Physics 1: Mechanics and Thermodynamics # Raymond A. Serway and W. Jewett, 2014

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PHYSICS 1 PHYSICS 1: MECHANICS AND THERMODYNAMICS PHYSICS 2: OSCILLATIONS, ELECTRICITY AND MAGNETISM PHYSICS 3: WAVES, OPTICS AND MODERN PHYSICS

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Introduction 1. Title: Physics 1 2. Credits: 3 3. Prerequisites: ❖ Analytics analysis 4. Course Description This course tends to give students opportunity to explore the basic concepts, laws and application of Mechanics and Thermodynamics, including: ❖ Properties and laws of motion of particle, rigid body ❖ Relationship among position, velocity and acceleration ❖ Laws of linear momentum, angular momentum and energy ❖ The kinetic theory of gases, thermodynamic quantities ❖ Laws of Thermodynamics

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Introduction 5. Textbook 1.

Raymond A. Serway and W. Jewett, Physics for Scientists and Engineers with Modern Physics (9th Edition), Cengage Learning, USA, 2014

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Trần Ngọc Hơi, Phạm Văn Thiều, Vật lý đại cương: Các nguyên lý và ứng dụng, Tập 1: Cơ học và Nhiệt học, NXB Giáo dục 2006

Reference Books 1.

Hugh D. Young and Roger A. Freedman, University Physics with Modern Physics (13th Edition), Pearson Education, USA, 2012

2.

Paul A. Tipler and Gene Mosca, Physics for Scientists and Engineers (6th Ed.), W. H. Freeman and Company, USA, 2008

3.

David Halliday, Cơ sở vật lý, tập 1, NXB Giáo dục, 2007

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Introduction 6. Lesson plan Week

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2

3

4

Lesson

Preparation of student

Introduction Chapter 1: Physics and measurement

Read the text book: 2-13

Chapter 2: Motion in one dimension

Read the text book: 21-47

Chapter 3: Montion in two dimensions

Read the text book: 78-98

Chapter 4: The laws of motion

Read the text book: 111-135

Solve problems of chapters 2, 3, 4

Prepare the solution of problems by group

Chapter 5: Circular motion and other applications of Newton’s laws

Read the text book: 150-167

Chapter 6: Energy of the system

Read the text book: 177-201

Chapter 7: Conservation of energy

Read the text book: 211-233

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Introduction 6. Lesson plan Week

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6

7

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Lesson

Preparation of student

Solve problems of chapters 5, 6, 7

Prepare the solution of problems by group

Chapter 8: Linear momentum and collisions

Read the text book: 247-279

Chapter 9: Rotation of rigid object about a fiexd axis

Read the text book: 293-321

Solve problems of chapters 8, 9

Prepare the solution of problems by group

Chapter 10: Angular momentum

Read the text book: 335-352

Chapter 11: Static equilibrium and elasticity, Universal gravitation, Fluid mechanics

Study by your self: 363-449

Introduction to the project of course

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Introduction 6. Lesson plan Week

Lesson

Preparation of student

Solve problems of chapters 10

Prepare the solution of problems by group

Chapter 12: Tempurature and the first law of Thermodynamics

Read the text book: 568-625

Chapter 13: The kinetic theory of gases

Read the text book: 626-652

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Chapter 14: Heat engines, entropy and the second law of Thermodynamics

Read the text book: 653-688

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Solve problems of chapters 12, 13, 14

Prepare the solution of problems by group

Represent the result of the project

Work in group to make the product of the project and to prepare a report of project

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Introduction 7. Assessment Plan Assessment Types

Assessment Components

Percentages

A1. Learning activities

A1.1. Attendance A1.2. Homework report A1.3. Project

20%

A2. Midterm Exam

A2.1. Written Test

20%

A3. Final Exam

A3.1. Written Test

60%

8. Student Responsibilities and Policies: â?–

Attendance: It is compulsory that students attend at least 80% of the course to be eligible for the final examination.

â?–

Missed tests: Students are not allowed to miss any of the tests. There are very few exceptions

PART 1: MECHANICS

CHAPTER 1: PHYSICS AND MEASUREMENT

PHYSICS AND MEASUREMENT CHAPTER 1

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CHAPTER 1: PHYSICS AND MEASUREMENT

Physics Physics, the most fundamental physical science, is concerned with the fundamental principles of the Universe. The study of physics can be divided into six main areas: ➢ Classical mechanics: concerning the motion of objects that are large

relative to atoms and move at speeds much slower than the speed of light

➢ Relativity: a theory describing objects moving at any speed, even speeds

approaching the speed of light

➢ Thermodynamics: dealing with heat, work, temperature, and the

statistical behavior of systems with large numbers of particles

➢ Electromagnetism: concerning electricity, magnetism, and electromagnetic

fields

➢ Optics: the study of the behavior of light and its interaction with materials ➢ Quantum mechanics: a collection of theories connecting the behavior of

matter at the submicroscopic level to macroscopic observations

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Physics and measurement Like all other sciences, physics is based on experimental observations and quantitative measurements. ➢ Objectives: to identify fundamental laws governing natural

phenomena and use them to develop theories

➢ Tool: Language of mathematics (a bridge between theory and

experiment) ➢ Classical

physics: includes the principles of classical mechanics, thermodynamics, optics, and electromagnetism developed before 1900 (Newton mechanics) ➢ Modern physics: a major revolution in physics began near the

end of the 19th century (theories of relativity and quantum mechanics)

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1.1. Standards of length, mass and time In 1960, an international committee established a set of standards for the fundamental quantities of science call SI (Système International)

Length ❖ The distance between two points in space ❖ Standard in SI: meter (m)

▪ 1960: 1m = the length of the meter was defined as the distance

between two lines on a specific platinum–iridium bar stored under controlled conditions in France ▪ 1960s-1970s: 1m = 1 650 763.73 wavelengths 1 of orange-red

light emitted from a krypton-86 lamp

▪ 1983: 1m = the distance traveled by light in vacuum during a

time of 1/299792458 second

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CHAPTER 1: PHYSICS AND MEASUREMENT

1.1. Standards of length, mass and time Mass Standard in SI: kilogram (kg) ▪ 1987: 1kg = the mass of a specific platinum–

iridium alloy cylinder kept at the International Bureau of Weights and Measures at Sèvres, France

Time Standard in SI: second (s) ▪ 1967: 1s = 9 192 631 770 times the period of

vibration of radiation from the cesium-133 atom (in an atomic clock)

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1.2. Dimentional analysis • The dimensions of length l, mass m, and time t are L, M,

and T, respectively. • Use brackets [ ] to denote the dimensions of a physical quantity. • Assume that a quantity f = g(z, y, z) is a function of quantities x, y and z with the corresponding dimensions [x], [y] and [z], the dimension of f: • [f ] = g([x ], [y ], [z ]) • Example: • The dimensions of speedv = l/t are written [v ] = L/T. • The dimensions of area A = l × l are [A ] = L × L = L2.

