge of slipping, f = (fs)max = µsn at both the floor

Topic 8

528

and the wall. From ΣFx = 0, we find f1 − n2 = 0 or

n2 = µs n1

[1]

Also, ΣFy = 0 gives n1 − w + µsn2 = 0 Using Equation [1], this becomes n 1 − w + µ s (µ s n 1 ) = 0

w w n1 = + 2 = 0.800 w 2 1 + µ 1 + 0.500 ( ) s !

[2]

Thus, Equation [1] gives n2 = 0.500(0.800w) = 0.400w

[3]

Choose an axis perpendicular to the page and passing through the lower end of the ladder. Then, Στ = 0 yields

⎛L ⎞ −w ⎜ cosθ ⎟ + n2 ( Lsin θ ) + f2 ( Lcosθ ) = 0 ⎠ ! ⎝2 Making the substitutions n2 = 0.400w and f2 = µsn2 = 0.200w this becomes

Topic 8

529

⎛L ⎞ −w ⎜ cosθ ⎟ + 0.400w Lsin θ + 0.200w Lcos θ = 0 ⎝2 ⎠

(

)(

) (

)(

)

⎛ 0.500 − 0.200 ⎞ and reduces to sin θ = ⎜ ⎟⎠ cos θ ⎝ 0.400 ! Hence, tan θ = 0.750 and θ = 36.9° 8.80

We treat each astronaut as a point object, of mass M, moving at speed v in a circle of radius r = d/2.

Figure P8.80 Then the total angular momentum is ⎡ ⎛ v⎞ ⎤ L = I1ω + I 2ω = 2 ⎢( Mr 2 ) ⎜ ⎟ ⎥ ⎝ r⎠⎦ ⎣ ⎛ d⎞ = 2Mvr = 2Mv ⎜ ⎟ = Mvd ⎝ 2⎠ !

(a) Li = Mvidi = (75.0 kg)(5.00 m/s)(10.0 m) = 3.75 × 103 kg ⋅ m2/s 1 1 KEi = m1v1i2 + m2 v2i2 2 2 (b) ⎛1 ⎞ = 2 ⎜ Mvi2 ⎟ = 75.0 kg 5.00 m/s 2 = 1.88 × 103 J = 1.88 kJ ⎝2 ⎠

(

)(

)

Topic 8

530

Lf 3.75 × 103 kg ⋅ m 2 /s = = 10.0 m/s (d) v f = Md f 75.0 kg ) ( 5.00 m ) ( !

⎛1 ⎞ 2 2 3 (e) KE f = 2 ⎜ Mv f ⎟ = 75.0 kg 10.0 m/s = 7.50 × 10 J = 7.50 kJ ⎝2 ⎠

(

)(

)

(f) Wnet = KEf − KEi = 5.62 kJ

8.81

⎡ ⎛ d⎞ ⎤ (a) Li = 2 ⎢ Mv ⎜ ⎟ ⎥ = Mvd ⎝ 2⎠ ⎦ ⎣ ! ⎛1 2⎞ 2 (b) KEi = 2 ⎜ Mvi ⎟ = Mv ⎝ ⎠ 2 ! (c) Lf = Li = Mvd

(d) v f = !

(

2 Mrf

)

Mvd = 2v 2M ( d/4 )

2 ⎛1 2⎞ 2 (e) KE f = 2 ⎜ Mv f ⎟ = M ( 2v ) = 4Mv ⎝ ⎠ 2 !

(f) Wnet = KEf − KEi = 3Mv2 8.82

Choose an axis that is perpendicular to the page and passing through the left end of the scaffold. Then Στ = 0 gives − (750 N)(1.00 m) − (345 N)(1.50 m) − (500 N)(2.00 m) −(1 000 N)(2.50 m) + TR(3.00 m) = 0 or

TR = 1.59 × 103 N = 1.59 kN

Topic 8

531

Then, ΣFy = 0 ⇒ TL = (750 + 345 + 500 + 1 000) N − 1.59 × 103 N = 1.01 kN 8.83

(a) Since the bar is in equilibrium, ΣFy = 0 giving Fs = mg − F1 − F2 and Fs = (2.35 kg)(9.80 m/s2) − 6.80 N − 9.50 N or

Fs = 6.73 N upward

(b) We require the sum of the torques about an axis perpendicular to the page and passing through the left end of the bar be zero. This gives Στ left = 0 + Fs ⋅ x − (mg)(/2) + F2 ⋅  = 0 ! end

Topic 8

532

⎡⎣( mg/2 ) − F2 ⎤⎦ ⋅  ⎡⎣( 2.35 kg ) ( 9.80 m s 2 ) 2 − 9.50 N ⎤⎦ (1.30 m ) x= = F 6.73 N s !

and 8.84

x = 0.389 m = 38.9 cm

(a) Taking PEg = 0 at the level of the horizontal axis passing through the center of the rod, the total energy of the rod in the vertical position is E = KE + PEg = 0 + m1g(+L) + m2g(−L) = (m1 − m2) gL

Figure P8.84 (b) In the rotated position of Figure P8.84b, the rod is in motion and the total energy is 1 E = KEr + PEg = I totalω 2 + m1 gy1 + m2 gy2 2 ! 1 (m1L2 + m2 L2 )ω 2 + m1 g ( +Lsin θ ) + m2 g ( −Lsin θ ) 2 !

E= !

(m1 + m2 ) L2ω 2 + m − m gLsin θ ( 1 2) 2

(c) In the absence of any nonconservative forces that do work on the rotating system, the total mechanical energy of the system is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 8

533

constant. Thus, the results of parts (a) and (b) may be equated to yield

(m1 + m2 ) L2ω 2 + m − m gLsin θ = m − m gL ( 1 2) ( 1 2) 2

and

ω=

(

)(

2 m1 − m2 g 1 − sin θ (m1 + m2 )L

)

(d) In the vertical position, the net torque acting on the system is zero,

τnet = 0. This is because the lines of action of both external gravitational forces (m1g and m2g) pass through the pivot, and hence have zero lever arms about the rotation axis. In the rotated position, the net torque (taking clockwise as positive) is

τnet = Στ = m1g(Lcos θ) − m2g(Lcos θ) = (m1 − m2)gLcos θ Note that the net torque is not constant as the system rotates. Thus, the angular momentum will change at a non-uniform rate, ΔL/Δt = τnet . (e) In the rotated position, the angular acceleration is

α= !

τ net ( m1 − m2 ) gLcos θ ( m1 − m2 ) gcos θ = = I m1L2 + m2 L2 (m1 + m2 ) L

The angular acceleration is a maximum in the horizontal position (θ = 0°), where the gravitational forces have maximum lever arms

Topic 8

534

and exert the maximum torque on the system. Also, note that α = 0 at θ = 90° This is understandable since the vertical orientation is a position of unstable equilibrium (τnet = 0). 8.85

The gymnast’s center of gravity is located at xcg =

ycg =

marms xcg,arms + mtorso xcg,torso + mthighs xcg,thighs + mlegs xcg,legs marms + mtorso + mthighs + mlegs

and

marms ycg,arms + mtorso ycg,torso + mthighs ycg,thighs + mlegs ycg,legs marms + mtorso + mthighs + mlegs

Use the figure and the provided data to locate xcg and ycg for each body part: Arms:

xcg,arms = 0

ycg,arms = rcg,arms = 0.239 m Torso:

xcg,torso = rcg,torso = 0.337 m ycg,torso = 0 Thighs: xcg,thighs = Ltorso + rcg,thighs cos60°

= 0.601 m + ( 0.151 m ) cos60° = 0.677 m

Topic 8

535

ycg,thighs = rcg,thighs sin 60°

= ( 0.151 m ) sin 60° = 0.131 m

Legs: xcg,legs = Ltorso + Lthighs cos60° + rcg,legs

= 0.601 m + ( 0.374 m ) cos60° + 0.227 m

=1.02 m

ycg,legs = Lthighs sin 60°

= ( 0.374 m ) sin 60° = 0.324 m

The x- and y-locations of the gymnast’s center of gravity are then xcg =

marms xcg,arms + mtorso xcg,torso + mthighs xcg,thighs + mlegs xcg,legs marms + mtorso + mthighs + mlegs

( 6.87 kg)(0) + ( 33.57 kg)(0.337 m ) + (14.07 kg)(0.677 m ) + (7.54 kg)(1.02 m ) 6.87 kg + 33.57 kg + 14.07 kg + 7.54 kg

= 0.460 m

and ycg =

marms ycg,arms + mtorso ycg,torso + mthighs ycg,thighs + mlegs ycg,legs marms + mtorso + mthighs + mlegs

( 6.87 kg)(0.239 m ) + ( 33.57 kg)([0]

ytorso

) + (14.07 kg)(0.131 m) + (7.54 kg)(0.324 m)

6.87 kg + 33.57 kg + 14.07 kg + 7.54 kg

= 9.55 × 10−2 m

8.86

(a) Since only conservative forces do work on the long rod, we use conservation of energy for this pure rotation about the fixed point

Topic 8

536

O. The rod starts from rest (ωi = 0) with the center of gravity at the level of point O. Choosing this level as the reference level for gravitational potential energy, we have KEf = KEi + PEg,i − PEg,f

and

⎛ L⎞ IOω 2f = 0 + 0 − Mg ⎜ − ⎟ 2 ⎝ 2⎠ 1

ωf =

MgL IO

MgL 1 ML2 3

3g L

Figure P8.47 The tangential speed of the center of mass when the rod reaches the vertical position is then 2

⎛ L ⎞ 3g ⎛ L ⎞ 3g vcg = rcgω f = ⎜ ⎟ = ⎜ ⎟ ⎝ 2⎠ L ⎝ 2⎠ L

vcg = 3gL/4 =

( )

3gL /2

(b) The tangential speed of the lowest point on the rod when it reaches the vertical position is

Topic 8

537

vlower = rlowerω f = (L) end

8.87

end

3g L

= (L)2

3g L

and

vlower = 3gL = 2vcg end

Choose an axis perpendicular to the page and passing through the center of the cylinder. Then, applying Στ = Iα to the cylinder gives

( 2T ) ⋅ R = ⎛⎜⎝ MR2 ⎞⎟⎠ α = ⎛⎜⎝ MR2 ⎞⎟⎠ ( at /R) ,or 2 2 1

1 T = Mat 4 !

[1]

Now apply ΣFy = may to the falling objects to obtain (2m)g − 2T = (2m)at, or T at = g − m !

[2]

(a) Substituting Equation [2] into [1] yields

Mg ⎛ M ⎞ T= −⎜ ⎟T 4 ⎝ 4m ⎠ !

Topic 8

538

Mmg which reduces to T = . M + 4m !

(b) From Equation [2] above, 1 ⎛ Mmg ⎞ Mg 4mg at = g − ⎜ = g− = ⎟ m ⎝ M + 4m ⎠ M + 4m M + 4m !

8.88

(a) A smooth (that is, frictionless) wall cannot exert a force parallel to its surface. Thus, the only force the vertical wall can exert on the upper end of the ladder is a horizontal normal force. (b) Consider the force diagram of the ladder given below.

If the rotation axis is perpendicular to the page and passing through the lower end of the ladder, the lever arm of the normal  force !n 2 that the wall exerts on the upper end of the ladder is

d2 = L sin θ

 (c) The lever arm of the force of gravity, !mg , acting on the ladder is d = ( L/2 ) cosθ = ( Lcosθ ) 2 !

Topic 8

539

(d) Refer to the force diagram given in part (b) of this solution and make use of the fact that the ladder is in both translational and rotational equilibrium.

(

)

ΣF = 0 ⇒ n1 − m g − mp g = 0!,!or!n1 = m + mp g ! y

When the ladder is on the verge of slipping,

(

)

f = ( f1 )max = µsn1 = µs m + mp g . !1

(

)

Then, ΣFx = 0 ⇒ n2 = f1, or n2 = µs m + mp g . ! Finally, Στ = 0 ⇒ n2 ( Lsin θ ) − m g ( L/2 ) cos θ − mp gxcos θ = 0 where ! x is the maximum distance the painter can go up the ladder before it will start to slip. Solving for x gives

x= !

⎛m ⎞ ⎛ m ⎞ n2 ( Lsin θ ) − m g ( L/2 ) cos θ = µs ⎜  + 1⎟ Ltan θ − ⎜  ⎟ L mp gcos θ ⎝ mp ⎠ ⎝ 2mp ⎠

and using the given numerical data, we find ⎡ 30 kg ⎤ ⎛ 30 kg ⎞ x = ( 0.45 ) ⎜ + 1⎟ ( 4.0 m ) tan 53° − ⎢ ⎥ ( 4.0 m ) = 2.5 m 2 ( 80 kg ) ⎦ ⎝ 80 kg ⎠ ⎣ !

8.89

The large mass (m1 = 60.0 kg) moves in a circular path of radius r1 = 0.140 m, while the radius of the path for the small mass (m2 = 0.120 kg) is r2 = ℓ − r1 = 3.00 m − 0.140 m = 2.86 m.

Topic 8

540

The system has maximum angular speed when the rod is in the vertical position as shown above. We take PEg = 0 at the level of the horizontal rotation axis and use conservation of energy to find

( )

1 ⎛1 2 2 ⎞ KE f + PEg = KEi + PEg ⇒ ⎜ I1ω max + I 2ω max ⎟⎠ + ( m2 gr2 − m1 gr1 ) = 0 + 0 f i ⎝ 2 2 ! Approximating the two objects as point masses, we have!I1 = m1r1 and 2

2 !I 2 = m2 r2 . The energy conservation equation then becomes

1 2 m1r12 + m2 r22 )ω max = ( m1r1 − m2 r2 ) g and yields ( !2

ω max =

(

)

2 m1r1 − m2 r2 g m1r12 − m2 r22

(

)(

) (

)(

2 ⎡ 60.0 kg 0.140 m − 0.120 kg 2.86 m ⎤ 9.80 m/s 2 ⎣ ⎦ = 2 2 60.0 kg 0.140 m + 0.120 kg 2.86 m

(

)(

) (

)(

)

Or ωmax = 8.56 rad/s. The maximum tangential speed of the small mass object is then (v2)max = r2ωmax = (2.86 m)(8.56 rad/s) = 24.5 m/s

Topic 8

8.90

541

(a) Note that the cylinder has both translational and rotational motion. The center of gravity accelerates downward while the cylinder rotates around the center of gravity. Thus, we apply both the translational and the rotational forms of Newton’s second law to the cylinder: ΣFy = may ⇒ T − mg = m(−a) or

T = m(g − a)

[1]

Στ = Iα ⇒ − Tr = I(−a/r)

1 2 For a uniform, solid cylinder, I = mr so our last result becomes ! 2 ⎛ mr 2 ⎞ ⎛ a ⎞ Tr = ⎜ ⎜ ⎟ ⎝ 2 ⎟⎠ ⎝ r ⎠ !

2T m

[2]

Substituting Equation [2] into Equation [1] gives T = mg − 2T, and solving for T yields T = mg/3. (b) From Equation [2] above,

Topic 8

542

⎛ 2g ⎞ vy = 0 + 2 ⎜ − ⎟ ( −h ) = ⎝ 3⎠ !

4gh/3

Using conservation of energy, with PEg = 0 at the final level of the cylinder, gives (KEt + KEr + PEg)f = (KEt + KEr + PEg)i , or

1 2 1 2 mvy + Iω + 0 = 0 + 0 + mgh 2 !2

1 2 Since ω = vy /r and I= mr , this becomes ! 2

(

)

⎞ 1⎛ 1 mv y2 + ⎜ m r 2 ⎟ v y2 r 2 = mgh, or 2 2⎝ 2 ⎠ 1

3 4 8.91

mv y2 = mgh yielding v y =

4gh/3 .

Considering the shoulder joint as the pivot, the second condition of equilibrium gives w Στ = 0 ⇒ (70 cm ) − ( Fm sin 45° ) ( 4.0 cm ) = 0 2 !

Fm =

Thus,

Fm =

(

w 70 cm

)

2 4.0 cm sin 45°

(750 N)(70 cm) = 9.3 × 10 N = 9.3 kN (8.0 cm)sin 45° 3

Topic 8

8.92

543

(a) Choosing the elbow (right end of the forearm) as the pivot, the second condition of equilibrium gives

⎛ ⎞ Στ = 0 + 0 + mg ⎜ ⎟ + 0 − (T sin θ )  = 0 ⎝ 2⎠ ! or

(b)

1.60 kg ) ( 9.80 m/s 2 ) ( mg T= = = 10.2 N 2sin θ 2sin 50.0° !

ΣFx = 0 ⇒ Rx = Tcos θ = (10.2 N)cos 50.0° = 6.56 N and ΣFy = 0 gives

⎛ mg ⎞ Ry = mg − T sin θ = mg − ⎜ ⎟ sin θ ⎝ 2sin θ ⎠ =

8.93

mg 2

(1.60 kg )(9.80 m/s ) = 7.84 N = 2

(a) Force diagrams for each block and the pulley are given below. Observe that the angular acceleration of the pulley will be clockwise in direction and has been given a negative sign.

Topic 8

544

Since Στ = Iα, the positive sense for torques and angular acceleration must be the same (counterclockwise). For m1: ΣFy = may ⇒ T1 − m1g = m1(−a), or T1 = m1(g − a)

[1]

For m2: ΣFx = max ⇒ T2 = m2a

[2]

For the pulley: Στ = Iα ⇒ T2r − T1r = I(−a/r), or

⎛ 1⎞ T1 − T2 = ⎜ 2 ⎟ a ⎝r ⎠ !

[3]

Substitute Equations [1] and [2] into Equation [3] and solve for a to obtain a = m1g/[(I/r)2 + m1 + m2], or

( 4.00 kg )( 9.80 m s ) 2

a= = 3.12 m/s 2 2 2 ( 0.500 kg ⋅ m ) (0.300 m) + 4.00 kg + 3.00 kg !

(b) Equation [1] above gives: T1 = (4.00 kg)(9.80 m/s2 − 3.12 m/s2) = 26.7 N

Topic 8

545

and Equation [2] yields: T2 = (3.00 kg)(3.12 m/s2 ) = 9.36 N. 8.94

(a)

(b) ΣFy = 0 ⇒ nF − 120 N − mmonkey g = 0, so nF = 120 N + (10.0 kg)(9.80 m/s2) = 218 N (c) When x = 2L/3, we consider the lower end of the ladder as our pivot and obtain ∑ τ lower = 0 end

⎛ L/ ⎞ ⎛ 2 L/ ⎞ / − 120 N ⎜ cos 60.0°⎟ − 98.0 N ⎜ cos 60.0°⎟ + nW Lsin 60.0° = 0 ⎝2 ⎠ ⎝ 3 ⎠

(

)

(

)

(

)

(

)

⎡ 60.0 N + 196/3 N ⎤ cos 60.0° ⎦ nW = ⎣ = 72.4 N sin 60.0° Then, ΣFx = 0 ⇒ T − nW = 0

T = nW = 72.4 N

Topic 8

546

(d) When the rope is ready to break, T = nW = 80.0 N. Then ∑ τ lower = 0 end

yields

⎛L ⎞ − (120 N ) ⎜ cos 60.0°⎟ − ( 98.0 N ) xcos 60.0° + ( 80.0 N ) ( Lsin 60.0° ) = 0 ⎝ ⎠ 2 ! or

(

)

(

)

⎡ 80.0 N sin 60.0° − 60.0 N cos 60.0° ⎤ L ⎦ x= ⎣ 98.0 N cos 60.0°

(

)

= 0.802 L = 0.802 3.00 m = 2.41 m (e) If the horizontal surface were rough and the rope removed, a horizontal static friction force directed toward the wall would act on the lower end of the ladder. Otherwise, the analysis would be much as what is done above. The maximum distance the monkey could climb would correspond to the condition that the friction force have its maximum value, µsnF, so you would need to know the coefficient of static friction to solve part (d). 8.95

(a) Choose the initial position of the block as the zero gravitational potential energy level. Then, conservation of energy from when the block is released from rest until it comes to rest momentarily after falling a distance h gives KEt,f + KEr,f + PEg,f + PEs,f = KEt,i + KEr,i + PEg,i + PEs,i

Topic 8

547

Since vf = vi = 0 and ωf = ωi = 0, this becomes 1 0 + 0 − mgh + kh2 != 0 + 0 + 0 + 0 , or 2 !

2mg 2 ( 3.2 kg ) ( 9.80 m/s ) h= = = 0.73 m k 86 N/m ! 2

(b) Here, we use conservation of energy from when the block is released from rest until it has dropped a distance h = 25 cm = 0.25 m. Recognize that if the string does not slip on the pulley, the angular speed of the pulley is given by ω = vt/r, where vt is the translational speed of the block and r is the radius of the pulley. KEt,f + KEr,f + PEg,f + PEs,f = KEt,i + KEr,i + PEg,i + PEs,i 2

1 2 1⎛ 1 1 ⎞⎛v ⎞ mvt + ⎜ Mr 2 ⎟ ⎜ t ⎟ + mg(−h) + kh2 = 0 + 0 + 0 + 0 ⎠⎝ r ⎠ 2⎝ 2 2 !2 which reduces to !12 ( m + M/2 ) vt = mgh − 12 kh , and 2

2 ( 3.2 kg ) ( 9.8 m/s 2 ) ( 0.25 m ) − ( 86 N/m ) ( 0.25 m ) 2mgh − kh2 vt = = m + M/2 3.2 kg + (1.8 kg ) 2 !

yielding vt = 1.6 m/s.

Topic 9

548

Topic 9 Fluids and Solids

QUICK QUIZZES

9.1

Choice (c). The mass that you have of each element is:

mgold = ρgoldVgold = (19.3 × 103 kg/m3)(1 m3) = 19.3 × 103 kg

msilver = ρsilverVsilver = (10.5 × 103 kg/m3)(2 m3) = 21.0 × 103 kg

maluminum = ρaluminumValuminum = (2.70 × 103 kg/m3)(6 m3) = 16.2 × 103 kg

9.2

Choice (a). At a fixed depth, the pressure in a fluid is directly proportional to the density of the fluid. Since ethyl alcohol is less dense than water, the pressure is smaller than P when the glass is filled with alcohol.

9.3

Choice (c). For a fixed pressure, the height of the fluid in a barometer is inversely proportional to the density of the fluid. Of the fluids listed in the selection, ethyl alcohol is the least dense.

Topic 9

9.4

549

Choice (b). The blood pressure measured at the calf would be larger than that measured at the arm. If we imagine the vascular system of the body to be a vessel containing a liquid (blood), the pressure in the liquid will increase with depth. The blood at the calf is deeper in the liquid than that at the arm and is at a higher pressure.

Blood pressures are normally taken at the arm because that is approximately the same height as the heart. If blood pressures at the calf were used as a standard, adjustments would need to be made for the height of the person, and the blood pressure would be different if the person were lying down.

9.5

Choice (c). The level of a floating ship is unaffected by the atmospheric pressure. The buoyant force results from the pressure differential in the fluid. On a high-pressure day, the pressure at all points in the water is higher than on a low-pressure day. Because water is almost incompressible, however, the rate of change of pressure with depth is the same, resulting in no change in the buoyant force.

9.6

Choice (b). Since both lead and iron are denser than water, both objects will be fully submerged and (since they have the same dimensions) will displace equal volumes of water. Hence, the buoyant forces acting on the

Topic 9

550

two objects will be equal.

9.7

Choice (a). When there is a moving air stream in the region between the balloons, the pressure in this region will be less than on the opposite sides of the balloons where the air is not moving. The pressure differential will cause the balloons to move toward each other. This is a demonstration of Bernoulli’s principle in action.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

9.2

We approximate the thickness of the atmosphere by using P = P0 + ρgh with P0 = 0 at the top of the atmosphere and P = 1 atm at sea level. This gives an approximation of

P − P0 105 Pa − 0 h= ~ = 10 4 m 3 1 2 ρg (1 kg/m )(10 m/s ) !

h  10 km

Because both the density of the air, ρ, and the acceleration of gravity, g, decrease with altitude, the actual thickness of the atmosphere will be greater than our estimate.

9.4

The two dams must have the same strength. The force on the back of each

Topic 9

551

dam is the average pressure of the water times the area of the dam. If the two reservoirs are equally deep, the forces exerted on the two dams by the water have equal magnitudes.

9.6

A fan driven by the motor removes air, and hence decreases the pressure inside the cleaner. The greater air pressure outside the cleaner pushes air in through the nozzle toward this region of lower pressure. This inward rush of air pushes or carries the dirt along with it.

9.8

The external pressure exerted on the chest by the water makes it difficult to expand the chest cavity and take a breath while under water. Thus, a snorkel will not work in deep water.

9.10

The water level on the side of the glass stays the same. The floating ice cube displaces its own weight of liquid water, and so does the liquid water into which it melts.

9.12

The higher the density of a fluid, the higher an object will float in it. Thus, an object will float lower in low-density alcohol.

9.14

The ski jumper gives her body the shape of an airfoil. She deflects the air stream downward as it rushes past and it deflects her upward in agreement with Newton’s third law. Thus, the air exerts a lift force on

Topic 9

552

her, giving a higher and longer trajectory.

9.16

Since most of the ice at the south pole is supported by land, it does not displace any seawater, and hence, does not contribute to the water level in the oceans. However, after this ice melts and ﬂows into the sea, it will signiﬁcantly add to the water level in the oceans. On the contrary, the ice at the north pole is currently displacing its own weight in water, just as it will after melting. Thus, the ice at the south pole will have the greater impact on sea levels as it melts and the correct choice is (b).

ANSWERS TO EVEN NUMBERED PROBLEMS

9.2

(a) 7.322 × 10−3 kg

(b) Vgold = 3.79 × 10−7 m3, Vcopper = 7.46 × 10−8 m3

9.4

24.8 kg

Topic 9

553

9.6

1.9 × 104 N

9.8

(a) 1.17 × 105 Pa

(b)

1.60 × 104 Pa

9.10

(a) 20.0 cm

(b)

0.490 cm

9.12

860 N

9.14

(a) 10.5 m

(b) No. Since water and alcohol are more volatile than mercury, more liquid will evaporate and degrade the vacuum above the liquid column inside the tube of this barometer.

9.16

2.3 lb

9.18

(a) 1.77 × 10−3 m3

(b)

17.3 N

(c)

196 N

(d)

9.20

(a) See Solution.

(b)

ΣFy = B − w − wr = 0

(c)

964 N

(d) 356 N

(e)

101 kg/m3

(f) 3.62 × 103 N

179 N

(g) 333 kg

9.22

(a) See Solution.

(b)

4.11 × 103 N

Topic 9

554

(d)

136 kg

(e) The balloon and its load accelerate upward.

(f) As the balloon rises, decreasing at atmospheric density decreases the buoyancy force.

9.24

(a) 23.2% after inhaling, 17.1% after exhaling

9.26

3.33 × 103 kg/m3

9.28

16.5 cm

9.30

(a) 8.57 × 103 kg/m3

9.32

6.25 m/s

9.34

(b)

715 kg/m3

(a) 0.471 m/s

(b)

4.24 m/s

9.36

(a) 11.0 m/s

(b)

2.64 × 104 Pa

9.38

4.4 × 10−2 Pa

9.40

1.41 m/s

9.42

(a) 2.02 m/s

(b)

8.08 m/s

(b)

See Solution.

(c)

5.71 × 10−3

Topic 9

555

m3/s

9.44

(a) 17.7 m/s

(b)

1.73 mm

9.46

(a) 15.1 MPa

(b)

2.95 m/s

9.48

(a) 1.91 m/s

(b)

8.64 × 10−4 m3/s

9.50

7.32 × 10−2 N/m

9.52

5.6 m

9.54

0.12 N

9.56

1.5 m/s

9.58

1.5 × 105 Pa

9.60

Yes

9.62

9.5 × 10−10 m2/s

9.64

1.02 × 103 kg/m3

9.66

2.11 × 10−3 m

9.68

2.58 × 10−8 m3

(c)

4.35 kPa

Topic 9

556

9.70

(a) 3.14 × 104 N

9.72

22 N toward the bottom of the page in Figure P9.14

9.74

(a) −5.4 × 10−2 m3

9.76

Yes, the average stress is 5.5 × 107 Pa, considerably less than 16 × 107 Pa.

9.78

4.74ρ0

9.80

(a) See Solution.

(b)

1.23 × 104 Pa

9.82

(a) 10.3 m

(b)

9.84

See Solution.

9.86

1.9 m

9.88

(a) 1.25 cm

(b)

13.8 m/s

9.90

1.71 cm

(b)

6.28 × 104 N

1.1 × 103 kg/m3

(b)

PROBLEM SOLUTIONS

9.1

(a) Apply the definition of density to find

Topic 9

557

ρ=

M M 81.5 kg → V= = = 8.27 × 10−2 m 3 3 V ρ 985 kg/m

(b) From the definition of pressure:

(

F mg ( 81.5 kg ) 9.80 m/s P= = = A A 4.50 × 10−2 m 2 9.2

) = 1.77 × 10 Pa 4

(a) The mass of gold in the coin is

mAu = !

(# karats ) m

total

22 11 mtotal = (7.988 × 10−3 kg ) = 7.322 × 10−3 kg 24 12

and the mass of copper is mCu = mtotal/12 = (7.988 × 10−3 kg)/12 =

6.657 × 10−4 kg

(b) The volume of the gold present is

m 7.322 × 10−3 kg VAu = Au = = 3.79 × 10−7 kg 3 3 3 ρ 19.3 × 10 kg/m Au ! and the volume of the copper is

m 6.657 × 10−4 kg VCu = Cu = = 7.46 × 10−8 m 3 3 3 ρCu 8.92 × 10 kg/m ! (c) The average density of the British sovereign coin is

Topic 9

558

m mtotal 7.988 × 10−3 kg ρCu = total = = = 1.76 × 10 4 kg/m 3 −7 3 −5 3 V V + V 3.79 × 10 m + 7.46 × 10 m total Au Cu ! 9.3

Apply the definition of pressure, taking A to be the surface area of a sphere of radius RE:

(

F w = A 4π RE2

→ w = P 4π RE2

(

)(

)

w = 1.01× 105 N/m 2 4π RE2 = 5.17 × 1019 N 9.4

Mbar = ρAuVbar = ρAu(l × w × h) = (19.3 × 103 kg/m3)[(0.045 0 m)(0.110 m)(0.260 m)]

9.5

Mbar = 24.8 kg

(a) If the particles in the nucleus are closely packed with negligible space between them, the average nuclear density should be approximately that of a proton or neutron. That is

ρ nucleus = !

mproton Vproton

mproton 4π r 3 3



3 (1.67 × 10−27 kg ) 4π (1 × 10

−15

= −4 × 1017 kg/m 3

(b) The density of iron is ρFe = 7.86 × 103 kg/m3 and the densities of other solids and liquids are on the order of 103 kg/m3. Thus, the nuclear density is about 1014 times greater than that of common solids and © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 9

559

liquids, which suggests that atoms must be mostly empty space. Solids and liquids, as well as gases, are mostly empty space.

9.6

Let the weight of the car be W. Then, each tire supports W/4, and the gauge pressure is P = F/A = (W/4)/A = W/4A. Thus,

W = 4AP = 4(0.024 m2)(2.0 × 105 Pa) = 1.9 × 104 N

9.7

(a) Fatm = PA = Patm(πr2) = (8.04 × 104 Pa)π(2.00 m)2 = 1.01 × 106 N

(b) Fg = mg = (ρV)g = ρ[(πr2)h]g

= (415 kg/m3)[π(2.00 m)2(10.0 m)](7.44 m/s2) = 3.88 × 105 N

(c) Now, consider the thin disk-shaped region 2.00 m in radius at the bottom end of the column of methane. The total downward force on it is the weight of the 10.0-meter tall column of methane plus the downward force exerted on the upper end of the column by the atmosphere. Thus, the pressure (force per unit area) on the diskshaped region located 10.0 meters below the ocean surface is

Ftotal Fatm + Fg 1.01 × 106 N + 3.88 × 105 N P= = = = 1.11 × 105 Pa 2 2 A πr π (2.00 m) !

Topic 9

9.8

560

(a) The absolute pressure at the base of a 0.120 m column of mercury is: P = P0 + ρ gh

(

)(

)

= 1.01 × 105 Pa + 13.6 × 103 kg/m 3 9.80 m/s2 ( 0.120 m ) = 1.17 × 105 Pa

(b) The corresponding gauge pressure is Pgauge = P − P0 = ρ gh = 1.60 × 10 4 Pa

9.9

(a)

P = P0 + pgh = 101.3 × 103 Pa + (1.00 × 103 kg/m 3 ) ( 9.80 m/s 2 ) ( 27.5 m ) !

= 3.71 × 105 Pa

(b) The inward force the water will exert on the window is

⎛ 35.0 × 10−2 m ⎞ F = PA = P (π r ) = ( 3.71 × 10 Pa ) π ⎜ = 3.57 × 10 4 N ⎟ ⎝ 2 ⎠ ! 2

9.10

(a) Consider the tube as shown in part (b) of the sketch below.

The volume of water in the right arm is

Topic 9

561

Vw =

ρw

100 g 1.00 g/cm

= 100 cm 3

But, we also know that Vw = A2hw, giving

Vw 100 cm 3 hw = = = 20.0 cm A2 5.00 cm 2 ! (b) From the sketch above, observe that the mercury that has been forced out of the right arm into the left arm is Vdisplaced = A1h = A2h2, so h2 = (A1/A2)h

⎛ 10.0 cm 2 ⎞ h2 = ⎜ h = 2.00h 2 ⎝ 5.00 cm ⎟⎠ !

The absolute pressure at the water-mercury interface in the right arm is

P = P0 + ρwghw

[1]

The absolute pressure at the same level in the left arm is

P = P0 + ρHgg(h + h2) = P0 + ρHgg(h + 2.00h)

P = P0 + 3.00ρHggh [2]

Topic 9

562

Since the pressure is the same at all points at a given level in a static fluid, we equate the pressures in Equations [1] and [2] to obtain P0 + 3.00ρHggh = P0 + ρwghw, which yields

⎡ ⎤ ⎛ ρ ⎞ 1.00 × 103 kg/m 3 w ⎢ ⎥ 20.0 cm = 0.490 cm h=⎜ ⎟h = ⎜⎝ 3.00 ρ Hg ⎟⎠ w ⎢ (3.00) 13.6 × 103 kg/m 3 ⎥ ⎣ ⎦

(

9.11

)(

)

The density of the solution is ρ = 1.02ρwater = 1.02 × 103 kg/m3. If the glucose solution is to flow into the vein, the minimum required gauge pressure of the fluid at the level of the needle is equal to the gauge pressure in the vein, giving

Pgauge = P − P0 = ρghmin = 1.33 × 103 Pa

and hmin = ! 9.12

1.33 × 103 Pa 1.33 × 103 Pa = = 0.133 m ρg (1.02 × 103 kg/m 3 )( 9.80 m/s2 )

Neglecting any difference in height between the input and output pistons, Pascal’s principle gives

Topic 9

563

Pin = Pout Fin F = out Ain Aout Fin =

9.13

→ Fin =

Ain A Fout = in wcar Aout Aout

(

)

0.050 m 2 1.2 × 10 4 N = 860 N 0.70 m 2

We first find the absolute pressure at the interface between oil and water.

P1 = P0 + ρoil ghoil !

= 1.013 × 105 Pa + (700 kg m 3 ) ( 9.80 m s 2 ) ( 0.300 m ) = 1.03 × 105 Pa

This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use P2 = P1 + ρwaterghwater, or

P2 = 1.03 × 105 Pa + (103 kg/m3)(9.80 m/s2)(0.200 m) = 1.05 × 105 Pa

9.14

(a) If we assume a vacuum (P = 0) inside the tube above the wine column and atmospheric pressure at the base of the column (that is, at the level of the wine in the open container), we start at the top of the liquid in the tube and calculate the pressure at depth h in the wine as Patmo = 0 + ρgh = ρgh. Thus,

P 1.013 × 105 Pa h = atmo = = 10.5 m ρ g ( 984 kg/m 3 ) ( 9.80 m/s 2 ) !

Topic 9

564

(b) No. Since water and alcohol are more volatile than mercury, more liquid will evaporate and degrade the vacuum above the liquid column inside the tube of this barometer.

9.15

The gauge pressure required to support a 0.155-m deep column of mercury is

Pgauge = P − P0 = ρ gh

(

)(

)

= 13.6 × 103 kg/m 3 9.80 m/s2 ( 0.115 m ) = 1.53 × 10 4 Pa 9.16

First, use Pascal’s principle, F1/A1 = F2/A2, to find the force piston 1 will exert on the handle when a 500-lb force pushes downward on piston 2.

⎛ π d2 4 ⎞ ⎛ d2 ⎞ ⎛A ⎞ 1 1 F1 = ⎜ ⎟ F2 = ⎜ 2 ⎟ F2 = ⎜ 12 ⎟ F2 ⎝ A2 ⎠ ⎝ π d2 4 ⎠ ⎝ d2 ⎠

(0.25 in ) (500 lb) = 14 lb = (1.5 in ) 2

Now, consider an axis perpendicular to the page, passing through the left end of the jack handle. Στ = 0 yields

Topic 9

565

+(14 lb)(2.0 in) − F(12 in) = 0,

9.17

F = 2.3 lb

When held underwater, the ball will have three forces acting on it: a downward gravitational force, mg = ρballVg = ρball(4πr3/3)g; an upward buoyant force, B = ρwaterVg = ρwater(4πr3/3)g; and an applied force, F. If the ball is to be in equilibrium, we have (taking upward as positive)

ΣFy = F + B − mg = 0

⎡ ⎛ 4π r 3 ⎞ ⎤ ⎛ 4π r 3 ⎞ ⎛ 4π r 3 ⎞ F = mg − B = ⎢ ρ ball ⎜ g − ρ g = ρ − ρ ( ball water ) ⎜⎝ 3 ⎟⎠ g ⎥ water ⎜ ⎝ 3 ⎟⎠ ⎦ ⎝ 3 ⎟⎠ ⎣ !

giving

4π ⎛ 0.038 0 m ⎞ 2 F = ⎡ 0.084 0 − 1.00 × 10 kg/m ⎤ ⎟ 9.80 m/s ⎣ ⎦ 3 ⎜⎝ 2 ⎠

(

)

(

)

= −0.258 N

! so the required applied force is F = 0.258 Ndirecteddownward .

9.18

(a) Use the definition of density to find

ρ lead =

Mlead Vlead

→ Vlead =

Mlead 20.0 kg = ρ lead 11.3 × 103 kg/m 3

Vlead = 1.77 × 10−3 m 3

Topic 9

566

(b) The buoyant force magnitude equals the weight of the displaced water. The lead is submerged, so the volume of displaced water equals the lead’s volume and:

B = mwater g = ρ waterVlead g

(

)(

)

B = 1.00 × 103 kg/m 3 1.77 × 10−3 m 3 9.80 m/s2 = 17.3 N (c) The lead’s weight is wlead = Mlead g = 196 N . (d) Three forces act on the lead (at rest): The normal and buoyant forces are directed up in the +y-direction and weight is directed down. Apply Newton’s second law to find: ΣFy = may n + B − wlead = 0 → n = wlead − B n = 196 N − 17.3 N = 179 N

9.19

The boat sinks until the weight of the additional water displaced equals the weight of the truck. Thus,

wtruck = [ ρwater ( ΔV )] g ! or

kg ⎞ m⎞ ⎛ ⎛ = ⎜ 103 3 ⎟ ⎡⎣( 4.00 m ) ( 6.00 m ) ( 4.00 × 10−2 m ) ⎤⎦ ⎜ 9.80 2 ⎟ ⎝ ⎝ m ⎠ s ⎠

wtruck = 9.41 × 103 N = 9.41 kN

Topic 9

9.20

567

(a)

(b) Since the system is in equilibrium,

(c)

( )

B = ρ w gVsubmerged = ρ w g d ⋅ A

(

)(

ΣFy = B − w − wr = 0

)(

)

= 1 025 kg m 3 9.80 m s 2 0.024 0 m 4.00 m 2 = 964 N

(d) From B - w − wr = 0,

wr = B − w = B − mg = 964 N − (62.0 kg)(9.80 m/s2) = 356 N

(e)

(f)

ρ force =

mr Vr

wr g t⋅A

356 N

(9.80 m s )(0.090 m)( 4.00 m ) 2

= 101 kg m 3

( ) = (1 025 kg m ) ( 9.80 m s ) ( 0.090 0 m ) ( 4.00 m ) = 3.62 × 10 N

Bmax = ρ w gVr = ρ w g t ⋅ A 3

(g) The maximum weight of survivors the raft can support is wmax =

mmaxg = Βmax − wr, so

mmax = !

Bmax − wr 3.62 × 103 N − 356 N = = 333 kg g 9.80 m/s 2

Topic 9

9.21

568

Three forces act on the neutrally buoyant balloon/basket: the weight of the balloon/basket (wBB), the weight of the hot air inside the balloon (whot), and the buoyant force (equal to the weight of the displaced ambient air, wair). Apply Newton’s second law to find: ΣFy = may B − mBB g − mhot g = 0

ρ airVBB g − mBB g − ρ hotVBB g = 0 Solve for the density of the hot air:

ρ hot = ρ air −

mBB 545 kg = 1.25 kg/m 3 − VBB 2.55 × 103 m 3

ρ hot = 1.04 kg/m 3 9.22

(a)

(b) Since the balloon is fully submerged in air, Vsubmerged = Vb = 325 m3, and © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 9

569

B = ρairgVb = (1.29 kg/m3)(9.80 m/s2)(325 m3) = 4.11 × 103 N

(

)

∑ Fy = B − wb − wHe = B − mb g − ρ He gVb = B − mb + ρ HeVb g (c)

(

)(

= 4.11 × 103 N − ⎡ 226 kg + 0.179 kg m 3 325 m 3 ⎤ 9.80 m s 2 ⎣ ⎦

)

= +1.33 × 103 N

Since ΣFy = may > 0, aywill be positive (upward), and the balloon rises.

(d) If the balloon and load are in equilibrium, ΣFy = (B − wb − wHe) − wload = 0 and wload = (B − wb − wHe) = 1.33 × 103 N. Thus, the mass of the load is

mload = !

wload 1.33 × 103 N = = 136 kg g 9.80 m/s 2

(e) If mload < 136 kg, the net force acting on the balloon + load system is upward and the balloon and its load will accelerate upward.

(f) As the balloon rises, decreasing atmospheric density decreases the buoyancy force. At some height the balloon will come to equilibrium and go no higher.

9.23

⎛ 4π r 3 ⎞ 3 ⎛ 4π ⎞ B = ρair gVballoon = ρair g ⎜ = (1.29 kg/m 3 ) ( 9.80 m/s 2 ) ⎜ 3.00 m ) ( ⎟ ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠ (a) = 1.43 × 103 N = 1.43 kN !

Topic 9

570

(b)

(

)(

ΣFy = B − wload = 1.43 × 103 N − 15.0 kg 9.80 m/s 2

)

= +1.28 × 103 N = 1.28 kN upward

(c) The balloon expands as it rises because the external pressure (atmospheric pressure) decreases with increasing altitude.

9.24

(a) To float, the buoyant force acting on the person must equal the weight of that person, or the weight of the water displaced by the person must equal the person’s own weight. Thus,

B = mg ⇒ (ρseaVsubmerged)g = (ρbodygVtotal)g

Vsubmerged ! Vtotal

ρ body ρsea

After inhaling,

Vsubmerged Vtotal

945 kg/m 3 1 230 kg/m 3

= 0.768 = 76.8%

leaving 23.2% above surface

After exhaling,

Vsubmerged Vtotal

1 020 kg m 3 1 230 kg m 3

= 0.829 = 82.9%

Topic 9

571

(b) In general, “sinkers” would be expected to be thinner with heavier bones, whereas “floaters” would have lighter bones and more fat.

9.25

⎡ ⎛ 4π 3 ⎞ ⎤ Btotal = 600 ⋅ Bsingle = 600 ρ air gVballoon = 600 ⎢ ρ air g ⎜ r ⎟⎥ 3 ⎝ ⎠ ⎥⎦ baloon ⎢⎣ (a) 3⎤ ⎡ 4π = 600 ⎢ 1.29 kg/m 3 9.80 m/s 2 0.50 m ⎥ = 4.0 × 103 N = 4.0 kN 3 ⎢⎣ ⎥⎦

(

)

)(

) (

)

(b) ΣFy = Btotal − mtotalg = 4.0 × 103 N − 600(0.30 kg)(9.8 m/s2) = 2.2 × 103 N = 2.2 kN

(c) Atmospheric pressure at this high altitude is much lower than at Earth's surface, so the balloons expanded and eventually burst.

9.26

The actual weight of the object is Fg,actual = mobjectg = 5.00 N, and its mass is mobject = 5.00 N/g. When fully submerged, the upward buoyant force (equal to the weight of the displaced water) and the upward force exerted on the object by the scale (Fg,apparent = 3.50 N) together support the actual weight of the object. That is,

ΣFy = 0 ⇒ B + Fg,apparent − Fg,actual = 0

and

B = Fg,actual − Fg,apparent = 5.00 N − 3.50 N = 1.50 N

Topic 9

572

Thus, B = ρwatergVobject gives Vobject = B/(ρwaterg) and the density of the object is

mobject ⎛ 5.00 N ⎞ ⎛ ρwater g ⎞ ρobject = =⎜ = 3.33 ρwater = 3.33 × 103 kg/m 3 ⎟ ⎜ ⎟ Vobject ⎝ g ⎠ ⎝ 1.50 N ⎠ !

9.27

(a) The wooden block sinks until the buoyant force (weight of the displaced water) equals the weight of the block. That is, when equilibrium is reached,

B = ρwaterg[(s − h)s2) = ρwoodg ⋅ s3, giving

⎛ρ ⎞ s − h = ⎜ wood ⎟ ⋅ s ⎝ ρwater ⎠ !

⎛ ⎞ ⎛ ρ 650 kg/m 3 ⎞ h = s ⋅ ⎜ 1 − wood ⎟ = 20.0 cm ⎜ 1 − ⎟ = 7.00 cm ρ water ⎠ 1 000 kg/m 3 ⎠ ⎝ ⎝

(

)

(b) When the upper surface of the block is level with the water surface, the buoyant force is

Topic 9

573

B = ρwatergVblock = ρwaterg ⋅ s3

This must equal the weight of the block plus the weight of the added lead, or mPbg + mblockg = B, and

mPb =

(

)

− mblock = ρ waterVblock − ρ woodVblock = ρ water − ρ wood ⋅ s3

mPb = (1 000 kg/m3 − 650 kg/m3)(0.200 m)3 = 2.80 kg

giving

9.28

At equilibrium, ΣFy = B − Fspring − mg = 0 so the spring force is

Fspring = B − mg =[(ρwaterVblock) − m]g, where

m 5.00 kg Vblock = = = 7.69 × 10−3 m 3 3 ρwood 650 kg/m ! Thus, Fspring = [(103 kg/m3)(7.69 × 10−3 m3) − 5.00 kg](9.80 m/s2) = 26.4 N.

The elongation of the spring is then

Δx = ! 9.29

Fspring k

26.4 N = 0.165 m = 16.5 cm 160 N/m

(a) The buoyant force is the difference between the weight in air and the apparent weight when immersed in the alcohol, or B = 300 N − 200 N

Topic 9

574

= 100 N. But, from Archimedes’ principle, this is also the weight of the displaced alcohol, so B = (ρalcoholV)g. Since the sample is fully submerged, the volume of the displaced alcohol is the same as the volume of the sample. This volume is

B 100 N V= = = 1.46 × 10−2 m 3 3 2 ρalcohol g (700 kg/m ) ( 9.80 m/s ) ! (b) The mass of the sample is

m= !

weight!in!air 300 N = = 30.6 kg g 9.80 m/s 2

and its density is

m 30.6 kg ρ= = = 2.10 × 103 kg/m 3 −2 3 V 1.46 × 10 m ! 9.30

The difference between the weight in air and the apparent weight when immersed is the buoyant force exerted on the object by the fluid.

(a) The mass of the object is

m= !

weight!in!air 300 N = = 30.6 kg g 9.80 m/s 2

The buoyant force when immersed in water is the weight of a volume © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 9

575

of water equal to the volume of the object, or Bwater = (ρwaterV)g. Thus, the volume of the object is

B 300 N − 265 N V = water = = 3.57 × 10−3 m 3 3 3 2 ρwater g (10 kg/m ) ( 9.80 m/s ) ! and its density is

m 30.6 kg ρobject = = = 8.57 × 103 kg/m 3 −3 3 V 3.57 × 10 m ! (b) The buoyant force when immersed in oil is equal to the weight of a volume V = 3.57 × 10−3 m3 of oil. Hence, Boil = (ρoilV)g, or the density of the oil is

B 300 N − 275 N ρoil = oil = = 715 kg/m 3 Vg ( 3.57 × 10−3 m 3 ) ( 9.80 m/s 2 ) ! 9.31

The volume of the iron block is

m 2.00 kg V = iron = = 2.54 × 10−4 m 3 3 3 ρiron 7.86 × 10 kg/m ! and the buoyant force exerted on the iron by the oil is

B = (ρoilV)g = (916 kg/m3)(2.54 × 10−4 m3)(9.80 m/s2) = 2.28 N

Topic 9

576

Applying ΣFy = 0 to the iron block gives the support force exerted by the upper scale (and hence the reading on that scale) as

Fupper = mirong − B = 19.6 N − 2.28 N = 17.3 N

From Newton’s third law, the iron exerts force B downward on the oil (and hence the beaker). Applying ΣFy = 0 to the system consisting of the beaker and the oil gives

Flower − B − (moil + mbeaker)g = 0

The support force exerted by the lower scale (and the lower scale reading) is then

Flower = B + (moil + mbeaker)g = 2.28 N + [(2.00 + 1.00) kg](9.80 m/s2) = 31.7 N

9.32

Use the equation of continuity to find:

A1 v1 = A2 v2 → v2 =

A1 πr 2 r2 v1 = 1 2 v1 = 12 v1 A2 π r2 r2

(0.250 m ) (1.00 m/s) = 6.25 m/s v = (0.100 m ) 2

Topic 9

9.33

577

Let y = 0 at the height of the faucet so the top of the tank is at height h = 3.00 m − 0.800 m = 2.20 m. Use Bernoulli’s equation along a streamline between the top of the tank and the faucet to find: P1 + 12 ρ v12 + ρ gy1 = P2 + 12 ρ v22 + ρ gy2 P0 + 0 + ρ gh = P0 + 12 ρ v22 + 0

(

)

v2 = 2gh = 2 9.80 m/s2 ( 2.20 m ) v2 = 6.57 m/s

9.34

(a) The cross-sectional area of the hose is A = π r2 = π d2/4 = π (2.74 cm)2/4, and the volume flow rate (volume per unit time) is Av = 25.0 L/1.50 min. Thus,

⎤ ⎛ 1 min ⎞ ⎛ 103 cm 3 ⎞ 25.0 L / 1.50 min ⎛ 25.0 L ⎞ ⎡ 4 =⎜ ⎢ 2 2⎥ A ⎝ 1.50 min ⎟⎠ ⎣ π ⋅(2.74) cm ⎦ ⎜⎝ 60 s ⎟⎠ ⎜⎝ 1 L ⎟⎠

⎛ 1m ⎞ = (47.1 cm/s) ⎜ 2 = 0.471 m/s 10 cm ⎟⎠ ⎝ ! 2

2 A2 ⎛ π d22 ⎞ ⎛ 4 ⎞ ⎛ d2 ⎞ ⎛ 1 ⎞ 1 =⎜ =⎜ ⎟ =⎜ ⎟ = (b) 2⎟ ⎟ ⎜ ⎝ ⎠ 3 9 ! A1 ⎝ 4 ⎠ ⎝ π d1 ⎠ ⎝ d1 ⎠

A2 =

A1 9

Then, from the equation of continuity, A2v2 = A1v1 and we find

⎛A ⎞ v2 = ⎜ 1 ⎟ v1 = 9 ( 0.471 m/s ) = 4.24 m/s ⎝ A2 ⎠ !

9.35

(a) The volume flow rate is Av, and the mass flow rate is

Topic 9

578

ρAv = (1.0 g/cm3)(2.0 cm2)(40 cm/s) = 80 g/s

(b) From the equation of continuity, the speed in the capillaries is

⎛ A ⎞ ⎛ 2.0 cm 2 ⎞ vcapiliaries = ⎜ aorta ⎟ vaorta = ⎜ 40 cm/s ) 3 2 ( vcapiliaries ⎠ ⎝ 3.0 × 10 cm ⎟⎠ ⎝ ! or

9.36

vcapiliaries = 2.7 × 10−2 cm/s = (0.27 mm/s)

(a) From the equation of continuity, the flow speed in the second section of the pipe is

⎛ 10.0 cm 2 ⎞ ⎛A ⎞ v2 = ⎜ 1 ⎟ v 1 = ⎜ 2.75 m/s ) = 11.0 m/s 2 ( ⎝ A2 ⎠ ⎝ 2.50 cm ⎟⎠ !

(b) Using Bernoulli’s equation and choosing y = 0 along the centerline of the pipe gives

1 P2 = P1 + ρ ( v12 − v22 ) 2 1 2 2 = 1.20 × 105 Pa + (1.65 × 103 kg/m 3 ) ⎡( 2.75 m/s ) − (11.0 m/s ) ⎤ ⎣ ⎦ 2 ! or

9.37

P2 = 2.64 × 104 Pa

From Bernoulli’s equation, choosing y = 0 at the level of the syringe and

Topic 9

579

1 2 1 2 needle, P2 + pv2 = P1 + pv1 , so the flow speed in the needle is 2 2 !

v2 = v12 + !

2 ( P1 − P2 ) ρ

In this situation,

F 2.00 N P1 − P2 = P1 − Patm = ( P1 )gauge = = = 8.00 × 10 4 Pa −5 2 A 2.50 × 10 m 1 ! Thus, assuming v1 ≈ 0,

2 ( 8.00 × 10 4 Pa )

v2 = 0 + = 12.6 m/s 1.00 × 103 kg/m 3 !

9.38

We apply Bernoulli’s equation, ignoring the very small change in vertical 1 1 3 2 2 2 2 2 position, to obtain P1 − P2 = ρ ( v2 − v1 ) = ρ ⎡⎣( 2v1 ) − v1 ⎤⎦ = ρ v1 , or 2 2 2 ! 2 3 ΔP = (1.29 kg/m 3 ) (15 × 10−2 m/s ) = 4.4 × 10−2 Pa 2 !

9.39

(a) Assuming the airplane is in level flight, the net lift (the difference in the upward and downward forces exerted on the wings by the air flowing over them) must equal the weight of the plane, or (Plower surface − Pupper surface)Awings = mg. This yields

Topic 9

580

(8.66 × 10 kg )( 9.80 m/s ) = 9.43 × 103 Pa mg Plower − Pupper = = Awings 90.0 m 2 surface surface ! 4

(b) Neglecting the small difference in altitude between the upper and lower surfaces of the wings, and applying Bernoulli’s equation, yields

1 1 2 2 Plower + ρair vlower = Pupper + ρair vupper 2 2 !

vupper = v !

2 lower

( +

2 Plower − Pupper

ρair

) = ( 225 m s) + 2 ( 9.43 × 10 Pa ) = 225 m s 3

1.29 kg m 3

(

)

1 2 2 smaller pressure difference, ΔP = ρair vupper − vlower . Beyond the 2 !

maximum operational altitude, the pressure difference can no longer support the aircraft.

9.40

In the equation of continuity, A1v1 = A2v2, the area A2 is the combined cross-sectional area of the two pipes after the divider. Substitute values to find:

Topic 9

581

A1 v1 = A2 v2 → π ( R1 ) v1 = 2π ( R2 ) v2 2

R2 ( 9.00 mm ) (1.25 m/s) = 1.41 m/s v2 = 1 2 v1 = 2 2R2 2 ( 6.00 mm ) 2

9.41

(a) Since the pistol is fired horizontally, the emerging water stream has initial velocity components of (v0x = vnozzle, v0y = 0). Then, Δy = v0yt + ayt2/2, with ay = −g, gives the time of flight as

t= !

2(Δy) 2(−1.50 m) = = 0.553 s ay −9.80 m/s 2

(b) With ax = 0 and v0x = vnozzle, the horizontal range of the emergent stream is Δx = vnozzlet, where t is the time of flight from above. Thus, the speed of the water emerging from the nozzle must be

Δx 8.00 m vnozzle = = = 14.5 m/s t 0.553 s ! (c) From the equation of continuity, A1v1 = A2v2, the speed of the water in the larger cylinder is v1 = (A2/A1)v2 = (A2/A1) vnozzle, or

⎛ π r2 ⎞ ⎛r ⎞ v1 = ⎜ 22 ⎟ v nozzle = ⎜ 2 ⎟ v nozzle ⎝ r1 ⎠ ⎝ π r1 ⎠ 2

⎛ 1.00 mm ⎞ =⎜ ⎟ 14.5 m/s = 0.145 m/s ⎝ 10.0 mm ⎠

(

)

Topic 9

582

(d) The pressure at the nozzle is atmospheric pressure, or

P2 = 1.013 × 105 Pa

(e) With the two cylinders horizontal, y1 ≈ y2 and gravity terms from Bernoulli’s equation can be neglected, leaving 2 2 !P1 + ρwater v1 / 2 = P2 + ρwater v2 / 2 so the needed pressure in the larger

cylinder is

P1 = P2 +

ρ water

(v − v ) 2 2 2

2 1

1.00 × 103 kg/m 3 ⎡ = 1.013 × 10 Pa + (14.5 m/s)2 − (0.145 m/s)2 ⎤ ⎣ ⎦ 2 5

P1 = 2.06 × 105 Pa

(f) To create an overpressure of ΔP = 2.06 × 105 Pa − 1.013 × 105 Pa = 1.05 × 105 Pa in the larger cylinder, the force that must be exerted on the piston is

( ) (

)(

)

F1 = (ΔP)A1 = (ΔP) π r12 = 1.05 × 105 Pa π 1.00 × 10−2 m = 33.0 N

9.42

(a) From Bernoulli’s equation,

Topic 9

583

P1 + !

ρwater v12 ρ v2 + ρwater gy1 = P2 + water 2 + ρwater gy2 2 2

⎡P − P ⎤ v22 − v12 = 2 ⎢ 1 2 − g ( y2 − y1 ) ⎥ ⎣ ρwater ⎦ !

and using the given data values, we obtain

⎡ 1.75 × 105 Pa − 1.20 × 105 Pa ⎤ v22 − v12 = 2 ⎢ − ( 9.80 m/s 2 ) (2.50 m) ⎥ 3 3 1.00 × 10 Kg/m ⎣ ⎦ !

and

2 2 2 2 !v2 − v1 = 61.0 m / s

From the equation of continuity,

⎛ 3.00 cm ⎞ ⎛A ⎞ ⎛ π r2 ⎞ ⎛r ⎞ v2 = ⎜ 1 ⎟ v1 = ⎜ 12 ⎟ v1 = ⎜ 1 ⎟ v1 = ⎜ v1 ⎝ A2 ⎠ ⎝ π r2 ⎠ ⎝ r2 ⎠ ⎝ 1.50 cm ⎟⎠ ! or

v2 = 4v1

Substituting Equation [2] into [1] gives !(16 − 1)v1 = 61.0 m / s , or 2

Topic 9

584

v1 =

61.0 m 2 / s2 15

= 2.02 m / s

(b) Equation [2] above now yields v2 = 4(2.02 m/s) = 8.08 m/s.

Looking at the lower point:

flow!rate = (π r12 ) v1 = π ( 3.00 × 10−2 m ) (2.02 m/s) = 5.71 × 10−3 m 3 /s 2

! 9.43

First, consider the path from the viewpoint of projectile motion to find the speed at which the water emerges from the tank. From 1 Δy = v0yt + ayt 2 with v0y = 0, Δy = −1.00 m, and ay = −g, we find the time of 2 !

flight as

t= !

2(Δy) 2.00 m = ay g

From the horizontal motion, the speed of the water coming out of the hole is

v2 = v0x =

Δx t

= (0.600 m)

g 2.00 m

(0.600 m)2 g 2.00 m

(1.80 × 10 m) g −1

Topic 9

585

We now use Bernoulli’s equation, with point 1 at the top of the tank and point 2 at the level of the hole. With P1 = P2 = Patm and v1 ≈ 0, this gives 1 ρ gy1 = ρ v22 + pgy2 , or 2 !

h = y1 − y2 =

9.44

v22 2g

(1.80 × 10 m) g = 9.00 × 10 m = 9.00 cm = −1

−2

(a) Apply Bernoulli’s equation with point 1 at the open top of the tank and point 2 at the opening of the hole. Then, P1 = P2 = Patm and we assume v1 ≈ 0. This gives

1 2 ρ v2 + pgy2 = ρ gy1 , or !2

v2 = 2g ( y1 − y2 ) = 2 ( 9.80 m/s 2 ) (16.0 m) = 17.7 m/s ! (b) The area of the hole is found from

A2 = !

flow!rate 2.50 × 10−3 m 3 / min ⎛ 1 min ⎞ = = 2.35 × 10−6 m 2 ⎜ ⎟ v2 17.7 m/s ⎝ 60 s ⎠

But, !A2 = π d2 / 4 and the diameter of the hole must be 2

d2 = ! 9.45

4A2 = π

4 ( 2.35 × 10−6 m 2 )

= 1.73 × 10−3 m = 1.73 mm

First, determine the flow speed inside the larger section from

Topic 9

586

v1 = !

flow!rate 1.80 × 10−4 m 3 /s = = 0.367 m/s 2 A1 π ( 2.50 × 10−2 m ) 4

The absolute pressure inside the large section on the left is P1 = Patm + ρgh1, where h1 is the height of the water in the leftmost standpipe. The absolute pressure in the constriction is P2 = Patm + ρgh2, so

P1 − P2 = ρg(h1 − h2) = ρg(5.00 cm)

The flow speed inside the constriction is found from Bernoulli’s equation with y1 = y2 (since the pipe is horizontal). This gives

2 v22 = v12 + ( P1 − P2 ) = v12 + 2g ( h1 − h2 ) ρ !

v = ! 2

( 0.367 m/s ) + 2 ( 9.80 m/s )( 5.00 × 10 m ) = 1.06 m/s 2

−2

The cross-sectional area of the constriction is then

A2 = !

flow!rate 1.80 × 10−4 m 3 / s = = 1.70 × 10−4 m 2 v2 1.06 m/s

and the diameter is

d2 = !

4A2 = π

4 (1.70 × 10−4 m 2 )

= 1.47 × 10−2 m = 1.47 cm

Topic 9

9.46

587

(a) For minimum input pressure so the water will just reach the level of the rim, the gauge pressure at the upper end is zero (i.e., the absolute pressure inside the upper end of the pipe is atmospheric pressure), and the flow rate is zero. Thus, Bernoulli’s equation,

⎛ ⎞ ⎛ ⎞ 1 2 1 2 P + ρ v + ρ gy = P + ρ v + ρ gy ⎜ ⎟ ⎜ ⎟ , becomes 2 2 ⎝ ⎠ river ⎝ ⎠ rim

(P ) + 0 = 1atm + 0 + ρ g( y − y ) = 1atm + ρ g( y − y ) river min

rim

river

⎛

rim

⎞⎛

river

[1]

⎞

(P ) = 1.013 × 10 Pa+ ⎜⎝ 10 mKg ⎟⎠ ⎜⎝ 9.80 ms ⎟⎠ (2 096 m − 564 m) 5

or,

river min

(P ) = (1.013 × 10 + 1.50 × 10 ) Pa = 1.51 × 10 Pa = 15.1 MPa 5

river min

(b) When volume flow rate is

⎛ π d2 ⎞ 3 flow rate = Av = ⎜ ⎟ v = 4 500 m /d ⎝ 4 ⎠

the velocity in the pipe is

4( flow rate)

π d2

4(4 500 m 3 /d) ⎛ 1d ⎞ = ⎜ ⎟ = 2.95 m/s π (0.150 m)2 ⎝ 86 400 s ⎠

Topic 9

588

(

)

1 2 Priver + 0 = ⎡1atm + ρ g yrim − yriver ⎤ + ρ vrim ⎣ ⎦ 2 or, using Equation [1] from above,

1⎛ kg ⎞ ⎛ m⎞ = Priver + ⎜ 103 2.95 ⎟⎜ ⎟ rim 2⎝ m3 ⎠ ⎝ s⎠

( ) + 2 ρv ( ) = ( P ) + 4.35 kPa

Priver = Priver

rim

2 rim

river rim

The additional pressure required to achieve the desired flow rate is

ΔP = 4.35 kPa

9.47

(a) For the upward flight of a water-drop projectile from geyser vent to fountain-top, !vy = v0y + 2ay (Δy) , with vy = 0 when Δy = Δymax, gives 2

v = 0 − 2a(Δy)max = −2 ( −9.80 m/s 2 ) (40.0 m) = 28.0 m/s ! 0y (b) Because of the low density of air and the small change in altitude, atmospheric pressure at the fountain top will be considered equal to that at the geyser vent. Bernoulli’s equation, with vtop = 0, then gives

Topic 9

589

(

)

1 2 ρvent = 0 + ρ g ytop − yvent ,!or !2

(

)

v = 2g ytop − yvent = 2 ( 9.80 m/s 2 ) (40.0 m) = 28.0 m/s ! vent

( P + 0 + ρ gy )

c hamber

1 2 = Patm + ρ vvent + ρ gyvent , or 2

⎡1 2 ⎤ P − Patm = ρ ⎢ vvent + g ( yvent − ychamber ) ⎥ ⎣2 ⎦ 2 ⎤ m⎞ ⎛ 3 kg ⎞ ⎡ ( 28.0 m s ) ⎛ = ⎜ 10 + ⎜ 9.80 2 ⎟ (175 m ) ⎥ = 2.11 × 106 Pa 3⎟ ⎢ ⎝ ⎝ m ⎠ ⎣⎢ 2 s ⎠ ⎥⎦

! or

9.48

Pgauge = P − Patm = 2.11 MPa = 20.8 atmospheres

(a) Since the tube is horizontal, y1 = y2 and the gravity terms in Bernoulli’s equation cancel, leaving

Topic 9

590 3 1 1 2(P1 − P2 ) 2 (1.20 × 10 Pa ) P1 + ρ v12 = P2 + ρ v22 or v22 − v12 = = 2 2 ρ 7.00 × 102 kg/m 3 ! !

and

2 2 2 2 !v2 − v1 = 3.43 m /s

[1]

From the continuity equation, A1v1 = A2v2, we find

⎛ 2.40 cm ⎞ ⎛A ⎞ ⎛r ⎞ v2 = ⎜ 1 ⎟ v1 = ⎜ 1 ⎟ v1 = ⎜ v1 or ⎝ A2 ⎠ ⎝ r2 ⎠ ⎝ 1.20 cm ⎟⎠ ! v2 = 4v1

[2]

Substituting Equation [2] into [1] yields !15v1 = 3.43 m /s , and v1 = 2

0.478 m/s.

Then, Equation [2] gives v2 = 4(0.478 m/s) = 1.91 m/s.

(b) The volume flow rate is

A1v1 = A2 v2 = (π r22 )ν 2 = π (1.20 × 10−2 m ) (1.91 m/s ) = 8.64 × 10−4 m 3 /s ! 2

9.49

From ΣFy = T − mg − Fy = 0, the balance reading is found to be T = mg + Fy where Fy is the vertical component of the surface tension force. Since this

Topic 9

591

is a two-sided surface, the surface tension force is F = γ(2 L) and its vertical component is Fy = γ(2L)cos φ where φ is the contact angle. Thus, T = mg + 2 γ Lcos φ.

T = 0.40 N when φ = 0°

⇒

mg + 2 γ L = 0.40 N

[1]

T= 0.39 N when φ = 180° ⇒

mg − 2 γ L = 0.39 N

[2]

Subtracting Equation [2] from [1] gives

γ = ! 9.50

0.40 N − 0.39 N 0.40 N − 0.39 N = = 8.3 × 10−2 N/m −2 4L 4 ( 3.0 × 10 m )

Because there are two edges (the inside and outside of the ring), we have

γ =

9.51

F Ltotal F 4π r

F 2(circumference) 1.61 × 10−2 N 4π (1.75 × 10 m) −2

= 7.32 × 10−2 N/m

The total vertical component of the surface tension force must equal the weight of the column of fluid, or F cos φ = γ(2πr) ⋅ cos φ = ρ(πr2)h ⋅ g. Thus,

Topic 9

592

γ =

hρ gr 2cos φ

(2.1 × 10 m)(1 080 kg/m )(9.80 m/s )(5.0 × 10 m) = −2

−4

2cos 0°

= 5.6 × 10−2 N/m

9.52

The blood will rise in the capillary until the weight of the fluid column equals the total vertical component of the surface tension force, or until

ρ(πr2)h ⋅ g = F cos φ = γ(2πr) ⋅ cos φ This gives

h= ! 9.53

2 ( 0.058 N/m ) cos 0° 2γ cos φ = = 5.6 m ρ gr (10.50 kg/m 2 )( 9.80 m/s2 )( 2.0 × 10−6 m )

From the definition of the coefficient of viscosity, η = Fd/Av, the required force is −3 2 η Av (1.79 × 10 N ⋅s/m ) ⎡⎣( 0.800 m ) (1.20 m ) ⎤⎦ ( 0.50 m/s ) F= = = 8.6 N −3 d 0.10 × 10 m !

9.54

From the definition of the coefficient of viscosity, η = Fd/Av, the required force is

Topic 9

593

(

)(

)

−3 2 ⎡ ⎤ η Av 1 500 × 10 N ⋅s/m ⎣ 0.010 m 0.040 m ⎦ 0.30 m/s F= = = 0.12 N d 1.5 × 10−3 m

9.55

Poiseuille’s law gives flow rate = (P1 − P2)πR4/8ηL, and P2 = Patm in this case. Thus, the desired gauge pressure is 2 −5 3 8η L( flow!rate) 8 ( 0.12 N ⋅s/m ) ( 50 m ) ( 8.6 × 10 m /s ) P1 − Patm = = 4 π R4 π ( 0.50 × 10−2 m ) !

9.56

P1 − Patm = 2.1 × 106 Pa = 2.1 MPa

From Poiseuille’s law, the flow rate in the artery is

ΔP ) π R 4 ( flow!rate = = 8η L

( 400 Pa )π ( 2.6 × 10 m ) −3

8 ( 2.7 × 10 N ⋅s/m ) ( 8.4 × 10 m ) −3

−2

= 3.2 × 10−5 m 3 /s

Thus, the flow speed is

flow!rate 3.2 × 10−5 m 3 /s v= = 2 = 1.5 m/s A π ( 2.6 × 10−3 m ) ! 9.57

If a particle is still in suspension after 1 hour, its terminal velocity must be less than

⎛ cm ⎞ ⎛ 1 h ⎞ ⎛ 1 m ⎞ −5 = 5.0 ⎟⎜ ⎟ = 1.4 × 10 m/s ⎜ ⎟⎜ t max h ⎠ ⎝ 3 600 s ⎠ ⎝ 100 cm ⎠ ⎝

(v )

Topic 9

594

Thus, from vt = 2r2g(ρ − ρf)/9η, we find the maximum radius of the particle:

(

max

2g ρ sphere − ρ water

)

( )( ) = 2.8 × 10 m = 2.8 µm 2 ( 9.80 m s ) ⎡(1 800 − 1 000 ) kg m ⎤ ⎣ ⎦

9 1.00 × 10−3 N ⋅s m 2 1.4 × 10−5 m s

9.58

( )

9ηwater v1

rmax =

−6

From Poiseuille’s law, the pressure difference required to produce a given volume flow rate of fluid with viscosity η through a tube of radius R and length L is

ΔP = !

8η L(ΔV / Δt) π R4

If the mass flow rate is (Δm/Δt) = 1.0 × 10−3 kg/s, the volume flow rate of the water is

ΔV Δm / Δt 1.0 × 10−3 kg/s = = = 1.0 × 10−6 m 3 /s 3 3 Δt ρ 1.0 × 10 kg/m ! and the required pressure difference is

ΔP = !

8 (1.0 × 10−3 Pa ⋅s ) ( 3.0 × 10−2 m ) (1.0 × 10−6 m 3 /s )

π ( 0.15 × 10 m ) −3

= 1.5 × 105 Pa

Topic 9

9.59

595

With the IV bag elevated 1.0 m above the needle and atmospheric pressure in the vein, the pressure difference between the input and output points of the needle is

ΔP = (Patm + ρgh) − Patm =ρgh = (1.0 × 103 kg/m3)(9.8 m/s2)(1.0 m)

= 9.8 × 103 Pa

The desired flow rate is

ΔV 500 cm (1 m / 10 cm ) = = 2.8 × 10−7 m 3 /s Δt 30 min ( 60 s 1 min ) ! 3

Poiseuille’s law then gives the required radius of the needle as

⎡ 8η L ( ΔV Δt ) ⎤ R=⎢ ⎥ ⎣ π ( ΔP ) ⎦ ! or

9.60

1/2

⎡ 8 (1.0 × 10−3 Pa ⋅s ) ( 2.5 × 10−2 m ) ( 2.8 × 10−7 m 3 s ) ⎤ =⎢ ⎥ π ( 9.8 × 103 Pa ) ⎢⎣ ⎥⎦

1/2

R = 2.1 × 10−4 m = 0.21 mm

The Reynolds number is

(

)(

)

3 −2 ρ vd 1 050 kg/m 0.55 m/s 2.0 × 10 m RN = = = 4.3 × 103 −3 2 η 2.7 × 10 N ⋅s/m

In this region (RN > 3 000), the flow is turbulent. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 9

9.61

596

The observed diffusion rate is (8.0 × 10–14) / (15 s) = 5.3 × 10–15 kg/s. Then, from Fick’s law, the difference in concentration levels is found to be

C2 − C1 = = !

9.62

( Diffusion!rate ) L DA ( 5.3 × 10−15 kg/s )( 0.10 m )

( 5.0 × 10−10 m 2 s )( 6.0 × 10−4 kg/m 2 )

= 1.8 × 10−3 kg/m 3

Fick’s law gives the diffusion coefficient as D = (Diffusion rate)/[A⋅(ΔC/L)], where ΔC/L is the concentration gradient.

Thus,

9.63

5.7 × 1015 kg/s D= = 9.5 × 10−10 m 2 /s −4 2 −2 4 2.0 × 10 m . 3.0 × 10 kg/m ( )( ) !

Stokes’s law gives the viscosity of the air as

Ft 3.0 × 10−13 N η= = = 1.4 × 10−5 N ⋅s m 2 −6 −4 6π rv 6π ( 2.5 × 10 m ) ( 4.5 × 10 m s ) !

9.64

Using vt = 2r2g(ρ − ρf)/9η, the density of the sphere is found to be ρsphere =

ρwater + 9ηwatervt/2r2g. Thus, if r = d/2 = 0.500 × 10−3 m and vt = 1.10 × 10−2 m/s when falling through 20°C water,

ρ sphere = 1.00 × 10

(

)( ) 2 ( 5.00 × 10 m ) ( 9.80 m/s )

−3 2 −2 kg 9 1.00 × 10 N.s/m 1.10 × 10 m/s

−4

Topic 9

9.65

597

Young’s modulus is defined as Υ = stress/strain = (F/A)(ΔL/L0) =

(F ⋅ L0)/(A ⋅ ΔL) Thus, the elongation of the wire is

)(

(

)(

)

⎡ 200. kg 9.80 m/s 2 ⎤ 4.00 m ⎦ ΔL = = ⎣ = 4.90 × 10−3 m = 4.90 mm −4 2 10 2 A ⋅Y 0.200 × 10 m 8.00 × 10 N/m F ⋅ L0

9.66

(

)(

)

In equilibrium, the tension in the cable equals the load’s weight so that F = mg: Apply the stress-strain relation to find: FL F ΔL =Y → ΔL = 0 = A L YA

(( 3.50 × 10 kg)( 9.80 m/s ))( 25.0 m) 3

( 20 × 10 N/m )( 2.03 × 10 m ) 10

−3

ΔL = 2.11 × 10−3 m

9.67

Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of the plank, separated by distance h = 2.00 m and each with area A = (2.00 cm)(15.0 cm) = 30.0 cm2, move a distance Δx = 5.00 × 10−2 m parallel to each other. The force causing this shearing effect in the plank is the weight of the man F = mg applied perpendicular to the length of the plank at its outer end. Since the shear modulus is given by, the shear modulus for the wood in this plank must be

(80.0 kg )( 9.80 m/s )( 2.00 m ) 2

S= !

( 5.00 × 10 m ) ⎡⎣( 30.0 cm )(1 m / 10 cm )⎤⎦ −2

= 1.05 × 107 Pa

Topic 9

9.68

598

From the definition of bulk modulus, the change in volume will be:

ΔP = −B

ΔV V

( ΔP )V = − (1.01× 10 Pa − 5.00 × 10 Pa )(1.00 × 10 m ) ΔV = − 5

−6

194 × 109 Pa

= 2.58 × 10−8 m 3

9.69

FL0 Using Y = with A = π d2/4 and F = mg, we get A(ΔL) !

Y= ! 9.70

4 ⎡⎣( 90 kg ) ( 9.80 m/s 2 ) ⎤⎦ (50 m)

π (1.0 × 10 m ) (1.6 m) −2

= 3.5 × 108 Pa

(a) The force needed to shear the bolt through its cross-sectional area is F = (A)(stress), or

F = π ( 5.00 × 10−3 m ) ( 4.00 × 108 N/m 2 ) = 3.14 × 10 4 N ! 2

(b) The area over which the shear occurs is equal to the circumference of the hole times its thickness.

Topic 9

599

Thus,

(

( )

)(

)

A = 2π r t = ⎡ 2π 5.00 × 10−3 m ⎤ 5.00 × 10−3 m = 1.57 × 10−4 m 2 ⎣ ⎦

The force required to punch a hole of this area in the 0.500-cm thick steel plate is then

F = (A)(stress) = (1.57 × 10−4 m2)(4.00 × 108 N/m2) = 6.28 × 104 N

9.71

From Y = stress/strain = (stress)(ΔL/L0), the maximum compression the femur can withstand before breaking is

ΔLmax = !

9.72

( stress)max ( Ln )

(1.60 × 10 Pa )( 0.50 m ) = 4.4 × 10 m = 4.4 mm = 6

−3

18 × 10 Pa 9

From Υ = FL0/A(ΔL), the tension needed to stretch the wire by 0.10 mm is

YA(ΔL)

( =

( )( )

Y π r 2 ΔL L0

)(

) (0.10 × 10 m) = 22 N (3.1 × 10 m)

18 × 1010 Pa π 0.11 × 10−3 m

−3

−2

Topic 9

600

The tension in the wire exerts a force of magnitude F on the tooth in each direction along the length of the wire as shown in the above sketch. The resultant force exerted on the tooth has an x-component of Rx = ΣFx = −F cos 30° + F cos 30° = 0, and a y-component of Ry = ΣFy = −F sin 30° − F sin 30° = −F = −22 N.

Thus, the resultant force is  !R = 22 N directed down the page in the diagram.

9.73

(a) When at rest, the tension in the cable equals the weight of the 800.-kg FL0 object, 7.84 × 103 N. Thus, from Y = , the initial elongation of A(ΔL) !

the cable is

(7.84 × 103 N)(25.0 m) = 2.5 × 10−3 m = 2.5 mm F ⋅ L0 ΔL = = A ⋅Y ( 4.00 × 10−4 m 2 ) ( 20 × 1010 Pa ) ! (b) When the load is accelerating upward, Newton’s second law gives the tension in the cable as

F − mg = may, or

F = m(g + ay)

[1]

If m = 800. kg and ay = +3.0 m/s2, the elongation of the cable will be

Topic 9

601

(

)(

)

(

⎡ 800 kg 9.80 + 3.0 m/s 2 ⎤ 25.0 m ⎦ ΔL = == ⎣ −4 2 A ⋅Y 4.00 × 10 m 20 × 1010 Pa F ⋅ L0

(

)(

= 3.2 × 10 m = 3.2 mm −3

)

Thus, the increase in the elongation has been

increase = (ΔL) − (ΔL)initial = 3.2 mm − 2.5 mm = 0.70 mm

Fmax = A ⋅ (stress)max = (4.00 × 10−4 m2)(2.2 × 108 Pa) = 8.8 × 104 N

Then, Equation [1] above gives the mass of the maximum load as

F 8.8 × 10 4 N mmax = max = = 6.9 × 103 kg 2 g + a 9.8 + 3.0 m/s ( ) ! 9.74

(a) From the definition of bulk modulus, the change in volume will be: ΔP = −B

ΔV V0

( ΔP )V = − (1.13 × 10 Pa − 1.01× 10 Pa )(1.00 m ) ΔV = − 8

0.21× 1010 Pa

= −5.4 × 10−2 m 3

(b) The density of seawater at the bottom of the Mariana Trench is

Topic 9

602

ρ=

M M = V V0 + ΔV

→ ρ=

M/V0 ρ0 = = 1.1 × 103 kg/m 3 1 + ΔV /V0 1 − ΔP/B

An alternative solution is

ρ=

9.75

M M 1030 kg = = = 1.1 × 103 kg/m 3 3 −2 3 V V0 + ΔV 1.00 m − 5.4 × 10 m

(

) (

)

The tension and cross-sectional area are constant through the entire length of the rod, and the total elongation is the sum of that of the aluminum section and that of the copper section.

⎡ F (L ) L) ⎤ ( ( ) F ⎢(L ) ⎥ ΔL = ΔL + ΔL = + = + F L0

rod

0 Cu

AYAl

AYCu

0 Al

A ⎢ YAl ⎣

0 Cu

YCu ⎥ ⎦

where A = πr2 with r = 0.20 cm = 2.0 × 10−3 m. Thus,

(5.8 × 10 N) ⎡⎢ 1.3 m + 2.6 m ⎤⎥ = 1.9 × 10 m = 1.9 cm ΔL = π ( 2.0 × 10 m ) ⎢⎣ 7.0 × 10 Pa 11 × 10 Pa ⎥⎦ 3

−2

rod

9.76

−3

The acceleration of the forearm has magnitude Δv 80 km/h (10 m / 1 km ) (1h / 3600 s) a= = = 4.4 × 103 m/s 2 −3 Δt 5.0 × 10 s ! 3

The compression force exerted on the arm is F = ma and the compressional stress on the bone material is

Topic 9

603 3 2 F ( 3.0 kg ) ( 4.4 × 10 m / s ) stress = = = 5.5 × 107 Pa 2 −4 2 2 A 2.4 cm (10 m 1 cm ) !

Since the calculated stress is less than the maximum stress bone material can withstand without breaking, the arm should survive.

9.77

(a) Both iron and aluminum are denser than water, so both blocks will be fully submerged. Since the two blocks have the same volume, they displace equal amounts of water and the buoyant forces acting on the two blocks are equal.

(b) Since the block is held in equilibrium, the force diagram below shows that

ΣFy = 0 ⇒ T = mg − B  The buoyant force !B is the same for the two blocks, so the spring  scale reading !T is largest for the iron block which has a higher

density, and hence weight, than the aluminum block. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 9

604

B = (ρwaterV)g = (1.00 × 103 kg/m3)(0.20 m3)(9.80 m/s2) = 2.0 × 103 N

For the iron block:

Tiron = (ρironV)g − B = (7.86 × 103 kg/m3)(0.20 m3)(9.8 m/s2) − B

Tiron = 1.5 × 104 N − 2.0 × 103 N = 13 × 103 N

For the aluminum block:

Taluminum = (ρaluminumV)g − B = (2.70 × 103 kg/m3)(0.20 m3)(9.8 m/s2) − B

9.78

Taluminum = 5.3 × 103 N − 2.0 × 103 N = 3.3 × 103 N

The average density of either of the two original worlds was

M M 3M ρ0 = = = 3 V 4π R / 3 4π R3 !

The average density of the combined world is

Mtotal 2M 42 (2M) 32M ρ= = = 3 = V′ π ( 32 ) R3 9π R3 4π ⎛ 3 ⎞ ⎜ R⎟ 3 ⎝4 ⎠ !

Topic 9

605

9.79

so !

ρ ⎛ 32M ⎞ ⎛ 4π R3 ⎞ 128 =⎜ = = 4.74 ⎟ ρ0 ⎝ 9π R3 ⎠ ⎜⎝ 3M ⎟⎠ 27

ρ = 4.74 ρ0

Three forces act on the suction cup in equilibrium: the weight equal to 230 times the fish’s body weight and the two pressure forces. The pressure inside the cup will exert a downward force and the external pressure will exert an upward force. Apply Newton’s second law to find:

ΣFy = may −230mg − Pcup Acup + Pambient Acup = 0

(

)(

)

230 30.0 × 10−3 kg 9.80 m/s 2 230mg Pcup /Pambient = 1− = 1− Pambient Acup 1.10 × 105 N/m 2 15.0 × 10−4 m 2

(

)(

)

Pcup /Pambient = 0.59 9.80

(a) Starting with P = P0 + ρgh, we choose the reference level at the level of the heart, so P0 = PH. The pressure at the feet, a depth hH below the reference level in the pool of blood in the body, is PF = PH + ρghH. The pressure difference between feet and heart is then PF − PH = ρghH .

(b) Using the result of part (a),

PF − PH = (1.05 × 103 kg/m3)(9.80 m/s2)(1.20 m) = 1.23 × 104 Pa

9.81

The cross-sectional area of the aorta is !A1 = π d1 / 4 and that of a single 2

Topic 9

606

capillary is !Ac = π d2 / 4 . If the circulatory system has N such capillaries, 2

the total cross-sectional area carrying blood from the aorta is 2 !A2 = NAc = Nπ d2 / 4 .

From the equation of continuity, A2 = (v1/v2)A1, or

Nπ d22 ⎛ v1 ⎞ π d12 =⎜ ⎟ ⎝ v2 ⎠ 4 ! 4

yielding

⎛ v1 ⎞ ⎛ d1 ⎞ ⎛ 1.0 m/s ⎞ ⎛ 0.50 × 10−2 m ⎞ N=⎜ ⎟⎜ ⎟ =⎜ = 2.5 × 107 −2 −6 ⎟ ⎜ ⎟ ⎝ v2 ⎠ ⎝ d2 ⎠ ⎝ 1.0 × 10 m/s ⎠ ⎝ 10 × 10 m ⎠ ! 9.82

(a) We imagine that a superhero is capable of producing a perfect vacuum above the water in the straw. Then P = P0 + ρgh, with the reference level at the water surface inside the straw and P being atmospheric pressure on the water in the cup outside the straw, gives the maximum height of the water in the straw as

hmax =

Patm − 0

ρ water g

Patm

water

(

1.013 × 105 N m 2

)(

1.00 × 103 kg m 3 9.80 m s 2

)

= 10.3 m

(b) The Moon has no atmosphere so Patm = 0, which yields hmax = 0.

Topic 9

9.83

607

(a) P = 160 mm of H 2 O = ρ H O g(160 mm) 2

kg ⎞ ⎛ m⎞ ⎛ = ⎜ 103 3 ⎟ ⎜ 9.80 2 ⎟ ( 0.160 m ) = 1.57 kPa m ⎠⎝ s ⎠ ! ⎝ ⎛ ⎞ 1 atm P = (1.57 × 103 Pa ) ⎜ = 1.55 × 10−2 atm 5 ⎟ 1.013 × 10 Pa ⎝ ⎠ !

The pressure is !P = ρH2O ghH2O = ρHg ghHg , so

⎛ ρH O ⎞ ⎛ 103 kg/m 3 ⎞ hHg = ⎜ 2 ⎟ hH2O = ⎜ 160 mm ) = 11.8 mm!of!Hg 3 3 ( ρHg ⎠ ⎝ 13.6 × 10 kg/m ⎟⎠ ⎝ ! (b) The fluid level in the tap should rise.

9.84

When the rod floats, the weight of the displaced fluid equals the weight of the rod, or (ρfVdisplaced)g = (ρ0Vrod)g. Assuming a cylindrical rod, Vrod =

πr2L. The volume of fluid displaced is the same as the volume of the rod that is submerged

Vdisplaced = π r2(L − h).

Thus, ρf[π r2(L − h)]g = ρ0[π r2L]g which reduces to ρf = ρ0L/(L − h).

Topic 9

9.85

608

Consider the diagram and apply Bernoulli’s equation to points A and B, taking y = 0 at the level of point B, and recognizing that vA ≈ 0. This gives

1 PA + 0 + ρw g(h − Lsin θ ) = PB + ρw vB2 + 0 2 !

Recognize that PA = PB = Patm since both points are open to the atmosphere. Thus, we obtain !vB = 2g(h − Lsin θ ) . Now the problem reduces to one of projectile motion with v0y = vB sin θ. At the top of the arc vy = 0, y = ymax, and !vy = v0y + 2ay (Δy) gives 2

ymax − 0 = !

2 vy2 − v0y

2ay

0 − vB2 sin 2 θ 2g (h − Lsin θ )sin θ = = 2(−g) 2g 2

ymax = [10.0 m − (2.00 m) sin 30.0°]sin2 30.0° = 2.25 m above the level of point B © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 9

9.86

609

When the balloon comes into equilibrium, the weight of the displaced air equals the weight of the filled balloon plus the weight of string that is above ground level. If ms and L are the total mass and length of the string, the mass of string that is above ground level is (h/L)ms. Thus, (ρairVballoon)g = mballoong + (ρheliumVballoon)g + (h/L)msg, which reduces to

⎡ ( ρ − ρ helium ) Vballoon − mballoon ⎤ h = ⎢ air ⎥L ms ⎣ ⎦ ! This yields

⎡ (1.29 kg/m 3 − 0.179 kg/m 3 ) ⎡ 4π ( 0.40 m )3 / 3 ⎤ − 0.25 kg ⎤ ⎣ ⎦ ⎥ ( 2.0 m ) = 1.9 m h=⎢ ⎢ ⎥ 0.050 kg ⎦ ! ⎣

9.87

Four forces are acting on the balloon: an upward buoyant force exerted by the surrounding air, B = (ρairVballoon)g; the downward weight of the balloon envelope, Fg, balloon = mg; the downward weight of the helium filling the balloon, Fg, He = (ρHeVballoon)g; and the downward spring force, Fs = k|Δx|. At equilibrium, |Δx| = L, and we have

ΣFy = 0 ⇒ B − Fs − Fg, balloon − Fg, He = 0

Topic 9

610

Fs = kL = B − Fg, balloon − Fg, He = (ρairVballoon)g − mg − (ρHeVballoon)g

⎡( ρ − ρHe ) Vballoon − m ⎤⎦ g and L = ⎣ air k ! This yields

{⎡⎣(1.29 − 0.179) kg/m ⎤⎦ ( 5.00 m ) − 2.00 × 10 kg}( 9.80 m/s ) L= 3

! or

−3

90.0 N/m

L = 0.605 m

9.88

(a) Consider the pressure at points A and B in part (b) of the figure by applying P = P0 + ρfgh. Looking at the left tube gives PA = Patm +

Topic 9

611

ρwaterg(L − h), and looking at the tube on the right, PB = Patm + ρoilgL. Pascal’s principle says that PB = PA. Therefore, Patm + ρoilgL = Patm +

ρwaterg(L − h), giving

⎛ ⎛ ρ ⎞ 750 kg/m 3 ⎞ h = ⎜ 1 − oil ⎟ L = ⎜ 1 − 5.00 cm = 1.25 cm 3⎟ ρ 1 000 kg/m ⎝ ⎠ ⎝ water ⎠

(

)

(b) Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid levels in the two tubes. First, apply Bernoulli’s equation to points A and B. This gives 1 1 PA + ρair v A2 + ρair gy A = PB + ρair vB2 + ρair gyB . Since yA = yB, vA = v, and 2 2 !

vB = 0, this reduces to

1 PB − PA = ρair v 2 2 !

[1]

Now use P = P0 + ρfgh to find the pressure at points C and D, both at the level of the oil−water interface in the right tube. From the left tube, PC = PA + ρwatergL, and from the right tube, PD = PB + ρoilgL.

Pascal’s principle says that PD = PC, and equating these two gives PB +

ρoilgL = PA + ρwatergL, or

Topic 9

612

PB − PA = (ρwater − ρoil)gL

[2]

Combining Equations [1] and [2] yields

9.89

(

)

2 ρ water − ρ oil gL

ρ air

(

)(

2(1 000 − 750) 9.80 m/s 2 5.00 × 10−2 m 1.29

) = 13.8 m/s

The pressure on the surface of the two hemispheres is constant at all points, and the force on each element of surface area is directed along the radius of the hemispheres. The applied force along the horizontal axis must balance the net force on the “effective” area, which is the crosssectional area of the sphere,

Aeffective = π R2

and

9.90

F = PgaugeAeffective = (P0 − P)π R2

Since the block is floating, the total buoyant force must equal the weight

Topic 9

613

of the block. Thus,

ρoil = ⎡⎣ A ( 4.00 cm − x ) ⎤⎦ g + ρwater [ A ⋅ x ] g !

= ρwood ⎡⎣ A ( 4.00 cm ) ⎤⎦ g

where A is the surface area of the top or bottom of the rectangular block.

Solving for the distance x gives

⎛ρ ⎛ 960 − 930 ⎞ − ρ oil ⎞ x = ⎜ wood ⎟ 4.00 cm = ⎜ ⎟ 4.00 cm = 1.71 cm ⎝ 1 000 − 930 ⎠ ⎝ ρ water − ρ oil ⎠

(

9.91

)

(

)

A water droplet emerging from one of the holes becomes a projectile with v0y = 0 and v0x = v. The time for this droplet to fall distance h to the floor is 1 2 found from Δy = v0yt + ayt !to!be!t = 2h / g 2 !

Topic 9

614

The horizontal range is !R = vt = v 2h / g . If the two streams hit the floor at the same spot, it is necessary that R1 = R2, or

v1 !

2h1 2h2 = v2 g g

With h1 = 5.00 cm and h2 = 12.0 cm, this reduces to v = v2 h2 / h1 = v2 12.0 cm / 5.00 cm , or !1

!v1 = v2 2.40

[1]

Apply Bernoulli’s equation to points 1 (the lower hole) and 3 (the surface of the water). The pressure is atmospheric pressure at both points and, if the tank is large in comparison to the size of the holes, v3 ≈ 0. Thus,

Topic 9

615

1 Patm + pv12 + ρ gh1 = Patm + 0 + ρ gh3 , or 2 !

v 2 = 2g ( h3 − h1 ) !1

[2]

Similarly, applying Bernoulli’s equation to point 2 (the upper hole) and

1 point 3 gives Patm + pv22 + ρ gh2 = ρ atm + 0 + ρ gh3 , or 2

v22 = 2g ( h3 − h2 ) !

[3]

Square Equation [1] and substitute from Equations [2] and [3] to obtain

2g(h3 − h1) = 2.40[2g(h3 − h2)]

Solving for h3 yields

h3 = !

2.40h2 − h1 2.40(12.0 cm) − 5.00 cm = = 17.0 cm 1.40 1.40

so the surface of the water in the tank is 17.0 cm above floor level.

Topic 10

616

Topic 10 Thermal Physics

QUICK QUIZZES 10.1

Choice (c). When two objects having different temperatures are in thermal contact, energy is transferred from the higher temperature object to the lower temperature object. As a result, the temperature of the hotter object decreases and that of the cooler object increases until thermal equilibrium is reached at some intermediate temperature.

10.2

Choice (b). The glass surrounding the mercury expands before the mercury does, causing the level of the mercury to drop slightly. The mercury rises after it begins to get warmer and approaches the temperature of the hot water, because its coefficient of expansion is greater than that for glass.

10.3

Choice (c). Gasoline has the highest coefficient of expansion so it undergoes the greatest change in volume per degree change in temperature.

10.4

Choice (c). A cavity in a material expands in exactly the same way as if the cavity were filled with the surrounding material. Thus, both spheres

Topic 10

617

will expand by the same amount. 10.5

Unlike land-based ice, ice floating in the ocean already displaces a quantity of liquid water whose weight equals the weight of the ice. This is the same situation as will exist after the ice melts, so the melting oceanbased ice will not change ocean levels much.

10.6

Choice (b). Since the two containers are at the same temperature, the average kinetic energy per molecule is the same for the argon and helium gases. However, helium has a lower molar mass than does argon, so the rms speed of the helium atoms must be higher than that of the argon atoms.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 10.2 (a) From the ideal gas law, pressure is linearly proportional to temperature. Therefore doubling the temperature results in a doubling of the pressure: Tf /Ti = Pf /Pi = 2 . (b) The average kinetic energy per molecule is also linearly proportional to the temperature, so doubling the temperature doubles the average molecular kinetic energy and Tf /Ti = KEf /KEi = 2 .

Topic 10

618

v 2 , is proportional to the square root of the temperature:

Tf /Ti = KE f /KEi = v 2f /vi2 =

2 .

(d) The internal energy of a monatomic gas is U = 32 nRT . The internal energy is proportional to the temperature so that doubling the temperature results in a doubling of the internal energy: Tf /Ti = Uf /Ui = 10.4

2 .

The pressure inside the balloon is greater than the ambient atmospheric pressure because the pressure inside must not only resist the external pressure, but also the force exerted by the elastic material of the balloon.

10.6

From the ideal gas law, the number of moles is proportional to the product of pressure and volume. If each of these doubles while the temperature is constant, the number of moles must have increased by a factor of 2 × 2 = 4 .

10.8

The measurements are too short. At 22°C the tape would read the width of the object accurately, but an increase in temperature causes the divisions ruled on the tape to be farther apart than they should be. This “too long” ruler will, then, measure objects to be shorter than they really

Topic 10

619

are. 10.10 (a) Average kinetic energy is proportional to the gas temperature. Therefore two different gas constituents at the same temperature have the same kinetic energy. (b) At temperature T, the rms speed varies as 1 over the square root of the mass. Nitrogen, the less massive constituent, will therefore have the higher rms speed. 10.12

Doubling the volume while reducing the pressure by half results in no change in the quantity PV that appears in the ideal gas law equation. Consequently, the temperature and hence the internal energy remain the same.

10.14

As the washer is heated, both the inner and outer radii expand. Therefore the correct answer is a.

ANSWERS TO EVEN NUMBERED PROBLEMS 10.2

(a)

−251°C

(b)

1.36 atm

10.4

(a)

56.7°C, −62.1°C

(b)

330 K, 211 K

10.6

(a)

−270°C

(b)

1.27 atm

(c)

1.74 atm

Topic 10

620

10.8

(a)

31.7°C

(b)

31.7 K

10.10

(a)

TR = TF + 459.67

(b)

5 TK = TR 9 !

10.12

(a)

L = 1.3 m − 0.49 mm (b)

The clock will run fast.

10.14

(a)

2.542 cm

(b)

3.0 × 102 °C

10.16

1.70 × 10−5 (°C)−1

10.18

(a) 396 N

(b)

−101°C

(c) The initial length of the wire cancels during the calculation, so the results are not affected by doubling the length of the bridge. 10.20

See Solution.

10.22

18.702 m

10.24 (a) (b)

1.5 km Accordion-like expansion joints are placed in the pipeline at periodic intervals.

10.26

(a)

27.7 kg

(b)

1.02 m3

(d)

27.2 kg

(e)

0.5 kg

(b)

5.52 × 1024

10.28

1.2 cm

10.30

(a) 9.17 mol

(c)

716 kg/m3

Topic 10

621

10.32

2.00 m3

10.34

(a)

3.95 atm = 400 kPa

(b)

4.43 atm = 449 kPa

10.36

(a)

3.00 mol

(b)

1.81 × 1024 molecules

10.38

0.131 kg/m3

10.40

0.704 atm

10.42

712 m/s

10.44

(a) 1.10 × 104 J

(b)

1.10 × 104 J

(d) 9.14 × 105 Pa

(e)

9.13 × 105 Pa

( ) = 1.03 v ( Cl )

(b)

The less massive atom, 35Cl, moves faster.

385 K

(b)

7.97 × 10−21 J

10.46

(a)

vmax 35 Cl 37

(c)

6.07 × 10−21 J

max

10.48 (a)

18 kPa

10.52

(a)

0.176 mm

(b)

8.78 µm

10.54

(a)

0.34%

(b)

0.48%

(c)

The moment of inertia for each of the shapes has the same

(c)

9.30 × 10−8 m3

Topic 10

622

mathematical form: the product of a constant, the mass, and the square of a length. 10.56

28 m

10.58

(a)

See Solution.

(c)

The equilibrium value of h increases as the temperature T increases.

(a)

292 K

(b)

7.83 mol

(c)

(d)

345 g

(e)

516 K, 5.98 mol

(f) Pf = (nfTf /niTi)Pi

10.60

(b)

greater than atmospheric pressure

44.0 g/mol

(g) 1.28 × 106 Pa 10.62

(a)

343 K

(b)

12.5%

10.64

(a)

16.9 cm

(b)

1.35 × 105 Pa

10.66

(a)

6.0 cm

(b)

See Solution.

1.15 atm

Topic 10

623

PROBLEM SOLUTIONS 10.1

9 9 (a) TF = TC + 32 = (−273.15) + 32 = −460°F 5 5 ! 5 5 (b) TC = (TF − 32) = (98.6 − 32) = 37°C 9 9 ! 9 9 9 (c) TF = TC + 32 = (TK − 273.15) + 32 = (−173.15) + 32 = −280°F 5 5 5 !

10.2

When the volume of a low-density gas is held constant, pressure and temperature are related by a linear equation P = AT + B, where A and B are constants to be determined. For the given constant-volume gas thermometer, P = 0.700 atm when T = 100°C ⇒ 0.700 atm = A(100°C) + B

[1]

P = 0.512 atm when T = 0°C ⇒ 0.512 atm = A(0) + B

[2]

From Equation [2], B = 0.512 atm. Substituting this result into Equation [1] yields

A= !

0.700 atm − 0.512 atm = 1.88 × 10−3 atm/°C 100°C

so, the linear equation for this thermometer is: P = (1.88 × 10−3 atm/°C)T + 0.512 atm

Topic 10

624

(a) If P = 0.040 0 atm, then

P−B A

0.040 0 atm − 0.512 atm 1.88 × 10−3 atm / °C

= −251°C

(b) If T = 450°C, then P = (1.88 × 10−3 atm/°C)(450°C) + 0.512 atm = 1.36 atm 10.3

(a) TC = TK − 273.15 = 20.3−273.15 = −253°C 9 9 (b) TF = TC + 32 = (−253) + 32 = −423°F 5 5 !

10.4

5 5 (a) TC = (TF − 32) = (134 − 32) = 56.7°C , and 9 9 5 TC = (−79.8 − 32) = −62.1°C 9 !

(b) T = TC + 273.15 = 56.7 + 273.15 = 330 K , and TK = TC + 273.15 = −62.1 + 273.15 = 211 K 10.5

Use the relations TC = T − 273.15 and TC = 59 (TF − 32 ) to see that ΔT = 59 ΔTF (where T is the Kelvin temperature and TF is the Fahrenheit temperature). Substitute ΔTF = TFf – TFi = 45.0°F −(−4.00°F) = 49.0°F to find that

ΔT = 59 ΔTF = 59 ( 49.0°F ) = 27.2 K 10.6

Since we have a linear graph, we know that the pressure is related to the

Topic 10

625

temperature as P = A + BTC, where A and B are constants. To find A and B, we use the given data: 0.900 atm = A + B(−78.5°C)

[1]

and 1.635 atm = A + B(78.0°C)

[2]

Solving Equations [1] and [2] simultaneously, we find: A = 127 atm, and Therefore,

B = 4.70 × 10−3 atm/°C

P = 1.27 atm + (4.70 × 10−3 atm/°C)TC

(a) At absolute zero the gas exerts zero pressure (P = 0), so

−1.27 atm TC = = −270°C −3 4.70 × 10 atm/°C ! (b) At the freezing point of water, TC = 0°C and P = 1.27 atm + 0 = 1.27 atm (c) At the boiling point of water, TC = 100°C, so P = 1.27 atm + (4.70 × 10−3 atm/°C)(100°C) = 1.74 atm 10.7

Use the relation TC = 59 (TF − 32 ) to realize that ΔTC = 59 ΔTF . For ΔTF = 3.0°F,

ΔTC = 1.67°C . Alternatively, calculate that 98.6°F = 37.0°C and 101.6°F =

Topic 10

626

38.67°C so that ΔTC = 1.67°C as before. 10.8

(a) As in the solution to Problem 10.7 above, 5 5 ΔTC = (ΔTF ) = (57.0)°C = 31.7°C 9 9 ! 5 (b) ΔTK = (TC ,out + 273.15 ) − (TC ,in + 273.15 ) = (TC ,out − TC ,in ) = ΔTC = 31.7 K 9 !

10.9

9 9 (a) TF = TC + 32 = (41.5) + 32 = 107°F 5 5 !

(b) Yes The normal body temperature is 98.6°F, so this patient has a high fever and needs immediate attention. 10.10

(a) Since temperature differences on the Rankine and Fahrenheit scales are identical, the temperature readings on the two thermometers must differ by no more than an additive constant (i.e., TR = TF + constant). To evaluate this constant, consider the temperature readings on the two scales at absolute zero. We have TR = 0°R at absolute zero, and 9 9 TF = TC + 32.00 = (−273.15) + 32.00 = −459.67°F 5 5 !

Substituting these temperatures in our Fahrenheit to Rankine conversion gives

Topic 10

627

0° = −459.67°+ constant or constant = 459.67° giving

TR = TF + 459.67

(b) We start with the Kelvin temperature and convert to the Rankine temperature in several stages, using the Fahrenheit to Rankine conversion from part (a) above.

(9 T − 32.00) + 273.15 = 59 ⎡⎣(T − 459.67 ) − 32.00⎤⎦ + 273.15 5 5 5 5 = (T − 491.67 ) + 273.15 = T − ( 491.67 ) + 273.15 = T − 273.15 + 273.15 9 9 9 9

TK = TC + 273.15 =

10.11

5 TK = TR 9 !

The increase in temperature is ∆T = 35°C−(−20°C) = 55°C. Thus, ΔL = αL0(ΔT) = [11 × 10−6 (°C)−1](518 m)(55°C) = 0.31 m = 31 cm

10.12 (a) As the temperature drops by 20°C, the length of the pendulum changes by ΔL = αL0(ΔT) = [19 × 10−6(°C)−1](1.3 m)(−20°C) or ΔL = −4.9 × 10−4 m = −0.49 mm Thus, the final length of the rod is L = 1.3 m − 0.49 mm. (b) From the expression for the period, T = 2π L/g , we see that as the

Topic 10

628

length decreases the period decreases. Thus, the pendulum will swing too rapidly and the clock will run fast. 10.13

We choose the radius as our linear dimension. Then, from ΔL = αL0(ΔT),

ΔT = TC − 20.0°C = !

L − L0 2.21 − 2.20 cm = = 35.0°C ΔL0 ⎡1.30 × 10−4 ( °C )−1 ⎤ ( 2.20 cm ) ⎣ ⎦

or TC = 55.0°C 10.14 (a) The diameter is a linear dimension, so we consider the linear expansion of steel: d = d0[1 + α(ΔT) = (2.540 cm)[1 + (11 × 10−6)(°C)−1(100.0°C − 25.00°C )] = 2.542 cm (b) If the volume increases by 1.000%, then ∆V = (1.000 × 10−2)V0. Then, using ∆V = βV0(∆T), where β = 3α is the volume expansion coefficient, we find

ΔT =

10.15

ΔV/V0

1.000 × 10−2 3[11 × 10 (°C) ] −6

−1

= 3.0 × 102 °C

From ∆L = L − L0 = αL0(∆T), the final value of the linear dimension is L = L0 + αL0(∆T). To remove the ring from the rod, the diameter of the ring

Topic 10

629

must be at least as large as the diameter of the rod. Thus, we require that LBrass = LAl, or (L0)Brass+αBrass(L0)Brass(∆T) = (L0)Al + αAl +(L0)Al (∆T)

(L0 )Al − (L0 )Brass ΔT = α Brass (L0 )Brass − α Al (L0 )Al !

This gives

(a) If (L0)Al = 10.01 cm,

ΔT = !

⎡19 × 10−6 ( °C ) ⎣

−1

10.01 − 10.00 = −199°C ⎤ (10.00 ) − ⎡ 24 × 10−6 ( °C )−1 ⎤ (10.01) ⎦ ⎣ ⎦

so T = T0 + ΔT = 20.0°C − 199°C = −179°C which is attainable (b) If (L0)Al = 10.02 cm

ΔT = !

⎡19 × 10 ⎣

−6

(°C)

−1

10.02 − 10.00 = −396°C ⎤ (10.00 ) − ⎡ 24 × 10−6 ( °C )−1 ⎤ (10.02 ) ⎦ ⎣ ⎦

and T = T0 + ΔT = −376°C which is below absolute zero and unattainable 10.16 Use the defining equation for linear expansion to find

ΔL = α L0 ΔT → α =

α = 1.70 × 10−5 ( °C )

ΔL 1.19 × 10−2 m = L0 ΔT ( 25.0 m ) ( 30.0°C − 2.00°C )

−1

4 3 10.17 The density of lead at 20.0°C is ρ 0 = M/V0 = 1.13 × 10 kg/m . As the

temperature increases from 20.0°C to 105°C, the volume will expand to V © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 10

630

= V0 + ΔV where ΔV = βV0ΔT. With the mass unchanged, the density will become

ρ=

M M M = = V V0 + ΔV V0 (1 + βΔT )

Substitute ρ0 = M/V0, β = 3α = 8.7 × 10−5 (°C)−1, and ΔT = 105°C − 20.0°C = 85°C to find

ρ=

ρ0 1.13 × 10 4 kg/m 3 = = 1.12 × 10 4 kg/m 3 1 + βΔT 1 + 8.7 × 10−5 ( °C )−1 ( 85°C )

(

)

Alternatively, the mass of lead is M = ρ0V0 = ρV. Substitute V = V0 + ΔV where ΔV = βV0ΔT to find

ρ 0V0 = ρ (V0 + ΔV ) = ρ (V0 + βV0 ΔT ) ρ 0V0 = ρV0 (1 + βΔT ) → ρ =

ρ0 1 + βΔT

as before. 10.18 (a) When a wire undergoes a decrease in temperature of magnitude |∆T|, it will attempt to contract by an amount |∆L| = αL0|∆T|. If the ends of the wire are held fixed, not allowing it to contract, the wire will develop a tension sufficient to stretch it by the amount of its normal contraction, |∆L|. This tension is

⎛ α L0 | ΔT|⎞ ⎛ ΔL ⎞ F = YA ⎜ ⎟ = YA ⎜ ⎟ = YAα |ΔT| L0 ⎝ L0 ⎠ ⎝ ⎠ ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 10

631

where Y is Young’s modulus for the wire material and A is the crosssectional area of the wire. For the given steel wire, with |ΔT| = |−10.0−35.0|°C, the tension that develops in the wire is F = (20.0 × 1010 N/m2)(4.00 × 10−6 m2)[11.0 × 10−6(°C)−1](45.0°C) = 396 N (b) As the temperature decreases while the wire is prevented from contracting, the stress that develops in the wire is stress = (F/A) = Yα|ΔT|. The decrease in temperature required to reach the elastic limit is

( stress)limit

ΔT =

Yα

=−

3.00 × 108 N / m 2 = −136°C ( 20.0 × 1010 N / m 2 ) ⎡⎣11 × 10−6 (°C)−1 ⎤⎦

Thus, the temperature of the wire when it reaches its elastic limit is T = T0 + ΔT = 35.0°C+(−136°C) = 101 °C. (c) Observe that the initial length of the wire cancels out in the calculations of parts (a) and (b). Thus, the results obtained would not be changed if the initial length of the wire were doubled. 10.19

The difference in Celsius temperature in the underground tank and the tanker truck is 5 5 ΔTC = (ΔTF ) = (95.0 − 52.0) = 23.9°C 9 9 !

Topic 10

632

If V52°F is the volume of gasoline that fills the tank at 52.0°F, the volume this quantity of gas would occupy on the tanker truck at 95.0°F is V95°F = V52°F + ΔV = V52°F + βV52°F(ΔT) = V52°F[1 + β(ΔT) = (1.00 × 103 gal)[1 + (9.6 × 10−4(°C)−1(23.9°C)] = 1.02 × 103 gal 10.20

Consider a regular solid with initial volume given by V0 = A0L0 at temperature T0. Here, A is the cross-sectional area and L is the length of the regular solid. When temperature undergoes a change ΔT = T − T0, the change in the cross-sectional area is ΔA = A − A0 = γA0(ΔT) = 2αA0(ΔT), giving A = A0 + 2αA0(ΔT). Similarly, the new length will be L = L0 + αL0(ΔT), so the new volume is V = [A0 + 2αA0(ΔT)][L0 + αL0(ΔT)] = A0L0 + 3αA0L0(ΔT) + 2α 2A0L0(ΔT)2 The term involving α 2 is negligibly small in comparison to the other terms, so V ≈ A0L0 + 3αA0L0(ΔT) = V0 + 3αV0(ΔT) This is of the form ΔV = V − V0 = βV0(ΔT) where β = 3α .

10.21 (a) The volume of turpentine that overflows as the temperature rises

Topic 10

633

equals the difference in the increase in the volume of the turpentine and the increase in the volume of the aluminum container. Voverflow = ΔVT − ΔVAl = βTV0(ΔT) − βAlV0(ΔT) = (βT − βAl)V0(ΔT) = (βT − 3αAl)V0(ΔT) = [9.0 × 10−4(°C−1) − 3(24 × 10−6(°C−1))](2.000 L)(80.0°C − 20.0°C) or

Voverflow = 9.9 × 10−2 L = 99 mL

(b) The volume of turpentine remaining in the cylinder at 80.0°C is the same as the volume of the aluminum cylinder at 80.0°C. This is V1 = V0[1 + 3αAl(ΔT)] = (2.000 L)[1 + 3(24 × 10−6(°C−1))(80.0 − 20.0) °C] = 2.000 L + 0.008 6 L = 2.009 L or

V1 = Vturpentine remaining = 2.009 L

(c) The volume of the aluminum cylinder when cooled back to 20.0°C will be V0 = 2.000 L. The volume of the remaining turpentine when cooled to 20.0°C will be V2 = V1[1 + βT(ΔT) = (2.009 L)[1 + (9.0 × 10−1(°C−1))(20.0 − 80.0)°C] = 2.009 L − 0.11 L = 1.90 L The fraction of the cylinder’s volume that is now empty will be

Topic 10

634

fraction!empty = !

V0 − V2 2.000 L − 1.90 L = = 5.00 × 10−2 V0 2.000 L

so, the empty height above the remaining turpentine at 20.0°C is hempty = h(fraction empty) = (20.0 cm)(5.00 × 10−2) = 1.00 cm 10.22

[Note to Instructor: Some rules concerning significant figures are deliberately violated in this solution to better illustrate the method of solution.] Let L be the final length of the aluminium column. This will also be the final length of the quantity of tape now stretching from one end of the column to the other. In order to determine what the scale reading now is, we need to find the initial length this quantity of tape had at 21.2°C (when the scale markings were presumably put on the tape). Thus, we let this initial length of tape be (L0)tape and require that L = (L0)tape[1 + αsteel(ΔT)] = (L0)column[1 + αAl(ΔT)] which gives

(L0 )tape

(L ) =

( L0 )tape = !

( ) ( )

⎡1 + α ΔT ⎤ Al ⎣ ⎦ 1 + α steel ΔT

0 column

(18.700 m ) ⎡⎣1 + ( 24 × 10 (°C) )( 29.4°C − 21.2°C)⎤⎦ −6

(

1 + 11 × 10−6 ( °C )

−1

)( 29.4°C − 21.2°C)

= 18.702 m

The measured length of the column, according to the markings on the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 10

635

tape, at 29.4°C is therefore 18.702 m. 10.23

If allowed to do so, the amount the band (with initial length L0)would contract as it cools to 37°C is ΔL = αL0|ΔT|. Since the band is not allowed to contract, it will develop a tensile stress given by

⎛ α L0 |ΔT|⎞ ⎛ ΔL ⎞ Stress = Y ⎜ ⎟ = Y ⎜ ⎟ = Yα |ΔT| L0 ⎝ L0 ⎠ ⎝ ⎠ ! If A = (height ⋅ thickness) = (4.0 mm)(0.50 mm) = 2.0 × 10−6 m2 is the crosssectional area of the band, the tension in the band will be F = A ⋅(Stress) ⎛ N⎞ = (2.0 × 10−6 m 2 ) ⎜ 18 × 1010 2 ⎟ 17.3 × 10−6 °C−1 80°C − 37°C m ⎠ ⎝

(

)(

)

= 2.7 × 102 N

10.24 (a) The expansion of the pipeline will be ΔL = αL0(ΔT), or ΔL = [11 × 10−6 (°C)−1](1 300 km)[35°C − (−73°C)] = 1.5 km (b) This is accommodated by accordion-like expansion joints placed in the pipeline at periodic intervals. 10.25

The drum and the carbon tetrachloride, both having an initial volume of V0 = 50.0 gal, expand at different rates as the temperature rises by ΔT = 20.0°C From ΔV = βV0(ΔT), with β = 3α as the coefficient of volume

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expansion for the steel drum, we obtain

⎛ ⎞ Vspillage = ΔVcarbon − ΔVsteel = ⎜ β carbon − 3α steel ⎟ V0 (ΔT) ⎝ tetrachloride ⎠ tetrachloride drum ! or

Vspillage = [5.81 × 10−4(°C)−1 − 3(11 × 10−6(°C)−1)](50.0 gal)(20.0°C) = 0.548 gal

10.26

⎛ 3.80 × 10−3 m 3 ⎞ 2 3 (a) m0 = ρ0V = (7.30 × 10 kg / m ) (10.0 gal ) ⎜ ⎟⎠ = 27.7 kg 1 gal ⎝ !

(b) V = V0 + ΔV = V0 + βV0(ΔT) = V0[1 + β (ΔT)] or

V = (1.000 m3)[1+ (9.60 × 10−4(°C)−1(20.0°C)] = 1.02 m3

(c) Gasoline having a mass of m = 7.30 × 102 kg occupies a volume of V0 = 1.000 m3 at 0°C and a volume of V0 = 1.02 m3 at 20.0°C. The density of gasoline at 20.0°C is then

m 7.30 × 102 kg ρ20 = = = 716 kg/m 3 3 V 1.02 m ! ⎛ 3.80 × 10−3 m 3 ⎞ 2 3 (d) m20 = ρ20V = (7.16 × 10 kg / m ) (10.0 gal ) ⎜ ⎟⎠ = 27.2 kg 1 gal ⎝ !

(e) ∆m = m0 − m20 = 27.7 kg − 27.2 kg = 0.5 kg 10.27 (a) The gap width is a linear dimension, so it increases in “thermal enlargement” as the temperature goes up. The gap expands in the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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same way the material removed to create the gap would have expanded. (b) At 190°C, the length of the piece of steel that is missing, or has been removed to create the gap, is L = L0 + ΔL = L0[1 + α(ΔT)]. This gives L = (1.600 cm)[1 + (11 × 10−6(°C−1))(190°C − 30.0°C)] = 1.603 cm 10.28

Each slab will undergo an increase in length of ΔL = αL0(ΔT), and the gap between successive slabs must be at least this wide to accommodate this expansion. Thus, the minimum gap size should be ΔL = αconcreteL0(T − T0) = (12 × 10−6(°C−1))(25.0 m)(50.0°C − 10.0°C) or

ΔL = 1.2 × 10−2 m = 1.2 cm

10.29 (a) Consult a periodic table to find the molar mass of copper is MCu = 63.546 g/mol. The number of moles n in a 12.5 g samples is then

12.5 g = 0.197 mol 63.546 g/mol

(b) The number of copper atoms is

(

N = nN A = ( 0.197 mol ) 6.02 × 10 23 atoms/mol

)

= 1.18 × 10 23 atoms 10.30 (a) The molecular mass of formaldehyde (CH2O) is Mmolecule = MC + 2MH + MO = 12.0 g/mol + 2.0 g/mol + 16.0 g/mol = 30.0 g/mol. The number of © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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moles in 275 g of formaldehyde is then

275 g = 9.17 mol 30.0 g/mol

(b) The number of CH2O molecules is

(

N = nN A = ( 9.17 mol ) 6.02 × 10 23 molecules/mol

)

= 5.52 × 10 24 molecules 10.31 (a) From the ideal gas law, PV = nRT, we find P/T = nR/V. Thus, if both n and V are constant as the gas is heated, the ratio P/T is constant, giving Pf /Tf = Pi /Ti, or ⎛ Pf ⎞ ⎛ 3P ⎞ Tf = Ti ⎜ ⎟ = ( 300 K ) ⎜ i ⎟ = 900 K = 627°C ⎝ Pi ⎠ ⎝ Pi ⎠ !

(b) If both pressure and volume double as n is held constant, the ideal gas law gives

( )

⎛ 2P (2V ) ⎞ ⎛PV ⎞ i i f f ⎟ = 4Ti = 4 300 K = 1 200 K = 927°C Tf = Ti ⎜ ⎟ = Ti ⎜ ⎜⎝ PiVi ⎟⎠ ⎜⎝ PiVi ⎟⎠

(

)

10.32 For the states before and after the valve is opened, PiVi = nRT and PfVf = nRT. The temperature T remains constant so that PiVi = PfVf and

Vf =

(

Pi 4.00 atm Vi = 0.500 m 3 Pf 1.00 atm

)

= 2.00 m 3

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10.33

639 5 −6 3 PV (1.013 × 10 Pa/atm ) (1.0 × 10 m ) n = = = 4.2 × 10−5 mol (a) RT (8.31 J/mol ⋅ K )( 293 K ) !

Thus,

molecules ⎞ ⎛ 19 N = n⋅ NA = ( 4.2 × 10−5 mol ) ⎜ 6.02 × 1023 ⎟⎠ = 2.5 × 10 molecules . ⎝ mol ! (b) Since both V and T are constant,

n /n = (P2 V2 / RT 2 ) / (P1 V1 / RT 1 ) = P2 /P1 , or ! 2 1 ⎛ 1.0 × 10−11 Pa ⎞ ⎛P ⎞ n2 = ⎜ 2 ⎟ n1 = ⎜ 4.2 × 10−5 mol ) = 4.1 × 10−21 mol ( 5 ⎟ P 1.013 × 10 Pa ⎝ 1⎠ ⎝ ⎠ !

10.34

(a) With P0 = 1.00 atm, T0 = 10.0°C = 283 K, T1 = 40.0°C = 313 K, and V1 = 0.280V0, we find ⎛V ⎞⎛T ⎞ P1V1 n1 RT1 ⎛ 1 ⎞ ⎛ 313 K ⎞ = ⇒ P1 = ⎜ 0 ⎟ ⎜ 1 ⎟ P0 = ⎜ (1.00 atm ) = 3.95 atm ⎝ 0.280 ⎟⎠ ⎜⎝ 283 K ⎟⎠ P0V0 n0 RT0 ⎝ V1 ⎠ ⎝ T0 ⎠ !

and P1 = (3.95 atm)(1.013 × 105 Pa/1 atm) = 4.00 × 105 Pa = 400 kPa. (b) If now, conditions inside the tire change so that Vf = 1.02V1 and Tf = 85.0°C = 358 K, we form a new ratio to find

Pf V f PV ! l l

⎛ V ⎞ ⎛ Tf ⎞ ⎛ 1 ⎞ ⎛ 358 K ⎞ ⇒ Pf = ⎜ l ⎟ ⎜ ⎟ Pl = ⎜ ( 3.95 atm ) = 4.43 atm ⎝ 1.02 ⎟⎠ ⎜⎝ 313 K ⎟⎠ nl RTl ⎝ V f ⎠ ⎝ Tl ⎠

n f RTf

⎛ 1.013 × 105 Pa ⎞ P = 4.43 atm ) ⎜ 1 atm ⎟ = 4.49 × 105 Pa = 449 kPa and f ( ⎝ ⎠ ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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10.35

640

The initial and final absolute temperatures are Ti = TC,i + 273 = (25.0 + 273)K = 298 K

and

Tf = TC,f + 273 = (75.0 + 273) K = 348 K The volume of the tank is assumed to be unchanged, or Vf = Vi. Also, twothirds of the gas is withdrawn, so nf = ni/3. Thus, from the ideal gas law,

Pf V f PV ! i i

⎛ n f ⎞ ⎛ Tf ⎞ ⎛ 1 ⎞ ⎛ 348 K ⎞ ⇒ Pf = ⎜ ⎟ ⎜ ⎟ Pi = ⎜ ⎟ ⎜ (11.0 atm ) = 4.28 atm ⎝ 3 ⎠ ⎝ 298 K ⎟⎠ ni RTi ⎝ ni ⎠ ⎝ Ti ⎠

n f RTf

10.36 (a) The volume of the gas is V = 8.00 L = 8.00 × 103 cm3 = 8.00 × 10−3 m3, and the absolute temperature is T = (20.0 + 273) K = 293 K. The ideal gas law then gives the number of moles present as 5 −3 3 PV ⎡⎣ 9.00 atm (1.013 × 10 Pa/1 atm ) ⎤⎦ ( 8.00 × 10 m ) n= = = 3.00 mol RT 8.31 J/mol ⋅ K ) ( 293K ) ( !

(b) The number of molecules present in the container is N = n⋅NA = (3.00 mol)(6.02 × 1023 molecules/mol = 1.81 × 1024 molecules 10.37

With n held constant, the ideal gas law gives V1 ⎛ P2 ⎞ ⎛ T1 ⎞ ⎛ 0.030 atm ⎞ ⎛ 300 K ⎞ = = = 4.5 × 10−2 V2 ⎜⎝ P1 ⎟⎠ ⎜⎝ T2 ⎟⎠ ⎜⎝ 1.0 atm ⎟⎠ ⎜⎝ 200 K ⎟⎠ !

Since the volume of a sphere is V = (4π/3)r3, V1/V2 = (r1/r2)3.

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⎛V ⎞ Thus, r1 = ⎜ 1 ⎟ ! ⎝ V2 ⎠

10.38

1/3

r2 = ( 4.5 × 10−2 )

1/3

( 20 m ) = 7.1 m

The mass of the gas in the balloon does not change as the temperature increases. Thus,

ρf ρ ! i

(m/V ) = V f

(m/Vi )

⎛V ⎞ ρf = ρi ⎜ i ⎟ ⎝ Vf ⎠

From the ideal gas law with both n and P constant, we find Vi/Vf = Ti/Tf , and now have

⎛T ⎞ ⎛ 273 K ⎞ 3 ρf = ρi ⎜ i ⎟ = ( 0.179 kg/m 3 ) ⎜ ⎟⎠ = 0.131 kg/m T 373 K ⎝ ⎝ f⎠ ! 10.39

The pressure 100 m below the surface is found, using P1 = Patm + ρgh, to be P1 = 1.013 × 105 Pa + (103 kg/m3)(9.80 m/s2)(100 m) = 1.08 × 106 Pa The ideal gas law, with both n and T constant, gives the volume at the surface as ⎛ 1.08 × 106 Pa ⎞ ⎛ P1 ⎞ ⎛ P1 ⎞ V2 = ⎜ ⎟ V1 = ⎜ V =⎜ 1.50 cm 3 ) = 160 cm 3 ( 5 ⎟ ⎟ ⎝ P2 ⎠ ⎝ Pα n ⎠ ⎝ 1.013 × 10 Pa ⎠ !

10.40 For the states before and after the chest is expanded, PiVi = nRT and PfVf = nRT The temperature T and number of moles remain constant so that PiVf = PfVf. With Vi = 1 250 mL, the final volume is Vf = 1 250 mL + 525 mL =

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1 775 mL and Pf =

Vi 1 250 mL Pi = (1.00 atm ) Vf 1 775 mL

= 0.704 atm

10.41

The average kinetic energy of the molecules of any ideal gas at 300. K is

1 3 3⎛ J⎞ KE = mv 2 = kBT = ⎜ 1.38 × 10−23 ⎟ 300. K = 6.21 × 10−21 J 2 2 2⎝ K⎠

(

)

10.42 Consult a periodic table to find that the molecular mass of methane (CH4) is M = MC + 4MH = 12.0 g/mol + 4.0 g/mol = 16.0 g/mol = 16.0 × 10−3 kg/mol. The rms speed is

vrms = v 2 =

3RT M

3 ( 8.31 J/mol ⋅ K ) ( 325 K ) = 712 m/s 16.0 × 10−3 kg/mol

10.43 (a) The average kinetic energy per molecule of any ideal gas at 275 K is

1 2

(

)

mv 2 = 32 kBT = 32 1.38 × 10−23 J/K (275 K) = 5.69 × 10−21 J

(b) The rms speed is vrms = 3kBT /m = 3RT / M where M is the molar mass in kg/mol. Substitute M = 39.948 × 10−3 kg/mol for argon to find

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vrms =

(

)

3 8.31 J/( mol ⋅ K ) ( 275 K ) 3RT = = 414 m/s M 39.948 × 10−3 kg/mol

(

)

U = 32 nRT = 32 ( 3 mol ) 8.31 J/( mol ⋅ K ) ( 275 K ) = 1.03 × 10 4 J

10.44 From the given information, V = (20.0 cm)3 = 8.00 × 10−3 m3, n = 3 mol, T = 20.0°C +273.15 = 293 K, and MNe = 20.180 g/mol = 20.180 × 10−3 kg/mol. (a) The internal energy of the gas is

U = 32 nRT = 32 ( 3 mol ) ( 8.31 J/mol ⋅ K ) ( 293 K ) = 1.10 × 10 4 J

(b) The internal energy of a monatomic gas is due entirely to its translational kinetic energy so that

KEtotal = U = 32 nRT = 1.10 × 10 4 J

1 2

KEtotal 1.01 × 10 4 J mv = = N ( 3 mol ) 6.02 × 1023 atoms/mol 2

(

)

= 6.07 × 10−21 J

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(d) Use Equation 10.13 to find the pressure is

(

)(

⎛ ( 3 mol ) 6.02 × 10 23 atoms/mol ⎞ ⎛ N⎞ 1 2 −21 2 P = ⎜ ⎟ 2 mv = 3 ⎜ ⎟ 6.07 × 10 J −3 3 ⎝V⎠ 8.00 × 10 m ⎝ ⎠ 2 3

(

)

= 9.14 × 105 Pa (e) From PV = nRT, P=

nRT ( 3 mol ) ( 8.31 J/mol ⋅ K ) ( 293 K ) = V 8.00 × 10−3 m 3

= 9.13 × 105 Pa

10.45

One mole of any substance contains Avogadro’s number of molecules and has a mass equal to the molar mass, M. Thus, the mass of a single molecule is m = M/NA. For helium, M = 4.00 g/mol = 4.00 × 10−3 kg/mol, and the mass of a helium molecule is

4.00 × 10−3 kg/mol m= = 6.64 × 10−27 kg/molecule 23 6.02 × 10 molecule/mol ! Since a helium molecule contains a single helium atom, the mass of a helium atom is matom = 6.64 × 10−27 kg 10.46 (a) The rms speed of molecules in a gas of molar mass M and absolute © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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temperature T is !vrms = 3RT /M . Since the molar masses of 35Cl and 37

Cl are 35.0 × 10−3 kg/mol and 37.0 × 10−3 kg/mol, respectively, the

desired ratio is

3RT /35.0 × 10−3 kg/mol vrms (35 Cl) 37.0 = = = 1.03 37 −3 vrms ( Cl) 35.0 3RT /37.0 × 10 kg/mol ! (b) Since the above ratio is larger than 1, the less massive atom, 35Cl, moves faster. 10.47 If vrms = vesc, we must have !vrms = 3kBT /m = vesc , where kB = 1.38 × 10−23 J/K is Boltzmann’s constant and m is the mass of a molecule (for helium, m = 6.64 × 10−27 kg. Thus, the required absolute temperature is !T = mvesc /3 kB . 2

(a) To have vrms = vesc on Earth where vesc = 1.12 × 104 m/s, the required temperature for the helium gas is

( 6.64 × 10 kg )(1.12 × 10 m/s ) = 2.01 × 10 K T= 3 (1.38 × 10 J/K ) ! −27

−23

(b) If vrms = vesc on the Moon where vesc = 2.37 × 103 m/s, the temperature must be

( 6.64 × 10 kg )( 2.37 × 10 m/s ) = 901 K 3 (1.38 × 10 J/K ) −27

T= !

−23

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646

(a) The volume occupied by this gas is V = 7.00 L = 7.00 × 103 cm3 = 7.00 × 10−3 m3 Then, the ideal gas law gives 6 −3 3 PV (1.60 × 10 Pa ) (7.00 × 10 m ) T= = = 385 K nR 3.50 mol ) ( 8.31 J/mol ⋅ K ) ( !

(b) The average kinetic energy per molecule in this gas is 3 3 KE molecule = kBT = (1.38 × 10−23 J/K ) ( 385 K ) = 7.97 × 10−21 J 2 2 !

(c) You would need to know the mass of the gas molecule to find its rms speed, which in turn requires knowledge of the molar mass of the gas. 10.49

Consider a time interval of 1.0 min = 60 s, during which 150 bullets bounce off Superman’s chest. From the impulse-momentum theorem, the magnitude of the average force exerted on Superman is

Fav = = ! 10.50

I 150 Δp bullet 150 m ( v − v0 ) = = Δt Δt Δt −3 150 ( 8.0 × 10 kg ) ⎡⎣( 400 m/s ) − ( −400 m/s ) ⎤⎦ 60 s

= 16 N

From the impulse-momentum theorem, the average force exerted on the wall is

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647

I N Δp molecule N m ( v − v0 ) Fav = = = Δt Δt Δt !

( 5.0 × 10 )( 4.68 × 10 23

Fav = !

−26

kg ) ⎡⎣( 300 m/s ) − ( −300 m/s ) ⎤⎦ 10 s

= 14 N

The pressure on the wall is then Fav 14 N ⎛ 10 4 cm 2 ⎞ P= = = 1.8 × 10 4 N/m 2 = 18 kPa 2 ⎜ 2 ⎟ A 8.0 cm ⎝ 1 m ⎠ !

10.51

As the pipe undergoes a temperature change ΔT = 46.5°C − 18.0°C = 28.5°C, the expansion toward the right of the horizontal segment is ΔLx = αL0x(ΔT) = [17 × 10−6 (°C)−1](28.0 cm)(28.5°C) = 1.4 × 10−2 cm = 0.14 mm The downward expansion of the vertical section is ΔLy = αL0y(ΔT) = [17 × 10−6(°C)−1](134 cm)(28.5°C) = 6.5 × 10−2 cm = 0.65 mm The total displacement of the pipe elbow is ΔL = ΔL2x + ΔL2y = !

( 0.14 mm ) + ( 0.65 mm ) = 0.66 mm 2

⎛ 0.65 mm ⎞ ⎛ ΔLy ⎞ θ = tan −1 ⎜ = tan −1 ⎜ = 78° ⎟ ⎝ ΔLx ⎠ ⎝ 0.14 mm ⎟⎠ !

 !ΔL = 0.66 mm toward the right at 78° below the horizontal

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10.52 (a) ΔL = αL0(ΔT) = [9.00 × 10−6(°C)−1](30.0 cm)(65.0°C) = 1.76 × 10−2 cm = 0.176 mm (b) ΔD = αD0(ΔT) = [9.00 × 10−6(°C)−1](1.50 cm)(65.0°C) = 8.78 × 10−4 cm = 8.78 µm (c) The initial volume is 2 ⎛ π D02 ⎞ π V0 = ⎜ L0 = (1.50 × 10−2 m ) ( 0.300 m ) = 5.30 × 10−5 m 3 ⎟ ⎝ 4 ⎠ 4 !

ΔV = β V0 ( ΔT ) !

10.53

−1 = 3α V0 ( ΔT ) = 3 ⎡⎣ 9.00 × 10−6 ( °C ) ⎤⎦ ( 5.30 × 10−5 m 3 ) ( 65.0°C ) = 9.30 × 10−8 m 3

The number of moles of CO2 present is n = m/M, where m = 6.50 g and M = 44.0 g/mol. Thus, at the given temperature (20.0°C = 293 K) and pressure (1.00 atm = 1.013 × 105 Pa, the volume will be nRT mRT ( 6.50g ) ( 8.31 J/mol ⋅ K ) ( 293 K ) V= = = = 3.55 × 10−3 m 3 = 3.55 L 5 P MP ( 44.0 g/mol )(1.013 × 10 Pa ) !

10.54

Note that the moment of inertia of each object in Table 8.1 is of the form 2 !I = CM , where C is a constant, M is the mass of the object, and  is

some length. As the temperature increases, the factors C and M are unchanged, while the value of the length measurement at temperature T is ! =  0 (1 + α ⋅ ΔT) . Therefore, the percentage increase in the moment of © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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inertia of the object is ⎧⎪ CM [  0 (1 + α ⋅ ΔT )]2 − CM20 ⎫⎪ I − I0 %change = × 100% = ⎨ ⎬ × 100% 2 I0 CM 0 ⎩⎪ ⎭⎪ !

% change = [(1+α ⋅ΔT)2 − 1] × 100%

(a) For copper, with α = 17 × 10−6 °C−1 and ΔT = 100°C % change = {[1 + (17 × 10−6(°C−1))(100°C)]2 −1} × 100% = 0.34% (b) For aluminum, with α = 24 × 10−6 (°C−1) and ΔT = 100°C, % change = {[1 + (24 × 10−6 (°C−1))(100°C)]2−1} × 100% = 0.48% (c) The moment of inertia for each of the shapes has the same mathematical form: the product of a constant, the mass, and the square of a length. 10.55

For a temperature change ΔTF = TF − TF,0 on the Fahrenheit scale, the corresponding temperature change on the Celsius scale is 5 5 5 5 ΔTC = TC − TC ,0 = (TF − 32 ) − (TF ,0 − 32 ) = (TF − TF ,0 ) = ( ΔTF ) 9 9 9 9 !

Therefore, if L0 = 35.000 m and ΔTF = 90.000°F − 15.000°F = 75.000°F, the final length of the beam is

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5 ⎡ ⎤ L = L0 [1 + α (ΔT)] = ( 35.000 m ) ⎢1 + (11 × 10−6 °C−1 ) (75.000 ) °C ⎥ = 35.016 m 9 ⎣ ⎦ ! 10.56

When air trapped in the tube is compressed, at constant temperature, into a cylindrical volume 0.40 m long, the ideal gas law gives its pressure as ⎛ 1.5 m ⎞ ⎛V ⎞ ⎛L ⎞ P2 = ⎜ 1 ⎟ P1 = ⎜ 1 ⎟ P1 = ⎜ 1.013 × 105 Pa ) = 3.8 × 105 Pa ( ⎟ ⎝ V2 ⎠ ⎝ L2 ⎠ ⎝ 0.40 m ⎠ !

This is also the water pressure at the bottom of the lake. Thus, P = Patm +

ρgh gives the depth of the lake as P2 − Patm 3.8 − 1.013 ) × 105 Pa ( h= = = 28 m ρg 103 kg/m 3 ) ( 9.80 m/s 2 ) ( !

10.57

The mass of CO2 produced by three astronauts in 7.00 days is m = 3(1.09 kg/d)(7.00 d) = 22.9 kg, and the number of moles of CO2 available is

m 22.9 kg n= = = 520 mol M 44.0 × 10−3 kg/mol ! The recycling process will generate 520 moles of methane to be stored. In a volume of V = 150 L = 0.150 m3 and at temperature T = −45.0°C = 228 K, the pressure of the stored methane is nRT ( 520 mol ) ( 8.31 J/mol ⋅ K ) ( 228 K ) P= = = 6.57 × 106 Pa= 6.57 MPa 3 V 0.150 m !

10.58 (a) The piston in this vertical cylinder has three forces acting on it. These

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are: (1) a downward gravitational force, mg, the piston’s own weight; (2) a downward pressure force, Fd = P0A, due to the atmospheric pressure above the piston; and (3) an upward pressure force, Fu = PA, due to the absolute pressure of the gas trapped inside the cylinder. Since the piston is in equilibrium, Newton’s second law requires ΣFy = 0 ⇒ Fu − mg − Fd = 0

PA = mg + P0A

[1]

From the ideal gas law, the absolute pressure of the trapped gas is nRT nRT P= = V Ah !

[2]

Substituting Equation [2] into [1] yields ⎛ nRT ⎞ ⎜⎝ A h ⎟⎠ A = mg + P0 A !

nRT mg + P0 A

(b) From Equation [1] above, the absolute pressure inside the cylinder is mg P= + P0 A !

where P0 is atmospheric pressure. This is greater than atmospheric pressure because mg/A>0. (c) Observe from the result of part (a) above, if the absolute temperature T increases, the equilibrium value of h also increases.

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10.59 (a) As the acetone undergoes a change in temperature ΔT = (20.0 − 35.0) °C = −15.0°C, the final volume will be Vƒ = V0 + ΔV = V0 + βV0(ΔT) = V0[1 + β(ΔT)] =(100 mL)[1 + (1.50 × 10−4 (°C)−1)(−15.0°C)] = 99.8 mL (b) When acetone at 35°C is poured into the Pyrex flask that was calibrated at 20°C, the volume of the flask temporarily expands to be larger than its calibration markings indicate. However, the coefficient of volume expansion for Pyrex [β = 3α = 9.6 × 10−6(°C)−1] is much smaller than that of acetone [β = 1.5 × 10−4(°C)−1]. Hence, the temporary increase in the volume of the flask will be much smaller than the change in volume of the acetone as the materials cool back to 20°C, and this change in volume of the flask has negligible effect on the answer. 10.60

(a) Ti = TC + 273.15 = (19.0 + 273.15) K = 292 K

ni = (b)

PiVi RTi

9.50 × 10 Pa ) ⎡( 20.0 L ) (10 cm / 1 L ) (1 m / 10 cm ) ⎤ ( ⎣ ⎦ = (8.31 J / mol ⋅ K )(292 K ) 5

= 7.83 mol

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(c)

g MCO2 = [12.0 + 2 (16.0 )] = 44.0 g / mol mol !

(d) mi = niMCO2 = (7.83 mol)(44.0 g/mol) = 345 g (e) Tf = Ti + ΔT = 292 K + 224 K = 516 K

nf =

mf M

= 2

mi − Δm M

345 g − 82.0 g 44.0 g/mol

= 5.98 mol

(f) Neglecting any change in volume of the tank, Vf ≈ Vi , and we have

Pf V f !

Pi Vi

⎛ n f Tf ⎞ ⇒ Pf = ⎜ Pi ni RTi ⎝ niTi ⎟⎠

n f RTf

⎛ 5.98 mol ⎞ ⎛ 516 K ⎞ ⎛ n f ⎞ ⎛ Tf ⎞ ( 9.50 × 105 Pa ) = 1.28 × 106 Pa (g) Pf = ⎜ ⎟ ⎜ ⎟ Pi = ⎜ ⎝ ni ⎠ ⎝ Ti ⎠ ⎝ 7.83 mol ⎟⎠ ⎜⎝ 292 K ⎟⎠ !

10.61 (a) The volume of the liquid expands by ΔVliquid = βV0(ΔT) and the volume of the glass flask expands by ΔVflask = (3α)V0(ΔT). The amount of liquid that must overflow into the capillary is Voverflow = ΔVliquid − ΔVflask = V0(β − 3α)(ΔT). The distance the liquid will rise into the capillary is then

Δh = !

Voverflow ⎛ V0 ⎞ = ⎜ ⎟ (β − 3α )(ΔT) ⎝ A⎠ A

(b) For a mercury thermometer, βHg = 1.82 × 10−4(°C)−1 and (assuming © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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654

Pyrex glass), 3αglass = 3(3.2 × 10−6 (°C)−1) = 9.6 × 10−6(°C)−1. Thus, the expansion of the mercury is almost 20 times the expansion if the flask, making it a rather good approximation to neglect the expansion of the flask. 10.62 (a) The initial absolute pressure in the tire is P1 = Patm + P1,gauge = 2.80 atm, and the final absolute pressure is P2 = Patm + P2,gauge = 3.20 atm. The ideal gas law, with both n and V constant, gives ⎛ 3.20 atm ⎞ ⎛P ⎞ T2 = ⎜ 2 ⎟ T1 = ⎜ ( 300 K ) = 343 K ⎝ P1 ⎠ ⎝ 2.80 atm ⎟⎠ !

(b) When the quantity of gas varies, while volume and temperature are constant, the ideal gas law gives

n3 P3 = . Thus, when air is released !n2 P2

to lower the absolute pressure back to 2.80 atm, we have

n3 2.80 atm = = 0.875 . n 3.20 atm 2 ! At the end, we have 87.5% of the original mass of air remaining, or 12.5% of the original mass was released. 10.63

After expansion, the increase in the length of one span is ΔL = αL0(ΔT)

Topic 10

655

= [12 × 10−6 (°C)−1](125 m)(20.0°C) = 3.0 × 10−2 m giving a final length of L = L0 + ΔL = 125 m + 3.0 × 10−2 m

From the Pythagorean theorem, y = L2 − L20 = !

10.64

[(125 + 0.030) m ] − (125 m ) = 2.7 m 2

(a) From the ideal gas law,

⎛ P ⎞⎛V ⎞ ⎛T ⎞ P2V2 P1V1 = ,!or! ⎜ 2 ⎟ ⎜ 2 ⎟ = ⎜ 2 ⎟ T1 ⎝ P1 ⎠ ⎝ V1 ⎠ ⎝ T1 ⎠ ! T2

The initial conditions are: P1 = 1 atm, V1 = 5.00 L = 5.00 × 10−3 m3, and T1 = 20.0°C = 293 K The final conditions are: F k ⋅h P2 = 1 atm+ = 1 atm+ ,!V2 = V1 + A ⋅ h,!and!T2 = 250°C = 523 K A A ! ⎛ k ⋅h ⎞ ⎛ A ⋅ h ⎞ ⎛ 523 K ⎞ ⎜ 1 + A 1 atm ⎟ ⎜⎝ 1 + V ⎟⎠ = ⎜⎝ 293 K ⎟⎠ ( )⎠ ⎝ 1 !

Thus,

(

)

( (

)

⎛ ⎞⎛ 2.00 × 103 N/m ⋅ h 0.010 0 m 2 ⋅ h ⎞ ⎛ 523 ⎞ ⎜1+ ⎟ ⎜1+ ⎟ =⎜ ⎟ ⎜⎝ 0.010 0 m 2 1.013 × 105 N/m 2 ⎟⎠ ⎜⎝ 5.00 × 10−3 m 3 ⎟⎠ ⎝ 293 ⎠

(

)(

)

Simplifying and using the quadratic formula yields © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 10

656

h = 0.169 m = 16.9 cm (b)

P2 = 1 atm+

k ⋅h A

(2.00 × 10 N/m)(0.169 m) = 1.35 × 10 Pa = 1.013 × 10 Pa+ 3

0.010 0 m

10.65 (a) The two metallic strips have the same length L0 at the initial temperature T0. After the temperature has changed by ΔT = T − T0, the lengths of the two strips are L1 = L0[1 + α1(ΔT)]

and

L2 = L0[1 + α2(ΔT)]

The lengths of the circular arcs are related to their radii by L1 = r1θ and L2 = r2θ, where θ is measured in radians.

Thus,

θ=

L L (α − α 1 ) L0 (ΔT) Δr = r2 − r1 = 2 − 1 = 2 , or θ θ θ !

(α 2 − α 1 ) L0 ( ΔT )

Δr

(b) As seen in the above result, θ = 0 if either ΔT = 0 or α1 = α2 (c) If ΔT < 0, then θ is negative so the bar bends in the opposite direction. 10.66 (a) If the bridge were free to expand as the temperature increased by ΔT = 20°C, the increase in length would be © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 10

657

ΔL = αL0(ΔT) = (12 × 10−6 (°C−1))(250 m)(20°C) = 6.0 × 10−2 m = 6.0 cm (b) When the bridge is not allowed to expand naturally, stress builds up in the bridge, effectively compressing it the distance ΔL that it would normally have expanded. Combining the defining equation for Young’s modulus, (Y = Stress/Strain = Stress/(ΔL/L), with the expression, ΔL = αL(ΔT), for the linear expansion when the temperature changes by ΔT yields

⎛ ΔL ⎞ ⎛ α L(ΔT) ⎞ Stress = Y ⎜ ⎟ = Y ⎜ ⎟ = α Y(ΔT) ⎝ ⎠ ⎝ L L ⎠ ! (c) When ΔT = 20°C, the stress in the specified bridge would be Stress = αY(ΔT) = (12 × 10−6 (°C−1)(2.0 × 1010 Pa)(20°C) = 4.8 × 106 Pa Since this is considerably less than the maximum stress, 2.0 × 107 Pa, that concrete can withstand, the bridge will not crumble. 10.67 (a) Yes, the angular speed will increase as the disk cools. Since no external torque acts on the disk, the angular momentum of the disk will be conserved. As the disk cools, its radius, and hence its moment of inertia will decrease. Then, in order to keep angular momentum (L = Iω) constant, the angular speed must increase.

Topic 10

658

(b) Since angular momentum is conserved, Iω = I0ω0 or ω = (I0/I) ω0. Thus,

⎛1 ⎞ 2 2 MR02 ⎛ ⎞ R0 ⎛ R0 ⎞ ⎜2 ⎟ ω =⎜ ⎟ ω 0 = ⎜⎝ R ⎟⎠ ω 0 = ⎜ R [1 + α (ΔT)] ⎟ ω 0 1 2 ⎝ 0 ⎠ ⎜⎝ MR ⎟⎠ 2 !

10.68

ω=

ω0

( )

⎡1 + α ΔT ⎤ ⎣ ⎦

25.0 rad/s

(

)(

)

⎡1 + 17 × 10−6 °C−1 20.0 − 850 °C ⎤ ⎣ ⎦

= 25.7 rad/s

Let container 1 be maintained at T1 = T0 = 0°C = 273 K, while the temperature of container 2 is raised to T2 = 100°C = 373 K. Both containers have the same constant volume, V, and the same initial pressures, (P0)2 = (P0)1 = P0. As the temperature of container 2 is raised, gas flows from one container to the other until the final pressures are again equal, P2 = P1 = P. The total mass of gas is constant, so n2 + n1 = (n0)2 + (n0)1

[1]

PV From the ideal gas law, n = , so Equation [1] becomes RT ! PV PV P0V P0V + = + , !RT2 RT1 RT0 RT0

⎛ 1 1 ⎞ 2P P⎜ + ⎟ = 0 ⎝ T2 T1 ⎠ T0

Thus, 2P ⎛ T T ⎞ 2 (1.00 atm ) ⎛ 273 ⋅ 373 ⎞ P= 0⎜ 1 2 ⎟ = ⎜⎝ 273 + 373 ⎟⎠ = 1.15 atm T0 ⎝ T1 + T2 ⎠ 273 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 11

659

Topic 11 Energy in Thermal Processes

QUICK QUIZZES

11.1

(a) Water, glass, iron. Because it has the highest specific heat (4 186 J/kg ⋅ °C), water has the smallest change in temperature. Glass is next (837 J/kg ⋅ °C), and iron (448 J/kg ⋅ °C) is last.

(b) Iron, glass, water. For a given temperature increase, the energy transfer by heat is proportional to the specific heat.

11.2

Choice (b). The slopes are proportional to the reciprocal of the specific heat, so larger specific heat results in a smaller slope, meaning more energy to achieve a given change in temperature.

11.3

Choice (c). The blanket acts as a thermal insulator, slowing the transfer of energy by heat from the air into the cube.

11.4

Choice (b). The rate of energy transfer by conduction through a rod is proportional to the difference in the temperatures of the ends of the rod. When the rods are in parallel, each rod experiences the full difference in the temperatures of the two regions. If the rods are

Topic 11

660

connected in series, neither rod will experience the full temperature difference between the two regions, and hence neither will conduct energy as rapidly as it did in the parallel connection.

11.5

(a) PA/PB = 4. From Stefan’s law, the power radiated from an object at absolute temperature T is proportional to the surface area of that object. Star A has twice the radius and four times the surface area of star B.

(b) PA/PB = 16. From Stefan’s law, the power radiated from an object having surface area A is proportional to the fourth power of the

( ) = 2 (σ AeT ) =16P . (c)

absolute temperature. Thus, PA = σ Ae 2TB

4 B

PA/PB = 64. When star A has both twice the radius and twice the absolute temperature of star B, the ratio of the radiated powers is

( )( ) = 2(R ) (2T ) = (2) (2) = 64 = = P σ A eT RT σ ( 4π R ) (1)T

σ AA eTA4

4 B

σ 4π RA2 1 TA4 2

2 4 B B

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

11.2

One of the ways that objects transfer energy is by radiation. The top of the mailbox is oriented toward the clear sky. Radiation emitted by the

Topic 11

661

top of the mailbox goes upward and into space. There is little radiation coming down from space to the top of the mailbox. Radiation leaving the sides of the mailbox is absorbed by the environment. Radiation from the environment (trees, houses, cars, etc.), however, can enter the sides of the mailbox, keeping them warmer than the top. As a result, the top is the coldest portion and frost forms there first.

11.4

Substances A and B have equal masses so the one that undergoes the lower temperature change has the higher specific heat. Substance A undergoes a 65°C temperature change while B undergoes a 15°C change. The correct answer is therefore (b) : Substance B has the higher specific heat.

11.6

The specific heats and temperature changes are the same for each sample so that the ratio of energy transfers is equal to the ratio of sample masses (e.g., a sample with twice the mass will require twice the energy transfer for the same values of specific heat and temperature change). The largest energy transfer is QB so the sample

B has the largest mass.

11.8

Yes, if you know the specific heat of zinc and copper, you can determine the relative fraction of each by heating a known weight of

Topic 11

662

pennies to a specific initial temperature, say 100°C, then dump them into a known quantity of water, at say 20°C. The equation for conservation of energy will be

mpennies[x ⋅ cCu + (1 − x)cZn](100°C − T) = mwatercwater(T − 20°C)

The equilibrium temperature, T, and the masses will be measured. The specific heats are known, so the fraction of metal that is copper, x, can be computed.

11.10

The final temperature T must lie between the highest and lowest sample temperatures, so the correct answer is ( e ) , TA > T > TC . It is not possible in general to determine whether the final temperature will be higher or lower than the intermediate sample temperature TB. Similarly, it is not possible in general to determine the sign or magnitude of the energy transfers to sample B.

11.12 (a) Radiated power is proportional to T4 so, other things being equal, TB = 4TA results in PB = 44PA. Therefore PB /PA = 256 . (b) For equal surface areas and radiated power, the product eT4 is constant so that an object with 4 times the emissivity of another must have 1/ 2 times the other’s temperature:

4eBTA4 = eBTB4 → TB /TA =

2 .

Topic 11

11.14

663

The black car absorbs more of the incoming energy from the Sun than does the white car, making it more likely to cook the egg.

11.16

With eA = eB, rA = 2rB, and TA = 2TB, the ratio of the power output of A to that of B is

4 PA σ AA e A TA 4 π rA2TA4 ⎛ rA ⎞ ⎛ TA ⎞ = = = ⎜ ⎟ ⎜ ⎟ = (2)2 (2)4 = (2)6 = 64 4 2 4 PB σ AB e B TB 4 π rB TB ⎝ rB ⎠ ⎝ TB ⎠

ANSWERS TO EVEN NUMBERED PROBLEMS

11.2

(a) 4.40 × 105 J

(b)

938 J

11.4

(a)

2.3 × 106 J

(b)

2.8 × 104 stairs

(c)

7.0 × 103 stairs

11.6

0.234 kJ/kg ⋅ °C

11.8

0.15 mm

(c)

47.7°C

11.10 (a) 25.8°C

(b) No. Gravitational potential energy is proportional to the mass, just like the temperature change.

Topic 11

664

11.12 0.105°C

11.14

11.16

11.18

−

(a)

9.9 × 10 3 °C

(b)

It is absorbed by the rough horizontal surface.

(a)

F Stress = = Y[α (ΔT)] A !

(b)

mc ⎛ F ⎞ Q= ⎜ ⎟ Yα ⎝ A ⎠ !

(c)

(d)

6.7 × 106 J

(e)

79°C

(f) ∼4 h

96.0 kg

467

11.20 29.4°C

11.22

29.6°C

11.24

47°C

11.26

1.18 × 103 J/kg ⋅ °C

11.28

4.62 × 106 J

11.30

49 kJ

11.32

0.12 MJ

11.34

(a) ice at −10.0°C to ice at 0°C; ice at 0°C to liquid water at 0°C; water at 0°C to water at T; aluminum at 20.0°C to aluminum at T; ethyl

Topic 11

665

alcohol at 30.0°C to ethyl alcohol at T.

(b) See Solution.

(d) 4.81°C

11.36

0.33 kg, 0.067 or 6.7%

11.38

403 cm3

11.40

11.1 W

11.42

(a)

6 × 103 W

11.44

(a)

Rskin = 5.0 × 10 2 m2 ⋅ K/W, Rfat = 2.5 × 10 2 m2 ⋅ K/W,

(b)

5 × 108 J

−

Rtissue = 6.4 × 10 2 m2 ⋅ K/W

Rtotal = 0.14 m2 ⋅ K/W

(b)

5.3 × 102 W

11.46

39 m3

11.48

(a)

52 W

(b)

2 kW

Topic 11

666 −

11.50

7.3 × 10 2 W/m ⋅ °C

11.52

(a) 408 K

11.54

1.1 × 10 5 m2

11.56

1.83 h

11.58

(a)

1.1 × 102 W

(b)

The positive sign indicates that the body is radiating energy

(b)

50.3 m2

(c)

2.54 × 104 W

−

away faster than it absorbs energy from the environment.

11.60

(a)

1.6 × 102 W

(b)

2.7 × 102 W

(d)

1.2 × 102 W

2.0 kW

(b)

4.5°C

(b)

7.84 ft2 ⋅ °F ⋅ h/Btu

11.62

45°C

11.64

(a)

11.66

28°C

11.68

(a)

2.03 × 103 J/s

11.70

(a)

0.457 kg

(c)

11 W

(b) If the samples and inner surface of the insulation are preheated,

Topic 11

667

nothing undergoes a temperature change during the test. Therefore, only the mass of the wax, which undergoes a change of phase, needs to be known.

11.72

0.9 kg

11.74

(a) The processes involved are the removal of energy to: (1) cool liquid water from 20.0 °C to 0°C, (2) convert liquid water at 0°C to solid water (ice) at 0°C, and (3) cool ice from 0°C to −8.00°C.

(b) 32.5 kJ

PROBLEM SOLUTIONS 11.1

(a) 1 kcal = 1.00 × 103 cal so that 3.50 × 103 cal = 3.50 kcal = 3.50 Cal . (b) Use the conversion 1 cal = 4.186 J to find

⎛ 4.186 J ⎞ 3.50 × 103 cal = 3.50 × 103 cal ⎜ = 1.47 × 10 4 J ⎝ 1 cal ⎟⎠ 11.2

(a) Use the conversion 1 Cal = 1 kcal = 4.186 × 103 J to find ⎛ 4.186 × 103 J ⎞ 5 105 Cal = 105 Cal ⎜ ⎟⎠ = 4.40 × 10 J 1 Cal ⎝

(b) Use the definition of kinetic energy to find

1 2

mv = 4.39 × 10 J → v = 2

(

2 4.40 × 105 J 1.00 kg

) = 938 m/s

Topic 11

668

Q = 4.40 × 105 J = mcΔT ΔT =

Q 4.40 × 105 J = = 27.7°C mc ( 3.79 kg ) 4 186 J/( kg ⋅°C )

(

)

Tf = Ti + ΔT = 20.0°C + 27.7°C = 47.7°C

11.3

(a)

(

)

1 1 Wnet = ΔKE = m v 2f − v02 = (75 kg ) ⎡⎣(11.0 m/s2 ) − 0 ⎤⎦ = 4.54 × 103 J → 4.5 × 103 J 2 2 !

Wnet 4.54 × 10−3 J P = = = 9.1 × 102 J/s = 910 W (b) Δt 5.0 s ! (c) If the mechanical energy is 25% of the energy gained from converting food energy, then Wnet = 0.25(ΔQ) and !P = 0.25(ΔQ)/Δt , so the food energy conversion rate is

ΔQ P ⎛ 910 J/s ⎞ ⎛ 1 Cal ⎞ = =⎜ = 0.87 Cal/s ⎟ Δt 0.25 ⎝ 0.25 ⎠ ⎜⎝ 4186 J ⎟⎠ !

(d) The excess thermal energy is transported by conduction and convection to the surface of the skin and disposed of through the evaporation of perspiration.

11.4

⎛ 103 cal ⎞ ⎛ 4.186 J ⎞ = 2.3 × 106 J (a) Q = 540 Cal ⎜ ⎝ 1 Cal ⎟⎠ ⎜⎝ 1 cal ⎟⎠ !

Topic 11

669

(b) The work done lifting her weight mg up one stair of height h is W1 = mgh. Thus, the total work done in climbing N stairs is W = Nmgh, and we have W = Nmgh = Q or

Q 2.3 × 106 J N= = = 2.8 × 10 4 stairs mgh ( 55 kg ) ( 9.80 m/s 2 ) ( 0.15 m ) !

⎛ Q ⎞ 0.25 Q N= = 0.25 ⎜ = 0.25 ( 2.8 × 10 4 stairs ) = 7.0 × 103 stairs ⎟ mgh ⎝ mgh ⎠ !

11.5

(a) Use the conversion 1 day = 86 400 s to find

7.00 × 106 J/day = 7.00 × 106

J ⎛ 1 day ⎞ = 81.0 W day ⎜⎝ 86 400 s ⎟⎠

(b) Use the conversion 1 Cal = 4.186 × 103 J:

7.00 × 106 J/day = 7.00 × 106

J ⎛ 1 Cal ⎞ ⎛ 1 day ⎞ 3 ⎟⎜ ⎜ day ⎝ 4.186 × 10 J ⎠ ⎝ 24 hr ⎟⎠

= 69.7 Cal/hr (c) Use PEg = mgy to calculate

81.0 J/s =

( ) = mgΔy

Δ PEg Δt

( 81.0 J/s) Δy = = 8.27 m/s 1 s (1.00 kg ) 9.80 m/s2

(

)

Topic 11

670

1.23 × 103 J

11.6

11.7

As mass m of water drops from the top to the bottom of the falls, the

m(ΔT)

(

)(

0.525 kg 10.0°C

)

= 234 J/kg ⋅°C = 0.234 kJ/kg ⋅°C

gravitational potential energy given up (and hence, the kinetic energy gained) is Q = mgh. If all of this goes into raising the temperature, the rise in temperature will be

ΔT =

Q mcwater

m gh m cwater

(9.80 m/s )(807 m) = 1.89°C = 2

4 186 J/kg ⋅°C

and the final temperature is Tf = Ti + ΔT = 15.0°C + 1.89°C = 16.9°C

11.8

The change in temperature of the rod is

Q 1.00 × 10 4 J ΔT = = = 31.7°C mc ( 0.350 kg ) ( 900 J/kg°C ) !

and the change in the length is

ΔL = αL0(ΔT)

−

= [24 × 10 6 (°C) 1](20.0 cm)(31.7°C) = 1.5 × 10 2 cm = 0.15 mm

11.9

The mass of water involved is

kg ⎞ ⎛ m = ρV = ⎜ 103 3 ⎟ ( 4.00 × 1011 m 3 ) = 4.00 × 1014 kg ⎝ m ⎠ ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 11

671

(a) Q = mc(ΔT) = (4.00 × 1014 kg)(4 186 J/kg⋅°C)(1.00°C) = 1.67 × 1018 J

(b) The power input is P = 1 000 MW = 1.00 × 109 J/s,

11.10

⎞ 1 yr Q 1.67 × 1018 J ⎛ t= = = 52.9 yr 9 7 ⎟ ⎜ P 1.00 × 10 J/s ⎝ 3.156 × 10 s ⎠ !

(a) Q = 0.600|ΔPEg| = 0.600(mg|h|) = 0.600 ⋅ m(9.80 m/s2)(50.0 m) or

Q = (294 m2/s2) ⋅ m

From Q = mc(ΔT) = mc(Tf − Ti), we find the final temperature as

( 294 m 2 /s2 ) ⋅ m = 25.8°C Q Tf = Ti + = 25.0°C+ mc m ( 387 J/kg ⋅°C ) ! (b) Observe that the mass of the coin cancels out in the calculation of part (a). Hence, the result is independent of the mass of the coin.

11.11

The mechanical energy transformed into internal energy of the bullet 1 1 ⎛ 1 2⎞ 1 2 is Q = ( KEi ) = ⎜ mvi ⎟ = mvi . Thus, the change in temperature of ⎠ 4 2 2⎝ 2 !

the bullet is

1 2 2 300 m/s ) ( Q 4 m vi ΔT = = = = 176°C mc m clead 4 (128 J/kg ⋅°C ) !

11.12

The internal energy added to the system equals the gravitational

Topic 11

672

potential energy given up by the 2 falling blocks, or Q = ΔPEg = 2mbgh. Thus,

ΔT =

11.13

Q mw cw

2mb gh mw cw

( )( )( ) = 0.105°C (0.200 kg )( 4 186 J/kg ⋅°C)

2 1.50 kg 9.80 m/s 2 3.00 m

The quantity of energy transferred from the water-cup combination in a time interval of 1 minute is

( ) (

( ) ( ) ) (

Q = ⎡ mc + mc ⎤ ΔT ⎢⎣ water cup ⎥ ⎦ ⎡ ⎛ ⎛ J ⎞ J ⎞⎤ = ⎢ 0.800 kg ⎜ 4 186 + 0.200 kg 900 ⎟ ⎜ ⎟ ⎥ 1.5°C kg ⋅°C ⎠ kg ⋅°C ⎠ ⎥⎦ ⎝ ⎝ ⎢⎣ 3 = 5.3 × 10 J

)

(

)

The rate of energy transfer is

Q 5.3 × 103 J J P= = = 88 = 88 W Δt 60 s S ! 11.14 (a) The mechanical energy converted into internal energy of the block ⎛ 1 2⎞ is Q = 0.85 KEi = 0.85 ⎜ mvi ⎟ . The change in temperature of the ⎝2 ⎠

( )

block will be

ΔT =

Q mcCu

⎛1 ⎞ 0.85 ⎜ mvi2 ⎟ ⎝2 ⎠ mcCu

(

) = 9.9 × 10 °C = 2 ( 387 J/kg ⋅°C ) 0.85 3.0 m/s

−3

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673

(b) The remaining energy is absorbed by the horizontal surface on which the block slides.