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1.2. Dimensional analysis • Quantities can be added or subtracted only if they have the same

dimensions. • The terms on both sides of an equation must have the same dimensions. Example 1.1. Analysis of an equation

Show that the expression � = ��, where � represents speed, � acceleration, and � an instant of time, is dimensionally correct. • Solve: • The dimensions of v: [v] = L/T • The dimensions of at: [at] =

L T T2

đ??ż

= � • Therefore, v = at is dimensionally correct because we have the same dimensions on both sides.

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1.3. Conversion of units Sometimes it is necessary to convert units from one measurement system to another or convert within a system. ▪ 1 km = 1000 m, 1 m = 10 dm = 100 cm =1000 mm ▪ 1 mile (mi)= 1609 m ▪ 1 feet (ft) = 0.3048 m, 1 m = 39.37 in. ▪ 1 inch (in.) = 0.0254 m, 1 m = 3.281 ft

Example 1.3 ▪ 36 km/h = ? m/s ▪ 200 mi/h = ? m/s

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1.4. Estimates and Order-of-Magnitude

Calculations Order of magnitude is a power often determined as follows Example 1.4 ▪ 0.0086 m ∼ 10−2m

1720 kg ∼ 103 kg

▪ 0.0021 s ∼ 10−3 s

3259 kg ∼ 103 kg

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1.5. Significant figures Significant figures is the number of numerical digits used to express the measurement + 1; 2 have one significant figure; 12; 2,3 have two SF; 123; 1,56 have three SF. + 1,01; 202 have three SF; 2016; 50,25 have four SF; 30001; 1,1001 have five SF + 0,1; 0,002 have one SF; 0,16; 0,0025 have two SF; 0,102; 0,123 have three SF

+ 12,00; 2,010 have four SF; 20; 2,0 have two SF; 0,100; 10,0 have three SF + Mass of 1500 g is an ambiguous value → should use scientific notation such as 1.5 × 103 g (2 SF), or 1.50 × 103 g (3 SF), or 1.500 × 103 g (4 SF) + The same rule holds for numbers less than 1. Ex: 2.3 × 10−5 (2 SF), 1.560 × 10−2 (4 SF)

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1.5. Significant figures The rule of determinating the number of significant figures When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the quantity having the smallest number of significant figures. The same rule applies to division Ex: Report the result of multiplications â–Ş The area of a carpet whose length is 15.24 m and whose width is

2.19 m đ??´ = 15.24 Ă— 2.19 = 33.4 m2

33.3756

â–Ş The area of the disc whose radius is 6.0 cm

đ??´ = đ?œ‹6.02 = 1.1 Ă— 102 cm2 (113.097 335 ‌ )

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1.5. Significant figures The rule of determinating the number of significant figures When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum or difference Ex: 23.5 + 6.174 = 29.7 (29.674) Note: The rule for rounding number ▪ The last digit retained is increased by 1 if the last digit dropped is greater

than 5 (Ex: 2.567 → 2.57) ▪ If the last digit dropped is less than 5, the last digit retained remains as it is

(Ex: 2.564 → 2.56) ▪ If the last digit dropped is equal to 5, the remaining digit should be rounded

to the nearest even number (Ex: 2.565 → 2.56, 2.555 → 2.56)

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CHAPTER 2: MOTION IN ONE DIMENSION

Introduction Kinematics: describe the motion of an object while ignoring the interactions with external agents â–Ş Motion in one dimension: motion of an object along a straight line â–Ş Particle model: describe the moving object as a particle regardless

of its size (a particle to be a point-like object) Physical terms http://www.conservapedia.com/Physical_Science_Terms motion, particle, kinematics, position, reference point, coordinate system, velocity, speed, average/instantaneous velocity/speed, derivative, acceleration, gravity, resistance, period, angular speed, centripetal acceleration, tangential and radial acceleration, relative velocity/acceleration

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CHAPTER 2: MOTION IN ONE DIMENSION

2.1. Position, velocity, speed Position A particle’s position � is the location of the particle with respect to a chosen reference point that we can consider to be the origin of a coordinate system Example

One dimension coordinate system Pictorial representation

Position-time graph Graphical representation

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2.1. Position, velocity, speed Displacement and distance The displacement � of a particle is its change in position in some time interval As the particle moves from an initial position �� to a final position ��, its displacement is given by

đ?šŤđ?’™ = đ?’™đ?’‡ − đ?’™đ?’Š Note: Displacement differs from distance Displacement (đ?šŤđ??ą)

Distance (d)

change in position

the length of a path

vector quantity

scalar quantity

Ex: After each period of motion of a particle moving in a circle of radius đ?‘&#x;: Δx = 0, d = 2đ?œ‹đ?‘&#x;

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CHAPTER 2: MOTION IN ONE DIMENSION

2.1. Position, velocity and speed Velocity and speed The average velocity ��,avg (�� ) of a particle is defined as the particle’s displacement Δ� divided by the time interval Δ� during which that displacement occurs:

Dimension: L/T Note 1: In one dimension motion, the average velocity can be positive or negative, depending on the sign of the displacement Note 2: Velocity differs from speed The average speed đ?‘Łavg of a particle is defined as

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CHAPTER 2: MOTION IN ONE DIMENSION

2.2. Instantaneous velocity and speed Instantaneous velocity The instantaneous velocity (or velocity for short) of a particle at a particular instant in time � equals the limiting value of the ratio Δ�/Δ� as Δ� approaches zero:

Note 1: The instantaneous velocity can be positive, negative, or zero Note 2: Velocity differs from speed The instantaneous speed (or speed for short) of a particle is defined as the magnitude of its velocity đ?‘Ł.

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CHAPTER 2: MOTION IN ONE DIMENSION

2.3. Particle under constant velocity Analysis model An analysis model is a common situation that occurs time and again when solving physics problems All of the analysis models that we will develop are based on four fundamental simplification models: particle, system, rigid object, wave Two basic steps to solve a problem: â–Ş Identify the analysis model that is appropriate for the problem â–Ş The model tells you which equation(s) to use for the mathematical

representation

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CHAPTER 2: MOTION IN ONE DIMENSION

2.3. Particle under constant velocity Analysis model: particle under constant velocity If the velocity of a particle is constant, its instantaneous velocity at any instant during a time interval is the same as the average velocity over the interval:

→ The position of the particle as a function of time is given by �� = �� + �� � Note: Particle under constant speed If a particle moves at a constant speed through a distance � along a straight line or a curved path in a time interval Δ�, its constant speed is

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CHAPTER 2: MOTION IN ONE DIMENSION

2.4. Acceleration Average acceleration The average acceleration of the particle is defined as the change in velocity divided by the time interval during which that change occurs:

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CHAPTER 2: MOTION IN ONE DIMENSION

2.4. Acceleration Instantaneous acceleration The instantaneous acceleration of the particle is defined as the limit of the average acceleration as Δ� approaches zero:

Dimension: L/T2 Acceleration is a vectorial quantity Note: For the case of motion in a straight line ▪ When the object’s velocity and acceleration are in the same

direction, the object is speeding up.

▪ when the object’s velocity and acceleration are in opposite

directions, the object is slowing down.

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CHAPTER 2: MOTION IN ONE DIMENSION

2.5. Particle under constant acceleration Analysis model: Particle under constant acceleration If a particle begins from position đ?‘Ľđ?‘– and initial velocity đ?‘Łđ?‘Ľđ?‘– and moves in a straight line with a constant acceleration đ?‘Žđ?‘Ľ , its subsequent position and velocity are described by the following kinematic equations: đ?‘Łđ?‘Ľđ?‘“ = đ?‘Łđ?‘Ľđ?‘– + đ?‘Žđ?‘Ľ đ?‘Ą

2 2 đ?‘Łđ?‘Ľđ?‘“ = đ?‘Łđ?‘Ľđ?‘– + 2đ?‘Žđ?‘Ľ (đ?‘Ľđ?‘“ − đ?‘Ľđ?‘– )

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CHAPTER 2: MOTION IN ONE DIMENSION

• EX2-1. In a 100m foot-race, you cover the first 50m with an average

velocity of 10m/s and the second 50m with an average velocity of 8m/s. What is your average velocity for the entire 100m. •

EX2-2 A car travels 80km in a straight line. If the first 40km is covered with an average velocity of 80km/h, and the total trip takes 1,2h, what was the average velocity during the second 40km?

• EX2-3

You run 100m in 12s, then turn around and jog 50m back toward the starting point in 30s. Calculate your average speed and your average velocity for the total trip.

• EX2-4

Two train 75 km apart approach each other on parallel tracks, each moving at 15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass each other. How far does the bird fly?

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CHAPTER 2: MOTION IN ONE DIMENSION

2.6. Freely falling object A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. If we neglect air resistance and assume the free-fall acceleration does not vary with altitude over short vertical distances, the motion of a freely falling object moving vertically is equivalent to the motion of a particle under constant acceleration in one dimension. → apply the particle under constant acceleration model (� = 9.80m/s2) Note: ▪ the motion is in the vertical direction (the � direction) rather than

in the horizontal direction

â–Ş if the direction of đ?‘‚đ?‘Ś axis is chosen upward, đ?‘Žđ?‘Ś = −đ?‘” = −9.80 m/s2

where the negative sign means that the acceleration of a freely falling object is downward

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CHAPTER 2: MOTION IN ONE DIMENSION

2.6. Freely falling object EX. 2.5. A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down. (a) Using đ?‘Ąđ??´ = 0 as the time the stone leaves the thrower’s hand at position A, determine the time at which the stone reaches its maximum height. (b) Find the maximum height of the stone. (c) Determine the velocity of the stone when it returns to the height from which it was thrown. (d) Find the velocity and position of the stone at đ?‘Ą = 5.00 s.

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2.6. Freely falling object • EX2-6

Upon graduation, a student throws his cap upward with an initial speed of 14,7m/s. Given that its acceleration is 9,8m/s2 downward. (we neglect air resistance). • a) How long does it take to reach its highest point? • b) What is the distance to the highest point? • c) What is the total time the cap is in the air?

x

Highest point v=0

a = -g

x=0

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2.6. Freely falling object • EX2-7

On a highway at night you see a stalled vehicle and brake your car to stop with an acceleration of magnitude 5m/s2. (deceleration). What is the car’s stopping distance if its initial speed is. • a) 15 m/s (about 54 km/h) • b) 30 m/s • E2-8

An electron in a cathode-ray tube accelerates from rest with a constant acceleration of 5,33 . 1012 m/s2 for 0,15 s. The electron then drifts with constant velocity for 0,2 m. Finally, it comes to rest with an acceleration of -0,67 . 1043 m/s2. How far does the electron travel ?

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CHAPTER 2: MOTION IN ONE DIMENSION

2.6. Freely falling object EX2-8 While standing in an elevator, you see a screw fall from the ceiling. The ceiling is 3m above the floor. a) If the elevator is moving upward with a constant speed of 2,2m/s, how long does it take for the screw to hit the floor? b) How long is the screw in the air if the elevator starts from rest when the screw falls, and moves upward with a constant acceleration of a = 4m/s2 ?

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.1. Position, velocity, acceleration vectors Position vector đ?‘&#x;: ÔŚ drawn from the origin of some coordinate system to the location of the particle in the đ?‘Ľđ?‘Ś plane Displacment vector: Δđ?‘&#x;ÔŚ = đ?‘&#x;ÔŚđ?‘“ − đ?‘&#x;ÔŚđ?‘–

Average velocity vector �Ԍavg during Δ�:

Instantaneous velocity:

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.1. Position, velocity, acceleration vectors Average acceleration is the change in its instantaneous velocity vector Δ�Ԍ divided by the time interval Δ� during which that change occurs

Instantaneous acceleration is the limiting value of the average acceleration as Δ� approaches zero:

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.2. Two-dimensional motion with Motion in two dimensions can be modeled as two independent motions in each of the two perpendicular directions associated with the x and y axes → we can write â–Ş the position vector: đ?‘&#x;ÔŚ = đ?‘ĽiƸ + đ?‘ŚjƸ â–Ş the velocity vector:

â–Ş the acceleration vector:

where iƸ and jƸ are the unit vectors of đ?‘Ľ and đ?‘Ś axes, respectively.