11.15 The swimming pool has volume V = (5.00 m) × (10.0 m) × (1.78 m) = 89.0 m3. (a) Use the definition of density to solve for the mass:

ρ=

(

)(

)

M → M = ρV = 1.00 × 103 kg/m 3 89.0 m 3 = 8.90 × 10 4 kg V

(b) From the definition of specific heat:

(

)

Q = mcΔT = 8.90 × 10 4 kg ( 4186 J/kg ⋅°C ) ( 26.5°C − 15.5°C ) = 4.10 × 109 J (c) The cost C in dollars for this amount of energy is

⎛ $0.100 ⎞ ⎛ 1 kW ⎞ ⎛ 1 h ⎞ C = 4.10 × 109 J ⎜ = $114 ⎝ kW ⋅ h ⎟⎠ ⎜⎝ 1 000 J/s ⎟⎠ ⎜⎝ 3 600 s ⎟⎠

11.16

(a) From the relation between compressive stress and strain, F/A = Y(ΔL/L0), where Y is Young’s modulus of the material. From the discussion on linear expansion, the strain due to thermal expansion can be written as (ΔL/L0) = α(ΔT), where α is the coefficient of linear expansion. Thus, the stress becomes

F/A = Y[α(ΔT)].

(b) If the concrete slab has mass m, the thermal energy required to

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674

produce a change in temperature ΔT is Q = mc(ΔT), where c is the specific heat of concrete. Using the result from part (a), the absorbed thermal energy required to produce compressive stress F/A is

⎛ F/A ⎞ Q = mc ⎜ ⎝ Yα ⎟⎠ !

mc ⎛ F ⎞ ⎜ ⎟ Yα ⎝ A ⎠

−

m = ρV = (2.40 × 103 kg/m3)[(4.00 × 10 2 m)(1.00 m)(1.00 m)]

= 96.0 kg

(d) If the maximum compressive stress concrete can withstand is F/A = 2.00 × 107 Pa, the maximum thermal energy this slab can absorb before starting to break up is found, using Table 10.1, to be

Qmax =

(

)( )

) ( ( )

96.0 kg 880 J/kg ⋅°C mc ⎛ F ⎞ = 2.00 × 107 Pa ⎜ ⎟ −1 ⎛ ⎞ Yα ⎝ A ⎠ max 2.1 × 1010 Pa ⎜ 12 × 10−6 °C ⎟ ⎝ ⎠

(

)

= 6.7 × 106 J

(e) The change in temperature of the slab as it absorbs the thermal energy computed above is

Q 6.7 × 106 J ΔT = = = 79°C mc ( 96.0 kg ) ( 880 J/kg ⋅°C ) ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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(f) The rate the slab absorbs solar energy is

Pabsorbed = 0.5Psolar = 0.5(1.00 × 103 W) = 5 × 102 J/s

so the time required to absorb the thermal energy computed in (d) above is

6.7 × 106 J ⎛ 1 h ⎞ t= = ⎜ ⎟ ~ 4h Pabsorbed 5 × 102 J/s ⎝ 3 600 s ⎠ Q

11.17

When thermal equilibrium is reached, the water and aluminum will have a common temperature of Tf = 65.0°C. Assuming that the wateraluminum system is thermally isolated from the environment, Qcold = −Qhot, so mwcw(Tf − Ti,w) = −mAlcAl(Tf − Ti,Al), or

mw =

(

−mAl cAl Tf − Ti ,Al

(

cw Tf − Ti ,w

)

) = − (1.85 kg)(900 J/kg ⋅°C)(65.0°C − 150°C) ( 4 186 J/kg ⋅°C)(65.0°C − 25.0°C)

= 0.845 kg

11.18

If N pellets are used, the mass of the lead is Nmpellet. Since the energy lost by the lead must equal the energy absorbed by the water, |Nmpelletc(ΔT)|lead = [mc(ΔT)]water or the number of pellets required is

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676

11.19

( )

mwcw ΔT

mpellet clead (ΔT)

lead

(0.500 kg )( 4 186 J/kg ⋅°C)(25.0°C − 20.0°C) = 467 (1.00 × 10 kg )(128 J/kg ⋅°C)(200°C − 25.0°C) −3

The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver sample. Thus, [mccAl + mscCu + mwcw](ΔT)w = [mc|ΔT|]Ag Solving for the mass of the cup gives

ΔT Ag ⎤ 1 ⎡ mc = − ms cCu − mw cw ⎥ , ⎢ mAg cAg cAl ⎣ ( ΔT )w ⎦ !

(

)

mc =

)( ) ((

) ( )( ) ( )

⎡ ⎤ 87 − 32 1 ⎢ 400 g 234 − 40 g 387 − 225 g 4 186 ⎥ = 80 g ⎥ 900 ⎢ 32 − 27 ⎣ ⎦

(

)(

)

11.20 By conservation of energy, the two air masses will exchange energy such that Qwarm + Qcool = 0. Substitute Q = mcΔT to find the final temperature:

Qwarm = −Qcool mwarm cair (T − Twarm ) = −mcoolcair (T − Tcool ) T ( mwarm + mcool ) = mwarmTwarm + mcoolTcool

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677

mwarmTwarm + mcoolTcool ( 975 kg ) ( 30.0°C ) + ( 50.0 kg ) (18.0°C ) = mwarm + mcool 975 kg + 50.0 kg

= 29.4°C

11.21

(a) The total energy given up by the copper and the unknown sample equals the total energy absorbed by the calorimeter and water. Hence,

mCucCu|ΔT|Cu + munkcunk|ΔT|unk = [mccAl + mwcw](ΔT)w

Solving for the specific heat of the unknown material gives

cunk = !

cunk =

[ m c + m c ]( ΔT ) − m c c Al

w w

Cu Cu

munk ΔT unk

ΔT Cu

, or

⎡(100 g ) ( 900 J/kg ⋅°C ) + ( 250 g ) ( 4 186 J/kg ⋅°C ) ⎤ (10.0°C ) { ⎦ (70.0 g )(80.0°C) ⎣ 1

(

)(

− 50.0 g 387 J/kg ⋅°C 60.0°C

)}

= 1.82 × 103 J/kg ⋅°C

(b) The unknown could be beryllium but other possibilities also exist.

11.22

Qcold = −Qhot ⇒ mwcw(Tf − Ti,w) = −mFecFe(Tf − Ti,Fe) or

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678

Tf = =

mw cwTi ,w + mFe cFeTi ,Fe mw cw + mFe cFe

(20.0 kg )( 4 186 J/kg ⋅°C)(25.0°C) + (1.50 kg )( 448 J/kg ⋅°C)(600°C) (20.0 kg )( 4 186 J/kg ⋅°C) + (1.50 kg )( 448 J/kg ⋅°C)

= 29.6°C

11.23

Since the temperature of the water and the steel container is unchanged, and neither substance undergoes a phase change, the internal energy of these materials is constant. Thus, all the energy given up by the copper is absorbed by the aluminum, giving mAlcAl(ΔT)Al = mCucCu|ΔT| Cu, or

⎛ c ⎞ ⎡ ΔT ⎤ Cu ⎥ mAl = ⎜ Cu ⎟ ⎢ mCu ⎢ ⎥ c (ΔT) ⎝ Al ⎠ Al ⎣ ⎦ ⎛ 387 ⎞ ⎛ 85°C − 25°C ⎞ 2 =⎜ ⎟⎜ ⎟ 200 g = 2.6 × 10 g = 0.26 kg ⎝ 900 ⎠ ⎝ 25°C − 5.0°C ⎠

(

11.24

)

The kinetic energy given up by the car is absorbed as internal energy by the four brake drums (a total mass of 32 kg of iron). Thus, ΔKE = Q = mdrumscFe(ΔT), or

(

)(

) )

2 1 1 mcar vi2 1 500 kg 30 m/s ΔT = 2 =2 = 47°C mdrums cFe 32 kg 448 J/kg ⋅°C

(

)(

11.25 By conservation of energy mwatercwater ΔTwater + mCucCu ΔTCu = 0 . Solve for the final temperature and substitute known values to find © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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679

mwatercwater (T − Twater ) = −mCucCu (T − TCu ) T=

mwater cwaterTwater + mCu cCuTCu mwater cwater + mCu cCu

( 0.275 kg )( 4 186 J/kg ⋅°C )( 25.0°C ) + ( 0.100 kg )( 387 J/kg ⋅°C )( 90.0°C ) ( 0.275 kg )( 4186 J/kg ⋅°C ) + ( 0.100 kg )( 387 J/kg ⋅°C )

= 27.1°C

11.26

Assuming that the unknown-water-calorimeter system is thermally isolated from the environment, −Qhot = Qcold, or −mxcx(Tf − Ti,x) = mwcw(Tf − Ti,w) + mAlcAl(Tf − Ti,Al) and, since Ti,w = Ti,Al = Tcold = 25.0°C, we have

cx = (mwcw + mAlcAl)(Tf − Tcold)/mx(Ti,x − Tf)

(

)(

) (

)(

)

⎡ 0.285 kg 4 186 J/kg ⋅°C + 0.510 kg 900 J/kg ⋅°C ⎤ 32.0 − 25.0 °C ⎦ cx = ⎣ 0.125 kg 95.0°C − 32.0°C

yielding

(

)(

)

cx = 1.18 × 103 J/kg ⋅ °C

11.27 (a) The energy required to melt 2.00 kg of ice into liquid water is

(

)

Qmelt = +mL f = ( 2.00 kg ) 3.33 × 105 J/kg = 6.66 × 105 J (b) The energy remaining to warm the water is Qwarm = Qtotal − Qmelt = 9.30 × 105 J − 6.66 × 105 J = 2.64 × 105 J .

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680

(c) Apply the equation for specific heat to find the water’s final temperature: Q = mcΔT = mc (T − 0°C ) Qwarm 2.64 × 105 J T= = = 31.5°C mc ( 2.00 kg )( 4186 J/kg ⋅°C )

11.28 The water is at the boiling point so the energy required to boil it into steam at the same temperature is

(

)

Qboil = +mLv = ( 2.00 kg ) 2.26 × 106 J/kg = 4.52 × 106 J The additional energy required to raise the temperature of the steam from 100°C to 125°C is

Qsteam = mcΔT = msteam csteam (T − 100°C )

= ( 2.00 kg ) ( 2 010 J/kg ⋅°C ) ( 25°C ) = 1.01× 105 J

The total required thermal energy is then Qtotal = Qboil + Qsteam =

4.62 × 106 J .

11.29

Remember that energy must be supplied to melt the ice before its temperature will begin to rise. Then, assuming a thermally isolated system, Qcold = −Qhot, or miceLf + micecwater(Tf − 0°C) = −mwatercwater(Tf − 25°C) and

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681

Tf =

(

) )c

mwater cwater 25°C − mice L f

(m + m (= 825 g )( 4 186 J/kg ⋅°C)(25°C) − (75 g )(3.33 × 10 J/kg) (75 g + 825 g )( 4 186 J/kg ⋅°C) ice

water

yielding

11.30

Tf = 16°C

The total energy input required is

Q = (energy to melt 50. g of ice)

+ (energy to warm 50. g of water to 100°C)

+ (energy to vaporize 5.0 g water)

= (50. g)Lf + (50. g)cwater(100°C − 0°C) + (5.0 g)Lv

⎛ J ⎞ Thus, Q = 0.050 kg ⎜ 3.33 × 105 ⎟ kg ⎠ ⎝ ⎛ J ⎞ + 0.050 kg ⎜ 4 186 ⎟ 100°C − 0°C kg ⋅°C ⎠ ⎝ ⎛ J ⎞ + 5.0 × 10−3 kg ⎜ 2.26 × 106 ⎟ kg ⎠ ⎝

(

)

(

)

(

)

which gives Q = 4.9 × 104 J = 49 kJ

11.31

The conservation of energy equation for this process is

(energy to melt ice) + (energy to warm melted ice to T) = (energy to

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682

cool water to T)

miceLf + micecw(T − 0°C) = mwcw(80°C − T)

This yields T = !

mw cw ( 80°C ) − mice L f

(mice + mw ) cw

(1.0 kg )( 4 186 J/kg ⋅°C)(80°C) − (0.100 kg )(3.33 × 10 J/kg ) = 65°C T= (1.1 kg )( 4 186 J/kg ⋅°C) 5

11.32

The energy required is the following sum of terms:

Q = (energy to reach melting point) + (energy to melt) + (energy to reach boiling point) + (energy to vaporize) + (energy to reach 110°C)

Mathematically,

Q = m[cice[0°C − (−10°C)] + Lf + cwater(100°C − 0°C) + Lv + csteam(110°C − 100°C)]

This yields

⎡⎛ ⎛ ⎞ J ⎞ 5 J Q = 40 × 10−3 kg ⎢⎜ 2 090 ⎟ 10°C + ⎜ 3.33 × 10 ⎟ kg ⋅°C ⎠ kg ⎠ ⎝ ⎢⎣⎝ ⎤ ⎛ ⎛ ⎞ ⎛ J ⎞ J ⎞ 6 J + ⎜ 4 186 ⎟ 100°C + ⎜ 2.26 × 10 ⎟ + ⎜ 2 010 ⎟ 10°C ⎥ kg ⋅°C ⎠ kg ⎠ ⎝ kg ⋅°C ⎠ ⎝ ⎝ ⎥⎦

(

)

(

)

(

)

Topic 11

683

11.33

Q = 1.2 × 105 J = 0.12 MJ

Assuming all work done against friction is used to melt snow, the energy balance equation is f ⋅ s = msnowLf. Since f = µk(mskierg), the distance traveled is

(1.0 kg )(3.33 × 10 J/kg ) = 2.3 × 10 m = 2.3 km s= = µ ( m g ) 0.20 (75 kg ) ( 9.80 m/s ) msnow L f

11.34

skier

(a) Observe that the equilibrium temperature will lie between the two extreme temperatures (−10.0°C and +30.0°C) of the mixed materials. Also, observe that a water-ice change of phase can be expected in this temperature range, but that neither aluminum nor ethyl alcohol undergoes a change of phase in this temperature range. The thermal energy transfers we can anticipate as the system comes to an equilibrium temperature are:

ice at −10.0°C to ice at 0°C; ice at 0°C to liquid water at 0°C; water at 0°C to water at T; aluminum at 20.0°C to aluminum at T; ethyl alcohol at 30.0°C to ethyl alcohol at T.

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684

(b)

m (kg) c (J/kg · °C) L (J/kg)

Tf (°C) Ti (°C)

Expression

Qice

1.00

–10.0

micecice[0 – (–10.0°C)]

Qmelt

1.00

3.33 × 105 0

miceLf

4 186

micecwater(T – 0)

Qwater 1.00

2 090

QAl

0.500

900

20.0

mAlcAl[T – 20.0°C]

Qalc

6.00

2 430

30.0

malccalc[T – 30.0°C]

(d)

T= !

mAl cAl ( 20.0°C ) malc calc ( 30.0°C) − mice ⎡⎣ cice (10.0°C ) + L f ⎤⎦ mice cwater + mAl cAl + malc calc

Substituting in numeric values from the table in (b) above gives

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685

0.500 ) ( 900 ) ( 20.0 ) + ( 6.00 ) ( 2 430 ) ( 30.0 ) − (1.00 ) ⎡( 2 090 ) (10.0 ) + 3.33 × 10 ⎤ ( ⎣ ⎦ T= (1.00)( 4 186) + (0.500)(900) + (6.00)(2 430) 5

and yields

11.35

T = 4.81°C

Assume that all the ice melts. If this yields a result T > 0, the assumption is valid, otherwise the problem must be solved again based on a different premise. If all ice melts, energy conservation (Qcold = −Qhot) yields mice[cice[0°C − (−78°C)]Lf + cw(T − 0°C)] = −(mwcw + mcalcCu)(T − 25°C)

T= !

With

(mw cw + mcal cCu )( 25°C) − mice ⎡⎣ cice (78°C) + L f ⎤⎦ (mw + mice ) cw + mcal cCu

mw = 0.560 kg, mcal = 0.080 g, mice = 0.040, cw = 4 186 J/kg ⋅ °C,

cCu = 387 J/kg ⋅ °C, cice = 2 090 J/kg ⋅ °C, and Lf = 3.33 × 105 J/kg this gives

(

)(

) (

)( ) (

) (

)(

)( ( )

)

⎡ 0.560 4 186 + 0.080 387 ⎤ 25°C − 0.040 ⎡ 2090 78°C + 3.33 × 105 ⎤ ⎦ ⎣ ⎦ T=⎣ 0.560 + 0.040 4 186 + 0.080 387

(

)(

)

or T = 16°C and the assumption that all ice melts is seen to be valid.

11.36

At a rate of 400 kcal/h, the excess internal energy that must be

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686

eliminated in a half-hour run is

cal ⎞ ⎛ 4.186 J ⎞ ⎛ Q = ⎜ 400 × 103 0.500 h ) = 8.37 × 105 J ( ⎟ ⎜ ⎟ ⎝ h ⎠ ⎝ 1 cal ⎠ !

The mass of water that will be evaporated by this amount of excess energy is

Q 8.37 × 105 J mevaporated = = = 0.33 kg Lv 2.5 × 106 J/kg ! The mass of fat burned (and thus, the mass of water produced at a rate of 1 gram of water per gram of fat burned) is

( 400 kcal/h )( 0.500 h ) = 22 g = 22 × 10−3 kg mproduced = 9.0 kcal/gram!of!fat ! so the fraction of water needs provided by burning fat is

22 × 10−3 kg f= = = 0.067 or 6.7% mevaporated 0.33 kg mproduced

−

11.37 (a) The mass of 2.0 liters of water is mw = ρV = (103 kg/m3)(2.0 × 10 3 m3) = 2.0 kg.

The energy required to raise the temperature of the water (and pot) up to the boiling point of water is

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687

Qboil = (mwcw + mAlcAl)(ΔT)

⎡ ⎛ ⎛ J ⎞ J ⎞⎤ Qboil = ⎢ 2.0 kg ⎜ 4 186 + 0.25 kg 900 ⎟ ⎜ ⎟ ⎥ 100°C − 20°C kg ⎠ kg ⎠ ⎥⎦ ⎝ ⎝ ⎢⎣

(

)

(

)

(

)

= 6.9 × 105 J

The time required for the 14 000 Btu/h burner to produce this much energy is

tboil =

Qboil 14 000 Btu/h

⎞ 6.9 × 105 J ⎛ 1 Btu ⎜ ⎟ 14 000 Btu/h ⎝ 1.054 × 103 J ⎠

= 4.7 × 10−2 h = 2.8 min

(b) Once the boiling temperature is reached, the additional energy required to evaporate all of the water is

Qevaporate = mwLv = (2.0 kg)(2.26 × 106 J/kg) = 4.5 × 106 J

and the time required for the burner to produce this energy is

tevaporate =

Qevaporate 14 000 Btu/h

⎞ 4.5 × 106 J ⎛ 1 Btu ⎜ ⎟ 14 000 Btu/h ⎝ 1.054 × 103 J ⎠

= 0.30 h = 18 min

11.38

In one hour, the energy dissipated by the runner is

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688

ΔE = P ⋅ t = (300 J/s)(3 600 s) = 1.08 × 106 J

Ninety percent, or Q = 0.900(1.08 × 106 J) = 9.72 × 105 J, of this is used to evaporate bodily fluids. The mass of fluid evaporated is

Q 9.72 × 105 J m= = = 0.403 kg 6 L 2.41 × 10 J/kg v ! Assuming the fluid is primarily water, the volume of fluid evaporated in one hour is

⎛ 106 cm 3 ⎞ V= = = 4.03 × 10 m ⎜ = 403 cm 3 3 3 ⎟ ρ 1 000 kg/m ⎝ 1m ⎠ m

11.39

0.403 kg

(

−4

)

The energy required to melt 50 g of ice is

Q1 = miceLf = (0.050 kg)(333 kJ/kg) = 17 kJ

The energy needed to warm 50 g of melted ice from 0°C to 100°C is

Q2 = micecw(ΔT) = (0.050 kg)(4.186 kJ/kg ⋅ °C)(100°C) = 21 kJ

(a) If 10 g of steam is used, the energy it will give up as it condenses is

Q3 = msteamLv = (0.010 kg)(2 260 kJ/kg) = 23 kJ

Since Q3 > Q1, all of the ice will melt. However, Q3 < Q1 + Q2, so the final temperature is less than 100°C. From conservation of energy, © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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689

we find

Qcold = −Qhot

mice = [Lf = cw(T − 0°C)] = −msteam[−Lv + cw(T − 100°C)]

T= !

msteam [ Lv + cw (100°C)] − mice L f

(mice + msteam ) cw

giving 10 g ) ⎡ 2.26 × 10 + ( 4 186 ) (100 ) ⎤ − ( 50 g ) ( 3.33 × 10 ) ( ⎣ ⎦ T= = 40°C (50 g + 10 g )( 4 186) 6

(b) If only 1.0 g of steam is used, then !Q3′ = msteam Lv = 2.26 kJ . The energy 1.0 g of condensed steam can give up as it cools from 100°C to 0°C is

−

Q4 = msteamcw(ΔT) = (1.0 × 10 3 kg)(4.186 kJ/kg ⋅ °C)(100°C) = 0.42 kJ

Since !Q3′ + Q4 is less than Q1, not all of the 50 g of ice will melt, so the final temperature will be 0°C. The mass of ice which melts as the steam condenses and the condensate cools to 0°C is

Q3′ + Q4 Lf

(2.26 + 0.42) kJ = 8.0 × 10 kg = 8.0 g −3

333 kJ/kg

Topic 11

11.40

690 −

First, we use the ideal gas law (with V = 0.600 L = 0.600 × 10 3 m3 and T = 37.0°C = 310 K) to determine the quantity of water vapor in each exhaled breath: 3 −3 3 PV ( 3.20 × 10 Pa ) ( 0.600 × 10 m ) PV = nRT ⇒ n = = = 7.45 × 10−4 mol RT (8.31 J/mol ⋅ K )( 310 K ) !

⎛ g ⎞ ⎛ 1 kg ⎞ −5 m = nMwater = 7.45 × 10−4 mol ⎜ 18.0 ⎟ ⎜ 3 ⎟ = 1.34 × 10 kg mol ⎠ ⎝ 10 g ⎠ ⎝ !

(

)

The energy required to vaporize this much water, and hence the energy carried from the body with each breath, is

−

Q = mLv = (1.34 × 10 5 kg)(2.26 × 106 J/kg) = 30.3 J

The rate of losing energy by exhaling humid air is then

J ⎞⎛ breaths ⎞ ⎛ 1 min ⎞ ⎛ P = Q ⋅ ( respiration!rate ) = ⎜ 30.3 22.0 = 11.1 W ⎟ ⎜ ⎟ ⎝ breath ⎠ ⎝ min ⎠ ⎜⎝ 60 s ⎟⎠ !

11.41

(a) The bullet loses all of its kinetic energy as it is stopped by the ice. Also, thermal energy is transferred from the bullet to the ice as the bullet cools from 30.0°C to the final temperature. The sum of these two quantities of energy equals the energy required to melt part of the ice. The final temperature is 0°C because not all of the ice

Topic 11

691

melts.

(b) The total energy transferred from the bullet to the ice is

1 Q = KEi + mbullet clead 0°C − 30.0°C = mbullet vi2 + mbullet clead ( 30.0°C ) 2 ⎡ ( 2.40 × 102 m/s )2 ⎤ −3 = ( 3.00 × 10 kg ) ⎢ + (128 J/kg ⋅°C ) ( 30.0°C ) ⎥ = 97.9 J 2 ⎢⎣ ⎥⎦ !

The mass of ice that melts when this quantity of thermal energy is absorbed is

m= ! 11.42

(L ) f

water

⎛ 103 g ⎞ 97.9 J −4 = 2.94 × 10 kg ⎜⎝ 1 kg ⎟⎠ = 0.294 g 3.33 × 105 J/kg

(a) The rate of energy transfer by conduction through a material of area A, thickness L, with thermal conductivity k, and temperatures Th > Tc on opposite sides is P = kA(Th − Tc)/L. For the given windowpane, this is

)(

)

25°C − 0°C ⎛ ⎞ J ⎡ 1.0 m 2.0 m ⎤ P = ⎜ 0.8 = 6 × 103 J/s = 6 × 103 W ⎟⎣ −2 ⎦ s ⋅ m ⋅ °C ⎠ 0.62 × 10 m ⎝

(

)(

(b) The total energy lost per day is

E = P ⋅ Δt = (6 × 103 J/s)(8.64 × 104 s) = 5 × 108 J

11.43

The rate of energy transfer by conduction through a material having

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692

thermal conductivity κ, cross-sectional area A, thickness L and a temperature change of Th − Tc across it is P = κA(Th − Tc)/L. Hence, with

κ = 0.6 J/s ⋅ m ⋅ °C for water, the rate of energy transfer by conduction to the bottom of the pond is

( 0.6 J/s ⋅ m ⋅°C)(820 m )( 25°C − 12°C) = 3 × 10 J/s = 3 × 10 W P= 2

2.0 m

11.44

(a) The R value of a material is R = L/ κ, where L is its thickness and κ is the thermal conductivity. The R values of the three layers covering the core tissues in this body are:

1.0 × 10−3 m Rskin = = 5.0 × 10−2 m 2 ⋅ K/W 0.020 W/m ⋅ K ! 0.50 × 10−2 m Rfat = = 2.5 × 10−2 m 2 ⋅ K/W 0.020 W/m ⋅ K !

and

3.2 × 10−2 m Rtissue = = 6.4 × 10−2 m 2 ⋅ K/W 0.50 W/m ⋅ K !

so the total R value of the three layers taken together is

(

)

Rtotal = ∑ Ri = 5.0 + 2.5 + 6.4 × 10−2 i=1

= 1.4 × 10−2

m2 ⋅ K W

= 0.14

m2 ⋅ K W

(b) The rate of energy transfer by conduction through these three © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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693

layers with a surface area of A = 2.0 m2 and temperature difference of ΔT = (37 − 0)°C = 37°C = 37 K is 2 A ( ΔT ) ( 2.0 m ) ( 37 K ) P= = = 5.3 × 102 W 2 Rtotal 0.14 m ⋅ K/W !

11.45

⎛ ΔT ⎞ P = κ A⎜ , with ⎝ L ⎟⎠ ! cal ⎛ 102 cm ⎞ ⎛ 4.186 J ⎞ 1 κ = 0.200 = 83.7 ⎜ ⎟ ⎜ ⎟ cm ⋅°C ⋅s ⎝ 1 m ⎠ ⎝ 1 cal ⎠ s ⋅ m ⋅°C !

Thus, the energy transfer rate is

⎛ ⎞ ⎛ ⎞ J ⎡ 8.00 m 50.0 m ⎤ 200°C − 20.0°C P = ⎜ 83.7 ⎟ ⎦ ⎜⎝ 1.50 × 10−2 m ⎟⎠ s ⋅ m ⋅°C ⎠ ⎣ ⎝

(

)(

)

= 4.02 × 108 J/s = 4.02 × 108 W = 402 MW

11.46

The total surface area of the house is

A = Aside walls + Aend walls + Agables + Aroof

where Aside walls = 2[(5.00 m) × (10.0 m)] = 100 m2

Aend walls = 2[(5.00 m) × (8.00 m)] = 80.0 m2

⎡1 ⎤ Agables = 2 ⎢ base × altitude ⎥ ⎢⎣ 2 ⎥⎦ ⎡1 ⎤ = 2 ⎢ 8.00 m × 4.00 m tan 37.0° ⎥ = 24.1 m 2 ⎢⎣ 2 ⎥⎦

( ) ( (

) (

)

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694

Aroof = 2[(10.0 m) × (4.00 m/cos 37.0°)] = 100 m2

Thus, A = 100 m2 + 80.0 m2 + 24.1 m2 + 100 m2 = 304 m2

With an average thickness of 0.210 m, average thermal conductivity of −

4.8 × 10 4 kW/m ⋅ °C, and a 25.0°C difference between inside and outside temperatures, the energy transfer from the house to the outside air each day is

( )( )

⎡κ A ΔT ⎤ ⎥ Δt E = P Δt = ⎢ ⎢ ⎥ L ⎣ ⎦ ⎡ 4.8 × 10−4 kW/m ⋅°C 304 m 2 25.0°C ⎤ ⎥ (86 400 s) =⎢ ⎢ ⎥ 0.210 m ⎣ ⎦

( )

(

)(

)

E = 1.5 × 106 kJ = 1.5 × 109 J

The volume of gas that must be burned to replace this energy is

11.47

E heat of combustion

1.5 × 109 J

(9 300 kcal/m )( 4 186 J/kcal)) 3

= 39 m 3

Because the two pots hold the same quantity of water at the same initial temperature, the same amount, Q, of thermal energy is required to boil away the water in each pot. The time required to do this for one of the pots is t = Q/P, where P is the rate of energy conduction from the stove to the water through the bottom of the pot. The ratio of the

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695

times required for the two pots is

tAl ⎛ Q ⎞ ⎛ PCu ⎞ κ Cu A ( ΔT )/ L κ Cu = = = tCu ⎜⎝ PAl ⎟⎠ ⎜⎝ Q ⎟⎠ κ Al A ( ΔT )/ L κ Al !

Note that in the above calculation everything except the thermal conductivities canceled because the two pots are identical except for the material making up the bottoms. Thus, the time required to boil away the water in the aluminum bottomed pot is

⎛κ ⎞ ⎛ 397 W/m ⋅°C ⎞ tAl = ⎜ Cu ⎟ tCu = ⎜ ⎟ 425 s = 709 s ⎝ 238 W/m ⋅°C ⎠ ⎝ κ Al ⎠

(

11.48

)

The rate of energy transfer through a compound slab is

A(ΔT) R

, where R = ΣLi /κ i

(a) For the thermopane, R = Rpane + Rtrapped air + Rpane = 2Rpane + Rtrapped air

Thus, ⎛ 0.50 × 10−2 m ⎞ 1.0 × 10−2 m m 2 ⋅°C m 2 ⋅°C R = 2⎜ + = 0.01 + 0.43 = 0.44 ( ) W W ⎝ 0.8 W/m ⋅°C ⎟⎠ 0.023 4 W/m ⋅°C !

(1.0 m )(23°C) = 52 W P= 2

and

0.44 m 2 ⋅°C/W

Topic 11

696

(b) For the 1.0 cm thick pane of glass,

2 1.0 × 10−2 m −2 m ⋅°C R= = 1 × 10 0.8 W/m ⋅°C W !

(1.0 m )(23°C) = 2 × 10 W = 2 kW , about 38 times so P = 2

1 × 10 m ⋅°C/W −2

greater.

11.49

When the temperature of the junction stabilizes, the energy transfer rate must be the same for each of the rods, or PCu = PAl. The crosssectional areas of the rods are equal, and if the temperature of the junction is 50°C, the temperature difference is ΔT = 50°C for each rod.

Thus,

⎛ ΔT ⎞ ⎛ ΔT ⎞ PCu = κ Cu A ⎜ = κ Al A ⎜ = PAl , which gives ⎟ ⎝ LAl ⎟⎠ ⎝ LCu ⎠ ! ⎛ 238 W/m ⋅°C ⎞ ⎛κ ⎞ LAl = ⎜ Al ⎟ LCu = ⎜ (15 cm ) = 9.0 cm ⎝ κ Cu ⎠ ⎝ 397 W/m ⋅°C ⎟⎠ !

11.50

The energy transfer rate is

(5.0 kg )(3.33 × 10 J/kg ) = 58 W P= = = Δt Δt (8.0 h )(3 600 s/1 h ) ΔQ

mice L f

⎛ ΔT ⎞ Thus, P = κ A ⎜ gives the thermal conductivity as ⎝ L ⎟⎠ !

Topic 11

697

(58 W)(2.0 × 10 m) = 7.3 × 10 W/m ⋅°C κ= = A(ΔT) ( 0.80 m ) ( 25°C − 5.0°C ) −2

P⋅L

−2

11.51

The window will consist of the glass pane and a stagnant air layer on each side (see Example 11.10 in text). From Tables 11.3 and 11.4, the Rvalues for these layers are

2 0.40 × 10−2 m −3 m ⋅°C Rpane = = = 5.0 × 10 κ glass 0.80 W/m ⋅°C W !

ft 2 ⋅°F ⎛ 1 Btu/h ⎞ ⎛ 1 m ⎞ ⎛ 1°C ⎞ m 2 ⋅°C Rair = 0.17 ⋅⎜ = 0.030 ⎜ ⎟ Btu/h ⎝ 0.293 W ⎟⎠ ⎜⎝ 3.281 ft ⎟⎠ ⎝ 9/5°F ⎠ W layer !

and

Thus,

Rtotal = ΣRi = ( 0.030 + 5.0 × 10−3 + 0.030 ) !

m 2 ⋅°C m 2 ⋅°C = 0.065 W W

The energy loss through the window in a 12 hour interval is then

( Q = P ⋅t =

A Th − Tc ΣRi

) ⋅t = (2.0 m )(20°C) ⋅ (12 h )⎛ 3 600 s ⎞ = 2.7 × 10 J 2

0.065 m ⋅°C/W

⎜ ⎝

⎟ 1h ⎠

11.52 (a) Use the conversion T = TC + 273.15 to find T = 135 + 273.15 = 408 K (b) The ball has the surface area of sphere: A = 4πR2 = 4π(2.00 m)2 =

50.3 m 2 (c) With T0 = 25.0°C = 298 K, the net radiated power is

Topic 11

698

(

Pnet = σ Ae T 4 − T04

(

)

)(

(

)

= 5.67 × 10−8 W/m 2 ⋅ K 4 50.3 m 2 ( 0.450 ) ( 408 K ) − ( 298 K ) 4

)

= 2.54 × 10 4 W

11.53

The absolute temperatures of the two stars are TX = 5 727 + 273 = 6 000 K and TY = 11 727 + 273 = 12 000 K. Thus, the ratio of their radiated powers is

PY σ AeTY4 ⎛ TY ⎞ = = ⎜ ⎟ = (2)4 = 16 4 P σ AeT ⎝ TX ⎠ X ! X 11.54

From Stefan’s law, the power radiated by an object at absolute −

temperature T and surface area A is P = σAeT4, where σ = 5.669 6 × 10 8 W/m2 ⋅ K and e is the emissivity. Thus, the surface area of the filament must be

11.55

σ eT

75 W

(5.669 6 × 10 W/m ⋅ K )(1.0)(3 300 K ) −8

= 1.1 × 10−5 m 2

At a pressure of 1 atm, water boils at 100°C. Thus, the temperature on the interior of the copper kettle is 100°C and the energy transfer rate through the bottom is

2 ⎤ ⎛ 102°C − 100°C ⎞ ⎛ ΔT ⎞ ⎛ W ⎞⎡ P = κ A⎜ = 397 π 0.10 m ⎟ ⎟ ⎜ ⎟⎢ ⎥⎜ m ⋅°C ⎠ ⎣ ⎦ ⎝ 2.0 × 10−3 m ⎠ ⎝ L ⎠ ⎝

(

)

Topic 11

11.56

699

The mass of the water in the heater is

⎛ 3.786 L ⎞ ⎛ 1 m 3 ⎞ kg ⎞ ⎛ m = ρV = ⎜ 103 3 ⎟ ( 50.0 gal ) ⎜ = 189 kg 3 3 ⎝ m ⎠ ⎝ 10 L ⎟⎠ ⎜⎝ 10 L ⎟⎠ !

The energy required to raise the temperature of the water from 20.0°C to 60.0°C is

Q = mc(ΔT) = (189 kg)(4 186 J/kg)(60.0°C − 20.0°C) = 3.16 × 107 J

The time required for the water heater to transfer this energy is

11.57

Q P

3.16 × 107 J ⎛ 1 h ⎞ ⎜ ⎟ = 1.83 h 4 800 J/s ⎝ 3 600 s ⎠

The energy conservation equation is Qcold = −Qhot, or

miceLf + [(mice + mw)cw + mcup + cCu](12°C − 0°C) = −mPbcPb(12°C − 98°C)

This gives

⎛ J ⎞ 5 mPb ⎜ 128 ⎟ 86°C = 0.040 kg 3.33 × 10 J/kg kg ⋅°C ⎠ ⎝ + ⎡ 0.24 kg 4 186 J/kg ⋅°C + 0.100 kg 387 J/kg ⋅°C ⎤ 120°C ⎣ ⎦

(

) (

(

11.58

)(

) (

)

)(

)

mPb = 2.3 kg

(a) The net rate of energy transfer by radiation between a body at

Topic 11

700

absolute temperature T and its surroundings at absolute temperature T0 is !Pnet = σ Ae (T − T0 ) . Hence, with T = 33.0°C = 4

306 K, T0 = 20.0°C = 293 K, emissivity= e = 0.95, and surface area A = 1.50 m2, the net power radiated is

4 ⎛ W ⎞ ⎡ ⎤ Pnet = ⎜ 5.669 6 × 10−8 2 4 ⎟ 1.50 m 2 0.95 ⎢ 306 K − 293 K )4 ⎥ m ⋅K ⎠ ⎣ ⎦ ⎝

(

)( ) (

) (

)

= +1.1 × 102 W

(b) The positive sign on the net power radiated means that the body is radiating energy away faster than it is absorbing energy from the environment.

11.59

Assuming the aluminum-water-calorimeter system is thermally isolated from the environment, Qcold = −Qhot, or

mAlcAl(Tf − Ti,Al) = −mwcw(Tf − Ti,w) − mcalccal(Tf − Ti,cal)

Since Tf = 66.3°C and Ti,cal = Ti,w = 70.0°C, this gives

cAl = !

(mw cw + mcal ccal )(Ti ,w − Tf )

(

mAl Tf − Ti ,Al

)

, or

Topic 11

cAl =

701

⎡ ⎛ ⎛ J ⎞ J ⎞⎤ + 0.040 kg 630 ⎢ 0.400 kg ⎜ 4 186 ⎟ ⎜ ⎟ ⎥ 70.0 − 66.3 °C kg ⋅°C ⎠ kg ⋅°C ⎠ ⎥⎦ ⎝ ⎝ ⎢⎣

(

)

(

)

(

)

(0.200 kg )(66.3 − 27.0)°C

= 8.00 × 102

J kg ⋅°C

The variation between this result and the value from Table 11.1 is

⎛ variation ⎞ ⎛ 800 − 900 J/kg ⋅°C ⎞ %=⎜ × 100% = ⎜ × 100% = 11.1% ⎟ 900 J/kg ⋅°C ⎟⎠ ⎝ accepted!value ⎠ ⎝ !

which is within the 15% tolerance.

11.60

(a) With a body temperature of T = 37°C + 273 = 310 K and surroundings at temperature T0 = 24°C + 273 = 297 K, the rate of energy transfer by radiation is

(

Pradiation = σ Ae T 4 − T04

)

4 4⎤ ⎛ W ⎞ ⎡ = ⎜ 5.669 6 × 10−8 2 4 ⎟ 2.0 m 2 0.97 ⎢ 310 K − 297 K ⎥ = 1.6 × 102 W m ⋅K ⎠ ⎣ ⎦ ⎝

(

)( ) (

) (

)

(b) The rate of energy transfer by evaporation of perspiration is

Pperspiration =

Q Δt

mLv.perspiration Δt

(0.40 kg )(2.43 × 10 kJ/kg )(10 J/kJ) = 2.7 × 10 W = 3

3 600 s

Topic 11

702

⎛ kJ ⎞ ⎛ 1 h ⎞ ⎛ 103 J ⎞ Plungs = ⎜ 38 ⎟ ⎜ ⎟ = 11 W ⎟⎜ ⎝ h ⎠ ⎝ 3 600 s ⎠ ⎝ 1 kJ ⎠

(d) The excess thermal energy that must be dissipated is

⎛ kJ ⎞ ⎛ 1 h ⎞ ⎛ 103 J ⎞ 2 Pexcess = 0.80Pmetabolic = 0.80 ⎜ 2.50 × 103 ⎟ ⎜ ⎟ = 5.6 × 10 W ⎟⎜ h ⎠ ⎝ 3 600 s ⎠ ⎝ 1 kJ ⎠ ⎝

so the rate energy must be transferred by conduction and convection is

Pc&c = Pexcess − (Pradiation + Pperspiration + Plungs) = (5.6 − 1.6 − 2.7 − .11) × 102 W = 1.2 × 102 W

11.61

The total energy needed is

Q = mLv = (2.00 kg)(2.00 × 104 J/kg) = 4.00 × 104 J

and the time required to supply this energy is

⎛ 1 min ⎞ Q 4.00 × 10 4 J t= = = 4.00 × 103 s ⎜ = 66.7 min P 10.0 J/s ⎝ 60.0 s ⎟⎠ !

11.62

The energy added to the air in one hour is

Q = (Ptotal)t = [10(200 W)](3 600 s) = 7.20 × 106 J

and the mass of air in the room is

Topic 11

703

m = ρV = (1.3 kg/m3)[(6.0 m)(15.0 m)(3.0 m)] = 3.5 × 102 kg

The change in temperature is Q 7.2 × 106 J ΔT = = = 25°C mc ( 3.5 × 102 kg ) ( 837 J/kg ⋅°C ) !

giving

11.63

T = T0 + ΔT = 20°C + 25°C = 45°C

⎛ 80.0°C − T ⎞ ⎛ T − 30.0°C ⎞ In the steady state, PAu = PAg, or κ Au A ⎜ ⎟⎠ = κ Ag A ⎜⎝ ⎟⎠ ⎝ L L !

This gives

T= !

11.64

κ Au ( 80.0°C ) +κ Ag ( 30.0°C ) 314 ( 80.0°C ) + 427 ( 30.0°C ) = = 51.2°C κ Au + κ Ag 314 + 427

(a) The rate work is done against friction is

P = f ⋅ v = (50 N)(40 m/s) = 2.0 × 103 J/s = 2.0 kW

(b) In a time interval of 10 s, the energy added to the 10 kg of iron is

Q = P ⋅ t = (2.0 × 103 J/s)(10 s) = 2.0 × 104 J

and the change in temperature is

Q 2.0 × 10 4 J ΔT = = = 4.5°C mc (10 kg ) ( 448 J/kg ⋅°C ) !

Topic 11

11.65

704

(a) The energy required to raise the temperature of the brakes to the melting point at 660°C is

Q = mc(ΔT) = (6.00 kg)(900 J/kg⋅°C)(660°C − 20.0°C) = 3.46 × 106 J

The internal energy added to the brakes on each stop is

(

)(

)

1 1 Q1 = ΔKE = mcar vi2 = 1 500 kg 25.0 m/s 2 = 4.69 × 105 J 2 2 The number of stops before reaching the melting point is

Q 3.46 × 106 J N= = = 7 stops 5 Q 4.69 × 10 J 1 ! (b) As the car stops, it transforms part of its kinetic energy into internal energy due to air resistance. As soon as the brakes rise above air temperature, they transfer energy by heat to the air. If they reach a high temperature, they transfer energy to the air very quickly.

11.66

When liquids 1 and 2 are mixed, the conservation of energy equation is

mc1(17°C − 10°C) = mc2(20°C − 17°C)

⎛ 7⎞ c2 = ⎜ ⎟ c1 ⎝ 3⎠ !

When liquids 2 and 3 are mixed, energy conservation yields © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 11

705

mc3(30°C − 28°C) = mc2(28°C − 20°C)

Thus, we now know that

c3 = 4(7c1/3)

c3 = 4c2

c3/c1 = 28/3

Mixing liquids 1 and 3 will give mc1(T − 10°C) = mc3(30°C − T)

T= !

c1 (10°C ) + c3 ( 30°C ) 10°C + ( 28/3 ) ( 30°C ) = = 28°C c1 + c3 1 + ( 28/3 )

11.67 The power radiated by Earth is Pradiated = σ AeT 4 . Equating this to the absorbed power Pabs and solving for the temperature (with e = 1) gives

(

)

σ AeT 4 = Pabs = 960 W/m 2 π RE2

( 960 W/m )π R ( 960 W/m ) T = = ( 5.67 × 10 W/m ⋅ K ) 4π R (1) 4 ( 5.67 × 10 W/m ⋅ K ) 2

2 E

−8

2 E

−8

T = 255 K

11.68

(a) The surface area of the stove is Astove = Aends + Acylindrical = 2(π r 2 ) + (2π rh) , or side !

Astove = 2π(0.200 m)2 + 2π(0.200 m)(0.500 m) = 0.880 m2 The temperature of the stove is

5 Tstove = (400°F − 32.0°F) = 204°C = 477 K while that of the air in 9 5 the room is Troom = (70.0°F − 32.0°F) = 21.1°C = 294 K . If the 9 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 11

706

emissivity of the stove is e = 0.920, the net power radiated to the room is

(

4 4 P = σ Astove e Tstove − Troom

(

)

)(

) (

)

4 4⎤ ⎡ = 5.669 6 × 10−8 W/m 2 ⋅ K 4 0.800 m 2 0.920 ⎢ 477 K − 294 K ⎥ ⎣ ⎦

or P = 2.03 × 103 W = 2.03 × 103 J/s (b) The total surface area of the walls and ceiling of the room is

A = 4Awall + Aceiling = 4[(8.00 ft)(25.0 ft)] + (25.0 ft)2 = 1.43 × 103 ft2 If the temperature of the room is constant, the power lost by conduction through the walls and ceiling must equal the power radiated by the stove. Thus, from thermal conduction equation, P = A(Th = Tc)/ΣRi, the net R value needed in the walls and ceiling is

( ΣR =

A Th − Tc

) = (1.43 × 10 ft )(70.0°F − 32.0°F) ⎛ 1 054 J ⎞ ⎛ 1 h ⎞ 3

2.03 × 103 J/s

⎜ ⎟⎜ ⎟ ⎝ 1 Btu ⎠ ⎝ 3 600 s ⎠

or ΣRi = 7.84 ft2 ⋅ °F ⋅ h/Btu

11.69

−

A volume of 1.0 L of water has a mass of m =ρV = (103 kg/m3)(1.0 × 10 3 m3) = 1.0 kg The energy required to raise the temperature of the water to 100°C and then completely evaporate it is Q = mc(ΔT) + mLv, or

Topic 11

707

Q = (1.0 kg)(4 186 J/kg ⋅ °C)(100°C − 20°C) + (1.0 kg)(2.26 × 106 J/kg)

= 2.59 × 106 J

The power input to the water from the solar cooker is

⎡ π ( 0.50 m )2 ⎤ P = ( efficiency ) IA = ( 0.50 ) ( 600 W/m ) ⎢ ⎥ = 59 W 4 ⎢ ⎥⎦ ⎣ ! 2

so the time required to evaporate the water is

11.70

Q P

2.59 × 106 J 59 J/s

⎛ 1h ⎞ = 4.4 × 10 4 s ⎜ ⎟ = 12 h ⎝ 3 600 s ⎠

(

)

(a) From the thermal conductivity equation, P =κA[(Th − Tc)/L], the total energy lost by conduction through the insulation during the 24-h period will be

(

) (

) κA⎡ = (37.0°C − 23.0°C) + (37.0°C − 16.0°C)⎤⎦ (12.0 h ) L ⎣

Q = P1 12.0 h + P2 12.0 h

(0.012 0 J/s ⋅ m°C)(0.490 m ) ⎡14.0°C + 21.0°C⎤ (12.0 h )⎛ 3 600 s ⎞ Q= 2

9.50 × 10−2 m

⎣

⎦

⎜ ⎝

⎟ 1h ⎠

= 9.36 × 10 4 J

The mass of molten wax which will give off this much energy as it © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 11

708

solidifies (all at 37°C) is

Q 9.36 × 10 4 J m= = = 0.457 kg L f 205 × 103 J/kg !

(b) If the test samples and the inner surface of the insulation are preheated to 37.0°C during the assembly of the box, nothing undergoes a temperature change during the test period. Thus, the masses of the samples and insulation do not enter into the calculation. Only the duration of the test, inside and outside temperatures, along with the surface area, thickness, and thermal conductivity of the insulation need to be known.

11.71

The total power radiated by the Sun is P = σAeT4 where σ = 5.669 6 × −

10 8 W/m2 ⋅ K4 the emissivity is e = 0.986, the surface area (a sphere) is A = 4πr2 and the absolute temperature is T = 5 800 K. Thus,

−

P = (5.669 6 × 10 8 W/m2 ⋅ K4)4π(6.96 × 108 m)2(0.986)(5 800 K)4

or P = 3.85 × 1026 W Thus, the energy radiated each second is

E = P ⋅ Δt = (3.85 × 1026 J/s)(1.00 s) = 3.85 × 1026 J

11.72

We approximate the latent heat of vaporization of water on the skin (at 37°C) by asking how much energy would be needed to raise the

Topic 11

709

temperature of 1.0 kg of water to the boiling point and evaporate it. The answer is

( )

(

)(

)

L37°C ≈ cwater ΔT + L100°C = 4 186 J/kg ⋅°C 100°C − 37°C + 2.26 × 106 J/kg v v

37°C 6 !Lv ≈ 2.5 × 10 J/kg

Assuming that you are approximately 2.0 m tall and 0.30 m wide, you will cover an area of A = (2.0 m)(0.30 m) = 0.60 m2 of the beach, and the energy you receive from the sunlight in one hour is

Q = IA(Δt) = (1 000 W/m2)(0.60 m2)(3 600 s) = 2.2 × 106 J

The quantity of water this much energy could evaporate from your body is

Q 2.2 × 106 J m = 37°C ≈ = 0.9 kg 6 L 2.5 × 10 J/kg v ! The volume of this quantity of water is

m 0.9 kg V= = 3 ≈ 10−3 m 3 = 1 L 3 ρ 10 kg/m ! Thus, you will need to drink almost a liter of water each hour to stay hydrated. Note, of course, that any perspiration that drips off your body does not contribute to the cooling process, so drink up! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 11

11.73

710

During the first 50 minutes, the energy input is used converting m kilograms of ice at 0°C into liquid water at 0°C. The energy required is Q1 = mLf = m(3.33 × 105 J/kg), so the constant power input must be m ( 3.33 × 10 J/kg ) Q1 P= = 50 min ( Δt )1 ! 5

During the last 10 minutes, the same constant power input raises the temperature of water having a total mass of (m + 10 kg) by 2.0°C. The power input needed to do this is

(m + 10 kg ) c ( ΔT ) = (m + 10 kg )( 4 186 J/kg ⋅°C)(2.0°C) 10 min ( Δt) ( Δt) Q2

Since the power input is the same in the two periods, we have m ( 3.33 × 105 J/kg ) !

50 min

(m + 10 kg )( 4186 J/kg ⋅°C)( 2.0°C) 10 min

10 kg = 1.4 kg . which simplifies to (8.0)m = m + 10 kg or m = 7.0 ! 11.74

(a) First, energy must be removed from the liquid water to cool it to 0°C. Next, energy must be removed from the water at 0°C to freeze it, which corresponds to a liquid-to-solid phase transition. Finally, once all the water has frozen, additional energy must be removed from the ice to cool it from 0°C to −8.00°C.

Topic 11

711

(b) The total energy that must be removed is

Q = Qcool!water + Qfreeze + Qcool!ice = mw cw 0°C − Ti + mw L f + mw cice Tf − 0°C to!0°C at!0°C to −8.00°C !

⎡⎛ ⎤ ⎛ J ⎞ J ⎞ 5 J Q = 7.5 × 10−3 kg ⎢⎜ 4 186 + ⎜ 2 090 ⎟ −20.0 °C + 3.33 × 10 ⎟ −8.00 °C ⎥ kg ⋅°C ⎠ kg ⎝ kg ⋅°C ⎠ ⎢⎣⎝ ⎥⎦

(

)

= 3.25 × 10 4 J = 32.5 kJ

11.75 (a) In steady state, the energy transfer rate is the same for each of the rods, or PAl = PFe. Thus, ⎛ 100°C − T ⎞ ⎛ T − 0°C ⎞ κ Al A ⎜ = κ Fe A ⎜ ⎟ ⎟ ⎝ ⎠ ⎝ L L ⎠ !

giving ⎛ κ Al ⎞ 238 ⎞ T=⎜ (100°C) = ⎛⎜⎝ ⎟ (100°C ) = 75.0°C ⎟ 238 + 79.5 ⎠ ⎝ κ Al + κ Fe ⎠ !

(b) If L = 15 cm and A = 5.0 cm2, the energy conducted in 30 min is

⎡⎛ ⎛ 100°C − 75.0°C ⎞ ⎤ W ⎞ −4 2 Q = PAl ⋅t = ⎢⎜ 238 ⎟ ⎥ 1 800 s ⎟ 5.0 × 10 m ⎜ m ⋅°C 0.15 m ⎝ ⎠ ⎝ ⎠ ⎥⎦ ⎢⎣

(

)

(

)

= 3.6 × 10 4 J = 36 kJ

Topic 12

712

Topic 12 The Laws of Thermodynamics

QUICK QUIZZES 12.1

Choice (b). The magnitude of the work done on a gas during a thermodynamic process is equal to the area under the curve on a PV diagram. Processes in which the volume decreases do positive work on the gas, while processes in which the volume increases do negative work on the gas. The work done on the gas in each of the four processes shown is Wa = −4.00 × 105 J, Wb = +3.00 × 105 J, Wc = −3.00 × 105 J, and Wd = +4.00 × 105 J Thus, the correct ranking (from most negative to most positive) is a, c, b, d.

12.2

A is isovolumetric, B is adiabatic, C is isothermal, D is isobaric.

12.3

Choice (c). The highest theoretical efficiency of an engine is the Carnot efficiency given by ec = 1 − Tc/Th. Hence, the theoretically possible efficiencies of the given engines are

eA = 1 −

700 K 1 000 K

= 0.300, eB = 1 −

500 K 800 K

= 0.375, and eC = 1 −

300 K 600 K

= 0.500

Topic 12

713

and the correct ranking (from highest to lowest) is C, B, A.

12.4

Choice (b). ΔS =

Qτ T

= 0 and Q = 0 in an adiabatic process. If the

process was reversible, but not adiabatic, the entropy of the system could undergo a non-zero change. However, in that case, the entropy of the system’s surroundings would undergo a change of equal magnitude but opposite sign, and the total change of entropy in the universe would be zero. If the process was irreversible, the total entropy of the universe would increase. 12.5

The number 7 is the most probable outcome because there are six ways this could occur: 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. The numbers 2 and 12 are the least probable because they could only occur one way each: either 1-1, or 6-6. Thus, you are six times more likely to throw a 7 than a 2 or 12.

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS 12.2

The true statement is given in (d). Choice (c) is not true because the area is unsigned while the work can be either positive or negative. Choices (a) and (b) are false because the path on a PV diagram can go in either direction.

Topic 12

12.4

714

Temperature = a measure of molecular motion. Heat = the process through which energy is transferred between objects by means of random collisions of molecules. Internal energy = energy associated with random molecular motions plus chemical energy, strain potential energy, and an object’s other energy not associated with center of mass motion or location.

12.6

The true statement is given in (c). Choice (a) is false because temperature can change due to work W. Choice (b) is false because temperature decreases during an adiabatic expansion (as negative work is done on the gas). Choice (d) is false because temperature increases during an adiabatic compression (as positive work is done on the gas). Choice (e) is false for the same reason that that choice (d) is false.

12.8

Choice (d) is impossible because it would require an efficiency greater than 1 in violation of energy conservation.

12.10 (a) Q = 0 and W < 0 so ΔU < 0 and U decreases. (b) W = 0 and Q < 0 so ΔU < 0 and U decreases. (c) W = 0 and Q = 0 so ΔU = 0 and U remains constant. (d) The internal energy of an ideal gas is proportional to its temperature so as temperature increases, internal energy

Topic 12

715

increases. 12.12 (a) Entropy change is proportional to Q so the cooler object will have an entropy increase as it absorbs energy and warms. The correct answer is object B. (b) Object A loses energy to B so its entropy decreases (QA < 0). The correct answer is object A. (c) QA = −QB and entropy change is inversely proportional to temperature, so the cooler object will have the greater magnitude of entropy change. TA > TB so the correct answer is object B. 12.14

Heat is not a form of energy. Rather, it is the process of transferring energy through microscopic collisions between atoms or molecules. An object having a large mass and/or a high specific heat may be at a low temperature and still contain more internal energy than does a higher temperature, low-mass object.

12.16

The Clausius statement of the second law of thermodynamics says that in any thermodynamics process, reversible or irreversible, the total entropy of the universe must either remain constant (reversible process) or increase (irreversible process). Thus, if in a thermodynamics process, the entropy of a system changes by −6 J/K, the entropy of the environment (i.e., the rest of the universe) must

Topic 12

716

increase by +6 J/K or more. The correct choice here is (e).

ANSWERS TO EVEN NUMBERED PROBLEMS (a)

6.1 × 102 J

(b)

(d)

(e)

+2.0 × 102 J

12.4

(a)

−3.00 × 105 J

(b)

4.00 × 105 J

12.6

(a)

−16.0 × 105 J

(b)

(d)

(e)

−12.0 × 105 J

(a)

1.25 MJ

(b)

−1.25 MJ

12.10 (a)

−12 MJ

(b)

+12 MJ

12.12 (a) −1.75 × 103 J

(b)

36.1 K

(a)

8.12 × 104 Pa

(b)

3.65 × 104 J

(c)

(d)

977 K

(e)

1.22 × 105 J

(f) 8.55 × 104 J

12.2

12.8

12.14

(c)

−4.1 × 102 J

4.00 × 105 J

−5.68 × 104 J

(g) 1.42 × 105 J

12.16

(h)

additional energy must be transferred to the gas

(i)

1.42 × 105 J

(a)

−2.7 × 105 J

(b)

energy is transferred from the sprinter to the environment by

(j)

Q = ΔU − W

Topic 12

717

heat 12.18

6P0V0

12.20

3 3 !W = + 2 P0V0 > 0, ΔU = 0, Q = − 2 P0V0 < 0

12.22

(a)

−395 J

(c)

37 J

(a)

See Solution.

(c)

See Solution.

(d)

No. From part (c), !Q = 52 Wenv , and Wenv is positive for an isobaric

12.24

(b)

285 K; at point A

(b)

See Solution.

expansion process. Thus, the energy transfer by heat must be directed into the gas. 12.26 (a) 41.7 mol (c) 246 K 12.28

(b)

5.68 × 104 Pa

(d)

−6.84 × 104 J

WBC = 0, QBC < 0, ΔUBC < 0; WCA > 0, QCA < 0, ΔUCA < 0; WAB < 0, QAB > 0, ΔUAB > 0

12.30

12.32

(a)

WIAF = −76.0 J, WIBF = −101 J, WIF = −88.7 J

(b)

QIAF = 167 J, QIBF = 192 J, QIF = 180 J

600 kJ

12.34 (a)

564°C

Topic 12

718

(b)

No. Any real heat engine operates in an irreversible manner and has an efficiency less than the Carnot efficiency.

12.36

(a)

4 200 J

12.38

3.35 hp

12.40

(a)

12.42

0.391

12.44

(a)

7.69 × 108 J

0.051 2 (or 5.12%)

(b)

0.43 (or 43%)

(b)

5.67 × 108 J

(b)

5.27 × 1012 J

(a)

9.09 kW

(b)

11.9 kJ

12.48

(a)

+79 J/K

(b)

−79 J/K

12.50

6.06 kJ/K

12.52

(a) 2.13 × 105 J

(b)

−724 J/K

12.54

(a)

−3.45 J/K

(b)

+8.06 J/K

(d)

Thermal energy always flows from (Qr < 0) the hot reservoir at

(c)

+461 J/K

Topic 12

719

Th and into (Qr > 0) the cooler reservoir at Tc. Thus, in the expression for the total change in entropy, ΔS = Qr/Tc + Qr/Th = |Qr|/Tc − |Qr|/Th, the positive term has the smaller denominator and dominates the sum. 12.56 (a)

ΔSh = −|Qh|/Th

(c)

(b)

ΔSc = +|Qh|/Tc

ΔSuniverse = −|Qh|/Th + |Qh|/Tc

12.58 (a)

6.54 h

(b)

0.717 h

(c)

2.35 × 103 J

(d)

16.6 lifts

(e)

No, the human body is only about 25% efficient in converting chemical energy to mechanical energy. −

12.60

(a) 3.10 × 10 2 L/s

12.62

3.01

12.64

0.55 kg

12.66

(a)

251 J

(c)

104 J done by the gas

(d) 12.68

(a)

12.70

8 × 106 J/K ⋅ s

(b)

14.8 breaths/min

(b)

314 J

104 J done on the gas

(e)

ΔU = 0(cyclic process)

4P0V0

(b)

4P0V0

(c)

9.07 kJ

Topic 12

720

(a)

1.00 × 105 Pa

(d)

9.28 × 103 K; 2.12 × 106 Pa

(e)

0.259 kg; 0.494 kg/m3

(f)

2.93 × 103 m/s

12.74

(a)

2.4 × 106 J

12.76

(a)

−6.00 × 105 J

(d)

1.60 × 106 J

12.72

1.31 × 105 J

(c)

3.48 × 106 J

(b)

1.6 × 106 J

(c)

2.8 × 102 J

(b)

5.50 × 102 K

(c)

1.00 × 106 J

(b)

PROBLEM SOLUTIONS 12.1

(a) The work done on the gas in this constant pressure process is ⎛ nRTf nRTi ⎞ W = −P(ΔV ) = −P ⎜ − ⎟⎠ = −nR(Tf − Ti ) ⎝ P P !

W = −(0.200 mol)(8.31 J/mol ⋅ K)(573 K − 293 K) = −465 J

(b) The negative sign for work done on the gas indicates that the gas does positive work on its surroundings.

(Wenv )ab = Pa (Vb − Va ) 12.2

⎛ 10−3 m 3 ⎞ = ⎡⎣ 3(1.013 × 105 Pa) ⎤⎦ (3.0 L − 1.0 L) ⎜ ⎝ 1 L ⎟⎠

(a)

= 6.1 × 102 J

Topic 12

721

(b) (Wenv)bc = P(Vc − Vb) = 0

(Wenv )cd = Pc (Vd − Vc ) (c)

⎛ 10−2 m 3 ⎞ = ⎡⎣ 2 (1.013 × 105 Pa ) ⎤⎦ (1.0 L − 3.0 L ) ⎜ = −4.1 × 102 J ⎟ 1 L ⎝ ⎠

(d) (Wenv)da = P(Va − Vd) = 0

(e)

(Wenv )dcb = (Wenv )ab + (Wenv )bc + (Wenv )cd + (Wenv )da !

= +6.1 × 102 J + 0 − 4.1 × 102 J + 0 = +2.0 × 102 J

Note that, to 2 significant figures, this equals the area enclosed within the process diagram given above. 12.3

The constant pressure is P = (1.5 atm)(1.013 × 105 Pa/atm) and the work done on the gas is W = −P(ΔV). (a) ΔV = +4.0 m3 and W = −P(ΔV) = −(1.5 atm)(1.013 × 105 Pa/atm)(+4.0 m3) = −6.1 × 105 J (b) ΔV = −3.0 m3, so W = −P(ΔV) = −(1.5 atm)(1.013 × 105 Pa/atm)(−3.0 m3) = +4.6 × 105 J

12.4

The area under the graph in a PV diagram is equal in magnitude to the

Topic 12

722

work done on the gas. Work on the gas is negative if the gas expands and positive if it is compressed. (a) The magnitude of work is equal to the sum of areas A1 and A2 given by

(

)(

)

A1 = height × width = 1.00 × 105 Pa 2.00 m 3 = 2.00 × 105 J

(

A2 = 12 base × height = 12 1.00 m

)( 2.00 × 10 Pa ) = 1.00 × 10 J 5

The gas expands so the work done on the gas is negative and

W = − ( A1 + A2 ) = −3.00 × 105 J (b) The magnitude of work is equal to the sum of areas A1 and A2 given by

( )( ) A = height × width = ( 2.00 × 10 Pa ) (1.00 m ) = 2.00 × 10 J A1 = height × width = 1.00 × 105 Pa 2.00 m 3 = 2.00 × 105 J 5

The gas is compressed so the work done on the gas is positive and

W = ( A1 + A2 ) = 4.00 × 105 J 12.5

In each case, the work done on the gas is given by the negative of the area under the path on the PV diagram. Along those parts of the path where volume is constant, no work is done. Note that 1 atm = 1.013 × −

105 Pa and 1 Liter = 10 3 m3.

(a)

(

)

WIAF = WIA + WAF = −PI VA − VI + 0

(

) (

)

= − ⎡ 4.00 1.013 × 105 Pa ⎤ ⎡ 4.00 − 2.00 × 10−3 m 3 ⎤ = −810 J ⎦ ⎣ ⎦⎣

Topic 12

723

(

) ( ) 1 1 = − ( P − P ) (V − V ) − P (V − V ) = − ( P + P ) (V − V ) 2 2 1⎡ = − ( 4.00 + 1.00 ) (1.013 × 10 Pa ) ⎤ ( 4.00 − 2.00 ) × 10 m ⎦ 2⎣

WIF = − triangular area − rectangular area (b)

−3

= −507 J

(c)

12.6

(

WIBF = WIB + WBF = 0 − PB VF − VB

(

) (

)

= − ⎡1.00 1.013 × 105 Pa ⎤ ⎡ 4.00 − 2.00 × 10−3 m 3 ⎤ = −203 J ⎦ ⎣ ⎦⎣

Calculate the area under each indicated path section. The work on the gas for each section is positive if the gas is compressed and negative if the gas expands. (a) The area under path AB is a rectangle of height 4.00 × 105 Pa and width 4.00 m3 so that

(

)(

)

AAB = height × width = 4.00 × 105 Pa 4.00 m 3 = 16.0 × 105 J

The gas expands on path AB so the work is negative and

WAB = −AAB = −16.0 × 105 J (b) Path BC is a vertical line with ΔVBC = 0 so that ABC = 0 and WBC = 0. (c) The area under path CD is a rectangle of height 1.00 × 105 Pa and width 4.00 m3 so that

(

)(

)

ACD = 1.00 × 105 Pa 4.00 m 3 = 4.00 × 105 J

The gas is compressed on path CD so the work is positive and

Topic 12

724

WCD = +ACD = 4.00 × 105 J (d) Path DA is a vertical line with ΔVDA = 0 so that ADA = 0 and WDA = 0. (e) The work WABCDA = WAB + WBC + WCD + WDA = −16.0 × 105 J + 0 + 4.00 × 105 J + 0 = −12.0 × 10 J . 5

12.7

With Pf = Pi = P, the ideal gas law gives PfVf − PiVi = P(ΔV) = nR(ΔT), so the work done by the gas is

⎛ m ⎞ Wenv = +P(ΔV ) = nR(ΔT) = ⎜ R Tf − Ti ⎝ MHe ⎟⎠ !

(

)

If Wenv = 20.0 J, the mass of helium in the gas sample is

( )=

(20.0 J)( 4.00 g/mol) R (T − T ) ( 8.31 J/mol ⋅ K ) ( 373 K − 273 K )

Wenv MHe f

= 0.096 3 g = 96.3 mg

12.8

(a) The work done by the gas as it expands from point A to point B is given by the area under the PV diagram between these points. Consider the sketch given below and observe that this area can be broken into two rectangular areas and two triangular areas. The total area is given by

Wenv = (100 kPa ) ( 5.00 m 3 ) + !

1 300 kpa ) ( 2.00 m 3 ) ( 2

1 100 kPa ) (1.00 m 3 ) + ( 200 kPa ) ( 2.00 m 3 ) ( 2

Topic 12

725

( )( )

(

)( )

Wenv = 1 250 kPa m 3 = 1.25 × 103 N m 2 m 3 = 1.25 MJ

(b) When the volume is decreasing, the work done by the gas is the negative of the area under the PV diagram. Thus, if the gas is compressed from point B to point A along the same path, Wenv = −1.25 MJ 12.9

(a) From the ideal gas law, nR = PVf/Tf = PVi/Ti. With pressure constant this gives

⎛ Vf ⎞ Tf = Ti ⎜ ⎟ = (273 K)(4) = 1.09 × 103 K ⎝ Vi ⎠ ! (b) The work done on the gas is

(

)

(

)

W = −P ( ΔV ) = − PV f − PVi = −nR Tf − Ti = −nR ( 4Ti − Ti ) ! 12.10

= − (1.00 mol ) ( 8.31 J/mol ⋅ K ) ⎡⎣ 3 ( 273 K ) ⎤⎦ = −6.81 × 103 J = −6.81 kJ

(a) The work done on the fluid is the negative (since Vf > Vi) of the area under the curve on the PV diagram. Thus,

Topic 12

726

{

1 Wif = − ( 6 × 106 !Pa − 0 ) ( 2!m 3 − 1!m 3 ) + ⎡⎣(6 − 2) × 106 !Pa ⎤⎦ (2!m 3 − 1!m 3 ) 2 !

}

+ ( 2.00 × 106 !Pa − 0 ) ( 4!m 3 − 2!m 3 ) Wif = −12 × 106 J = −12 MJ

(b) When the system follows the process curve in the reverse direction (with Vf < Vi), the work done on the fluid equals the area under the process curve, which is the negative of that computed in (a). Thus, Wfi = −Wif = +12 MJ 12.11 (a) Substitute the values Q = 925 J and W = −102 J (the work done on the gas is negative because it expands to a larger volume) into the first law of thermodynamics to find

ΔU = Q + W = 925 J − 102 J= 823 J (b) Use the relation ΔU = nCvΔT to find the change in temperature. Here, n = 5.00 moles and C = 3 R (for helium, a monatomic gas) to v

find

ΔU ΔU = nCv ΔT = 32 nRΔT → ΔT = 3 2 nR 823 J ΔT = 3 = 13.2 K 2 ( 5.00 mol ) ( 8.31 J/mol ⋅ K ) 12.12 (a) From the problem statement, Q = 1.25 × 103 J and W = −PΔV = −

−(1.50 × 105 Pa)(2.00 × 10 2 m3) = −3.00 × 103 J. Substitute values into © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

727

the first law of thermodynamics to find ΔU = Q + W = 1.25 × 103 J − 3.00 × 103 J ΔU = −1.75 × 103 J

(b) From the ideal gas law, PΔV = nRΔT. Solve for ΔT and substitute values to find

(

)(

1.50 × 105 Pa 2.00 × 10−2 m 3 PΔV ΔT = = nR (10.0 mol )( 8.31 J/mol ⋅ K )

)

= 36.1 K

12.13 From kinetic theory, the average kinetic energy per molecule is 3 3⎛ R ⎞ KE molecule = kBT = ⎜ ⎟T 2 2 ⎝ NA ⎠

For a monatomic ideal gas containing N molecules, the total energy associated with random molecular motions is 3⎛ N ⎞ 3 U = N ⋅ KE molecule = ⎜ RT = nRT ⎟ 2 ⎝ NA ⎠ 2 !

Since PV = nRT for an ideal gas, the internal energy of a monatomic ideal gas is found to be given by U = 32 PV . ! 12.14 (a) The initial absolute temperature is Ti = 20.0° + 273.15 = 293 K, so the initial pressure is nRTi (10.0 mol) ( 8.31 J/mol ⋅ K ) (293 K) Pi = = = 8.12 × 10 4 Pa 3 Vi 0.300 m ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

728

(b) For a monatomic ideal gas, the internal energy is U = 3nRT/2. Thus,

3 3 Ui = nRTi = (10.0 mol) ( 8.31 J/mol ⋅ K ) (293 K) = 3.65 × 10 4 J 2 2 ! (c) The work done on the gas in this isobaric expansion is W = −P(ΔV) = −(−8.12 × 104 Pa)(1.000 m3 − 0.300 m3) = −5.68 × 104 J

Pf V f

(8.12 × 10 )(1.00 m ) 4

= = 977 K (d) Tf = nR (10.0 mol ) ( 8.31 J/mol ⋅ K ) ! 3 3 5 (e) U f = nRTf = (10.0 mol) ( 8.31 J/mol ⋅ K ) (977 K) = 1.22 × 10 J 2 2 ! (f) ΔU = Uf − Ui = 1.22 × 105 J − 3.65 × 104 J = 8.55 × 104 J (g) ΔU − W = 8.55 × 104 J − (−5.68 × 104 J) = +1.42 × 105 J (h) Since ΔU − W > 0, the increase in the internal energy of the gas exceeds the energy transferred to the gas by work. Thus, additional energy must be transferred to the gas by heat. (i) The additional energy that must be transferred to the gas by heat is ΔU − W = +1.42 × 105 J

[See part (g) above.]

(j) The suggested relationship is Q = ΔU − W, which is a statement of the first law of thermodynamics.

Topic 12

12.15

729

(a) Along the direct path IF, the work done on the gas is

W = − ( −area under curve ) 1 ⎡ ⎤ = − ⎢(1.00 atm)(4.00 L − 2.00 L) + (4.00 atm − 1.00 atm)(4.00 L − 2.00 L) ⎥ 2 ⎣ ⎦ !W = −5.00 atm ⋅ L Thus, ΔU = Q + W = 418 J − 5.00 atm ⋅ L ⎛ 1.013 × 103 Pa ⎞ ⎛ 10−4 m 4 ⎞ = 418 J − (5.00 atm ⋅ L) ⎜ ⎟⎜ ⎟ = −88.5 J 1 atm ⎝ ⎠⎝ 1L ⎠

(b) Along path IAF, the work done on the gas is

⎛ 1.013 × 105 Pa ⎞ ⎛ 10−3 m 3 ⎞ W = − ( 4.00 atm ) ( 4.00 L − 2.00 L ) ⎜ ⎟⎠ ⎜⎝ 1 L ⎟⎠ = −810 J 1 atm ⎝ ! From the first law, Q = ΔU − W = −88.5 J − (−810 J) = 722 J 12.16

(a) We treat the sprinter’s body as a thermodynamic system and apply the first law of thermodynamics, ΔU = Q + W. Then, with ΔU = −7.5 × 105 J and W = −4.8 × 105 J (negative because the sprinter does work on the environment), the energy absorbed as heat is Q = ΔU − W = −7.5 × 105 J − (−4.8 × 105 J) = −2.7 × 105 J (b) The negative sign in the answer to part (a) means that energy is transferred from the sprinter to the environment by heat.

Topic 12

12.17

730

⎛ 1.013 × 103 Pa ⎞ ⎛ 10−3 m 3 ⎞ W = −P ( ΔV ) = − ( 0.800 atm ) ( −7.00 L ) ⎜ ⎟⎠ ⎜⎝ 1 L ⎟⎠ 1 atm ⎝ (a) = 567 J ! (b) ΔU = Q + W = −400 J + 567 J = 167 J

12.18

The work done on the gas is the negative (since Vf > Vi) of the area under the curve on the PV diagram, or

1 3 ⎡ ⎤ W = − ⎢ P0 ( 2V0 − V0 ) + ( 2P0 − P0 ) ( 2V0 − V0 ) ⎥ = − P0V0 2 2 ⎣ ⎦ ! From the result of Problem 11 (given as a hint in this problem),

( )( )

3 3 3 3 9 ΔU = Pf V f − PiVi = 2P0 2V0 − P0V0 = P0V0 2 2 2 2 2 9 ⎛ 3 ⎞ Thus, from the first law, Q = ΔU − W = P0V0 − ⎜ − P0V0 ⎟ = 6P0V0 ⎝ ⎠ 2 2 ! 12.19

(a) The change in the volume occupied by the gas is

(

)

ΔV = V f − Vi = A L f − Li = ( 0.150 m 2 ) ( −0.200 m ) = −3.00 × 10−2 m 3 ! and the work done by the gas is

W = +P ( ΔV ) = ( 6 000 Pa ) ( −3.00 × 10−2 m 3 ) = −180 J ! env (b) The first law of thermodynamics is ΔU = Q + W = −Qoutput − Wenv. Thus, if ΔU = −8.00 J, the energy transferred out of the system by heat is Qoutput = −ΔU − Wenv = −(−8.00 J) − (−180 J) = +188 J © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

12.20

731

The work done on the gas is the negative (since Vf < Vi) of the area under the curve on the PV diagram,

1 3 ⎡ ⎤ W = − ⎢ P0 (V0 − 2V0 ) + ( 2P0 − P0 ) (V0 − 2V0 ) ⎥ = + P0V0 , or W > 0 2 2 ⎣ ⎦ !

The change in the internal energy of this monatomic ideal gas is

3 3 3 3 ΔU = Pf V f − PiVi = ( 2P0 ) (V0 ) − ( P0 ) ( 2V0 ) = 0 2 2 2 2 ! 3 3 Then, from the first law, Q = ΔU − W = 0 − P0V0 = − P0V0 , or Q < 0 ! 2 2 ! 12.21

(a) W = F ⋅ d = (25.0 × 102 N)⋅(0.130 m) = 3.25 × 103 J = 3.25 kJ (b) Since the internal energy of an ideal gas is a function of temperature alone, the change in the internal energy in this isothermal process is ΔU = 0. (c) From the first law of thermodynamics, Q = ΔU − W = 0 − (3.25 kJ) = −3.25 kJ (d) If the energy exchanged as heat is Q = 0 while the work done on the gas is positive (W > 0), the first law of thermodynamics, ΔU = Q + W = 0 + W > 0, tells us that the internal energy of the system must increase. Since the internal energy of an ideal gas is directly proportional to the absolute temperature, the temperature must increase.

Topic 12

12.22

732

(a) The work done on the gas as it goes from point A to point C is the negative of the area under the PV diagram between these points. Consider the sketch given below and observe that this area can be broken into a rectangular area and a triangular area. The work done on the gas is

W = − [(Area 1) + (Area 2)]

1 = −(0.500 atm)(6.00 L) − (0.300 atm)(6.00 L) 2 ⎛ 1.013 × 105 N/m 2 ⎞ ⎛ 10−3 m 3 ⎞ = −3.90 atm ⋅ L ⎜ ⎟⎠ ⎜⎝ 1 L ⎟⎠ = −395 J 1 atm ⎝

(b) The absolute temperature of an ideal gas is given by T = PV/nR. Thus, the lowest temperature occurs where the product PV is the smallest. This is seen to be at point A, and the temperature at this point is

TA =

PAVA nR

(0.300 atm)(2.00 L) (0.025 6 mol)(0.082 1 L⋅atm/mol ⋅ K)

= 285 K

(a) In an isothermal process involving an ideal gas, the work done on

Topic 12

733

the gas is W = −Wenv = −nRT ln (Vf/Vi). But, when temperature is constant, the ideal gas law gives PiVi = PfVf = nRT and we may write the work done on the gas as

⎛V ⎞ f W = −PiVi ln ⎜ ⎟ ⎜⎝ Vi ⎟⎠ ⎛ 1.25 m 3 ⎞ = − 1.00 × 103 Pa 0.500 m 3 ln ⎜ = −4.58 × 10 4 J 3⎟ ⎝ 0.500 m ⎠

(

)(

)

(b) The change in the internal energy of an ideal gas is ΔU = nCv(ΔT), and for an isothermal process, we have ΔU = 0. Thus, from the first law of thermodynamics, the energy transfer by heat in this isothermal expansion is Q = ΔU − W = 0 − (−4.85 × 104 J) = +4.58 × 104 J (c) ΔU = 0 12.24

[See part (b) above.]

(a) From the ideal gas law, PiVi = nRTi and PfVf = nRTf. Thus, if Pi = Pf = P, subtracting these two expressions gives PVf − PVi = nRTf − nRTi, or P(ΔV) = nR(ΔT). (b) For a monatomic, ideal gas containing N gas atoms, the internal

( ) ( )( k T ) = nRT . Thus, the change in

2 energy is U = N 12 mv = nN A

3 2

internal energy of this gas in a thermodynamic process is 3 !ΔU = 2 nR(ΔT) . But, using the result of part (a) above, we have,

Topic 12

734

for an isobaric process involving an monatomic ideal gas, 3 3 3 ΔU = nR(ΔT) = P(ΔV ) = Wenv 2 2 2 !

(c) We recall that the work done on the gas is W = −Wenv, and use the first law of thermodynamics to find that the energy transferred to the gas by heat to be 3 Q = ΔU − W = ΔU + Wenv = Wenv + Wenv 2 !

5 Q = Wenv 2

(d) In an isobaric expansion (⇒ ΔV > 0), the work done on the environment is Wenv = P(ΔV) > 0. Thus, from the result of part (c) above, the energy transfer as heat is Q > 0, meaning that the energy flow is into the gas. Therefore, it is impossible for the gas to exhaust thermal energy in an isobaric expansion 12.25 (a) The work done on a gas at constant pressure is W = −PΔV. Substitute values to find

(

)(

)

W = − 1.50 × 105 Pa 0.500 m 3 − 1.25 m 3 = 1.13 × 105 J (b) The relation ΔU = nCvΔT holds for all ideal gases, whether or not the volume is constant. Here the gas is monatomic so Cv = 32 R . From the ideal gas law applied to a constant pressure process, nΔT = PΔV/R so that

Topic 12

735

⎛ PΔV ⎞ 3 ΔU = Cv ( nΔT ) = ( 32 R ) ⎜ = PΔV ⎝ R ⎟⎠ 2

(

)(

)

ΔU = 32 1.50 × 105 Pa 0.500 m 3 − 1.25 m 3 = −1.69 × 105 J (c) Use the first law of thermodynamics to find

ΔU = Q + W Q = ΔU − W = −1.69 × 105 J − 1.13 × 105 J = −2.82 × 105 J (d) The initial temperature is Ti = 425 K. To find the final temperature, take the ratio of the ideal gas law in the initial and final states and solve for Tf :

PVi = nRTi and PV f = nRTf PVi nRTi = PV f nRTf

⎛ Vf ⎞ ⎛ 0.500 m 3 ⎞ → Tf = ⎜ ⎟ Ti = ⎜ 425 K ) = 1.70 × 10 2 K 3 ⎟( ⎝ 1.25 m ⎠ ⎝ Vi ⎠

12.26 The gas expands adiabatically with Vi = 0.750 m3, Vf = 1.50 m3, Pi = 1.50 × 105 Pa, and Ti = 325 K. (a) Use the ideal gas law to find the number of moles:

PV = nRT

(

)(

)

1.50 × 105 Pa 0.750 m 3 PV PiVi n= = = = 41.7 mol RT RTi ( 8.31 J/mol ⋅ K )( 325 K ) (b) For an adiabatic process, PiViγ = Pf V fγ where γ = Cp/Cv = 7/5 is the adiabatic index for a diatomic gas. Solve for the final gas pressure to find

Topic 12

736

⎛V ⎞ ⎛ 0.750 m 3 ⎞ 5 Pf = Pi ⎜ i ⎟ = 1.50 × 105 Pa ⎜ = 5.68 × 10 4 Pa 3 ⎟ V 1.50 m ⎝ ⎠ ⎝ f⎠

(

)

(c) To find the final gas temperature, take the ratio of the ideal gas law for quantities in the final and initial states, and solve for Tf :

Pf V f PiVi

nRTf nRTi

( (

)( )(

) )

⎛ 5.68 × 10 4 Pa 1.50 m 3 ⎞ ⎛ Pf V f ⎞ Tf = ⎜ Ti = ⎜ ( 325 K ) = 246 K 5 3 ⎟ 1.50 × 10 Pa 0.750 m ⎝ PiVi ⎟⎠ ⎝ ⎠ (d) From the first law of thermodynamics, the work done on a diatomic gas during an adiabatic (Q = 0) process is

ΔU = Q + W → W = ΔU = nCv ΔT W = ( 41.7 mol ) ( 52 R ) ( 246 K − 325 K ) = −6.84 × 10 4 J 12.27 (a) From PV = nRT, and with Vf = Vi = V = 0.200 m3, we have ⎛ 1.013 × 105 N m 2 ⎞ 5.00 atm ⎜ 0.200 m 3 ) ( ⎟ 1 atm ⎝ ⎠ PV n= i = = 40.6 mol RTi 8.31 J/mol ⋅ K ) ( 300 K ) ( !

(b) The total heat capacity of the gas is Ctotal = nCv = n(3R/2) = (40.6 mol)(1.50)(8.31 J/mol⋅K) = 506 J/K (c) The work done by the gas during this constant volume process is W=0 (d) From the first law of thermodynamics, ΔU = Q + W = 16.0 kJ + 0 =

Topic 12

737

16.0 kJ. (e) The change in internal energy of a monatomic ideal gas is ΔU = nCv(ΔT) = CtotalΔT so the change in temperature is

ΔU 16.0 kJ 16.0 × 103 J ΔT = = = = 31.6 K C 506 J/K 506 J/K total ! (f) The final temperature is Tf = Ti + ΔT = 300 K + 31.6 K = 332 K (g) From the ideal gas law, the final pressure of the gas is

Pf =

! 12.28

nRTf V

(40.6 mol)(8.31 J/mol ⋅ K)(332 K) 0.200 m 3

⎛ ⎞ 1 atm = 5.60 × 103 Pa ⎜ = 5.53 atm 5 ⎝ 1.013 × 10 Pa ⎟⎠

Volume is constant in process BC, so WBC = 0. Given that QBC < 0, the first law shows that ΔUBC = QBC + WBC = QBC + 0. Thus, ΔUBC < 0. For process CA, ΔVCA = VA − VC < 0, so W = −P(ΔV) shows that WCA > 0. Then, given that ΔUCA < 0, the first law gives QCA = ΔUCA − WCA and QCA < 0. In process AB, the work done on the system is W = −(area under curve AB) where

1 (area under curve!AB) = PA (VB − VA ) + (PB − PA )(VB − VA ) > 0 2 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

738

Hence, WAB < 0. For the cyclic process, ΔU = ΔUAB + ΔUBC + ΔUCA = 0, so ΔUAB = (ΔUBC + ΔUCA). This gives ΔUAB > 0, since both ΔUBC and ΔUCA are negative. Finally, from the first law, Q = ΔU − W shows that QAB > 0, since both ΔUAB and −WAB are positive. 12.29 (a) The original volume of the 738 aluminium is V0 = m/ ρAl, and the change in volume is ΔV = βV0(ΔT) = (3αAl)(m/ρAl)(ΔT). The work done by the 738 luminium is then

( ) ( )( )( ) ⎞ 5.0 kg ⎡ ⎤⎛ = (1.013 × 10 Pa ) 3 ⎢ 24 × 10 ( °C ) ⎥ ⎜ ⎟ (70°C ) ⎣ ⎦ ⎝ 2.70 × 10 kg m ⎠

Wenv = +P ΔV = P 3α Al m ρ Al ΔT 3

−6

−1

= 0.95 J (b) The energy transferred by heat to the aluminum is

Q = mcAl (ΔT) = (5.0 kg)(900 J/kg ⋅°C)(70°C) = 3.2 × 105 J (c) The work done on the aluminum is W = −Wenv = −0.95 J, so the first law gives ΔU = Q + W = 3.2 × 105 J − 0.95 J = 3.2 × 105 J 12.30

(a) The work done on the gas in each process is the negative of the area under the process curve on the PV diagram. For path IAF, WIAF = WIA + WAF = 0 + WAF, or

Topic 12

739

⎡ ⎛ 1.013 × 105 Pa ⎞ ⎤ ⎡ ⎛ 10−3 m 3 ⎞ ⎤ WIAF = − ⎢(1.50 atm ) ⎜ 0.500 L ) ⎜ 1 L ⎟ ⎥ = −76.0 J ⎟⎠ ⎥ ⎢( 1 atm ⎝ ⎝ ⎠⎦ ⎣ ⎦⎣ ! For path IBF, WIBF = WIB + WBF = WIB + 0, or

⎡ ⎛ 1.013 × 105 Pa ⎞ ⎤ ⎡ ⎛ 10−3 m 3 ⎞ ⎤ WIBF = − ⎢( 2.00 atm ) ⎜ 0.500 L ( ) ⎟⎠ ⎥ ⎢ ⎜⎝ 1 L ⎟⎠ ⎥ = −101 J 1 atm ⎝ ⎣ ⎦ ⎣ ⎦ ! For path IF, WIF = WAF − (triangular area), or 5 ⎤⎡ ⎛ 10−3 m 3 ⎞ ⎤ 1⎡ 3 ⎛ 1.013 × 10 Pa ⎞ WIF = −76.0 J − ⎢ 0.500 m ⎜ ⎟⎠ ⎥ ⎢( 0.500 L ) ⎜⎝ 1 L ⎟⎠ ⎥ = −88.7 J 2⎣ 1 atm ⎝ ⎦⎣ ⎦ !

(b) Using the first law, with ΔU = UF − UA = (182 − 91.0) J = 91.0 J for each process gives QIAF = ΔU − WIAF = 9.10 J − (−76.0 J) = 167 J QIBF = ΔU − WIBF = 91.0 J − (−101 J) = 192 J QIF = ΔU − WIF = 91.0 J − (−88.7 J) = 180 J 12.31

The net work done by a heat engine operating on the cyclic process shown in Figure P12.31 equals the triangular area enclosed by this process curve. Thus,

1 6.00 atm − 2.00 atm ) ( 3.00 m 2 − 1.00 m 2 ) ( 2 ⎛ 1.013 × 105 Pa ⎞ = 4.00 atm ⋅ m 3 ⎜ = 4.05 × 105 J ⎟ 1 atm ⎝ ⎠

Wenv =

= 405 × 103 J = 405 kJ

Topic 12

740

FIGURE P12.31 12.32

The net work done by a heat engine operating on the cyclic process shown in Figure P12.32 equals the triangular area enclosed by this process curve. This is 1 (base )( altitude ) 2 1 = ⎡⎣( 4.00 − 1.00 ) m 3 ⎤⎦ ⎡⎣( 6.00 − 2.00 ) × 105 Pa ⎤⎦ 2 = 6.00 × 105 Pa = 600 kJ

Weng =

FIGURE P12.32 12.33

The maximum possible efficiency for a heat engine operating between reservoirs with absolute temperatures of Tc = 25° + 273 = 298 K and Th = 375° + 273 = 648 K is the Carnot efficiency

T 298 K ec = 1 − c = 1 − = 0.540 or 54.0% T 648 K h !

Topic 12

12.34

741

(a) The absolute temperature of the cold reservoir is Tc = 20° + 273 = 293 K. If the Carnot efficiency is to be eC = 0.65, it is necessary that

T 1 − c = 0.65 ! Th Thus,

Tc = 0.35 Th

293 K Th = = 837 K 0.35 !

and

Th =

Tc 0.35

Th = 837 − 273 = 564°C

(b) No. Any real heat engine will have an efficiency less that the Carnot efficiency because it operates in an irreversible manner. 12.35

(a) The efficiency of a heat engine is e = Weng/|Qh|, where Weng is the work done by the engine and |Qh| is the energy absorbed from the higher temperature reservoir. Thus, if Weng = |Qh|/4, the efficiency is e = 1/4 = 0.25 or 25%. (b) From conservation of energy, the energy exhausted to the lower temperature reservoir is |Qc| = |Qh| − Weng. Therefore, if Weng = |Qh|/4, we have |Qc| = 3|Qh|/4 or |Qc|/|Qh| = 3/4.

12.36

(a) The work done by a heat engine equals the net energy absorbed by the engine, or Weng = |Qh|−|Qc|. Thus, the energy absorbed from the high temperature reservoir is |Qh| = Weng +|Qc| = 1 800 J + 2 400 J = 4 200 J (b) The efficiency of the heat engine is

Topic 12

742

e≡

12.37

Weng

1 800 J 4 200 J

= 0.43 or

43%

(a) The maximum efficiency possible is that of a Carnot engine operating between reservoirs having absolute temperatures of Th = 1 870 + 273 = 2 143 K and Tc = 430 + 273 = 703 K.

ec =

Th − Tc Th

= 1−

Tc Th

= 1−

703 K 2 143 K

= 0.672 (or 67.2%)

Weng (b) From e = , we find Weng = e|Qh| = 0.420(1.40 × 105 J) = Qh !

5.88 × 104 J

Weng t

5.88 × 10 4 J 1.00 s

= 5.88 × 10 4 W = 58.8 kW

12.38 The engine’s horsepower rating equals the work per second done by the engine, converted to from J/s to hp. 4 Each second the engine ejects energy Qc = 1.00 × 10 J to the cold

reservoir while operating with an efficiency of 0.200. Use the definition of efficiency to find the energy Qh per second extracted from the hot reservoir:

e = 1−

Qc Qh

→

Qc = 1− e Qh

Qc 1.00 × 10 4 J Qh = = = 1.25 × 10 4 J 1− e 1 − 0.200 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

743

From the first law of thermodynamics applied to heat engines, each second the engine does work

Weng = Qh − Qc = 1.25 × 10 4 J − 1.00 × 10 4 J = 2.50 × 103 J so the engine’s horsepower rating is ⎛ 1 hp ⎞ Peng = 2.50 × 103 J/s ⎜ = 3.35 hp ⎝ 746 J/s ⎟⎠

(

12.39 (a) e =

Weng Qh

= 1−

)

Qc Qh

= 1−

1.20 × 103 J 1.70 × 103 J

= 0.294

(or 29.4%)

(b) Weng = |Qh|−|Qc| = 1.70 × 103 J − 1.20 × 103 J = 5.00 × 102 J

(c)

12.40

Weng t

5.00 × 102 J 0.300 s

= 1.67 × 103 W = 1.67 kW

(a) The coefficient of performance of a heat pump is COP = |Qh|/W, where |Qh| is the thermal energy delivered to the warm space and W is the work input required to operate the heat pump. Therefore,

Qh = W ⋅ COP = (P ⋅ Δt)⋅ COP ⎡⎛ ⎛ 3600 s ⎞ ⎤ J⎞ 6 = ⎢⎜ 7.03 × 103 ⎟ 8.00 h ⎜ ⎟ ⎥ 3.80 = 7.69 × 10 J ⎜ ⎟ ⎢⎝ s⎠ ⎝ 1 h ⎠ ⎥⎦ ⎣

(

)

(b) The energy extracted from the cold space (outside air) is

Qc = Qh − W = Qh −

Qh 1 ⎞ ⎛ = Qh ⎜ 1 − ⎟ ⎝ COP COP ⎠

Topic 12

744

1 ⎞ ⎛ 8 Qc = (7.69 × 108 J ) ⎜ 1 − ⎟⎠ = 5.67 × 10 J ⎝ 3.80 !

12.41

(a)

⎛ ⎞ kWh ⎞ ⎛ 3.60 × 106 J ⎞ ⎛ 1 y 6 W = P ⋅ Δt = ⎜ 457 ⎟⎜ ⎜ ⎟ ⋅(1 d) = 4.50 × 10 J ⎟ y ⎠⎝ 1d ⎠ ⎝ 365.242 d ⎠ ⎝ ! (b) From the definition of the coefficient of performance for a refrigerator, (COP)R = |Qc|/W, the thermal energy removed from the cold space each day is |Qc| = (COP)R ⋅ W = 6.30(4.50 × 106 J) = 2.84 × 107 J (c) The water must be cooled 20.0°C before it will start to freeze, so the thermal energy that must be removed from mass m of water to freeze it is |Qc| = mcw|ΔT| + mLf. The mass of water that can be frozen each day is then

Qc cw ΔT + L f

(

2.84 × 102 J

)(

)

4 186 J/kg ⋅°C 20.0°C + 3.33 × 105 J/kg

12.42 The overall efficiency is e =

= 68.2 kg

WA + WB , where WA = 275 J is known. QhA

Engine B absorbs 75.0% of the energy ejected by engine A so that QhB = 0.750 QcA . Also, eA = eB = 0.250.

First, use the definition of efficiency to find QhA :

Topic 12

745

eA =

WA QhA

→

QhA =

WA 275 J = = 1.10 × 103 J eA 0.250

Now find the energy ejected by engine A to the cold reservoir:

WA = QhA − QcA QcA =

→ QcA = QhA − WA

⎛ 1 − eA ⎞ WA − WA = WA ⎜ = 825 J eA ⎝ e A ⎟⎠

The energy absorbed by engine B is therefore:

⎛ ⎛ 1 − eA ⎞ ⎞ QhB = 0.750 QcA = 0.750 ⎜ WA ⎜ = 619 J ⎝ e A ⎟⎠ ⎟⎠ ⎝ The work done by engine B is

WB = e B QhB = ( 0.750 ) (1 − e A ) WA = ( 0.750 ) (1 − 0.250 ) ( 275 J ) = 155 J or, alternatively,

WB = e B QhB = ( 0.250 ) ( 619 J ) = 155 J

Substitute values or expressions into the definition of overall efficiency to find e=

WA + WB 275 J + 155 J = = 0.391 QhA 1.10 × 103 J

or, alternatively, e=

12.43

WA + ( 0.750 ) (1 − e A ) WA WA /e

= e A ⎡⎣1 + ( 0.750 ) (1 − e A ) ⎤⎦ = 0.391

The actual efficiency of the engine is

Q 300 J e = 1− c = 1− = 0.400 Qh 500 J ! If this is 60.0% of the Carnot efficiency, then

Topic 12

746

eC =

e 0.600

0.400 0.600

2 3

Thus, from !eC = 1 − Tc /Th , we find Tc Th

12.44

= 1 − eC = 1 −

2 3

1 3

(a) The Carnot efficiency represents the maximum possible efficiency. With Th = 20.0°C = 293 K and Tc = 5.00°C = 278 K, this efficiency is given by

eC = 1 −

Tc Th

= 1−

278 K 293 K

= 0.051 2 (or 5.12%)

(b) The efficiency of an engine is e = Weng/|Qh|, so the minimum energy input by heat each hour is

min

Weng emax

P ⋅ Δt emax

(75.0 × 10 J/S)(3 600 s) = 5.27 × 10 J = 6

0.051 2

(c) As fossil-fuel prices rise, this could be an attractive way to use solar energy. However, the potential environmental impact of such an engine would require serious study. The energy output, |Qc|=|Qh|− Weng = |Qh|(1 − e), to the low temperature reservoir (cool water deep in the ocean) could raise the temperature of over a million cubic meters of water by 1°C every hour. 12.45 (a) We treat the power plant as a heat engine and compute its © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

747

efficiency as

Weng Qh

435 MW 1 420 MW

= 0.306 or 30.6%

(b) The work done by a heat engine equals the net energy absorbed by the engine, or Weng = |Qh|−|Qc|. Thus, the energy expelled to the low temperature reservoir is |Qc|=|Qh|− Weng = 1 420 MW − 435 MW = 985 MW 12.46

(a) With reservoirs at absolute temperatures of Tc = 80.0 + 273 = 353 K and Th = 350 + 273 = 623 K, the Carnot efficiency is

T 353 K eC = 1 − c = 1 − = 0.433 (or 43.3%) T 623 K h ! so the maximum power output is

Pmax =

(W ) = e Q = 0.433(21.0 kJ) = 9.09 kW eng

max

1.00 s

(b) From e = 1 − |Qc|/|Qh|, the energy expelled by heat each cycle is |Qc|=|Qh|(1 − e) = (21.0 kJ)(1 − 0.433) = 11.9 kJ 12.47

The thermal energy transferred to the room by the water as the water cools from 1.00 × 102°C to 20.0°C is Q = mcw|ΔT| = (0.125 kg)(4 186 J/kg⋅°C)(80.0°C) = 4.19 × 104 J

Topic 12

748

If the room has a constant absolute temperature of T = 20.0° + 273 = 293 K, the increase in the entropy of the room is

Q 4.19 × 10 4 J ΔS = = = 143 J/K T 293 K ! 12.48

(a) The energy transferred to the ice cube as it melts at a constant temperature of T = 0.0°C = 273 K is Qice = +miceLf and the change in entropy of the ice cube is

mice L f (0.065 kg) ( 3.33 × 10 J/kg ) Q ΔSenv = ice = = = + 79 J/K T T 273 K ! 5

(b) The energy transferred to the environment during this melting process is Qenv = −Qice = −miceLf and the change in entropy of the environment is

−mice L f Q ΔSenv = env = = −79 J/K T T ! 12.49

The energy transferred from the water by heat, and absorbed by the freezer, is

( )

Q = mL f = ρV L f ⎛ J ⎞ = ⎡ 103 kg/m 3 1.0 × 10−3 m 3 ⎤ ⎜ 3.33 × 103 ⎟ = 3.3 × 105 J ⎣ ⎦⎝ kg ⎠

(

)(

)

Thus, the change in entropy of the water is

(a)

ΔSwater = !

( ΔQr )water T

−3.3 × 103 J J = −1.2 × 103 = −1.2 kJ/K 273 K K

Topic 12

749

and that of the freezer is

(b) ΔSfreezer = ! 12.50

( ΔQr )freezer T

+3.3 × 103 J = +1.2 kJ/K 273 K

The energy added to the water by heat is

ΔQr = mLv = (1.00 kg)(2.26 × 106 J/ kg) = 2.26 × 106 J so the change in entropy is

ΔQr 2.26 × 106 J J ΔS = = = 6.06 × 103 = 6.06 kJ/K T 373 K K ! 12.51

The potential energy lost by the log is transferred away by heat, so the energy transferred from the log to the reservoir is ΔQr = mgh. The change in entropy of the reservoir (universe) is then

ΔQc mgh (70.0 kg) ( 9.80 m/s ) (25.0 m) ΔS = = = = 57.2 J/K T T 300 K ! 2

12.52 (a) The energy absorbed by the nitrogen will first boil the nitrogen to a gas at 77.3 K and then warm the nitrogen gas to the room temperature of 21.0°C + 273.15 = 294 K: Qtotal = Qboil + Qwarm. Assume the nitrogen gas is warmed at constant pressure (the ambient pressure in the room) so that Qwarm = mcv ΔT where

c v = cN 2 = 1.04 × 103 J/kg ⋅ K . Substitute expressions for these energies to find

Topic 12

750

Qtotal = mLv + mc p ΔT

(

)

(

)

= ( 0.500 kg ) 2.01 × 105 J/kg + ( 0.500 kg ) 1.04 × 103 J/kg ⋅ K ( 294 K − 77.3 K ) = 2.13 × 105 J

(b) The energy expelled by the room is Qroom = −Qtotal. Assuming the room’s temperature is constant at 294 K, its change in entropy is

ΔS =

Qroom −2.13 × 105 J = = −724 J/K T 294 K

12.53 A quantity of energy, of magnitude Q, is transferred from the Sun and

−Q +Q added to Earth. Thus, ΔSSun = and ΔSEarth = , so the total T T Sun Earth ! ! change in entropy is

ΔStotal = ΔSEarth + ΔSSun =

Q TEarth

−

Q TSun

⎛ 1 ⎞ 1 = (1 000 J) ⎜ − ⎟ = +3.27 J/K ⎝ 290 K 5 700 K ⎠

12.54

The change in entropy of a reservoir is ΔS = Qr/T, where Qr is the energy absorbed (Qr > 0) or expelled (Qr < 0) by the reservoir, and T is the absolute temperature of the reservoir.

(a) For the hot reservoir: ΔSh = !

−2.50 × 103 J = −3.45 J/K 725 K

(b) For the cold reservoir: ΔSc = !

+2.50 × 103 J = +8.06 J/K 310 K

Topic 12

751

(c) For the Universe: ΔSU = ΔSh + ΔSc = −3.45 J/K + 8.06 J/K = +4.61 J/K (d) The magnitudes of the thermal energy transfers, appearing in the numerators, are the same for the two reservoirs, but the cold reservoir necessarily has a smaller denominator. Hence, its positive change dominates. 12.55 (a) The table is shown below. On the basis of the table, the most probable result of a toss is 2H and 2T.

End

Possible Tosses

Result

Total Number of Same Result

All H

HHHH

1T, 3H

HHHT, HHTH, HTHH, THHH

2T, 2H

HHTT, HTHT, THHT, HTTH,

THTH, TTHH

3T, 1H

TTTH, TTHT, THTT, HTTT

All T

TTTT

Topic 12

752

(b) The most ordered state is the least likely. This is seen to be all H or all T. (c) The least ordered state is the most likely. This is seen to be 2H and 2T. 12.56

The change in entropy of a reservoir is ΔS = Qr/T, where Qr is the energy absorbed (Qr > 0) or expelled (Qr < 0) by the reservoir, and T is the absolute temperature of the reservoir.

− Qh (a) For the hot reservoir, Qr = −|Qh|, and ΔSh = Th ! + Qh (b) For the cold reservoir, Qr = +|Qh|, and ΔSc = Tc !

! 12.57

Qh Qh + Th Tc

Using the metabolic rates from Table 12.4, we find the change in the body’s internal energy (i.e., the energy consumed) for each activity, and the total change for the day, as: Metabolic Activity

Sleeping

Rate (W)

Internal Energy Change

ΔU1 = −(80 J/s)(8.0 h)(3 600 s/1 h) =

Topic 12

753

− 8.0 h

−2.3 × 106 J

Light Chores −

ΔU2 = −(230 J/s)(3.0 h)(3 600 s/1 h) = −2.5 × 230

3.0 h

106 J

Slow Walk −

ΔU3 = − (230 J/s)(1.0 h)(3 600 s/1 h) = −8.3 230

1.0 h

× 105 J ΔU4 = − (465 J/s)(0.50 h)(3 600 s/1 h) = −8.4

Running

465

− 0.5 h

× 105 J ΔU2 + ΔU3 + ΔU4 = −6.5 × 106 J = −6.5 MJ

Total

12.58 (a) At the sleeping rate of 80.0 W, the time required for the body to use the 450 Cal of energy supplied by the bagel is

Δt =

ΔU P

450 Cal ⎛ 4 186 J ⎞ ⎛ 1 h ⎞ ⎜ ⎟⎜ ⎟ = 6.54 h 80.0 J/s ⎝ 1 Cal ⎠ ⎝ 3 600 s ⎠

(b) The metabolic rate while working out is P = 80.0 W + 650 W = 7.30 × 102 W, and the time to use the energy from the bagel at this rate is

Δt =

ΔU P

⎛ 4 186 J ⎞ ⎛ 1 h ⎞ ⎜ ⎟⎜ ⎟ = 0.717 h 7.30 × 102 J/s ⎝ 1 Cal ⎠ ⎝ 3 600 s ⎠ 450 Cal

= F ⋅ Δx = mgh = (120 kg ) ( 9.80 m s 2 ) ( 2.00 m ) = 2.35 × 103 J (c) Wper ! ltr

Topic 12

754

(d) The additional energy consumed in 1 minute while working out (instead of sleeping) is ΔU = Pincrease(1 min) = (650 J/s)(60.0 s) = 3.90 × 104 J. The number of barbell lifts this should allow in 1 minute is ΔU 3.90 × 10 4 J N= = = 16.6 Wper 2.35 × 103 J ltr !

(e) No. The body is only about 25% efficient in converting chemical energy to mechanical energy. 12.59

The maximum rate at which the body can dissipate waste heat by sweating is

⎞⎛ ⎛ ⎛ Δm ⎞ kg ⎞ ⎛ 1 h ⎞ 3 J 3 ⎜ ⎟ ⎜ ⎟ =⎜ L = 1.5 2 430 × 10 ⎜ ⎟ = 1.0 × 10 W ⎟ v ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Δt ⎝ Δt ⎠ h ⎠⎝ kg ⎠ ⎝ 3 600 s ⎠ ⎝

ΔQ

If this represents 80% of the maximum sustainable metabolic rate [i.e., ΔQ/Δt = 0.80(ΔU/Δt)max], then that maximum rate is

ΔQ/Δt 1.0 × 10 W ⎛ ΔU ⎞ = = = 1.3 × 105 W ⎜⎝ ⎟⎠ 0.80 0.80 ! Δt max 3

12.60 The metabolic rate equation requires a metabolic rate in units of kcal/s. Convert 625 W to kcal/s: ⎛ 1 kcal ⎞ 625 W = 625 J/s ⎜ = 0.149 kcal/s ⎝ 4186 J ⎟⎠

(a) Her volume rate of oxygen consumption is then

Topic 12

755

ΔVO 2 ΔU = 4.8 Δt Δt

→

ΔVO 2 Δt

1 ΔU 1 = (0.149 kcal/s) = 3.10 × 10−2 L/s 4.8 Δt 4.8

(b) Finding the respiratory rate R in breaths/min requires a unit conversion from L/s to breaths/min:

R = 3.10 × 10−2

LO 2 ⎛ 1 breath ⎞ ⎛ 1 Lair ⎞ ⎛ 60 s ⎞ = 14.8 breaths/min ⎜ ⎟ s ⎜⎝ 0.600 Lair ⎟⎠ ⎝ 0.209 LO 2 ⎠ ⎜⎝ 1 min ⎟⎠

12.61 (a) The 78.0-kg athlete consumes oxygen at a rate of ΔVO 2

⎛ ⎛ 10−3 L ⎞ ⎛ 1 min ⎞ mL ⎞ = ⎜ 70.0 78.0 kg = 9.10 × 10−2 L/s ( ) ⎜ ⎟ ⎜ ⎟ ⎟ Δt min⋅ kg ⎠ ⎝ 1 mL ⎠ ⎝ 60 s ⎠ ⎝

with a metabolic rate of

ΔVO 2 ΔU = 4.8 = ( 4.8 ) 9.10 × 10−2 L/s = 0.437 kcal/s = 26.2 kcal/min Δt Δt

(

)

(b) Treating the athlete as a heat engine operating for 30.0 min with an efficiency of 20.0%, Q = ΔU = 26.2 kcal/min 30.0 min = 786 kcal . ( ) h Applying the definition of efficiency gives the energy released to the environment (the cold reservoir):

e = 1−

ΔU → Qc = (1 − e ) ΔU Qc

Qc = (1 − 0.200 ) ( 786 kcal ) = 629 kcal 12.62

Operating between reservoirs having temperatures of Th = 100°C = 373 K and Tc = 20°C = 293 K, the theoretical efficiency of a Carnot engine is

eC = 1 −

Tc Th

= 1−

293 K 373 K

= 0.214

Topic 12

756

If the temperature of the hotter reservoir is changed to

!Th′ = 550°C!=!823 K , the theoretical efficiency of the Carnot engine increases to

T 293 K eC′ = 1 − c = 1 − = 0.644 T 823 K ′ h ! The factor by which the efficiency has increased is

eC′ 0.644 = = 3.01 e 0.214 C ! 12.63

The work output from the engine in an interval of one second is Weng = 1 500 kJ. Since the efficiency of an engine may be expressed as Weng Weng e= = Qh Weng + Qc !

the exhaust energy each second is

⎛1 ⎞ ⎛ 1 ⎞ Qc = Weng ⎜ − 1⎟ = (1 500 kJ) ⎜ − 1⎟ = 4.5 × 103 kJ ⎝e ⎠ ⎝ 0.25 ⎠ The mass of water flowing through the cooling coils each second is −

m = ρV = (103 kg/m3)(60 L)(10 3 m3/1 L) = 60 kg so the rise in the temperature of the water is

ΔT =

12.64

Qc mcwater

4.5 × 106 J (60 kg)(4 186 J/kg ⋅°C)

= 18°C

The energy exhausted from a heat engine is

Topic 12

757

Qc = Qh − Weng = !

⎛1 ⎞ − Weng = Weng ⎜ − 1⎟ ⎝e ⎠ e

Weng

where Qh is the energy input from the high temperature reservoir, Weng is the useful work done, and e = Weng/Qh is the efficiency of the engine. For a Carnot engine, the efficiency is eC = 1 − Tc/Th = (Th − Tc)/Th

so we now have

⎛ Th ⎞ ⎛ Tc ⎞ Qc = Weng ⎜ − 1⎟ = Weng ⎜ ⎝ Th − Tc ⎠ ⎝ Th − Tc ⎟⎠ !

Thus, if Th = 100°C = 373 K and Tc = 20°C = 293 K, the energy exhausted when the engine has done 5.0 × 104 J of work is

⎛ ⎞ 293 K Q = (5.0 × 10 4 J) ⎜ = 1.83 × 105 J ⎝ 373 K − 293 K ⎟⎠ ! The mass of ice (at 0°C) this exhaust energy could melt is Q 1.83 × 105 J m= = = 0.55 kg L f ⋅water 3.33 × 105 J/kg !

12.65

(a) The work done by the system in process AB equals the area under this curve on the PV diagram. Thus, Wenv = (triangular area) + (rectangular area), or

Wenv = ⎡⎣ 12 ( 4.00 atm ) ( 40.0 L ) Pa ⎞ ⎛ 10−3 m 3 ⎞ ⎛ !! + (1.00 atm ) ( 40.0 L ) ⎤⎦ ⎜ 1.013 × 103 ⎟ ⎝ atm ⎠ ⎜⎝ L ⎟⎠ !

= 1.22 × 10 4 J = 12.2 kJ

Note that the work done on the system is WAB = −Wenv = −12.2 kJ © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

758

for this process. (b) The work done on the system (that is, the work input) for process BC is the negative of the area under the curve on the PV diagram, or −3 3 ⎛ 5 Pa ⎞ ⎛ 10 m ⎞ WBC = − ⎡⎣(1.00 atm ) (10.0 L − 50.0 L ) ⎤⎦ ⎜ 1.013 × 10 ⎟ ⎝ atm ⎠ ⎜⎝ 1 L ⎟⎠

= 4.05 kJ

(

Qcycle = ΔUcycle − Wcycle = 0 − WAB + WBC + WCA

(

)

= 0 − −12.2 kJ + 4.05 kJ + 0 = 8.15 kJ

12.66 (a) From the first law, ΔU1→3 = Q123 + W123 = +418 J + (−167 J) = 251 J. (b) The difference in internal energy between states 1 and 3 is independent of the path used to get from state 1 to state 3. Thus, and

ΔU1→3 = Q143 + W143 = 251 J, Q143 = 251 J − W143 = 251 J −(−63.0 J) = 314 J

(c) W12341 = W123 + W341 + W123 + (−W143) = −167 J – (−63.0 J) = −104 J or 104 J of work is done by the gas in the cyclic process 12341. (d) W14321 = W143 + W321 = W143 + (−W123) = −63.0 J −(−167 J) = +104 J

Topic 12

759

or 104 J of work is done on the gas in the cyclic process 14321. (e) The change in internal energy is zero for both parts (c) and (d) since both are cyclic processes. 12.67

(a) The change in length, due to linear expansion, of the rod is −

−

ΔL = αL0(ΔT) = [11 × 10 6 (°C) 1](2.0 m)(40°C − 20°C) = 4.4 × 10 4 m The load exerts a force F = mg = (6 000 kg)(9.80 m/s2) = 5.88 × 104 N on the end of the rod in the direction of movement of that end. Thus, the work done on the rod is −

W = F ⋅ ΔL = (5.88 × 104 N)(4.4 × 10 4 m) = 26 J (b) The energy added by heat is

⎛ J ⎞ Q = mc(ΔT) = (100 kg) ⎜ 488 (20°C) = 9.0 × 105 J kg ⋅°C ⎟⎠ ⎝ ! (c) From the first law, ΔU = Q + W = 9.0 × 105 J + 26 J = 9.0 × 105 J 12.68 (a) The work done by the gas during each full cycle equals the area enclosed by the cycle on the PV diagram. Thus Wenv = (3P0 − P0)(3V0 − V0) = 4P0V0 (b) Since the work done on the gas is W = −Wenv = −4P0V0 and ΔU = 0 for any cyclic process, the first law gives Q = ΔU − W = 0 − (−4P0V0) = 4P0V0 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

760

J ⎞ ⎛ Wenv = 4nRT0 = 4 (1.00 mol ) ⎜ 8.31 ⎟ ( 273 K ) ⎝ mol ⋅ K ⎠ ! 12.69

= 9.07 × 103 J = 9.07 kJ

(a) The energy transferred to the gas by heat is

⎛ J ⎞ 3 Q = mc(ΔT) = (1.00 mol) ⎜ 20.79 ⎟ (120 K) = 2.49 × 10 J = 2.49 kJ mol ⋅ K ⎠ ⎝ (b) Treating the neon as a monatomic ideal gas, Equation 12.3b gives the change in internal energy as !ΔU = 32 nR(ΔT) , or

⎛ 3 J ⎞ 3 ΔU = (1.00 mol) ⎜ 8.31 ⎟ (120 K) = 1.50 × 10 J = 1.50 kJ 2 mol ⋅K ⎠ ⎝ (c) From the first law, the work done on the gas is W = ΔU − Q = 1.50 × 103 J − 2.49 × 103 J = −990 J 12.70

Assuming the gravitational potential energy given up by the falling water is transformed into thermal energy when the water hits the bottom of the falls, the rate of thermal energy production is

ΔQ ⎛ Δm ⎞ ⎛ ΔV ⎞ =⎜ ⎟⎠ gh = ρw ⎜⎝ ⎟ gh ⎝ Δt Δt ⎠ ! Δt Then, if the absolute temperature of the environment is TK = 20.0° + 273 = 293 K, the rate of entropy production is

Topic 12

761

ΔS ΔQ / Δt ρw ( ΔV / Δt ) gh = = Δt T TK K ! ΔS or

Δt

3 ⎞⎛ ⎛ 3 kg ⎞ ⎛ m⎞ 3 m 10 5 × 10 9.8 2 ⎟ (50.0 m) ⎜ ⎟ ⎜ ⎟ ⎜ 293 K ⎝ s ⎠⎝ s ⎠ m3 ⎠ ⎝

= 8 × 109

and

12.71

(kg ⋅ m/s 2 )⋅ m K ⋅s

ΔS (N ⋅ m) = 8 × 106 = 8 × 106 J/K ⋅s Δt K ⋅s !

3 3 PAVA (10.0 × 10 Pa ) (1.00 m ) T = = (a) A nR (10.0 mol )(8.31 J/mol ⋅ K ) !

= 1.20 × 102 K 3 3 PBVB (10.0 × 10 Pa ) ( 6.00 m ) TB = = nR (10.0 mol ) ( 8.31 J/mol ⋅ K ) !

= 722 K

(b) As it goes from A to B, the gas is expanding and hence, doing work on the environment. The magnitude of the work done equals the area under the process curve from A to B. We subdivide this area into 2 rectangular and 2 triangular parts:

Topic 12

762

Wenv = ⎡⎣(10.0 × 103 Pa ) ( 6.00 − 1.00 ) m 3 ⎤⎦ + ⎡⎣( 40.0 − 10.0 ) × 103 Pa ⎤⎦ (1.00 m 3 ) !

⎡1 ⎤ +2 ⎢ (1.00 m 3 )(40.0 − 10.0) × 103 Pa ⎥ = 1.10 × 105 J ⎣2 ⎦ (c) The change in the internal energy of a monatomic, ideal gas is

3 ΔU = nR(ΔT) , so 2 ! 3 3 ΔU A→B = nR (TB − TA ) = (10.0 mol) ( 8.31 J/mol ⋅ K ) (722 K − 120 K) = 7.50 × 10 4 J 2 2 !

(d) From the first law of thermodynamics, Q = ΔU − W, where W is the work done on the gas. In this case, W = −Wenv = −1.10 × 105 J, and Q = ΔU − W = 7.50 × 104 J − (−1.10 × 105 J) = 1.85 × 105 J 12.72

(a) The constant volume occupied by the gases is 3 3 3 4 !V = 3 π r = 4π (0.500 m) /3 = 0.524 m and the initial absolute

temperature is Ti = 20.0° + 273 = 293 K. To determine the initial pressure, we treat each component of the mixture as an ideal gas and compute the pressure it would exert if it occupied the entire volume of the container. The total pressure exerted by the mixture is then the sum of the partial pressures exerted by the components of the mixture. This gives PH2 = !

nH2 RT V

PO2 = !

nO2 RT V

Topic 12

763

and

Pi = PH + PO = 2

nH RT 2

nO RT 2

(= n + n ) RT H2

Pi =

(14.4 mol + 7.2 mol)(8.31 J/mol ⋅ K)(293 K) 0.524 m

= 1.00 × 103 Pa

(b) Treating both the hydrogen and oxygen as ideal gases, each with internal energy U = nCvT where Cv is the molar specific heat at constant volume, we use Table 12.1 and find the initial internal energy of the mixture as

(

)

U = UH2 ,i + Uo2 ,i = nH2 Cv ,H2 + no2 Cv ,o2 Ti ! or Ui = [(14.4 mol)(20.4 J/mol ⋅ K) + (7.2 mol)(21.1 J/mol ⋅ K)](293 K) = 1.31 × 105 J (c) During combustion, this mixture produces 14.4 moles of water (1 mole of water for each mole of hydrogen used) with a conversion of 241.8 kJ of chemical potential energy per mole. Since the volume is constant, no work is done and the additional internal energy generated in the combustion is ΔU = (14.4 mol)(241.8 kJ/mol) = 3.48 × 103 kJ = 3.48 × 106 J (d) After combustion, the internal energy of the system is Uf = Ui + ΔU = 1.31 × 105 J + 3.48 × 106 J = 3.61 × 106 J © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

764

Treating the steam as an ideal gas, so U = nCvTK, and obtaining the molar heat capacity for water vapor (a polyatomic gas) from Table 12.1, we find Uf 3.61 × 106 J Tf = = = 9.28 × 103 K nwaterCv ,water (14.4 mol)(27.0 J/mol ⋅ K) !

and the final pressure is

Pf = !

Nwater RTf V

(14.4 mol)(8.31 J/mol ⋅ K)(9.28 × 103 K) = = 2.12 × 106 Pa 3 0.524 m

(e) The total mass of steam present after combustion is

m = nsteam Mwater = (14.4 mol)(18.0 × 10−3 kg/mol) = 0.259 kg ! steam

and its density is

(f)

ρs =

msteam V

0.259 kg 0.524 m

= 0.494 kg/m 3

Assuming the steam is essentially at rest within the container (v1 ≈ 0), P2 = 0 (since the steam spews into a vacuum), and y2 = y1, we use the pressure from part (d) above and Bernoulli’s equation

P + 1 ρ v 2 + ρs gy2 = P1 + 12 ρsv12 + ρs gy1 ) to find the exhaust speed as !( 2 2 s 2

2 ( 2.12 × 106 Pa ) 2P1 v2 = = = 2.93 × 103 m/s 3 ρs 0.494 kg/m ! 12.73

The work that you have done is

Topic 12

765

Weng = mg(Δh) ⎡ ⎛ 4.448N ⎞ ⎤ ⎡⎛ ⎛ step ⎞ in ⎞ ⎛ 2.54 × 10−2 m ⎞ ⎤ = ⎢(150 lb) ⎜ 90.0 (30.0 min) 8.00 ⎢ ⎥ ⎟⎥ ⎟ ⎜ ⎜ ⎟⎜ ⎟ min ⎠ step ⎠ ⎝ 1 in ⎝ ⎝ 1 lb ⎠ ⎥⎦ ⎢⎣⎝ ⎠ ⎥⎦ ⎢⎣ or

Weng = 3.66 × 105 J

⎛ 4 186 J ⎞ 6 If the energy input by heat was Qh = (600 kcal) ⎜ ⎟ = 2.51 × 10 J , ⎝ 1 kcal ⎠ your efficiency has been Weng 3.66 × 105 J e= = = 0.146 or 14.6% Qh 2.51 × 106 J !

If the actual efficiency was e = 0.180 or 18.0%, the actual energy input was

12.74

actual

Weng eactual

3.66 × 105 J 0.180

⎛ 1 kcal ⎞ = (2.03 × 106 J) ⎜ ⎟ = 485 kcal ⎝ 4 186 J ⎠

(a) The energy transferred from the water by heat as it cools is

Qh = mc ΔT = ( ρV ) c ΔT ⎡⎛ ⎛ 103 cm 3 ⎞ ⎤ ⎛ g ⎞ cal ⎞ ⎛ 4.186 J ⎞ = ⎢⎜ 1.0 1.0 L 1.0 ( 570°C − 4.0°C) ( ) ⎥ ⎟ 3 ⎜ ⎟ ⎜ cm ⎠ g ⋅°C ⎟⎠ ⎜⎝ 1 cal ⎟⎠ ⎝ 1 L ⎠ ⎦⎝ ⎣⎝ !

Qh = 2.4 × 106 J (b) The maximum efficiency of a heat engine is the Carnot efficiency. Thus,

T (4.0 + 273) K 277 K eC = 1 − c = 1 − = 1− = 0.67 T (570 + 273) K 843 K h !

Topic 12

766

The maximum useful work output is then (Weng)max = eC|Qh| = (0.67)(2.4 × 106 J) = 1.6 × 106 J (c) The energy available from oxidation of the hydrogen sulfide in 1.0 L of this water is

mol ⎞ J ⎞ ⎡⎛ ⎤⎛ 2 U = n(310 kJ/mol) = ⎢⎜ 0.90 × 10−3 (1.0 L) ⎥ ⎜ 310 × 103 ⎟ ⎟⎠ = 2.8 × 10 J ⎝ ⎠ ⎝ L mol ⎣ ⎦ ! 12.75

(a) With an overall efficiency of e = 0.15 and a power output of Pout = 150 MW, the required power input (from burning coal) is

Pout 150 × 106 W Pin = = = 1.0 × 109 J/s e 0.15 ! The coal used each day is 1.0 × 109 J/s ) ( 8.64 × 10 4 s/d ) ( Δm Pin kg = = = 2.6 × 106 3 Δt heat of combustion ⎛ d 3 cal ⎞ ⎛ 10 g ⎞ ⎛ 4.186 J ⎞ ⎜⎝ 7.8 × 10 g ⎟⎠ ⎜⎝ 1 kg ⎟⎠ ⎜⎝ 1 cal ⎟⎠ !

Δm ⎛ kg ⎞ ⎛ 1 metric ton ⎞ = ⎜ 2.6 × 106 ⎟ ⎜ = 2.6 × 103 metric ton/d Δt ⎝ d ⎠ ⎝ 103 kg ⎟⎠ !

(b) The annual fuel cost is cost = (coal used yearly)⋅(rate), or cost = (2.6 × 103 ton/d)(365.242 d/y)($8.0/ton) = $7.6 × 106/y (c) The rate of energy transfer to the river by heat is Pexhuast = Pin − Pout = 1.0 × 109 W − 150 × 106 W = 8.5 × 108 W Thus, the flow required if the maximum rise in temperature is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 12

767

5.0°C is

flow rate =

12.76

Δmwater Δt

Pexhaust cwater (ΔT)

8.5 × 10 4 J/s (4 186 J/kg ⋅°C)(5.0°C)

= 4.1 × 10 4 kg/s

(a) The work done on the gas is the negative of the area under the process curve in a PV diagram. From the sketch below, observe that this area consists of a triangular area sitting atop a rectangular area, or

W = − [ 12 ( base ) (altitude) !

+(width)(height)]

Thus,

⎡1 ⎤ W = − ⎢ (3.00 − 1.00)m 3 (4.00 − 2.00) × 105 Pa + (3.00 − 1.00) m 3 (4.00 − 2.00) × 105 Pa ⎥ ⎢⎣ 2 ⎥⎦ = −6.00 × 105 J

(b)

PV P V P ( ΔV ) ( 2.00 × 10 Pa ) ( 3.00 − 1.00 ) m ΔT = TB − TA = B B − A A = A = = 5.50 × 10 K nR nR nR 87.5 mol ) ( 8.31 J/mol ⋅ K ) ( ! 5

Topic 12

768

(c) The change in internal energy of an ideal gas is ΔU = nCvΔT. From Table 12.1, which gives the molar specific heats of diatomic gases, we take the average of the first four entries to obtain Cv ≈ 20.8 J/mol ⋅ K. This then yields ΔU = UB − UA = nCΔT = (87.5 mol)(20.8 J/mol ⋅ K)(550 K) = 1.00 × 106 J (d) From the first law of thermodynamics, ΔU = Q + W, we find Q = ΔU − W = 1.00 × 106 J − (−6.00 × 105 J) = 1.60 × 106 J

769

Topic 13

Topic 13 Vibrations and Waves

QUICK QUIZZES 13.1

Choice (d). To complete a full cycle of oscillation, the object must travel distance 2A to position x = −A and then travel an additional distance 2A returning to the original position at x = +A.

13.2

Choice (c). The force producing harmonic oscillation is always directed toward the equilibrium position, and hence, directed opposite to the displacement from equilibrium. The acceleration is in the direction of the force. Thus, it is also always directed opposite to the displacement from equilibrium.

13.3

Choice (b). In simple harmonic motion, the force (and hence, the acceleration) is directly proportional to the displacement from equilibrium. Therefore, force and acceleration are both at a maximum when the displacement is a maximum.

13.4

Choice (a). The period of an object-spring system is T = 2π m k . ! Thus, increasing the mass by a factor of 4 will double the period of oscillation.

13.5

Choice (c). The total energy of the oscillating system is equal to 12 kA 2 , !

770

Topic 13

where A is the amplitude of oscillation. Since the object starts from rest at displacement A in both cases, it has the same amplitude of oscillation in both cases. 13.6

Choice (d). The expressions for the total energy, maximum speed, and maximum acceleration are E = 12 kA 2 ,!vmax = A k/m ,!and!amax = A ( k m) , ! where A is the amplitude. Thus, all are changed by a change in amplitude. The period of oscillation is T = 2π m k and is unchanged ! by altering the amplitude.

13.7

Choices (c) and (b). An accelerating elevator is equivalent to a gravitational field. Thus, if the elevator is accelerating upward, this is equivalent to an increased effective gravitational field magnitude g, and the period will decrease. Similarly, if the elevator is accelerating downward, the effective value of g is reduced and the period increases. If the elevator moves with constant velocity, the period of the pendulum is the same as that in the stationary elevator.

13.8

Choice (a). The clock will run slow. With a longer length, the period of the pendulum will increase. Thus, it will take longer to execute each swing, so that each second according to the clock will take longer than an actual second.

13.9

Choice (b). Greater. The value of g on the Moon is about one-sixth the value of g on Earth, so the period of the pendulum on the Moon will

771

Topic 13

be greater than the period on Earth.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 13.2

Friction. This includes both air resistance and damping within the spring.

13.4

The spring force and the spring potential energy are each proportional to the spring constant k. As the spring constant doubles, (a) the force doubles so that Fn/F0 = 2 and (b) the spring potential energy doubles so that PEsn/PEs0 = 2.

13.6

Each half-spring will have twice the spring constant of the full spring, as shown by the following argument. The force exerted by a spring is proportional to the separation of the coils as the spring is extended. Imagine that we extend a spring by a given distance and measure the distance between coils. We then cut the spring in half. If one of the half-springs is now extended by the same distance, the coils will be twice as far apart as they were for the complete spring. Thus, it takes twice as much force to stretch the half-spring, from which we conclude that the half-spring has a spring constant which is twice that of the complete spring.

13.8

(a) False. When x = ±A the block’s velocity is v = 0.

772

Topic 13

(b) True. (c) False. When v > 0, a could be positive or negative depending on whether the block is moving toward or away from equilibrium. (d) True. This follows from Hooke’s law and Newton’s second law: ΣFx = −kx = ma. (e) False. When x > 0, the velocity v can be positive or negative depending on whether the block is moving toward or away from equilibrium. 13.10

Shorten the pendulum to decrease the period between ticks.

13.12

The speed of the pulse is !v = F µ , so increasing the tension F in the hose increases the speed of the pulse. Filling the hose with water increases the mass per unit length µ, and will decrease the speed of the pulse.

13.14

As the temperature increases, the length of the pendulum will increase due to thermal expansion, and with a greater length, the period of the pendulum increases. Thus, it takes longer to execute each swing, so that each second according to the clock will take longer than an actual second. Consequently, the clock will run slow.

773

Topic 13

ANSWERS TO EVEN NUMBERED PROBLEMS 13.2

1.54 cm

13.4

0.242 kg

13.6

(a)

327 N

(b)

1.25 × 103 N

13.8

(a)

0.625 J

(b)

0.791 m/s

13.10

(a)

575 N/m

(b)

46.0 J

13.12

2.23 m/s

13.14

(a)

2 1 !E = 2 kA

(b)

!2 mv = kx

(c)

!x = ±A

13.16

(a)

144 N/m

(b)

1.13 J

(c)

2.60 m/s

13.18

(a)

0.28 m/s

(b)

0.26 m/s

(c)

0.26 m/s

(d)

3.5 cm

13.20

(a)

2.00 Hz

(b)

158 N/s

(c)

2.72 m/s

13.22

(a)

0.628 m/s

(b)

0.500 Hz

(c)

3.14 rad/s

13.24

(a)

1.89 Hz

(b)

33.7 N/m

(c)

0.118 m

13.26

(a)

3.9 × 105 N/m

(b)

2.2 Hz

13.28

(a)

7.00 × 10 2 m/s

(b)

2.45 × 10 2 m/s2

(c)

0.153 m

−

774

Topic 13

−

(d)

−4.51 × 10 2 m/s

13.30

(a)

!±A 3 2

(b)

!±A

13.32

(a)

250 N/m

(b)

22.4 rad/s, 3.56 Hz , 0.281 s

(c)

0.313 J

(d)

5.00 cm

(e)

1.12 m/s, 25.0 m/s2

(f)

0.919 cm

(g)

+1.10 m/s, −4.59 m/s2

13.34

(a)

59.6 m

(b)

37.5 s

13.36

1.001 5

13.38

1.66 × 10 2 kg ⋅ m2

13.40

(a)

3.65 s

(b)

6.41 s

(c)

4.24 s

13.42

(a)

2.00 cm

(b)

4.00 s

(c)

π/2 rad/s

(d)

π cm/s

(e)

4.93 cm/s2

(f)

x = (2.00 cm)sin(πt/2)

13.44

(a)

0.357 Hz

13.46

5.72 mm

13.48

(a)

13.50

(a)

(e)

−1.87 × 10 2 m/s2 2

−

(b)

0.985 m/s

0.20 Hz

(b)

0.25 Hz

22.5 m/s

(b)

20.2 m/s

13.52 (a) The units of the first T are seconds, the units of the second are

775

Topic 13

newtons. (b)

The first T is period of time; the second is force of tension.

13.54

1.64 m/s2

13.56

7.07 m/s

13.58

586 m/s

13.60

(a)

v = 2nL/t

(b)

F = 4n2ML/t2

13.62

(a)

0.25 m

(b)

0.47 N/m

(d)

0.12 m/s

13.64

(a)

5.10 ms

(b)

1.75 m

13.66

0.990 m

13.68

(a)

100 m/s

(b)

374 J

13.70

(a)

19.8 m/s

(b)

8.95 m

13.72

(a) and (b)

13.74

1.3 cm/s

13.76

(a)

ΣFy = −kyf − mg = mv2/(L − yf)

(b)

mv 2 = 2mg L − y f − ky 2f (c) !

(d)

greater than

(c)

0.23 m

See Solution for proofs.

(

)

yf = −0.110 m

776

Topic 13

PROBLEM SOLUTIONS 13.1

(a) Taking to the right as positive, the spring force acting on the block at the instant of release is Fs = −kA = −(130 N/m)(+0.13 m) = −17 N or 17 N to the left (b) At this instant, the acceleration is

F −17 N a= s = = −28 m/s 2 ! m 0.60 kg 13.2

a = 28 m/s2 to the left

The force compressing the spring is the weight of the object. Thus, the spring will be compressed a distance of

F mg ( 2.30 kg ) ( 9.80 m s ) x= = = = 1.54 × 10−2 m = 1.54 cm 3 k k 1.46 × 10 N m ! 2

13.3

Assuming the spring obeys Hooke’s law, the magnitude of the force required to displace the end a distance |x| from the equilibrium position (by either compressing or stretching the spring) is |F| = k|x|, where k is the force constant of the spring. (a) If x = −4.80 cm, the required force is |F| = k|x| = −

(137 N/m)(4.80 × 10 2 m) = 6.58 N (b) If x = +7.36 cm, the required force is |F| = k|x| = −

(137 N/m)(7.36 × 10 2 m) = 10.1 N 13.4

When the system is in equilibrium, the tension in the spring F = k|x|

777

Topic 13

must equal the weight of the object. Thus, −2 k x ( 47.5 N/m ) ( 5.00 × 10 m ) = = 0.242 kg k|x| = mg giving m = 2 g 9.80 m/s !

13.5

Two equal spring forces act vertically upward on the sample and its weight acts vertically downward. Apply Newton’s second law to the sample in equilibrium, solve for the spring constant, and substitute values to find:

ΣFy = 0

(

mg ( 0.725 kg ) 9.80 m/s 2kx − mg = 0 → k = = 2x 2 ( 0.200 m ) 13.6

) = 17.8 N/m

(a) The free-body diagram of the point in the center of the string is given below. From this, we see that

ΣFx = 0 ⇒ F − 2T sin 35.0° = 0

F 2sin 35.0°

375 N 2sin 35.0°

= 327 N

(b) Since the bow requires an applied horizontal force of 375 N to hold the string at 35.0° from the vertical, the tension in the spring must be 375 N when the spring is stretched 30.0 cm. Thus, the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

778

Topic 13

spring constant is

F 375 N k= = = 1.25 × 103 N/m x 0.300 m ! 13.7

(a) When the block comes to equilibrium, ΣFy = −ky0 − mg = 0 giving

(10.0 kg )( 9.80 m s ) = −0.206 m mg y0 = − =− k 475 N/m ! 2

or the equilibrium position is 0.206 m below the unstretched position of the lower end of the spring. (b) When the elevator (and everything in it) has an upward acceleration of a = 2.00 m/s2, applying Newton’s second law to the block gives ΣFy = −k(y0 + y) − mg = may

ΣFy = −(ky0 − mg) − ky = may

where y = 0 at the equilibrium position of the block. Since −ky0 − mg = 0 [see part (a)], this becomes −ky = ma and the new position of the block is

may (10.0 kg )( +2.00 m/s2 ) = −4.21 × 10−2 m = −4.21 cm y=− =− −k −475 N/m ! or

4.21 cm below the equilibrium position.

(c) When the cable breaks, the elevator and its contents will be in free-fall with ay = −g. The new “equilibrium” position of the block

779

Topic 13

is found from ∑ Fy = −ky0′ − mg = m ( −g ) , which yields !y0′ = 0 . ! When the cable snapped, the block was at rest relative to the elevator at distance y0 + y = 0.206 m + 0.042 1 m = 0.248 m below the new “equilibrium” position. Thus, while the elevator is in free-fall, the block will oscillate with amplitude = 0.248 m about the new “equilibrium” position, which is the unstretched position of the spring’s lower end. 13.8

(a) The work required to stretch the spring equals the elastic potential energy of the spring in the stretched condition, or 2 1 1 W = kx 2 = ( 5.00 × 102 N/m ) ( 5.00 × 10−2 m ) = 0.625 J 2 2 !

(b) In the initial condition, the spring-block system is at rest (KEi = 0) with elastic potential energy of PEs,i = 0.625 J. Since the spring force is conservative, conservation of energy gives KEf + PEs,f = KEi + PEs,i = 0.625 J. Thus, when the block is at the equilibrium 2 position (PEs,f = 0), we have !KE f = 12 mv f = 0.625 J , or

vf = ! 13.9

2 ( 0.625 J ) 2 ( 0.625 J ) = = 0.791 m/s m 2.00 kg

(a) Assume the rubber bands obey Hooke’s law. Then, the force constant of each band is

780

Topic 13

F 15 N k= s = = 1.5 × 103 N/m −2 x 1.0 × 10 m ! Thus, when both bands are stretched 0.20 m, the total elastic potential energy is 2 ⎛1 ⎞ PEs = 2 ⎜ kx 2 ⎟ = (1.5 × 103 N/m ) ( 0.20 m ) = 60 J ⎝2 ⎠ !

(b) Conservation of mechanical energy gives (KE + PEs)f = (KE + PEs)i, or

1 2 mv + 0 = 0 + 60 J 2 ! 13.10

2 ( 60 J ) = 49 m/s 50 × 10−3 kg

Fmax 230 N = = 575 N/m (a) k = x 0.400 m max !

(

)(

)

2 1 1 (b) work done = PEs = kx 2 = 575 N/m 0.400 = 46.0 J 2 2

13.11 In the absence of nonconservative forces, apply conservation of mechanical energy with y = 0 at ground-level:

( KE + PE + PE ) = ( KE + PE + PS ) g

0 + 0 + kx = 0 + mgy f + 0 1 2

2 i

Solve for the block’s maximum height yf and substitute xi = −0.125 m, m = 1.50 kg, and k = 575 N/m to find:

781

Topic 13

kxi2 ( 575 N/m ) ( −0.125 m ) yf = = = 0.306 m 2mg 2 (1.50 kg ) 9.80 m/s2 2

(

13.12

)

Conservation of mechanical energy, (KE + PEg + PEs)i = (KE + PEg + PEs)f, gives 1 2 1 mvi + 0 + 0 = 0 + 0 + kx 2f , 2 !2

13.13

vi =

k m

xi =

5.00 × 106 N/m 1 000 kg

(3.16 × 10 m) = 2.23 m/s −2

An unknown quantity of mechanical energy is converted into internal energy during the collision. Thus, we apply conservation of momentum from just before to just after the collision and obtain mvi = M(0) = (M + m)V, or the speed of the block and embedded bullet just after collision is V = (m/M + m)vi. We now use conservation of mechanical energy, (KE + PEs)f = (KE + PEs)i, from just after collision 2 2 until the block comes to rest. This gives !0 + 12 kx f = 12 (M + m)V + 0 , or

xf = v !

M+m mvi ⎛ m ⎞ M+m = vi ⎜ = ⎟ ⎝ M + m⎠ k k (M + m)k

(10.0 × 10 kg )( 300 m/s ) = 0.478 m −3

yielding

xf = !

13.14

( 2.01 kg )(19.6 N/m )

(a) At either of the turning points, x = ±A, the constant total energy of the system is momentarily stored as elastic potential energy in the

782

Topic 13

spring. Thus, E = 12 kA 2 . ! (b) When the object is distance x from the equilibrium position, the elastic potential energy is PEs = kx2/2 and the kinetic energy is KE = mv2/2. At the position where KE = 2PEs, it is necessary that 1 2 ⎛1 ⎞ mv = 2 ⎜ kx 2 ⎟ ⎝ ⎠ 2 !2

1 2 mv = kx 2 2

⎛1 ⎞ kA 2 = 3 ⎜ kx 2 ⎟ 2 ⎝2 ⎠ 1

13.15

⇒

x=±

k A2 / 2 3k /2

x=±

A 3

(a) At maximum displacement from equilibrium, all of the energy is in the form of elastic potential energy, giving !E = 12 kxmax , and 2

2 ( 47.0 J ) 2E 3 k= 2 = 2 = 1.63 × 10 N/m xmax ( 0.240 m ) ! (b) At the equilibrium position (x = 0), the spring is momentarily in its relaxed state and PEs = 0, so all of the energy is in the form of kinetic energy. This gives 1 2 KE x=0 = mvmax = E = 47.0 J 2 !

783

Topic 13

block is

2 ( 47.0 J ) 2E m= 2 = 2 = 7.90 kg vmax ( 3.45 m/s ) ! (d) At any position, the constant total energy is 2 2 1 1 !E = KE + PEs = 2 mv + 2 kx , so at x = 0.160 m

2 ( 47.0 J ) (1.63 × 103 N/m ) ( 0.160 m ) 2E − kx 2 v= = = 2.57 m/s m 7.90 kg ! 2

(e) At x = 0.160 m, where v = 2.57 m/s, the kinetic energy is 1 1 2 KE = mv 2 = (7.90 kg ) ( 2.57 m/s ) = 26.1 J 2 2 !

(f) At x = 0.160 m, where KE = 26.1 J, the elastic potential energy is PEs = E − KE = 47.0 J − 26.1 J = 20.9 J or alternately: 1 1 2 PEs = kx 2 = (1.63 × 103 N/m ) ( 0.160 m ) = 20.9 J 2 2 !

(g) At the first turning point (for which x < 0 since the block started from rest at x = +0.240 m and has passed through the equilibrium at x = 0), all of the remaining energy is in the form of elastic potential energy, so 1 2 kx = E − Eloss = 47.0 J − 14.0 J = 33.0 J 2 !2

784

Topic 13

x=− !

and

2(33.0 J) 2(33.0 J) =− = −0.201 m k 1.63 × 103 N/m

13.16 The given values are: m = 0.250 kg, A = 0.125 m, and vmax = 3.00 m/s (at x = 0). (a) Conservation of energy gives the spring constant k: 2 mvmax (0.250 kg)( 3.00 m/s) = 144 N/m → k= = 2 A (0.125 m )2 2

1 2

kA = mv 2

1 2

2 max

(b) The total energy of the block-spring system is:

Etotal = PEs,max = KEmax so that Etotal = 12 kA 2 = 12 (144 N/m ) ( 0.125 m ) = 1.13 J 2

or 2 Etotal = 12 mvmax = 12 ( 0.250 kg ) ( 3.00 m/s) = 1.13 J 2

(c) Apply conservation of energy with y = 0 at the block’s height to find the speed when x = A/2. Take the initial point where x = A and v = 0 and the final point where x = A/2:

( KE + PE + PE ) = ( KE + PE + PE ) g

0 + 0 + kA = mv + 0 + kx 1 2

1 2

785

Topic 13

kA 2 − kx 2 kA − k ( A/2 ) 3kA 2 v = = = m m 4m 2

3 (144 N/m ) ( 0.125 m ) 3kA 2 v= = = 2.60 m/s 4m 4 ( 0.250 kg ) 2

13.17 (a) The block is released from rest, so the oscillation amplitude will equal its initial distance from equilibrium: A = 8.00 × 10−2 m . (b) Substitute values into the definition of angular frequency:

ω = k/m =

( 425 N/m )/( 2.00 kg) = 14.6 rad/s

2.32 Hz . (d) Substitute values into the expression for period T to find:

T = 2π

m 2.00 kg = 2π = 0.431 s k 425 N/m

Alternatively, use the relation f = 1/T to find the same result. (e) There are many ways to find the block’s maximum velocity (which is equal to the block’s maximum speed). For example, use conservation of energy to find

786

Topic 13

1 2

2 kA 2 = 12 mvmax

vmax =

kA 2 = m

( 425 N/m )( 8.00 × 10 m ) −2

= 1.17 m/s

2.00 m

Alternatively, realize that the maximum value of v = −Aω sin(2π ft) is vmax = Aω = (8.00 × 10-2 m)(14.6 rad/s) = 1.17 m/s. (f) There are many ways to find the block’s maximum acceleration. For example, realize that the maximum value of a = −Aω2cos(2πft) −

is amax = Aω2 = A(k/m) = (8.00 × 10 2 m)(425 N/m)/(2.00 kg) =

17.0 m/s2 . Alternatively, use Hooke’s law and Newton’s second law to find −kxmin = kA = mamax so that, as before, amax = A(k/m). 13.18

From conservation of mechanical energy, (KE + PEg + PEs)f = (KE + PEg + PEs)i, we have !12 mv + 0 + 12 kx = 0 + 0 + 12 kA , or 2

k 2 v= A − x2 ) ( m ! (a) The speed is a maximum at the equilibrium position, x = 0.

k 2 vmax = A = m !

(19.6 N/m ) ( 0.040 m ) = 0.28 m/s ( 0.40 kg ) 2

(b) When x = −0.015 m,

v= !

(19.6 N/m ) ⎡( 0.040 m ) − −0.015 m ⎤ = 0.26 m/s ( )⎦ ( 0.40 kg ) ⎣ 2

787

Topic 13

v= !

(19.6 N/m ) ⎡( 0.040 m ) − +0.015 m ⎤ = 0.26 m/s ( )⎦ ( 0.40 kg ) ⎣ 2

(d) If !v = 12 vmax , then

k 2 1 k 2 A − x2 ) = A ( 2 m ! m This gives !A − x = 14 A , or 2

3 3 x=A = ( 4.0 cm ) = 3.5 cm 2 2 ! 13.19

The maximum speed occurs at the equilibrium position and is !vmax = A k/m . Thus,

kA 2 (16.0 N/m ) ( 0.200 m ) m= 2 = = 4.00 kg 2 vmax 0.400 m/s) ( ! 2

and

F = mg = ( 4.00 kg ) ( 9.80 m s 2 ) = 39.2 N !g 13.20 (a) The oscillation frequency is given by f = 1/T = 1/(0.500 s) = 2.00 Hz .

(b) The angular frequency is

ω = 2π f = k/m . Solve for the spring

constant k and substitute values to find

788

Topic 13

k = 4π 2 f 2 m = 4π 2 ( 2.00 Hz ) (1.00 kg ) = 158 N/m 2

1 2

kA 2 = 12 mv 2 + 12 kx 2 →

1 2

kA 2 = 12 mv 2 + 12 k ( A/2 )

3 (158 N/m ) ( 0.250 m ) 3kA 2 v= = = 2.72 m/s 4m 4 (1.00 kg ) 2

13.21

(

)(

)

2 1 1 (a) PEs,i = kxi2 = 850 N m 6.00 × 10−2 m = 1.53 J 2 2

(b) Since the surface is frictionless, the total energy of the blockspring system is constant. Thus, KE + PEs = KEi + PEs,i = 0 + 1.53 J. At the equilibrium position, PEs = 0, so the kinetic energy must be KE = 1 mv 2 = 1.53 J , which yields ! max 2 max

vmax = !

2 (1.53 J ) 2KEmax = = 1.75 m/s m 1.00 kg

the conservation of energy gives KE + PEs = E, or !12 mv + 12 kx = E 2

and

v= !

13.22

2E = kx 2 = m

2(1.53 J) − (850 N/m) ( 3.00 × 10−2 m ) 1.00 kg

= 1.51 m/s

2π r 2π ( 0.200 m ) = = 0.628 m/s (a) v f = T 2.00 s !

789

Topic 13

(b)

1 1 = = 0.500 Hz T 2.00 s

2π 2π = = 3.14 rad/s (c) ω = T 2.00 s ! 13.23 The position, velocity, and acceleration as functions of time are given by

x = Acos(ω t)

v = −Aω sin (ω t )

a = −Aω 2 cos (ω t )

Substitute values to find:

(

)

(a) x(1.15 s) = ( 0.250 m ) cos (12.0 rad/s) (1.15 s) = 8.27 × 10−2 m

(b)

(

)

(

)

v(1.15 s) = − ( 0.250 m ) (12.0 rad/s) sin (12.0 rad/s) (1.15 s) = −2.83 m/s

a(1.15 s) = − ( 0.250 m ) (12.0 rad/s) cos (12.0 rad/s) (1.15 s) 2

(c)

13.24

= −11.9 m/s2

(a)

1 1 = = 1.89 Hz T 0.528 s

(b) The period of oscillation of an object-spring system is !T = 2π m/k , so the force constant is 2 4π 2 m 4π ( 0.238 kg ) k= = = 33.7 N/m 2 T2 0.528 s ) ( !

790

Topic 13

(c) At the turning points (x = ±A) in the oscillation, all of the energy is temporarily stored as elastic potential energy, or E = kA2/2. Thus,

2 ( 0.234 J ) 2E = = 0.118 m k 33.7 N/m

A= ! 13.25

The spring constant is found from

Fs x

mg x

(0.010 kg )(9.80 m s ) = 2.5 N m = 2

3.9 × 10−2 m

When the object attached to the spring has mass m = 25 g, the period of oscillation is

T = 2π ! 13.26

m 0.025 kg = 2π = 0.63 s k 2.5 N/m

(a) The springs compress 0.80 cm when supporting an additional load of m = 320 kg. Thus, the spring constant is

ΔF mg ( 320 kg ) ( 9.80 m s ) k= x = = = 3.9 × 105 N m −2 Δx Δx 0.80 × 10 m ! 2

(b) When the empty car, M = 2.0 × 103 kg, oscillates on the springs, the frequency will

1 1 be f = = T 2π ! 13.27

k 1 = M 2π

3.9 × 105 N/m = 2.2 Hz 2.0 × 103 kg

The general expression for the position as a function of time for an object undergoing simple harmonic motion with x = 0 at t = 0 is x =

791

Topic 13

Asin(ωt). Thus, if x = (5.2 cm)sin(8.0π⋅t), we have that the amplitude is A = 5.2 cm and the angular frequency is ω = 8.0π rad/s. (a) The period is

2π 2π T= = = 0.25 s −1 ω 8.0 π s ! (b) The frequency of motion is

f= !

1 1 = = 40 s −1 = 4.0 Hz T 0.25 s

(c) As discussed above, the amplitude of the motion is A = 5.2 cm. (d) Note: For this part, your calculator should be set to operate in radians mode. If x = 2.6 cm, then

⎛ 2.6 cm ⎞ ⎛ x⎞ ω t = sin −1 ⎜ ⎟ = sin −1 ⎜ = sin −1 (0.50) = 0.52 radians ⎟ ⎝ A⎠ 5.2 cm ⎝ ⎠ ! and

0.52 rad

0.52 rad 8.0π rad/s

= 2.1 × 102 s = 21 × 10−3 s = 21 ms

13.28 The position of a harmonic oscillator as a function of time is given by

x = Acos(ω t) . Match this pattern with the values given in x(t) = (0.200 m) cos (0.350t) to recognize that A = 0.200 m and ω = 0.350 rad/s. (a) The maximum value of v = −Aω sin(ω t) is vmax = Aω = (0.200

792

Topic 13

m)(0.350 rad/s) = 7.00 × 10−2 m/s . (b) The maximum value of a = −Aω 2 cos (ω t ) is amax = Aω2 = (0.200 m)(0.350 rad/s)2 = 2.45 × 10−2 m/s2 .

In parts (c)−(e), make sure your calculator is set to radians mode and substitute values to find:

(c)

(

x ( 2.00 s) = Acos (ω t ) = ( 0.200 m ) cos ( 0.350 rad/s) ( 2.00 s)

)

= 0.153 m

v ( 2.00 s) = −Aω sin (ω t )

(

= − ( 0.200 m ) ( 0.350 rad/s) sin ( 0.350 s) ( 2.00 s)

(d)

)

= −4.51 × 10−2 m/s a ( 2.00 s) = −Aω 2 cos (ω t )

(

= − ( 0.200 m ) ( 0.350 rad/s) cos ( 0.350 s) ( 2.00 s) 2

(e)

)

= −1.87 × 10−2 m/s2

13.29

(a) At the equilibrium position, the total energy of the system is in the form of kinetic energy and !12 mvmax = E , so the maximum 2

speed is

vmax = !

2 ( 5.83 J ) 2E = = 5.98 m/s m 0.326 kg

(b) The period of an object-spring system is !T = 2π m/k , so the force constant of the spring is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

793

Topic 13

4π 2m 4π ( 0.326 kg ) k= = = 206 N/m 2 T2 0.250 s ) ( ! 2

(c) At the turning points, x = ±A, the total energy of the system is in 2 the form of elastic potential energy, or !E = 12 KA , giving the

amplitude as

A= ! 13.30

2 ( 5.83 J ) 2E = = 0.238 m k 206 N/m

For a system executing simple harmonic motion, the total energy may 2 be written as E = KE + PEs = 12 mvmax = 12 KA 2 , where A is the amplitude !

and vmax is the speed at the equilibrium position. Observe from this 2 2 expression, that we may write vmax = kA m .

2 (a) If v = 12 vmax , then E = 12 mv 2 + 12 kx 2 = 12 mvmax becomes ! !

2 ⎞ 1 2 1 2 1 ⎛ vmax m⎜ ⎟⎠ + 2 kx = 2 mvmax 2 ⎝ 4 !

and gives 2 3 ⎛ m⎞ 2 3 ⎛ m⎞ ⎡ k ⎤ 3A x 2 = ⎜ ⎟ vmax = ⎜ ⎟ ⎢ A2 ⎥ = 4⎝ k ⎠ 4 ⎝ k ⎠ ⎣m ⎦ 4 !

x=±

A 3 2

x=±

A 2

(b) If the elastic potential energy is PEs = 12 E , we have !

1 2 1 ⎛ 1 2⎞ kx = ⎜ kA ⎟ ⎠ 2 2⎝ 2 ! 13.31

x2 =

A2 2

and

Note: Your calculator must be in radians mode for part (a) of this

794

Topic 13

problem. (a) The angular frequency of this oscillation is !ω = k/m and the displacement at time t is x = Acosωt. At t = 3.50 s, the spring force will be F = −kx = −kA cos(ωt), or

⎡⎛ 5.00 N/m ⎞ ⎤ N⎞ ⎛ F = − ⎜ 5.00 ⎟ ( 3.00 m ) cos ⎢⎜ 3.50 s ( ) ⎥ = −11.0 N , ⎟ ⎝ m⎠ ⎝ 2.00 kg ⎠ ⎢ ⎥⎦ ⎣ ! or

F = 11.0 N directed to the left

(b) The time required for one complete oscillation is !T = 2π /ω = 2π m/k . Hence, the number of oscillations made in

3.50 s is

Δt Δt 3.50 s 5.00 N/m N= = = = 0.881 T 2π m/k 2π 2.00 kg ! 13.32

F 7.50 N = 250 N/m (a) k = = −2 x 3.00 × 10 m ! k 250 N/m = = 22.4 rad/s , (b) ω = m 0.500 kg ! f= ! and

ω 1 = 2π 2π

k 1 = m 2π

250 N/m = 3.56 Hz , 0.500 kg

1 1 T= = = 0.281 s f 3.56 Hz ! −

795

Topic 13

oscillator is 1 1 1 E = KE + PEs = mv 2 + kx 2 = 0 + (250 N/m)(5.00 × 10−2 m)2 = 0.313 J 2 2 2 ! 1 2 (d) When x = A, v = 0!so!E = KE + PEs = 0 + KA . 2 !

Thus,

(

2 0.313 J

−2

250 N/m

1 2 (e) At x = 0, KE = mvmax =E, 2

vmax = !

) = 5.00 × 10 m = 5.00 cm

2 ( 0.313 J ) 2E = = 1.12 m/s m 0.500 kg

amax =

Fmax m

kA m

(250 N m)(5.00 × 10 m) = 25.0 m s = −2

0.500 kg

Note: To solve parts (f) and (g), your calculator should be set in radians mode. (f) At t = 0.500 s, Equation 13.14a gives the displacement as

⎡ 250 N/m ⎤ x = Acos (ω t ) = Acos t k/m = ( 5.00 cm ) cos ⎢( 0.500 s ) ⎥ = 0.919 cm 0.500 kg ⎣ ⎦ !

(

)

(g) From Equation 13.14b, the velocity at t = 0.500 s is

(

v = −Aω sin (ω t ) = −A k/m sin t k/m = − ( 5.00 × 10−2 m ) !

)

⎡ 250 N/m 250 N/m ⎤ sin ⎢( 0.500 s ) ⎥ = +1.10 m/s 0.500 kg 0.500 kg ⎦ ⎣

796

Topic 13

and from Equation 13.14c, the acceleration at this time is

(

a = −Aω 2 cos (ω t ) = −A ( k/m) cos ( t k/m

! 13.33

)

⎡ ⎛ 250 N/m ⎞ 250 N/m ⎤ 2 = − ( 5.00 × 10−2 m ) ⎜ cos ⎢( 0.500 s ) ⎥ = −4.59 m/s ⎟ 0.500 kg 0.500 kg ⎝ ⎠ ⎣ ⎦ k 2 A − x 2 ) = ± ω 2 ( A2 − x 2 ) From Equation 13.6, v = ± ( m ! Hence,

v = ±ω A 2 − A 2 cos 2 (ω t ) = ±ω A 1 − cos 2 (ω t ) = ±ω Asin (ω t ) !

k 2 2 From Equation 13.2, a = − x = −ω [ Acos (ω t )] = −ω Acos (ω t ) m !

13.34

(a) The height of the tower is almost the same as the length of the pendulum. From !T = 2π L/g , we obtain

gT 2 ( 9.80 m/s ) (15.5 s ) L= = = 59.6 m 4π 2 4π 2 ! 2

(b) On the Moon, where g = 1.67 m/s2, the period will be

L 59.6 m T = 2π = 2π = 37.5 s g 1.67 m s 2 ! 13.35 (a) The period is the time for one complete oscillation. Hence,

T= !

2.00 min ⎛ 60 s ⎞ 120 s ⎜⎝ 1 min ⎟⎠ = 82.0 82

T = 1.46 s

(b) The period of oscillation of a simple pendulum is !T = 2π /g , so the local acceleration of gravity must be

797

Topic 13 2 4π 2  4π ( 0.520 m ) 2 g= 2 = 2 = 9.59 m/s T (120 s 82.0) !

13.36

LT The period in Tokyo is TT = 2π and the period in Cambridge is gT ! L TC = 2π C . gC ! We know that TT = TC = 2.000 s, from which we see that

LT LC = g gC T !

gC LC 0.994 2 = = = 1.001 5 gT LT 0.992 7

13.37 If the clock kept perfect time, it would oscillate 86 400 times per day (the number of seconds in a day). Instead, the clock runs slow by 1.500 minutes/day = 90.00 s/day. The number of oscillations per day is then 86 400 − 90.00 = 86 310. The clock’s period is then:

86 400 s/day = 1.001 s/oscillation 86 310 oscillations/day

(a) Solve for the local value of g using

T = 2π

L glocal

4π 2 L 4π ( 0.248 2 m ) → glocal = = = 9.779 m/s 2 2 2 T (1.001 s ) 2

(b) Substitute values to find

T = 2π

L glocal

(

)

9.779 m/s (1.000 s) g T2 → L = local 2 = = 0.247 7 m 4π 4π 2 2

798

Topic 13

13.38

The coat hanger acts as a physical pendulum and its period of oscillation is T = 2π I /mgd , where d is the distance from the pivot to ! the center of mass. Thus, the moment of inertia about the axis perpendicular to the plane of oscillation and passing through the pivot must be 2

⎛ T ⎞ ⎛ 1.25 s ⎞ −2 2 I = mgd ⎜ ⎟ = ( 0.238 kg ) ( 9.80 m s 2 ) ( 0.180 m ) ⎜ ⎟⎠ = 1.66 × 10 kg ⋅ m ⎝ ⎠ ⎝ 2π 2π ! 13.39

gT 2 From T = 2π L/g , the length of a pendulum with period T is L = . ! 4π 2 !

(9.8 m s )(1.0 s) = 0.25 m = 25 cm (a) On Earth, with T = 1.0 s, L = 2

4π 2

(3.7 m/s )(1.0 s) = 0.094 m = 9.4 cm (b) If T = 1.0 s on Mars, L = 2

4π 2

(c) and (d) The period of an object on a spring is !T = 2π m/k , which is independent of the local free-fall acceleration. Thus, the same mass will work on Earth and on Mars. This mass is

KT 2 (10 N m ) (1.0 s ) m= = = 0.25 kg 4π 2 4π 2 ! 2

13.40

The apparent free-fall acceleration is the vector sum of the actual freefall acceleration and the negative of the elevator’s acceleration. To see this, consider an object that is hanging from a vertical string in the elevator and appears to be at rest to the elevator passengers. These

799

Topic 13

passengers believe the tension in the string is the negative of the    object’s weight, or T = −mg apparent where g apparent is the apparent free! !

fall acceleration in the elevator. An observer located outside the elevator applies Newton’s second law      to this object by writing ∑ F = T + mg = ma e where !a e is the acceleration !

of the elevator and all its contents. Thus,        T = m ( a e − g ) = −mg apparent , which gives g apparent = g − a e . ! !

(a) If we choose downward as the positive direction, then  a = −5.00 m s 2 in this case and ! e  g = ( 9.80 + 5.00 ) m s 2 = +14.8 m/s 2 (downward). The period ! apparent

of the pendulum is

L 5.00 m T = 2π = 2π = 3.65 s gapparent 14.8 m/s 2 !  (b) Again choosing downward as positive, a e = 5.00 m/s 2 and !  g apparent = ( 9.80 − 5.00 ) m s 2 = +4.80 m/s 2 (downward) !

in this case. The period is now given by

L 5.00 m T = 2π = 2π = 6.41 s gapparent 4.80 m/s 2 !     (c) If a e = 5.00 m s 2 horizontally, the vector sum g apparent = g − a e ! !

800

Topic 13

is as shown in the sketch above. The magnitude is g = ! apparent

( 5.00 m s ) + ( 9.80 m s ) = 11.0 m/s , 2 2

2 2

and the period of the pendulum is

L 5.00 m T = 2π = 2π = 4.24 s gapparent 11.0 m/s 2 ! 13.41

(a) The distance from the bottom of a trough to the top of a crest is twice the amplitude of the wave. Thus, 2A = 8.26 cm and A = 4.13 cm.

FIGURE P13.41 (b) The horizontal distance from a crest to a trough is a half wavelength. Hence,

λ/2 = 5.20 cm and λ = 10.4 cm (c) The period is

801

Topic 13

1 1 T= = = 5.56 × 10−2 s −1 f 18.0 s ! (d) The wave speed is

v = λ f = (10.4 cm ) (18.0 s −1 ) = 187 cm s = 1.87 m s ! 13.42

(a) The amplitude is the magnitude of the maximum displacement from equilibrium (at x = 0). Thus, A = 2.00 cm

FIGURE P13.42 (b) The period is the time for one full cycle of the motion. Therefore, T = 4.00 s (c) The period may be written as T = 2π/ω, so the angular frequency is 2π 2π π ω= = = rad/s T 4.00 s 2 ! 2 2 (d) The total energy may be expressed as !E = 12 mvmax = 12 KA . Thus,

v = A k /m , and since !ω = k/m , this becomes vmax = ωA and ! max

yields ⎛π ⎞ vmax = ω A = ⎜ rad/s⎟ ( 2.00 cm ) = π cm/s ⎝ ⎠ 2 !

802

Topic 13

(e) The spring exerts maximum force, |F| = k|x|, when the object is at maximum distance from equilibrium, i.e., at |x| = A = 2.00 cm. Thus, the maximum acceleration of the object is

F kA ⎛π ⎞ amax = max = = ω 2 A = ⎜ rad s⎟ ( 2.00 cm ) = 4.93 cm s 2 ⎝2 ⎠ m m ! 2

(f) The general equation for position as a function of time for an object undergoing simple harmonic motion with t = 0 when x = 0 is x = Asin(ωt). For this oscillator, this becomes

! 13.43

⎛π ⎞ x = ( 2.00 cm ) sin ⎜ t ⎟ ⎝2 ⎠

(a) The speed of propagation for a wave is the product of its frequency and its wavelength, v = λf. Thus, the frequency must be

f= !

v 3.00 × 108 m/s = = 5.45 × 1014 Hz λ 5.50 × 10−7 m

1 1 = 1.83 × 10−15 s . (b) The period is T = = 14 f 5.45 × 10 Hz ! 13.44

(a) The frequency of a transverse wave is the number of crests that pass a given point each second. Thus, if 5.00 crests pass in 14.0 seconds, the frequency is

f= !

5.00 = 0.357 s −1 = 0.357 Hz 14.0 s

(b) The wavelength of the wave is the distance between successive © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

803

Topic 13

maxima or successive minima. Thus, λ = 2.76 m and the wave speed is −

v = λf = (2.76 m)(0.357 s 1) = 0.985 m/s 13.45

The speed of the wave is

Δx 425 cm v= = = 42.5 cm/s Δt 10.0 s ! and the frequency is number of vibrations occurring each second, or f = 40.0 vib/30.0 s.

Thus,

13.46

( 42.5 cm/s )( 30.0 s ) = 31.9 cm v 42.5 cm/s λ= = = f 40.0 vib/30.0 s 40.0 vib !