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.2. Two-dimensional motion with In the case of constant acceleration, đ?‘ŽÔŚ = const, its components đ?‘Žđ?‘Ľ and đ?‘Žđ?‘Ś also are constants. Therefore, we can model the particle as a particle under constant acceleration independently in each of the two directions Along đ?‘Ľ direction đ?‘Žđ?‘Ľ = const đ?‘Łđ?‘Ľđ?‘“ = đ?‘Łđ?‘Ľđ?‘– + đ?‘Žđ?‘Ľ đ?‘Ą 1 đ?‘Ľđ?‘“ = đ?‘Ľđ?‘– + đ?‘Łđ?‘Ľđ?‘– đ?‘Ą + đ?‘Žđ?‘Ľ đ?‘Ą 2 2

Along đ?‘Ś direction đ?‘Žđ?‘Ś = const đ?‘Łđ?‘Śđ?‘“ = đ?‘Łđ?‘Śđ?‘– + đ?‘Žđ?‘Ś đ?‘Ą 1 đ?‘Śđ?‘“ = đ?‘Śđ?‘– + đ?‘Łđ?‘Śđ?‘– đ?‘Ą + đ?‘Žđ?‘Ś đ?‘Ą 2 2

đ?’—đ?’‡ = đ?‘Łđ?‘Ľ iƸ + đ?‘Łđ?‘Ś jƸ = đ?‘Łđ?‘Ľđ?‘– + đ?‘Žđ?‘Ľ đ?‘Ą iƸ + đ?‘Łđ?‘Śđ?‘– + đ?‘Žđ?‘Ś jƸ = đ?’—đ?’Š + đ?’‚đ?’•

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.2. Two-dimensional motion with Example 3.1. A particle moves in the đ?‘Ľđ?‘Ś plane, starting from the origin at đ?‘Ą = 0 with an initial velocity having an đ?‘Ľ component of 20 m/s and a đ?‘Ś component of −15 m/s. The particle experiences an acceleration in the đ?‘Ľ direction, given by đ?‘Ž = 4. 0 m/s2 . (a) Determine the total velocity vector at any time. (b) Calculate the velocity and speed of the particle at đ?‘Ą = 5.0 s and the angle the velocity vector makes with the đ?‘Ľ axis. (c) Determine the đ?‘Ľ and đ?‘Ś coordinates of the particle at any time đ?‘Ą and its position vector at this time.

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.3. Projectile motion Considering projectile motion of a particle with two assumptions: (1) the free-fall acceleration is constant over the range of motion and is directed downward, (2) the effect of air resistance is negligible. Along đ?‘Ľ direction đ?‘Žđ?‘Ľ = 0, đ?‘Łđ?‘Ľđ?‘– = đ?‘Łđ?‘– cos đ?œƒđ?‘– đ?‘Łđ?‘Ľđ?‘“ = đ?‘Łđ?‘Ľđ?‘– đ?‘Ľđ?‘“ = đ?‘Ľđ?‘– + đ?‘Łđ?‘Ľđ?‘– đ?‘Ą

Along đ?‘Ś direction đ?‘Žđ?‘Ś = −g, đ?‘Łđ?‘Śđ?‘– = đ?‘Łđ?‘– sin đ?œƒđ?‘– đ?‘Łđ?‘Śđ?‘“ = đ?‘Łđ?‘Śđ?‘– − đ?‘”đ?‘Ą 1 2 đ?‘Śđ?‘“ = đ?‘Śđ?‘– + đ?‘Łđ?‘Śđ?‘– đ?‘Ą − đ?‘”đ?‘Ą 2

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3.3. Projectile motion Maximum height

Horizontal Range

→ The maximum value of �

when đ?œƒđ?‘– = 45°

CHAPTER 3: MOTION IN TWO DIMENSIONS

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.4. Particle in uniform circular motion

If a particle moves in a circular path of radius đ?‘&#x; at a constant speed đ?‘Ł, the magnitude of its centripetal acceleration is

and the period of the particle’s motion is given by

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.5. Tangential and radial acceleration As the particle moves along the curved path, the total acceleration vector đ?‘ŽÔŚ can be written as the vector sum of the component vectors: đ?‘ŽÔŚ = đ?‘ŽÔŚđ?‘&#x; + đ?‘ŽÔŚđ?‘Ą â–Ş The tangential acceleration component đ?‘ŽÔŚđ?‘Ą causes a change in the speed đ?‘Ł of the particle, is parallel to the instantaneous velocity, and its magnitude is given by â–Ş The

radial acceleration component đ?‘ŽÔŚđ?‘&#x; arises from a change in direction of the velocity vector, is toward the center of the circle representing the radius of curvature đ?‘ŽÔŚđ?‘&#x; ≥ đ?‘ŽÔŚđ?‘? .

â–Ş Since đ?‘ŽÔŚđ?‘&#x; ⊼ đ?‘ŽÔŚđ?‘Ą , đ?‘Ž =

đ?‘Žđ?‘&#x;2 + đ?‘Žđ?‘Ą2.

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CHAPTER 3: MOTION IN TWO DIMENSIONS

3.6. Relative velocity and acceleration Galilean transformation equations â–Ş Relative velocity: The velocity đ?‘˘đ?‘ƒđ??´ of a particle measured in a fixed frame of reference Sđ??´ can be related to the velocity đ?‘˘đ?‘ƒđ??ľ of the same particle measured in a moving frame of reference đ?‘†B by đ?‘˘đ?‘ƒđ??´ = đ?‘˘đ?‘ƒđ??ľ + đ?‘ŁÔŚđ??ľđ??´ where đ?‘ŁÔŚđ?‘Šđ?‘¨ is velocity of đ?‘†đ??ľ relative to đ?‘†đ??´ . â–Ş Relative acceleration

If đ?‘ŁÔŚđ??ľđ??´ is constant, therefore đ?‘ŽÔŚđ?‘ƒđ??´ = đ?‘ŽÔŚđ?‘ƒđ??ľ .

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CHAPTER 4: THE LAWS OF MOTION

4.1. The concept of force The force refers to an interaction with an object by means of muscular activity and some change in the object’s velocity. Note: 1. Forces do not always cause motion. 2. Forces have been verified to behave as vectors

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CHAPTER 4: THE LAWS OF MOTION

4.2. Newton’s first law and inertial frame Newton’s first law of motion If an object does not interact with other objects, it is possible to identify a reference frame in which the object has zero acceleration. The reference frame mentioned in Newton’s first law is call an inertial frame of reference. The Newton’s first law of motion can be more practically statemented as follows: In the absence of external forces and when viewed from an inertial reference frame, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line).