From v = λf, the wavelength (and size of smallest detectable insect) is

λ=

v f

343 m/s 3

60.0 × 10 Hz

= 5.72 × 10−3 m = 5.72 mm

13.47 The wavelength is related to the wave speed and frequency by v = λf where v = 343 m/s. For two waves at 440 Hz and 442 Hz, the magnitude of the difference in their wavelengths is

Δλ = λ 440 − λ 442 =

13.48

v f 440

−

⎛f −f ⎞ = ⎜ 442 440 ⎟ v = 3.53 × 10−3 m f 442 ⎝ f 440 f 442 ⎠ v

(a) When the boat is at rest in the water, the speed of the wave relative to the boat is the same as the speed of the wave relative to the water, v = 4.0 m/s. The frequency detected in this case is

804

Topic 13

f= !

v 4.0 m/s = = 0.20 Hz λ 20 m

   (b) Taking eastward as positive, !v wave,boat = v wave,water − v boat,water (see the discussion of relative velocity in Topic 3 of the textbook) gives

(

)

! v wave,boat = +4.0 m/s − −1.0 m s = +5.0 m s

and

! v boat,wave = v wave,boat = 5.0 m/s

Thus,

f= ! 13.49

v boat,wave 5.0 m/s = = 0.25 Hz λ 20 m

The down and back distance is 4.00 m + 4.00 m = 8.00 m. The speed is then 4 ( 8.00 m ) d v = total = = 40.0 m/s = F/µ t 0.800 s !

Now,

m 0.200 kg µ= = = 5.00 × 10−2 kg/m L 4.00 m ! so

F = µv 2 = ( 5.00 × 10−2 kg/m ) ( 40.0 m/s ) = 80.0 N ! 2

13.50 The wave speed is given by v = F/ µ where F is the cable tension and µ is the linear density. Here, the linear density is µ = mcable/Lcable =

805

Topic 13

12.0 kg/20.0 m = 0.600 kg/m and, for a cable hanging vertically in equilibrium, the tension at any point equals underlying supported weight. (a) At the cable’s midpoint, the underling mass equals the 25.0-kg mass and half of the cable’s 12.0-kg mass so that munder = 25.0 kg + 6.0 kg = 31.0 kg and Tmiddle = munderg = (31.0 kg)(9.80 m/s2) = 304 N. The wave speed is then

vmiddle =

Tmiddle 304 N = = 22.5 m/s µ 0.600 kg/m

(b) At the bottom of the cable, the underlying mass is only the 25.0-kg mass so that Tbottom = (25.0 kg)(9.80 m/s2) = 245 N. The wave speed is then

vbottom =

Tbottom 245 N = = 20.2 m/s µ 0.600 kg/m

13.51

v= !

F 1 350 N = = 5.20 × 102 m/s µ 5.00 × 10−3 kg/m

13.52

(a)

f= !

v= !

1 1 1 1 → T = → [T ] = = 3 =T T f [f] T

units!are!seconds

T M L2 ML → T = µv 2 → [T ] = [ µ ] ⎡⎣ v 2 ⎤⎦ = ⋅ 2 = 2 µ L T T

units!are!newtons

(b) The first T is period of time; the second is force of tension.

806

Topic 13

13.53

(a) The mass per unit length is

µ=

m L

0.060 0 kg 5.00 m

= 1.20 × 10−2 kg/m

From !v = F/µ , the required tension in the string is

F = v 2 µ = ( 50.0 m s ) (1.20 × 10−2 kg/m ) = 30.0 N ! 2

(b) v = ! 13.54

F 8.00 N = = 25.8 m/s µ 1.20 × 10−2 kg/m

The mass per unit length of the wire is

m 4.00 × 10−3 kg µ= = = 2.50 × 10−3 kg/m , L 1.60 m ! and the speed of the pulse is

L Δt

1.60 m 0.036 1 s

= 44.3 m/s

The tension in the wire is F = mg = µv2, so the lunar acceleration of gravity must be −3 v 2 µ ( 44.3 m/s ) ( 2.50 × 10 kg/m ) g= = = 1.64 m s 2 m 3.00 kg ! 2

13.55

The period of the pendulum is T = 2π L/g , so the length of the string ! is

gT 2 ( 9.80 m s ) ( 2.00 s ) L= = = 0.993 m 4π 2 4π 2 ! 2

807

Topic 13

The mass per unit length of the string is then

m 0.060 0 kg kg µ= = = 6.04 × 10−2 L 0.993 m m ! When the pendulum is vertical and stationary, the tension in the string is F = Mball g = (5.00 kg)(9.80 m/s2) = 49.0 N and the speed of transverse waves in it is

v= ! 13.56

F 49.0 N = = 28.5 m/s µ 6.04 × 10−2 kg/m

If µ1 = m1/L is the mass per unit length for the first string, then µ2 = m2/L = m1/2L = µ1/2 is that of the second string. Thus, with F2 = F1 = F, the speed of waves in the second string is

⎛ F⎞ F 2F v2 = = = 2⎜ = 2v1 = 2 ( 5.00 m/s ) = 7.07 m/s µ2 µ1 µ1 ⎟⎠ ⎝ ! 13.57

(a) The tension in the string is F = mg = (3.00 kg)(9.80 m/s2) = 29.4 N. Then, from !v = F/µ , the mass per unit length is

F 29.4 N −2 µ= 2 = kg/m 2 = 5.10 × 10 v 24.0 m/s ) ( ! (b) When m = 2.00 kg, the tension is F = mg = (2.00 kg)(9.80 m/s2) = 19.6 N

808

Topic 13

and the speed of transverse waves in the string is

F 19.6 N = = 19.6 m/s µ 5.10 × 10−2 kg/m

v= ! 13.58

If the tension in the wire is F, the tensile stress is Stress = F/A, so the speed of transverse waves in the wire may be written as

v= !

F = µ

A ⋅Stress Stress = m/L m ( A ⋅ L)

But A ⋅ L = V = volume, so m/(A ⋅ L) = ρ = density. Thus, v = Stress/ρ . ! Taking the density of steel to be equal to that of iron, the maximum speed of waves in the wire is

vmax =

13.59

(Stress) ρ steel

max

2.70 × 109 Pa 7.86 × 103 kg m 3

= 586 m s

(a) The speed of transverse waves in the line is !v = F/µ , with

µ = m/L being the mass per unit length. Therefore,

v= !

F F = = µ m/L

FL = m

(12.5 N)( 38.0 m ) = 13.4 m/s 2.65 kg

(b) The worker could throw an object, such as a snowball, at one end of the line to set up a pulse, and use a stopwatch to measure the time it takes a pulse to travel the length of the line. From this measurement, the worker would have an estimate of the wave

809

Topic 13

speed, which in turn can be used to estimate the tension. 13.60

(a) In making n round trips along the length of the line, the total distance traveled by the pulse is Δx = n(2L) = 2nL. The wave speed is then Δx 2nL v= = t t !

(b) From !v = F/µ as the speed of transverse waves in the line, the tension is 2 2 2 ⎛ M ⎞ ⎛ 2nL ⎞ ⎛ M ⎞ ⎛ 4n L ⎞ 4n ML F = µv 2 = ⎜ ⎟ ⎜ = = ⎝ L ⎠ ⎝ t ⎟⎠ ⎜⎝ L ⎟⎠ ⎜⎝ t 2 ⎟⎠ t2 ! 2

13.61

(a) Constructive interference produces the maximum amplitude

A′ = A1 + A2 = 0.30 m + 0.20 m = 0.50 m ! max (b) Destructive interference produces the minimum amplitude

A′ = A1 − A2 = 0.30 m − 0.20 m = 0.10 m ! min 13.62

We are given that x = Acos(ωt) = (0.25 m)cos(0.4πt). (a) By inspection, the amplitude is seen to be A = 0.25 m (b) The angular frequency is ω = 0.4π rad/s. But !ω = k/m , so the spring constant is k = mω2 = (0.30 kg)(0.4π rad/s)2 = 0.47 N/m

810

Topic 13

(c) Note: Your calculator must be in radians mode for part (c). At t = 0.30 s, x = (0.25 m)cos[(0.4π rad/ s)(0.30 s)] = 0.23 m (d) From conservation of mechanical energy, the speed at displacement x is given by 2 2 !v = ω A − x . Thus, at t = 0.30 s, when x = 0.23 m, the speed is

v = ( 0.4π rad/s ) !

( 0.25 m ) − ( 0.23 m ) = 0.12 m/s 2

13.63 (a) The period of a vibrating object-spring system is !T = 2π /ω = 2π m/k , so the spring constant is 2 4π 2m 4π ( 2.00 kg ) k= = = 219 N/m 2 T2 0.600 s ) ( !

(b) If T = 1.05 s for mass m2, this mass is

kT 2 ( 219 N m ) (1.05 s ) m2 = = = 6.12 kg 4π 2 4π 2 ! 2

13.64

(a) The period is the reciprocal of the frequency, or

1 1 T= = = 5.10 × 10−3 s = 5.10 ms −1 f 196 s ! (b) λ = ! 13.65

vsound 343 m/s = = 1.75 m f 196 s −1

(a) The period of a simple pendulum is !T = 2π /g , so the period of the first system is

811

Topic 13

 0.700 m T1 = 2π = 2π = 1.68 s g 9.80 m s 2 ! (b) The period of an object-spring system is !T = 2π m/k , so if the period of the second system is T2 = T1, then 2π m/k = 2π /g and ! the spring constant is

mg (1.20 kg ) ( 9.80 m s ) k= = = 16.8 N/m  0.700 m ! 2

13.66

Since the spring is “light,” we neglect any small amount of energy lost in the collision with the spring, and apply conservation of mechanical energy from when the block first starts until it comes to rest again. This gives 1 2 (KE + PEg + PEs)f = (KE + PEg + PEs)i, or 0 + 0 + kxmax = 0 + 0 + mghi 2 !

2 ( 0.500 kg ) ( 9.80 m s 2 ) ( 2.00 m ) 2mghi Thus, xmax = = = 0.990 m k 20.0 N/m ! 13.67

Choosing PEg = 0 at the initial height of the 3.00-kg object, conservation of mechanical energy gives (KE + PEg + PEs)f = (KE + PEg + PEs)i, or 12 mv 2 + mg(−x) + 12 kx 2 = 0 , where v is the speed of the object ! after falling distance x. (a) When v = 0, the non-zero solution to the energy equation from above gives !12 kxmax = mgxmax , or 2

812

Topic 13

2mg 2 ( 3.00 kg ) ( 9.80 m s ) k= = = 588 N/m x 0.100 m max ! 2

(b) When x = 5.00 cm = 0.500 0 m, the energy equation gives

v = 2gx − kx 2 m , or !

( 588 N/m )( 0.050 0 m ) = 0.700 m/s v = 2 9.80 m s ( 0.050 0 m ) −

(

)

3.00 kg

! 13.68

(a) We apply conservation of mechanical energy from just after the collision until the block comes to rest. Conservation of energy gives (KE + PEs)f = (KE + PEs)i or 0 + 12 kx 2f = 12 MV 2 + 0 . The speed of ! the block just after the collision is then

V= !

kx 2f M

( 900 N/m )( 0.050 0 m ) = 1.50 m/s = 2

1.00 kg

Now, we apply conservation of momentum from just before impact to immediately after the collision. This gives m(vbullet)i + 0 = m(vbullet)f + MV, or

⎛

⎞

( vbullet ) f = ( vbullet )i − ⎛⎜⎝ m ⎞⎟⎠ V = 400 m/s − ⎜ 5.00 × 10−1 kg ⎟ (1.5 m/s ) = 100 m/s M

1.00 kg

⎝

⎠

(b) The mechanical energy converted into internal energy during the collision is ΔE = KEi − ΣKE f = 12 m ( v bullet )i − 12 m ( v bullet ) f − 12 MV , or ! 2

813

Topic 13

1 1 2 2 2 ΔE = ( 5.00 × 10−3 kg ) ⎡( 400 m/s ) − (100 m/s ) ⎤ − (1.00 kg ) (1.50 m/s ) ⎣ ⎦ 2 2 !

ΔE = 374 J ! 13.69

The maximum acceleration of the oscillating system is

amax = ω2A = (2πf)2 A The friction force, fs, acting between the two blocks must be capable of accelerating block B at this rate. When block B is on the verge of slipping, fs = (fs)max = µsn = µsmg = mamax and we must have amax = (2πf)2 A = µsg

( 0.600)( 9.80 m s ) µs g = 6.62 × 102 m = 6.62 cm Thus, A = 2 = 2 ⎡⎣ 2π (1.50 Hz ) ⎤⎦ ( 2π f ) ! 2

13.70

(a) When the gun is fired, the energy initially stored as elastic potential energy in the spring is transformed into kinetic energy of the bullet. Assuming no loss of energy, we have !12 mv = 12 kxi , 2

k 9.80 N/m v = xi = ( 0.200 m ) = 19.8 m/s m 1.00 × 10−2 kg ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

814

Topic 13

(b) From Δy = v0yt + 12 ayt 2 , the time required for the pellet to drop 1.00 ! m to the floor, starting with v0y = 0, is

t= !

2 ( Δy ) 2 ( −1.00 m ) = = 0.452 s ay −9.80 m/s 2

The range (horizontal distance traveled during the flight) is then Δx = v0xt = (19.8 m/s)(0.452 s) = 8.95 m 13.71

(a) The force diagram below shows the forces acting on the balloon when it is displaced distance s = Lθ along the circular arc it follows.

The net force tangential to this path is Fnet = ΣFx = −B sin θ + mg sin θ = −(B − mg) sin θ For small angles, sin θ ≈ θ = s/L. Also, mg = (ρHeV)g and the buoyant force is B = (ρairV)g. Thus, the net restoring force acting on the balloon is

815

Topic 13

(

)

⎡ ρ − ρ Vg ⎤ air He ⎥s Fnet ≈ − ⎢ ⎢ ⎥ L ⎣ ⎦ Observe that this is in the form of Hooke’s law, F = −ks, with k = (ρair − ρHe)Vg/L. Thus, the motion will be simple harmonic. (b) The period of this simple harmonic motion is given by

1 2π m T= = = 2π = 2π f ω k !

⎛ ρHe ⎞ L ρHeV = 2π ⎜ ⎝ ρair − ρHe ⎟⎠ g ( ρair − ρHe )Vg L

This yields ⎛ ⎞ ( 3.00 m ) 0.179 kg m 3 T = 2π ⎜ = 1.40 s 3 3⎟ ⎝ 1.29 kg m − 0.179 kg m ⎠ ( 9.80 m s 2 ) !

13.72

(a) When the object is given some small upward displacement, the net restoring force exerted on it by the rubber bands is y Fnet = ΣFy = −2F sin θ, where tan θ = L !

For small displacements, the angle θ will be very small. Then sin θ ≈ tan θ = y/L, and the net restoring force is

⎛ y⎞ ⎛ 2F ⎞ Fnet = −2F ⎜ ⎟ = − ⎜ ⎟ y ⎝ L⎠ ⎝ L⎠ ! (b) The net restoring force found in part (a) is in the form of Hooke’s law F = −ky, with k = 2F/L. Thus, the motion will be simple harmonic, and the angular frequency is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

816

Topic 13

k ω= = m ! 13.73

2F mL

Newton’s law of gravitation is

F=− !

GMm ⎛4 ⎞ ,!where!M = ρ ⎜ π r 3 ⎟ 2 ⎝ ⎠ r 3

Thus,

⎛4 ⎞ F = − ⎜ πρGm⎟ r ⎝3 ⎠

which is of Hooke’s law form, F = −kr, with 4 k = πρGm 3 !

13.74

The inner tip of the wing is attached to the end of the spring and always moves with the same speed as the end of the vibrating spring. Thus, its maximum speed is

k 4.7 × 10−4 N m vinner,!max = vspring,!max = A = ( 0.20 cm ) = 0.25 cm s m 0.30 × 10−3 ! Treating the wing as a rigid bar, all points in the wing have the same angular velocity at any instant in time. As the wing rocks on the

817

Topic 13

fulcrum, the inner tip and outer tips follow circular paths of different radii. Since the angular velocities of the tips are always equal, we may write

v v ω = outer = inner . The maximum speed of the outer tip is then router rinner ! ⎛ 15.0 mm ⎞ ⎛r ⎞ vouter,max = ⎜ outer ⎟ vinner,max = ⎜ ( 0.25 cm/s ) = 1.3 cm/s ⎝ rinner ⎠ ⎝ 3.00 mm ⎟⎠ ! k 500 N/m = = 15.8 rad/s 13.75 (a) ω = m 2.00 kg ! (b) Apply Newton’s second law to the block while the elevator is accelerating: ΣFy = Fs − mg = may With Fs = kx and ay = g/3, this gives kx = m(g + g/3), or 2 4mg 4 ( 2.00 kg ) ( 9.80 m s ) x= = = 5.23 × 102 m = 5.23 cm 3k 3 ( 500 N/m ) !

13.76

(a) Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius L − yf. Also, observe that the y-coordinate of the object at this point must be negative (yf < 0) so the spring is stretched and exerting an upward tension force of magnitude greater than the object’s weight. This is necessary so the object experiences a net force toward the pivot to

818

Topic 13

supply the needed centripetal acceleration in this position. This is summarized by Newton’s second law applied to the object at this point, stating

ΣFy = −ky f − mg = !

mv 2 L − yf

(b) Conservation of energy requires that E = KEi + PEg,i + PEs,i = KEf + PEg,f + PEs,f, or 1 1 E = 0 + mgL + 0 = mv 2 + mgy f + ky 2f 2 2 !

reducing to

(

)

mv 2 = 2mg L − y f − ky 2f

(c) From the result of part (a), observe that mv2 = −(L − yf)(kyf + mg) Substituting this into the result from part (b) gives

(

) (

)(

)

2mg L − y f = − L − y f ky f + mg + ky 2f ! After expanding and regrouping terms, this becomes

( 2k ) y 2f + ( 3mg − kL) y f + ( −3mgL) = 0

which is a quadratic equation !ay f + by f + c = 0 , with 2

a = 2k = 2(1 250 N/m) = 2.50 × 103 N/m b = 3mg − kL = 3(5.00 kg)(9.80 m/s2) − (1 250 N/m)(1.50 m) = © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

819

Topic 13

−1.73 × 103 N and c = −3mgL = −3(5.00 kg)(9.80 m/s2)(1.50 m) = −221 N ⋅ m Applying the quadratic formula, keeping only the negative solution [see the discussion in part (a)] gives

−b − b2 − 4ac − ( −1.73 × 10 ) − yf = = 2a ! 3

( −1.73 × 10 ) − 4 ( 2.50 × 10 )( −221) 2 ( 2.50 × 10 ) 3 2

yf = −0.110 m

(d) Because the length of this pendulum varies and is longer throughout its motion than a simple pendulum of length L, its period will be greater than that of a simple pendulum.

Topic 14

820

Topic 14 Sound

QUICK QUIZZES 14.1

Choice (c). The speed of sound in air is given by v = (331 m/s) T /273 K . Thus, increasing the absolute temperature, T,

will increase the speed of sound. Changes in frequency, amplitude, or air pressure have no effect on the speed of sound. 14.2

Choice (c). The distance between you and the buzzer is increasing. Therefore, the intensity at your location is decreasing. As the buzzer falls, it moves away from you with increasing speed. This causes the detected frequency to decrease.

14.3

Choice (b). The speed of sound increases in the warmer air, while the speed of the sound source (the plane) remains constant. Therefore, the ratio of the speed of the source to that of sound (that is, the Mach number) decreases.

14.4

Choices (b) and (e). A string fastened at both ends can resonate at any integer multiple of the fundamental frequency. Of the choices listed, only 300 Hz and 600 Hz are integer multiples of the 150 Hz fundamental frequency.

Topic 14

14.5

821

Choice (d). In the fundamental mode, an open pipe has a node at the center and antinodes at each end. The fundamental wavelength of the open pipe is then twice the length of the pipe and the fundamental frequency is fopen = v/2L. When one end of the pipe is closed, the fundamental mode has a node at the closed end and an antinode at the open end. In this case, the fundamental wavelength is four times the length of the pipe and the fundamental frequency is fclose = v/4L.

14.6

Choice (a). The change in the length of the pipe, and hence the fundamental wavelength, is negligible. As the temperature increases, the speed of sound in air increases and this causes an increase in the fundamental frequency, f0 = v/λ0.

14.7

Choice (b). Since the beat frequency is steadily increasing, you are increasing the difference between the frequency of the string and the frequency of the tuning fork. Thus, your action is counterproductive and you should reverse course by loosening the string.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 14.2

Increasing the sound intensity by a factor of 2n increases the decibel level by 3⋅n dB. Because 2.0 = 21, 4.0 = 22, 8.0 = 23, and 16 = 24, the decibel levels are: (a) 85 dB + 1⋅(3.0 dB) = 88 dB, (b) 85 dB + 2⋅(3.0 dB) =

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822

91 dB, (c) 85 dB + 3⋅(3.0 dB) = 94 dB, and (d) 85 dB + 4⋅(3.0 dB) = 97 dB. 14.4

The speed of light is so high that the arrival of the flash is practically simultaneous with the lightning discharge. Thus, the delay between the flash and the arrival of the sound of thunder is the time sound takes to travel the distance separating the lightning from you. By counting the seconds between the flash and thunder and knowing the approximate speed of sound in air, you have a rough measure of the distance to the lightning bolt.

14.6

A vibrating string is not able to set very much air into motion when vibrated alone. Thus it will not be very loud. If it is placed on the instrument, however, the string’s vibration sets the sounding board of the guitar into vibration. A vibrating piece of wood is able to move a lot of air, and the note is louder.

14.8

The answers follow directly from definitions related to standing waves on a string: (i) The harmonic number is equal to the number of antinodes . (ii) The distance from a node to its adjacent antinode is

always equal to a quarter wavelength. (iii) The fundamental frequency has harmonic number one . 14.10

Consider the level of fluid in the bottle to be adjusted so that the air column above it resonates at the first harmonic. This is given by f =

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823

v/4L. This equation indicates that as the length L of the column increases (fluid level decreases), the resonant frequency decreases.

ANSWERS TO EVEN NUMBERED PROBLEMS 14.2

1 × 1011 Pa

14.4

0.196 s

14.6

λ = 17 m for f = 20 Hz; λ = 1.7 cm for f = 20 000 Hz

14.8

18.6 m

14.10

(a) 98.8 dB

(b)

6.12 × 10−6 W/m2

(a) 1.0 × 103 W/m2 (b) The intensity level is 1 000 times that at the threshold of pain.

14.14

150 dB

14.16

(a) 5.0 × 10−17 W

(b)

5.0 × 10−5 W

14.18

(a) 0.316 W/m2

(b)

1.58 W/m2

(e)

1.26 × 106 m

(f) The sound intensity falls as the sound wave travels farther from

Topic 14

824

the source until it is much lower than the ambient noise level and is drowned out. 14.20

(a) 1.32 × 10−4 W/m2

(b)

81.2 dB

14.22

(a) 7.96 × 10−2 W/m2

(b)

109 dB

14.24

(a) IA/IB = 2

(b)

IA/IC = 5

14.26

(a) 10.7 m/s

(b)

415 Hz

14.28

(a)

10.0 kHz

(b)

3.33 kHz

14.30

(a) behind the car

(b)

32.0 m/s

14.32

(a) 3.29 m/s (b)

14.34

14.36

2.82 m

Yes, the bat gains on the insect at a rate of 1.71 m/s.

(a) 2.16 × 10−2 m/s (c)

(c)

(b)

2 000 029 Hz

(b)

the yellow submarine

2 000 058 Hz

(a) fO = fs[(v + vO)/(v − vs)] (c)

the red submarine

(d)

increases the time (period), decreases the frequency

(e)

negative

(f) decreases the time (period), increases the frequency (g) positive

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825

(h) 5.30 × 103 Hz 14.38

(a) 0.227 m

14.40

1.43 m

14.42

823.8 N

14.44

1.00 cm toward the nut

14.46

(a) 4

14.48

n (a) f A = A 2LA !

TA µA

(b)

0.454 m

(b)

1.00 × 103 Hz 2

⎛ n ⎞ (b) fB = fA/2 (c) TB = ⎜ A ⎟ TA ⎝ nA + 1 ⎠ !

(d) TB/TA = 9/16 14.50

(a) µ = 4.9 × 10−3 kg/m

(b)

9.00 kHz

14.54

(a) 229 Hz and 457 Hz

(b)

114 Hz and 343 Hz

14.56

(a) 536 Hz

(b)

4.29 cm

14.58

(a) fn = n(0.085 8 Hz) n = 1, 2, 3, … (b) Yes. The tunnel can resonate at many closely spaced frequencies, and the sound would be greatly amplified.

14.60

(a) f1 = 50.0 Hz

(b)

open at only one end

Topic 14

826

14.62

523 Hz and 527 Hz

14.64

29.7 cm

14.66

3.98 Hz

14.68

2.95 cm

14.70

∼1000 mosquitoes

14.72

(a) 65.0 dB

14.74

21.4 m

14.76

1 204 Hz

14.78

(a) 0.655 m

14.80

1.95 m/s

14.82

(a) 32.2°C

(b)

67.8 dB

(b)

13.4°C

(b)

3.6 × 102 Hz

(c)

69.6 dB

PROBLEM SOLUTIONS 14.1

(a) We ignore the time required for the lightning flash to arrive. Then, the distance to the lightning stroke is d = vsound ⋅ Δt = (343 m/s)(16.2 s) = 5.56 × 103 m = 5.56 km (b) No. Since vlight >> vsound, the time required for the flash of light to reach the observer is negligible in comparison to the time

Topic 14

827

required for the sound to arrive, and knowledge of the actual value of the speed of light is not needed. 14.2

The speed of longitudinal waves in a fluid is !v = B/ρ . Considering the Earth’s crust to consist of a very viscous fluid, our estimate of the average bulk modulus of the material in Earth’s crust is B = ρv2 = (2 500 kg/m3)(7 × 103 m/s)2 = 1 × 1011 Pa.

14.3

5 5 The Celsius temperature is TC = (TF − 32 ) = (114 − 32 ) = 45.6°C and 9 9 !

the speed of sound in the air is

T 45.6 v = ( 331 m/s ) 1 + C = ( 331 m/s ) 1 + = 358 m/s 273 273 ! 14.4

The speed of sound in seawater at 25°C is 1 533 m/s. Therefore, the time for the sound to reach the sea floor and return is

14.5

2d v

(

2 150 m

) = 0.196 s

1 533 m/s

Since the sound had to travel the distance between the hikers and the mountain twice, the time required for a one-way trip was 1.50 s. The distance the sound traveled to the mountain was d = (343 m/s)(1.50 s) = 515 m

14.6

At T = 27°C, the speed of sound in air is

Topic 14

828

T 27 v = ( 331 m/s ) 1 + C = ( 331 m/s ) 1 + = 347 m/s 273 273 ! The wavelength of the 20 Hz sound is

v 347 m/s λ= = = 17 m f 20 Hz ! and that of the 20 000 Hz is

λ=

347 m/s 20 000 Hz

= 1.7 × 10−2 m = 1.7 cm

Thus, range of wavelengths of audible sounds at 27°C is 1.7 cm to 17 m. 14.7

The reflected percentage of an incident ultrasound wave is 2

⎛ ρ − ρt ⎞ PR = ⎜ i × 100 ⎝ ρ + ρ ⎟⎠ i

where ρi is the ambient material (muscle tissue, in this case) and ρt is the reflecting material (bone, in this case). Substitute the given values to find

⎛ ρ − ρt ⎞ ⎛ 1.06 × 103 kg/m 3 − 1.90 × 103 kg/m 3 ⎞ PR = ⎜ i × 100 = ⎜⎝ 1.06 × 103 kg/m 3 + 1.90 × 103 kg/m 3 ⎟⎠ × 100 ⎝ ρ + ρ ⎟⎠ i

= 8.05% 14.8

At a temperature of T = 10.0°C the speed of sound in air is

Topic 14

829

T 10.0 v = ( 331 m/s ) 1 + C = ( 331 m/s ) 1 + = 337 m/s 273 273 ! The elapsed time between when the stone was released and when the sound is heard is the sum of the time t1 required for the stone to fall distance h and the time t2 required for sound to travel distance h in air on the return up the well. That is, t1 + t2 = 2.00 s. The distance the stone

gt 2 falls, starting from rest, in time t1 is h = 1 . 2 ! Also, the time for the sound to travel back up the well is h t2 = = 2.00 s − t1 v !

Combining these two equations yields !(g/2v)t1 = 2.00 s − t1 . 2

With v = 337 m/s and g = 9.80 m/s2, this becomes −2 −1 2 !(1.45 × 10 s )t1 + t1 − 2.00 s = 0

Applying the quadratic formula yields one positive solution of t1 = 1.95 s, so the depth of the well is

gt12 (9.80 m/s 2 )(1.95 s)2 h= = = 18.6 m 2 2 ! 14.9

(a) Because the speed of sound in air is vair = 343 m/s while its speed in the steel rail is vsteel = 5 950 m/s, the pulse traveling in the steel rail arrives first.

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830

(b) The difference in times when the two pulses reach the microphone at the opposite end of the rail is

Δt =

L vair

−

⎛ ⎞ 1 1 −2 = 8.50 m ⎜ − ⎟ = 2.34 × 10 s = 23.4 ms vsteel ⎝ 343 m/s 5 950 m/s ⎠ L

(

)

14.10 (a) The intensity level corresponding to an intensity of 7.50 × 10−3 W/m2 is ⎛ I⎞ ⎛ 7.50 × 10−3 W/m 2 ⎞ β = 10log ⎜ ⎟ = 10log ⎜ = 98.8 dB ⎝ 1.00 × 10−12 W/m 2 ⎟⎠ ⎝I ⎠ 0

(b) The ratio of wave intensities on spherical surfaces of different radii is I1 /I 2 = r22 /r12 . Solve for I2 and substitute values to find the intensity at a distance of r2 = 35.0 m:

r2 (1.00 m ) 7.50 × 10−3 W/m 2 = 6.12 × 10−6 W/m 2 I 2 = 12 I1 = r2 ( 35.0 m )2 2

(

)

(c) The intensity level corresponding to an intensity of 6.12 × 10−6 W/m2 is ⎛I ⎞ ⎛ 6.12 × 10−6 W/m 2 ⎞ β 2 = 10log ⎜ 2 ⎟ = 10log ⎜ = 67.9 dB ⎝ 1.00 × 10−12 W/m 2 ⎟⎠ ⎝ I0 ⎠

14.11 (a) Solve for the intensity at a distance of 1.00 m using the definition of decibel level:

Topic 14

831

⎛ ⎞ β −12⎟ ⎛ I1 ⎞ 10.5−12 ) ⎝⎜ 10 ⎠ 10 β = 10log ⎜ ⎟ → I1 = I0 10 = 10 = 10( = 3.16 × 10−2 W/m 2 ⎝I ⎠ 0

(b) For sound propagating as a spherical wave, the ratio of wave intensities on spherical surfaces of different radii is I1 /I 2 = r22 /r12 . Solve for I2 and substitute values:

r2 (1.00 m ) 3.16 × 10−2 W/m 2 = 7.91 × 10−5 W/m 2 I 2 = 12 I1 = r2 ( 20.0 m )2 2

(

)

(c) Multiply the sound intensity from one cicada by the number of cicadas (in this case, 100) to find the total intensity: I100 = 7.91 × 10−3 W/m2. Finally, apply the definition of decibel level to find ⎛I ⎞ ⎛ 7.91 × 10−3 W/m 2 ⎞ β100 = 10log ⎜ 100 ⎟ = 10log ⎜ = 99.0 dB ⎝ 1.00 × 10−12 W/m 2 ⎟⎠ ⎝ I0 ⎠

14.12

(a) The decibel level, β, of a sound is given β = 10 log(I/I0), where I is the intensity of the sound, and I0 = 1.0 × 10−12 W/m2 is the reference intensity. Therefore, if β = 150 dB, the intensity is I = I0 × 10β/10 = (1.0 × 10−12 W/m2) × 1015 = 1.0 × 103 W/m2 (b) The threshold of pain is I = 1 W/m2 and the answer to part (a) is 1 000 times greater than this, explaining why some airport employees must wear hearing protection equipment.

Topic 14

14.13

832

If the intensity of this sound varied inversely with the square of the distance from the source (I = constant/r2), the ratio of the intensities at distances r1 = 161 km and r2 = 4 800 km from the source is given by 2

⎛ constant ⎞ ⎛ ⎞ ⎛r ⎞ ⎛ 161 km ⎞ r12 1 =⎜ = = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ I1 ⎝ r22 ⎝ 4 800 km ⎠ ⎠ ⎝ constant ⎠ ⎝ r2 ⎠

The difference in the decibel levels at distances r1 and r2 from this source was then 2 ⎛ ⎞ ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ 161 km ⎞ I2 I0 2 1 2 β 2 − β1 = 10 ⋅ log ⎜ ⎟ − 10 ⋅ log ⎜ ⎟ = 10 ⋅ log ⎜ ⋅ ⎟ = 10 ⋅ log ⎜ ⎟ = 10 ⋅ log ⎜ ⎟ ⎜ I I1 ⎟ I1 ⎠ 4 800 km ⎠ ⎝ ⎝ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ 0 ⎠

β2 − β1 = −29.5 dB. This gives the decibel level on Rodriguez Island

β2 = β1 − 29.5 dB = 180 dB − 29.5 dB ≈ 151 dB 14.14

The decibel level due to the first siren is ⎛ 100.0 W/m 2 ⎞ β1 = 10 ⋅ log ⎜ = 140 dB −12 2 ⎝ 1.0 × 10 W/m ⎟⎠ !

Thus, the decibel level of the sound from the ambulance is

β2 = β1 + 10 dB = 140 dB + 10dB = 150 dB 14.15

In terms of their intensities, the difference in the decibel level of two sounds is

Topic 14

833

⎛ I I0 ⎞ ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ β 2 − β1 = 10 ⋅ log ⎜ 2 ⎟ − 10 ⋅ log ⎜ 1 ⎟ = 10 ⋅ log ⎜ 2 ⋅ ⎟ = 10 ⋅ log ⎜ 2 ⎟ ⎝ I1 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ I 0 I1 ⎠ ! Thus,

I2 = 10( β2 −β1 ) 10 !I1

I 2 = I1 × 10( β2 −β1 ) 10

If β2 − β1 = 30.0 dB and I1 = 3.0 × 10−11 W/m2, then I2 = (3.0 × 10−11 W/m2) × 103.00 = 3.0 × 10−8 W/m2 14.16

The sound power incident on the eardrum is P = IA, where I is the intensity of the sound and A = 5.0 × 10−5 m2 is the area of the eardrum. (a) At the threshold of hearing, I = 1.0 × 10−12 W/m2, and P = (1.0 × 10−12 W/m2)(5.0 × 10−5 m2) = 5.0 × 10−17 W (b) At the threshold of pain, I = 1.0 W/m2, and P = (1.0 W/m2)(5.0 × 10−5 m2) = 5.0 × 10−5 W

14.17

The decibel level β = 10log(I/I0), where I0 = 1.00 × 10−12 W/m2. (a) If β = 100 dB, then log(I/I0) = 10, giving I = 1010 I0 = 1.00 × 10−2 W/m2 (b) If all three toadfish sound at the same time, the total intensity of the sound produced is I′ = 3I = 3.00 × 10−2 W/m2, and the decibel level is

Topic 14

834

⎛ 3.00 × 10−2 W/m 2 ⎞ β ′ = 10log ⎜ −12 2 ⎝ 1.00 × 10 W/m ⎟⎠ !

14.18

= 10log ⎡⎣( 3.00 ) (1010 ) ⎤⎦ = 10 [ log ( 3.00 ) + 10 ] = 105 dB

(a) From the defining equation of the decibel level, β = 10 ⋅ log(I/I0), we solve for the intensity as I = I0⋅10β/10 and find that I = (1.0 × 10−12 W/m2)⋅10115/10 = 1.0 × 10−12+11.5 W/m2 = 10−0.5 W/m2 = 0.316 W/m2 (b) If 5 trumpets are sounded together, the total intensity of the sound is I5 = 5I1 = 5(0.316 W/m2) = 1.58 W/m2 (c) If the sound propagates uniformly in all directions, the intensity varies inversely as the square of the distance from the source, I = constant/r2, and we find that

⎛r ⎞ =⎜ 1⎟ I1 ⎝ r2 ⎠

⎛r ⎞ ⎛ 1.0 m ⎞ −2 2 or I 2 = I1 ⎜ 1 ⎟ = 1.58 W/m 2 ⎜ ⎟ = 2.47 × 10 W/m ⎝ 8.0 m ⎠ ⎝ r2 ⎠

(

)

⎛ 2.47 × 10−2 W / m 2 ⎞ ⎛I ⎞ β = 10 ⋅ log = 10 ⋅ log (d) ⎜⎝ I ⎟⎠ ⎜⎝ 1.0 × 10−12 W / m 2 ⎟⎠ = 104 dB 0 !

(e) The intensity of sound at the threshold of hearing is I0 = 1.0 × 10−12 W/m2, and from the discussion and result of part (c), we have I0/I2 = (r2/r0)2, and with the intensity being I2 = 2.47 × 10−2 W/m2 at

Topic 14

835

distance r2 = 8.0 m, the distance at which the intensity would be I0 = 1.0 × 10−12 W/m2 is

I 2.47 × 10−2 W/m 2 r0 = r2 2 = ( 8.0 m ) = 1.26 × 106 m I0 1.0 × 10−12 W/m 2 ! (f) The sound intensity level falls as the sound wave travels farther from the source until it is much lower than the ambient noise level and is drowned out. 14.19

The intensity of a spherical sound wave at distance r from a point source is I = Pav/4πr2, where Pav is the average power radiated by the source. Thus, at distances r1 = 5.0 m and r2 = 10 km = 104 m, the intensities of the sound wave radiating out from the elephant are I = Pav /4π r12 and I 2 = Pav /4π r22 giving I2 = (r1/r2)2I1. From the defining !1

equation, β = 10 log(I/I0), the intensity level of the sound at distance r2 from the elephant is seen to be 2 ⎡⎛ r ⎞ 2 I ⎤ ⎛ I2 ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ r1 ⎞ ⎛r ⎞ 1 1 β 2 = 10log ⎜ ⎟ = 10log ⎢⎜ ⎟ ⎥ = 10log ⎜ ⎟ + 10log ⎜ 1 ⎟ = 20log ⎜ 1 ⎟ + 10log ⎜ 1 ⎟ ⎝ r2 ⎠ ⎝ r2 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎢⎣⎝ r2 ⎠ I 0 ⎥⎦ !

14.20

⎛ 5.0 m ⎞ β 2 = 20log ⎜ 4 ⎟ + β1 = −66 dB + 103 dB = 37 dB ⎝ 10 m ⎠ !

(a) The intensity of the musical sound (β = 80 dB) is Imusic = I010β/10 = I0(108.0), and that produced by the crying baby (β = 75 dB) is Ibaby =

Topic 14

836

I0(107.5). Thus, the total intensity of the sound engulfing you is

I = I music + I baby = I 0 (108.0 + 107.5 ) !

= (1.0 × 10−12 W/m 2 ) (1.32 × 108 ) = 1.32 × 10−4 W/m 2

(b) The combined sound level is

β = 10 log(I/I0) = 10 log(1.32 × 108) = 81.2 dB 14.21 (a) The intensity of sound at 10 km from the horn (where β = 50 dB) is I = I010β/10 = (1.0 × 10−12 W/m2)105.0 = 1.0 × 10−7 W/m2 Thus, from I = P/4πr2, the power emitted by the source is P = 4πr2I = 4π(10 × 103 m)2(1.0 × 10−7 W/m2) = 4π × 101 W = 1.3 × 102 W (b) At r = 50 m, the intensity of the sound will be P 1.3 × 102 W −3 I= = W/m 2 2 = 4.1 × 10 2 4π r 4π ( 50 m ) !

and the sound level is ⎛ 4.1 × 10−7 W/m 2 ⎞ ⎛ I⎞ β = 10log ⎜ ⎟ = 10log ⎜ = 10log ( 4.1 × 109 ) = 96 dB −12 2⎟ ⎝ I0 ⎠ ⎝ 1.0 × 10 W/m ⎠ !

14.22

P 100 W −2 W/m 2 2 = 7.96 × 10 (a) I = 4π r 2 = 4π (10.0 m ) !

Topic 14

837

⎛ I⎞ ⎛ 7.96 × 10−2 W/m 2 ⎞ (b) β = 10log ⎜ ⎟ = 10log ⎜ −12 2⎟ ⎝ 1.00 × 10 W/m ⎠ ⎝ I0 ⎠

(

)

= 10log 7.96 × 1010 = 109 dB (c) At the threshold of pain (β = 120 dB), the intensity is I = 1.00 W/m2. Thus, from I = P/4πr2, the distance from the speaker is

r= ! 14.23

P = 4π I

100 W = 2.82 m 4π (1.00 W/m 2 )

The sound level for intensity I is β = 10log(I/I0). Therefore,

⎛ I I0 ⎞ ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ β 2 − β1 = 10log ⎜ 2 ⎟ − 10log ⎜ 1 ⎟ = 10log ⎜ 2 ⋅ ⎟ = 10log ⎜ 2 ⎟ ⎝ I1 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎝ I 0 I1 ⎠ ! Since I = P/4πr2 = (P/4π)/r2, the ratio of intensities is I 2 ⎛ P/4π ⎞ ⎛ r12 ⎞ r12 =⎜ 2 ⎟⎜ ⎟⎠ = r 2 I r ⎝ P/4 π ⎝ ⎠ 1 2 2 ! 2

⎛ r12 ⎞ ⎛ r1 ⎞ ⎛ r1 ⎞ β − β = 10log = 10log = 20log Thus, 2 1 2 ⎜⎝ r ⎟⎠ ⎜⎝ r ⎟⎠ ⎜⎝ r ⎟⎠ 2 2 2 !

14.24

P ( P/4π ) = The intensity at distance r from the source is I = 2 r2 ! 4π r

I r 2 (100 m ) + (100 m ) = 2 (a) A = B2 = 2 I B rA 100 m ) ( ! 2

Topic 14

838

I r 2 (100 m ) + ( 200 m ) = 5 (b) A = C2 = 2 I r (100 m) C A ! 2

14.25 Here the sound source is stationary and the observer is moving away at 6.50 m/s so that vO = −6.50 m/s and vS = 0. The boy will hear a frequency lower than f0. Take the speed of sound to be v = 343 m/s and substitute values into the Doppler shift equation:

⎛ v + vO ⎞ ⎛ 343 m/s − 6.50 m/s ⎞ fO = fS ⎜ = 1250 Hz ⎜ = 1.23 × 103 Hz ⎟ ⎟ ⎝ ⎠ 343 m/s ⎝ v − vS ⎠ 14.26 The sound source at frequency fS is stationary (vS = 0) and there are two moving observers. The observer at the back of the train is moving toward the source and has velocity +vO while the observer at the front has velocity −vO. (a) Write the Doppler shift equation for each observer and divide the two equations, canceling the unknown source frequency:

⎛ v + vO ⎞ ⎛ v − vO ⎞ f back = fS ⎜ and ffront = fS ⎜ → ⎟ ⎝ v ⎠ ⎝ v ⎟⎠

f back ⎛ v + vO ⎞ = ffront ⎜⎝ v − vO ⎟⎠

Topic 14

839

Solve for the observer’s speed vO to find

f back ⎛ v + vO ⎞ = → ffront ⎜⎝ v − vO ⎟⎠

f back ( v − vO ) = v + vO → ffront ( v + vO ) = fback ( v − vO ) ffront

⎛f −f ⎞ vO ( ffront + f back ) = v ( f back − ffront ) → vO = v ⎜ back front ⎟ ⎝ f back + ffront ⎠ Substitute fback = 428 Hz and ffront = 402 Hz to find the train’s speed vO:

⎛f −f ⎞ ⎛ 428 Hz − 402 Hz ⎞ vO = v ⎜ back front ⎟ = 343 m/s ⎜ = 10.7 m/s ⎝ 428 Hz + 402 Hz ⎟⎠ ⎝ f back + ffront ⎠

(b) Substitute values into the Doppler shift equation for either observer:

⎛ v ⎞ 343 m/s ⎛ v + vO ⎞ ⎛ ⎞ f back = fS ⎜ → f = f = 428 Hz ⎜ = 415 Hz S back ⎜ ⎟ ⎟ ⎝ 343 m/s + 10.7 m/s ⎟⎠ ⎝ v ⎠ ⎝ v + vO ⎠

14.27

When a stationary observer (vO = 0) hears a moving source, the observed frequency is

⎛ v+v ⎞ ⎛ v ⎞ O fO = fS ⎜ ⎟ = fS ⎜ ⎟ ⎝ v − vS ⎠ ⎝ v − vS ⎠ (a) When the train is approaching, vS = +40.0 m/s, and

⎛ ⎞ 343 m/s = 320 Hz ) ⎜ ⎟ = 362 Hz O approach ⎝ 343 m/s − 40.0 m/s ⎠

(f )

(

)

Topic 14

840

After the train passes and is receding, vS = −40.0 m/s, and

⎡ ⎤ 343 m/s = 320 Hz ⎢ ⎥ = 287 Hz O recede ⎢⎣ 343 m/s − (−40.0 m/s) ⎥⎦

(f )

(

)

Thus, the frequency shift that occurs as the train passes is ΔfO = (fO)recede − (fO)approach = −75 Hz, or it is a 75 Hz drop. (b) As the train approaches, the observed wavelength is

λ=

(f )

O approach

14.28

343 m/s 362 Hz

= 0.948 m

The general expression for the observed frequency of a sound when the source and/or the observer are in motion is

⎛ v+v ⎞ O fO = fS ⎜ ⎟ ⎝ v − vS ⎠ Here, v is the velocity of sound in air, vO is the velocity of the observer, vS is the velocity of the source, and fS is the frequency that would be detected if both the source and observer were stationary. (a) If fS = 5.00 kHz and the observer is stationary (vO = 0), the frequency detected when the source moves toward the observer at half the speed of sound (vS = +v/2) is

⎛ v+0 ⎞ fO = 5.00 kHz ⎜ ⎟ = 5.00 kHz 2 = 10.0 kHz ⎝ v − v/2 ⎠

(

)

(

)( )

Topic 14

841

(b) When fS = 5.00 kHz and the source moves away from a stationary observer at half the speed of sound (vS = −v/2), the observed frequency is

⎛ v+0 ⎞ ⎛ 2⎞ fO = 5.00 kHz ⎜ ⎟ = 5.00 kHz ⎜ ⎟ = 3.33 kHz ⎝ v + v/2 ⎠ ⎝ 3⎠

(

14.29

)

(

)

⎛ v+v ⎞ O Both source and observer are in motion, so fO = fS ⎜ ⎟ . Since each v − v ⎝ S ⎠ train moves toward the other, vO > 0 and vS > 0. The speed of the source (train 2) is

vS = 90.0

km ⎛ 1 000 m ⎞ ⎛ 1 h ⎞ ⎜ ⎟⎜ ⎟ = 25.0 m/s h ⎝ 1 km ⎠ ⎝ 3 600 s ⎠

and that of the observer (train 1) is vO = 130 km/h = 36.1 m/s. Thus, the observed frequency is

⎛ 343 m/s + 36.1 m/s ⎞ fO = 500 Hz ⎜ ⎟ = 596 Hz ⎝ 343 m/s − 25.0 m/s ⎠

(

14.30

)

(a) Since the observer hears a reduced frequency, the source and observer are getting farther apart. Hence, the cyclist is behind the car. (b) With the cyclist (observer) behind the car (source) and both moving in the same direction, the observer moves toward the source (vO > 0) while the source moves away from the

Topic 14

842

⎡ v + v /3 ⎤ ⎛ v + v /3 ⎞ ⎛ v+v ⎞ car car ⎢ ⎥= f ⎜ O ⎟, fO = fS ⎜ = f ⎟ S⎢ S ⎥ v − v ⎜ ⎟ v + v v − − v ⎝ S ⎠ ⎝ car ⎠ car ⎥ ⎣⎢ ⎦

(

)

giving

⎛ 343 m/s + vcar /3 ⎞ 415 Hz = ( 440 Hz ) ⎜ ⎟ ⎝ 343 m/s + vcar ⎠ !

and

vcar = 32.0 m/s

and |vcar| = 32.0 m/s 14.31

With the train approaching the stationary observer (vO = 0) at speed |vt|, the source velocity is vS = +|vt| and the observed frequency is ⎛ ⎞ 343 m/s ⎟ 465 Hz = fS ⎜ ⎜⎝ 343 m/s − vt ⎟⎠

[

As the train recedes, the source velocity is vS = −|vt| and the observed frequency is ⎛ ⎞ 343 m/s ⎟ 441 Hz = fS ⎜ ⎜⎝ 343 m/s + vt ⎟⎠

Dividing Equation [1] by [2] gives 465 343 m/s + vt = 441 343 m/s − vt ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

[

Topic 14

843

and solving for the speed of the train yields |vt| = 9.09 m/s 14.32

(a) We let the speed of the insect be |vbug| and the speed of the bat be |vbat| = 5.00 m/s, and break the action into two steps. In the first step, the bat is the sound source flying toward the observer (the insect), so vS = +|vbat|, while the insect (observer) is flying away from the source, making vO = −|vbug|. If f0 is the actual frequency sound emitted by the bat, the frequency detected (and reflected) by the moving insect is

(

⎡v + − v ⎛ v+v ⎞ bug ⎢ O f reflect = fO ⎜ ⎟ = fO ⎢ ⎝ v − vS ⎠ ⎢⎣ v − + v bat

(

) ⎤⎥ or f

) ⎥⎥⎦

⎛ v− v ⎞ bug ⎟ = fO ⎜ reflect ⎜ v− v ⎟ bat ⎠ ⎝

In the second step of the action, the insect acts as a sound source, reflecting a wave of frequency freflect back to the bat which acts as a moving observer. Since the source (insect) is moving away from the observer, vS = −|vbug|, and the observer (bat) is moving toward the source (insect) giving vO = +|vbat|. The frequency of the return sound received by the bat is then

(

⎡v+ + v ⎛ v+v ⎞ bat ⎢ O f return = f reflect ⎜ ⎟ = f reflect ⎢ ⎝ v − vS ⎠ ⎢⎣ v − − v bug

(

) ⎤⎥ or f

) ⎥⎥⎦

⎛ v+ v ⎞ bat ⎟ = f reflect ⎜ return ⎜ v+ v ⎟ bug ⎠ ⎝

Combing the results of the two steps gives

Topic 14

844

⎛ v− v ⎞⎛ v+ v ⎞ bug bat ⎟ ⎟⎜ f return = fO ⎜ ⎜ v− v ⎟⎜ v+ v ⎟ bat ⎠ ⎝ bug ⎠ ⎝ or

⎛ 343 m/s − v bug ⎞ ⎛ 343 + 5.00 ⎞ 40.4 kHz = ( 40.0 kHz )) ⎜ ⎟ ⎟⎜ ⎜⎝ 343 − 5.00 ⎟⎠ ⎜⎝ 343 m/s + v bug ⎟⎠ ! This reduces to

(

⎛ 40.0 ⎞ ⎛ 348 ⎞ 343 m/s + v bug = ⎜ 343 m/s − v bug ⎝ 40.4 ⎟⎠ ⎜⎝ 338 ⎟⎠ !

)

⎡⎛ 40.0 ⎞ ⎛ 348 ⎞ ⎤ ⎡⎛ 40.0 ⎞ ⎛ 348 ⎞ ⎤ ⎢⎜⎝ 40.4 ⎟⎠ ⎜⎝ 338 ⎟⎠ + 1⎥ v bug = ( 343 m/s ) ⎢⎜⎝ 40.4 ⎟⎠ ⎜⎝ 338 ⎟⎠ − 1⎥ ⎦ ⎣ ⎦ !⎣ and yields |vbug| = 3.29 m/s. (b) Yes, the bat is gaining on the insect at a rate of 5.00 m/s − 3.29 m/s = 1.71 m/s. 14.33

For a source receding from a stationary observer,

⎛ ⎞ ⎛ ⎞ v v ⎜ ⎟ ⎟ fO = fS = fS ⎜ ⎜ v− − v ⎟ ⎜⎝ v + vS ⎟⎠ S ⎠ ⎝

( )

Thus, the speed the falling tuning fork must reach is

⎛ f ⎞ ⎛ 512 Hz ⎞ vS = v ⎜ S − 1⎟ = 343 m/s ⎜ − 1⎟ = 19.1 m/s f 485 Hz ⎝ ⎠ ⎝ O ⎠

(

)

Topic 14

845

The distance it has fallen from rest before reaching this speed is

(19.1 m/s) − 0 = 18.6 m Δy = = 2a 2 ( 9.80 m/s ) 2

vS2 − 0

The time required for the 485 Hz sound to reach the observer is

Δy1 v

18.6 m 343 m/s

= 0.054 2 s

During this time the fork falls an additional distance

(

) (

)(

)

2 1 1 Δy2 = vSt + ayt 2 = 19.1 m/s 0.054 2 s + 9.80 m/s 2 0.054 2 s = 1.05 m 2 2

The total distance fallen before the 485 Hz sound reaches the observer is Δy = Δy1 + Δy2 = 18.6 m + 1.05 m = 19.7 m

14.34

⎛ 115/ min ⎞ (a) ω = 2π f = 2π ⎜ = 12.0 rad/s , and for harmonic motion, ⎝ 60.0 s/min ⎟⎠ !

(

)(

)

vmax = ω A = 12.0 rad/s 1.80 × 10−3 m = 2.16 × 10−2 m/s (b) The heart wall is a moving observer (vO = +|vmax|) and the detector a stationary source, so the maximum frequency reflected by the heart wall is

( ) f wall

⎛ v+ v ⎞ ⎛ 1 500 + 0.021 6 ⎞ max ⎟ = (2000 000 Hz) ⎜ = fS ⎜ ⎟ = 2 000 029 Hz max v 1 500 ⎜⎝ ⎟⎠ ⎝ ⎠

Topic 14

846

(c) Now, the heart wall is a moving source (vS = +|vmax|) and the detector a stationary observer. The observed frequency of the returning echo is ⎛ ⎞ ⎛ ⎞ v 1 500 ⎜ ⎟ = 2 000 029 Hz ⎜ f echo = f wall ⎟ = 2 000 058 Hz max ⎜ 1 500 − 0.021 6 ⎟ ⎝ ⎠ v − v ⎝ max ⎠

( )

14.35

(a)

(

)

For a plane traveling at Mach 3.00, the half-angle of the conical wave front is

⎛v ⎞ ⎛ 1 ⎞ θ = sin −1 ⎜ sound ⎟ = sin −1 ⎜ ⎝ 3.00 ⎟⎠ vplane ⎠ ⎝ !

FIGURE P14.35 The distance the plane has moved when the wave front reaches the observer is x = h/tan θ, or

20.0 km x= = 56.6 km tan ⎡⎣sin −1 (1/3.00 ) ⎤⎦ ! The time required for the plane to travel this distance, and hence the time when the shock wave reaches the observer, is

Topic 14

847

x x 56.6 × 101 m t= = = = 56.3 s vplane 3.00vsound 3.00 ( 335 m/s ) !

(b) The plane is 56.6 km farther along as computed above. 14.36

(a) From Equation 14.12 in the textbook,

⎛ v+v ⎞ O fO = fS ⎜ ⎟ ⎝ v − vS ⎠ where fS is the frequency emitted by the source, fO is the frequency detected by the observer, v is the speed of the wave in the propagating medium, vO is the velocity of the observer relative to the medium, and vS is the velocity of the source relative to the propagating medium. (b) The yellow submarine is the source or emitter of the sound waves. (c) The red submarine is the observer or receiver of the sound waves. (d) The motion of the observer away from the source tends to increase the time observed between arrivals of successive pressure maxima. This effect tends to cause an increase in the observed period and a decrease in the observed frequency. (e) In this case, the sign of vO should be negative to decrease the numerator in Equation 14.12, and thereby decrease the calculated © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

848

observed frequency. (f) The motion of the source toward the observer tends to decrease the time between the arrival of successive pressure maxima, decreasing the observed period, and increasing the observed frequency. (g) In this case, the sign of vS should be positive to decrease the denominator in Equation 14.12, and thereby increase the calculated observed frequency. (h)

( (

) )

⎡ 1 533 m/s + −3.00 m/s ⎤ ⎛ v+v ⎞ 3 O ⎥ = 5.30 × 103 Hz fO = fS ⎜ ⎟ = 5.27 × 10 Hz ⎢ ⎢ 1 533 m/s − +11.0 m/s ⎥ ⎝ v − vS ⎠ ⎣ ⎦

(

)

14.37 Taking the speed of sound as v = 343 m/s, the wavelength at a frequency of 625 Hz is λ = v/f = (343 m/s)/(625 Hz) = 0.549 m. The bicyclist is at a point of constructive interference, so the difference in path lengths between the two horns must be an integer number of wavelengths (and, because she is an equal distance from each horn, the horns must sound in phase with each other). To reach a point of destructive interference, she must move one quarter of a wavelength backwards. Doing this, the wave arriving

Topic 14

849

from behind will travel one quarter of a wavelength less far and the wave arriving from in front will travel one quarter of a wavelength farther than to her previous location. The difference in their path lengths will then be 2(λ/4) = λ/2 as required for destructive interference. To quantify this, let her move a distance d backwards and set the path difference to λ/4:

λ λ 0.549 m m + d ) − ( 4.50 m − d ) = → d= = = 0.137 m (!4.50 # #"## $ !# #"## $ 2 4 4

distance to front horn

14.38

distance to back horn

The wavelength of the sound emitted by the speaker is

v 343 m/s λ= = = 0.454 m , and a half wavelength is λ/2 = 0.227 m. f 756 Hz ! (a) If a condition of constructive interference currently exists, this can be changed to a case of destructive interference by adding a distance of λ/2 = 0.227 m to the path length through the upper arm. (b) To move from a case of constructive interference to the next occurrence of constructive interference, one should increase the path length through the upper arm by a full wavelength, or by λ = 0.454 m. 14.39

At point D, the distance of the ship from point A is

Topic 14

850

(

ℓ = d 2 + 800 m

) = (600 m) + (800 m) = 1 000 m 2

Since destructive interference occurs for the first time when the ship reaches D, it is necessary that ! − d = λ /2 , or

l = 2(ℓ − d) = 2(1 000 m − 600 m) = 800 m 14.40

The speakers emit sound of wavelength

v 343 m/s λ= = = 0.762 m f 450 Hz ! so λ/2 = 0.381 m Initially, Δy = 0, and

r = r = (1.50 m)2 + (8.00 m)2 = 8.14 m !1 2 To create destructive interference at point O, we move the top speaker upward distance Δy from its original location until we have r1 − r2 = © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

851

λ/2. Since this did not change r2, we must now have r1 = r2 + λ/2 = 8.14 m + 0.381 m = 8.52 m But, after moving the speaker, this gives

r = (1.50 m + Δy)2 + (8.00 m)2 = 8.52 m !1 2 2 2 2 !(1.50 m + Δy) = (8.52 m) − (8.00 m) = 8.59 m

or Thus, 14.41

Δy = 8.59 m 2 − 1.50 m = 1.43 m . !

The wavelength of the sound is

v 343 m/s λ= = = 0.500 m f 686 Hz !

FIGURE P14.37 (modified) (a) At the first relative maximum (constructive interference), r = x + λ = x + 0.500 m Using the Pythagorean theorem, r2 = x2 + d2, or (x + 0.500 m)2 = x2 + (0.700 m)2 giving x = 0.240 m. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

852

(b) At the first relative minimum (destructive interference), r = x + λ/2 = x + 0.250 m Therefore, the Pythagorean theorem yields (x + 0.250 m)2 = x2 + (0.700 m)2 or 14.42

x = 0.855 m

In the fundamental mode of vibration, the wavelength of waves in the wire is

λ = 2L = 2(0.700 0 m) = 1.400 m If the wire is to vibrate at f = 261.6 Hz, the speed of the waves must be v = λf = (1.400 m)(261.6 Hz) = 366.2 m/s The mass per unit length of the wire is

µ=

m L

4.300 × 10−3 kg 0.700 0 m

= 6.143 × 10−4 kg/m

and the required tension is given by !v = F/µ as F = v2µ = (366.2 m/s)7(6.143 × 10−3 kg/m = 823.8 N 14.43

In the third harmonic, the string of length L forms a standing wave of three loops, each of length λ/2 = L/3.

Topic 14

853

The wavelength of the wave is then 2L 16.0 m λ= = ≈ 5.33 m 3 3 !

(a) The nodes in this string, fixed at each end, will occur at distances of 0, λ/2 = 2.67 m, λ = 5.33 m, and 3λ/2 = 8.00 m from the end. Antinodes occur halfway between each pair of adjacent nodes, or at distances of λ/4 = 1.33 m, 3λ/4 = 4.00 m, and 5λ/4 = 6.67 m from the end. (b) The linear density is

m 40.0 × 10−3 kg µ= = = 5.00 × 10−3 kg/m L 8.00 m ! and the wave speed is

v= !

F 49.0 N = = 99.0 m/s µ 5.00 × 10−3 kg/m

Thus, the frequency is

! 14.44

v 99.0 m/s = = 18.6 Hz λ 5.33 m

In the fundamental mode, the distance from the finger of the cellist to the far end of the string is one-half of the wavelength for the transverse waves in the string. Thus, when the string resonates at 449 Hz,

λ = 2(68.0 cm − 20.0 cm) = 96.0 cm

Topic 14

854

The speed of transverse waves in the string is therefore v = λf = (0.960 m)(449 Hz) = 431 m/s For a resonance frequency of 440 Hz, the wavelength would be

v 431 m/s λ′ = = = 0.980 m = 98.0 cm f 440 Hz ′ ! To produce this tone, the cellist should position her finger at a distance of

λ 98.0 cm x = L − = 68.0 cm − = 19.0 cm 2 2 ! from the nut. Thus, she should move her finger 1.00 cm toward the nut. 14.45

When the string vibrates in the fifth harmonic (i.e., in five equal segments) at a frequency of f5 = 630 Hz, we have L = 5(λ5/2) or the wavelength is λ5 = 2L/5. The speed of transverse waves in the string is then v = λ5f5 = (2L/5)f5 For the string to vibrate in three segments (i.e., third harmonic), the wavelength must be such that L = 3(λ3/2) or λ3 = 2L/3. The new frequency would then be

f4 = !

v ( 2L/5 ) f5 3 3 = = f5 = ( 630 Hz ) = 378 Hz λ3 2L/3 5 5

Topic 14

855

14.46 (a) The string has a length of 20.0 cm and there are 5.00 cm between nodes. See the figure below to see that there are 5 nodes and 4 antinodes in the standing wave.

The harmonic number n equals the number of antinodes so that n= 4 .

(b) Given the string’s mass and length, its linear mass density is λ = M/L = (1.75 × 10–2 kg)/(0.200 m) = 8.75 × 10−2 kg/m. The frequency of the nth mode (where n = 4) is:

14.47

fn = nf1 =

n F 2L µ

f 4 = 4 f1 =

4 F 4 875 N = = 1.00 × 103 Hz −2 2L µ 2 ( 0.200 m ) 8.75 × 10 kg/m

(a) The linear density is

m 25.0 × 10−3 kg µ= = = 1.85 × 10−2 kg/m L 1.35 m !

(b) In a string fixed at both ends, the fundamental mode has a node at

Topic 14

856

each end and a single antinode in the center, so that L = λ/2, or λ = 2L = 2(1.10 m) = 2.20 m. Then, the desired wave speed in the wire is v = λf = (2.20 m)(41.2 Hz) = 90.6 m/s. (c) The speed of transverse waves in a string is !v = F/µ , so the required tension is

F = µv 2 = (1.85 × 10−2 kg/m)(90.6 m/s)2 = 152 N ! (d) λ = 2L = 2(1.10 m) = 2.20 m

[See part (b) above.]

(e) The wavelength of the longitudinal sound waves produced in air by the vibrating string is

v 343 m/s λair = air = = 8.33 m f 41.2 Hz ! 14.48

(a) A string fixed at each end forms standing wave patterns with a node at each end and an integer number of loops, each loop of length λ/2, with an antinode at its center. Thus, L = n(λ/2) or λ = 2L/n. If the string has tension T and mass per unit length µ, the speed of transverse waves is !v = λ f = T /µ . Thus, when the string forms a standing wave of n loops (and hence n antinodes), the frequency of vibration is

Topic 14

857

T /µ 2L/n

2L µ

⇒

fA =

2LA

µA

(b) Assume the length is doubled, LB = 2LA, and a new standing wave is formed having nB = nA and TB = TA. Then

fB =

2LB

µA

( )

2 2LA

⎛ n = ⎜ A µ A 2 ⎜⎝ 2LA

TA ⎞ f A ⎟= µ A ⎟⎠ 2

(c) Solving the general result obtained in part (a) for the tension in the string gives T = 4µf2L2/n2. Thus, if fB = fA, LB = LA, and nB = nA + 1, we find

TB =

4 µ A f B2 L2B nB2

⎛ 4 µ f 2 L2 ⎞ ⎛ n ⎞ nA2 A A A A = = = T = ⎟ ⎜ ⎟ TA A 2 2 ⎜ 2 n + 1 n n + 1 ⎝ ⎠ ⎝ ⎠ A A n +1 n +1 A 4 µ A f A2 L2A

(

nA2

) (

(

)

(d) If now we have fB = 3fA, LB = LA/2, and nB = 2nA, then

TB =

14.49

4 µ A f B2 L2B 2 B

( )(

4 µ A 9 f A2 L2A 4 4n

2 A

) = 9 ⎛⎜ 4µ f L ⎞⎟ = 9 T or T = 9 2 2

16 ⎝

A A 2 A

⎠

(a) From the sketch below, notice that when d = 2.00 m,

Topic 14

858

L= !

⎛ d/2 ⎞ 5.00 m − d = 1.50 m , and θ = sin −1 ⎜ ⎟ = 41.5° 2 ⎝ L ⎠

Then evaluating the net vertical force on the lowest bit of string, ∑Fy = 2F cos θ − mg = 0 gives the tension in the string as

(12.0 kg )( 9.80 m/s ) = 78.9 N mg F= = 2cos θ 2cos ( 41.8° ) ! 2

(b) The speed of transverse waves in the string is

78.9 N 0.001 00 kg/m

= 2.81 × 102 m/s

For the pattern shown, 2d 4.00 m 3(λ / 2) = d,!so!λ = = 3 3 !

Thus, the frequency is v 3 ( 2.81 × 10 m/s ) f= = = 2.11 × 102 Hz λ 4.00 m ! 2

14.50

(a) For a standing wave of 6 loops, 6(λ/2) = L, or λ = L/3 = (2.0 m)/3. The speed of the waves in the string is then

⎛ 2.0 m ⎞ 2 v=λf =⎜ ⎟⎠ (150 Hz ) = 1.0 × 10 m/s ⎝ 3 ! Since the tension in the string is F = mg = (5.0 kg)(9.80 m/s2) = 49 N, !v = F/µ gives © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

859

µ= !

F 49 N −4 = kg/m 2 = 4.9 × 10 2 2 v 1.0 × 10 m ( )

(b) If m = 45 kg, then F = (45 kg)(9.80 m/s2) = 4.4 × 102 N, and

4.4 × 102 N = 3.0 × 102 m/s −3 4.9 × 10 kg/m

v= !

Thus, the wavelength will be

v 3.0 × 102 m/s λ= = = 2.0 m f 150 Hz ! and the number of loops is

L 2.0 m n= = = 2 λ /2 1.0 m !

v= !

v 1.4 × 102 m/s = 0.93 m Then, λ = = f 150 Hz !

L 2.0 m = and n = is not an integer, λ /2 0.47 m ! so no standing wave will form. 14.51

The speed of transverse waves in the string is

50.000 N 1.000 0 × 10−2 kg/m

= 70.711 m/s

The fundamental wavelength is λ1 = 2L = 1.200 0 m and its frequency © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

860

f1 =

λ1

70.711 m/s 1.200 0 m

= 58.926 Hz

The harmonic frequencies are then fn = nf1 = n(58.926 Hz), with n being an integer The largest one under 20 000 Hz is f339 = 19 976 Hz = 19.976 kHz. 14.52

The distance between adjacent nodes (and between adjacent antinodes) is one-quarter of the circumference.

λ 20.0 cm dNN = dAA = = = 5.00 cm 2 4 ! so λ = 10.0 cm = 0.100 m, and

f= !

v 900 m/s = = 9.00 × 103 Hz = 9.00 kHz λ 0.100 m

The singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it. 14.53 The spring’s natural oscillation frequency is

ω = k/m =

(1.00 × 10 N/m)/( 30.0 kg) = 57.5 rad/s which 5

corresponds to a period of T = 2π/ω = 0.109 s. To drive the tire into resonance, it should hit a bump (spaced d = 0.750 m) once per period.

Topic 14

861

The car’s minimum speed is then v = d/T = (0.750 m)/(0.109 s) =

6.88 m/s (approximately 15.4 mph).

14.54 (a) The harmonic frequencies of a pipe open at both ends are given by

fn = nf1 = n

v 2L

n = 1, 2,3, . . .

where v =343 m/s is the sound speed and L = 0.750 m is the pipe’s length. Substitute values to find

f1 = (1)

v 343 m/s = = 229 Hz 2L 2 ( 0.750 m )

f 2 = 2 f1 = 457 Hz (b) The harmonic frequencies of a pipe closed at one end are given by

fn = nf1 = n

v 4L

n = 1,3,5, . . .

Substitute values to find

f1 = (1)

v 343 m/s = = 114 Hz 4L 4 ( 0.750 m )

f3 = 3 f1 = 343 Hz 14.55

Assuming an air temperature of T = 37°C = 310 K, the speed of sound inside the pipe is

Topic 14

862

(

v = 331 m/s

= 353 m/s ) 273 K = (331 m/s) 310 273

In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is λ = 4L. Thus, for the whooping crane

λ = 4(5.0 ft) = 2.0 × 101 ft

and !

14.56

v ( 353 m/s ) ⎛ 3.281 ft ⎞ = = 58 Hz λ 2.0 × 101 ft ⎜⎝ 1 m ⎟⎠

(a) In the fundamental resonant mode of a pipe open at both ends, the distance between antinodes is dAA = λ/2 = L. Thus, λ = 2L = 2(0.320 m) = 0.640 m

and

(b) dAA =

14.57

v 343 m/s = = 536 Hz λ 0.640 m

λ 1 ⎛ v ⎞ 1 ⎛ 343 m/s ⎞ −2 = ⎜ ⎟= ⎜ ⎟ = 4.29 × 10 m = 4.29 cm 2 2 ⎝ f ⎠ 2 ⎝ 4 000 Hz ⎠

Hearing would be best at the fundamental resonance, so λ = 4L = f=

4(2.8 cm) and !

14.58

v 343 m/s ⎛ 100 cm ⎞ = = 3.1 × 101 Hz = 3.1 kHz λ 4 ( 2.8 cm ) ⎜⎝ 1 m ⎟⎠

(a) To form a standing wave in the tunnel, open at both ends, one must have an antinode at each end, a node at the middle of the tunnel, and the length of the tunnel must be equal to an integral number of half-wavelengths [L = n(λ/2) or λ = 2L/n]. The

Topic 14

863

resonance frequencies of the tunnel are then

fn =

vsound in air

λn

⎛ ⎞ 343 m/s ⎟ = n 0.085 8 Hz n = 1, 2, 3,… = n⎜ 2L/n ⎜⎝ 2 2.00 × 103 m ⎟⎠

343 m/s

(

)

(

)

(b) It would be good to make such a rule. Any car horn would produce several closely spaced resonance frequencies of the air in the tunnel, so the sound would be greatly amplified. Other drivers might be frightened directly into dangerous behavior or might blow their horns also. 14.59

(a) The fundamental wavelength of the pipe open at both ends is λ1 = 2L = v/f1. Since the speed of sound is 331 m/s at 0°C, the length of the pipe is

v 331 m/s L= = = 0.552 m 2 f1 2 ( 300 Hz ) ! (b) At T = 30°C = 303 K,

TK 303 v = ( 331 m/s ) = ( 331 m/s ) = 349 m/s 273 273 ! and

f1 = ! 14.60

v v 349 m/s = = = 316 Hz λ1 2L 2 ( 0.552 m )

(a) Observe from Equations 14.18 and 14.19 in the textbook that the difference between successive resonance frequencies is constant,

Topic 14

864

regardless of whether the pipe is open at both ends or is closed at one end. Thus, the resonance frequencies of 650 Hz or less for this pipe must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, and 50.0 Hz, with the lowest or fundamental frequency being f1 = 50.0 Hz. (b) Note, from the list given above, the resonance frequencies are only the odd multiples of the fundamental frequency. This is a characteristic of a pipe that is open at only one end and closed at the other. (c) The length of a pipe with an antinode at the open end and a node at the closed end is one-quarter of the wavelength of the fundamental frequency, so the length of this pipe must be

14.61

λ1 4

vsound 4 f1

343 m/s 4(50.0 Hz)

= 1.72 m

The beat frequency is f b = f 2 − f1 = 196 Hz − 199 Hz = 3 Hz .

14.62 The beat frequency equals the absolute value of the difference between two frequencies. Therefore a frequency 2 Hz above and below 525 Hz will result in the required beat frequency of 2 Hz:

f = 523 Hz and 527 Hz because

2 Hz = 525 Hz − 523 Hz and 2 Hz = 525 Hz − 527 Hz . © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

14.63

865

In a string fixed at both ends, the length of the string is equal to a halfwavelength of the fundamental resonance frequency, so λ1 = 2L. The fundamental frequency may then be written as

f1 = !

v 1 F = = λ1 2L µ

F 4L2 µ

If a second identical string with tension F′ < F is struck, the fundamental frequency of vibration would be

f1′ = !

⎛ F ⎞ F′ F′ F′ = ⎜ 2 ⎟ = f1 2 ⎝ 4L µ ⎠ F 4L µ F

When the two strings are sounded together, the beat frequency heard will be

14.64

⎛ ⎛ F′ ⎞ 5.40 × 102 N ⎞ 2 f beat = f1 − f1′ = f1 ⎜ 1 − = (1.10 × 10 Hz ) ⎜ 1 − ⎟ = 5.64 beats/s F ⎟⎠ 6.00 × 102 N ⎠ ⎝ ⎝

By shortening her string, the second violinist increases its fundamental frequency. Thus, ! f1′ = f1 + f beat = (196 + 2.00) Hz = 198 Hz . Since the tension and the linear density are both identical for the two strings, the speed of transverse waves, !v = F/µ , has the same value for both strings. Therefore, !λ1′ f1′ = λ1 f1 , or !λ1′ = λ1 ( f1 /f1′) . The fundamental wavelength of a string fixed at both ends is λ = 2L, and this yields

Topic 14

866

⎛f ⎞ ⎛ 196 ⎞ L′ = L ⎜ 1 ⎟ = ( 30.0 cm ) ⎜ = 29.7 cm ⎝ 198 ⎟⎠ ⎝ f1′ ⎠ !

14.65

The commuter, stationary relative to the station and the first train, hears the actual source frequency (fO,1 = fS = 180 Hz) from the first train. The frequency the commuter hears from the second train, moving relative to the station and commuter, is given by fO,2 = fS ± fbeat = 180 Hz ± 2 Hz = 178 Hz or 182 Hz This stationary observer (vO = 0) hears the lower frequency (fO,2 = 178 Hz) if the second train is moving away from the station (vS = −|vS|), so fO = fS[(v + vO)/v − vS)] gives the speed of the receding second train as ⎛ ⎞ ⎛ 343 m/s + 0 ⎞ 343 m/s + 0 ⎟ = 180 Hz ⎜ 178 Hz = 180 Hz ⎜ ⎟ ⎜⎝ 343 m/s − −|vS | ⎟⎠ ⎝ 343 m/s +|vS |⎠

(

)

(

) (

)

or ⎡⎛ 180 Hz ⎞ ⎤ ⎛ 180 Hz ⎞ 343 m/s +|vS |= 343 m/s ⎜ and |v | = 343 m/s ⎢⎜ ⎟ ⎟ − 1⎥ = 3.85 m/s S ⎝ 178 Hz ⎠ ⎢⎣⎝ 178 Hz ⎠ ⎥⎦

(

)

(

)

so one possibility for the second train is vS = 3.85 m/s away from the station. The other possibility is that the second train is moving toward the station (vS = +|vs|) and the commuter is detecting the higher of the

Topic 14

867

possible frequencies (fO,2 = 182 Hz). In this case, fO = fS[(v + vO)/(v − vS)] yields ⎛ ⎞ 343 m/s + 0 ⎟ 182 Hz = 180 Hz ⎜ ⎜⎝ 343 m/s − vS ⎟⎠

(

)

⎛ 180 Hz ⎞ and 343 m/s − vS = 343 m/s ⎜ ⎟ ⎝ 182 Hz ⎠

(

)

⎡ ⎛ 180 Hz ⎞ ⎤ vs = ( 343 m/s ) ⎢1 − ⎜ ⎥ = 3.77 m/s ⎝ 182 Hz ⎟⎠ ⎦ ⎣ !

In this case, the velocity of the second train is vS = 3.77 m/s toward the station. 14.66

The temperatures of the air in the two pipes are T1 = 27°C = 300 K and T2 = 32°C = 305 K. The speed of sound in the two pipes is T1 T2 v1 = ( 331 m/s ) and!v2 = ( 331 m/s ) 273 K 273 K !

Since the pipes have the same length, the fundamental wavelength, λ = 4L, is the same for them. Thus, from f = v/λ, the ratio of their fundamental frequencies is seen to be f2/f1 = v2/v1, which gives f2 = f1(v2/v1). The beat frequency produced is then

⎛ T ⎞ ⎛v ⎞ f beat = f2 − f1 = f1 ⎜ 2 − 1⎟ = f1 ⎜ 2 − 1⎟ ⎝ v1 ⎠ ⎝ T1 ⎠ !

Topic 14

868

! 14.67

⎛ 305 K ⎞ f beat = ( 480 Hz ) ⎜ − 1⎟ = 3.98 Hz ⎝ 300 K ⎠

(a) First consider the wall a stationary observer receiving sound from an approaching source having velocity va. The frequency received and reflected by the wall is freflect = fS[v/(v − va)]. Now consider the wall as a stationary source emitting sound of frequency freflect to an observer approaching at velocity va. The frequency of the echo heard by the observer is

⎛ v ⎞⎛ v+v ⎞ ⎛ v+v ⎞ ⎛ v+v ⎞ a a a f echo = f reflect ⎜ ⎟⎜ ⎟ ⎟ = fS ⎜ ⎟ = fS ⎜ v v − v v v − v ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ a a⎠ Thus, the beat frequency between the tuning fork and its echo is

⎛ v+v ⎞ ⎛ 2v ⎞ ⎛ 2(1.33) ⎞ a a f beat = f echo − fS = fS ⎜ − 1⎟ = fS ⎜ ⎟ = 256 Hz ⎜ ⎟ = 1.99 Hz ⎝ 343 − 1.33 ⎠ ⎝ v − va ⎠ ⎝ v − va ⎠

(

)

(b) When the student moves away from the wall, va changes sign so the frequency of the echo heard is fecho = fS[(v − |va|)/(v + |va|)]. The beat frequency is then

⎛ ⎛ 2|v | ⎞ v−|va |⎞ a f beat = fS − f echo = fS ⎜ 1 − ⎟ = fS ⎜ ⎟ v+|v | v+|v | ⎝ ⎠ ⎝ a a ⎠

giving

va =

vf beat 2 fS − f beat

Topic 14

869

The receding speed needed to observe a beat frequency of 5.00 Hz is

va = !

14.68

( 343 m/s )( 5.00 Hz ) = 3.38 m/s 2 ( 256 Hz ) − 5.00 Hz

The extra sensitivity of the ear at 3 000 Hz appears as downward dimples on the curves in Figure 14.30 of the textbook. At T = 37°C = 310 K the speed of sound in air is

TK 310 v = ( 331 m/s ) = ( 331 m/s ) = 353 m/s 273 273 ! Thus, the wavelength of 3 000 Hz sound is

λ=

v f

353 m/s 3 000 Hz

= 0.118 m

For the fundamental resonant mode in a pipe closed at one end, the length required is

14.69

λ 0.118 m = = 0.029 5 m = 2.95 cm 4 4

At normal body temperature of T = 37°C, the speed of sound in air is

T 37.0 v = ( 331 m/s ) 1 + C = ( 331 m/s ) 1 + 273 273 ! and the wavelength of sound having a frequency of f = 20 000 Hz is

Topic 14

870

λ=

(331 m/s) 1 + 37.0 = 1.76 × 10 m = 1.76 cm f ( 20 000 Hz ) 273

−2

Thus, the diameter of the eardrum is 1.76 cm 14.70

From the defining equation for the decibel level, β = 10log(I/I0), the intensity of sound having a decibel level β is I = (10β/10)I0 . Thus, the intensity of a 40 dB sound is I40 = (104.0)I0, while that of a 70 dB sound is I70 = (107.0)I0. Since the combined intensity of sound from a swarm of n mosquitoes is Iswarm = nI40, we must require that Iswarm = nI40 = I70

(10 ) I = 10 = 1 000 n= = I (10 ) I I70

7.0

3.0

4.0

We conclude that the swarm should contain ∼1 000 mosquitoes to yield a 70 dB sound. 14.71

(a) With a decibel level of 103 dB, the intensity of the sound at 1.60 m from the speaker is found from β = 10 ⋅ log(I/I0) as I = I0⋅10β/10 = (1.00 × 10−12 W/m2) ⋅ 1010.3 = 1.00 × 10−1.7 W/m2 If the speaker broadcasts equally well in all directions, the intensity (power per unit area) at 1.60 m from the speaker is uniformly distributed over a spherical wave front of radius r =

Topic 14

871

1.60 m centered on the speaker. Thus, the power radiated is P = IA = I(4πr2) = (1.00 × 10−1.7 W/m2)4π(1.60 m)2 = 0.642 W

(b) efficiency =

14.72

Poutput Pinput

0.642 W 150 W

= 0.004 3 or 0.43%

(a) At point C, the distance from speaker A is

r = !A

( 3.00 m ) + ( 4.00 m ) = 5.00 m 2

and the intensity of the sound from this speaker is

IA = !

PA 1.00 × 10−3 W = 2 4π rA2 4π ( 5.00 m )

= 3.18 × 10−6 W/m 2

The sound level at C due to speaker A alone is then ⎛ 3.18 × 10−6 W/m 2 ⎞ ⎛I ⎞ β A = 10 ⋅ log ⎜ A ⎟ = 10 ⋅ log ⎜ = 65.0 dB −12 2 ⎝ I0 ⎠ ⎝ 1.00 × 10 W/m ⎟⎠ !

(b) The distance from point C to speaker B is r = !B

( 2.00 m ) + ( 4.00 m ) = 4.47 m and the intensity of the sound 2

Topic 14

872

P 1.50 × 10−3 W −6 IB = B 2 = W/m 2 2 = 5.97 × 10 4π rB 4π ( 4.47 m ) !

The sound level at C due to speaker B alone is therefore ⎛ 5.97 × 10−6 W/m 2 ⎞ ⎛I ⎞ β B = 10 ⋅ log ⎜ B ⎟ = 10 ⋅ log ⎜ = 67.8 dB −12 2 ⎝ I0 ⎠ ⎝ 1.00 × 10 W/m ⎟⎠ !

(c) If both speakers are sounded together, the total sound intensity at point C is IAB = IA + IB = 3.18 × 10−6 W/m2 + 5.97 × 10−6 W/m2 = 9.15 × 10−6 W/m2 and the total sound level in decibels is ⎛ 9.15 × 10−6 W/m 2 ⎞ ⎛ I AB ⎞ β AB = 10 ⋅ log ⎜ = 10 ⋅ log ⎜ = 69.6 dB −12 2 ⎝ I 0 ⎟⎠ ⎝ 1.00 × 10 W/m ⎟⎠ !

14.73

We assume that the average intensity of the sound is directly proportional to the number of cars passing each minute. If the sound level in decibels is β = 10 ⋅ log(I/I0), the intensity of the sound is I = I0 ⋅ 10β/10, so the average intensity in the afternoon, when 100 cars per minute are passing, is

(

)

I100 = I 0 ⋅1080.0/10 = 1.00 × 10−12 W / m 2 ⋅108.00 = 1.00 × 10−4 W / m 2

The expected average intensity at night, when only 5 cars pass per minute, is given by the ratio I5/I100 = 5/100 = 1/20, or

I100 1.00 × 10−4 W/m 2 I5 = = = 5.00 × 10−6 W/m 2 20 20 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

873

and the expected sound level in decibels is ⎛ 5.00 × 10−6 W/m 2 ⎞ ⎛ I5 ⎞ β 5 = 10 ⋅ log ⎜ ⎟ = 10 ⋅ log ⎜ = 67.0 dB −12 2 ⎝ I0 ⎠ ⎝ 1.00 × 10 W/m ⎟⎠ !

14.74

The well will act as a pipe closed at one end (the bottom) and open at the other (the top). The resonant frequencies are the odd integer multiples of the fundamental frequency, or fn = (2n − 1)f1, where n = 1, 2, 3, …. Thus, if fn and fn+1 are two successive resonant frequencies, their difference is

(

)

f − f = [ 2 ( n + 1) − 1] f1 − ( 2n − 1) f1 = 2n + 2 − 1 − 2n + 1 f1 = 2 f1 ! n+1 n In this case, we have 60.0 Hz − 52.0 Hz = 2f1, giving the fundamental frequency for the well as f1 = 4.00 Hz. In the fundamental mode, the well (pipe closed at one end) forms a standing wave pattern with a node at the bottom and the first antinode at the top, making the depth of the well

14.75

⎞ 1 ⎛ 343 m/s ⎞ 1⎛ v = ⎜ sound ⎟ = ⎜ ⎟ = 21.4 m 4 4 ⎝ f1 ⎠ 4 ⎝ 4.00 Hz ⎠

λ1

If r1 and r2 are the distances of the two observers from the speaker, the intensities of the sound at their locations are

P I1 = 4π r12 !

and

I2 =

P 4π r22

Topic 14

874

where P is the power output of the speaker. The difference in the decibel levels for the two observers is 2

⎛I ⎞ ⎛I ⎞ ⎛I ⎞ ⎛ r2 ⎞ ⎛r ⎞ ⎛r ⎞ β1 − β 2 = 10log ⎜ 1 ⎟ − 10log ⎜ 2 ⎟ = 10log ⎜ 2 ⎟ = 10log ⎜ 22 ⎟ = 10log ⎜ 2 ⎟ = 20log ⎜ 2 ⎟ ⎝ I1 ⎠ ⎝ r1 ⎠ ⎝ r1 ⎠ ⎝ r1 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ !

Since β1 = 80 dB and β2 = 60 dB, we find that 80 − 60 = 20log(r2/r1). This yields log(r2/r1) = 1.0 We also know that

and

r2/r1 = 10

r2 = 10r1

r1 + r2 = 36.0 m

[1] [2]

Substituting Equation [1] into [2] gives: 11r1 = 36.0 m or r1 = 36.0 m/11 ≈ 3.3 m Then, Equation [2] yields r2 = 36.0 m − r1 = 36.0 m − 3.3 m = 32.7 m. 14.76

We take toward the east as the positive direction, so the velocity of the

 sea water relative to Earth is !v WE = –10.0 km/h. The velocity of the trailing ship, which is the sound source (S), relative the propagation medium (sea water) is then ⎛ 0.278 m/s ⎞    v SW = v SE − v WE = +64.0 km/h − ( −10.0 km/h ) = +74.0 km/h ⎜ = 20.6 m/s ⎝ 1 km/h ⎟⎠ !

The velocity of the leading ship, the observer (O) in this case, relative to the water is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

875

⎛ 0.278 m/s ⎞    v OW = v OE − v WE = +45.0 km/h − ( −10.0 km/h ) = +55.0 km/h ⎜ = 15.3 m/s ⎝ 1 km/h ⎟⎠ !

With the source moving toward the observer, but the observer moving away from the source, the frequency detected by the observer is given by Equation 14.12 as

⎛ v+v ⎞ ⎛ v + (−| v! |) ⎞ O OW fO = fS ⎜ ⎟ = fS ⎜ ⎟ ! ⎝ v − vS ⎠ ⎝ v − (+| v SW |) ⎠ The speed of sound in sea water is v = 1 533 m/s (Table 14.1) and the frequency emitted by the source is fS = 1 200.0 Hz, so the observed frequency should be

⎛ 1 533 m/s − 15.3 m/s ⎞ fO = 1 200.0 Hz ⎜ ⎟ = 1 204 Hz ⎝ 1 533 m/s − 20.6 m/s ⎠

(

14.77

)

On the weekend, there are one-fourth as many cars passing per minute as on a weekday. Thus, the intensity, I2, of the sound on the weekend is one-fourth of that, I1, on a weekday. The difference in the decibel levels is therefore ⎛I ⎞ ⎛I ⎞ ⎛I ⎞ β1 − β 2 = 10log ⎜ 1 ⎟ − 10log ⎜ 2 ⎟ = 10log ⎜ 1 ⎟ = 10log ( 4 ) = 6 dB ⎝ I2 ⎠ ⎝ I0 ⎠ ⎝ I0 ⎠ !

so, β2 = β1 − 6 dB = 70 dB − 6 dB = 64 dB 14.78

(a) At T = 20°C = 293 K, the speed of sound in air is

Topic 14

876

T 20.0 v = ( 331 m/s ) 1 + C = ( 331 m/s ) 1 + = 343 m/s 273 273 ! The first harmonic or fundamental of the flute (a pipe open at both ends) is given by

v 343 m/s λ1 = 2L = = = 1.31 m f 261.6 Hz 1 ! Therefore, the length of the flute is

λ 1.31 m L= 1 = = 0.655 m 2 2 ! (b) In the colder room, the length of the flute, and hence its fundamental wavelength, is essentially unchanged (that is,

!λ1′ = λ1 = 1.31 m ). However, the speed of sound, and thus the frequency of the fundamental, will be lowered. At this lower temperature, the frequency must be

! f1′ = f1 − f beat = 261.6 Hz − 3.00 Hz = 258.6 Hz The speed of sound in this room is v′ = λ1′ f1′ = (1.31 m ) ( 258.6 Hz ) = 339 m/s !

From !v = (331 m/s) 1 + TC /273 , the temperature in the colder room is given by

Topic 14

877

2 ⎡⎛ ⎤ ⎡⎛ 339 m/s ⎞ 2 ⎤ ⎞ v T = ( 273°C ) ⎢⎜ ⎟ − 1⎥ = ( 273°C ) ⎢⎜⎝ 331 m/s ⎟⎠ − 1⎥ = 13.4°C ⎢⎣⎝ 331 m/s ⎠ ⎥⎦ ⎢⎣ ⎥⎦ !

14.79

The maximum speed of the oscillating block and speaker is k 20.0 N/m vmax = Aω = A = ( 0.500 m ) = 1.00 m/s m 5.00 kg !

(a) When the speaker moves away from the stationary observer, the source velocity is vS = −vmax and the minimum frequency heard is

⎛

⎞

⎛

⎞

343 m/s ( f ) = f ⎜⎝ v + vv ⎟⎠ = ( 440 Hz)⎜⎝ 343 m/s ⎟ = 439 Hz + 1.00 m/s ⎠ O min

max

(b) When the speaker (sound source) moves toward the stationary observer, then vS = +vmax and the maximum frequency heard is

⎛ ⎛ ⎞ 343 m/s v ⎞ = f = 440 Hz ⎜ ⎟ ⎜ ⎟ = 441 Hz O max S v − v 343 − 1.00 m/s m/s ⎝ ⎠ ⎝ max ⎠

(f ) 14.80

(

)

The speed of transverse waves in the wire is

F = µ

vT = !

F⋅L = m

( 400 N)( 0.750 m ) = 365 m/s 2.25 × 10−3 kg

When the wire vibrates in its third harmonic, λ = 2L/3 = 0.500 m, so the frequency of the vibrating wire and the sound produced by the wire is

fS =

365 m/s 0.500 m

= 730 Hz

Since both the wire and the wall are stationary, the frequency of the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 14

878

wave reflected from the wall matches that of the waves emitted by the wire. Thus, as the student approaches the wall at speed |vO|, he approaches one stationary source and recedes from another stationary source, both emitting frequency fS = 730 Hz. The two frequencies that will be observed are

⎛ v +|v |⎞ ⎛ v −|v |⎞ O O = fS ⎜ and f = f ⎟ ⎟ O 2 S⎜ 1 v v ⎝ ⎠ ⎝ ⎠

( ) fO

( )

The beat frequency is

(

)

⎛ v+ v − v− v ⎞ 2f v O O ⎟= S O f beat = fO − fO = fS ⎜ 1 2 ⎜ ⎟ v v ⎝ ⎠

( ) ( )

14.81

⎡ ⎤ ⎛f ⎞ 8.30 Hz ⎥ beat ⎢ vO = ⎜ 343 m/s = 1.95 m/s ⎟v= ⎢ 2 730 Hz ⎥ ⎝ 2 fS ⎠ ⎣ ⎦

(

)(

)

The speeds of the two types of waves in the rod are

vlong = !

Y Y Y(A ⋅ L) = = !and!vtrans = ρ m/V m

Thus, if vlong = 8vtrans, we have

F F = = µ m/L

F⋅L m

Y(A ⋅ L) ⎛ F ⋅ L⎞ = 64 ⎜ , or the required ⎝ m ⎟⎠ ! m

tension is 2 10 2 ⎡ −2 ⎤ Y ⋅ A ( 6.80 × 10 N/m ) ⎣π ( 0.200 × 10 m ) ⎦ F= = = 1.34 × 10 4 N 64 64 !

14.82

(a) For the fundamental mode of a pipe open at both ends, L = λ1/2 or

Topic 14

879

the wavelength of the waves traveling through the air in the pipe is λ1 = 2L = 2(0.500 m) = 1.00 m. If the frequency of this fundamental mode is f1 = 350 Hz, the speed of sound waves within the pipe must be v = λ1f1 = (1.00 m)(350 Hz) = 350 m/s From !v = ( 331 m/s ) 1 + TC /273 , the Celsius temperature of the air in the pipe is 2 ⎡⎛ ⎤ ⎡⎛ 350 m/s ⎞ 2 ⎤ ⎞ v TC = ( 273°C ) ⎢⎜ ⎟ − 1⎥ = ( 273°C ) ⎢⎜⎝ 331 m/s ⎟⎠ − 1⎥ = 32.2°C ⎢⎣⎝ 331 m/s ⎠ ⎥⎦ ⎢⎣ ⎥⎦ !

(b) If the temperature rises to T′ = T + 20.0°C = 52.2°C, the speed of sound in the air will be

v′ = ( 331 m/s ) 1 + TC′ / 273 = ( 331 m/s ) 1 + 52.2 / 273 , and the new ! length of the pipe will be L′ = L0[1 + α(ΔT)] = (0.500 m)[1 + (19 × 10−6 °C−1)(20.0°C)] The new fundamental wavelength is !λ1′ = 2 L′ , and the new fundamental resonance frequency will be

f1′ = !

( 331 m/s ) 1 + 52.2 / 273 v′ = = 3.6 × 102 Hz λ1′ 2 ( 0.500 m ) ⎡⎣1 + (19 × 10−6 °C−1 ) ( 20.0°C ) ⎤⎦

Topic 15

880

Topic 15 Electric Forces and Fields

QUICK QUIZZES 15.1

Choice (b). Object A must have a net charge because two neutral objects do not attract each other. Since object A is attracted to positively-charged object B, the net charge on A must be negative.

15.2

Choice (b). By Newton’s third law, the two objects will exert forces having equal magnitudes but opposite directions on each other.

15.3

Choice (c). The electric field at point P is due to charges other than the test charge. Thus, it is unchanged when the test charge is altered. However, the direction of the force this field exerts on the test change is reversed when the sign of the test charge is changed.

15.4

Choice (a). If a test charge is at the center of the ring, the force exerted on the test charge by charge on any small segment of the ring will be balanced by the force exerted by charge on the diametrically opposite segment of the ring. The net force on the test charge, and hence the electric field at this location, must then be zero.

Topic 15

15.5

881

Choices (c) and (d). The electron and the proton have equal magnitude charges of opposite signs. The forces exerted on these particles by the electric field have equal magnitude and opposite directions. The electron experiences an acceleration of greater magnitude than does the proton because the electron’s mass is much smaller than that of the proton.

15.6

Choice (a). The field is greatest at point A because this is where the field lines are closest together. The absence of lines at point C indicates that the electric field there is zero.

15.7

Choice (c). When a plane area A is in a uniform electric field E, the flux through that area is ΦE = EA cos θ, where θ is the angle the electric field makes with the line normal to the plane of A. If A lies in the xy-plane and E is in the z-direction, then θ = 0° and ΦE = EA = (5.00 N/C)(4.00 m2) = 20.0 N ⋅ m2/C.

15.8

Choice (b). If θ = 60° in Quick Quiz 15.7 above, then ΦE = EA cos θ which yields ΦE = (5.00 N/C)(4.00 m2)cos (60°) = 10.0 N ⋅ m2/C.

15.9

Choice (d). Gauss’s law states that the electric flux through any closed surface is equal to the net enclosed charge divided by the permittivity of free space. For the surface shown in Figure 15.28, the net enclosed charge

Topic 15

882

is Q = −6 C, which gives ΦE = Q/e0 = − (6 C)/e0. 15.10

Choices (b) and (d). Since the net flux through the surface is zero, Gauss’s law says that the net change enclosed by that surface must be zero as stated in (b). Statement (d) must be true because there would be a net flux through the surface if more lines entered the surface than left it (or viceversa).

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 15.2

(a) False. Electric charge can only be transferred in multiples of the fundamental charge +e. (b) True. (c) False. All electrons have an electric charge of −e.

15.4

(a) Removing charge −Q from object A leaves it with a net positive charge of +Q. Object B acquires a net charge −Q so that A and B are oppositely charged and the Coulomb force, proportional to QAQB =

−Q2, is attractive. (b) After another transfer of −Q from A to B, QA = +2Q and QB = −2Q so that QAQB = −4Q2 and Fnew/F0 = 4.

Topic 15

883

No. Object A might have a charge opposite in sign to that of B, but it also might be neutral. In this latter case, object B causes object A to be polarized, pulling charge of the sign opposite the change on B toward the near face of A and pushing an equal amount of charge of the same sign as that on B toward the far face. Then, due to difference in distances, the force of attraction exerted by B on the induced charge of opposite sign is slightly larger than the repulsive force exerted by B on the induced charge of like sign. Therefore, the net force on A is toward B.

15.8

(a)

Yes. The positive charges create electric fields that extend in all directions from those charges. The total field at point A is the vector sum of the individual fields produced by the charges at that point.

(b) No, because there are no field lines emanating from or converging on point A.

Topic 15

884

Electric field lines start on positive charges and end on negative charges. Thus, if the fair-weather field is directed into the ground, the ground must have a negative charge.

15.12

To some extent, a television antenna will act as a lightning rod on the house. If the antenna is connected to the Earth by a heavy wire, a lightning discharge striking the house may pass through the metal support rod and be safely carried to the Earth by the ground wire.

15.14 (a)

If the charge is tripled, the flux through the surface is also tripled because the net flux is proportional to the charge inside the surface.

(b) The flux remains constant when the volume changes because the surface surrounds the same amount of charge, regardless of its volume. (c)

The flux does not change when the shape of the closed surface changes.

(d) The flux through the closed surface remains unchanged as the charge inside the surface is moved to another location inside that surface. (e)

The flux is zero because the charge inside the surface is zero. All of

Topic 15

885

these conclusions are arrived at through an understanding of Gauss’s law. 15.16

All of the constituents of air are nonpolar except for water. The polar water molecules in the air quite readily “steal” charge from a charged object, as any physics teacher trying to perform electrostatics demonstrations in the summer well knows. As a result—it is difficult to accumulate large amounts of excess charge on an object in a humid climate. During a North American winter, the cold, dry air allows accumulation of significant excess charge, giving the possibility for shocks caused by static electricity sparks.

ANSWERS TO EVEN NUMBERED PROBLEMS 15.2

1.57 N directed to the left

15.4

(a)

15.6

2.25 × 10−9 N/m

15.8

4.33keq2/a2 to the right and 45° above the horizontal

15.10

F(6 µC) = 46.7 N to the left; F(1.5 µC) = 157 N to the right; F(−2 µC) = 111

0.115 N

(b)

1.25 cm

N to the left

Topic 15

886

15.12

5.15 × 103 N/m

15.14

16.7 µC

15.16

(a)

(b)

30.0 N

(c)

(d)

17.3 N

(e)

−13.0 N

(f) 17.3 N

(g) 17.0 N

(h)

! FR = 24.3 N at 44.5° above the + x-axis

21.6 N

15.18

(a)

2.00 × 107 N/C to the right

15.20

(a)

5.27 × 1013 m/s2

(b)

5.27 × 105 m/s

15.22

(a)

See Solution.

(b)

1.4 × 103 N/C

(c)

7.5 × 10−2 N

(a)

2.19 × 105 N/C at 85.2° below the +x-axis

(b)

The electric field would be unchanged, but the force would double

15.24

(b)

40.0 N to the left

in magnitude. 15.26 (a) 1.93 × 105 m/s2

(b)

8.69 × 105 m

15.28 (a) −3.80 m/s2

(b)

−12.6 m/s

15.30 (a) 1.80 × 104 N/C to the right 15.32

(b)

9.00 × 10−5 N to the left

1.88 × 103 N/C at 4.40° below the +x-axis

Topic 15

887

15.34 (a)

q1/q2 = −1/3

15.36 (a), (b), and (c) 15.38 (a) (c) 15.40 (a) (c)

(b)

See Solution for Sketches.

See Solution.

(b)

 !E = 0 at the center of the triangle.

1.73keq/a2

(b)

positive y-dirtection

zero

(b)

1.8 × 106 N/m

1.1 × 105 N/m

15.42

r ∼ 10−6 m

15.44

1.38 × 103 N⋅m2/C

15.46 (a) (b)

q1 < 0, q2 > 0

3.38 nC The sphere contains a net positive charge in a spherically symmetric distribution.

15.48 (a)

6.55 × 105 N ⋅ m2/C

(b)

1.64 × 105 N ⋅ m2/C

15.50 (a)

1.59 × 10−8 C/m2

(b)

898 N/C

15.52 (a)

1.92 × 107 N ⋅ m2/C

(b)

3.20 × 106 N ⋅ m2/C

(c)

The answer to (b) would change because of the loss of symmetry; however, the answer to (a) would be unchanged since Qinside is unaltered.

Topic 15

15.54

888

The electric field is everywhere parallel to the z-axis with the following magnitudes:

15.56

(a)

Ez = + σ/2e0

(b)

Ez = + 3σ/2e0

(c)

Ez = − σ/2e0

(a)

2.94 × 105 N/C

(b)

The magnitude of the field is independent of distance from the sheet, provided that distance is small compared to the dimensions of the sheet.

15.58

+5.25 µC

15.60

E = K/ed in the direction of the electron’s motion.

15.62 (a) (b) 15.64 (a)

0.307 s Yes, neglecting gravity causes a 2.28% error. For 0 < x < 1.00, the field contributions made by the two charges are in the same direction and cannot add to zero.

(b)

Any point for x < 0 is nearer the larger charge, so the field contributions made by the two charges cannot have equal magnitudes and add to zero.

(c) Points for x > 1.00 m are nearer the smaller charge and the field © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 15

889

contributions by the two charges are in opposite directions. Thus, it is possible for them to add to zero. (d) E = 0 at x = +9.47 m 15.66

0.951 m

15.68

1.98 µC

15.70

(a)

37.0° or 53.0°

(b)

1.67 × 10−7 s for θ = 37.0°; 2.21 × 10−7 s for θ = 53.0°

PROBLEM SOLUTIONS 15.1

QQ (a) From Coulomb’s law, F = ke 1 2 2 , we have r !

(7.50 × 10 C)( 4.20 × 10 C) = 8.74 × 10 N F = ( 8.99 × 10 N ⋅ m C ) −9

(1.80 m )

−9

−8

(b) Since these are like charges (both positive), the force is repulsive. 15.2

Particle A exerts a force toward the right on particle B. By Newton’s third law, particle B will then exert a of equal magnitude force toward the left on particle A. The ratio of the final magnitude of the force to the original magnitude of the force is

Topic 15

890

⎛ ri ⎞ = = F ke q1q2 ri2 ⎜⎝ rf ⎟⎠ ! i ke q1q2 rf2

⎛r⎞ ⎛ 13.7 mm ⎞ Ff = Fi ⎜ i ⎟ = ( 2.62 N ) ⎜ = 1.57 N ⎝ 17.7 mm ⎟⎠ ⎝ rf ⎠

The final vector force that B exerts on A is 1.57 N directed to the left. 15.3

Solve Coulomb’s law for the separation r and substitute values to find F = ke

q1 q2 r2

→ r = ke

q1 q2 F

(1.60 × 10 r = ( 8.99 × 10 N ⋅ m /C ) 9

−19

1.00 × 10

15.4

(a)

−14

) = 1.52 × 10 m 2

−7

The gravitational force exerted on the upper sphere by the lower one is negligible in comparison to the gravitational force exerted by the Earth and the downward electrical force exerted by the lower sphere. Therefore, ΣFy = 0 ⇒ T − mg − Fe = 0

T = mg + !

ke q1 q2 d2

9 2 2 −9 −9 m ⎞ ( 8.99 × 10 N ⋅ m C ) ( 32.0 × 10 C ) ( 58.0 × 10 C ) ⎛ T = (7.50 × 10 kg ) ⎜ 9.80 2 ⎟ + 2 ⎝ s ⎠ 2.00 × 10−2 m ) ( ! −3

giving

T = 0.115 N

Topic 15

891

(b) ∑ Fy = 0 → Fe = !

ke q1 q2 k q q = T − mg,!and!d = e 1 2 2 d T − mg

Thus, if T = 0.180 N,

(8.99 × 10 N ⋅ m C )( 32.0 × 10 C)( 58.0 × 10 C) = 1.25 × 10 m = 1.25 cm d= m⎞ ⎛ 0.180 N − (7.50 × 10 kg ) ⎜ 9.80 ⎟ ⎝ s ⎠ 9

−9

−2

−3

15.5

2 9 2 2 ⎡ −19 ⎤ 2 ke ( 2e ) ( 8.99 × 10 N ⋅ m C ) ⎣ 4 (1.60 × 10 C ) ⎦ (a) F = = = 36.8 N 2 r2 5.00 × 10−15 m ) ( !

(b)

The mass of an alpha particle is m = 4.002 6 u, where 1 u = 1.66 × 10−27 kg is the unified mass unit. The acceleration of either alpha particle is then

15.6

F m

(

36.8 N

4.002 6 1.66 × 10

−27

)

= 5.54 × 1027 m s 2

The attractive force between the charged ends tends to compress the molecule. Its magnitude is

Topic 15

892 2 2 1.60 × 10−19 C ) ke (1e ) ⎛ 9 N⋅m ⎞ ( −17 F= = ⎜ 8.99 × 10 N 2 = 4.89 × 10 2 2 ⎟ −6 r ⎝ C ⎠ ( 2.17 × 10 m ) ! 2

The compression of the “spring” is x = (0.010 0)r = (0.010 0)(2.17 × 10−6 m) = 2.17 × 10−8 m,

so the spring constant is k =

15.7

F x

4.89 × 10−17 N 2.17 × 10 m −8

= 2.25 × 10−9 N/m .

The charge q of N = 3.50 × 1012 electrons is q = −eN = −(1.60 × 10–19 C)(3.50 × 1012 electrons) = 5.60 × 10−7 C. One sphere will have charge q and the other will have charge −q. Substitute values into Coulomb’s law to find

)(

5.60 × 10−7 C q q F = ke 1 2 2 = 8.99 × 109 N ⋅ m 2 /C r ( 2.00 m )2

(

15.8

) = 7.05 × 10 N 2

−4

See the sketch below. The magnitudes of the forces are q ( 2q ) q ( 3q ) q2 q2 F1 = F2 = ke 2 = 2ke 2 !and!F3 = ke 2 = 1.50ke 2 a a a a 2 !

(

)

The components of the resultant force on charge q are

Topic 15

893

q2 q2 Fx = F1 + F3 cos 45° = ( 2 + 1.50cos 45° ) ke 2 = 3.06ke 2 a a ! and

q2 q2 Fy = F1 + F3 sin 45° = ( 2 + 1.50sin 45° ) ke 2 = 3.06ke 2 a a !

The magnitude of the resultant force is ⎛ q2 ⎞ q2 Fe = Fx2 + Fy2 = 2 ⎜ 3.06ke 2 ⎟ = 4.33ke 2 a ⎠ a ⎝ ⎛F ⎞ and it is directed at θ = tan −1 ⎜ y ⎟ = tan −1 (1.00 ) = 45°above the horizontal . ⎝ Fx ⎠ !

15.9

(a)

The spherically symmetric charge distributions behave as if all charge was located at the centers of the spheres. Therefore, the magnitude of the attractive force is

12 × 10−9 C ) (18 × 10−9 C ) ke q1 q2 9 2 2 ( F= = ( 8.99 × 10 N ⋅ m C ) = 2.2 × 10−5 N 2 2 r ( 0.30 m ) ! (b)

When the spheres are connected by a conducting wire, the net charge qnet = q1 + q2 = − 6.0 × 10−9 C will divide equally between the two identical spheres. Thus, the force is now −9 k ( q 2) ⎛ N ⋅ m 2 ⎞ ( −6.0 × 10 C ) F = e net2 = ⎜ 8.99 × 109 2 r ⎝ C2 ⎟⎠ 4 ( 0.30 m ) ! 2

F = 9.0 × 10−7 N (repulsion)

Topic 15

15.10

894

The forces are as shown in the sketch below.

F1 = !

2 6.00 × 10−6 C ) (1.50 × 10−6 C ) ke q1q2 ⎛ 9 N⋅m ⎞ ( = 8.99 × 10 = 89.9 N 2 ⎜⎝ −2 r122 C2 ⎟⎠ 3.00 × 10 m ( )

2 6.00 × 10−6 C ) ( 2.00 × 10−6 C ) ke q1 q3 ⎛ 9 N⋅m ⎞ ( F2 = = ⎜ 8.99 × 10 = 43.2N r132 ⎝ C2 ⎟⎠ (5.00 × 10−2 m)2 ! 2 1.50 × 10−6 C ) ( 2.00 × 10−6 C ) ke q2 q3 ⎛ 9 N⋅m ⎞ ( F3 = = ⎜ 8.99 × 10 = 67.4 N r232 ⎝ C2 ⎟⎠ (2.00 × 10−2 m)2 !

The net force on the 6 µC charge is F = F1 − F2 = 46.7 N to the left. The net force on the 1.5 µC charge is F = F1 − F3 = 157 N to the right. The net force on the −2 µC charge is F = F2 + F3 = 111 N to the left. 15.11

In the sketch below, FR is the resultant of the forces F6 and F3 that are exerted on the charge at the origin by the 6.00 nC and the −3.00 nC charges, respectively.

Topic 15

895 −9 −9 ⎛ N ⋅ m 2 ⎞ ( 6.00 × 10 C ) ( 5.00 × 10 C ) F6 = ⎜ 8.99 × 109 = 3.00 × 10−6 N 2 2 ⎟ ⎝ C ⎠ ( 0.300 m )

2 3.00 × 10−9 C ) ( 5.00 × 10−9 C ) ⎛ 9 N⋅m ⎞ ( F3 = ⎜ 8.99 × 10 = 1.35 × 10−5 N 2 2 ⎟ ⎝ C ⎠ ( 0.100 m ) !

The resultant is FR = !

or 15.12

⎛F ⎞ −1 3 θ = tan at ⎜ ⎟ = 77.5° ( F6 ) + ( F3 ) = 1.38 × 10 N ⎝ F6 ⎠ 2

−5

! FR = 1.38 × 10−5 N at 77.5° below the − x-axis

At equilibrium, ΣFx = Fe − Fs = 0 or Fs = Fe

Thus,

kx = !

ke q1q2 and the force constant of the spring is d2

9 2 2 −6 −6 ke q1q2 ( 8.99 × 10 N ⋅ m C ) ( 2.70 × 10 C ) ( 8.60 × 10 C ) k= = 2 xd 2 5.00 × 10−3 m ) ( 9.00 × 10−2 m ) ( !

k = 5.15 × 103 N/m

15.13

Please see the sketch below.

Topic 15

896

(8.99 × 10 N ⋅ m C )( 2.00 × 10 C)( 4.00 × 10 C) F = 9

( 0.500 m )

! or

−6

F1 = 0.288 N

(8.99 × 10 N ⋅ m C )( 2.00 × 10 C)(7.00 × 10 C) F = 9

( 0.500 m )

! or

−6

F2 = 0.503 N

The components of the resultant force acting on the 2.00 µC charge are: Fx = F1 − F2 cos 60.0° = 0.288 N − (0.503 N)cos 60.0° = 3.65 × 10−2 N and Fy = −F2 sin 60.0° = −(0.503 N)sin60.0° = −0.436 N The magnitude and direction of this resultant force are

F = Fx2 + Fy2 =

(

) ( 2

0.036 5 N + 0.436 N

) = 0.438 N 2

⎛F ⎞ ⎛ −0.436 N ⎞ y θ = tan ⎜ ⎟ = tan −1 ⎜ ⎟ = −85.2° or 85.2° below the +x-axis ⎜⎝ Fx ⎟⎠ ⎝ 0.036 5 N ⎠ −1

Topic 15

15.14

897

At equilibrium, Fe = Fs or

keQ 2 = kx = k ( L − Li ) 2 ! L

Thus,

k ( L − Li ) L2 Q= = ke !

(100 N/m )( 0.500 m − 0.400 m )( 0.500 m )

8.99 × 109 N ⋅ m 2 C2

Q = 1.67 × 10−5 C = 16.7 µC 15.15

Consider the free-body diagram of one of the spheres given below. Here, T is the tension in the string and Fe is the repulsive electrical force exerted by the other sphere.

ΣFy = 0 ⇒ T cos 5.0° = mg,

mg T= cos 5.0° !

ΣFx = 0 ⇒ Fe = T sin 5.0° = mg tan 5.0°

Topic 15

898

At equilibrium, the distance separating the two spheres is r = 2L sin 5.0°.

Thus, Fe = mg tan 5.0° becomes

q = ( 2Lsin 5.0° )

15.16

ke q 2

2 Lsin 5.0° ) !(

= mg! tan 5.0° and yields

mg tan 5.0° ke

= ⎡⎣ 2 ( 0.300 m ) sin 5.0° ⎤⎦

( 0.20 × 10 kg )( 9.80 m s ) tan 5.0° = 7.2 nC −3

8.99 × 109 N ⋅ m 2 C2

In the sketch below,

r = ! BC

( 4.00 m ) + ( 3.00 m ) = 5.00 m 2

⎛ 3.00 m ⎞ −1 and θ = tan ⎜ ⎟ = 36.9° ⎝ 4.00 m ⎠ (a) (FAC)x = 0

Topic 15

899

(b)

( ) FAC

= ke

qA qC 2 rAC

(3.00 × 10 C)(1.00 × 10 C) = 30.0 N = ( 8.99 × 10 N ⋅ m C ) (3.00 m) −4

−4

(c)

FBC = ke

qB qC 2 rBC

(6.00 × 10 C)(1.00 × 10 C) = 21.6 N = ( 8.99 × 10 N ⋅ m C ) (5.00 m) −4

−4

(d) (FBC)x = |FBC|cosθ = (21.6 N)cos(36.9°) = 17.3 N (e) (FBC)y = −|FBC|sinθ = −(21.6 N)sin(36.9°) = −13.0 N (f) (FR)x = (FAC)x + (FBC)x = 0 + 17.3 N = 17.3 N (g) (FR)y = (FAC)y + (FBC)y = 30.0 − 13.0 N = 17.0 N

(h) FR = !

( FR )2x + ( FR )2y = (17.3 N)2 + (17.0 N)2 = 24.3 N

and

⎡ ( FR )y ⎤ ⎛ 17.0 N ⎞ ϕ = tan −1 ⎢ = 44.5° ⎥ = tan −1 ⎜ ⎝ 17.3 N ⎟⎠ ⎢⎣ ( FR )x ⎥⎦ !

! FR = 24.3 N at 44.5° above the +x-axis

Topic 15

15.17

900

In order to suspend the object in the electric field, the electric force exerted on the object by the field must be directed upward and have a magnitude equal to the weight of the object. Thus, Fe = qE = mg, and the magnitude of the electric field must be

(

)

3.80 g ( 9.80 m s ) ⎛ 1 kg ⎞ mg 3 E= = ⎜ 103 g ⎟ = 2.07 × 10 N/C q 18 × 10−6 C ⎝ ⎠ ! 2

The electric force on a negatively charged object is in the direction opposite to that of the electric field. Since the electric force must be directed upward, the electric field must be directed downward. 15.18 (a) Taking to the right as positive, the resultant electric field at point P is given by ER = E1 + E3 − E2 =

ke q1 r12

k e q3 r32

−

ke q2 r22

⎡ ⎤ −6 −6 −6 2⎞ ⎛ ⎢ 6.00 × 10 C 2.00 × 10 C 1.50 × 10 C ⎥ 9 N⋅m = ⎜ 8.99 × 10 + − ⎟ 2 2⎥ C2 ⎠ ⎢⎢ 0.020 0 m 2 ⎝ 0.030 m 0.010 0 m ⎥ ⎣ ⎦

(

) (

)

Topic 15

901

 E R = 2.00 × 107 N/C!to!the!right !

  7 (b) !F = qER = ( −2.00 × 10 N C ) = −40.0 N or 15.19

 F = 40.0 N!to!the!left !

The force on a negative charge is opposite to the direction of the electric field and has magnitude F = |q|E. Thus, F = |− 6.00 × 10−6 C|(5.25 × 105 N/C) = 3.15 N

 and F = 3.15 N!due!the!north ! 15.20

(a) The magnitude of the force on the electron is F = |q|E = eE, and the acceleration is −19 F eE (1.60 × 10 C ) ( 300 N/C ) a= = = = 5.27 × 1013 m s 2 −31 m m 9.11 × 10 kg e e !

13 2 −8 5 (b) v = v0 + at = 0 + ( 5.27 × 10 m s ) (1.00 × 10 s ) = 5.27 × 10 m/s !

15.21 See the figure below illustrating the electric fields from charges q1 and q2.

q1 +

x1= 0

E1 x2–x

x x

The net electric field at a position 0 < x < 2 is given by:

Topic 15

902

⎛q q2 ⎞ E = ke ⎜ 12 − ⎟ ⎜⎝ x ( x − x )2 ⎟⎠ 2 Set this expression equal to zero and solve for the position x: ⎛q q2 ⎞ ke ⎜ 12 − ⎟ =0 → ⎜⎝ x ( x − x )2 ⎟⎠ 2

(

(x − x ) q − x q = 0 x (x − x )

)

q1 x 2 + x22 − 2xx2 − x 2 q2 = 0 →

( q − q ) x − ( 2x q ) x + q x = 0 2

2 1

2 1 2

Substitute values, suppressing units and zeros for clarity, and noticing that the factor 10−9 (from the nC) cancels out of each term:

(1 − 3) x − ( 2 ( 2)(1)) x + (1)( 2) = 0 2

−2x 2 − 4x + 4 = 0 x 2 + 2x − 2 = 0 Apply the quadratic formula, solving for the values of x that lies between 0 and 2.00 m: −b ± b 2 − 4ac −2 ± 4 − 4 (1) ( −2 ) x= = = −1 ± 1 + 2 = −1 ± 3 2a 2

Choosing the positive root gives: x = −1 + 3 = 0.732 m . 15.22 (a)

Topic 15

903

(b) ΣFy = 0 ⇒ Fe sin θ − mg = 0

mg Fe = sin θ !

Since Fe = qE, this gives

5.8 × 10−3 kg ) ( 9.8 m s 2 ) ( mg E= = = 1.4 × 103 N/C −6 qsin θ ( 68 × 10 C) sin 37° ! ⎛ mg ⎞ mg cos θ = (c) ΣFx = 0 ⇒ Fe cosθ − T = 0 or T = Fe cos θ = ⎜ ⎟ tan θ ⎝ sin θ ⎠ !

( 5.8 × 10 kg )( 9.8 m s ) = 7.5 × 10 N and T = −3

−2

tan 37°

15.23

−19 F qE (1.60 × 10 C ) ( 640 N/C ) = = 6.12 × 1010 m s 2 (a) a = = m mp 1.673 × 10−27 kg !

Δv 1.20 × 106 m s = = 1.96 × 10−5 s = 19.6 µs (b) t = 10 2 a 6.12 × 10 m s !

(c)

Δx = !

v 2f − v02 2a

(1.20 × 10 m s ) − 0 = 11.8 m = 2 ( 6.12 × 10 m s ) 2

2 1 1 2 −27 6 −15 (d) KE f = mp v f = (1.673 × 10 kg ) (1.20 × 10 m s ) = 1.20 × 10 J 2 2 !

15.24 (a)

Please refer to the solution of Problem 15.13 earlier in this chapter. There it is shown that the resultant electric force experienced by the ! 2.00 µC located at the origin is F = 0.438 N at 85.2° below the +x-axis.

Since the electric field at a location is defined as the force per unit © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 15

904

charge experienced by a test charge placed in that location, the electric field at the origin in the charge configuration of Figure P15.13 is

  F 0.438 N E= = at − 85.2° = 2.19 × 105 N/C!at!85.2°!below!the + x7axis −6 q0 2.00 × 10 C ! (b)

The electric force experienced by the charge at the origin is directly proportional to the magnitude of that charge. Thus, doubling the magnitude of this charge would double the magnitude of the electric force. However, the electric field is the force per unit charge and the field would be unchanged if the charge was doubled. This is easily seen in the calculation of part (a) above. Doubling the magnitude of the charge at the origin would double both the numerator and the  denominator of the ratio !F q0 , but the value of the ratio (i.e., the

electric field) would be unchanged. 15.25 Let the square have sides of length d = 2.00 cm. (a) With equal charges on each corner of a square, the electric field at the square’s center is E = 0 by symmetry. (b) Now let three of the charges be positive and one be negative, each with a magnitude of 3.20 nC. If charges q1 and q2 are two of three

Topic 15

905

positive charges and are located on opposite corners of the square, ! ! notice that E1 + E2 = 0 by symmetry. If q3 > 0 and q4 < 0, again note ! ! ! ! from symmetry that E3 + E4 = 2E3 = 2E4 . The net electric field

magnitude at the center is therefore E = 2ke

q3 r2

where the

Pythagorean theorem gives r2 = d2/2. Substitute values to find

E = 2ke

q3 2

d /2

(

= 4ke

q3 d2

= 4 8.99 × 10 N ⋅ m /C 9

)

3.20 × 10−9 C

( 2.00 × 10 m) −2

= 2.88 × 105 N/C

! ! 15.26 (a) The electric force is F = qE . Substitute values and apply Newton’s

second law to find the x-component of the nucleus’ acceleration:

! ! ! F = qE = ma

(

)(

)

6.41 × 10−19 C 2.00 × 10−3 N/C qEx ax = = = 1.93 × 105 m/s2 m 6.64 × 10−27 kg (b) The nucleus has constant acceleration, so apply a kinematic equation with v0 = 0 to find Δx = v0t + 12 axt 2

(

)

= 0 + 12 1.93 × 105 m/s2 ( 3.00 s) = 8.69 × 105 m 2

Topic 15

906

15.27 (a) The dust particle is held in equilibrium so the two vertical forces acting on it (gravity and electric forces) must sum to zero. Solve Newton’s second law for the charge q and substitute values:

ΣFy = qE − mg = 0

(

)(

)

−10 2 mg 7.50 × 10 kg 9.80 m/s = = 1.55 × 10−11 C E 475 N/C

(b) Neutralizing the dust particle’s charge requires that q − Ne = 0 where N is the number of required electrons and e is the magnitude of the electron charge. Solve for N to find

(

)( )

)

7.50 × 10−10 kg 9.80 m/s2 q mg N= = = = 9.67 × 107 e eE 1.60 × 10−19 C ( 475 N/C )

(

15.28 (a) Apply Newton’s second law and solve for ay, the y-component of the particle’s acceleration:

ΣFy = may qE − mg = may → ay =

qE − mg m

Substitute values to find:

( 3.00 × 10 a =

−12

)(

) (

)(

C 2.00 × 103 N/C − 1.00 × 10−9 kg 9.80 m/s2 1.00 × 10

−9

)

= −3.80 m/s2 (b) The particle’s acceleration is constant so its final velocity can be found using the first kinematic equation: © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 15

907

v y = v0 y + ayt

(

)

= −5.00 m/s + −3.80 m/s2 ( 2.00 s) = −12.6 m/s

15.29

From the figure below, observe that 2r cos 45.0° + d = 2d. Thus, r= !

d d d = = 2cos 45.0° 2 2 2 2

(

)

k Q k Q ⎛ 2 ⎞ keQ k Q ⎛ 2⎞ kQ Ex = E2 − E1 cos 45.0° = e 2 − e 2 ⎜ = 2 − 2e ⎜ = 1− 2 e2 ⎟ ⎟ d r ⎝ 2 ⎠ d d ( d 2) ⎝ 2 ⎠ !

(

Ey = E1 sin 45.0° + 0 =

15.30

k eQ ⎛ 2 ⎞ k eQ ⎛ 2 ⎞ = ⎜ ⎟ ⎜ ⎟= r 2 ⎜⎝ 2 ⎟⎠ d 2 2 ⎜⎝ 2 ⎟⎠

( )

)

k eQ d2

(a) Observe the figure below:

k q kQ E1 = E2 = e 2 !and!E3 = e 2 r r ! Ey = E1 sin 30.0° − E2 sin 30.0° = 0 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 15

908

Ex = E1 cos 30.0° + E2 cos 30.0° − E3

(

)

k q k 2keQ cos 30.0° − e 2 = 2e = 2Qcos 30.0° − q 2 r r r 9 2 2 (8.99 × 10 N ⋅ m C ) ⎡ 2 ( 3.00 × 10−9 C) cos 30.0° − 2.00 × 10−9 C ⎤ = 2 ⎦ ( 4.00 × 10−2 m ) ⎣

Ex =

Ex = +1.80 × 104 N/C. Then, E = Ex2 + Ey2 = 1.80 × 10 4 N/C and !  E = 1.80 × 10 4 N/C to!the!right !

(b) Fe = q′E = (−5.00 × 10−9 C)(1.80 × 104 N/C) = −9.00 × 10−5 N

! or Fe = 9.00 × 10−5 N to the left . 15.31

If the resultant field is zero, the contributions from the two charges must be in opposite directions and also have equal magnitudes. Choose the line connecting the charges as the x-axis, with the origin at the −2.5 µC charge. Then, the two contributions will have opposite directions only in the regions x < 0 and x > 1.0 m. For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the point of zero resultant field is on the x-axis at x < 0.

Topic 15

909

Requiring equal magnitudes gives

ke q1 ke q2 = 2 r12 r2 ! Thus,

2.5 µC 6.0 µC = 2 d (1.0 m + d )2

=d (1.0 m + d ) 2.5 6.0

Solving for d yields d = 1.8 m, 15.32

1.8 m to the left of the −2.5 µC charge

The altitude of the triangle is

h = (0.500 m) sin 60.0° = 0.433 m and the magnitudes of the fields due to each of the charges are 9 2 2 −9 ke q1 ( 8.99 × 10 N ⋅ m C ) ( 3.00 × 10 C ) E1 = 2 = h ( 0.433 m )2

= 144 N/C

9 2 2 −9 ke q2 ( 8.99 × 10 N ⋅ m C ) ( 8.00 × 10 C ) E2 = 2 = = 1.15 × 103 N/C 2 r2 ( 0.250 m ) !

Topic 15

910 9 2 2 −9 ke q3 ( 8.99 × 10 N ⋅ m C ) ( 5.00 × 10 C ) E = = = 719 N/C and 3 2 r32 0.250 m ) ( !

ΣEx = E2 + E3 = 1.87 × 103 N/C

Thus,

and

ΣEy = −E1 = −144 N/C

giving

E = ! R

( ∑ Ex )2 + ( ∑ Ey ) = 1.88 × 103 N/C 2

and θ = tan−1(ΣEy/ΣEx) = tan−1(−0.076 9) = −4.40° Hence 15.33

 E R = 1.88 × 103 N/C at!4.40° below!the!+x4axis

From the symmetry of the charge distribution, students should recognize

 that the resultant electric field at the center is E R = 0 . !

If one does not recognize this intuitively, consider:     E = E R 1 + E2 + E3 !

k q k q Ex = E1x − E2x = e 2 cos 30° − e 2 cos 30° = 0 r r !

Topic 15

911

and

k q k q k q Ey = E1y + E2y − E3 = e 2 sin 30° + e 2 sin 30° − e 2 = 0 r r r !

Thus, ER = Ex2 + Ey2 = 0 ! 15.34

The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines emerge from q2 as enter q1. |q2| = 3|q1| (a) Then,

q1/q2 = −1/3

(b) q2 > 0 because lines emerge from it, and q1 < 0 because lines terminate on it. 15.35

Note in the sketches below that electric field lines originate on positive charges and terminate on negative charges. The density of lines is twice as great for the − 2q charge in (b) as it is for the 1q charge in (a).

15.36

Rough sketches for these charge configurations are shown below.

Topic 15

912

15.37 (a) The sketch for (a) is shown below. Note that four times as many lines should leave q1 as emerge from q2 although, for clarity, this is not shown in this sketch.

(b) The field pattern looks the same here as that shown for (a) with the exception that the arrows are reversed on the field lines. 15.38 (a) The electric field in the plane of the charges has the general appearance shown below:

Topic 15

913

(b) It is zero at the center of the triangle, where (by symmetry) one can see that the three charges individually produce fields that cancel out. In addition to the center of the triangle, the electric field lines in the second figure below indicate three other points near the middle of each leg of the triangle where E = 0, but they are more difficult to find mathematically.

(c)

Using the sketch below, observe that E1 = E2 = keq/a2 and the resultant

   field at point P is !E = E1 + E2 .

Topic 15

914

Thus, Ex = E1 cos60.0° − E2 cos60.0° = 0

and

2k q Ey = E1 sin 60.0° + E2 sin 60.0° = 2e sin 60.0° a !

The magnitude of the resultant field is then

E = Ex2 + Ey2 = 0 + Ey2 = Ey = !

1.73ke q a2

(d) Since Ex = 0 and Ey > 0, the resultant field at point P is directed vertically upward in the positive y-direction. 15.39

(a) Zero net charge on each surface of the sphere. (b) The negative charge lowered into the sphere repels −5µC on the outside surface, and leaves +5µC on the inside surface of the sphere. (c) The negative charge lowered inside the sphere neutralizes the inner surface, leaving zero charge on the inside. This leaves repels −5µC on the outside surface of the sphere. (d) When the object is removed, the sphere is left with repels −5.00 µC on the outside surface and zero charge on the inside.

15.40

(a) The dome is a closed conducting surface. Therefore, the electric field

Topic 15

915

is zero everywhere inside it. At the surface and outside of this spherically symmetric charge distribution, the field is as if all the charge were concentrated at the center of the sphere. (b) At the surface, 9 2 2 −4 ke q ( 8.99 × 10 N ⋅ m C ) ( 2.0 × 10 C ) E= 2 = = 1.8 × 106 N/C 2 R (1.0 m ) !

(8.99 × 10 N ⋅ m C )( 2.0 × 10 C) = 1.1 × 10 N/C E= 9

−4

( 4.0 m )

! 15.41

For a uniformly charged sphere, the field is strongest at the surface. Thus,

kq Emax = e max , R2 !

6 R2 Emax ( 2.0 m ) ( 3.0 × 10 N C ) = = 1.3 × 10−3 C or qmax = 9 2 2 ke 8.99 × 10 N ⋅ m C ! 2

15.42

If the weight of the drop is balanced by the electric force, then mg = |q|E = eE, so the mass of the drop must be

eE g

(1.6 × 10 C)(3 × 10 N C) ≈ 5 × 10 kg = 4

−19

−16

9.8 m s

Topic 15

916

But m = ρV = ρ(4πr3/3) and the radius of the drop is r = [3m/4πρ]1/3, which gives ⎡ 3 ( 5 × 10−16 kg ) ⎤ r=⎢ 3 ⎥ 4 π 858 kg/m ( ) ⎥⎦ ⎢ ! ⎣

15.43 (a)

1/3

= 5.2 × 10−7 m or r ∼ 1µm

Fe = ma = (1.67 × 10−27 kg)(1.52 × 1012 m/s2) = 2.54 × 10−15 N in the direction of the acceleration, or radially outward.

(b) The direction of the field is the direction of the force on a positive charge (such as the proton). Thus, the field is directed radially outward. The magnitude of the field is

Fe 2.54 × 10−15 N E= = = 1.59 × 10 4 N/C −19 q 1.60 × 10 C ! 15.44

When an electric field of magnitude E is incident on a surface of area A, the flux through the surface is ΦE = EAcosθ

 where θ is the angle between !E and the line normal to the surface. Thus, in this case,

θ = 90.0° − 65.0° = 25.0° and ΦE = (435 N/C)(3.50 m2)cos25.0° = 1.38 × 103 N ⋅ m2/C

Topic 15

917

15.45 The area of the rectangular plane is A = (0.350 m)(0.700 m) = 0.245 m2. (a) When the plane is parallel to the yz-plane, its normal line is parallel to the x-axis and makes an angle θ = 0° with the direction of the field. The flux is then ΦE = EAcosθ = (3.50 × 103 N/C)(0.245 m2)cos 0° = 858 N ⋅ m2/C (b) When the plane is parallel to the x-axis, θ = 90° and ΦE = 0 (c) ΦE = EAcos θ = (3.50 × 103 N/C)(0.245 m2)cos 40.0° = 657 N ⋅ m2/C 15.46 (a) Gauss’s law states that the electric flux through any closed surface equals the net charge enclosed divided by e0. We choose to consider a closed surface in the form of a sphere, centered on the center of the charged sphere and having a radius infinitesimally larger than that of the charged sphere. The electric field then has a uniform magnitude and is perpendicular to our surface at all points on that surface. The flux through the chosen closed surface is therefore ΦE = EA = E(4πr2),

Topic 15

918

and Gauss’s law gives Qinside = e0ΦE = 4πe0Er2 = 4π(8.85 × 10−12 C2/N ⋅ m2)(575 N/C)(0.230 m)2 = 3.38 × 10−9 C = 3.38 nC (b) Since the electric field displays spherical symmetry, you can conclude that the charge distribution generating that field is spherically symmetric. Also, since the electric field lines are directed outward away from the sphere, the sphere must contain a net positive charge. 15.47

From Gauss’s law, the electric flux through any closed surface is equal to the net charge enclosed divided by e0.

Thus, the flux through each surface (with a positive flux coming outward from the enclosed interior and a negative flux going inward toward that

Topic 15

919

interior) is For S1:

ΦE = Qnet/e0 = (+Q − 2Q)/e0 = −Q/e0

For S2:

ΦE = Qnet/e0 = (+Q − Q)/e0 = 0

For S3:

ΦE = Qnet/e0 = (−2Q + Q − Q)/e0 = − 2Q/e0

For S4:

ΦE = Qnet/e0 = (0)/e0 = 0

15.48 (a)

From Gauss’s law, the total flux through this closed surface is

ΦE =

Qenclosed

+5.80 × 10−6 C 8.85 × 10

−12

C N⋅m

= 6.55 × 105 N ⋅ m 2 C

(b) When the charge is located in the center of this tetrahedron, we have total symmetry and the flux is the same through each of the four surfaces. Thus, the flux through any one face is

Φ 6.55 × 105 N ⋅ m 2 C Φ1face = E = = 1.64 × 105 N ⋅ m 2 C 4 4 ! 15.49 (a) For the top surface, the normal direction is upward (out of the closed cylinder), in the direction of the constant electric field so that the angle between the normal and the electric field is θ = 0. Apply the definition of electric flux and substitute values to find:

Topic 15

920

(

)(

Φ E,top = EAcosθ = EA = 3.50 × 103 N/C 2.00 m 2

)

= 7.00 × 103 N ⋅ m 2 /C (b) For the bottom surface, the normal direction is downward (out of the closed cylinder), in the direction opposite the constant electric field so that the angle between the normal and the electric field is θ = 180°. Apply the definition of electric flux and substitute values to find: Φ E,bottom = EAcosθ = EAcos (180° ) = −EA

(

)(

)

= − 3.50 × 103 N/C 2.00 m 2 = −7.00 × 103 N ⋅ m 2 /C

(c) The electric flux is everywhere perpendicular to the cylinder’s lateral surface, so Φ lateral = 0 and the total electric flux is Φ E = Φ E,top + Φ E,bottom + Φ E,lateral = 0 . Apply Gauss’ law to show there is

zero net charge inside the closed cylinder:

ΦE = =

Qinside = 0 → Qinside = 0 É0

15.50 (a) The surface charge density on the disk is σ = q/A where A = πR2 is the disk’s surface area. Substitute values to find

σ=

q q 2.00 × 10−9 C = = = 1.59 × 10−8 C/m 2 2 2 A πR π ( 0.200 m )

Topic 15

921

(b) Place a Gaussian cylinder of cross-sectional area A0 at the center of the disk. Assuming the Gaussian cylinder is small compared to the charged disk so that edge effects can be neglected, Gauss’ law gives

Φ E = Φ E,top + Φ E,bottom + Φ E,lateral = =

Qinside É0

Recognize that Φ E,top = Φ E,bottom = EA0 and Φ lateral = 0 , and that Qinside =

σA0. Substitute values into Gauss’ law to find Φ E = 2Φ E,top = 2EA0 = E=

= 15.51

σ A0 É0

σ 1.59 × 10−8 C/m 2 = 2É 0 2 8.85 × 10−12 C 2 / N ⋅ m 2

(

))

= 898 N/C

We choose a spherical Gaussian surface, concentric with the charged spherical shell and of radius r. Then, ΣEA cosθ = E(4πr2)cos0° = 4πr2E. (a)

For r > a (that is, outside the shell), the total charge enclosed by the Gaussian surface is Q = + q − q = 0. Thus, Gauss’s law gives 4πr2E = 0, or E = 0.

(b) Inside the shell, r < a, and the enclosed charge is Q = +q.

Therefore, from Gauss’s law, 4π r E =

or E =

q 4π! 0 r

= 2

ke q r2

 2 The field for r < a is E = ke q r directed!radially!outward . ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 15

922

15.52 (a)

The surface of the cube is a closed surface which surrounds a total charge of Q = 1.70 × 102 µC. Thus, by Gauss’s law, the electric flux through the full surface of the cube is

ΦE =

Qinside

1.70 × 10−4 C 8.85 × 10

−12

C N⋅m

= 1.92 × 107 N ⋅ m 2 C

(b) Since the charge is located at the center of the cube, the six faces of the cube are symmetrically positioned around the location of the charge. Thus, one-sixth of the flux passes through each of the faces, or

Φ 1.92 × 107 N ⋅ m 2 C Φ face = E = = 3.20 × 106 N ⋅ m 2 C 6 6 ! (c)

The answer to part (b) would change because the charge could now be at different distances from each face of the cube, but the answer to part (a) would be unchanged because the flux through the entire closed surface depends only on the total charge inside the surface.

15.53

Note that with the point charge −2.00 nC positioned at the center of the spherical shell, we have complete spherical symmetry in this situation. Thus, we can expect the distribution of charge on the shell, as well as the electric fields both inside and outside of the shell, to also be spherically

Topic 15

923

symmetric. (a)

We choose a spherical Gaussian surface, centered on the center of the conducting shell, with radius r = 1.50 m < a as shown below. Gauss’s law gives

(

)

Φ E = EA = E 4π r 2 =

Qinside

or E =

Qinside

keQcenter

4π! 0 r

= 2

(8.99 × 10 N ⋅ m C )( −2.00 × 10 C) E= 9

−9

(1.50 m )

and E = −7.99 N/C. The negative sign means that the field is radial inward.

(b) All points at r = 2.20 m are in the range a < r < b, and hence are located within the conducting material making up the shell. Under conditions of electrostatic equilibrium, the field is E = 0 at all points inside a conducting material.

Topic 15

(c)

924

If the radius of our Gaussian surface is r = 2.50 m > b, Gauss’s law (with total spherical symmetry) leads to E =

Qinside 4π! 0 r

keQinside r2

just as

in part (a). However, now Qinside = Qshell + Qcenter = +3.00 nC − 2.00 nC = +1.00 nC. Thus, we have

(8.99 × 10 N ⋅ m C )( +1.00 × 10 C) = +1.44 N C E= 9

−9

( 2.50 m )

with the positive sign telling us that the field is radial outward at this location. (d) Under conditions of electrostatic equilibrium, all excess charge on a conductor resides entirely on its surface. Thus, the sum of the charge on the inner surface of the shell and that on the outer surface of the shell is Qshell = +3.00 nC. To see how much of this is on the inner surface, consider our Gaussian surface to have a radius r that is infinitesimally larger than a. Then, all points on the Gaussian surface lie within the conducting material, meaning that E = 0 at all points and the total flux through the surface is ΦE = 0. Gauss’s law then states that Qinside = Qinner + Qcenter = 0 , or surface !

Topic 15

925

Qinner = −Qcenter = − ( −2.00 nC ) = +2.00 nC ! surface The charge on the outer surface must be

Qouter = Qshell − Qinner = 3.00 nC − 2.00 nC = +1.00 nC surface ! surface 15.54

Please review Example 15.8 in your textbook. There it is shown that the electric field due to a nonconducting plane sheet of charge parallel to the xy-plane has a constant magnitude given by Ez = |σsheet|/2e0, where σsheet is the uniform charge per unit area on the sheet. This field is everywhere perpendicular to the xy-plane, is directed away from the sheet if it has a positive charge density, and is directed toward the sheet if it has a negative charge density. In this problem, we have two plane sheets of charge, both parallel to the xy-plane and separated by a distance of 2.00 cm. The upper sheet has charge density σsheet = −2σ, while the lower sheet has σsheet = +σ. Taking upward as the positive z-direction, the fields due to each of the sheets in the three regions of interest are:

Topic 15

926

Lower sheet (at z = 0)

Upper sheet (at z = 2.00 cm)

Region

z<0

0 < z < 2.00 cm

z > 2.00 cm

Electric Field

Ez = −

Ez = +

+σ 2! 0

Electric Field

=−

σ 2! 0

Ez = +

σ 2! 0

Ez = +

σ 2! 0

Ez = −

−2σ 2! 0

σ !0

=−

σ !0

When both plane sheets of charge are present, the resultant electric field in each region is the vector sum of the fields due to the individual sheets for that region.

(a) For z < 0:

Ez = Ez ,lower + Ez ,upper = −

σ σ σ + = + 2! 0 ! 0 2! 0

Topic 15

15.55

927

(b) For 0 < z < 2.00 cm:

Ez = Ez ,lower + Ez ,upper = +

σ σ 3σ + = + 2! 0 ! 0 2! 0

Ez = Ez ,lower + Ez ,upper = +

σ σ σ − = − 2! 0 ! 0 2! 0

The radius of each sphere is small in comparison to the distance to the nearest neighboring charge (the other sphere). Thus, we shall assume that the charge is uniformly distributed over the surface of each sphere and, in its interaction with the other charge, treat it as though it were a point charge. In this model, we then have two identical point charges, of magnitude 35.0 mC, separated by a total distance of 310 m (the length of the cord plus the radius of each sphere). Each of these charges repels the other with a force of magnitude 35.0 × 10−3 C ) Q2 9 2 2 ( Fe = ke 2 = ( 8.99 × 10 N ⋅ m C ) 2 = 115 N r 3.10 × 102 m ) ( ! 2

Thus, to counterbalance this repulsion and hold each sphere in equilibrium, the cord must have a tension of 115 N, so it will exert a 115 N on that sphere, directed toward the other sphere. 15.56 (a)

As shown in Example 15.8 in the textbook, the electric field due to a nonconducting plane sheet of charge has a constant magnitude of E =

Topic 15

928

σ/2e0, where σ is the uniform charge per unit area on the sheet. The direction of the field at all locations is perpendicular to the plane sheet and directed away from the sheet if σ is positive, and toward the sheet if σ is negative. Thus, if σ = +5.20 µC/m2, the magnitude of the electric field at all distances greater than zero from the plane (including the distance of 8.70 cm) is

+5.20 × 10−6 C m 2 σ = = 2.94 × 105 N C −12 2 2 2! 0 2 8.85 × 10 C N ⋅ m

(

)

(b) The field does not vary with distance as long as the distance is small compared with the dimensions of the sheet. 15.57

The three contributions to the resultant electric field at the point of interest are shown in the sketch below.

The magnitude of the resultant field is ER = −E1 + E2 + E3

⎡ q ke q ke q ke q q q ⎤ ER = 2 1 + 2 2 + 2 3 = ke ⎢ − 21 + 22 + 23 ⎥ r1 r2 r3 r2 r3 ⎦ ⎣ r1 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 15

929

⎛ N ⋅ m 2 ⎞ ⎡ 4.0 × 10−9 C 5.0 × 10−9 C 3.0 × 10−9 C ⎤ ER = ⎜ 8.99 × 109 = ⎢− + + ⎥ 2 2 2 ⎝ C2 ⎟⎠ ⎢⎣ ( 2.5 m ) 2.0 m ) 1.2 m ) ⎥⎦ ( ( !

 ER = + 24 N/C or E R = 24 N/C!in!the! + x,direction ! 15.58

Consider the force diagram shown below.

mg ΣFy = 0 ⇒ T cos θ = mg or T = cos θ ! ⎛ mg ⎞ ∑ Fx = 0 ⇒ Fe = Tsin θ = ⎜ sin θ = mg tan θ ⎝ cos θ ⎟⎠ !

Since Fe = qE, we have qE = mg tan θ, or q = mg tan θ/E

( 2.00 × 10 kg )( 9.80 m s ) tan 15.0° = +5.25 × 10 C = +5.25 µC −3

q= !

−6

1.00 × 10 N C 3

15.59 (a) Apply Gauss’ to the top and bottom plates to find the electric field E in the region between the plates:

(

)(

(

))

E = σ /É 0 = 4.20 × 10−9 C/m 2 / 8.854 × 10−12 C 2 / N ⋅ m 2 = 475 N/C =

Topic 15

930

(b) The electric force on the proton has magnitude

(

)

F = eE = 1.60 × 10−19 C (σ /É 0 ) = 7.59 × 10−17 N

(c) The proton’s horizontal acceleration is ax = 0 so vx = v0 = 1.50 × 106 m/s and it spends a time Δt = d/v0 between the plates. The electric force is directed down so that ay = −F/mp (where mp is the proton’s mass) is constant. Apply kinematics to find ⎛ eE ⎞ ⎛ d ⎞ Δy = v0 y Δt + ay ( Δt ) = ⎜ − ⎟ ⎜ ⎟ ⎝ mp ⎠ ⎝ v0 ⎠ 1 2

(

1 2

)

⎛ 1.60 × 10−19 C ( 474 N/C ) ⎞ ⎛ 2.00 × 10−2 m ⎞ 2 = 12 ⎜ − ⎟⎜ 6 ⎟ 1.67 × 10−27 kg ⎝ ⎠ ⎝ 1.50 × 10 m/s ⎠ = −4.05 × 10−6 m

15.60

It is desired that the electric field exert a retarding force on the electrons, slowing them down and bringing them to rest. For the retarding force to have maximum effect, it should be anti-parallel to the direction of the electron’s motion. Since the force an electric field exerts on negatively charged particles (such as electrons) is in the direction opposite to the field, the electric field should be in the direction of the electron’s motion. The work a retarding force of magnitude Fe = |q|E = eE does on the electrons as they move distance d is W = Fed cos180° = −Fed = − eEd. The

Topic 15

931

work-energy theorem (W = ΔKE) then gives −eEd = KEf − KEi = 0 − K and the magnitude of the electric field required to stop the electrons in distance d is −K E= −ed !

15.61

E = K/ed

Consider the sketch below and observe:

⎛ d⎞ θ = tan −1 ⎜ ⎟ = 45.0° ⎝ d⎠ ! and

2 2 !r = d + d = d 2

Thus,

Fx = F1x + F2x = +

k Q2 ⎛ 2 ⎞ sin 45.0° + 0 = + e 2 ⎜ ⎟ 2d ⎜⎝ 2 ⎟⎠

k eQ 2 ⎛ k eQ ⎞ ⎜ 2 ⎟ = +0.354 2 4 ⎝ d ⎠ d 2

ke −Q Q

Topic 15

15.62

932

and

k (Q ) ( 2Q ) k −Q Q keQ 2 ⎛ 2 ⎞ 2keQ 2 Fy = F1y + F2y = − e 2 cos 45.0° + e = − + r d2 2d 2 ⎜⎝ 2 ⎟⎠ d2 !

⎛ 2 ⎞ keQ 2 keQ 2 Fy = ⎜ 2 − = +1.65 4 ⎟⎠ d 2 d2 ⎝ !

(a) The downward electrical force acting on the ball is Fe = qE = (2.00 × 10−6 C)(1.00 × 105 N/C) = 0.200 N The total downward force acting on the ball is then

F = Fe + mg = 0.200 N + (1.00 × 10−3 kg ) ( 9.80 m s 2 ) = 0.210 N ! Thus, the ball will behave as if it were in a modified gravitational field where the effective free-fall acceleration is

F 0.210 N " g" = = = 210 m s 2 −3 m 1.00 × 10 kg ! The period of the pendulum will be L 0.500 m T = 2π = 2π = 0.307 s # g# 210 m s 2 !

(b) Yes. The force of gravity is a significant portion of the total downward force acting on the ball. Without gravity, the effective acceleration would be

Topic 15

933

F 0.200 N " g" = e = = 200 m s 2 −3 m 1.00 × 10 kg ! 0.500 m = 0.314 s giving T = 2π 200 m/s 2 !

a 2.28% difference from the correct value with gravity included. 15.63

The sketch below gives a force diagram of the positively charged sphere. Here, F1 = ke|q|2/r2 is the attractive force exerted by the negatively charged sphere, and F2 = qE is exerted by the electric field.

mg ΣFy = 0 ⇒ T cos10° = mg or T = cos 10° ! 2

k q ΣFx = 0 ⇒ F2 = F1 + T sin 10° or qE = e 2 + mg tan 10° r ! At equilibrium, the distance between the two spheres is r = 2(L sin10°). Thus,

Topic 15

934

ke q mg tan 10° + 4 ( Lsin 10° ) q

(8.99 × 10 N ⋅ m C )( 5.0 × 10 C) + ( 2.0 × 10 kg )( 9.80 m s ) tan 10° = ( 5.0 × 10 C) 4 ⎡⎣( 0.100 m ) sin 10° ⎤⎦ ! 9

−8

−3

−8

or the needed electric field strength is E = 4.4 × 105 N/C 15.64 (a)

At any point on the x-axis in the range 0 < x < 1.00 m, the contributions made to the resultant electric field by the two charges are both in the positive x-direction. Thus, it is not possible for these contributions to cancel each other and yield a zero field.

(b) Any point on the x-axis in the range x < 0 is located closer to the larger magnitude charge (q = 5.00 µC) than the smaller magnitude charge (|q| = 4.00 µC). Thus, the contribution to the resultant electric field by the larger charge will always have a greater magnitude than the contribution made by the smaller charge. It is not possible for these contributions to cancel to give a zero resultant field. (c)

If a point is on the x-axis in the region x > 1.00 m, the contributions made by the two charges are in opposite directions. Also, a point in this region is closer to the smaller magnitude charge than it is to the larger charge. Thus, there is a location in this region where the contributions of these charges to the total field will have equal

Topic 15

935

magnitudes and cancel each other. (d) When the contributions by the two charges cancel each other, their magnitudes must be equal. That is,

ke !

( 5.00 µC) = k ( 4.00 µC ) !or!x − 1.00 m = +x 4/5 x ( x − 1.00 m ) 2

Thus, the resultant field is zero at

15.65

1.00 m x= = +9.47 m . ! 1− 4 / 5

The two spheres have charges q1 and q2 = 2q1, so the repulsive force that one exerts on the other has magnitude.!Fe = ke q1q2 r = 2ke q1 r 2

From Figure P15.59 in the textbook, observe that the distance separating the two spheres is r = 3.00 cm + 2[(5.00 cm)sin10.0°] = 4.74 cm = 0.047 4 m From the force diagram of one sphere given above, observe that ΣFy 0 ⇒ T cos10.0° = mg or T = mg/cos10.0°

Topic 15

936

⎛ mg ⎞ ΣFx = 0 ⇒ Fe = T sin 10.0° = ⎜ sin 10.0° = mg tan 10.0° ⎝ cos 10.0° ⎟⎠ !

and

2 2 !2ke q1 r = mg! tan 10.0°

Thus,

mgr 2 tan 10.0°

q1 =

giving

2ke

(0.015 0 kg )(9.80 m s )(0.047 4 m) tan 10.0° = 2 ( 8.99 × 10 N ⋅ m C ) 2

q1 = 5.69 × 10−8 C as the charge on one sphere,

and q2 = 2q1 = 1.14 × 10−7 C as the charge on the other sphere. 15.66

Consider the sketch below. The electrical forces acting on the third charge, Q, have magnitudes

k q Q 3k qQ F1 = e 12 = e 2 x x !

and

kqQ k qQ F2 = e 2 2 = e 2 (d − x) (d − x) !

The bead with charge Q is in equilibrium when F1 = F2, or 3 ke qQ ! x

ke qQ (d − x)2

Topic 15

937

giving

and

15.67

3(d − x)2 = x2

! 3 (d − x) = x

3 (1.50 m ) 3d x= = = 0.951 m 1+ 3 ! 1+ 3

Because of the spherical symmetry of the charge distribution, any electric field present will be radial in direction. If a field does exist at distance R from the center, it is the same as if the net charge located within r ≤ R were concentrated as a point charge at the center of the inner sphere. Charge located at r > R does not contribute to the field at r = R.

(a) At r = 1.00 cm, E = 0 since static electric fields cannot exist within conducting materials. (b) The net charge located at r ≤ 3.00 cm is Q = +8.00 µC. Thus, at r = 3.00 cm,

Topic 15

938

E= = !

keQ r2 (8.99 × 109 N ⋅ m 2 C2 )(8.00 × 10−6 C)

( 3.00 × 10 m ) −2

= 7.99 × 107 N / C (outward)

(c) At r = 4.50 cm, E = 0, since this is located within conducting materials. (d) The net charge located at r ≤ 7.00 cm is Q = +4.00 µC. Thus, at r = 7.00 cm, E= = !

15.68

keQ r2 (8.99 × 109 N ⋅ m 2 C2 )( 4.00 × 10−6 C)

(7.00 × 10 m ) −2

= 7.34 × 106 N/C (outward)

Consider the free-body diagram of the rightmost charge given below.

ΣFy = 0 ⇒ T cos θ = mg or T = mg/cos θ and ΣFx = 0 ⇒ Fe = Tsin θ = (mg/cos θ)sinθ = mg tan θ

But,

Thus,

k q2 k q2 ke q 2 ke q 2 5ke q2 Fe = e 2 + e 2 = + = r1 r2 ( Lsin θ )2 ( 2Lsin θ )2 4L2 sin 2 θ !

5ke q2 = mg tan θ 4L2 sin 2 θ !

4L2mgsin 2 θ tan θ 5ke

Topic 15

939

If θ = 45.0°, m = 0.100 kg, and L = 0.300 m, then

4 ( 0.300 m ) ( 0.100 kg ) ( 9.80 m s 2 ) sin 2 ( 45.0° ) tan ( 45.0° ) 2

q= ! or 15.69 (a)

5 ( 8.99 × 109 N ⋅ m 2 C2 )

q = 1.98 × 10−6 C = 1.98 µC When an electron (negative charge) moves distance Δx in the direction of an electric field, the work done on it is W = Fe(Δx)cos θ = eE(Δx)cos180° = −eE(Δx) From the work-energy theorem (Wnet = KEf − KEi) with KEf = 0, we have −eE(Δx) = −KEi,

KEi 1.60 × 10−17 J E= = = 1.00 × 103 N/C −19 e ( Δx ) (1.60 × 10 C ) ( 0.100 m ) !

(b) The magnitude of the retarding force acting on the electron is Fe = eE, and Newton’s -second law gives the acceleration as a = −Fe/m = −eE/m. Thus, the time required to bring the electron to rest is

t= !

2m ( KEi ) v − v0 0 − 2 ( KEi ) m = = a −eE m eE

Topic 15

940

(c)

t= !

2 ( 9.11 × 10−31 kg ) (1.60 × 10−17 J )

(1.60 × 10

−19

C ) (1.00 × 10 N C ) 3

= 3.37 × 10−8 s = 33.7 ns

After bringing the electron to rest, the electric force continues to act on it, causing the electron to accelerate in the direction opposite to the field at a rate of −19 3 eE (1.60 × 10 C ) (1.00 × 10 N C ) a= = = 1.76 × 1014 m s 2 −31 m 9.11 × 10 kg !

15.70

(a) The acceleration of the protons is downward (in the direction of the field) and

ay = !

−19 Fe eE (1.60 × 10 C ) (720 N C ) = = = 6.90 × 1010 m s 2 −27 m m 1.67 × 10 kg

The time of flight for the proton is twice the time required to reach the peak of the arc, or

⎛ v ⎞ 2v sin θ 0y t = 2tpeak = 2 ⎜ ⎟= 0 ⎜⎝ ay ⎟⎠ ay !

The horizontal distance traveled in this time is

Topic 15

941

⎛ 2v sin θ ⎞ v 2 sin 2θ R = v0xt = ( v0 cos θ ) ⎜ 0 ⎟= 0 ⎜⎝ ⎟⎠ ay ay ! Thus, if R = 1.27 × 10−3, we must have

(6.90 × 10 m s )(1.27 × 10 m) = 0.961 sin 2θ = = v (9 550 m s ) ay R

2 0

−3

giving 2θ = 73.9° or 2θ = 180° − 73.9° = 106°. Hence, θ = 37.0° or 53.0° (b) The time of flight for each possible angle of projection is:

For θ = 37.0°:

For θ = 53.0°:

2v0 sin θ ay

(

)

2 9 550 m s sin 37.0° 10

6.90 × 10 m s

(

)

2 9 550 m s sin 53.0° 10

6.90 × 10 m s

= 1.67 × 10−7 s

= 2.21 × 10−7 s

Topic 16

942

Topic 16 Electrical Energy and Capacitance

QUICK QUIZZES 16.1

Choice (b). The field exerts a force on the electron, causing it to accelerate in the direction opposite to that of the field. In this process, electrical potential energy is converted into kinetic energy of the electron. Note that the electron moves to a region of higher potential, but because the electron has negative charge this corresponds to a decrease in the potential energy of the electron.

16.2

Choice (a). The electron, a negatively charged particle, will move toward the region of higher electric potential. Because of the electron’s negative charge, this corresponds to a decrease in electrical potential energy.

16.3

Choice (b). Charged particles always tend to move toward positions of lower potential energy. The electrical potential energy of a charged particle is PE = qV and, for positively-charged particles, this decreases as V decreases. Thus, a positively-charged particle located at x = A would move toward the left.

Topic 16

16.4

943

Choice (d). For a negatively-charged particle, the potential energy (PE = qV) decreases as V increases. A negatively charged particle would oscillate around x = B which is a position of minimum potential energy for negative charges.

16.5

Choice (d). If the potential is zero at a point located a finite distance from charges, negative charges must be present in the region to make negative contributions to the potential and cancel positive contributions made by positive charges in the region.

16.6

Choice (c). Both the electric potential and the magnitude of the electric field decrease as the distance from the charged particle increases. However, the electric flux through the balloon does not change because it is proportional to the total charge enclosed by the balloon, which does not change as the balloon increases in size.

16.7

Choice (a). From the conservation of energy, the final kinetic energy of either particle will be given by KEf = KEi + (PEi − PEf) = 0 + qVi − qVf = −q(Vf − Vi) = −q(ΔV) For the electron, q = −e and ΔV = +1 V giving KEf = −(−e)(+1 V) = +1 eV. For the proton, q = +e and ΔV = −1 V, so KEf = −(e)(−1 V) = +1 eV, the same as that of the electron.

Topic 16

16.8

944

Choice (c). The battery moves negative charge from one plate and puts it on the other. The first plate is left with excess positive charge whose magnitude equals that of the negative charge moved to the other plate.

16.9

(a)

C decreases.

(c)

E stays the same.

(d)

ΔV increases.

(b)

Q stays the same.

(e)

The energy stored increases.

Because the capacitor is removed from the battery, charges on the plates have nowhere to go. Thus, the charge on the capacitor plates remains the same as the plates are pulled apart. Because E = σ/e0 = (Q/A)/e0, the electric field is constant as the plates are separated. Because ΔV = Ed and E does not change, ΔV increases as d increases. Because the same charge is stored at a higher potential difference, the capacitance (C = Q/ΔV) has decreased. Because energy stored = Q2/2C and Q stays the same while C decreases, the energy stored increases. The extra energy must have been transferred from somewhere, so work was done. This is consistent with the fact that the plates attract one another, and work must be done to pull them apart. 16.10

(a)

C increases.

(c)

E stays the same.

(b)

Q increases.

Topic 16

945

(d)

ΔV remains the same.

(e)

The energy stored increases.

The presence of a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant. Since the battery holds the potential difference constant while the capacitance increases, the charge stored (Q = CΔV) will increase. Because the potential difference and the distance between the plates are both constant, the electric field (E = ΔV/d) will stay the same. The battery maintains a constant potential difference. With ΔV constant while capacitance increases, the stored energy

⎡ energy!stored = 12 C ( ΔV )2 ⎤ will increase. ⎦ !⎣ 16.11 Choice (a). Increased random motions associated with an increase in temperature make it more difficult to maintain a high degree of polarization of the dielectric material. This has the effect of decreasing the dielectric constant of the material, and in turn, decreasing the capacitance of the capacitor.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 16.2

(a) I. By conservation of energy, − eΔV + ΔKE = 0 . ΔKE > 0 so −eΔV < 0 and ΔV > 0.

Topic 16

946

(b) I. ΔPE = −eΔV so ΔPE is proportional to ΔV. (c) U. Total energy is conserved. 16.4

The potential energy between a pair of point charges separated by distance R is PE = keq1q2/R. Thus, the potential energy for each of the four systems is:

(a)

PEa = ke !

Q ( 2Q ) Q2 = 2ke r r

(b)

PEb = ke !

(−Q)(−Q) Q2 = ke r r

(c)

PEc = ke !

Q(−Q) 1 Q2 = − ke 2r 2 r

(d)

PEd = ke !

(−Q)(−2Q) Q2 = ke 2r r

Therefore, the correct ranking from largest to smallest is (a) > (b) = (d) > (c). 16.6

(a) Capacitance is proportional to the dielectric constant, so Cnew/C0 = 2. (b) The potential difference is inversely proportional to the dielectric constant, so ΔVnew/ΔV0 = 1/2. (c) The energy stored in a capacitor is inversely proportional to the dielectric constant, so PEC,new/PEC,0 = 1/2.

16.8

A sharp point on a charged conductor would produce a large electric field in the region near the point. An electric discharge could most easily take place at the point.

Topic 16

16.10

947

There are eight different combinations that use all three capacitors in the circuit. These combinations and their equivalent capacitances are:

⎛ 1 1 1⎞ + ⎟ All three capacitors in series: Ceq = ⎜ + ⎝ C1 C2 C3 ⎠ ! All three capacitors in parallel:

−1

Ceq = C1 + C2 + C3

One capacitor in series with a parallel combination of the other two: −1

−1

⎛ 1 ⎛ 1 ⎛ 1 1⎞ 1⎞ 1⎞ Ceq = ⎜ + ⎟ ,!Ceq = ⎜ + ⎟ ,!Ceq = ⎜ + ⎟ ⎝ C1 + C2 C3 ⎠ ⎝ C3 + C1 C2 ⎠ ⎝ C2 + C3 C1 ⎠ !

−1

One capacitor in parallel with a series combination of the other two: ⎛ CC ⎞ ⎛ CC ⎞ ⎛ CC ⎞ Ceq = ⎜ 1 2 ⎟ + C3 ,!Ceq = ⎜ 3 1 ⎟ + C2 ,!Ceq = ⎜ 2 3 ⎟ + C1 ⎝ C1 + C2 ⎠ ⎝ C3 + C1 ⎠ ⎝ C2 + C3 ⎠ !

16.12

(a) If the wires are disconnected from the battery and not allowed to touch each other or another object, the charge on the plates is unchanged. (b) If, after being disconnected from the battery, the wires are connected to each other, electrons will rapidly flow from the negatively charged plate to the positively charged plate to leave the capacitor uncharged with both plates neutral.

16.14

The primary choice would be the dielectric. You would want to choose a

Topic 16

948

dielectric that has a large dielectric constant and dielectric strength, such as strontium titanate, where k ≈ 233 (Table 16.1). A convenient choice could be thick plastic or Mylar. Secondly, geometry would be a factor. To maximize capacitance, one would want the individual plates as close as possible, since the capacitance is proportional to the inverse of the plate separation—hence the need for a dielectric with a high dielectric strength. Also, one would want to build, instead of a single parallel-plate capacitor, several capacitors in parallel. This could be achieved through “stacking” the plates of the capacitor. For example, you can alternately lay down sheets of a conducting material, such as aluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their nearest neighbors, connect every other plate together as illustrated in the figure below.

This technique is often used when “home-brewing” signal capacitors for radio applications, as they can withstand huge potential differences without flashover (without either discharge between plates around the

Topic 16

949

dielectric or dielectric breakdown). One variation on this technique is to sandwich together flexible materials such as aluminum roof flashing and thick plastic, so the whole product can be rolled up into a “capacitor burrito” and placed in an insulating tube, such as a PVC pipe, and then filled with motor oil (again to prevent flashover). 16.16

The material of the dielectric may be able to withstand a larger electric field than air can withstand before breaking down to pass a spark between the capacitor plates.

ANSWERS TO EVEN NUMBERED PROBLEMS 16.2

(a)

6.16 × 10−17 N

(b)

3.69 × 1010 m/s2 in the direction of the electric field

(c)

7.38 cm

16.4

1.67 × 106 N/C

16.6

(a)

1.10 × 10−2 N to the right

(b)

1.98 × 10−3 J

(c)

−1.98 × 10−3 J

Topic 16

16.8

950

(d)

−49.5 V

(a)

−2.31 kV

(b)

Protons would require a greater potential difference.

(c)

ΔVp/ΔVe = −mp/me

16.10

40.2 kV

16.12

(a)

16.14

−9.08 J

16.16

(a)

16.18

(b)

+10.8 kV

See Solution.

(b)

3keq/a

(a)

See Solution.

(b)

⎛ 1 2⎞ V = ( 22.5 V ⋅ m ) ⎜ − ⎟ ⎝ 1.20 m − x x ⎠ !

(c)

−37.5 V

(d)

x = 0.800 m

16.20 (a)

+5.39 kV

(c)

See Solution.

Conservation of energy alone yields one equation with two unknowns.

(b)

Conservation of linear momentum

(c)

vp = 1.05 × 107 m/s, vα = 2.64 × 106 m/s

16.22

5.4 × 105 V

16.24

(a) !V = 4 2keQ/a

(b)

!W = 4 2ke qQ/a

Topic 16

951

16.26

(a)

−2.50 × 102 V

(b)

−2.40 × 10−16 J

16.28

(a)

2.19 × 106 m/s

(b)

5.29 × 10−11 m

16.30

(a)

3.00 µF

(b)

36.0 µC

16.32

(a)

5.31 × 10−9 C

(b)

7.08 × 10−5 V

16.34

31.0 Å

16.36

1.23 kV

16.38

(a)

17.0 µF

(b)

9.00 V

(c)

45.0 µC on C1, 108 µC on C2

16.40

3.00 pF and 6.00 pF

16.42

(a)

6.00 µF

(d)

Q4 = 144 µC, Q2 = 72.0 µC, Qrightmost = 216 µC

(e)

Q24 = Q8 = 216 µC

16.44

(b)

12.0 µF

(c)

432 µC

branch

(f) 9.00 V

(g) 27.0 V

(a)

(b)

(c)

ΔV1 > ΔV2 = ΔV3

(d)

Q1 and Q3 increase, Q2 decreases

Q1 > Q3 > Q2

Topic 16

16.46

952

(a)

6.04 µF

16.48 (a)

5.96 µF

(b)

83.6 µC

(b) 87.4 µC on the 20.0 µF capacitor, 63.0 µC on the 6.00 µF capacitor, 26.3 µC on the 15.0 µF capacitor, and 26.3 µC on the 3.00 µF capacitor 16.50 (a) Ceq = 12.0 µF, Estored,total = 8.64 × 10−4 J (b) Estored,1 = 5.76 × 10−4 J, Estored,2 = 2.88 × 10−4 J It will always be true that Estored,1 + Estored,2 = Estored,total (c) 5.66 V; C2, with the largest capacitance, stores the most energy. 16.52 (a) 12.0 V

(b)

3.60 × 10−4 J

16.54

(a)

13.3 nC

(b)

272 nC

16.56

(a)

3.10 × 10−9 F

(b)

4.88 × 10−9 F

16.58

0.443 mm

16.60

(a)

13.5 mJ

(b)

Estored,2 = 3.60 mJ, Estored,3 = 5.40 mJ, Estored,4 = 1.80 mJ, Estored,6 = 2.70 mJ

(c)

The energy stored in the equivalent capacitance equals the sum of

(c)

7.28 × 10−9 F

the energies stored in the individual capacitors.

Topic 16

953

16.62

1 1 2 1 1 2 C1 = Cp ± Cp − CpCs ,!C2 = Cp  Cp − CpCs 2 4 2 4 !

16.64

(a)

1.8 × 104 V

(c)

−1.8 × 104 V

(d)

−5.4 × 10−2 J

16.66

(a)

ab C= ke (b − a) !

16.68

k = 2.33

16.70

(a)

2ke q !d 5

(b)

−3.6 × 104 V

(b)

See Solution.

(b)

4ke q2 !d 5

(b)

4.4 mm

(c)

4ke q2 !d 5

8ke q2 (d) ! md 5 16.72

(a)

0.1 mm

PROBLEM SOLUTIONS 16.1

(a) Because the electron has a negative charge, it experiences a force in the direction opposite to the field and, when released from rest, will move in the negative x-direction. The work done on the electron by the field is

Topic 16

954

W = Fx(Δx) = (qEx) Δx = (−1.60 × 10−19 C)(375 N/C)(−3.20 × 10−2 m) = 1.92 × 10−18 J (b) The change in the electric potential energy is the negative of the work done on the particle by the field. Thus, ΔPE = −W = −1.92 × 10−18 J (c) Since the Coulomb force is a conservative force, conservation of energy gives ΔKE + ΔPE = 0, or !KE f = 12 me v f = KEi − ΔPE = 0 − ΔPE , 2

and

vf =

16.2

( ) = −2(−1.92 × 10 J) = 2.05 × 10 m/s in the − x-direction −18

−2 ΔPE me

9.11 × 10

−31

(a) F = qE = (1.60 × 10−19 C)(385 N/C) = 6.16 × 10−17 N

(b)

F mp

6.16 × 10−17 N 1.67 × 10−27 kg

= 3.69 × 1010 m/s 2 in the direction of the electric field 2 1 1 Δx = v0t + at 2 = 0 + ( 3.69 × 1010 m/s ) ( 2.00 × 10−6 s ) 2 2 (c) −2 = 7.38 × 10 m = 7.38 cm !

16.3

The work done by the agent moving the charge out of the cell is Winput = −Wfield = −(−ΔPEe) = +q(ΔV)

Topic 16

955

= (1.60 × 10−19 C)(+90 × 10−3 J/C) = 1.4 × 10−20 J 16.4

Solve for the electric field magnitude using the relationship between potential difference and a 1-D, constant, electric field:

ΔV 25.0 × 103 V ΔV = −Ex Δx → Ex = − = = 1.67 × 106 N/C −2 Δx 1.50 × 10 m 16.5

(a) The change in the proton’s kinetic energy is ΔKE = KEf − KEi where KEi = 0 because the proton starts from rest so that vi = 0. Substitute values to find

(

)(

)

ΔKE = KE f = 12 mp v 2f = 12 1.67 × 10−27 kg 1.50 × 105 m/s = 1.88 × 10−17 J (b) Apply either conservation of energy (ΔKE + ΔPE = 0) or the workkinetic energy theorem (Wnet = −ΔPE = ΔKE) and substitute values to find:

ΔPE = −ΔKE = −1.88 × 10−17 J (c) Calculate the electric field magnitude from the work done on the proton: 2

W = Fe Δx = eEx Δx →

1 W W 2 mp v f Ex = = net = eΔx eΔx eΔx

(1.67 × 10 kg)(1.50 × 10 m/s) = 58.7 N/C E = (1.60 × 10 C )( 2.00 m) 1 2

−27

−19

Topic 16

16.6

956

(a)

(

)(

! ! F = qE = +40.0 × 10−6 C +275 N/C

)

= 1.10 × 10−2 N directed toward the right

(b) WAB = F(Δx)cos θ = (1.10 × 10−2 N)(0.180 m)cos 0° = 1.98 × 10−3 J (c) ΔPE = −WAB = −1.98 × 10−3 J

(d) ΔV = VB − VA = ! 16.7

ΔPE −1.98 × 10−3 J = = −49.5 V q +40.0 × 10−6 C

ΔV 600 J/C = = 1.13 × 105 N/C (a) E = −3 d 5.33 × 10 m !

(b) F = q E = !

−19 q ΔV (1.60 × 10 C ) ( 600 J/C ) = = 1.80 × 10−14 N d 5.33 × 10−3 m

(a) Using conservation of energy, ΔKE + ΔPE = 0, with KEf = 0 since the particle is “stopped,” we have 2 1 1 ⎛ ⎞ ΔPE = −ΔKE = − ⎜ 0 − me vi2 ⎟ = + ( 9.11 × 10−31 kg ) ( 2.85 × 107 m/s ) ⎝ ⎠ 2 2

= +3.70 × 10−16 J

The required stopping potential is then

ΔPE +3.70 × 10−16 J ΔV = = −2.31 × 103 V = −2.31 kV −19 q −1.60 × 10 C ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

957

(b) Being more massive than electrons, protons traveling at the same initial speed will have more initial kinetic energy and require a greater magnitude stopping potential. (c) Since ΔVstopping = ΔPE/q = (−ΔKE)/q = (−mv2/2)/q, the ratio of the stopping potential for a proton to that for an electron having the same initial speed is ΔVp

ΔVe

16.9

−mp vi2 2(+e)

( )

− me vi2 2 −e

= −mp me

The mass of an ionized oxygen molecule is mO2 = 5.32 × 10−26 kg and its charge is q = +e. (a) Apply conservation of energy to find Ex:

ΔKE + ΔPE = 0 mO 2 v 02

( 0 − m v ) + −eE Δx = 0 → E = − eΔx 1 2

2 0

1 2

( 5.32 × 10 kg )( 2.00 × 10 m/s) = −887 V/m E =− (1.60 × 10 C )( 0.750 × 10 m ) 1 2

−26

−19

−3

(b) The potential difference between points A and B is

(

)

ΔV = −Ex Δx = − ( −887 V/m ) 0.750 × 10−3 m = 0.665 V

16.10

2 2 Using !Δy = v0yt + 12 ayt for the full flight gives !0 = v0yt f + 12 ayt f , or ay =

Topic 16

958

−2v0y/tf, where tf is the full time of the flight. Then, using !vy = v0y + 2ay (Δy) 2

for the upward part of the flight gives

( Δy ) !

= max

2 0 − v0y

2ay

(

2 −v0y

2 −2v0y t f

)

v0yt f 4

( 20.1 m/s )( 4.10 s ) = 20.6 m 4

From Newton’s second law,

ΣFy −mg − qE qE ⎞ ⎛ ay = = = −⎜ g + ⎟ ⎝ m m m⎠ ! Equating this to the earlier result gives ay = −(g + qE/m) = −2v0y/tf, so the electric field strength is

⎤ ⎛ 2.00 kg ⎞ ⎡ 2 ( 20.1 m/s ) ⎤ ⎛ m ⎞ ⎡ 2v0y E=⎜ ⎟⎢ − g⎥ = ⎜ − 9.80 m s 2 ⎥ = 1.95 × 103 N/C ⎢ −6 ⎟ ⎝ q ⎠ ⎢⎣ t f ⎥⎦ ⎝ 5.00 × 10 C ⎠ ⎣ 4.10 s ⎦ ! Thus, (ΔV)max = (Δymax)E = (20.6 m)(1.95 × 103 N/C) = 4.02 × 104 V = 40.2 kV

16.11

9 2 2 −19 ke q ( 8.99 × 10 N ⋅ m C ) ( −1.60 × 10 C ) V = = = −5.75 × 10−7 V (a) A −2 rA 0.250 × 10 m ! 9 2 2 −19 ke q ( 8.99 × 10 N ⋅ m C ) ( −1.60 × 10 C ) V = = = −1.92 × 10−7 V (b) B −2 rB 0.750 × 10 m !

ΔV = VB − VA = −1.92 × 10−7 V−(–5.75 × 10−7 V) = +3.83 × 10−7 V (c) No. The original electron will be repelled by the negatively charged particle which suddenly appears at point A. Unless the electron is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

959

fixed in place, it will move in the opposite direction, away from points A and B, thereby lowering the potential difference between these points. 16.12

(a) kq VA = ∑ e i ri i

⎛ −15.0 × 10−9 C 27.0 × 10−9 C ⎞ = 8.99 × 109 N ⋅ m 2 C2 ⎜ + ⎟ = +5.39 kV −2 −2 ⎝ 2.00 × 10 m 2.00 × 10 m ⎠

(

)

(b) kq VB = ∑ e i ri i

⎛ −15.0 × 10−9 C 27.0 × 10−9 C ⎞ = 8.99 × 109 N ⋅ m 2 C2 ⎜ + ⎟ = +10.8 kV −2 −2 ⎝ 1.00 × 10 m 1.00 × 10 m ⎠

(

16.13

)

(a) Calling the 2.00 µC charge q3

⎛q q ⎞ kq q V = ∑ e i = ke ⎜ 1 + 2 + 2 3 2 ⎟ ri r1 + r2 ⎠ i ⎝ r1 r2 2 ⎛ −6 ⎛ C 4.00 × 10−6 C 9 N ⋅ m ⎞ 8.00 × 10 ⎜ = ⎜ 8.99 × 10 + + ⎝ C2 ⎟⎠ ⎜ 0.060 0 m 0.030 0 m ⎝

⎞ ⎟ ( 0.060 0)2 + ( 0.030 0)2 m ⎟⎠ 2.00 × 10−6 C

= V = 2.67 × 106 V

(b) Replacing 2.00 × 10−6 C by −2.00 × 10−6 in part (a) yields V = 2.13 × 106 V © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

16.14

960

W = q(ΔV) = q(Vf − Vi), and Vf = 0 since the final location of the 8.00 µC is an infinite distance from other charges. The potential, due to the other charges, at the initial location of the 8.00 µC is Vi = ke (q1/r1 + q2/r2). Thus, the required energy for the move is

⎡ ⎛ q q ⎞⎤ W = q ⎢ 0 − ke ⎜ 1 + 2 ⎟ ⎥ ⎝ r1 r2 ⎠ ⎦ ⎣ 2 ⎛ −6 ⎛ C 9 N ⋅ m ⎞ 2.00 × 10 ⎜ = − ( 8.00 × 10 C ) ⎜ 8.99 × 10 + 2 ⎟ ⎝ C ⎠ ⎜ 0.030 0 m ⎝ −6

⎞ ⎟ 2 2 ( 0.030 0) + ( 0.060 0) m ⎟⎠ 4.00 × 10−6 C

W = −9.08 J !

16.15

2 −9 ke qi ⎛ C 3.00 × 10−9 C ⎞ 9 N ⋅ m ⎞ ⎛ 5.00 × 10 V = = 8.99 × 10 − = 103 V (a) ∑ r ⎜⎝ C2 ⎟⎠ ⎜⎝ 0.175 m 0.175 m ⎟⎠ i i !

PE =

(b)

k e qi q 2 r12

(

)(

)

−9 2 ⎞ 5.00 × 10 C −3.00 × 10−9 C ⎛ 9 N⋅m = ⎜ 8.99 × 10 = −3.85 × 10−7 J ⎟ 2 0.350 m C ⎠ ⎝

The negative sign means that positive work must be done to separate the charges by an infinite distance (that is, bring them up to a state of zero potential energy). 16.16

(a) At the center of the triangle, each of the identical charges produce a field contribution of magnitude E1 = keq/a2. The three contributions are oriented as shown below and the components of the resultant

Topic 16

961

field are: Ex = ΣEx = +E1 cos 30° − E1 cos 30° = 0 Ey = ΣEy = +E1 sin 30° − E1 + E1 sin 30° = 0

Thus, the resultant field has magnitude

E = Ex2 + Ey2 = 0 ! (b) The total potential at the center of the triangle is

k q k q k q k q 3ke q V = ∑Vi = ∑ e i = e + e + e = ri a a a a ! (c) Imagine a test charge placed at the center of the triangle. Since the field is zero at the center, the test charge will experience no electrical force at that point. The fact that the potential is not zero at the center means that work would have to be done by an external agent to move a test charge from infinity to the center. 16.17

The Pythagorean theorem gives the distance from the midpoint of the

Topic 16

962

base to the charge at the apex of the triangle as r = !3

( 4.00 cm ) − (1.00 cm ) = 15 cm = 15 × 10 m 2

−2

Then, the potential at the midpoint of the base is V = ∑ ke qi /ri , or i !

2 ⎛ −7.00 × 10−9 C ) ( −7.00 × 10−9 C ) ( +7.00 × 10−9 C ) ⎞ ⎛ 9 N⋅m ⎞ ( V = ⎜ 8.99 × 10 + + ⎜ ⎟ ⎝ C2 ⎟⎠ ⎝ 0.010 0 m 0.010 0 m 15 × 10−2 m ⎠

16.18

= −1.10 × 10 4 V = −11.0 kV

(a) See the sketch below:

(b) At the point (x, 0), where 0 < x < 1.20 m, the potential is ⎛ k q k ( −2q ) ke q 1 2⎞ V =∑ e i = e + = ke q ⎜ − ⎟ ri x 1.20 m − x ⎝ 1.20 m − x x ⎠ i !

Topic 16

963

2⎞ ⎛ ⎛ 1 2⎞ 9 N⋅m −9 V = ⎜ 8.99 × 10 2.50 × 10 C − ⎜ ⎟ ⎟ C2 ⎠ ⎝ ⎝ 1.20 m − x x ⎠

(

)

⎛ 1 2⎞ = 22.5 V ⋅ m ⎜ − ⎟ ⎝ 1.20 m − x x ⎠

(

)

(c) At x = +0.600 m, the potential is ⎛ ⎞ 1 2 22.5 V ⋅ m V = ( 22.5 V ⋅ m ) ⎜ − =− = −37.5 V ⎟ 0.600 m ⎝ 1.20 m − 0.600 m 0.600 m ⎠ !

(d) When 0 < x < 1.20 m and V = 0, we have 1/(1.20 m − x) − 2/x = 0, or x = 2.40 m − 2x. This yields x = 2.40 m/3 = 0.800 m. 16.19 (a) When the charge configuration consists of only the two protons (q1 and q2 in the sketch), the potential energy of the configuration is 9 2 2 −19 k q q ( 8.99 × 10 N ⋅ m C ) (1.60 × 10 C ) PEa = e 1 2 = r12 6.00 × 10−15 m !

PEa = 3.84 × 10−14 J

(b) When the alpha particle (q3 in the sketch) is added to the configuration, there are three distinct pairs of particles, each of which possesses potential energy. The total potential energy of the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

964

configuration is now

PEb = !

⎛ ke ( 2e 2 ) ⎞ ke q1q2 ke q1q3 ke q2 q3 + + = PEa + 2 ⎜ ⎟ r12 r13 r23 ⎝ r13 ⎠

where use has been made of the facts that q1q3 = q2q3 = e(2e) = 2e2 and r =r = ! 13 23

( 3.00 fm ) + ( 3.00 fm ) = 18.0 fm = 18.0 × 10 2

−15

m . Also,

note that the first term in this computation is just the potential energy computed in part (a). Thus,

PEb = PEa +

4ke e 2 r13

= 3.84 × 10 !

−14

J+

4 ( 8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10−19 C )

= 2.55 × 10−13 J

18.0 × 10−15 m

(c) If we start with the three-particle system of part (b) and allow the alpha particle to escape to infinity [thereby returning us to the twoparticle system of part (a)], the change in electric potential energy will be ΔPE = PEa − PEb = 3.84 × 10−14 J − 2.55 × 10−13 J = −2.17 × 10−13 J (d) Conservation of energy, ΔKE + ΔPE = 0, gives the speed of the alpha particle at infinity in the situation of part (c) as !12 mα vα − 0 = −ΔPE , or 2

Topic 16

965

−2 ( −2.17 × 10−13 J ) −2 ( ΔPE ) vα = = = 8.08 × 106 m/s mα 6.64 × 10−27 kg ! (e) When, starting with the three-particle system, the two protons are both allowed to escape to infinity, there will be no remaining pairs of particles and hence no remaining potential energy. Thus, ΔPE = 0 − PEb = −PEb, and conservation of energy gives the change in kinetic energy as ΔKE = −ΔPE = +PEb. Since the protons are identical particles, this increase in kinetic energy is split equally between them giving !KEproton = 12 mp vp = 12 ( PEb ) , 2

16.20

PEb 2.55 × 10−13 J vp = = = 1.24 × 107 m/s −27 mp 1.67 × 10 kg !

(a) If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and allowed to recede to infinity, the final speeds of the two particles will differ because of the difference in the masses of the particles. Thus, attempting to solve for the final speeds by use of conservation of energy alone leads to a situation of having one equation with two unknowns, and does not permit a solution. (b) In the situation described in part (a) above, one can obtain a second equation with the two unknown final speeds by using conservation

Topic 16

966

of linear momentum. Then, one would have two equations which could be solved simultaneously for both unknowns. (c) From conservation of energy:

(

)

⎡ 12 mα vα2 + 12 mp vp2 − 0 ⎤ + ⎡ 0 − ke qα qp /ri ⎤ = 0 , or ⎦ ⎦ ⎣ !⎣ mα vα2 + mp vp2 = !

2ke qα qp ri

2 ( 8.88 × 109 N ⋅ m 2 C2 ) ( 3.20 × 10−19 C ) (1.60 × 10−19 C ) 4.00 × 10−15 m

yielding !mα vα + mp vp = 2.30 × 10 2

−13

[1]

From conservation of linear momentum,

mα vα + mp vp = 0 !

⎛ mp ⎞ vα = ⎜ vp ⎝ mα ⎟⎠

[2]

Substituting Equation [2] into Equation [1] gives 2

⎛ mp ⎞ 2 ⎛ mp ⎞ 2 −13 mα ⎜ v + m v = 2.30 × 10 J or + 1 mp vp2 = 2.30 × 10−13 J p p p ⎟ ⎜ ⎟ ⎝ mα ⎠ ⎝ mα ⎠ ! and

vp =

2.30 × 10−13 J

(m m + 1) m p

(1.67 × 10

2.30 × 10−13 −27

)(

6.64 × 10−27 + 1 1.67 × 10−27 kg

)

= 1.05 × 107 m/s

Topic 16

967

Then, Equation [2] gives the final speed of the alpha particle as

16.21

⎛ 1.67 × 10−27 kg ⎞ ⎛ mp ⎞ vα = ⎜ vp = ⎜ 1.05 × 107 m/s ) = 2.64 × 106 m/s ( −27 ⎟ ⎟ ⎝m ⎠ ⎝ 6.64 × 10 kg ⎠ α

(a) Conservation of energy gives

⎛1 1⎞ KE f = KEi + PEi − PE f = 0 + ke q1q2 ⎜ − ⎟ ⎝ ri rf ⎠ !

(

)

With q1 = +8.50 nC, q2 = −2.80 nC, ri = 1.60 µm, and rf = 0.500 µm, this becomes ⎛ ⎞ ⎛ N ⋅ m2 ⎞ 1 1 KE f = ⎜ 8.99 × 109 8.50 × 10−9 C ) ( −2.80 × 10−9 C ) ⎜ − ( 2 −6 −6 ⎟ ⎝ C ⎠ ⎝ 1.60 × 10 m 0.500 × 10 m ⎟⎠ !

yielding

KEf = 0.294 J

(b) When r = rf = 0.500 µm and KE = KEf = 0.294 J, the speed of the sphere having mass m = 8.00 mg = 8.00 × 10−6 kg is

vf = !

16.22

(

2 KE f m

2 ( 0.294 J ) = 271 m/s 8.00 × 10−6 kg

The excess charge on the metal sphere will be uniformly distributed over its surface. In this spherically symmetric situation, the electric field and the electric potential outside the sphere is the same as if all the excess

Topic 16

968

charge were concentrated as a point charge at the center of the sphere. Thus, for points outside the sphere, Q E = ke 2 r !

and

V = ke

Q = E⋅r r

If the sphere has a radius of r = 18 cm = 0.18 m and the air breaks down when E = 3.0 × 106 V/m, the electric potential at the surface of the sphere when breakdown occurs is V = (3.0 × 106 V/m)(0.18 m) = 5.4 × 105 V 16.23

From conservation of energy, (KE + PEe)f = (KE + PEe)i, which gives 0 + keQq rf = 12 mα vi2 + 0 , or ! 2k Qq 2k (79e ) ( 2e ) rf = e 2 = e mα vi mα vi2 !

rf =

2 ( 8.99 × 109 N ⋅ m 2 C2 ) (158 ) (1.60 × 10−19 C )

( 6.64 × 10

16.24

−27

kg ) ( 2.00 × 10 m/s ) 7

= 2.74 × 10−14 m

(a) The distance from any one of the corners of the square to the point at the center is one half the length of the diagonal of the square, or

r= !

diagonal a2 + a2 a 2 a = = = 2 2 2 2

Since the charges have equal magnitudes and are all the same

Topic 16

969

distance from the center of the square, they make equal contributions to the total potential. Thus,

kQ kQ Q Vtotal = 4Vsingle = 4 e = 4 e = 4 2ke r a a 2 charge ! (b) The work required to carry charge q from infinity to the point at the center of the square is equal to the increase in the electric potential energy of the charge, or

Q⎞ qQ ⎛ W = PEcenter − PE∞ = qVtotal − 0 = q ⎜ 4 2ke ⎟ = 4 2ke ⎝ ⎠ a a ! 16.25 The energy 1 eV = 1.60 × 10−19 J. (a) Use the definition of kinetic energy to find the electron’s speed:

KE = m v 1 2

→ ve =

2 e e

2 ( KE) = me

(

2 1.60 × 10−19 J 9.11 × 10

−31

) = 5.93 × 10 m/s 5

(b) Use the definition of kinetic energy to find the proton’s speed:

KE = mp v 1 2

→ vp =

2 p

2 ( KE) = mp

(

2 1.60 × 10−19 J 1.67 × 10

−27

) = 1.38 × 10 m/s 4

(

)(

)

kBT = 32 1.38 × 10−23 J/K 3.00 × 10 2 K = 6.21× 10−21 J . Convert to eV

to find

Topic 16

970

1 eV ⎛ ⎞ 6.21 × 10−21 J ⎜ = 3.88 × 10−2 eV −19 ⎟ ⎝ 1.60 × 10 J ⎠ 16.26 Convert the given work to find 1.50 × 103 eV = 2.40 × 10−16 J. The work done by the electric field is W = −ΔPE = −qΔV. (a) Solve for the potential difference and substitute values to find

W 2.40 × 10−16 J ΔV = − =− = −2.50 × 10 2 V −19 q 9.61 × 10 C (b) The change in electric potential energy is ΔPE = −W = −2.40 × 10−16 J . 16.27

The work is W = −qΔV = −q(VB – VA) = −(3.20 × 10−19 C)(5.80 − 3.60) × 103 J/C = –7.04 × 10−16 J. Convert to eV using 1 eV = 1.60 × 10–19 J to find W =

−4.40 × 103 eV . 16.28 Convert the given energies to joules to find KEe = 13.6 eV = 2.18 × 10-18 J and PEe = −27.2 eV = −4.35 × 10−18 J. (a) Use the definition of kinetic energy to find

KE = mv 1 2

→ ve =

2 ( KEe ) me

(

2 2.18 × 10−18 J 9.11 × 10

−31

) = 2.19 × 10 m/s 6

(b) Use the potential energy of a pair of charges to solve for the orbital radius r:

Topic 16

971

qq PE = ke 1 2 r

)(

(

⎛ C2 ⎞ 1.0 × 10 m 16.29 (a) C = ! 0 = ⎜ 8.85 × 10−12 ⎟ d ⎝ 800 m N ⋅ m2 ⎠ A

(b)

)

1.60 × 10−19 C e2 → re = −ke = − 8.99 × 109 N ⋅ m 2 /C 2 = 5.29 × 10−11 m −18 PEe −4.35 × 10 J

( )

Qmax = C ΔV

max

(

)

= C Emax d = ! 0

(

)(

A d

) = 1.1 × 10 F −8

(E d ) = ! AE max

)(

max

)

= 8.85 × 10−12 C2 N ⋅ m 2 1.0 × 106 m2 3.0 × 106 N/C = 27 C

Q 27.0 µC = = 3.00 µF 16.30 (a) C = ΔV 9.00 V ! (b) Q = C(ΔV) = (3.00 µF)(12.0 V) = 36.0 µC 16.31 (a) The capacitance of this air-filled (dielectric constant, κ = 1.00) parallelplate capacitor is

C= =

k! 0 A d 1.00 8.85 × 10−12 C2 N ⋅ m 2 2.30 × 10−4 m 2

( )(

)(

1.50 × 10 m −3

) = 1.36 × 10 F = 1.36 pF −12

(b) Q = C(ΔV) = (1.36 × 10−12 F)(12.0 V) = 1.63 × 10−11 C = 16.3 × 10−12 C = 16.3 pC

ΔV 12.0 V = = 8.00 × 103 V/m = 8.00 × 103 N/C (c) E = −3 d 1.50 × 10 m !

Topic 16

972

ÉA Q 16.32 (a) An air-filled parallel-plate capacitor has capacitance C = 0 = d ΔV = where A is the plate area and d is the distance between the plates. Assuming the electric field between the plates is constant, E = ΔV/d so that ΔV = Ed. Substitute that expression and the known values to find

É0 A Q Q = → É0 A = → Qmax = É 0 AEmax d ΔV E

(

)(

Qmax = 8.85 × 10−12 C 2 /N ⋅ m 2 2.00 × 10−4 m 2 3.00 × 106 V/m

)

−9 =Qmax = 5.31 × 10 C

(b) Use the definition of capacitance to find

Q ΔV

→ ΔV =

Qmax 5.31 × 10−9 C = = 7.08 × 10−5 V −6 C 75.0 × 10 F

ΔV 20.0 V = = 1.11 × 10 4 V/m = 11.1 kV/m toward the 16.33 (a) E = −3 d 1.80 × 10 m ! negative plate

(b) =

!0 A

(

d 8.85 × 10−12 C2 N ⋅ m 2 7.60 × 10−4 m 2

)(

1.80 × 10 m −3

) = 3.74 × 10 F = 3.74 pF −12

Topic 16

16.34

973

C = e0A/d, so

!0 A C

(8.85 × 10 C N ⋅ m )(21.0 × 10 m ) = 3.10 × 10 m = −12

−12

−9

60.0 × 10

−15

⎛ 1Å ⎞ d = ( 3.10 × 10−9 m ) ⎜ −10 ⎟ = 31.0 Å ⎝ 10 m ⎠ !

16.35

(a) Assuming the capacitor is air-filled (κ = 1), the capacitance is

!0 A d

(8.85 × 10 C N ⋅ m )(0.200 m ) = 5.90 × 10 F = −12

−10

3.00 × 10 m −3

(b) Q = C(ΔV) = (5.90 × 10−10 F)(6.00 V) = 3.54 × 10−9 C

(c)

ΔV 6.00 V E= = = 2.00 × 103 V/m = 2.00 × 103 N/C −3 d 3.00 × 10 m !

Q 3.54 × 10−9 C = 1.77 × 10−8 C/m 2 (d) σ = = 2 A 0.200 m ! (e) Increasing the distance separating the plates decreases the capacitance, the charge stored, and the electric field strength between the plates. This means that all of the previous answers will be decreased.

16.36

mg ΣFy = 0 ⇒ T cos 15.0° = mg or T = cos 15.0° !

Topic 16

974

ΣFx = 0 ⇒ qE = T sin 15.0° = mg tan 15.0°

E= !

mg tan 15.0° q

ΔV = Ed = !

mgd tan 15.0° q

(350 × 10 kg )(9.80 m s )(0.040 0 m) tan 15.0° = 1.23 × 10 V = 1.23 kV ΔV = −6

30.0 × 10 C −9

16.37

(a) Capacitors in a series combination store the same charge, Q = Ceq(ΔV), where Ceq is the equivalent capacitance and ΔV is the potential difference maintained across the series combination. The equivalent capacitance for the given series combination is 1/Ceq = 1/C1 + 1/C2, or Ceq = C1C2/(C1 + C2), giving

( 2.50 µF)( 6.25 µF) = 1.79 µF Ceq = 2.50 µF + 6.25 µF ! and the charge stored on each capacitor in the series combination is

Topic 16

975

Q = Ceq ( ΔV ) = (1.79 µF ) ( 6.00 V ) = 10.7 µC !

(b) When connected in parallel, each capacitor has the same potential difference, ΔV = 6.00 V, maintained across it. The charge stored on each capacitor is then

16.38

For C1 = 25.0 µF:

Q1 = C1(ΔV) = (2.50 µF)(6.00 V) = 15.0 µC

For C2 = 6.25 µF:

Q2 = C2(ΔV) = (6.25 µF)(6.00 V) = 37.5 µC

(a) Ceq = C1 + C2 = 5.00 µF + 12.0 µF = 17.0 µF (b) In a parallel combination, the full potential difference maintained between the terminals of the battery exists across each capacitor. Thus, ΔV1 = ΔV2 = ΔVbattery = 9.00 V (c) Q1 = C1(ΔV1) = (5.00 µF)(9.00 V) = 45.0 µC Q2 = C2(ΔV2) = (12.0 µF)(9.00 V) = 108 µC

16.39

(a) First, we replace the parallel combination between points b and c by its equivalent capacitance, Cbc = 2.00 µF + 6.00 µF = 8.00 µF. Then, we have three capacitors in series between points a and d. The equivalent capacitance for this circuit is therefore

Topic 16

976

1 1 1 1 3 = + + = C Cab C bc Ccd 8.00 µF ! eq

giving

Ceq = !

8.00 µF = 2.67 µF 3

(b) The charge stored on each capacitor in the series combination is Qab = Qbc = Qcd = Ceq(ΔVad) = (2.67 µF)(9.00 V) = 24.0 µC Then, note that ΔVbc = Qbc/Cbc = 24.0 µC/8.00 µF = 3.00 V. The charge on each capacitor in the original circuit is: On the 8.00 µF between a and b:

Q8 = Qab = 24.0 µC

On the 8.00 µF between c and d:

Q8 = Qcd = 24.0 µC

On the 2.00 µF between b and c: Q2 = C2(ΔVbc) = (2.00 µF)(3.00 V) = 6.00 µC On the 6.00 µF between b and c: Q6 = C6(ΔVbc) = (6.00 µF)(3.00 V) = 18.0 µC

Topic 16

977

(c)

Note that ΔVab = Qab/Cab = 24.0 µC/8.00 µF = 3.00 V, and that ΔVcd = Qcd/Ccd = 24.0 µC/8.00 µF = 3.00 V. We earlier found that ΔVbc = 3.00 V, so we conclude that the potential difference across each capacitor in the circuit is ΔV8 = ΔV2 = ΔV6 = ΔV8 = 3.00 V

16.40

Cparallel = C1 + C2 = 9.00 pF

1 !Cseries

1 1 + C1 C2

⇒ C1 = 9.00 pF − C2

⇒ Cseries =

Thus, using Equation [1],

[1]

C1C2 = 2.00 pF C1 + C2

( 9.00 pF − C2 )C2 = 2.00 pF Cseries = which 9.00 pF − C2 ) + C2 ( !

reduces to 2 2 !C2 − (9.00 pF)C2 + 18.0 (pF) = 0,!or!(C2 − 6.00 pF)(C2 − 3.00 pF) = 0

Therefore, either C2 = 6.00 pF and, from Equation [1], C1 = 3.00 pF or

C2 = 3.00 pF

and C1 = 6.00 pF.

We conclude that the two capacitances are 3.00 pF and 6.00 pF. 16.41 (a) The equivalent capacitance of the series combination in the upper branch is 1 C ! upper

1 1 2+1 + = 3.00 µF 6.00 µF 6.00 µF

Topic 16

978

Cupper = 2.00 µF

Likewise, the equivalent capacitance of the series combination in the lower branch is

1 !Clower

1 1 2+1 + = 2.00 µF 4.00 µF 4.00 µF

Clower = 1.33 µF

These two equivalent capacitances are connected in parallel with each other, so the equivalent capacitance for the entire circuit is Ceq = Cupper + Clower = 2.00 µF + 1.33 µF = 3.33 µF (b) Note that the same potential difference, equal to the potential difference of the battery, exists across both the upper and lower branches. The charge stored on each capacitor in the series combination in the upper branch is Q3 = Q6 = Qupper = Cupper (ΔV) = (2.00 µF)(90.0 V) = 180 µC and the charge stored on each capacitor in the series combination in

Topic 16

979

the lower branch is Q2 = Q4 = Qlower = Clower (ΔV) = (1.33 µF)(90.0 V) = 120 µC (c) The potential difference across each of the capacitors in the circuit is:

Q 120 µC ΔV2 = 2 = = 60.0 V C 2.00 µ F 2 !

ΔV4 =

Q4 120 µC = = 30.0 V C4 4.00 µF

Q 180 µC ΔV3 = 3 = = 60.0 V C 3.00 µ F 3 !

ΔV6 =

Q6 180 µC = = 30.0 V C6 6.00 µF

16.42 (a) The equivalent capacitance of the series combination in the rightmost branch of the circuit is 1 C ! right

1 1 1+ 3 + = 24.0 µF 8.00 µF 24.0 µF

Cright = 6.00 µF

Figure P16.38 (b) The equivalent capacitance of the three capacitors now connected in parallel with each other and with the battery is C = 4.00 µF + 2.00 µF + 6.00 µF = 12.0 µF ! eq © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

980

Diagram 1 (c) The total charge stored in this circuit is Qtotal = Ceq (ΔV) = (12.0 µF)(36.0 V) or

Qtotal = 432 µC

Diagram 2 (d) The charges on the three capacitors shown in Diagram 1 are: Q4 = C4(ΔV) = (4.00 µF)(36.0 V) = 144 µC Q2 = C2(ΔV) = (2.00 µF)(36.0 V) = 72 µC Qright = Cright(ΔV) = (6.00 µF)(36.0 V) = 216 µC Yes. Q4 + Q2 + Qright = Qtotal as it should. (e) The charge on each capacitor in the series combination in the rightmost branch of the original circuit (Figure P16.38) is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

981

Q24 = Q8 = Qright = 216 µC

Q24 216 µC = = 9.00 V (f) ΔV24 = C 24.0 µ F 24 ! Q8 216 µC = = 27.0 V (g) ΔV8 = C 8.00 µ F 8 !

Note that ΔV8 + ΔV24 = ΔV = 36.0 V as it

should. 16.43

The circuit may be reduced in steps as shown above. Using Figure 3,

Qac = (4.00 µF)(24.0 V) = 96.0 µC

Qac 96.0 µC = = 16.0 V Then, in Figure 2, (ΔV )ab = Cab 6.00 µF ! and (ΔV)bc = (ΔV)ac − (ΔV)ab = 24.0 V − 16.0 V = 8.00 V Finally, using Figure 1, Q1 = C1(ΔV)ab = (1.00 µF)(16.0 V) = 16.0 µC,

Q5 = (5.00 µF)(V)ab = 80.0 µC

Topic 16

982

Q8 = (8.00 µF)(ΔV)bc = 64.0 µC, 16.44

and Q4 = (4.00 µF)(ΔV)bc = 32.0 µC

(a) Consider the simplification of the circuit as shown below:

Since C2 and C3 are connected in parallel, Cab = C2 + C3 = C + 5C = 6C. Now observe that C1 and Cab are connected in series, giving 1 1 1 = + C C1 Cab ! eq

or Ceq =

C1Cab (3C)(6C) = = 2C C1 + Cab 3C + 6C

(b) Since capacitors C1 and Cab are connected in series, Q1 = Qab = Qeq = Ceq (ΔV) = 2C(ΔV)

Then,

Q 2C(ΔV ) ΔV C ( ΔV ) ΔVab = ab = = ,!giving!!!Q2 = C2 ( ΔVab ) = Cab 6C 3 3 !

Also,

Q3 = C3 ( ΔVab ) = !

5C(ΔV ) . 3

Therefore,

Q1 > Q3 > Q2

Also, with capacitors C2 and C3 in parallel between points a and b, © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

983

ΔV ΔV2 = ΔV3 = ΔVab = 3 !

Thus,

ΔV1 > ΔV2 = ΔV3

(d) Consider the following steps: (i) Increasing C3 while C1 and C2 remain constant will increase Cab = C2 + C3. ⎛ Cab ⎞ Therefore, the equivalent capacitance, Ceq = C1 ⎜ , will ⎝ C1 + Cab ⎟⎠ !

increase. (ii)

Since C1 and Cab are in series, Q1 = Qab = Ceq (ΔV). Thus, Q1 will increase as Ceq increases. Also, Qab experiences the same increase.

(iii)

Because ΔV1 = Q1/C1, an increase in Q1 causes ΔV1 to increase and causes ΔVab = ΔV − ΔV1 to decrease. Thus, since Q2 = C2(ΔVab), Q2 will decrease.

(iv) With capacitors C2 and C3 in parallel between points a and b, we have Qab = Q2 + Q3 or Q3 = Qab − Q2. Thus, with Qab increasing [see Step (ii)] while Q2 is decreasing [see Step (iii)], we see that Q3 will increase.

Topic 16

16.45

984

(a) From Q = C(ΔV), Q25 = (25.0 µF)(50.0 V) = 1.25 × 103 µC = 1.25 mC and Q40 = (40.0 µF)(50.0 V) = 2.00 × 103 µC = 2.00 mC (b) Since the negative plate of one capacitor was connected to the positive plate of the other, the net charge stored in the new parallel combination is Q = Q40 − Q25 = 2.00 × 103 µC − 1.25 × 103 µC = 750 µC The two capacitors, now in parallel, have a common potential difference ΔV across them. The new charges on each of the capacitors

′ = C1 (ΔV ) and !Q40 ′ = C2 (ΔV ) . Thus, are !Q25 ⎛ 25 µF ⎞ C 5 Q25 Q40 ′ = 1 Q40 ′ =⎜ ′ = Q40 ′ ⎟ C2 8 ⎝ 40 µF ⎠ !

and the total change now stored in the combination may be written as 5 13 Q = Q40 ′ + Q25 ′ = Q40 ′ + Q40 ′ = Q40 ′ = 750 µC 8 8 ! 8 ′ = (750 µC) = 462 µC and giving Q40 13 ! Q′ = Q − Q40 ′ = (750 − 462)µC = 288 µC ! 25

(c) The potential difference across each capacitor in the new parallel © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

985

combination is Q Q 750 µC ΔV = = = = 11.5 V Ceq C1 + C2 65.0 µF !

16.46

(a) The original circuit reduces to a single equivalent capacitor in the steps shown below.

⎛ 1 1⎞ Cs = ⎜ + ⎟ ⎝ C1 C2 ⎠ !

−1

⎛ 1 1 ⎞ =⎜ = ⎝ 5.00 µF 10.0 µF ⎟⎠

−1

= 3.33 µF

Cp1 = Cs + C3 + Cs = 2(3.33 µF) + 2.00 µF = 8.66 µF Cp2 = C2 + C2 = 2(10.0 µF) = 20.0 µF ⎛ 1 1 ⎞ Ceq = ⎜ + ⎟ ⎝ Cp1 Cp2 ⎠ !

−1

⎛ 1 1 ⎞ =⎜ + ⎝ 8.66 µF 20.0 µF ⎟⎠

−1

= 6.04 µF

(b) The total charge stored between points a and b is Qtotal = Ceq (ΔV)ab = (6.04 µF)(60.0 V) = 362 µC

Topic 16

986

Then, looking at the third figure, observe that the charges of the series capacitors of that figure are Qp1 = Qp2 = Qtotal = 362 µC. Thus, the potential difference across the upper parallel combination shown in the second figure is

( ΔV )p1 = !

Qp1 Cp1

362 µC = 41.8 V 8.66 µF

Finally, the charge on C3 is Q = C3 ( ΔV )p1 = (2.00 µF)(41.8 V) = 83.6 µC ! 3

16.47

From Q = C(ΔV), the initial charge of each capacitor is Q1 = (1.00 µF)(10.0 V) = 10.0 µC

and Q2 = (2.00 µF)(0) = 0

After the capacitors are connected in parallel, the potential difference across one is the same as that across the other. This gives

Q1′ Q2′ ΔV = = 1.00 µF 2.00 µF !

Q2′ = 2Q1′

[1]

From conservation of charge, !Q1′ + Q2′ = Q1 + Q2 = 10.0 µC . Then, substituting from Equation [1], this becomes

!Q1′ + 2Q1′ = 10.0 µC , giving Finally, from Equation [1],

Q′ = 10 µC/3 = 3.33 µC ! 1

Q′ = 20 µC/3 = 6.67 µC ! 2

Topic 16

16.48

987

(a) We simplify the circuit in stages as shown below:

1 1 1 = + !Cs 15.0 µF 3.00 µF

or Cs =

(15.0 µF)( 3.00 µF) = 2.50 µF 15.0 µF + 3.00 µF

Cp = Cs + 6.00 µF = 2.50 µF + 6.00 µF = 8.50 µF 1 1 1 = + C Cp 20.0 µF ! eq

or Ceq =

(8.50 µF)(20.0 µF) = 5.96 µF 8.50 µF + 20.0 µF

(b) Q20 = QCp = Qeq = Ceq (ΔVab ) = (5.96 µ F) ( 15.0 V ) = 89.4 µC QC 89.4 µC ΔVCp = p = = 10.5 V Cp 8.50 µF !

Q6 = C6(ΔVCp) = (6.00 µF)(10.5 V) = 63.0 µC

and

Q15 = Q3 = Qs = Cs (ΔVCp ) = (2.50 µ F) ( 10.5 V ) = 26.3 µC

The charges are 89.4 µC on the 20 µF capacitor, 63.0 µC on the 6 µF capacitor, and 26.3 µC on both the 15 µF and 3 µF capacitors.

16.49

Q2 1 1 2 2 Energy!stored!= = C ( ΔV ) = ( 4.50 × 10−6 F ) (12.0 V ) = 3.24 × 10−4 J 2C 2 2 !

Topic 16

16.50

988

(a) The equivalent capacitance of a series combination of C1 and C2 is 1 1 1 2+1 = + = C 18.0 µF 36.0 µF 36.0 µF ! eq

Ceq = 12.0 µF

When this series combination is connected to a 12.0-V battery, the total stored energy is 1 1 2 2 Total!energy!stored!= Ceq ( ΔV ) = (12.0 × 10−6 F ) (12.0 V ) = 8.64 × 10−4 J 2 2 !

(b) The charge stored on each of the two capacitors in the series combination is Q1 = Q2 = Qtotal = Ceq (ΔV) = (12.0 µF)(12.0 V) = 144 µC = 1.44 × 10−4 C and the energy stored in each of the individual capacitors is −4 Q 2 (1.44 × 10 C ) Energy!stored!in!C1 = 1 = = 5.76 × 10−4 J −6 2C1 2 (18.0 × 10 F ) ! 2

−4 Q22 (1.44 × 10 C ) Energy!stored!in!C = = = 2.88 × 10−4 J and 2 −6 2C2 2 ( 36.0 × 10 F ) ! 2

Energy stored in C1 + Energy stored in C2 = 5.76 × 10−4 J + 2.88 × 10−4 J = 8.64 × 10−4 J, which is the same as the total stored energy found in part (a). This must be true if the computed equivalent capacitance is truly equivalent to the original combination.

Topic 16

989

(c) If C1 and C2 had been connected in parallel rather than in series, the equivalent capacitance would have been Ceq =C1 + C2 = 18.0 µF + 36.0 2 µF = 54.0 µF. If the total energy stored ⎡⎣ 12 Ceq (ΔV ) ⎤⎦ in this parallel !

combination is to be the same as was stored in the original series combination, it is necessary that 2 ( 8.64 × 10−4 J ) 2(Total!energy!stored) ΔV = = = 5.66 V Ceq 54.0 × 10−6 F !

Since the two capacitors in parallel have the same potential difference across them, the energy stored in the individual capacitors ⎡ 1 C(ΔV )2 ⎤⎦ is directly proportional to their capacitances. The larger !⎣ 2

capacitor, C2, stores the most energy in this case. 16.51

(a) The energy initially stored in the capacitor is

Qi2 1 1 (Energy!stored)1 = = Ci (ΔV )i2 = (3.00 µF)(6.00 V)2 = 54.0 µ J 2Ci 2 2 ! (b) When the capacitor is disconnected from the battery, the stored charge becomes isolated with no way off the plates. Thus, the charge remains constant at the value Qi as long as the capacitor remains disconnected. Since the capacitance of a parallel-plate capacitor is C =

ke0A/d, when the distance d separating the plates is doubled, the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

990

capacitance is decreased by a factor of 2 (Cf = Ci/2 = 1.50 µF). The stored energy (with Q unchanged) becomes ⎛ Q2 ⎞ Q2 Qi2 (Energy!stored)2 = i = = 2 ⎜ i ⎟ = 2(Energy!stored)1 = 108 µ J 2C f 2(Ci /2) ⎝ 2Ci ⎠ !

(c) When the capacitor is reconnected to the battery, the potential difference between the plates is reestablished at the original value of ΔV = (ΔV)i = 6.00 V, while the capacitance remains at Cf = Ci/2 = 1.50

µF. The energy stored under these conditions is 1 1 2 (Energy!stored)3 = C f ( ΔV )i = (1.50 µF)(6.00 V)2 = 27.0 µ J 2 2 !

16.52 (a) Use the definition of capacitance to find

Q ΔV

→ ΔV =

Q 60.0 × 10−6 C = = 12.0 V C 5.00 × 10−6 F

(b) The energy stored in the capacitor is

(

)

PEC = 12 QΔV = 12 60.0 × 10−6 C (12.0 V ) = 3.60 × 10−4 J 16.53

(a) Note that the charge on the plates remains constant at the original value, Q0, as the dielectric is inserted. Thus, the change in the potential difference, ΔV = Q/C, is due to a change in capacitance alone. The ratio of the final and initial capacitances is

Topic 16

991

Cf Ci

κ! 0 A/d ! 0 A/d

=κ

and

Cf Ci

( ΔV ) = ( ΔV ) = 85.0 V = 3.40 25.0 V Q ( ΔV ) ( ΔV )

Thus, the dielectric constant of the inserted material is κ = 3.40, and the material is probably nylon (see Table 16.1). (b) If the dielectric only partially filled the space between the plates, leaving the remaining space air-filled, the equivalent dielectric constant would be somewhere between κ = 1.00 (air) and κ = 3.40. The resulting potential difference would then lie somewhere between (ΔV)i = 85.0 V and (ΔV)f = 25.0 V. 16.54

(a) If the maximum electric field that can exist between the plates before breakdown (i.e., the dielectric strength) is Emax, the maximum potential difference across the plates is ΔVmax = Emax ⋅ d, where d is the plate separation. The maximum change on either plate then has magnitude Q = C ( ΔVmax ) = C ( Emax ⋅ d ) ! max

Since the capacitance of a parallel-plate capacitor is C = ke0A/d, the maximum charge is

Topic 16

992

(

)

⎛ âÉ A ⎞ Qmax = ⎜ 0 ⎟ Emax ⋅ d = âÉ 0 AEmax ⎝ d ⎠ = The area of each plate is A = 5.00 cm2 = 5.00 × 10−4 m2, and when air is the dielectric, k = 1.00 and Emax = 3.00 × 106 V/m (see Table 16.1). Thus, Qmax = (1.00)(8.85 × 10−12 C/N ⋅ m2)(5.00 × 10−4 m2)(3.00 × 106 V/m) = 1.33 × 10−8 C = 13.3 nC (b) If the dielectric is now polystyrene (k = 2.56 and Emax = 24.0 × 106 V/m), then Qmax = (2.56)(8.85 × 10−12 C/N ⋅ m2)(5.00 × 10−4 m2)(24.0 × 106 V/m) = 2.72 × 10−7 C = 272 nC 16.55 (a) The dielectric constant for Teflon® is κ = 2.1, so the capacitance is

C= =

âÉ 0 A d

(2.1)(8.85 × 10 C N ⋅ m )(175 × 10 m ) = −12

−4

0.040 0 × 10−3 m

C = 8.1 × 10−9 F = 8.1 nF (b) For Teflon®, the dielectric strength is Emax = 60 × 106 V/m, so the maximum voltage is ΔVmax = Emaxd = (60 × 106 V/m)(0.040 0 × 10−3 m)

Topic 16

993

ΔVmax = 2.4 × 103 V = 2.4 kV 16.56 (a) The capacitance of an air-filled parallel-plate capacitor is

(

É A 8.85 × 10 C= 0 = d =

−12

)(

C 2 /N ⋅ m 2 7.00 × 10−2 m 2 2.00 × 10

−4

) = 3.10 × 10 F −9

(b) This capacitor is equivalent to two capacitors in series, one having a dielectric constant k, and each with a plate separation of d/2. The equivalent capacitance of the two capacitors in series is

⎛ d ⎞⎛1 ⎞ 1 1 1 d d = + = + =⎜ +1 Ceq C1 C2 2âÉ 0 A 2É 0 A ⎝ 2É 0 A ⎟⎠ ⎜⎝ â ⎟⎠ ⎞⎛ 1 1 ⎛ 2.00 × 10−4 m ⎞ =⎜ + 1⎟ −12 2 2 −2 2 ⎟⎜ Ceq ⎝ 2 8.85 × 10 C /N ⋅ m 7.00 × 10 m ⎠ ⎝ 3.70 ⎠

(

)(

)

C = 4.88 × 10−9 F = eq (c) This capacitor is equivalent to two capacitors in parallel, one having a dielectric constant k, and each with a plate area equal to A/2. The equivalent capacitance of two capacitors in parallel is and

Ceq = C1 + C2 =

âÉ 0 A É 0 A É 0 A + = (â + 1) 2d 2d 2d

( 8.85 × 10 C = eq

16.57

(a) V =

−12

)(

C 2 /N ⋅ m 2 7.00 × 10−2 m 2

(

2 2.00 × 10

1.00 × 10−12 kg 1 100 kg m

−4

)

) ( 3.70 + 1) = 7.28 × 10 F −9

= 9.09 × 10−16 m 3

Topic 16

994

Since V = 4πr3/3, the radius is r = [3V/4π]1/3, and the surface area is ⎡ 3V ⎤ A = 4π r = 4π ⎢ ⎣ 4π ⎥⎦ !

2/3

(b) C= =

âÉ 0 A d

⎡ 3 ( 9.09 × 10−16 m 3 ) ⎤ = 4π ⎢ ⎥ 4π ⎢⎣ ⎥⎦

2/3

= 4.54 × 10−10 m 2

(5.00)(8.85 × 10 C N ⋅ m )( 4.54 × 10 m ) = 2.01 × 10 F = 2

−12

−10

−13

100 × 10 m −9

Q 2.01 × 10−14 C n= = = 1.26 × 105 −19 e 1.60 × 10 C ! 16.58

For a parallel-plate capacitor, C = ke0 A/d and Q = σA = C(ΔV). Thus,

(

)( )

σ A = âÉ 0 A /d ΔV , and d = (ke0/σ)(ΔV). With air as the dielectric = material (k = 1.00), the separation of the plates must be

(1.00)(8.85 × 10−12 C2 N ⋅ m 2 )(150 V )

d= = 4.43 × 10−4 m = 0.443 mm −10 2 4 2 2 ( 3.00 × 10 C cm )(10 cm 1 m ) ! 16.59

Since the capacitors are in series, the equivalent capacitance is given by

1 Ceq

1 C1

1 C2

1 C3

!0 A !0 A !0 A

d1 + d2 + d3

!0 A

Topic 16

995

Ceq =

!0 A d

where d = d1 + d2 + d3

16.60 (a) Please refer to the solution of Problem 16.37 where the following results were obtained: Ceq = 3.33 µF

Q3 = Q6 = 180 µC

Q2 = Q4 = 120 µC

The total energy stored in the full circuit is then

(Energy!stored )

total

1 1 2 2 = Ceq ( ΔV ) = ( 3.33 × 10−6 F ) ( 90.0 V ) 2 2 −2 = 1.35 × 10 J = 13.5 × 10−3 J = 13.5 mJ

(b) The energy stored in each individual capacitor is For 2.00 µF:

120 × 10 C (Energy stored) = 2C = 2( (2.00 × 10 F) ) = 3.60 × 10 J = 3.60 mJ Q22

−6

−3

−6

For 3.00 µF:

Topic 16

996

180 × 10 C (Energy stored) = 2C = 2( (3.00 × 10 F) ) = 5.40 × 10 J = 5.40 mJ Q32

−6

−3

−6

For 4.00 µF:

120 × 10 C (Energy stored) = 2C = 2( ( 4.00 × 10 F) ) = 1.80 × 10 J = 1.80 mJ 4

−6

Q42

−3

−6

For 6.00 µF:

180 × 10 C (Energy stored) = 2C = 2( (6.00 × 10 F) ) = 2.70 × 10 J = 2.70 mJ Q62

−6

−3

−6

(c) The total energy stored in the individual capacitors is Energy stored = (3.60 + 5.40 + 1.80 + 2.70) mJ = 13.5 mJ = (Energy stored)total Thus, the sums of the energies stored in the individual capacitors equals the total energy stored by the system. 16.61

In the absence of a dielectric, the capacitance of the parallel-plate capacitor is C0 = e0A/d.

Topic 16

997

With the dielectric inserted, it fills one-third of the gap between the plates as shown in sketch (a) above. We model this situation as consisting of a pair of capacitors, C1 and C2, connected in series as shown in sketch (b) above. In reality, the lower plate of C1 and the upper plate of C2 are one and the same, consisting of the lower surface of the dielectric shown in sketch (a). The capacitances in the model of sketch (b) are given by

âÉ A 3âÉ 0 A C1 = 0 = d/3 d =

and C2 =

É0 A 2d/3

3É 0 A 2d

The equivalent capacitance of the series combination is

Topic 16

998

1 Ceq

d 3âÉ 0 A

2d 3É 0 A

⎛1 ⎞⎛ d ⎞ = ⎜ + 2⎟ ⎜ ⎟ ⎝â ⎠ ⎝ 3É 0 A ⎠ ⎛ 2κ + 1 ⎞ d =⎜ ⎟ ⎝ κ ⎠ 3É 0 A ⎛ 2â + 1 ⎞ d =⎜ ⎟ ⎝ 3â ⎠ É 0 A

⎛ 2â + 1 ⎞ 1 =⎜ ⎟ ⎝ 3â ⎠ C0

and Ceq = [3k/(2k + 1)]C0 . 16.62

For the parallel combination: CP = C1 + C2 which gives C2 = CP − C1

[1]

For the series combination:

1 1 1 1 1 1 C1 − Cs = + !or! = − = C C C C C C CsC1 s 1 2 2 s 1 !

CsC1 Thus, we have C2 = and equating this to Equation [1] above gives C1 − Cs ! CC Cp − C1 = s 1 C1 − Cs ! We write this result as:

CpC1 − CpCs − C12 + CsC1 = CsC1

C 2 − CpC1 + CpCs = 0 ! 1

and use the quadratic formula to obtain

1 1 2 C1 = Cp ± Cp − CpCs 2 4 !

Topic 16

999

Then, Equation [1] gives

16.63

1 1 2 C2 = C p ∓ C − C pCs 2 4 p

For a parallel-plate capacitor with plate separation d,

ΔVmax = Emax ⋅ d

ΔVmax d= Emax !

The capacitance is then

âÉ 0 A d

⎛ E ⎞ = âÉ 0 A ⎜ max ⎟ ⎝ ΔVmax ⎠

and the needed area of the plates is A = C ⋅ ΔVmax/ke0 Emax, or

( 0.250 × 10−6 F)( 4.00 × 103 V ) A= = 0.188 m 2 −12 2 2 8 ( 3.00)(8.85 × 10 C N ⋅ m )( 2.00 × 10 V/m ) ! 16.64 (a) The 1.0-µC is located 0.50 m from point P, so its contribution to the potential at P is ⎛ 1.0 × 10−6 C ⎞ q V1 = ke 1 = ( 8.99 × 109 N ⋅ m 2 C2 ) ⎜ = 1.8 × 10 4 V ⎟ r1 0.50 m ⎝ ⎠ !

(b) The potential at P due to the −2.0-µC charge located 0.50 m away is −6 q2 C⎞ 9 2 2 ⎛ −2.0 × 10 V2 = ke = ( 8.99 × 10 N ⋅ m C ) ⎜ = −3.6 × 10 4 V ⎟ r2 ⎝ 0.50 m ⎠ !

Topic 16

1000

(c) The total potential at point P is VP = V1 + V2 = (+1.8 − 3.6) × 104 V = −1.8 × 104 V (d) The work required to move a charge q = 3.0 µC to point P from infinity is W = qΔV = q(VP − V∞) = (3.0 × 10−6 C)(−1.8 × 104 V − 0) = −5.4 × 10−2 J 16.65

The stages for the reduction of this circuit are shown below.

Thus, 16.66

Ceq = 6.25 µF

(a) Due to spherical symmetry, the charge on each of the concentric spherical shells will be uniformly distributed over that shell. Inside a spherical surface having a uniform charge distribution, the electric field due to the charge on that surface is zero. Thus, in this region, the potential due to the charge on that surface is constant and equal to the potential at the surface. Outside a spherical surface having a

Topic 16

1001

uniform charge distribution, the potential due to the charge on that surface is given by V = keq/r, where r is the distance from the center of that surface and q is the charge on that surface. In the region between a pair of concentric spherical shells, with the inner shell having charge +Q and the outer shell having radius b and charge −Q, the total electric potential at distance r from the center is given by

k Q k (−Q) ⎛ 1 1⎞ V = Vdue!to + Vdue!to = e + e = keQ ⎜ − ⎟ ⎝ r b⎠ r b inner!shell outer!shell ! The potential difference between the two shells is therefore ⎛ 1 1⎞ ⎛ 1 1⎞ ⎛ b − a⎞ ΔV = V r=a − V r=b = keQ ⎜ − ⎟ − keQ ⎜ − ⎟ = keQ ⎜ ⎝ ⎠ ⎝ ⎠ ⎝ ab ⎟⎠ a b b b !

The capacitance of this device is given by

Q ab C= = ΔV ke (b − a) ! (b) When b >> a, then b − a ≈ b. Thus, in the limit as b → ∞, the capacitance found above becomes

C→

ab ke (b)

a ke

= 4π! 0 a

Topic 16

16.67

1002

The energy stored in a charged capacitor is !Estored = 12 C(ΔV ) . Hence, 2

ΔV = ! 16.68

2 ( 300 J ) 2Estored = = 4.47 × 103 V = 4.47 kV C 30.0 × 10−6 F

From Q = C(ΔV), the capacitance of the capacitor with air between the plates is Q 150 µC C0 = 0 = ΔV ΔV !

After the dielectric is inserted, the potential difference is held to the original value, but the charge changes to Q = Q0 + 200 µC = 350 µ. Thus, the capacitance with the dielectric slab in place is Q 350 µC C= = ΔV ΔV !

The dielectric constant of the dielectric slab is therefore

â= = 16.69

⎛ 350 µC ⎞ ⎛ ΔV ⎞ 350 =⎜ = 2.33 ⎟= ⎟⎜ C0 ⎝ ΔV ⎠ ⎝ 150 µC ⎠ 150 C

The charges initially stored on the capacitors are Q1 = C1(ΔV)i = (6.0 µF)(250 V) = 1.5 × 103 µC and Q2 = C2(ΔV)i = (2.0 µF)(250 V) = 5.0 × 102 µC When the capacitors are connected in parallel, with the negative plate of

Topic 16

1003

one connected to the positive plate of the other, the net stored charge is Q = Q1 − Q2 = 1.5 × 103 µC − 5.0 × 102 µC = 1.0 × 103 µC The equivalent capacitance of the parallel combination is Ceq = C1 + C2 = 8.0 µF. Thus, the final potential difference across each of the capacitors is

Q 1.0 × 103 µC (ΔV )′ = = = 125 V Ceq 8.0 µF ! and the final charge on each capacitor is Q′ = C1 (ΔV )′ = (6.0 µF)(125 V) = 750 µC = 0.75 mC ! 1

and 16.70

Q′ = C2 (ΔV )′ = (2.0 µF)(125 V) = 250 µC = 0.25 mC ! 2

(a) The distance from the charge 2q to either of the charges on the y-axis is r = d 2 + (2d)2 = 5 d . Thus, !

kq kq kq 2ke q V =∑ e i = e + e = ri 5d 5d 5d i !

ke q1q3 ke q2 q3 ke q(2q) ke q(2q) 4ke q2 + = + = (b) PE2q = r1 r2 5d 5d 5d ! (c) From conservation of energy with PE = 0 at r = ∞,

KE f = KEi + PEi − PE f = 0 + !

4ke q2 4ke q2 −0= 5d 5d

Topic 16

1004

(d) v f =

(

2 KE f m

16.71

) = 2 ⎛ 4k q ⎞ = ⎛ 8k q ⎞ 2

m ⎜⎝ 5 d ⎟⎠

1 2

⎜⎝ 5 md ⎟⎠

When excess charge resides on a spherical surface that is far removed from any other charge, this excess charge is uniformly distributed over the spherical surface, and the electric potential at the surface is the same as if all the excess charge were concentrated at the center of the spherical surface. In the given situation, we have two charged spheres, initially isolated from each other, with charges and potentials of QA = +6.00 µC and VA = keQA/RA, where RA = 12.0 cm, QB = −4.00 µ, and VB = keQB/RB, with RB = 18.0 cm. When these spheres are then connected by a long conducting thread, the charges are redistributed (yielding charges of !QA′ and !QB′ , respectively) until the two surfaces come to a common potential

(V ′ = kQ′ R = V ′ = kQ′

RB ) . When equilibrium is established, we have:

From conservation of charge: !Q′A + QB′ = QA + QB ⇒

!Q′A + QB′ = +2.00 µC

[1]

From equal potentials:

Topic 16

1005

kQ′A kQB′ = R RB A !

⎛R ⎞ ⇒ QB′ = ⎜ B ⎟ Q′A ⎝ RB ⎠

or QB′ = 1.50 Q′A

[2]

Q′A = !

+2.00 µC = 0.800 µC 2.50

Substituting Equation [2] into [1] gives

Q′ = 1.50(0.800 µC) = 1.20 µC ! B

Then, Equation [2] gives 16.72

The electric field between the plates is directed downward with magnitude

Ey = !

ΔV 100 V = = 5.0 × 10 4 N/m −3 D 2.0 × 10 m

Since the gravitational force experienced by the electron is negligible in comparison to the electrical force acting on it, the vertical acceleration is Fy

qEy

ay = = m me e !

(a)

( −1.60 × 10 =

−19

C ) ( −5.0 × 10 4 N/m )

9.11 × 10−31 kg

= +8.8 × 1015 m s 2

At the closest approach to the bottom plate, vy = 0. Thus, the vertical 2 2 displacement from point O is found from !vy = v0y + 2ay (Δy) as

2 − ⎡⎣ − ( 5.6 × 106 m/s ) sin!45° ⎤⎦ 0 − ( v0 sin θ 0 ) Δy = = = −8.9 × 10−4 m = −0.89 mm 15 2 2ay 2 ( 8.8 × 10 m/s ) ! 2

The minimum distance above the bottom plate is then D d = + Δy = 1.0 mm − 0.89 mm = 0.1 mm 2 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 16

1006

(b) The time for the electron to go from point O to the upper plate (Δy = 2 +1.0 mm) is found from !Δy = v0yt + 12 ayt as

m⎞ m⎞ ⎡ ⎛ ⎤ 1⎛ +1.0 × 10−3 m = ⎢ − ⎜ 5.6 × 106 ⎟ sin 45° ⎥ t + ⎜ 8.8 × 1015 2 ⎟ t 2 ⎝ ⎠ ⎝ s s ⎠ ⎣ ⎦ 2 !

Solving for t gives a positive solution of t = 1.1 × 10−9 s. The horizontal displacement from point O at this time is Δx = v0xt = [(5.6 × 106 m/s)cos 45°](1.1 × 10−9 s) = 4.4 mm

Topic 17

1007

Topic 17 Current and Resistance

QUICK QUIZZES 17.1

Choice (d). Negative charges moving in one direction are equivalent to positive charges moving in the opposite direction. Thus, Ia, Ib, Ic, and Id are equivalent to the movement of 5, 3, 4, and 2 charges respectively, giving Id < Ib < Ic < Ia .

17.2

Choice (b). Under steady-state conditions, the current is the same in all parts of the wire. Thus, the drift velocity, given by vd = I/nqA, is inversely proportional to the cross-sectional area.

17.3

Choices (c) and (d). Neither circuit (a) nor circuit (b) applies a difference in potential across the bulb. Circuit (a) has both lead wires connected to the same battery terminal. Circuit (b) has a low resistance path (a “short”) between the two battery terminals as well as between the bulb terminals.

17.4

Choice (b). The slope of the line tangent to the curve at a point is the reciprocal of the resistance at that point. Note that as ΔV increases, the slope (and hence 1/R) increases. Thus, the resistance decreases.

Topic 17

1008

17.5

Choice (b). From Ohm’s Law, R = ΔV/I = 120 V/6.00 A = 20.0 Ω.

17.6

L L Choice (b). Consider the expression for resistance: R = ρ = ρ 2 . A πr !

Doubling all linear dimensions increases the numerator of this expression by a factor of 2 but increases the denominator by a factor of 4. Thus, the net result is that the resistance will be reduced to one-half of its original value. 17.7

Choice (a). The resistance of the shorter wire is half that of the longer wire. The power dissipated, P = (ΔV)2/R, (and hence the rate of heating) will be greater for the shorter wire. Consideration of the expression P = I2R might initially lead one to think that the reverse would be true. However, one must realize that the currents will not be the same in the two wires.

17.8

Choice (b). Ia = Ib > Ic = Id > Ie = If. Charges constituting the current Ia leave the positive terminal of the battery and then split to flow through the two bulbs; thus, Ia = Ic + Ie. Because the potential difference ΔV is the same across the two bulbs and because the power delivered to a device is P = I(ΔV), the 60-W bulb with the higher power rating must carry the greater current, meaning that Ic > Ie. Because charge does not accumulate in the

Topic 17

1009

bulbs, all the charge flowing into a bulb from the left has to flow out on the right; consequently Ic = Id and Ie = If. The two currents leaving the bulbs recombine to form the current back into the battery, If + Id = Ib. 17.9

Choice (a). The power dissipated by a resistor may be expressed as P = I2R, where I is the current carried by the resistor of resistance R. Since resistors connected in series carry the same current, the resistor having the largest resistance will dissipate the most power.

17.10

Choice (c). Increasing the diameter of a wire increases the cross-sectional area. Thus, the cross-sectional area of A is greater than that of B, and from R = ρL/A, we see that RA < RB. Since the power dissipated in a resistance may be expressed as P = (ΔV)2/R, the wire having the smallest resistance dissipates the most power for a given potential difference.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 17.2

(a) D. Current is inversely proportional to resistance. (b) U. The voltage across the device will equal the battery voltage. (c) D. From P = I ΔV, a decrease in current with constant voltage results in a power decrease.

Topic 17

17.4

1010

(a) The number of cars would correspond to charge Q. (b) The rate of flow of cars past a point would correspond to current.

17.6

(a) The 25 W bulb has the higher resistance. Because R = (ΔV)2/P, and both operate from 120 V, the bulb dissipating the least power has the higher resistance. (b) When the voltage is constant, the current and power are directly proportional to each other, P = (ΔV)I = (constant)I. Thus, the higher power bulb (100 W) carries more current.

17.8

An electrical shock occurs when your body serves as a conductor between two points having a difference in potential. The concept behind the admonition is to avoid simultaneously touching points that are at different potentials.

17.10 (a) D. Current is inversely proportional to resistance. (b) U. The voltage across the device will equal the battery voltage. (c) D. From P = IΔV, a decrease in current with constant voltage results in a power decrease.

Topic 17

1011

ANSWERS TO EVEN NUMBERED PROBLEMS 17.2

(a)

4.28 × 10−5 m/s

17.4

80.0 A

17.6

159 mA

17.8

0.130 mm/s

17.10

8.89 Ω

(b)

2.23 × 10−5 m/s

17.12 (a)

5.43 × 103 A/m2

(b)

0.109 A

17.14 (a)

1.8 m

(b)

0.28 mm

(b)

1.8 × 107 A

17.16

3.22 × 10−8 Ω ⋅ m

17.18

3.4 × 1021 electrons

17.20 (a)

2.8 × 108A

17.22 (a)

5.0 × 105 V

(b)

Rubber gloves and soles can increase resistance to current passing through the body.

17.24

6.3 Ω

Topic 17

1012

17.26 (a) (b)

1.6 × 102 °C The expansion of the cross-sectional area contributes slightly more than the expansion of the length of the wire, so the answer would be slightly reduced.

17.28 2.3 × 102 °C 17.30 1.1 × 10−3 (°C)−1 17.32 (a) Yes, the design goal can be met with R0,Carbon = 4.4 Ω and R0,Nichrome = 5.6 Ω. (b) Lcarbon = 0.89 m, LNichrome = 26 m 17.34 63.3°C 17.36 (a)

$0.29

(b)

$2.6

17.38 (a)

72.0 W

(b)

29.0 min

17.40 (a)

0.263 A

(b)

1.80 × 104 J

17.42 (a)

184 W

(b)

461°C

17.44 (a)

0.66 kWh

(b)

$0.079 = 7.9 cents

17.46 (a)

$1.61

(b)

0.582 cents

(c)

$0.416 = 41.6 cents

Topic 17

1013

17.48 (a) (d) 17.50 (a)

3.5 × 102 W

(b)

41 Ω

d = 4 ρ0 L π R0 !

(e)

0.38 mm

2.7 × 10−3 Ω

(b)

1.45 × 103 K

(c)

4.0 × 101 Ω

(c)

1.7 × 10−2 Ω

(d)

1.1 V

(e)

They radiate only a small portion of the energy consumed as visible light.

17.52

15.0 h

17.54

1.5 × 102 °C

17.56

90 µV

17.58 (a)

9.3 m

(b)

0.93 mm

17.60 (a)

18 C

(b)

3.6 A

17.62

9.1 Ω

(a)

(b) We assume the temperature coefficient of resistivity for Nichrome remains constant over this temperature range. 17.64

No, the fuse should limit the current to 3.87 A or less.

17.66 (a)

See Solution.

(b)

1.420 Ω versus 1.418 Ω

Topic 17

1014

17.68 (a)

470 W

(c)

(b)

1.60 mm or more

2.93 mm or more

PROBLEM SOLUTIONS 17.1

(a) The charge that moves past the cross section is ΔQ = I(Δt), and the number of electrons is

n= =

ΔQ I(Δt) = |e| |e| (80.0 × 10−3 C/s )[(10.0 min)(60.0!s/min)]

1.60 × 10−19 C

= 3.00 × 1020 electrons

(b) The negatively charged electrons move in the direction opposite to the conventional current flow.

17.2

The wire’s cross-sectional area is A = πr2 = π(1.25 × 10−3 m)2 = 4.91 × 10−6 m2. (a) Solve for the drift speed from the current expression I = nqvdA:

vd =

I 3.70 A = 29 3 nqA 1.10 × 10 electrons/m 1.60 × 10−19 C 4.91 × 10−6 m 2

(

)(

)

= 4.28 × 10−5 m/s (b) Substitute the new value of n to find

Topic 17

1015

vd =

I 3.70 A = 29 3 nqA 2.11 × 10 electrons/m 1.60 × 10−19 C 4.91 × 10−6 m 2

(

)(

)

= 2.23 × 10−5 m/s 17.3

The period of the electron in its orbit is T = 2πr/v, and the current represented by the orbiting electron is I = ΔQ/Δt = |e|/T = v|e|/2πr, or

( 2.19 × 10 m/s )(1.60 × 10 I= 2π ( 5.29 × 10 m ) ! 6

−19

−11

17.4

= 1.05 × 10−3 C/s = 1.05 mA

Use the definition of average current to find

(

)(

)

20 −19 ΔQ 1.00 × 10 electrons 1.60 × 10 C/electron I av = = = 80.0 A Δt 0.200 s

17.5

The resistance of a wire of length L and diameter d is R = ρL/A =

ρL/(πd2/4), giving d2 = 4ρL/πR. Using Table 17.1, the needed diameter is found to be

d= !

17.6

4ρL = πR

4 ( 2.82 × 10−8 Ω ⋅ m ) ( 32.0 m )

π ( 2.50 Ω )

= 6.78 × 10−4 m = 0.678 mm

The mass of a single gold atom is matom = M/NA, where M is the molecular weight of gold and NA is Avogadro’s number. Thus, the number of atoms deposited, and hence the number of ions moving to the negative electrode, is

Topic 17

1016 −3 23 m mNA ( 3.25 × 10 kg ) ( 6.02 × 10 atoms/mol ) n= = = = 9.93 × 1021 −3 matom M (197 g/mol )(10 kg/g ) !

The current in the cell is then

(9.93 × 10 )(1.60 × 10 C) = 0.159 A= 159 mA I= = = Δt Δt (2.78 h )(3 600 s/1 h ) ΔQ

17.7

−19

The drift speed of electrons in the line is vd = I/nqA = I/n|e|(πd2/4). The time to travel the length of the 200-km line is then Δt = L/vd = Ln|e|(πd2)/4I, or

( 200 × 10 m )(8.5 × 10 m )(1.6 × 10 C)π ( 0.02 m ) = 27 yr Δt = 4 (1 000 A ) ( 3.156 × 10 s/yr ) ! 3

−19

17.8

The mass of a single aluminum atom is matom = M/NA, where the molecular weight (mass per mole) of aluminum is M = 27.0 g/mol, and Avogadro’s number (number of atoms per mole) is NA = 6.02 × 1023/mol. If each aluminum atom contributes one conduction electron, the number of conduction electrons per unit volume is

n= !

mass!per!unit!volume density ρ NA = = mass!per!atom matom M

and the drift speed is vd = I/nqA = MI/ρNAqA. Thus, for the given case, we find

Topic 17

1017

vd = ! or 17.9

( 27.0 g / mol )( 5.00 C / s )

g ⎞ ⎛ 10 cm ⎞ ⎛ 6.02 × 1023 ⎞ ⎛ 1.60 × 10−19 C ) ( 4.00 × 10−6 m 2 ) ( ⎜⎝ 2.70 3⎟⎜ 3 ⎟⎜ ⎟ ⎠ cm ⎝ 1 cm ⎠ ⎝ mol ⎠ 6

vd = 1.30 × 10−4 m/s = 0.130 mm/s.

(a) Using the periodic table on the inside back cover of the textbook, we find MFe = 55.85 g/mol = (55.85 g/mol)(1 kg/103 g) = 55.85 × 10−3 kg/mol (b) From Table 9.1, the density of iron is ρFe = 7.86 × 103 kg/m3, so the molar density is

ρFe 7.86 × 103 kg m 3 molar!density = = ( )Fe M 55.85 × 10−3 kg mol = 1.41 × 105 mol m 3 Fe ! (c) The density of iron atoms is

density of atoms = N A (molar density)

⎛ ⎞ atoms ⎞ ⎛ 5 mol = ⎜ 6.02 × 1023 1.41 × 10 ⎟⎜ ⎟ mol ⎠ ⎝ m3 ⎠ ⎝ = 8.49 × 1028

atoms m3

(d) With two conduction electrons per iron atom, the density of charge carriers is

Topic 17

1018

n = (charge!carriers/atom)(density!of!atoms) !

atoms ⎞ ⎛ electrons ⎞ ⎛ 29 3 = ⎜2 8.49 × 1028 ⎟ ⎜ ⎟ = 1.70 × 10 electrons m ⎝ atom ⎠ ⎝ m3 ⎠

(e) With a current of I = 30.0 A and cross-sectional area A = 5.00 × 10−6 m2, the drift speed of the conduction electrons in this wire is

I 30.0 C/s vd = = = 2.21 × 10−4 m/s 29 −3 nqA (1.70 × 10 m ) (1.60 × 10−19 C ) ( 5.00 × 10−6 m 3 ) ! 17.10

ΔV 1.20 × 102 V From Ohm’s law, R = = = 8.89 Ω I 13.5 A !

17.11 (a) Solve for the resistance R using Ohm’s law to find ΔV = IR → R =

ΔV 115 V = = 2.20 × 10 2 Ω I 0.522 A

(b) The wire’s cross-sectional area is A = πr2 = π(2.30 × 10−5 m)2 = 1.66 × 10−9 m2. Solve for the resistivity using

R=ρ

ℓ A

(

RA ΔV π R ρ= = ℓ Iℓ

) = (115 V )π ( 2.30 × 10 m) −5

(0.522 A )(0.600 m )

= 6.10 × 10−7 Ω ⋅ m 17.12 (a) Substitute R = ρℓ/A into Ohm’s law and solve for the current density given by J = I/A:

Topic 17

1019

ΔV = IR = I

ρℓ = Jρℓ A

I ΔV 5.00 V = = = 5.43 × 103 A/m 2 A ρ ℓ ( 0.460 Ω ⋅ m ) 2.00 × 10−3 m

(

)

(b) Assuming the current density is constant across the area, the current is I = JA = (5.43 × 103 A/m2)(2.00 × 10−5 m2) = 0.109 A . 17.13

(ΔV)max = ImaxR = (80 × 10−6 A)R Thus, if R = 4.0 × 105 Ω, (ΔV)max = 32 V and if R = 2 000 Ω, (ΔV)max = 0.16 V

17.14

The volume of the copper is

m 1.00 × 10−3 kg V= = = 1.12 × 10−7 m 3 3 3 density 8.92 × 10 kg/m ! Since V = A ⋅ L, this gives A⋅L = 1.12 × 10−7 m3.

[1]

ρL (a) From R = , where ρ is the resistivity of copper, we find that A ! ⎛ 1.7 × 10−8 Ω ⋅ m ⎞ ⎛ ρ⎞ A=⎜ ⎟L=⎜ L = ( 3.4 × 10−8 m ) L ⎟ ⎝ R⎠ 0.500 Ω ⎝ ⎠ !

Inserting this expression for A into Equation [1] gives (3.4 × 10−8 m)L2 = 1.12 × 10−7 m3, which yields L = 1.8 m

Topic 17

1020

π d 2 1.12 × 10−7 m 3 = (b) From Equation [1], A = , or 4 L !

(

4 1.12 × 10−7 m 3

πL

) = 4(1.12 × 10 m ) π (1.8 m ) −7

= 2.8 × 10−4 m = 0.28 mm ΔV 1.20 × 102 V R = = = 13.0 Ω 17.15 (a) From Ohm’s law, I 9.25 A ! (b) Using R = ρL/A and data from Table 17.1, the required length is found to be 2 −3 RA R (π r ) (13.0 Ω ) π ( 0.791 × 10 m ) L= = = = 17.0 m −8 ρ ρ 150 × 10 Ω ⋅ m ! 2

17.16

From R = ρL/A = ρL/(πd2/4), the resistivity is −3 π Rd 2 π (1.60 Ω ) ( 0.800 × 10 m ) ρ= = = 3.22 × 10−8 Ω ⋅ m 4L 4 ( 25.0 m ) ! 2

17.17

ΔV 12 V = = 30 Ω (a) R = I 0.40 A ! (b) From R = ρL/A, 2 −2 ⎡ ⎤ R ⋅ A ( 30 Ω ) ⎣π ( 0.40 × 10 m ) ⎦ ρ= = = 4.7 × 10−4 Ω ⋅ m L 3.2 m !

Topic 17

1021

17.18 Find the total charge using the definition of average current and the conversion 1 hr = 3 600 s: ΔQ = IΔt = (0.15 A)(3 600 s) = 5.40 × 102 C. The number N of electrons is then

ΔQ 5.40 × 10 2 C ΔQ = Ne → N = = = 3.4 × 10 21 −19 e 1.60 × 10 C/electron 17.19

From Ohm’s law, R = ΔV/I, and R = ρL/A = ρL/(πd2/4), the resistivity is

π Rd 2 π (ΔV )d 2 π (9.11 V)(2.00 × 10−3 m)2 ρ= = = = 1.59 × 10−8 Ω ⋅ m 4L 4IL 4(36.0 A)(50.0 m) ! Then, from Table 17.1, we see that the wire is made of silver. 17.20

With different orientations of the block, three different values of the ratio L/A are possible. These are:

10 cm 1 1 ⎛ L⎞ = = ⎜⎝ ⎟⎠ = , A 1 ( 20 cm ) ( 40 cm ) 80 cm 0.80 m ! 20 cm 1 1 ⎛ L⎞ = = ⎜⎝ ⎟⎠ = , A 2 (10 cm ) ( 40 cm ) 20 cm 0.20 m ! and

40 cm 1 1 ⎛ L⎞ = = ⎜⎝ ⎟⎠ = ! A 3 (10 cm)(20 cm) 5.0 cm 0.050 m

ΔV ΔV (60 V)(0.80 m) = = = 2.8 × 108 A (a) I max = −8 R ρ L/A 1.7 × 10 Ω ⋅ m ( ) min min ! ΔV ΔV (6.0 V)(0.050 m) = = 1.8 × 107 A (b) I min = R = ρ L/A −8 1.7 × 10 Ω ⋅ m ( ) max max ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 17

1022

17.21 The wire’s cross-sectional area is A = πr2 = π(1.29 × 10−3 m)2 = 5.23 × 10−6 m2. (a) Its resistance is

(

)

−8 ρ L 1.70 × 10 Ω ⋅ m (1.00 m ) R= = = 3.25 × 10−3 Ω −6 2 A 5.23 × 10 m

(b) The wire’s volume is V = AL = (5.23 × 10−6 m2)(1.00 m) = 5.23 × 10−6 m 3 (c) For the same volume of wire, its new cross-sectional area is Anew = V/Lnew = (5.23 × 10-6 m3)/(2.00 m) = 2.62 × 10−6 m2 and the new resistance is

(

)

−8 ρ Lnew 1.70 × 10 Ω ⋅ m ( 2.00 m ) Rnew = = = 1.30 × 10−2 Ω −6 2 Anew 2.62 × 10 m

Alternatively, substitute the relationships symbolically to find: Rnew =

17.22

ρ ( 2L ) −2 ρ Lnew ρL ⎛ L⎞ = = 4ρL ⎜ ⎟ = 4 = 4R = 1.30 × 10 Ω ⎝V⎠ Anew V /( 2L ) A

(a) From Ohm’s law, ΔV = IR = (500 × 10−3 A)(1.0 × 106 Ω) = 5.0 × 105 V (b) Rubber-soled shoes and rubber gloves can increase the resistance to current and help reduce the likelihood of a serious shock.

17.23

If a conductor of length L has a uniform electric field E maintained within it, the -potential -difference between the ends of the conductor is ΔV = EL. But, from Ohm’s law, the -relation between the potential difference across a conductor and the current through it is ΔV = IR, where R = ρL/A.

Topic 17

1023

Combining these relations, we obtain ΔV = EL = IR = I(ρL/A) 17.24

E = ρ(I/A) = ρJ

Using R = R0[1 + α(T − T0)], with R0 = 6.00 Ω at T0 = 20.0°C and αsilver = 3.8 × 10−3 (°C)−1 (from Table 17.1 in the textbook), the resistance at T = 34.0°C is R = (6.00 Ω)[1 + 3.8 × 10−3 (°C)−1(34.0°C − 20.0°C)] = 6.3 Ω

17.25 Solve for the temperature coefficient using the relation

R = R0 ⎡⎣1 + α (T − T0 ) ⎤⎦

α=

17.26

(a)

R − R0 2.00 × 10 2 Ω −1 = = 5.08 × 10−3 ( °C ) R0 (T − T0 ) ( 75.0 Ω ) ( 525°C )

Given: Aluminum wire with α = 3.9 × 10−3 (°C)−1 (see Table 17.1 in textbook), and R0 = 30.0 Ω at T0 = 20.0°C. If R = 46.2 Ω at temperature T, solving R = R0[1 + α(T − T0)] gives the final temperature as

T = T0 + !

( R/R0 ) − 1 = 20.0°C + ( 46.2 Ω/30.0 Ω) − 1 = 1.6 × 102 °C α

3.9 × 10−3 (°C)−1

(b) The expansion of the cross-sectional area contributes slightly more than the expansion of the length of the wire, so the answer would be slightly reduced.

Topic 17

17.27

1024

The volume of the gold wire may be written as V = A ⋅ L = m/ρd, where ρd is the density of gold. Thus, the cross-sectional area is A = m/ρdL. The resistance of the wire is R = ρeL/A, where ρe is the electrical resistivity. Therefore, −8 3 3 3 ρe L ρ ρ L2 ( 2.44 × 10 Ω ⋅ m ) (19.3 × 10 kg m ) ( 2.40 × 10 m ) R= = e d = m ρd L m 1.00 × 10−3 kg !

giving 17.28

R = 2.71 × 106 Ω = 2.71 MΩ

For aluminum, the resistivity at room temperature is ρ0,Al = 2.82 × 10−8 Ω⋅m, and the temperature coefficient of resistivity is αAl = 3.9 × 10−3 (°C)−1. Thus, if at some temperature, the aluminum has a resistivity of ρAl, solving ρAl = ρ0,Al [1 + αAl(T − T0)] for that temperature gives T = T0 + {[(ρAl/ρ0,Al) − 1]/αAl}, where T0 = 20°C. When ρAl = 3ρ0,Cu = 3(1.7 × 10−8 Ω ⋅ m) = 5.1 × 10−8 Ω ⋅ m, the temperature must be

T = 20°C + ! 17.29

⎛ 5.1 × 10−8 Ω ⋅ m ⎞ ⎜⎝ 2.82 × 10−8 Ω ⋅ m ⎟⎠ − 1 −3

3.9 × 10 (°C)

−1

= 2.3 × 102 °C

At 80°C,

Topic 17

1025

I= !

ΔV ΔV 5.0 V = = R R0 ⎡⎣1 + α (T − T0 ) ⎤⎦ ( 2.0 × 102 Ω ) ⎡⎣1 + ( −0.5 × 10−3 °C−1 ) ( 80°C − 20°C ) ⎤⎦

or 17.30

I = 2.6 × 10−2 A = 26 mA

If R = 41.0 Ω at T = 20.0°C and R = 41.4 Ω at T = 29.0°C, then R = R0[1 + α(T − T0)] gives the temperature coefficient of resistivity of the material making up this wire as R − R0 41.4 Ω − 41.0 Ω α= = = 1.1 × 10−3 (°C)−1 R0 (T − T0 ) ( 41.0 Ω ) ( 29.0°C − 20.0°C ) !

17.31

(a) The resistance at 20.0°C is

R0 = ρ !

−8 L (1.7 × 10 Ω ⋅ m ) ( 34.5 m ) = = 3.0 Ω 2 A π ( 0.25 × 10−3 m )

ΔV 9.0 V = = 3.0 A and the current will be I = R0 3.0 Ω ! (b) At 30.0°C, R = R0[1 + α(T − T0)] = (3.0 Ω)[1 + (3.9 × 10−3 (°C)−1)(30.0°C − 20.0°C)] = 3.1 Ω

ΔV 9.0 V = = 2.9 A . Thus, the current is I = R 3.1 Ω ! 17.32

We call the carbon resistor 1 and the Nichrome resistor 2. Then,

Topic 17

1026

connecting the resistors end to end, the total resistance is R = R1 + R2 = R01[1 + α1 ⋅ ΔT] + R02[1 + α2 ⋅ ΔT] = R01 + R02 + (R01α1 + R02α2) ΔT If this is not to vary with temperature, it is necessary that R01α1 + R02α2 = 0, or

⎛ 0.4 × 10−3 ( °C )−1 ⎞ ⎛α ⎞ R01 = − ⎜ 2 ⎟ R02 = − ⎜ −1 R02 = 0.8R02 ⎝ α1 ⎠ −0.5 × 10−3 ( °C ) ⎟⎠ ⎝ ! If the constant total resistance is to be R = 10.0 Ω, it is necessary that R01 + R02 = 0.8R02 + R02 = 10.0 Ω

R02 = !

10.0 Ω = 5.6 Ω 1.8

and R01 = 10.0 Ω − R02 = 4.4 Ω

(a) Yes it is possible for her to meet the design goal with this method. (b) From R = ρL/A = ρL/πr2, we have L = (πr2) ⋅ R/ρ. This yields 2 R 4.4 Ω L1 = (π r 2 ) 01 = π (1.50 × 10−3 m ) = 0.89 m (Carbon) −5 ρ 3.5 × 10 Ω ⋅ m 01 !

and 2 R 5.6 Ω L2 = (π r 2 ) 02 = π (1.50 × 10−3 m ) = 26 m (Nichrome) ρ02 150 × 10−8 Ω ⋅ m !

17.33

(a) From R = ρL/A, the initial resistance of the mercury is

Topic 17

1027

(9.4 × 10 Ω ⋅ m)(1.000 0 m) = 1.2 Ω R = = A π (1.00 × 10 m ) 4 −7

ρ Li

−3

(b) Since the volume of mercury is constant, V = Af ⋅ Lf = Ai ⋅ Li gives the final cross-sectional area as Af = Ai ⋅ (Li/Lf). Thus, the final resistance is 2 given by !R f = ρ L f A f = ρ L f Ai ⋅ Li . The fractional change in the

resistance is then

R f − Ri

Δ= !

Rf Ri

−1=

ρ L2f ( Ai ⋅ Li ) ρ Li Ai

⎛ Lf ⎞ −1= ⎜ ⎟ −1 ⎝L ⎠ i

⎛ 100.04 ⎞ −4 Δ=⎜ ⎟⎠ − 1 = 8.0 × 10 or a 0.080% increase ⎝ 100.00 !

17.34

The resistance at the reference temperature of 20.0°C is

R 200.0 Ω R0 = = = 217 Ω −3 1 + α (T − T0 ) 1 + ⎡⎣ 3.92 × 10 (°C)−1 ⎤⎦ ( 0°C − 20.0°C ) ! Solving R = R0[1 + α(T − T0)] for T gives the temperature of the melting potassium as T = T0 + !

R − R0 253.8 Ω − 217 Ω = 20.0°C + = 63.3°C α R0 ⎡ 3.92 × 10−3 ( °C )−1 ⎤ ( 217 Ω ) ⎣ ⎦

17.35 (a) The maximum power delivered by the power supply is Pmax = ImaxΔV = (10.0 A)(5.00 V) = 50.0 W .

Topic 17

1028

(b) The number N of cell phone chargers, each of power Pcharger, is given by: Pmax = N(Pcharger). Solve for the number N to find

17.36

Pmax 50.0 W = = 25.0 Pcharger 2.00 W

(a) The energy used by a 100-W bulb in 24 h is E = P ⋅ Δt = (100 W)(24 h) = (0.100 kW)(24 h) = 2.4 kWh and the cost of this energy, at a rate of $0.12 per kilowatt-hour is cost = E ⋅ rate = (2.4 kWh)($0.12/kWh) = $0.29 (b) The energy used by the oven in 5.0 h is

⎡ ⎛ 1 kW ⎞ ⎤ E = P ⋅ Δt = [ I ( ΔV )] ⋅ Δt = ⎢( 20.0 C/s ) ( 220 J/C ) ⎜ 3 ⎟ ⎥ ( 5.0 h ) = 22 kWh ⎝ 10 J/s ⎠ ⎦ ⎣ ! and the cost of this energy, at a rate of $0.12 per kilowatt-hour is cost = E ⋅ rate = (22 kWh)($0.12/kWh) = $2.6

17.37

−8 ρCu L 4 ρCu L 4 (1.7 × 10 Ω ⋅ m ) (1.00 m ) = = = 5.2 × 10−3 Ω (a) RCu = 2 −2 A π d2 π ( 0.205 × 10 m ) !

and PCu = I2RCu = (20.0 A)2(5.2 × 10−3 Ω) = 2.1 W −8 ρ Al L 4 ρ Al L 4 ( 2.82 × 10 Ω ⋅ m ) (1.00 m ) = = = 8.54 × 10−3 Ω (b) RAl = 2 −2 A π d2 π ( 0.205 × 10 m ) !

Topic 17

1029

and PAl = I2RAl = (20.0 A)2(8.54 × 10−3 Ω) = 3.42 W (c) No, the aluminum wire would not be as safe. If surrounded by thermal insulation, it would get much hotter than the copper wire. 17.38 (a) The consumed power is P = ( ΔV ) /R = (12.0 V ) /2.00 Ω = 72.0 W . 2

(b) The energy required to heat the coffee is Q = mcΔT = (1.00 kg)(4184 J/kg⋅°C)(50.0°C − 20.0°C) = 1.26 × 105 J. Use the definition of average power to solve for the time Δt: Pav =

17.39

Q Q mcΔT → Δt = = = 1.74 × 103 s = 29.0 min Δt Pav 72.0 W

The energy required to bring the water to the boiling point is E = mc(ΔT) = (0.500 kg)(4 186 J/kg⋅°C)(100°C − 23.0°C) = 1.61 × 105 J The power input by the heating element is Pinput = (ΔV)I = (120 V)(2.00 A) = 240 W = 240 J/s Therefore, the time required is ⎛ 1 min ⎞ E 1.61 × 105 J t= = = 671 s ⎜ = 11.2 min Pinput 240 J/s ⎝ 60 s ⎟⎠ !

17.40 (a) From P = IΔV, the average current is I = P/ΔV = (1.00 W)/(3.80 V) =

0.263 A .

Topic 17

1030

(b) By conservation of energy, the energy stored in a fully-charged battery will equal the energy consumed when the battery is “dead” and requires charging. This energy is E = PΔt = (1.00 W)(5.00 hr⋅3600 s/hr) = 1.80 × 10 4 J . 2 ΔV ) (120 V ) ( = = 576 Ω (a) R = 2

17.41

25.0 W

2 ΔV ) (120 V ) ( R = = = 144 Ω 2

and

100 W

⎛ 576 Ω ⎞ Q ⎛R ⎞ = Q ⎜ A ⎟ = (1.00 C ) ⎜ = 4.80 s (b) ΔtA = ⎝ ΔV ⎠ IA ⎝ 120 V ⎟⎠ !

(d) P =

W Δt

ΔtA =

W PA

1.00 J 25.0 J/s

= 0.040 0 s

(e) Energy enters the bulb by electrical transmission and leaves by heat and radiation.

⎡ ⎛ 1 kW ⎞ ⎤ ⎡ ⎛ 24.0 h ⎞ ⎤ (f) E = PA ( Δt ) = ⎢( 25.0 W ) ⎜ 3 ⎟ ⎥ ⎢( 30.0 d ) ⎜ ⎟⎠ ⎥ = 18.0 kWh 10 W 1 d ⎝ ⎠ ⎝ ⎣ ⎦⎣ ⎦ ! and cost = E × rate = (18.0 kwh)($0.110/kWh) = $1.98

Topic 17

17.42

1031

(a) At the operating temperature, P = (ΔV)I = (120 V)(1.53 A) = 184 W (b) From R = R0[1 + α(T − T0)], the temperature T is given by T = T0 + (R − R0)/αR0. The resistances are given by Ohm’s law as

R= !

( ΔV ) = 120 V and R = ( ΔV )0 = 120 V 0 I

1.53 A

1.80 A

Therefore, the operating temperature is

T = 20.0°C + ! 17.43

(120/1.53) − (120/1.80) = 461°C ( 0.400 × 10−3 (°C)−1 )(120/1.80)

The power loss per unit length of the cable is P/L = (I2R)/L = I2(R/L). Thus, the resistance per unit length of the cable is

R P/L 2.00 W/m = 2 = = 2.22 × 10−5 Ω/m 2 I (300 A) !L From R = ρL/A, the resistance per unit length is also given by R/L = ρ/A. Hence, the cross-sectional area is πr2 = A = ρ/(R/L), and the required radius is

17.44

( )

π R/L

1.7 × 10−8 Ω ⋅ m

(

π 2.22 × 10−5 Ω/m

)

= 0.016 m = 1.6 cm

(a) The rating of the 12-V battery is I ⋅ Δt = 55 A ⋅ h. Thus, the stored

Topic 17

1032

energy is Energy stored = P ⋅ Δt = (ΔV)I ⋅ Δt = (12 V)(55 A ⋅ h) = 660 W ⋅ h = 0.66 kWh (b) value = (0.66 kWh)($0.12/kWh) = $0.079 = 7.9 cents 17.45

P = (ΔV)I = (75.0 × 10−3 V)(0.200 × 10−3 A) = 1.50 × 10−5 W = 15.0 × 10−6 W = 15.0 µW

17.46

(a) E = P ⋅ t = (40.0 W)(14.0 d)(24.0 h/d) = 1.34 × 104 Wh = 13.4 kWh cost = E ⋅ (rate) = (13.4 kWh)($0.120/kWh) = $1.61 (b) E = P ⋅ t = (0.970 kW)(3.00 min)(1 h/60 min) = 4.85 × 10−2 kWh cost = (4.85 × 10−2 kWh)($0.120/kWh) = $0.005 82 = 0.582 cents (c) E = P ⋅ t = (5.20 kW)(40.0 min)(1 h/60 min) = 3.47 kWh cost = E ⋅ (rate) = (3.47 kWh)($0.120/kWh) = $0.416 = 41.6 cents

17.47

Total length of transmission lines: L = 2(50.0 m) = 100 m. Thus, the resistance of these lines is R = (0.108 Ω/300 m)(100 m) = 3.60 × 10−2 Ω. (a) The total potential drop along the transmission lines is (ΔV)lines = IR, giving

Topic 17

1033

(ΔV)house = (ΔV)source − (ΔV)lines = 120 V − (110 A)(3.60 × 10−2 Ω) = 116 V (b) Pdelivered = I(ΔV)house = (110 A)(116 V) = 1.28 × 104 W = 12.8 kW 17.48

(a) The thermal energy needed to raise the water temperature by ΔT is E = mc(ΔT). If ΔT = 100°C − 20°C = 80°C and this energy is to be delivered in 4.00 min, the average power required is E mc ( ΔT ) ( 0.250 kg ) ( 4186 J/kg ⋅°C ) (100°C − 20°C ) P= = = = 3.5 × 102 W Δt Δt 4.00 min 60 s/1!min ( )( ) !

(b) The required resistance (at 100°C) of the heating element is then

( ΔV )2 = (120 V ) R= !

3.5 × 102 W

= 41 Ω

R 41 Ω R0 = = = 4.0 × 101 Ω −3 1 + α (T − T0 ) 1 + ( 0.4 × 10 °C−1 ) (100°C − 20°C ) ! (d) We find the needed dimensions of a Nichrome wire for this heating element from R0 = ρ0L/A = ρ0L/(πd2/4) = 4ρ0L/πd2, where L is the length of the wire and d is its diameter. This gives the diameter as

d = 4 ρ0 L π R0 .

(e) If L = 3.00 m, the required diameter of the wire is

Topic 17

1034 2 −8 ⎡ 4 ρ0 L ⎤ 2 ⎡ 4 (150 × 10 Ω ⋅ m ) ( 3.00 m ) ⎤ −4 d=⎢ =⎢ ⎥ = 3.8 × 10 m = 0.38 mm ⎥ 1 π R0 ⎦ ⎢⎣ π ( 4.0 × 10 Ω ) ⎥⎦ ! ⎣ 1

17.49

The power dissipated in a conductor is P = (ΔV)2/R, so the resistance may be written as R = (ΔV)2/P. Hence,

RB RA

(ΔV)2 PB

⋅

PA (ΔV)2

PA PB

RB = 3RA

Since R = ρL/A = ρL/(πd2/4), this result becomes

⎛ 4ρ L ⎞ = 3⎜ ⎟ ⎜⎝ π d 2 ⎟⎠ π dB2 A

4ρ L

dA2 dB2

and yields dA dB = 3 . 17.50

(a) For tungsten, Table 17.1 from the textbook gives the resistivity at T0 = 20.0°C = 293 K as ρ0 = 5.6 × 10−8 Ω and the temperature coefficient of resistivity as α = 4.5 × 10−3 (°C)−1 = 4.5 × 10−3 K−1. Thus, for a tungsten wire having a radius of 1.00 mm and a length of 15.0 cm, the resistance at T0 = 293 K is

15.0 × 10−2 m ) ( L L −8 −3 R0 = ρ0 = ρ0 = ( 5.6 × 10 Ω ⋅ m ) Ω 2 = 2.7 × 10 2 −3 A π r ( ) π (1.00 × 10 m ) ! (b) From Stefan’s law (see Section 11.5 of the textbook), the radiated © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 17

1035

power is P = σAeT4, where A is the area of the radiating surface. Note that since we are computing the radiated power, not the net energy gained or lost as a result of radiation, the ambient temperature is not needed here. In the case of a wire, this is the cylindrical surface area A = 2πrL. The temperature of the wire when it is radiating a power of P = 75.0 W is given by T = [P/σAe]1/4 as

⎡ ⎤ 75.0 W ⎥ T=⎢ ⎢ 5.669 × 10−8 W/m 2 ⋅ K 4 2π 1.00 × 10−3 m 0.150 m 0.320 ⎥ ⎣ ⎦

(

) (

)(

1/ 4

)

= 1.45 × 103 K (c) Assuming a linear temperature variation of resistance, the resistance of the wire at this temperature is R = R0[1 + α(T − T0)] = (2.7 × 10−3 Ω)[1 + (4.5 × 10−3 K−1)(1.45 × 103 K − 293 K)] giving

R = 1.7 × 10−2Ω

(d) The voltage drop across the wire when it is radiating 75.0 W and has the resistance found in part (c) above is given by P = (ΔV)2/R as

ΔV = R ⋅ P = !

(1.7 × 10 Ω)(75.0 W ) = 1.1 V −2

(e) Tungsten bulbs radiate little of the energy they consume in the form © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 17

1036

of visible light, making them inefficient sources of light. 17.51

The battery is rated to deliver the equivalent of 60.0 amperes of current (i.e., 60.0 C/s) for 1 hour. This is Q = I ⋅ Δt = (60.0 A)(1 h) = (60.0 C/s)(3 600 s) = 2.16 × 105 C

17.52

The energy available in the battery is Energy stored = P ⋅ t = (ΔV)I ⋅ t = (12.0 V)(90.0 A ⋅ h) = 1.08 × 103 W ⋅ h The two head lights together consume a total power of P = 2(36.0 W) = 72.0 W, so the time required to completely discharge the battery is

Energy stored 1.08 × 103 W ⋅ h Δt = = = 15.0 h P 72.0 W !

17.53

−8 ρL ρL 4 ρ L 4 ( 2.82 × 10 Ω ⋅ m ) (15.0 m ) = = = = 1.50 Ω (a) R = 2 A π d2 4 π d2 π ( 0.600 × 10−3 m ) !

ΔV 9.00 V = = 6.00 A (b) I = R 1.50 Ω ! 17.54

Using chemical symbols to denote the two different metals, the resistances are equal when R 1 + α Cu ( ΔT )] = R0 W [1 + α W ( ΔT )] ! 0Cu [

⎡ ⎤ ⎛ R0 ⎞ − 1 = ⎢α W − ⎜ Cu ⎟ α Cu ⎥ ( ΔT ) R ⎝ R0W ⎠ ⎢⎣ ⎥⎦ ! 0W R0Cu

Topic 17

1037

ΔT = T − 20.0 °C =

Thus,

17.55

0Cu

(R R ) − 1 α − (R R )α 0Cu

T = 20.0 °C +

0Cu

T = 20.0 °C +

(R R ) − 1 α − ( R R )α 0W

0Cu

( 5.00/4.75) − 1 −1 4.5 × 10−3 ( °C ) − ( 5.00/4.75 ) ( 3.9 × 10−3 ( °C) ) −1

= 20.0 °C + 1.3 × 102 °C = 1.5 × 102 °C

From P = (ΔV)2/R, the total resistance needed is R = (ΔV)2/P = (20 V)2/48 W = 8.3 Ω. Thus, from R = ρL/A, the length of wire required is

L= !

17.56

−6 2 R ⋅ A ( 8.3 Ω ) ( 4.0 × 10 m ) = = 1.1 × 103 m = 1.1 km −8 ρ 3.0 × 10 Ω ⋅ m

The resistance of the 4.0 cm length of wire between the feet is −8 ρ L (1.7 × 10 Ω ⋅ m ) ( 0.040 m ) R= = = 1.8 × 10−6 Ω 2 A π ( 0.022 m ) 4 !

so the potential difference is ΔV = IR = (50 A)(1.8 × 10−6 Ω) = 9.0 × 10−5 V = 90 µV 17.57

Ohm’s law gives the resistance as R = (ΔV)/I. From R = ρL/A, the resistivity is given by ρ = R ⋅ (A/L). The results of these calculations for

Topic 17

1038

each of the three wires are summarized in the table below.

L (m)

R (Ω)

ρ (Ω ⋅ m)

0.540

10.4

1.41 × 10−6

1.028

21.1

1.50 × 10−6

1.543

31.8

1.50 × 10−6

The average value found for the resistivity is Σρ ρav = i = 1.47 × 10−6 Ω ⋅ m 3 !

which differs from the value of ρ = 150 × 10−8 Ω ⋅ m = 1.50 × 10−6 Ω ⋅ m given in Table 17.1 by 2.0 %. 17.58

The volume of the material is mass 50.0 g ⎛ 1 m 3 ⎞ V= = = 6.36 × 10−6 m 3 3 ⎜ 6 3⎟ density 7.86 g cm ⎝ 10 cm ⎠ !

Since V = A ⋅ L, the cross-sectional area of the wire is A = V/L. (a) From R = ρL/A = ρL/(V/L) = ρL2/V, the length of the wire is given by

L= !

R ⋅V = ρ

(1.5 Ω)( 6.36 × 10 m ) = 9.3 m −6

11 × 10−8 Ω ⋅ m

(b) The cross-sectional area of the wire is A = πd2/4 = V/L. Thus, the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 17

1039

diameter is

4V d= = πL !

17.59

4 ( 6.36 × 10−6 m 3 )

π ( 9.3 m )

= 9.3 × 10−4 m = 0.93 mm

(a) The total power you now use while cooking breakfast is P = (1 200 + 500) W = 1.70 kW The cost to use this power for 0.500 h each day for 30.0 days is

⎡ ⎤ ⎛ h ⎞ cost = [ P × ( Δt )] × rate = ⎢(1.70 kW ) ⎜ 0.500 30.0 days ) ⎥ ( $0.120 kWh ) = $3.06 ( ⎟ day ⎠ ⎝ ⎢⎣ ⎥⎦ ! (b) If you upgraded, the new power requirement would be P = (2 400 + 500) W = 2 900 W, and the required current would be I = P/ΔV = 2 900 W/110 V = 26.4 A > 20 A. No, your present circuit breaker cannot handle the upgrade. 17.60

(a) The charge passing through the conductor in the interval 0 ≤ t ≤5.0 s is represented by the area under the I vs. t graph given in Figure P17.60. This area consists of two rectangles and two triangles. Thus, ΔQ = Arectangle!1 + Arectangle! 2 + Atriangle!1 + Atriangle! 2

= ( 5.0 s − 0 ) ( 2.0 A − 0 ) + ( 4.0 s − 3.0 s ) ( 6.0 A − 2.0 A )

1 1 3.0 s − 2.0 s ) ( 6.0 A − 2.0 A ) + ( 5.0 s − 4.0 s ) ( 6.0 A − 2.0 A ) ( 2 2 ΔQ = 18 C ! +

Topic 17

1040

(b) The constant current that would pass the same charge in 5.0 s is

ΔQ 18 C I= = = 3.6 A Δt 5.0 s ! 17.61 (a) The power input to the motor is Pinput = (ΔV)I = Poutput/efficiency, so the required current is

(2.50 hp)(746 W 1 hp) = 17.3 A ( ΔV )(efficiency) (120 V )(0.900) Poutput

Poutput

( ) efficiency

(b) E = Pinput Δt =

yielding

2.50 hp ) ( 0.746 kW 1 hp ) ( ( ) (3.00 h ) Δt =

0.900

⎛ 3.60 MJ ⎞ E = 6.22 kWh = ( 6.22 kWh ) ⎜ = 22.4 MJ ⎝ 1 kWh ⎟⎠ !

(c) cost = E × rate = (6.22 kWh)($0.110/kWh) = $0.684 17.62 (a) R = R0[1 + α(ΔT)] = (8.0 Ω)[1 + (0.4 × 10−3 °C−1)(350°C − 20°C )] = 9.1 Ω (b) We assume that the temperature coefficient of resistivity for

Topic 17

1041

Nichrome remains constant over this temperature range. 17.63 The current in the wire is I = ΔV/R = 15.0 V/0.100 Ω = 150 A. Then, from the expression for the drift velocity, vd = I/nqA, the density of free electrons is n= !

or 17.64

I 150 A = 2 2 vd e (π r ) ( 3.17 × 10−4 m/s ) (1.60 × 10−19 C ) π ( 5.00 × 10−3 m )

n = 3.77 × 1028/m3

Each speaker has a resistance of R = 4.00 Ω and can handle a maximum power of 60.0 W. From P = I2R, the maximum safe current is

I max = !

Pmax 60.0 W = = 3.87 A R 4.00 Ω

Thus, the system is not adequately protected by a 4.00 A fuse. 17.65

The cross-sectional area of the conducting material is !A = π ( router − rinner ) . 2

Thus,

3.5 × 105 Ω ⋅ m ) ( 4.0 × 10−2 m ) ( ρL R= = = 3.7 × 107 Ω = 37 MΩ 2 2 −2 −2 A π ⎡(1.2 × 10 m ) − ( 0.50 × 10 m ) ⎤ ⎣ ⎦ ! 17.66

ρL (a) At temperature T, the resistance is R = , where A !

Topic 17

1042

ρ = ρ0[1 + α(T − T0)], L = L0[1 + α′(T − T0)], and A = A0[1 + 2α′(T − T0)] Thus,

⎛ ρ L ⎞ ⎡1 + α (T − T0 ) ⎤⎦ ⋅ ⎡⎣1 + α ′ (T − T0 ) ⎤⎦ R0 ⎡⎣1 + α (T − T0 ) ⎤⎦ ⋅ ⎡⎣1 + α ′ (T − T0 ) ⎤⎦ R=⎜ 0 0⎟ ⎣ = ⎡⎣1 + 2α ′ (T − T0 ) ⎤⎦ ⎡⎣1 + 2α ′ (T − T0 ) ⎤⎦ ⎝ A0 ⎠ ! −8 ρ0 L0 (1.7 × 10 Ω ⋅ m ) ( 2.00 m ) R = = = 1.082 Ω (b) 0 2 A0 π ( 0.100 × 10−3 ) !

Then R = R0[1 + α(T − T0)] gives R = (1.082 Ω)[1 + (3.9 × 10−3 °C−1)(80.0°C)] = 1.420 Ω The more complex formula gives

(1.420 Ω) ⋅ ⎡⎣1 + (17 × 10 °C )(80.0°C)⎤⎦ −6

R= !

−1

⎡⎣1 + 2 (17 × 10−6 °C−1 ) ( 80.0°C ) ⎤⎦

= 1.418 Ω

Note: Some rules for handing significant figures have been deliberately violated in this solution in order to illustrate the very small difference in the results obtained with these two expressions. 17.67

The power the beam delivers to the target is P = (ΔV)I = (4.0 × 106 V)(25 × 10−3 A) = 1.0 × 105 W The mass of cooling water that must flow through the tube each second if

Topic 17

1043

the rise in the water temperature is not to exceed 50°C is found from P = (Δm/Δt)c(ΔT) as

Δm Δt 17.68

P c(ΔT)

(

1.0 × 105 J/s

)(

4 186 J/kg ⋅°C 50°C

)

= 0.48 kg/s

Note: All potential differences in this solution have a value of ΔV = 120 V. First, we shall do a symbolic solution for many parts of the problem and then enter the specified numeric values for the cases of interest. From the marked specifications on the cleaner, its internal resistance (assumed constant) is

Rvac !

2 ΔV ) ( =

where P1 = 535 W

[1]

If each of the two conductors in the extension cord has resistance Rc, the total resistance in the path of the current (outside of the power source) is Rt = Rvac + 2Rc

[2]

so the current which will exist is I = ΔV/Rt, and the power that is delivered to the cleaner is

⎛ ΔV ⎞ ⎛ ΔV ⎞ ( ΔV )2 ( ΔV )4 Pdelivered = I Rvac = ⎜ Rvac = ⎜ = 2 P1 Rt P1 ⎝ Rt ⎟⎠ ⎝ Rt ⎟⎠ ! 2

[3]

The resistance of a copper conductor of length L and diameter d is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 17

1044

L L 4ρ L Rc = ρCu = ρCu = Cu2 2 A (π d 4 ) π d ! Thus, if Rc, max is the maximum allowed value of Rc, the minimum acceptable diameter of the conductor is 4 ρCu L dmin = π Rc ,! max !

[4]

(a) If Rc = 0.900 Ω, then from Equations [2] and [1], 2 (120 V ) + 1.80 Ω ΔV ) ( Rt = Rvac + 2 ( 0.900 Ω ) = + 1.80 Ω = 2

535 W

and, from Equation [3], the power delivered to the cleaner is

Pdelivered = !

(120 V )

⎡ (120 V )2 ⎤ + 1.80 Ω ⎥ ( 535 W ) ⎢ ⎢⎣ 535 W ⎥⎦ 2

= 470 W

If the minimum acceptable power delivered to the cleaner is Pmin, then the maximum allowable total resistance is given by Equations [2] and [3] as 4 2 ΔV ) ΔV ) ( ( Rt ,! max = Rvac + 2Rc ,! max = =

Pmin P1

Topic 17

1045

2 ⎤ 1 ⎡ ( ΔV )2 ( ΔV )2 ⎤ ( ΔV )2 ⎡ 1 1 ⎡ ( ΔV ) 1⎤ Rc ,! max = ⎢ − Rvac ⎥ = ⎢ − − ⎥ ⎥= ⎢ 2 ⎣ Pmin P1 P1 ⎦ 2 ⎣ Pmin P1 P1 ⎦ ⎦ 2 ⎣ Pmin P1 !

(b) When Pmin = 525 W, then

(120 V ) ⎡⎢ = 2

Rc ,! max

( 525 W )( 535 W )

⎢⎣

−

1 ⎤ ⎥ = 0.128 Ω 535 W ⎥ ⎦

and, from Equation [4], 4 (1.7 × 10−8 Ω ⋅ m ) (15.0 m )

dmin = !

π ( 0.128 Ω )

= 1.60 mm

⎡ 120 V ) ⎢ ( = 2

Rc , max

and

dmin =

(

⎢ ⎢⎣

⎤ ⎥ − ⎥ = 0.037 9 Ω 535 W 532 W 535 W ⎥⎦ 1

)(

)

)(

) = 2.93 mm

4 1.7 × 10−8 Ω ⋅ m 15.0 m

(

π 0.037 9 Ω

)

Topic 18

1046

Topic 18 Direct-Current Circuits

QUICK QUIZZES

18.1

True. When a battery is delivering a current, there is a voltage drop within the battery due to internal resistance, making the terminal voltage less than the emf.

18.2

Because of internal resistance, power is dissipated into the battery material, raising its temperature.

18.3

Choice (b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases.

18.4

Choice (a). When the switch is opened, resistors R1 and R2 are in series, so the total circuit resistance is larger than and the current through R1 is less with the switch open than when it is closed. Since the power delivered to R1 is P = I2R1, Po < Pc.

Topic 18

18.5

1047

Choice (a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the current increases.

18.6

Choice (b). Observe that the potential difference across R1 equals the terminal potential difference of the battery. If the battery has negligible internal resistance, the terminal potential difference is the same with the switch open or closed. Under these conditions, the power delivered to R1, equal to P = (ΔV)2/R1, is unchanged when the switch is closed.

18.7

The voltage drop across each bulb connected in parallel with each other and across the battery equals the terminal potential difference of the battery. As more bulbs are added, the current supplied by the battery increases. However, if the internal resistance is negligible, the terminal potential difference is constant and the current through each bulb is the same regardless of the number of bulbs connected. Under these conditions: (a) The brightness of a bulb, determined by the current flowing in the bulb, is unchanged as bulbs are added. (b) The individual currents in the bulbs, I = ΔV/R, are constant as bulbs are added since ΔV does not change. (c) The total power delivered by the battery increases by an amount (ΔV)2/R each time a bulb is added. (d) With the total delivered

Topic 18

1048

power increasing, energy is drawn from the battery at an increasing rate and the battery’s lifetime decreases.

18.8

Adding bulbs in series with each other and the battery increases the total load resistance seen by the battery. This means that the current supplied by the battery decreases with each new bulb that is added. (a) The brightness of a bulb is determined by the power delivered to that bulb, Pbulb = I2R, which decreases as bulbs are added and the current decreases. (b) For a series connection, the individual currents in the bulbs are the same and equal to the total current supplied by the battery. This decreases as bulbs are added. (c) The total power delivered by the battery is given by Ptotal = (ΔV)I, where ΔV is the terminal potential difference of the battery and I is the total current supplied by the battery. With negligible internal resistance, ΔV is constant. Thus, with I decreasing as bulbs are added, the total delivered power decreases. (d) With the delivered power decreasing, energy is drawn from the battery at a decreasing rate, which increases the lifetime of the battery.

18.9

Choice (c). After the capacitor is fully charged, current flows only around the outer loop of the circuit. This path has a total resistance of 3 Ω, so the 6-V battery will supply a current of 2 amperes.

Topic 18

1049

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

18.2

18.4

A short circuit can develop when the last bit of insulation frays away between the two conductors in a lamp cord. Then the two conductors touch each other, creating a low resistance path in parallel with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A very large current will be produced in the power source, the house wiring, and the wire in the lamp cord up to and through the short. The circuit breaker will interrupt the circuit quickly but not before considerable heating and sparking is produced in the short-circuit path.

18.6

A wire or cable in a transmission line is thick and made of material with very low resistivity. Only when its length is very large does its resistance become significant. To transmit power over a long distance it is most efficient to use low current at high voltage. The power loss per unit length of the transmission line is Ploss/L = I2(R/L), where R/L is the

Topic 18

1050

resistance per unit length of the line. Thus, a low current is clearly desirable, but to transmit a significant amount of power P = (ΔV)I with low current, a high voltage must be used.

18.8

The bulbs of set A are wired in parallel. The bulbs of set B are wired in series, so removing one bulb produces an open circuit with infinite resistance and zero current.

18.10 (a) I. The charging time increases as the time constant τ = RC increases.

(

)

(b) U. The ratio q/Q = ΔV / ε = 90% = 1 − e −t/RC only depends on the time constant, not on the battery voltage. (c) I. Adding a second resistor in series will increase the equivalent R, increasing the time constant and the time required to reach 90% of its final charge.

18.12

Compare two runs in series to two resistors connected in series. Compare three runs in parallel to three resistors connected in parallel. Compare one chairlift followed by two runs in parallel to a battery followed immediately by two resistors in parallel.

The junction rule for ski resorts says that the number of skiers coming into a junction must be equal to the number of skiers leaving. The loop

Topic 18

1051

rule would be stated as the total change in altitude must be zero for any skier completing a closed path.

18.14

Because water is a good conductor, if you should become part of a short circuit when fumbling with any electrical circuit while in a bathtub, the current would follow a pathway through you, the water, and to ground. Electrocution would be the obvious result.

ANSWERS TO EVEN NUMBERED PROBLEMS

18.2

(a)

1.13 A

(b)

9.17 Ω

18.4

(a)

1.64 A

(b)

14.7 W

18.6

(a)

27 Ω

(b)

0.44 A

18.8

(a)

35.0 Ω/3 = 11.7 Ω

(c)

3.0 Ω, 1.3 A

(b) 1.00 A in the 12.0-Ω and 8.00-Ω resistors, 2.00 A in the 6.00-Ω and 4.00-Ω resistors, and 3.00 A in the 5.00-Ω resistor

18.10 (a)

3.33 Ω

(b)

7.33 Ω

(c)

2.13 Ω

Topic 18

1052

(d)

4.13 Ω

(e)

1.94 A

(g) 4.12 V

(h)

1.37 A

18.12 (a)

(b)

(f) 3.88 V

ΔV1 = ε/3, ΔV2 = 2ε/9, ΔV3 = 4ε/9, ΔV4 = 2ε/3

I1 = I, I2 = I3 = I/3, I4 = 2I/3

18.14 (a)

1.00 kΩ

(b)

2.00 kΩ

(c)

3.00 kΩ

18.16 (a)

Yes, Req = 8.00 Ω

(b)

2.25 A

(c)

40.5 W

18.18 (a)

I6 = 3.0 A, I12 = 2.0 A, I24 = 1.0 A

(b)

18.20 (a)

(b)

Answers are the same as for part (a).

0.909 A

|ΔVab| = 1.82 V with point b at the lower potential.

18.22 I2 = 2.0 A flowing from b toward a, I3 = 1.0 A in direction shown

18.24 (a)

60.0 Ω

(d)

0.50 W

(b)

0.20 A

(c)

2.4 W

Topic 18

1053

18.28 (a)

1.7 × 102 A

(b)

0.20 A

18.30 (a)

18.0I12 + 13.0I18 = 30.0

(b)

5.00I36 − 18.0I12 = 24.0

(c)

I18 = I12 + I36

(d)

I36 = I18 − I12

(e)

5.00I18 − 23.0I12 = 24.0

(f) I12 = −0.416 A, I18 = 2.88 A

(g) I36 = 3.30 A

(h)

I12 flows opposite to the assumed direction and has magnitude 0.416 A.

18.32

See Solution.

18.34 (a)

2.00 ms

(b)

180 µC

18.36 (a)

5.21 × 104 Ω

(b)

30.9 A

18.38 (a)

0.164 A

(b)

6.02 × 10−2 A

18.40 (a)

12.5 Ω

(b)

1 250 W

18.42 (a)

Icofee = 10 A,!I toaster = 9.2 A,!I waffle = 12 A maker ! maker

(b)

Itotal = 31 A

(c)

114 µC

No, Itotal > 15 A.

Topic 18

1054

18.44 (a)

18.46

4.1 × 10−11 J

(b)

0.56 µA

Sixteen distinct resistances are possible. See Solution for how these are produced.

18.48 (a)

2.0 kΩ

(b)

15 V

(c)

9.0 V

(d) Assumed negligible current through the voltmeter and negligible resistance in the ammeter.

18.50 (a)

40.0 W

(b)

ΔV1 = 80.0 V, ΔV2 = ΔV3 = 40.0 V

18.52 (a)

0.708 A

(b)

2.51 W

(c) Only the circuit of Figure P18.52c. In the other circuits, the batteries can be combined into a single effective battery while the 5.00-Ω and 8.00-Ω resistors remain in parallel with each other.

(d) The power is lowest in Figure P18.52c. The circuits in Figures P18.52b and P18.52d have in effect 30-V batteries driving the current.

18.54

14 Ω

18.56

14 s

Topic 18

1055

18.58

R = 20 Ω or R = 98 Ω

18.60

See Solution.

18.62

q1 = (240 µC)(1 − e−1 000 t/6.0 s), q2 = (360 µC)(1 − e−1 000 t/6.0 s)

18.64 (a)

Req = 0.099 9 Ω, I R 1 = 50.0 A, I R 2 = I R3 = I100 = 45.0 mA

(b)

Req = 1.09 Ω, I R 1 = I R2 = 4.55 A, I R3 = I100 = 45.5 mA

(c)

Req = 9.99 Ω, I R 1 = I R2 = I R3 = 0.450 A , I100 = 50.0 mA

18.66 (a) In case 1, the two bulbs have the same current, power supplied, and brightness.

(b) In case 2, the two bulbs have the same current, power supplied, and brightness.

(d) In case 1, both go out when one bulb fails. In case 2, the other bulb remains lit with unchanged brightness when one bulb fails.

18.68 (a)

61.6 mA

(b)

0.235 µC

(c)

1.96 A

Topic 18

1056

PROBLEM SOLUTIONS

18.1

From ΔV = I (R + r), the internal resistance is

ΔV 9.00 V r= −R= − 72.0 Ω = 4.92 Ω I 0.117 A ! 18.2

(a) When the 8.00-Ω resistor is connected across the 9.00-V terminal potential difference of the battery, the current through both the resistor and the battery is

ΔV 9.00 V I= = = 1.3 A R 8.00 Ω ! (b) The relation between the emf and the terminal potential difference of a battery supplying current I is ΔV = ε − Ir, where r is the internal resistance of the battery. Thus, if the battery has r = 0.15 Ω and maintains a terminal potential difference of ΔV = 9.00 V while supplying the current found above, the emf of this battery must be

ε = ΔV + Ir = 9.00 V + (1.13 A)(0.15 Ω) = (9.00 + 0.17) Ω = 9.17 Ω

Topic 18

18.3

1057

(a) The terminal voltage is related to the emf ε and internal resistance r by ΔV = ε − Ir. Solve for the internal resistance and substitute values to find: r=

ε − ΔV = 12.0 V − 11.5 V = 0.167 Ω I

3.00 A

(b) The load resistance R is given by ε = IR + Ir. Solve and substitute values to find:

R= 18.4

ε − Ir = 12.0 V − ( 3.00 A )(0.167 Ω ) = 3.83 Ω I

3.00 A

(a) The total power supplied by the battery is I ε = I(IR + Ir) = 15.0 W, where r = 0.100 Ω. Substitute IR = ε − Ir = ΔV and rearrange to find I

ε = I(ΔV + Ir) = 15.0 W. Solve for the current I: IΔV + I 2 r = 15.0 W 15.0 W ⎛ ΔV ⎞ I2 + ⎜ I− =0 ⎟ ⎝ r ⎠ r Substitute known values, suppressing units for clarity, and apply the quadratic equation, choosing the positive root:

15.0 W ⎛ 9.00 V ⎞ I2 + ⎜ I− =0 ⎟ ⎝ 0.100 Ω ⎠ 0.100 Ω I 2 + ( 90.0 ) I − 150. = 0

−90.0 ±

( 90.0) − 4 (1)( −150.) → 2

I = 1.64 A

Topic 18

1058

(b) The power delivered to the load resistor is PR = I2R = I ε − I2r. Substitute values to find:

PR = Iε − I 2 r = 15.0 W − (1.64 A ) ( 0.100 Ω ) = 14.7 W 2

Alternatively, use PR = I(ΔV) to find the same result to within roundoff error. 18.5

(a) Two resistors connected in series have the equivalent resistance Req = R1 + R2. Substitute values to find Req = (2.00 Ω) + (4.00 Ω) = 6.00 Ω . (b) Two resistors connected in parallel have the equivalent resistance given by 1/Req = 1/R1 + 1/R2. Substitute values to find 1/Req = 1/(2.00 Ω) + 1/(4.00 Ω) → Req = 1.33 Ω .

18.6

(a) When the three resistors are in series, the equivalent resistance of the circuit is Req = R1 + R2 + R3 = 3(9.0 Ω) = 27 Ω.

(b) The terminal potential difference of the battery is applied across the series combination of the three 9.0-Ω resistors, so the current supplied by the battery and the current through each resistor in the series combination is

ΔV Req

12 V 27 A

− 72.0 Ω = 0.44 A

Topic 18

1059

1 1 1 1 3 = + + = R 9.0 Ω 9.0 Ω 9.0 Ω 9.0 Ω ! eq

Req =

9.0 Ω = 3.0 Ω 3

When this parallel combination is connected to the battery, the potential difference across each resistor in the combination is ΔV = 12 V, so the current through each of the resistors is

ΔV 12 V I= = = 1.3 A R 9.0 Ω ! 18.7

(a) The equivalent resistance of the two parallel resistors is

⎛ 1 1 ⎞ RP = ⎜ + ⎝ 7.00 Ω 10.0 Ω ⎟⎠ !

−1

= 4.12 Ω.

Thus, Rab = R4 + Rp + R9 = (4.00 + 4.12 + 9.00) Ω = 17.1 Ω

(b) I ab = !

( ΔV )ab 34.0 V Rab

17.1 Ω

= 1.99 A , so I4 = I9 = 1.99 A

Topic 18

18.8

1060

Also,

(ΔV)p + IabRp = (1.99 A)(4.12 Ω) = 8.20 V

Then,

I7 =

and

( ΔV )p 8.20 V I10 = = = 0.820 A R 10.0 Ω 10 !

( ΔV ) = 8.20 V = 1.17 A p

7.00 Ω

(a) The parallel combination of the 6.00-Ω and 12.0-Ω resistors has an equivalent resistance of 1 1 1 2+1 = + = R 6.00 Ω 12.0 Ω 12.0 Ω ! p1

Rp1 =

12.0 Ω = 4.00 Ω 3

Similarly, the equivalent resistance of the 4.00-Ω and 8.00-Ω parallel combination is 1 1 1 2+1 = + = R 4.00 Ω 8.00 Ω 8.00 Ω ! p2

Rp2 =

8.00 Ω 3

The total resistance of the series combination between points a and b is then

8.00 35.0 Rab = Rp1 + 5.00 Ω + Rp2 = 4.00 Ω + 5.00 Ω + Ω= Ω 3 3 ! (b) If ΔVab = 35.0 V, the total current from a to b is

Topic 18

1061

Iab = ΔVab/Rab = 35.0 V/(35.0 Ω/3) = 3.00 A

and the potential differences across the two parallel combinations are

ΔVp1 = IabRp1 = (3.00 A)(4.00 Ω) = 12.0 V, and

⎛ 8.00 ⎞ ΔVp2 = I ab Rp2 = ( 3.00 A ) ⎜ Ω⎟ = 8.00 V ⎝ 3 ⎠ !

The individual currents through the various resistors are:

and

18.9

I12 = ΔVp1/12.0 Ω = 1.00 A;

I6 = ΔVp1/6.00 Ω = 2.00 A;

I5 = Iab = 3.00 A;

I8 = ΔVp2/8.00 Ω = 1.00 A;

I4 = ΔVp2/4.00 Ω = 2.00 A

When connected in series, we have R1 + R2 = 690 Ω

[1]

which we may rewrite as R2 = 690 Ω − R1

[1a]

When in parallel,

1 1 1 + = !R1 R2 150 Ω

R1R2 = 150 Ω R1 + R2

[2]

Substitute Equations [1] and [1a] into Equation [2] to obtain:

Topic 18

1062

R1 ( 690 Ω − R1 ) = 150 Ω 690 Ω !

R12 − ( 690Ω ) R1 + ( 690 Ω ) (150 Ω ) = 0

Using the quadratic formula to solve Equation [3] gives

R1 = !

690 Ω ±

( 690 Ω) − 4 ( 690Ω)(150Ω) 2

with two solutions of R1 = 470 Ω

and

Then Equation [1a] yields R2 = 220 Ω

R1 = 220 Ω

R2 = 470 Ω

Thus, the two resistors have resistances of 220 Ω and 470 Ω.

18.10

(a) The equivalent resistance of this first parallel combination is

1 1 1 = + R 10.0 Ω 5.00 Ω ! p1

Rp1 = 3.33 Ω

(b) For this series combination,

[3]

Topic 18

1063

Rupper = Rp1 + 4.00 Ω = 7.33 Ω

1 1 1 1 1 = + = + R Rupper 3.00 Ω 7.33 Ω 3.00 Ω ! p2

Rp2 = 2.13 Ω

(d) For the second series combination (and hence the entire resistor network)

Rtotal = 2.00 Ω + Rp2 = 2.00 Ω + 2.13 Ω = 4.13 Ω

(e) The total current supplied by the battery is

ΔV 8.00 V I total = = = 1.94 A R 4.13 Ω total ! (f) The potential drop across the 2.00 Ω resistor is

ΔV2 = R2Itotal = (2.00 Ω)(1.94 A) = 3.88 V

(g) The potential drop across the second parallel combination must be

ΔVp2 = ΔV − ΔV2 = 8.00 V − 3.88 V = 4.12 V

(h) So the current through the 3.00 Ω resistor is

Topic 18

1064

ΔVp2 4.12 V I total = = = 1.37 A R 3.00 Ω 3 ! 18.11

(a) Using the rules for combining resistors in series and parallel, the circuit reduces as shown below:

From the figure of Step 3, observe that

25.0 V I= = 1.93 A and ΔVab = I(2.94 Ω) = (1.93 A)(2.94 Ω) 10.0 Ω + 2.49 Ω ! = 5.67 V

(b) From the figure of Step 1, observe that

ΔVab 5.67 V I 20 = = = 0.227 A 25.0 Ω 25.0 Ω ! 18.12

(a) The figures below show the simplification of the circuit in stages:

Topic 18

1065

Note that R2 and R3 are in series with equivalent resistance R23 = R2 + R3 = 6R. Then, R4 and R23 are in parallel with equivalent resistance

RR (3R)(6R) R234 = 4 23 = = 2R R + R 3R + 6R 4 23 ! The total current supplied by the battery is then

ε R1 + R234

ε R + 2R

ε 3R

⎛ ε ⎞ The potential difference across R1 is ΔV1 = 1R1 = ⎜ ⎟R= ε /3 ⎝ 3R ⎠

⎛ ε ⎞ and that across R4 is ΔV4 = ΔVab = IR234 = ⎜ ⎟ (2R) = 2ε / 3 ⎝ 3R ⎠

The current through R2 and R3 is I23 = ΔVab/R23 = (2ε/3R)/6R = ε/9R.

so the potential difference across R2 is

Topic 18

1066

⎛ ε ⎞ ΔV2 = I 23 R2 = ⎜ ⎟ (2R) = 2ε / 9 ⎝ 9R ⎠

⎛ ε ⎞ and that across R3 is ΔV3 = I 23 R3 = ⎜ ⎟ (4R) = 4ε / 9 ⎝ 9R ⎠

(b) From above, we have I1 = I and I2 = I3 = I23 = ε/9R = I/3.

The current through R4 is I4 = ΔV4/R4 = (2ε/3)/3R = 2ε/9R = 2I/3.

18.13 The equivalent resistance is Req = R + Rp, where Rp is the total resistance of the three parallel branches:

⎛ 1 ⎞ 1 1 Rp = ⎜ + + ⎝ 120 Ω 40 Ω R + 5.0 Ω ⎟⎠ !

Thus,

75 Ω = R +

−1

⎛ 1 ⎞ 1 =⎜ + ⎝ 30 Ω R + 5.0 Ω ⎟⎠

(30 Ω)(R + 5.0 Ω) R + 35 Ω

−1

(30 Ω)(R + 5.0 Ω) R + 35 Ω

R 2 + (65 Ω)R + 150 Ω 2 R + 35 Ω

which reduces to

R2 − (10 Ω)R − 2 475 Ω2 = 0 or (R − 55 Ω)(R + 45 Ω) = 0.

Only the positive solution is physically acceptable, so R = 55 Ω.

18.14

The sketch below shows the equivalent circuit when the switch is in the

Topic 18

1067

open position. For this simple series circuit,

R1 + R2 + R3 =

6.00 V ε = IO 1.00 × 10−3 A

R1 + R2 + R3 = 6.00 kΩ

[1]

When the switch is closed in position a, the equivalent circuit is shown in Figure 2. The equivalent resistance of the two parallel resistors, R2, is Rp = R2/2 and the total resistance of the circuit is Ra = R1 + (R2/2) + R3. Thus,

R1 +

R2 2

+ R3 =

6.00 V ε = I a 1.20 × 10−3 A

R R1 + 2 + R3 = 5.00 kΩ 2 !

[2]

When the switch is closed in position b, resistor R3 is shorted out, leaving © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

1068

R1 and R2 in series with the battery as shown in Figure 3. This gives

ε 6.00 V R1 + R2 = = Ib 2.00 × 10−3 A !

and

R1 + R2 = 3.00 kΩ

[3]

Substitute Equation [3] into Equation [1] to obtain

3.00 kΩ + R3 = 6.00 kΩ

and

R3 = 3.00 kΩ

Now, Equation [1] minus Equation [2] gives R2/2 = 1.00 kΩ or R2 = 2.00 kΩ

Finally, Equation [3] tells that R1 + 2.00 kΩ = 3.00 kΩ, or R1 = 1.00 kΩ

In summary, we have

(a) R1 = 1.00 kΩ,

(b) R2 = 2.00 kΩ,

and

18.15 The resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance of Rad = (63/11) Ω.

Topic 18

1069

Figure 1

Figure 2

Figure 3

From Figure 5,

Figure 4

I= !

Then, from Figure 4,

( ΔV )ad Rad

Figure 5

18 V = 3.1 A (63 / 11) Ω

(ΔV)bd = IRbd = (3.1 A)(30/11 Ω) = 8.5 V

Now, look at Figure 2 and observe that

( ΔV )bd 8.5 V I2 = = = 1.7 A 3.0 Ω + 2.0 Ω 5.0 Ω ! so (ΔV)be = I2Rbe = (1.7 A)(3.0 Ω) = 5.1 V

Topic 18

1070

Finally, from Figure 1, I12 = !

( ΔV )be R12

5.1 V = 0.43 A 12 Ω

18.16 (a) The resistor network connected to the battery in Figure P18.16 can be reduced to a single equivalent resistance in the following steps. The equivalent resistance of the parallel combination of the 3.00 Ω and 6.00 Ω resistors is

1 1 1 3 = + = R 3.00 Ω 6.00 Ω 6.00 Ω ! p

Rp = 2.00 Ω

Figure P18.16

This resistance is in series with the 4.00 Ω and the other 2.00 Ω resistor, giving a total equivalent resistance of Req = 2.00 Ω + Rp + 4.00 Ω = 8.00 Ω.

(b) The current in the 2.00 Ω resistor is the total current supplied by the battery and is equal to

I total =

ΔV Req

18.0 V 8.00 Ω

= 2.25 A

Topic 18

1071

P = (ΔV)Itotal = (18.0 V)(2.25 A) = 40.5 W

18.17

(a) Connect two 50-Ω resistors in parallel to get 25 Ω. Then connect that parallel combination in series with a 20-Ω resistors for a total resistance of 45 Ω.

(b) Connect two 50-Ω resistors in parallel to get 25 Ω.

Also, connect two 20-Ω resistors in parallel to get 10 Ω.

Then, connect these two parallel combinations in series to obtain

35 Ω.

18.18

(a) The equivalent resistance of the parallel combination between points b and e is

1 1 1 = + !Rbe 12 Ω 24 Ω

Rbe = 8.0 Ω

Topic 18

1072

The total resistance between points a and e is then

Rae = Rab + Rbe = 6.0 Ω + 8.0 Ω = 14 Ω

The total current supplied by the battery (and also the current in the 6.0-Ω resistor) is

I total = I 6 =

ΔVae Rae

42 V 14 Ω

= 3.0 A

The potential difference between points b and e is

ΔVbe = RbeItotal = (8.0 Ω)(3.0 A) = 24 V

ΔVbe 24 V I12 = = = 2.0 A R 12 Ω bce !

and

I 24 =

(b) Applying the junction rule at point b yields

ΔVbe 24 V = = 1.0 A Rbde 24 Ω

I6 − I12 − I24 = 0 [1]

Using the loop rule on loop abdea gives +42 − 6I6 − 24I24 = 0

I6 = 7.0 − 4I24

[2]

and using the loop rule on loop bcedb gives −12I12 + 24I24 = 0

I12 = 2I24

[3]

Topic 18

1073

Substituting Equations [2] and [3] into [1] yields 7I24 = 7.0

I24 = 1.0 A

Then, Equations [2] and [3] yield

I6 = 3.0 A

and

I12 = 2.0 A

18.19 (a) With the bridge balanced, no charge flows through the ammeter and the wire through the ammeter is an equipotential. Taking I1 as the current in the left branch and I2 as the current in the right branch (both directed down in the figure), apply Kirchhoff’s loop rule, moving clockwise around the top and bottom halves of the bridge to find:

− (1.00 Ω ) I 2 + ( 3.00 Ω ) I1 = 0 and −RI 2 + ( 9.00 Ω ) I1 = 0 →

1.00 Ω 3.00 Ω = R 9.00 Ω

R = 3.00 Ω (b) Apply Kirchhoff’s loop rule to the outside perimeter of the bridge, moving clockwise to find

9.00 V − (1.00 Ω ) I1 − ( 3.00 Ω ) I1 = 0 → I1 = 2.25 A

18.20

(a) Applying Kirchhoff’s junction rule at a gives I2 = I4 − I6

[1]

Topic 18

1074

Going counterclockwise around the lower loop, and applying Kirchhoff’s loop rule, we obtain

+ 8.00 V − (6.00 Ω)I6 + (2.00 Ω)I2 = 0

4 1 I6 = + I2 3 3 !

[2]

Applying the loop rule as we go counterclockwise around the upper loop:

−(2.00 Ω)I2 − (4.00 Ω)I4 + 12.0 V = 0

1 I 4 = 3.00 − I 2 2 !

Substituting Equations [2] and [3] into Equation [1] yields

1 1⎞ 4 ⎛ ⎜⎝ 1 + + ⎟⎠ I 2 = 3.00 − 2 3 3 !

and I2 = 0.909 A

(b) The potential difference between points a and b is

ΔVab = −I2 (2.00 Ω) = −(0.909 A)(2.00 Ω) = −1.82 V

[3]

Topic 18

1075

|ΔVab| = −1.82 V with point b at the lower potential.

18.21 Consider the circuit diagram below, in which Kirchhoff’s junction rule has already been applied at points a and e.

Applying the loop rule around loop abca gives

ε − R(I − I) − 4RI = 0 1

⎞ 1⎛ε I1 = ⎜ + I ⎟ 5⎝ R ⎠

[1]

Next, applying the loop rule around loop cedc gives

−3RI2 + 2ε − 2R(I2 + I) = 0

⎞ 2⎛ε I2 = ⎜ − I ⎟ 5⎝ R ⎠

[2]

Finally, applying the loop rule around loop caec gives

−4RI1 + 3RI2 = 0

4I1 = 3I2

[3]

Topic 18

1076

Substituting Equations [1] and [2] into Equation [3] yields I =

ε .

Thus, if ε = 250 V and R = 1.00 kΩ = 1.00 × 103 Ω, the current in the wire between a and e is

(

250 V 3

5 1.00 × 10 Ω

)

= 50.0 × 10−3 A

= 50.0 mA flowing from a toward e.

18.22 Following the path of I1 from a to b and recording changes in potential gives

Vb − Va = +24 V − (6.0 Ω)(3.0 A) = + 6.0 V

Now, following the path of I2 from a to b and recording changes in potential gives

Vb − Va = −(3.0 Ω)I2 = + 6.0 V, or I2 = −2.0 A

Topic 18

1077

Thus, I2 is directed from b toward a and has magnitude of 2.0 A.

Applying Kirchhoff’s junction rule at point a gives

I3 = I1 + I2 = 3.0 A + (−2.0 A) = 1.0 A

18.23 Starting from the circuit’s leftmost branch, define the current in each branch to be positive in the upward direction: I200 in the 200-Ω resistor; I80 in the 80.0-Ω resistor; I20 in the 20.0-Ω resistor; and I70 in the 70.0-Ω resistor. From Kirchhoff’s node rule, I200 + I80 + I20 = I70. (a) Apply Kirchhoff’s loop rule three times, one for each of the inner loops, moving clockwise to find (suppressing units and sig fig notation for clarity):

−200I 200 + 80I 80 + 0 = 40 0 − 80I 80 + 20I 20 = −400 −70I 200 − 70I 80 − 90I 20 = 440 One method of solving this system of equations is to multiply the middle equation by 9/2 and add all three together, eliminating I20. Then, between the resulting sum and the first equation, eliminate I200 to find I 80 = 3.00 A (up) . Knowing I80, substitute into the top equation to find I 200 = 1.00 A (up) . Next substitute into the middle

Topic 18

1078

equation to find I20 = −8.00 A so that I20 = 8.00 A (down) . Finally, apply Kirchhoff’s node rule to find I70 = 4.00 A (up) . (b) The potential difference across the 200.-Ω resistor is ΔV200 = I200R = (1.00 A)(200. Ω) = 2.00 × 10 2 V .

(c) A battery delivers power P = IV. Identifying each battery using a subscript equal to its emf, the delivered powers are P40 = I80(40.0 V) = (3.00 A)(40.0 V) = 1.20 × 10 2 W , P360 = I20(360. V) = 2 880 W , and P80 = I70(80.0 V) = 3.20 × 10 2 W .

18.24

(a) The 30.0-Ω and 50.0-Ω resistors in the upper branch are in series, and add to give a total resistance of Rupper = 80.0 Ω for this path. This 80.0Ω resistance is in parallel with the 80.0-Ω resistance of the middle branch, and the rule for combining resistors in parallel yields a total resistance of Rab = 40.0 Ω between points a and b. This resistance is in series with the 20.0-Ω resistor, so the total equivalent resistance of the circuit is

Req = 20.0 Ω + Rab = 20.0 Ω + 40.0 Ω = 60.0 Ω

Topic 18

1079

(b) The current supplied to this circuit by the battery is ΔV 12 V I total = = = 0.20 A . Req 60.0 Ω !

(d) The potential difference between points a and b is

ΔVab = RabItotal = (40.0 Ω)(0.20 A) = 8.0 V

ΔVab 8.0 V = = 0.10 A , and the current in the upper branch is I upper = Rupper 80.0 Ω !

so the power delivered to the 50.0 Ω resistor is

2 P = R50 I upper = (50.0 Ω)(0.10 A)2 = 0.50 W ! 50

18.25

(a) We name the currents I1, I2, and I3 as shown.

Topic 18

1080

Applying Kirchhoff’s loop rule to loop abcfa gives +ε1 − ε2 − R2I2 − R1I1 = 0 or

3I2 + 2I1 = 10.0 mA

and I1 = 5.00 mA − 1.50I2

[1]

Applying the loop rule to loop edcfe yields +ε3 − R3I3 − ε2 − R2I2 = 0

3I2 + 4I3 = 20.0 mA

and I3 = 5.00 mA − 0.750I2

[2]

Finally, applying Kirchhoff’s junction rule at junction c gives

I2 = I1 + I3

[3]

Substituting Equations [1] and [2] into [3] yields

I2 = 5.00 mA − 1.50I2 + 5.00 mA − 0.750I2

3.25I2 = 10.0 mA

Topic 18

1081

This gives I2 = (10.0 mA)/3.25 = 3.08 mA. Then, Equation [1] yields

⎛ 10.0 mA ⎞ I1 = 5.00 mA − 1.50 ⎜ = 0.385 mA ⎝ 3.25 ⎟⎠ ! ⎛ 10.0 mA ⎞ = 2.69 mA and, from Equation [2], I 3 = 5.00 mA − 0.750 ⎜ ⎝ 3.25 ⎟⎠ !

(b) Start at point c and go to point f, recording changes in potential to obtain Vf − Vc = −ε2 − R2I2 = −60.0 V − (3.00 × 103 Ω)(3.08 × 10−3 A) = −69.2 V

|ΔV|cf = 69.2 V and point c is at the higher potential.

18.26 Sum voltage drops around the circuit, moving clockwise and assuming the current is clockwise: ε − IR1 − IR2 = 0. Substitute ε = 5.00 V and IR2 = ΔV = 1.50 V to find the current I: I=

ε − ΔV = 5.00 V − 1.50 V = 3.50 × 10 A −3

1.00 × 10 Ω 3

From Ohm’s law, the required resistance is ΔVout = IR2 ΔVout 1.50 V = I 3.50 × 10−3 A = 429 Ω

R2 =

Topic 18

18.27

1082

(a) No. Some simplification could be made by recognizing that the 2.0-Ω and 4.0-Ω resistors are in series, adding to give a total of 6.0 Ω; and the 5.0-Ω and 1.0-Ω resistors form a series combination with a total resistance of 6.0 Ω. The circuit cannot be simplified any further, and Kirchhoff’s rules must be used to analyze the circuit.

(b) Applying Kirchhoff’s junction rule at junction a gives

I1 = I2 + I3

[1]

Using Kirchhoff’s loop rule on the upper loop yields

+24 V − (2.0 + 4.0)I1 − (3.0)I3 = 0

I3 = 8.0 A − 2I1

For the lower loop:

[2]

+12 V + (3.0)I3 − (1.0 + 5.0)I2 = 0

Using Equation [2], this reduces to

Topic 18

1083

I2 = !

12 V + 3.0 ( 8.0 A − 2I1 ) 6.0

I2 = 6.0 A − I1

Substituting Equations [2] and [3] into [1] gives I1 = 3.5 A.

Then, Equation [3] gives

I2 = 2.5 A, and Equation [2] yields

I3 = 1.0 A.

18.28 Using Kirchhoff’s loop rule on the outer perimeter of the circuit gives

+12 V − (0.01)I1 − (0.06)I3 = 0

I1 = 1.2 × 103 A − 6.0I3

[1]

For the rightmost loop, the loop rule gives +10 V + (1.00)I2 − (0.06)I3 = 0

I2 = 0.06I3 − 10 A

[2]

Applying Kirchhoff’s junction rule at either junction gives

I1 = I2 + I3

[3]

Topic 18

1084

(a) Substituting Equations [1] and [2] into [3] yields

7.1I3 = 1.2 × 103 A and I3 = 1.7 × 102 A (in starter)

(b) Then, Equation [2] gives I2 = 0.20 A (in dead battery).

18.29

(a) No. This multi-loop circuit does not contain any resistors in series (i.e., connected so all the current in one must pass through the other) nor in parallel (connected so the voltage drop across one is always the same as that across the other). Thus, this circuit cannot be simplified any further, and Kirchhoff’s rules must be used to analyze it.

(b) Assume currents I1, I2, and I3 in the directions shown above. Then, using Kirchhoff’s junction rule at junction a gives

I3 = I1 + I2

Applying Kirchhoff’s loop rule on the lower loop,

[1]

Topic 18

1085

+10.0 V − (5.00)I2 − (20.0)I3 = 0

I2 = 2.00 A − 4I3

[2]

and for the loop around the perimeter of the circuit,

20.0 V − 30.0I1 − 20.0I3 = 0

I1 = 0.667 A − 0.667I3

Substituting Equations [2] and [3] into [1]:

I3 = 0.667 A − 0.667I3 + 2.00 A − 4I3

which reduces to 5.67I3 = 2.67 A and gives I3 = 0.471 A.

Then, Equation [2] gives I2 = 0.116 A, and from Equation [3], I1 = 0.353 A.

All currents are in the directions indicated in the circuit diagram given above.

18.30 (a) Going counterclockwise around the upper loop, Kirchhoff’s loop rule gives

[3]

Topic 18

1086

−11.0I12 + 12.0 − 7.00I12 − 5.00I18 + 18.0 − 8.00I18 = 0

18.0I12 + 13.0I18 = 30.0

[1]

(b) Going counterclockwise around the lower loop:

−5.00I36 + 36.0 − 7.00I12 − 12.0 + 11.0I12 = 0

5.00I36 − 18.0I12 = 24.0

[2]

I18 = I12 + I36

[3]

(d) Solving Equation [3] for I36 yields I36 = I18 − I12 . [4]

(e) Substituting Equation [4] into [2] gives 5.00(I18 − I12) − 18.0I12 = 24.0, or

5.00I18 − 23.0I12 = 24.0

[5]

Topic 18

1087

(f) Solving Equation [5] for I18 yields I18 = (24.0 + 23.0I12)/5.00. Substituting this into Equation [1] and simplifying gives 389I12 = −162, and I12 = −0.416 A. Then, from Equation [1], I18 = (30.0 − 18.0I12)/13.0 which yields I18 = 2.88 A.

(g) Equation [4] gives I36 = 2.88 A − (−0.416 A), or

I36 = 3.30 A.

(h) The negative sign in the answer for I12 means that this current flows in the opposite direction to that shown in the circuit diagram and assumed during this solution. That is, the actual current in the middle branch of the circuit flows from right to left and has a magnitude of 0.416 A.

18.31 Applying Kirchhoff’s junction rule at junction a gives

I3 = I1 + I2

Using Kirchhoff’s loop rule on the leftmost loop yields

[1]

Topic 18

1088

−3.00 V − (4.00)I3 − (5.00)I1 + 12.0 V = 0

so I1 = (9.00 A − 4.00I3)/5.00

I1 = 1.80 A − 0.800I3

[2]

For the rightmost loop,

−3.00 V − (4.00)I3 − (3.00 + 2.00)I2 + 18.0 V = 0

and I2 = (15.0 A − 4.00I3)/5.00

I2 = 3.00 A − 0.800I3

[3]

Substituting Equations [2] and [3] into [1] and simplifying gives 2.60I3 = 4.80 and I3 = 1.846 A. Then Equations [2] and [3] yield I1 = 0.323 A and I2 = 1.523 A.

Therefore, the potential differences across the resistors are

ΔV2 = I2(2.00 Ω) = 3.05 V, ΔV3 = I2(3.00 Ω) = 4.57 V

ΔV4 = I3(4.00 Ω) = 7.38 V, and ΔV5 = I1(5.00 Ω) = 1.62 V

18.32

The time constant is τ = RC. Considering units, we find

⎛ Volts ⎞ ⎛ Coulombs ⎞ ⎛ Coulombs ⎞ RC → ( Ohms ) ( Farads ) = ⎜ ⎜ ⎟= ⎝ Amperes ⎟⎠ ⎝ Volts ⎠ ⎜⎝ Amperes ⎟⎠

Coulombs ⎛ ⎞ =⎜ = Second ⎝ Coulombs/Second ⎟⎠

Topic 18

1089

18.33 (a)

(b)

τ = RC has units of time.

The time constant is: τ = RC = (75.0 × 103 Ω)(25.0 × 10−6 F) = 1.88 s.

At t = τ, q = 0.632Qmax = 0.632(Cε) = 0.632(25.0 × 10−6 F)(12.0 V) = 1.90 × 10−4 C.

18.34 (a) τ = RC = (100 Ω)(20.0 × 10−6 F) = 20.0 × 10−3 F = 2.00 ms

(b) Qmax = Cε = (20.0 × 10−6 F)(9.00 V) = 1.80 × 10−4 C = 180 µC

1⎞ ⎛ −t/τ − τ /τ (c) Q = Qmax (1 − e ) = Qmax (1 − e ) = Qmax ⎜ 1 − ⎟ = 114 µC ⎝ e⎠ !

18.35 (a) The time constant of an RC circuit is τ = RC. Thus,

τ = (1.00 × 10−6 Ω)(5.00 × 10−6 F) = 5.00 s

(b) Qmax = Cε = (5.00 µF)(30.0 V) = 150 µC

(c) To obtain the current through the resistor at time t after the switch is closed, recall that the charge on the capacitor at that time is q = Cε (1 τ

− e−t/ ) and the potential difference across a capacitor is Vc = q/C. Thus,

Topic 18

1090

Vc =

(

Cε 1 − e −t/τ C

) = ε (1 − e ) −t/τ

Then, considering switch S to have been closed at time t = 0, apply Kirchhoff’s loop rule around the circuit shown above to obtain

+ε − iR − Vc = 0

ε − ε (1 − e R

−t/τ

)

The current in the circuit at time t after the switch is closed is then i = (ε/R)e−t/τ), so the current in the resistor at t = 10.0 s is

⎛ 30.0 V ⎞ − 5.00s i=⎜ e = ( 30.0 µA ) e −2.00 = 4.06 µA 6 ⎟ 1.00 × 10 Ω ⎝ ⎠ ! 10.0s

(

)

18.36 (a) The charge on a charging capacitor increases as q = Q 1 − e −t/τ where

τ is the RC time constant and Q is the maximum charge. At t = 15.0 s the capacitor holds 90.0% of its final charge so that q/Q = 0.900. Substitute these values and solve for the time constant τ :

Topic 18

1091

(

)

q − 15.0 s /τ − 15.0 s /τ = 1 − e ( ) = 0.900 → e ( ) = 0.100 Q −15.0 s = ln ( 0.100 ) → τ = 6.51 s τ Solve for the resistance R from the definition of the time constant: τ = RC: R=

τ 6.51 s = = 5.21 × 10 4 Ω C 125 × 10−6 F

(b) The charge stored when q/Q = 0.900 is q = 0.900(C ε ) = 3.09 × 10−2 C. If this charge is delivered in 1.00 ms, the average current is Iav =

q 3.09 × 10−2 C = = 30.9 A Δt 1.00 × 10−3 s

18.37 (a) Use the relation q = CΔV(t) to find the voltage across a charging

(

)

capacitor as a function of time: ΔV ( t ) = ε 1 − e −t/τ . First, substitute known values to find the time constant τ :

ΔV

ε (

= 1 − e −t/τ

) → e(

− 1.50 s)/τ

= 1−

775 V = 0.380 1250 V

−1.50 s = ln ( 0.380 ) → τ = 1.55 s τ Use the definition τ = RC to find the unknown resistance: R=

τ 1.55 s = = 1.94 × 105 Ω −6 C 8.00 × 10 F

Topic 18

1092

(b) The voltage across the discharging capacitor will decrease as ΔV ( t ) = ΔVmax e −t/τ where τ = RtorsoC = (1250 Ω)(8.00 µF) = 0.0100 s.

Substitute this time constant and the value ΔVmax = 775 V to find the voltage at t = 5.00 ms:

ΔV t=5.00 ms = ( 775 V ) e

(

)

− 5.00×10−3 s /( 0.0100 s)

= 4.70 × 10 2 V

18.38 (a) The capacitor is uncharged at t = 0 (q = CΔV = 0). Kirchhoff’s loop rule gives the current as 9.00 V ε − IR = 0 → I = εR = 55.0 = 0.164 A Ω

(b) At t = τ, the potential difference across the charging capacitor is

(

) → ΔV (t ) = ε (1 − e )

(

)

q ( t ) = CΔVcap ( t ) = Q 1 − e −t/τ

−t/τ

cap

ΔVcap (τ ) = ( 9.00 V ) 1 − e −τ /τ = 5.69 V Kirchhoff’s loop rule then gives the current as

ε − IR − ΔV = 0 → I = cap

18.39

ε − ΔV

cap

9.00 V − 5.69 V = 6.02 × 10−2 A 55.0 Ω

The current drawn by a single 75-W bulb connected to a 120-V source is I1 = P/ΔV = 75 W/120 V. Thus, the number of such bulbs that can be connected in parallel with this source before the total current drawn will equal 30.0 A is

Topic 18

1093

n= !

⎛ 120 V ⎞ 30.0 A = (30.0 A) ⎜ = 48 I1 ⎝ 75 W ⎟⎠

18.40 (a) The toaster’s power is Ptoaster = IΔV = I2R = (ΔV)2/R = 1 150 W. Substitute ΔV = 120. V to find

( ΔV ) = (120 V ) = 12.5 Ω R= 2

Ptoaster

1 150 W

(b) The maximum total power is Pmax = ImaxΔV = (20.0 A)(120. V) = 2.40 × 103 W. If the toaster consumes Ptoaster = 1 150 W, the maximum power available to the microwave is

Pmax = Ptoaster + Pmicrowave,max Pmicrowave,max = Pmax − Ptoaster = 2.40 × 103 W − 1 150 W = 1250 W

18.41 From P = (ΔV)2/R, the resistance of the element is

( ΔV ) = (240 V ) = 19.2 Ω R= 2

3 000 W

When the element is connected to a 120-V source, we find that

ΔV 120 V = = 6.25 A , and (a) I = R 19.2 Ω ! (b) P = (ΔV)I = (120 V)(6.25 A) = 750 W © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

18.42

1094

(a) The current drawn by each appliance operating separately is

Coffee Maker:

Toaster:

Waffle Maker:

P ΔV

1 200 W 120 V

1 100 W

P ΔV

120 V

= 10 A

= 9.2 A

1 400 W 120 V

= 12 A

(b) If the three appliances are operated simultaneously, they will draw a total current of Itotal = (10 + 9.2 + 12) A = 31 A.

(c) No. The total current required exceeds the limit of the circuit breaker, so they cannot be operated simultaneously. In fact, with a 15 A limit, no two of these appliances could be operated at the same time without tripping the breaker.

18.43

(a) The area of each surface of this axon membrane is

A = L(2πr) = (0.10 m)[2π(10 × 10−6 m)] = 2π × 10−6 m2

and the capacitance is

Topic 18

1095

C = κ! 0

A d

(

= 3.0 8.85 × 10

−12

⎛ 2π × 10−6 m 2 ⎞ −8 C N⋅m ⎜ ⎟ = 1.67 × 10 F −8 ⎝ 1.0 × 10 m ⎠ 2

)

In the resting state, the charge on the outer surface of the membrane is

Qi = C(ΔV)i = (1.67 × 10−8 F) (70 × 10−3 V) = 1.17 × 10−9 C → 1.2 × 10−9 C

The number of potassium ions required to produce this charge is

Qi 1.17 × 10−9 C NK + = = = 7.3 × 109 K + ions −19 e 1.6 × 10 C ! and the charge per unit area on this surface is

⎞ ⎛ 10−20 m 2 ⎞ Q 1.17 × 10−9 C ⎛ 1e 1e 1e σ= i= = = −6 2 ⎜ −19 2 4 2 ⎜ ⎟ ⎟ A 2π × 10 m ⎝ 1.6 × 10 C ⎠ ⎝ 1 Å ⎠ 8.6 × 10 Å (290 Å)2 ! This corresponds to a low charge density of one electronic charge per square of side 290 Å, compared to a normal atomic spacing of one atom every few Å.

(b) In the resting state, the net charge on the inner surface of the membrane is −Qi = −1.17 × 10−9 C, and the net positive charge on this surface in the excited state is

Topic 18

1096

Qf = C(ΔV)f = (1.67 × 10−8 F) (+30 × 10−3 V) = +5.0 × 10−10 C

The total positive charge which must pass through the membrane to produce the excited state is therefore ΔQ = Q f − Qi !

= +5.0 × 10−10 C − ( −1.17 × 10−9 C ) = 1.67 × 10−9 C → 1.7 × 10−9 C

corresponding to

ΔQ 1.67 × 10−9 C NNa+ = = = 1.0 × 1010 Na + ions −19 + e 1.6 × 10 C Na ion ! (c) If the sodium ions enter the axon in a time of Δt = 2.0 ms, the average current is

ΔQ 1.67 × 10−9 C I= = = 8.3 × 10−7 A = 0.83 µA −3 Δt 2.6 × 10 s ! (d) When the membrane becomes permeable to sodium ions, the initial influx of sodium ions neutralizes the capacitor with no required energy input. The energy input required to charge the now neutral capacitor to the potential difference of the excited state is

2 1 1 2 W = C ( ΔV ) f = (1.67 × 10−8 F ) ( 30 × 10−3 V ) = 7.5 × 10−12 J 2 2 !

Topic 18

1097

18.44 The capacitance of the 10 cm length of axon was found to be C = 1.67 × 10−8 F in the solution of Problem 18.43.

(a) When the membrane becomes permeable to potassium ions, these ions flow out of the axon with no energy input required until the capacitor is neutralized. To maintain this outflow of potassium ions and charge the now neutral capacitor to the resting action potential requires an energy input of

2 1 1 2 W = C ( ΔV ) = (1.67 × 10−8 F ) (70 × 10−3 V ) = 4.1 × 10−11 J 2 2 !

(b) As found in the solution of Problem 18.43, the charge on the inner surface of the membrane in the resting state is −1.17 × 10−9 C, and the charge on this surface in the excited state is +5.0 × 10−10 C. Thus, the positive charge which must flow out of the axon as it goes from the excited state to the resting state is

ΔQ = 5.0 × 10−10 C + 1.17 × 10−9 C = 1.67 × 10−9 C

and the average current during the 3.0 ms required to return to the resting state is

Topic 18

1098

ΔQ 1.67 × 10−9 C I= = = 5.6 × 10−7 A = 0.56 µA −3 Δt 3.0 × 10 s ! 18.45 From Figure 18.29, the duration of an action potential pulse is 4.5 ms. From the solution of Problem 18.43, the energy input required to reach the excited state is W1 = 7.5 × 10−12 J. The energy input required during the return to the resting state is found in Problem 18.44 to be W2 = 4.1 × 10−11 J. Therefore, the average power input required during an action potential pulse is

Wtotal W1 + W2 7.5 × 10−12 J + 4.1 × 10−11 J P= = = = 1.1 × 10−8 W = 11 nW −3 Δt Δt 4.5 × 10 s ! 18.46 Using a single resistor → 3 distinct values:

R1 = 2.0 Ω, R2 = 4.0 Ω, R3 = 6.0 Ω

2 resistors in Series → 2 additional distinct values:

R4 = 2.0 Ω + 6.0 Ω = 8.0 Ω, and R5 = 4.0 Ω + 6.0 Ω = 10.0 Ω.

Note: 2.0 Ω and 4.0 Ω in series duplicates R3 above.