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CHAPTER 4: THE LAWS OF MOTION

4.3. Newton’s second law Newton’s second law of motion When viewed from an inertial reference frame, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass:

Ex1. A hockey puck having a mass of 0.30 kg slides on the friction. The force đ??šÔŚđ?&#x;? has a magnitude of 5.0 N, and is directed at đ?œƒ = 20° below the đ?‘Ľ axis. The force đ??šÔŚđ?&#x;? has a magnitude of 8.0 N and its direction is đ?œ™ = 60° above the đ?‘Ľ axis. Determine both the magnitude and the direction of the puck’s acceleration

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4.4. The gravitational force and weight All objects are attracted to the Earth. The attractive force exerted by the Earth on an object is called the gravitational force đ??šÔŚg. This force is directed toward the center of the Earth, and its magnitude is called the weight of the object. â–Ş The gravitational force đ??šÔŚg = đ?‘šg â–Ş The weight of an object

đ??šg = đ?‘šg Note: Inertial mass vs gravitational mass â–Ş Mass in Newton’s second law is call inertial mass â–Ş Mass in the expression of the weight is the gravitational mass The experiments in Newtonian dynamics conclude that gravitational mass and inertial mass have the same value.

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4.5. Analysis models using Newton’s second law Particle in equilibrium

Particle under a net force

If it has several forces acting on it so that the forces all cancel, giving a net force of zero, the object will have an acceleration of zero. This condition is mathematically described as ∑đ??šÔŚ = 0

If it has one or more forces acting on it so that there is a net force on the object, it will accelerate in the direction of the net force. The relationship between the net force and the acceleration is ∑đ??šÔŚ = đ?‘šđ?‘ŽÔŚ

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4.5. Analysis models using Newton’s second law Ex. 5.2. A traffic light weighing 122 N hangs from a cable tied to two other cables fastened to a support as in the below figure. The upper cables make angles of đ?œƒ1 = 37.0° and đ?œƒ2 = 53.0° with the horizontal. These upper cables are not as strong as the vertical cable and will break if the tension in them exceeds 100 N. Does the traffic light remain hanging in this situation, or will one of the cables break?

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4.5. Analysis models using Newton’s second law Ex. 5.3. A car of mass m is on an icy driveway inclined at an angle đ?œƒ as in beside figure. (a) Find the acceleration of the car, assuming the driveway is frictionless. (b) Suppose the car is released from rest at the top of the incline and the distance from the car’s front bumper to the bottom of the incline is d. How long does it take the front bumper to reach the bottom of the hill, and what is the car’s speed as it arrives there?

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4.5. Analysis models using Newton’s second law Ex. 5.4. Two blocks of masses đ?‘š1 and đ?‘š2, with đ?‘š1 > đ?‘š2, are placed in contact with each other on a frictionless, horizontal surface as in beside figure. A constant horizontal force đ??šÔŚ is applied to đ?‘š1 as shown. (a) Find the magnitude of the acceleration of the system. (b) Determine the magnitude of the contact force between the two blocks.

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4.5. Analysis models using Newton’s second law Ex. 5.5. A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice.

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5 5.1. Extending the article in Uniform Circular Motion Model 5.2. Non-uniform Circular Motion 5.3 Motion in Accelerated Frames 5.4 Motion Presence of Forces

in the Resistive

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CHAPTER 5: CIRCULAR MOTION AND APPLICATIONS OF NEWTON’S LAW

5.1. Extending the Particle in Uniform Circular Motion Model Applying Newton’s second law along the radial direction, the net force causing the centripetal acceleration can be related to the acceleration as follows:

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5.1. Extending the Particle in Uniform Circular Motion Model Ex. 6.1. A small ball of mass đ?‘š is suspended from a string of length đ??ż. The ball revolves with constant speed đ?‘Ł in a horizontal circle of radius đ?‘&#x; . (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) Find an expression for đ?‘Ł. Ans. đ?‘Ł =

đ??żđ?‘” sin đ?œƒ tan đ?œƒ

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5.2. Non-uniform Circular Motion The total force exerted on the particle can be written in terms of РѕЉ­Юљ╣нд = РѕЉ­Юљ╣нд­ЮЉЪ + РѕЉ­Юљ╣нд­ЮЉА , where Рќф РѕЉ­Юљ╣нд­ЮЉЪ = ­ЮЉџ­ЮЉјнд ­ЮЉЪ :radial component directed toward the center of the circle Рќф РѕЉ­Юљ╣нд­ЮЉА = ­ЮЉџ ­ЮЉјнд­ЮЉА : tangential component representing a change in the particleРђЎs speed with time. Remind:

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5.2. Non-uniform Circular Motion Ex. 6.6. A small sphere of mass đ?‘š is attached to the end of a cord of length đ?‘… and set into motion in a vertical circle about a fixed point O as illustrated in the beside figure. Determine the tangential acceleration of the sphere and the tension in the cord at any instant when the speed of the sphere is đ?‘Ł and the cord makes an angle đ?œƒ with the vertical. Ans.

cos đ?œƒá‰

đ?‘Žđ?‘Ą = đ?‘” sin đ?œƒ , đ?‘‡ =

�2 �� ቀ + ��

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5.3. Motion in Accelerated Frames An observer in a noninertial (accelerating) frame of reference introduces fictitious forces when applying Newton’s second law in that frame. The fictitious force has the form of đ??šÔŚđ?‘“ = −đ?‘šđ??´ÔŚ with đ?‘š is the mass of a particle, đ??´ÔŚ acceleration of the noninertial frame.

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5.3. Motion in Accelerated Frames Ex. 6.7. A small sphere of mass đ?‘š hangs by a cord from the ceiling of a boxcar that is accelerating to the right as shown in the below figure. Both the inertial observer on the ground and the noninertial observer on the train agree that the cord makes an angle đ?œƒ with respect to the vertical. The noninertial observer claims that a force, which we know to be fictitious, causes the observed deviation of the cord from the vertical. How is the magnitude of this force related to the boxcar’s acceleration measured by the inertial observer?

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5.4. Motion in the Presence of Resistive Forces Model 1: Resistive Force Proportional to Object Velocity If we model the resistive force acting on an object moving through a liquid or gas as proportional to the object’s velocity, the resistive force can be expressed as đ?‘š = −đ?’ƒđ?’— (b = const) Ex. 6.8. A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil, where it experiences a resistive force proportional to its speed. The sphere reaches a terminal speed of 5.00 cm/s. Determine the time constant t and the time at which the sphere reaches 90.0% of its terminal speed.

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5.4. Motion in the Presence of Resistive Forces Model 2: Resistive Force Proportional to Object Speed Squared For objects moving at high speeds through air, such as airplanes, skydivers, cars, and baseballs, the resistive force is reasonably well modeled as proportional to the square of the speed. In these situations, the magnitude of the resistive force can be expressed as đ?&#x;? đ?‘š = đ?‘Ťđ??†đ?‘¨đ?’—đ?&#x;? đ?&#x;? where đ??ˇ is a dimensionless empirical quantity called the drag coefficient, đ?œŒ is the density of air, and đ??´ is the cross-sectional area of the moving object measured in a plane perpendicular to its velocity.

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6

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CHAPTER 6: ENERGY OF A SYSTEM

6.1. Work Done by a Constant Force The work đ?‘Š done on a system by an agent exerting a constant force on the system is the product of the magnitude đ??š of the force, the magnitude Δđ?‘&#x; of the displacement of the point of application of the force, and cos đ?œƒ, where đ?œƒ is the angle between the force and displacement vectors: đ?‘Š = đ??šÔŚ . Δđ?‘&#x;ÔŚ = đ??šÎ”đ?‘&#x; cos đ?œƒ Note: â–Ş If đ?œƒ = 90° then đ?‘Š = 0 â–Ş If đ?œƒ = 0° then đ?‘Š = đ??šÎ”đ?‘&#x; â–Ş The sign of the work also depends on the direction of đ??šÔŚ relative to Δđ?‘&#x;: ÔŚ đ?‘Š > 0 if đ?œƒ < 90° đ?‘Š < 0 if đ?œƒ > 90°

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6.2. Work Done by a Varying Force For the general case of a net force ∑đ??šÔŚ whose magnitude and direction may both vary, the total work done on the system is ∑đ?‘Š = đ?‘Šđ?‘’đ?‘Ľđ?‘Ą = ŕśą ∑đ??šÔŚ . đ?‘‘đ?‘&#x;ÔŚ Note: Work Done by a Spring â–Ş Hooke’s law đ??šđ?‘ = −đ?‘˜đ?‘Ľ (đ?‘˜: spring constant) â–Ş If the block undergoes an arbitrary displacement from đ?‘Ľ = đ?‘Ľđ?‘– to đ?‘Ľ = đ?‘Ľđ?‘“ , the work done by the spring force on the block is

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6.2. Work Done by a Varying Force Ex. 7.5. A common technique used to measure the force constant of a spring is demonstrated by the setup in the Figure. The spring is hung vertically, and an object of mass m is attached to its lower end. Under the action of the “loadâ€? mg, the spring stretches a distance d from its equilibrium position. (a) If a spring is stretched 2.0 cm by a suspended object having a mass of 0.55 kg, what is the force constant of the spring? (b) How much work is done by the spring on the object as it stretches through this distance? Ans. đ?‘˜ = 2.7 Ă— 102N/m đ?‘Šđ?‘ = −5.4 Ă— 10−2J

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6.3. Kinetic Energy and the Work–Kinetic Energy Theorem The net work done on the block through a displacement Δđ?‘&#x;ÔŚ = Δđ?‘Ľ đ?‘–Ƹ = đ?‘Ľđ?‘“ − đ?‘Ľđ?‘– đ?‘–:Ƹ 1 1 2 đ?‘Šđ?‘’đ?‘Ľđ?‘Ą = đ?‘šđ?‘Łđ?‘“ − đ?‘šđ?‘Łđ?‘–2 2 2 â–Ş Kinetic energy

1 đ??ž = đ?‘šđ?‘Ł 2 2 â–Ş Work-kinetic energy theorem When work is done on a system and the only change in the system is in its speed, the net work done on the system equals the change in kinetic energy of the system, as expressed by đ?‘Šđ?‘’đ?‘Ľđ?‘Ą = đ??žđ?‘“ − đ??žđ?‘– = Δđ??ž

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6.3. Kinetic Energy and the Work–Kinetic Energy Theorem Ex. 7.6. A 6.0-kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force of magnitude 12 N. Find the block’s speed after it has moved through a horizontal distance of 3.0 m. Ans. �� = 3.5 m/s

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6.4. Potential Energy of a System Conservative forces have these two equivalent properties 1. The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle. 2. The work done by a conservative force on a particle moving through any closed path is zero. Note: The work đ?‘Šđ?‘–đ?‘›đ?‘Ą done by a conservative force on an object that is a member of a system as the system changes from one configuration to another is equal to the initial value of the potential energy of the system minus the final value: đ?‘žđ??˘đ??§đ??­ = đ?‘źđ?’Š − đ?‘źđ?’‡ = đ?šŤđ?‘ź â–Ş The gravitational potential energy of the particle–Earth system is đ?‘ˆđ?‘” = đ?‘šđ?‘”đ?‘Ś + đ??ś â–Ş The elastic potential energy stored in a spring of force constant k is 1 2 đ?‘ˆđ?‘ = đ?‘˜đ?‘Ľ + đ??ś 2

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6.4. Potential Energy of a System Relationship Between Conservative Forces and Potential Energy The đ?‘Ľ component of a conservative force acting on a member within a system equals the negative derivative of the potential energy of the system with respect to đ?‘Ľ đ?’…đ?‘ź đ?‘­đ?’™ = − đ?’…đ?’™ Mechanical energy of the system: đ?‘Ź =đ?‘˛+đ?‘ź â–Ş K: kinetic energy of all moving members of the system â–Ş U: all types of potential energy in the system

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CHAPTER

7 7.1. Non-isolated system 7.2. Isolated system 7.3. Situations involving Kinetic friction 7.4. Changes in mechanical energy for non-conservative forces 7.5. Power

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Energy transfer mechanisms

7.1. Non-isolated system

CHAPTER 7: CONSERVATION OF ENERGY

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7.1. Non-isolated system If the total amount of energy in a system changes, it can only be because energy has crossed the boundary of the system by a transfer mechanism. đ?šŤđ?‘Źđ??Źđ??˛đ??Źđ??­đ??žđ??Ś = ∑đ?‘ť Δđ??ž + Δđ?‘ˆ + Δđ??¸int = đ?‘Š + đ?‘„ + đ?‘‡MW + đ?‘‡MT + đ?‘‡ET + đ?‘‡ER â–Ş đ??¸system: total energy of the system including đ??ž (kinetic energy), đ?‘ˆ

(potential energy), đ??¸int (internal energy) â–Ş đ?‘‡: amount of energy transferred across the system boundary by some mechanism including đ?‘Š (work), đ?‘„ (heat), đ?‘‡MW (mechanical waves), đ?‘‡MT (matter transfer), đ?‘‡ET (electrical transmission), đ?‘‡ER (electromagnetic radiation) Note: Non-isolated system is a system for which energy crosses the boundary of the system during some time interval due to an interaction with the environment.