2 resistors in Parallel → 3 additional distinct values:

Topic 18

1099

R6 = 2.0 Ω and 4.0 Ω in parallel = 1.3 Ω

R7 = 2.0 Ω and 6.0 Ω in parallel = 1.5 Ω

R8 = 4.0 Ω and 6.0 Ω in parallel = 2.4 Ω

3 resistors in Series → 1 additional distinct value:

R9 = 2.0 Ω + 4.0 Ω + 6.0 Ω = 12 Ω

3 resistors in Parallel → 1 additional distinct value:

R10 = 2.0 Ω, 4.0 Ω and 6.0 Ω in parallel = 1.1 Ω

1 resistor in Parallel with Series combination of the other 2: → 3 additional values:

R11 = (Rp = 2.0 Ω; 4.0 Ω and 6.0 Ω in series) = 1.7 Ω

R12 = (Rp = 4.0 Ω; 2.0 Ω and 6.0 Ω in series) = 2.7 Ω

R13 = (Rp = 6.0 Ω; 2.0 Ω and 4.0 Ω in series) = 3.0 Ω

Topic 18

1100

1 resistor in Series with Parallel combination of the other 2: → 3 additional values:

R14 = (Rs = 2.0 Ω; 4.0 Ω and 6.0 Ω in parallel) = 4.4 Ω

R15 = (Rs = 4.0 Ω; 2.0 Ω and 6.0 Ω in parallel) = 5.5 Ω

R16 = (Rs = 6.0 Ω; 2.0 Ω and 4.0 Ω in parallel) = 7.3 Ω

Thus, 16 distinct values of resistance are possible using these three resistors.

18.47 Since the circuit is open at points a and b, no current flows through the 4.00-V battery or the 10.0-Ω resistor. A current I will flow around the closed path through the 2.00-Ω resistor, 4.00-Ω resistor, and the 12.0-V battery as shown in the sketch below. This current has magnitude ΔV 12.0 V I= = = 2.00 A Rpath 2.00 Ω + 4.00 Ω !

Topic 18

1101

Along the path from point a to point b, the change in potential that occurs is given by

ΔVab = +ε4 − IR4 = +4.00 V − (2.00 A)(4.00 Ω) = −4.00 V

(a) The potential difference between points a and b has magnitude

|ΔVab| = 4.00 V

(b) Since the change in potential in going from a to b was negative, we conclude that point a is at the higher potential.

18.48

ΔV 6.0 V = = 2.0 × 103 Ω = 2.0 kΩ (a) R = −3 I 3.0 × 10 A !

(b) The resistance in the circuit consists of a series combination with an equivalent resistance of Req = 2.0 kΩ + 3.0 kΩ = 5.0 kΩ. The emf of the

Topic 18

1102

battery is then

ε = IR = (3.0 × 10 A)(5.0 × 10 Ω) = 15 V eq

−3

(d) In this solution, we have assumed that we have ideal devices in the circuit. In particular, we have assumed that the battery has negligible internal resistance, the voltmeter has an extremely large resistance and draws negligible current, and the ammeter has an extremely low resistance and a negligible voltage drop across it.

18.49 The voltage across one of the series-connected resistors is ΔVseries = IseriesR = (1 A)(1 Ω) = 1 V. The current through each parallel-connected resistor is Iparallel = 0.5 A so that ΔVparallel = IparallelR = (0.5 A)(1 Ω) = 0.5 V. (a) The ratio ΔVseries/ΔVparallel is then ΔVseries/ΔVparallel = 2 . (b) The power dissipated by a resistor is P = IΔV so that

Pseries = Iseries ΔVseries = (1 A ) (1 V ) = 1 W

Pparallel = I parallel ΔVparallel = ( 0.5 A ) ( 0.5 V ) = 0.25 W Pseries 1W = = 4 Pparallel 0.25 W

Topic 18

18.50

1103

(a) From P = (ΔV)2/R, the resistance of each of the three bulbs is given by 2 ΔV ) (120 V ) ( R= = = 240 Ω 2

60.0 W

As connected, the parallel combination of R2 and R3 is in series with R1. Thus, the equivalent resistance of the circuit is

⎛ 1 1⎞ Req = R1 + ⎜ + ⎟ ⎝ R2 R3 ⎠ !

−1

⎛ 1 1 ⎞ = 240 Ω + ⎜ + ⎝ 240 Ω 240 Ω ⎟⎠

−1

= 360 Ω

The total power delivered to the circuit is

( ΔV )2 = (120 V ) = 40.0 W P= 2

Req

360 Ω

ΔV 120 V 1 = = A . Thus, (b) The current supplied by the source is I = Req 360 Ω 3 !

the potential difference across R1 is

( ΔV )1 = I R1 = ⎛⎜⎝ A⎞⎟⎠ ( 240 Ω ) = 80.0 V 3 ! 1

Topic 18

1104

The potential difference across the parallel combination of R2 and R3 is then

( ΔV ) = ( ΔV ) = ( ΔV ) 2

source

( ) = 120 V − 80.0 V = 40.0 V

− ΔV

18.51 When the two resistors are connected in series, the equivalent resistance is Rs = R1 + R2 and the power delivered when a current I = 5.00 A flows through the series combination is

Ps = I2Rs = (5.00 A)2(R1 + R2) = 225 W

Thus,

225 W R1 + R2 = 25.0 A 2 !

giving

R1 + R2 = 9.00 Ω [1]

When the resistors are connected in parallel, the equivalent resistance is Rp = R1R2/(R1 + R2) and the power delivered by the same current (I = 5.00 A) is

⎞ 2⎛ R R Pp = I 2 Rp = ( 5.00 A ) ⎜ 1 2 ⎟ = 50.0 W ⎝ R1 + R2 ⎠ !

giving

50.0 W Rp = 25.0 A 2 !

R1R2 = 2.00 Ω R1 + R2

[2]

Substituting Equation [1] into Equation [2] yields

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1105

R ( 9.00 Ω − R1 ) R1R2 = 1 = 2.00 Ω R + R 9.00 Ω 1 2 ! 2 2 or !R1 − (9.00 Ω)R1 + 18.0 Ω = 0 . This quadratic equation factors as

(R1 − 3.00 Ω)(R1 − 6.00 Ω) = 0

Thus, either R1 = 3.00 Ω or R1 = 6.00 Ω, and from Equation [1], we find that either R2 = 6.00 Ω or R2 = 3.00 Ω. Therefore, the pair contains one 3.00 Ω resistor and one 6.00 Ω resistor.

18.52

(a) Recognize that the 5.00-Ω and the 8.00-Ω resistors are connected in parallel and that the effective resistance of this parallel combination is

(5.00 Ω)(8.00 Ω) Rp = = 3.08 Ω 5.00 Ω + 8.00 Ω ! This resistance is in series with the 10.0-Ω resistor, giving a total resistance for the circuit of Req = 10.0 Ω + 3.08 = 13.1 Ω. Thus, the current supplied by the battery is

I total =

ε = 15.0 V = 1.15 A

Req

13.1 Ω

Topic 18

1106

and the potential difference across the parallel combination is

ΔVp = ItotalRp = (1.15 A)(3.08 Ω) = 3.54 V

The current through the 5.00-Ω in this parallel combination is then

ΔVp 3.54 V I5 = = = 0.708 A R 5.00 Ω 5 ! (b) The power delivered to the 5.00-Ω resistor is

P5 = I 52 R5 = ( 0.708 A ) ( 5.00 Ω ) = 2.51 W ! 2

(c) Only the circuit in Figure P18.52c requires the use of Kirchhoff’s rules for solution. In the other circuits, the batteries can be combined into a single effective battery while the 5.00-Ω and 8.00-Ω resistors remain in parallel with each other.

(d) The power delivered is lowest in Figure 18.52c. The circuits in Figures P18.52b and P18.52d have in effect 30.0-V batteries driving current through the 10.0-Ω resistor, thus delivering more power than the circuit in Figure 18.52a. In Figure 18.52c, the two 15.0-V batteries tend to oppose each other’s efforts to drive current through the 10.0Ω resistor, making them less effective than the single 15.0-V battery © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

1107

of Figure 18.52a.

18.53

(a) When switch S is open, all three bulbs are in series and the equivalent resistance is Req !

open

= R + R + R = 3R .

When the switch is closed, bulb C is shorted across and no current will flow through that bulb. This leaves bulbs A and B in series with an equivalent resistance of Req !

closed

= R + R+ = 2R .

(b) With the switch open, the power delivered by the battery is

Popen =

= open

Req

, and with the switch closed,

closed Pclosed = ε 2 Req = ε 2 2R .

(c) When the switch is open, the three bulbs have equal brightness. When S is closed, bulb C goes out, while A and B remain equal at a greater brightness. than they had when the switch was open.

18.54 With the switch open, the circuit may be reduced as follows: © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

1108

With the switch closed, the circuit reduces as shown below:

Since the equivalent resistance with the switch closed is one-half that when the switch is open, we have

1 R + 18 Ω = ( R + 50 Ω ) , which yields R = 14 Ω 2 !

18.55

(a) Note the assumed directions of the three distinct currents in the circuit diagram below. Applying the junction rule at point c gives

I1 = I3 − I2

Applying Kirchhoff’s loop rule to loop gbcfg gives

[1]

Topic 18

1109

+800 V + 3.00I2 − 5.00I1 = 0

5.00I1 − 3.00I2 = 8.00 V

[2]

Finally, applying Kirchhoff’s loop rule to loop fcdef yields

−3.00I2 − 5.00I3 + 4.00 V = 0

5.00I3 + 3.00I2 = 4.00 V

Substituting Equation [1] into Equation [2] gives

5.00I3 − 8.00I2 = 8.00 V

and subtracting this result from Equation [3] yields I2 = −4.00 V/11.0

Equation [3] then gives the current in the 4.00-V battery as

I3 = !

4.00 V − 3.00(−4.00 V / 11.0) = +1.02 A 5.00

I3 = 1.02 A down

(b) From above, the current in the 3.00-Ω resistor is

4.00 V I2 = = −0.364 A 11.0 !

I2 = 0.364 A down

I1 = 1.02 A − (−0.364 A) = +1.38 A

I1 = 1.38 A up

(d) Once the capacitor is charged, the current in the 3.00-V battery is I = 0

[3]

Topic 18

1110

because of the open circuit between the plates of the capacitor.

(e) To obtain the potential difference between the plates of the capacitor, we start at the negative plate and go to the positive plate (noting the changes in potential) along path hgbak. The result is

ΔVhk = +8.00 V + 3.00 V = 11.0 V

so the charge on the capacitor is

Qhk = Chk(ΔVhk) = (6.00 µF)(11.0 V) = 66.0 µC

18.56 At time t, the charge on the capacitor will be q = Qmax(1 − e−t/τ), where

τ = RC = (2.0 × 106 Ω)(3.0 × 10−6 F) = 6.0 s

When q = 0.90Qmax, this gives 0.90 = 1 − e−t/τ or e−t/τ = 0.10. Thus, −t/τ = ln(0.10), giving t = −(6.0 s)ln(0.10) = 14 s.

18.57 (a)

For the first measurement, the equivalent circuit is as shown in 1 Figure 1. From this, Rab = R1 = Ry + Ry = 2 Ry so Ry = R1 2 !

[1]

Topic 18

1111

Figure 1

For the second measurement, the equivalent circuit is shown in 1 Figure 2. This gives Rac = R2 = Ry + Rx 2 !

[2]

Figure 2

Substitute [1] into [2] to obtain

1⎛ 1 ⎞ 1 R2 = ⎜ R1 ⎟ + Rx ,!or! Rx = R2 − R1 ⎝ ⎠ 2 2 4 ! (b) If R1 = 13 Ω and R2 = 6.0 Ω, then Rx = 2.8 Ω.

Since this exceeds the limit of 2.0 Ω, the antenna is inadequately grounded.

18.58 Assume a set of currents as shown in the circuit diagram below.

Topic 18

1112

Applying Kirchhoff’s loop rule to the leftmost loop gives

+75 − (5.0)I − (30)(I − I1) = 0

7I − 6I1 = 15

[1]

For the rightmost loop, the loop rule gives

− (40 + R)I1 + (30)(I − I1) = 0,

⎛7 R ⎞ I = ⎜ + ⎟ I1 ! ⎝ 3 30 ⎠

[2]

Substituting Equation [2] into [1] and simplifying gives

310I1 + 7(I1R) = 450

[3]

Also, it is known that PR = I1 R = 20 W,!so!!I1R = ! 2

20 W I1

[4]

Substitution of Equation [4] into [3] yields

140 310I1 + = 450 I 1 !

310I12 − 450I1 + 140 = 0

Topic 18

1113

Using the quadratic formula: I1 = !

− ( −450 ) ±

( −450)2 − 4 ( 310)(140) , 2(310)

yielding I1 = 1.0 A and I1 = 0.452 A. Then, from R = !

20 W , we find two I12

possible values for the resistance R. These are:

R = 20 Ω or R = 98 Ω.

18.59 When connected in series, the equivalent resistance is Req = R1 + R2 + … + Rn = nR. Thus, the current is Is = (ΔV)/Req = (ΔV)/nR, and the power consumed by the series configuration is

2 2 ΔV ) ΔV ) ( ( Ps = I Req = 2 ( nR ) = nR nR ( ) ! 2 s

For the parallel connection, the power consumed by each individual resistor is P1 = (ΔV)2/R, and the total power consumption is

Pp = nP1 = !

n(ΔV )2 R

Ps (ΔV )2 R 1 1 = ⋅ = 2 !or! Ps = 2 Pp Therefore, 2 P nR n(ΔV ) n n ! p

18.60 Consider a battery of emf ε connected between points a and b as below. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

1114

Applying Kirchhoff’s loop rule to loop acbea gives −(1.0)I1 − (1.0)(I1 − I3) + ε = 0 or

I3 = 2I1 − ε

[1]

Applying the loop rule to loop adbea gives −(3.0)I2 − (5.0)(I2 + I3) + ε = 0 or

8I2 + 5I3 = ε

[2]

I1 + I 3 3

[3]

For loop abca, the loop rule yields

−(3.0)I2 + (1.0)I3 + (1.0)I1 = 0 or I 2 = !

Substituting Equation [1] into [3] gives I2 = I1 − ε/3

Now, substitute Equations [1] and [4] into [2] to obtain 18I1 =

reduces to I1 =

13 27

[4]

26 3

ε , which

ε.

Topic 18

1115

⎡ 13 9 ⎤ 4 1 Then, Equation [4] gives I 2 = ⎢ − ⎥ ε = ε , and [1] yields I 3 = ε . 27 ⎢⎣ 27 27 ⎥⎦ 27

Then, applying Kirchhoff’s junction rule at junction a gives

I = I1 + I 2 =

Therefore, Rab =

ε= I

ε + 4 ε = 17 ε . 27

ε (17 ε / 27)

27 17

18.61 (a) and (b): With R the value of the load resistor, the current in a series circuit composed of a 12.0 V battery, an internal resistance of 10.0 Ω, and a load resistor is

12.0 V I= ! R + 10.0 Ω

and the power delivered to the load resistor is

(144 V ) R P =I R= 2

( R + 10.0 Ω)

Topic 18

1116

Some typical data values for the graph are

R (Ω)

PL (W)

1.00

1.19

5.00

3.20

10.0

3.60

15.0

3.46

20.0

3.20

25.0

2.94

30.0

2.70

Topic 18

1117

The curve peaks at PL = 3.60 W at a load resistance of R = 10.0 Ω.

18.62 The total resistance in the circuit is

⎛ 1 1⎞ R=⎜ + ⎟ ⎝ R1 R2 ⎠ !

−1

⎛ 1 1 ⎞ =⎜ + ⎝ 2.0 kΩ 3.0 kΩ ⎟⎠

−1

= 1.2 kΩ

and the total capacitance is C = C1 + C2 = 2.0 µF + 3.0 µF = 5.0 µF.

Thus,

Qmax = Cε = (5.0 µF)(120 V) = 600 µC

and

6.0 s τ = RC = (1.2 × 103 Ω ) ( 5.0 × 10−6 F ) = 6.0 × 10−3 s = 1 000 !

The total stored charge at any time t is then

q = q1 + q2 = Qmax(1 − e−t/τ)

q1 + q2 = (600 µC)(1 − e−1 000 t/6.0 s)

[1]

Topic 18

1118

Since the capacitors are in parallel with each other, the same potential difference exists across both at any time.

Therefore,

q C1

q C2

( ΔV )c = 1 = 2 , !

⎛C ⎞ or q2 = ⎜ 2 ⎟ q1 = 1.5q1 ⎝ C1 ⎠

[2]

Substituting Equation [2] into [1] gives

2.5q1 = (600 µC)(1 − e−1 000 t/6.0 s)

and

q1 = (240 µC)(1 − e−1 000 t/6.0 s)

Then, Equation [2] yields

q2 = 1.5(240 µC)(1 − e−1 000 t/6.0 s) = (360 µC)(1 − e−1 000 t/6.0 s)

18.63

(a) With 4.0 × 103 cells, each with an emf of 150 mV, connected in series, the total terminal potential difference is

ΔV = (4.0 × 103)(150 × 10−3 V) = 6.0 × 102 V

When delivering a current of I = 1.0 A, the power output is

P = I(ΔV) = (1.0 A)(6.0 × 102 V) = 6.0 × 102 W

(b) The energy released in one shock is

E1 = P(Δt)1 = (6.0 × 102 W)(2.0 × 10−3 s) = 1.2 J © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

1119

(c) The energy released in 300 such shocks is Etotal = 300E1 = 300(1.2 J) = 3.6 × 102 J. For a 1.0-kg object to be given a gravitational potential energy of this magnitude, the height the object must be lifted above the reference level is

PEg 3.6 × 102 J h= = = 37 m mg (1.0 kg ) ( 9.80 m/s 2 ) !

18.64

(a) When the power supply is connected to points A and B, the circuit reduces as shown below to an equivalent resistance of Req = 0.099 9 Ω.

5.00 V = 50.0 A From the center figure above, observe that I R1 = I1 = 0.100 Ω !

and

I R = I R = I100 = 2

5.00 V 111 Ω

= 0.045 0A = 45.0 mA

(b) When the power supply is connected to points A and C, the circuit

Topic 18

1120

reduces as shown below to an equivalent resistance of Req = 1.09 Ω.

From the center figure above, observe that

5.00 V I R1 = I R2 = I1 = = 4.55 A 1.10 Ω !

and

I R = I100 = 3

5.00 V 110 Ω

= 0.045 5A = 45.5 mA

(c) When the power supply is connected to points A and D, the circuit reduces as shown below to an equivalent resistance of Req = 9.99 Ω.

From the center figure above, observe that

5.00 V I R1 = I R2 = I R3 = I1 = = 0.450 A 11.1 Ω ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

1121

and

I100 =

5.00 V 100 Ω

= 0.050 0A = 50.0 mA

18.65 In the circuit diagram below, note that all points labeled a are at the same potential and equivalent to each other. Also, all points labeled c are equivalent.

To determine the voltmeter reading, go from point e to point d along the path ecd, keeping track of all changes in potential to find

ΔVed = Vd − Ve = −4.50 V + 6.00 V = +1.50 V

Apply Kirchhoff’s loop rule around loop abcfa to find

−(6.00 Ω)I + (6.00 Ω)I3 = 0

I3 = I

[1]

Apply Kirchhoff’s loop rule around loop abcda to find

−(6.00 Ω)I + 6.00 V − (10.0 Ω)I2 = 0 or I2 = 0.600 A − 0.600I [2]

Topic 18

1122

Apply Kirchhoff’s loop rule around loop abcea to find

−(6.00 Ω)I + 4.50 V − (5.00 Ω)I1 = 0 or

I1 = 0.900 A − 1.20I

[3]

Finally, apply Kirchhoff’s junction rule at either point a or point c to obtain I + I3 = I1 + I2

[4]

Substitute Equations [1], [2], and [3] into Equation [4] to obtain the current through the ammeter. This gives

I + I = 0.900 A − 1.20I + 0.600 A − 0.600I

3.80I = 1.50 A

and

I = 1.50 A/3.80 = 0.395 A

18.66 In the figure given below, note that all bulbs have the same resistance, R.

(a) In the series situation, Case 1, the same current I1 flows through both bulbs. Thus, the same power, !P1 = I1 R , is supplied to each bulb. Since 2

the brightness of a bulb is proportional to the power supplied to it,

Topic 18

1123

they will have the same brightness. We conclude that the bulbs have the same current, power supplied, and brightness.

(b) In the parallel case, Case 2, the same potential difference ΔV is maintained across each of the bulbs. Thus, the same current I2 = ΔV/R will flow in each branch of this parallel circuit. This means that, again, the same power !P2 = I 2 R is supplied to each bulb, and the two 2

bulbs will have equal brightness.

(c) The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the current in this case is I1 = ΔV/2R. Note that this is one half of the current I2 that flows through each bulb in the parallel circuit (Case 2). Since the power supplied is proportional to the square of the current, the power supplied to each bulb in Case 2 is four times that supplied to each bulb in Case 1. Thus, the bulbs in Case 2 are much brighter than those in Case 1.

(d) If either bulb goes out in Case 1, the only conducting path of the circuit is broken and all current ceases. Thus, in the series case, the other bulb must also go out. If one bulb goes out in Case 2, there is still a continuous conducting path through the other bulb. Neglecting

Topic 18

1124

any internal resistance of the battery, the battery continues to maintain the same potential difference ΔV across this bulb as was present when both bulbs were lit. Thus, in the parallel case, the second bulb remains lit with unchanged current and brightness when one bulb fails.

18.67

(a) The equivalent capacitance of this parallel combination is

Ceq = C1 + C2 = 3.00 µF + 2.00 µF = 5.00 µF

When fully charged by a 12.0-V battery, the total stored charge before the switch is closed is

Q0 = Ceq(ΔV) = (5.00 µF)(12.0 V) = 60.0 µC

Once the switch is closed, the time constant of the resulting RC circuit is

τ = RCeq = (5.00 × 102 Ω)(5.00 µF) = 2.50 × 10−3 s = 2.50 ms © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 18

1125

Thus, at t = 1.00 ms after closing the switch, the remaining total stored charge is

q = Q0e−t/τ = (60.0 µC)e−1.00 ms/2.50 ms = (60.0 µC)e−0.400 = 40.2 µC

The potential difference across the parallel combination of capacitors is then q 40.2 µC ΔV = = = 8.04 V Ceq 5.00 µF !

and the charge remaining on the 3.00-µF capacitor will be

q3 = C3(ΔV) = (3.00 µF)(8.04 V) = 24.1 µC

(b) The charge remaining on the 2.00-µF capacitor at this time is

q2 = q − q3 = 40.2 µC − 24.1 µC = 16.1 µC

or alternately,

(c)

q2 = C2(ΔV) = (2.00 µF)(8.04 V) = 16.1 µC

Since the resistor is in parallel with the capacitors, it has the same potential difference across it as do the capacitors at all times. Thus, Ohm’s law gives

Topic 18

1126

ΔV 8.04 V I= = = 1.61 × 10−2 A = 16.1 mA 2 R 5.00 × 10 Ω ! 18.68 (a)

If the switch S in the circuit below is closed at t = 0, the charge remaining on the capacitor at time t is q = Q0e−t/τ, where Q0 = 5.10 µC is the initial charge and

τ = RC = (1.30 × 103 Ω)(2.00 × 10−9 F) = 2.60 µs

is the time constant of the circuit.

The potential difference across the capacitor, ΔvC, at time t is

ΔvC =

q C

Q0 e −t/τ C

Applying Kirchhoff’s loop rule to the above circuit gives ΔvC − iR = 0, or

ΔvC R

⎛Q ⎞ = ⎜ 0 ⎟ e −t/τ RC ⎝ τ ⎠

Q0 e −t/τ

Topic 18

1127

The current through the resistor at time t = 9.00 µs is then

9.00 µs

⎛ 5.10 µC ⎞ − 2.60 µs ⎛Q ⎞ i = ⎜ 0 ⎟ e −t/τ = ⎜ e = 6.16 × 10−2 C/s = 61.6 mA ⎟ ⎝ τ ⎠ 2.60 µ s ⎝ ⎠ ! (b) The charge remaining on the capacitor at t = 8.00 µs is

q = Q0 e !

−t/τ

= (5.10 µC)e

−

8.00 µs 2.60 µs

= 0.235 µC

5.10 µ C Q ⎛Q ⎞ i0 = ⎜ 0 ⎟ e −0 = 0 = = 1.96 A ⎝ τ ⎠ τ 2.60 µ s !

Topic 19

1128

Topic 19 Magnetism

QUICK QUIZZES

19.1

Choice (b). The force that a magnetic field exerts on a charged particle moving through it is given by F = qvB sin θ = qvB⊥, where B⊥ is the component of the field perpendicular to the particle’s velocity. Since the particle moves in a straight line, the magnetic force (and hence B⊥, since

!qv ≠ 0 ) must be zero. 19.2

Choice (c). The magnetic force exerted on a charge by a magnetic field is proportional to the charge’s velocity relative to the field. If the charge is stationary, as in this situation, there is no magnetic force.

19.3

Choice (a). The magnetic force acting on the particle is always perpendicular to the velocity of the particle and hence to the displacement the particle is undergoing. Under these conditions, the force does no work on the particle, and the particle’s kinetic energy remains constant.

Topic 19

19.4

1129

Choice (c). The torque that a planar current loop will experience when it is in a magnetic field is given by τ = BIA sin θ. Note that this torque depends on the strength of the field, the current in the coil, the area enclosed by the coil, and the orientation of the plane of the coil relative to the direction of the field. However, it does not depend on the shape of the loop.

19.5

Choices (a) and (c). The magnitude of the force per unit length between two parallel current carrying wires is F/ = ( µ0 I1I 2 ) ( 2π d ) . The magnitude ! of this force can be doubled by doubling the magnitude of the current in either wire. It can also be doubled by decreasing the distance between them, d, by half. Thus, both choices (a) and (c) are correct.

19.6

Choice (b). The two forces are an action-reaction pair. They act on different wires and have equal magnitudes but opposite directions.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

19.2

It should point straight down toward the surface of the Earth.

19.4

No. The force that a constant magnetic field exerts on a charged particle is

Topic 19

1130

dependent on the velocity of that particle. If the particle has zero velocity, it will experience no magnetic force and cannot be set in motion by a constant magnetic field.

19.6

Yes. Regardless of which pole is used, the magnetic field of the magnet induces magnetic poles in the nail, with each end of the nail taking on a magnetic polarity opposite to that of the pole of the magnet nearest to it. These opposite magnetic poles then attract each other.

19.8

The magnet causes domain alignment in the iron, inducing magnetic poles such that the iron becomes magnetic and attracted to the original magnet. Now that the iron is magnetic, it can produce an identical effect in another piece of iron.

19.10

No. The magnetic field created by a single current loop resembles that of a bar magnet—strongest inside the loop, and decreasing in strength as you move away from the loop. Neither is the field uniform in direction— the magnetic field lines form closed paths that curve around the conductor forming the current loop and pass through the area enclosed by that current loop.

19.12

Near the poles the magnetic field of Earth points almost straight

Topic 19

1131

downward (or straight upward), in the direction (or opposite to the direction) the charges are moving. As a result, there is little or no magnetic force exerted on the charged particles at the pole to deflect them away from Earth.

19.14

The loop can be mounted on an axle so it is free to rotate. The current loop will tend to rotate in a manner that brings the plane of the loop perpendicular to the direction of the field. As the current in the loop is increased, the torque causing it to rotate will increase in magnitude.

19.16

(a) The blue magnet experiences an upward magnetic force equal to its weight. The yellow magnet is repelled by the red magnets with a force whose magnitude equals the weight of the yellow magnet plus the magnitude of the reaction force exerted on this magnet by the blue magnet.

(b) The rods prevent motion to the side and prevent the magnets from rotating under their mutual torques. Its constraint changes unstable equilibrium into stable.

(c) Most likely, the disks are magnetized perpendicular to their flat faces, making one face a north pole and the other a south pole. The

Topic 19

1132

yellow magnet has a pole on its lower face which is the same as the pole on the upper faces of the red magnets. The pole on the lower face of the blue magnet is the same as that on the upper face of the yellow magnet.

(d) If the upper magnet were inverted, the yellow and blue magnets would attract each other and stick firmly together. The yellow magnet would continue to be repelled by and float above the red magnets.

19.18 Choices (a) and (c) are correct. At point B, each current produces a magnetic field directed out of the page; at point D, each current produces a magnetic field into the page. Because the fields at each point are in the same direction, they add together to produce a stronger net field. (At points A and C, the two currents produce oppositely-directed fields at each point. These fields cancel to produce a relatively weak net field.)

ANSWERS TO EVEN NUMBERED PROBLEMS

19.2

(a)

a′) toward the left

b′) into the page

c′) out of the page

Topic 19

1133

d′) toward top of page

e′) into the page

f′) out of the page

(b) The answer for each subpart is opposite to that given in part (a) above.

19.4

(a) toward top of page

(b)

out of page (q < 0)

(c)

zero force

(d) into the page

19.6

 −2 B ! =!2.09 × 10 !T in the negative y-direction

19.8

(a)

19.10

807 N

(b)

3.01 × 105 m/s2

(b)

7.49 × 1013 m/s2

(b)

1.36 × 106

19.20 (a) !rd = 2 ⋅ rp

(b)

r = 2 ⋅ rp !α

19.22 (a)

(b)

into the page

19.12 (a)

8.00 × 10−21 N

1.25 × 10−13 N

19.14 See solution.

19.16 3.11 cm

19.18 (a)

4.88 × 10−5 T

toward the left

(c)

out of the page

Topic 19

1134

(d)

19.24 (a)

toward top of page

(e)

into the page

(f) out of the page

0.12 N

(b) Both the direction of the field and the direction of the current must be known before the direction of the force can be determined.

 19.26 !Bmin = 0.245!T eastward 19.28 (a)

(b)

19.30 (a)

9.0 × 10−3 N at 15° above the horizontal in the northward direction

2.3 × 10−3 N horizontal and due west

9.80 A

19.32

B = µkmg/Id

19.34

5.8 N into the page

19.36

4.33 × 10−3 N ⋅ m

19.38 (a)

(c)

19.40 (a)

(b)

11.7 Ω

0.150 N in the +(z-direction)

(b)

0.150 N in the −(z-direction)

(d)

(e)

1.80 × 10−2 N ⋅ m

4.00 × 10−2 A

(b)

786

(c)

7.50 × 10−7 N ⋅ m

Topic 19

1135

19.42 (a)

+x-direction, zero torque about x-axis

(b)

−x-direction, zero torque about x-axis

(c) No. The two forces are equal in magnitude and opposite in direction, canceling each other, and can have no effect on the motion of the loop.

(d) in the yz-plane at 130° counterclockwise from +y-direction. Torque is counterclockwise about x-axis.

(e)

counterclockwise about x-axis

(f) 0.135 A ⋅ m2

19.44 (a) 5.00 × 10−7 T

(g) 130°

(h)

0.155 N · m

(b)

(c)

5.00 × 10−7 T

+(y-direction)

(d) +(x-direction)

19.46 675 A downward

19.48 (a)

40.0 µT into the page

(b)

5.00 µT out of the page

(c)

1.67 µT out of the page

Topic 19

1136

 19.50 (a) !Bset = 4.00!µ T toward the bottom of the page

(b)

19.52 (a)

(c)

 B ! set = 6.67!µ T upward at 77.0° to the left of vertical

in the −y-direction

(b)

upward, in the positive z-direction

The magnitude of the upward magnetic force must equal that of the downward gravitational force.

(d)

19.54 (a)

(c)

d = qvµ0I/2πmg

(e)

5.40 cm

in the +y-direction

(b)

BP = µ0Ix/π(x2 + d2)

BP|x=0 = 0 This is as expected since the two field contributions have equal magnitudes and opposite directions at the point midway between the wires.

19.56 (a)

8.0 A

(b)

opposite directions

(c) Reversing the direction of one current changes the force from repulsion to attraction. Doubling the magnitude of one current doubles the magnitude of the force.

19.58 2.70 × 10−5 N toward the left

Topic 19

1137

19.60 4.8 × 104 turns

19.62

27.9 turns/m

19.64 (a)

B = mg/NIw directed out of the page

(b) The forces on the two sides of the loop are equal in magnitude and opposite in direction, canceling each other, and do not affect the state of balance of the system.

(c)

19.66 (a)

B = 0.26 T

5.00 cm

(b)

8.78 × 106 m/s

19.68 3.92 × 10−2 T

19.70 (a)

6.2 m/s2

(b)

0.40 s

19.72 (a)

opposite directions

(b)

68 A

19.76 (a) 1.00 × 10−5 T

(b)

8.00 × 10−5 N directed toward wire 1

(d)

8.00 × 10−5 N directed toward wire 2

19.74 0.59 T

Topic 19

1138

PROBLEM SOLUTIONS

19.1

Remember that the direction of the magnetic force exerted on the negatively charged electron is opposite to the direction predicted by right-hand rule number 1. The magnetic field near the Earth’s equator is horizontal and directed toward the north. The magnetic force experienced by a moving charged particle is always perpendicular to the plane formed by the vectors representing the magnetic field and the particle’s velocity.

(a) When the velocity of a positively charged particle is downward, right-hand rule number 1 predicts a magnetic force toward the east. Hence, the force experienced by the negatively charged electron (and also the deflection of its velocity) is directed toward the west.

(b) When the particle moves northward, its velocity is parallel to the magnetic field, and it will experience zero force and zero deflection.

(c) The direction of the force on the negatively charged electron (and the deflection of its velocity) will be vertically upward.

Topic 19

1139

(d) The direction of the force on the negatively charged electron (and the deflection of its velocity) will be vertically downward.

19.2

(a) For a positively charged particle, the direction of the force is that predicted by the right-hand rule number one. These are:

(a′)

in plane of page and to left

(b′)

(c′)

out of the page

(d′)

in plane of page and toward the top

(e′)

into the page

(f′)

into the page

out of the page

(b) For a negatively charged particle, the direction of the force is exactly opposite what the right-hand rule number 1 predicts for positive charges. Thus, the answers for part (b) are reversed from those given in part (a).

19.3

Since the particle is positively charged, use the right-hand rule number 1.  In this case, start with the fingers of the right hand in the direction of !v

 and the thumb pointing in the direction of !F . As you start closing the  hand, the fingers point in the direction of !B after they have moved 90°.

Topic 19

1140

The results are

19.4

(a)

into the page

(b)

(c)

toward bottom of page

toward the right

 Hold the right hand with the fingers in the direction of !v so that as you

 close your hand, the fingers move toward the direction of !B . The thumb will point in the direction of the force (and hence the deflection) if the particle has a positive charge. The results are

19.5

(a)

toward top of page

(b)

out of the page, since the charge is negative

(c)

θ = 180° ⇒ zero force

(d)

into the page

(a) The proton experiences maximum force when it moves perpendicular to the magnetic field, and the magnitude of this maximum force is

Fmax = qvB sin 90° = (l.60 × 10−19 C)(6.00 × 106 m/s)(1.50 T)(1) = 1.44 × 10−12 N

Topic 19

1141

Fmax 1.44 × 10−12 N = = 8.62 × 1014 m s 2 (b) amax = mp 1.67 × 10−27 kg !

(c) Since the magnitude of the charge of an electron is the same as that of a proton, the force experienced by the electron would have the same magnitude but would be in the opposite direction due to the negative charge of the electron.

(d) The acceleration of the electron would have a much greater magnitude than that of the proton because of the mass of the electron is much smaller.

19.6

Since the acceleration (and hence the magnetic force) is in the positive xdirection, the magnetic field must be in the negative y-direction (see sketch below) according to right-hand rule number 1.

The magnitude of the magnetic field is found from Fm = qvB sin θ as

Topic 19

1142

(1.67 × 10−27 kg )( 2.00 × 1013 m s2 ) = 2.09 × 10−2 T B = or 1.60 × 10−19 C ) (1.00 × 107 m/s ) sin 90.0° ( ! yielding 19.7

 B = 2.09 × 10−2 T!in!the!negative!y0direction .

(a) The proton’s velocity is perpendicular to Earth’s magnetic field, so −

the magnetic force has magnitude F = qvB = (1.60 × 10 19 C)(4.00 × 105 −

m/s)(3.00 × 10 8 T) = 1.92 × 10−21 N . (b) The direction of the magnetic force on a proton is given by right-hand ! rule number 1: point the right-hand fingers along v in the positive x-

! direction and curl them toward B in the positive z-direction. The thumb points in the direction of the magnetic force on a positive charge, in the − ( y-direction ) . (c) The electron’s velocity is perpendicular to Earth’s magnetic field, so −

the magnetic force has magnitude F = qvB= (−1.60 × 10 19 C)(4.00 × 105 −

m/s)(3.00 × 10 8 T) = 1.92 × 10−21 N . (d) The direction of the magnetic force on an electron is given by right! hand rule number 1: point the right-hand fingers along v in the

! positive x-direction and curl them toward B in the positive zdirection. The thumb points in the direction of the magnetic force on

Topic 19

1143

a positive charge, in the − ( y-direction ) . The magnetic force on a negative charge is in the opposite direction, in the + ( y-direction ) . 19.8

(a) The ion’s velocity is perpendicular to the magnetic field, so the magnetic force has magnitude F = qvB = (1.60 × 10–19 C)(2.50 × 103 m/s)(2.00 × 10−5 T) = 8.00 × 10−21 N . (b) The ion’s acceleration has magnitude a = F/m where m is the mass of an oxygen ion (O+). Oxygen’s molar mass is 15.999 g/mol so that one −

ion has a mass m = (15.999 × 10 3 kg/mol)(1 mol/6.02 × 1023 ions/mol) = −

2.66 × 10 26 kg. The ion’s acceleration magnitude is then

19.9

F 8.00 × 10−21 N = = 3.01 × 105 m/s2 m 2.66 × 10−26 kg

The magnetic force experienced by a moving charged particle has magnitude Fm = qvB sin θ, where θ is the angle between the directions of the particle’s velocity and the magnetic field. Thus,

F 8.20 × 10−13 N sin θ = m = = 0.754 qvB (1.60 × 10−19 C ) ( 4.00 × 106 m/s ) (1.70 T ) !

and

19.10

θ = sin−1 (0.754) = 48.9°

θ = 180° − 48.9° = 131° .

The force on a single ion is

Topic 19

1144

F1 = qvB sin θ = (1.60 × 10−19 C)(0851 m/s)(0.254 T) sin (51.0°) = 2.69 × 10−20 N

The total number of ions present is

ions ⎞ ⎛ N = ⎜ 3.00 × 1020 100 cm 3 ) = 3.00 × 1022 3⎟( ⎝ ⎠ cm ! Thus, assuming all ions move in the same direction through the field, the total force is

F = N ⋅ F1 = (3.00 × 1022)(2.69 × 10−20 N) = 807 N

19.11

Gravitational force:

Fg = mg = (9.11 × 10−31 kg)(9.80 m/s2) = 8.93 × 10−30 N downward

Electric force:

Fe = qE = (−1.60 × 10−19 C)(−100 N/C) = 1.60 × 10−17 N upward

Magnetic force:

Fm = qvB sin θ = (−1.60 × 10−19 C)(6.00 × 106 m/s)(50.0 × 10−16 T) sin (90.0°) = 4.80 × 10−17 N in direction opposite right hand rule prediction

Topic 19

1145

Fm = 4.80 × 10−17 N downward

19.12

(a) F = qvB sin θ = (1.60 × 10−19 C)(5.02 × 106 m/s)(0.180 T) sin (60.0°)

= 1.25 × 10−13 N

F 1.25 × 10−13 N = 7.49 × 1013 m/s 2 (b) a = = −27 m 1.67 × 10 kg ! 19.13

2 (a) From !KE = 12 me v , the speed of the electron is

2 ( 3.30 × 10−19 J ) 2 ( KE ) v= = = 8.51 × 105 m me 9.11 × 10−31 kg ! (b) The magnetic force acting on the electron must provide the necessary 2 centripetal acceleration. Thus, !me v r = qvB sin θ , which gives

9.11 × 10−31 kg ) ( 8.51 × 105 m/s ) ( me v r= = qB sin θ (1.60 × 10−19 C ) ( 0.235 T ) sin 90.0° ! 19.14

= 2.06 × 10−5 m = 20.6 × 10−6 m = 20.6 µm

For the particle to pass through with no deflection, the net force acting on it must be zero. Thus, the magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives

Topic 19

1146

Fm = Fe, or qvB = qE, which reduces to v = E/B

19.15

The speed of the particles emerging from the velocity selector is v = E/B (see Problem 14). In the deflection chamber, the magnetic force supplies the centripetal acceleration, so

mv 2 qvB = r !

mv m(E/B) mE = = 2 qB qB qB

Using the given data, the radius of the path is found to be

( 2.18 × 10 kg )( 950 V/m ) = 1.50 × 10 m = 0.150 mm (1.60 × 10 C)( 0.930 T ) −26

r= ! 19.16

−4

−19

Since the centripetal acceleration is furnished by the magnetic force

mv mv 2 acting on the ions, qvB = , or the radius of the path is r = . Thus, r ! ! qB the distance between the impact points (that is, the difference in the diameters of the paths followed by the U238 and the U235 isotopes) is

Δd = 2 ( r238 − r235 ) = = ! or

2v (m238 − m235 ) qB

2 ( 3.00 × 105 m/s )

(1.60 × 10

−19

⎡ ⎞⎤ ( 238 u − 235 u ) ⎛⎜⎝ 1.66 × 10−27 kg ⎟ ⎢ u ⎠ ⎥⎦ C ) ( 0.600 T ) ⎣

Δd = 3.11 × 10−2 m = 3.11 cm

Topic 19

1147

19.17 (a) Convert units to find 75.0 eV = 75.0 eV(1.60 × 10−19 J/1 eV) = 1.20 × 10-17 J. From KE = 12 mv 2 , the sulfur ion’s speed is then v = 2 ( KE) /m =

(

)(

)

2 1.20 × 10−17 J / 5.32 × 10−26 kg = 2.12 × 10 4 m/s . The maximum

magnetic force is

(

)(

Fmax = qvB = 1.60 × 10−19 C 2.12 × 10 4 m/s 4.28 × 10−4 T

)

= 1.45 × 10−18 N

(b) From the cyclotron equation, the ion’s circular path has radius

(

)( )(

)

5.32 × 10−26 kg 2.12 × 10 4 m/s mv r= = = 16.5 m qB 1.60 × 10−19 C 4.28 × 10−4 T

(

)

19.18 (a) Use the cyclotron equation to solve for the magnitude of Earth’s magnetic field:

(

)( )(

9.11 × 10−31 kg 6.00 × 105 m/s mv mv r= → B= = qB qr 1.60 × 10−19 C 7.00 × 10−2 m

(

)

= 4.88 × 10−5 T (b) Because the electron’s velocity is perpendicular to the magnetic field, the number of times, N, it circles the magnetic field each second is

(

)

6.00 × 10 m/s (1 s) N ( 2π r ) v v= → N= = = 1.36 × 106 −2 1s 2π r 2π 7.00 × 10 m

19.19

(

)

In the perfectly elastic, head-on collision between the α-particle and the initially stationary proton, conservation of momentum requires that mpvp

Topic 19

1148

+ mαvα = mαv0 while conservation of kinetic energy also requires that v0 − 0 = −(vα − vp) or vp = vα + v0. Using the fact that mα = 4mp and combining these equations gives

mp(vα + v0) + (4mp)vα = (4mp)v0

vα = 3v0/5

and vp = (3v0/5) + v0 = 8v0/5

Thus,

3 3⎛ 5 ⎞ 3 vα = v0 = ⎜ vp ⎟ = vp 5 5⎝ 8 ⎠ 8 !

After the collision, each particle follows a circular path in the horizontal plane with the magnetic force supplying the centripetal acceleration. If the radius of the proton’s trajectory is R, and that of the alpha particle is r, we have

vp2 qp vp B = mp R !

and

19.20

v2 qα vα B = mα α r !

mp vp qp B

mp vp eB

(

)(

)

4mp 3vp 8 mα vα 3 ⎛ mp vp ⎞ 3 = = ⎜ = R qα B 4 ⎝ eB ⎟⎠ 4 ( 2e ) B

Consider a particle of mass m and charge q accelerated from rest through a potential difference ΔV = Vi − Vf, Applying conservation of energy gives

Topic 19

1149

KEf + PEf = 0 + PEi

1 KE f = mv 2 = PEi − PE f = qVi − qV f = q ( ΔV ) 2 !

or the speed given the particle is v = 2q ( ΔV ) m . ! If the particle now enters a magnetic field of strength B, moving perpendicular to the direction of the field, it will follow a circular path of radius

mv m 2q ( ΔV ) r= = qB qB m !

which reduces to

r= !

2m ( ΔV ) [1] qB2

For a proton (mass mp and charge q = e), Equation [1] gives

2mp ( ΔV )

rp = !

eB2

(a) For the deuteron (mass md = 2mb and charge qd = e), Equation [1] gives

rd = !

( )

⎛ 2mp ( ΔV ) ⎞ 2 2mp ( ΔV ) 2md ( ΔV ) = = 2 ⎜ ⎟= qd B2 eB2 eB2 ⎠ ⎝

2 ⋅ rp

(b) For the alpha particle (mass mα = 4me and charge qα = 2e), we find

Topic 19

1150

rα = ! 19.21

( )

⎛ 2mp ( ΔV ) ⎞ 2 2mp ( ΔV ) 2mα ( ΔV ) = = 2 ⎜ ⎟= qα B2 eB2 ⎠ ( 2e ) eB2 ⎝

2 ⋅ rp

(a) Within the velocity selector, the electric and magnetic fields exert forces in opposite directions on charged particles passing through. For particles having a particular speed, these forces have equal magnitudes, and the particles pass through without deflection. The selected speed is found from Fe = qE = qvB = Fm, giving v = E/B. In the deflection chamber, the selected particles follow a circular path having a diameter of d = 2r = 2mv/qB. Thus, the mass to charge ratio for these particles is

m q

Bd 2v

( ) 2E

2 E/B

(

) (0.396 m) = 2.08 × 10 kg/C 2 ( 8 250 V/m )

0.093 1 T

−7

(b) If the particle is doubly ionized (i.e., two electrons have been removed from the neutral atom), then q = 2e, and the mass of the ion is

⎛ m⎞ m = ( 2e ) ⎜ ⎟ = 2 (1.60 × 10−19 C ) ( 2.08 × 10−7 kg/C ) = 6.66 × 10−26 kg ⎝ q⎠ !

(c) Assuming this is an element, the mass of the ion should be © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

1151

approximately equal to the atomic weight multiplied by the atomic mass unit (see Table C.5 in Appendix C of the textbook). This would give the atomic weight as

m 6.66 × 10−26 kg At.!wt. ≈ = = 40.1 , suggesting that the element −27 1 u 1.66 × 10 kg ! is calcium.

19.22

Hold the right hand with the fingers in the direction of the current so, as you close the hand, the fingers move toward the direction of the magnetic field. The thumb then points in the direction of the force. The results are

19.23

(a)

to the left

(b)

into the page

(c)

out of the page

(d)

toward top of page

(e)

into the page

(f)

out of the page

From F = BIL sin θ, the magnetic field is

F/L 0.12 N/m B= = = 8.0 × 10−3 T I sin θ (15 A)sin 90° !

  The direction of !B must be the +z-direction to have !F in the −y-direction  when !I is in the +x-direction.

Topic 19

19.24

1152

(a) F = BIL sin θ = (0.28 T)(3.0 A)(0.14 m) sin 90° = 0.12 N

(b) Neither the direction of the magnetic field nor that of the current is given. Both must be known before the direction of the force can be determined. In this problem, you can only say that the force is perpendicular to both the wire and the field.

19.25

Use the right-hand rule number 1, holding your right hand with the fingers in the direction of the current and the thumb pointing in the direction of the force. As you close your hand, the fingers will move toward the direction of the magnetic field. The results are

19.26

(a)

into the page

(b)

toward the right

(c)

toward the bottom of the page

In order to just lift the wire, the magnetic force exerted on a unit length of the wire must be directed upward and have a magnitude equal to the weight per unit length. That is, the magnitude is

F ⎛ m⎞ = BI sin θ = ⎜ ⎟ g ⎝ ⎠ !

giving

⎛ m⎞ g B=⎜ ⎟ ⎝  ⎠ I sin θ

Topic 19

1153

To find the minimum possible field, the magnetic field should be perpendicular to the current (θ = 90.0°). Then,

⎡ g g ⎛ 1 kg ⎞ ⎛ 102 cm ⎞ ⎤ 9.80 m/s 2 ⎛ m⎞ Bmin = ⎜ ⎟ = ⎢ 0.500 = 0.245 T ⎥ ⎝  ⎠ I sin 90.0° ⎣ cm ⎜⎝ 103 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ ⎦ ( 2.00 A ) (1) ! To find the direction of the field, hold the right hand with the thumb pointing upward (direction of the force) and the fingers pointing southward (direction of current). Then, as you close the hand, the fingers point eastward. The magnetic field should be directed eastward.

19.27

F = BIL sin θ = (0.300 T)(10.0 A)(5.00 m) sin (30.0°) = 7.50 N

19.28

(a) The magnitude is

F = BIL sin θ = (0.60 × 10−4 T)(15 A)(10.0 m) sin (90°) = 9.0 × 10−3 N

  !F is perpendicular to !B . Using the right-hand rule number 1, the  orientation of !F is found to be 15° above the horizontal in the northward direction.

(b) F = BIL sin θ = (0.60 × 10−4 T)(15 A)(10.0 m) sin (165°) = 2.3 × 10−3 N

and, from the right-hand rule number 1, the direction is horizontal

Topic 19

1154

and due west.

19.29

 For minimum field, !B should be perpendicular to the wire. If the force is to be northward, the field must be directed downward.

To keep the wire moving, the magnitude of the magnetic force must equal that of the kinetic friction force. Thus, BIL sin 90° µk(mg), or

µ ( m/L) g ( 0.200 ) (1.00 g/cm ) ( 9.80 m s ) ⎛ 1 kg ⎞ ⎛ 102 cm ⎞ B= k = ⎜⎝ 103 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ = 0.131 T I sin 90° 1.50 A ) (1.00 ) ( ! 2

19.30 (a) The mass is hanging in equilibrium so that ΣFy = 0 and BI ℓ sinθ − mg = 0, where θ = 90°. Solve for the current I to find

(

(b) From Ohm’s law,

ε = IR, the required resistance is

19.31

)

mg (1.00 kg ) 9.80 m/s = = 9.80 A Bℓ ( 2.00 T )(0.500 m )

ε = 115 V = 11.7 Ω I

9.80 A

(a) The magnetic force must be directed upward and its magnitude must equal mg, the weight of the wire. Then, the net force acting on the wire will be zero, and it can move upward at constant speed.

(b) The magnitude of the magnetic force must be BIL sin θ = mg and for

Topic 19

1155

minimum field θ = 90°. Thus,

mg ( 0.015 kg ) ( 9.80 m s ) Bmin = = = 0.20 T IL 5.0 A ) ( 0.15 m ) ( ! 2

For the magnetic force to be directed upward when the current is

 toward the left, !B must be directed out of the page. (c) If the field exceeds 0.20 T, the upward magnetic force exceeds the downward force of gravity, so the wire accelerates upward.

19.32

As shown in end view in the sketch below, the rod (and hence, the current) is horizontal, while the magnetic field is vertical. For the rod to move with constant velocity (zero acceleration), it is necessary that

∑ Fx = Fm – f k = BIℓsin 90.0° − µk n = 0

µk n µ (mg) B= = k Isin 90.0° Id(1.00) !

reducing to

B = µk mg Id !

Topic 19

19.33

1156

For each segment, the magnitude of the force is given by F = BIL sin θ, and the direction is given by the right-hand rule number 1. The results of applying these to each of the four segments, with B = 0.020 0 T and I = 5.00 A, are summarized below.

Segment

L (m)

F (N)

Direction

0.400

180°

0.400

90.0°

0.040 0 negative x

!0.400 2

45.0°

0.040 0 negative z

no direction

parallel to xz-

!0.400 2

90.0°

0.056 6

plane at 45° to both +x- and +z-directions

19.34

The magnitude of the force is

F = BIL sin θ = (5.0 × 10−5 N)(2.2 × 103 A)(58 m) sin 65° = 5.8 N

Topic 19

1157

and the right-hand rule number 1 shows its direction to be into the page.

19.35

The torque exerted on a single turn current loop in a magnetic field is given by τ = BIA sin θ, where A is the area enclosed by the loop and θ is the angle between the line normal to the plane of the loop and the direction of the magnetic field B. For maximum torque, θ = 90.0°, so we have

⎛ π d2 ⎞ 1 2 τ max = BI ⎜ sin 90.0° = ( 3.00 × 10−3 T ) ( 5.00 A ) π ( 0.100 m ) (1.00 ) ⎟ ⎝ 4 ⎠ 4 !

19.36

= 1.18 × 10−4 N ⋅ m = 118 × 10−6 N ⋅ m = 118 µN ⋅ m

The magnitude of the torque is τ = NBIA sin θ, where θ is the angle between the field and the line perpendicular to the plane of the loop. The circumference of the loop is 2πr = 2.00 m, so the radius is r = !

1.00 m and π

1 2 2 the area is A = π r = m . π !

Thus,

19.37

⎛1 ⎞ τ = (1) ( 0.800 T ) (17.0 × 10−3 A ) ⎜ m 2 ⎟ sin 90.0° = 4.33 × 10−3 N ⋅ m ⎝ ⎠ π !

The area of the loop is A = πab. where a = 0.200 m and b = 0.150 m. Since the field is parallel to the plane of the loop, θ = 90.0° and the magnitude of the torque is

Topic 19

1158

τ = NBIA sin θ = 8(2.00 × 10−4 T)(6.00 A)[π (0.200 m)(0.150 m)] sin 90.0°

= 9.05 × 10−4 N ⋅ m

The torque is directed to make the left-hand side of the loop move toward you and the right-hand side move away. 19.38 (a) For the left segment, θ = 90° and Fleft = BIb = 0.150 N in the + ( z-direction ) found using the right-hand rule: direct the right-hand

fingers along the current and curl them toward the magnetic field. The thumb then points in the direction of the magnetic force (along the +z-direction). (b) For the right segment, θ = 90° and Fright = BIb = 0.150 N in the − ( z-direction ) found using the right-hand rule: direct the right-hand

fingers along the current and curl them toward the magnetic field. The thumb then points in the direction of the magnetic force (along the −z-direction). (c) Along the top segment, the current is antiparallel to the magnetic field so that θ = 180° and Ftop = 0.

Topic 19

1159

(d) Along the bottom segment, the current is parallel to the magnetic field so that θ = 0 and Fbottom = 0. (e) Taking advantage of symmetry, the net torque about the axle is

⎛ a⎞ ⎛ 0.120 m ⎞ τ = 2 ⎜ ⎟ Fleft = 2 ⎜ 0.150 N ) = 1.80 × 10−2 N ⋅ m ( ⎟ ⎝ 2⎠ ⎝ ⎠ 2 19.39 (a) A current-carrying circular coil of area A = πr2 has magnetic moment

(

µ = IAN = (1.60 A ) π ( 0.200 m )

)(6.00) = 1.21 A ⋅ m

(b) For θ = 60.0°, the torque is

(

)

τ = BIAN sin θ = µBsin θ = 1.21 A ⋅ m 2 ( 0.200 T ) sin ( 60.0° ) = 0.209 N ⋅ m

19.40 (a) Use the relation P = I

ε to find the current: I = P/ ε = 0.200 W/5.00 V

= 4.00 × 10−2 A . (b) Find the number of turns, N, using the definition of magnetic moment: µ = IAN:

µ 0.020 0 A ⋅ m 2 N= = = 786 IA 4.00 × 10−2 A 6.36 × 10−4 m 2

(

)(

)

(

)(

)

τ max = µB = 0.020 0 A ⋅ m 2 3.75 × 10−5 T = 7.50 × 10−7 N ⋅ m

19.41

(a) Let θ be the angle the plane of the loop makes with the horizontal as

Topic 19

1160

shown in the sketch below. Then, the angle it makes with the vertical is !φ = 90.0° − θ . The number of turns on the loop is

L 4.00 m N= = = 10.0 circumference 4(0.100 m) !

⎛s ⎞ The torque about the z-axis due to gravity is τ g = mg ⎜ cos θ ⎟ where ⎝2 ⎠ ! s = 0.100 m is the length of one side of the loop. This torque tends to rotate the loop clockwise. The torque due to the magnetic force tends to rotate the loop counterclockwise about the z-axis and has magnitude !τ m = NBIA sin θ . At equilibrium, τ m = τ g , or NBI(s2) sin θ = ! mg(s cos θ)/2. This reduces to

0.100 kg ) ( 9.80 m s 2 ) ( mg tan θ = = = 14.4 2NBIs 2 (10.0 ) ( 0.100 0 T ) ( 3.40 A ) ( 0.100 m ) !

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1161

Since tan θ = tan (90.0° − φ) = cot φ, the angle the loop makes with the vertical at equilibrium is φ = cot−1(14.4) = 3.97°.

(b) At equilibrium,

τm =NBI(s2) sin θ = (10.0)(0.010 0 T)(3.40 A)(0.100 m) sin (90.0° − 3.97°) = 3.39 × 10−3 N ⋅ m 19.42

(a) The current in segment ab is in the +y-direction. Thus, by right-hand rule 1, the magnetic force on it is in the +x-direction. This force is parallel to the x-axis and therefore has zero torque about that axis.

(b) The current in segment cd is in the −y-direction, and the right-hand rule 1 gives the direction of the magnetic force as the −x-direction. This force is parallel to the x-axis and can exert no torque about that axis.

(c) No. These two forces are equal in magnitude and opposite in directions, so their sum is zero. Further, each force has zero torque © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

1162

about the axis the loop is hinged on. Since the two forces cancel each other and are both parallel to the rotation axis, they can have no effect on the motion of the loop.

(d) The magnetic force is perpendicular to both the direction of the current in bc (the x-axis) and the magnetic field. As given by righthand rule 1, this places it in the yz-plane at 130° counterclockwise from the +y-axis. The force acting on segment bc tends to rotate it counterclockwise about the x-axis, so the torque is in the +x direction.

(e) The loop tends to rotate counterclockwise about the x-axis.

(f) µ = IAN = (0.900 A)[(0.500 m)(0.300 m)](1) = 0.135 A ⋅ m2

(g) The magnetic moment vector is perpendicular to the plane of the loop (the xy-plane) and is therefore parallel to the z-axis. Because the current flows clockwise around the loop, the magnetic moment vector is directed downward, in the negative z-direction. This means that the angle between it and the direction of the magnetic field is θ = 90.0° + 40.0° = 130°.

Topic 19

1163

(h) τ = µB sin θ = (0.135 A ⋅ m2)(1.50 T) sin (130°) = 0.155 N ⋅ m

19.43

Treat the lightning bolt as a long, straight conductor. Then, the magnetic field is −7 4 µ0 I ( 4π × 10 T ⋅ m A ) (1.00 × 10 A ) B= = = 2.00 × 10−5 T = 20.0 µ T 2π r 2π (100 m ) !

19.44 (a) Use Ampere’s law to find

(

)

4π × 10−7 T ⋅ m/A ( 3.00 A ) µ0 I = = 5.00 × 10−7 T 2π r 2π (1.20 m )

(b) Use right-hand rule number 2 to find that the magnetic field is in the + ( y-direction ) : Point the right-hand thumb up, in the +z-direction.

On the +x axis, the fingers point in the +y-direction. (c) At the same distance from the current, the magnetic field will have the same magnitude so that, as before, B = 5.00 × 10−7 T . (d) Use right-hand rule number 2 to find that the magnetic field is in the + ( x-direction ) : Point the right-hand thumb up, in the +z-direction.

On the −y axis, the fingers point in the +x-direction.

Topic 19

19.45

1164

The magnetic field at distance r from a long conducting wire is B =

µ0I/2πr. Thus, if B = 1.0 × 10−15 T at r = 4.0 cm, the current must be 2π rB 2π ( 0.040 m ) (1.0 × 10 I= = µ 4π × 10−7 T ⋅ m/A 0 !

−15

19.46

= 2.0 × 10−10 A

Model the tornado as a long, straight, vertical conductor and imagine grasping it with the right hand so the fingers point northward on the western side of the tornado (that is, at the observatory’s location.) The thumb is directed downward, meaning that the conventional current is downward or negative charge flows upward.

The magnitude of the current is found from B = µ0I/2πr as 3 −8 2π rB 2π ( 9.00 × 10 m ) (1.50 × 10 T ) I= = = 675 A µ0 4π × 10−7 T ⋅ m/A !

19.47

From B = µ0I/2πr, the required distance is −7 µ0 I ( 4π × 10 T ⋅ m/A ) ( 20 A ) r= = = 2.4 × 10−3 m = 2.4 mm 2π B 2π (1.7 × 10−3 T ) !

19.48

Assume that the wire on the right is wire 1 and that on the left is wire 2. Also, choose the positive direction for the magnetic field to be out of the

Topic 19

1165

page and negative into the page.

(a) At the point half way between the two wires,

⎡µ I µI ⎤ µ ⎡ I I ⎤ µ ⎡ 4I ⎤ Bnet = −B1 − B2 = − ⎢ 0 1 + 0 2 ⎥ = − 0 ⎢ + =− 0 ⎢ ⎥ ⎥ 2π ⎣ d/2 d/2 ⎦ 2π ⎣ d ⎦ ⎣ 2π r1 2π r2 ⎦

( 4π × 10 T ⋅ m/A ) ⎡ 4(5.00 A) ⎤ = −4.00 × 10 T −7

=− ! or

2π

⎢ ⎥ ⎣ 0.100 m ⎦

−5

Bnet = 40.0 µT into the page

µ ⎡ I I ⎤ µ ⎡ I I ⎤ µ ⎡ I ⎤ (b) At point P1, Bnet = B1 − B2 = 0 ⎢ + 1 − 2 ⎥ = 0 ⎢ + − ⎥ = 0 ⎢ + ⎥ 2π ⎣ r1 r2 ⎦ 2π ⎣ d 2d ⎦ 2π ⎣ 2d ⎦ !

Bnet = !

4π × 10−7 T ⋅ m/A ⎡ 5.00 A ⎤ −6 ⎢ ⎥ = 5.00 × 10 T = 5.00 µ T!out!of!page 2π 2 0.100 m )⎦ ⎣ (

µ ⎡ I I ⎤ µ ⎡ I I ⎤ µ ⎡ I ⎤ (c) At point P2, Bnet = −B1 + B2 = 0 ⎢ − 1 + 2 ⎥ = 0 ⎢ − + ⎥ = 0 ⎢ ⎥ 2π ⎣ r1 r2 ⎦ 2π ⎣ 3d 2d ⎦ 2π ⎣ 6d ⎦ !

Bnet = !

19.49

4π × 10−7 T ⋅ m/A ⎡ 5.00 A ⎤ −6 ⎢ ⎥ = 1.67 × 10 T = 1.67 µ T!out!of!page 2π ⎣ 6(0.100 m) ⎦

The distance from each wire to point P is given by r = 12 s2 + s2 = s !

where s = 0.200 m.

At point P, the magnitude of the magnetic field produced by each of the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

1166

wires is −7 µ I ( 4π × 10 T! ⋅ !m/A ) ( 5.00 A ) 2 B= 0 = = 7.07 µ T 2π r 2π ( 0.200 m ) !

Carrying currents into the page, the field A produces at P is directed to the left and down at −135°, while B creates a field to the right and down at − 45°. Carrying currents toward you, C produces a field downward and to the right at − 45°, while D’s contribution is down and to the left at −135°. The horizontal components of these equal magnitude contributions cancel in pairs, while the vertical components all add. The total field is then

Bnet = 4(7.07 µT) sin 45.0° = 20.0 µT toward the bottom of the page

19.50

Wire 1 has current I1 = 3.00 A and wire 2 has I2 = 5.00 A, with both currents directed out of the page. Choose the line running from wire 1 to

Topic 19

1167

wire 2 as the positive x-direction.

(a) At the point midway between the wires, the field due to each wire is parallel to the y-axis, and the net field is B = +B1y − B2y = µ0 ( I1 − I 2 ) ( 2π d 2 ) ! net

( 4π × 10 T! ⋅ !m/A ) = ( 3.00 A − 5.00 A ) = −4.00 × 10 T B = −7

Thus,

−6

2π ( 0.100 m )

net

 Bnet = 4.00 µ T!toward!the!bottom!of!the!page !

(b) At point P, B1 is directed at θ1 = +135°.

The magnitude of B1 is

4π × 10−7 T ⋅ m/A ) ( 3.00 A ) ( µ 0 I1 µ 0 I1 B1 = = = = 2.12 µ T 2π r1 2π d 2 2π 0.200 2 m !

(

)

(

)

Topic 19

1168

The contribution from wire 2 is in the −x-direction and has magnitude −7 µ0 I 2 µ0 I 2 ( 4π × 10 T ⋅ m/A ) ( 5.00 A ) B2 = = = = 5.00 µ T 2π r2 2π d 2π ( 0.200 m ) !

Therefore, the components of the net field at point P are Bx = B1 cos 135° + B2 cos 180° = (2.12 µT) cos 135° + (5.00 µT) cos 180° = −6.50 µT and By = Bt sin 135° + B2 sin 180° = (2.12 µT) sin 135° + 0 = +1.50 µT

Thus, Bnet = Bx2 + By2 = 6.67 µ T at !

⎛B ⎞ ⎛ 6.50 µ T ⎞ θ = tan− 1 ⎜ x ⎟ = tan −1 ⎜ = 77.0° By ⎠ ⎝ 1.50 µ T ⎟⎠ ⎝ ! or

 Bnet = 6.67 µ T!upward!at!77.0°!to!the!left!of!vertical !

Topic 19

19.51

1169

Call the wire along the x-axis wire 1 and the other wire 2. Also, choose the positive direction for the magnetic fields at point P to be out of the page.

µI µI µ ⎛I I ⎞ At point P, Bnet = +B1 − B2 = 0 1 − 0 2 = 0 ⎜ 1 − 2 ⎟ 2π r1 2π r2 2π ⎝ r1 r2 ⎠ !

( 4π × 10 T ⋅ m/A ) ⎛ 7.00 A − 6.00 A ⎞ = +1.67 × 10 T B = −7

net

2π

⎜⎝ 3.00 m

4.00 m ⎟⎠

−7

 Bnet = 0.167 µ T!out!of!the!page ! 19.52

(a) Imagine the horizontal xy-plane being perpendicular to the page, with the positive x-axis coming out of the page toward you and the positive y-axis toward the right edge of the page. Then, the vertically upward positive z-axis is directed toward the top of the page. With the current in the wire flowing in the positive x-direction, the righthand rule 2 gives the direction of the magnetic field above the wire as being toward the left, or in the −y-direction.

(b) With the positively charged proton moving in the −x-direction (into the page), right-hand rule 1 gives the direction of the magnetic force on the proton as being directed toward the top of the page or © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

1170

upward, in the positive z-direction.

(c) Since the proton moves with constant velocity, a zero net force acts on it. Thus, the magnitude of the magnetic force must equal that of the gravitational force.

(d) ΣFz = maz = 0 ⇒ Fm = Fg or qvB = mg, where B = µ0I/2πd. This gives qvµ0I/2πd = mg, or the distance the proton is above the wire must be

d = qvµ0I/2πmg.

qvµ0 I (1.60 × 10 = (e) d = 2π mg !

−19

C ) ( 2.30 × 10 4 m/s ) ( 4π × T ⋅ m/A ) (1.20 × 10−6 A ) 2π (1.67 × 10−27 kg ) ( 9.80 m s 2 )

d = 5.40 × 10−2 m = 5.40 cm

19.53

(a) From B = µ0I/2πr, observe that the field is inversely proportional to the distance from the conductor. Thus, the field will have one-tenth its original value if the distance is increased by a factor of 10. The required distance is then r′ = 10r = 10(0.400 m) = 4.00 m.

(b) A point in the plane of the conductors and 40.0 cm from the center of the cord is located 39.85 cm from the nearer wire and 40.15 cm from

Topic 19

1171

the far wire. Since the currents are in opposite directions, so are their contributions to the net field. Therefore, Bnet = B1 − B2, or

(

)(

)

4π × 10−7 T⋅ m/A 2.00 A ⎛ ⎞ µ0 I ⎛ 1 1 ⎞ 1 1 Bnet = − ⎜ − ⎟= ⎜ ⎟ 2π ⎝ r1 r2 ⎠ 2π ⎝ 0.398 5 m 0.401 5 m ⎠ = 7.50 × 10−9 T = 7.50 nT

(c) Call r the distance from cord center to field point P and 2d = 3.00 mm the distance between centers of the conductors.

µ I⎛ 1 1 ⎞ µ0 I ⎛ 2d ⎞ Bnet = 0 ⎜ − ⎟= ⎜ ⎟ 2π ⎝ r − d r + d ⎠ 2π ⎝ r 2 − d 2 ⎠ !

( 4π × 10 T ⋅ m/A )( 2.00 A ) ⎛ −7

7.50 × 10−10 T = !

2π

3.00 × 10−3 m ⎞ ⎜⎝ r 2 − 2.25 × 10−6 m 2 ⎟⎠

so r = 1.26 m

The field of the two-conductor cord is weak to start with and falls off rapidly with distance.

Topic 19

1172

(d) The cable creates zero field at exterior points, since a loop in Ampère’s law encloses zero total current.

19.54

(a) Point P is equidistant from the two wires which carry identical   currents. Thus, the contributions of the two wires, !Bupper and !Blower ,

to the magnetic field at P will have equal magnitudes. The horizontal components of these contributions will cancel, while the vertical components add. The resultant field will be vertical in the +ydirection.

(b) The distance of each wire from point P is !r = x 2 + d 2 , and the cosine   of the angle that Bupper and Blower make with the vertical is cos θ = x/r. ! !   The magnitude of either Bupper or Blower is !Bwire = µ0 I 2π r , and the ! !

vertical components of either of these contributions have values of

µI x

µ Ix

( Bwire )y = ( Bwire ) cos θ = ⎛⎜⎝ 2π0 r ⎞⎟⎠ r = 2π0 r 2

Topic 19

1173

The magnitude of the resultant field at point P is then

BP = 2 ( Bwire )y = !

µ0 Ix µ0 I x = 2 πr π ( x2 + d2 )

(c) The point midway between the two wires is the origin (0, 0). From the above result for part (b), the resultant field at this midpoint is BP

x=0

= 0 . This is as expected, because right-hand rule 2 shows that

at the midpoint the field due to the upper wire is toward the right, while that due to the lower wire is toward the left. Thus, the two fields cancel, yielding a zero resultant field.

19.55

(a) The magnetic force per unit length on each of two parallel wires separated by the distance d and carrying currents I1 and I2 has the magnitude F µ 0 I 1I 2 = 2π d !

In this case, we have −7 F ( 4π × 10 T ⋅ m/A ) (1.25 A ) ( 3.50 A ) = = 3.50 × 10−5 N/m −2  2π ( 2.50 × 10 m ) !

Topic 19

1174

(b) The magnetic forces two parallel wires exert on each other are attractive if their currents are in the same direction and repulsive if the currents flow in opposite directions. In this case, the currents in the two wires are in opposite directions, so the forces are repulsive.

19.56

(a) The force per unit length that parallel conductors exert on each other is !F/ = µ0 I1I 2 2π d . Thus, if F/ = 2.0 × 10−4 N/m , I1 = 5.0 A, and d = 4.0 ! cm, the current in the second wire must be

2π ( 4.0 × 10−2 m ) 2π d ⎛ F ⎞ I2 = 2.0 × 10−4 N/m ) = 8.0 A ( ⎜⎝ ⎟⎠ = −7 µ 0 I1  ( 4π × 10 T ⋅ m/A )( 5.0 A ) ! (b) Since parallel conductors carrying currents in the same direction attract each other (see Section 19.8 in the textbook), the currents in these conductors which repel each other must be in opposite directions.

(c) The result of reversing the direction of either of the currents would be that the force of interaction would change from a force of repulsion to an attractive force. The expression for the force per unit length, !F/ = µ0 I1I 2 2π d , shows that doubling either of the currents would double the magnitude of the force of interaction. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

19.57

1175

In order for the system to be in equilibrium, the repulsive magnetic force per unit length on the top wire must equal the weight per unit length of this wire.

!F/ = µ0 I1I 2 2π d = 0.080 N/m , and the distance between the wires

Thus, will be

4π × 10−7 T ⋅ m/A ) ( 60.0 A ) ( 30.0 A ) ( µ 0 I 1I 2 d= = 2π ( 0.080 N/m ) 2π ( 0.080 N/m ) ! 19.58

= 4.5 × 10−3 m = 4.5 mm

The magnetic forces exerted on the top and bottom segments of the rectangular loop are equal in magnitude and opposite in direction. Thus, these forces cancel, and we only need consider the sum of the forces exerted on the right and left sides of the loop. Choosing to the left (toward the long, straight wire) as the positive direction, the sum of these two forces is

Fnet =

µ 0 I 1I 2 ℓ 2π c

−

µ 0 I 1I 2 ℓ

( )

2π c + a

µ 0 I 1I 2 ℓ ⎛ 1 1 ⎞ − ⎜ ⎟ 2π ⎝ c c + a ⎠

Topic 19

1176

( 4π × 10 T ⋅ m/A)(5.00 A)(10.0 A)(0.450 m) ⎛ F = −7

net

2π

⎞ 1 1 − ⎜ ⎟ ⎝ 0.100 m 0.250 m ⎠

= +2.70 × 10−5 N = 2.70 × 10−5 N to the left

19.59

The magnetic field inside a solenoid which carries current I is given by B = µ0nI, where !n = N/ is the number of turns of wire per unit length. Thus, the current in the windings of this solenoid must be −4 B B (1.00 × 10 T ) ( 0.400 m ) I= = = = 3.18 × 10−2 A = 31.8 mA −7 3 µ0 n µ0 N ( 4π × 10 T ⋅ m/A ) (10 ) !

19.60

The magnetic field inside of a solenoid is B = µ0nI = µ0(N/L)I. Thus, the number of turns on this solenoid must be

( 9.0 T )( 0.50 m ) BL N= = = 4.8 × 10 4 turns −7 µ0 I ( 4π × 10 T ⋅ m/A ) (75 A ) ! 19.61

(a) From R = ρL/A, the required length of wire to be used is 2 −3 ⎡ ⎤ R ⋅ A ( 5.00 Ω ) ⎣π ( 0.500×10 m ) 4 ⎦ L= = = 58 m −8 ρ 1.7 × 10 Ω ⋅ m !

The total number of turns on the solenoid (that is, the number of times this length of wire will go around a 1.00-cm radius cylinder) is

Topic 19

1177

L 58 m N= = = 9.2 × 102 = 920 −2 2π r 2π (1.00 × 10 m ) ! (b) From B = µ0nI, the number of turns per unit length on the solenoid is

B 4.00 × 10−2 T n= = = 7.96 × 103 turns/m −7 µ0 I ( 4π × 10 T ⋅ m/A ) ( 4.00 A ) !

Thus, the required length of the solenoid is

N 9.2 × 102 turns = = = 0.12 m = 12 cm n 7.96 × 103 turns/m ! 19.62 The magnetic field of a long solenoid is B = µ0nI where n is the number of turns per meter. For B = 3.50 × 10−5 T, the value of n is

B 3.50 × 10−5 T = = 27.9 turns/m µ0 I 4π × 10−7 T ⋅ m/A (1.00 A )

(

)

19.63 (a) The magnetic force supplies the centripetal acceleration, so qvB = mv2/r. The magnetic field inside the solenoid is then found to be −31 4 mv ( 9.11 × 10 kg ) (1.0 × 10 m/s ) B= = = 2.8 × 10−6 T = 2.8 µ T −19 −2 qr (1.60 × 10 C)( 2.0 × 10 m ) !

(b) From B = µ0nI. the current is the solenoid is found to be

Topic 19

1178

I= !

19.64

B 2.8 × 10−6 T = µ0 n ( 4π × 10−7 T ⋅ m/A ) ⎡⎣( 25 turns/cm ) (100 cm 1 m ) ⎤⎦

= 8.9 × 10−4 A = 0.89 mA

(a) When switch S is closed , a total current NI (current I in a total of N conductors) flows toward the right through the lower side of the coil. This results in a downward force of magnitude Fm = B(NI)w being exerted on the coil by the magnetic field, with the requirement that the balance exert a upward force F′ = mg on the coil to bring the system back into balance.

In order for the magnetic force to be downward, the right-hand rule number 1 shows that the magnetic field must be directed out of the page toward the reader. For the system to be restored to balance, it is necessary that

Fm = F′

B(NI)w = mg, giving B = mg/NIw

(b) The magnetic field exerts forces of equal magnitudes and opposite directions on the two sides of the coil. These forces cancel each other © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

1179

and do not affect the balance of the coil. Hence the dimension of the sides is not needed. −3 2 mg ( 20.0 × 10 kg ) ( 9.80 m s ) = = 0.26 T (c) B = NIw ( 50 ) ( 0.30 A ) ( 5.0 × 10−2 m ) !

19.65

(a) The magnetic field at the center of a circular current loop of radius R and carrying current I is B = µ0I/2R. The direction of the field at this center is given by right-hand rule number 2. Taking out of the page (toward the reader) as positive, the net magnetic field at the common center of these coplanar loops has magnitude

( 4π × 10 T ⋅ m/A ) 3.00 A − 5.00 A µI µI Bnet = B2 − B1 = 0 2 − 0 1 = 2r2 2r1 2 9.00 × 10−2 m 12.0 × 10−2 m −7

= −5.24 × 10−6 T = 5.24 µT

(b) Since we chose out of the page as the positive direction, and now find that Bnet < 0, we conclude the net magnetic field at the center is into the page.

⎛ 3.00 A ⎞ ⎛I ⎞ r2 = ⎜ 2 ⎟ r1 = ⎜ (12.0 cm ) = 7.20 cm ⎝ I1 ⎠ ⎝ 5.00 A ⎟⎠ ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

19.66

1180

The angular momentum of a point mass moving in a circular path is

⎛ v⎞ L = Iw = ( mr 2 ) ⎜ ⎟ = mvr ⎝ r⎠ ! where m is the mass of the particle, v is its speed, and r is the radius of its path.

(a) The magnetic force experienced by the moving electron supplies the needed centripetal acceleration, so

⎛ v2 ⎞ m ⎜ ⎟ = qvBsin 90.0° ! ⎝ r⎠

mv = qBr

Thus, L = mvr = (qBr)r = qBr2, and the radius of the path must be

L r= = qB !

4.00 × 10−25 kg ⋅ m 2 s = 5.00 × 10−2 m = 5.00 cm −19 −3 (1.60 × 10 C)(1.00 × 10 T )

(b) The speed of the electron may now be found from L = mvr as

L 4.00 × 10−25 k ⋅gm 2 s v= = = 8.78 × 106 m/s −31 −2 mr ( 9.11 × 10 kg ) ( 5.00 × 10 m ) !

19.67

Assume wire 1 is along the x-axis and wire 2 is along the y-axis.

(a) Choosing out of the page as the positive field direction, the field at © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 19

1181

point P is

(

)

4π × 10−7 T ⋅ m/A ⎛ 5.00 A µ0 ⎛ I1 I 2 ⎞ 3.00 A ⎞ B = B1 − B2 = − ⎜ − ⎟= ⎜ ⎟ 2π ⎝ r1 r2 ⎠ 2π ⎝ 0.400 m 0.300 m ⎠ = 5.00 × 10−7 T = 0.500 µ T out of the page

(b) At 30.0 cm above the intersection of the wires, the field components are as shown below, where

µ 0 I1 2π r ( 4π × 10−7 T ⋅ m/A )( 5.00 A )

By = −B1 = − =− !

2π ( 0.300 m )

= −3.33 × 10−6 T

−7 µ0 I 2 ( 4π × 10 T ⋅ m/A ) = = 2.00 × 10−6 T and Bx = B2 = 2π r 2π ( 0.300 m ) !

With Bz = 0, the resultant field is parallel to the xy-plane and

⎛ By ⎞ B = Bx2 + By2 = 3.88 × 10−6 T at θ = tan −1 ⎜ ⎟ = −59.0° ⎝ Bx ⎠ !

Topic 19

1182

! B = 3.88 µ T parallel to the xy - plane and 59.0 clockwise from the + x-direction

19.68

For the rail to move at constant velocity, the net force acting on, it must be zero. Thus, the magnitude of the magnetic force must equal that of the friction force, giving BIL = µk(mg), or

B= ! 19.69

µk ( mg ) ( 0.100 ) ( 0.200 kg ) ( 9.80 m s 2 ) = = 3.92 × 10−2 T IL (10.0 A )( 0.500 m )

(a) Since the magnetic field is directed from N to S (that is, from left to right within the artery), positive ions with velocity in the direction of the blood flow experience a magnetic deflection toward electrode A. Negative ions will experience a force deflecting them toward electrode B. This separation of charges creates an electric field directed from A toward B. At equilibrium, the electric force caused by this field must balance the magnetic force, so

qvB = qE = q(ΔV/d)

ΔV 160 × 10−6 V v= = = 1.33 m/s Bd ( 0.040 0 T ) ( 3.00 × 10−3 m ) !

Topic 19

1183

(b) The magnetic field is directed from N to S. If the charge carriers are  negative moving in the direction of !v , the magnetic force is directed

toward point B. Negative charges build up at point B, making the potential at A higher than that at B. If the charge carriers are positive  moving in the direction of !v , the magnetic force is directed toward

A, so positive charges build up at A. This also makes the potential at A higher than that at B. Therefore the sign of the potential difference does not depend on the charge of the ions.

19.70

(a) The magnetic force acting on the wire is directed upward and of magnitude !Fm = BIL sin 90° = BI .

∑ Fy Fm − mg BIL − mg BI ay = = = = − g , or m m m (m/L) !

Thus,

( 4.0 × 10 T )( 2.0 A ) − 9.80 m s = 6.2 m s a = −3

−4

5.0 × 10 kg/m

2 (b) Using !Δy = v0yt + 12 ayt , with v0y = 0, gives

t= ! 19.71

2 ( Δy ) 2 ( 0.50 m ) = = 0.40 s ay 6.2 m s 2

Label the wires 1, 2, and 3, as shown in Figure 1. Also, let B1, B2, and B3

Topic 19

1184

respectively, represent the magnitudes of the fields produced by the currents in those wires, and observe that θ = 45°.

(

)

At point A, B1 = B2 = µ0 I 2π a 2 , or !

( 4π × 10 T ⋅ m/A )( 2.0 A ) = 28 µT −7

B1 = B2 = !

2π ( 0.010 m ) 2

4π × 10−7 T ⋅ m/A ) ( 2.0 A ) ( µ0 I = = 13 µ T and B3 = 2π ( 3a) 2π ( 0.030 m ) ! These field contributions are oriented as shown in Figure 2.

  Observe that the horizontal components of B1 and B2 cancel while their ! !  vertical components add to B3 . The resultant field at point A is then !

Topic 19

1185

BA = (B1 + B2)cos 45° + B3 = 53 µT, or

 BA = 53 µ T!directed!toward!the!bottom!of!the!page ! −7 µ0 I ( 4π × 10 T ⋅ m/a ) ( 2.0 A ) = = 40 µ T . At point B, B1 = B2 = 2π a 2π ( 0.010 m ) !

µ0 I and B3 = = 20 µ T . These contributions are oriented as shown in 2 π 2a ( ) ! Figure 3.

Thus, the resultant field at B is

  BB = B3 = 20 µ T!directed!toward!the!bottom!of!the!page !

(

)

At point C, B1 = B2 = µ0 I 2π a 2 , while B3 = µ0I/2πa. These contributions ! are oriented as shown in Figure 4.

Topic 19

1186

  Observe that the horizontal components of B1 and B2 cancel, while their ! !  vertical components add to oppose B3 . The magnitude of the resultant !

field at C is

(

)

BC = B1 + B2 sin 45° − B3 =

19.72

µ0 I ⎛ 2sin 45° ⎞ − 1⎟ = 0 ⎜ 2π a ⎝ ⎠ 2

(a) Since one wire repels the other, the currents must be in opposite directions.

(b) Consider a free-body diagram of one of the wires as shown below.

ΣFy = 0 ⇒ T cos 8.0° = mg

mg T= cos 8.0° ! ⎛ mg ⎞ ∑ Fx = 0 ⇒ Fm = T sin 8.0° = ⎜ sin 8.0° ⎝ cos 8.0° ⎟⎠ !

Topic 19

1187

Fm = (mg) tan 8.0°. Thus,

I= !

µ0 I 2 L = ( mg ) tan 8.0° , which gives ! 2π d

d [( m/L) g ] tan 8.0° µ0 2π

Observe that the distance between the two wires is

d = 2 ⎡⎣( 6.0 cm ) sin 8.0° ⎤⎦ = 1.7 cm , so !

(1.7 × 10 m )( 0.040 kg/m )( 9.80 m s ) tan 8.0° = 68 A I= −2

2.0 × 10−7 T ⋅ m/A

! 19.73

Note: We solve part (b) before part (a) for this problem.

(b) Since the magnetic force supplies the centripetal acceleration for this particle, qvB = mv2/r, or the radius of the path is r = mv/qB = p/qB, where p = mv = 2m ( KE ) = 2 (1.67 × 10−27 kg ) ( 5.00 × 106 eV ) (1.60 × 10−19 J/eV ) ! = 5.17 × 10

−20

kg ⋅ m/s

Consider the circular path shown below and observe that the desired angle is

Topic 19

1188

⎡ (1.00 m ) qB ⎤ ⎛ 1.00 m ⎞ α = sin −1 ⎜ = sin −1 ⎢ ⎟ ⎥ ⎝ r ⎠ p ⎣ ⎦ !

(

)(

)

⎡ 1.00 m 1.60 × 10−19 C 0.050 0 T ⎤ ⎥ = 8.90° α = sin ⎢ ⎢ ⎥ 5.17 × 10−20 kg ⋅ m/s ⎣ ⎦ −1

(a) The linear momentum of the particle has constant magnitude p = mv, and its vertical component as the particle leaves the field is py = −p sin

α, or

py = −(5.17 × 10−20 kg ⋅ m/s) sin (8.90°) = −8.00 × 10−21 kg ⋅ m/s

19.74

The force constant of the spring system is found from the elongation produced by the weight acting alone.

F mg k= = ! x1 x1

where

x1 = 0.50 cm

Topic 19

1189

The total force stretching the springs when the field is turned on is

∑ F = Fm + mg = kxtotal ! y

where

xtotal = x1 + 0.30 cm = 0.80 cm

Thus, the downward magnetic force acting on the wire is

⎛ mg ⎞ ⎛x ⎞ Fm = kxtotal − mg = ⎜ xtotal − mg = ⎜ total − 1⎟ mg ⎟ ⎝ x1 ⎠ ⎝ x1 ⎠

⎛ 0.80 cm ⎞ =⎜ − 1⎟ (10.0 × 10−3 kg ) ( 9.80 m s 2 ) = 5.9 × 10−2 N ⎝ 0.50 cm ⎠

Since the magnetic force is given by Fm = BIL sin 90°, the magnetic field is

12 Ω ) ( 5.9 × 10−2 N ) ( Fm Fm B= = = = 0.59 T IL ( ΔV /R ) L ( 24 V ) ( 5.0 × 10−2 m ) ! 19.75

The magnetic force is very small in comparison to the weight of the ball, so we treat the motion as that of a freely falling body. Then, as the ball approaches the ground, it has velocity components with magnitudes of

vx = v0x = 20.0 m/s, and

2 v = v0y + 2ay ( Δy ) = 0 + 2 ( −9.80 m s 2 ) ( −20.0 m ) = 19.8 m/s ! y

The velocity of the ball is perpendicular to the magnetic field and, just

Topic 19

1190

before it reaches the ground, has magnitude v = vx2 + vy2 = 28.1 m/s . Thus, ! the magnitude of the magnetic force is

Fm = qvB sin θ

= (5.00 × 10−6 C)(28.1 m/s)(0.010 0 T) sin 90.0° = 1.41 × 10−6 N

19.76

−7 µ0 I1 ( 4π × 10 T ⋅ m/A ) ( 5.00 A ) = = 1.00 × 10−5 T (a) B1 = 2π d 2π ( 0.100 m ) !

(b) F21 = B1I 2 = (1.00 × 10−5 T ) ( 8.00 A ) = 8.00 × 10−5 N!directed!toward!wire!1 ! −7 µ0 I 2 ( 4π × 10 T ⋅ m/A ) ( 8.00 A ) = = 1.60 × 10−5 T (c) B2 = 2π d 2π ( 0.100 m ) !

(d) F12 = B2 I1 = (1.60 × 10−5 T ) ( 5.00 A ) = 8.00 × 10−5 N!directed!toward!wire!2 !

Topic 20

1191

Topic 20 Induced Voltages and Inductance

QUICK QUIZZES 20.1

b, c, a. At each instant, the magnitude of the induced emf is proportional to the magnitude of the rate of change of the magnetic field (hence, proportional to the absolute value of the slope of the curve shown on the graph).

20.2

Choice (c). Taking downward as the normal direction, the north pole approaching from above produces an increasing positive flux through the area enclosed by the loop. To oppose this, the induced current must generate a negative flux through the interior of the loop, or the induced magnetic field must point upward through the enclosed area of the loop. Imagine gripping the wire of the loop with your right hand so the fingers curl upward through the area enclosed by the loop. You will find that your thumb, indicating the direction of the induced current, is directed counterclockwise around the loop as viewed from above.

20.3

Choice (b). If the positive z-direction as chosen as the normal direction, the increasing counterclockwise current in the left-hand loop produces an

Topic 20

1192

increasing positive flux through the area enclosed by this loop. The magnetic field lines due to this current curl around and pass through the area of the xy-plane outside the left-hand loop in the negative direction. Thus, the right-hand loop has an increasing negative flux through it. To counteract this effect, the induced current must produce positive flux, or generate a magnetic field in the positive z-direction, through the area enclosed by the right-hand loop. Imagine gripping the wire of the righthand loop with the right hand so the fingers point in the positive zdirection as they penetrate the area enclosed by this loop. You should find that the thumb is directed counterclockwise around the loop as viewed from above the xy-plane. 20.4

Choice (a). All charged particles within the metal bar move straight downward with the bar. According to right-hand rule number 1, positive changes moving downward through a magnetic field that is directed northward will experience magnetic forces toward the east. This means that the free electrons (negative charges) within the metal will experience westward forces and will drift toward the west end of the bar, leaving the east end with a net positive charge.

20.5

Choice (b). According to Equation 20.3, when B and v are constant, the

Topic 20

1193

emf depends only on the length of the wire cutting across the magnetic field lines. Thus, you want the long dimension of the rectangular loop perpendicular to the velocity vector. This means that the short dimension is parallel to the velocity vector, and (b) is the correct choice. From a more conceptual point of view, you want the rate of change of area in the magnetic field to be the largest, which you do by thrusting the long dimension into the field. 20.6

Choice (b). When the iron rod is inserted into the solenoid, the inductance of the coil increases. As a result, more potential difference appears across the coil than before. Consequently, less potential difference appears across the bulb, and its brightness decreases.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 20.2

Consider the copper tube to be a large set of rings stacked one on top of the other. As the magnet falls toward or falls away from each ring, a current is induced in the ring. Thus, there is a current in the copper tube around its circumference.

20.4

(a) The flux is calculated as ΦB = BA cos θ = B⊥A. The flux is therefore maximum when the magnetic field vector is perpendicular to the

Topic 20

1194

plane of the loop. (b) The flux is zero when the magnetic field is parallel to the plane of the loop (and hence, θ = 90°, making B⊥ = B cos θ = 0). 20.6

As water falls, it gains velocity and kinetic energy. It then pushes against the blades of a turbine, transferring this energy to the rotor or coil of a large generator. The rotor moves in a strong external magnetic field and a voltage is induced in the coil. This induced emf is the voltage source for the current in our electric power lines.

20.8

No. Once the bar is in motion and the charges are separated, no external force is necessary to maintain the motion. During the initial acceleration of the bar, an external applied force will be necessary to overcome both the inertia of the bar and a retarding magnetic force exerted on the bar.

20.10 The current at time t is I =

ε 1− e R

(

−t/τ

) where τ = L/R.

(a) Doubling both ε and R has the net effect of reducing the time constant by a factor of 2 (because ε /R = 2 ε /2R) so that the current will be greater than Iref. (b) Doubling the inductance L doubles the time constant so that the current will be less than Iref.

Topic 20

1195

(c) Doubling L, R, and ε has no net effect on the circuit ( ε /R = 2ε /2R and τ = L/R = 2L/2R ) and the current will therefore equal Iref. 20.12

The increasing counterclockwise current in the solenoid coil produces an upward magnetic field that increases rapidly. The increasing upward flux of this field through the ring induces an emf to produce a clockwise current in the ring. At each point on the ring, the field of the solenoid has a radially outward component as well as an upward component. This radial field component exerts an upward force on the current at each point in the ring. The resultant magnetic force on the ring is upward and exceeds the weight of the ring. Thus, the ring accelerates upward off of the solenoid.

20.14

As the magnet moves at high speed past the fixed coil, the magnetic flux through the coil changes very rapidly, increasing as the magnet approaches the coil and decreasing as the magnet moves away. The rapid

Topic 20

1196

change in flux through the coil induces a large emf, large enough to cause a spark across the gap in the spark plug.

ANSWERS TO EVEN NUMBERED PROBLEMS 1.00 × 10−7 T ⋅ m2

(b)

8.66 × 10–8 T ⋅ m2 (c)

1.69 m

(b)

10.8 Wb

(b)

no induced current

(b)

6.27 × 10–8 V

(b)

clockwise

20.2

(a)

20.4

zero

20.6

(a)

20.8

0.10 mV

20.10

34 mV

20.12

4.74 V

20.14

into the page

20.16

(a)

left to right

(c)

right to left

20.18

(a)

1.88 × 10–7 T ⋅ m2

20.20

8.8 A

20.22 (a) | ε | = NB0πr2/t (c)

I = B0πr2/tR

Topic 20

20.24

1197

1.20 mV, west end is positive

20.26 2.36 × 103 V 20.28 (a)

6.0 µT

(b) The magnitude and direction of the Earth’s field varies from one location to another, so the induced voltage in the wire will change. 20.30 (a)

2.00 m/s

(d)

6.00 W

(b)

6.00 W

(c)

20.32

13 mV

20.34

(a)

7.5 kV

(b)

when the plane of the coil is parallel to the magnetic field

20.36

(a)

8.0 A

(b)

3.2 A

20.38

(a)

1.32 × 10−8 H

(b)

6.34 × 10−6 V

20.40

See Solution.

20.42

1.92 × 10–5 T ⋅ m2

20.44

(a)

1.00 kΩ

(b)

3.00 ms

20.46

(a)

(b)

3.8 V

(c)

6.0 V

(d)

2.2 V

(c)

3.00 N

60 V

Topic 20

20.48

1198

(a)

2.00 ms

(b)

0.176 A

(c)

1.50 A

(d)

3.22 ms

20.50

(a)

4.44 mH

(b)

0.555 mJ

20.52

(a)

1.3 Ω

(b)

4.8 × 102 turns

(c)

(d)

0.76 mH

(e)

0.46 ms

(f) 3.6 A

(g) 3.2 ms

(h)

4.9 mJ

20.54

(a)

increasing

(b)

62.2 mT/s

20.56

(a)

0.73 m/s, counterclockwise

(b)

0.65 mW

(c)

Work is being done on the bar by an external force to maintain

0.48 m

constant speed. 20.58

(a)

2.1 × 106 m/s

(c)

1.7 × 1010 V

(b)

from side to side

(d) The very large induced emf would lead to powerful spontaneous electrical discharges. The strong electric and magnetic fields would disrupt the flow of ions in their bodies. 20.60

0.158 mV

Topic 20

1199

20.62

0.120 A, clockwise

20.64

1.60 A

20.66

(a)

ε = NBℓv

(b)

!I = NBv/R

(c)

P = N 2 B2 ℓ 2 v 2 R

(d)

2 2 2 !F = N B  v R

(e)

clockwise

(f) toward the left

PROBLEM SOLUTIONS 20.1

The angle between the direction of the constant field and the normal to the plane of the loop is θ = 0°, so ΦB = BA cos θ = (0.50 T)[(8.0 × 10–2 m)(12 × 10–2 m)]cos 0° = 4.8 × 10–3 T ⋅ m2

20.2

The magnetic flux through the loop is given by ΦB = BA cos θ, where B is the magnitude of the magnetic field, A is the area enclosed by the loop, and θ is the angle the magnetic field makes with the normal to the plane of the loop. Thus,

⎡ ⎛ 10−2 m ⎞ ⎤ −7 2 Φ B = BAcos θ = ( 5.00 × 10−5 T ) ⎢ 20.0 cm 2 ⎜ ⎥ cos θ = (1.00 × 10 T ⋅ m ) cos θ ⎟ 1 cm ⎝ ⎠⎦ ⎣ !

 (a) When !B is perpendicular to the plane of the loop, θ = 0° and ΦB =

Topic 20

1200

1.00 × 10–7 T ⋅ m2 (b) If θ = 30.0°, then ΦB = (1.00 × 10–7 T ⋅ m2) cos 30.0° = 8.66 × 10–8 T ⋅ m2 . (c) If θ = 90.0°, then ΦB = (1.00 × 10–7 T⋅m2)cos 90.0° = 0. 20.3

The magnetic flux is Φ B = BAcosθ where θ is the angle between the magnetic field and the normal to the plane of the loop. The loop area is A = ℓ 2 = (0.250 m)2 = 0.0625 m2.

(

)

(a) Here θ = 0° so that Φ B = BAcos0° = ( 2.00 T ) 0.062 5 m 2 = 0.125 Wb . (b) When the plane of the loop makes an angle of 60.0° with the magnetic field, the loop’s normal makes an angle θ = 30.0° so that

(

)

Φ B = BAcos30° = ( 2.00 T ) 0.0625 m 2 cos ( 30.0° ) = 0.108 Wb .

θ = 90° and Φ B = 0 (because cos(90°) = 0). 20.4

The magnetic field lines are everywhere parallel to the surface of the cylinder, so no magnetic field lines penetrate the cylindrical surface. The total flux through the cylinder is zero.

20.5

(a) Every field line that comes up through the area A on one side of the wire goes back down through area A on the other side of the wire. Thus, the net flux through the coil is zero.

Topic 20

1201

(b) The magnetic field is parallel to the plane of the coil, so θ = 90.0°. Therefore, ΦB = BA cos θ = BA cos 90.0° = 0. 20.6

(a) The angle between the magnetic field and the loop normal is θ = 0° so that, with a loop area of A = π R2,

(

Φ B = BA = B π R 2

Wb = 1.69 m ) → R = π2.70 (0.300 T )

(b) When the loop radius is doubled, the area increases by a factor of 4. Because the flux is proportional to the area, the flux also increases by a factor of 4 so that Φ B,new = 4Φ B = 4 ( 2.70 Wb ) = 10.8 Wb . 20.7

(a) The magnetic flux through an area A may be written as ΦB = (B cos θ)A = (component of B perpendicular to A).A

Thus, the flux through the shaded side of the cube is ΦB = Bx ⋅ A = (5.0 T) ⋅ (2.5 × 10–2 m)2 = 3.1 × 10–3 T ⋅ m2

Topic 20

1202

(b) Unlike electric field lines, magnetic field lines always form closed loops, without beginning or end. Therefore, no magnetic field lines originate or terminate within the cube and any line entering the cube at one point must emerge from the cube at some other point. The net flux through the cube, and indeed through any closed surface, is zero. 20.8

ε=

ΔΦ B Δt

⎡ ⎤ ( ΔB) Acos θ (1.5 T − 0) ⎢⎣π (1.6 × 10 m ) ⎥⎦ cos 0° −3

Δt

120 × 10−3 s

= 1.0 × 10−4 V = 0.10 mV 20.9

(a) As loop A moves parallel to the long straight wire, the magnetic flux through loop A does not change. Hence, there is no induced current in this loop. (b) As loop B moves to the left away from the straight wire, the magnetic flux through this loop is directed out of the page, and is decreasing in magnitude. To oppose this change in flux, the induced current flows counterclockwise around loop B producing a magnetic flux directed out of the page through the area enclosed by loop B. (c) As loop C moves to the right away from the straight wire, the magnetic flux through this loop is directed into the page and is decreasing in magnitude. In order to oppose this change in flux, the

Topic 20

1203

induced current flows clockwise around loop C producing a magnetic flux directed into the page through the area enclosed by loop C. 20.10

ε=

ΔΦ B

Δt 2 ⎡ ⎤ 0.15 T ⎢π 0.12 m − 0 ⎥ cos 0° ⎣ ⎦ = = 3.4 × 10−2 V = 34 mV 0.20 s

(

Δt

B ΔA cos θ

) (

)

20.11 The average induced emf is ε = −N

ΔΦB . As the magnetic field decreases Δt

from its maximum value to zero, ΔΦ B = −BA so that

Δt =

NBA

(15)(1.00 × 10 T )( 3.00 × 10 −3

−4

5.00 V

) = 9.00 × 10 s −7

20.12 Each coil has an area A = πR2 = π(1.50 × 10−2 m)2 = 7.07 × 10–4 m2. The Topic 19 result gives the maximum magnetic field as

(

)

4π × 10−7 T ⋅ m/A (10.0 A ) µ0 I = (10 ) = 4.19 × 10–3 T Bmax = N −2 2R 2 1.50 × 10 m

(

)

The average induced emf is then

(

)(

)

4.19 × 10−3 T 7.07 × 10−4 m 2 ΔΦB −BA ε = −N Δt = −N Δt = 10 = 4.74 V 6.25 × 10−6 s 20.13

The magnetic field changes from zero to a magnitude of 2.5 T, directed at 45° to the plane of the band, in a time of Δt = 0.18 s. If the band has a

Topic 20

1204

diameter of 6.5 cm, the magnitude of the average induced emf in the metal band is

ε=

ΔΦ B Δt

ΔB Δt

(2.5 T − 0) π (6.5 × 10 m ) cos 45° Acos θ = −2

0.18 s

= 3.3 × 10−2 V = 33 mV 20.14

When the switch is closed, the magnetic field due to the current from the battery will be directed toward the left along the axis of the cylinder. To oppose this increasing leftward flux, the induced current in the other loop must produce a field directed to the right through the area it encloses. Thus, the induced current is directed out of the page through the resistor.

20.15

(a) When the magnet moves to the left, the flux through the interior of the coil is directed toward the right and is decreasing in magnitude. To oppose this change in flux, the magnetic field generated by the induced current should be directed to the right along the axis of the coil. The current must then be left to right through the resistor. (b) When the magnet moves to the right, the flux through the interior of the coil is directed toward the right and is increasing in magnitude. To oppose this increasing flux, the magnetic field generated by the

Topic 20

1205

induced current should be directed toward the left along the axis of the coil. The current must then be right to left through the resistor. 20.16

When the switch is closed, the current from the battery produces a magnetic field directed toward the right along the axis of both coils. (a) As the battery current is growing in magnitude, the induced current in the rightmost coil opposes the increasing rightward directed field by generating a field toward to the left along the axis. Thus, the induced current must be left to right through the resistor. (b) Once the battery current, and the field it produces, have stabilized, the flux through the rightmost coil is constant, and there is no induced current (c) As the switch is opened, the battery current and the field it produces rapidly decrease in magnitude. To oppose this decrease in the rightward directed field, the induced current must produce a field toward the right along the axis, so the induced current is right to left through the resistor.

20.17

(a) The current is zero. The magnetic flux the current produces through the right side of the loop is directed into the page, and is equal in

Topic 20

1206

magnitude to the outward-directed flux the current produces through the left side of the loop. Thus, the net flux through the loop has a constant value of zero and does not induce a current. (b) The flux through the loop due to the long wire is directed out of the page and is increasing in magnitude. To oppose this increasing outward flux, the induced current must generate a magnetic field that is directed into the page through the area enclosed by the loop. Thus, the induced current in the loop must be clockwise. 20.18

The initial magnetic field inside the solenoid is ⎛ 100 ⎞ ⎛ N⎞ B = µ0 nI = µ0 ⎜ ⎟ I = ( 4π × 10−7 T ⋅ m/A ) ⎜ ( 3.00 A ) = 1.88 × 10−3 T ⎝ ⎠ ⎝ 0.200 m ⎟⎠ !

(a) ΦB = BAloop cos θ = (1.88 × 10–3 T)(1.00 × 10–2 m)2 cos 0° = 1.88 × 10–7 T ⋅ m2 (b) When the current is zero, the flux through the loop is ΦB = 0, and the average induced emf has been

ε=

ΔΦ B Δt

0 − 1.88 × 10−7 T ⋅ m 2 3.00 s

= 6.27 × 10−8 V

20.19 (a) The initial field inside the solenoid is

Topic 20

1207

⎛ 300 ⎞ Bi = µ0 nIi = ( 4π × 10−7 T ⋅ m/A ) ⎜ ( 2.00 A ) = 3.77 × 10−3 T ⎝ 0.200 m ⎟⎠ !

(b) The final field inside the solenoid is ⎛ 300 ⎞ B f = µ0 nI f = ( 4π × 10−7 T ⋅ m/A ) ⎜ 5.00 A ) = 9.42 × 10−3 T ( ⎟ ⎝ 0.200 m ⎠ !

(c) The 4-turn coil encloses an area A = πr2 = π(1.50 × 10–2 m)2 = 7.07 × 10–4 m2 (d) The change in flux through each turn of the 4-turn coil during the 0.900-s period is ΔΦB = (ΔB)A = (9.42 × 10–3 T – 3.77 × 10–3 T)(7.07 × 10–4 m2) = 3.99 × 10–6 Wb (e) The average induced emf in the 4-turn coil is ⎛ ΔΦ ⎞ ⎛ 3.99 × 10−6 Wb ⎞ −5 B = 4 ⎟ ⎜ ⎟ = 1.77 × 10 V 2 0.900 s ⎝ ⎠ ⎝ Δt ⎠

ε =N ⎜

Since the current increases at a constant rate during this time interval, the induced emf at any instant during the interval is the same as the average value given above. (f) The induced emf is small, so the current in the 4-turn coil will also be very small. This means that the magnetic field generated by this

Topic 20

1208

current will be negligibly small in comparison to the field generated by the solenoid. 20.20

The magnitude of the average emf is

ε= =

N ΔΦ B

(

NBA Δ cos θ

)

Δt Δt 200 1.1 T 100 × 10−4 m 2 cos 180° − cos 0°

(

)(

0.10 s

)

Therefore, the average induced current is I =

20.21

ε R

= 44 V

44 V 5.0 Ω

= 8.8 A .

If the magnetic field makes an angle of 28.0° with the plane of the coil, the angle it makes with the normal to the plane of the coil is θ = 62.0°. Thus,

ε=

( ) = NB( ΔA)cos θ

N ΔΦ B

Δt Δt −6 200 50.0 × 10 T ⎡ 39.0 cm 2 1 m 2 10 4 cm 2 ⎤ cos 62.0° ⎣ ⎦ = 1.80 s

(

)(

)

= 1.02 × 10−5 V = 10.2 µV

20.22

With the magnetic field perpendicular to the plane of the coil, the flux through each turn of the coil is ΦB = BA = B(πr2). Since the area remains constant, the change in flux due to the changing magnitude of the magnetic field is ΔΦB = (ΔB)πr2.

Topic 20

1209

⎡ B − 0 π r 2 ⎤ NB π r 2 0 0 ⎥= = N⎢ (a) The induced emf is ε = N . ⎢ t−0 ⎥ Δt t ⎣ ⎦ ΔΦ

(b) When looking down on the coil from a location on the positive z-axis, the magnetic field (in the positive z-direction) is directed up toward you and increasing in magnitude. This means the change in the flux through the coil is directed upward. In order to oppose this change in flux, the induced current must produce a magnetic field directed downward through the area enclosed by the coil. Thus, the current must flow clockwise as seen from your viewing location. (c) Since the turns of the coil are connected in series, the total resistance of the coil is Req = NR. Thus, the magnitude of the induced current is

20.23

ε Req

NB0π r 2 t NR

B0π r 2 tR

The motional emf induced in a metallic object of length  moving through a magnetic field at speed v is given by ε = B⊥ vℓ , where B⊥ is the component of the magnetic field perpendicular to the velocity of the object. Thus,

ε = (35.0 × 10 T)(25.0 m/s)(15.0 m) = 1.31 × 10 V= 13.1 mV –6

–2

Topic 20

20.24

1210

The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of the wire. Thus, the magnitude of the motional emf induced in the wire is

ε = B ℓv = ( 40.0 × 10 T )( 2.00 m )(15.0 m/s ) = 1.20 × 10 V = 1.20 mV −6

−3

⊥

Imagine holding your right hand horizontal with the fingers pointing north (the direction of the wire’s velocity), such that when you close your hand the fingers curl downward (in the direction of B⊥). Your thumb will then be pointing westward. By right-hand rule number 1, the magnetic force on charges in the wire would tend to move positive charges westward. Thus, the west end of the wire will be positive relative to the east end. 20.25

The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of the metallic truck body. Thus, the motional emf induced across the width of the truck is ⎡

⎞⎤ −3 ⎟ ⎥ 37 m/s = 2.6 × 10 V = 2.6 mV ⎝ 39.37 in ⎠ ⎥⎦ ⎛

ε = B ℓv = ( 35 × 10 T ) ⎢(79.8 in ) ⎜ −6

⊥

⎢⎣

(

)

20.26 The motional emf is ΔV = Bℓv = (1.50 × 10−5 T)(20.0 × 103 m)(7.86 × 103 m/s) = 2.36 × 103 V .

Topic 20

20.27

1211

(a) Observe that only the horizontal component, Bh, of Earth’s magnetic field is effective in exerting a vertical force on charged particles in the antenna. For the magnetic force, Fm = qvBh sin θ, on positive charges in the antenna to be directed upward and have maximum magnitude (when θ = 90°), the car should move eastward through the northward horizontal component of the magnetic field. (b)

ε = B ℓv , where B is the horizontal component of the magnetic field. h

⎡⎛

⎞⎛

⎞⎤

ε = ⎡⎣( 50.0 × 10 T ) cos 65.0° ⎤⎦ (1.20 m ) ⎢⎜ 65.0 km ⎟ ⎜ 0.278 m/s ⎟ ⎥ −6

⎢⎣⎝

h ⎠ ⎝ 1 km/h ⎠ ⎥⎦

= 4.58 × 10−4 V 20.28 (a) Since ε = B⊥ ℓv , the magnitude of the vertical component of the Earth’s magnetic field at this location is

Bvertical = B⊥ =

ε=

ℓv

0.45 V

(25 m )(3.0 × 10 m/s) 3

= 6.0 × 10−6 T = 6.0 µ T

(b) Yes. The magnitude and direction of the Earth’s field varies from one location to the other, so the induced voltage in the wire changes. Further, the voltage will change if the tether cord changes its orientation relative to the Earth’s field. 20.29

The metal bar of length  and moving at speed v through the magnetic

Topic 20

1212

field experiences an induced emf of magnitude ε = B⊥ ℓv , where B⊥ = B cos θ is the component of the magnetic field perpendicular to the velocity of the bar as shown in figure (a) below.

As the bar slides down the rails, the magnetic flux through the conducting path formed by the bar, the rails, and the resistor R is directed downward and is increasing in magnitude. Thus, the induced current must flow counterclockwise around the conducting path to generate an upward flux, opposing the increase in flux due to the field B. This current flows into the page as indicated in figure (b) and has magnitude I = ε /R = Bℓvcos θ /R .

According to right-hand rule number 1, the bar will experience a  magnetic force !Fm directed horizontally toward the left as shown in

figure (b). The magnitude of this force is

B  vcos θ ⎛ Bvcos θ ⎞ Fm = BI = B ⎜ ⎟⎠  = ⎝ R R ! 2 2

Topic 20

1213

 Now, consider the free-body diagram of the bar in figure (b), where !n is the normal force exerted on the bar by the rails. If the bar is to move with constant velocity (i.e., be in equilibrium), it is necessary that

ΣFy = 0

and ΣFx = 0

⇒

n cos θ = mg

⇒ Fm = n sin θ or

mg n= ! cos θ

B2 2 vcos θ ⎛ mg ⎞ =⎜ sin θ = mg tan θ R ⎝ cos θ ⎟⎠ !

Thus, the equilibrium speed of the bar is

(mg tan θ ) R = ( 0.200 kg )( 9.80 m s ) tan 25.0° (1.00 Ω) = 2.80 m/s v= B  cos θ ( 0.500 T ) (1.20 m ) cos 25.0° ! 2

2 2

20.30 (a) The motional emf is ΔV = Bℓv. Substitute this result into Ohm’s law and solve for the speed v to find I=

ΔV Bℓv IR (1.00 A ) ( 6.00 Ω ) = → v= = = 2.00 m/s R R Bℓ ( 2.50 T ) (1.20 m )

(b) The power delivered to the resistor is P = I2R = (1.00 A)2(6.00 Ω) = 6.00 W .

(c) The force on a current-carrying wire has magnitude F = BI ℓ sinθ. Here, θ = 90° so that F = (2.50 T)(1.00 A)(1.20 m) = 3.00 N .

Topic 20

1214

(d) The bar moves at constant speed in the same direction so that a = 0 and F − Fapp = 0 so that F = Fapp and P = Fappv = (3.00 N)(2.00 m/s) = 6.00 W .

(Alternatively, apply conservation of energy, equating the power dissipated by the resistor to the work done each second by Fapp on the bar.) 20.31

(a) In the initial orientation of the coil, the magnitude of the flux passing through the loop is |ΦB| = NBA, where A is the area enclosed by the loop and N is the number of turns on the loop. After the loop has rotated 90°, the magnetic field is now parallel to the plane of the loop and the flux through the loop is zero. The average emf induced in the loop as it rotates is

ε=

ΔΦ B Δt

( NBA − 0 = =

)(

28 1.25 T 2.80 × 10−2 m

Δt

0.335 s

)

= 8.19 × 10−2 V = 81.9 mV

(b) The average induced current is I =

20.32

ε = 81.9 mV = 105 mA . R

0.780 Ω

Note that the vertical component of the magnetic field is always parallel to the plane of the coil, and can never contribute to the flux through the

Topic 20

1215

coil. The maximum induced emf in the coil is then

2 ⎡⎛ rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎤ −5 = NB A ω = 100 2.0 × 10 T 0.20 m 1 500 ⎢ ⎟⎜ ⎟⎥ ⎜ ⎟⎜ max horizontal min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠ ⎥⎦ ⎢⎣⎝

(

)(

)

= 1.3 × 10−2 V = 13 mV 20.33

Note the similarity between the situation in this problem and a generator. In a generator, one normally has a loop rotating in a constant magnetic field so the flux through the loop varies sinusoidally in time. In this problem, we have a stationary loop in an oscillating magnetic field, and the flux through the loop varies sinusoidally in time. In both cases, a sinusoidal emf ε = ε max sin ω t , where ε max = NBAω, is induced in the loop. The loop in this case consists of a single band (N = 1) around the perimeter of a red blood cell with diameter d = 8.0 × 10–6 m. The angular frequency of the oscillating flux through the area of this loop is ω = 2πf = 2π(60 Hz) = 120π rad/s. The maximum induced emf is then

(

)(

1.0 × 10−3 T 8.0 × 10−6 m ⎛ π d2 ⎞ ε max = NBAω = B ⎜ ⎟ ω = 4 ⎝ 4 ⎠

) (120π s ) 2

−1

= 1.9 × 10−11 V 20.34

(a) Using ε max = NBAω, we find

Topic 20

1216

⎡⎛ rev ⎞ ⎛ 2π rad ⎞ ⎤ 2 = 1 000 0.20 T 0.10 m ⎢⎜ 60 ⎟⎥ ⎟⎜ max s ⎠ ⎝ 1 rev ⎠ ⎥⎦ ⎢⎣⎝

)(

(

)

= 7.5 × 103 V = 7.5 kV (b) The maximum induced emf occurs when the flux through the coil is changing the most rapidly. This is when the plane of the coil is parallel to the magnetic field. 20.35 The emf of a generator is ε = NBA sinωt. Here, ε = 6.00 V is the maximum emf obtained when sin ωtmax = 1. (a) Solve for the loop area A to find

NBω

6.00 V = 5.00 × 10−3 m 2 100 0.600 T 20.0 rad/s ( )( )( )

(b) The maximum emf is obtained when sin ωtmax = 1 and the minimum is obtained when sin ωtmin = −1. One way to find the interval |tmax − tmin| is to realize that it equals half of period:

tmax − tmin = 12 T = 12

2π 1 2π =2 = 0.157 s ω 20.0 rad/s

Alternatively, see that ωtmax = π/2 and ωtmin = 3π/2 when sin ωtmax = 1 and sin ωtmin = −1, respectively. Solve for |tmax − tmin| to find

tmax − tmin =

π 3π 1 2π − = = 0.157 s 2ω 2ω 2 ω

Topic 20

20.36

1217

(a) Immediately after the switch is closed, the motor coils are still stationary and the back emf is zero. Thus,

ε = 240 V = 8.0 A R

30 Ω

(b) At maximum speed, ε back = 145 V and

ε −ε R

back

240 V − 145 V 30 Ω

= 3.2 A

= ε – IR = 240 V – (6.0 A)(30 Ω) = 60 V

back

(a) When a coil having N turns and enclosing area A rotates at angular frequency ω in a constant magnetic field, the emf induced in the coil is

ε =ε

back

sin ωt, where ε max = NB⊥ Aω

Here, B⊥ is the magnitude of the magnetic field perpendicular to the rotation axis of the coil. In the given case, B⊥ = 55.0 µT; A = πab, where a = (10.0 cm)/2 and b = (4.00 cm)/2; and rev ⎞ ⎛ 1 min ⎞ ⎛ ω = 2π f = 2π ⎜ 100 = 10.5 rad/s ⎟ ⎝ min ⎠ ⎜⎝ 60.0 s ⎟⎠ !

Thus,

Topic 20

1218

⎡π ⎤ −6 = 10.0 55.0 × 10 T 0.100 m 0.040 0 m ⎢ ⎥ 10.5 rad/s max ⎢⎣ 4 ⎥⎦

( )(

) (

max

)(

) (

)

= 1.81×10–5 V = 18.1 µV

(b) When the rotation axis is parallel to the field, then B⊥ = 0, giving ε max = 0. It is easily understood that the induced emf is always zero in this case if you recognize that the magnetic field lines are always parallel to the plane of the coil, and the flux through the coil has a constant value of zero. 20.38 (a) The solenoid’s inductance is

(

)

(

)

−7 −9 2 µ0 N 2 A 4π × 10 T ⋅ m/A ( 475 ) 2.80 × 10 m L= = = 1.32 × 10−8 H ℓ 6.00 × 10−2 m

(

)

(b) The emf across the solenoid is A ε L = −L ΔI = − (1.32 × 10−8 H ) −2.00 A − 2.00 = 6.34 × 10−6 V −3 Δt

20.39

From | ε | = L|ΔI/Δt|, we have

20.40

8.33 × 10

ε ΔI /Δt

ε ( Δt ) Δt

(12 × 10 V )(0.50 s) = 4.0 × 10 H = 4.0 mH = −3

−3

2.0 A − 3.5 A

The units of NΦB/I are T ⋅ m2/A. From the force on a moving charged particle, F = qvB the magnetic field is B = F/qv, and we find that

Topic 20

1219

N N ⋅s 1T=1 =1 C ⋅(m/s) C⋅m ! (N ⋅ m)⋅s ⎛ J ⎞ ⎛ N ⋅s ⎞ 2 2 = ⎜ ⎟ ⋅s = V ⋅s , Thus, T ⋅ m = ⎜ ⎟⎠ ⋅ m = ⎝ ⎝ C⎠ C⋅m C ! and T ⋅ m2/A = V ⋅ s/A. The units of ε /(ΔI/Δt) are V/(A/s) = V ⋅ s/A, the same as the units of NΦB/I.

(

20.41

)( ) (

)

2⎤ 2⎡ 4π × 10−7 T ⋅ m/A 400 ⎢π 2.5 × 10−2 m ⎥ µN A ⎣ ⎦ L= 0 = (a) ℓ 0.20 m 2

= 2.0 × 10−3 H = 2.0 mH

(b) From | ε |= L(ΔI/Δt),

20.42

ΔI Δt

ε L

75 × 10−3 V 2.0 × 10−3 H

= 38 A/s

From | ε | = L(ΔI/Δt), the self-inductance is

ε ΔI/Δt

24.0 × 10−3 V 10.0A/s

= 2.40 × 10−3 H

Then, from L = NΦB/I, the magnetic flux through each turn is −3 L⋅ I ( 2.40 × 10 H ) ( 4.00 A ) ΦB = = = 1.92 × 10−5 T ⋅ m 2 N 500 !

20.43

(a) In the series circuit of Figure P20.43, maximum current occurs after

Topic 20

1220

the switch has been closed for a very long time, when current has stabilized and the back emf due to the inductance has decreased to zero. This maximum current is given by

I max =

ε = 24.0 V = 5.33 A R

4.50 Ω

(b) The time constant of the RL circuit is

L 12.0 H 12.0 Ω ⋅s τ= = = = 2.67 s R 4.50 Ω 4.50 Ω ! (c) If the switch in the RL circuit is closed at time t = 0, the current as a τ

function of time is given by I = Imax (1 – e–t/ ). τ

Thus, e–t/ = 1 – I/Imax, or t = –τ ln(1 – I/Imax). With τ = 2.67 s, the current in this circuit will be I = 0.950Imax at time t = –(2.67 s)ln(1 – 0.950) = 8.00 s 20.44

(a) The time constant of the RL circuit is τ = L/R, and that of the RC circuit is τ = RC. If the two time constants have the same value, then RC = L/R, or L R= = C !

3.00 H = 1.00 × 103 Ω = 1.00 kΩ −6 3.00 × 10 F

(b) The common value of the two time constants is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 20

1221

L 3.00 H τ= = = 3.00 × 10−3 s = 3.00 ms 3 ! R 1.00 × 10 Ω 20.45 (a) The maximum current is Imax = ε /R = 2.00 A. With ε = 9.00 V, solve for the resistance R to find R = 4.50 Ω.

The current at thalf = 0.100 s is half the maximum value so that I(thalf)/( ε /R) = ½. Solve for the time constant from the expression for current in an RL circuit and substitute values to find I(t) =

ε 1− e τ

( R

−t/

) 1 = = 1− e ) → I(t ε /R 2 ( half

−thalf/τ

) →e

−thalf/τ

= 1 − 12 = 12

t 0.100 s − half = ln ( 12 ) → τ = − = 0.144 s τ ln ( 12 )

(b) Apply Kirchhoff’s loop rule, moving clockwise around the circuit at t = 0.100 s when I = 1.00 A to find

ε − IR − ΔV = 0 → ΔV = ε − IR L

ΔVL = 9.00 V − (1.00 A ) ( 4.50 Ω ) = 4.50 V

(c) At the instant t = 0 when the switch is closed, I = 0 and Kirchhoff’s loop rule gives ε − ΔVL = 0 so that ΔVL = ε = 9.00 V .

20.46

The current in the RL circuit at time t is I =

ε 1− e

( R

−t/τ

) . The potential

difference across the resistor is ΔVR = RI = ε (1 – e–t/ ), and from τ

Topic 20

1222

Kirchhoff’s loop rule, the potential difference across the inductor is ΔVL = ε – ΔVR = ε [1 –(1 – e–t/ )] = ε e–t/ τ

(a) At t = 0, ΔVR = ε (1 – e–0) = ε (1 – 1) = 0. (b) At t = τ, ΔVR = ε (1 – e–1) = (6.0 V)(1 – 0.368) = 3.8 V. (c) At t = 0, ΔVL = ε e–0 = ε = 60 V. (d) At t = τ, ΔVL = ε e–1 = (6.0 V)(0.368) = 2.2 V. 20.47

From I = Imax(1 – e–t/ ), we obtain e–t/ = 1 –I/Imax. If I/Imax = 0.900 at t = 3.00 s, then τ

e–(3.00 s)/ = 0.100

−3.00 s τ= = 1.30 s ! ln(0.100)

Since the time constant of an RL circuit is τ = L/R, the resistance is

L 2.50 H R= = = 1.92 Ω τ 1.30 s ! L 8.00 mH = 2.00 ms 20.48 (a) τ = = R 4.00 Ω ! (b) I =

ε 1− e

( R

−t /τ

V ) = ⎜⎝ 6.00 ⎟ (1 − e 4.00 Ω ⎠ ⎛

⎞

−250×10−6 s/2.00×10-3 s

) = 0.176 A

ε = 6.00 V = 1.50 A R

4.00 Ω

Topic 20

1223

(d) I = Imax(1 – e–t/ ) yields e–t/ = 1 – I/Imax, and t = –τ ln(1 – I/Imax) = –(2.00 ms)ln (1 – 0.800) = 3.22 ms 20.49 (a) The energy stored by an inductor is !PEL = 12 LI , so the self inductance 2

is −3 2 ( PEL ) 2 ( 0.300 × 10 J ) L= = = 2.08 × 10−4 H! = ! 0.208 mH 2 2 I (1.70 A ) !

(b) If I = 3.0 A, the stored energy will be

1 1 2 PEL = LI 2 = ( 2.08 × 10−4 H ) ( 3.0 A ) = 9.36 × 10−4 J = 0.936 mJ 2 2 ! 20.50 (a) The inductance of a solenoid is given by !L = µ0 N A/ , where N is the 2

number of turns on the solenoid, A is its cross-sectional area, and  is its length. For the given solenoid,

L= !

µ0 N 2 (π r 2 ) 

( 4π × 10 T ⋅ m/A )( 300) π ( 5.00 × 10 m ) = −7

−2

0.200 m

= 4.44 × 10−3 H = 4.44 mH

(b) When the solenoid described above carries a current of I = 0.500 A, the stored energy is

1 1 2 PEL = LI 2 = ( 4.44 mH ) ( 0.500 A ) = 0.555 mJ 2 2 !

Topic 20

20.51

1224

The current in the circuit at time t is I = Imax (1 – e–t/ ), where Imax = ε /R, τ

and the energy stored in the inductor is !PRL = 12 LI . 2

(a) As t→∝, I → I max =

ε = 24 V = 3.0 A , and R

8.0 Ω

1 2 1 2 PEL → LI max = ( 4.0 H ) ( 3.0 A ) = 18 J 2 2 ! (b) At t = τ, I = Imax (1 – e–1) = (3.0 A)(1 – 0.368) = 1.9 A, and

1 2 PEL = ( 4.0 H ) (1.9 A ) = 7.2 J 2 ! 20.52 (a) Use Table 17.1 to obtain the resistivity of the copper wire and find −8 ρCu L ρCu L (1.7 × 10 Ω ⋅ m ) ( 60.0 m ) Rwire = = 2 = = 1.3 Ω 2 Awire π rwire π ( 0.50 × 10−3 m ) !

(b)

L L 60.0 m N= = = = 4.8 × 102 turns circumference!of!a!loop 2π rsolenoid 2π ( 2.0 × 10−2 m ) !

 = N ( diameter!of!wire ) = N ( 2rwire ) = ( 480 ) 2 ( 0.50 × 10−3 m ) = 0.48 m !

Topic 20

1225

(d) 2 4π × 10−7 T ⋅ m/A ) ( 480 ) π ( 2.0 × 10−2 m ) ( µ0 N 2 Asolenoid µ0 N 2π rsolenoid L= = =   0.48 m ! 2

giving

L = 7.6 × 10–4 H = 0.76 mH

L L 7.6 × 10−4 H = = = 4.6 × 10−4 s = 0.46 ms (e) τ = R R + r 1.3 Ω + 0.350 Ω total wire internal ! (f) I max =

ε =

Rtotal

6.0 V 1.3 Ω + 0.350 Ω τ

= 3.6 A

(g) I = Imax (1 – e–t/ ), so when I = 0.999Imax, we have e–t/ = 1 – 0.999 = 0.001. Thus, –t/τ = ln(0.001), or t = –τ ⋅ ln(0.001) = –(0.46 ms) ⋅ ln(0.001) = 3.2 ms.

1 2 1 2 −4 −3 (h) ( PEL )max = LI max = (7.6 × 10 H ) ( 3.6 A ) = 4.9 × 10 J = 4.9 mJ 2 2 ! 20.53

The flux due to the current in loop 1 passes from left to right through the area enclosed by loop 2. As loop 1 moves closer to loop 2, the magnitude of this flux through loop 2 is increasing. The induced current in loop 2 generates a magnetic field directed toward the left through the area it encloses in order to oppose the increasing flux from loop 1. This means that the induced current in loop 2 must flow counterclockwise as viewed

Topic 20

1226

from the left end of the rod. 20.54

(a) The clockwise induced current in the loop produces a flux directed into the page through the area enclosed by the loop. Since this flux opposes the change in flux due to the external field, the outwarddirected flux due to the external field must be increasing in magnitude. This means that the magnitude of the external field itself must be increasing in time. (b) The induced emf in the loop must be

ε = IR = (2.50 mA)(0.500 Ω) = 1.25 mV Since ε =

ΔΦ B Δt

Δ(BAcos 0°) Δt

⎛ ΔB ⎞ =⎜ ⎟ A , the rate of change of the field ⎝ Δt ⎠

is ΔB Δt

20.55

ε ε = = = A

πr

1.25 × 10−3 V

(

π 8.00 × 10−2 m

)

= 6.22 × 10−2 T/s = 62.2 mT/s

(a) After the right end of the coil has entered the field, but the left end has not, the flux through the area enclosed by the coil is directed into the page and is increasing in magnitude. This increasing flux induces an emf of magnitude

Topic 20

1227

ε=

ΔΦ B Δt

NB(ΔA) Δt

= NBwv

in the loop. Note that in the above equation, ΔA = wv is the area enclosed by the coil that enters the field in time Δt. This emf produces a counterclockwise current in the loop to oppose the increasing inward flux. The magnitude of this current is I = | ε |/R = NBwv/R. The right end of the loop is now a conductor, of length Nw, carrying a current toward the top of the page through a field directed into the page. The field exerts a magnetic force of magnitude NBwv ⎛ NBwv ⎞ F = BI(Nw) = B ⎜ (Nw) = !directed! toward!the!left ⎟ ⎝ R ⎠ R ! 2

on this conductor, and hence, on the loop. (b) When the loop is entirely within the magnetic field, the flux through the area enclosed by the loop is constant. Hence, there is no induced emf or current in the loop, and the field exerts zero force on the loop.

Topic 20

1228

(c) After the right end of the loop emerges from the field, and before the left end emerges, the flux through the loop is directed into the page and is decreasing. This decreasing flux induces an emf of magnitude | ε | = NBwv in the loop, which produces an induced current directed clockwise around the loop so as to oppose the decreasing flux. The current has magnitude I = ⎪ ε ⎪/R = NBwv/R. This current flowing upward, through conductors of total length Nw, in the left end of the loop, experiences a magnetic force given by NBwv ⎛ NBwv ⎞ F = BI(Nw) = B ⎜ (Nw) = !directed! toward!the!left ⎟ ⎝ R ⎠ R ! 2

20.56

(a) The motional emf induced in the bar must be ε = IR, where I is the current in this series circuit. Since ε = B⊥ ℓv , the speed of the moving bar must be

ε = IR = (8.5 × 10 A )( 9.0 Ω) = 0.73 m/s v= −3

B⊥ ℓ

(0.30 T )(0.35 m )

The flux through the closed loop formed by the rails, the bar, and the resistor is directed into the page and is increasing in magnitude. To oppose this increasing inward flux, the induced current must generate a magnetic field directed out of the page through the area

Topic 20

1229

enclosed by the loop. This means the current will flow counterclockwise. (b) The rate at which energy is delivered to the resistor is P = I2R = (8.5 × 10–3 A)2 (9.0 Ω) = 6.5 × 10–4 W = 0.65 mW (c) An external force directed to the right acts on the bar to balance the magnetic force to the left. Hence, work is being done by the external force, which is transformed into internal energy within the resistor. 20.57

(a) The current in the solenoid reaches Isol = 0.632Imax in a time of t = τ = L/R, where −7 −4 2 µ0 N 2 A ( 4π × 10 T ⋅ m/A ) (12 500 ) (1.00 × 10 m ) L= = = 0.280 H  7.00 × 10−2 m ! 2

Thus, t = (0.280 H)/(14.0 Ω) = 2.00 × 10–2 s = 20.0 ms. (b) The change in the solenoid current during this time is ⎛ 60.0 V ⎞ ⎛ ΔV ⎞ ΔIsol = 0.632I max − 0 = 0.632 ⎜ = 0.632 ⎜ = 2.71 A ⎟ ⎝ R ⎠ ⎝ 14.0 Ω ⎟⎠ !

so the average back emf is

⎛ ΔI ⎞ ⎛ 2.71 A ⎞ sol = L = (0.280 H) = 37.9 V ⎜ ⎟ ⎜ back −2 ⎟ ⎜⎝ Δt ⎟⎠ ⎝ 2.00 × 10 s ⎠

Topic 20

1230

(c) The change in the magnitude of the magnetic field at the location of the coil is one-half the change in the magnitude of the field at the center of the solenoid. Thus, ΔBcoil = 12 ⎡⎣ µ0 nsol ( ΔIsol ) ⎤⎦ , and the average ! rate of change of flux through each turn of the coil is

( ΔB)coil Acoil 12 ⎡⎣ µ0nsol ( ΔIsol )⎤⎦ Acoil µ0Nsol ( ΔIsol ) Acoil ⎛ ΔΦ B ⎞ = = ⎜⎝ ⎟⎠ = Δt coil Δt Δt 2 sol ⋅ ( Δt )

( 4π × 10 T ⋅ m/A )(12 500)( 2.71 A )(1.00 × 10 m ) = 1.52 × 10 V = 2 (7.00 × 10 m ) ( 2.00 × 10 s ) −7

−4

−3

−2

(d)

I coil =

ε coil Rcoil

(

Ncoil ΔΦ B Δt Rcoil

−2

) = (820)(1.52 × 10 V ) −3

coil

24.0 Ω

= 0.051 9 A = 51.9 mA 20.58

(a) The gravitational force exerted on the ship by the pulsar supplies the centripetal acceleration needed to hold the ship in orbit. Thus, Fg = !

v= !

GMpulsarmship 2 rorbit

GMpulsar rorbit

mship v 2 rorbit

, giving

( 6.67 × 10 =

−11

N ⋅ m kg 2 ) ( 2.0 × 1030 kg ) 3.0 × 10 m 7

= 2.1 × 106 m/s

(b) The magnetic force acting on charged particles moving through a magnetic field is perpendicular to both the magnetic field and the velocity of the particles (and therefore perpendicular to the ship’s length). Thus, the charged particles in the materials making up the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 20

1231

spacecraft experience magnetic forces directed from one side of the ship to the other, meaning that the induced emf is directed from side to side within the ship. (c)

ε = B ℓv , where ! = 2r ⊥

ship

= 0.080 km = 80 m is the side-to-side

dimension of the ship. This yields

ε = (1.0 × 10 T)(80 m)(2.1 × 10 m/s) = 1.7 × 10 V 2

(d) The very large induced emf would lead to powerful spontaneous electric discharges. The strong electric and magnetic fields would disrupt the flow of ions in their bodies. 20.59

(a) To move the bar at uniform speed, the magnitude of the applied force must equal that of the magnetic force retarding the motion of the bar. Therefore, Fapp = BI . The magnitude of the induced current ! is

ε R

( ΔΦ /Δt) = B( ΔA/Δt) = Bℓv B

2 so the field strength is !B = IR/ , giving !Fapp = ( IR/v ) I = I R v , and

the current is

Topic 20

1232

I= !

Fapp ⋅ v R

(1.00 N)( 2.00 m/s ) = 0.500 A 8.00 Ω

(b) Pdissipated = I2 R = (0.500 A)2(8.00 Ω) = 2.00 W (c) Pinput = Fapp ⋅ v = (1.00 N)(2.00 m/s) = 2.00 W 20.60

Since the magnetic field outside the solenoid is negligible in comparison to the field inside the solenoid, we shall assume that the flux through the single-turn square loop is the same as that through each turn of the solenoid. Then, the induced emf in the square loop is ⎡ ⎤ Δ ⎢ Binside Asolenoid ⎥ ⎛ ΔI ⎞ ΔΦ ⎥⎦ Δ µ0 nIsolenoid Asolenoid ε = B = ⎢⎣ solenoid = = µ0 n ⎜ solenoid ⎟ Asolenoid ⎜⎝ Δt ⎟⎠ Δt Δt Δt

(

ε = µ n(ΔI 0

solenoid

)

/Δt)πr2, which gives

ε = (4π × 10 T ⋅ m/A)(3.50 × 10 /m)(28.5 A/s) π (2.00 × 10 m) –7

–2

= 1.58 × 10–4 V = 0.158 × 10–3 V = 0.158 mV 20.61

If d is the distance from the lightning bolt to the center of the coil, then

Topic 20

1233

N ( ΔB) A N ⎡⎣ µ ( ΔI ) 2π d ⎤⎦ A N µ ( ΔI ) A ( ) ε = = = = Δt Δt Δt 2π d ( Δt ) ⎡ ⎤ 100 ( 4π × 10 T ⋅ m/A ) ( 6.02 × 10 A − 0 ) ⎢π ( 0.800 m ) ⎥ ⎣ ⎦ = 2π ( 200 m ) (10.5 × 10 s ) N ΔΦ B

−7

−6

= 1.15 × 105 V = 115 kV 20.62

When A and B are 3.00 m apart, the area enclosed by the loop consists of four triangular sections, each having hypotenuse of 3.00 m, altitude of 1.50 m, and base of

( 3.00 m ) − (1.50 m ) = 2.60 m . The decrease in the ! 2

enclosed area has been 2 ⎡1 ⎤ ΔA = Ai − A f = ( 3.00 m ) − 4 ⎢ (1.50 m ) ( 2.60 m ) ⎥ = 1.20 m 2 ⎣2 ⎦ !

The average induced current has been

ε I = av

2 ΔΦ B / Δt ) B ( ΔA / Δt ) ( 0.100 T )( 1.20 m / 0.100 s ) ( = = = = 0.120 A

10.0 Ω

As the enclosed area decreases, the flux due to the external field (directed into the page) through this area also decreases. Thus, the induced current will be directed clockwise around the loop to create additional flux © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 20

1234

directed into the page through the enclosed area.

ε 20.63

= av

(a)

ΔΦ B Δt

(

)

B ⎡ π d2 4 − 0⎤ ⎦ = ⎣ Δt Δt

B ΔA

(25.0 mT )π (2.00 × 10 m ) = 0.157 mV = 4 ( 50.0 × 10 s ) 2

−2

−3

As the inward-directed flux through the loop decreases, the induced current goes clockwise around the loop to create additional inward flux through the enclosed area. With positive charges accumulating at B, point B is positive relative to A.

( ΔB) A = ⎡⎣(100 − 25.0) mT ⎤⎦ π (2.00 × 10 m ) = 5.89 mV (b) ε = = Δt Δt 4 ( 4.00 × 10 s ) −2

ΔΦ B

−3

As the inward-directed flux through the enclosed area increases, the induced current goes counterclockwise around the loop in to create flux directed outward through the enclosed area. With positive charges now accumulating at A, point A is positive relative to B. 20.64

The induced emf in the ring is

( ΔB) A

( ΔB

)

⎛ ΔI ⎞⎤ 1⎡ ⎢ µ n ⎜ solenoid ⎟ ⎥ A Δt Δt Δt 2 ⎢⎣ 0 ⎜⎝ Δt ⎟⎠ ⎥⎦ solenoid 2⎞ ⎤ 1⎡ ⎛ = ⎢ 4π × 10−7 T ⋅ m/A 1 000 270 A/s ⎜ π ⎡ 3.00 × 10−2 m ⎤ ⎟ ⎥ = 4.80 × 10−4 V ⎦ ⎠ ⎝ ⎣ 2⎣ ⎦

= av

ΔΦ B

(

solenoid

)(

solenoid

2 Asolenoid

)(

)

Topic 20

1235

Thus, the induced current in the ring is

I ring =

20.65

4.80 × 10−4 V 3.00 × 10−4 Ω

= 1.60 A

(a) As the rolling axle (of length ! = 1.50 m ) moves perpendicularly to the uniform magnetic field, an induced emf of magnitude ε = Bℓv will exist between its ends. The current produced in the closed-loop circuit by this induced emf has magnitude

( ΔΦ Δt) = B( ΔA/Δt) = Bℓv = (0.800 T )(1.50 m )(3.00 m/s) B

0.400 Ω

= 9.00 A

(b) The induced current through the axle will cause the magnetic field to exert a retarding force of magnitude !Fr = BI on the axle. This force is

 directed opposite to the velocity !v to oppose the motion of the axle. If the axle is to continue moving at constant speed, an applied force in

 the direction of !v , and having magnitude Fapp = Fr, must be exerted on the axle.

F = BI = (0.800 T)(9.00 A)(1.50 m) = 10.8 N ! app (c) Using right-hand rule number 1, observe that positive charges within the moving axle experience a magnetic force toward the rail © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 20

1236

containing point b, and negative charges experience a force directed toward the rail containing point a. Thus, the rail containing b is positive relative to the other rail, so b is at the higher potential.

  (d) No. Both the velocity !v of the rolling axle and the magnetic field !B are unchanged. Thus, the polarity of the induced emf in the moving axle is unchanged, and the current continues to be directed from b to a through the resistor R. 20.66

(a) The time required for the coil to move distance  and exit the field is

!t = /v , where v is the constant speed of the coil. Since the speed of the coil is constant, the flux through the area enclosed by the coil decreases at a constant rate. Thus, the instantaneous induced emf is the same as the average emf over the interval t seconds in duration, or 2

ε = −N ΔΦ = −N (0 − BA) = N Bℓ = NBℓ = NBℓv Δt

t−0

ℓ/v

(b) The current induced in the coil is I = ε /R = NBℓv/R . (c) Since the coil moves with constant velocity, the power delivered to the coil must equal the power being dissipated within the coil. This is given by !P = I 2 R , or © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 20

1237

⎛ N 2 B2  2 v 2 ⎞ N 2 B2  2 v 2 P=⎜ ⎟⎠ R = ⎝ R2 R !

(d) The rate that the applied force does work must equal the power delivered to the coil, so Fapp ⋅ v = P, or P N 2 B2  2 v 2 R N 2 B2  2 v Fapp = = = v v R !

(e) As the coil is emerging from the field, the flux through the area it encloses is directed into the page and is decreasing in magnitude. The induced current must flow clockwise around the coil to generate a magnetic field directed into the page through the area enclosed by the coil, opposing the decrease in the inward flux. (f) As the coil is emerging from the field, the left side of the coil carries an induced current directed toward the top of the page through a magnetic field that is directed into the page. Right-hand rule number 1 then shows that this side of the coil will experience a magnetic force directed to the left, opposing the motion of the coil. 20.67

(a) As the bottom conductor of the loop falls, it cuts across the magnetic field lines coming out of the page. This induces an emf of magnitude | ε | = Bwv in this conductor, with the left end at the higher

Topic 20

1238

potential. As a result, an induced current of magnitude

ε R

Bwv R

flows clockwise around the loop. The field then exerts an upward force of magnitude

Bwv ⎛ Bwv ⎞ Fm = BIw = B ⎜ w= ⎟ ⎝ R ⎠ R ! 2

on this current-carrying conductor forming the bottom of the loop. If the loop is falling at terminal speed, the magnitude of this force must equal the downward gravitational force acting on the loop. That is, when v = vt, we must have B2 w 2 vt Fm = = Mg R !

vt =

MgR B2 w 2

(b) A larger resistance would make the current smaller, so the loop must

Topic 20

1239

reach higher speed before the magnitude of the magnetic force will equal the gravitational force. (c) The magnetic force is proportional to the product of the field and the current, while the current itself is proportional to the field. If B is cut in half, the speed must become four times larger to compensate and yield a magnetic force with magnitude equal to that of the gravitational force.

Topic 21

1240

Topic 21 Alternating Current Circuits and Electromagnetic Waves

QUICK QUIZZES 21.1

Choice (c). The average power is proportional to the rms current which is non-zero even though the average current is zero. (a) is only valid for an open circuit, for which R → ∞. (b) and (d) can never be true because iav = 0 for AC currents.

21.2

Choice (b). Choices (a) and (c) are incorrect because the unaligned sine curves in Figure 21.9 mean the voltages are out of phase, and so we cannot simply add the maximum (or rms) voltages across the elements. (In other words, ΔV ≠ ΔVR + ΔVL + ΔVC even though Δv = ΔvR + ΔvL + ΔvC.)

21.3

Choice (b). Closing switch A replaces a single resistor with a parallel combination of two resistors. Since the equivalent resistance of a parallel combination is always less than the lowest resistance in the combination, the total resistance of the circuit decreases, which causes the impedance 2 Z = Rtotal + ( X L − XC ) to decrease. ! 2

21.4

Choice (a). Closing switch A replaces a single resistor with a parallel

Topic 21

1241

combination of two resistors. Since the equivalent resistance of a parallel combination is always less than the lowest resistance in the combination, the total resistance of the circuit decreases, which causes the phase angle,

φ = tan−1[(XL − XC)/R], to increase. 21.5

Choice (a). Closing switch B replaces a single capacitor with a parallel combination of two capacitors. Since the equivalent capacitance of a parallel combination is greater than that of either of the individual capacitors, the total capacitance increases, which causes the capacitive reactance XC = 1/2πfC to decrease. Thus, the net reactance, XL − XC, increases causing the phase angle, φ = tan−1[(XL − XC)/R], to increase.

21.6

Choice (b). Inserting an iron core in the inductor increases both the selfinductance and the inductive reactance, XL = 2πfL. This means the net 2 reactance, XL − XC, and hence the impedance, Z = Rtotal + ( X L − XC ) , ! 2

increases, causing the current (and therefore, the bulb’s brightness) to decrease. 21.7

Choices (b) and (c). Since pressure is force per unit area, changing the size of the area without altering the intensity of the radiation striking that area will not cause a change in radiation pressure. In (b), the smaller disk absorbs less radiation, resulting in a smaller force. For the same reason,

Topic 21

1242

the momentum in (c) is reduced. 21.8

Choices (b) and (d). The frequency and wavelength of light waves are related by the equation λf = v or f = v/λ, where the speed of light v is a constant within a given medium. Thus, the frequency and wavelength are inversely proportional to each other; when one increases the other must decrease.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 21.2

The phase angle in an RLC series circuit is given by

⎛ X − XC ⎞ ⎛ 2π fL − 1/2π fC ⎞ φ = tan −1 ⎜ L = tan −1 ⎜ ⎟ ⎟⎠ ⎝ ⎝ R ⎠ R ! which is clearly frequency dependent. As f → 0, φ → tan−1 (−∞) = −90° and as f → ∞, φ → tan−1 (+∞) = +90°. At resonance, XL = XC, and φ = tan−1 (0) = 0°. 21.4

An antenna that is a conducting line responds to the electric field of the electromagnetic wave—the oscillating electric field causes an electric force on electrons in the wire along its length. The movement of electrons along the wire is detected as a current by the radio and is amplified. Thus, a line antenna must have the same orientation as the broadcast

Topic 21

1243

antenna. A loop antenna responds to the magnetic field in the radio wave. The varying magnetic field induces a varying current in the loop (by Faraday’s law), and this signal is amplified. The loop should be in the vertical plane containing the line of sight to the broadcast antenna, so the magnetic field lines go through the area of the loop. 21.6

(a) In an electromagnetic wave, electric and magnetic fields oscillate at right angles to each other and perpendicular to the direction of propagation of the wave. (b) Energy is the quantity transported by the oscillating electric and magnetic fields of the electromagnetic wave.

21.8

Comparing the wavelengths for the different types of electromagnetic waves, the proper ranking from shortest to longest wavelength is choice (c): gamma rays, x-rays, visible light, radio waves.

21.10

The brightest portion of your face shows where you radiate the most. Your nostrils and the openings of your ear canals are particularly bright. Brighter still are the pupils of your eyes.

21.12

The changing magnetic field of the solenoid induces eddy currents in the conducting core. This is accompanied by I2R conversion of electricallytransmitted energy into internal energy in the conductor.

Topic 21

21.14

1244

The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° in phase ahead of the current in the circuit.

21.16

When a power source, AC or DC, is ﬁrst connected to a RL combination, the presence of the inductor impedes the buildup of a current in the circuit. The value of the current starts at zero and increases as the back emf induced across the inductor decreases somewhat in magnitude. Thus, the correct choice is (c).

21.18

At resonance of the RLC series circuit, XL = XC, and the impedance becomes

Z = R 2 + (X L − X C )2 = R 2 + 0 = R so choice (c) is correct.

ANSWERS TO EVEN NUMBERED PROBLEMS 21.2

(a)

1.70 × 102 V

(b)

2.40 × 102 Ω

Topic 21

1245

(c)

The 100-W bulb will have the lower resistance.

(a)

I1,rms = I2,rms = 1.25 A, I3,rms = 0.833 A

(b)

R1 = R2 = 96.0 Ω, R3 = 144 Ω

(a)

170 V

(b)

30.0 Hz

(c)

(d)

6.00 A

(e)

8.49 A

(f) 720 W

(g) 6.9 A

(h)

radians

21.8

(a)

2.65 × 10−6 F

(b)

3.18 × 10−6 F

21.10

(a)

221 Ω

(b)

0.163 A

(d)

No. The charge is a maximum when the voltage is a maximum, but

21.4

21.6

(c)

120 V

0.231 A

the voltage across a capacitor is 90° out of phase with the current. 21.12

(a)

69.3 V

(b)

40 Hz

(c)

0.354 A

(d)

196 Ω

(e)

20.3 µF

21.14

(a)

1.8 × 104 Ω

(b)

6.8 × 10−3 A

21.16

(a)

15.0 Hz

(b)

84.9 V

(c)

47.1 Ω

(d)

1.80 A

(e)

2.55 A

(f) 0

(g) i = (2.55 A)sin(30πt − π/2) 21.18

(a)

57.5 Ω

(b)

(h)

20.9 ms

1.39 A

Topic 21

1246

21.20

(a)

194 V

(b)

The current leads the voltage by 49.9°.

21.22

(a)

138 V

(b)

104 V

(d)

641 V

(e)

See Solution.

(a)

0.11 A

(b)

ΔVR, max = 1.3 × 102 V, ΔVL, max = 1.2 × 102 V

(c)

ΔvR = 1.3 × 102 V, ΔvL = 0, Δvsource = 1.3 × 102 V

(d)

ΔvR = 0, ΔvL = 1.2 × 102 V, Δvsource = 1.2 × 102 V

(a)

88.4 Ω

(d)

The voltage lags the current by 55.8°.

(e)

It changes the impedance, and therefore the current, in the circuit.

21.24

21.26

(b)

107 Ω

(c)

729 V

1.12 A

21.28

2.79 kHz

21.30

(a)

0.11 A

(c)

ΔvR = 0, ΔvC = 1.2 × 102 V, Δvsource = 1.2 × 102 V, qC = 300 µC

(d)

ΔvR = 1.3 × 102 V, ΔvC = 0, Δvsource = 1.3 × 102 V, qC = 0

21.32

(a)

66.8 Ω

21.34

88.0 W

21.36

(a)

(b)

ΔVR, max = 1.3 × 102 V, ΔVC, max = 1.2 × 102 V

0.953 A

(c)

45.4 W

No, the algebraic sum of the rms voltage drops is 20.9 V.

Topic 21

1247

(b) 21.38

to the resistor

(c)

3.2 W

(b)

1.15 × 10−8 F

2.31 kHz

21.40 (a) 6.57 × 105 Hz 21.42 (a)

480 W

(b)

0.192 W

(c)

30.7 mW

(d)

0.192 W

(e)

30.7 mW; Maximum power is delivered at the resonance frequency.

21.44

(a)

9.23 V

(b)

30.0 W

21.46

(a)

278 W

(b)

2.41 A

21.48

(a)

29.0 kW

(b)

0.577%

(c)

See Solution.

21.50

(a)

6.80 × 102 y

(b)

8.31 min

(c)

2.56 s

21.52

8.854 187 8 × 10−12 C2/(N ⋅ m2)

21.54

0.80 or 80%

21.56

(a)

I = mgc/2A

(b)

1.46 × 109 W/m2

(c)

Propulsion by light pressure in a significant gravity field is impractical because of the enormous power requirements. In addition, no material is perfectly reflecting, so the absorbed energy would melt the reflecting surface.

21.58

(a)

8.89 × 10−8 W/m2

(b)

11.4 MW

Topic 21

1248

21.60

11.0 m

21.62

The radio listeners hear the news 8.4 ms before the studio audience because radio waves travel so much faster than sound waves. 6.003 6 × 1014 Hz

(b)

3.6 × 1011 Hz

0.63 pF

(b)

8.4 mm

(a)

6.0 Ω

(b)

12 mH

(a)

6.2 mW/cm2, 24% higher than the maximum allowed leakage from a

21.64

(a)

21.66

1.1 × 107 m/s

21.68

∼106 J

21.70

XC,i = 3R

21.72

(a)

21.74 21.76

(c)

25 Ω

microwave (b)

0.024 mW/cm2

PROBLEM SOLUTIONS 21.1

For an AC circuit containing only resistance (the filament of the lightbulb), the power dissipated is

(

2 2 P = I rms R = ( ΔVrms R ) R = ΔVrms R = ΔVmax ! 2

) R. 2

Topic 21

1249

(

) = 193 Ω

(

) = 145 Ω

2 170 V/ 2 ΔVrms (a) R = = P 75.0 W !

2 170 V/ 2 ΔVrms (b) R = = P 100.0 W !

21.2

2 2 (a) ΔVR,max = 2 ( ΔVR,rms ) = 2 (1.20 × 10 V ) = 1.70 × 10 V !

(b) Pav = I !

2 2 (1.20 × 10 V ) = 2.40 × 102 Ω ΔVrms ΔVrms R= ⇒R= = R Pav 60.0 W 2

2 rms

2 ΔVrms R = (c) Because (see above), if the bulbs are designed to operate at Pav !

the same voltage, the 100 W will have the lower resistance. 21.3

For a voltage source with an rms voltage of 120 V, the maximum voltage is ΔVmax =

2 ΔVrms = 170 V.

(a) The maximum voltage across a resistor is ΔVR,max = ImaxR =

(

ΔVmax R

) R = ΔV

max

= 170 V .

(b) Solve for the maximum current using Ohm’s law:

I max =

ΔVR ,max 170 V = = 0.11 A R 1.5 × 103 Ω

I rms =

ΔVR ,rms 120 V = = 8.0 × 10−2 A R 1.5 × 103 Ω

Topic 21

1250

(d) The average power dissipated by the resistor is

(

2 Pav = I rms R = 8.0 × 10−2 A

21.4

) (1.5 × 10 Ω ) = 9.6 W 2

All lamps are connected in parallel with the voltage source, so ΔVrms = 120 V for each lamp. Also, for each bulb, the current is Irms = Pav/ΔVrms and the resistance is R = ΔVrms/Irms. 150 W I1,rms = I 2,rms = = 1.25 A 120 V !

(a) For bulbs 1 and 2:

100 W I 3,rms = = 0.833 A 120 V !

For bulb 3:

120 V = 96.0 Ω (b) R1 = R2 = 1.25 A !

21.5

and

R3 =

120 V = 144 Ω 0.833 A

(a) Solve for the rms current from the expression for average power dissipated by a resistor: 2 Pav = I rms R → I rms =

Pav = R

0.600 W = 5.00 × 10−3 A 3 24.0 × 10 Ω

(b) Apply Kirchhoff’s loop rule to find ΔVrms − IrmsR = 0 so that

(

)(

)

ΔVrms = I rmsR = 5.00 × 10−3 A 24.0 × 103 Ω = 1.20 × 10 2 V

21.6

The general form of the generator voltage is Δv = (ΔVmax)sin (ωt), so by inspection

Topic 21

1251

(a) ΔVR, max = 170 V

(c)

ΔVR,rms = !

(d) I rms = !

and

ω 60π rad/s = = 30.0 Hz (b) f = 2 π 2 π rad !

ΔVR,max 170 V = = 120 V 2 2

ΔVR,rms 120 V = = 6.00 A R 20.0 Ω

(e) I max = 2I rms = 2(6.00 A) = 8.49 A (f)

2 Pav = I rms R = (6.00 A)2 (20.0 Ω) = 720 W !

(g) At t = 0.005 0 s, the instantaneous current is Δv (170 V) (170 V) i= = sin [(60π rad/s)(0.005 0 s)] = sin (0.94 rad) = 6.9 A R 20.0 Ω 20.0 Ω !

(h) The argument of the sine function has units of [ωt] = (rad/s)(s) = radians. 21.7

(a) The expression for capacitive reactance is XC = 1/2πfC. Thus, if XC < 175 Ω, it is necessary that

f= !

1 1 > 2π CXC 2π ( 22.0 × 10−6 F ) (175 Ω )

f > 41.3 Hz

(b) For C1, the reactance is XC,1 = 1/2πfC1, while for C2, XC, 2 = 1/2πfC2. Thus, for the same frequencies, the ratio of the reactance for the two

Topic 21

1252

capacitors is

⎛ 1 ⎞ ⎛ 2π f C1 ⎞ C1 =⎜ ⎟⎜ ⎟= XC , 1 ⎝ 2π f C2 ⎠ ⎝ 1 ⎠ C2 ! XC , 2

⎛C ⎞ XC , 2 = ⎜ 1 ⎟ XC , 1 ⎝ C2 ⎠

If C1 = 22.0µF, C2 = 44.0 µF, and XC,1 < 175 Ω, we have

⎛ 22.0 µF ⎞ XC , 2 < ⎜ (175 Ω) ⎝ 44.0 µF ⎟⎠ ! 21.8

21.9

XC,2 < 87.5 Ω

The capacitive reactance is XC = 1/(2πfC). Solve for C to find (a) C =

1 1 = = 2.65 × 10−6 F 2π fXC 2π ( 60.0 Hz ) 1.00 × 103 Ω

(b) C =

1 1 = = 3.18 × 10−6 F 3 2π fXC 2π ( 50.0 Hz ) 1.00 × 10 Ω

( (

) )

ΔVrms I rms = = 2π fC ( ΔVrms ) , so XC !

f= ! 21.10

I rms 0.30 A = = 4.0 × 102 Hz −6 2π C ( ΔVrms ) 2π ( 4.0 × 10 F ) ( 30 V )

1 1 = = 221 Ω (a) XC = 2π fC 2π ( 60.0 Hz ) (12.0 × 10−6 F ) ! (b) I rms = ! (c)

ΔVC ,!rms 36.0 V = = 0.163 A XC 221 Ω

I = 2 I rms = 2 ( 0.163 A ) = 0.231 A ! max

Topic 21

1253

(d) No. The charge is a maximum when the voltage is a maximum, but the voltage across a capacitor is 90° out of phase with the current. 21.11

ΔVmax I max = = 2π fC ( ΔVmax ) = 2π ( 90.0 Hz ) ( 3.70 × 10−6 F ) ( 48.0 V ) XC !

21.12

Imax = 1.00 × 10−1 A = 100 mA

(a) By inspection, ΔVC,max = 98.0 V, so ΔVC,rms = !

ΔVC ,max 98.0 V = = 69.3 V 2 2

ω 80π rad/s = = 40 Hz (b) Also by inspection, ω = 80π rad/s, so f = 2 π 2 π rad ! (c)

I 0.500 A I rms = max = = 0.354 A 2 2 !

(d) XC = !

ΔVC ,max 98.0 V = = 196 Ω I max 0.500 A

(e)

XC =

21.13

1 2π fC

ωC

ω XC

1 (80π rad/s)(196 Ω)

= 2.03 × 10−5 F = 20.3 µF

54.0 Ω X L = 2π (60.0 Hz)L = 54.0 Ω ⇒ 2π L = = 0.900 Ω ⋅s −1 60.0 s !

Then, when ΔVrms = 100 V and f = 50.0 Hz, the maximum current will be

Topic 21

1254

2 ( ΔVrms ) 2 (100 V ) ΔVmax I max = = = = 3.14 A XL 2π L) f 0.900 Ω ⋅s ) ( 50.0 Hz ) ( ( ! 21.14 (a) The inductive reactance is XL = 2πfL. Substitute values to find XL = 2π ( 60.0 Hz ) ( 47 H ) = 1.8 × 10 4 Ω = 18 kΩ

(b) The voltage across an inductor is ΔVL = IrmsXL. In a purely inductive circuit, Kirchhoff’s loop rule gives ΔVrms = ΔVL so that ΔVrms = I rmsXL → I rms =

I rms =

ΔVrms XL

ΔVrms 120 V = = 6.8 × 10−3 A 2π fL 2π ( 60.0 Hz ) ( 47 H )

(Note that substituting XL = 1.8 × 104 Ω results in Irms = (120 V)/(1.8 × 104 Ω) = 6.7 × 10−3 A, the difference from the answer above being due to rounding error.) 21.15

ΔVmax ΔVmax = , so (a) I max = X 2 π fL L ! ΔVmax 100 V L= = = 4.24 × 10−2 H = 42.4 mH 2π fI max 2π (50.0 Hz)(7.50 A) !

(b) Imax = ΔVmax/XL = ΔVmax/ωL, or Imax is inversely proportional to ω. Thus, Imax,1/Imax,2 = ω2/ω1, or

Topic 21

1255

⎛I ⎞ ⎛ 7.50 A ⎞ ⎡ 2π ( 50.0 Hz ) ⎤⎦ = 942 rad/s ω 2 = ⎜ max,1 ⎟ ω 1 = ⎜ I max,2 ⎠ ⎝ 2.50 A ⎟⎠ ⎣ ⎝ !

21.16

Given: vL = (1.20 × 102 V)sin (30πt) and L = 0.500 H.

ω 30π rad/s (a) By inspection, ω = 30π rad/s, so f = = = 15.0 Hz 2π 2π ! (b) Also by inspection, ΔVL,max = 1.20 × 102 V, so ΔVL,max 1.20 × 102 V ΔVL,rms = = = 84.9 V 2 2 !

(d) I rms = !

ΔVL,rms XL

84.9 V = 1.80 A 47.1 Ω

(e) I max = 2 I rms = 2(1.80 A) = 2.55 A ! (f) The phase difference between the voltage across an inductor and the current through the inductor is φL = 90°, so the average power delivered to the inductor is PL,av = IrmsΔVL,rms cos φL = IrmsΔVL,rms cos (90°) = 0 (g) When a sinusoidal voltage with a peak value ΔVL,max is applied to an inductor, the current through the inductor also varies sinusoidally in time, with the same frequency as the applied voltage, and has a © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 21

1256

maximum value of Imax = ΔVL,max/XL. However, the current lags behind the voltage in phase by a quarter-cycle, or π/2 radians. Thus, if the voltage is given by ΔvL = ΔVL,max sin (ωt), the current as a function of time is i = Imax sin (ωt − π/2). In the case of the given inductor, the current through it will be i = (2.55 A)sin (30πt − π/2). (h) When i = +1.00 A, we have sin (30πt − π/2) = (1.00 A/2.55 A), or 30πt − π/2 = sin−1(1.00 A/2.55 A) = sin−1(0.392) = 0.403 rad t= !

and

21.17

π /2 rad + 0.403 rad = 2.09 × 10−2 s = 20.9 ms 30π rad/s

From L = NΦB/I (see Section 20.5 in the textbook), the total flux through the coil is ΦB,total = NΦB = L ⋅ I, where ΦB is the flux through a single turn on the coil. Thus,

(Φ

)

B,total max

⎡ ΔVmax ⎤ = L ⋅ I max = L ⋅ ⎢ ⎥ ⎣ XL ⎦ =L

21.18

2 ( ΔVrms ) 2 (120 V ) = = 0.450 T ⋅ m 2 2π fL 2π (60.0 Hz)

(a) The applied voltage is Δv = ΔVmax sin(ωt) = (80.0 V)sin (150t), so ΔVmax = 80.0 V and ω = 2πf = 150 rad/s. The impedance for the circuit is

Z = R2 + ( X L − XC ) = R2 + (2π fL − 1/2π fC)2 = R2 + (ω L − 1/ω C)2 ! 2

Topic 21

1257

⎡ ⎤ 1 Z = ( 40.0 Ω ) + ⎢(150 rad/s ) ( 80.0 × 10−3 H ) − ⎥ (150 rad/s )(125 × 10−6 F) ⎥⎦ ⎢⎣

= 57.5 Ω

(b) I max = !

ΔVmax Z

80.0 V = 1.39 A 57.5 Ω

21.19 (a) When XL = XC, the expression for impedance, Z = R 2 − ( XL − XC ) , 2

shows that Z = R (the impedance is purely resistive when the inductive reactance equals the capacitive reactance). Apply XL = XC and solve for the capacitance C to find XL = XC → 2π fL =

1 1 → C= 2 2 2π fC 4π f L

4π ( 60.0 Hz ) ( 0.500 H ) 2

= 1.41 × 10−5 F

(b) In a purely resistive circuit, the current and voltage are in phase so that φ = 0 . Alternatively, solve for the phase angle from

tan φ =

XL − XC 0 = → tan φ = 0 R R

φ = tan −1 ( 0 ) = 0 21.20

(a) XL = 2πfL = 2π(50.0 Hz)(400 × 10−3 H) = 126 Ω

Topic 21

1258

1 1 XC = = = 719 Ω 2π fC 2π ( 50.0 Hz ) ( 4.43 × 10−6 F ) !

Z = R2 + ( X L − XC ) = ! 2

( 500 Ω) + (126 Ω − 719 Ω) = 776 Ω 2

ΔVmax = ImaxZ = (0.250 A)(776 Ω) = 194 V

−1 ⎛ X L − XC ⎞ −1 ⎛ 126 Ω − 719 Ω ⎞ (b) φ = tan ⎜⎝ ⎟⎠ = tan ⎜ ⎟⎠ = −49.9° R 500 Ω ⎝ !

Thus, the current leads the voltage by 49.9°. 21.21

(a) XL = 2πfL = 2π(240 Hz)(2.50 H) = 3.77 × 103 Ω

1 1 XC = = = 2.65 × 103 Ω −6 2π fC 2π ( 240 Hz ) ( 0.250 × 10 F ) !

Z = R2 + ( X L − XC ) = ! 2

(b) I max =

ΔVmax Z

( 900 Ω) + ⎡⎣( 3.77 − 2.65) × 10 Ω ⎤⎦ = 1.44 kΩ

140 V 1.44 × 103 Ω

= 0.097 2 A

3 −1 ⎛ X L − XC ⎞ −1 ⎡ (3.77 − 2.65) × 10 Ω ⎤ = tan (c) φ = tan ⎜⎝ ⎟ ⎢ ⎥ = 51.2° R ⎠ 900 Ω ⎣ ⎦ !