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7.1. Non-isolated system

Conservation of energy

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7.2. Isolated system The total energy of an isolated system (no interact with its environment) is conserved: Δđ??¸system = 0 Especially, if no non-conservative forces act in an isolated system, the mechanical energy of the system is conserved: đ?šŤđ?‘Źđ?’Žđ?’†đ?’„đ?’‰ = đ?šŤđ?‘˛ + đ?šŤđ?‘ź = đ?&#x;Ž Examples â–Ş Mechanical energy of an object alone by the gravitational force is conserved: 1 1 2 đ?‘šđ?‘”đ?‘§đ??´ + đ?‘šđ?‘Łđ??´ = đ?‘šđ?‘”đ?‘§đ??ľ + đ?‘šđ?‘Łđ??ľ2 2 2 â–Ş Mechanical energy of an object attached to a horizontal is conserved if if the surface is frictionless: 1 2 1 1 2 1 2 đ?‘˜đ?‘Ľđ??´ + đ?‘šđ?‘Łđ??´ = đ?‘˜đ?‘Ľđ??ľ + đ?‘šđ?‘Łđ??ľ2 2 2 2 2

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7.3. Situations involving kinetic friction If a friction force of magnitude đ?‘“đ?‘˜ acts over a distance đ?‘‘ within a system, the change in internal energy of the system is Δđ??¸int = đ?‘“đ?‘˜ đ?‘‘ Example 8.4. A block of mass 1.6 kg is attached to a horizontal spring that has a force constant of 1000 N/m as shown in the below Figure. The spring is compressed 2.0 cm and is then released from rest . (a) Calculate the speed of the block as it passes through the equilibrium position đ?‘Ľ = 0 if the surface is frictionless. (b) Calculate the speed of the block as it passes through the equilibrium position if a constant friction force of 4.0 N retards its motion from the moment it is released.

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7.4. Changes in mechanical energy Reminding that: â–Ş Change in kinetic energy: Δđ??ž = ∑đ?‘Šđ??š0 = đ?‘Šđ??šđ?‘? + đ?‘Šđ?‘“đ?‘˜ +đ?‘Šother â–Ş Change in potential energy: đ?›Ľđ?‘ˆ = −đ?‘Šđ??šđ?‘? â–Ş Change in internal energy: Δđ??¸int = đ?‘“đ?‘˜ đ?‘‘ = −đ?‘Šđ?‘“đ?‘˜

→ or

Δđ??¸system = Δđ??ž + Δđ?‘ˆ + Δđ??¸int = đ?‘Šother đ?šŤđ?‘Źđ?’Žđ?’†đ?’„đ?’‰ = đ?œ&#x;đ?‘˛ + đ?œ&#x;đ?‘ź = đ?‘žđ?’?đ?’•đ?’‰đ?’†đ?’“ − đ?šŤđ?‘Źđ?’Šđ?’?đ?’• = đ?‘žđ?’?đ?’•đ?’‰đ?’†đ?’“ + đ?‘žđ?’‡đ?’Œ

Note: â–Ş đ??šđ?‘? : conservative force â–Ş đ?‘“đ?‘˜ : friction force â–Ş đ?‘Šđ?‘œđ?‘Ąâ„Žđ?‘’đ?‘&#x; is the work of all forces acting into the system excepting for conservative force and friction force.

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7.4. Changes in mechanical energy Ex. 8.7. A 3.00-kg crate slides down a ramp. The ramp is 1.00 m in length and inclined at an angle of 30.0° as shown in the Figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. (a) Use energy methods to determine the speed of the crate at the bottom of the ramp. (b) How far does the crate slide on the horizontal floor if it continues to experience a friction force of magnitude 5.00 N?

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7.5. Power The time rate of energy transfer is called the instantaneous power đ?‘ƒ and is defined as

In the case of energy transfer by work:

â–Ş Unit of power: 1đ?‘Š = 1đ??˝/đ?‘ â–Ş Relation of power, force and velocity

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7.5. Power Ex. 8.11. An elevator car has a mass of 1 600 kg and is carrying passengers having a combined mass of 200 kg. A constant friction force of 4 000 N retards its motion. (a) How much power must a motor deliver to lift the elevator car and its passengers at a constant speed of 3.00 m/s? (b) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1.00 m/s2?

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8 8.1. Linear momentum 8.2. Analysis model: Isolated system 8.3. Analysis model: nonisolated system 8.4. Collisions in 1D 8.5. Collisions in 2D 8.6. The center of mass 8.7. Systems of many particles 8.8. Rocket propulsion

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8.1. Linear momentum The other expression of Newton’s second law â–Ş According to Newton’s 2nd law đ??šÔŚ = đ?‘šđ?‘ŽÔŚ â–Ş Replacing the acceleration with its definition, we get

â–Ş If đ?‘š = const then

The linear momentum of a particle or an object that can be modeled as a particle of mass đ?‘š moving with a velocity đ?‘ŁÔŚ is defined to be the product of the mass and velocity of the particle: đ?’‘ = đ?’Žđ?’—

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8.2. Analysis model: momentum of an isolated system Momentum version of the isolated system model If no external forces act on the system ( ∑đ??šÔŚ = 0), the total momentum of the system is constant đ?šŤđ?‘ˇđ??­đ??¨đ??­ = đ?&#x;Ž Note: â–Ş The total momentum of a system Δđ?‘ƒtot = đ?‘ƒ1 + đ?‘ƒ2 + â‹Ż = đ?‘š1đ?‘ŁÔŚ1 + đ?‘š2đ?‘ŁÔŚ2 + â‹Ż â–Ş The equation Δđ?‘ƒtot = 0 can be written

đ?‘ƒtot,đ?‘– = đ?‘ƒtot,đ?‘“

or

đ?‘ƒ1đ?‘– + đ?‘ƒ2đ?‘– + â‹Ż = đ?‘ƒ1đ?‘“ + đ?‘ƒ2đ?‘“ + â‹Ż

â–Ş The momentum conservation in one direction

If đ??šđ?‘Ľ = 0 (note: ∑đ??šÔŚ is not necessary to be 0), the x-component of the total momentum of the system is conservative đ?‘ˇđ?&#x;?đ?’™,đ?’Š + đ?‘ˇđ?&#x;?đ?’™,đ?’Š + â‹Ż = đ?‘ˇđ?&#x;?đ?’™,đ?’‡ + đ?‘ˇđ?&#x;?đ?’™,đ?’‡ + â‹Ż

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8.2. Analysis model: momentum of an isolated system Ex. 9.1. A 60-kg archer stands at rest on frictionless ice and fires a 0.030-kg arrow horizontally at 85 m/s. With what velocity does the archer move across the ice after firing the arrow?

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8.3. Analysis model: momentum of a nonisolated system Impulse-momentum theorem for a particle According to the second form of Newton’s 2nd law, we have ÔŚ đ?‘‘đ?‘ƒ = ∑đ??šđ?‘‘đ?‘Ą Integrating the above expression to get the chance in momentum:

đ?‘Ą ÔŚ called the impulse of the net force ∑đ??šÔŚ acting on a where đ??źÔŚ = ‍đ?‘Ąđ?‘‘đ??šâˆ‘ đ?‘“ đ?‘Ą×Źâ€Ź đ?‘–

particle over the time interval Δđ?‘Ą = đ?‘Ąđ?‘“ − đ?‘Ąđ?‘–.

→ The change in the momentum of a particle is equal to the

impulse of the net force acting on the particle: �� = �Ԍ

Note: If ∑đ??šÔŚ = const, đ??źÔŚ = ∑đ??šÔŚ Δt, therefore đ?šŤđ??? = ∑đ?‘­ đ?šŤđ??­ ≥ ∑đ?‘­đ?’‚đ?’—đ?’ˆđ?šŤđ??­

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8.3. Analysis model: momentum of a nonisolated system Ex. 9.3. In a particular crash test, a car of mass 1500 kg collides with a wall. The initial and final velocities of the car are đ?‘Łđ?‘– = − 15.0đ?‘–Ƹ m/s and đ?‘ŁÔŚđ?‘“ = 2.60đ?‘–Ƹ m/s , respectively. If the collision lasts 0.150s, find the impulse caused by the collision and the average net force exerted on the car.

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8.4. Collisions in one dimension Categorization of collisions Elastic collision

Inelastic collision

Total momentum of two objects is conserved đ?’Žđ?&#x;? đ?’—đ?&#x;?đ?’Š + đ?’Žđ?&#x;?đ?’—đ?&#x;?đ?’Š = đ?’Žđ?&#x;?đ?’—đ?&#x;?đ?’‡ + đ?’Žđ?&#x;?đ?’—đ?&#x;?đ?’‡

Total kinetic energy of the system is constant đ?‘˛đ?&#x;?đ?’Š + đ?‘˛đ?&#x;?đ?’Š = đ?‘˛đ?&#x;?đ?’‡ + đ?‘˛đ?&#x;?đ?’‡

Perfectly inelastic collision Conservation of momentum đ?‘š1đ?‘ŁÔŚ1đ?‘– + đ?‘š2đ?‘ŁÔŚ2đ?‘– = đ?‘š1 + đ?‘š2 đ?‘ŁÔŚđ?‘“

Total kinetic energy of the system is not conserved đ?šŤđ?‘˛ = ∑đ?‘˛đ?’‡ − ∑đ?‘˛đ?’Š < đ?&#x;Ž

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8.4. Collisions in one dimension Elastic collision Momentum: đ?‘?đ?‘– = đ?‘?đ?‘“ → đ?‘š1 đ?‘Ł1đ?‘– + đ?‘š2 đ?‘Ł2đ?‘– = đ?‘š1 đ?‘Ł1đ?‘“ + đ?‘š2 đ?‘Ł2đ?‘“ 2 2 2 2 Kinetic energy: đ??žđ?‘– = đ??žđ?‘“ → đ?‘š1 đ?‘Ł1đ?‘– + đ?‘š2 đ?‘Ł2đ?‘– = đ?‘š1 đ?‘Łđ?‘–đ?‘“ + đ?‘š2 đ?‘Ł2đ?‘“

→ Solutions of these equations are

Special cases â–Ş If đ?‘š1 = đ?‘š2 then đ?‘Ł1đ?‘“ = đ?‘Ł2đ?‘– and đ?‘Ł2đ?‘“ = đ?‘Ł1đ?‘– â–Ş If đ?‘Ł2đ?‘– = 0, đ?‘š1 ≍ đ?‘š2 then đ?‘Ł1đ?‘“ ≈ đ?‘Ł1đ?‘– and đ?‘Ł2đ?‘“ ≈ 2đ?‘Ł1đ?‘– â–Ş If đ?‘Ł2đ?‘– = 0, đ?‘š1 ≪ đ?‘š2 then đ?‘Ł1đ?‘“ ≈ −đ?‘Ł1đ?‘– and đ?‘Ł2đ?‘“ ≈ 0

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8.5. Collisions in two dimension Conservation of momentum

Example We have đ?‘ŁÔŚ1đ?‘– = đ?‘Ł1đ?‘– đ?‘–,Ƹ đ?‘ŁÔŚ2i = 0 đ?‘ŁÔŚ1đ?‘“ = đ?‘Ł1đ?‘“ cos đ?œƒ đ?‘–Ƹ + đ?‘Ł1đ?‘“ sin đ?œƒ đ?‘—Ƹ đ?‘ŁÔŚ2đ?‘“ = đ?‘Ł2đ?‘“ cos đ?œ™ đ?‘–Ƹ − đ?‘Ł2đ?‘“ sin đ?œ™ đ?‘—Ƹ Conservation of momentum đ?‘Ł1đ?‘– = đ?‘Ł1đ?‘“ cos đ?œƒ + đ?‘Ł2đ?‘“ cos đ?œ™

0 = đ?‘Ł1đ?‘“ sin đ?œƒ − đ?‘Ł2đ?‘“ sin đ?œ™

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8.5. Collisions in two dimension Ex. 9.8.

A 1 500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2 500-kg truck traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

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8.6. Center of mass The position vector of the center of mass of a system of particles is defined as

where đ?‘€ = ∑đ?‘– đ?‘šđ?‘– is the total mass of the system and đ?‘&#x;ÔŚđ?‘– is the position vector of the đ?‘– th particle. Note. The position vector of the center of mass of an extended object can be obtained from the integral expression đ?‘&#x;ÔŚđ??śđ?‘€

1 = ŕśąđ?‘&#x;ÔŚ đ?‘‘đ?‘š đ?‘€

Ex. A system consists of three particles (đ?‘š1 = đ?‘š2 = 1 kg, đ?‘š3 = 2 kg) located as shown in the figure. Find the center of mass of the system.

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8.7. System of many particles â–Ş The velocity of the center of mass for a system of particles is

â–Ş The total momentum of a system of particles equals the total mass

multiplied by the velocity of the center of mass:

▪ Newton’s second law applied to a system of particles is

where đ?‘ŽÔŚđ??śđ?‘€ is the acceleration of the center of mass and the sum is over all external forces. → The center of mass moves like an imaginary particle of mass M under the influence of the resultant external force on the system.

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8.8. Rocket propulsion ▪ At the time �: M + Δm � where � is

the speed of the rocket relative to the Earth.

▪ Over a short time intervalΔ� , the

rocket ejects fuel of mass Δ� to reach the speed � + Δ�.

▪ The fuel is ejected with a speed �

relative to the rocket.

→ the final speed �� of the rocket when its mass is �� relates to the initial speed �� of the rocket at the mass �� as follows

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