Topic 21

21.22

1259

XL = 2πfL = 2π(60.0 Hz)(0.100 H) = 37.7 Ω

1 1 XC = = = 265 Ω 2π fC 2π ( 60.0 Hz ) (10.0 × 10−6 F ) !

Z = R2 + ( X L − XC ) = ! 2

( 50.0 Ω) + ( 37.7 Ω − 265 Ω) = 233 Ω 2

(a) ΔVR,rms = IrmsR = (2.75 A)(50.0 Ω) = 138 V (b) ΔVL,rms = IrmsXL = (2.75 A)(37.7 Ω) = 104 V (c) ΔVC = IrmsXC = (2.75 A)(265 Ω) = 729 V (d) ΔVrms = IrmsZ = (2.75 A)(233 Ω) = 641 V

(e)

21.23 Applying Ohm’s law and solving for the maximum current gives

I max =

ΔVL,max ΔVC ,max ΔVR ,max 145 V = = = = 12.1 A XL XC R 12.0 Ω

(a) Solve for ΔVL,max :

ΔVL,max = I max XL = (12.1 A ) ( 30.0 Ω ) = 363 V (b) Solve for ΔVC,max : © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 21

1260

ΔVC ,max = I max XC = (12.1 A ) ( 20.0 Ω ) = 242 V (c) As found above, the maximum current is Imax = 12.1 A . (d) The impedance is

Z = R 2 + ( XL − XC ) = 2

21.24

(12.0 Ω ) + ( 30.0 Ω − 20.0 Ω ) = 15.6 Ω 2

XL = 2πfL = 2π(60 Hz)(2.8 H) = 1.1 × 103 Ω

Z = R2 + ( X L − XC ) = ! 2

(a) I max = !

(1.2 × 10 Ω) + (1.1 × 10 Ω − 0) = 1.6 × 10 Ω 3

ΔVmax 170 V = = 0.11 A Z 1.6 × 103 Ω

(b) ΔVR,max = ImaxR = (0.11 A)(1.2 × 103 Ω) = 1.3 × 102 V ΔVL,max = ImaxXL = (0.11 A)(1.1 × 103 Ω) = 1.2 × 102 V (c) When the instantaneous current is a maximum (i = Imax), the instantaneous voltage across the resistor is ΔvR = iR = ImaxR = ΔVR,max = 1.3 × 102 V. The instantaneous voltage across an inductor is always 90° or a quarter cycle out of phase with the instantaneous current. Thus, when i = Imax, ΔvL = 0. Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit Δvsource = ΔvR + ΔvL,

Topic 21

1261

and at this instant when i = Imax, we have Δvsource = ImaxR + 0 = 1.3 × 102 V. (d) When the instantaneous current i is zero, the instantaneous voltage across the resistor is ΔvR = iR = 0. Again, the instantaneous voltage across an inductor is a quarter cycle out of phase with the current. Thus, when i = 0, we must have ΔvL = ΔVL,max = 1.2 × 102 V. Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit at the instant when i = 0 gives Δvsource = ΔvR + ΔvL = 0 + ΔVL, max = 1.2 × 102 V.

21.25

1 1 XC = = = 1.33 × 108 Ω −12 2π fC 2π ( 60.0 Hz ) ( 20.0 × 10 F ) ! 2 ZRC = Rbody + XC2 = !

and

ΔV ( I =

( 50.0 × 10 Ω) + (1.33 × 10 Ω) = 1.33 × 10 Ω 3

secondary

rms

ZRC

) = 5 000 V = 3.76 × 10 A −5

rms

1.33 × 10 Ω

Therefore, ΔVbody, rms = Irms Rbody = (3.76 × 10−5 A)(50.0 × 103 Ω) = 1.88 V.

21.26

1 1 = = 88.4 Ω (a) XC = 2π fC 2π ( 60.0 Hz ) ( 30.0 × 10−6 F ) ! (b) Z = R2 + ( 0 − XC ) = R2 + XC2 = ! 2

( 60.0 Ω) + (88.4 Ω) = 107 Ω 2

Topic 21

1262

(c)

I max = !

ΔVmax 1.20 × 102 V = = 1.12 A Z 107 Ω

(d) The phase angle in this RC circuit is

⎛ 0 − 88.4 ⎞ ⎛ X − XC ⎞ φ = tan −1 ⎜ L = tan −1 ⎜ = −55.8° ⎟ ⎝ R ⎠ ⎝ 60.0 Ω ⎟⎠ ! Since φ < 0, the voltage lags behind the current by 55.8°. (e) Adding an inductor will change the impedance and the current in the circuit. If the added inductive reactance is XL < 2XC, the impedance will be decreased and the current will increase. However, if XL > 2XC, the impedance will be increased and the current will decrease. 21.27

(a) XL = 2πfL = 2π(50.0 Hz)(0.150 H) = 47.1 Ω

1 1 = = 637 Ω (b) XC = 2π fC 2π ( 50.0 Hz ) ( 5.00 × 10−6 F ) ! ΔVmax 240 V = = 2.40 × 103 Ω = 2.40 kΩ (c) Z = −1 I max 1.00 × 10 A !

(d) Z = R2 + ( X L − XC ) !

R= !

R = Z 2 − ( X L − XC )

( 2.40 × 10 Ω) − ( 47.1 Ω − 637 Ω) = 2.33 × 10 Ω = 2.33 kΩ 3

Topic 21

1263

⎛ 47.1 Ω − 637 Ω ⎞ ⎛ X − XC ⎞ (e) φ = tan −1 ⎜ L = tan −1 ⎜ = −14.2° ⎟ 3 ⎝ R ⎠ ⎝ 2.33 × 10 Ω ⎟⎠ ! 21.28

XL = ωL and XC = 1/ωC, where ω = 2πf. At resonance (i.e., when XL = XC), we have

1 ω 0 = 2π f0 = LC !

f0 =

1 2π LC

With L = 57.0 µH and C = 57.0 µF, this resonance frequency is f0 = !

21.29

2π

( 57.0 × 10 H )( 57.0 × 10 F) −6

−6

= 2.79 × 103 Hz = 2.79 kHz

XL = 2πfL = 2π(50.0 Hz)(0.185 H) = 58.1 Ω

1 1 XC = = = 49.0 Ω 2π fC 2π ( 50.0 Hz ) ( 65.0 × 10−6 F ) !

Zad = R2 + ( X L − XC ) = ! 2

(

( 40.0 Ω) + ( 58.1 Ω − 49.0 Ω) = 41.0 Ω 2

)

ΔVmax 2 ΔVrms 150 V = = = 2.59 A and I rms = Zad Zad 41.0 Ω ) 2 ( ! (a) Zab = R = 40.0 Ω, so (ΔVrms)ab = Irms Zab = (2.59 A)(40.0 Ω) = 104 V (b) Zbc = XL = 58.1 Ω, and (ΔVrms)bc = Irms Zbc = (2.59 A)(58.1 Ω) = 150 V (c) Zcd = XC = 49.0 Ω, and (ΔVrms)cd = Irms Zcd = (2.59 A)(49.0 Ω) = 127 V

Topic 21

1264

(d) Zbd = |XL − XC| = 9.10 Ω, so (ΔVrms)bd = Irms Zbd = (2.59 A)(9.10 Ω) = 23.6 V

21.30

1 1 XC = = = 1.1 × 103 Ω −6 2π fC 2π ( 60 Hz ) ( 2.5 × 10 F ) !

Z = R2 + ( X L − XC ) = ! 2

(a) I max = !

(1.2 × 10 Ω) + ( 0 − 1.1 × 10 Ω) = 1.6 × 10 Ω 3

ΔVmax 170 V = = 0.11 A Z 1.6 × 103 Ω

(b) ΔVR,max = ImaxR = (0.11 A)(1.2 × 103 Ω) = 1.3 × 102 V ΔVC,max = ImaxXC = (0.11 A)(1.1 × 103 Ω) = 1.2 × 102 V (c) When the instantaneous current i is zero, the instantaneous voltage across the resistor is |ΔvR| = iR = 0. The instantaneous voltage across a capacitor is always 90°, or a quarter cycle, out of phase with the instantaneous current. Thus, when i = 0, |ΔvC| = ΔVC,max = 1.2 × 102 V, and qC = C(ΔvC) = (2.5 × 10−6 F)(1.2 × 102 V) = 3.0 × 10−4 C = 300 µC Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path. Thus, for this series circuit, Δvsource + ΔvR + ΔvC = 0, and at this instant when i = 0, we have |Δvsource| = 0 + |ΔvC| = ΔVC,max = 1.2 × 102 V. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 21

1265

(d) When the instantaneous current is a maximum (i = Imax), the instantaneous voltage across the resistor is |ΔvR| = |iR| = ImaxR = ΔVR,max = 1.3 × 102 V. Again, the instantaneous voltage across a capacitor is a quarter cycle out of phase with the current. Thus, when i = Imax, we must have |ΔvC| = 0 and qC = C|ΔvC| = 0. Then, applying Kirchhoff’s loop rule to the instantaneous voltages around the series circuit at the instant when i = Imax and |ΔvC| = 0 gives Δvsource + ΔvR + ΔvC = 0 ⇒

21.31

|Δvsource| = |ΔvR| = ΔVR,max = 1.3 × 102 V

ΔVrms 104 V = = 208 Ω (a) Z = I 0.500 A rms !

Pav 10.0 W 2 (b) Pav = I rms R!gives!R = 2 = 2 = 40.0 Ω I rms ( 0.500 A ) ! (c) Z = R2 + X L2 ,!so!X L = Z 2 − R2 = (208 Ω)2 − (40.0 Ω)2 = 204 Ω ! and

21.32

X 204 Ω L= L = = 0.541 H 2 π f 2 π (60.0 Hz) !

Given v = ΔVmax sin (ωt) = (90.0 V) sin (350t), observe that ΔVmax = 90.0 V and ω = 350 rad/s. Also, the net reactance is XL − XC = 2πfL − 1/2πfC = ωL − 1/ωC. (a)

Topic 21

1266

1 1 X L − XC = ω L − = ( 350 rad/s ) ( 0.200 H ) − = −44.3 Ω ωC 350 rad/s ) ( 25.0 × 10−6 F ) ( !

so the impedance is

Z = R2 ( X L − XC ) = ! 2

(b) I rms = !

ΔVrms ΔVmax = Z Z

( 50.0 Ω) + ( −44.3 Ω) = 66.8 Ω 2

90.0 V = 0.953 A 2 ( 66.8 Ω )

⎛ −44.3 Ω ⎞ ⎛ X − XC ⎞ φ = tan −1 ⎜ L = tan −1 ⎜ = −41.5° ⎟ ⎝ R ⎠ ⎝ 50.0 Ω ⎟⎠ ! so the average power delivered to the circuit is ⎛ ΔV ⎞ ⎛ 90.0 V ⎞ Pav = I rms ΔVrms cos φ = I rms ⎜ max ⎟ cos φ = ( 0.953 A ) ⎜ ⎟ cos ( −41.5° ) ⎝ ⎝ 2 ⎠ 2 ⎠ = 45.4 W

21.33 (a) From cos φ = 0.707, the phase angle φ = ±cos−1(0.707). Choose the negative value so that XC is positive: tan φ =

XL − XC R

(

)

XC = XL − R tan φ = 175 Ω − ( 225 Ω ) tan − cos−1 ( 0.707 ) = 400 Ω

(b) For cos φ = 1.00, φ = 0° so that, as before,

Topic 21

1267

XC = XL − R tan φ = 175 Ω − ( 225 Ω ) tan 0° = 175 Ω (c) For cos φ = 1.00 × 10−2, the phase angle φ = ±(1.00 × 10−2). Choose the negative value so that XC is positive:

(

))

(

XC = XL − R tan φ = 175 Ω − ( 225 Ω ) tan − cos−1 1.00 × 10−2 = 2.27 × 10 4 Ω

21.34

The rms current in the circuit is I rms = !

ΔVrms 160 V = = 2.00 A Z 80.0 Ω

and the average power delivered to the circuit is

(

)

(

)

2 Pav = I rms ΔVrms cosφ = I rms ΔVR,rms = I rms I rms R = I rms R

(

= 2.00 A

21.35

) (22.0 Ω) = 88.0 W 2

Pav 14 W 2 = = 0.28 A (a) !Pav = I rms R = I rms ( I rms R ) = I rms ( ΔVR,rms ) , so I rms = ΔV 50 V R,rms !

Thus,

R= !

ΔVR,rms 50 V = = 1.8 × 102 Ω I rms 0.28 A

(b) Z = R2 + X L2 , which yields ! 2

⎛ 90 V ⎞ ⎛ ΔV ⎞ 2 X L = Z − R = ⎜ rms ⎟ − R2 = ⎜ − (1.8 × 102 Ω ) = 2.7 × 102 Ω ⎟ ⎝ I rms ⎠ ⎝ 0.28 A ⎠ ! 2

XL 2.7 × 102 Ω L = = = 0.72 H and 2π f 2π ( 60 Hz ) ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 21

21.36

1268

XL = 2πfL = 2π(600 Hz)(6.0 × 10−3 H) = 23 Ω

1 1 XC = = = 11 Ω 2π fC 2π ( 600 Hz ) ( 25 × 10−6 F ) !

Z = R2 + ( X L − XC ) = ! 2

( 25 Ω) + ( 23 Ω − 11 Ω) = 28 Ω 2

⎛ 10 V ⎞ ⎛ ΔVrms ⎞ ( 25 Ω) = 8.9 V (a) ΔVR,rms = I rms R = ⎜⎝ ⎟⎠ R = ⎜ Z ⎝ 28 Ω ⎟⎠ ! ⎛ 10 V ⎞ ⎛ ΔV ⎞ ΔVL,rms = I rms X L = ⎜ rms ⎟ X L = ⎜ ( 23 Ω) = 8.2 V ⎝ Z ⎠ ⎝ 28 Ω ⎟⎠ ! ⎛ 10 V ⎞ ⎛ ΔV ⎞ ΔVC ,rms = I rms XC = ⎜ rms ⎟ XC = ⎜ (11 Ω) = 3.9 V ⎝ Z ⎠ ⎝ 28 Ω ⎟⎠ ! No, ΔVR,rms + ΔVL,rms + ΔVC,rms = 8.9 V + 8.2 V + 3.8 V = 20.9 V ≠ 10 V (b) The power delivered to the resistor is the greatest. No power losses occur in an ideal capacitor or inductor. 2

(c)

21.37

Pav = I !

2 rms

2 ⎛ 10 V ⎞ ⎛ ΔVrms ⎞ R=⎜ R=⎜ ( 25 Ω) = 3.2 W ⎟ ⎝ Z ⎠ ⎝ 28 Ω ⎟⎠

(

)

The resonance frequency of a series RLC circuit is f0 = 1 2π LC . Thus, ! if L = 1.40 µH and the desired resonance frequency is f0 = 99.7 MHz, the needed capacitance is

Topic 21

1269

C= ! 21.38

1 1 = = 1.82 × 10−12 F = 1.82 pF 2 2 2 6 −6 4π f0 L 4π ( 99.7 × 10 Hz ) (1.40 × 10 H ) 2

(

)

The resonance frequency of a series RLC circuit is f0 = 1 2π LC . Thus, ! the ratio of the resonance frequencies when the same inductance is used with two different capacitances in the circuit is

⎞ ⎛ 2π LC1 ⎞ f0,2 ⎛ 1 C1 =⎜ = ⎟ ⎜ ⎟ f0,1 ⎝ 2π LC2 ⎠ ⎝ 1 C2 ⎠

If f0,1 = 2.84 kHz when C1 = 6.50 µF, the resonance frequency when the capacitance is C2 = 9.80 µF will be

f0,2 = f0,1 ! f0 =

21.39 !

1 2π LC

C1 6.50 µF = ( 2.84 kHz ) = 2.31 kHz C2 9.80 µF

,!so!C =

1 4π f02 L 2

For f0 = (f0)min = 500 kHz = 5.00 × 105 Hz

C = Cmax = !

4π ( 5.00 × 10 Hz ) ( 2.0 × 10 H ) 2

−6

= 5.1 × 10−8 F = 51 nF

For f0 = (f0)max = 1 600 kHz = 1.60 × 106 Hz

C = Cmin = !

4π (1.60 × 10 Hz ) ( 2.0 × 10 H ) 2

−6

= 4.9 × 10−9 F = 4.9 nF

Topic 21

1270

21.40 (a) The resonant frequency of an RLC circuit is f0 =

1 2π LC

2π

(1.25 × 10 H )( 47.0 × 10 F ) −6

−9

= 6.57 × 105 Hz

(b) Solve for the required capacitance to find f0 =

21.41

1 2π LC 1 1 = = 1.15 × 10−8 F 2 2 2 2 −6 6 4π Lf0 4π 1.25 × 10 H 1.33 × 10 Hz

(

)(

)

(

)

The resonance frequency of an LC circuit is f0 = 1 2π LC . If the ! capacitance remains constant while the inductance increases by 1.000%, the new resonance frequency will be

f 0′ = !

1 1 f0 ⎛ 1 ⎞ =⎜ = ⎟ 2π (1.010L)C ⎝ 1.010 ⎠ 2π LC 1.010

The beat frequency detected in this case will be ! f beat = f0 − f0′, or 1 ⎞ 1 ⎞ ⎛ ⎛ f beat = ⎜ 1 − f0 = ⎜ 1 − ⎟ ⎟⎠ (725 kHz ) = 3.60 kHz ⎝ ⎠ ⎝ 1.010 1.010 !

21.42

The resonance frequency is !ω 0 = 2π f0 = 1

LC . Also, XL = ωL and XC =

1/ωC. (a) At resonance,

Topic 21

1271

⎛ 1 ⎞ 3.00 H L XC = X L = ω 0 L = ⎜ = = 1 000 Ω. ⎟L= C 3.00 × 10−6 F ⎝ LC ⎠ Thus, !Z = R2 + 02 = R , Irms = ΔVrms/Z = 120 V/30.0 Ω = 4.00 A, and 2 Pav = I rms R = ( 4.00 A ) ( 30.0 Ω ) = 480 W . ! 2

1 1⎛ ⎞ ⎛ ⎞ (b) At ω = ω 0 ; X L = ⎜ X L ⎟ = 500 Ω, XC = 2 ⎜ XC ⎟ = 2 000 Ω, ω0 ⎠ ⎝ 2 2 ⎝ ω0 ⎠

(

Z = R 2 + X L − XC

and I rms =

120 V 1 500 Ω

(

)

(

) ( 2

30.0 Ω + 500 Ω − 2 000 Ω

) = 1 500 Ω 2

= 0.080 0 A. Thus,

2 Pav = I rms R = 0.080 0 A

) (30.0 Ω ) = 0.192 W . 2

1 1⎛ ⎞ ⎛ ⎞ (c) At ω = ω 0 ; X L = ⎜ X L ⎟ = 250 Ω, XC = 4 ⎜ XC ⎟ = 4 000 Ω, ω0 ⎠ ⎝ 4 4 ⎝ ω0 ⎠

Z = 3 750 Ω, and I rms =

120 V 3 750 Ω

= 0.032 0 A.

Therefore,

(

2 Pav = I rms R = 0.032 0 A

) (30.0 Ω ) = 3.07 × 10 W = 30.7 mW . 2

−2

1⎛ ⎛ ⎞ ⎞ (d) At ω = 2ω0; X L = 2 ⎜ X L ⎟ = 2 000 Ω, XC = ⎜ XC ⎟ = 500 Ω, ω0 ⎠ ⎝ ω0 ⎠ 2⎝

Topic 21

1272

Z = 1 500 Ω, and I rms =

(

2 Pav = I rms R = 0.080 0 A

120 V 1 500 Ω

= 0.080 0 A. Then,

) (30.0 Ω ) = 0.192 W . 2

1⎛ ⎛ ⎞ ⎞ (e) At ω = 4ω0; X L = 4 ⎜ X L ⎟ = 4 000 Ω, XC = ⎜ XC ⎟ = 250 Ω, ω0⎠ ⎝ ω0⎠ 4⎝

Z = 3 750 Ω, and I rms =

(

120 V 3 750 Ω

2 Pav = I rms R = 0.032 0 A

= 0.032 0 A. Hence,

) (30.0 Ω ) = 3.07 × 10 W = 30.7 mW 2

−2

The power delivered to the circuit is a maximum when the rms current is a maximum. This occurs when the frequency of the source is equal to the resonance frequency of the circuit. 21.43

The maximum output voltage (ΔVmax)2 is related to the maximum input

(

)

N2 ( ΔVmax )1 , where N1 and voltage (ΔVmax)1 by the expression ΔVmax 2 = N 1 !

N2 are the number of turns on the primary coil and the secondary coil, respectively. Thus, for the given transformer,

500 ( ΔV ) = 1250 (170 V ) = 1.02 × 10 V 3

max 2

and the rms voltage across the secondary is

Topic 21

1273

( ΔVrms )2 =

21.44

( ΔVmax )2 = 1.02 × 103 V = 721 V . 2

(a) The output voltage of the transformer is

⎛N ⎞ ⎛ 1⎞ ΔV2,rms = ⎜ 2 ⎟ ΔV1,rms = ⎜ ⎟ (120 V ) = 9.23 V ⎝ 13 ⎠ ⎝ N1 ⎠ ! (b) Assuming an ideal transformer, Poutput = Pinput, and the power delivered to the CD player is (Pav)2 = (Pav)1 = I1,rms(ΔV1,rms) = (0.250 A)(120 V) = 30.0 W 21.45

The power input to the transformer is (Pav)input = (ΔV1,rms)I1,rms = (3 600 V)(50 A) = 1.8 × 105 W For an ideal transformer, (Pav)output = (ΔV2,rms)I2,rms = (Pav)input, so the current in the long-distance power line is

I 2,rms

P ) ( 1.8 × 10 W = = = 1.8 A ( ΔV ) 100 000 V 5

av input 2,rms

The power dissipated as heat in the line is then

P = I 2 R = (1.8 A ) (100 Ω ) = 3.2 × 102 W ! lost 2,rms line 2

The percentage of the power delivered by the generator that is lost in the line is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 21

1274

⎛ 3.2 × 102 W ⎞ P % Lost = lost × 100% = ⎜ × 100% = 0.18% 5 Pinput ⎝ 1.8 × 10 W ⎟⎠ ! 21.46 (a) The transformer’s power is P = IinputΔVinput = IoutputΔVoutput. Substitute values to find

(

)

P = Iout ΔVout = 30.0 × 10−3 A ( 9250 V ) = 278 W

(b) Solve for the input current to find

(

)

30.0 × 10−3 A ( 9250 V ) Iout ΔVout I input = = = 2.41 A ΔVinput 115 V 21.47

(a) At 90% efficiency, (Pav)output = 0.90(Pav)input. Thus, if (Pav)output = 1 000 kW, the input power to the primary is

( ) Pav

(b) I1,rms =

(c)

21.48

I 2,rms =

input

(P )

av input

ΔV1,rms

(P )

av output

0.90

av output

ΔV2,rms

1 000 kW 0.90

1.1 × 103 kW

ΔV1,rms

1 000 kW ΔV2,rms

= 1.1 × 103 kW

1.1 × 106 W 3 600 V

1.0 × 106 W 120 V

= 3.1 × 102 A

= 8.3 × 103 A

Rline = (4.50 × 10−4 Ω/m)(6.44 × 105 m) = 290 Ω (a) The power transmitted is (Pav)transmitted = (ΔVrms)Irms, so

Topic 21

1275

I rms = !

( Pav )transmitted ΔVrms

5.00 × 106 W = 10.0 A 500 × 103 V

2 4 Thus, ( Pav )loss = I rms Rline = (10.0 A ) ( 290 Ω ) = 2.90 × 10 W = 29.0 kW . ! 2

(b) The power input to the line is (Pav)input = (Pav)transmitted + (Pav)loss = 5.00 × 106 W + 2.90 × 104 W = 5.03 × 106 W and the fraction of input power lost during transmission is

(P ) fraction = (P )

av loss

av input

2.90 × 10 4 W 5.03 × 106 W

= 0.005 77 or 0.577%

(c) It is impossible to deliver the needed power at the generator voltage of 4.50 kV. The maximum line current with an input voltage of 4.50 kV to the line occurs when the line is shorted out at the customer’s end, and this current is

(I )

rms max

ΔVrms Rline

4 500 V 290 Ω

= 15.5 A

The maximum input power is then (Pinput)max = (ΔVrms)(Irms)max = (4.50 × 103 V)(15.5 A) = 6.98 × 104 W = 69.8 kW © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 21

1276

This is far short of meeting the customer’s request, and all of this power is lost in the transmission line. 21.49

From v = λf, the wavelength is

v 3.00 × 108 m/s λ= = = 4.00 × 106 m = 4!000 km f 75 Hz ! The required length of the antenna is then L = λ/4 = 1 000 km, or about 621 miles. Not very practical at all.

21.50

⎞ 1y d 6.44 × 1018 m ⎛ = 6.80 × 102 y (a) t = = 8 7 ⎟ ⎜ c 3.00 × 10 m/s ⎝ 3.156 × 10 s ⎠ ! (b) From Table C.4 (in Appendix C of the textbook), the average EarthSun distance is d = 1.496 × 1011 m, giving the transit time as

d 1.496 × 1011 m ⎛ 1 min ⎞ t= = = 8.31 min c 3.00 × 108 m/s ⎜⎝ 60 s ⎟⎠ ! (c) Also from Table C.4, the average Earth-Moon distance is d = 3.84 × 108 m, giving the time for the round trip as 8 2d 2 ( 3.84 × 10 m ) t= = = 2.56 s 8 c 3.00 × 10 m/s !

21.51

(a) The solar energy incident each second on 1.00 m2 of the surface of Earth’s atmosphere is

Topic 21

1277

U total = 1 370

J/s

N⋅m

m 2 ⋅s

= 1 370 2

= 1 370

N m ⋅s

Of this, 38.0% is reflected and 62.0% is absorbed. From Equation 21.29 and Equation 21.30 in the textbook, the radiation pressures P1 due to the reflected radiation and P2 due to the absorbed radiation are given by

P1 = !

and

2 Ureflected 2 ( 0.380 Utotal ) 0.760 Utotal = = c c c

P2 = !

Uabsorbed 0.620 Utotal = c c

The total radiation pressure is then

Prad = P1 + P2 = !

Prad =

(

( 0.760 + 0.620)Utotal c

1.38 1 370 N/m ⋅s 8

3.00 × 10 m/s

) = 6.30 × 10 N m = 6.30 × 10 Pa −6

−6

(b) In comparison, atmospheric pressure at the surface of the Earth is

Patm 101 × 103 Pa = = 1.60 × 1010 times!greater!than!the!radiation!pressure −6 P 6.30 × 10 Pa ! rad 21.52 Solve the definition of c for the permittivity of free space to find

Topic 21

1278

µ0É 0

→ É0 =

1 1 = 2 c µ0 ( 299 792 458 m/s) 4π × 10−7 N ⋅s2 /C 2

(

É 0 = 8.854 187 8 × 10−12 C 2 / N ⋅ m 2 = 21.53

)

(a) The frequency of an electromagnetic wave is f = c/λ, where c is the speed of light and λ is the wavelength of the wave. The frequencies of the two light sources are then

fred =

Red:

c 3.00 × 108 m s = = 4.55 × 1014 Hz −9 λred 660 × 10 m

and

c 3.00 × 108 m/s fIR = = = 3.19 × 1014 Hz −9 λIR 940 × 10 m !

Infrared:

(b) The intensity of an electromagnetic wave is proportional to the square of its amplitude. If 67% of the incident intensity of the red light is absorbed, then the intensity of the emerging wave is (100% − 67%) = 33% of the incident intensity, or If = 0.33Ii. Hence, we must have Emax, f !

21.54

Emax,i

If Ii

= 0.33 = 0.57

If I0 is the incident intensity of a light beam, and I is the intensity of the

Topic 21

1279

beam after passing through length L of a fluid having concentration C of absorbing molecules, the Beer-Lambert law states that log10 (I/I0) = −eCL, where e is a constant. For 660-nm light, the absorbing molecules are oxygenated hemoglobin. Thus, if 33% of this wavelength light is transmitted through blood, the concentration of oxygenated hemoglobin in the blood is

CHBO = 2

− log10 (0.33)

[1]

The absorbing molecules for 940-nm light are deoxygenated hemoglobin, so if 76% of this light is transmitted through the blood, the concentration of these molecules in the blood is

CHB =

( )

− log 10 0.76

[2]

Dividing Equation [2] by Equation [1] gives the ratio of deoxygenated hemoglobin to oxygenated hemoglobin in the blood as CHB CHBO

= 2

log 10 (0.76) log 10 (0.33)

= 0.25 or CHB = 0.25CHBO

Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is necessary that CHB + CHBO = 1.00 , and we now have 2

Topic 21

1280

0.25CHBO + CHBO = 1.0 . The fraction of hemoglobin that is oxygenated in 2

this blood is then

CHBO = 2

1.0 1.0 + 0.25

= 0.80

80%

Someone with only 80% oxygenated hemoglobin in the blood is probably in serious trouble, needing supplemental oxygen immediately.

21.55

From Intensity = !

2 Emax Emax Bmax = c, we find Intensity = c Bmax . and 2 µ0 2 µ0 !Bmax !

Thus,

Bmax =

( ) (1 370 W/m ) = 3.39 × 10 T Intensity ) = ( c 3.00 × 10 m/s

2 µ0

2 4π × 10−7 T ⋅ m/A

−6

and Emax = Bmaxc = (3.39 × 10−6 T)(3.00 × 108 m/s) = 1.02 × 103 V/m 21.56

(a) To exert an upward force on the disk, the laser beam should be aimed vertically upward, striking the lower surface of the disk. To just levitate the disk, the upward force exerted on the disk by the beam should equal the weight of the disk. The momentum that electromagnetic radiation of intensity I, incident normally on a perfectly reflecting surface of area A, delivers to that surface in time Δt is given by Equation 21.30 in the textbook as Δp =

Topic 21

1281

2U/c = 2(IAΔt)/c. Thus, from the impulse-momentum theorem, the average force exerted on the reflecting surface is F = Δp/Δt = 2IA/c. To just levitate the surface, F = 2IA/c = mg, and the required intensity of the incident radiation is I = mgc/2A. (b) −3 2 8 mgc mgc ( 5.00 × 10 kg ) ( 9.80 m s ) ( 3.00 × 10 m s ) I= = = = 1.46 × 109 W m 2 2 2 −2 2A 2π r 2π ( 4.00 × 10 m ) !

(c) Propulsion by light pressure in a significant gravity field is impractical because of the enormous power requirements. In addition, no material is perfectly reflecting, so the absorbed energy would melt the reflecting surface. 21.57

The distance between adjacent antinodes in a standing wave is λ/2. Thus, λ = 2(6.00 cm) = 12.0 cm = 0.120 m, and c = λf = (0.120 m)(2.45 × 109 Hz) = 2.94 × 108 m/s

21.58

(a) The intensity of the radiation at distance r = 20.0 ly from the star is

P 4π r

= 2

(

4.00 × 1028 W

)(

)

4π ⎡ 20.0 1y 9.461 × 1015 m/ly ⎤ ⎣ ⎦

= 8.89 × 10−8 W m 2

(b) The power of the starlight intercepted by Earth, with cross-sectional © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 21

1282

area of !Acs = π RE , is 2

P = IAcs = I ⋅ π RE2 = ( 8.89 × 10−8 W m 2 ) π ( 6.38 × 106 m ) ! or

21.59

P = 1.14 × 107 W = 11.4 MW

c 3.00 × 108 m/s = 6.00 × 10−12 m = 6.00 pm (a) λ = = 19 f 5.00 × 10 Hz ! c 3.00 × 108 m/s = 7.50 × 10−2 m = 7.50 cm (b) λ = = 9 f 4.00 × 10 Hz !

21.60

c 3.00 × 108 m/s λ= = = 11.0 m 6 f 27.33 × 10 Hz !

21.61

(a) For the AM band,

c 3.00 × 108 m/s λmin = = = 188 m 3 f 1600 × 10 Hz max ! c 3.00 × 108 m/s λmax = = = 556 m 3 f 540 × 10 Hz min ! (b) For the FM band,

c 3.00 × 108 m/s λmin = = = 2.78 m 6 f 108 × 10 Hz max ! c 3.00 × 108 m/s λmax = = = 3.4 m 6 f 88 × 10 Hz min ! 21.62

The transit time for the radio wave is

Topic 21

1283

d 100 × 103 m tR = R = = 3.33 × 10−4 s = 0.333 ms 8 c 3.00 × 10 m/s ! and that for the sound wave is d 3.0 m ts = s = = 8.7 × 10−3 s = 8.7 ms vsound 343 m/s !

Thus, the radio listeners hear the news 8.4 ms before the studio audience because radio waves travel so much faster than sound waves. 21.63 Use the relation c = λ f to find f = c/λ and (a) λ = 655 nm → f = c/ λ = 4.58 × 1014 Hz (b) λ = 515 nm → f = c/ λ = 5.82 × 1014 Hz (c) λ = 475 nm → f = c/ λ = 6.31 × 1014 Hz 21.64 (a) Since the space station and the ship are moving toward one another, the frequency after being Doppler shifted is fO = fS(1 + u/c), where u is the relative speed between the observer and source. Thus,

⎡ 1.800 0 × 105 m/s ⎤ 14 fO = 6.000 0 × 10 Hz ⎢1 + ⎥ = 6.003 6 × 10 Hz 8 ⎢⎣ 3.000 0 × 10 m/s ⎥⎦

(

)

(b) The change in frequency is Δf = fO − fS = 6.003 6 × 1014 Hz − 6.000 0 × 1014 = 3.6 × 1011 Hz

Topic 21

1284

21.65 Convert units to find u = 65 mph = 29.1 m/s. The moving vehicle receives

u⎞ ⎛ waves with a frequency fvehicle ≈ fS ⎜ 1 + ⎟ and then acts as a moving ⎝ c⎠ source, reradiating the signal so that the stationary police radar receives the waves at a frequency 2

u⎞ u⎞ 2u ⎞ ⎛ ⎛ ⎛ fO ≈ fvehicle ⎜ 1 + ⎟ = fS ⎜ 1 + ⎟ ≈ fS ⎜ 1 + ⎟ ⎝ ⎝ ⎝ c⎠ c⎠ c ⎠

The frequency shift is then

Δf = fO − fS = 2 fS

u ⎛ 29.1 m/s ⎞ = 2 24.0 × 109 Hz ⎜ = 4.66 × 103 Hz ⎟ ⎝ ⎠ c c

(

)

Alternatively, realize that the radar signal undergoes two frequency shifts. One occurs when it is absorbed by the car and the second occurs when it is reradiated by a moving source. Each frequency shift is approximately fS(u/c) for a total shift of 2 fS(u/c). 21.66

The driver was driving toward the warning lights, so the correct form of the Doppler shift equation is fO = fS(1 + u/c). The frequency emitted by the yellow warning light is

fS =

λS

3.00 × 108 m/s 580 × 10 m −9

= 5.17 × 1014 Hz

and the frequency the driver claims that she observed is

Topic 21

1285

fO =

λO

3.00 × 108 m/s 560 × 10 m −9

= 5.36 × 1014 Hz

The speed with which she would have to approach the light for the Doppler effect to yield this claimed shift is

⎛f ⎞ ⎛ 5.36 × 1014 Hz ⎞ 8 O u = c ⎜ − 1⎟ = 3.00 × 10 m/s ⎜ − 1⎟ = 1.1 × 107 m/s 14 ⎝ 5.17 × 10 Hz ⎠ ⎝ fS ⎠

(

21.67

)

The energy incident on the mirror in time Δt is U = Plaser ⋅ Δt, where Plaser is the power transmitted by the laser beam Plaser = 25.0 × 10−3 W = 25.0 × 10−3 J/s = 25.0 × 10−3 N ⋅ m/s From Equation 21.30 in the textbook, the rate of change in the momentum of the mirror as the beam reflects from it is Δp 2U/c 2Plaser = = Δt c ! Δt

The impulse-momentum theorem then gives the force exerted on the mirror as F = Δp/Δt = 2Plaser/c, and the radiation pressure on the mirror is F 2P c 2P Prad = = laser = laser A A cA !

where A = πr2 = πd2/4 is the area of the mirror illuminated (i.e., the crosssectional area of the laser beam). Thus,

Topic 21

1286

Prad = !

Prad = 5.31 × 10−5 N/m2 = 5.31 × 10−5 Pa

or 21.68

8 ( 25.0 × 10−3 N ⋅ m/s ) 2Plaser 8Plaser = = 2 c (π d 2 4 ) π cd 2 π ( 3.00 × 108 m/s ) ( 2.00 × 10−3 m )

Suppose you cover a 1.7 m-by-0.3 m section of beach blanket. Suppose the elevation angle of the Sun is 60°. Then the target area you fill in the Sun’s field of view is (1.7 m)(0.3 m) cos 30° = 0.4 m2. The intensity the radiation at Earth’s surface is Isurface = 0.6Iincoming, and only 50% of this is absorbed. Since I = Pav/A = (ΔE/Δt)/A, the absorbed energy is

) ( ) ( ) ( ) = (0.5)(0.6) (1 370 W/m ) ( 0.4 m ) ( 3 600 s ) = 6 × 10 J (

ΔE = 0.5I surface A Δt = ⎡⎢ 0.5 0.6I incoming ⎤⎥ A Δt ⎣ ⎦ 2

Z = R2 + ( XC ) = R2 + ( 2π fC ) 2

21.69

∼ 106 J

−2

( 200 Ω) + ⎡⎣ 2π ( 60 Hz )( 5.0 × 10 F)⎤⎦ 2

−6

−2

= 5.7 × 102 Ω

Thus, 2

Pav = I !

2 rms

2 ⎛ 120 V ⎞ ⎛ ΔVrms ⎞ R=⎜ R=⎜ ( 200 Ω) = 8.9 W = 8.9 × 10−3 kW ⎟ 2 ⎝ Z ⎠ ⎝ 5.7 × 10 Ω ⎟⎠

and cost = ΔE ⋅ (rate) = Pav ⋅ Δt ⋅ (rate) = (8.9 × 10−3 kW)(24 h)(8.0 cents/kWh) = 1.7 cents 21.70

For a parallel-plate capacitor, C = eA/d: cutting the plate separation d in

Topic 21

1287

half will double the capacitance (Cf = 2Ci). Since the capacitive reactance is XC = 1/(2πfC), doubling the capacitance reduces the capacitive reactance by a factor of 2, giving XC,f = XC,i/2. The current in the circuit is Irms = ΔVrms/Z, so the impedance must be cut in half (Zf = Zi/2) if the current doubles while the applied voltage remains constant. With an inductive reactance equal to the resistance (XL = R), we then have

(

R2 + R − XC , f

) = 12 R + ( R − X ) 2

C ,i

4[R2 + (R − XC,i/2)2] = R2 + (R − XC,i)2

Simplifying gives 6R2 = 2RXC,i, and XC,i = 6R2/2R = 3R.

21.71

R= !

( ΔV )DC I DC

12.0 V = 19.0 Ω 0.630 A

ΔVrms 24.0 V Z = R2 + (2π fL)2 = = = 42.1 Ω I rms 0.570 A !

(42.1 Ω)2 − (19.0 Ω)2 Z 2 − R2 = = 9.97 × 10−2 H = 99.7 mH Thus, L = 2 π f 2 π (60.0 Hz) !

21.72

(a) The frequency of a 3.0-cm radar wave, and hence the desired resonance frequency of the tuning circuit, ! f0 = 1/2π LC , is

f0 = f = !

c 3.00 × 108 m/s = = 1.0 × 1010 Hz λ 3.00 × 10−2 m

Topic 21

1288

Therefore, the required capacitance is

( 2π f ) L ( 2π × 10 Hz ) ( 400 × 10 2

! (b) C =

ℓ=

!0 A d

C ⋅d

! 0ℓ2 d

−12

= 6.3 × 10−13 F = 0.63 pF

, so

(6.3 × 10 F)(1.0 × 10 m) = 8.4 × 10 m = 8.4 mm = −13

−3

8.85 × 10

−12

C N⋅m

21.73

Emax Emax 0.20 × 10−6 V/m = c,so!B = = = 6.7 × 10−16 T (a) max 8 c 3.00 × 10 m/s !Bmax (b) Intensity = !

−6 −16 Emax Bmax ( 0.20 × 10 V/m ) ( 6.7 × 10 T ) = = 5.3 × 10−17 W m 2 −7 2 µ0 2 ( 4π × 10 T ⋅ m/A )

⎡ π d2 ⎤ (c) Pav = (Intensity) ⋅ A = (Intensity) ⎢ ⎥ !⎣ 4 ⎦ = ( 5.3 × 10 ! 21.74

−17

⎡ π ( 20.0 m )2 ⎤ W m )⎢ ⎥ = 1.7 × 10−14 W 4 ⎢⎣ ⎥⎦ 2

ΔVrms 12 V = = 6.0 Ω (a) Z = I 2.0 A rms ! ΔVDC 12 V = = 4.0 Ω (b) R = I 3.0 A DC !

Topic 21

1289

From Z = R2 + X L2 = R2 + ( 2π fL) , we find ! 2

Z 2 − R2 L= = 2π f ! 21.75

( 6.0 Ω) − ( 4.0 Ω) = 1.2 × 10 H = 12 mH 2π ( 60 Hz ) 2

−2

(a) From Equation 21.30 in the textbook, the momentum imparted in time Δt to a perfectly reflecting sail of area A by normally incident radiation of intensity I is Δp = 2U/c = 2(IAΔt)/c. From the impulse-momentum theorem, the average force exerted on the sail is then

Fav =

Δp Δt

( =

(b) aav =

2 IAΔt

Fav m

Δt

) c = 2IA = 2(1 340 W m )(6.00 × 10 m ) = 0.536 N

0.536 N 6 000 kg

3.00 × 108 m/s

= 8.93 × 10−5 m s 2

1 2 (c) From Δx = v0t + at , with v0 = 0, the time is 2 !

2 ( 3.84 × 108 m ) ⎛ ⎞ 2 ( Δx ) 1d t= = = ( 2.93 × 106 s ) ⎜ = 33.9 d −5 2 4 ⎟ aav 8.93 × 10 m s ⎝ 8.64 × 10 s ⎠ ! 21.76

(a) The intensity of radiation at distance r from a point source, which radiates total power P, is I = P/A = P/4πr2. Thus, at distance r = 2.0 in from a cell phone radiating a total power of P = 2.0 W = 2.0 × 103 mW,

Topic 21

1290

the intensity is

I= !

2.0 × 103 mW

4π ⎡⎣( 2.0 in ) ( 2.54 cm/1 in ) ⎤⎦

= 6.2 m W cm 2

This intensity is 24% higher than the maximum allowed leakage from a microwave at this distance of 2.0 inches. (b) If when using a Bluetooth headset (emitting 2.5 mW of power) in the ear at distance rh = 2.0 in = 5.1 cm from the brain, the cell phone (emitting 2.0 W of power) is located in the pocket at distance rp = 1.0 m = 1.0 × 102 cm from the brain, the total radiation intensity at the brain is

I total = I phone + I headset = !

2.0 × 103 mW

4π (1.0 × 102 cm )

2.5 mW

4π ( 5.1 cm )

I total = 1.6 × 10−2 = 2.4 × 10

mW 2

cm −2 mW cm

+ 7.6 × 10−3

mW cm 2

= 0.024 mW cm 2

Topic 22

1291

Topic 22 Reflection and Refraction of Light

QUICK QUIZZES 22.1

Choice (a). In part (a), you can see clear reflections of the headlights and the lights on the top of the truck. The reflection is specular. In part (b), although bright areas appear on the roadway in front of the headlights, the reflection is not as clear, and no separate reflection of the lights from the top of the truck is visible. The reflection in part (b) is mostly diffuse.

22.2

Beams 2 and 4 are reflected; beams 3 and 5 are refracted.

22.3

Choice (b). When light goes from one material into one having a higher index of refraction, it refracts toward the normal line of the boundary between the two materials. If, as the light travels through the new material, the index of refraction continues to increase, the light ray will refract more and more toward the normal line.

22.4

Choice (c). Both the wave speed and the wavelength decrease as the index of refraction increases. The frequency is unchanged.

Topic 22

1292

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 22.2

(a) D. Snell’s law is n1 sinθ1 = n2 sinθ2. As n increases, the angle θ decreases to keep the product n sinθ constant. (b) D. The speed of light in a medium is v = c/n = λf. As n increases, the speed and the wavelength decrease (while the frequency remains unchanged). (c) U. As light travels from one medium to another, its frequency doesn’t change. (d) D. The speed of light in a medium is v = c/n. As n increases, the speed decreases. (e) U. The photon energy is E = hf. Because the frequency is unchanged, the photon energy is unchanged.

22.4

Moving from A to B, the ray bends toward the normal so that nB > nA. Moving from A to C, the ray bends away from the normal so that nA > nC. The correct ranking is then nB > nA > nC as in choice (d).

22.6

The upright image of the hill is formed by light that has followed a direct path from the hill to the eye of the observer. The second image is a result

Topic 22

1293

of refraction in the atmosphere. Some light is reflected from the hill toward the water. As this light passes through warmer layers of air directly above the water, it is refracted back up toward the eye of the observer, resulting in the observation of an inverted image of the hill directly below the upright image. 22.8.

The index of refraction of water is 1.333, quite different from that of air, which has an index of refraction of about 1. The boundary between the air and water is therefore easy to detect, because of the differing refraction effects above and below the boundary. (Try looking at a glass half full of water.) The index of refraction of liquid helium, however, happens to be much closer to that of air. Consequently, the refractive differences above and below the helium-air boundary are harder to see.

22.10 Total internal reflection requires that nB < nA so that (a) is true and (b) is false. (c) is false because only rays at the critical angle travel along the interface (rays at a more glancing angle are reflected back into medium A). (d) is true because total internal reflection can only occur for rays traveling into a region with a lower index of refraction. Collecting the true statements gives the answer as: (a) and (d).

Topic 22

22.12

1294

With no water in the cup, light rays from the coin do not reach the eye because they are blocked by the side of the cup. With water in the cup, light rays are bent away from the normal as they leave the water so that some reach the eye. In figure (a) below, ray a is blocked by the side of the cup so it cannot enter the eye, and ray b misses the eye. In figure (b), ray a is still blocked by the side of the cup, but ray b refracts at the water’s surface so that it reaches the eye. Ray b seems to come from position B, directly above the coin at position A.

(a)

(b)

22.14 Water and air have different indices of refraction, with nwater ≈ 4nair/3. In passing from one of these media into the other, light will be refracted (deviated in direction) unless the angle of incidence is zero (in which case, the angle of refraction is also zero). Thus, rays B and D cannot be correct. In refraction, the incident ray and the refracted ray are never on the same side of the line normal to the surface at the point of contact, so ray A © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 22

1295

cannot be correct. Also in refraction, the ray makes a smaller angle with the normal in the medium having the highest index of refraction. Therefore, ray E cannot be correct, leaving only ray C as a likely path. Choice (c) is the correct answer.

ANSWERS TO EVEN NUMBERED PROBLEMS 22.2

22.4.

(a)

1.99 × 10−15 J

(b)

12.4 keV

(c)

decreased

(d)

increased

(a)

λ = hc/E

(b)

Higher energy photons have shorter

wavelengths. 22.6

(a)

2.25 × 108 m/s

(b)

1.97 × 108 m/s

22.8

(a)

1.94 m

(b)

50.0° above the horizontal

22.10

See Solution.

22.12

(a) The angle of refraction increases with wavelength, so the longest

(c)

1.24 × 108 m/s

wavelength deviates the least from the original path. (b) 22.14

80.0°

22.16

(a)

λ = 400 nm, θr = 16.0°; λ = 500 nm, θr = 16.1°; λ = 650 nm, θr = 16.3°

(b)

A, B, and C

Topic 22

22.18

1296

(a)

θr,top = 19.5°

(b)

θi,bottom = 19.5°, θr,bottom = 30.0°

(c)

0.386 cm

(d)

2.00 × 108 m/s

(e)

1.06 × 10−10 s

(f) The angle of incidence affects the distance the ray travels in the glass and hence affects the travel time. 22.20

1.22

22.22

6.30 cm

22.24

(a)

22.26

See Solution.

22.28

0.39°

22.30

θred = 34.9°, θviolet = 34.5°

22.32

(a)

19.1°

(b)

20.4°

22.34

(a)

43.3°

(b)

42.2°

22.36

48.5°

22.38

67.3°

22.40

(a)

34.2°

(b)

34.2°

1.97 × 108 m/s

(a)

24.42°

(b)

388 nm

(c)

3.38 × 10−19 J

(c)

40.4°

Neither thickness nor index of refraction affects the result. (b)

See Solution.

Topic 22

1297

(c)

33.44°

(d)

Total internal reflection still occurs.

(e)

rotate clockwise (f) 2.9°

22.44

2.55 m

22.46

(a)

23.7°

(b)

θr → θi = 30.0°

22.48

(a)

θincidence ≤ 90.0°

(b)

θincidence ≤ 29.9°

(c)

Total internal reflections is not possible since npolystyrene < ncarbon disulfide

22.50

77.5°

22.52

3.50 × 10−4 m

22.54

φ = 8.0°

22.56

82 complete reflections

22.58

1.33

22.60

23.1°

22.62

4.5°

(c)

θr > θi = 30.0°

PROBLEM SOLUTIONS 22.1

The total distance the light travels is

Topic 22

1298

⎛ ⎞ Δd = 2 ⎜ Dcenter!to − REarth − RMoon ⎟ ⎝ center ⎠ !

= 2 ( 3.84 × 108 − 6.38 × 106 − 1.76 × 106 ) m = 7.52 × 108 m

Therefore,

22.2

Δd 7.52 × 108 m v= = = 3.00 × 108 m/s Δt 2.51 s !

(a) The energy of a photon is E = hf = hc/λ, where Planck’s constant is h = 6.63 × 10−34 J ⋅ s and the speed of light in vacuum is c = 3.00 × 108 m/s. If λ = 1.00 × 10−10 m,

( 6.63 × 10 E=

−34

J ⋅s ) ( 3.00 × 108 m/s )

1.00 × 10−10 m

= 1.99 × 10−15 J

⎛ ⎞ 1 eV −15 = 1.24 × 10 4 eV = 12.4 keV (b) E = (1.99 × 10 J ) ⎜ −19 ⎟ 1.602 × 10 J ⎝ ⎠ ! (c) and (d)

For the x-rays to be more penetrating, the photons should be

more energetic. Since the energy of a photon is directly proportional to the frequency and inversely proportional to the wavelength, the wavelength should decrease, which is the same as saying the frequency should increase.

Topic 22

22.3

1299

(a)

⎛ ⎞ 1 eV E = hf = 6.63 × 10−34 J ⋅s 5.00 × 1017 Hz ⎜ −19 ⎟ ⎝ 1.60 × 10 J ⎠

(

)(

)

= 2.07 × 103 eV = 2.07 keV −34 8 hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m/s ) ⎛ 1 nm ⎞ −19 = (b) E = hf = ⎜⎝ 10−9 m ⎟⎠ = 6.63 × 10 J λ 3.00 × 102 nm !

⎛ ⎞ 1 eV E = 6.63 × 10−19 J ⎜ = 4.14 eV −19 ⎟ ⎝ 1.60 × 10 J ⎠ ! 22.4

(a) The energy of a photon is

⎛ c ⎞ hc E = hf = h ⎜ ⎟ = ⎝ λ⎠ λ ! Thus,

λ = hc/E

(b) Higher energy photons have shorter wavelengths. 22.5

(a) The speed of light is c = 3.00 × 108 m/s and the Sun-Earth distance is 1 AU = d = 1.50 × 1011 m. The travel time in minutes is then

Δt =

d 1.50 × 1011 m = = 500 s = 8.33 min c 3.00 × 108 m/s

(b) Use the conversion 1 eV = 1.60 × 10−19 J to find

(

)(

6.63 × 10−34 J ⋅s 3.00 × 108 m/s hc E = hf = = λ 558 × 10−9 m = 3.56 × 10−19 J = 2.23 eV

)

Topic 22

1300

(

)(

6.63 × 10−34 J ⋅s 3.00 × 108 m/s hc hc E= → λ= = λ E 1.60 × 10−19 J

)

= 1.24 × 10−6 m 22.6

The speed of light in a medium with index of refraction n is v = c/n where c is its speed in vacuum.

(a) For water, n = 1.333 and v = !

3.00 × 108 m/s = 2.25 × 108 m/s . 1.333

(b) For crown glass, n = 1.52 and v = ! (c) For diamond, n = 2.419 and v = ! 22.7

3.00 × 108 m/s = 1.97 × 108 m/s . 1.52

3.00 × 108 m/s = 1.24 × 108 m/s . 2.419

From Snell’s law, n2 sin θ2 = n1 sin θ1. Thus, when θ1 = 45.0° and the first medium is air (n1 = 1.00), we have sin θ2 = (1.00) sin 45.0°/n2. −1 ⎛ (1.00)sin 45.0° ⎞ (a) For quartz, n2 = 1.458 and θ 2 = sin ⎜ ⎟⎠ = 29.0° . ⎝ 1.458 !

(b) For carbon disulfide, n2 = 1.628, and ⎛ (1.00 ) sin 45.0° ⎞ θ 2 = sin −1 ⎜ ⎟⎠ = 25.7° . ⎝ 1.628 ! −1 ⎛ (1.00 ) sin 45.0° ⎞ (c) For water, n2 = 1.333, and θ 2 = sin ⎜ ⎟⎠ = 32.0° . ⎝ 1.333 !

Topic 22

1301

22.8

(a) From geometry, 1.25 m = d sin 40.0° so d = 1.94 m. (b) 50.0° above horizontal or parallel to the incident ray 22.9

n1 sin θ1 = n2 sin θ2 sin θ1 = 1.333 sin 45.0°

θ1 = sin−1(1.333 sin 45.0°) = 70.5° Thus, the sun appears to be 19.5° above the horizontal.

22.10

In the sketch below, observe that the law of reflection is obeyed as the ray reflects from each of the mirrors. Also, note that the normal lines to the

Topic 22

1302

two mirrors intersect at a right angle since the two mirrors are perpendicular to each other.

Considering the right triangle formed by the two normal lines and the ray, and recalling that the sum of the interior angles of any triangle is 180°, we find that

θ1 +θ2 + 90.0° = 180°

θ2 = 90.0° − θ1

Looking at the point where the incident ray strikes mirror 1, we see that α = 90.0° − θ1. Thus, both the incident ray and the final outgoing reflected ray are at angle 90.0° − θ1 above the horizontal and hence, parallel to each other.

22.11

n sin θ 1 (1.00 ) sin 30.0° = = 1.52 . (a) From Snell’s law, n2 = 1 sin θ sin 19.24° 2 !

λ0 632.8 nm = = 416 nm (b) λ2 = n 1.52 2 ! f=

(c)

c 3.00 × 108 m/s = = 4.74 × 1014 Hz in the air and also in the −9 λ0 632.8 × 10 m

Topic 22

1303

solution

c 3.00 × 108 m/s = = 1.97 × 108 m/s (d) v2 = n2 1.52 ! 22.12

(a) When light refracts from air (n1 = 1.00) into the crown glass, Snell’s law gives the angle of refraction as

θ2 = sin−1 (sin 25.0°/ncrown glass) For first quadrant angles, the sine of the angle increases as the angle increases. Thus, from the above equation, note that θ2 will increase when the index of refraction of the crown glass decreases. From Figure 22.13, we see that the angle of refraction will increase with wavelength, so the longer wavelengths deviate the least from the original path.

Topic 22

1304

(b) From Figure 22.13, observe that the index of refraction of crown glass for the given wavelengths is

λ = 400 nm, ncrown glass = 1.53;

λ = 500 nm, ncrown glass = 1.52;

and λ = 650 nm, ncrown glass = 1.51 Thus, Snell’s law gives

λ = 400 nm, θ2 = sin−1 (sin 25.0°/1.53) = 16.0° λ = 500 nm, θ2 = sin−1 (sin 25.0°/1.52) = 16.1° λ = 650 nm, θ2 = sin−1 (sin 25.0°/1.51) = 16.3° 22.13

From Snell’s law,

⎡ n sin θ 1 ⎤ ⎡ (1.00 ) sin 40.0° ⎤ θ 2 = sin −1 ⎢ 1 = sin −1 ⎢ ⎥ ⎥ = 29.4° n 1.309 ⎣ ⎦ ⎣ ⎦ 2 ! and from the law of reflection, λ = 40.0 nm, φ = θ1 = 40.0°. Hence, the angle between the reflected and refracted rays is

α = 180° − θ2 − φ = 180.0° − 29.4° − 40.0° = 110.6°

Topic 22

22.14

1305

Consider the sketch below and note that the incident horizontal ray is parallel to the surface of mirror 2. Thus, the angle the incident ray makes with mirror 1 must be α = θ1 = 50.0°.

Since the ray must obey the law of reflection at mirror 1, the angle β must be β = α = 50.0°. Recalling that the sum of the interior angles of a triangle is always 180.0°, we find that

γ = 180.0° − θ1 − β = 180.0° − 50.0° − 50.0° = 80.0° Hence, in order to obey the law of reflection at mirror 2, the angle the outgoing reflected ray makes with the surface of mirror 2 must be θ2 = γ = 80.0°. 22.15

The index of refraction of zircon is n = 1.923.

c 3.00 × 108 m/s = 1.56 × 108 m/s (a) v = = n 1.923 !

λ 632.8 nm (b) The wavelength in the zircon is λn = 0 = = 329.1 nm . n 1.923 !

Topic 22

1306

v c 3.00 × 108 m/s (c) The frequency is f = = = = 4.74 × 1014 Hz . −9 λ λ 632.8 × 10 m n 0 ! 22.16

The sketch below shows the path of the ray inside the glass slab.

Considering the reflections at points A and B, the law of reflection tells us that α = θA and β = θB. Also, we observe that

θB = 90° − α = 90° − θA and θC = 90° − β = 90° − θB = 90° − (90° − θA) = θA Thus, if θA = 55° then θB = 35° and θC = 55°. In order for part of the ray to leave the glass slab and enter the surrounding medium at a reflection point, the angle of incidence at that point must be less than the critical angle, θC = sin−1(n2/nglass) = sin−1(n2/1.52). (a) If the surrounding medium is air, then n2 = 1.00 and θC = sin−1(1.00/1.52) = 41.1°. Thus, we see that total internal reflection will occur at points A and C, but part of the ray can refract into the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 22

1307

surrounding air at point B. (b) When the surrounding medium is carbon disulfide, n2 = 1.628 > nglass. Thus, the critical angle does not exist, and total internal reflection will not occur when the ray attempts to go from the glass into the carbon disulfide. This means that part of the ray will enter the carbon disulfide at points A, B, and C. 22.17

The incident light reaches the left-hand mirror at distance d/2 = (1.00 m) tan 5.00° = 0.087 5 m above its bottom edge. The reflected light first reaches the right-hand mirror at height d = 2(0.087 5 m) = 0.175 m It bounces between the mirrors with distance d between points of contact with a given mirror. Since the full 1.00 length of the right-hand mirror is available for reflections, the number of reflections from this mirror will be

1.00 m Nright = = 5.71 → 5 full!reflections 0.175 m ! Since the first reflection from the left-hand mirror occurs at a height of d/2

Topic 22

1308

= 0.087 5 m. The total number of that can occur from this mirror is

Nleft = 1 + !

22.18

1.00 m − 0.087 5 m = 6.21 → 0.175 m

6 full!reflections

(a) From Snell’s law, the angle of refraction at the first surface is

⎡ n sin θ 1 ⎤ −1 ⎡ (1.00 ) sin 30.0° ⎤ θ 2 = sin −1 ⎢ air ⎥ = sin ⎢ ⎥ = 19.5° nglass ⎥⎦ 1.50 ⎣ ⎦ ⎢ ⎣ ! (b) Since the upper and lower surfaces are parallel, the normal lines where the ray strikes these surfaces are parallel. Hence, the angle of incidence at the lower surface will be θ2 = 19.5°. The angle of refraction at this surface is then

⎡ nglass sin θ glass ⎤ ⎡ (1.50 ) sin 19.5° ⎤ θ 3 = sin −1 ⎢ = sin −1 ⎢ ⎥ ⎥ = 30.0° nair 1.00 ⎣ ⎦ ⎣ ⎦ ! Thus, the light emerges traveling parallel to the incident beam. (c) Consider the sketch below, and let h represent the distance from

Topic 22

1309

point a to c (that is, the hypotenuse of triangle abc).

Then,

2.00 cm 2.00 cm h= = = 2.12 cm cos θ 2 cos 19.5° ! Also, α = θ1 − θ2 = 30.0° − 19.5° = 10.5°, so d = h sin α = (2.12 cm) sin 10.5° = 0.386 cm (d) The speed of the light in the glass is

c 3.00 × 108 m/s v= = = 2.00 × 108 m/s nglass 1.50 ! (e) The time required for the light to travel through the glass is

c 2.12 cm ⎛ 1 m ⎞ t= = = 1.06 × 10−10 s 8 2 ⎜ ⎟ v 2.00 × 10 m/s ⎝ 10 cm ⎠ ! (f) Changing the angle of incidence will change the angle of refraction and therefore the distance h the light travels in the glass. Thus, the travel time will also change. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 22

22.19

1310

From Snell’s law, the angle of incidence at the air-oil interface is

⎡ n sin θ oil ⎤ ⎡ (1.48 ) sin 20.0° ⎤ θ = sin −1 ⎢ oil = sin −1 ⎢ ⎥ ⎥ = 30.4° nair 1.00 ⎣ ⎦ ⎣ ⎦ ! and the angle of refraction as the light enters the water is

⎡ n sin θ oil ⎤ ⎡ (1.48 ) sin 20.0° ⎤ θ ′ = sin −1 ⎢ oil = sin −1 ⎢ ⎥ ⎥ = 22.3° nwater ⎦ 1.333 ⎣ ⎦ ⎣ ! 22.20

Since the light ray strikes the first surface at normal incidence, it passes into the prism without deviation. Thus, the angle of incidence at the second surface (hypotenuse of the triangular prism) is θ1 = 45.0° as shown in the sketch below.

The angle of refraction is

θ2 = 45.0° + 15.0° = 60.0° and Snell’s law gives the index of refraction of the prism material as n sin θ 2 (1.00 ) sin (60.0°) n1 = 2 = = 1.22 sin θ sin (45.0°) 1 !

Topic 22

22.21

1311

From the figure geometry, tan θ 2 =

2.50 m so that θ2 = 39.8°. Taking the 3.00 m

index of refraction of water to be n2 = 1.333, apply Snell’s law to find θ1: n1 sin θ1 = n2 sin θ 2 → sin θ1 =

n2 1.333 sin θ1 = sin ( 39.8° ) = 0.853 n1 1.000

θ1 = 58.6°

22.22

⎛ nmedium ⎞ sin 50.0°. From Snell’s law, sin θ = ⎜ ⎝ nliver ⎟⎠ ! But

nmedium c vmedium v = = liver = 0.900 c vliver vmedium ! nliver

so θ = sin−1[(0.900)sin 50.0°] = 43.6° From the law of reflection,

d= !

22.23

12.0 cm = 6.00 cm 2

and

d 6.00 cm = = 6.30 cm tan θ tan (43.6°)

(a) Before the container is filled, the ray’s path is as shown in figure (a) below. From this figure, observe that

sin θ 1 = !

d d = = 2 s1 h + d2

(h d) + 1 2

Topic 22

1312

After the container is filled, the ray’s path is shown in figure (b). From this figure, we find that

d/2 d/2 = = 2 s2 h2 + ( d / 2 )

sin θ 2 = !

1 4(h / d) + 1 2

From Snell’s law, nair sin θ1 = n sin θ2, or

1.00

(h / d) + 1 2

n 4(h / d) + 1 2

and 4(h/d)2 + 1 = n2(h/d)2 + n2

Simplifying, this gives (4 − n2)(h/d)2 = n2 − 1

! (b) If d = 8.00 cm

and

n2 − 1 4 − n2

n = nwater = 1.333, then

(1.333)2 − 1 = 4.73 cm 4 − (1.333)2

h = ( 8.00 cm ) ! 22.24

h = d

(a) The speed of light in a medium with index of refraction n is v=

c c = = 1.97 × 108 m/s n 1.52

(b) The wavelength is

λn =

λ0 589 nm = = 3.88 × 10−7 m = 388 nm n 1.52

Topic 22

1313

E = hf =

hc hv = λ λn

( 6.63 × 10 E=

−34

)(

J ⋅s 3.00 × 108 m/s

589 × 10

−9

) = 3.38 × 10

−19

(rounding differences can lead to slight variations from this result). 22.25

As shown below, θ1 + β + θ2 = 180.0°. When β = 90.0°, this gives θ2 = 90° −

θ1. Then, from Snell’s law

sin θ 1 = !

ng sin θ 2 nair

= ng sin ( 90° − θ 1 ) = ng cos θ 1

Thus, when β = 90.0°,

22.26

sin θ 1 = tan θ 1 = ng or θ1 = tan−1(ng). cos θ 1 !

The index of refraction of the atmosphere decreases with increasing altitude because of the decrease in density of the atmosphere with increasing altitude. As indicated in the ray diagram below, the Sun located at S below the horizon appears to be located at S΄.

Topic 22

22.27

1314

When the Sun is 28.0° above the horizon, the angle of incidence for sunlight at the air-water boundary is

θ1 = 90.0° − 28.0° = 62.0° Thus, the angle of refraction is

⎡ n sin θ 1 ⎤ θ 2 = sin −1 ⎢ air ⎥ ⎣ nwater ⎦ ⎡ (1.00)sin 62.0° ⎤ = sin −1 ⎢ ⎥⎦ = 41.5° 1.333 ⎣ !

3.00 m 3.00 m The depth of the tank is then h = = = 3.39 m . tan θ tan (41.5°) 2 !

22.28

The angles of refraction for the two wavelengths are

Topic 22

1315

⎛ n sin θ 1 ⎞ ⎛ 1.000sin 30.00° ⎞ θ red = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 18.03° ⎟ ⎝ 1.615 ⎝ nred ⎠ ! and

⎛ n sin θ 1 ⎞ ⎛ 1.000sin 30.00° ⎞ θ blue = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 17.64° ⎟ ⎝ 1.650 ⎝ nblue ⎠ !

Thus, the angle between the two refracted rays is Δθ = θred − θblue = 18.03° − 17.64° = 0.39° 22.29

Using Snell’s law gives

⎛ n sin θ i ⎞ ⎛ (1.000)sin 83.00° ⎞ (a) θ blue = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 47.79° ⎟ ⎝ 1.340 ⎝ nblue ⎠ ! ⎛ n sin θ i ⎞ ⎛ (1.000)sin 83.00° ⎞ (b) θ red = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 48.22° ⎟ ⎝ 1.331 ⎝ nred ⎠ ! 22.30

Using Snell’s law gives

⎛ n sin θ i ⎞ ⎛ (1.00)sin 60.0° ⎞ θ red = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 34.9° ⎟ ⎝ 1.512 ⎝ nred ⎠ ! and

22.31

⎛ n sin θ i ⎞ ⎛ (1.00)sin 60.0° ⎞ θ violet = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 34.5° ⎟ ⎝ 1.530 ⎝ nviolet ⎠ !

Using Snell’s law gives

⎛ n sin θ i ⎞ ⎛ (1.000)sin 50.00° ⎞ θ red = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 31.77° ⎟ ⎝ 1.455 ⎝ nred ⎠ !

Topic 22

1316

and

⎛ n sin θ i ⎞ ⎛ (1.000)sin 50.00° ⎞ θ violet = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 31.45° ⎟ ⎝ 1.468 ⎝ nviolet ⎠ !

Thus, the dispersion is θred − θviolet = 31.77° − 31.45° = 0.32°. 22.32

The necessary geometry is shown in the diagram below, where the incoming ray can be either blue or red and the blue lines indicate the prism surfaces:

30.0

30.0 60.0 60.0 60.0

With n1 = 1, apply Snell’s law to the prism’s first surface to find ⎛ 0.500 ⎞ . n1 sin 30.0° = nsin β → β = sin −1 ⎜ ⎝ n ⎟⎠

The angle φ can be found from the largest triangle inside the prism, for ⎛ 0.500 ⎞ which β + 60°+90°+φ = 180° so that φ = 30°−β = 30° − sin −1 ⎜ . The ⎝ n ⎟⎠

angle of deviation is then given by Snell’s law as

Topic 22

1317

nsin φ = (1) sin δ → δ = sin −1 ( nsin φ )

(a) For red light with an index of refraction nred = 1.60, substitute numbers to find ⎛ 0.500 ⎞ φred = 30.0° − sin −1 ⎜ = 11.8° ⎝ 1.60 ⎟⎠

and

δ red = sin −1 ((1.60 ) sin (11.8° )) = 19.1° (b) In the same way for blue light using nblue = 1.64: ⎛ 0.500 ⎞ φblue = 30.0° − sin −1 ⎜ = 12.25° ⎝ 1.64 ⎟⎠

and

δ blue = sin −1 ((1.64 ) sin (12.25° )) = 20.4° 22.33

(a) The angle of incidence at the first surface is θ1i = 30°, and the angle of refraction is

⎛ n sin θ 1i ⎞ θ 1r = sin −1 ⎜ air ⎟ ⎝ nglass ⎠ !

⎛ 1.0sin 30° ⎞ = sin −1 ⎜ ⎟⎠ = 19° ⎝ 1.5

Also, α = 90° − θ1r = 71° and β = 180.0° − 60.0° − α = 49°. Therefore, the angle of incidence at the second surface is θ2i = 90° − β

Topic 22

1318

= 41°. The angle of refraction at this surface is

⎛ nglass sin θ 2i ⎞ ⎛ 1.5sin 41° ⎞ θ 2r = sin −1 ⎜ = sin −1 ⎜ ⎟⎠ = 80° ⎟ ⎝ nair 1.0 ⎝ ⎠ !

(b) The angle of reflection at each surface equals the angle of incidence at that surface. Thus,

(θ1 )reflection = θ1i = 30° ,!and! (θ 2 )reflection = θ 2i = 41°

! 22.34

As light goes from a medium having a refractive index n1 to a medium with refractive index n2 < n1, the critical angle is given by the relation sin

θc = n2/n1. Table 22.1 gives the refractive index for various substances at λ0 = 589 nm. (a) For fused quartz surrounded by air, n1 = 1.458 and n2 = 1.00, giving θc = sin−1 (1.00/1.458) = 43.3°. (b) In going from polystyrene (n1 = 1.49) to air, θc = sin−1 (1.00/1.49) = 42.2°.

Topic 22

1319

When light is coming from a medium of refractive index n1 into water (n2 = 1.333), the critical angle is given by θc = sin−1 (1.333/n1). (a) For fused quartz, n1 = 1.458, giving θc = sin−1 (1.333/1.458) = 66.1°. (b) In going from polystyrene (n1 = 1.49) to water, θc = sin−1 (1.333/1.49) = 63.5°. (c) From sodium chloride (n1 = 1.544) to water, θc = sin−1 (1.333/1.544) = 59.7°.

22.36

Using Snell’s law, the index of refraction of the liquid is found to be nliquid = (nair sinθi)/sin θr Thus, the critical angle for light going from this liquid into air is

⎡ ⎤ ⎛ n ⎞ nair ⎡ sin 22.0° ⎤ ⎥ = sin −1 ⎢ θ c = sin −1 ⎜ air ⎟ = sin −1 ⎢ ⎥ = 48.5° sin 30.0° ⎦ ⎢ nair sin θ i sin θ r ⎥ ⎝ nliquid ⎠ ⎣ ⎣ ⎦ !

(

22.37

)

When light attempts to cross a boundary from one medium of refractive index n1 into a new medium of refractive index n2 < n1, total internal reflection will occur if the angle of incidence exceeds the critical angle

Topic 22

1320

given by θc = sin−1 (n2/n1). −1 ⎛ 1.00 ⎞ (a) If n1 = 1.53!and!n2 = nair = 1.00,!then!θ c = sin ⎜ ⎟⎠ = 40.8° . ⎝ 1.53 !

−1 ⎛ 1.333 ⎞ = 60.6° . (b) If n1 = 1.53 and n2 = nwater = 1.333,then!θ c = sin ⎜ ⎝ 1.53 ⎟⎠ !

22.38

The critical angle for this material in air is

⎛ n ⎞ ⎛ 1.00 ⎞ θ c = sin −1 ⎜ air ⎟ = sin −1 ⎜ = 47.3° ⎝ 1.36 ⎟⎠ npipe ⎠ ⎝ ! Thus,

θr = 90.0° − θc = 42.7° and from Snell’s law,

⎛ npipe sin θ r ⎞ ⎛ (1.36)sin 42.7° ⎞ θ i = sin −1 ⎜ = sin −1 ⎜ ⎟⎠ = 67.3° ⎟ ⎝ nair 1.00 ⎝ ⎠ !

22.39

The angle of incidence at each of the shorter faces of the prism is 45°, as shown in the figure below.

Topic 22

1321

For total internal reflection to occur at these faces, it is necessary that the critical angle be less than 45°. With the prism surrounded by air, the critical angle is given by sin θc = nair/nprism = 1.00/nprism, so it is necessary that 1.00 sin θ c = < sin 45° nprism !

22.40

1.00 1.00 nprism > = = sin 45° 2 2 !

(a) The minimum angle of incidence for which total internal reflection occurs is the critical angle. At the critical angle, the angle of refraction is 90°, as shown in the figure below. From Snell’s law, ng sin θi = na sin 90.0°, the critical angle for the glass-air interface is found to be

⎛ n sin 90° ⎞ ⎛ 1.00 ⎞ θ i = θ c = sin −1 ⎜ a = sin −1 ⎜ = 34.2° ⎟ ⎝ 1.78 ⎟⎠ ng ⎝ ⎠ !

(b) When the slab of glass has a layer of water on top, we want the angle of incidence at the water-air interface to equal the critical angle for that combination of media. At this angle, Snell’s law gives

Topic 22

1322

nw sin θc = na sin 90.0° = 1.00 and

sin θc = 1.00/nw

Now, considering the refraction at the glass-water interface, Snell’s law gives ng sin θi = ng sin θc. Combining this with the result for sin θc from above, we find the required angle of incidence in the glass to be

⎛ n sin θ c ⎞ ⎛ n (1.00 nw ) ⎞ ⎛ 1.00 ⎞ ⎛ 1.00 ⎞ θ i = sin −1 ⎜ w = sin −1 ⎜ w = sin −1 ⎜ = sin −1 ⎜ = 34.2° ⎟ ⎟ ⎟ ⎝ 1.78 ⎟⎠ ng ⎠ ng ng ⎠ ⎝ ⎝ ⎠ ⎝ ! (c) and (d)

Observe in the calculation of part (b) that all the physical

properties of the intervening layer (water in this case) canceled, and the result of part (b) is identical to that of part (a). This will always be true when the upper and lower surfaces of the intervening layer are parallel to each other. Neither the thickness nor the index of refraction of the intervening layer affects the result. 22.41

(a) Snell’s law can be written as sin θ1/sin θ2 = v1/v2. At the critical angle of incidence (θ1 = θc), the angle of refraction is 90°, and Snell’s law

Topic 22

1323

becomes sin θc = v1/v2. At the concrete-air boundary, ⎛v ⎞ ⎛ 343 m/s ⎞ θ c = sin −1 ⎜ 1 ⎟ = sin −1 ⎜ ⎟ = 10.7° ⎝ 1 850 m/s ⎠ ⎝ v2 ⎠

(b) Sound can be totally reflected only if it is initially traveling in the slower medium. Hence, at the concrete-air boundary, the sound must be traveling in air. (c) Sound in air falling on the wall from most directions is 100% reflected, so the wall is a good mirror. 22.42

(a) The index of refraction for diamond is ndiamond = 2.419, and the critical angle at a diamond-air boundary is

⎛ nair ⎞ ⎛ 1.000 ⎞ θ c = sin −1 ⎜ = sin −1 ⎜ = 24.42° ⎟ ⎝ 2.419 ⎟⎠ ⎝ ndiamond ⎠ ! (b) With the face of the diamond tilted up 35.0° from the horizontal, the normal line to this face at point P is tipped over 35.0° from the vertical. Thus, the angle of incidence at point P is θi = 35.0° > θc = 24.42°. Since this angle of incidence exceeds the critical angle, total internal reflection occurs at point P. (c) If the diamond is immersed in water (nwater = 1.333), the critical angle

Topic 22

1324

⎛ n ⎞ ⎛ 1.333 ⎞ θ c = sin −1 ⎜ water ⎟ = sin −1 ⎜ = 33.44° ⎝ 2.419 ⎟⎠ ⎝ ndiamond ⎠ ! (d) The light continues to enter the top surface of the diamond at normal incidence, so the angle of incidence at point P continues to be θi = 35.0° > θc = 33.44°. Since this angle of incidence (barely) exceeds the critical angle for the diamond-water boundary, total internal reflection still occurs at point P. (e) To have light exit the diamond at point P, we need to decrease the angle of incidence at this point. Thus, we should rotate the diamond clockwise, thereby bringing the normal line closer to the incident ray at point P. (f) Rotating the diamond clockwise by an angle θ changes the angle of refraction at point A where the ray enters the diamond. To find what this new angle of refraction will be, we extend the line of the top of the diamond and the line of the face containing point P until they intersect at point B as shown below.

Topic 22

1325

Summing the interior angles in triangle ABP gives

α + β + 35.0° = 180° or

(90.0° − θr) + (90.0° − φ) + 35.0° = 180°

and θr = 35.0° − φ If the ray is to exit the diamond and enter the water at point P, the angle of incidence at this point must be less than or equal to the critical angle found in part (c). Thus, we require φ ≤ 33.44° and see that we must have

θr ≥ 35.0° − 33.44° = 1.6° Applying Snell’s law at point A gives the minimum required rotation as

⎛n sin θ r ⎞ ⎡ 2.419sin (1.6°) ⎤ θ min = sin −1 ⎜ diamond = sin −1 ⎢ ⎟ ⎥⎦ = 2.9° nwater 1.333 ⎝ ⎠ ⎣ !

Topic 22

22.43

1326

If θc = 42.0° at the boundary between the prism glass and the surrounding medium, then sin θc = n2/n1 gives nm = sin 42.0° nglass !

From the geometry shown in the figure below,

α = 90.0° − 42.0° = 48.0°, β = 180.0° − 60.0° − α = 72.0° and θr = 90.0° − β = 18.0°. Thus, applying Snell’s law at the first surface gives

⎛ sin θ r ⎞ ⎛ sin 18.0° ⎞ ⎛ nglass sin θ r ⎞ −1 θ 1 = sin −1 ⎜ = sin = sin −1 ⎜ = 27.5° ⎜ ⎟ ⎟ nm nm nglass ⎠ ⎝ ⎠ ⎝ sin 42.0° ⎟⎠ ⎝ !

22.44

The maximum distance d occurs when the underwater ray reaches the critical angle, θc, so that nwater sin θ c = nair sin 90° = 1 and θ c = 48.6° . From the diagram geometry with θc being the angle between the surface normal and the underwater ray,

Topic 22

1327

tan θ c =

22.45

d → d = 2.55 m 2.25 m

At the air-ice boundary, Snell’s law gives the angle of refraction in the ice as

sin θ 1r = !

nair sin θ 1i nice

Since the sides of the ice layer are parallel, the angle of incidence at the ice-water boundary is θ2i = θ1r. Then, from Snell’s law, the angle of refraction in the water is ⎡ nice ⎛ n sin θ ⎞ ⎤ ⎛ n sin θ 2i ⎞ −1 ⎛ nice sin θ 1r ⎞ −1 air 1i θ 2r = sin −1 ⎜ ice = sin = sin ⎢ ⎜ ⎟⎥ ⎟ ⎜ ⎟ nice ⎝ nwater ⎠ ⎝ nwater ⎠ ⎢⎣ nwater ⎝ ⎠ ⎥⎦ !

⎛ n sin θ 1i ⎞ −1 ⎛ (1.00)sin 30.0° ⎞ θ 2r = sin −1 ⎜ air = sin ⎜⎝ ⎟⎠ = 22.0° 1.333 ⎝ nwater ⎟⎠ !

Note that all of the properties of the ice canceled out in the above calculation, and the result is the same as if the ice had not been present. This will always be true when the intermediate medium has parallel sides. 22.46

When light coming from the surrounding medium is incident on the surface of the glass slab, Snell’s law gives ng sin θr = ns sin θi, or

Topic 22

1328

sin θr = (ns/ng) sin θi (a) If θi = 30.0° and the surrounding medium is water (ns = 1.333), the angle of refraction is

⎡ 1.333sin (30.0°) ⎤ θ r = sin −1 ⎢ ⎥⎦ = 23.7° 1.66 ⎣ !

(b) From Snell’s law given above, we see that as ns → ng we have sin θr → sin θi, or the angle of refraction approaches the angle of incidence,

θr → θi = 30.0°. (c) If ns > ng, then sin θr = (ns/ng) sin θi > sin θi, or θr > θi . 22.47

(a) Given that the angle θ shown in the figure below is 30.0°, the maximum distance the observer can be from the pool and continue to see the lower edge on the opposite side of the pool is d = (1.85 m) tan 30.0° = 1.07 m

Topic 22

1329

(b) If the pool is now completely filled with water, the light ray coming to the observer’s eye from the lower opposite edge of the pool will refract at the surface of the water as shown in the second figure below.

The angle of refraction is

⎛n sin 30.0° ⎞ ⎡ (1.333)sin 30.0° ⎤ φ = sin −1 ⎜ water = sin −1 ⎢ ⎟ ⎥⎦ = 41.8° nair 1.00 ⎝ ⎠ ⎣ ! The maximum distance the observer can now be from the pool and still see the same boundary is d' = (1.85 m) tan φ = (1.85 m) tan 41.8° = 1.65 m 22.48

(a) For polystyrene surrounded by air, total internal reflection at the left vertical face requires that

Topic 22

1330

⎛n ⎞ ⎛ 1.00 ⎞ θ 3 ≥ θ c = sin −1 ⎜ air ⎟ = sin −1 ⎜ = 42.2° ⎝ 1.49 ⎟⎠ np ⎠ ⎝ ! From the geometry shown in the figure below,

θ2 = 90.0° − θ3 ≤ 90.0° − 42.2° = 47.8°

Thus, use of Snell’s law at the upper surface gives

sin θ 1 = !

np sin θ 2 nair

≤

(1.49)sin 47.8° = 1.10 1.00

so it is seen that any angle of incidence ≤ 90.0° at the upper surface will yield total internal reflection at the left vertical face. (b) Repeating the steps of part (a) with the index of refraction of air replaced by that of water yields θ3 ≥ 63.5°, θ2 ≤ 26.5°, sin θ1 ≤ 0.499°, and θ1 ≤ 29.9°. (c) Total internal reflection is not possible since npolystyrene < ncarbon disulfide .

Topic 22

22.49

1331

The angle between the horizontal surface and the underwater portion of the line extending to the fish’s apparent position is equal to θ1, given by

tan θ1 = 2.00 m 1.25 m so that θ1 = 58.0. Applying Snell’s law with n1 = 1.000 and n2 = 1.333 gives ⎛ 1.000 ⎞ n1 sin θ1 = n2 sin θ 2 → θ 2 = sin −1 ⎜ sin ( 58.0° )⎟ = 39.5° ⎝ 1.333 ⎠

The fish’s actual depth d is then

tan θ 2 =

22.50

2.00 m 2.00 m → d= = 2.43 m d tan ( 39.5° )

Applying Snell’s law to this refraction, recognizing that nair = 1.00, gives nglass sin θ2 = nair sin θ1 = sin θ1 If θ1 = 2θ2, this becomes nglass sin θ2 = sin (2θ2) = 2sin θ2 cos θ2

nglass cos θ 2 = 2 !

and

⎛ nglass ⎞ θ 2 = cos −1 ⎜ ⎝ 2 ⎟⎠

Then, the desired angle of incidence is

⎛ nglass ⎞ ⎛ 1.56 ⎞ θ 1 = 2θ 2 = 2cos −1 ⎜ = 2cos −1 ⎜ = 77.5° ⎟ ⎝ 2 ⎟⎠ ⎝ ⎠ 2 ! 22.51

In the figure below, observe that β = 90° − θ1 and α = 90° − θ1.

Topic 22

1332

Thus, β = α. Similarly, on the right side of the prism, δ = 90° − θ2 and γ = 90° − θ2, giving δ = γ. Next, observe that the angle between the reflected rays is B = (α + β) + (γ +

δ), so B = 2(α + γ). Finally, observe that the left side of the prism is sloped at angle α from the vertical, and the right side is sloped at angle γ. Thus, the angle between the two sides is A = α + γ, and we obtain the result B = 2(α + γ) = 2A. 22.52

Total internal reflection occurs within the fiber when θ ≥ θc or sin θ ≥ sin

θc. From the figure’s geometry, sin θ =

sin θ c =

R−d . The critical angle is given by R

1 R−d 1 . Substituting these results gives the condition ≥ . As R n R n

decreases, the ratio

R−d decreases (for positive d) so that R

Topic 22

1333

Rmin − d 1 = → Rmin ( n − 1) = nd Rmin n Rmin = 22.53

(1.40)1.00 × 10 nd = n−1 1.40 − 1

−4

= 3.50 × 10−4 m

Consider light which leaves the lower end of the wire and travels parallel to the wire while in the benzene. If the wire appears straight to an observer looking along the dry portion of the wire, this ray from the lower end of the wire must enter the observer’s eye as he sights along the wire. Thus, the ray must refract and travel parallel to the wire in air. The angle of refraction is then θ2 = 90.0° − 30.0° = 60.0°. From Snell’s law, the angle of incidence was

⎛ n sin θ 2 ⎞ ⎛ (1.00)sin 60.0° ⎞ θ 1 = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 35.3° ⎟ ⎝ 1.50 ⎝ nbenzene ⎠ ! and the wire is bent by angle θ = θ2 − θ1 = 60.0° − θ1 = 60.0° − 35.3° = 24.7°.

22.54

In the sketch below, the angle of incidence at A is the same as the prism angle at point O. This is true because tipping the line OA up angle θ from

Topic 22

1334

the horizontal necessarily tips its normal line over at angle θ from the vertical. Given that θ = 60.0°, application of Snell’s law at point A gives 1.50 sin β = (1.00) sin 60.0° or

β = 35.3°

From triangle AOB, we calculate the angle of incidence and reflection, γ at point B:

θ + (90.0° − β) + (90.0° − γ) = 180°

γ = θ − β = 60.0° − 35.3° = 24.7°

Now, we find the angle of incidence at point C using triangle BCQ: (90.0° − γ) + (90.0° − δ) + (90.0° − θ) = 180° or

δ = 90.0° − (θ + γ) = 90.0° − 84.7° = 5.3°

Finally, application of Snell’s law at point C gives (1.00) sin φ = (1.50) sin (5.3°), or φ = sin−1(1.50 sin 5.3°) = 8.0°. 22.55

The path of a light ray during a reflection and/or refraction process is always reversible. Thus, if the emerging ray is parallel to the incident ray, the path which the light follows through this cylinder must be symmetric

Topic 22

1335

about the center line as shown below.

−1 ⎛ d / 2 ⎞ −1 ⎛ 1.00 m ⎞ = 30.0° Thus, θ 1 = sin ⎜⎝ ⎟⎠ = sin ⎜ R ⎝ 2.00 m ⎟⎠ !

Triangle ABC is isosceles, so γ = α and β = 180° − α − γ = 180° − 2α. Also, β = 180° − θ1, which gives α = θ1/2 = 15.0°. Then, from applying Snell’s law at point A,

ncylinder = !

nair sin θ 1 (1.00)sin 30.0° = = 1.93 sin α sin 15.0°

22.56

The angle of refraction as the light enters the left end of the slab is

⎛ n sin θ 1 ⎞ ⎛ (1.00)sin 50.0° ⎞ θ 2 = sin −1 ⎜ air = sin −1 ⎜ ⎟⎠ = 31.2° ⎟ ⎝ 1.48 ⎝ nslab ⎠ ! Observe from the figure that the first reflection occurs at x = d, the second

Topic 22

1336

reflection is at x = 3d, the third is at x = 5d, and so forth. In general, the Nth reflection occurs at x = (2N − 1)d, where

d= !

(0.310 cm / 2) 0.310 cm = = 0.256 cm tan θ 2 2 tan 31.2°

Therefore, the number of reflections made before reaching the other end of the slab at x = L = 42 cm is found from L = (2N − 1)d to be

⎞ 1 ⎛ L ⎞ 1 ⎛ 42 cm N = ⎜ + 1⎟ = ⎜ + 1⎟ = 82.5 or 82 complete reflections 2 ⎝ d ⎠ 2 ⎝ 0.256 cm ⎠ ! 22.57

(a) If θ1 = 45.0°, application of Snell’s law at the point where the beam enters the plastic block gives (1.00) sin 45.0° = n sin φ

[1]

Application of Snell’s law at the point where the beam emerges from the plastic, with θ2 = 76.0°, gives n sin (90.0° − φ) = (1.00) sin 76.0°

(1.00) sin 76.0° = n cos φ

Dividing Equation [1] by Equation [2], we obtain

sin 45.0° tan φ = = 0.729 sin 76.0° ! Thus, from Equation [1],

n= !

and

φ = 36.1°

sin 45.0° sin 45.0° = = 1.20 . sin φ sin 36.1°

[2]

Topic 22

1337

(b) Observe from the figure above that sin φ = L/d. Thus, the distance the light travels inside the plastic is d = L/ sin φ, and if L = 50.0 cm = 0.500 m, the time required is

Δt =

d v

L / sin φ c/n

nL csin φ

(

1.20 0.500 m

)

(3.00 × 10 m/s)sin 36.1° 8

= 3.39 × 10−9 s = 3.39 ns 22.58

Snell’s law would predict that nair sin θi = nwater sin θr, or since nair = 1.00, sin θi = nwater sin θr Comparing this equation to the equation of a straight line, y = mx + b, shows that if Snell’s law is valid, a graph of sin θi versus sin θr should yield a straight line that would pass through the origin if extended and would have a slope equal to nwater.

Topic 22

1338

θ i (deg)

θ r (deg)

sin θ i

sin θ r

10.0

7.50

0.174

0.131

20.0

15.1

0.342

0.261

30.0

22.3

0.500

0.379

40.0

28.7

0.643

0.480

50.0

35.2

0.766

0.576

60.0

40.3

0.866

0.647

70.0

45.3

0.940

0.711

80.0

47.7

0.985

0.740

The straightness of the graph line and the fact that its extension passes through the origin demonstrates the validity of Snell’s law. Using the end points of the graph line to calculate its slope gives the value of the index of refraction of water as

Topic 22

1339

0.985 − 0.174 nwater = slope = = 1.33 0.740 − 0.131 !

22.59

Applying Snell’s law at points A, B, and C gives 1.40 sin α = 1.60 sin θ1

[1]

1.20 sin β = 1.40 sin α

[2]

and 1.00 sin θ2 = 1.20 sin β

[3]

Combining Equations [1], [2], and [3] yields sin θ2 = 1.60 sin θ1 Note that Equation [4] is exactly what Snell’s law would yield if the second and third layers of this “sandwich” were ignored. This will always be true if the surfaces of all the layers are parallel to each other.

(a) If θ1 = 30.0°, then Equation [4] gives θ2 = sin−1 (1.60 sin 30.0°) = 53.1°. (b) At the critical angle of incidence on the lowest surface, θ2 = 90.0°.

[4]

Topic 22

1340

Then, Equation [4] gives

⎛ sin θ 2 ⎞ ⎛ sin 90.0° ⎞ θ 1 = sin −1 ⎜ = sin −1 ⎜ ⎟ ⎟⎠ = 38.7° ⎝ ⎠ ⎝ 1.60 1.60 ! 22.60

For the first placement, Snell’s law gives

n sin 26.5° n2 = 1 sin 31.7° !

In the second placement, application of Snell’s law yields

⎛ n sin 26.5° ⎞ n sin 36.7° n3 sin 26.5° = n2 sin 36.7° = ⎜ 1 sin 36.7°,!or!n3 = 1 ⎟ sin 31.7° ⎝ sin 31.7° ⎠ ! Finally, using Snell’s law in the third placement gives

sin θ R = !

⎛ sin 31.7° ⎞ n1 sin 26.5° = ( n1 sin 26.5° ) ⎜ = 0.392 n3 ⎝ n1 sin 36.7° ⎟⎠

and θR = 23.1° 22.61

Light from all angles above the water’s surface will refract below the surface and light from the most extreme angle (horizontal to the water’s

Topic 22

1341

surface) will be refracted to the critical angle. Occurring for all directions to create Snell’s window, the cone’s full angle equals twice the critical ⎛ 1 ⎞ angle so that θ = 2θc = 2sin −1 ⎜ = 97.2° . ⎝ 1.333 ⎟⎠

22.62

For the violet light, nglass = 1.66, and

⎛ n sin θ 1i ⎞ θ 1r = sin −1 ⎜ air ⎟ ⎝ nglass ⎠ !

⎛ 1.00sin 50.0° ⎞ = sin −1 ⎜ ⎟⎠ = 27.5° ⎝ 1.66

α = 90° − θ1r = 62.5°, β = 180.0° − 60.0° − α = 57.5°, and θ2i = 90.0° − β = 32.5° The final angle of refraction of the violet light is

⎛ nglass sin θ 2i ⎞ ⎛ 1.66sin 32.5° ⎞ θ 2r = sin −1 ⎜ = sin −1 ⎜ ⎟⎠ = 63.1° ⎟ ⎝ nair 1.00 ⎝ ⎠ ! Following the same steps for the red light (nglass = 1.62) gives

θ1r = 28.2°, α = 61.8°, β = 58.2°, θ2i = 31.8° and θ2r = 58.6° Thus, the angular dispersion of the emerging light is

Dispersion = θ 2r violet −θ 2r red = 63.1° − 58.6° = 4.5° !

Topic 22

1342

Topic 23

1343

Topic 23 Mirrors and Lenses

QUICK QUIZZES 23.1

At C.

23.2

Choice (c). Since nwater > nair, the virtual image of the fish formed by refraction at the flat water surface is closer to the surface than is the fish. See Equation 23.9 in the textbook.

23.3

(a) False. A concave mirror forms an inverted image when the object distance is greater than the focal length. (b) False. The magnitude of the magnification produced by a concave mirror is greater than 1 if the object distance is less than the radius of curvature.

Topic 23

1344

Choice (b). In this case, the index of refraction of the lens material is less than that of the surrounding medium. Under these conditions, a biconvex lens will be divergent.

23.5

Although a ray diagram only uses 2 or 3 rays (those whose direction is easily determined using only a straight edge), an infinite number of rays leaving the object will always pass through the lens.

23.6

(a) False. A virtual image is formed on the left side of the lens if p < f. (b) True. An upright, virtual image is formed when p < f while an inverted, real image is formed when p > f. (c) False. A magnified, real image is formed if 2 f > p > f and a magnified, virtual image is formed if p < f.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 23.2

Three rules are helpful here: 1) Rays traveling through (or virtually extending through) the focus are reflected in a direction parallel to the principal axis. 2) Rays traveling through (or virtually extending through) the center are reflected back on themselves. 3) Rays may be traced either

Topic 23

1345

forward or backward. Applying these three rules gives the correct rays as: (a) B (b) R (c) R (d) R (e) R (f) B. 23.4

See the ray diagrams below. (a) True (b) False – the image magnification increases as the object moves from C to F. (c) True

23.6

Light rays diverge from the position of a virtual image just as they do from an actual object. Thus, a virtual image can be as easily photographed as any object can. Of course, the camera would have to be placed near the axis of the lens or mirror in order to intercept the light rays.

23.8

Actually no physics is involved here. The design is chosen so your eyelashes will not brush against the glass as you blink. A reason involving a little physics is that with this design, when you direct your gaze near the outer circumference of the lens, you receive a ray that has

Topic 23

1346

passed through glass with more nearly parallel surfaces of entry and exit. Then the lens minimally distorts the direction to the object you are looking at. 23.10

Both words are inverted. However, OXIDE looks the same right-side-up and upside-down. LEAD does not.

23.12

(a) No. The screen is needed to reflect the light toward your eye. (b) Yes. The light is traveling toward your eye and diverging away from the position of the image, the same as if the object were located at that position.

23.14

Light rays reflecting from the fish, and passing through the flat refracting surface of the water (nwater > nair) on the way to the fisherman’s eye, form a virtual image located closer to the surface than the fish’s actual depth (see Example 23.6 in the textbook). Thus, the fisherman should aim below the apparent location of the fish, and (b) is the correct choice.

ANSWERS TO EVEN NUMBERED PROBLEMS 23.2

1.50 m

23.4

4.58 m

Topic 23

1347

23.6

(a)

R = + 2.22 cm

(b)

M = + 10.0

23.8

R = −0.790 cm

23.10

The mirror is concave with R ≈ 60 cm and f ≈ 30 cm.

23.12

(a)

p = +15.0 cm

(b)

|R| = 60.0 cm

23.14

(a)

2.50 m

(b)

3.46 m

(d)

real

(e)

inverted

23.16

(a)

−24.0 m

(b)

0.375 m

23.18

50.0 cm

23.20

(a) As the ball moves from p = 3.00 m to p = 0.500 m, the image moves

(c)

−0.384

from q = +0.600 m to q = +∞. As the ball moves from p = 0.500 m to p = 0 the image moves from q = −∞ to q = 0. (b) at t = 0.639 s when p = 1.00 m, and at t = 0.782 s when p = 0 23.22

in the water, 9.00 cm inside the wall of the bowl

23.24

(a)

1.50 m

(b)

1.75 m

23.26

(a)

−0.656 m

(b)

0.874

23.28

See Solution.

23.30

(a)

(b)

diverging

−13.8 cm

(c)

−13.8 cm

Topic 23

23.32

23.34

1348

(d)

−3.67 cm

(e)

−10.8 cm

(a)

See Solution.

(b)

40 cm behind the mirror

(d)

The mirror equation yields q = −40.0 cm and M = +2.00.

(a)

See Solution.

(b)

(c)

See Solution.

(c) Graph (a) yields an upright, virtual image located 13.3 cm from the lens and one-third the size of the object. Graph (b) yields an upright, virtual image located 6.7 cm from the lens and two-thirds the size of the object. Both agree with the algebraic solutions. 23.36

f = +5.68 cm

23.38

(a)

12.3 cm to the left of the lens

(c)

See Solution.

(a)

q = pf/(p − f)

(c)

q > 0 only if p > f when f > 0

23.40

(b)

M = +0.615

q < 0 for all values of p > 0 and f < 0

23.42. (a)

4.91 cm

(b)

0.727

23.44

(a)

13.3 cm

(b)

M = −5.91

23.46

(a)

q1 = +30.0 cm

(b)

20.0 cm beyond the second lens

(c)

p2 = −20.0 cm

(d)

q2 = +4.00 cm

(c)

(e)

inverted

M1 = −2.00

Topic 23

23.48

23.50

23.52

1349

(f) M2 = +0.200

(g) Mtotal = −0.400

(a)

q = 5f/4

(b)

(c)

real, inverted, opposite side

(a)

f = −12.0 cm

(b)

q = f = −12.0 cm

(d)

q = −6.00 cm

(e)

q = −4.00 cm

(a)

25.3 cm to the right of the mirror

(b)

virtual

23.54

Δx = 21.3 cm

23.56

See the Solution.

23.58

See the Solution.

23.60

(a)

p = 4f/3

(c)

Ma = −3, Mb = +4

(h)

real, inverted

(c)

q = −9.00 cm

(d)

M = +8.05

(b)

See Solution.

at the 30.0-cm mark (c)

f = −20.0 cm

M = −1/4

(c)

upright

(b)

p = 3f/4

23.62

fmirror = 11.7 cm

23.64

(a)

0.708 cm in front or the spherical ornament

23.66

(a)

convex

(b)

Topic 23

1350

PROBLEM SOLUTIONS 23.1

(a) Due to the finite value of the speed of light, the light arriving at your eye must have reflected from your face at a slightly earlier time. Thus, the image viewed in the mirror shows you younger than your actual age. (b) If you stand 40 cm in front of the mirror, the time required for light scattered from your face to travel to the mirror and back to your eye is 2 ( 0.40 m ) 2d Δt = = = 2.7 × 10−9 s or ∼10−9 s 8 c 3.0 × 10 m/s !

23.2

For a flat mirror, p = |q|. The camera focuses on the image a distance d = p + |q| = 2p from the object. The object distance is then p = d/2 = 3.00 m/2 = 1.50 m .

23.3

(1) The first image in the left-hand mirror is 5.00 ft behind the mirror or 10.0 ft from the person. (2) The first image in the right-hand mirror serves as an object for the left-hand mirror. It is located 10.0 ft behind the right-hand mirror,

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1351

which is 25.0 ft from the left-hand mirror. Thus, the second image in the left-hand mirror is 25.0 ft behind the mirror or or 30.0 ft from the person. (3) The first image in the left-hand mirror serves as an object for the right-hand mirror. It is located 20.0 ft in front of the right-hand mirror and forms an image 20.0 ft behind that mirror. This image then serves as an object for the left-hand mirror. The distance from this object to the left-hand mirror is 35.0 ft. Thus, the third image in the left-hand mirror is 35.0 ft behind the mirror or 40.0 ft from the person. 23.4

The virtual image is as far behind the mirror as the choir is in front of the mirror. Thus, the image is 5.30 m behind the mirror.

The image of the choir is 0.800 m + 5.30 m = 6.10 m © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 23

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from the organist. Using similar triangles gives

h" 6.10 m = !0.600 m 0.800 m or

23.5

⎛ 6.10 m ⎞ h" = ( 0.600 m ) ⎜ = 4.58 m ⎝ 0.800 m ⎟⎠ !

When an object is in front of a plane mirror, that mirror forms an upright, virtual image that is the same size as the object and as far behind the mirror as the object is in front of the mirror. This statement is true even if the mirror is rotated, as shown in the ray diagrams given below. In figure (a), a real object O1 is distance p1 in front of the upper mirror in the periscope. This mirror forms the virtual image I1 at distance p1 behind the mirror. As shown in figure (b), this image serves as the object for the lower mirror in the periscope, and is distance p2 = p1 + h in front of the lower mirror. The lower mirror then forms the final image I2, an upright, virtual image, located distance p2 = p1 + h behind this mirror.

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1353

(a) As shown in the above ray diagrams, the final image is distance |q2| = p1 + h behind the lower mirror, where p1 is the distance from the original object to the upper mirror and h is the vertical distance between the two mirrors in the periscope. (b) The final image is behind the mirror and is virtual. (c) As seen from the ray diagrams, the final image I2 is oriented the same way as the original object, and is therefore upright. (d) The overall magnification is ⎛ q ⎞⎛ q ⎞ ⎛ p ⎞⎛ p ⎞ M = M1M2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ( +1) ( +1) = +1 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p1 ⎠ ⎝ p2 ⎠ !

The final image is therefore upright and the same size as the original object. (e) No. The images formed by plane mirrors are upright in both

Topic 23

1354

directions. Just as plane mirrors do not reverse up and down, neither do they reverse left and right. 23.6

(a) Since the object is in front of the mirror, p > 0 With the image behind the mirror, q < 0 The mirror equation gives the radius of curvature as

2 1 1 1 1 10 − 1 = + = − = !R p q 1.00 cm 10.0 cm 10.0 cm or

⎛ 10.0 cm ⎞ R = 2⎜ ⎟ = +2.22 cm ⎝ 9 ⎠ !

q ( −10.0 cm ) = +10.0 (b) The magnification is M = − = p 1.00 cm !

23.7

(a) The center of curvature of a convex mirror is behind the mirror. Therefore, the radius of curvature, and hence the focal length f = R/2 is negative. With the image behind the mirror, the image is virtual and q = –10.0 cm. The mirror equation then gives

( −10.0 cm )( −15.0 cm ) = +30.0 cm qf p= = q − f −10.0 cm − ( −15.0 cm ) ! The object should be placed 30.0 cm in front of the mirror. (b) The magnification of the mirror is

Topic 23

1355

Therefore, the image is upright and one-third the size of the object. 23.8

The lateral magnification is given by M = −q/p. Therefore, the image distance is q = −Mp = −(0.013 0)(30.0 cm) = −0.390 cm

The mirror equation,

R= !

2 1 1 2pq = + ,!or!R = ,!gives R p q p + q !

2 ( 30.0 cm ) ( −0.390 cm ) = −0.790 cm 30.0 cm − 0.390 cm

The negative sign tells us that the surface is convex, as expected. 23.9

(a) The center of curvature of a concave mirror is in front of the mirror. Therefore, both the radius of curvature and the focal length, f = R/2 are positive. Since the image is virtual, the image distance is negative and q = −20.0 cm. With R = +40.0 cm and f = +20.0 cm, the mirror equation gives

( −20.0 cm )( +20.0 cm ) = +10.0 cm qf p= = q − f −20.0 cm − ( +20.0 cm ) ! Thus, the object should be placed 10.0 cm in front of the mirror. (b) The magnification of the mirror is

Topic 23

1356

( −20.0 cm ) = +2.00 q M=− =− p +10.0 cm ! Therefore, the image is upright and twice the size of the object. 23.10

The image was initially upright but became inverted when Dina was more than 30 cm from the mirror. From this information, we know that the mirror must be concave, because a convex mirror will form only upright, virtual images of real objects. When the object is located at the focal point of a concave mirror, the rays leaving the mirror are parallel, and no image is formed. Since Dina observed that her image disappeared when she was about 30 cm from the mirror, we know that the focal length must be f ≈ 30 cm. Also, for spherical mirrors, R = 2f. Thus, the radius of curvature of this concave mirror must be R ≈ 60 cm.

23.11

The enlarged, virtual images formed by a concave mirror are upright, so M > 0.

q h" 5.00 cm = +2.50 , giving Thus, M = − = = p h 2.00 cm ! q = −2.50p = −2.50(+3.00 cm) = −7.50 cm The mirror equation then gives

Topic 23

1357

f= !

23.12

( 3.00 cm )( −7.50 cm ) = +5.00 cm pq = p+q 3.00 cm − 7.50 cm

Realize that the magnitude of the radius of curvature, |R|, is the same for both sides of the hubcap. For the convex side, R = −|R| and for the concave side, R = +|R|. The object distance p is positive (real object) and has the same value in both cases. Also, we write the virtual image distance as q = −|q| in each case. The mirror equation then gives:

For the convex side,

1 2 1 = − −q −R p !

For the concave side,

1 −q

2 R

−

1 p

Rp R + 2p

[1]

[2]

R − 2p

Comparing Equations [1] and [2], we observe that the smaller magnitude image distance, |q| = 10.0 cm, occurs with the convex side of the mirror. Hence, we have

1 2 1 = − !−10.0 cm − R p

[3]

and for the concave side, |q| = 30.0 cm gives

1 2 1 = − !−30.0 cm R p

[4]

Topic 23

1358

(a) Adding Equations [3] and [4] yields

(b) Subtracting [3] from [4] gives

23.13

2 3+1 = , or p = +15.0 cm. !p 30.0 cm

4 3−1 = , or |R| = 60.0 cm. ! R 30.0 cm

The image is upright, so M > 0 and we have

q M = − = +2.0 or q = −2.0p = −2.0(25 cm) = −50 cm p ! The radius of curvature is then found to be

2 1 1 1 1 2−1 ⎛ 0.50 m ⎞ = + = − = !or!R = 2 ⎜ = 1.0 m ⎝ +1 ⎟⎠ R p q 25 cm 50 cm 50 cm ! 23.14

(a) The mirror’s focal length is f = R/2 = 2.50 m . (b) Applying the sign conventions, p = +9.00 m and f = +2.50 m. Substitute into the mirror equation to find the image distance:

( 2.50 m )( 9.00 m ) = 3.46 m 1 1 1 fp + = → q= = p q f p− f 9.00 m − 2.50 m (c) The mirror’s magnification is M=−

q 3.46 m =− = −0.384 m p 9.00 m

(d) The image distance q > 0 so, by the sign convention, the image is real . This result is confirmed by a ray diagram.

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1359

(e) The magnification is negative, so the image is inverted . A ray diagram confirms this result. 23.15

The focal length of the mirror may be found from the given object and image distances as 1/f = 1/p + 1/q or

f= !

(152 cm )(18.0 cm ) = +16.1 cm pq = p + q 152 cm + 18.0 cm

For an upright image twice the size of the object, the magnification is M = −q/p = +2.00, giving q = −2.00p Then, using the mirror equation again, 1/p + 1/q = 1/f becomes

1 1 1 1 2−1 1 + = − = = !p q p 2.00 p 2.00 p f or 23.16

f 16.1 cm p= = = 8.05 cm 2.00 ! 2.00

Applying the sign convention, find that p = +40.0 cm and q = −15.0 cm. (a) The focal length is given by the mirror equation: 1 1 1 + = p q f

→

1 1 1 − = 40.0 cm 15.0 cm f

→ f = −24.0 cm

(b) The mirror’s magnification is

M=−

q −15.0 cm =− = 0.375 p 40.0 cm

Topic 23

23.17

1360

(a) Since the mirror is convex, R < 0. Thus, R = −0.550 m and f = R/2 = −0.275 m. The mirror equation then yields

( +10.0 m )( −0.275 m ) = −0.268 m = −26.8 cm pf q= = p − f +10.0 m − ( −0.275 m ) ! The image is located 26.8 cm behind the mirror. (b) The magnification of the image is M = −q/p. Since p > 0 and q < 0 in this case, we see that M > 0. Therefore, the image is upright

(c)

23.18

M=−

q p

=−

(−26.8 cm) = − (−26.8 cm) = +2.68 × 10 = +0.026 8 −2

10.0 m

10.0 × 10 cm

A spherical mirror has a focal length equal to half its radius: f = R/2 so that R = 2f = 50.0 cm .

23.19

(a) An image formed on a screen is a real image. Thus, the mirror must be concave since, of mirrors, only concave mirrors can form real images of real objects. (b) The magnified, real images formed by concave mirrors are inverted, so M < 0 and

q q 5.0 m M = − = −5,!or!p = = = 1.0 m p 5 5 ! The object should be 1.0 m in front of the mirror. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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1361

(a—revisited) Now that we have both the object distance and image distance, we can use the mirror equation to be more specific about the required mirror. The needed focal length of the concave mirror is given by

f= !

23.20

(a) From

(1.0 m ) + ( 5.0 m ) = 0.83 m pq = p+q 1.0 m + 5.0 m

1 1 2 (1.00 m ) p Rp + = = , we find q = . !p q R% ! 2p − R 2p − 1.00 m

The table gives the image position at a few critical points in the motion. Between p = 3.00 m and p = 0.500 m, the real image moves from 0.600 m to positive infinity. From p = 0.500 cm to p = 0, the virtual image moves from negative infinity to 0.

Object Distance, p

Image Distance, q

3.00 m

0.600 m

0.500 m

±∞

Note the “jump” in the image position as the ball passes through the focal point of the mirror. (b) The ball and its image coincide when p = 0 and when 1/p + 1/p = 2/p =

Topic 23

1362

2/R, or p = R = 1.00 m. From Δy = v0yt + 12 ayt 2 , with v0y = 0, the times for the ball to fall from y ! = +3.00 m to these positions are found to be

23.21

p = 1.00 m position

t= !

p = 0 position

t= !

From

2 ( Δy ) 2 ( −2.00 m ) = = 0.639 s and ay −9.80 m s 2 2 ( −3.00 m ) = 0.782 s −9.80 m s 2

n1 n2 n2 − n1 + = , with R → ∞ the image position is found to be p q R !

n ⎛ 1.00 ⎞ q = − 2 p = −⎜ ( 50.0 cm ) = −38.2 cm ⎝ 1.309 ⎟⎠ n 1 ! or the virtual image is 38.2 cm below the upper surface of the ice. 23.22

The location of the image formed by refraction at this spherical surface is described by Equation 23.7 from the textbook, which states n1/p + n2/q = (n2 − n1)/R and uses the sign convention of Table 23.2. As light crosses this surface, passing from the water into air, we have n1 = nwater = 1.333, n2 = nair = 1.00, p = +10.0 cm (object is in front of the surface), and R = −15.0 cm (center of curvature is in front of the surface). Thus, the image distance is found from

Topic 23

1363

n2 n2 − n1 n1 = − R p !q

1.00 1.00 − 1.333 1.333 +0.666 − 3.999 −3.333 = − = = q −15.0 cm +10.0 cm 30.0 cm 30.0 cm

This yields q = 30.0 cm/(−3.333) = −9.00 cm. Thus, the goldfish appears to be in the water, 9.00 cm inside the wall of the bowl.

23.23

Since the center of curvature of the surface is on the side the light comes from, R < 0 giving R = −4.0 cm. Then,

n1 n2 n2 − n1 + = becomes R !p q

1.00 1.00 − 1.50 1.50 = − , or q = −4.0 cm −4.0 cm 4.0 cm ! q ⎛n ⎞q h" Thus, the magnification M = = − ⎜ 1 ⎟ gives h ⎝ n2 ⎠ p !

⎛ n q⎞ 1.50 ( −4.0 cm ) h" = − ⎜ 1 ⎟ h = − ( 2.5 mm ) = 3.8 mm 1.00 ( 4.0 cm ) ⎝ n2 q ⎠ !

23.24

For a plane refracting surface, R → ∞ and

n1 n2 n2 − n1 + = becomes p q R !

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1364

n q=− 2 p. n1 ! (a) Considering the bottom of the pool as the object, p = 2.00 m when the pool is full, and

⎛ 1.00 ⎞ q = −⎜ ⎟⎠ ( 2.00 m ) = −1.50 m ⎝ 1.333 ! or the pool appears to be 1.50 m deep. (b) If the pool is half filled, then p = 1.00 m and q = −0.750 m. Thus, the bottom of the pool appears to be 0.75 m below the water surface or 1.75 m below ground level. 23.25

As parallel rays from the Sun (object distance, p → ∞) enter the transparent sphere from air (n1 = 1.00) the center of curvature of the surface is on the side the light is going toward (back side). Thus, R > 0 It is observed that a real image is formed on the surface opposite the Sun, giving the image distance as q = +2R. Then,

n1 n2 n2 − n1 + = p q R !

becomes

n n − 1.00 = 2R R

which reduces to n = 2n − 2.00 and gives n = 2.00. 23.26

(a) For an image formed by refraction,

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n1 n2 n2 − n1 + = p q R

where n1 is the index of refraction for the object’s medium and n2 is index of refraction on the other side of the surface. Here, n1 = 1.333 and n2 = 1.000. Substitute values to find the image distance q: 1.000 1.000 − 1.333 1.333 = − → q = −0.656 m q 1.75 m 1.00 m

(b) The magnification is

M=−

23.27

n1q (1.333)( −0.656 m ) = 0.874 =− n2 p (1.000)(1.00 m )

In the drawing below, object O (the jellyfish) is located distance p in front of a plane water-glass interface. Refraction at that interface produces a virtual image I' at distance |q′| in front it. This image serves as the object for refraction at the glass-air interface. This object is located distance p′ = |q′| + t in front of the second interface, where t is the thickness of the layer of glass. Refraction at the glass-air interface produces a final virtual image, I, located distance |q| in front of this interface.

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From n1/p + n2/q = (n2 − n1)/R with R → ∞ for a plane, the relation between the object and image distances for refraction at a flat surface is q = −(n2/n1)p. Thus, the image distance for the refraction at the water-glass interface is q′ = −(ng/nw)p. This gives an object distance for the refraction at the glass-air interface of p′ = (ng/nw)p + t and a final image position (measured from the glass-air interface) of

⎡⎛ n ⎞ ⎛n ⎞ ⎤ ⎤ n n ⎡⎛ ng ⎞ q = − a p' = − a ⎢⎜ ⎟ p + t ⎥ = − ⎢⎜ a ⎟ p + ⎜ a ⎟ t ⎥ . ng ng ⎣⎝ nw ⎠ ⎝ ng ⎠ ⎥⎦ ⎢⎣⎝ nw ⎠ ⎦ ! (a) If the jellyfish is located 1.00 m (or 100 cm) in front of a 6.00-cm thick pane of glass, then !p = +100 cm,!t = 6.00 cm , and the position of the final image relative to the glass-air interface is

⎡⎛ 1.00 ⎞ ⎤ ⎛ 1.00 ⎞ q = − ⎢⎜ 100 cm ) + ⎜ 6.00 cm ) ⎥ = −79.0 cm ( ( ⎟ ⎟ ⎝ 1.50 ⎠ ⎣⎝ 1.333 ⎠ ⎦ ! It appears to be in the water, 79.0 cm back of the outer surface of the

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1367

glass pane. (b) If the thickness of the glass is negligible (t → 0) the distance of the final image from the glass-air interface is

⎤ ⎛n ⎞ n ⎡⎛ ng ⎞ ⎛ 1.00 ⎞ q = − a ⎢⎜ ⎟ p + 0 ⎥ = − ⎜ a ⎟ p = − ⎜ (100 cm ) = −75.0 cm ⎝ 1.333 ⎟⎠ ng ⎣⎝ nw ⎠ ⎝ nw ⎠ ⎦ ! Now, it appears to be 75.0 cm back of the outer surface of the glass pane. (c) Comparing the results of parts (a) and (b), we see that the 6.00-cm thickness of the glass in part (a) made a 4.00-cm difference in the apparent position of the jellyfish. We conclude that the thicker the glass, the greater the distance between the final image and the outer surface of the glass. 23.28

We assume the ray shown in the diagram below is a paraxial ray so θ1 and θ2 are both sufficiently small to allow us to write Snell’s law as n1 tan θ1 = n2 tan θ2

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1368

Also, note that we are assuming transverse distances measured upward from the axis OPI are positive, while those measured downward from this axis are negative. Looking at the right triangle having distance !OP as its base, we see that tan θ1 = h/p, and looking at the triangle having !PI as a base, we have tan θ2 = |h′|/q = −h′/q. Thus, Snell’s law becomes n1(h/p) = −n2(h′/q) and the magnification becomes

M= ! 23.29

h$ nq =− 1 h n1p

With R1 = +2.00 cm and R2 = +2.50 cm, the lens maker’s equation gives the focal length as ⎛ 1 ⎛ 1⎞ 1 1 ⎞ −1 = n − 1 ⎜ − ⎟ = 1.50 − 1 ⎜ − ⎟ = +0.050 0 cm f ⎝ 2.00 cm 2.50 cm ⎠ ⎝ Rt R2 ⎠

( )

23.30

1 0.050 0 cm −1

(

)

= +20.0 cm

(a) Solve for the focal length using the lens-maker’s equation:

⎛ 1 1 1⎞ 1 1 ⎛ ⎞ = ( n − 1) ⎜ − ⎟ = (1.67 − 1) ⎜ − ⎝ −12.0 cm 40.0 cm ⎟⎠ f ⎝ R1 R2 ⎠ f = −13.8 cm

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1369

(b) The focal length f < 0 so, from the sign convention, the lens is

diverging . Apply the thin-lens equation to find the image distances:

23.31

(c)

1 1 1 1 1 = − = − → q = −13.8 cm q f p −13.8 cm ∞

(d)

1 1 1 1 1 = − = − → q = −3.67 cm q f p −13.8 cm 5.00 cm

(e)

1 1 1 1 1 = − = − → q = −10.8 cm q f p −13.8 cm 50.0 cm

The focal length of a converging lens is positive, so f = +10.0 cm. The thinp (10.0 cm ) pf = lens equation then yields an image distance of q = . ! p − f p − 10.0 cm

( 20.0 cm )(10.0 cm ) = +20.0 cm (a) When p = + 20.0 cm q = , and 20.0 cm − 10.0 cm ! q M = − = −1.00 , so the image is located 20.0 cm beyond the lens is p ! real (q > 0), is inverted (M < 0), and is the same size as the object (|M| = 1.00). (b) When p = f = +10.0 cm, the object is at the focal point and no image is formed. Instead, parallel rays emerge from the lens.

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( 5.00 cm )(10.0 cm ) = −10.0 cm (c) When p = 5.00 cm, q = , and 5.00 cm − 10.0 cm ! q M = − = +2.00 , so the image is located 10.0 cm in front of the lens, is p ! virtual (q < 0), is upright (M > 0), and is twice the size of the object (|M| = 2.00). 23.32

(a) To approximate paraxial rays, the rays should be drawn so they reflect at the vertical plane that passes through the vertex of the mirror, rather than at the mirror’s surface as is generally done in the textbook. For this reason, the concave surface of the mirror appears flat in the ray diagram given below.

(b) In the diagram above, each square of the grid represents 5 cm. Thus, we see that the image I is located 40 cm behind the mirror, or q = −40 cm. (c) From the above ray diagram, the height h′ of the upright image I is

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1371

seen to be twice the height h of the object O. Therefore, the magnification is M = h′/h = +2. (d) From the mirror equation, the image distance is q = pf/(p − f), or

q= !

( +20.0 cm )( +40.0 cm ) = −40.0 cm 20.0 cm − 40.0 cm

and the magnification is

23.33

( −40.0 cm ) = +2.00 q M=− =− p +20.0 cm !

For a divergent lens, the focal length is negative. Hence, f = −20.0 cm in this case. The thin-lens equation gives the image distance as q = pf/(p − f) and the magnification is given by M = −q/p. (i) (a)

( 40.0 cm )( −20.0 cm ) = −13.3 cm ⇒ q= 40.0 cm − ( −20.0 cm ) ! (b)

q < 0 ⇒ virtual image

(c)

M = − q/p > 0 ⇒ upright image

(d)

( −13.3 cm ) = +0.333 M=− +40.0 cm !

13.3 cm!in!front!of!the!lens .

(ii) (a)

20.0 cm( − 20.0 cm) q= = −10.0 cm ⇒ ! 20.0 cm − (−20.0 cm)

10.0 cm!in!front!of!the!lens .

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1372

(b)

q < 0 ⇒ virtual image

(c)

M = −q/p > 0

(d)

( −10.0 cm ) = +0.500 M=− +20.0 cm !

⇒ upright image

(iii) (a)

(10.0 cm)(−20.0 cm) q=− = −6.67 cm ⇒ 10.0 cm − (−20.0 cm) !

23.34

(b)

q<0

(c)

M = −q/p > 0

(d)

( −6.67 cm ) = +0.667 M=− +10.0 cm !

(a) and (b)

6.67 cm!in!front!of!the!lens

virtual image

⇒ ⇒

upright image

Your scale drawings should look similar to those given

below:

Figure (a)

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1373

Figure (b) A carefully drawn-to-scale version of Figure (a) should yield an upright, virtual image located 13.3 cm in front of the lens and onethird the size of the object. Similarly, a carefully drawn-to-scale version of Figure (b) should yield an upright, virtual image located 6.7 cm in front of the lens and two-thirds the size of the object. (c) The results of the graphical solution are consistent with the algebraic answers found in Problem 23.33, allowing for small deviances due to uncertainties in measurement. Graphical answers may vary, depending on the size of the graph paper and accuracy of the drawing. 23.35

(a) The real image case is shown in the ray diagram below. Notice that p + q = 12.9 cm or q = 12.9 cm − p. The thin-lens equation, with f = 2.44 cm, then gives

1 1 1 + = !p 12.9 cm − p 2.44 cm

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1374

p2 − (12.9 cm)p + 31.5 cm2 = 0

Using the quadratic formula to solve gives p = 9.63 cm or p = 3.27 cm Both are valid solutions for the real image case. (b) The virtual image case is shown in the second diagram. Note that in this case, q = −(12.9 cm + p) so the thin-lens equation gives

1 1 1 + = !p 12.9 cm + p 2.44 cm or

p2 + (12.9 cm)p − 31.5 cm2 = 0

The quadratic formula then gives p = 2.10 cm or p = −15.0 cm.

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1375

Since the object is real, the negative solution must be rejected, leaving p = 2.10 cm. 23.36

We must first realize that we are looking at an upright, enlarged, virtual image. Thus, we have a real object located between a converging lens and its front-side focal point, so q < 0, p > 0, and f > 0.

q The magnification is M = − = +2 giving q = −p. Then, from the thin-lens p ! equation,

23.37

1 1 1 1 − =+ = or f = 2 p = 2(2.84 cm) = 5.68 cm. p 2p 2p f !

(a) The lens is converging so that f = +10.0 cm. Apply the thin-lens equation to find the image distance q: 1 1 1 1 1 1 = − → = − → q = 16.7 cm q f p q 10.0 cm 25.0 cm

(b) The magnification is M=−

q 16.7 cm =− = −0.668 p 25.0 cm

(c) The image height is h’ = Mh = (−0.668)(8.00 cm) = −5.34 cm . (d) The image distance q > 0 so, from the sign convention, the image is real . A ray diagram confirms that real rays converge on the image.

(e) The image height is negative so, from the sign convention, the image

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is inverted . 23.38

(a) This is a real object, so the object distance is p = +20.0 cm. The thinlens equation gives the image distance as

( 20.0 cm )( −32.0 cm ) = −12.3 cm pf q= = p − f 20.0 cm − ( −32.0 cm ) !

so the image is 12.3 cm to the left of the lens.

( −12.3 cm ) = +0.615 q (b) The magnification is M = − = − p +20.0 cm ! (c) The ray diagram for this arrangement is shown above. 23.39

Since the light rays incident to the first lens are parallel, p1 = ∞, and the thin-lens equation gives q1 = f1 = −10.0 cm. The virtual image formed by the first lens serves as the object for the second lens, so p2 = 30.0 cm + |q1| = 40.0 cm. If the light rays leaving the second lens are parallel, then q2 = ∞, and the thin-lens equation gives f2 = p2 = 40.0 cm.

Topic 23

23.40

1377

(a) Solving the thin-lens equation for the image distance q gives

1 1 1 p− f = − = q f p pf !

pf p− f

(b) For a real object, p > 0 and p = |p|. Also, for a diverging lens, f < 0 and f = −|f|. The result of part (a) then becomes

(

)

p − f p f q= =− p+ f p− − f !

(

)

Thus, we see that q < 0 for all numeric values of |p|and |f|. Since negative image distances mean virtual images, we conclude that a diverging lens will always form virtual images of real objects. (c) For a real object, p > 0 and p = |p|. Also, for a converging lens, f > 0 and f = |f|. The result of part (a) then becomes p f q= > 0! if p− f !

p − f >0

Since q must be positive for a real image, we see that a converging lens will form real images of real objects only when |p| > |f| (or p > f since both p and f are positive in this situation). 23.41

The thin-lens equation gives the image position for the first lens as

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1378

p f (30.0 cm)(15.0 cm) q1 = 1 1 = = +30.0 cm p − f 30.0 cm − 15.0 cm 1 1 ! q 30.0 cm = −1.00 . and the magnification by this lens is M1 = − 1 = − p 30.0 cm 1 ! The real image formed by the first lens serves as the object for the second lens, so p2 = 40.0 cm − q1 = +10.0 cm. Then, the thin-lens equation gives

p f (10.0 cm)(15.0 cm) q2 = 2 2 = = −30.0 cm p2 − f2 10.0 cm − 15.0 cm ! and the magnification by the second lens is

q (−30.0 cm) M2 = − 2 = − = +3.00 p 10.0 cm 2 ! Thus, the final, virtual image is located 30.0 cm in front of the seconds lens, and the overall magnification is M = M1M2 = (−1.00)(+3.00) = −3.00. 23.42

(a) For a combination of thin lenses, the image formed by the first lens is treated as the object for the second lens. Here, light from the object first reaches the converging lens so that pC = +2.00 cm and fC = +4.00 cm (where the subscript C indicates quantities associated with the converging lens). Find the object distance using the thin-lens equation:

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1379

1 1 1 = − qC fC pC

→ qC = −4.00 cm

This negative object distance indicates the object is virtual and located 4.00 cm to the left of the converging lens, at x = −4.00 cm. This virtual object becomes the image for the diverging lens (with D subscripts) so that pD = d − qC = 10.0 cm − (−4.00 cm) = 14.0 cm. With fD = −8.00 cm, apply the thin-lens equation to find the final image distance: 1 1 1 = − qD fD pD

→ qD = −5.09 cm

The negative object distance indicates the object is virtual and located 5.09 cm to the left of the diverging lens. The x-location of the image is then

ximage = 10.0 cm − 5.09 cm = 4.91 cm (b) The image’s overall magnification is the product of the magnifications from each successive lens:

⎛ q ⎞ ⎛ q ⎞ ⎛ −4.00 cm ⎞ ⎛ −5.09 cm ⎞ M = MC MD = ⎜ − C ⎟ ⎜ − D ⎟ = ⎜ − ⎟⎜− ⎟ = 0.727 ⎝ pC ⎠ ⎝ pD ⎠ ⎝ 2.00 cm ⎠ ⎝ 14.0 cm ⎠

23.43

p f (4.00 cm)(8.00 cm) = −8.00 cm From the thin-lens equation, q1 = 1 1 = p − f 4.00 cm − 8.00 cm 1 1 !

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1380

q (−8.00 cm) = +2.00 . The magnification by the first lens is M1 = − 1 = − p 4.00 cm 1 ! The virtual image formed by the first lens is the object for the second lens, so p2 = 6.00 cm + |q1| = +14.0 cm, and the thin-lens equation gives

p f (14.0 cm)(−16.0 cm) q2 = 2 2 = = −7.47 cm p2 − f2 14.0 cm − (−16.0 cm) ! q (−7.47 cm) = +0.533 , The magnification by the second lens is M2 = − 2 = − p2 14.0 cm ! so the overall magnification is M = M1M2 = (+2.00)(+0.533) = +1.07. The position of the final image is 7.47 cm in front of the second lens, and its height is h′ = Mh = (+1.07)(1.00 cm) = 1.07 cm. Since M > 0 the final image is upright and since q2 < 0 this image is virtual. 23.44

(a) We start with the final image and work backward. From Figure P23.44, observe that q2 = −(50.0 cm − 31.0 cm) = −19.0 cm. The thinlens equation then gives

q f (−19.0 cm)(20.0 cm) p2 = 2 2 = = +9.74 cm q − f −19.0 cm − 20.0 cm 2 2 ! The image formed by the first lens serves as the object for the second lens and is located 9.74 cm in front of the second lens. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Thus, the image distance for the first lens is q1 = 50.0 cm − 9.74 cm = 40.3 cm, and the thin-lens equation gives

q f (40.3 cm)(10.0 cm) p1 = 1 1 = = +13.3 cm q1 − f 1 40.3 cm − 10.0 cm ! The original object should be located 13.3 cm in front of the first lens. (b) The overall magnification is ⎛ q ⎞ ⎛ q ⎞ ⎛ 40.3 cm ⎞ ⎛ (−19.0 cm) ⎞ M = M1M2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − − = −5.91 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 13.3 cm ⎟⎠ ⎜⎝ 9.74 cm ⎟⎠ !

Note: Final answers to this problem are highly sensitive to round-off error. To avoid this, we retain extra digits in intermediate answers and round only the final answers to the correct number of significant figures. Since the final image is to be real and in the film plane, q2 = +d.

q f d(13.0 cm) Then, the thin-lens equation gives p2 = 2 2 = . q − f d − 13.0 cm 2 2 ! The object of the second lens (L2). is the image formed by the first lens (L1), so ⎛ 13.0 cm ⎞ d2 q1 = (12.0 cm − d ) − p2 = 12.0 cm − d ⎜ 1 + = 12.0 cm − d − 13 cm ⎟⎠ d − 13.0 cm ⎝ !

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1382

If d = 5.00 cm, then q1 = + 15.125 and when d = 10.0 cm, q1 = +45.3 cm.

q f q (15.0 cm) From the thin-lens equation, p1 = 1 1 = 1 . q1 − f1 q1 − 15.0 cm ! When q1 = +15.125 cm (d = 5.00 cm) then p1 = 1.82 × 103 cm = 18.2 m. When q1 = +45.3 cm (d = 10.0 cm), then p1 = 22.4 cm = 0.224 m. Thus, the range of focal distances for this camera is 0.224 m to 18.2 m. 23.46

(a) From the thin-lens equation, the image distance for the first lens is

p f (15.0 cm)(10.0 cm) q1 = 1 1 = = +30.0 cm p − f 15.0 cm − 10.0 cm 1 1 ! (b) With q1 = +30.0 cm, the image of the first lens is located 30.0 cm in back of that lens. Since the second lens is only 10.0 cm beyond the first lens, this means that the first lens is trying to form its image at a location 20.0 cm beyond the second lens. (c) The image the first lens forms (or would form if allowed to do so) serves as the object for the second lens. Considering the answer to part (b) above, we see that this will be a virtual object, with object distance p2 = −20.0 cm. (d) From the thin-lens equation, the image distance for the second lens is

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1383

p f (−20.0 cm)(5.00 cm) q2 = 2 2 = = +4.00 cm p − f −20.0 cm − 5.00 cm 2 2 ! q1 30.0 cm = −2.00 (e) M1 = − = − p 15.0 cm 1 ! (f)

q 4.00 cm M2 = − 2 = − = +0.200 p (−20.0 cm) 2 !

(g) Mtotal = M1M2 = (−2.00)(+0.200) = −0.400 (h) Since q2 > 0, the final image is real, and since Mtotal < 0, that image is inverted relative to the original object. 23.47

Since q = +8.00 cm when p = +10.0 cm, we find that

1 1 1 1 1 18.0 = + = + = ! f p q 10.0 cm 8.00 cm 80.0 cm Then, when p = 20.0 cm,

1 1 1 18.0 1 18.0 − 4.00 14.0 = − = − = = 80.0 cm 80.0 cm !q f p 80.0 cm 20.0 cm or

q= !

80.0 cm = +5.71 cm 14.0

Thus, a real image is formed 5.71 cm in front of the mirror. 23.48

(a) We are given that p = 5f, with both p and f being positive. The thinlens equation then gives

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1384

pf (5 f ) f 5f q= = = 4 ! p− f 5f − f q (5 f /4) 1 = − (b) M = − = − p 5f 4 !

(c) Since q > 0 the image is real. Because M < 0 the image is inverted. Since the object is real, it is located in front of the lens, and with q > 0 the image is located in back of the lens. Thus, the image is on the opposite side of the lens from the object. 23.49

In the sketch below, the center of curvature of the left side of the biconcave lens is on the side of the incoming light. Thus, by the convention of Table 23.2, the radius of curvature of this side is negative while the radius of curvature of the right side is positive.

If R1 = −32.5 cm and R2 = 42.5 cm, the lens maker’s equation gives the focal length of the lens as

⎛ 1 ⎞ 1 1 = ( n − 1) ⎜ − = (1 − n) ( 5.43 × 10−2 cm −1 ) ⎟ f −32.5 42.5 cm ⎝ ⎠

(a) For a very distant object (p → ∞) the thin-lens equation gives the

Topic 23

1385

image distance as q = f. Thus, if the index of refraction of the lens material is n = 1.53 for violet light,

1 q= f = = −34.7 cm 1 − 1.53 ) ( 5.43 × 10−2 cm −1 ) ( ! and the image of violet light is formed 34.7 cm to the left of the lens. (b) If n = 1.51 for red light, the image distance for very distance source emitting red light is

1 q= f = = −36.1 cm 1 − 1.53 ) ( 5.43 × 10−2 cm −1 ) ( ! The image of the very distant red light source is formed 36.1 cm to the left of the lens. 23.50

(a) Using the sign convention from Table 23.2, the radii of curvature of the surfaces are R1 = −15.0 cm and R2 = +10.0 cm. The lens maker’s equation then gives

⎛ ⎛ 1 1 1⎞ 1 1 ⎞ = ( n − 1) ⎜ − ⎟ = (1.50 − 1) ⎜ − or f ⎝ R1 R2 ⎠ ⎝ −15.0 cm 10.0 cm ⎟⎠

f = −12.0 cm (b) If p → ∞ then q = f = − 12.0 cm.

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1386

p ( −12.0 cm ) pf = The thin-lens equation gives q = and the p − f p + 12.0 cm !

following results: (c) If p = 3|f| = +36.0 cm, q = −9.00 cm. (d) If p = |f| = +12.0 cm, q = −6.00 cm. (e) If p = |f|/2 = +6.0 cm, q = −4.00 cm. 23.51

As light passes left to right through the lens, the image position is given by

p f (100 cm)(80.0 cm) q= 1 1 = = +400 cm p − f 100 cm − 80.0 cm 1 1 ! This image serves as an object for the mirror with an object distance of p2 = 100 cm − q1 = −300 cm (virtual object). From the mirror equation, the position of the image formed by the mirror is

p f (−300 cm)(−50.0 cm) q2 = 2 2 = = −60.0 cm p − f −300 cm − (−50.0 cm) 2 2 ! This image is the object for the lens as light now passes through it going right to left. The object distance for the lens is p3 = 100 cm − q2 = 100 cm − (−60.0 cm), or p3 = 160 cm. From the thin lens equation,

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1387

p f (160 cm)(80.0 cm) q3 = 3 3 = = +160.0 cm p − f 160 cm − 80.0 cm) 3 3 ! Thus, the final image is located 160 cm to the left of the lens. ⎛ q ⎞⎛ q ⎞⎛ q ⎞ The overall magnification is M = M1M2 M3 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ , or ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p3 ⎠ ! ⎛ 400 cm ⎞ ⎡ (−60.0 cm) ⎤ ⎛ 160 cm ⎞ M = ⎜− = −0.800 ⎢− ⎥ − ⎝ 100 cm ⎟⎠ ⎣ (−300 cm) ⎦ ⎜⎝ 160 cm ⎟⎠ !

Since M < 0 the final image is inverted. 23.52

(a) Since the object is midway between the lens and mirror, the object distance for the mirror is p1 = +12.5 cm. The mirror equation gives the image position as

1 2 1 2 1 5− 4 1 = − = − = = !q1 R p1 +20.0 cm 12.5 cm 50.0 cm 50.0 cm or

q1 = + 50.0 cm

This image serves as the object for the lens, so p2 = 25.0 cm − q1 = −25.0 cm. Note that since p2 < 0 this is a virtual object. The thin-lens equation gives the image position for the lens as

p f (−25.0 cm)(−16.7 cm) q2 = 2 2 = = −50.3 cm p − f −25.0 cm − (−16.7 cm) 2 2 !

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1388

Since q2 < 0 this image is located 50.3 cm in front of (to the right of) the lens or 25.3 cm behind (to the right of) the mirror. (b) With q2 < 0 the final image is a virtual image. (c) and (d)

The overall magnification is

⎛ q ⎞ ⎛ q ⎞ ⎛ 50.0 cm ⎞ ⎡ (−50.3 cm) ⎤ M = M1M2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − ⎢− ⎥ = +8.05 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ 12.5 cm ⎟⎠ ⎣ (−25.0 cm) ⎦ !

Since M > 0, the final image is upright. 23.53

A hemisphere is too thick to be described as a thin lens. The light is undeviated on entry into the flat face. We next consider the light’s exit from the curved surface, for which R = −6.00 cm.

The incident rays are parallel, so p = ∞.

Then,

n1 n2 n2 − n1 1.00 1.00 − 1.56 + = = becomes 0 + from which p q R q −6.00 cm ! !

q = 10.7 cm. 23.54

The diagram below shows a light ray traveling parallel to the principle

Topic 23

1389

axis of a plano-convex lens at distance h off the axis. This ray strikes the plane surface at normal incidence and passes into the glass undeviated.

The angle of incidence at the spherical surface (of radius R) is θ1 where sin θ1 = h/R. If the index of refraction of the lens material is n, Snell’s law gives the angle of refraction at the spherical surface as ⎛ nsin θ1 ⎞ ⎛ nh ⎞ θ 2 = sin −1 ⎜ = sin −1 ⎜ ⎟ ⎟ ⎝ R⎠ ⎝ nair ⎠ !

The distance x from the center of curvature of the spherical surface to the point where the refracted ray crosses the principle axis of the lens is

h x = R2 − h2 + tan φ !

where

φ = θ2 − θ1

If R = 20.0 cm, n = 1.60, and the first ray is distance h1 = 0.500 cm off the axis, we find ⎛ 0.500 cm ⎞ θ 1 = sin −1 ⎜ = 1.43° ⎝ 20.0 cm ⎟⎠ !

and

⎛ 1.60(0.500 cm) ⎞ θ 2 = sin −1 ⎜ ⎟⎠ = 2.29° 20.0 cm ⎝

Topic 23

1390

0.500 cm x1 = (20.0 cm)2 − (0.500 cm)2 + = 53.3 cm tan (2.29° − 1.43°) !

If the second ray is distance h2 = 12.0 cm from the axis, then ⎛ 12.0 cm ⎞ θ1 = sin −1 ⎜ = 36.9° ⎝ 20.0 cm ⎟⎠ !

giving

and

⎡ 1.60(12.0 cm) ⎤ θ1 = sin −1 ⎢ ⎥ = 73.7° ⎣ 20.0 cm ⎦

12.0 cm x2 = (20.0 cm 2 ) − (12.0 cm)2 + = 32.0 cm tan (73.7° − 36.9°) !

The distance between the points where these two rays cross the principle axis is then Δx = x1 − x2 = 53.3 cm − 32.0 cm = 21.3 cm 23.55

(a) With light going through the piece of glass from left to right, the radius of the first surface is positive and that of the second surface is negative according to the sign convention of Table 23.2. Thus, R1 = +2.00 cm, and R2 = −4.00 cm.

Applying

n1 n2 n2 − n1 + = to the first surface gives p q R !

1.00 1.50 1.50 − 1.00 + = , 1.00 cm q +2.00 cm 1 ! which yields q1 = −2.00 cm.. The first surface forms a virtual image 2.00 cm to the left of that surface, and 16.0 cm to the left of the second

Topic 23

1391

surface. The image formed by the first surface is the object for the second surface, so p2 = +16.0 cm, and

n1 n2 n2 − n1 + = gives p q R !

1.50 1.00 1.00 − 1.50 + = or q2 = +32.0 cm. q2 −4.00 cm !16.0 cm The final image is located 32.0 cm to the right of the second surface. (b) Since q2 > 0 the final image formed by the piece of glass is a real image. 23.56

Consider an object O1 at distance p1 in front of the first lens. The thin-lens equation gives the image position for this lens as

1 1 1 = − . q f1 p1 1 !

The image, I1, formed by the first lens serves as the object, O2, for the second lens. With the lenses in contact, this will be a virtual object if I1 is real and will be a real object if I1 is virtual. In either case, if the thicknesses of the lenses may be ignored,

Topic 23

1392

p2 = −q1

and

1 1 1 1 =− =− + q1 f1 p1 !p2

Applying the thin-lens equation to the second lens,

1 1 1 1 − + + = ! f1 p1 q2 f2

1 1 1 + = becomes p q1 f2 2 !

1 1 1 1 + = + p1 q2 f1 f2

Observe that this result is a thin-lens-type equation relating the position of the original object, O1 and the position of the final image, I2, formed by this two-lens combination. Thus, we see that we may treat two thin lenses in contact as a single lens having a focal length, f, given by

! 23.57

1 1 1 = + . f f1 f 2

From the thin-lens equation, the image distance for the first lens is

p f (40.0 cm)(30.0 cm) q1 = 1 1 = = +120 cm p − f 40.0 cm − 30.0 cm 1 1 ! q 120 cm and the magnification by this lens is M1 = − 1 = − = −3.00 . p 40.0 cm 1 ! The real image formed by the first lens serves as the object for the second lens, with object distance of p2 = 110 cm − q1 = −10.0 cm (a virtual object). The thin-lens equation gives the image distance for the second lens as

p f (−10.0 cm) f2 q2 = 2 2 = p2 − f2 −10.0 cm − f2 !

Topic 23

1393

(a) If f2 = −20.0 cm, then q2 = +20.0 cm, and the magnification by the second lens is M2 = −q2/p2 = −(20.0 cm)/(−10.0 cm) = +2.00. The final image is located 20.0 cm to the right of the second lens, and the overall magnification is M = M1M2(−3.00)(+2.00) = −6.00. (b) Since M < 0 the final image is inverted. (c) If f2 = +20.0 cm, then q2 = +6.67 cm, and

q 6.67 cm M2 = − 2 = − = +0.667 . p (−10.0 cm) 2 ! The final image is 6.67 cm to the right of the second lens, and the overall magnification is M = M1M2 = (−3.00)(+0.667) = −2.00. Since M < 0 the final image is inverted. 23.58

The diagram below gives a ray diagram showing how this mirror system forms a final image I2 located just above the opening in the upper mirror. The focal point of each mirror is at the center of the opposite mirror, so their focal length f is the vertical distance between the centers of the mirrors. The original object O1 sits just above the surface at the center of the lower mirror. This places it on the principle axis and just inside the

Topic 23

1394

focal point of the upper mirror.

Therefore, the object distance for the upper mirror is p1 = f − ε, where ε is a very small number approaching zero. The mirror equation then gives

p f ( f − ε )f f2 −ε f q1 = − 1 = =− = −δ p − f f − ε − f ε 1 !

and δ → ∞. when ε → 0

Thus, the upper mirror forms a virtual image, on the principle axis, far above this mirror. This image serves as the object for the lower mirror and is located far in front of that mirror. With !p2 ≈ q1 = δ  f , the image distance for the lower mirror is

p f δf δf q2 = 2 = ≈+ =+f p − f δ − f δ 2 ! Since q2 > 0, the final image I2 is a real image located just above the focal point of the lower mirror (at the opening in the upper mirror). The overall magnification for this mirror system is ⎛ q ⎞ ⎛ q ⎞ ⎛ −δ ⎞ ⎛ f ⎞ M = M1M2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ≈ ⎜ − ⎟ ⎜ − ⎟ = +1 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ f ⎠ ⎝ δ ⎠ ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 23

1395

This means that the final image is upright and the same size as the original object. When a flashlight beam is aimed at the final image, it passes through the opening in the upper mirror and reflects as shown in the ray diagram below. It emerges from the mirror as if it had reflected from that image, even obeying the law of reflection as it did so!

23.59

(a) The lens maker’s equation,

⎛ 1 1 1⎞ = (n − 1) ⎜ − ⎟ gives ⎝ R1 R2 ⎠ !f

⎛ ⎞ 1 1 1 = (n − 1) ⎜ − 5.00 cm ⎝ 9.00 cm −11.0 cm ⎟⎠ !

1 ⎛ 99.0 ⎞ which simplifies to n = 1 + ⎜⎝ ⎟ = 1.99 . 5.00 11.0 + 9.00 ⎠ ! (b) As light passes from left to right through the lens, the thin-lens equation gives the image distance as

p f (8.00 cm)(5.00 cm) q1 = 1 = = +13.3 cm p − f 8.00 cm − 5.00 cm 1 !

Topic 23

1396

This image formed by the lens serves as an object for the mirror with object distance p2 = 20.0 cm − q1 = +6.67 cm. The mirror equation then gives

p2 R (6.67 cm)(8.00 cm) q2 = = = +10.0 cm 2p2 − R 2(6.67 cm) − 8.00 cm ! This real image, formed 10.0 cm to the left of the mirror, serves as an object for the lens as light passes through it from right to left. The object distance is p3 = 20.0 cm − q2 = +10.0 cm, and the thin-lens equation gives

p f (10.0 cm)(5.00 cm) q3 = 1 = = +10.0 cm p − f 10.0 cm − 5.00 cm 3 ! The final image is located 10.0 cm to the left of the lens and its overall magnification is ⎛ q ⎞ ⎛ q ⎞ ⎛ q ⎞ ⎛ 13.3 ⎞ ⎛ 10.0 ⎞ ⎛ 10.0 ⎞ M = M1M2 M2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ = ⎜ − ⎟⎜− ⎟⎜− ⎟ = −2.50 ⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p3 ⎠ ⎝ 8.00 ⎠ ⎝ 6.67 ⎠ ⎝ 10.0 ⎠ !

23.60

qf From the thin-lens equation, the object distance is p = . ! q− f (4 f ) f 4f = (a) If q = +4f then p = or 1.33 f. 4f − f 3 !

Topic 23

1397

(−3 f ) f = 3 f /4 or 0.750 f . (b) When q = −3f, we find p = −3 f − f ! q 4f = −3 , and in case (b), (c) In case (a), M = − = − p 4 f /3 ! q −3 f M=− =− = +4 . p 3 f /3 ! 23.61

If R1 = −3.00 m and R2 = −6.00 m, the focal length is given by

⎞⎛ 1 ⎛ n1 1 1 ⎞ ⎛ n2 − n1 ⎞ ⎛ −1 ⎞ = ⎜ − 1⎟ ⎜ + = f ⎝ n2 ⎠ ⎝ −3.00 m 6.00 m ⎟⎠ ⎜⎝ n2 ⎟⎠ ⎜⎝ 6.00 m ⎟⎠

(6.00 m)n2 n2 − n1

(a) If n1 = 1.50 and n2 = 1.00, then f = !

[1]

(6.00 m)(1.00) = −12.0 m 1.00 − 1.50

pf (10.0 m)(−12.0 m) = = −5.45 m . The thin-lens equation gives q = p − f 10.0 m + 12.0 m ! A virtual image is formed 5.45 m to the left of the lens. (b) If n1 = 1.50 and n2 = 1.33, the focal length is

(6.00 m)(1.33) = −46.9 m , and 1.33 − 1.50

pf (10.0 m)(−46.9 m) q= = = −8.24 m p − f 10.0 m + 46.9 m ! The image is located 8.24 m to the left of the lens. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 23

1398

(6.00 m)(2.00) = +24.0 m and 2.00 − 1.50

pf (10.0 m)(24.0 m) q= = = −17.1 m p − f 10.0 m − 24.0 m ! The image is 17.1 m to the left of the lens. (d) Observe from Equation [1] that f < 0 if n1 > n2 and f > 0 when n1 < n2. Thus, a diverging lens can be changed to converging by surrounding it with a medium whose index of refraction exceeds that of the lens material. 23.62

The inverted image is formed by light that leaves the object and goes directly through the lens, never having reflected from the mirror. For the formation of this inverted image, we have

q M = − 1 = −1.50 p1 !

giving

q1 = +1.50p1

The thin-lens equation then gives

1 1 1 + = !p1 1.50p1 10.0 cm

1 ⎞ ⎛ p1 = (10.0 cm) ⎜ 1 + = 16.7 cm ⎝ 1.50 ⎟⎠

The upright image is formed by light that passes through the lens after reflecting from the mirror. The object for the lens in this upright image formation is the image formed by the mirror. In order for the lens to form

Topic 23

1399

the upright image at the same location as the inverted image, the image formed by the mirror must be located at the position of the original object (so the object distances, and hence image distances, are the same for both the inverted and upright images formed by the lens). Therefore, the object distance and the image distance for the mirror are equal, and their common value is qmirror = pmirror = 40.0 cm − p1 = 40.0 cm − 16.7 cm = +23.3 cm

The mirror equation,

1 ! fmirror 23.63

1 !pmirror

1 qmirror

1 1 +2 + = 23.3 cm 23.3 cm 23.3 cm

2 1 = , then gives R fmirror

fmirror = +

23.3 cm = +11.7 cm 2

(a) The lens-maker’s equation for a lens made of material with refractive index n1 = 1.55 and immersed in a medium having refractive index n2 is ⎞⎛ 1 1 ⎛ n1 1 ⎞ ⎛ 1.55 − n2 ⎞ ⎛ 1 1⎞ = ⎜ − 1⎟ ⎜ − ⎟ = ⎜ − ⎟ ⎟ ⎜ n2 ⎠ ⎝ R1 R2 ⎠ ⎝ ⎠ ⎝ R1 R2 ⎠ ! f ⎝ n2 1 ⎛ 1.55 − 1.00 ⎞ ⎛ 1 1⎞ =⎜ − ⎟ ⎟ ⎜ ⎝ 1.00 ⎠ ⎝ R1 R2 ⎠ ! fair

[1]

1⎞ ⎛ 1.55 − 1.33 ⎞ ⎛ 1 =⎜ − ⎟ ⎟ ⎜ fwater ⎝ 1.33 ⎠ ⎝ R1 R2 ⎠

[2]

Thus, when the lens is in air, we have

and when it is immersed in water, !

Topic 23

1400

Dividing Equation [1] by Equation [2] gives

fwater ⎛ 1.33 ⎞ ⎛ 1.55 − 1.00 ⎞ =⎜ = 3.33 ⎝ 1.00 ⎟⎠ ⎜⎝ 1.55 − 1.33 ⎟⎠ f air ! If fair = 79.0 cm, the focal length when immersed in water is fwater = 3.33(79.0 cm) = 263 cm (b) The focal length for a mirror is determined by the law of reflection, which is independent of the material of which the mirror is made and of the surrounding medium. Thus, the focal length depends only on the radius of curvature and not on the material making up the mirror or the surrounding medium. This means that, for the mirror, fwater = fair = 79.0 cm 23.64

(a) The spherical ornament serves as a convex mirror, forming an image that is three-fourths the size of the object. Since convex mirrors only form upright, virtual images for real objects, the magnification is positive. Thus,

h$ q 3 M= =− =+ h p 4 !

and

−3p 4

With R < 0 (convex mirror) and a diameter of 8.50 cm, we have R = −(8.50 cm)/2 = −4.25 cm © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 23

1401

The mirror equation, 1/p + 1/q = 2/R then gives

1 4 2 − = !p 3p R

p=−

R (−4.25 cm) =− = +0.708 cm 6 6

Hence, the object is 0.708 cm in front of the spherical ornament. (b) The ray diagram below shows the formation of the upright, virtual image by this convex mirror.

23.65

The diagram below shows a bubble O, located 5.00 cm above the center of a glass sphere having radius |R| = 15.0 cm. This bubble is an actual distance of p = 10.0 cm below the refracting surface separating the glass and air. Refraction at this surface forms a virtual image I at distance |q| below the surface as shown.

The object distance p and image distance q are related by

Topic 23

1402

n1 n2 n2 − n1 + = p q R !

n2 n2 − n1 n1 = = q R p

Thus, with R < 0 for the concave surface the light crosses going from glass into the air, we have R = − 15.0 cm, and

1.00 q or

1.00 − 1.50 −15.0 cm q=

−

30.0 cm −3.50

1.50 10.0 cm

+1.00 − 4.50 30.0 cm

−3.50 30.0 cm

= −8.57 cm

Therefore, the bubble has an apparent depth of 8.57 cm below the glass surface. 23.66

(a) The only upright images that spherical mirrors can form of real objects are virtual images. Therefore, we know that the image described is virtual and diminished in size (4.00 cm high in comparison to the 10.0-cm height of the object). Since the virtual images formed by concave mirrors are always enlarged, we must conclude that this image is formed by a convex mirror. (b) The sketch below shows an upright, diminished, virtual image formed by a convex mirror. The distance between the object and this virtual image in this case is known to be p + |q| = 42.0 cm.

Topic 23

1403

The image distance for a virtual image is negative, so q < 0 and |q| = −q. Thus, we have p − q = 42.0 cm

[1]

We also know that the magnification is positive (upright image) and

h$ q 4.00 cm M= =− = = 0.400 h p 10.0 cm !

q = − 0.400p

Substituting Equation [2] into Equation [1] gives

p − (−0.400 0)p = 42.0 cm

p= !

42.0 cm = 30.0 cm 1.40

With the object at the zero end of the meter stick, the mirror is at the 30.0-cm mark. (c) The image distance is now seen to be q = −0.400p = 0.400(30.0 cm) = −12.0 cm. The mirror equation (1/p + 1/q = 2/R = 1/f)) then gives the focal length of this mirror as

f= !

pq (30.0 cm)(−12.0 cm) = = −20.0 cm p+q 30.0 cm − 12.0 cm

[2]

Topic 24

1404

Topic 24 Wave Optics

QUICK QUIZZES 24.1

Choice (c). The fringes on the screen are equally spaced only at small angles, where tan θ ≈ sin θ is a valid approximation.

24.2

Choice (c). The screen locations of the dark fringes of order m are given by !( ydark )m = ( λ L/d ) ( m + 12 ) , with m = 0 corresponding to the first dark fringe on either side of the central maximum. The width of the central maximum is then 2 ( ydark )0 = 2 ( λ L/d ) ( 12 ) = λ L/d . Thus, doubling the ! distance d between the slits will cut the width of the central maximum in half.

24.3

Choice (c). The screen locations of the bright fringes of order m are given by (ybright)m = (λL/d)m and the distance between successive bright fringes for a given wavelength is Δybright = (ybright)m+1 − (ybright)m = (λL/d)m+1 − (λL/d)m = λL/d Observe that this spacing is directly proportional to the wavelength. Thus, arranged from smallest to largest spacing between bright fringes,

Topic 24

1405

the order of the colors will be blue, green, red. 24.4

Choice (b). The space between successive bright fringes is proportional to the wavelength of the light. Since the wavelength in water is less than that in air, the bright fringes are closer together in the second experiment.

24.5

Choice (b). The outer edges of the central maximum occur where sin θ = ±λ/a. Thus, as the width of the slit, a, becomes smaller, the width of the central maximum will increase.

24.6

The compact disc. The tracks of information on a compact disc are much closer together than on a phonograph record. As a result, the diffraction maxima from the compact disc will be farther apart than those from the record.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 24.2

The fringe spacing in a double-slit interference pattern is Δybright = λL/d. Therefore, as the distance L between the screen and the plane of the slits is increased, the spacing between the bright fringes will increase, and (a) is the correct choice.

24.4

The correct choice is (b)—the distance would decrease. The spacing between double-slit interference fringes on a screen is proportional to the

Topic 24

1406

wavelength of the light (Δybright = λL/d). Light of wavelength λ0 in vacuum has wavelength λ0/nwater = λ0/1.333 in water. Because λ0/1.333 < λ0, the distance between bright fringes will decrease when the double-slit experiment in air is repeated under water. 24.6

For a ray traveling in a medium with index of refraction n, a 180° phase reversal occurs upon reflection from a medium with index of refraction nreflect > n. No phase reversal occurs if nreflect < n. (a) 0. For both ray 1 and ray 2, nreflect < n and there are no phase reversals. (b) 1. For ray 1, nreflect > n (giving a phase reversal). For ray 2, nreflect < n (giving no phase reversal). (c) 2. For both ray 1 and ray 2, nreflect > n. Each ray undergoes a 180° phase reversal.

24.8

The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When the finger touches a smooth surface, the oils from the skin will be deposited on the surface in the pattern of the closely spaced ridges. The clear spaces between the lines of deposited oil can serve as the slits in a crude diffraction grating and produce a colored spectrum of the light passing through or reflecting from the glass surface.

24.10

Suppose the index of refraction of the coating is intermediate between vacuum and the glass. When the coating is very thin, light reflected from

Topic 24

1407

its top and bottom surfaces will interfere constructively, so you see the surface white and brighter. Once the thickness reaches one-quarter of the wavelength of violet light in the coating, destructive interference for violet light will make the surface look red. Then other colors in spectral order (blue, green, yellow, orange, and red) will interfere destructively, making the surface look red, magenta, and then blue. As the coating gets thicker, constructive interference is observed for violet light and then for other colors in spectral order. Even thicker coatings give constructive and destructive interference for several visible wavelengths, so the reflected light starts looking white again. 24.12

The reflected light is partially polarized, with the component parallel to the reflecting surface being the most intense. Therefore, the polarizing material should have its transmission axis oriented in the vertical direction in order to minimize the intensity of the reflected light from horizontal surfaces.

24.14

Due to gravity, the soap film tends to sag in its holder, being quite thin at the top and becoming thicker as one moves toward the bottom of the holding ring. Because light reflecting from the front surface of the film experiences a phase change, and light reflecting from the back surface of

Topic 24

1408

the film does not (see Figure 24.7 in the textbook), the film must be a minimum of a half wavelength thick before it can produce constructive interference in the reflected light. Thus, the light must be striking the film at some distance from the top of the ring before the thickness is sufficient to produce constructive interference for any wavelength in the visible portion of the spectrum. 24.16

The wavelength of light is extremely small in comparison to the dimensions of your hand, so the diffraction of light around obstacles the size of your hand is totally negligible. However, sound waves have wavelengths that are comparable to the dimensions of the hand or even larger. Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.

ANSWERS TO EVEN NUMBERED PROBLEMS 24.2

(a)

1.77 µm

24.4

2.40 µm

24.6

(a)

24.8

2.9 µm

1.52 cm

(b)

1.47 µm

(b)

2.13 cm

Topic 24

1409

24.10

(a)

2.53 × 10−6 m

(b)

2.22 × 10−6 m

24.12

(a)

2.3°, 4.6°, and 6.9°

(c)

At small angles, θ ≈ sin θ. The approximation breaks down at larger

(b)

1.1°, 3.4°, and 5.7°

angles. 24.14

24.16

(a)

See Solution.

(b)

(c)

0.144°, 2.51 × 10−3, sin θ ≈ tan θ only when θ is small

(d)

603 nm

(a)

640 nm

(e)

tan θ = 2.51 × 10−3

0.720°

(f) 2.26 cm

(b) make use of a higher order constructive interference (i.e., a larger value of m) (c)

360 nm (m = 1), 600 nm (m = 2)

24.18

(a)

512 nm

24.20

233 nm

24.22

(a)

(b)

2.5m1 = 2m2 + 1

541 nm

(b) 406 nm; The most strongly transmitted wavelengths are those which suffer destructive interference in reflection. 24.24

(a)

1 phase reversal

(b)

3.07 × 10−5 m

Topic 24

1410

24.26

6.5 × 102 nm

24.28

99.6 nm

24.30

(a)

97.8 nm

(b)

Yes, some of which are 293 nm, 489 nm, and 685 nm.

24.32

(a)

0.0725°

24.34

91.1 cm

24.36

0.227 mm

24.38

659 nm

24.40

(a)

2 complete orders

24.42

(a)

Three maxima (for m = −1, m = 0, and m = +1) will be observed.

(b)

for m = −1, θ = −45.2°; for m = 0, θ = 0°; and for m = +1, θ = +45.2°

(b)

3.16 mm

(b)

24.44

7.35°

24.46

(a)

39.9°

(b)

450 nm

24.48

(a)

10.8°

(b)

24.50

78.1 nm and 469 nm

24.52

(a)

(b)

0.164

24.54

36.9°

0.336

(c)

10.9°

69.7°

Topic 24

1411

24.56

60.5°

24.58

(a)

45.0°

(b)

60.0°

(c)

65.9°

24.60

(a)

30.0°

(b)

45.0°

(c)

60.0°

(d)

90.0°

24.62

632.8 nm

24.64

(a)

6m1 = 5m2

(b)

m1 = 5, m2 = 6; 2.52 cm from the central maximum

24.66

maxima at 0°, ±29.1°, and ±76.4° minima at ±14.1° and ±46.8°

24.68

113 dark fringes (counting the m = 0 order at the line of contact)

24.70

(a)

24.72

See Solution.

24.74

α = 20.0 × 10−6 (°C)−1

If /Ii = 0

(b)

If /Ii = 0.25

PROBLEM SOLUTIONS 24.1

The location of the bright fringe of order m (measured from the position of the central maximum) is (ybright)m = (λL/d)m, m = 0, ±1, ±2, … Thus, the spacing between successive bright fringes is

Topic 24

1412

Δybright = (ybright)m+1 − (ybright)m = (λL/d)(m + 1) − (λL/d)m = λL/d The wavelength of the laser light must be

( Δy ) d = (1.58 × 10 m )( 0.200 × 10 m ) = 6.32 × 10 m = 632 nm λ= −2

−3

bright

24.2

−7

5.00 m

(a) For a bright fringe of order m, the path difference is δ = mλ, where m = 0, 1, 2,…. At the location of the third order bright fringe, m = 3 and

δ = 3λ = 3(589 nm) = 1.77 × 103 nm = 1.77 µm 1⎞ ⎛ (b) For a dark fringe, the path difference is δ = ⎜ m + ⎟ λ where 2⎠ ! ⎝ m = 0, 1, 2,…. At the third dark fringe, m = 2 and

1⎞ 5 ⎛ δ = ⎜ 2 + ⎟ λ = ( 589 nm ) = 1.47 × 103 nm = 1.47 µm 2⎠ 2 ! ⎝ 24.3

(a) The condition for double-slit constructive interference (bright fringes) is mλ = d sinθbright with m = 0, ±1, ±2, …. Here, m = 1 for the first bright fringe next to the central maximum. Solve for the angle and substitute values:

(

−9 mλ 1 633 × 10 m sin θ bright = = d 1.45 × 10−5 m

) → θ

bright

= 2.50°

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1413

(b) The condition for double-slit destructive interference (dark fringes) is (m + ½)λ =d sinθdark with m = 0, ±1, ±2, …. Here, m = 1 for the second dark fringe (the first dark fringe occurs where m = 0 with a path difference of λ/2). Solve for the angle and substitute values:

sin θ dark

( m + ) λ = (1.5)( 633 × 10 m ) = 3.75° = −9

1 2

1.45 × 10−5 m

(c) The viewing screen is a distance ℓ = 2.00 m from the slits and the angle to the first bright fringe is θbright = 2.50°. Construct a triangle to find the distance y to the first bright fringe: tan θ bright =

24.4

ybright ℓ

→ ybright = ( 2.00 m ) tan 2.50° = 8.73 cm = 8.73 × 10−2 m

In the double-slit interference pattern, the bright fringe of order m is found where d sin θ = mλ with m = 0, ±1, ±2,…. Here, θ is the angle between the line to the central maximum and the line to the location of the bright fringe of interest. Thus, if θ = 15.0° for the m = 1 bright fringe of light having wavelength λ = 620 nm the spacing between the slits must be

(1)( 620 × 10 m ) mλ d= = = 2.40 × 10−6 m = 2.40 µm sin θ sin 15.0° ! −9

24.5

(a) From d sin θ = mλ the angle for the m = 1 maximum for the sound waves is

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1414

⎡m ⎛ v ⎡ ⎞⎤ ⎛ 354 m/s ⎞ ⎤ 1 ⎛m ⎞ θ = sin −1 ⎜ λ ⎟ = sin −1 ⎢ ⎜ sound ⎟ ⎥ = sin −1 ⎢ ⎜⎝ 2 000 Hz ⎟⎠ ⎥ = 36.2° ⎝d ⎠ d f 0.300 m ⎝ ⎠ ⎣ ⎦ ⎣ ⎦ !

(b) For 3.00-cm microwaves, the required slit spacing is mλ (1)(3.00 cm) d= = = 5.08 cm sin (36.2°) ! sin θ

24.6

(1)( 3.00 × 108 m/s ) c mc f= = = = 5.08 × 1014 Hz λ dsin θ (1.00 × 10−6 m ) sin 36.2°

In a double-slit interference pattern, the screen position of the mth order maximum for wavelength λ is ym = (λL/d)m. The separation between the maxima of orders m1 and m2 is then Δy = (λL/d)(m2 − m1). (a) If λ = 588 nm while m2 = 1 and m1 = 0, the separation is −9 λL λ L ( 588 × 10 m ) (1.50 m ) Δy = 1 − 0 = = = 1.52 × 10−2 m = 1.52 cm ( ) −3 d d 0.058 0 × 10 m !

(b) If λ = 412 nm, m2 = 4, and m1 = 2, the spacing between these maxima is

( 412 × 10 m )(1.50 m ) ( 4 − 2) = 2.13 × 10 m = 2.13 cm Δy = −9

−2

−3

! 24.7

0.058 0 × 10 m

As indicated in the sketch below, the path difference between waves from

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1415

the two antennas that travel toward the car is given by δ = d sin θ When δ = mλ where m = 0, 1, 2,…., constructive interference produces the maximum of order m.

Destructive interference produces the minimum of order m when 1 !δ = ( m + 2 ) λ .

(a) At the m = 2 maximum, δ = d sin θ = 2λ or

λ=

d ⋅sin θ m

⎤ ⎞ (300 m) ⎡ 400 m d⎛ y ⎢ ⎥ = 55.7 m = ⎜ ⎟= ⎢ (1 000 m)2 + (400 m)2 ⎥ 2 ⎜⎝ x 2 + y 2 ⎟⎠ 2 ⎣ ⎦

(b) The next minimum encountered is the m = 2 minimum, and it occurs where δ = 5λ/2 or

⎡ 5(55.7 m) ⎤ ⎛ 5λ ⎞ θ = sin −1 ⎜ ⎟ = sin −1 ⎢ ⎥ = 27.7° ⎝ 2d ⎠ 2(300 m) ⎦ ⎣ ! At this point, y = x tan θ = (1 000 m) tan 27.7° = 525 m, so the car must travel an additional 125 m 24.8

The angular position of the bright fringe of order m is given by d sin θ =

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1416

mλ. Thus, if the m = 1 bright fringe is located at θ = 12° when λ = 6.0 × 102 nm = 6.0 × 10−7 m the slit spacing is

(1)( 6.0 × 10−7 m ) mλ d= = = 2.9 × 10−6 m = 2.9 µm sin θ sin12° ! 24.9

The screen position of the mth order bright fringe of wavelength λ in a double-slit interference pattern is ym = (λL/d)m. Then, Δy = (λL/d)(m2 − m1) is the separation between the bright fringes of orders m1 and m2. If Δy = 0.552 mm for the first and second order bright fringes when L = 1.75 m and the slit separation is d = 0.552 mm the wavelength incident upon the set of slits is

( Δy ) d

λ= L ( m2 − m1 ) ! 24.10

( 0.552 × 10 m )( 2.10 × 10 m ) = 6.62 × 10 m = 662 nm = −3

−3

−7

(1.75 m )( 2 − 1)

(a) The path difference to the mth bright fringe is δm,bright = mλ. For m = 4,

(

)

δ 4 ,bright = 4 633 × 10−9 m = 2.53 × 10−6 m (b) The path difference to the mth dark fringe is δ m,dark = ( m + 12 ) λ . The fourth dark fringe occurs where m = 3 (the first dark fringe occurs where m = 0 with a path difference of λ/2) so that

(

)

δ 4 ,dark = ( 3.5 ) 633 × 10−9 m = 2.22 × 10−6 m 24.11

The distance between the central maximum (position of A) and the first

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1417

minimum is

λL ⎛ 1⎞ λL y= = ⎜⎝ m + ⎟⎠ d 2 m=0 2d !

λ L ( 3.00 m ) (150 m ) = = 11.3 m . Thus, d = 2y 2 ( 20.0 m ) ! 24.12

(a) The angular position of the bright fringe of order m is given by d sin

θ = mλ or θm = sin−1 (mλ/d). If d = 25λ the first three bright fringes are found at

⎛ 1⎞ ⎛ 2⎞ ⎛ 3⎞ θ 1 = sin −1 ⎜ ⎟ = 2.3° ,!θ 2 = sin −1 ⎜ ⎟ = 4.6° ,!!and!!θ 3 = sin −1 ⎜ ⎟ = 6.9° ⎝ 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ ! (b) The angular position of the dark fringe of order m is given by dsin θ = ( m + 12 ) λ !or!θ m = sin −1 ⎡⎣( m + 12 ) λ /d ⎤⎦ ,!m = 0,±1,±2,…. If d = 25λ !

the first three dark fringes are found at

⎛ 1 ⎞ ⎛3 ⎞ ⎛5 ⎞ θ 0 = sin −1 ⎜ 2 ⎟ = 1.1° , θ 1 = sin −1 ⎜ 2 ⎟ = 3.4° , and θ 2 = sin −1 ⎜ 2 ⎟ = 5.7° ⎝ 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ ! (c) The answers are evenly spaced because the angles are small and θ ≈ sin θ . At larger angles, the approximation breaks down and the spacing isn’t so regular. 24.13

As shown in the figure below, the path difference in the waves reaching

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1418

the telescope is δ = d2 − d1 = d2(1 − sin α). If the first minimum (δ = λ/2) occurs when θ = 25.0°, then

α = 180° − (θ + 90.0° + θ) = 40.0° and δ λ /2 (250 m/2) d2 = = = = 350 m 1 − sin α 1 − sin α 1 − sin 40.0° ! Thus, h = d2 sin 25.0° = 148 m.

24.14

(a)

y1 4.52 × 10−3 m = 2.51 × 10−3 (b) tan θ 1 = = L 1.80 m ! (c) θ1 = tan−1 (2.51 × 10−3) = 0.144° and sin θ1 = sin (0.144°) = 2.51 × 10−3 . The sine and the tangent are very nearly the same but only because

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1419

the angle is small. (d) From δ = d sin θ = mλ for the order m bright fringe, −4 dsin θ 1 ( 2.40 × 10 m ) sin ( 0.144° ) λ= = = 6.03 × 10−7 m = 603 nm 1 1 !

⎡ 5 ( 6.03 × 10−7 m ) ⎤ ⎛ 5λ ⎞ −1 ⎥ = 0.720° (e) θ 5 = sin ⎜⎝ ⎟⎠ = sin ⎢ −4 d 2.40 × 10 m ⎢ ⎥⎦ ⎣ ! −1

(f) y5 = L tan θ5 = (1.80 m) tan (0.720°) = 2.26 × 10−2 m = 2.26 cm 24.15

The path difference in the two waves received at the home is δ = 2d where d is the distance from the home to the mountain. Neglecting any phase change upon reflection, the condition for destructive interference is

1⎞ ⎛ δ = ⎜ m + ⎟ λ with m = 0, 1, 2,… 2⎠ ! ⎝ so

24.16

δ 1 ⎞ λ λ 300 m ⎛ dmin = min = ⎜ 0 + ⎟ = = = 75.0 m ⎝ 2 2⎠ 2 4 4 !

(a) With a phase change in the reflection at the outer surface of the soap film and no change on reflection from the inner surface, the condition for constructive interference in the light reflected from the soap bubble is !2nfilmt = ( m + 12 ) λ where m = 0, 1, 2,…. For the lowest order reflection, m = 0 so the wavelength is

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1420

2nwatert λ= = 4(1.333)(120 nm) = 6.4 × 102 nm = 640 nm (0 + 1/2) ! (b) To strongly reflect the same wavelength light, a thicker film will need to make use of a higher order constructive interference, i.e., use a larger value of m. (c) The next greater thickness of soap film that can strongly reflect 640 nm light corresponds to m = 1 giving

t1 = !

(1 + 1/2)λ 3 ⎡ 640 nm ⎤ = ⎢ = 3.6 × 102 nm = 360 nm ⎥ 2nfilm 2 ⎣ 2(1.333) ⎦

and the third such thickness (corresponding to m = 2) is

t2 = ! 24.17

( 2 + 1/2) λ = 5 ⎡ 640 nm ⎤ = 6.0 × 102 nm = 600 nm 2nfilm

2 ⎢⎣ 2(1.333) ⎥⎦

Light reflecting from the first (glass-iodine) interface suffers a phase change, but light reflecting at the second (iodine-glass) interface does not have a phase change. Thus, the condition for constructive interference in the reflected light is !2nfilmt = ( m + 12 ) λ with m = 0, 1, 2,…. The smallest film thickness capable of strongly reflecting the incident light is

tmin = ! 24.18

(mmin + 1/2) λ = ( 0 + 1/2) λ = 6.00 × 102 nm = 85.4 nm 2nfilm

2nfilm

4(1.756)

(a) Phase changes are experienced by light reflecting at either surface of

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1421

the oil film, a upper air-oil interface and a lower oil-water interface. Under these conditions, the requirement for constructive interference is 2nfilmt = m1λ, with m = 0, 1, 2,…., and the requirement for destructive interference is !2nfilmt = ( m2 + 12 ) λ , with m2 = 0, 1, 2,…. To have the thinnest film that produces simultaneous constructive interference of λ1 = 640 nm and destructive interference of λ2 = 512 nm it is necessary that

1⎞ ⎛ 2nfilmt = m1 (640 nm) = ⎜ m2 + ⎟ ( 512 nm ) ⎝ 2⎠ ! where both m1 and m2 are the smallest integers for which this is true. It is found that m1 = m2 = 2 are the smallest integer values that will satisfy this condition, giving the minimum acceptable film thickness as

(m + 1 ) λ 2(640 nm) = (2.5)(512 nm) = 512 nm mλ tmin = 1 1 = 2 2 2 = 2nfilm 2nfilm 2(1.25) 2(1.25) ! (b) From the discussion above, it is seen that in order to have a film thickness that produces simultaneous constructive interference of 640-nm light and destructive interference of 512-nm light, it is necessary that

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1422

1⎞ ⎛ m1 (640 nm) = ⎜ m2 + ⎟ (512 nm) ⎝ 2⎠ !

⎛ 640 nm ⎞ 1 m1 ⎜ = m2 + ⎟ 2 ⎝ 512 nm ⎠

1 This gives (1.25)m1 = m2 + ,!or! 2.5 m1 = 2m2 + 1 . 2 !

24.19

With nfilm = 1.52, > nair light reflecting from the upper surface (air to glass transition) experiences a phase change, while light reflecting from the lower surface (a glass to air transition) does not experience such a change. Under these conditions, the requirement for constructive interference of normally incident waves reflecting from these two surfaces is 1 !2nfilmt = ( m + 2 ) λ where m is any integer value. Thus, if the thickness of

the film is t = 0.420 µm = 420 nm the wavelengths that will strongly reflect from the film are

2nfilmt 4(1.52)(420 nm) λ= = m + 1/2 2m + 1 !

m = 0, 1, 2,…

yielding m = 0 ⇒ λ = 2.55 × 103 nm (infrared); m = 2 ⇒ λ = 511 nm (visible);

m = 1 ⇒ λ = 851 nm (infrared);

m = 3 ⇒ λ = 365 nm (ultraviolet),

with the wavelengths for all higher values of m being even shorter. Thus, the only visible light wavelength that can strongly reflect from this film is 511 nm © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 24

24.20

1423

Since nair < noil < nwater light reflected from both top and bottom surfaces of the oil film experiences a phase change, resulting in zero net phase difference due to reflections. Therefore, the condition for constructive interference in reflected light is

λ 2t = mλn = m nf ilm !

⎛ λ ⎞ t = m⎜ where m = 0, 1, 2,… ⎝ 2nfilm ⎟⎠

Assuming that m = 1 the thickness of the oil slick is

λ 600 nm t = (1) = = 233 nm . 2nfilm 2(1.29) ! 24.21

There will be a phase change of the radar waves reflecting from both surfaces of the polymer, giving zero net phase change due to reflections. The requirement for destructive interference in the reflected waves is then 2t = ( m + 12 ) λn !

t = (2m + 1) =

λ where m = 0, 1, 2,… 4nfilm

If the film is as thin as possible, then m = 0 and the needed thickness is

λ 3.00 cm t= = = 0.500 cm 4n 4(1.50) film ! This anti-reflectance coating could be easily countered by changing the wavelength of the radar—to 1.50 cm—now creating maximum reflection! 24.22

(a) With nair < nwater < noil, reflections at the air-oil interface experience a

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1424

phase change, while reflections at the oil-water interface experience no phase change. With one phase change at the surfaces, the condition for constructive interference in the light reflected by the film is !2nfilmt = ( m + 12 ) λ , where m is any positive integer. Thus,

2n t 2(1.45)(280 nm) 812 nm λm = film1 = = m+ 2 m + 12 m + 12 !

m = 0, 1, 2, 3,…

The possible wavelengths are: λ0 = 1.62 × 103 nm, λ1 = 541 nm, λ2 = 325 nm,…. Of these, only λ1 = 541 nm (green) is in the visible portion of the spectrum. (b) The wavelengths that will be most strongly transmitted are those that suffer destructive interference in the reflected light. With one phase change at the surfaces, the condition for destructive interference in the light reflected by the film is 2nfilmt = mλ where m is any positive, nonzero integer. The possible wavelengths are

λm = ! or

2nfilmt 2(1.45)(280 nm) 812 nm = = m m m

m = 1, 2, 3,…

λ1 = 812 nm, λ2 = 406 nm, λ3 = 271 nm,…

of which only λ2 = 406 nm (violet) is in the visible spectrum.

Topic 24

24.23

1425

(a) For maximum transmission, we want destructive interference in the light reflected from the front and back surfaces of the film. If the surrounding glass has refractive index greater than 1.378, light reflected from the front surface suffers no phase change, and light reflected from the back does undergo a phase change. Under these conditions, the condition for destructive interference in the reflected light is 2d = mλn = m λ/nfilm where m is a positive integer. Thus, for minimum nonzero film thickness, we require that m = 1 and find

λ 656.3 nm d= = = 238 nm 2(1.378) ! 2nfilm (b) The filter will expand. As d increases in 2nfilm d = λ so does λ increase. (c) Destructive interference of order 2 will occur for reflected light when 2d = 2λ/nfilm or for the wavelength

λ = nfilm d = (1.378)(238 nm) = 328 nm (near ultraviolet) 24.24

(a) Consider the n = 1.000 air layer with glass (n = 1.55) above and below. A phase reversal occurs on the bottom side of this layer but not on the top side. There is 1 phase reversal. (b) Given the spacer’s thickness and the glass plate’s length, the top plate is inclined at an angle θ = sin−1(2.90 × 10−4 m/3.00 × 10−2 m) =

Topic 24

1426

0.554° above the bottom plate. For n = 1 in the air gap and one phase reversal, adjacent dark bands occur when the air gap thickness changes by t = λ/2. The distance d between these bands is given by sin θ = t/d so that

24.25

t λ 594 × 10−9 m = = = 3.07 × 10−5 m sin θ 2sin θ 2sin ( 0.554° )

When the hair is inserted between one end of the glass plates, which have length L = 14.0 cm a wedge-shaped air film is created as shown below. The maximum thickness of this air wedge is equal to the diameter d of the hair.

If the bright interference fringe of order m occurs at point A (where the air wedge has thickness tm) and the adjacent bright fringe of order m + 1 occurs at B (where the thickness is tm+1), we may use the properties of similar triangles to write

d Δt tm+1 − tm = = Δx !L Δx

⎛ Δt ⎞ d=⎜ ⎟L ⎝ Δx ⎠

[1]

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1427

Since nair < nglass light waves reflecting at the upper surface of the air film do not undergo a phase change, but those reflecting from the lower surface do experience such a change. Thus, the condition for constructive interference to produce a bright fringe is !2nairt = ( m + 12 ) λ or the thickness of the film at the location of the bright fringe of order m is 1 !tm = ( m + 2 ) λ 2nair . The change in thickness between the locations of

adjacent bright fringes is then

Δt = tm+1 − tm = !

[(m + 1) + 1/2] λ − [m + 1/2] λ = λ = λ 2nair

2nair

If the observed spacing between adjacent bright fringes is Δx = 0.580 mm when using light of wavelength λ = 650 nm Equation [1] gives the diameter of the hair as

( 650 × 10−9 m )(14.0 × 10−2 m ) = 7.84 × 10−5 m = 78.4 µm λL ⎛ λ /2 ⎞ d=⎜ L = = ⎝ Δx ⎟⎠ 2(Δx) 2 ( 0.580 × 10−3 m ) ! 24.26

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1428

From the geometry shown in the figure, R2 = h2 + r2 = (R − t)2 + r2 or

t = R − R2 − r 2 = 3.0 m − (3.0 m)2 − (9.8 × 10−3 m)2 = 1.6 × 10−5 m ! With a phase change upon reflection at the lower surface of the air layer but no change with reflection from the upper surface, the condition for a bright fringe is 1⎞ 1⎞ λ ⎛ 1⎞ ⎛ ⎛ 2t = ⎜ m + ⎟ λn = ⎜ m + ⎟ = ⎜ m + ⎟ λ , where m = 0, 1, 2,… ⎝ ⎝ 2⎠ 2 ⎠ nair ⎝ 2⎠ !

At the 50th bright fringe, m = 49 and the wavelength is found to be

2 (1.6 × 10−5 m ) 2t λ= = = 6.5 × 10−7 m = 6.5 × 102 nm m + 1/2 49.5 ! 24.27

(a) There is one phase reversal (from the oil’s top surface) so that constructive interference occurs for 2nt = ( m + 12 ) λ . In the wavelength range from 400 nm to 600 nm, the constructive interference condition is satisfied when m = 2 so that

λconstructive =

2 (1.45 ) ( 425 nm ) 2nt = = 493 nm 1 m+ 2 2.50

(b) Similarly, the condition for destructive interference is 2nt = mλ and is satisfied when m = 3 so that

λdestructive =

2nt 2 (1.45 ) ( 425 nm ) = = 411 nm m 3

Topic 24

24.28

1429

With a phase change due to reflection at each surface of the magnesium fluoride layer, there is zero net phase difference caused by reflections. The condition for destructive interference is then 1⎞ 1⎞ λ ⎛ ⎛ 2t = ⎜ m + ⎟ λn = ⎜ m + ⎟ , where m = 0, 1, 2,… ⎝ ⎝ 2⎠ 2 ⎠ nfilm !

For minimum thickness, m = 0 and the thickness is

( 550 × 10 m ) = 9.96 × 10−8 m = 99.6 nm λ t = (2m + 1) = (1) 4nfilm 4(1.38) ! −9

24.29

Since the thin film has air on both sides of it, there is a phase change for light reflecting from the first surface but no change for light reflecting from the second surface. Under these conditions, the requirement to be met if waves reflecting from the two sides are to produce constructive interference and a strong reflection is

1⎞ ⎛ 2nfilmt = ⎜ m + ⎟ λ ⎝ 2⎠ !

λ=

4nfilmt 2m + 1

m = 0, 1, 2,…

With nfilm = 1.473 and t = 542 nm the wavelengths that produce strong reflections are given by λ = 4(1.473)(524 nm)/(2m + 1), which yields m = 0: λ = 3.09 × 103 nm m = 1: λ = 1.03 × 103 nm m = 2: λ = 617 nm m = 0: λ = 441 nm m = 4: λ = 343 nm m = 5: λ = 281 nm

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1430

and many other shorter wavelengths. Of these, the only ones in the range 300 nm to 700 nm are 617 nm, 441 nm, and 343 nm. 24.30

With nair < nfilm < nglass light undergoes a phase change when it reflects from either of the two surfaces of the film. Therefore, the condition for destructive interference of normally incident light reflecting from these two surfaces is

1⎞ ⎛ 2nfilmt = ⎜ m + ⎟ λ ⎝ 2⎠ !

⎛ λ ⎞ t = (2m + 1) ⎜ ⎝ 4nfilm ⎟⎠

where!m = 0, 1, 2,… [1]

(a) The minimum thickness film that will minimize the light reflected from the lens is given by m = 0 in Equation [1], and has the value

λ 540 nm tmin = = = 97.8 nm 4n 4(1.38) film ! (b) Yes. From Equation [1], we see than any film thickness given by t = (2m + 1)tmin where tmin = (λ/4nfilm) = 97.8 nm will produce destructive interference and hence minimum reflected light for wavelength λ = 540 nm. For m = 0, 1, 2,…, we obtain thicknesses of 293 nm, 489 nm, 685 nm,…. 24.31

In a single-slit diffraction pattern, with the slit having width a, the dark fringe of order m occurs at angle θm where sin θm = m(λ/a) and m = 0, ±1,

Topic 24

1431

±2, ±3,…. The location, on a screen located distance L from the slit, of the dark fringe of order m (measured from y = 0 at the center of the central maximum) is (ydark)m = L tan θm ≈ L sin θm = mλ(L/a). (a) The central maximum extends from the m = −1 dark fringe to the m = +1 dark fringe, so the width of this central maximum is ⎛ λL⎞ ⎛ λ L ⎞ 2λ L Central!max.!width = ( ydark )1 − ( ydark )−1 = 1 ⎜ ⎟ − ( −1) ⎜ ⎟ = ⎝ a ⎠ ⎝ a ⎠ a =

2 ( 5.40 × 10−7 m ) (1.50 m ) 0.200 × 10−3 m

= 8.10 × 10−3 m = 8.10 mm

(b) The first order bright fringe extends from the m = 1 dark fringe to the m = 2 dark fringe, or

( Δy ) = ( y ) − ( y ) = 2 ⎛⎜⎝ λaL ⎞⎟⎠ − 1⎛⎜⎝ λaL ⎞⎟⎠ = λaL bright

dark 2

dark 1

( 5.40 × 10 m )(1.50 m ) = 4.05 × 10 m = 4.05 mm = −7

−3

0.200 × 10 m

Note that the width of the first order bright fringe is exactly one-half the width of the central maximum. 24.32

(a) The condition for destructive interference for a single slit is

sin θ dark = m

λ . Here, m = ±1 for the first minimum so that (choosing a

m = +1)

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1432

⎛ 633 × 10−9 m ⎞ θ dark = sin −1 ⎜ 1 = 0.072 5° ⎝ 0.500 × 10−3 m ⎟⎠ (b) Take the central maximum width, w, to be the distance between the first minimums in the diffraction pattern. For a screen located 1.25 m beyond the slit,

tan θ dark =

w/2 1.25 m

w = ( 2.50 m ) tan ( 0.072 5° ) = 3.16 × 10−3 m = 3.16 mm 24.33

(a) Dark bands (minima) occur where sin θ = m(λ/a) For the first minimum, m = 1, and the distance from the center of the central maximum is y1 = L tan θ ≈ L sin θ = L(λ/a). Thus, the needed distance to the screen is

⎛ 0.75 × 10−3 m ⎞ ⎛ a⎞ L = y1 ⎜ ⎟ = ( 0.85 × 10−3 m ) ⎜ = 1.1 m −9 ⎝ λ⎠ ⎝ 587.5 × 10 m ⎟⎠ ! (b) The width of the central maximum is 2y1 = 2(0.85 mm) = 1.7 mm. 24.34

Note: The small angle approximation does not work well in this situation. Rather, you should proceed as follows. At the first order minimum, sin θ1 = (1) λ/a or

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1433

⎛ 5.00 cm ⎞ ⎛ λ⎞ θ 1 = sin −1 ⎜ ⎟ = sin −1 ⎜ = 7.98° ⎝ a⎠ ⎝ 36.0 cm ⎟⎠ ! Then, y1 = L tan θ1 = (6.50 m) tan 7.98° = 0.911 m = 91.1 cm

24.35

With the screen locations of the dark fringe of order m at (ydark)m = L tan θm ≈ L sin θm = m(λL/a)

for

m = ±1, ±2, ±3…

the width of the central maximum is

Δycentral = ( ydark )m=+1 − ( ydark )m=−1 = 2 ( λ L/a) , so ! maximum ⎛ ⎞ a ⎜ Δycentral ⎟ −3 −3 ⎝ maximum ⎠ ( 0.600 × 10 m ) ( 2.00 × 10 m ) λ= = = 4.62 × 10−7 m = 462 nm 2L 2(1.30 m) ! 24.36

At the positions of the minima, sin θm = m(λ/a) and ym = L tan θm ≈ L sin θm = m[L(λ/a)]. Thus, y3 − y1 = (3 − 1)[L(λ/a)] = 2[L(λ/a)] and

Topic 24

1434

2(0.500 m) ( 680 × 10−9 m ) 2Lλ a= = = 2.27 × 10−4 m = 0.227 mm −3 3.00 × 10 m ! y3 − y1 24.37

The locations of the dark fringes (minima) mark the edges of the maxima, and the widths of the maxima equals the spacing between successive minima. At the locations of the minima, sin θm = m(λ/a) and

⎡ ⎛ 500 × 10−9 m ⎞ ⎤ ym = Ltan θ m ≈ Lsin θ m = m ⎡⎣ L ( λ /a ) ⎤⎦ = m ⎢(1.20 m ) ⎜ −3 ⎟⎠ ⎥ = m (1.20 mm ) 0.500 × 10 m ⎝ ⎣ ⎦ !

Then, Δy = Δm(1.20 mm) and for successive minima, Δm = 1. Therefore, the width of each maximum, other than the central maximum, in this interference pattern is width = Δy = (1)(1.20 mm) = 1.20 mm 24.38

In a single-slit diffraction pattern, minima are found where ⎛ λ⎞ sin θ dark = m ⎜ ⎟ ⎝ a⎠ !

where!!m = ±1, ± 2,…

or, using the small angle approximation y = L tan θ ≈ L sin θ, at screen positions ym = m(λL/a). Thus, if the m = 2 minimum is seen at y = 1.40 mm on a screen located distance L = 85.0 cm from a single slit of width a = 0.800 mm the wavelength of the light passing through the slit must be

Topic 24

1435 −3 −3 ya (1.40 × 10 m ) ( 0.800 × 10 m ) λ= = = 6.59 × 10−7 m = 659 nm −2 mL 2 ( 85.0 × 10 m ) !

24.39

The grating spacing is d = (1/3 660)cm = (1/3.66 × 105) m, and bright lines are found where d sin θ = mλ. (a) The wavelength observed in the first-order spectrum is λ = d sin θ1 or 9 ⎛ 10 4 nm ⎞ ⎛ 1 m ⎞ ⎛ 10 nm ⎞ λ=⎜ sin θ = 1 ⎜⎝ 3.66 ⎟⎠ sin θ 1 ⎝ 3.66 × 105 ⎟⎠ ⎜⎝ 1 m ⎟⎠ !

This yields: at θ1 = 10.1°, λ = 479 nm; at θ1 = 13.7°, λ = 647 nm; and at

θ1 = 14.8°, λ = 698 nm (b) In the second order, m = 2. The second order images for the above wavelengths will be found at angles θ2 = sin−1 (2λ/d) = sin−1 [2sin θ1] This yields: for λ = 479 nm, θ2 = 20.5°; for λ = 647 nm, θ2 = 28.3°; and for λ = 698 nm, θ2 = 30.7°. 24.40

(a) The longest wavelength in the visible spectrum is 700 nm, and the grating spacing is d = 1 mm/600 = 1.67 × 10−3 mm = 1.67 × 10−6 m. The

Topic 24

1436

maximum viewing angle is θmax = 90.0° and maxima are found where mλ = d sin θ. Thus, −6 dsin 90.0° (1.67 × 10 m ) sin 90.0° mmax = = = 2.38 λred 700 × 10−9 m !

so 2 complete orders will be observed. (b) From λ = d sin θ the angular separation of the red and violet edges in the first order will be −9 ⎡ 700 × 10−9 m ⎤ m⎤ ⎡λ ⎤ ⎡λ ⎤ −1 ⎡ 400 × 10 Δθ = sin −1 ⎢ red ⎥ − sin −1 ⎢ violet ⎥ = sin −1 ⎢ − sin ⎥ ⎢ ⎥ −6 −6 ⎣ d ⎦ ⎣ d ⎦ ⎣ 1.67 × 10 m ⎦ ⎣ 1.67 × 10 m ⎦ !

24.41

Δθ = 10.9°

1.00 cm ⎛ 1.00 m ⎞ 1.00 m = . The grating spacing is d = 2 4 500 ⎜⎝ 10 cm ⎟⎠ 4.50 × 105 ! From d sin θ = mλ the angular separation between the given spectral lines will be Δθ = sin−1 [mλred/d] − sin−1[mλviolet/d], or

⎡ m ( 656 × 10−9 m ) ( 4.50 × 105 ) ⎤ ⎡ m ( 434 × 10−9 m ) ( 4.50 × 105 ) ⎤ −1 Δθ = sin ⎢ ⎥ − sin ⎢ ⎥ 1.00 m 1.00 m ⎢ ⎥ ⎢ ⎥⎦ ⎣ ⎦ ⎣ ! −1

The results obtained are: for m = 1, Δθ = 5.91°; for m = 2, Δθ = 13.2°; and

Topic 24

1437

for m = 3, Δθ = 26.5°. Complete orders for m ≥ 4 are not visible. 24.42

The array of wires will act as a diffraction grating for the ultrasound waves, with maxima located at those angles given by the grating equation d sin θ = mλ, where m = 0, ±1, ±2,…. The spacing between adjacent slits in this grating is d = 1.30 cm and the wavelength of these ultrasound waves is

λ= !

vsound 343 m/s = = 9.22 × 10−3 m = 9.22 mm 3 f 37.2 × 10 Hz

(a) The order number of the maximum found at angle θ is m = (d/λ) sin θ, so the maximum order that can be found will be the largest integer satisfying the relation

1.30 × 10−2 m ⎛ d⎞ ⎛ d⎞ m ≤ ⎜ ⎟ sin θ max = ⎜ ⎟ sin 90.0° = = 1.41 ⎝ λ⎠ ⎝ λ⎠ 9.22 × 10−3 m

Thus, three maxima corresponding to m = −1, m = 0, and m = +1 can be found. (b) We use θ = sin−1 (mλ/d) to find the direction for each maximum to be: for m = −1, θ = −45.2°; for m = 0, θ = 0°; and for m = +1, θ = +45.2°. 24.43

For diffraction by a grating, the angle at which the maximum of order m occurs is given by d sin θm = mλ, where d is the spacing between adjacent

Topic 24

1438

slits on the grating. Thus,

(1)( 632.8 × 10−9 m ) mλ d= = = 1.81 × 10−6 m = 1.81 µm sin θ sin 20.5° m ! 24.44

With 2 000 lines per centimeter, the grating spacing is 1 d= cm = 5.00 × 10−4 cm = 5.00 × 10−6 m ! 2 000

Then, from d sin θ = mλ, the location of the first order for the red light is

⎛ 1( 640 × 10−9 m ) ⎞ ⎛ mλ ⎞ −1 θ = sin −1 ⎜ = sin ⎜ ⎟ = 7.35° −6 ⎝ d ⎟⎠ ⎝ 5.00 × 10 m ⎠ ! 24.45

The spacing between adjacent slits on the grating is

1 cm 10−2 m d= = ! 5 310 slits 5 310 The maximum of order m is located where d sin θm = mλ so

⎛ d d ⎛ ym ⎞ 10−2 m ⎜ λ = sin θ m = ⎜ 2 ⎟= m m ⎝ L + ym2 ⎠ (1)(5 310) ⎜ ⎝ !

⎞ ⎟ 2 2 (1.72 m ) + ( 0.488) ⎟⎠ 0.488 m

= 5.14 × 10−7 m = 514 nm

Topic 24

24.46

1439

(a) The diffraction grating equation is mλ = d sinθ where θ is the angle to the mth-order maximum for wavelength λ. For a grating with 475 lines/mm, the distance between lines is d = (1 × 10−3 m)/(475 lines). Substitute values to find

(

)

−9 mλ 2 675 × 10 m ( 475 ) sin θ r 2 = = → θ r 2 = 39.9° d 10−3 m

(b) A shorter wavelength, λ3, will have a third-order maximum at the same angle when

sin θ 3 = sin θ r 2 m3 λ3 m2 λr 2 = d d 3 λ3 2 λr 2 = d d

λ3 =

24.47

2 2 λr 2 = ( 675 nm ) = 450 nm 3 3

1 cm 10−2 m The grating spacing is d = = = 3.64 × 10−6 m . 3 2 750 2.75 × 10 ! From d sin θ = mλ or θ = sin−1 (mλ/d) the angular positions of the red and violet edges of the second-order spectrum are found to be

⎛ 2 (700 × 10−9 m ) ⎞ ⎛ 2λred ⎞ −1 θ r = sin ⎜ = sin ⎜ ⎟ = 22.6° −6 ⎝ d ⎟⎠ ⎝ 3.64 × 10 m ⎠ ! −1

Topic 24

1440

⎛ 2 ( 400 × 10−9 m ) ⎞ ⎛ 2λviolet ⎞ −1 θ v = sin ⎜ = sin ⎜ ⎟ = 12.7° −6 ⎝ d ⎟⎠ ⎝ 3.64 × 10 m ⎠ ! −1

and

Note from the sketch below that yr = L tan θr and yv = L tan θv, so the width of the spectrum on the screen is Δy = L(tan θr − tan θv).

Since it is given that δ = (m + 1/2)λ, the distance from the grating to the screen must be

L= ! 24.48

Δy 1.75 cm = tan θ r − tan θ v tan (22.6°) − tan (12.7°)

= 9.17 cm

(a) For a grating with 325 lines/mm, the distance d between lines is d = (10−3 m)/325 so that 1/d = 325 × 103 m−1. Use the diffraction grating equation to find the angle to the first-order maximum:

sin θ1 =

mλ = 1 577 × 10−9 m 325 × 103 m −1 d

(

)(

) → θ = 10.8° 1

(b) The maximum possible angle of diffraction is θ = 90° so that sinθ ≤ 1. The highest order mmax is therefore the largest integer satisfying

Topic 24

1441

mλ = dsin θ ≤ d → mmax ≤

d 1 = = 5.33 3 −1 λ 325 × 10 m 577 × 10−9 m

(

)(

)

mmax = 5 (c) The angle to the mmax = 5 maximum is

((

⎛ mλ ⎞ θ max = sin −1 ⎜ = sin −1 5 577 × 10−9 m 325 × 103 m −1 ⎝ d ⎟⎠

)(

))

θ max = 69.7° 24.49

(a) For a diffraction grating having distance d between adjacent slits, primary maxima for wavelength λ are produced at angles satisfying the grating equation, d sin θ = mλ. If the m = 3 maximum for λ = 500 nm is observed at θ = 32.0°, the grating spacing is 3 ( 500 × 10 m ) mλ d= = = 2.83 × 10−6 m = 2.83 × 10−4 cm sin 32.0° ! sin θ −9

The number of rulings per centimeter on the grating is then n= !

1 cm 1 cm = = 3.53 × 103 grooves/cm −4 d 2.83 × 10 cm

(b) The order number of a maximum appearing at angle θ is m = d sin θ/λ.The magnitude of the largest order numbers visible is the largest integer satisfying the condition

Topic 24

1442

d 2.83 × 10−6 m ⎛ d⎞ ⎛ d⎞ m ≤ ⎜ ⎟ sin θ max = ⎜ ⎟ sin 90° = = = 5.66 ⎝ λ⎠ ⎝ λ⎠ λ 500 × 10−9 m

Thus, 11 maxima for wavelength λ = 500 nm will be observable with this grating. These are the maxima corresponding to m = 0, ± 1, ± 2, ± 3, ± 4, and ±5. 24.50

The grating spacing is d = 1 cm/1 200 = 8.33 × 10−4 cm = 8.33 × 10−6 m. Using sin θ = mλ/d and the small angle approximation, the distance from the central maximum to the maximum of order m for wavelength λ is ym = L tan θ ≈ L sin θ = (λL/d)m. Therefore, the spacing between successive maxima is Δy = ym+1 − ym = λL/d. The longer wavelength in the light is found to be

λlong = !

−3 −6 (Δy)d ( 8.44 × 10 m ) ( 8.33 × 10 m ) = = 4.69 × 10−7 m = 469 nm L 0.150 m

Since the third order maximum of the shorter wavelength falls halfway between the central maximum and the first order maximum of the longer wavelength, we have 3λshort L ⎛ 0 + 1 ⎞ λlong L =⎜ ⎝ 2 ⎟⎠ d ! d

24.51

⎛ 1⎞ λshort = ⎜ ⎟ ( 469 nm ) = 78.1 nm ⎝ 6⎠

(a) From Brewster’s law, the index of refraction is

Topic 24

1443

n = tan θ p = tan (48.0°) = 1.11 ! 2 (b) From Snell’s law, n2 sin θ2 = n1 sin θ1 the angle of refraction when θ1 =

θp is ⎛ n1 sin θ p ⎞ ⎛ (1.00)sin 48.0° ⎞ θ 2 = sin −1 ⎜ = sin −1 ⎜ ⎟⎠ = 42.0° ⎟ ⎝ 1.11 ⎝ n2 ⎠ ! Note that when θ1 = θp, θ2 = 90.0° − θp as it should. 24.52

(a) From Malus’s law, the fraction of the incident intensity of the unpolarized light that is transmitted by the polarizer is I′ = I0 (cos2 θ)av = I0 (0.500) The fraction of this intensity incident on the analyzer that will be transmitted is I = I′ cos2 (35.0°) = [0.500I0]cos2 (35.0°) = 0.336I0 Thus, the fraction of the incident unpolarized light transmitted is I/I0 = 0.336. (b) The fraction of the original incident light absorbed by the analyzer is I ′ − I 0.500I 0 − 0.336I 0 = = 0.164 I0 ! I0

24.53

The more general expression for Brewster’s angle is given in problem

Topic 24

1444

P24.57 as tan θp = n2/n1 −1 ⎛ n ⎞ −1 ⎛ 1.52 ⎞ = 56.7° . (a) When n1 = 1.00 and n2 = 1.52, θ p = tan ⎜ 2 ⎟ = tan ⎜ ⎝ 1.00 ⎟⎠ n ⎝ ⎠ 1 !

−1 ⎛ n ⎞ −1 ⎛ 1.52 ⎞ = 48.8° . (b) When n1 = 1.333 and n2 = 1.52, θ p = tan ⎜ 2 ⎟ = tan ⎜ ⎝ 1.333 ⎟⎠ ⎝ n1 ⎠ !

24.54

The polarizing angle for light in air striking a water surface is

⎛n ⎞ ⎛ 1.333 ⎞ θ p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = 53.1° ⎝ 1.00 ⎟⎠ n ⎝ ⎠ 1 ! This is the angle of incidence for the incoming sunlight (that is, the angle between the incident light and the normal to the surface). The altitude of the Sun is the angle between the incident light and the water surface. Thus, the altitude of the Sun is

α = 90.0° − θp = 90.0° − 53.1° = 36.9° 24.55

(a) Brewster’s angle (or the polarizing angle) is

⎛ nquartz ⎞ ⎛n ⎞ ⎛ 1.458 ⎞ θ p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = tan −1 ⎜ = 55.6° ⎟ ⎝ 1.000 ⎟⎠ ⎝ n1 ⎠ ⎝ nair ⎠ ! (b) When the angle of incidence is the polarizing angle, θp, the angle of refraction of the transmitted light is θ2 = 90.0° − θp. Hence, θ2 = 90.0° −

Topic 24

1445

55.6° = 34.4°. 24.56

The critical angle for total internal reflection is θc = sin−1 (n2/n1). Thus, if θc = 34.4° as light attempts to go from sapphire into air, the index of refraction of sapphire is n2 1.00 nsapphire = n1 = = = 1.77 sin θ sin 34.4° c !

Then, when light is incident on sapphire from air, the Brewster’s angle is

⎛n ⎞ ⎛ 1.77 ⎞ θ p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜ = 60.5° ⎝ 1.00 ⎟⎠ ⎝ n1 ⎠ ! 24.57

From Snell’s law, the angles of incidence and refraction are related by n1 sin θ1 = n2 sin θ2. If the angle of incidence is the polarizing angle (that is, θ1 = θp), the refracted ray is perpendicular to the reflected ray (see Figure 24.28 in the textbook), and the angles of incidence and refraction are also related by θp + 90° + θ2 = 180°, or θ2 = 90° − θp. Substitution into Snell’s law then gives n1 sin θp = n2 sin (90° − θp) = n2 cos θp

24.58

sin θp/cosθp = tan θp = n2/n1

From Malus’s law, the intensity of the light transmitted by the polarizer is

Topic 24

1446

I = I0 cos2 θ, where I0 is the intensity of the incident light, and θ is the angle between the direction of the plane of polarization of the incident light and the transmission axis of the polarizing disk. Thus,

θ = cos −1 !

24.59

( I/I ). 0

(a)

I 1 = I 2.00 !0

⇒

⎛ 1 ⎞ θ = cos −1 ⎜ = 45.0° ⎝ 2.00 ⎟⎠

(b)

I 1 = I 4.00 !0

⇒

⎛ 1 ⎞ θ = cos −1 ⎜ = 60.0° ⎝ 4.00 ⎟⎠

(c)

I 1 = I 6.00 !0

⇒

⎛ 1 ⎞ θ = cos −1 ⎜ = 65.9° ⎝ 6.00 ⎟⎠

From Malus’s law, the intensity of the light transmitted by the first polarizer is I1 = Ii cos2 θ1. The plane of polarization of this light is parallel to the axis of the first plate and is incident on the second plate. Malus’s law gives the intensity transmitted by the second plate as I2 = I1 cos2 (θ2 −

θ1) = Ii cos2 θ1 cos2 (θ2 − θ1). This light is polarized parallel to the axis of the second plate and is incident upon the third plate. A final application of Malus’s law gives the transmitted intensity as If = I2 cos2 (θ3 − θ2) = Ii cos2 θ1 cos2 (θ2 − θ1) cos2 (θ3 − θ2) With θ1 = 20.0°, θ2 = 40.0°, and θ3 = 60.0°, this result yields © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 24

1447

If = (10.0 units) cos2 (20.0°) cos2 (20.0°) cos2 (20.0°) = 6.89 units 24.60

In each case, solve for the angle θ using Malus’ law, I = I0 cos2θ or cos2θ = I/I0. (a) cos2 θ = 0.750 → θ = 30.0° (b) cos2 θ = 0.500 → θ = 45.0° (c) cos2 θ = 0.250 → θ = 60.0° (d) cos2 θ = 0 → θ = 90.0°

24.61

(a) If light has wavelength λ in vacuum, its wavelength in a medium of refractive index n is λn = λ/n. Thus, the wavelengths of the two components in the specimen are

λ 546.1 nm λn1 = = = 413.7 nm n1 1.320 ! and

λ 546.1 nm λn2 = = = 409.7 nm n2 1.333 !

(b) The numbers of cycles of vibration each component completes while passing through the specimen of thickness t = 1.000 µm are

t 1.000 × 10−6 m N1 = = = 2.417 λn1 413.7 × 10−9 m !

Topic 24

1448

and

t 1.000 × 10−6 m N2 = = = 2.441 λn2 409.7 × 10−9 m !

Thus, when they emerge, the two components are out of phase by N2 − N1 = 0.024 cycles. Since each cycle represents a phase angle of 360°, they emerge with a phase difference of Δφ = (0.024 cycles)(360°/cycles) = 8.6° 24.62

In a single-slit diffraction pattern, the first dark fringe occurs where sin(θdark)1 = (1)λ/a. If no diffraction minima are to be observed, the maximum width the slit can have is that which would place the first dark fringe at the maximum viewable angle of 90.0°. That is, when a = amax, we will have sin (θdark)1 = (1) λ/amax = sin 90.0° = 1.00, yielding amax = λ = 632.8 nm

24.63

The light has passed through a single slit since the central maximum is twice the width of other maxima (the space between the centers of successive dark fringes). In a double-slit pattern, the central maximum has the same width as all other maxima (compare Figures 24.1(b) and 24.18(b) in the textbook).

Topic 24

1449

Figure P24.63 In single-slit diffraction the width of the central maximum on the screen is given by ⎛ (1)λ ⎞ ⎛ λL⎞ Δycentral = 2Ltan (θ dark )m=1 ≈ 2Lsin (θ dark )m=1 = 2L ⎜ = 2⎜ ⎟ ⎟ ⎝ a ⎠ ⎝ a ⎠ ! maximun

The width of the slit is then

a= !

24.64

2λ L Δycentral

maximun

2 ( 632.8 × 10−9 m ) ( 2.60 m ) ⎛ 102 cm ⎞ = 1.2 × 10−4 m = 0.12 mm ⎜ ⎟ 10.3 − 7.6 cm 1 m ( ) ⎝ ⎠

(a) In a double-slit interference pattern, bright fringes on the screen occur where (ybright)m = m(λL/d). Thus, if bright fringes of the wavelengths λ1 = 540 nm and λ2 = 450 nm are to coincide, it is necessary that

m1 !

( 540 nm ) L = m ( 450 nm ) L d

or, dividing both sides by 90 nm, 6m1 = 5m2 , where both m1 and m2 are integers. (b) The condition found above may be written as m2 = (6/5)m1. Trial and

Topic 24

1450

error reveals that the smallest nonzero integer value of m1 that will yield an integer value for m2 is m1 = 5 yielding m2 = 6. Thus, the first overlap of bright fringes for the two given wavelengths occurs at the screen position (measured from the central maximum)

y bright = ! 24.65

5 ( 540 × 10−9 m ) (1.40 m ) 0.150 × 10−3 m

6 ( 450 × 10−9 m ) (1.40 m ) 0.150 × 10−3 m

= 2.52 × 10−2 m = 2.52 cm

Dark fringes (destructive interference) occur where d = sin θ = (m + 1/2)λ for m = 0, 1, 2,…. Thus, if the second dark fringe (m = 1) occurs at θ = (18.0 min)(1.00°/60.0 min) = 0.300°, the slit spacing is −9 1⎞ λ ⎛ ⎛ 3 ⎞ ( 546 × 10 m ) d = ⎜m+ ⎟ =⎜ ⎟ = 1.56 × 10−4 m= 0.156 mm ⎝ ⎠ ⎝ ⎠ 2 sin θ 2 sin (0.300°) !

24.66

The wavelength is λ = vsound/f = (340 m/s)/(2 000 Hz) = 0.170 m, and maxima occur where d sinθ = mλ, or θ = sin−1 [m(λ/d)] for m = 0, ± 1, ± 2,…. Since d = 0.350 m, λ/d = 0.486, which gives θ = sin−1[m(0.486)]. For m = 0, ± 1, and ± 2, this yields maxima at 0°, ± 29.1°, and ± 76.4°. No solutions exist for |m| ≥ 3, since that would imply sin θ > 1. Minima occur where d sin θ = (m + 1/2)λ, or θ = sin−1[(2m + 1)λ/2d] for m = 0, ± 1, ± 2,…. With λ/d = 0.486, this becomes θ = sin−1[(2m + 1)(0.243)]. For

Topic 24

1451

m = 0 and ± 1, we find minima at ± 14.1° and ± 46.8°. No solutions exist for |m| ≥ 2, since that would imply sin θ > 1. 24.67

The source and its image, located 1.00 cm below the mirror, act as a pair of coherent sources. This situation may be treated as double-slit interference, with the slits separated by 2.00 cm, if it is remembered that the light undergoes a phase change upon reflection from the mirror. The existence of this phase change causes the conditions for constructive and destructive interference to be reversed. Therefore, dark bands (destructive interference) occur where y = m(λL/d) for m = 0, 1, 2,…. The m = 0 dark band occurs at y = 0 (that is, at mirror level). The first dark band above the mirror corresponds to m = 1 and is located at −9 ⎛ λ L ⎞ ( 500 × 10 m ) (100 m ) y = (1) ⎜ ⎟ = = 2.50 × 10−3 m = 2.50 mm −2 ⎝ d ⎠ 2.00 × 10 m !

24.68

Assuming the glass plates have refractive indices greater than that of both air and water, there will be a phase change at the reflection from the lower surface of the film but no change from reflection at the top of the film. Therefore, the condition for a dark fringe is 2t = mλn = m(λ/nfilm) for m = 0, 1, 2,… If the highest order dark band observed is m = 84 (a total of 85 dark bands

Topic 24

1452

counting the m = 0 order at the edge of contact), the maximum thickness of the air film is

tmax = !

mmax ⎛ λ ⎞ 84 ⎛ λ ⎞ = ⎜ ⎟ = 42λ 2 ⎜⎝ nfilm ⎟⎠ 2 ⎝ 1.00 ⎠

When the film consists of water, the highest order dark fringe appearing will be

⎛n ⎞ ⎛ 1.333 ⎞ mmax = 2tmax ⎜ film ⎟ = 2 ( 42 λ ) ⎜ = 112 ⎝ λ ⎠ ⎝ λ ⎟⎠ ! Counting the m = 0 order, a total of 113 dark fringes are now observed. 24.69

In the figure below, observe that the path difference between the direct and the indirect paths is

δ = 2x − d = 2 h2 + ( d/2 ) − d ! 2

With a phase change (equivalent to a half-wavelength shift) occurring upon reflection at the ground, the condition for constructive interference is δ = (m + 1/2)λ, and the condition for destructive interference is δ = mλ. In both cases, the possible values of the order number are m = 0, 1, 2,…. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 24

1453

δ (a) The wavelengths that will interfere constructively are λ = . m + 1/2 ! The longest of these is for the m = 0 case and has a value of

λ = 2δ = 4 h2 + ( d/2 ) − 2d = 4 ( 50.0 m ) + ( 300 m ) − 2 ( 600 m ) = 16.6 m ! 2

(b) The wavelengths that will interfere destructively are λ = δ/m, and the largest finite one of these is for the m = 1 case. That wavelength is

λ = δ = 2 h2 + ( d/2 ) − d = 2 ( 50.0 m ) + ( 300 m ) − 600 m = 8.28 m ! 2

24.70

θ1) = Ii cos2 θ1 cos2 (θ2 − θ1). This light is polarized parallel to the axis of the second plate and is incident upon the third plate. A final application of Malus’s law gives the transmitted intensity as If = I2 cos2 (θ3 − θ2) = Ii cos2 θ1 cos2 (θ2 − θ1) cos2 (θ3 − θ2) (a) If θ1 = 45°, θ2 = 90°, and θ3 = 0°, then If/Ii = cos2 45° cos2 (90° − 45°) cos2 (0° − 90°) = 0

Topic 24

1454

(b) If θ1 = 0°, θ2 = 45°, and θ3 = 90°, then If/Ii = cos2 0° cos2 (45° − 0°) cos2 (90° − 45°) = 0.25 24.71

If the signal from the antenna to the receiver station is to be completely polarized by reflection from the water, the angle of incidence where it strikes the water must equal the polarizing angle from Brewster’s law. This is given by

⎛n ⎞ θ p = tan −1 ⎜ water ⎟ = tan −1 (1.333) = 53.1° ⎝ nair ⎠ !

From the triangle RST in the sketch above, the horizontal distance from the point of refection, T, to shore is given by x = (90.0 m) tan θp = (90.0 m)(1.33) = 120 m and from triangle ABT, the horizontal distance from the antenna to point T is y = (5.00 m) tan θp = (5.00 m)(1.333) = 6.67 m © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 24

1455

The total horizontal distance from ship to shore is then x + y = 120 m +6.67 m = 127 m. 24.72

There will be a phase change associated with the reflection at one surface of the film but no change at the other surface of the film. Therefore, the condition for a dark fringe (destructive interference) is

⎛ λ ⎞ 2t = mλn = m ⎜ ⎟⎠ where m = 0, 1, 2,… n ⎝ film ! From the figure below, note that R2 = r2 + (R − t)2 = r2 + R2 − 2Rt + t2, which reduces to r2 = 2Rt − t2. Since t will be very small in comparison to either r or R, we may neglect the term t2, leaving !r ≈ 2Rt.

mλ For a dark fringe, t = , so the radii of the dark rings will be ! 2nfilm

⎛ mλ ⎞ r ≈ 2R ⎜ = ⎝ 2nfilm ⎟⎠ ! 24.73

mλ R nfilm

for m = 0, 1, 2,…

In the single-slit diffraction pattern, destructive interference (or minima)

Topic 24

1456

occur where sin θ = m(λ/a) for m = 0, ±1, ±2,…. The screen locations, measured from the center of the central maximum, of these minima are at ym = L tan θm ≈ L sin θm = m(λL/a) If we assume the first order maximum is halfway between the first and second order minima, then its location is y= !

y1 + y2 (1 + 2)(λ L/a) 3λ L = = 2 2 2a

and the slit width is −9 3λ L 3 ( 500 × 10 m ) (1.40 m ) a= = = 3.50 × 10−4 m = 0.350 mm −3 2y 2 ( 3.00 × 10 m ) !

24.74

As light emerging from the glass reflects from the top of the air layer, there is no phase change produced. However, the light reflecting from the end of the metal rod at the bottom of the air layer does experience a phase change. Thus, the condition for constructive interference in the reflected light is !2t = ( m + 12 ) λ nair = ( m + 12 ) λ . As the metal rod expands, the thickness of the air layer decreases. The increase in the length of the rod is given by

λ λ λ ΔL = Δt = ( mi + 12 ) = m f + 12 = Δm 2 2 2 !

(

)

Topic 24

1457

The order number changes by one each time the film changes from bright to dark and back to bright. Thus, during the expansion, Δm = 200, and the measured change in the length of the rod is 500 × 10−9 m ) ( λ ΔL = ( 200 ) = ( 200 ) = 5.00 × 10−5 m 2 2 !

From ΔL = L0α(ΔT), the coefficient of linear expansion of the rod is

ΔL 5.00 × 10−5 m −1 α= = = 20.0 × 10−6 ( °C ) L0 ( ΔT ) ( 0.100 m ) ( 25.0°C ) !

Topic 25

1458

Topic 25 Optical Instruments

QUICK QUIZZES 25.1

Choice (c). The corrective lens for a farsighted eye is a converging lens, while that for a nearsighted eye is a diverging lens. Since a converging lens is required to form a real image of the Sun on the paper to start a fire, the campers should use the glasses of the farsighted person.

25.2

Choice (a). We would like to reduce the minimum angular separation for two objects below the angle subtended by the two stars in the binary system. We can do that by reducing the wavelength of the light—this in essence makes the aperture larger, relative to the light wavelength, increasing the resolving power. Thus, we would choose a blue filter.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 25.2

The f-number is given by f-number = f/D where, in this case, the focal length f is constant. Increasing the aperture D, which decreases the fnumber, decreases the depth of field, lets in more light, and decreases the

Topic 25

1459

appropriate exposure time. The correct answers are therefore: (a) I (b) D (c) I (d) D 25.4

There will be an effect on the interference pattern—it will be distorted. The high temperature of the flame will change the index of refraction of air for the arm of the interferometer in which the match is held. As the index of refraction varies randomly, the wavelength of the light in that region will also vary randomly. As a result, the effective difference in length between the two arms will fluctuate, resulting in a wildly varying interference pattern.

25.6

The aperture of a camera is a close approximation to the iris of the eye. The retina of the eye corresponds to the film of the camera, and a close approximation to the cornea of the eye is the lens of the camera.

25.8

For a farsighted person, the near point is always located farther than 25 cm from the eye and the corrective lens is converging.

25.10

Under low ambient light conditions, a photoflash unit is used to ensure that light entering the camera lens will deliver sufficient energy for a proper exposure to each area of the film. Thus, the most important criterion is the additional energy per unit area (product of intensity and the duration of the flash, assuming this duration is less than the shutter

Topic 25

1460

speed) provided by the flash unit. 25.12

The angular magnification produced by a simple magnifier is m = (25 cm)/f. Note that this is proportional to the optical power of a lens, P = 1/f, where the focal length f is expressed in meters. Thus, if the power of the lens is doubled, the angular magnification will also double.

25.14

In a nearsighted person the image of a distant object focuses in front of the retina. The cornea needs to be flattened so that its focal length is increased.

ANSWERS TO EVEN NUMBERED PROBLEMS 25.2

2.5 mm to 46 mm

25.4

6.48 m

25.6

(a)

25.8

1.05 m to 6.30 m

25.10

(a)

farsighted

(d)

virtual image, negative

(f )

+2.92 diopters

See Solution.

(b)

1/100 s

(b)

18.0 cm

(c)

38.0 cm

(e)

34.2 cm

Topic 25

1461

(g) p = 20.0 cm, q = −40.0 cm, f = +40.0 cm, P = +2.50 diopters 25.12

(a)

+50.8 diopters ≤ P ≤ +60.0 diopters

(b)

–0.800 diopters, diverging

25.14

(a)

−6.67

25.16

(a)

Yes, by using a bifocal or progressive lens.

(b)

+1.78 diopters

(c)

−1.18 diopters

25.18

(a)

−0.67 diopters

(b)

+0.67 diopters

25.20

(a)

−25.0 cm

(b)

nearsighted

25.22

(a)

10.0 cm

(b)

−3.33 cm

25.24

(a)

+5.0

(b)

+6.0

25.26

(a)

mmax = 2.39

(b)

m = 1.39

25.28

m = −588

25.30

0.809 µm

25.32

(a)

46.0 cm

(b)

30.7

25.34

(a)

L = fo[(m + 1)/m] (b)

25.36

1.6 × 102 mi

25.38

(a)

m = 7.50

(b)

diverging

(c)

−3.70 diopters

(c)

+4.2 cm

2.00 cm toward the objective lens

0.944 m

Topic 25

25.40

1462

(a)

virtual image

(b)

The final image is an infinite distance in front of the telescope.

(c)

fe = −5.00 cm, fo = 15.0 cm

25.42

1.7 m

25.44

(a)

25.46

2.2 × 1011 m

25.48

38 cm

25.50

(a)

3.6 × 103 lines

25.52

(a)

449 nm

(b)

smaller, because the wavelength is longer

2.29 × 10−4 rad

(b)

43.7 m

(b)

1.8 × 103 lines

25.54

39.6 µm

25.56

1.000 5

25.58

θmin ≈ 2.0 × 10−3 radians

25.60

(a)

−4.3 diopters

(b)

−4.0 diopters, 44 cm

25.62

(a)

1.96 cm

(b)

3.27

25.64

5.07 mm

(c)

9.80

Topic 25

25.66

1463

m = 10.7

PROBLEM SOLUTIONS 25.1

The f-number (or focal ratio) of a lens is defined to be the ratio of focal length of the lens to its diameter. Therefore, the f-number of the given lens is

f "number = ! 25.2

f 28 cm = = 7.0 D 4.0 cm

If a camera has a lens with focal length of 55 mm and can operate at fnumbers that range from f/1.2 to f/22, the aperture diameters for the camera must range from

f 55 mm Dmin = = = 2.5 mm f %number )max 22 ( ! to

f 55 mm Dmax = = = 46 mm f %number )min 1.2 ( ! 25.3

(a) A camera’s f-number = f/D where f is the focal length and D is the aperture diameter. Given the focal length and f-number, the aperture diameter is

Topic 25

1464

f 105 mm = = 37.5 mm f − number 2.80

(b) Use the thin-lens equation to solve for the image distance q, which must equal the distance between the lens and the CCD sensor:

1 1 1 1 1 1 + = → = − p q f q f p The minimum value of q occurs when the object is located at p = ∞ so that 1/p → 0 and qmin = f = 105 mm = 10.5 cm . (c) The maximum value of q occurs when the object is located at the camera’s near-point of p = 30.0 cm:

1 qmax 25.4

1 1 − → qmax = 16.2 cm 10.5 cm 30.0 cm

The person’s height h will be the limiting factor (the person’s image will fill the CCD vertically before their width fills the CCD horizontally). When the person fills the CCD so that the image height, h’, equals the CCD height, the object distance takes its minimum value and the magnification is

h' 14.0 × 10−3 m M=− = =− = −7.77 × 10−3 pmin h 1.80 m q

Substituting q = −Mpmin into the thin-lens equation and solving for pmin gives

Topic 25

1465

1 1 1 + = p q f

→

1 pmin

−

1 1 = Mpmin f

(

M −1 M = pmin f

→

)(

)

−3 −3 f ( M − 1) 50.0 × 10 m −7.77 × 10 − 1 pmin = = = 6.48 m M −7.77 × 10−3

25.5

The exposure time is being reduced by a factor of

t2 1/125 s 15 3 = = = !t1 1/15 s 125 25 Thus, to maintain correct exposure, the intensity of the light reaching the film should be increased by a factor of 25/3. This is done by increasing the area of the aperture by a factor of 25/3, so in terms of the diameter,

π D2 4 = ( 25/3 ) (π D12 4 ) ,!or!D2 = D1 25/3. ! 2 The new f-number will be

( f "number ) = Df = D f25/3 = ( f "number )

25.6

3 4.0 3 = = 1.4 25 5

f/1.4

(a) The intensity is a measure of the rate at which energy is received by the film per unit area of the image, or I ∝ 1/Aimage. Consider an object with horizontal and vertical dimensions hx and hy as shown below. If the vertical dimension intercepts angle θ, the vertical dimension of the image is hy′ = qθ ,!or!hy′ ∝ q. !

Topic 25

1466

Similarly for the horizontal dimension, !hx′ ∝ q, and the area of the image is Aimage = hx′ hy′ ∝ q2 . Assuming a very distant object, q ≈ f, so ! Aimage ∝ f 2, and we conclude that I ∝ 1/f 2. The intensity of the light reaching the film is also proportional to the cross-sectional area of the lens and hence to the square of the diameter of that lens, or I ∝ D2. Combining this with our earlier conclusion gives

D2 1 I∝ 2 = 2 f f /D) ( !

I∝

( f 'number )

(b) The total light energy delivered to the film is proportional to the product of intensity and exposure time, It. Thus, to maintain correct exposure, this product must be kept constant, or I2t2 = I1t1, giving 2 ⎡ ( f2 $number )2 ⎤ ⎛ I1 ⎞ ⎛ 4.0 ⎞ ⎛ 1 ⎞ t2 = ⎜ ⎟ t1 = ⎢ s ≈ 1/100 s 2 ⎥ t1 = ⎜ ⎝ 1.8 ⎟⎠ ⎜⎝ 500 ⎟⎠ ⎝ I2 ⎠ f1 $number ) ⎥⎦ ( ⎢ ⎣ !

25.7

Since the exposure time is unchanged, the intensity of the light reaching

Topic 25

1467

the film must be doubled if the energy delivered is to be doubled. Using the result of Problem 25.6 (part a), we obtain ⎛I ⎞

⎛ ⎞

( f -number) = ⎜⎝ I ⎟⎠ ( f -number) = ⎜⎝ 12 ⎟⎠ (11) = 61, 2

f2 -number = 16 = 7.8

Thus, you should use the f/8.0 setting on the camera. 25.8

The image must always be focused on the film, so the image distance is the distance between the lens and the film. From the thin-lens equation, 1/p + 1/q = 1/f, the object distance is p = qf/(q − f), and the range of object distances this camera can work with is from

pmin =

qmax f qmax − f

(210 mm)(175 mm) 210 mm − 175 mm

= 1.05 × 103 mm = 1.05 m

pmax =

25.9

qmin f qmin − f

(180 mm)(175 mm) 180 mm − 175 mm

= 6.30 × 103 mm = 6.30 m

The corrective lens must form an upright, virtual image at the near point of the eye (i.e., q = −60.0 cm in this case) for objects located 25.0 cm in front of the eye (p = +25.0 cm). From the thin-lens equation, 1/p + 1/q = 1/f, the required focal length of the corrective lens is

Topic 25

1468

f= !

pq (25.0 cm)( − 60.0 cm) = = +42.9 cm p+q 25.0 cm − 60.0 cm

and the power (in diopters) of this lens will be

1 1 P= = = +2.33 diopters f +0.429 m in!meters ! 25.10

(a) The person is farsighted, able to see distant objects but unable to focus on objects at the normal near point for a human eye. (b) With the corrective lens 2.00 cm in front of the eye, the object distance for an object 20.0 cm in front of the eye is p = 20.0 cm − 2.00 cm = 18.0 cm. (c) The upright, virtual image formed by the corrective lens will serve as the object for the eye, and this object must be 40.0 cm in front of the eye. With the lens 2.00 cm in front of the eye, the magnitude of the image distance for the lens will be |q| = 40.0 cm − 2.00 cm = 38.0 cm. (d) The image must be located in front of the corrective lens, so it is a virtual image, and the image distance is negative. Thus, q = −38.0 cm. (e) From the thin-lens equation, 1/p + 1/q = 1/f, the required focal length of the corrective lens is

Topic 25

1469

f= !

pq (18.0 cm)( − 38.0 cm) = = +34.2 cm p+q 18.0 cm − 38.0 cm

(f) The power of the corrective lens is then

1 1 P= = = +2.92 diopters f +0.342 m in!meters ! (g) With a contact lens, the lens-to-eye distance would be zero, so we would have p = 20.0 cm, and q = −40.0 cm, giving a required focal length of

f= !

pq (20.0 cm)( − 40.0 cm) = = +40.0 cm p+q 20.0 cm − 40.0 cm

and a power in diopters of

1 1 P= = = +2.50 diopters f +0.400 m in!meters ! 25.11

His lens must form an upright, virtual image of a very distant object (p ≈ ∞) at his far point, 80.0 cm in front of the eye. Therefore, the focal length is f = q = −80.0 cm. If this lens is to form a virtual image at his near point (q = −18.0 cm), the object distance must be

qf (−18.0 cm)( − 80.0 cm) p= = = 23.2 cm ! q − f −18.0 cm − (−80.0 cm)

Topic 25

25.12

1470

(a) When the child clearly sees objects at her far point (pmax = 125 cm), the lens-cornea combination has assumed a focal length suitable for forming the image on the retina (q = 2.00 cm). The thin-lens equation gives the optical power under these conditions as

Pfar =

1 f in meters

1 p

1 q

1 1.25 m

1 0.020 0 m

= +50.8 diopters

When the eye is focused (q = 2.00 cm) on objects at her near point (pmin = 10.0 cm), the optical power of the lens-cornea combination is

1 1 1 1 1 Pnear = = + = + = +60.0 diopters f p q 0.100 m 0.020 0 m in!meters ! Therefore, the range of the power of the lens-cornea combination is +50.8 diopters ≤ P ≤ +60.0 diopters (b) If the child is to see very distant objects (p → ∞) clearly, her eyeglass lens must form an erect virtual image at the far point of her eye (q = −125 cm). The optical power of the required lens is

1 1 1 1 P= = + = 0+ = −0.800 diopters f p q −1.25 m in!meters ! Since the power, and hence the focal length, of this lens is negative, it is diverging.

Topic 25

25.13

1471

(a) The lens should form an upright, virtual image at the far point (q = −50.0 cm) for very distant objects (p ≈ ∞). Therefore, f = q = −50.0 cm, and the required power is

1 1 P= = = −2.00 diopters f −0.500 m ! (b) If this lens is to form an upright, virtual image at the near point of the unaided eye (q = −13.0 cm), the object distance should be

qf (−13.0 cm)( − 50.0 cm) p= = = 17.6 cm q − f −13.0 cm − (−50.0 cm) ! 25.14

(a) The required lens will form an image at q = −15.0 cm for an object located at infinity so that

1 1 1 1 1 + = =P → P= + → P = −6.67 p q f ∞ −15.0 cm (b) The required lens has a negative power (and thus a negative focal length) so, by the sign convention, the lens is diverging. 25.15

(a) The person cannot see nearby objects clearly so they are farsighted. (b) The required contact lens will form a virtual image (on the same side of the lens as the object) at 35.0 cm for an object located at 20.0 cm from the eye. The object distance is p = 20.0 cm and the object

Topic 25

1472

distance is q = −35.0 cm (negative because the image is virtual). The required lens power is then

1 1 1 1 1 + = =P → P= + = 2.14 −2 p q f 20.0 × 10 m −35.0 × 10−2 m (c) If eyeglasses are to be prescribed and the lenses are 2.00 cm from the eye, the object minimum object distance changes to p = 18.0 cm (from the lens) and the image distance changes to q = −33.0 cm. The required lens power is then

1 1 1 1 1 + = =P → P= + = 2.53 −2 p q f 18.0 × 10 m −33.0 × 10−2 m (d) The lens power (and thus the focal length) is positive so, by the sign convention, the lens is converging. 25.16

(a) Yes, a single lens can correct the patient’s vision. The patient needs corrective action in both the near vision (to allow clear viewing of objects between 45.0 cm and the normal near point of 25 cm) and the distant vision (to allow clear viewing of objects more than 85.0 cm away). A single lens solution is for the patient to wear a bifocal or progressive lens. Alternately, the patient must purchase two pairs of glasses, one for reading, and one for distant vision. (b) To correct the near vision, the lens must form an upright, virtual

Topic 25

1473

image at the patient’s near point (q = −45.0 cm) when a real object is at the normal near point (p = +25.0 cm). The thin-lens equation gives the needed focal length as

f= !

pq (25.0 cm)( − 45.0 cm) = = +56.3 cm p+q 25.0 cm − 45.0 cm

so the required power in diopters is

1 1 P= = = +1.78 diopters f 0.563 m in!meters ! (c) To correct the distant vision, the lens must form an upright, virtual image at the patient’s far point (q = −85.0 cm) for the most distant objects (p → ∞). The thin-lens equation gives the needed focal length as f = q = −85.0 cm, so the needed power is

1 1 P= = = −1.18 diopters f −0.850 m in!meters ! 25.17

Considering the image formed by the cornea as a virtual object for the implanted lens, the object distance for this lens is p = −5.33 cm, and the image distance is q = +2.80 cm. The thin-lens equation then gives the focal length of the implanted lens as

f= !

pq (−5.33 cm)(2.80 cm) = = +5.90 cm p + q −5.33 cm + 2.80 cm

Topic 25

1474

1 1 = +17.0 diopters . so the power is P = = f +0.059 0 m ! 25.18

(a) The upper portion of the lens should form an upright, virtual image of very distant objects (p ≈ ∞) at the far point of the eye (q = −1.5 m). The thin-lens equation then gives f = q = −1.5 m, so the needed power is

1 1 P= = = −0.67 diopters f −1.5 m ! (b) The lower part of the lens should form an upright, virtual image at the near point of the eye (q = −30 cm) when the object distance is p = 25 cm. From the thin-lens equation,

f= !

pq (25 cm)( − 30 cm) = = +1.5 × 102 cm = +1.5 m p+q 25 cm − 30 cm

1 1 = +0.67 diopters . Therefore, the power is P = = f +1.5 m ! 25.19

The corrective lens should form an upright, virtual image at the woman’s far point (q = −40.0 cm) for a very distant object (p → ∞). The thin-lens equation gives the required focal length as f = q = −40.0 cm = −0.400 m. Since f < 0, it is a diverging lens, and the required power is

1 1 P= = = −2.50 diopters f −0.400 m in!meters ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 25

25.20

1475

(a)

1 1 = = −0.250 m = −25.0 cm P −4.00 diopters

(b) The corrective lens forms virtual images of very distant objects (p → ∞) at q = f = −25.0 cm. Thus, the person must be very nearsighted, unable to see objects clearly when they are more than (25.0 + 2.00) cm = 27.0 cm from the eye. (c) If contact lenses are to be worn, the far point of the eye will be 27.0 cm in front of the lens, so the needed focal length will be f = q = −27.0 cm, and the power is

1 1 P= = = −3.70 diopters fin!meters −0.270 m ! 25.21

(a) The simple magnifier (a converging lens) is to form an upright, virtual image located 25 cm in front of the lens (q = −25 cm). The thinlens equation then gives

qf (−25 cm)(7.5 cm) p= = = +5.8 cm ! q − f −25 cm − 7.5 cm so the stamp should be placed 5.8 cm in front of the lens. (b) When the image is at the near point of the eye, the angular magnification produced by the simple magnifier is

Topic 25

1476

m = mmax = 1 + ! 25.22

25 cm 25 cm = 1+ = 4.3 f 7.5 cm

(a) Solve for the focal length using the expression for a simple magnifier’s maximum magnification:

mmax = 1 +

25 cm 25 cm 25 cm → f= = = 10.0 cm f mmax − 1 3.50 − 1

(b) The surface closest to the object has a radius of curvature R1 → ∞ . Given the water drop’s focal length and index of refraction (n = 1.333 for water), the top surface has a curvature R2 given by ⎛ 1 ⎛ 1⎞ 1 1⎞ = ( n − 1) ⎜ − ⎟ = ( n − 1) ⎜ − ⎟ f ⎝ R1 R2 ⎠ ⎝ R2 ⎠ R2 = − ( n − 1) f = − (1.333 − 1) (10.0 cm ) = −3.33 cm

25.23

(a) From the thin-lens equation,

f= !

pq (3.50 cm)( − 25.0 cm) = = +4.07 cm p+q 3.50 cm − 25.0 cm

(b) With the image at the normal near point, the angular magnification is

m = mmax = 1 + ! 25.24

25.0 cm 25.0 cm = 1+ = +7.14 f 4.07 cm

(a) When the object is at the focal point of the magnifying lens, a virtual image is formed at infinity and parallel rays emerge from the lens.

Topic 25

1477

Under these conditions, the eye is most relaxed and the magnification produced is

m= !

25 cm 25 cm = = +5.0 f 5.0 cm

(b) When the object is positioned so the magnifier forms a virtual image at the near point of the eye (q = −25 cm), maximum magnification is produced and this is

mmax = 1 + !

25 cm 25 cm = 1+ = +6.0 f 5.0 cm

(c) From the thin-lens equation, the object distance needed to yield the maximum magnification computed in part (b) above is

qf (−25 cm)(5.0 cm) p= = = +4.2 cm q − f −25 cm 5.0 cm − ! 25.25

(a) From the thin-lens equation, a real inverted image is formed at an image distance of

pf (71.0 cm)(39.0 cm) q= = = +86.5 cm p − f 71.0 cm − 39.0 cm ! so the lateral magnification produced by the lens is

h′ q 86.5 cm M= =− =− = −1.22 h p 71.0 cm !

Topic 25

1478

and the magnitude is |M| = 1.22. (b) If |h| is the actual length of the leaf, the small-angle approximation gives the angular width of the leaf when viewed by the unaided eye from a distance of d = 126 cm + 71.0 cm = 197 cm as

h h θ≈ = d 197 cm ! The length of the image formed by the lens is |h′| = |M h| = 1.22|h|, and its angular width when viewed from a distance of d′ = 126 cm − q = 39.5 cm is

h′ 1.22 h θ′ ≈ = d′ 39.5 cm ! The angular magnification achieved by viewing the image instead of viewing the leaf directly is

θ ′ 1.22 h 39.5 cm 1.22(197 cm) ≈ = = 6.08 θ h 197 cm 39.5 cm ! 25.26

(a) When a converging lens is used as a simple magnifier, maximum magnification is obtained when the upright, virtual image is formed at the near point of the eye (|q| = 25.0 cm for a normal eye). For the given lens, this maximum magnification is

Topic 25

1479

mmax = 1 + !

25.0 cm 25.0 cm = 1+ = 2.39 f 18.0 cm

(b) When this simple magnifier is positioned for relaxed viewing (virtual image formed at infinity), the magnification produced is

mrelaxed = ! 25.27

25.0 cm 25.0 cm = = 1.39 f 18.0 cm

The overall magnification is m = M1me = M1(25 cm/fe), where M1 is the lateral magnification produced by the objective lens. Therefore, the required focal length for the eyepiece is

25.28

fe =

M1 (25 cm) (−12)(25 cm) = = 2.1 cm m −140

The approximate overall magnification of a compound microscope is given by m = −(L/fo)(25.0 cm/fe), where L is the distance between the objective and eyepiece lenses, while fo and fe are the focal lengths of the objective and eyepiece lenses, respectively. Thus, the described microscope should have an approximate overall magnification of ⎛ 20.0 cm ⎞ ⎛ 25.0 cm ⎞ L ⎛ 25.0 cm ⎞ m=− ⎜ = − ⎜⎝ 0.500 cm ⎟⎠ ⎜⎝ 1.70 cm ⎟⎠ = −588 fo ⎝ fe ⎟⎠ !

25.29

The angular magnification produced by a telescope is m = fo/fe, where fo is the focal length of the objective element, and fe is that of the eyepiece lens.

Topic 25

1480

Also, the optical power of a lens is P = 1/fmeters, where fmeters is the focal length of that lens, expressed in meters. Thus, fo = 1/Po and fe = 1/Pe, so the angular magnification of this telescope is

f 1 Po Pe 35.0 diopters m= o = = = = 12.7 f 1 P P 2.75 diopters e e o ! 25.30

It is specified that the final image the microscope forms of the blood cell is 29.0 cm in front of the eye and that the diameter of this image intercepts an angle of θ = 1.43 mrad. The diameter of this final image must then be he = rθ = (29.0 × 10−2 m)(1.43 × 10−3 rad) = 4.15 × 10−4 m At this point, it is tempting to use Equation 25.7 from the textbook for the overall magnification of a compound microscope and compute h = he/m as the size of the blood cell serving as the object for the microscope. However, the derivation of that equation is based on several assumptions, one of which is that the eye is relaxed and viewing a final image located an infinite distance in front of the eyepiece. This is clearly not true in this case, and the use of Equation 25.7 would introduce considerable error. Instead, we shall return to basics and use the thin-lens equation to find the size of the original object.

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1481

The image formed by the objective lens is the object for the eyepiece, and we label the size of this image as h′. The lateral magnification of the objective lens is M1 = h′/h = −q1/p1 and that of the eyepiece is Me = he/h′ = −qe/pe. The overall magnification produced by the microscope is

h ⎛ h′ ⎞ ⎛ h ⎞ Mtotal = e = ⎜ ⎟ ⎜ e ⎟ h ⎝ h ⎠ ⎝ h′ ⎠ ! which gives the size of the original object as h = he/|Mtotal|. From the thin-lens equation, the required object distance for the eyepiece is

q f (−29.0 cm)(0.950 cm) pe = e e = = 0.920 cm q − f −29.0 cm − 0.950 cm e e ! and the magnification produced by the eyepiece is

q (−29.0 cm) Me = − e = − = +31.5 pe 0.920 cm ! The image distance for the objective lens is then q1 = L − pe = 29.0 cm − 0.920 cm = 28.1 cm and the object distance for this lens is

q f (28.1 cm)(1.622 cm) p1 = 1 o = = 1.72 cm q − f 28.1 cm − 1.622 cm 1 o !

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1482

The magnification by the objective lens is

q (28.1 cm) M1 = − 1 = − = −16.3 p 1.72 cm 1 ! and the overall lateral magnification is !Mtotal = M1Me = (−16.3)(+31.5) = −513.

The actual diameter of the red blood cell serving as the original object is found to be he 4.15 × 10−4 m h= = = 8.09 × 10−7 m = 0.809 µm Mtotal 513 !

25.31

(a) The overall magnification of a compound microscope is m = Mlme where Ml = −10.0 is the lateral magnification and me is the eyepiece magnification. Solve for me to find

me =

m −115 = = 11.5 M1 −10.0

(b) The eyepiece focal length is given by

me =

25 cm 25 cm 25 cm → fe = = = 2.17 cm fe me 11.5

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1483

Ml ≈ −

25.32

L fo

→ fo ≈ −

L 20.0 cm =− = 2.00 cm Ml −10.0

(a) The length L of a refracting telescope tube equals the sum of the focal lengths so that L = fo + fe. Solve for the object lens focal length, fo, to find

f o = L − fe = 47.5 cm − 1.50 cm = 46.0 cm (b) The angular magnification is

25.33

f o 46.0 cm = = 30.7 fe 1.50 cm

The sketch below shows an image of the nebula formed by an objective lens of diameter Do and focal length fo. The diameter of the image of the nebula formed on the film is given by h′ = fo⋅θ

where θ is the angular width of the nebula and its image as shown. If the intensity of the light arriving at the objective lens from the nebula is I, the energy entering the lens during an exposure time Δt is

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1484

⎛ π Do2 ⎞ E = IAobjective ( Δt ) = I ⎜ ⎟⎠ Δt ⎝ 4 !

The energy per unit area deposited on the film during the exposure time is then 2

E 4E 4 ⎛ π IDo2 Δt ⎞ ⎛ I ⎞ ⎛ Do ⎞ = = = =⎜ ⎟ Δt 2 2 Aimage π ( h′ ) 4 π ( fo ⋅ θ ) πθ 2 ⎜⎝ 4 fo2 ⎟⎠ ⎝ θ 2 ⎠ ⎜⎝ fo ⎟⎠ E

Telescope 1 has an objective diameter Do,1 = 200 mm, focal length fo,1 = 2 000 mm, and uses an exposure time (Δt)1 = 1.50 min. If telescope 2, with Do,2 = 60.0 mm, and fo,2 = 900 mm, is to deposit the same energy per unit area on the film as does telescope 1, it is necessary that 2

E ⎛ I ⎞⎛D ⎞ ⎛ I ⎞⎛D ⎞ = ⎜ 2 ⎟ ⎜ o,2 ⎟ ( Δt )2 = ⎜ 2 ⎟ ⎜ o,1 ⎟ ( Δt )1 = ⎝ θ ⎠ ⎝ fo,1 ⎠ Aimage ,2 ⎝ θ ⎠ ⎝ fo,2 ⎠ Aimage ,1 E

The required exposure time for the second telescope is therefore I/θ ) ( D ( ( Δt) = (I/θ ) (D 2

o,1

o,2

) Δt = ⎛ D ⎞ ⎛ f ⎞ Δt ( ) ⎜⎜⎝ f ⎟⎟⎠ ⎜⎜⎝ D ⎟⎟⎠ ( ) f ) f o,1

o,2

o,1

o,2

o,1

o,2

⎛ 200 mm ⎞ ⎛ 900 mm ⎞ =⎜ ⎟ ⎜ ⎟ (1.50 min) = 3.38 min ⎝ 2 000 mm ⎠ ⎝ 60.0 mm ⎠

25.34

(a) For a refracting telescope, the overall length is L = fo + fe, and the magnification produced is m = fo/fe, where fo and fe are the focal

Topic 25

1485

lengths of the objective element and the eyepiece, respectively. Thus, we may write fe = fo/m to obtain f 1⎞ ⎛ ⎛ m + 1⎞ L = fo + o = fo ⎜ 1 + ⎟ = fo ⎜ ⎝ ⎠ ⎝ m ⎟⎠ m m !

(b) Using the result of part (a), the required change in the length of the telescope will be

⎛ m′ + 1 m + 1 ⎞ ⎛ 101 51.0 ⎞ ΔL = fo ⎜ − = (2.00 m) ⎜ − = −2.00 × 10−2 m = −2.00 cm ⎟ ⎟ ⎝ m′ ⎝ 100 50.0 ⎠ m ⎠ ! or the telescope must be shortened by moving the eyepiece 2.00 cm forward toward the objective lens. 25.35

(a) The magnification of a telescope is m = fo/fe, where fo and fe are the focal lengths of the objective and eyepiece lenses, respectively. Thus, if m = 34.0, while fo = 86.0 cm, the focal length of the eyepiece must be

fe =

fo 86.0 cm = = 2.53 cm m 34.0

(b) When the telescope is adjusted for relaxed-eye viewing, the distance between the objective and eyepiece lenses is L = fo + fe = 86.0 cm + 2.53 cm = 88.5 cm 25.36

The Moon may be considered an infinitely distant object (p → ∞) when

Topic 25

1486

viewed with this lens, so the image distance will be q = fo = 1 500 cm.

Considering the rays that pass undeviated through the center of this lens as shown in the sketch above, observe that the angular widths of the image and the object are equal. Thus, if w is the linear width of an object forming a 1.00-cm-wide image, then

θ=

25.37 (a)

w 8

3.8 × 10 m

1.0 cm fo

1.0 cm 1 500 cm

⎛ 1.0 cm ⎞ ⎛ 1 mi ⎞ 2 w = 3.8 × 108 m ⎜ ⎟⎜ ⎟ = 1.6 × 10 mi ⎝ 1 500 cm ⎠ ⎝ 1 609 m ⎠

(

)

From the thin-lens equation, q = pf/(p − f), so the lateral magnification by the objective lens is M = h′/h = − q/p = − f/(p − f). Therefore, the image size will be

fh fh h′ = M h = − = p− f f −p ! fh . (b) If !p  f , then f − p ≈ − p and h′ ≈ − p !

Topic 25

1487

fh (4.00 m)(108.6 m) h′ ≈ − =− = −1.07 × 10−3 m = −1.07 mm 3 p 407 × 10 m ! 25.38

Use the larger focal length (lowest power) lens as the objective element and the shorter focal length (largest power) lens for the eyepiece. The focal lengths are

fo = !

1 1 = +0.833 m,!and!fe = = +0.111 m +1.20 diopters +9.00 diopters

(a) The angular magnification (or magnifying power) of the telescope is then

f +0.833 m m= o = = 7.50 fe +0.111 m ! (b) The length of the telescope is L = fo + fe = 0.833 m + 0.111 m = 0.944 m 25.39

The lens for the left eye forms an upright, virtual image at qL = −50.0 cm when the object distance is pL = 25.0 cm, so the thin-lens equation gives its focal length as

fL = !

pLqL (25.0 cm)( − 50.0 cm) = = 50.0 cm pL + qL 25.0 cm − 50.0 cm

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1488

Similarly for the other lens, qR = −100 cm when pR = 25.0 cm, and fR = 33.3 cm. (a) Using the lens for the left eye as the objective,

f f 50.0 cm m= o = L = = 1.50 fe f R 33.3 cm ! (b) Using the lens for the right eye as the eyepiece and, for maximum magnification, requiring that the final image be formed at the normal near point (qe = −25.0 cm) gives the object distance for the eyepiece as

q f (−25.0 cm)(33.3 cm) pe = e e = = +14.3 cm q − f −25.0 cm − 33.3 cm e e ! The maximum magnification by the eyepiece is then

me = 1 + !

25.0 cm 25.0 cm = 1+ = +1.75 fe 33.3 cm

and the image distance for the objective is q1 = L − pe = 10.0 cm − 14.3 cm = −4.3 cm The thin-lens equation then gives the object distance for the objective as

q f (−4.3 cm)(50.0 cm) p1 = 1 1 = = +4.0 cm q − f −4.3 cm − 50.0 cm 1 1 !

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1489

The magnification by the objective is then

q (−4.33 cm) M1 = − 1 = − = +1.1 p 4.0 cm 1 ! and the overall magnification is m = M1me = (+1.1)(+1.75) = 1.9. 25.40

Note: We solve part (b) before answering part (a) in this problem. (b) The objective forms a real, inverted image, diminished in size, of a very distant object at q1 = fo. This image is a virtual object for the eyepiece at pe = −|fe|, giving

1 1 1 1 1 = − = + =0 qe pe fe − fe fe ! and

qe → ∞

(a) Parallel rays emerge from the eyepiece, so the eye observes a virtual image.

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1490

fo (c) The angular magnification is m = f = 3.00, giving fo = 3.00|fe|. Also, e ! the length of the telescope is L = fo + fe = 3.00|fe| − |fe| = 10.0 cm, giving

25.41

fe = − fe = −

10.0 cm = −5.00 cm !and!fo = 3.00 fe = 15.0 cm 2.00

The angular separation of the lights is θ = d/h, where d = 1.00 m is their linear separation, and h is the altitude of the satellite. If the lights are just resolved according to the Rayleigh criterion, then θ = θmin = 1.22(λ/D), where l is the wavelength of the light, and D is the diameter of the lens. Thus, the altitude of the satellite must be

d d d ⋅ D (1.00 m)(0.300 m) h= = = = = 4.92 × 105 m = 492 km −9 θ 1.22( λ /D) 1.22 λ 1.22(500 × 10 m) ! 25.42

The angular separation of two objects seen on the ground is θ = d/h, where d is their linear separation and h is your altitude. If the objects are just resolved according to the Rayleigh criterion, then θ = θmin = 1.22(λ/D), where λ is the wavelength of the light, and D is the diameter of the aperture (your pupil). Thus, the minimum separation of objects you can

Topic 25

1491

distinguish is 3 −9 1.22 ⋅ h ⋅ λ 1.22 ( 9.50 × 10 m ) ( 575 × 10 m ) d = h ⋅ θ min = = = 1.7 m −3 D 4.0 × 10 m !

25.43

λ The limit of resolution in air is θ min air = 1.22 = 0.60 µrad . D ! In oil, the limiting angle of resolution will be

( λ noil ) = ⎛ 1.22 λ ⎞ 1 λ θ min oil = 1.22 oil = 1.22 ⎜⎝ ⎟ D D D ⎠ noil ! θ min air 0.60 µrad θ min oil = = = 0.40 µrad n 1.5 oil !

25.44

(a) The wavelength of the light within the eye is λn = λ/n. Thus, the limiting angle of resolution for light passing through the pupil (a circular aperture with diameter D = 2.00 mm) is

500 × 10−9 m ) ( λn λ θ min = 1.22 = 1.22 = 1.22 = 2.29 × 10−4 rad −3 D nD (1.33)( 2.00 × 10 m ) ! (b) From s = rθ, the distance from the eye that two points separated by a distance s = 1.00 cm will intercept this minimum angle of resolution is

s 1.00 cm r= = = 4.37 × 103 cm = 43.7 m −4 θ 2.29 × 10 rad min !

Topic 25

25.45

1492

The angular separation of the headlights when viewed from a distance of r = 10.0 km is

s 2.00 m θ= = = 2.00 × 10−4 rad 3 r 10.0 × 10 m ! If the headlights are to be just resolved, this separation must equal the limiting angle of resolution for the circular aperture, θmin = 1.22λ/D, so the diameter of the aperture is −9 1.22λ 1.22λ 1.22 ( 885 × 10 m ) D= = = = 5.40 × 10−3 m = 5.40 mm −4 θ min θ 2.00 × 10 rad !

25.46

The angular separation of the two stars is θ = d/r, where d is their linear separation and r is their distance from Earth. If the stars are just resolved according to the Rayleigh criterion, then θ = θmin = 1.22(λ/D), where λ is the wavelength of the light, and D is the diameter of the aperture (telescope objective). Thus, the linear separation of the stars must be

d = r ⋅ θ min =

25.47

r ⋅1.22λ D

(

)(

)

(

)

⎡ 23 ly 9.461 × 1015 m/ly ⎤ 1.22 575 × 10−9 m ⎦ =⎣ = 2.2 × 1011 m 0.68 m

λ If just resolved, the angular separation of the objects is θ = θ min = 1.22 D ! ⎡ ⎛ 500 × 10−9 m ⎞ ⎤ 7 s = r θ = 8.0 × 10 km 1.22 ( ) and ⎢ ⎜⎝ 5.00 m ⎟⎠ ⎥ = 9.8 km ⎣ ⎦ !

Topic 25

25.48

1493

λ If just resolved, the angular separation of the objects is θ = θ min = 1.22 D ! ⎡ ⎛ 550 × 10−9 m ⎞ ⎤ and s = r θ = ( 200 × 10 m ) ⎢1.22 ⎜ ⎥ = 0.38 m = 38 cm ⎝ 0.35 m ⎟⎠ ⎦ ⎣ ! 3

25.49

(a) A diffraction grating with N lines illuminated has an mth order resolving power given by R = Nm. Solve for N to find

R 1250 = = 625 m 2

(b) Solve the definition of resolving power for the wavelength difference Δλ to find

25.50

λ λ 525 nm → Δλ = = = 0.420 nm Δλ R 1250

The resolving power of a diffraction grating is R = λ/Δλ = Nm. (a) The number of lines the grating must have to resolve the Hα line in the first order is

R λ /Δλ 656.2 nm N= = = = 3.6 × 103 lines m (1) 0.18 nm ! R 656.2 nm = 1.8 × 103 lines . (b) In the second order (m = 2), N = = 2 2(0.18 nm) ! 25.51

The grating spacing is d = 1 cm/6 000 = 1.67 × 10−4 cm = 1.67 × 10−6 m, and the highest order of 600 nm light that can be observed is

Topic 25

1494

mmax = !

−6 dsin 90° (1.67 × 10 m ) (1) = = 2.78 → 2 orders λ 600 × 10−9 m

The total number of slits is N = (15.0 cm)(6 000 slits/cm) = 9.00 × 104, and the resolving power of the grating in the second order is Ravailable = Nm = (9.00 × 104)2 = 1.80 × 105 The resolving power required to separate the given spectral lines is

λ 600.000 nm Rneeded = = = 2.0 × 105 Δ λ 0.003 nm ! These lines cannot be separated with this grating. 25.52

(a) When the central spot in the interferometer pattern goes through a full cycle from bright to dark and back to bright, two fringe shifts have occurred, and the movable mirror has moved a distance of 2(λ/4) = λ/2. Thus, if Ncycles = 1 700 such cycles are observed as the mirror moves distance d = 0.382 mm, it must be true that d = Ncycles(λ/2), or λ = 2d/Ncycles. The wavelength of the light illuminating the interferometer is therefore

λ= !

2 ( 0.382 × 10−3 m ) 1700

= 4.49 × 10−7 m = 449 nm

which is in the blue region of the visible spectrum.

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1495

(b) Red light has a longer wavelength than blue light, so fewer wavelengths would cover the given displacement. Hence, Ncycles would be smaller. 25.53

A fringe shift occurs when the mirror moves distance λ/4. Thus, if the mirror moves distance ΔL = 0.180 mm, the number of fringe shifts observed is −3 ΔL 4(ΔL) 4 ( 0.180 × 10 m ) Nshifts = = = = 1.31 × 103 −9 λ /4 λ 550 × 10 m !

25.54

A fringe shift occurs when the mirror moves distance λ/4. Thus, the distance the mirror moves as 250 fringe shifts are counted is ⎛ 632.8 × 10−9 m ⎞ ⎛ λ⎞ ΔL = Nshifts ⎜ ⎟ = 250 ⎜ = 3.96 × 10−5 m = 39.6 µm ⎟ ⎝ 4⎠ ⎝ 4 ⎠ !

25.55

A fringe shift occurs when the mirror moves distance λ/4. Thus, the distance moved (length of the bacterium) as 310 shifts occur is ⎛ 650 × 10−9 m ⎞ ⎛ λ⎞ −5 ΔL = Nshifts ⎜ ⎟ = 310 ⎜ ⎟⎠ = 5.04 × 10 m = 50.4 µm ⎝ 4⎠ ⎝ 4 !

25.56

The wavelength of light within the tube decreases from λ to λn = λ/ngas as the tube fills with gas. Thus, the number of wavelengths that will fit in the length L of the tube increases from L/λ to ngasL/λ. Four fringe shifts

Topic 25

1496

occur for each additional wavelength that fits within the tube, so the number of fringes shifts to be seen as the tube fills with gas is ⎡ ngas L L ⎤ 4L Nshifts = 4 ⎢ − ⎥= n −1 λ ⎦ λ gas ⎣ λ !

(

Hence,

25.57

)

⎡ ⎤ ⎛ λ ⎞ 600 × 10−9 m ⎥ ngas = 1 + ⎜ ⎟ Nshifts = 1 + ⎢ (160) = 1.000 5 ⎢ 4 5.00 × 10−2 m ⎥ ⎝ 4L ⎠ ⎣ ⎦

(

)

When the optical path in one arm of a Michelson’s interferometer increases by one wavelength, four fringe shifts will occur (one shift for every quarter-wavelength change in path length). The number of wavelengths (in a vacuum) that fit in a distance equal to a thickness t is Nvac = t/λ. The number of wavelengths that fit in this thickness while traveling through the transparent material is Nn = t/λn = t/(λ/n) = nt/λ. Thus, the change in the number of wavelengths that fit in the path down this arm of the interferometer is t ΔN = Nn − Nvac = (n − 1) λ !

and the number of fringe shifts that will occur as the thin sheet is inserted will be

Topic 25

1497

⎛ 15.0 × 10−6 m ⎞ t #fringe!shifts = 4(ΔN) = 4(n − 1) = 4(1.40 − 1) ⎜ = 40 −9 λ ⎝ 600 × 10 m ⎟⎠ !

25.58

When viewed from a distance of 50 meters, the angular length of a mouse (assumed to have an actual length of ≈10 cm) is

s 0.10 m θ= = = 2.0 × 10−3 radians r 50 m ! Thus, the limiting angle of resolution of the eye of the hawk must be

θmin ≈ 2.0 × 10−3 radians 25.59

(a) For a refracting telescope, the magnification is m = fo/fe, where fo and fe are the focal lengths of the objective lens and the eyepiece, respectively. Thus, when the Yerkes telescope uses an eyepiece with fe = 2.50 cm, the magnification is

f 20.0 m m= o = = 8.00 × 102 = 800 −2 fe 2.50 × 10 m ! (b) Standard astronomical telescopes form inverted images. Thus, the observer Martian polar caps are upside down. 25.60

(a) Since this eye can already focus on objects located at the near point of a normal eye (25 cm), no correction is needed for near objects. To correct the distant vision, a corrective lens (located 2.0 cm from the

Topic 25

1498

eye) should form virtual images of very distant objects at 23 cm in front of the lens (or at the far point of the eye). Thus, we must require that q = −23 cm when p → ∞. This gives

1 1 1 1 P = = + = 0+ = −4.3 diopters f p q −0.23 m ! (b) A corrective lens in contact with the cornea should form virtual images of very distant objects at the far point of the eye. Therefore, we require that q = −25 cm when p → ∞, giving

1 1 1 1 P = = + = 0+ = −4.0 diopters f p q −0.25 m !

1 ⎛ ⎞ When the contact lens ⎜ f = = −25 cm ⎟ is in place, the object ⎝ ⎠ P ! distance which yields a virtual image at the near point of the eye (that is, q = −16 cm) is given by

qf (−16 cm)( − 25 cm) p= = = 44 cm q − f −16 cm − (−25 cm) ! 25.61

With 485 lines equally spaced in a height ℓ, the distance separating adjacent lines is !d = /485. When the screen is viewed from a distance L, the angular separation between adjacent lines is θ = d/L. If the individual lines are not to be seen (i.e., the lines are to be unresolved), this angular

Topic 25

1499

separation must be less than the minimum angle of resolution, θmin = 1.22(λ/D) by the Rayleigh criterion. That is, we must have d /485 1.22 ⋅ λ θ= = < θ min = L L D !

25.62

L D 5.00 × 10−3 m > = = 15.4  485(1.22 ⋅ λ ) 485 (1.22 ) ( 550 × 10−9 m ) !

(a) The image must be formed on the back of the eye (retina), so we must have q = 2.00 cm when p = 1.00 m = 100 cm. The thin-lens equation gives the required focal length as

f= !

pq (100 cm)(2.00 cm) = = 1.96 cm p + q 100 cm + 2.00 cm

(b) The f-number of a lens aperture is the focal length of the lens divided by the diameter of the aperture. Thus, the smallest f-number occurs with the largest diameter of the aperture. For the typical eyeball focused on objects 1.00 m away, this is

( f "number ) !

min

f Dmax

1.96 cm = 3.27 0.600 cm

Topic 25

1500

( f "number ) ! 25.63

= max

f Dmin

1.96 cm = 9.80 0.200 cm

(a) The lens should form an upright, virtual image at the near point of the eye, q = −75.0 cm, when the object distance is p = 25.0 cm. The thin-lens equation then gives

f= !

pq (25.0 cm)( − 75.0 cm) = = 37.5 cm = 0.375 m p+q 25.0 cm − 75.0 cm

1 1 = +2.67 diopters . so the needed power is P = = f 0.375 m ! (b) If the object distance must be p = 26.0 cm to position the image at q = −75.0 cm, the actual focal length is

f= !

pq (26.0 cm)( − 75.0 cm) = = 0.398 m p+q 26.0 cm − 75.0 cm

1 1 = +2.51 diopters and P = = f 0.398 m ! The error in the power is ΔP = (2.67 − 2.51) diopters = 0.16 diopters too low

25.64

We use

n1 n2 n2 − n1 + = , with p → ∞, and q equal to the cornea to retina p q R !

distance.

Topic 25

1501

⎛n −n ⎞ ⎛ 1.34 − 1.00 ⎞ Then, R = q ⎜ 2 1 ⎟ = (2.00 cm) ⎜ ⎟⎠ = 0.507 cm = 5.07 mm . ⎝ n 1.34 ⎝ ⎠ 2 !

25.65

(a) The implanted lens should give an image distance of q = 22.4 mm for distant (p → ∞) objects. The thin-lens equation then gives the focal length as f = q = 22.4 mm, so the power of the implanted lens should be

1 1 Pimplant = = = +44.6 diopters −3 f 22.4 × 10 m ! (b) When the object distance is p = 33.0 cm, the corrective lens should produce parallel rays (q → ∞). Then the implanted lens will focus the final image on the retina. From the thin-lens equation, the required focal length is f = p = 33.0 cm, and the power of this lens should be

1 1 Pcorrective = = = +3.03 diopters f 0.330 m ! 25.66

The angular magnification is m = θ/θ0, where θ is the angle subtended by the final image, and θ0 is the angle subtended by the object as shown in the figure below. When the telescope is adjusted for minimum eyestrain, the rays entering the eye are parallel. Thus, the objective lens must form its image at the focal point of the eyepiece.

Topic 25

1502

From triangle ABC, θ0 ≈ tan θ0 = h′/q1, and from triangle DEF, θ ≈ tan θ =

θ h′ fe q1 = . h′/fe. The angular magnification is then m = = θ h q fe ′ 0 1 ! From the thin-lens equation, the image distance of the objective lens in this case is

p f (300 cm)(20.0 cm) q1 = 1 1 = = 21.4 cm p − f 300 cm − 20.0 cm 1 1 ! With an eyepiece of focal length fe = 2.00 cm, the angular magnification for this telescope is

q 21.4 cm m= 1 = = 10.7 f 2.00 cm e ! 25.67

When a converging lens forms a real image of a very distant object, the image distance equals the focal length of the lens. Thus, if the scout started a fire by focusing sunlight on kindling 5.00 cm from the lens, f = q = 5.00 cm. (a) When the lens is used as a simple magnifier, maximum

Topic 25

1503

magnification is produced when the upright, virtual image is formed at the near point of the eye (q = −15 cm in this case). The object distance required to form an image at this location is

qf (−15 cm)(5.0 cm) 15 cm p= = = q − f −15 cm − 5.0 cm 4.0 ! and the lateral magnification produced is

q −15 cm M=− =− = +4.0 . Note that adapting Equation 25.5 for p 15 cm/4.0 ! use with this “abnormal” eye would give an angular magnification of mmax = 1 + |q|/f = 1 + 15 cm/5.0 cm = +4.0. (b) When the object is viewed directly while positioned at the near point of the eye, its angular size is θ0 = h/15 cm. When the object is viewed by the relaxed eye while using the lens as a simple magnifier (with the object at the focal point so parallel rays enter the eye), the angular size of the upright, virtual image is θ = h/f. Thus, the angular magnification gained by using the lens in this manner is

θ h/f 15 cm 15 cm m= = = = = +3.0 θ h/15 cm f 5.0 cm 0 !

Topic 26

1504

Topic 26 Relativity

QUICK QUIZZES 26.1

False. One of Einstein’s two basic postulates in his special theory of relativity is the constancy of the speed of light. This postulate states that the speed of light in a vacuum has the same value in all inertial reference frames, regardless of the velocity of the observer or the velocity of the source emitting the light.

26.2

Choice (a). Less time will have passed for you in your frame of reference than for your employer back on Earth. Thus, to maximize your paycheck, you should choose to have your pay calculated according to the elapsed time on a clock on Earth.

26.3

False. Clocks, including biological clocks, in the observer’s own reference frame appear to run at normal speeds. It is only clocks in reference frames moving relative to the observer that the observer will judge to be running slower than normal.

26.4

No. From your perspective you are at rest with respect to the cabin, so

Topic 26

1505

you will measure yourself as having your normal length and will require a normal-sized cabin. 26.5

(i) Choices (a) and (e). The outgoing rocket will appear to have a shorter length and a slower clock. (ii) Choices (a) and (e). The answers for the incoming rocket are the same as for the outgoing rocket. Length contraction and time dilation depend only on the magnitude of the relative velocity, not on the direction.

26.6

False. According to the mass-energy equivalence equation, the particle’s total energy is E = mc 2 !

1 − v 2 c 2 , and it is seen that as v → c, the total

energy becomes infinitely large. As the total energy becomes infinitely large, so will the kinetic energy given by KE = E − mc2 and the relativistic momentum given by p = E 2 − ( mc 2 ) c . ! 2

26.7

(a) False

(b) False

(d) False

A reflected photon does exert a force on the surface. Although a photon has zero mass, a photon does carry momentum. When it reflects from a surface, there is a change in the momentum, just like the change in momentum of a ball bouncing off a wall. According to the momentum interpretation of Newton’s second law, a change in momentum results in

Topic 26

1506

a force on the surface. This concept is used in theoretical studies of space sailing. These studies propose building non-powered spacecraft with huge reflective sails oriented perpendicularly to the rays from the Sun. The large number of photons from the Sun reflecting from the surface of the sail will exert a force which, although small, will provide a continuous acceleration. This would allow the spacecraft to travel to other planets without fuel.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 26.2

The proper time interval is measured by an observer at rest relative to the location where the two events occur.

26.4

Because γ for an object in relative motion is always greater than 1, a proper time interval Δtp is always shorter than the time interval Δt and a proper length Lp is always longer than the length L.

26.6

The two observers will agree on the speed of light and on the speed at which they move relative to one another.

26.8

Special relativity describes inertial reference frames—that is, reference frames that are not accelerating. General relativity describes all reference

Topic 26

1507

frames. 26.10

You would see the same thing that you see when looking at a mirror when at rest. The theory of relativity tells us that all experiments will give the same results in all inertial frames of reference.

26.12

(a) Attempts to apply special relativity to this situation would lead one to say that each observer (one on Earth and one in orbit) would find the other’s clock to run slow relative to their own clock. However, the observer in orbit is not in an inertial reference frame, and special relativity does not apply here. According to general relativity, the clock in orbit and undergoing an acceleration (the centripetal acceleration) will run slower. (b) When the moving clock returns to Earth, they are again in inertial reference frames and subject to the same laws of physics. Thus, the identical clocks should tick at the same rate once again. However, they will not be synchronized. The clock that underwent the accelerations (i.e., the orbiting clock) will have permanently lost time (or will have “aged less”) and will be behind the clock that remained on Earth.

26.14

The 8 light-years represents the proper length of a rod from Earth to

Topic 26

1508

Sirius, measured by an observer seeing both the rod and Sirius nearly at rest. The astronaut sees Sirius coming toward her at 0.8c but also sees the distance contracted to d = 8 ly ) 1 − ( v / c ) = ( 8 ly ) 1 − ( 0.8 ) = 5 ly ! ( 2

d 5 ly (5 yr)c = = 6 yr. So the travel time measured on her clock is t = = 0.8c ! v 0.8c

26.16

Relativistic momentum is related to the total energy through the equation

(

E2 = p 2c 2 + mc 2

) or p c = E − ( mc ) . For a given E, relativistic 2

2 2

momentum p decreases with increasing mass. Among a set of particles with the same energy E, the most massive particle has the least momentum and the least massive particle has the greatest momentum. For a photon, proton, and electron with the same E, the correct ranking is given in choice (e): pproton < pelectron < pphoton.

ANSWERS TO EVEN NUMBERED PROBLEMS 26.2

(a)

172 m

26.4

(a)

0.436 m

(b)

less than 0.436 m by an undetectable amount

(a)

70 beats/min

26.6

(b)

2.78 × 10−7 s

(c)

6.38 × 10−7 s

30 beats/min

Topic 26

1509

26.8

(a)

65.0 beats/min

(b)

10.6 beats/min

26.10

(a)

2.94 × 108 m/s

(b)

15.3 yr

26.12

26.0 min + 1.06 × 10−11 s

26.14

(a) rectangular box (b)

26.16

1.25 × 108 m/s

26.18

(a)

(c)

3.06 yr

(b)

3.52 × 10−18 kg ⋅ m/s

1.2 m, 2.0 m, 2.0 m

4.96 × 10−19 kg ⋅ m/s

0.94c toward the left

26.22

+0.696c

26.24

(a)

26.26

0.161 Hz

26.28

(a)

939 MeV

26.30

(a)

KE ≈ γ ER = γ mpc2

26.32

(a)

0.582 MeV

26.34

2.51 × 10−28 kg with speed 0.987c, and 8.84 × 10−28 kg with speed 0.868c.

2.73 × 108 m/s

(b)

72.6 m

(b)

3.01 GeV (b)

(b)

(c)

2.07 GeV

0.999.99c

2.45 MeV

Topic 26

26.36

26.38

26.40

1510

(a)

KEi + mi c2 = KEf + mf c2

(c)

235.861 612 u

(a)

0.0.23 6c = 7.08 × 103 km/s

(b)

(6.17 × 10−4)c = 185 km/s

(d)

0.190 976 u

(b)

236.052 588 u

(e)

177.893 MeV

(a) Yes. As the spring is compressed, positive work is done on it or energy is added to it. Since mass and energy are equivalent, mass has been added to the spring. (b)

Δm = kx2/2c2

(c)

2.5 × 10−17 kg

26.42

(a)

2.50 MeV/c

(b)

4.60 GeV/c

26.44

1.2 × 10−13 m

26.46

(a)

2d t= ! c+v

(b)

2d c − v t′ = c c+v !

26.48

(a)

∼108 km

(b)

∼102 s

26.50

(a)

See Solution.

(b)

0.943c

26.52

Rg = 1.47 km for the Sun

26.54

(a)

21.0 yr

(b)

14.7 ly

(c)

10.5 ly

(d)

35.7 yr

Topic 26

26.56

1511

(a)

0.946c

(b)

0.160 ly

(c)

0.114 yr

(d)

7.51 × 10−22 J

PROBLEM SOLUTIONS 26.1

(a) Proper time is measured by an observer at rest with respect with the clock. In this case, the astronaut on the ship measures proper time so that Δtp = 2.00 s. (b) As observed by a person on Earth, the “clock” (a blinking light) is moving and therefore runs more slowly than an identical stationary clock. In fact, the time between blinks is

Δt = γ Δtp =

26.2

1 1 − vc 2

Δtp =

1 1 − ( 0.750 )

( 2.00 s) = 3.02 s

(a) Proper length is measured by an observer at rest relative to the object. In this case, an astronaut on the spaceship would measure the proper length. The length measured by an observer on Earth is L = 75.0 m so that

→ Lp = γ L =

1 v2 1− 2 c

1 1 − ( 0.900 )

(75.0 m ) = 172 m

Topic 26

1512

(b) As measured by a person on Earth, the spaceship passes in a time interval (in fact, it’s the proper time interval because the two events defining the interval occur at the same location for the Earth-bound observer) Δtp =

L 75.0 m = = 2.78 × 10−7 s v 0.900c

Δt = γ Δtp =

26.3

1 1 − ( 0.900 )

( 2.78 × 10 s) = 6.38 × 10 s −7

−7

(a) Observers on Earth measure the time for the astronauts to reach Alpha Centauri as ΔtE = 4.42 yr. But these observers are moving relative to the astronaut’s internal biological clock and hence experience a dilated version of the proper time interval Δtp measured on that clock. From ΔtE = γ Δtp, we find

Δt = ΔtE γ = ΔtE 1 − ( v / c ) = ( 4.42yr ) 1 − ( 0.950 ) = 1.38yr ! p 2

(b) The astronauts are moving relative to the span of space separating Earth and Alpha Centauri. Hence, they measure a length-contracted version of the proper distance, Lp = 4.20 ly. The distance measured by the astronauts is

Topic 26

1513

L = Lp γ = Lp 1 − ( v / c ) = ( 4.20ly ) 1 − ( 0.950 ) = 1.31ly ! 2

26.4

(a) The length of the meterstick measured by the observer moving at speed v = 0.900c relative to the meterstick is

L = Lp γ = Lp 1 − ( v / c ) = (1.00m ) 1 − ( 0.900 ) = 0.436 m ! 2

(b) If the observer moves relative to Earth in the direction toward the meterstick, the velocity of the observer relative to the meterstick is greater than that in part (a). The measured length of the meterstick will be less than 0.436 m under these conditions, but at the speed the observer could run, the effect would be too small to detect. 26.5

The contracted length of the ship, L = Lp − ΔL, as measured by the Earthbased observer is L = Lp − ΔL = Lp 1 − ( v / c ) . This yields ! 2

(

)

1 − ( v / c ) = 1 − ΔL Lp , and solving for the speed v of the ship, we find

(

v = c 1 − 1 − ΔL Lp !

) . Thus, if the proper length of the ship is L = 28.0 m, 2

and the observed contraction is ΔL = 0.150 m, the speed of the ship must be 2

⎛ 0.150 m ⎞ v = c 1− ⎜1− = 0.103c 28.0 m ⎟⎠ ⎝ !

Topic 26

26.6

1514

(a) The time for 70 beats, as measured by the astronaut and any observer at rest with respect to the astronaut, is Δtp = 1.0 min. The observer in the ship then measures a rate of 70 beats/min. (b) The observer on Earth moves at v = 0.90c relative to the astronaut and measures the time for 70 beats as

Δt = γΔtp = !

Δtp 1 − (v / c)

= 2

1.0min 1 − (0.90)2

= 2.3 min

This observer then measures a beat rate of 70 beats 70 beats/2.3 min = 30 beats/min.

26.7

Δtp 2.6 × 10−8 s Δt = γ Δt = = = 1.3 × 10−7 s (a) p 2 2 1 − (0.98) 1 − (v / c) ! (b) d = v(Δt) = [0.98(3.0 × 108 m/s)](1.3 × 10−7 s) = 38 m (c) d' = v(Δtp) = [0.98(3.0 × 108 m/s)](2.6 × 10−8 s) = 7.6 m

26.8

(a) As measured by observers in the ship (that is, at rest relative to the astronaut), the time required for 75.0 beats is Δtp = 1.00 min. The time interval required for 75.0 beats as measured by the Earth observer is Δt = γΔtp = (1.00 min ) !

1 − (0.500)2 , so the Earth observer

measures a pulse rate of

Topic 26

1515

75.0 75.0 1 − (0.500)2 rate = = = 65.0/ min Δt 1.00 min !

1.00 min (b) If v = 0.990c, then Δt = γΔtp = 1 − (0.990)2 ! and the pulse rate observed on Earth is 75.0 75.0 1 − (0.990)2 rate = = = 10.6/ min Δt 1.00 min !

That is, the life span of the astronaut (reckoned by the total number of his heartbeats) is much longer as measured by an Earth clock than by a clock aboard the space vehicle. 26.9

(a) To the observer on Earth, the muon appears to have a lifetime of

d 4.60 × 103 m Δt = = = 1.55 × 10−5 s v 0.990 ( 3.00 × 108 m/s ) ! 1 1 = = 7.09 (b) γ = 2 2 1 − (0.990) 1 − (v / c) ! (c) To an observer at rest with respect to the muon, its proper lifetime is

Δt 1.55 × 10−5 s Δtp = = = 2.19 × 10−6 s γ 7.09 ! (d) The muon is at rest relative to the observer traveling with the muon. Thus, the muon travels zero distance as measured by this observer.

Topic 26

1516

However, during the observed lifetime of the muon, this observer sees Earth move toward the muon a distance of d' = v(Δtp) = [0.990(3.00 × 108 m/s)](2.19 × 10−6 s) = 6.50 × 102 m = 650 m (e) As the third observer travels toward the incoming muon, his speed relative to the muon is greater than that of the observer at rest on Earth. Thus, his observed gamma factor (Δt = γΔtp) is higher, and he measures the muon's lifetime as longer than that measured by the observer at rest with respect to Earth. 26.10

(a) Given the relation L = Lp /γ with L = 3.00 ly and Lp = 15.0 ly, γ = 5.00 so the required speed is

γ =

= 5 → 1 − vc 2 = 2

1 − vc 2

1 → v = 2.94 × 108 m/s 25

(b) A person on Earth would measure the proper distance Lp to the star and a time interval Δt given by

Lp = vΔt

(

)

15.0 ly 9.461 × 1015 m/ly ⎛ 1 yr ⎞ Δt = = = 15.3 yr ⎜ 8 7 ⎝ 3.156 × 10 s ⎟⎠ v 2.94 × 10 m/s Lp

Topic 26

1517

Δtp =

26.11

Δt 15.3 yr = = 3.06 yr γ 5

The proper length of the faster ship is three times that of the slower ship (Lpf = 3Lps), yet they both appear to have the same contracted length, L. Thus,

( ) 1 − ( v c) ,!or!1 − ( v c) = 9 − 9 ( v c)

L = Lps 1 − ( vs c) = 3Lps ! 2

This gives

vf = !

26.12

c 8 + ( vs c ) 3

8 + (0.350)2 c = 0.950c 3

The driver is the observer at rest with respect to the clock measuring the 26.0-min time interval. Thus, this observer measures the proper time Δtp, and the Earth-based observer measures the dilated time −

2 Δt = γΔtp = ⎡⎣1 − ( v / c ) ⎤⎦ 2 ⋅ Δtp In this case, v = 35.0 m/s  c, so we use the !

1 − 1 binominal expansion [1 − x] 2 ≈ 1 + x!!when!!x  1 . This gives the time 2 !

measured by the observer fixed on Earth as 1

1 2 − 2⎞ ⎛ Δt = (26.0 min) ⎡⎣1 − ( v / c ) ⎤⎦ 2 ≈ (26.0 min) ⎜ 1 + ( v / c ) ⎟ ⎝ ⎠ 2 ! or

Topic 26

1518

( 26.0 min ) ⎛ Δt ≈ 26.0 min+ 2

26.13

35.0 m/s ⎞ ⎛ 60.0 s ⎞ −11 ⎜⎝ 3.00 × 108 m/s ⎟⎠ ⎜⎝ 1 min ⎟⎠ = 26.0 min+ 1.06 × 10 s

The trackside observer sees the supertrain length-contracted as L = Lp 1 − ( v / c ) = (100 m) 1 − (0.95)2 = 31 m ! 2

The supertrain appears to fit in the tunnel with 50 m − 31 m = 19 m to spare. 26.14

Length contraction occurs only in the dimension parallel to the motion. (a) The sides labeled L2 and L3 in the figure below are unaffected, but the side labeled L1 will appear contracted, giving the box a rectangular shape, or more formally, the shape of a rectangular parallelepiped.

(b) The dimensions of the box, as measured by the observer moving at v = 0.80c relative to it, are L2 = L2p = 2.0 m, L3 = L3p = 2.0 m, and © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 26

1519

L1 = L1p 1 − ( v / c ) = (2.0 m) 1 − (0.80)2 = 1.2 m ! 2

26.15

(a) For a speed v = 0.900c, γ = 1/

(1 − (0.900) ) = 2.29 . The proton’s 2

momentum is p = γ mv = (2.29)(1.67 × 10−27 kg)(0.900c) = 1.03 × 10−18 kg ⋅ m/s .

(b) This occurs where γ = 2 so that

γ =

= 2 → 1− 2

1 − vc 2

v2 1 = → v = 2.60 × 108 m/s c2 4

26.16 The required condition is γ mv − mv = (0.100)mv so that

γ − 1 = 0.100 → γ = 1.100 =

1 1 − vc 2

→ 1−

v2 1 = 2 c (1.100)2

v = 1.25 × 108 m/s 26.17

(a) The classical expression for linear momentum is pclasscial = mv, while the relativistic expression is p = γmv, where γ = 1 !

1 − ( v / c ) . Thus, if 2

p = 3pclasscial, it is necessary that the gamma factor have a value γ = 3. Solving for the speed of the particle gives

1 1 c 8 2c 2 v = c 1− 2 = c 1− = = = 0.943c γ 9 3 3 ! (b) Observe that the calculation above did not depend on the mass of the

Topic 26

1520

particle involved. Thus, the result is the same for a proton or any other particle. 26.18

(a) Classically, p = mv = m(0.990c) = (1.67 × 10−27 kg)(0.990)(3.00 × 108 m/s) = 4.96 × 10−19 kg ⋅ m/s (b) By relativistic calculations,

mv 1 − (v / c)

m(0.990c) 1 − (0.990)2

(1.67 × 10 =

−27

kg ) ( 0.990 ) ( 3.00 × 108 m/s ) 1 − (0.990)2

= 3.52 × 10−18 kg ⋅ m/s

Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus, their magnitudes must be equal, or γ2m2v2 =

γ1m1v1. This gives

(1.67 × 10 !

−27

kg ) v

1 − (v / c)

( 2.50 × 10 =

−28

kg ) ( 0.893c )

1 − (0.893)2

Topic 26

1521 2 ⎡ (1.67 × 10−27 kg ) 1 − (0.893)2 ⎤ ⎛ v ⎞ ⎛ v⎞ ⎥ = 1 − and reduces to ⎢ ⎜ ⎟ ⎜⎝ ⎟⎠ , or −28 c ⎢⎣ ( 2.50 × 10 kg ) ( 0.893 ) ⎥⎦ ⎝ c ⎠ !

12.3(v/c)2 = 1 and yields v = 0.285c. 26.20

We take to the right as the positive direction. Then, the velocities of the two ships relative to Earth are vRE = +0.70c and vLE = −0.70c. The velocity of ship L relative to ship R is given by the relativistic relative velocity relation (Equation 26.7 in the textbook) as

vLR = !

26.21

vLE − vRE (−0.70c) − 0.70c −1.40c = = = −0.94c = 0.94c!toward!the!left vLEvRE (−0.70c)(0.70c) 1 + 0.49 1− 1− c2 c2

Taking to the right as the positive direction, the velocity of the electron relative to the laboratory is vEL = +0.90c, and the velocity of the proton relative to the electron is vPE = −0.70c. Thus, the relativistic addition of velocities (Equation 26.8 in the textbook) gives the velocity of the proton relative to the laboratory as

vPE + vEL (−0.70c) + 0.90c +0.20c = = = +0.54c = 0.54c!toward!the!right vPEvEL (−0.70c)(0.90c) 1 + 0.63 1+ 1 + c2 c2

26.22

We choose the direction of the spaceship’s motion relative to Earth as the

vPL =

positive direction. Then, the spaceship’s velocity relative to Earth is vSE = +0.750c. It is desired to have the velocity of the rocket relative to Earth be © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 26

1522

vRE = +0.950c. The relativistic relative velocity relation (Equation 26.7 in the textbook) then gives the required velocity of the rocket relative to the ship as vRS = !

26.23

vRE − vSE 0.950c − 0.750c = = +0.696c vREvSE (0.950c)(0.750c) 1− 1 − c2 c2

Assign subscript S for the spaceship, E for the Earth, and R for the rocket. The relativistic addition of velocities gives the rocket’s speed relative to Earth:

vRE =

26.24

vRS + vSE 0.500c + 0.900c = = 2.90 × 108 m/s vRS vSE 1 + ( 0.500 ) ( 0.900 ) 1+ c2

(a) Taking the direction of spaceship A’s velocity as the positive direction, apply the equation for relativistic relative velocity to find

vBA =

vBE − vAE −0.750c − 0.500c = = 2.73 × 108 m/s vBE vAE 1 − ( −0.750 ) ( 0.500 ) 1− c2

(b) For v = 2.73 × 108 m/s, γ = 2.41. From L = Lp /γ , find that

26.25

175 m = 72.6 m 2.41

Taking away from Earth as the positive direction, the velocity of ship A relative to Earth is vAE = +0.800c, and the velocity of ship B relative to

Topic 26

1523

Earth is vBE = +0.900c. The relativistic relative velocity relation (Equation 26.7 in the textbook) gives the velocity of ship B relative to ship A (and hence, the speed with which B is overtaking A) as vBA = !

26.26

vBE − vAE 0.900c − 0.800c = = +0.357c vBEvAE (0.900c)(0.800c) 1− 1 − c2 c2

We first determine the velocity of the pulsar relative to the rocket. Taking toward the Earth as the positive direction, the velocity of the pulsar relative to Earth is vPE = +0.950c, and the velocity of the rocket relative to Earth is vRE = −0.995c. The relativistic relative velocity relation (Equation 26.7 in the textbook) gives the velocity of the pulsar relative to the rocket as

vPR =

vPE − vRE 1−

vPEvRE c2

0.950c − (−0.995c) (0.950c)(−0.995c) 1− c2

= +999 87c

The time interval between successive pulses in the pulsar’s own reference frame (i.e., on a clock at rest with respect to the event being timed) is Δtp = 1/10.0 Hz = 0.100 s. The duration of this interval in the rocket’s frame of reference is given by the time dilation relation as

Topic 26

1524

Δt = γΔtp =

Δtp

(

1 − vPR c

)

0.100 s 1 − (0.999 87)

= 6.20 s

and the frequency of the pulses as observed in the rocket’s reference frame is f= !

26.27

1 1 = = 0.161 Hz Δt 6.20 s

Taking to the right as positive, it is given that the velocity of the rocket relative to observer A is vRA = +0.92c. If observer B observes the rocket to have a velocity vRB = −0.95c, the velocity of observer B relative to the rocket is vBR = +0.95c. The relativistic velocity addition relation then gives the velocity of B relative to the stationary observer A as

vBA = !

26.28

vBR + vRA +0.95c + 0.92c = = +0.998c!!!or!!! 0.998c!toward!the!right vBR vRA (0.95c)(0.92c) 1− 1− c2 c2

2⎛ 1 MeV ⎞ 2 −27 8 = 939 MeV (a) ER = mc = (1.67 × 10 kg ) ( 3.00 × 10 m/s ) ⎜ −13 ⎝ 1.60 × 10 J ⎟⎠ !

Topic 26

1525

(b)

E = γ mc 2 = γ ER = = !

939 MeV 1 − (0.950)

ER 1 − (v / c)

= 3.01 × 103 MeV = 3.01 GeV

(a) The total energy is 400 times the rest energy, or E = γER = 400 ER, so it is necessary that γ = 400. But γ = 1 !

1 − ( v / c ) , and solving for the 2

speed gives

v = c 1−

= c 1−

1 (400)2

= 0.999 997c

(b) The kinetic energy is KE = E − ER = 400ER − ER = 399ER. For a proton, ER = 938.3 MeV. Thus, KE = 399(938.3 MeV) = 3.74 × 105 MeV 26.30

(a) The rest energy of a proton is ER = 938.3 MeV. If the kinetic energy is KE = 175 GeV >> ER, we have KE = (γ − 1) >> ER, or γ − 1 >> 1, and γ − 1 >> γ. Thus, we may write the approximate relation KE = (γ − 1)ER ≈

γEE = γmpc2 .

Topic 26

1526

(b) The gamma factor is γ = 1 !

1 − ( v / c ) , so solving for the speed gives 2

2 . Therefore, when KE >> ER and we may use the !v = c 1 − γ

approximation KE ≈ γER, we have γ ≈ER/KE. Then, for the proton having KE = 175 GeV, 2

⎛E ⎞ ⎛ 938.3 MeV ⎞ v = c 1−γ 2 ≈ c 1− ⎜ R ⎟ = c 1− ⎜ ⎟ = 0.999 99c 3 ⎝ 175 × 10 MeV ⎠ ⎝ KE ⎠

26.31

1 2 The nonrelativistic expression for kinetic energy is KE = mv , while the 2 !

relativistic expression is KE = E − ER = (γ − 1)ER = (γ − 1)mc2, where

γ =1 !

1 − ( v / c ) . Thus, when the relativistic kinetic energy is twice the 2

predicted nonrelativistic value, we have ⎛ ⎞ 1 ⎛1 ⎞ − 1⎟ mc 2 = 2 ⎜ mv 2 ⎟ ⎜ 2 ⎝2 ⎠ ⎜⎝ 1 − ( v / c ) ⎟⎠ !

2 2 1 = ⎡⎣1 + ( v / c ) ⎤⎦ 1 − ( v / c )

Squaring both sides of the last result and simplifying gives (v/c)2[(v/c)4 + (v/c)2 − 1] = 0 Ignoring the trivial solution v/c = 0, we must have (v/c)4 + (v/c)2 − 1 = 0. This is a quadratic equation of the form x2 + x − 1 = 0, with x = (v/c)2.

(

)

Applying the quadratic formula gives x = −1 ± 5 2 . Since x = (v/c)2, we ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 26

1527

ignore the negative solution and find

−1 + 5 ⎛ v⎞ x=⎜ ⎟ = = 0.618 2 ! ⎝ c⎠ 2

which yields v = c 0.618 = 0.786 c . ! 26.32

The energy input to the electron will be W = Ef − Ei = (γ f − γ i)ER, or ⎛ ⎞ 1 1 ⎟E W =⎜ − 2 2 ⎟ R ⎜ 1 − ( vi c ) ⎟⎠ ⎜⎝ 1 − v f c !

(

where

)

ER = 0.511 MeV

(a) If vf = 0.900c and vi = 0.500c, then ⎛ ⎞ 1 1 W =⎜ − ⎟ ( 0.511 MeV ) = 0.582 MeV 2 ⎜⎝ 1 − ( 0.900 )2 1 − ( 0.500 ) ⎟⎠ !

(b) When vf = 0.990c and vi = 0.900c, we have ⎛ ⎞ 1 1 W =⎜ − ⎟ ( 0.511 MeV ) = 2.45 MeV 2 ⎜⎝ 1 − ( 0.990 )2 1 − ( 0.900 ) ⎟⎠ !

26.33

(a) The mass of a helium nucleus is mHe = 6.64424 × 10−27 kg so that

(

)

Δm = 4mp − mHe = 4 1.672 62 × 10−27 kg − 6.644 24 × 10−27 kg = 4.62 × 10−29 kg

(b) The energy released in MeV is 2⎛ 1 MeV ⎞ E = Δmc 2 = 4.62 × 10−29 kg 3 × 108 m/s ⎜ = 26.0 MeV ⎝ 1.60 × 10−13 J ⎟⎠

(

)(

)

Topic 26

26.34

1528

Let m1 be the mass of the fragment moving at v1 = 0.987c, and m2 be the mass having a speed of v2 = 0.868c. From conservation of mass-energy, E1 + E2 = ER,i, or 2 2 2 !γ 1m1 c + γ 2m2 c = mi c , giving

m1 ! 1 − (0.987)

m2 1 − (0.868)2

= mi

and

6.22m1 + 2.01m2 = 3.34 × 10−27 kg

[1]

Since the original particle was at rest, the momenta of the two fragments after decay must add to zero. Thus, the magnitudes must be equal, giving P2 = P1, or γ 2m2v2 = γ 1m1v1, and yielding 2.01m2(0.868c) = 6.22m1(0.987c)

m2 = 3.52m1

Substituting Equation [2] into [1] gives (6.22 + 7.08)m1 = 3.34 × 10−27 kg, or m1 = 2.51 × 10−28 kg Equation [2] then yields m2 = 3.52(2.51 × 10−28 kg) = 8.84 × 10−28 kg. 26.35

The relativistic total energy is E = γ ER = γ mc2, and the momentum is p =

γmv, where γ = 1 !

1 − v 2 c 2 . Thus, E2/c2 = γ 2m2c2, and p2 = γ 2m2v2, so

subtracting yields

[2]

Topic 26

1529

⎛ ⎞ E2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 − p = γ m c − v = γ m c 1 − v c = ⎜ ⎟ m c 1− v c = m c ( ) ( ) ⎜⎝ 1 − v 2 c 2 ⎟⎠ c2 !

(

)

Rearranging, this becomes

E2 = p 2 + m2 c 2 2 c ! 26.36

E2 = p2c2 + m2c4

(a) Since the total relativistic energy of a particle is E = KE + ER = KE + mc2, requiring that total energy be conserved (i.e., total energy before reaction = total energy after reaction) gives KEi + mi c2 = KEf + mf c2 , where mi is the total mass of the particles present before the reaction, KEi is the total kinetic energy before the reaction, mf is the total mass of the particles present after the reaction, and KEf is the final total kinetic energy. (b) Using atomic masses from the tables in Appendix B of the textbook, the total mass of the initial particles is found to be

m = m 235 U + m 1 n = 235.043 923 u + 1.008 665 u = 236.052588 u 92 0 ! i (c) The total mass of the particles present after the reaction is m f = m148 La + m87 Br + m 1 n 57

= 147.932 236 u + 86.920 7119 u + 1.008 665 u = 235.861 612 u

Topic 26

1530

(d) The mass that must have been converted to energy in the reaction is Δm = mi − mf = (236.052 588 − 235.861 612)u = 0.190 976 u (e) Assuming that KEi = 0, the total kinetic energy of the product particles is then

⎛ 931.494 MeV / c 2 ⎞ KE f = Δmc 2 + KEi = (0.190 976 u)c 2 ⎜ ⎟ + 0 = 177.893 MeV 1u ⎝ ⎠

26.37

E 20.0 GeV ⎛ 103 MeV ⎞ γ = = = 3.91 × 10 4 (a) ⎜ ⎟ ER 0.511 MeV ⎝ 1 GeV ⎠ ! Lp 3.00 × 103 m = = 7.67 × 10−2 m = 7.67 cm (b) L = 4 γ 3.91 × 10 !

26.38

(a) For an electron moving at vc = 0.750c, the gamma factor is

γ =1 ! e

1 − ( ve c ) = 1 2

1 − (0.750)2 = 1.51

The kinetic energy of a particle is KE = E − ER = (γ − 1)ER, so if the kinetic energy of a proton (ER,p = 938.3 MeV) equals that of the electron (ER,e = 0.511 MeV), we must have (γp − 1)ER,p = (γe − 1)ER,e or

⎛ 0.511 MeV ⎞ E γ p = 1 + (γ e − 1) R,e = 1 + (1.51 − 1) ⎜ = 1 + 2.78 × 10−4 ⎟ ER,p 938.3 GeV ⎝ ⎠ ! But γ p = 1 !

(

)

(

)

1 − vp c ,!so! vp c = 1 − 1 γ p2 and the speed of the

Topic 26

1531

proton must be

vp = c 1 − !

1 1 3 = c 1− 2 = 0.023 6c = 7.08 × 10 km/s 2 −4 γp (1 + 2.78 × 10 )

(b) As above, γe = 1.51 for an electron having speed ve = 0.750c. The momentum of a particle is

p = γ mv = !

γ mc 2 ( v / c ) γ ER ( v / c ) = c c

pc = γER(v/c)

If the momentum of a proton equals that of the electron, then ppc = pec, or

γpER,p(vp/c) = γeER,e(ve/c) and

( v c) = γ E ⎛ v ⎞ = (1.51) ⎛ 0.511 MeV ⎞ (0.750) = 6.17 × 10 ⎜ ⎟ ⎜⎝ 938.3 MeV ⎟⎠ E ⎝ c⎠ 1 − (v c) ! p

R,e

R,p

( v c) = 6.17 × 10 !

Thus,

−4

(

1 − vp c

) , and 2

(vp/c)2 = (6.17 × 10−4)2 − (6.17 × 10−4)2 − (6.17 × 10−4)2(vp/c)2, which yields 6.17 × 10 ( v c) = 1 +( 6.17 × 10 ) = 3.81 × 10 ( ) ! 2

−4 2

−7

−4 2

and the speed of the proton must be

Topic 26

1532

vp = c 3.81 × 10−7 = ( 3.00 × 108 m/s ) ( 6.17 × 10−4 ) = 1.85 × 105 m/s = 185 km/s !

26.39

KE = E − ER = (γ − 1)ER

γ = 1+ !

KE 1 = 2 ER 1 − (v / c)

giving

v = c 1−

(1 + KE / ER )2

(a) The speed of an electron having KE = 2.00 Mev will be

ve = c 1 −

(1 + KE / E )

= c 1−

R ,e

(1 + 2.00 / 0.511)

= 0.979c

(b) For a proton with KE = 2.00 MeV, the speed is

vp = c 1 −

(1 + KE E )

= c 1−

R,p

1 = 0.065 2c (1 + 2.00 / 938)2

(a) Yes. As the spring is compressed, positive work is done on it or energy is added to it. Since mass and energy are equivalent, mass has been added to the spring. 1 ΔE ΔPE kx 2 kx 2 (b) Δm = 2 R = 2 s = 2 2 = c c c 2c 2 !

( 2.0 × 10 N/m )( 0.15 m ) = 2.5 × 10 (c) Δm = 2 ( 3.00 × 10 m/s ) ! 2

−17

Topic 26

26.41

1533

An observer who moves at speed v relative to an object (or span of space) having proper length Lp sees a contracted length given by L = Lp γ = Lp 1 − ( v / c ) . Thus, if the proper distance to the star is Lp = ! 2

5.00 ly, and this length is to have a contracted value of L = 2.00 ly in the reference frame of the spacecraft, the speed of the spacecraft relative to the star must be 2

⎛ L⎞ ⎛ 2.00 ly ⎞ v = c 1− ⎜ ⎟ = c 1− ⎜ = 0.917c ⎝ 5.00 ly ⎟⎠ ⎝ Lp ⎠ !

26.42

2 2 2 From !E = (pc) + ER , with E = 5ER, we find that !p = ER 24 c .

(a) For an electron, p = ! (b) For a proton, p = ! 26.43

(0.511 MeV) 24 = 2.50 MeV/c . c

(938.3MeV) 24 MeV = 4.60 × 103 = 4.60 GeV/c . c c

(a) Observers on Earth measure the distance to Andromeda to be d = 2.00 × 106 ly = (2.00 × 106 yr)c. The time for the trip, in Earth’s frame of

( )

reference, is Δt = γ Δtp = 30.0 yr !

1 − ( v / c ) . The required speed is 2

then

2.00 × 106 yr ) c ( d v= = Δt 30.0 yr 1 − ( v / c )2 !

Topic 26

1534

which gives (1.50 × 10−5 ) ( v / c ) = 1 − ( v / c ) . Squaring both sides of ! 2

this equation and solving for v/c yields v / c = 1 ! Then, the approximation !1

1 + 2.25 × 10−10 .

1 + x = 1 − x / 2 gives

v 2.25 × 10−10 ≈ 1− = 1 − 1.13 × 10−10 c 2 ! (b) KE = (γ − 1)mc2, and

γ = !

1 1 − (v / c)

1 1 − (1 − 1.13 × 10

)

−10 2

1 2.26 × 10−10

Thus,

⎛ ⎞ 2 1 KE = ⎜ − 1 1.00 × 106 kg ) ( 3.00 × 108 m/s ) = 5.99 × 1027 J ( ⎟ −10 ⎝ 2.26 × 10 ⎠ ! (c) cost = KE × rate

⎡ ⎛ 1 kWh ⎞ ⎤ 27 $0.13/kWh ) = $2.16 × 1020 ⎢( 5.99 × 10 J ) ⎜ 6 ⎟ ⎥( ⎝ 3.60 × 10 J ⎠ ⎦ !⎣ 26.44

The clock, at rest in the ship’s frame of reference, will measure a proper time of Δtp = 10 h before sounding. Observers on Earth move at v = 0.75c relative to the clock and measure an elapsed time of

( )

Δt = γ Δtp = !

Δtp 1 − (v / c)

10 h 1 − ( 0.75 )

= 15 h

Topic 26

1535

The observers on Earth see the clock moving away at 0.75c and compute the distance traveled before the alarm sounds as

⎛ 3 600 s ⎞ 13 d = v(Δt) = ⎡ 0.75 3.0 × 108 m/s ⎤ 15 h ⎜ ⎟ = 1.2 × 10 m ⎣ ⎦ ⎝ 1h ⎠

(

26.45

)( )

(a) Since Dina and Owen are at rest in the same frame of reference (S'), they both see the ball traveling in the negative x' direction with speed !v′ball = 0.800c . Note that the velocity of the ball relative to Dina is vBD = −0.800c. (b) The distance between Dina and Owen, measured in their own rest frame, is Lp = 1.80 × 1013 m. Therefore, the time required for the ball to reach Dina, measured on her own clock, is

1.80 × 1012 m 2.25 × 1012 m Δtp = = = = 7.50 × 103 s 8 v′ball 0.800c 3.00 × 10 m/s ! Lp

(c) Ed sees a contracted length for the distance separating Dina and Owen. According to him, they are separated by a distance L = Lp γ = Lp 1 − ( v / c ) , where v = 0.600c is the speed of the S' frame ! 2

relative to Ed’s reference frame, S. Thus, according to Ed, the ball must travel a distance

Topic 26

1536

L = (1.80 × 1012 m ) 1 − (0.600)2 = 1.44 × 1012 m !

(d) The ball has a velocity of vBD = −0.800c relative to Dina, and Dina moves at velocity vDE = +0.600c relative to Ed. The relativistic velocity addition relation (Equation 26.8 from the textbook) then gives the velocity of the ball relative to Ed as vBE = !

vBD + vDE −0.800c + 0.600c −0.200c = = = −0.385c vBDvDE 1 + (−0.800)(0.600) 0.520 1+ c2

Thus, the speed of the ball according to Ed is vball = |vBE| = 0.385c. 26.46 (a)

As seen by an observer at rest relative to the mirror in frame S, the light must travel distance d before it strikes the mirror and then a distance d − d1 back to the ship after reflection. Here, distance d1 = vt is the distance the ship moves toward the mirror in the time t between when the pulse was emitted from the ship and when the reflected pulse was received by the ship. Since all observers agree that light travels at speed c, the total travel time for the light is

t= !

d + ( d − d1 ) 2d − d1 2d ⎛ v ⎞ = = −⎜ ⎟t c c c ⎝ c⎠

v⎞ 2d ⎛ ⎜⎝ 1 + ⎟⎠ t = c c

Solving for the travel time t of the light gives t = 2d/(c + v).

Topic 26

1537

(b) From the viewpoint of observers in the spacecraft, the spacecraft is at rest, and the mirror moves toward it at speed v. At the time the pulse starts toward the mirror, these observers see a contracted initial distance d′ = d 1 − ( v / c ) to the mirror. If the total time of travel for ! 2

the light is t', during the time t'/2 while the light is traveling toward the mirror, the mirror moves a distance Δx = vt'/2. closer to the ship. Thus, the light must travel a distance d' − Δx before reflection and the same distance d' − Δx back to the stationary ship after reflection. The total distance traveled by the light is then D = 2(d' − Δx), and the time required (traveling at speed c) is

t′ =

D c

( ) = 2d ′ − ⎛ v ⎞ t ′

2d ′ − 2v t ′ / 2 c

⎜ ⎟ ⎝ c⎠

( )

⎛ v⎞ 2d ′ 2d 1 − v / c 1 + t = = ′ ⎜ ⎟ c⎠ c c ⎝

Solving for the travel time t' measured by observers in the spacecraft then gives

Topic 26

1538

c3 − v 2

(c + v)t ′ = 2d

and

26.47

t′ =

c2 2d (c + v)(c − v) c

(c + v)

2d c − v c

c+v

The length of the space ship, as measured by observers on Earth, is L = Lp 1 − ( v / c ) . In Earth’s frame of reference, the time required for the ! 2

ship to pass overhead is L Lp 1 − ( v / c ) 1 1 Δt = = = Lp 2 − 2 v v v c ! 2

Thus, 2

2 ⎛ 0.75 × 10−6 s ⎞ 1 1 ⎛ Δt ⎞ 1 −17 s = = 2 +⎜ ⎟ = 1.74 × 10 m 2 v 2 c 2 ⎜⎝ Lp ⎟⎠ 3.00 × 108 m/s ) ⎝ 300 m ⎠ ( !

1 1.74 × 10−17

! 26.48

s2 m2

⎞ m⎞⎛ c ⎛ = ⎜ 2.4 × 108 ⎟ ⎜ = 0.80c ⎝ s ⎠ ⎝ 3.00 × 108 m/s ⎟⎠

From KE = (γ − 1)ER, we find

KE 1013 MeV γ = 1+ = 1+ = 1.07 × 1010 ER 938 MeV ! or γ ∼ 1010. Thus, the speed of the proton (and hence, the speed of the galaxy as seen by the proton) is v = c 1 − 1 γ 2 ~ c 1 − 10−20 ≈ c . ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 26

1539

(a) The diameter of the galaxy, as seen in the proton’s frame of reference, is

L = Lp ! or

105 ly 1 − (v / c) = ~ = 10−5 ly 10 γ 10 2

⎛ 9.461 × 1015 m ⎞ ⎛ 1 km ⎞ L = 10−5 ly ⎜ = 9.461 × 107 km 3 ⎟ ⎜ ⎟ 1 ly 10 m ⎠ ⎝ ⎠⎝ !

~ 108 km

(b) The proton sees the galaxy rushing by at v ≈ c. The time, in the proton’s frame of reference, for the galaxy to pass is

L 108 km 1011 m Δt = ~ = = 333 s 8 v c 3.00 × 10 m/s ! 26.49

~ 102 s

The difference between the relativistic momentum, p = γ mv, and the classical momentum, mv is Δp = γ mv − mv = (γ − 1)mv. (a) The error is 1.00% when Δp/p = 0.010 0, or (γ − 1)mv = 0.010 0γ mv. This gives γ = 1/0.990, or 1 − (v/c)2 = (0.990)2, and yields v = 0.141c. (b) When the error is 10.0%, we have γ = 1/0.900, and 1 − (v/c)2 = (0.900)2. In this case, the speed of the particle is v = c 1 − (0.900)2 = 0.436c . !

26.50

The kinetic energy gained by the electron will equal the loss of potential energy, so KE = q(ΔV) = e(1.02 MV) = 1.02 MeV

Topic 26

1540

(a) If Newtonian mechanics remained valid, then !KE = 12 mv 2 , and the speed attained would be

2(1.02 MeV) (1.60 × 10−13 J/MeV ) 2KE = = 5.99 × 108 m/s ≈ 2c −31 m 9.11 × 10 kg

v= !

KE 1.02 MeV = 1+ = 3.00 (b) KE = (γ − 1)ER, so γ = 1 + ER 0.511 MeV !

The actual speed attained is

v = c 1 − 1 γ 2 = c 1 − 1 (3.00)2 = 0.943c ! 26.51

(a) When at rest, muons have a mean lifetime of Δtp = 2.2 µs. In a frame of reference where they move at v = 0.95c, the dilated mean lifetime of the muons will be

( )

τ = γ Δtp = !

Δtp 1 − (v / c)

= 2

2.2 µs 1 − (0.95)2

= 7.0 µs

(b) In a frame of reference where the muons travel at v = 0.95c, the time required to travel 3.0 km is

d 3.0 × 103 m t= = = 1.05 × 10−5 s = 10.5 µs 8 v 0.95 ( 3.00 × 10 m/s ) ! If N0 = 5.0 × 104 muons started the 3.0 km trip, the number remaining at the end is © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 26

1541

N = N0e−t/τ = (5.0 × 104)e−10.5 µs/7.0 µs = 1.1 × 104 26.52

The work required equals the increase in the gravitational potential energy, or W = GMSunm/Rg. If this is to equal the rest energy of the mass removed, then

mc 2 =

Rg =

GMSun m

26.53

( 6.67 × 10

−11

Rg =

GMSun c2

N ⋅ m 2 kg 2 ) (1.99 × 1030 kg )

( 3.00 × 10 m/s ) 8

= 1.47 × 103 m = 1.47 km

According to Earth-based observers, the times required for the two trips are:

For Speedo:

Lp 20.0 ly 20.0 yr Δts = = = = 21.1 yr v 0.950 c 0.950 s !

Lp 20.0 ly 20.0 yr = = = 26.7 yr For Goslo: ΔtG = v 0.750 c 0.750 G ! Thus, after Speedo lands, he must wait and age at the same rate as planetbased observers, for an additional ΔTage,S = ΔtG − ΔtS = (26.7 − 21.1) yr = 5.6 yr before Goslo arrives. The time required for the trip according to Speedo’s internal biological clock (which measures the proper time for his aging process during the

Topic 26

1542

trip) is Δt 2 Tage ,S = Δtp,S = S = ΔtS 1 − ( vs c ) = (21.1 yr) 1 − (0.950)2 = 6.59 yr γS !

When Goslo arrives, Speedo has aged a total of ΔTS = Tage,S + ΔTage,S = 6.59 yr + 5.6 = 12.2 yr The time required for the trip according to Goslo’s internal biological clock (and hence the amount he ages) is Δt 2 TG = Δtp,G = G = ΔtG 1 − ( vG c ) = (26.7 yr) 1 − (0.750)2 = 17.7 yr γG !

Thus, we see that when he arrives, Goslo is older than Speedo, having aged an additional TG − TS = 17.7 yr − 12.2 yr = 5.5 yr 26.54

(a) The proper lifetime is measured in the ship’s reference frame, and Earth-based observers measure a dilated lifetime of

Δt = γΔtp = !

Δtp 1 − (v / c)

= 2

15.0 yr 1 − (0.700)2

= 21.0 yr

(b) d = v(Δt) = [0.700 c](21.0 yr) = [(0.700)(1.00 ly/yr)](21.0 yr) = 14.7 ly (c) Looking out the rear window, the astronauts see Earth recede at a rate of v = 0.700 c. The distance it has receded, as measured by the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 26

1543

astronauts, when the batteries fail is d' = v(Δtp) = (0.700c)(15.0 yr) = [(0.700)(1.00 ly/yr)](15.0 yr) = 10.5 ly (d) Mission control gets signals for 21.0 yr while the battery is operating and then for 14.7 yr after the battery stops powering the transmitter, 14.7 ly away. The total time that signals are received is 21.0 yr + 14.7 yr = 35.7 yr. 26.55

Note: Excess digits are retained in some steps given below to more clearly illustrate the method of solution. We are given that L = 2.00 m and θ = 30.0° (both measured in the observer’s rest frame). The components of the rod’s length as measured in the observer’s rest frame are Lx = L cos θ = (2.00 m)cos 30.0° = 1.732 m and Ly = L sin θ = (2.00 m)sin 30.0° = 1.00 m

Topic 26

1544

The component of length parallel to the motion has been contracted, but the component perpendicular to the motion is unaltered. Thus, Lpy = Ly = 1.00 m and

Lpx = !

Lx 1 − (v / c)

1.732 m 1 − (0.995)2

= 17.34 m

(a) The proper length of the rod is then Lp = L2px + L2py = (17.34 m)2 + (1.00 m)2 = 17.4 m !

(b) The orientation angle in the rod’s rest frame is

⎛ Lpy ⎞ ⎛ 1.00 m ⎞ θ p = tan −1 ⎜ = tan −1 ⎜ = 3.30° ⎟ Lpx ⎠ ⎝ 17.34 m ⎟⎠ ⎝ ! 26.56

(a) Taking toward Earth as the positive direction, the velocity of the ship relative to Earth is vSE = +0.600c, and the velocity of the lander relative to the ship is vLS = +0.800c. The relativistic velocity addition relation (Equation 26.8 in the textbook) then gives the velocity of the lander relative to Earth as vLE = !

vLS + vSE 0.800c + 0.600c 1.40c = = = +0.946c vLS vSE 1 + (0.800)(0.600) 1.48 1+ c2

(b) Observers at rest on Earth measure the proper length between the

Topic 26

1545

ship and Earth as Lp = 0.200 ly. The contracted distance measured by observers at rest relative to the spaceship is

L = Lp 1 − ( vSE c ) = (0.200 ly) 1 − (0.600)2 = 0.160 ly ! 2

(c) Observers on the ship see the lander start toward Earth from an initial distance of L = 0.160 ly. They see this distance diminish as the lander nibbles into it from one end at vLS = 0.800c, and the Earth (as it appears to approach them) reducing it from the other end at vSE = 0.600c. The time they compute it will take the lander and Earth to meet is 0.160 ly 0.160 yr L t= = = = 0.114 yr v + v 1.40c 1.40 LS SE !

(d) The kinetic energy of the lander as observed in the Earth reference frame is KE = E − ER = (γ − 1)mc2, where γ = 1 !

1 − ( vLE c ) . This gives 2

⎛ ⎞ 2 1 5 8 KE = ⎜ − 1 4.00 × 10 kg 3.00 × 10 m/s = 7.51 × 1022 J ( ) ( ) ⎟ 2 ⎝ 1 − (0.946) ⎠ !

Topic 27

1546

Topic 27 Quantum Physics

QUICK QUIZZES 27.1

True. When a photon scatters off an electron that was initially at rest, the electron must recoil to conserve momentum. The recoiling electron possesses kinetic energy that was gained from the photon in the scattering process. Thus, the scattered photon must have less energy than the incident photon.

27.2

Choice (b). Some energy is transferred to the electron in the scattering process. Therefore, the scattered photon must have less energy (and hence, lower frequency) than the incident photon.

27.3

Choice (c). Conservation of energy requires the kinetic energy given to the electron be equal to the difference between the energy of the incident photon and that of the scattered photon.

27.4

False. The de Broglie wavelength of a particle of mass m and speed v is λ = h/p = h/mv. Thus, the wavelength decreases when the momentum increases.

Topic 27

27.5

1547

Choice (c). Two particles with the same de Broglie wavelength will have the same momentum p = mv = h/λ. If the electron and proton have the same momentum, they cannot have the same speed because of the difference in their masses. For the same reason, remembering that KE = p2/2m, they cannot have the same kinetic energy. Because the kinetic energy is the only type of energy an isolated particle can have, and we have argued that the particles have different energies, the equation f = E/h tells us that the particles do not have the same frequency.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 27.2

A microscope can see details no smaller than the wavelength of the waves it uses to produce images. Electrons with kinetic energies of several electron volts have wavelengths of less than a nanometer, which is much smaller than the wavelength of visible light (having wavelengths ranging from about 400 to 700 nm). Therefore, an electron microscope can resolve details of much smaller size as compared to an optical microscope.

27.4

Measuring the position of a particle implies having photons reflect from it. However, collisions between photons and the particle will alter the

Topic 27

1548

velocity of the particle. Thus, the more accurately you try to measure one quantity, the more uncertainty you create in your knowledge of the other. 27.6

Light has both wave and particle characteristics. In Young’s double-slit experiment, light behaves as a wave. In the photoelectric effect, it behaves like a particle. Light can be characterized as an electromagnetic wave with a particular wavelength or frequency, yet at the same time, light can be characterized as a stream of photons, each carrying a discrete energy, hf.

27.8

Ultraviolet light has a shorter wavelength and higher photon energy, Ephoton = hf = hc/λ than visible light.

27.10

Increasing the temperature of the substance increases the average kinetic energy of the electrons inside the material. This makes it slightly easier for an electron to escape from the material when it absorbs a photon.

27.12

Most stars radiate nearly as blackbodies. Of the two stars, Vega has the higher surface temperature and, in agreement with Wien’s displacement law, radiates more intensely at shorter wavelengths and has a bluish appearance, while Arcturus has a more reddish appearance.

27.14

No, the crystal cannot produce diffracted beams of visible light. The angles where one could expect to observe diffraction maxima are given

Topic 27

1549

by Bragg’s law, 2d sin θ = mλ, where m is a nonzero integer. Because sin θ ≤ 1, this equation gives m = (2d/λ) sin θ ≤ 2d/λ. Since the wavelengths of visible light (λ ∼ 102 nm) are much larger than the distances between atomic planes in crystals (d ∼ 1 nm), there are no nonzero integer values for m that can satisfy Bragg’s law for visible light. 27.16

Any object of macroscopic size—including a grain of dust—has an undetectably small wavelength, so any diffraction effects it might exhibit are very small, effectively undetectable. Recall historically how the diffraction of sound waves was at one time well known, but the diffraction of light was not.

ANSWERS TO EVEN NUMBERED PROBLEMS 27.2

(a)

∼100 nm, ultraviolet

(b)

∼10−1 nm, gamma

(b)

4.23 × 103 K

rays 27.4

(a)

5.78 × 103 K

27.6

(a)

2.90 × 10−19 J/photon

(b)

(d)

501 nm

5.69 × 1044 photon/s

5.7 × 103 photon/s

Topic 27

1550

27.10

(a)

288 nm

(b)

1.04 × 1015 Hz

27.12

(a)

only lithium

(b)

0.81 eV

27.14

(a)

1.90 eV

(b)

0.216 V

27.16

(a)

8.29 × 10−11 m

(b)

1.24 × 10−11 m

27.18

1.07 kV

27.20

0.455 nm

27.22

m = 2, θ = 25.8°; m = 3, θ = 40.8°; m = 4, θ = 60.8°

(c)

1.19 eV

(c)

λmin decreases

Higher orders cannot be found since sinθ cannot exceed a value of 1.00. 27.24

1.03 × 10−3 nm

27.26

(a)

5.6 pm

(b)

29.9 pm

27.28

(a)

0.101 nm

(b)

80.9°

27.30

(a)

1.99 × 10−11 m

(b)

1.98 × 10−14 m

27.32

(a)

5.79 × 10−10 m

(b)

2.26 × 10−25 m

27.34

(a)

p = 2mqΔV

(b)

λ =h

(c)

The proton, with the larger mass, will have the shorter wavelength.

27.36

(c)

2.76 × 10−7 m

2mqΔV

23 m/s

Topic 27

1551

27.38

(a)

23.2 m/s

(b)

1.26 × 10−2 m/s

27.40

(a)

See Solution.

(b)

5.3 MeV

27.42

v=c 2 2

27.44

(a)

2.49 × 10−11 m

(b)

0.29 nm

27.46

(a)

λ = 2.82 × 10−37 m

(b)

ΔE ≥ 1.06 × 10−32 J

(c)

2.88 × 10−35 %

27.48

0.14 keV

27.50

0.785 eV

27.52

(a) 148 days (b) This result is totally contrary to observations of the photoelectric effect

PROBLEM SOLUTIONS 27.1

From Wien’s displacement law,

(a) T =

0.289 8 × 10−2 m ⋅ K

λmax

0.289 8 × 10−2 m ⋅ K = 2.99 × 103 K, or ≈ 3 000 K −9 970 × 10 m

Topic 27

1552

(b)

0.289 8 × 10−2 m ⋅ K

λmax

0.289 8 × 10−2 m ⋅ K 145 × 10−9 m

= 2.00 × 10 4 K, or ≈ 20 000 K 27.2

Using Wien’s displacement law,

0.289 8 × 10−2 m ⋅ K = 2.898 × 10−7 m ~ 100 nm ultraviolet (a) λmax = 4 10 K 0.289 8 × 10−2 m ⋅ K = 2.898 × 10−10 m ~ 10−1 nm γ − rays (b) λmax = 7 10 K 27.3

From Wien’s displacement law,

λmax =

0.289 8 × 10 −2 m ⋅ K 0.289 8 × 10 −2 m ⋅ K = = 9.47 × 10 −6 m = 9.47 µ m(infrared) T 306 K

27.4

(a) The power radiated by an object with surface area A and absolute temperature T is given by Stefan’s law (see Chapter 11 of the textbook) as P = σAeT4, where σ is the Stefan-Boltzmann constant equal to 5.669 6 × 10−8 W/m2 ⋅ K4. For an object that approximates a blackbody, the emissivity is e = 1. Thus, the temperature of our Sun’s surface should be

Topic 27

1553

⎤4 P ⎡ P ⎤4 ⎡ T=⎢ = ⎢ ⎥ 2 ⎣ σ Ae ⎥⎦ ⎢⎣ σ ( 4π RSun ) e ⎥⎦ 1

⎡ ⎤4 3.85 × 1026 W ⎥ = 5.78 × 103 K =⎢ 2 −8 2 4 8 ⎢⎣ 4π ( 5.669 6 × 10 W/m ⋅ K ) ( 6.96 × 10 m ) (1) ⎥⎦ (b) Wien’s displacement law gives the peak wavelength of the radiation from the Sun as

0.289 8 × 10 −2 m ⋅ K 0.289 8 × 10 −2 m ⋅ K λmax = = = 5.01 × 10 −7 m = 501 nm 3 T 5.78 × 10 K 27.5

Wien’s displacement law gives the wavelength of peak emission on a blackbody curve. Solving for this wavelength gives

λmax =

27.6

0.2898 × 10−2 m ⋅ K 0.2898 × 10−2 m ⋅ K = = 1.01 × 10−5 m T 287 K

(a) EPeak =

(b) T =

λmax

(6.63 × 10 =

0.289 8 × 10 −2 m ⋅ K

λmax

−34

)(

J ⋅ s 3.00 × 10 8 m/s −9

685 × 10 m

) = 2.90 × 10

−19

J/photon

0.289 8 × 10 −2 m ⋅ K = 4.23 × 10 3 K 685 × 10 −9 m

(c) The surface area of the spherical star is A = 4πr2, and (considering it to be a blackbody) its emissivity is e = 1. Stefan’s law (see Chapter 11 of the textbook) gives the radiated power as P = σAeT4 = σ(4πr2)eT4 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 27

1554

= 4π(5.669 6 × 10−8 W/m2 ⋅ K4)(8.50 × 108 m)2 (1)(4.23 × 103 K)4 = 1.65 × 1026 W (d) Assuming the average energy of an emitted photon equals the energy of photons at the peak of the radiation distribution, the number of photons emitted by the star each second is

1.65 × 10 26 J/s ΔN P = = = 5.69 × 10 44 photon/s −19 Δt Epeak 2.90 × 10 J/photon 27.7

The energy of a photon having frequency f is given by Ephoton = hf, where Planck’s constant has a value of h = 6.63 × 10−34 J ⋅ s. This energy may be converted to units of electron volts by use of the conversion factor 1 eV = 1.60 × 10−19 J.

⎛ ⎞ 1 eV −34 12 = 2.57 eV (a) Ephoton = hf = 6.63 × 10 J ⋅ s 620 × 10 Hz ⎜ −19 ⎟ 1.60 × 10 J ⎝ ⎠

(

)(

)

(b)

⎛ ⎞ 1 eV Ephoton = hf = 6.63 × 10−34 J ⋅s 3.10 × 109 Hz ⎜ −19 ⎟ ⎝ 1.60 × 10 J ⎠

(

)(

)

= 1.28 × 10−5 eV

Topic 27

1555

(c)

⎛ ⎞ 1 eV Ephoton = hf = 6.63 × 10−34 J ⋅s 46.0 × 106 Hz ⎜ −19 ⎟ ⎝ 1.60 × 10 J ⎠

(

)(

)

= 1.91 × 10−7 eV 27.8

The energy entering the eye each second is P = I ⋅ A, where A is the area of the pupil. If the light has wavelength λ, the energy of a single photon is Ephoton = hc/λ. Hence, the number of photons entering the eye in time Δt is given by N = ΔE/Ephoton = P ⋅ (Δt)/Ephoton, or N = IA(Δt)/(hc/λ) = IA(Δt)λ/hc. With I = 4.0 × 10−11 W/m2 , λ = 500 nm, Δt = 1.0 s, and the opening of the pupil having an area of A = πd2/4, where d is the pupil diameter, we find

N = Δt 27.9

( 4.0 × 10

−11

) (

)

(

)

2 J/s ⋅ m 2 ⎡π 8.5 × 10 −3 m 4 ⎤ 500 × 10 −9 m ⎢⎣ ⎥⎦ = 5.7 × 10 3 photons/s −34 8 6.63 × 10 J ⋅ s 3.00 × 10 m/s

(

)(

)

(a) From the photoelectric effect equation, the work function is φ = hc/λ − KEmax, or

( 6.63 × 10 φ=

−34

)(

)

J ⋅ s 3.00 × 10 8 m/s ⎛ ⎞ 1 eV ⎜ ⎟ − 1.31 eV = 2.24 eV −9 −19 350 × 10 m ⎝ 1.60 × 10 J ⎠

Topic 27

1556

(b)

λc =

(6.63 × 10 J ⋅s)(3.00 × 10 m/s) ⎛⎜ = 8

−34

2.24 eV

⎞ −19 ⎟ ⎝ 1.60 × 10 J ⎠ 1 eV

= 5.55 × 10−7 m = 555 nm

(c)

27.10

3.00 × 108 m/s fc = = = 5.41 × 1014 Hz −9 λc 555 × 10 m c

(a) At the cutoff wavelength, KEmax = 0, so the photoelectric effect equation (KEmax = hc/λ − φ) gives the cutoff wavelength as

λc =

( 6.63 × 10 =

−34

)(

)

J ⋅ s 3.00 × 108 m/s ⎛ ⎞ 1 ev −7 ⎜ ⎟ = 2.88 × 10 m = 288 nm −19 4.31 eV 1.60 × 10 J ⎝ ⎠

(b) The lowest frequency of light that will free electrons from the material is fc = c/λc, where λc is the cutoff wavelength (calculated above). Thus,

3.00 × 108 m/s fc = = = 1.04 × 1015 Hz −7 λc 2.88 × 10 m c

(c) The photoelectric effect equation may be written as KEmax = Ephoton − φ. Therefore, if the photons incident on this surface have energy Ephoton = 5.50 eV, the maximum kinetic energy of the ejected electrons is KEmax = Ephoton − φ = 5.50 eV − 4.31 eV = 1.19 eV

Topic 27

27.11

1557

⎛ 1.60 × 10−19 J ⎞ −18 (a) φ = 6.35 eV ⎜ ⎟ = 1.02 × 10 J 1 eV ⎝ ⎠ (b) The energy of a photon having the cutoff frequency or cutoff wavelength equals the work function of the surface, or Ephoton = hfc = hc/λc = φ. Thus, the cutoff frequency of a surface having a work function of φ = 6.35 eV is

fc =

φ h

⎛ 1.60 × 10−19 J ⎞ 6.35 eV 15 ⎜ ⎟ = 1.53 × 10 Hz −34 1 eV 6.63 × 10 J ⋅ s ⎝ ⎠

λc =

c 3.00 × 108 m/s = = 1.96 × 10−7 m = 196 nm 15 fc 1.53 × 10 Hz

(d) KEmax = Ephoton − φ = 8.50 eV − 6.35 eV = 2.15 eV (e) eVs = KEmax, so the stopping potential is Vs = KEmax/e = 2.15 eV/e = 2.15 V 27.12

(a) The energy of the incident photons is

Ephoton =

(6.63 × 10 J ⋅s)(3.00 × 10 m/s) ⎛ = −34

400 × 10−9 m

⎞ 1 eV = 3.11 eV ⎜ −19 ⎟ ⎝ 1.60 × 10 J ⎠

For photoelectric emission to occur, it is necessary that Ephoton ≥ φ. Thus, of the three metals given, only lithium will exhibit the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 27

1558

photoelectric effect. (b) For lithium, KEmax = Ephoton − φ = 3.11 eV − 2.30 eV = 0.81 eV. 27.13

(a) By conservation of energy, the maximum kinetic energy equals eΔV, the highest potential energy achieved by the photoelectrons. The maximum kinetic energy is therefore 2.50 eV . (b) The maximum speed is found from the definition of kinetic energy:

(

)

2 KE = 12 me vmax = 2.50 eV 1.60 × 10−19 J/eV = 4.00 × 10−19 J

vmax =

(

2 4.00 × 10−19 J 9.11 × 10

−31

) = 9.37 × 10 m/s 5

φsilver = 4.73 eV. Substitute f = c/λ and solve for the wavelength λ to find hc = KEmax + φsilver λ

(

)( (

)

6.63 × 10−34 J ⋅s 3.00 × 108 m/s hc λ= = KEmax + φsilver ( 2.50 eV +4.73 eV ) 1.60 × 10−19 J/eV

)

λ = 1.72 × 10−7 m = 172 nm 27.14

(a) The maximum kinetic energy of the ejected electrons is related to the stopping potential by the expression KEmax = e(ΔVs). Thus, if the

Topic 27

1559

stopping potential is Vs = 0.376 V when the incident light has wavelength λ = 546.1 nm, the photoelectric effect equation gives the work function of this metal as

φ=

− KEmax =

(6.63 × 10 =

−34

− e ( ΔVs )

)(

)

J ⋅ s 3.00 × 10 8 m/s ⎛ ⎞ 1 eV ⎜ ⎟ − 0.376 eV = 1.90 eV −19 546.1 × 10 −9 m 1.60 × 10 J ⎝ ⎠

(b) If light of wavelength λ = 587.5 nm is incident on this metal, the maximum kinetic energy of the ejected electrons is

e ( ΔVs ) = KEmax =

(6.63 × 10 =

−φ

−34

)(

)

J ⋅ s 3.00 × 10 8 m/s ⎛ ⎞ 1 eV ⎜ ⎟ − 1.90 eV = 0.216 eV −9 −19 587.5 × 10 m 1.60 × 10 J ⎝ ⎠

Thus, the stopping potential will be

ΔVs = 27.15

KEmax 0.216 eV = = 0.216 V e e

Assuming the electron produces a single photon as it comes to rest, the energy of that photon is Ephoton = (KE)i = e(ΔV). The accelerating voltage is then

ΔV =

Ephoton e

(

)(

)

6.63 × 10 −34 J ⋅ s 3.00 × 108 m/s 1.24 × 10 −6 V ⋅ m hc = = = eλ λ 1.60 × 10 −19 C λ

(

)

Topic 27

1560

For λ = 1.0 × 10−8 m, V =

1.24 × 10−6 V ⋅ m = 1.2 × 102 V −8 1.0 × 10 m

and for λ = 1.0 × 10−13 m, V =

27.16

1.24 × 10−6 V ⋅ m = 1.2 × 107 V −13 1.0 × 10 m

A photon of maximum energy and minimum wavelength is produced when the electron gives up all its kinetic energy in a single collision, or

λmin = hc/KE = hc/e ΔV. (a) If ΔV = 15.0 kV,

( 6.63 × 10 J ⋅ s )( 3.00 × 10 m/s ) = 8.29 × 10 = (1.60 × 10 C )(15.0 × 10 V ) −34

λmin

−8

−19

−11

(b) If ΔV = 100 kV,

( 6.63 × 10 J ⋅s )( 3.00 × 10 m/s ) = 1.24 × 10 = (1.60 × 10 C)(100 × 10 V ) −34

λmin

−19

−11

A photon of maximum energy or minimum wavelength is produced when the electron gives up all of its kinetic energy in a single collision within the target. Thus, Emax = hc/λmin = KEe = e ΔV. If λmin = 70.0 pm = 70.0 × 10−12 m, the required accelerating voltage is

Topic 27

1561

ΔV =

27.18

hc eλmin

( 6.63 × 10 J ⋅ s )( 3.00 × 10 m/s ) = 1.78 × 10 V = 17.8 KV = (1.60 × 10 C )(70.0 × 10 m ) −34

−19

−12

By inspection, a minimum potential difference magnitude of ΔV = 1.07 kV will provide the energy (eΔV) necessary to produce a 1.07 keV

x-ray photon. 27.19

(a) Convert 75.0 keV to joules to find 75.0 keV = 1.20 × 10−14 J. First solve for the electron’s speed using the nonrelativistic definition of kinetic energy:

KE = me v 1 2

2 e

→ ve =

2 ( KE) = me

(

2 1.20 × 10−14 J 9.11 × 10

−31

) = 1.62 × 10 m/s 8

This result is more than half the speed of light with a relativistic factor γ = 1.19, indicating that the relativistic kinetic energy equation should be used. That equation gives

KE = mc 2 (γ − 1)

(

)(

−14 −31 8 KE + mc 2 1.20 × 10 J + 9.11 × 10 kg 3.00 × 10 m/s γ = = 2 mc 2 9.11 × 10−31 kg 3.00 × 108 m/s

(

γ =

1 1−

2 min 2

v c

)(

)

) = 1.15 2

= 1.15 → vmin = 1.48 × 108 m/s

(b) Find the wavelength from the energy of a photon:

Topic 27

1562

(

)(

)

6.63 × 10−34 J ⋅s 3.00 × 108 m/s hc hc E = hf = → λ= = = 1.66 × 10−11 m −14 λ E 1.20 × 10 J 27.20

From the Bragg equation, 2d sin θ = mλ, the interplanar spacing when 0.129-nm x-rays produce a first order diffraction maximum at θ = 8.15° is

27.21

Using Bragg’s law, the wavelength is found to be

λ= 27.22

(1)( 0.129 nm ) = 0.455 nm mλ = 2 sin θ 2 sin 8.15°

2d sin θ 2 ( 0.296 nm ) sin 7.6° = = 0.078 nm m 1

From the Bragg equation, 2d sin θ = mλ, the angle at which the diffraction maximum of order m will be found when the first order maximum is at θ1 = 12.6°, is given by sin θm = m(λ/2d) = m sin θ1 = m sin (12.6°) Thus, additional orders will be found at m = 2:

sin θ2 = 2 sin 12.6° = 0.436 and θ2 = sin−1 (0.436) = 25.8°

m = 3:

sin θ3 = 3 sin 12.6° = 0.654 and θ3 = sin−1 (0.654) = 40.8°

m = 4:

sin θ4 = 4 sin 12.6° = 0.873 and θ4 = sin−1 (0.873) = 60.8°

m = 5:

sin θ5 = 5 sin 12.6° = 1.09 Impossible, since sin θ ≤ 1 for all θ.

Topic 27

1563

No orders higher than m = 4 will be seen because it is mathematically impossible for the sin θ to be greater than one. 27.23

The interplanar spacing in the crystal is given by Bragg’s law as

(1)( 0.140 nm ) = 0.281 nm mλ = 2 sin θ 2 sin 14.4°

27.24

From the Compton shift equation, the wavelength shift of the scattered x−rays is 6.63 × 10 −34 J ⋅ s h Δλ = (1 − cos θ ) = (1 − cos 55.0 ) me c 9.11 × 10 −31 Kg 3.00 × 10 8 m/s

(

)(

)

⎛ 1 nm ⎞ = 1.03 × 10 −12 m ⎜ −9 ⎟ = 1.03 × 10 −3 nm ⎝ 10 m ⎠

27.25

If the scattered photon has energy equal to the kinetic energy of the recoiling electron, the energy of the incident photon is divided equally between them. Thus,

Ephoton =

(E )

0 photon

⇒

hc , so λ = 2λ0 and Δλ = 2λ0 − λ0 = λ0 = 0.001 60 nm 2λ0

The Compton scattering formula, Δλ = λC (1 − cos θ), then gives the scattering angle as

⎛ 1 − Δλ ⎞ 0.001 60 nm ⎞ −1 ⎛ ⎟ = cos ⎜ 1 − ⎟ = 70.0° λ 0.002 43 nm ⎝ ⎠ ⎝ c ⎠

θ = cos−1 ⎜

Topic 27

27.26

1564

(a) The Compton shift formula is λ = λ0 +

h (1 − cosθ ) . Use this mec

formula to find the first shifted wavelength, λ’:

λ ' = 25.0 pm +

(

6.63 × 10−34 J ⋅s 1 − cos ( 40.0° ) = 25.6 pm 9.11 × 10−31 kg 3.00 × 108 m/s

)(

)

(

)

(b) Apply the Compton shift formula again, with θ = 140° so that the final photon’s direction is directly opposite the first photon (40.0° + 140° = 180°):

λ ′′ = 25.6 pm +

(

6.63 × 10−34 J ⋅s 1 − cos (140° ) = 29.9 pm 9.11 × 10−31 kg 3.00 × 108 m/s

)(

)

(

)

(Alternatively, the wavelength λ’’ can be directly found by using an incoming photon of wavelength 25.0 pm and a scattered angle of 180°.) 27.27

The figure below shows the situation before and after the scattering process. Note that the scattering angle is θ = 180°, so the Compton equation gives

λ′ − λ =

h 2h 1 − cos 180° ) = ( me c mec

(

)

2 6.63 × 10 −34 J ⋅ s 2h −9 λ′ = λ + = 0.110 × 10 m + = 0.115 × 10 −9 m −31 8 me c 9.11 × 10 Kg 3.00 × 10 m/s

(

)(

)

Topic 27

1565

(a) The momentum of the incident photon is p = h/λ, while that of the scattered photon is p′ = − h/λ′ (the negative sign is included since momentum is a vector quantity and the scattered photon travels in the negative x − direction). Thus, conservation of momentum gives pe − h/λ′ = h/λ + 0, or the momentum of the recoiling electron is

pe =

⎛ ⎞ h 1 1 = 6.63 × 10 −34 J ⋅ s ⎜ + ⎟ −9 −9 λ′ ⎝ 0.110 × 10 m 0.115 × 10 m ⎠

(

)

= 1.18 × 10 −23 kg ⋅ m/s (b) Assuming the recoiling electron is nonrelativistic, its kinetic energy is given by

(

1.18 × 10 −23 Kg ⋅ m/s pe2 KEe = = 2me 2 9.11 × 10 −31 Kg

(

)

) = 7.64 × 10 2

−17

⎛ ⎞ 1 eV J⎜ ⎟ = 478 eV −19 ⎝ 1.60 × 10 J ⎠

Note that this energy is very small in comparison to the rest energy of an electron. Thus, our assumption that the recoiling electron

Topic 27

1566

would be non-relativistic is seen to be valid. 27.28

First, observe that v = 2.18 × 106 m/s << c. Thus, the recoiling electron is nonrelativistic, and its kinetic energy is KEe = 21 me v2 , while its momentum is given by pe = mev. (a) The Compton equation gives the wavelength of the scattered photon in terms of that of the incident photon as

λ = λ0 +

h (1 − cos θ ) me c

[1]

Conservation of energy gives the kinetic energy of the recoiling electron as 21 me v2 = hc /λ0 − hc /λ = hc(λ − λ0 )/λ0 λ . Substituting Equation [1] into this yields hc ⎡⎣( h me c )(1 − cos θ )⎤⎦ 1 me v 2 = 2 λ0 ⎡⎣λ0 + ( h me c )(1 − cos θ )⎤⎦ 2

⎛ h ⎞ λ0 ⎡⎣λ0 + ( h me c )(1 − cos θ )⎤⎦ = 2 ⎜ ⎟ (1 − cos θ ) ⎝ me v ⎠

Simplifying, this equation is seen to be of the form aλ02 + bλ0 + c = 0, where a = 1, and the constants b and c are as computed below:

(

) )(

6.63 × 10 −34 J ⋅ s (1 − cos 17.4° ) h b= = 1.11 × 10 −13 m (1 − cos θ ) = −31 8 me c 9.11 × 10 Kg 3.00 × 10 m/s

(

)

Topic 27

1567

(

)

−2 6.63 × 10 −34 J ⋅ s (1 − cos 17.4° ) ⎛ h ⎞ c = −2 ⎜ = −1.02 × 10 −20 m 2 ⎟ (1 − cos θ ) = 2 2 ⎝ me v ⎠ 9.11 × 10 −31 Kg 2.18 × 10 6 m/s 2

(

)(

)

2 From the quadratic formula, λ0 = ⎡⎢ −b ± b − 4ac ⎤⎥ 2a , using only the

⎣

⎦

upper sign since λ0 must be positive, we find λ0 = 1.01 × 10−10 m = 0.101 nm. (b) Choosing the x-axis to be the direction of the incident photon’s motion, the initial momentum in the y-direction is zero. Thus, conserving momentum in the y-direction gives pe sin φ = p sin θ

where θ = 17.4°, pe = mev, and p = h/λ. From Equation [1] and the result of part (a), the wavelength of the scattered photon is

(

) )(

6.63 × 10 −34 J ⋅ s (1 − cos 17.4° ) h −10 λ = λ0 + (1 − cos θ ) = 1.01× 10 m + me c 9.11 × 10 −31 Kg 3.00 × 10 8 m/s

(

)

or λ = 1.011 × 10−10 m. Thus, the scattering angle for the electron is φ =

Topic 27

1568

sin−1 (p sin θ/pe) = sin−1 (h sin θ/λmev), which gives

(

)

⎡ ⎤ 6.63 × 10−34 J ⋅ s sin 17.4° ⎥ = 80.9° φ = sin ⎢ ⎢ 1.011 × 10−10 m 9.11 × 10 −31 Kg 2.18 × 106 m/s ⎥ ⎣ ⎦ −1

27.29

(

)(

)

(a) From λ = h/p = h/mv, the speed is

6.63 × 10−34 J ⋅ s h v= = = 1.46 × 103 m/s = 1.46 Km/s −31 −7 me λ 9.11 × 10 Kg 5.00 × 10 m

(

(b) λ =

27.30

)(

)

6.63 × 10 −34 J ⋅ s h = = 7.28 × 10 −11 m −31 7 me v 9.11 × 10 Kg 1.00 × 10 m/s

(

)(

)

The de Broglie wavelength of a particle of mass m is λ = h/p, where the momentum is given by p = γ mv = mv

1 − (v / c)2 . Note that when the

particle is not relativistic, then γ ≈ 1, and this relativistic expression for momentum reverts back to the classical expression. (a) For a proton moving at speed v = 2.00 × 104 m/s, v << c and γ ≈ 1, so

λ=

6.63 × 10 −34 J ⋅ s h = = 1.99 × 10−11 m −27 4 mp v 1.67 × 10 Kg 2.00 × 10 m/s

(

)(

)

(b) For a proton moving at speed v = 2.00 × 107 m/s,

Topic 27

1569

λ=

2 h h = 1 − (v / c) λ mp v mp v

⎛ 2.00 × 10 m/s ⎞ (6.63 × 10 J ⋅ s ) = 1−⎜ ⎟ = 1.98 × 10 (1.67 × 10 Kg )( 2.00 × 10 m/s ) ⎝ 3.00 × 10 m/s ⎠ −34

−27

27.31

E2 − ER2

For relativistic particles, p =

−14

and λ =

h = p

hc E2 − ER2

For 3.00 MeV electrons, E = KE + ER = 3.00 MeV + 0.511 MeV = 3.51 MeV, so

(6.63 × 10 λ=

)(

)

J ⋅ s 3.00 × 108 m/s ⎛ 1 Mev ⎞ −13 ⎜ ⎟ = 3.58 × 10 m −13 2 2 1.60 × 10 J ⎠ ( 3.51 MeV ) − ( 0.511 MeV ) ⎝

27.32

−34

Convert units to find 4.50 eV = 7.20 × 10−19 J. Use the kinetic energy to find each particle’s speed or realize that KE = p2/2m to write the de Broglie wavelength as λ =

2m ( KE)

(a) The electron’s de Broglie wavelength is

λe =

(

6.63 × 10−34 J ⋅s

)(

2 9.11 × 10−31 kg 7.20 × 10−19 J

)

= 5.79 × 10−10 m

(b) The bowling ball’s de Broglie wavelength is

λb =

6.63 × 10−34 J ⋅s

(

2 ( 6.00 kg ) 7.20 × 10

−19

)

= 2.26 × 10−25 m

Topic 27

1570

(

)(

)

6.63 × 10−34 J ⋅s 3.00 × 108 m/s hc λp = = = 2.76 × 10−7 m −19 E 7.20 × 10 J 27.33

(a) The required electron momentum is

6.63 × 10−34 J ⋅ s ⎛ 1 keV ⎞ keV ⋅ s = 4.1× 10−7 ⎜ −11 −16 ⎟ m 1.0 × 10 m ⎝ 1.60 × 10 J ⎠

and the total energy is

E = p 2c 2 + ER2 2

2 2 ⎛ keV.s ⎞ = ⎜ 4.1 × 10−7 3.00 × 108 m/s + ( 511 keV ) = 526 keV ⎟ m ⎠ ⎝

(

)

The kinetic energy is then KE = E − ER = 526 keV − 511 keV = 15 keV (b)

Ephoton =

(6.63 × 10 J ⋅s)(3.00 × 10 m/s) ⎛⎜ = −34

1.0 × 10−11 m

⎞ −16 ⎟ ⎝ 1.60 × 10 J ⎠ 1 keV

= 1.2 × 102 keV

27.34

(a) From conservation of energy, the increase in kinetic energy must equal the decrease in potential energy, or KE − 0 = qΔV Since the particle is nonrelativistic, its kinetic energy and momentum are related by the expression KE = p2/2m. Thus, the momentum is p2 =

Topic 27

1571

2m(KE) = 2mqΔV, and p = 2mqΔV . (b) The de Broglie wavelength is λ = h/p. Therefore, using the result of part (a) gives λ = h

2mqΔV .

(c) The electron and proton have the same magnitude charge. Thus, when both are accelerated through the same magnitude potential difference, the resulting de Broglie wavelengths are inversely proportional to the square root of the masses of the particles. The proton (with the larger mass) will have the shorter wavelength 27.35

From the uncertainty principle, the minimum uncertainty in the momentum of the electron is

Δpx =

6.63 × 10 −34 J ⋅ s h = = 5.3 × 10 −25 kg ⋅ m/s −9 4π ( Δx ) 4π 0.10 × 10 m

(

)

so the uncertainty in the speed of the electron is

Δpx 5.3 × 10 −25 kg ⋅ m/s Δvx = = = 5.8 × 10 5 m/s or ~ 106 m/s −31 me 9.11 × 10 kg If the speed is on the order of the uncertainty in the speed, then v ∼ 106 m/s. 27.36

Assuming the electron is nonrelativistic, the uncertainty in its momentum

Topic 27

1572

is Δp = me Δv. From the uncertainty principle, ΔxΔpx ≥ h/4π, so if there is an uncertainty of Δx = 2.5 µm = 2.5 × 10−6 m in the position of the electron, the minimum uncertainty in its speed is

( Δvx )min = 27.37

( Δp )

x min

6.63 × 10 −34 J ⋅ s h = = 23 m/s 4π me ( Δx )max 4π 9.11 × 10 −31 kg 2.5 × 10-6 m

(

)(

)

The uncertainty in the magnitude of the velocity of each particle is Δvx = vx ⋅(0.010 0%) = (500 m/s)(0.010 0 × 10−2) = 5.00 × 10−2 m/s If the mass is known precisely, the uncertainty in momentum is Δpx = m(Δvx) and the minimum uncertainty in position is Δxmin = h/4π(Δpx) = h/4πm(Δvx). For the electron:

Δxmin =

4π me ( Δvx )

(

6.63 × 10 −34 J ⋅ s

)(

4π 9.11 × 10 −31 kg 5.00 × 10 −2 m/s

)

= 1.16 × 10−3 m = 1.16 mm For the bullet:

6.63 × 10 −34 J ⋅ s h Δxmin = = = 5.28 × 10 −32 m −2 4π m ( Δvx ) 4π ( 0.020 0 kg ) 5.00 × 10 m/s

(

27.38

)

(a) The x-component of the electron’s non-relativistic momentum is found using Heisenberg’s uncertainty principle:

Topic 27

1573

ΔxΔpx ≥

h 4π

Δvx ,min =

h 6.63 × 10−34 J ⋅s = = 23.2 m/s 4π ( me ) ( Δx ) 4π 9.11 × 10−31 kg 2.50 × 10−6 m

→ Δx ( me Δvx ) ≥

h 4π

(

)(

)

(b) Similarly for the proton:

ΔxΔpx ≥

h 4π

(

) 4hπ

→ Δx mp Δvx ≥

6.63 × 10−34 J ⋅s Δvx ,min = = = 1.26 × 10−2 m/s −27 −6 4π mp ( Δx ) 4π 1.67 × 10 kg 2.50 × 10 m h

( )

27.39

(

)(

)

The maximum time one can use in measuring the energy of the particle is equal to the lifetime of the particle, or Δtmax ≈ 2 µs. One form of the uncertainty principle is ΔE Δt ≥ h/4π. Thus, the minimum uncertainty one can have in the measurement of a muon’s energy is

6.63 × 10 −34 J ⋅ s h ΔEmin = = = 3 × 10 −29 J 4π Δtmax 4π 2 × 10 −6 s

(

27.40

)

2 2 (a) For a nonrelativistic particle, KE = 21 mv = ( mv ) 2m = p 2m . 2

(b) From the uncertainty principle,

Δpx ≥

6.63 × 10 −34 J ⋅ s h = = 5.3 × 10 −20 kg ⋅ m/s 4π ( Δx ) 4π 1.0 × 10 −15 m

(

)

Since the momentum must be at least as large as its own uncertainty,

Topic 27

1574

the minimum kinetic energy is

(

) )

2 5.3 × 10 −20 kg ⋅ m/s ⎛ 1 MeV ⎞ pmin KEmin = = = 5.3 MeV ⎜ −13 ⎟ 2m 2 1.67 × 10 −27 kg ⎝ 1.60 × 10 J ⎠

27.41

(a) p =

(b)

λ c

(

6.63 × 10−34 J ⋅ s = 2.21× 10−32 kg ⋅ m/s −2 3.00 × 10 m

3.00 × 108 m/s = 1.00 × 1010 Hz −2 3.00 × 10 m

(c)

(

)(

E = hf = 6.63 × 10 −34 J ⋅ s 1.00 × 1010 Hz

)

⎛ ⎞ 1 eV −5 = 6.63 × 10 −24 J ⎜ ⎟ = 4.14 × 10 eV −19 ⎝ 1.60 × 10 J ⎠ 27.42

The de Broglie wavelength is λ = h/p, and the Compton wavelength is λc = h/mec. Thus, if λ = λc, it is necessary that p = mec. The relativistic expression for the momentum of an electron is p = γmev, so if p = mec, we must have

γ me v = me c , or v / c = 1/ γ = 1 − ( v / c ) . 2

Squaring both sides of this result gives (v/c)2 = 1 − (v/c)2, or 2(v/c)2 = 1, and

27.43

c 2

c 2 2

The total mechanical energy of this object is

Topic 27

1575

Emech = KEf = PEi = mgyi = (2.0 kg)(9.80 m/s2)(5.0 m) = 98 J A photon of light having wavelength λ = 5.0 × 10−7 m has an energy of

Ephoton =

(6.63 × 10 =

−34

)(

J ⋅ s 3.00 × 108 m/s −7

5.0 × 10 m

) = 4.0 × 10

−19

Thus, if all the initial gravitational potential energy of the object were converted to light with λ = 5.0 × 10−7 m, the number of photons that would be produced is

27.44

Emech 98 J = = 2.5 × 1020 photons −19 Ephoton 4.0 × 10 J/photon

(a) Minimum wavelength photons are produced when an electron gives up all its kinetic energy in a single collision. Then, Ephoton = 50 000 eV and

λmin =

hc Ephoton

(6.63 × 10 J ⋅ s )( 3.00 × 10 m/s ) = 2.49 × 10 = ( 5.00 × 10 eV )(1.60 × 10 J/eV ) −34

−19

−11

(b) From Bragg’s law, the interplanar spacing is

(

−11

)

(1) 2.49 × 10 m mλ d= = = 2.9 × 10−10 m = 0.29 nm 2 sin θ 2 sin (2.5°) 27.45

(a) The peak radiation occurs at approximately 560 nm wavelength. From Wien’s displacement law,

Topic 27

1576

0.289 8 × 10−2 m ⋅ K

λmax

0.289 8 × 10 −2 m ⋅ K ≈ 5 200 K 560 × 10 −9 m

(b) Clearly, a firefly is not at this temperature, so this is not blackbody radiation. 27.46

(a) From v = v0 + 2ay ( Δy ) Johnny’s speed just before impact is 2

v = 2 g Δy = 2 ( 9.80 m/s )( 50.0 m ) = 31.3 m/s and his de Broglie wavelength is

λ=

6.63 × 10 −34 J ⋅ s h = = 2.82 × 10 −37 m mv ( 75.0 kg )( 31.3 m/s )

(b) The energy uncertainty is

ΔE ≥

6.63 × 10−34 J ⋅ s h = = 1.06 × 10 −32 J 4π ( Δt ) 4π 5.00 × 10−3 s

(

)

(c)

(1.60 × 10 J )(100%) % error = 100%) ≥ ( mg Δy (75.0 kg )(9.80 m s )(50.0 m ) ΔE

−32

= 2.88 × 10−35 %

27.47

The magnetic force supplies the centripetal acceleration for the electrons, so m(v2/r) = qvB, or p = mv = qrB. The maximum kinetic energy is then

Topic 27

1577

KEmax = p2/2m = q2r2B2/2m, or

(1.60 × 10 J ) (0.200 m ) ( 2.00 × 10 T ) = 2.25 × 10 = 2 ( 9.11 × 10 kg ) −29

KEmax

−5

−19

−31

The work function of the surface is given by φ = Ephoton − KEmax = hc/λ − KEmax, or

( 6.63 × 10 φ=

−34

)(

J ⋅ s 3.00 × 10 8 m/s −9

450 × 10 m

) − 2.25 × 10

−19

⎛ ⎞ 1 eV = 2.17 × 10 −19 J ⎜ ⎟ = 1.36 eV −19 ⎝ 1.60 × 10 J ⎠

27.48

The energy of the incident photon is E0 = 6.20 keV, and its wavelength is

λ0 = hc/E0. In a head-on collision with an electron, the photon is back scattered, or the scattering angle is θ = 180°. The Compton equation then gives the wavelength of the scattered photon as λ = λ0 + λc(1 − cos 180°) = λ0 + 2λc, where λc = 2.43 × 10−12 m is the Compton wavelength. Thus,

(

)(

) ( )

6.63 × 10 −34 J ⋅ s 3.00 × 10 8 m/s hc λ = + 2λc = + 2 2.43 × 10 −12 m = 2.05 × 10 −10 m −16 E0 ( 6.20 keV ) 1.60 × 10 J/1 keV

(

)

The energy of the scattered photon is E = hc/λ, and from conservation of energy, the kinetic energy of the recoiling electron is

Topic 27

1578

( 6.63 × 10 KE = E − E = 6.20 keV − e

27.49

−34

)(

)

J ⋅ s 3.00 × 10 8 m/s ⎛ 1 keV ⎞ ⎜ ⎟ −10 −16 2.05 × 10 m ⎝ 1.60 × 10 J ⎠

KEe = 6.20 keV − 6.06 keV = 0.14 keV

From the photoelectric effect equation, KEmax = Ephoton − φ =

For λ = λ0, KEmax = 1.00 eV, so

For λ =

λ0 2

1.00 eV=

− φ.

−φ

λ0

, KEmax = 4.00 eV, giving 4.00 eV =

2 hc

λ0

[1]

−φ

[2]

Multiplying Equation [1] by a factor of 2 and subtracting the result from Equation [2] gives the work function as φ = 2.00 eV. 27.50

From the photoelectric effect equation, KEmax = Ephoton − φ = hc/λ − φ.

For λ = 670 nm, KEmax = E1, so

E1 =

hc −φ 670 nm

For λ = 520 nm, KEmax = 1.50 E1,

giving

1.50 E1 =

[1]

hc −φ 520 nm

Multiplying Equation [1] by a factor of − 1.50 and adding the result to Equation [2] gives

⎛ −1.50 1.00 ⎞ 0 = hc ⎜ + ⎟ + (1.50 − 1.00)φ ⎝ 670 nm 520 nm ⎠

[2]

Topic 27

1579

The work function for the material is then

(6.63 × 10 φ= or 27.51

−34

)(

)

J ⋅ s 3.00 × 108 m/s ⎛ 1.50 ⎞ 1 eV 1.00 ⎞ ⎛ 1 nm ⎞⎛ − ⎟ ⎜ ⎟ ⎜ −9 ⎟⎜ −19 0.50 ⎝ 670 nm 520 nm ⎠⎝ 10 m ⎠⎝ 1.60 × 10 J ⎠

φ = 0.785 eV.

(a) If the single photon produced has wavelength λ = 1.00 × 10−8 m, the kinetic energy of the electron was

KEe =

(6.63 × 10 =

−34

)(

J ⋅ s 3.00 × 10 8 m/s −8

1.00 × 10 m

) = 1.99 × 10

−17

J = 124 eV

This electron is nonrelativistic and its speed is given by

2 ( KEe ) me

(

)

2 1.99 × 10−17 J ⎛ ⎞ c = ⎜ ⎟ = 0.022 0c −31 8 9.11 × 10 kg ⎝ 3.00 × 10 m/s ⎠

(b) When the single photon produced has wavelength λ = 1.00 × 10−13 m,

KEe =

( 6.63 × 10 =

−34

)(

J ⋅ s 3.00 × 10 8 m/s

1.00 × 10

−13

) =⎛

1 MeV ⎞ ⎜ ⎟ = 12.4 MeV −13 1.60 × 10 J ⎝ ⎠

the electron is highly relativistic and KE = (γ − 1)ER, giving

γ = 1+

12.4 MeV KE = 1+ = 25.3 ER 0.511 MeV

Then, v = c 1 − 1 γ 2 = c 1 − 1 ( 25.3 ) = 0.999 2c . 2

Topic 27

27.52

1580

(a) If light were a classical wave, the time required for a surface of area A = πr2 to absorb energy E when illuminated with radiation of intensity I would be given by E = p ⋅ (Δt) = IA ⋅ (Δt) = I(πr2)(Δt). Using the given data values, we find the time to absorb energy E = 1.00 eV to be

Δt =

( )

I πr

(

)

1.00 eV 1.60 × 10 −19 J/1 eV ⎛ 1 day ⎞ = ⎟ = 148 days 2 ⎜ 4 −15 8.64 × 10 s ⎝ ⎠ 500 J/s π 2.82 × 10 m ( )

(

)

(b) This result is totally contrary to observations of the photoelectric effect. There, when a surface is illuminated with light above the threshold frequency of the surface, photoelectrons are seen to be ejected with a delay of less than 10−9 s.

Topic 28

1581

Topic 28 Atomic Physics

QUICK QUIZZES 28.1

Choice (b). The allowed energy levels in a one-electron atom may be expressed as En = −Z2 (13.6 eV)/n2, where Z is the atomic number. Thus, the ground state (n = 1 level) in helium, with Z = 2, is lower than the ground state in hydrogen, with Z = 1.

28.2

(a) For n = 5, there are 5 allowed values of ℓ, namely ! = 0,!1,!2,!3, and 4. (b) Since !m ranges from − to + in integer steps, the largest allowed value of  (! = 4 in this case) permits the greatest range of values for

!m . For n = 5, there are 9 possible values for !m : −4, −3, −2, −1, 0, +1, +2, +3, and +4. (c) For each value of !, there are !2 + 1 possible values of !m . Thus, there is 1 distinct pair with ! = 0, 3 distinct pairs with ! = 1, 5 distinct pairs with ! = 2, 7 distinct pairs with ! = 3, and 9 distinct pairs with ! = 4. This yields a total of 25 distinct pairs of  and !m that are possible when n = 5.

Topic 28

28.3

1582

Choice (d). Krypton has a closed configuration consisting of filled n = 1, n= 2, and n = 3 shells as well as filled 4s and 4p subshells. The filled n = 3 shell (the next to outer shell in krypton) has a total of 18 electrons, 2 in the 3s subshell, 6 in the 3p subshell and 10 in the 3d subshell.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 28.2

Neon signs do not emit a continuous spectrum, as can be determined by observing the light from the sign through a spectrometer. The discrete wavelengths a neon gas can produce are determined by the differences in the energies of the allowed states within the neon atom. The specific wavelengths and intensities account for the color of the sign.

28.4

In a neutral helium atom, one electron can be modeled as moving in an electric field created by the nucleus and the other electron. According to Gauss’s law, if the electron is above the ground state it moves in the electric field of a net charge of +2e − 1e = +1e. We say the nuclear charge is screened by the inner electron. The electron in a He+ ion moves in the field of the unscreened nuclear charge of 2 protons. Then, the potential energy function for the electron is about double that of one electron in the neutral atom.

Topic 28

28.6

1583

Classically, the electron can occupy any energy state. That is, all energies would be allowed. Therefore, if the electron obeyed classical mechanics, its spectrum, which originates from transitions between states, would be continuous rather than discrete.

28.8

Fundamentally, three quantum numbers describe an orbital wave function because we live in three-dimensional space. They arise mathematically from boundary conditions on the wave function, expressed as a product of a function of r, a function of θ, and a function of

φ. 28.10

In both cases the answer is yes. Recall that the ionization energy of hydrogen is 13.6 eV. The electron can absorb a photon of energy less than 13.6 eV by making a transition to some intermediate state such as one with n = 2. It can also absorb a photon of energy greater than 13.6 eV, but in doing so, the electron would be separated from the proton and have some residual kinetic energy.

28.12

It replaced the simple circular orbits in the Bohr theory with electron clouds. More important, quantum mechanics is consistent with Heisenberg’s uncertainty principle, which tells us about the limits of accuracy in making measurements. In quantum mechanics, we talk about

Topic 28

1584

the probabilistic nature of the outcome of a measurement on a system, a concept which is incompatible with the Bohr theory. Finally, the Bohr theory of the atom contains only one quantum number n, while quantum mechanics provides the basis for additional quantum numbers to explain the finer details of atomic structure. 28.14

Each of the given atoms has a single electron in an ! = 0 (or s) state outside a fully closed-shell core, shielded from all but one unit of the nuclear charge. Since they reside in very similar environments, one would expect these outer electrons to have nearly the same electrical potential energies and hence nearly the same ionization energies. This is in agreement with the given data values. Also, since the distance of the outer electron from the nuclear charge should tend to increase with Z (to allow for greater numbers of electrons in the core), one would expect the ionization energy to decrease somewhat as atomic number increases. This is also in agreement with the given data.

ANSWERS TO EVEN NUMBERED PROBLEMS 28.2

(a)

1 875 nm, 1 281 nm, 1 094 nm

(b)

All are in the infrared region of the spectrum.

Topic 28

1585

28.4

1.94 µm

28.6

45 fm

28.8

(a)

−3.40 eV

(d)

0.476 nm

28.10

(a)

28.12

(b)

−1.51 eV

1.13 eV

(b)

1.10 µm

(a)

2.19 × 106 m/s

(b)

28.14

(a)

1 281 nm

28.16

(a)

Transition II

(c)

Transitions II and III

28.18

(a)

12.1 eV

28.20

See Solution.

28.22

(a)

6 distinct wavelengths

(c)

Paschen series

(a)

2.89 × 1034 kg ⋅ m2/s

(c)

7.27 × 10−69

28.24

28.26

(a) !KEn = − 12 PEn

28.28

(a)

(c)

0.212 nm

13.6 eV

(c)

−27.2 eV

(b)

2.34 × 1014 Hz

(c)

0.970 eV

(b)

Transition I

(b)

12.1 eV, 189 eV, 10.2 eV

ΔPE = 2E

(b)

1.88 × 103 nm

(b)

2.75 × 1068

(c)

ΔKE = −E

En = −(122 eV)/n2

Topic 28

1586

(b)

−7.63 eV

(c)

−30.5 eV

(d)

22.9 eV = 3.66 × 10−18 J

(e)

5.52 × 1015 Hz, 54.3 nm

(f) deep ultraviolet region 28.30

(a)

See Solution.

(b)

See Solution.

28.32

(a)

(b)

28.34

(a)

4: ! = 0,!1,!2,!and!3

(b)

7: !m = −3,! − 2,! − 1,!0, + 1, + 2,!and! + 3.

28.36

28.38

(a)

(c)

1s2 2s2 2p3

(b) !( n = 1,! = 0,!m = 0,!ms = ± 12 ) ;! ( n = 2,! = 0,!m = 0,!ms = ± 12 ) ;

( n = 2,! = 1,!m = −1,!ms = ± 12 ) ;! ( n = 2,! = 1,!m = 0,!ms = ± 12 ) ;

( n = 2,! = 1,!m = 1,!ms = ± 12 )

28.40

(a) 30 allowed states—see Solution for tables (b) 36 possible combinations—six possible states for each electron independently

Topic 28

1587

28.42

0.031 1 nm

28.44

germanium

28.46

137

28.48

(a)

4.20 mm

(c)

8.84 × 1016 photons/mm3

(a)

See Solution.

28.50

(b)

1.05 × 1019 photons

2.54 × 1074

(c)

1.18 × 10−63 m

(d) This number is much smaller than the radius of an atomic nucleus (∼10−15 m) so the distance between quantized orbits of the Earth is too small to observe. 28.52

(a)

135 eV

(b)

≈10 times the magnitude of the ground state energy

28.54

Topic 28

1588

PROBLEM SOLUTIONS 28.1

(a) The wavelengths in the Lyman series of hydrogen are given by 1/λ = RH (1 − 1/n2), where n = 2, 3, 4, …, and the Rydberg constant is RH = 1.097 373 2 × 107 m−1. This can also be written as λ = (1/RH)[n2/(n2 − 1)], so the first three wavelengths in this series are

⎛ 22 ⎞ 1 −7 λ1 = 7 −1 ⎜ 2 ⎟⎠ = 1.215 × 10 m = 121.5 nm 1.097 373 2 × 10 m ⎝ 2 − 1 ! ⎛ 32 ⎞ 1 λ2 = = 1.025 × 10−7 m = 102.5 nm 7 −1 ⎜ 2 ⎟ 1.097 373 2 × 10 m ⎝ 3 − 1 ⎠ ! ⎛ 42 ⎞ 1 λ3 = = 9.720 × 10−8 m = 97.20 nm 7 −1 ⎜ 2 ⎟ 1.097 373 2 × 10 m ⎝ 4 − 1 ⎠ ! (b) These wavelengths are all in the far ultraviolet region of the spectrum. 28.2

(a) The wavelengths in the Paschen series of hydrogen are given by 1/λ = RH (1/32 − 1/n2), where n = 4, 5, 6,…, and the Rydberg constant is RH = 1.097 373 2 × 107 m−1. This can also be written as

λ = (1/RH)[9n2/(n2 − 9)], so the first three wavelengths in this series are

Topic 28

1589

⎛ 9(4)2 ⎞ −6 λ1 = ⎟ = 1.875 × 10 m = 1 875 nm 7 −1 ⎜ 2 1.097 373 2 × 10 m ⎝ 4 − 9 ⎠ 1

⎛ 9(5)2 ⎞ 1 −6 λ2 = 7 −1 ⎜ 2 ⎟⎠ = 1.281 × 10 m = 1 281 nm 1.097 373 2 × 10 m ⎝ 5 − 9 !

λ3 =

⎛ 9(6)2 ⎞ −6 ⎟ = 1.094 × 10 m = 1 094 nm 7 −1 ⎜ 2 1.097 373 2 × 10 m ⎝ 6 − 9 ⎠ 1

(b) These wavelengths are all in the infrared region of the spectrum. 28.3

(a) From Coulomb’s law, 9 2 2 −19 ke q1q2 ( 8.99 × 10 N ⋅ m C ) (1.60 × 10 C ) F = = 2 r2 (1.0 × 10−10 m )

= 2.3 × 10−8 N

(b) The electrical potential energy is 9 2 2 −19 −19 ke q1q2 ( 8.99 × 10 N ⋅ m C ) ( −1.60 × 10 C ) (1.60 × 10 C ) PE = = r 1.0 × 10−10 m

! 28.4

⎛ ⎞ 1 eV = −2.3 × 10−18 J ⎜ = −14 eV −19 ⎝ 1.60 × 10 J ⎟⎠

If E2, E5, and E6, are the energies of an electron in the second, fifth, and sixth excited states within this atom, respectively, the energies of the photons produced are: (Ephoton)1 = hc/λ1 = E2 − E5, (Ephoton)2 = hc/λ2 = E2 − E6, and (Ephoton)3 = hc/λ3 = E5 − E6. Thus, we have hc/λ3 = E5 − E6 = (E2 − E6) − (E2 − E5), or hc/λ3 = hc/λ2 − hc/λ1. This reduces to

Topic 28

1590

λλ (520 nm)(410 nm) λ3 = 1 2 = = 1.94 × 103 nm = 1.94 µm λ1 − λ2 520 nm − 410nm ! 28.5

(a)

The electrical force supplies the centripetal acceleration of the electron, so mev2/r = kee2/r2 or v = ke e 2 me r. !

(8.99 × 10 N ⋅ m )(1.60 × 10 C) = 1.6 × 10 m/s v= ( 9.11 × 10 kg )(1.0 × 10 m ) ! 9

−19

−31

−10

v 1.6 × 106 m/s = 5.3 × 10−3 << 1, so the electron is not (b) No. = 8 !c 3.00 × 10 m/s relativistic. (c) The de Broglie wavelength for the electron is λ = h/p = h/mev, or

6.63 × 10−34 J ⋅s λ= = 4.5 × 10−10 m = 0.45 nm −31 6 ( 9.11 × 10 kg )(1.6 × 10 m/s ) ! (d) Yes. The wavelength and the atom are roughly the same size. 28.6

Assuming a head-on collision, the α-particle comes to rest momentarily at the point of closest approach. From conservation of energy, KEf + PEf = KEi + PEi or 0+ !

ke (2e)(79e) k (2e)(79e) = KEi + e rf ri

Topic 28

1591

With ri → ∞, this gives the distance of closest approach as 9 2 2 −19 158ke e 2 158 ( 8.99 × 10 N ⋅ m C ) (1.60 × 10 C) rf = = KEi 5.0 MeV (1.60 × 10−13 J/MeV )

! 28.7

= 4.5 × 10−14 m = 45 fm

(a) The energy of a Lyman-α photon is ⎛ 1 1⎞ Eα = E2 − E1 = −13.6 eV ⎜ 2 − 2 ⎟ = 10.2 eV ⎝2 1 ⎠

(b) The Lyman-α wavelength is

28.8

(

)(

)

6.63 × 10−34 J ⋅s 3.00 × 108 m/s hc → λα = = 122 nm λ 10.2 eV 1.60 × 10−19 J/eV

(

)

The hydrogen atom energy levels are given by En = −

13.6 eV . Substitute n2

values for n to find (a) n = 2: E2 = −

13.6 eV = −3.40 eV 22

(b) n = 3: E3 = −

13.6 eV = −1.51 eV 32

The nth orbital radius is rn = n2a0 where a0 = 0.052 9 nm is the Bohr radius. Substitute values to find (c) n = 2: r2 = 2 ( 0.052 9 nm ) = 0.212 nm 2

(d) n = 3: r2 = 3 ( 0.052 9 nm ) = 0.476 nm 2

Topic 28

28.9

1592

Electron energy levels in a hydrogen-like atom are given by

En = −

Z 2 (13.6 eV ) . For helium, Z = 2 so the required energy is n2

⎛ 1 1⎞ E = E2 − E1 = − 2 2 (13.6 eV ) ⎜ 2 − 2 ⎟ = 40.8 eV ⎝2 1 ⎠

( )

28.10

(a) The photon energy is equal to the energy difference between the n = 6 and the n = 3 levels: ⎛ 1 1⎞ E = E6 − E3 = −13.6 eV ⎜ 2 − 2 ⎟ = 1.13 eV ⎝6 3 ⎠

(b) The wavelength of a photon with this energy is

hc λ

(

)(

)

6.63 × 10−34 J ⋅s 3.00 × 108 m/s hc λ= = = 1.10 × 10−6 m = 1.10 µm −19 E 1.13 eV 1.60 × 10 J/eV

28.11

(

)

(a) rn = n2a0 yields r2 = 4(0.052 9 nm) = 0.212 nm (b) With the electrical force supplying the centripetal acceleration, 2 2 2 !me vn rn = ke e rn , giving

vn = ke e 2 me rn ,!and!pn = me vn = me ke e 2 rn . !

Thus,

Topic 28

1593

me ke e 2 p2 = = r2 ! (c)

( 9.11 × 10

−31

kg ) ( 8.99 × 109 N ⋅ m 2 C2 ) (1.6 × 10−19 C )

0.212 × 10−9 m

= 9.95 × 10−25 kg ⋅ m/s

⎛ 6.63 × 10−34 J ⋅s ⎞ ⎛ h ⎞ −34 Ln = n ⎜ ⎟ → L2 = 2 ⎜ ⎟⎠ = 2.11 × 10 J ⋅s ⎝ 2π ⎠ ⎝ 2 π !

(d)

( 9.95 × 10−25 kg ⋅ m/s ) 1 p2 KE2 = mv22 = 2 = 2 2me 2 ( 9.11 × 10−31 kg )

⎛ ⎞ 1 eV = 5.44 × 10−19 J ⎜ = 3.40 eV −19 ⎟ ⎝ 1.60 × 10 J ⎠ (e)

8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10−19 C ) ( ke (−e)e PE2 = =− r2 ( 0.212 × 10−9 m ) !

= −1.09 × 10−18 J = −6.80 eV

(f) E2 = KE2 + PE2 = 3.40 eV − 6.80 eV = −3.40 eV 28.12

(a) With the electrical force supplying the centripetal acceleration, 2 2 2 2 2 2 !me vn rn = ke e rn , giving !vn = ke e me rn , where rn = n a0 = n (0.052 9

nm). Thus,

ke e 2 v1 = = me r1 !

(8.99 × 10 N ⋅ m C )(1.60 × 10 C) = 2.19 × 10 m/s ( 9.11 × 10 kg )( 0.052 9 × 10 m ) 9

−19

−31

−9

Topic 28

1594

(b)

(

)(

)

2 1 1 KE1 = me v12 = 9.11 × 10−31 kg 2.19 × 106 m/s 2 2 ⎛ ⎞ 1 eV = 2.18 × 10−18 J ⎜ = 13.6 eV −19 ⎟ ⎝ 1.60 × 10 J ⎠

(c)

PE1 =

ke (−e)e r1

(8.99 × 10 N ⋅ m C )(1.60 × 10 C) =− (0.052 9 × 10 m) 9

−19

−9

= −4.35 × 10−18 J = −27.2 eV 28.13

Since the electrical force supplies the centripetal acceleration,

me vn2 ke e 2 = 2 r rn n !

vn2 =

k2 e 2 me rn

From !Ln = me rnvn = n, we have!rn = n me vn , so

k e2 ⎛ m v ⎞ vn2 = 2 ⎜ e n ⎟ me ⎝ n ⎠ ! 2 which reduces to vn = ke e n . !

28.14

(

)

2 2 (a) The Rydberg equation is 1/λ = RH 1 n f − 1 ni , or !

2 2 1 ⎛ ni n f ⎞ λ= RH ⎜⎝ ni2 − n2f ⎟⎠ !

With ni = 5 and nf = 3,

Topic 28

1595

1 ⎡ (25)(9) ⎤ λ= = 1.281 × 10−6 m = 1 281 nm 7 −1 ⎢ ⎥ 1.097 373 2 × 10 m ⎣ 25 − 9 ⎦ !

(b)

c 3.00 × 108 m/s = = 2.34 × 1014 Hz λ 1 281 × 10−9 m

−34 8 ⎞ hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m/s ) ⎛ 1 eV = = 0.970 eV (c) Ephoton = −9 −19 ⎟ ⎜ λ 1 281 × 10 m ⎝ 1.60 × 10 J ⎠ !

28.15

The energy of the emitted photon is −34 8 ⎞ hc ( 6.626 × 10 J ⋅s ) ( 2.998 × 10 m/s ) ⎛ 1 eV Ephoton = = = 1.89 eV −9 −19 ⎟ ⎜ λ 656 × 10 m ⎝ 1.60 × 10 J ⎠ !

This photon energy is also the difference in the electron’s energy in its initial and final orbits. The energies of the electron in the various allowed orbits within the hydrogen atom are En = − !

13.6 eV n2

where

n = 1, 2, 3, …

giving E1 = −13.6 eV, E2 = −3.40 eV, E3 = −1.51 eV, E4 = −0.850 eV,…. Observe that Ephoton = E3 − E2, so the transition was from the n = 3 orbit to the n = 2 orbit. 28.16

The change in the energy of the atom is

⎛ 1 1⎞ ΔE = E f − Ei = 13.6 eV ⎜ 2 − 2 ⎟ ⎝ ni n f ⎠ !

Topic 28

1596

⎛1 1 ⎞ Transition I: ΔE = 13.6 eV ⎜ − ⎟ = 2.86 eV (absorption) ⎝ 4 25 ⎠ ! ⎛ 1 1⎞ Transition II: ΔE = 13.6 eV ⎜ − ⎟ = −0.967 eV (emission) ⎝ 25 9 ⎠ ! 1⎞ ⎛ 1 Transition III: ΔE = 13.6 eV ⎜ − ⎟ = −0.572 eV (emission) ⎝ 49 16 ⎠ ! 1⎞ ⎛ 1 Transition IV: ΔE = 13.6 eV ⎜ − ⎟ = 0.572 eV (absorption) ⎝ 16 49 ⎠ ! hc hc = , (a) Since λ = E −ΔE transition II emits the shortest wavelength photon !

photon. (b) The atom gains the most energy in transition I. (c) The atom loses energy in transition II and III. 28.17

The energy absorbed by the atom is

Ephoton = E f − Ei = !

⎛ 1 −13.6 eV (−13.6 eV) 1⎞ − = +13.6 eV ⎜ 2 − 2 ⎟ 2 2 nf ni ⎝ ni n f ⎠

⎛ 1 1⎞ (a) Ephoton = 13.6 eV ⎜ 2 − 2 ⎟ = 2.86 eV ⎝2 5 ⎠ ! ⎛ 1 1⎞ (b) Ephoton = 13.6 eV ⎜ 2 − 2 ⎟ = 0.472 eV ⎝4 6 ⎠ ! 28.18

(a) The energy absorbed is

Topic 28

1597

ΔE = E f − Ei = !

−13.6 eV (−13.6 eV) ⎛ 1 1⎞ = = 13.6 eV ⎜ − ⎟ = 12.1 eV 2 2 ⎝ 1 9⎠ nf ni

(b) Three transitions are possible as the electron returns to the ground state. These transitions and the emitted photon energies are

28.19

ni = 3 → nf = 1:

⎛ 1 1⎞ ΔE = 13.6 eV ⎜ 2 − 2 ⎟ = 12.1 eV ⎝1 3 ⎠ !

ni = 3 → nf = 2:

⎛ 1 1⎞ ΔE = 13.6 eV ⎜ 2 − 2 ⎟ = 1.89 eV ⎝2 3 ⎠ !

ni = 2 → nf = 1:

⎛ 1 1⎞ ΔE = 13.6 eV ⎜ 2 − 2 ⎟ = 10.2 eV ⎝1 2 ⎠ !

(a) The photon of longest wavelength is produced in the transition for which the atom gives up the smallest amount of energy. From Figure P28.15, this is seen to be the n = 3 to n = 2 transition, for which Ephoton = E3 − E2 = −1.512 eV − (−3.401 eV) = 1.889 eV (b)

λ= !

hc Ephoton

( 6.63 × 10 =

J ⋅s ) ( 3.00 × 108 m/s ) ⎛ ⎞ 1 eV −19 ⎟ ⎜ 1.889 eV ⎝ 1.60 × 10 J ⎠

−34

= 6.58 × 10−7 m = 658 nm

(c) The photon of shortest wavelength is produced in the transition for which the atom gives up the greatest amount of energy. From Figure P28.15, this is seen to be the n = 6 to n = 2 transition, for which © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 28

1598

Ephoton = E6 − E2 = −0.378 eV − (−3.401 eV) = 3.02 eV (d)

λ= !

hc Ephoton

( 6.63 × 10

−34

J ⋅s ) ( 3.00 × 108 m/s ) ⎛ ⎞ 1 eV −19 ⎜⎝ 1.60 × 10 J ⎟⎠ 3.02 eV

= 4.12 × 10−7 m = 412 nm

(e) The shortest wavelength is produced for the n = ∞ to n = 2 transition. The photon energy for this transition is Ephoton = 0 − (−3.401 eV) = 3.401 eV, and its wavelength is

λ= ! 28.20

hc Ephoton

( 6.63 × 10

−34

J ⋅s ) ( 3.00 × 108 m/s ) ⎛ ⎞ 1 eV −19 ⎜⎝ 1.60 × 10 J ⎟⎠ 3.401 eV

= 3.66 × 10−7 m = 366 nm

The magnetic force supplies the centripetal acceleration, so mv2/r = qvB, or r = mv/qB. If angular momentum is quantized according to

2n Ln = mvnrn = 2n,!then!mvn = rn ! and the allowed radii of the path are given by

rn = !

28.21

1 ⎛ 2n ⎞ qB ⎜⎝ rn ⎟⎠

rn =

2n qB

(a) The energy of the emitted photon is

Topic 28

1599

⎛ 1 1⎞ Ephoton = E4 − E2 = −13.6 eV ⎜ 2 − 2 ⎟ = 2.55 eV ⎝4 2 ⎠ ! This photon has a wavelength of

λ= Ephoton !

( 6.63 × 10 =

−34

J ⋅s ) ( 3.00 × 108 m/s )

( 2.55 eV )(1.60 × 10

−19

J/eV )

= 4.88 × 10−7 m = 488 nm

(b) Since momentum must be conserved, the photon and the atom go in opposite directions with equal magnitude momenta. Thus, p = matomv = h/λ, or

hc 6.63 × 10−34 J ⋅s v= = = 0.814 m/s matom λ (1.67 × 10−27 kg ) ( 4.88 × 10−7 m ) ! 28.22

(a) Starting from the n = 4 state, there are 6 possible transitions as the electron returns to the ground (n = 1) state. These transitions are: n = 4 → n = 1, n = 4 → n = 2, n = 4 → n = 3, n = 3 → n = 1, n = 3 → n = 2, and n = 2 → n = 1. Since there is a different change in energy associated with each of these transitions, there will be 6 distinct wavelengths observed in the emission spectrum of these atoms. (b) The longest observed wavelength is produced by the transition involving the smallest change in energy. This is the n = 4 → n = 3 transition, and the wavelength is

Topic 28

1600

( 6.63 × 10−34 J ⋅s )( 3.00 × 108 m/s ) ⎛ 1 eV ⎞ ⎛ 1 nm ⎞ hc λmax = = ⎜⎝ 1.60 × 10−19 J ⎟⎠ ⎜⎝ 10−9 m ⎟⎠ E4 − E3 −13.6 eV (1/42 − 1/32 ) ! or

λmax = 1.88 × 103 nm.

(c) Since the transition producing this wavelength terminates on the n = 3 level, this is part of the Paschen series. 28.23

(a) For the absorption of the photon to ionize the hydrogen atom, the electron in this atom must be in an excited state having an ionization energy less than or equal to the photon energy. That is, we must have Eionization = −En ≤ Ephoton = 2.28 eV, or En ≥ −2.28 eV. The state with the smallest value of n meeting this requirement is the n = 3 state, with E3 = −1.51 eV. (b) After the electron spends 1.51 eV of energy to escape from the atom, it will retain the remaining absorbed photon energy (2.28 eV − 1.51 eV = 0.77 eV) as kinetic energy. Its speed will then be

v= !

28.24

2KE 2(0.77 eV) ⎛ 1.60 × 10−19 J ⎞ 5 = ⎟⎠ = 5.2 × 10 m/s = 520 km/s mc 9.11 × 10−31 kg ⎜⎝ 1 eV

7.36 × 1022 kg ) 2π ( 3.84 × 108 m ) ( ⎛ 2π r ⎞ L = MM vr = MM ⎜ r= ⎝ T ⎟⎠ 2.36 × 106 s (a)

= 2.89 × 1034 kg ⋅ m 2 s

Topic 28

1601 34 2 L 2π ( 2.89 × 10 kg ⋅ m s ) n = = = 2.75 × 1068 (b) −34  1.05 × 10 J ⋅s !

(c) The gravitational force supplies the centripetal acceleration, so MMv2/r = GMEMM/r2 or rv2 = GME. Then, from !Ln = MM vnrn = n or

!vn = n MM rn , we have

r = ( n MM rn ) = GME , which gives rn = n2 (  2 GME MM2 ) = n2 r1 . ! !n 2

Therefore, when n increases by 1, the fractional change in the radius is

Δr rn+1 − rn (n + 1)2 r1 − n2 r1 2n + 1 2 = = = ≈ !for!n >> 1 2 2 r r n r n n n 1 ! Δr 2 ≈ = 7.27 × 10−69 68 ! r 2.75 × 10

28.25

2 2 2 (a) rn = n a0 = n (0.052 9 nm) ⇒ r3 = 3 (0.052 9 nm) = 0.476 nm !

(b) In the Bohr model, the circumference of an allowed orbits must be an integral multiple of the de Broglie wavelength for the electron in that orbit, or 2πrn = nλ. Thus, the wavelength of the electron when in the n = 3 orbit in hydrogen is 2π r3 2π (0.476 nm) λ= = = 0.997 nm 3 3 !

28.26

(a) The Coulomb force supplies the necessary centripetal force to hold

Topic 28

1602

2 2 2 2 2 the electron in orbit, so !me vn rn = ke e rn or !me vn = ke e rn . But 2 2 !me vn = 2KEn , and kee /rn = −PEn, where PEn is the electrical potential

energy of the electron-proton system when the electron is in an orbit of radius rn. We then have 2KEn = −PEn, or KEn = − 12 PEn . ! (b) When the atom absorbs energy E, and the electron moves to a higher level, both the kinetic and potential energies will change. Conservation of energy requires that E = ΔKE + ΔPE. However, from the result of part (a), !ΔKE = − 12 ΔPE, and we have 1 1 E = − ΔPE + ΔPE = + ΔPE 2 2 !

(c)

28.27

1 1 ΔKE = − ΔPE = − (2E) 2 2 !

ΔPE = 2E

ΔKE = −E

2 2 2 2 The radii for atomic number Z are !rn = n (  me ke e ) Z = n a0 Z , so r1 =

a0/Z, where a0 = 0.052 9 nm is the radius of the first Bohr orbit in hydrogen. (a) For He+, Z = 2 and r1 = 0.052 9 nm/2 = 0.026 5 nm. (b) For Li2+, Z = 3 and r1 = 0.052 9 nm/3 = 0.017 6 nm. (c) For Be3+, Z = 4 and r1 = 0.052 9 nm/4 = 0.013 2 nm.

Topic 28

28.28

1603

(a) The energy levels in a single electron atom with nuclear charge +Ze are En = −Z2 (13.6 eV)/n2 For doubly ionized lithium, Z = 3, giving En = −(122 eV)/n2. (b) E4 = !

−122 eV = −7.63 eV 42

E4 = !

−122 eV = −30.5 eV 22

(c)

(d) Ephoton = Ei − Ef = −7.63 eV − (−30.5 eV) = 22.9 eV

⎛ 1.60 × 10−19 J ⎞ Ephoton = (22.9 eV) ⎜ = 3.66 × 10−18 J ⎟ 1 eV ⎝ ⎠ ! f=

(e)

Ephoton h

3.66 × 10−18 J = 5.52 × 1015 Hz −34 6.63 × 10 J ⋅s

c 3.00 × 108 m/s λ= = = 5.43 × 10−8 m = 54.3 nm 15 f 5.52 × 10 Hz ! (f) This wavelength is in the deep ultraviolet region of the spectrum. 28.29

The charge of the electron is q1 = −e, and that of the nucleus is q2 = Ze. The energy levels for the one-electron atoms and ions can then be written as 2 4 2 En = −kµZ2/n2, where !k = ke e 2 = constant. The energy of the photon

emitted in the n = 3 to n = 2 transition becomes Ephoton = E3 − E2 = kµZ2(1/22 − 1/32) = µZ2 (5k/36), and the wavelength is

Topic 28

1604

hc 1 ⎛ 36hc ⎞ λ= = ⎜ ⎟ Ephoton µZ 2 ⎝ 5k ⎠ ! For hydrogen, ZH = 1 and the reduced mass is µH = mpme/(mp + me). Since the proton mass is very large in comparison to the electron mass, mp + me ≈ mp and µH ≈ me. (a) For positronium, Zp = 1 and the reduced mass is µp = meme/(me + me) = me/2. Thus, the ratio of the wavelength for the n = 3 to n = 2 transition in positronium to that for the same transition in hydrogen is

λp λH !

(1 µ Z ) ( 36hc 5k ) = µ Z = m ⋅1 = p

2 p

2 H

(1 µ Z ) ( 36hc 5k )

µ pZ

2 H 2 p

(m 2) ⋅1

giving λp = 2λH = 2(656.3 nm) = 1 313 nm = 1.313 µm (infrared region). (b) For singly ionized helium, ZHe = 2, and the reduced mass is µHe ≈ 2mpme/(2mp + me). Again, mp >> me and we find that µHe ≈ me. The ratio of the wavelength for the n = 3 to n = 2 transition in singly ionized helium to that for the same transition in hydrogen is then

λHe µ Z2 m ⋅12 1 = H H2 = e 2 = 4 ! λH µHeZHe me ⋅ 2 λ 656.3 nm = 164.1 nm!(ultraviolet!region) . giving λHe = H = 4 4 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 28

28.30

1605

(a) For standing waves in a string fixed at both ends, L = nλ/2, or λ = 2L/n. According to the de Broglie hypothesis, p = h/λ. Combining these expressions gives p = mv = nh/2L. 2 2 (b) Using !E = 12 mv = p 2m , with p as found in (a) above:

n2 h 2 En = 2 or 4L (2m) !

28.31

En = n2 E0

where E0 =

h2 8mL

As a consequence of the allowable values of ℓ , mℓ, and ms, there exist N = 2n2 distinct states in hydrogen’s nth energy level. Substitute values to find (a) N = 2 (1) = 2 2

(b) N = 2 ( 2 ) = 8 2

29.32

(a) By inspection, an electron in a 3d state has the principle quantum number n = 3 . (b) Similarly, the letter d indicates the orbital quantum number is ℓ = 2 . ( ℓ = 0, 1, 2, … are indicated by the letters s, p, d, …). (c) For an electron in the n = 3, ℓ = 2 state, the magnetic quantum number can vary from − ℓ to + ℓ in steps of 1 for a total of 5 possible mℓ values (−2, −1, 0, +1, +2).

Topic 28

28.33

28.34

1606

In the 3d subshell, n = 3 and ! = 2. The 10 possible quantum states are n=3

1 ! = 2 !m = +2 !ms = + 2

n=3

1 ! = 2 !m = +2 !ms = − 2

n=3

1 ! = 2 !m = +1 !ms = + 2

n=3

1 ! = 2 !m = +1 !ms = − 2

n=3

1 ! = 2 !m = +0 !ms = + 2

n= 3

1 ! = 2 !m = +0 !ms = − 2

n=3

1 ! = 2 mℓ = −1 !ms = + 2

n=3

1 ! = 2 mℓ = −1 !ms = − 2

n=3

1 ! = 2 !m = −2 !ms = + 2

n=3

1 ! = 2 !m = −2 !ms = − 2

(a) For a given value of the principal quantum number n, the orbital quantum number  varies from 0 to n − 1 in integer steps. Thus, if n = 4, there are 4 possible values of ! : ! = 0,!1,!2,!and!3. (b) For each possible value of the orbital quantum number !, the orbital magnetic quantum number !m ranges from !−!to! +  in integer steps. When the principal quantum number is n = 4, and the largest allowed value of the orbital quantum number is ! = 3, and there are 7 distinct possible values for !m . These values are:

!m = −3,! − 2,! − 1,! + 0,! + 1,! + 2,!and! + 3.

Topic 28

28.35

1607

The 3d subshell has n = 3 and ! = 2 . For ρ-mesons, we also have s = 1. Thus, there are 15 possible quantum states, as summarized in the table below.



!m

+2 +2 +2 +1 +1 +1

−1

−2

−1 +1 0 −1

−1

ms +1

28.36

−1 +1

The Zeeman effect is a splitting of one energy level into several closelyspaced levels according to their mℓ values. For an electron in the 4f subshell, ℓ = 3 so there are 7 possible mℓ values (−3, −2, −1, 0, +1, +2, +3) and therefore 7 unique energy levels.

28.37

(a) By the Pauli exclusion principle, 2 electrons can occupy the n = 3, ℓ = 2, mℓ = −1 state (one each with ms = ±(1/2 )). (b) By the Pauli exclusion principle, 6 electrons can occupy the n = 3, ℓ = 1 state (mℓ = −1, 0, 1 and, for each of those three states, one each in ms = ±(1/2 ), for a total of 3 × 2 = 6). (c) The nth energy level has a total of 2n2 discrete states so, by the Pauli exclusion principle, the n = 4 state could be occupied by a maximum

Topic 28

1608

of 2(4)2 = 32 electrons. 28.38

(a) The electronic configuration for nitrogen (Z = 7) is 1s22s22p3 . (b) The quantum numbers for the 7 electrons can be:

1s states

2s states

n=1

n=2

! = 0

!m = 0

!m = −1

2p states

n=2

! = 1

!m = 0

!m = 1

28.39

1 !ms = + 2 1 !ms = − 2 1 !ms = + 2 1 !ms = + 2 1 !ms = + 2 1 !ms = − 2 1 !ms = + 2 1 !ms = − 2 1 !ms = + 2 1 !ms = − 2

In the table of electronic configurations (Table 28.4), or the periodic table on the inside back cover of the text, look for the element whose last electron is in a 3p state and which has three electrons outside a closed shell. Its electron configuration will then end in 3s23p1. You should find that the element is aluminum.

Topic 28

28.40

1609

(a) For Electron #1 and also for Electron #2, n = 3 and ! = 1. The other quantum numbers for each of the 30 allowed states are listed in the tables below.

!m

Electron #1

1 !+ 2

1 !− 2

Electron #2

1 !− 2

1 !± 2

−1

1 !± 2

1 !+ 2

1 !± 2

−1

1 !± 2

!m

Electron #1

1 !+ 2

1 !− 2

Electron #2

1 !± 2

1 !− 2

−1

1 !± 2

1 !+ 2

−1

1 !± 2

!m

Electron #1

−1

1 !+ 2

−1

1 !+ 2

−1

1 !+ 2

−1

1 !− 2

−1

1 !− 2

−1

1 !− 2

Electron #2

1 !± 2

−1

1 !− 2

1 !± 2

−1

1 !+ 2

There are 30 allowed states, since Electron #1 can have any of three possible values of !m for both spin up and spin down, totaling six possible states. For each of these states, Electron #2 can be in either of the remaining five states. (b) Were it not for the exclusion principle, there would be 36 possible

Topic 28

1610

states, six for each electron independently. 28.41

(a) Observe the electron configurations given in the periodic table on the inside back cover of the textbook. Zirconium, with 40 electrons, has 4 electrons outside a closed krypton core. The krypton core, with 36 electrons, has all states up through the 4p subshell filled. Normally, one would expect the next 4 electrons to go into the 4d subshell. However, an exception to the rule occurs at this point, and the 5s subshell fills (with 2 electrons) before the 4d subshell starts filling. The two remaining electrons in zirconium are in an incomplete 4d subshell. Thus, n = 4 and ℓ = 2 for each of these electrons. (b) For electrons in the 4d subshell, with ! = 2 , the possible values of !m are mℓ = 0, ± 1, ± 2 , and those for ms are ms = ±1 / 2 . ! (c) We have 40 electrons, so the electron configuration is: 1s22s22p63s23p63d104s24p64d25s2 = [Kr]4d25s2

28.42

For electrons accelerated through a potential difference of 40.0 kV, starting from rest, the kinetic energy is KE = e(ΔV) = e(40.0 kV) = 40.0 keV. When these electrons strike the tungsten target, the shortest wavelength possible is produced when the electron loses all its kinetic energy and is

Topic 28

1611

brought to rest in a single collision, producing a single photon. The wavelength of the photon produced is λmin = hc/(Ephoton)max = hc/KE = hc/e(ΔV). This gives

(6.63 × 10 J ⋅s)(3.00 × 10 m/s) = 3.11 × 10 m = 0.031 1 nm λ = (1.60 × 10 C)( 40.0 × 10 V ) −34

−11

min

28.43

−19

(a) Note that Z = 83 for bismuth. With a vacancy in the L-shell, an electron in the M-shell is shielded from the nuclear charge by a total of 9 electrons, 2 electrons in the filled K-shell and 7 electrons in the partially filled L-shell. Thus, our estimate for the energy of this electron while it is in the M-shell is

EM ≈ !

−(Z − 9)2 (13.6 eV) −(74)2 (13.6 eV) = = −8.27 × 103 eV = −8.27 eV 32 9

When this electron drops down to fill the vacancy in the L-shell, it will continue to be shielded from the nuclear charge by the 2 electrons in the filled K-shell. Our estimate for the energy of this electron in the L-shell is then

−(Z − 2)2 (13.6 eV) −(81)2 (13.6 eV) EL ≈ = = −2.23 × 10 4 eV = −22.3 keV 2 2 4 ! Therefore, the estimate for the transitional energy, and hence the energy of the photon produced, in a M- to L-shell transition in © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 28

1612

bismuth is Ephoton = EM − EL ≈ −8.27 keV −(−22.3 keV) = 14 keV (b) The wavelength of the photon produced in the M- to L-shell transition should be approximately hc

λ= Ephoton ! 28.44

( 6.63 × 10 ≈

−34

J ⋅s ) ( 3.00 × 108 m/s ) ⎛ ⎞ 1 keV = 8.9 × 10−11 m −16 ⎟ ⎜ 14 eV 1.60 × 10 J ⎝ ⎠

The energies in the K- and M-shells are

Ek ≈ − !

(Z − 1)2 (13.6 eV) (Z − 9)2 (13.6 eV) !and!E ≈ − M (1)2 (3)2

Thus, Ephoton = EM − EK ≈ (13.6 eV)[−(Z − 9)2/9 + (Z − 1)2] = (13.6 eV)(8Z2/9 − 8) and Ephoton = hc/λ gives Z2 = 9[8 + hc/(13.6 eV)λ]/8 or 9 ( 6.63 × 10−34 J ⋅s ) ( 3.00 × 108 m/s ) ⎛ ⎞ 1 eV Z ≈ 9+ = 3.20 −19 ⎟ −9 ⎜ 8 (13.6 eV ) ( 0.101 × 10 m ) ⎝ 1.60 × 10 J ⎠ !

The element is germanium. 28.45

The transitions that produce the three longest wavelengths in the K series are shown below.

Topic 28

1613

The energy of the K-shell is EK = −69.5 keV. Thus, the energy of the L-shell is

hc EL = EK + λ3 !

( 6.63 × 10 E = −69.5 keV + L

J ⋅s ) ( 3.00 × 108 m/s ) ⎛ ⎞ 1 keV −9 −16 ⎟ ⎜ 0.0215 × 10 m ⎝ 1.60 × 10 J ⎠ −34

= −69.5 keV + 57.8 keV = −11.7 keV

Similarly, the energies of the M- and N-shells are

( 6.63 × 10−34 J ⋅s )( 3.00 × 108 m/s ) = −10.0 keV hc EM = EK + = −69.5 keV + λ2 ( 0.020 9 × 10−9 m )(1.60 × 10−16 J/keV ) ! and

6.63 × 10−34 J ⋅s ) ( 3.00 × 108 m/s ) ( hc EN = EK + = −69.5 keV+ = −2.30 keV λ1 0.018 5 × 10−9 m ) (1.60 × 10−16 J/keV ) ( ! The ionization energies of the L-, M-, and N-shells are 11.7 keV, 10.0 keV, and 2.30 keV, respectively 28.46

According to the Bohr model, the radii of the electron orbits in hydrogen

Topic 28

1614

are given by rn = n2a0, , with a0 = 0.052 9 nm = 5.29 × 10−11 m Then, if rn ≈ 1.00 µm = 1.00 × 10−6 m, the quantum number is

n= !

28.47

1.00 × 10−6 m ≈ 137 5.29 × 10−11 m

rn = a0

(a) ΔE = E2 − E1 = −13.6 eV/(2)2 − (−13.6 eV/(1)2) = 10.2 eV (b) The average kinetic energy of the atoms must equal or exceed the needed excitation energy, or 32 kBT ≥ ΔE, which gives −19 2(ΔE) 2(10.2 eV) (1.60 × 10 J/eV ) T≥ = = 7.88 × 10 4 K −23 3kB 3 (1.38 × 10 J/K ) !

28.48

(a) L = c(Δt) = (3.00 × 108 m/s)(14.0 × 10−12 s) = 4.20 × 10−3 m = 4.20 mm

(b) N =

Epulse Ephoton

Epulse hc/λ

Epulse λ hc

( 3.00 J)( 694.3 × 10 m ) −9

( 6.63 × 10

! (c)

−34

J ⋅s ) ( 3.00 × 10 m/s ) 8

= 1.05 × 1019 photons

N N 4N n= = = 2 V L (π d 4 ) L (π d 2 ) !

4 (1.05 × 1019 photons )

= = 8.84 × 1016 photons/mm 3 2 (4.20 mm) π (6.00 mm) ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 28

28.49

1615

(a) With one vacancy in the K-shell, an electron in the L-shell has one electron shielding it from the nuclear charge, so Zeff = Z −1 = 24 − 1 = 23. The estimated energy the atom gives up during a transition from the L-shell to the K-shell is then

ΔE ≈ Ei − E f = !

2 Zeff (13.6 eV ) − ⎡ − Zeff2 (13.6 eV ) ⎤ = Z 2 (13.6 eV ) ⎡ 1 − 1 ⎤ ⎢ ⎥ ⎢ 2 eff 2⎥ ni2 n2f ⎢⎣ ⎥⎦ ⎢⎣ n f ni ⎥⎦

⎡1 1⎤ ΔE ≈ (23)2 (13.6 eV) ⎢ 2 − 2 ⎥ = 5.40 × 103 eV = 5.40 keV ⎣1 2 ⎦ ! (b) With a vacancy in the K-shell, we assume that Z − 2 = 24 − 2 = 22 electrons shield the outermost electron (in a 4s state) from the nuclear charge. Thus, for this outer electron, Zeff = 24 − 22 = 2, and the estimated energy required to remove this electron from the atom is

⎡ Z 2 (13.6 eV) ⎤ 22 (13.6 eV) Eionization = E f − Ei = 0 − Ei ≈ − ⎢ − eff 2 = 3.40 eV ⎥= ni 42 ⎣ ⎦ ! (c) KE = ΔE − Eionization = 5.40 keV − 3.40 eV ≈ 5.40 keV 28.50

(a) Requiring the angular momentum associated with the orbital motion of Earth to satisfy Bohr’s postulate gives !ME vr = n, or the speed of

Topic 28

1616

the Earth in the nth allowed orbit would be !vn = n ME rn . The gravitational force between Earth and the Sun supplies the needed centripetal acceleration as Earth moves in its orbit. Thus, MEv2/r = GMSME/r2, or v2 = GMS/r. Substituting for the speed in the nth orbit

(

from above gives n! ME rn

) = GM r , which reduces to 2

rn = n2  2 GMS ME2 .

(b) From Table 7.3 and the back flysheet of the textbook, MS = 1.991 × 1030 kg, ME = 5.98 × 1024 kg, r = 1.496 × 1011 m, G = 6.67 × 10−11 N ⋅ −34 m2/kg2, and ! = 1.05 × 10 J ⋅s. We then find the quantum number for

Earth’s orbit to be

ME rnGMS 

( 5.98 × 10 kg ) (1.496 × 10 m )( 6.67 × 10 = 24

−11

N ⋅ m 2 kg 2 ) (1.991 × 1030 kg )

1.05 × 10−34 J ⋅s

! or

n = 2.54 × 1074

⎡( n + 1)2 − n2 ⎤  2 [ 2n + 1]  2 2n 2 ⎦ = Δr = rn+1 − rn = ⎣ ≈ GMS ME2 GMS ME2 GMS ME2 ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 28

1617

Using the values from part (b) above gives

2 ( 2.54 × 1074 ) (1.05 × 10−34 J ⋅s ) 2n Δr ≈ − = 2 GMs ME2 ( 6.67 × 10−11 N ⋅ m 2 kg 2 ) (1.991 × 1030 kg ) ( 5.98 × 1024 kg ) 2

= 1.18 × 10−63 m

(d) The result computed in part (d) above is much smaller than the radius of an atomic nucleus (∼10−15), so the distance between quantized orbits of the Earth is too small to observe.

28.51

−3 −9 P ( ΔE/Δt ) 4 ( 3.00 × 10 J/100 × 10 s ) = = 4.24 × 1015 W/m 2 (a) I = = 2 −6 A π d2 4 π ( 3.00 × 10 m ) !

(b) 2⎤ W ⎞ ⎡π ⎛ E = IA(Δt) = ⎜ 4.24 × 1015 2 ⎟ ⎢ ( 0.600 × 10−9 m ) ⎥ (1.00 × 10−9 s ) ⎝ m ⎠⎣4 ⎦

= 1.20 × 10−12 J

28.52

(a) Given that the de Broglie wavelength of the electron is λ = 2a0, its momentum is p = h/λ = h/2a0. The kinetic energy of this nonrelativistic electron is

KE =

p2 h2 = 2me 8me a02

( 6.63 × 10 J ⋅s ) (1 eV/1.60 × 10 J ) = 135 eV = 8 ( 9.11 × 10 kg ) ( 0.052 9 × 10 m ) −34

−31

−19

−9

(b) The kinetic energy of this electron is ≈10 times the magnitude of the © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 28

1618

ground state energy of the hydrogen atom, which is −13.6 eV. 28.53

In the Bohr model,

f= !

En − En−1 1 ⎡ −me ke2 e 4 ⎛ 1 1 ⎞ ⎤ 4π 2me ke2 e 4 ⎡ 1 1⎤ = ⎢ − ⎢ 2⎟⎥= 2 − 2⎥ 2 2 3 ⎜ h h ⎢⎣ 2 ⎝ n ( n − 1) ⎠ ⎥⎦ 2h ⎣ ( n − 1) n ⎦

which reduces to f = !

28.54

2π 2me ke2 e 4 ⎛ 2n − 1 ⎞ ⎜⎝ (n − 1)2 n2 ⎟⎠ . h3

−34 8 hc ( 6.626 × 10 J ⋅s ) ( 2.998 × 10 m/s ) 1 240 eV ⋅ nm Ephoton = = = = ΔE λ λ λ (1.602 × 10−19 J/eV ) (10−9 m/nm ) !

For λ = 310.0 nm, ΔE = 4.000 eV; λ = 400.0 nm, ΔE = 3.100 eV; and λ = 1 378 nm, ΔE = 0.900 0 eV. The ionization energy is 4.100 eV. The energy level diagram having the smallest number of levels, and consistent with these energy differences, is shown below.

Topic 29

1619

Topic 29 Nuclear Physics

QUICK QUIZZES 29.1

False. In a sample containing a very large number of identical radioactive atoms, 50% of the original atoms remain after 1 half-life has elapsed. During the second half-life, half of the remaining 50% of the atoms decay, leaving 25% of the original atoms still present. Thus, after 2 half-lives have elapsed, only 75% of the original radioactive atoms have decayed. An individual radioactive atom may exist an indefinite time before decaying.

29.2

Choice (c). At the end of the first half-life interval, half of the original sample has decayed and half remains. During the second half-life interval, half of the remaining portion of the sample decays, leaving onequarter of the original sample. During the third half-life, half of that quarter will decay. The total fraction of the sample that has decayed during the three half-lives is 1 1 ⎛ 1⎞ 1 ⎛ 1⎞ 7 + ⎜ ⎟+ ⎜ ⎟= !2 2 ⎝ 2 ⎠ 2 ⎝ 4 ⎠ 8

Topic 29

29.3

1620

Choice (c). The half-life of a radioactive material is !T1/2 = ln 2/λ where λ is the decay constant for that material. Thus, if λA = 2λB, we have

(T ) = lnλ 2 = 2lnλ2 =

! 29.4

1/ 2 A

(T ) = 4 h = 2 h 1/2 B

Choices (a) and (b). Reactions (a) and (b) both conserve total charge and total mass number as required. Reaction (c) violates conservation of mass number with the sum of the mass numbers being 240 before the reaction and being only 223 after the reaction.

29.5

Choice (b). The reactant nuclei in this endothermic reaction must supply |Q| = 2.17 MeV of energy to be converted into mass during the reaction. However, the reactant nuclei must also supply the emerging particles with sufficient kinetic energy to allow momentum to be conserved during the reaction. Thus, the threshold kinetic energy for the reaction (minimum total kinetic energy of the particles going into the reaction) must exceed the value |Q| = 2.17 MeV.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 29.2

By convention, 18 X contains 18 protons and has a mass number of 40 so 40

that 40 = Nprotons + Nneutrons and Nneutrons = 22. By charge neutrality, the atom © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 29

1621

contains 18 electrons. 29.4

Extra neutrons are required to overcome the increasing electrostatic repulsion of the protons. The neutrons participate in the net attractive effect of the nuclear force but feel no Coulomb repulsion.

29.6

An alpha particle is a doubly positive charged helium nucleus, is very massive, and does not penetrate very well. A beta particle is a singly negative charged electron, is very low mass, and only slightly more difficult to shield from. A gamma ray is a high-energy photon or high frequency electromagnetic wave and has high penetrating ability.

29.8

The activity of a sample is defined to be the number of radioactive decays per second. (a) I. If the number of radioactive nuclei in the sample is doubled, the activity will double. (b) D. Doubling the half-life will decrease the number of decays per second. (c) I. The activity is proportional to the decay constant. (d) D. After two half-lives, only 1/4 of the original sample remains. Fewer radioactive nuclei corresponds to lower activity.

Topic 29

29.10

1622

With a very small mass in comparison to alpha particles, beta particles have greater penetrating ability than do alpha particles.

29.12

The alpha particle, with a mass about 7 000 times that of the beta particle, is much more difficult to deflect. Even under the influence of an electrical force whose magnitude is double that acting on the beta particle, the alpha particle is deviated from its original path by a much smaller amount.

ANSWERS TO EVEN NUMBERED PROBLEMS 29.2

(a) 2.6567 × 10−26 kg (b)

1.4903 × 104 MeV/c

29.4

(a)

1.9 fm

(b)

7.4 fm

29.6

(a)

4.8 fm

(b)

4.6 × 10−43 m3

29.8

(a)

r12C = 7.89 cm, r13C = 8.21 cm m12

(b)

m13

29.10

16 km

29.12

(a)

12 u 13 u

= 0.961 and

1.11 MeV/nucleon

r12 r13

7.89 cm 8.21 cm

(c)

2.3 × 1017 kg/m3

= 0.961 .

(b)

7.07 MeV/nucleon

Topic 29

1623

8.79 MeV/nucleon

(d)

7.57 MeV/nucleon.

59 Co , Eb/A = For !25 Mn , Eb/A = 8.765 MeV. For !26 Fe , Eb/A = 8.790 MeV. For !27 55

8.768 MeV. This gives us finer detail than is shown in Figure 29.4. 29.16

7.93 MeV

29.18

8.7 × 103 Bq

29.20

(a)

0.755

(b)

0.570

(c)

9.77 × 10−4

(d) No. The decay model depends on large numbers of nuclei. After some long but finite time, only one undecayed nucleus will remain. It is likely that the decay of this final nucleus will occur before infinite time. 29.22

1.72 × 104 yr

29.24

1.7 × 103 yr

29.26

(a)

29.28

(a) ! 6 C

29.30

4.28 MeV

29.32

(a)

66 28

29.34

(a)

230

3.54 × 109 12

(b)

7.72 × 108

(b)

!2 He

Ni → 66 Cu + −10 e + ν e 29 (b)

(c)

14 6

(b)

186 keV

Topic 29

1624

29.36

1.00 MeV

29.38

(a)

8.023 829 u

(d)

mpv = (mBe + mn)V

(b)

8.025 594 u

(e) !12 mp v = 12 ( mBe + mn ) V − Q 2

29.40

29.42

(c)

−1.64 MeV

(f) 1.88 MeV

(a)

6.258 MeV, exothermic; 5.494 MeV, exothermic

(b)

the first reaction releases more energy

(c)

1.88 × 10−15 m

(a) !2 He + 7 N → 1 H + 8 O 4

(b)

Li + 11 H → 42 He + 42 He

29.44

The equivalent dose is 5.0 rad of heavy ions.

29.46

(a)

2.00 J/kg

29.48

(a)

2.5 × 10−3 rem/x-ray

(b)

~38 times background levels

29.50

(a)

10 h

29.52

22.0 MeV

29.54

156 keV

(b)

4.78 × 10−4 °C

3.2 m

Topic 29

1625

29.56

12 mg

29.58

(a)

60.6 Bq/L

(b)

40.6 days

29.60

(a)

N0 = 2.5 × 1024

(b)

R0 = 2.3 × 1012 Bq

29.62

2.81 × 104 yr

(c)

1.1 × 106 yr

PROBLEM SOLUTIONS 29.1

By convention, the symbol 26 Fe indicates an iron atom containing (a) 26 56

electrons, (b) 26 protons, and (c) 56 − 26 = 30 neutrons. 29.2

(a) Use the conversion factor 1 u = 1.660559 × 10−27 kg to find

⎛ 1.660559 × 10−27 kg ⎞ 15.999 u = 15.999 u ⎜ = 2.6567 × 10−26 kg ⎟ 1u ⎝ ⎠ (b) Use the conversion factor 1 u = 931.494 MeV/c2 to find

⎛ 931.494 MeV/c 2 ⎞ 15.999 u = 15.999 u ⎜ = 1.4903 × 10 4 MeV/c 2 ⎟ 1u ⎝ ⎠ 29.3

The average nuclear radii are r = r0A1/3, where r0 = 1.2 × 10−15 m = 1.2 fm and A is the mass number. (a) For !21 H

r = (1.2 fm)(2)1/3 = 1.5 fm

60 Co , (b) For !27

r = (1.2 fm)(60)1/3 = 4.7 fm

Topic 29

1626

r = (1.2 fm)(197)1/3 = 7.0 fm

(d) For ! 94 Pu , r = (1.2 fm)(239)1/3 = 7.4 fm 239

29.4

(a) For !42 He

r = r0A1/3 = (1.2 × 10−15 m)(4)1/3 = 1.9 × 10−15 m = 1.9 fm

(b) For ! 92 U

r = r0A1/3 = (1.2 × 10−15 m)(238)1/3 = 7.4 × 10−15 m = 7.4 fm

238

29.5

From ME = ρnuclearV = ρnuclear (4πr3/3), we find ⎛ 3ME ⎞ r=⎜ ⎝ 4πρ nuclear ⎟⎠ !

29.6

1/3

⎡ 3 ( 5.98 × 1024 kg ) ⎤ =⎢ 17 3 ⎥ ⎢⎣ 4π ( 2.3 × 10 kg/m ) ⎥⎦

1/3

= 1.8 × 102 m

(a) r ≈ r0A1/3 = (1.2 fm)(65)1/3 = 4.8 fm 3 4 3 4 −15 −43 3 (b) V = π r ≈ π ( 4.8 × 10 m ) = 4.6 × 10 m 3 3 !

m 65 u 65 (1.66 × 10 kg ) = = 2.3 × 1017 kg m 3 (c) ρ = ≈ −43 3 V V 4.6 × 10 m ! −27

29.7

2 9 2 2 ⎡ −19 ⎤ ke q1q2 ( 8.99 × 10 N ⋅ m C ) ⎣( 2 ) ( 6 ) (1.60 × 10 C ) ⎦ (a) Fmax = 2 = = 27.6 N 2 −14 rmin 1.00 × 10 m ( ) !

Fmax 27.6 N = = 4.16 × 1027 m s 2 (b) amax = −27 mα 6.64 × 10 kg !

Topic 29

1627

(c)

PEmax =

ke q1q2

(

rmin

) ( )( )(

)

2⎤ ⎡ 8.99 × 109 N ⋅ m 2 C2 ⎢ 2 6 1.60 × 10−19 C ⎥ ⎣ ⎦ ⎛ 1 MeV ⎞ = ⎜ −13 ⎟ 1.00 × 10−14 m ⎝ 1.60 × 10 J ⎠

yielding PEmax = 1.73 MeV. 29.8

(a) From conservation of energy, ΔKE = − ΔPE, or !12 mv = q(ΔV ) Also, 2

the centripetal acceleration is supplied by the magnetic force, so mv2/r = qvB or v = qBr/m. The energy equation then yields r = 2m(ΔV ) qB2 . Applying this to the two isotopes of carbon in this !

case gives

r12 = C

(

)(

)

2 ⎡12 1.66 × 10−27 kg ⎤ 1 000 V ⎣ ⎦ = 7.89 × 10−2 m = 7.89 cm 2 1.60 × 10−19 C 0.200 T

(

)(

)

and

r13 = C

(

)(

)

2 ⎡13 1.66 × 10−27 kg ⎤ 1 000 V ⎣ ⎦ = 8.21 × 10−2 m = 8.21 cm 2 1.60 × 10−19 C 0.200 T

(

)(

)

(b) From above, we see that the radii of the paths followed by singly 2 ionized atoms of the two isotopes should be r12C = 2m12C ΔV eB and

r = 2m13C ΔV eB2 Thus, the ratio of these radii will be ! 13C © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 29

1628

r12C = 2m12C ΔV eB2 r = 2m13C ( ΔV ) eB ! 13C

= 2

m12C m13C

12 u = 0.961 13 u

and the ratio of our computed radii is

r r = 7.89 cm/8.21 cm = 0.961 , so they do agree. ! 12C 13C 29.9

(a) At the point of closest approach, PEf = KEi , so k 2e 79e r = mα vi2 2 or ! e ( ) ( ) min

vi = !

2ke (2e)(79e) mα rmin

(

)(

316 8.99 × 109 N ⋅ m 2 C2 1.60 × 10−19 C

(6.64 × 10 kg )(3.2 × 10 m) −27

)

−14

= 3.42 × 107 m/s = 1.9 × 107 m/s

(b)

2 9 2 2 ⎡ −19 ⎤ ke q1q2 ( 8.99 × 10 N ⋅ m C ) ⎣( 2 ) (79 ) (1.60 × 10 C ) ⎦ KEi = PE f = = rmin 3.2 × 10−14 m

29.10

⎛ 1 MeV ⎞ = 1.14 × 10−12 J ⎜ = 7.1 MeV −13 ⎝ 1.60 × 10 J ⎟⎠

If a star with a mass of two solar masses collapsed into a gigantic nucleus by converting all of its mass into neutrons, the total number of nucleons (all neutrons), and hence the atomic number, would be

Topic 29

1629

m mn

2mSun mn

(

2 1.99 × 1030 kg 1.67 × 10

−27

) = 2.38 × 10

and its approximate radius would be r ≈ r0A1/3 = (1.2 × 10−15 m)(2.38 × 1057)1/3 = 1.6 × 104 m = 16 km 29.11

(a) The total binding energy for !12 Mg is 24

(

)

Eb = ( Δm) c 2 = 12m1H + 12mn − m24Mg c 2 , and the average binding ! energy per nucleon is

(

) (

)

(

⎡12 1.007 825 u + 12 1.008 665 u − 23.985 042 u ⎤ 931.5 MeV/u ⎦ =⎣ A 24 = 8.26 MeV/nucleon

(

)

(b) For !37 Rb , Eb != ( Δm) c = 37m1H + 48mn − m85Rb c , yielding ! 85

(

)

(

)

(

⎡ 37 1.007 825 u + 48 1.008 665 u − 84.911 789 u ⎤ 931.5 MeV/u ⎦ =⎣ A 85 = 8.70 MeV/nucleon

29.12

)

(a) For !21 H , Δm = 1(1.007 825 u) + 1(1.008 665 u) − (2.014 102 u) = 0.02 388 u, and

Eb ( Δm) c 2 ( 0.002 388 u ) ( 931.5 MeV/u ) = = = 1.11 MeV/nucleon A 2 !A (b) For !42 He , Δm = 2(1.007 825 u) + 2(1.008 665 u) − (4.002 603 u) = © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 29

1630

0.030 377 u, and

Eb A

(Δm)c 2 A

(0.030 377 u)(931.5 MeV/u) 4

= 7.07 MeV/nucleon

0.528 458 u, and Eb (Δm)c 2 (0.528 458 u)(931.5 MeV/u) = = = 8.79 MeV/nucleon A 56 !A 238

(d) For 92 U , Δm = 92(1.007 825 u) + 146(1.008 665 u) − (238.050 783) = 1.934 207 u, and Eb (Δm)c 2 (1.934 207 u)(931.5 MeV/u) = = = 7.57 MeV/nucleon A 238 !A

29.13

For ! 8 O , Δm = 8(1.007 825 u) + 7(1.008 665 u) − (15.003 065) = 0.120 190 u, 15

and E = (Δm)c2 = (0.120 190 u)(931.5 MeV/u) = 112.0 MeV ! b 15O

For !157 N , Δm = 7(1.007 825 u) + 8(1.008 665 u) − (15.000 109) = 0.123 986 u, and E = (Δm)c2 = (0.123 986 u)(931.5 MeV/u) = 115.5 MeV ! b 15N

Therefore,

ΔEb = Eb |15N −Eb |15O = 3.5 MeV

Topic 29

29.14

1631

Δm = Zm1 H + (A − Z)mn − m and Eb/A = Δm(931.5 MeV/u)/A 1 !

Nucleus

(A − Z)

m(in u)

Δm (in u)

Eb/A (in MeV)

54.938 050

0.517 525

8.765

55.934 942

0.528 458

8.790

58.933 200

0.555 355

8.768

!25 Mn !26 Mn

!27 Co

Therefore, !26 Fe has a greater binding energy per nucleon than its 56

neighbors. This gives us finer detail than is shown in Figure 29.4. 29.15

(a) For !23 11 Na Δm = 11(1.007 825 u) + 12(1.008 665 u) − (22.989 770 u) = 0.200 285 u, and for !12 Mg Δm = 12(1.007 825 u) + 11(1.008 665 u) − 23

(22.994 127 u) = 0.195 088 u. The difference in the binding energy per nucleon for these two isobars is then 2 ΔEb ⎡⎣( Δm) 2311 Na − ( Δm) 2312 Mg ⎤⎦ c [ 0.200 285 u − 0.195 088 u ]( 931.5 MeV/u ) = = A A 23 = 0.210 MeV/nucleon !

(b) The binding energy per nucleon is greater by 0.210 MeV/nucleon in

!11 Na This is attributable to less proton repulsion in !11 Na than in 23

!12 Mg . 23

Topic 29

29.16

1632

The sum of the mass of !20 Ca plus the mass of a neutron exceeds the mass 42

of !20 Ca . This difference in mass must represent the mass equivalence of 43

the energy spent removing the last neutron from !20 Ca to produce !20 Ca 43

plus a free neutron. Thus,

(

)

E = m 42 Ca + mn − m 43 Ca c 2 = ( 41.958 622 u + 1.008 665 u − 42.958 770 u) c 2 20 20 ! or E = (0.008 517 u)(931.5 MeV/u) = 7.93 MeV 29.17

The mass of radon present at time t is equal to m = matom = N = matom N0e−λt = m0 e−λt, where matom is the mass of a single radon atom, N is the number or radon nuclei (and hence, atoms) present, and N0 is the number present at time t = 0, making m0 the mass of radon present at t = 0. The decay constant for radon is λ = ln 2/T1/2 = ln2/(3.83 d), yielding m = m0 e − λt = m0 e

29.18

−(t/T1/2 )ln2

= ( 3.00 g ) e −(1.50 d/3.83 d)ln2 = 2.29 g

The activity is R = λN = λN0e−λt = R0e−λt, where R0 is the activity at time t = 0, and the decay constant is λ = ln 2/T1/2 Thus,

R = R0e −(t/T1/2 )ln2 = (1.1 × 10 4 Bq)e −(2.0 h/6.05 h)ln 2 = 8.7 × 103 Bq

Topic 29

29.19

1633

⎛ 8.64 × 10 4 s ⎞ 5 T = 8.04 d (a) 1/ 2 ⎜⎝ ⎟⎠ = 6.95 × 10 s 1d !

ln 2 ln 2 = = 9.97 × 10−7 s −1 (b) λ = 5 T1/2 6.95 × 10 s ! ⎛ 3.7 × 1010 Bq ⎞ = 1.9 × 10 4 Bq (c) R = 0.500 µCi = ( 0.500 × 10 Ci ) ⎜ ⎟ 1 Ci ⎝ ⎠ ! −6

(d) From R = λN, the number of radioactive nuclei in a 0.500 µCi of 131I is

R 1.9 × 10 4 S −1 N= = = 1.9 × 1010 nuclei −7 −1 λ 9.97 × 10 S ! (e) The number of half-lives that have elapsed is n = t/T1/2 = 40.2 d/8.04 d = 5.00, so the remaining activity of the sample is R R0 6.40 mCi R = n0 = 5.00 = = 0.200 mCi 2 2 32.0 !

29.20

(a) From Equation (29.4a) in the textbook, the fraction remaining at t = 5.00 yr will be

N N0

= e − λt = e

(b) At t = 10.0 yr,

(

) = e −(5.00 yr/12.33 yr) ln 2 = 0.755

− t T1/2 ln 2

N = e − λt = e −(t/T1/2 )ln 2 = e −(10.0 yr/12.33 yr)ln 2 = 0.570 . N0

N = e − λt = e −(t/T1/2 )ln 2 = e −(123.3 yr/12.33 yr)ln 2 = 9.77 × 10−4 . N0

Topic 29

1634

From R = λN = λN0e−λt = R0e−λt, with R = (0.842) R0, we find e−λt = R/R0, and

λt = −ln(R/R0). Since the half-life may be expressed as T1/2 = ln 2/λ, this yields

( 2.00 d) ln 2 = 8.06 d t ln 2 T1/2 = =− ln ( R/R0 ) ln(0.842) ! 29.22

Using R = λN = λN0e−λt, with R/R0 = 0.125, gives λt = − ln(R/R0), or

( t=−

ln R/R0

29.23

) = −T ⎡⎢ ln (R/R ) ⎤⎥ = −(5 730 yr) ⎡⎢ ln(0.125) ⎤⎥ = 1.72 × 10 yr 0

1/2

⎢ ⎣

ln 2

⎥ ⎦

⎢⎣

ln 2

⎥⎦

(a) The initial activity is R0 = 10.0 mCi, and at t = 4.00 h, R = 8.00 mCi. Then, from R = λN = λNe−λt = R0e−λt, the decay constant is

λ=− !

ln ( R / R0 ) ln(0.800) =− = 5.58 × 10−2 h −1 t 4.00 h

ln 2 ln 2 = = 12.4 h and the half-life is T1/2 = −2 −1 λ 5.58 × 10 h !

Topic 29

1635 −3 10 −1 R0 (10.0 × 10 Ci ) ( 3.70 × 10 s 1 Ci ) N = = = 2.39 × 1013 nuclei (b) 0 −2 −1 λ ( 5.58 × 10 h )(1 h 3 600 s ) !

R = R0e − λt = (10.0! mCi ) e −(5.58×10 h )(30!h) = 1.9 mCi ! −2

−1

The number of !38 Sr nuclei initially present is 90

total!mass 5.0 kg N0 = = = 3.4 × 1025 mass!per!nucleus ( 89.907 7 u ) (1.66 × 10−27 kg/u ) ! The half-life of !38 Sr is T1/2 = 29.1 yr (Appendix B), so the initial activity is 90

3.4 × 1025 ) ln 2 ( N0 ln 2 R0 = λΝ 0 = = = 2.6 × 1016 count/s 7 T1/2 29.1 yr 3.156 × 10 s/yr ( )( ) ! From R = R0e−λt, the time when the activity will be R = 10.0 counts/min is

t=− !

ln ( R/R0 ) ln ( R/R0 ) = − (T1/2 ) λ ln 2

⎡ 10.0 min −1 ⎛ 1 min ⎞ ⎤ ln ⎢ ⎥ 2.6 × 1016 s −1 ⎜⎝ 60 s ⎟⎠ ⎦ ⎣ = − ( 29.1 yr ) = 1.7 × 103 yr ln 2 ! 29.25

(a) First, use the half-life to calculate the decay constant, λ:

T1/2 = 27.7 days = 2.39 × 106 s T1/2 =

ln 2 = 2.39 × 106 s → λ = 2.90 × 10−7 s−1 λ

A radioactive sample’s activity is R = λN where N is the number of radioactive nuclei. Solve for N to find (note the conversion to the SI unit of becquerels along the way) © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 29

1636

⎛ 3.7 × 1010 decays/s ⎞ 2.00 × 10−6 Ci ⎜ ⎟⎠ 7.40 × 10 4 Bq 1 Ci ⎝ R N= = = = 2.55 × 1011 −7 −1 −7 −1 λ 2.90 × 10 s 2.90 × 10 s (b) Because the activity is proportional the number of radioactive nuclei n

⎛ 1⎞ N, and because N = N 0 ⎜ ⎟ , where n is the number of half-lives, the ⎝ 2⎠ activity after 1 year (365.24 days, or n = 365.24/27.7 half-lives) is n

365.24 days

⎛ 1⎞ ⎛ 1 ⎞ 27.7 days R = R0 ⎜ ⎟ = 7.40 × 10 4 Bq ⎜ ⎟ = 7.94 Bq ⎝ 2⎠ ⎝ 2⎠ 29.26

(

)

The half-lives of 137 Cs and 134 Cs are T1/2,137 = 1.10 × 104 days and T1/2,134 = 734 days, respectively. From λ =

0.693 , the respective decay constants T1/2

are λ137 = 6.30 × 10−5 (days)−1 = 7.29 × 10−10 (s)−1 and λ134 = 9.44 × 10−4 (days)−1 = 1.09 × 10−8 (s)−1. The decay constant is related to the sample’s activity R by R = λ N = λ N 0 e − λt where N is the number of radioactive nuclei in the

sample. In a 1 m3 seawater sample obtained 1 730 days after the meltdown, the combined activity was 11.0 Bq so that

11.0 Bq = R137 + R134 = λ137 N 0 e − λ137t + λ134 N 0 e − λ134t where N0 is the number of nuclei (of each type) in the sample when it was first released and t = 1 730 days. Solve for N0 to find

Topic 29

1637

(

)

11.0 Bq = λ137 e − λ137t + λ134 e − λ134t N 0 N0 =

(7.29 × 10

−10

) (

s e

− 6.30×10

11.0 Bq

−5

(days ) )(1 730 days ) −1

) (

(

+ 1.09 × 10−8 s e

− 9.44×10−4 ( days )

−1

)(1 730 days)

N 0 = 3.95 × 109 nuclei Having found the initial number of radioactive nuclei in a 1 m3 seawater sample, a straightforward decay calculation gives the number of each isotope 1 730 days later:

) (

(

N137 = N 0 e − λt = 3.95 × 109 nuclei e

(a)

− 6.30×10−5 ( days )

−1

)(1 730 days)

= 3.54 × 109 nuclei N134 = N 0 e

(b)

− λt

( = 3.95 × 10 nuclei e

(

)

− 9.44×10−4 ( days )

−1

)(1 730 days)

= 7.72 × 108 nuclei 29.27

We identify the missing nuclide in each case by requiring that both the total mass number and the total charge number be the same on the two sides of the decay equation. (a)

(b)

212 83

95 36

Bi → 20881Tl + 42 He

95 Kr → 37 Rb + −10 e + ν

144

! 60

Nd → 42 He + 140 58 Ce

We complete the decay formula in each case by requiring that both the total mass number and the total charge number be the same on the two

Topic 29

1638

sides of the equation. (a)

(b)

12 5

B → 126 C + −10 e + ν

234

! 90 14

! 6

4 Th → 230 88 Ra + 2 He

C → 147 N + −10 e + ν

56 Co decays into the less massive !26 Fe . To conserve The more massive !27 56

charge, the charge of the emitted particle must be +1e. Since the parent and the daughter have the same mass number, the emitted particle must have essentially zero mass. Thus, the decay must be positron emission or e+ decay. The decay equation is 29.30

56 27

0 Co → 56 26 Fe + +1 e + ν e .

The energy released in the decay ! 92 U → 2 He + 90Th is 238

( )

(

)

234

Q = Δm c 2 = ⎡ m238 − m4 + m234 ⎤ c 2 ⎥ U He Th ⎦ ⎣⎢ = ⎡ 238.050 783 u − 4.002 603 u + 234.043 583 u ⎤ 931.5 MeV/u ⎣ ⎦ = 4.28 MeV 29.31

(

)(

)

The Q-value of a decay, Q = (Δm)c2, is the amount of energy released in the decay. Here, Δm is the difference between the mass of the original nucleus and the total mass of the decay products. If Q > 0, the decay may occur spontaneously.

Topic 29

1639

(a) For the decay !20 Ca → e + 19 K the masses of the electrons do not 40

automatically cancel. Thus, we add 20 electrons to each side of the decay to yield neutral atoms and obtain !( 20

− − Ca → 20e − ) → e + + ( 40 19 K + 19e ) + e

Caatom → e+ + 40Katom + e−

Then,

(

)

Q = m 40 Ca − m 40 K − 2me c 2 = ⎡⎣ 39.962 591 u − 39.963 999 u − 2 ( 0.000 549 u ) ⎤⎦ c 2 20 atom 20 atom !

Q = (− 0.002 506 u)c2 < 0

so the decay cannot occur

spontaneously. (b) In the decay !60 Nd → 2 He + 58 Ce , we may add 60 electrons to each 144

140

side, forming all neutral atoms, and use masses from Appendix B to find

(

)

Q = m144 Nd − m 4 He − m140 Ce c 2 = (143.910 083 u − 4.002 603u − 139.905 434 u ) c 2 60 2 52 ! or 29.32

(a)

66 28

Q = (+ 0.002 046 u)c2 > 0 so the decay can occur spontaneously. 0 Ni → 66 29 Cu + −1 e + ν e

(b) Because of the mass differences, neglect the kinetic energy of the

Topic 29

1640

recoiling daughter nucleus in comparison to that of the other decay products. Then, the maximum kinetic energy of the beta particle occurs when the antineutrino is given zero energy. That maximum is

(

)

KEmax = m66Ni − m66Cu c 2 = ( 65.929 1 u − 65.928 9 u) ( 931.5 MeV/u ) = 0.186 MeV = 186 keV

! 29.33

In the decay !31 H → 32 He + −10 e + ν e , the antineutrino is massless. Adding 1 electron to each side of the decay gives 3 1

( H + e ) → ( H + 2e ) + ν or 3 1

−

3 2

−

H atom → 32 H atom + ν e . Therefore, using neutral atomic masses from

Appendix B, the energy released is

(

( )

E = Δm c 2 = m3 − m3 H

) c = (3.016 049 u − 3.016 029 u) − (931.5 MeV/u) 2

= 0.018 6 MeV = 18.6 keV 29.34

(a) Balancing the mass numbers on each side of the reaction gives 234 = A + 4 → A = 230

(b) Balancing atomic numbers gives 90 = Z + 2 → Z= 88

29.35

From R = λN = λN0e−λt = R0e−λt, and T1/2 = 5 730 yr for 14C (Appendix B), the age of the sample is

t=−

(

ln R R0

) = −T ln (R R ) = − (5 730 yr) ln (0.600) = 4.22 × 10 yr 0

1/2

ln 2

Topic 29

29.36

1641

The energy released in the reaction is given by

( )

Q = Δm c 2 = ⎛ m1 + m 27 − m 27 − mn ⎞ c 2 Al Si ⎝ 1H ⎠ 13 14 = ⎡⎣1.007 825 u + 26.981 539 u − 26.986 721 u − 1.008 665 u ⎤⎦ 931.5 MeV/u = − 5.61 MeV

(

)

If we require total energy to be conserved and ignore the kinetic energy of the recoiling product nucleus, the kinetic energy of the emerging neutron must be KEf = KEi + Q = 6.61 MeV − 5.61 MeV = 1.00 MeV 29.37

We identify the missing particles by requiring that both the total mass number and the total charge number be the same on the two sides of the equation and by remembering that some form of neutrino always accompanies the emission of a beta particle or positron. (a)

(b)

! 10 235 92

1 Ne + 42 He → 24 12 Mg + 0 n

144 1 U + 01 n → 90 38 Sr + 54 Xe + 2 0 n

2 11 H → 21 H + +10 e + ν e !

(a) mi = m11 H + m73 Li = 1.007 825 u + 7.016 004 u = 8.023 829 u ! (b) m f = m74 Be + mn = 7.016 929 u + 1.008 665 u = 8.025 594 u !

Topic 29

1642

(c) Q = (mi − mf)c2 = [8.023 829 u − 8.025 594 u](931.5 MeV/u) = −1.64 MeV (d) mpv = (mBe + mn)V (e) From conservation of energy, the total kinetic energy before the reaction, plus the energy released during the reaction, must equal the total kinetic energy after the reaction. That is, KEp + Q = KEBe + KEn, or

(f)

1 2

mp v 2 = 12 ( mBe + mn ) V 2 − Q .

⎛ mp ⎞ ⎛ m⎞ 1.007 825 u ⎞ ⎛ KEmin = ⎜ 1 + ⎟ Q = ⎜ 1 + Q = ⎜1+ −1.64 MeV ⎟ ⎝ M⎠ m7 ⎠ 7.016 004 u ⎟⎠ ⎝ ⎝ 3 Li

= 1.88 MeV

29.39

We determine the product nucleus by requiring that both the total mass number and the total charge number be the same on the two sides of the reaction equation. The completed reaction equations are given below: (a)

29.40

10 5

B + 42 He → 11 H + 136 C

(b)

13 6

C + 11 H → 42 H + 105 B

(a) For the first reaction: !n + 21 H → 31 H

(

Q1 = ⎛ mn + m 2 − m 3 ⎞ c 2 = ⎡⎣1.008 665 u + 2.014 102 u − 3.016 049 u ⎤⎦ 931.5 MeV/u H H⎠ ⎝ 1 1

)

= 6.258 MeV > 0 ⇒ exothermic For the second reaction:

!1 H + 1 H → 2 He 1

Topic 29

1643

Q2 = ⎛ m1 + m 2 − m 3 ⎞ c 2 H He ⎠ ⎝ 1H 1 2 = ⎡⎣1.007 825 u + 2.014 102 u − 3.016 029 u ⎤⎦ 931.5 MeV/u

(

= 5.494 MeV > 0 ⇒

)

exothermic

(b) Since Q1 > Q2 the first reaction released more energy. One reason for this is that the product nucleus in the second reaction contains 2 protons. Some energy had to be left stored as electrical potential energy of this system, leaving less energy to be released as kinetic energy of the product nucleus. (c) Assuming the electrical potential energy of the 2 protons in !32 He fully accounts for the difference in the Q-values of the two reactions, we have ΔQ = PEe = kee2/r, where r is the distance separating the 2 protons. Thus, 9 2 2 −19 k e 2 ( 8.99 × 10 N ⋅ m C ) (1.60 × 10 C ) ⎛ 1 MeV ⎞ r= e = = 1.88 × 10−15 m −13 ⎟ ⎜ ΔQ ( 6.258 − 5.494) MeV ⎝ 1.60 × 10 J ⎠ ! 2

29.41

(a) Requiring that both charge and the number of nucleons (atomic mass number) be conserved, the reaction is found to be 197

! 79

0 Au + 01 n → 198 80 Hg + −1 e + ν e . Note that the antineutrino has been

included to conserve electron-lepton number, which will be discussed in the next chapter. © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 29

1644

(b) We add 79 electrons to both sides of the reaction equation given above to produce neutral atoms so we may use mass values from Appendix B. This gives ! 79 Au atom + 0 n → 80 Hg atom + ν e , and, 197

198

remembering that the antineutrino is massless, the Q-value is found to be

( )

Q = Δm c 2

(

)(

= ⎛ m197 + mn − m198 ⎞ c 2 196.966 552 u + 1.008 665 u − 197.966 750 u 931.5 MeV/u Hg ⎠ ⎝ 79 Au 80 = 7.89 MeV

The kinetic energy carried away by the daughter nucleus is negligible. Thus, the energy released may be split in any manner between the electron and antineutrino, with the maximum kinetic energy of the electron being 7.89 MeV. 29.42

We complete the reaction equation in each case by requiring that both the total mass number and the total charge number be the same on the two sides of the equation. (a)

29.43

He + 147 N → 11 H + 178 O (b)

7 3

Li + 11 H → 24 He + 24 He

(a) Determine the product of the reaction by requiring that both the total mass number and the total charge number be the same on the two

)

Topic 29

1645

sides of the equation. The completed reaction equation is 7 3

Li + 42 He → 105 B + 01 n .

(b)

(

) (

)

Q = Δmc 2 = ⎡ m7 Li + m 4 He − m10 B + m 1 n ⎤ c 2 2 5 0 ⎣ 3 ⎦

= ⎡⎣(7.016 004u + 4.002 603 u ) − (10.012 937 u + 1.008 665 u ) ⎤⎦ ( 931.5 MeV/u )

29.44

= −2.79 MeV

For equal amounts of biological damage, the two doses (in rem units) must be equal, or (heavy ion dose in rad) × RBEheavy ions = (x-ray dose in rad) × RBEx-rays or

ion dose in rad =

29.45

( x-ray dose in rad) × RBE

x-rays

REE heavy ions

(100 rad)(1.0) = 5.0 rad 20

For each rad of radiation, 10−2 J of energy is delivered to each kilogram of absorbing material. Thus, the total energy delivered in this whole body dose to a 75.0-kg person is J/kg ⎞ ⎛ E = ( 25.0 rad ) ⎜ 10−2 ⎟ (75.0 kg ) = 18.8 J ⎝ rad ⎠ !

29.46

(a) Each rad of radiation delivers 10−2 J of energy to each kilogram of

Topic 29

1646

absorbing material. Thus, the energy delivered per unit mass with this dose is E J/kg ⎞ ⎛ = ( 200 rad ) ⎜ 10−2 ⎟⎠ = 2.00 J/kg ⎝ m rad !

(b) From E = Q = mc(ΔT), the expected temperature rise with this dosage is

ΔT =

29.47

E/m c

2.00 J/kg 4 186 J/kg ⋅°C

= 4.78 × 10−4°C

The rate of delivering energy to each kilogram of absorbing material is J/kg ⎛ E/m ⎞ ⎛ −2 J/kg ⎞ ⎜⎝ ⎟⎠ = (10 rad/s ) ⎜⎝ 10 ⎟⎠ = 0.10 rad s ! Δt

The total energy needed per unit mass is

⎛ J ⎞ 5 E/m = c ΔT = ⎜ 4 186 ⎟ 50°C = 2.1 × 10 J/kg kg ⋅°C ⎠ ⎝

( )

(

)

so the required time will be

⎛ ⎞ energy!needed 2.1 × 105 J/kg 1d Δt = = = 2.1 × 106 s ⎜ = 24 d 4 ⎟ delivery!rate 0.10 J/kg ⋅s ⎝ 8.64 × 10 s ⎠ ! 29.48

(a) The number of x-rays taken per year is production = (8 x-ray/d)(5 d/week)(50 weeks/yr) = 2.0 × 103 x-ray/yr

Topic 29

1647

so the exposure per x-ray taken is

5.0 rem/yr exposure exposure!rate = = = 2.5 × 10−3 rem/x.ray 3 production 2.0 × 10 x.ray/yr ! (b) The exposure due to background radiation is 0.13 rem/yr. Thus, the work-related exposure of 5.0 rem/yr is

5.0 rem/yr ≈ 38 times!background!levels 0.13 rem/yr ! 29.49

(a) From N = R / λ = R0 e

− λt

/ λ = (T1/2 R0 / ln2)e −tln2/T1/2 the number of

decays occurring during the 10-day period is

(

)

⎛T R ⎞ −t ln 2 T1/2 ΔN = N0 − N = ⎜ 1/2 0 ⎟ 1 − e ⎜⎝ ln 2 ⎟⎠ ⎡ 14.3 d 1.31 × 106 decay/s ⎛ ⎤ 8.64 × 10 4 s ⎞ ⎛ −(10.0 d ) ln 2/14.3 d ⎞ ⎢ ⎥ = ⎜ ⎟ ⎝1 − e ⎠ ⎢ ⎥ ln 2 1d ⎝ ⎠ ⎣ ⎦

(

)(

)

= 8.97 × 1011 decays, and one electron is emitted per decay (b) The total energy deposited is found to be

⎛ ⎛ 1.60 × 10−16 J ⎞ keV ⎞ 11 E = ⎜ 700 (8.97 × 10 decays ) ⎜⎝ 1 keV ⎟⎠ = 0.100 J decay ⎟⎠ ⎝ ! (c) The total absorbed dose (measured in rad) is given by

Topic 29

1648

dose = ! 29.50

energy!deposited!per!unit!mass ( 0.100 J/0.100 kg ) = = 100 rad ⎛ −2 J/kg ⎞ energy!deposition!per!rad ⎜⎝ 10 ⎟ rad ⎠

(a) The dose (in rem) received in time Δt is given by rad ⎞ ⎤ rem ⎞ ⎡⎛ ⎛ dose = (dose in rad) × RBE = ⎢⎜ 100 × 10−13 ⎟⎠ Δt ⎥ × (1.00) = ⎜⎝ 0.100 ⎟ Δt ⎝ h h ⎠ ⎣ ⎦ !

If this dose is to be 1.0 rem, the required time is

1.0 rem Δt = = 10 h 0.100 rem/h ! (b) Assuming the radiation is emitted uniformly in all directions, the intensity of the radiation is given by I = I0/4πr2 Therefore,

(1.0 m ) Ir I 0 4π r 2 = 2 = I r2 I 0 4π (1.0 m ) !1 and

29.51

I 100 mrad/h r = (1.0 m ) 1 = (1.0 m ) = 3.2 m Ir 10 mrad/h !

(a) The mass of a ! 6 C atom is 11

matom = (11.011 u)(1.661 × 10−27 kg/u) = 1.829 × 10−26 kg = 1.829 × 10–23 g and the mass of 1 mole of ! 6 C is 11

M = matom = NA = (1.829 × 10−23 g/atom)(6.022 × 1023 atoms/mol) = 11.01 g/mol

Topic 29

1649

The number of moles in a 3.50 µg sample is then

n= !

msample M

3.50 × 10−6 g = 3.18 × 10−7 mol 11.01 g/mol

(b) The number of nuclei in the original sample is N0 = nNA = (3.18 × 10−7 mol)(6.022 × 1023 atoms/mol) = 1.91 × 1017 atoms or alternatively,

3.50 × 10−6 g N0 = = = 1.91 × 1017 atoms . −23 matom 1.829 × 10 g/atom ! msample

(

)(

)

= 1.08 × 1014 Bq

(d) The activity after an elapsed time of t = 8.00 h = 480 min will be

R = R0e − λt = R0e −tln2/T1/2 = (1.08 × 1014 Bq)e −( 480 min)ln2/(20.4 min) = 8.92 × 106 Bq 29.52

We must first calculate the Q value, given by

(

) (

)

Q = ( Δm) c 2 = ⎡ m 1 n + m 4 He − m 2 H + m 3 H ⎤ c 2 . 2 1 1 ⎣ 0 ⎦ ! Q = [(1.008 665 u + 4.002 603 u) − (2.014 102 u + 3.016 049 u)](931.5 MeV/u)

Topic 29

1650

= −17.6 MeV The threshold kinetic energy for the incident neutron is then

⎛ m1 ⎞ ⎛ 1.008 665 u ⎞ n KEmin = ⎜ 1 + 0 ⎟ Q = ⎜ 1 + ⎟ 17.6 MeV = 22.0 MeV ⎜ M4 ⎟ 4.002 603 u ⎠ ⎝ He ⎠ ⎝ 2

(

29.53

From R = R0e−λt, the elapsed time is

t=− ! 29.54

)

ln ( R/R0 ) ln ( R/R0 ) ln ( 20.0 mCi 200 mCi ) = −T1/2 = − (14.0 d ) = 46.5 d λ ln 2 ln 2

To compute the Q value for the reaction ! 6 C → 7 N + e + ν , we add 6 14

−

electrons to each side of the equation to obtain C + 6e ) → ( N + 7e ) + ν , or C → N ! !( 14 6

−

14 7

−

14 6

14 7

atom

+ ν . Now, we use the atomic

masses from Appendix B of the textbook. Since the neutrino is a massless

(

)

particle, Q = m146 Catom − m147 Natom c , giving ! 2

Q = (14.003 242 u − 14.003 074 u)(931.5 MeV/u) = 0.156 MeV = 156 keV 29.55

(a) If we assume all the 87Sr nuclei in a gram of material came from the decay of 87Rb nuclei, the original number of 87Rb nuclei was N0 = 1.82 × 1010 + 1.07 × 109 = 1.93 × 1010. Then, from N = N0e−λt the elapsed time is

Topic 29

1651

× 10 ⎞ ( 4.8 × 10 yr ) ln ⎛⎜⎝ 1.82 1.93 × 10 ⎟⎠ 10

t=− !

ln ( N N0 ) T ln ( N N0 ) = − 1/2 =− λ ln 2

ln 2

= 4.1 × 109 yr

(b) It could be no older. It could be younger if some 87Sr were initially present. (c) We have assumed that all the 87Sr present came from the decay of 87

29.56

Rb.

From R = λN = N0e−λt, if we have an activity of R after time t has passed, the number of unstable nuclei that must have been present initially is given by N0 = Reλt/λ. With R = 10 Ci = 10 Ci = 10(3.7 × 1010 Bq) = 3.7 × 1011 decay/s, λ = ln 2/T1/2 = ln 2/5.2 yr, and t = 30 months = 2.5 yr, this yields

(

N0 = 3.7 × 10 s

−1

)(

)

⎡ 5.2 yr 3.156 × 107 s/l yr ⎤ ⎢ ⎥ e ( ln2 5.2 yr )( 2.5 yr ) = 1.2 × 1020 ⎢ ⎥ ln 2 ⎣ ⎦

Thus, the initial mass of 60C required is found to be (using Appendix B from the text) m = N0matom = (1.2 × 1020)(59.933 822 u) = 7.2 × 1021 u

⎛ 1.66 × 10−27 kg ⎞ ⎛ 106 mg ⎞ or m = 7.2 × 10 u ⎜ ⎟⎠ ⎜⎝ 1 kg ⎟⎠ = 12 mg 1u ⎝ ! 21

29.57

The total activity of the working solution at t = 0 was

Topic 29

1652

(R0)total = (2.5 mCi/mL)(10 mL) = 25 mCi Therefore, the initial activity of the 5.0-mL sample, drawn from the 250mL working solution, was

⎛ 5.0 mL ⎞

⎝

( R0 )sample = ( R0 )total ⎜ 250 mL ⎟ = ( 25 mCi ) ⎜ 250 mL ⎟ = 0.50 mCi = 5.0 × 10−4 Ci

⎠

With a half-life of 14.96 h for 24Na (Appendix B), the activity of the sample after 48 h is R = R0 e − λt = R0 e −t ln 2 T1/2 = ( 5.0 × 10−4 Ci ) e (

− 48 h ) ln 2/(14.96 h)

29.58

= 5.4 × 10−5 Ci = 54 µCi

(a) The mass of a single 40 K atom is m = (39.964 u)(1.66 × 10−27 kg/u) = 6.63 × 10−26 kg = 6.63 × 10−23 g Therefore, the number of 40K nuclei in a liter of milk is

total mass of 40 K present mass per atom

(2.00 g/L)(0.011 7 / 100) = 3.53 × 10 L 18

6.63 × 10

−23

and the activity due to potassium is

( 3.53 × 1018 /L ) ln 2 N ln 2 R = λΝ = = = 60.6 Bq/L T1/2 1.28 × 109 yr ) ( 3.156 × 107 s/yr ) ( ! (b) Using R = R0e−λt, the time required for the 131I activity to decrease to

Topic 29

1653

the level of the potassium is given by

t=−

29.59

(

ln R R0

) = − T ln (R R ) = − (8.04 d) ln (60.6/2 000) = 40.6 d 1/2

ln 2

The decay constant for 235U is λ235 = ln2/T1/2 = ln 2/0.70 × 109 yr = 9.9 × 10−10 yr−1, while that for 238U is λ238 = ln2/4.47 × 109 yr = 1.55 × 10−10 yr−1, Assuming there were N0 nuclei of each isotope present initially, the number of each type still present should be N 235 = N 0 e

− λ235t

and

N 238 = N 0 e − λ238t . With the currently observed ratio of 235U to 238U being

0.007, we have !N235 N238 = e ( 235 − λ

− λ238 )t

= 0.007 , or our estimate of the

elapsed time since the release of the elements forming our Earth is

ln(0.007) ln(0.007) t=− =− = 5.9 × 109 yr −10 −1 −10 −1 λ235 − λ236 9.9 × 10 yr − 1.55 × 10 yr ! 29.60

(a) Obtaining the mass of a single ! 94 Pu atom from the table of 239

Appendix B in the text, we find

msample 1.0 kg N0 = = = 2.5 × 1024 matom ( 239.052 156 u ) (1.66 × 10−27 kg/u ) ! (b) The initial activity of this sample is

Topic 29

1654

R0 = λ N 0

⎛ ln 2 ⎞ =⎜ ⎟N ⎜⎝ T ⎟⎠ 0 1/2 ⎡ ⎤ ln 2 ⎢ ⎥ 2.5 × 1024 = 7 ⎢ 24 000 yr 3.156 × 10 s/yr ⎥ ⎣ ⎦

)(

(

)

= 2.3 × 1012 Bq (c) From R = R0e−λt, the time that must elapse before this sample will have a “safe” activity level of 0.10 Bq is

t=− =− =−

(

ln R R0

)

λ T1/2 ln R R0

(

)

ln 2 24 000 yr ln 0.10 2.3 × 1012

) (

ln 2

)

= 1.1 × 10 yr n

29.61

⎛ 1⎞ Here we must solve for n using N = N 0 ⎜ ⎟ . ⎝ 2⎠ n

⎛ 1⎞ (a) If N/N0 = 10.0% = 0.100, then ⎜ ⎟ = 0.100 or 2n = 10.0 . Solving for n ⎝ 2⎠ would be easy if calculators had a button for the base-2 logarithm. Because this function isn’t typically available, the problem can be most easily solved by using a property of logarithms:

log a ( b ) =

logc ( b ) so that logc ( a)

Topic 29

1655

( )

n = log 2 2n = log 2 (10.0 ) =

log10 (10.0 ) = 3.32 log10 ( 2 ) n

⎛ 1⎞ (b) If N/N0 = 5.00% = 0.050 0, then ⎜ ⎟ = 0.050 0 or 2 n = 20.0 . Using the ⎝ 2⎠ same property of logarithms,

( )

n = log 2 2n = log 2 ( 20.0 ) =

log10 ( 20.0 ) = 4.32 log10 ( 2 ) n

⎛ 1⎞ (c) If N/N0 = 1.00% = 0.010 0, then ⎜ ⎟ = 0.010 0 or 2 n = 100. Again ⎝ 2⎠ using the same property of logarithms,

( )

n = log 2 2n = log 2 (100.) =

29.62

log10 (100.) = 6.64 log10 ( 2 )

The initial activity of the 1.00-kg carbon sample would have been

⎛ 15.0 counts/min ⎞ R0 = (1.00 × 103 g ) ⎜ = 1.50 × 10 4 min −1 ⎟ 1.00 g ⎝ ⎠ ! From R = λN = λN0e−λt = R0e−λt, and T1/2 = 5 730 yr for 14C (Appendix B), the age of the sample is

t=− !

ln ( R R0 ) ln ( R R0 ) = −T1/2 λ ln 2

(

)(

)

ln ⎡ 5.00 × 102 min −1 1.50 × 10 4 min −1 ⎤ ⎣ ⎦ = 2.81 × 10 4 yr = − 5 730 yr ln 2

(

)

Topic 30

1656

Topic 30 Nuclear Energy and Elementary Particles

QUICK QUIZZES 30.1

Choice (a). This reaction fails to conserve charge and also fails to conserve baryon number. For each of these reasons, it cannot occur.

30.2

Choice (b). This reaction fails to conserve charge and cannot occur.

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 30.2

Choice (b). The electron and the positron have, as a system, approximately zero momentum. Conservation of momentum therefore requires zero momentum for the system of reaction products. Because photons carry momentum, the reaction product must involve pairs of equal-energy photons traveling in opposite directions.

30.4

Notice in the fusion reactions discussed in the text that the most commonly formed by-product of the reactions is helium, which is inert and not radioactive.

30.6

They are hadrons. Such particles decay into other strongly interacting particles such as p, n, and π with very short lifetimes. In fact, they

Topic 30

1657

decay so quickly that they cannot be detected directly. Decays which occur via the weak force have lifetimes of 10−13 s or longer; particles that decay via the electromagnetic force have lifetimes in the range of 10−16 s to 10−19 s. 30.8

Each flavor of quark can have three colors, designated as red, green, and blue. Antiquarks are colored antired, antigreen, and antiblue. Baryons consist of three quarks, each having a different color. Mesons consist of a quark of one color and an antiquark with a corresponding anticolor. Thus, baryons and mesons are colorless or white.

30.10

The decays of the neutral pion, eta, and neutral sigma occur by the electromagnetic interaction. These are the three shortest lifetimes in the table. All produce photons, which are the quanta of the electromagnetic force, and all conserve strangeness.

30.12

A neutron inside a nucleus is stable because it is in a lower energy state than a free neutron and lower in energy than it would be if it decayed into a proton (plus electron and antineutrino). The nuclear force gives it this lower energy by binding it inside the nucleus and by favoring pairing between neutrons and protons.

Topic 30

1658

ANSWERS TO EVEN NUMBERED PROBLEMS 30.2

(a)

30.4

192 MeV

30.6

9.4 × 102 g

30.8

2.63 kg

30.10

(a)

30.12

3.33 × 10−11 J

(b)

−3.70 × 10−28 kg

3.1 × 1010 g

(b)

1.3 × 108 mol; 7.8 × 1031 atoms

(c)

2.6 × 1021 J

(d)

5.5 yr

(e)

Fission alone cannot meet the world’s energy needs.

(a)

4 × 1015 g

(c)

The uranium comes from dissolving rock and minerals. Rivers

(b)

5 × 103 yr

carry such solutes into the oceans, but the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half-million years. (b)

8.99 × 1037

(c)

(a) !7 N 13

(b)

!6 C

(d)

(e) !7 N

30.14

(a)

4.27 × 10−12 J

30.16

5.49 MeV

30.18

!8 O

4.27 × 109 kg

(f)

!6 C

Topic 30

30.20

1659

(a)

8.68 MeV

(b) The proton and the boron nucleus must each have enough kinetic energy to overcome the repulsive electrical force one exerts on the other. 30.22

118 MeV

30.24

(a)

30.26

(a) may occur via the weak interaction

(b)

(c)

(b) violates conservation of baryon number − cannot occur by any interaction 30.28

(a) cannot occur, violates baryon number (b) can occur, no laws violated (c) can occur, no laws violated (d) can occur, no laws violated (e) cannot occur, violates baryon number and muon-lepton number

30.30

(a) See Solution. (b) The second reaction violates conservation of strangeness and cannot occur via the strong or electromagnetic interactions. (c) If a neutral kaon, K0, was produced in addition to those shown, the second reaction could occur without violating any

Topic 30

1660

conservation laws. Because of the greater total mass of the product particles, the total kinetic energy of the incident particles in the this reaction would have to be greater than in the first reaction. 30.32

See Solution.

30.34

See Solution.

30.36

(a)

30.38

(a) conservation of charge

Q = −e, antiproton

(b)

Q = 0, antineutron

(b) conservation of electron-lepton number and conservation of muon-lepton number (c)

conservation of baryon number

30.40

ν µ and νe

30.42

a neutron

30.44

(a)

7.85 × 1013 J

(b)

$1.74 × 106

(c)

$1.50 × 103

(d) At a 10% profit margin, the profit is $1.74 × 105 on each kilogram of deuterium. 3.61 × 1030 J

(b)

1.63 × 108 yr

30.46

(a)

30.48

The first reaction has a net 1u and 2d quarks both before and after the reaction. The second reaction has a net 1u and 2d quarks before the

Topic 30

1661

reaction and 1u, 3d, and 1 s quark present afterwards. 30.50

0.827 9c

PROBLEM SOLUTIONS 30.1

The number of uranium-235 nuclei is

(

)(

)

N = 0.720 × 10−2 2.50 × 10 28 nuclei = 1.80 × 10 26 nuclei 30.2

(a) Use the conversion factor 1 MeV = 1.60 × 10−13 J to find: ⎛ 1.60 × 10−13 J ⎞ 208 MeV = 208 MeV ⎜ = 3.33 × 10−11 J ⎝ 1 MeV ⎟⎠

(b) Find the change in mass using E = Δmc2:

E = Δmc 2 → Δm =

30.3

E −3.33 × 10−11 J = = −3.70 × 10−28 kg 2 2 8 c 3.00 × 10 m/s

(

)

The energy consumed by a 100-W lightbulb in a 1.0-h time period is

⎛ 3 600 s ⎞ 5 E = P ⋅ Δt = (100 J/s)(1.0 h) ⎜ ⎟ = 3.6 × 10 J ⎝ 1h ⎠ The number of fission events, yielding an average of 208 MeV each, required to produce this quantity of energy is

30.4

E 208 MeV

3.6 × 105 J ⎛ 1 MeV ⎞ = 1.1 × 1016 ⎜ −13 ⎟ 208 MeV ⎝ 1.60 × 10 J ⎠ 1

235

135

The energy released in the reaction 0 n + 92 U → 40 Zr + 52 Te + 30 n is

Topic 30

1662

Q = (Δm)c 2 = [m 235 − 2mn − m 98 − m135 ]c 2 92

= [235.043 923 u − 2(1.008 665 u) − 97.912 0 u − 134.908 7 u](931.5 MeV/u) = 192 MeV

30.5

235

136

The energy released in the reaction 0 n + 92 U → 38 Sr + 54 Xe + 120 n is

Q = (Δm)c 2 = [m 235 − 11mn − m88 − m136 ]c 2 92

= [235.043 923 u − 11(1.008 665 u) − 87.905 614 u − 135.907 220 u](931.5 MeV/u) = 126 MeV

30.6

The total energy released was

⎛ 4.0 × 109 J ⎞ ⎛ 1 MeV ⎞ 26 E = (20 × 10 ton TNT ) ⎜ ⎟⎜ −13 ⎟ = 5.0 × 10 MeV 1 ton TNT 1.60 × 10 J ⎠ ⎝ ⎠⎝ ! 3

The number of 235U nuclei that must have undergone fission to yield this energy is

E 5.0 × 1026 MeV n= = = 2.4 × 1024 nuclei 208 MeV/nucleus 208 MeV/nucleus ! The mass of 235U which will contain this number of atoms is ⎛ n ⎞ ⎛ 2.4 × 1024 atoms ⎞ ⎛ g ⎞ 2 m=⎜ M = ⎟ mol ⎜ ⎟ ⎜ 235 ⎟ = 9.4 × 10 g 23 mol ⎠ ⎝ 6.02 × 10 atoms/mol ⎠ ⎝ ⎝ NA ⎠

30.7

(a) With a specific gravity of 4.00, the density of soil is ρ = 4.00 × 103 kg/m3. Thus, the mass of the top 1.00 m of soil is

Topic 30

1663

2 ⎛ 1m ⎞ ⎤ ⎛ ⎞⎡ 3 kg ⎢ 2 7 ⎥ m = ρV = ⎜ 4.00 × 10 (1.00 m)(43 560ft ) ⎜ ⎟ ⎥ = 1.62 × 10 kg 3⎟ ⎢ m ⎠ ⎝ ⎝ 3.281 ft ⎠ ⎣ ⎦

At a rate of 1 part per million, the mass of uranium in this soil is then

m 1.62 × 107 kg mU = 6 = = 16.2 kg 10 106 ! (b) Since 0.720% of naturally occurring uranium is !92 U the mass of 235

235

!92

U in the soil of part (a) is

m 235 = (7.20 × 10−3 )mU = (7.20 × 10−3 )(16.2 kg) = 0.117 kg = 117 g 92

30.8

With a power output of Pout = 1.00 × 109 W and efficiency of e = 0.400, the total energy input required each day is

Einput = !

9 Pout ⋅ Δt (1.00 × 10 J /s)(1.00 d ) ⎛ 1 MeV ⎞ ⎛ 8.64 × 10 4 s ⎞ = = 1.35 × 1027 MeV ⎜ ⎟ ⎜ ⎟ e 0.400 1d ⎠ ⎝ 1 J ⎠⎝

At 200 MeV per fission event, the number of 235U atoms consumed each day is

Einput 1.35 × 1027 MeV n= = = 6.75 × 1024 atoms 200 MeV/atom 200 MeV/atom ! The mass of 235U which will contain this number of atoms is

Topic 30

1664

⎛ n ⎞ ⎛ 6.75 × 1024 atoms ⎞ ⎛ g ⎞ 3 m=⎜ 235 ⎟ Mmol = ⎜ ⎟⎜ ⎟ = 2.63 × 10 g = 2.63 kg 23 mol ⎠ ⎝ 6.02 × 10 atoms/mol ⎠ ⎝ ⎝ NA ⎠

30.9

(a) For a sphere of radius a, the surface area is A = 4πa2, and the volume is V = 4πa3/3. Thus, the surface-to-volume ratio is

A 4π a2 ratiosphere = = = 3/a V 4π a3 /3 ! (b) For a cube of side s, A = 6 ⋅ Aface = 6s2 and V = s ⋅ s ⋅ s = s3. The surface-to-volume ratio for a cube is then

ratiocube =

A V

6s2

= 3

If the cube has the same volume as the sphere, then s3 = 4πa3/3 or 3 !s = (4π /3) a , and the surface-to-volume ratio for the cube becomes 1

ratiocube =

A V

6s2 s

6 1

(4π 3) 3 a

= 3.72 / a

(a) The mass of 235U in the reserve is kg ⎞ ⎛ 3 g ⎞ ⎛ 0.70 ⎞ ⎛ m 235 U = ⎜ (4.4 × 106 metric ton) ⎜ 103 = 3.1 × 1010 g ⎟ ⎟ 10 ⎝ 100 ⎠ ⎝ ton ⎠ ⎜⎝ kg ⎟⎠ !

(b) The number of moles is n = m/Mmol = (3.1 × 1010 g)/(235 g/mol) =

Topic 30

1665

1.3 × 108 mol and the number of 235U atoms in this reserve is N = nNA = (1.3 × 108 mol)(6.02 × 1023 atoms/mol) = 7.8 × 1031 atoms (c) Assuming all atoms undergo fission and all released energy captured, the total energy available is MeV ⎞ MeV ⎞ ⎛ 1.60 × 10−13 J ⎞ ⎛ ⎛ 31 E = N ⎜ 208 = (7.87 × 10 atoms) 208 = 2.6 × 1021 J ⎟⎠ ⎜⎝ ⎟⎠ ⎜ ⎟ ⎝ atom atom ⎝ 1 MeV ⎠ !

(d) At a consumption rate of 1.5 × 1013 J/s, the maximum time this energy supply could last is ⎞ 1 yr E 2.6 × 1021 J ⎛ t= = = 5.5 yr P 1.5 × 1013 J ⎜⎝ 3.156 × 107 s ⎟⎠ !

(e) Fission alone cannot meet the world's energy needs at a price of $130 or less per kilogram of uranium. 30.11

The total energy required for one year is E = (2 000 kWh/month)(3.60 × 106 J/kWh)(12.0 months) = 8.64 × 1010 J The number of fission events needed will be

E 8.64 × 1010 J N= = = 2.60 × 1021 −13 E (208 MeV)(1.60 × 10 J/MeV) event ! and the mass of this number of 235U atoms is ⎛ N⎞ ⎛ 2.60 × 1021 atoms ⎞ m=⎜ M = ⎟ mol ⎜ ⎟ (235 g/mol) = 1.01 g 23 N 6.02 × 10 atoms/mol ⎝ ⎠ ⎝ A⎠

Topic 30

30.12

1666

(a) At a concentration of c = 3 mg/m3 = 3 × 10−3 g/m3 the mass of uranium dissolved in the oceans covering two-thirds of Earth’s surface to an average depth of hav = 4 km is

m = cV = c( 23 A)⋅ hav = c ⎡⎣ 23 (4π RE2 ) ⎤⎦ ⋅ hav or ! U

g ⎞ ⎛ 2⎞ ⎛ mU = ⎜ 3 × 10−3 3 ⎟ ⎜ ⎟ 4π (6.38 × 106 m)2 (4 × 103 m) = 4 × 1015 g ⎝ m ⎠ ⎝ 3⎠ ! (b) Fissionable 235U makes up 0.7% of all uranium, so

m235U = 0.70mU /100 . If we assume all of the 235U is collected and caused to undergo fission, with the release of about 200 MeV per event, the potential energy supply is

⎤⎛ MeV ⎞ ⎡⎛ 0.7mU /100 ⎞ MeV ⎞ ⎛ E = (number of 235 U atoms) ⎜ 200 = ⎢⎜ NA ⎥ ⎜ 200 ⎟ ⎟ ⎟ ⎝ ⎝ atom ⎠ ⎣⎝ Mmol ⎠ atom ⎠ ⎦ ! and at a consumption rate of P = 1.5 × 1013 J/s the time this could supply the world’s energy needs is t = E/P, or ⎡ 0.7 ⎛ m ⎞ ⎤ (200 MeV/atom) U t=⎢ ⎜ ⎟⎥ P ⎢⎣ 100 ⎝ Mmol ⎠ ⎥⎦ ⎡ 0.7 ⎛ 4 × 1015 g ⎞ ⎛ ⎞ ⎞ ⎤ 200 MeV/atom ⎛ 1.6 × 10−13 J ⎞ ⎛ 1 yr 23 atoms =⎢ 6.02 × 10 ⎜ ⎟ ⎜ ⎟⎜ ⎜ ⎟ ⎟⎥ mol ⎠ ⎥⎦ 1.5 × 1013 J/s ⎝ 1 MeV ⎠ ⎝ 3.156 × 107 s ⎠ ⎢⎣ 100 ⎝ 235 g/mol ⎠ ⎝ = 5 × 103 yr

(c) The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, but the Earth’s supply of

Topic 30

1667

uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half-million years. 30.13 Use Lawson’s criterion to solve for the required minimum density:

nτ ≥ 1016 s/cm 3 → nmin =

30.14

1016 s/cm 3 = 6.67 × 1015 cm −3 1.50 s

(a) Use the conversion factor 1 MeV = 1.60 × 10−13 J to find ⎛ 1.60 × 10−13 J ⎞ 26.7 MeV = 26.7 MeV ⎜ = 4.27 × 10−12 J ⎟ ⎝ 1 MeV ⎠

(b) Each second the Sun releases 3.84 × 1026 J of energy. The number of 26.7 MeV reaction cycles responsible for this energy release is

3.84 × 10 26 J = 8.99 × 1037 reaction cycles each second −12 4.27 × 10 J

(c) According to Einstein’s mass-energy relation, ΔE = Δm c2. Solve for the change in mass and substitute numbers to find

Δm =

30.15

ΔE 3.84 × 10 26 J = = 4.27 × 109 kg 2 8 c2 3.00 × 10 m/s

(

(a)

4 2

He + 42 He → 84 Be + γ

(b)

8 4

Be + 24 He → 12 C +γ 6

)

Topic 30

1668

= [3(4.002 603 u)−12.000 000 u](931.5 MeV/u) = 7.27 MeV 30.16

The energy released in the reaction 1 H + 1 H → 2 He + γ is

Q = (Δm)c 2 = ⎡ m1 + m 2 − m 3 ⎤ c 2 H He ⎥ ⎢⎣ 1 H ⎦ 1 2 = [1007 825 u + 2.014 102 u − 3.016 029 u](931.5 MeV/u)= 5.49 MeV

30.17

The energy released in the reaction 1 H + 1 H →1 H + 1 H is Q = ⎡ 2m 2 H − m 3 H − m1 H ⎤ c 2 , or ⎣ 1 ⎦ 1 1 !

Q = [2(2.014 102 u) − 3.016 049 u − 1.007 825 u](931.5 MeV/u) = 4.03 MeV 30.18

(a)

(c)

(e) 30.19

13 H + 12 6 C→ 7 N +γ

(b)

13 6

C + 11H + 14 7 N +γ

(d)

0 O + 15 7 H → +1 e + ν

(f)

1 1

13 7

0 N + 13 6 C + +1e + ν

14 7

N + 11H → 15 O +γ 8

4 N + 11H → 12 6 C + 2 He

With the deuteron and triton at rest, the total momentum before the reaction is zero. To conserve momentum, the neutron and the alpha particle must move in opposite directions with equal magnitude momenta after the reaction, or pα = pn. Neglecting relativistic effects, we use the classical relationship between momentum and kinetic energy, KE = p2/2m, and write ! 2mα KEα = 2mn KEn , or KEα = (mn/mα)KEn.

Topic 30

1669

To conserve energy, it is necessary that the kinetic energies of the reaction products satisfy the relation KEn + KEα = Q = 17.6 MeV. Then, using the result from above, we have KEn + (mn/mα)KEn = 17.6 MeV, or the kinetic energy of the emerging neutron must be

17.6 MeV KEn = = 14.1 MeV 1 + (1.008 665 u)/(4.002603 u) ! 30.20

(a) The energy released in the reaction 1 H + 5 B → 3(2 He) is Q = (Δm)c 2 = ⎡ m1 + m11 − 3m 4 ⎤ c 2 B He ⎥ ⎢⎣ 1 H 5 2 ⎦ = [1.007 825 u + 11.009 306 u − 3(4.002 603 u)](931.5 MeV/u) = 8.68 MeV

(b) The proton and the boron nucleus both have positive charges but must come very close to one another in order for fusion to occur. Thus, they must have sufficient kinetic energy to overcome the repulsive Coulomb force one exerts on the other. 30.21

Note that pair production cannot occur in a vacuum. It must occur in the presence of a massive particle which can absorb at least some of the momentum of the photon and allow total linear momentum to be conserved. When a particle-antiparticle pair is produced by a photon having the minimum possible frequency, and hence minimum possible energy, the nearby massive particle absorbs all the momentum of the photon,

Topic 30

1670

allowing both components of the particle-antiparticle pair to be left at rest. In such an event, the total kinetic energy afterwards is essentially zero, and the photon need only supply the total rest energy of the pair produced. The minimum photon energy required to produce a proton-antiproton pair is Ephoton = 2(ER)photon. Thus, the minimum photon frequency is

Ephoton h

and

30.22

2(ER )photon

λ=

c f

2(938.3 MeV)(1.60 × 10−13 J/MeV) = 4.53 × 1023 Hz 6.63 × 10−34 J ⋅s

3.00 × 108 m/s 23

4.53 × 10 Hz

= 6.62 × 10−16 m = 0.662 fm

The total kinetic energy after the pair production is KEtotal = Ephoton − 2(ER)photon = 2.09 × 103 MeV − 2(938.3 MeV) = 213 MeV The kinetic energy of the antiproton is then

KEp = KEtotal − KEp = 213 MeV − 95.0 MeV = 118 MeV 30.23

The total rest energy of the π0 is converted into energy of the photons. Since the total momentum was zero before the decay, the two photons must go in opposite directions with equal magnitude momenta (and hence equal energies). Thus, the rest energy of the π0 is split equally between the two photons, giving for each photon

Topic 30

1671

Ephoton =

pPhoton = !

and

30.24

R ,π 0

Ephoton c

Ephoton h

135 MeV 2

= 67.5 MeV

= 67.5 MeV/c

⎛ 1.60 × 10−13 J ⎞ 22 = =⎜ ⎟ = 1.63 × 10 Hz −34 6.63 × 10 J ⋅s ⎝ 1 MeV ⎠ 67.5 MeV

(a) A proton’s baryon number is +1 and an antiproton’s baryon’s number is −1. A gamma ray’s baryon number is 0. Because 1 + (−1) = 0, the reaction’s baryon number is 0 . (b) The Ω− has a baryon number of +1. On the RHS, the Λ0 has a baryon number of +1 and the K− has a baryon number of 0. Therefore the reaction’s baryon number is 1 . (c) The Ω− has an electro-lepton number of 0, as do the particles on the RHS of the reaction. Therefore the electron-lepton number is 0.

30.25

(a) The µ− muon has a muon-lepton number of +1. Only the muonneutrino on the RHS of the reaction has a muon-lepton number (of +1) so that +1 = 0 + 0 + 1 = 1 , the muon-lepton number. (b) Both the π − and the proton have a strangeness of zero. Strangeness values for the Λ 0 and the K0 are −1 and +1, respectively, so that 0 + 0 = −1 + 1 = 0.

Topic 30

30.26

1672

Observe that the given reactions involve only mesons and baryons. With no leptons before or after the reactions, we do not have to consider the conservation laws concerning the various lepton numbers. All interactions always conserve both charge and baryon numbers. The strong interaction also conserves strangeness. Conservation of strangeness may be violated by the weak interaction but never by more than one unit. With these facts in mind consider the given interactions: K0→π+ + π−

total K

Charge

total +

before

−

after

−1

Baryon number

Strangeness

This reaction conserves both charge and baryon number but does violate strangeness by one unit. Thus, it can occur via the weak

Topic 30

1673

interaction. but not other interactions. Λ0 → π+ + π− total Λ

Charge

before

Strangeness

π−

total after

−1

Baryon number

π+

This reaction fails to conserve baryon number and cannot occur via any interaction. 30.27 Reaction

Conservation Law Violated

(a)

+ !p + p → µ + e

Le: (0 + 0 → 0 + 1); and Lµ: (0 + 0 → −1 + 0)

(b)

π− + p → p + π+

Charge, Q: (−1 + 1 → +1 + 1)

(c)

p + p → p + π+

Baryon Number, B: (1 + 1 → 1 + 0)

(d)

p+p→p+p+

Baryon Number, B: (1 + 1 → 1 + 1 + 1)

(e)

n n + π0 γ+p→

Charge, Q: (0 + 1 → 0 + 0)

Topic 30

1674

30.28 Reaction

Conservation Law Violated

(a)

p → π+ + π0

Baryon Number, B: (1 → 0 + 0)

(b)

p + p → p + p + π0

No violations − The reaction can occur

(c)

π+ → µ+ + νµ

No violations − The reaction can occur

(d)

n → p + e– + ν e

No violations − The reaction can occur

(e)

π+ → µ+ + n

Baryon Number, B: (0 → 0 + 1) Muon-Lepton Number, Lµ : (0 → −1 + 0)

30.29 Reaction

Conservation Law

Conclusion

Baryon Number - violated: (a)

(b)

π + p → 2η −

K− + n → Λ0 + π−

(0 + 1 → 0)

Cannot Occur

All conservation laws

May occur via the

observed.

Strong Interaction Can occur via

Strangeness violated by 1 unit

(c)

(Strangeness violated by −1 → 0

K →π +π −

−

0 + 0). All other conservation laws observed.

Weak Interaction, but not by Electromagnetic or Strong Interactions.

Topic 30

1675

Can occur via Strangeness violated by 1 unit Weak Interaction, (d)

Ω− → Ξ− + π0

(−3 → −2 + 0). All other

but not by

conservation laws observed.

Electromagnetic or Strong Interactions. Allowed via all interactions, but photons are the mediator of the

(e)

All conservation laws

electromagnetic

observed.

interaction and the

η → 2γ

lifetime of the η0 is consistent with decay by that interaction.

30.30

(a) π+ + p → K+ + Σ+ Baryon number, B:

0+1→0+1

ΔB = 0

Charge, Q:

1+1→1+1

ΔQ = 0

Topic 30

1676

Baryon number, B: Charge, Q:

0+1→0+1

ΔB = 0

1+1→0+1

ΔQ = 0

(b) Strangeness is conserved in the first reaction: Strangeness, S:

0+0→1−1

ΔS = 0

The second reaction does not conserve strangeness: Strangeness, S:

0+0→0−1

ΔS = −1

The second reaction cannot occur via the strong or electromagnetic interactions. (c) If one of the neutral kaons were also produced in the second reaction, giving π+ + p → π+ + Σ+ + K0, then strangeness would no longer be violated: Strangeness, S:

0+0→0−1+1

ΔS = 0

Because the total mass of the product particles in this reaction would be greater than that in the first reaction [see part (a)], the total incident energy of the reacting particles would have to be greater for this reaction than for the first reaction. 30.31

(a) Λ0 → p + π− Strangeness, S:

−1 → 0 + 0 ⇒ ΔS ≠ 0

Not Conserved

Topic 30

1677

Strangeness, S:

0 + 0 → −1 + 1 ⇒ ΔS = 0

Conserved

0 + 0 → + 1 – 1 ⇒ ΔS = 0

Conserved

0 0 (c) !p + p → Λ + Λ

Strangeness, S: (d) π− + p → π− + Σ+ Strangeness, S:

0 + 0 → 0 – 1 ⇒ ΔS ≠ 0

Not Conserved

− 2 → −1 + 0 ⇒ ΔS ≠ 0

Not Conserved

−2 → 0 + 0 ⇒ ΔS ≠ 0

Not Conserved

(e) Ξ− → Λ0 + π− Strangeness, S: (f) Ξ0 → p + π− Strangeness, S: 30.32

strangeness baryon

proton

total

1 !3

2e/3

−e/3

number charge

Topic 30

1678

strangeness baryon number charge

30.33

neutron

total

1 !3

2e/3

−e/3

The number of water molecules in one liter (mass = 1 000 g) of water is

⎛ 1 000 g ⎞ 23 25 N= ⎜ ⎟ (6.02 × 10 molecule/mol) = 3.34 × 10 molecules ⎝ 18.0 g/mol ⎠ Each molecule contains 10 protons, 10 electrons, and 8 neutrons. Thus, there are Ne = 10 N = 3.34 × 1026 electrons Np = 10 N = 3.344 × 1026 protons, and Nn = 8 N = 2.68 × 1026 neutrons Each proton contains 2 up quarks and 1 down quark, while each neutron has 1 up quark and 2 down quarks. Therefore, there are Nu = 2 Np + Nn = 9.36 × 1026 up quarks, and Nd = Np = 2 Nn = 8.70 × 1026 down quarks.

Topic 30

1679

30.34

K0 Particle

strangeness

total

1/3

−1/3

−e/3

e/3

baryon number

charge

Λ0 Particle

Λ0

total

strangeness

−1

baryon number

1/3

charge

2e/3

−e/3

Topic 30

30.35

30.36

1680

Compare the given quark states to the entries in Table 30.4: (a) suu = Σ+

(b)

ud = π −

(c)

(d)

ssd = Ξ−

0 !sd = K

⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ (a) !uud charge = ⎜ − e ⎟ + ⎜ − e ⎟ + ⎜ + e ⎟ = −e . This is the ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ antiproton.

⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ (b) !uud charge = ⎜ − e ⎟ + ⎜ + e ⎟ + ⎜ + e ⎟ = 0 . This is the ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ antineutron. 30.37

The reaction is Σ0 + p → Σ+ + γ + X, or on the quark level, uds + uud → uus + 0 + x. The left side has a net 3u, 2d, and 1s. The right side has 2u, 0d, and 1s, plus the quark composition of the unknown particle. To conserve the total number of each type of quark, the composition of the unknown particle must be x = udd Therefore, the unknown particle must be a neutron.

30.38

(a) π− + p → Σ+ + π0 is forbidden by conversation of change. (b) µ− → π− + νe is forbidden by both conversation of electron-lepton number and conversation of muon-lepton number.

Topic 30

1681

To the reaction for nuclei, 1 H + 2 He → 2 He + +1 e + ν e , we add three electrons to both sides to obtain 1 1

H atom + 32 He atom → 24 He atom + +10 e + +10 e + ν e . Then we use the masses of

the neutral atoms from Appendix B of the textbook to compute Q = (Δm)c 2 = ⎡ m1 + m 3 − m 4 − 2me ⎤ c 2 He He ⎢⎣ 1 He ⎥⎦ 2 2 = [1.007 825 u + 3.016 029 u − 4.002 603 u − 2(0.000 549 u)](931.5 MeV = 18.8 MeV

30.40

For the particle reaction, µ+ + e− → 2ν, the lepton numbers before the event are Lµ = −1 and Le = +1. These values must be conserved by the reaction so one of the emerging neutrinos must have Lµ = −1 while the other has Le = +1. The emerging particles are ν µ and νe .

30.41

(a) K+ + p → X + p Since this occurs via the strong interaction, all conservation rules must be observed.

Baryon Number:

0+1→B+1

ΔB = 0 ⇒ B = 0

so X is not a baryon.

Strangeness:

+1+0→S+0

ΔS = 0 ⇒ S = +1

or X has S = +1

Charge:

+e+e→Q+e

ΔQ = 0 ⇒ Q = +e

or X has Q = +e

Lepton Numbers:

0+0→L+0

ΔL = 0 ⇒ L = 0

or X has Le = Lµ = Lτ = 0

Topic 30

1682

Of the particles in Table 30.2, the only non-baryon with S = +1 and Q = +e is the positive kaon, K+ Thus, this is an elastic scattering process. The weak interaction observes all conservation rules except strangeness, and ΔS = ±1. (b) Ω− → X + π− Baryon Number:

1→B+0

ΔB = 0 ⇒ B = +1

or X has B = +1

Strangeness:

−3 → S + 0

ΔS = +1 ⇒ S = −2

or X has S = −2

Charge:

−e → Q − e

ΔQ = 0 ⇒ Q = 0

or X has Q = 0

Lepton Numbers:

0→L+0

ΔL = 0 ⇒ L = 0

or X has Le = Lµ = Lτ = 0

The particle must be a neutral baryon with strangeness S = −2 Thus, it is the Ξ0 (c) K+ → X + µ+ + νµ Baryon Number:

0→B+0+0

ΔB = 0 ⇒ B = 0

so X is not a baryon.

Strangeness:

+1 → S + 0 + 0 ΔS = ±1 ⇒ S = 0, +2

Charge:

+e → Q + e + 0

ΔQ = 0 ⇒ Q = 0

or X has Q = 0

Lepton Numbers:

0 → Le + 0 + 0.

ΔLe = 0 ⇒ Le = 0

or X has Le = 0

0 → Lµ − 1 + 1

ΔLµ = 0 ⇒ Lµ + 0

or X has Lµ = 0

0 → Lτ + 0 + 0

ΔLτ = 0 ⇒ Lτ = + 0

or X has Lτ = 0

or X has S = 0 or +2

Topic 30

1683

The particle must be a neutral non-baryon with strangeness S = 0 or S = +2 and Le = Lµ = Lτ = 0 This is the neutral pion, π0 . 30.42

Assuming a head-on collision, the total momentum is zero both before and after the reaction p + p → p + π+ + X. Therefore, since the proton and the pion are at rest after the reaction, particle X must also be left at rest. Particle X must be a neutral baryon in order to conserve charge and baryon number in the reaction. Conservation of energy requires that the rest energy this particle be

(

)

ER,X = 2 ER,p + 70.4 MeV − ER,p − ER,π + + = ER,p − ER,π + + 140.8 MeV or

ER,X = 938.3 MeV − 139.6 MeV + 140.8 MeV = 939.5 MeV

Particle X is a neutron 30.43

If a neutron starts with kinetic energy KEi = 2.0 MeV and loses onehalf of its kinetic energy in each collision with a moderator atom, its kinetic energy after n collisions will be KEf = KEi/2n. The average kinetic energy associated with particles in a gas at temperature T = 20.0°C = 293 K. (see Chapter 10 of the textbook) is ⎛ ⎞ 1eV 3 3 KE f = kBT = (1.38 × 10−23 J/K)(293 K) ⎜ ⎟ = 0.037 9 eV −19 2 2 ⎝ 1.60 × 10 J ⎠

Topic 30

1684

Thus, the number of collisions the neutron must make before it reaches the energy associated with a room temperature gas is n ln 2 = ln(KEi/KEf), or

⎛ 1 ⎞ ⎛ 2.0 × 106 eV ⎞ n=⎜ ⎟ = 26 ⎟ ln ⎜ ⎝ ln 2 ⎠ ⎝ 0.037 9 eV ⎠ 30.44

(a) The number of deuterons in one kilogram of deuterium is

m 1 kg N= = = 3.00 × 1026 −27 m (2.01 u)(1.66 × 10 kg/u) atom ! Each occurrence of the reaction !1 D + 1 D → 2 He + 0 n consumes 2

two deuterons and releases Q = 3.27 MeV of energy. The total energy available from the one kilogram of deuterium is then ⎛ 1.60 × 10−13 ⎞ ⎛ 3.00 × 1026 ⎞ ⎛ N⎞ E=⎜ ⎟Q=⎜ (3.27 Mev) = 7.85 × 1013 J ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠ ⎝ 1 MeV ⎠ !

(b) At a rate of eight cents per kilowatt-hour, the value of this energy is ⎛ $0.08 ⎞ ⎛ 1 kWh ⎞ value = E × rate = (7.85 × 1013 J) ⎜ = $1.74 × 106 = $1 740 000 ⎝ kWh ⎟⎠ ⎜⎝ 3.60 × 106 J ⎟⎠ !

(c) Deuterium makes up four-twentieths or one-fifth of the mass of a heavy water molecule. Thus, five kilograms of heavy water are necessary to obtain one kilogram of deuterium. The cost for this water (5 kg)($300/kg) = $1 500.

Topic 30

1685

(d) Whether it would be cost-effective depends on how much it cost to fuse the deuterium and how much net energy was produced. If the cost is nine-tenths of the value of the energy produced, each kilogram of deuterium would still yield a profit of $174 000. 30.45

(a) The number of moles in 1.0 gal of water is

m ρV n= = = M M !

⎛ 3.786 L ⎞ ⎛ 103 cm 3 ⎞ g ⎞ ⎛ ⎜⎝ 1.0 ⎟ (1.0 gal) ⎜ cm 3 ⎠ ⎝ 1 gal ⎟⎠ ⎜⎝ 1 L ⎟⎠ 18 g/mol

= 2.1 × 102 mol

so the number of hydrogen atoms (2 per water molecule) will be NH = 2(nNA) = 2(2.1 × 102 mol)(6.02 × 1023 mol−1) = 2.5 × 1026 and one of every 6 500 of these contains a deuteron. Thus, the number of deuterons contained in 1.0 gal of water is ND = NH/6 500 = 2.5 × 1026/6 500 = 3.8 × 1022 deuterons and the available energy is E = (3.8 × 1022 deuterons)(1.64 MeV/deuteron)(1.6 × 10−13 J/MeV) = 1.0 × 1010 J (b) At a consumption rate of P = 1.0 × 104 J/s, the time that this could supply a person’s energy needs is E 1.0 × 1010 J ⎛ 1 d ⎞ Δt = = = 12d P 1.0 × 10 4 J/s ⎜⎝ 86 400 s ⎟⎠ !

Topic 30

30.46

1686

(a) The number of water molecules in the oceans is

⎛ mocean ⎞ ⎛ 1.32 × 1021 kg ⎞ water ⎟ NA = NH 2 O = ⎜ (6.02 × 1023 molecules/mol) −3 ⎜ ⎟ ⎝ 18.0 × 10 kg/mol ⎠ ⎜⎝ Mmol ⎟⎠ !

= 4.41 × 10 46 molecules Since there are 2 hydrogen atoms per water molecule, and the fraction of all hydrogen atoms which contain deuterons is 1.56 × 10−4, the number of deuterons in the oceans is

(

)

N D = 2N H 2O × 1.56 × 10− 4 ) = 2 (4.41 × 10 46 ) (1.56 × 10− 4 ) = 1.38 × 10 43 Each fusion event consumes 2 deuterons, so the number of fusion events possible is !Nevents = 12 ND . The reaction !1 D + 1 D → 2 He + 0 n 2

releases 3.27 MeV per event, so the total energy that could be released is E = (3.27 MeV)Nevents = (3.27 MeV)( 12 ND )

⎛ 1.38 × 10 43 ⎞ ⎛ 1.60 × 10−13 J ⎞ = (3.27 MeV) ⎜ = 3.61 × 1030 J ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 1 MeV ⎠

(b) One hundred times the current world electric power consumption is P = 100(7.00 × 1012 W) = 7.00 × 1014 J/s At this rate, the time the energy computed in part (a) would last is ⎞ 1 yr E 3.64 × 1030 J ⎛ t= = = 1.63 × 108 yr 14 7 ⎜ P 7.00 × 10 J/s ⎝ 3.156 × 10 s ⎟⎠ ! © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 30

30.47

1687 − − Conserving energy in the decay π → µ + ν µ , with the π− initially at

rest gives Eµ + Eν = ER ,π − , or Eµ + Eν = 139.6 MeV

[1]

Since the total momentum was zero before the decay, conservation of momentum requires the muon and antineutrino to recoil in opposite directions with equal magnitude momenta, or pµ = pν. The relativistic relation between total energy and momentum of a particle gives for the antineutrino: Eν = pνc, or pν = Eν/c. The same relation applied to the

( ) + E . Since we must have p = p , this may be

muon gives Eµ2 = pµ c !

2 R,µ

2 rewritten as Eµ2 = ( pν c ) + ER, µ , and using the fact that ! 2

2 2 2 pνc = Eν, we have !Eµ − Eν + ER,µ .

Rearranging and factoring then gives

(

)(

)

2 2 Eµ2 − Eν2 = Eµ + Eν Eµ − Eν = ER, µ = (105.7 MeV) !

and

Eµ − Eν = !

(105.7 MeV)2 Eµ + Eν

[2]

Substituting Equation [1] into [2] gives Eµ − Eν = (105.7 MeV)2/139.6 MeV, or Eµ − Eν = 80.0 MeV

[3]

Topic 30

1688

Subtracting Equation [3] from Equation [1] yields 2Eν = 59.6 MeV, and Eν = 29.8 MeV 30.48

The reaction π-− + p → K0 + Λ is !ud + uud → !ds + uds at the quark level. There is a net 1 u and 2d quarks both before and after the reaction. This reaction conserves the net number of each type quark. For the reaction π− + p → K0 + n, or !ud + uud → ds + udd , there is a net 1u and 2d before the reaction and 1u, 3d, and !1s quark afterwards. This reaction does not conserve the net number of each type quark.

30.49

To compute the energy released in each occurrence of the reaction 4p + 2e− → α + 2νe + 6γ we add two electrons to each side to produce neutral atoms and obtain !4(1 H atom ) → 2 He atom + 2ν e + 6γ . Then, recalling that the 1

neutrino and the photon both have zero rest mass, and using the neutral atomic masses from Appendix B in the textbook gives ⎤ c2 Q = (Δm)c 2 = ⎡ 4m1 − m4 He ⎢⎣ 1 Hatom ⎥ 2 atom ⎦ = ⎡ 4 1.007 825 u − 4.002 603 u ⎤ 931.5 MeV/u ⎣ ⎦ = 26.7 Mev

(

)

(

)

Each occurrence of this reaction consumes four protons. Thus, the energy released per proton consumed is E1 = 26.4 MeV/4 protons = © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Topic 30

1689

6.68 MeV/proton Therefore, the rate at which the Sun must be fusing protons to provide its power output is

rate =

30.50

P 3.85 × 1026 J/s ⎛ 1 MeV ⎞ = = 3.60 × 1038 proton/s E1 6.68 MeV/proton ⎝ 1.60 × 10−13 J ⎠

With the kaon initially at rest, the total momentum was zero before the decay. Thus, to conserve momentum, the two pions must go in opposite directions and have equal magnitudes of momentum after the decay. Since the pions have equal mass, this means they must have equal speeds and hence, equal energies. The rest energy of the kaon is then split equally between the two pions, and the energy of each is Eπ = ER ,K 0 / 2 = 497.7 MeV / 2 = 248.9 MeV

The total energy of one of the pions is related to its rest energy by Eπ = γER,π, where γ = 1 / 1 − ( v / c)2 ) . Therefore, the speed of each of the ! pions after the decay will be 2

⎛ 139.6 MeV ⎞ ⎛E ⎞ v = c 1 − ⎜ R,π ⎟ = c 1 − ⎜ = 0.827 9c ⎝ Eπ ⎠ ⎝ 248.9 MeV ⎟⎠ !