Cracking the AP Physics C Exam 2018

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ACKNOWLEDGMENTS

The Princeton Review would like to give a special thanks to Nick Owen and Scott Sennello for their careful reviews of the 2018 edition. Our gratitude also goes out to this book’s diligent production editors, Jim Melloan and Harmony Quiroz, and its extremely talented production artist, Deborah A. Silvestrini.

Cover

Title Page

Copyright

Acknowledgments

Part I: Using This Book to Improve Your AP Score

Preview: Your Knowledge, Your Expectations

Your Guide to Using This Book

How to Begin

Part II: Practice Test 1

Practice Test 1

Practice Test 1: Answers and Explanations

Part III: About the AP Physics C Exam

The Structure of the AP Physics C Exam

How the AP Physics C Exam Is Scored

Overview of Content Topics

How AP Exams Are Used

Other Resources

Designing Your Study Plan

Part IV: Test-Taking Strategies for the AP Physics C Exam

Preview

1 How to Approach Multiple-Choice Questions

2 How to Approach Free-Response Questions

3 Using Time Effectively to Maximize Points

Reflect

Part V: Content Review for the AP Physics C Exam

4 Vectors

Introduction

Definition

Vector Addition (Geometric)

Scalar Multiplication

Vector Subtraction (Geometric)

Standard Basis Vectors

Vector Operations Using Components

Magnitude of a Vector

Direction of a Vector

The Dot Product

The Cross Product

5 Kinematics

Introduction

Position, Distance, and Displacement

Speed and Velocity

Acceleration

Uniformly Accelerated Motion and the Big Five

Kinematics with Graphs

Free Fall

Projectile Motion

Chapter 5 Drill

Summary

6 Newton’s Laws

Introduction

The First Law

The Second Law

The Third Law

Weight

The Normal Force

Friction

Pulleys

Inclined Planes

Uniform Circular Motion

Chapter 6 Drill

Summary

7 Work, Energy, and Power

Introduction

Work

Kinetic Energy

The Work–Energy Theorem

Potential Energy

Conservation of Mechanical Energy

Potential Energy Curves

Power

Chapter 7 Drill

Summary

8 Linear Momentum

Introduction

Impulse

Conservation of Linear Momentum

Collisions

Center of Mass

Chapter 8 Drill

Summary

9 Rotational Motion

Introduction

Rotational Kinematics

The Big Five for Rotational Motion

Rotational Dynamics

Kinetic Energy of Rotation

Work and Power

Angular Momentum

Conservation of Angular Momentum

Equilibrium

Chapter 9 Drill

Summary

10 Laws of Gravitation

Kepler’s Laws

Newton’s Law of Gravitation

The Gravitational Attraction Due to an Extended Body

Gravitational Potential Energy

Proving the Equation

Orbits of the Planets

Chapter 10 Drill

Summary

11 Oscillations

Introduction

Simple Harmonic Motion (SHM): The Spring–Block Oscillator

The Kinematics of Simple Harmonic Motion

The Spring–Block Oscillator: Vertical Motion

The Sinusoidal Description of Simple Harmonic Motion

Pendula

Chapter 11 Drill

Summary

12 Electric Forces and Fields

Introduction

Coulomb’s Law

The Electric Field

Conductors and Insulators

Gauss’s Law

Chapter 12 Drill

Summary

13 Electric Potential and Capacitance

Introduction

Electrical Potential Energy

The Potential of a Sphere

The Potential of a Cylinder

Chapter 13 Drill

Summary

14 Direct Current Circuits

Introduction

Electric Current

Electric Circuits

Resistance–Capacitance (RC) Circuits

Chapter 14 Drill

Summary

15 Magnetic Forces and Fields

Introduction

The Magnetic Force on a Moving Charge

The Magnetic Force on a Current-Carrying Wire

Magnetic Fields Created by Current-Carrying Wires

The Biot–Savart Law

Ampere’s Law

Chapter 15 Drill

Summary

16 Electromagnetic Induction

Introduction

Motional Emf

Faraday’s Law of Electromagnetic Induction

Inductance

Maxwell’s Equations

Chapter 16 Drill

Summary

17 Chapter Drills: Answers and Explanations

Part VI: Practice Test 2

Practice Test 2

Practice Test 2: Answers and Explanations

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Part I

Using This Book to Improve Your AP Score

• Preview: Your Knowledge, Your Expectations

• Your Guide to Using This Book

• How to Begin

PREVIEW: YOUR KNOWLEDGE, YOUR EXPECTATIONS

Your route to a high score on the AP Physics C Exam will depend on how you plan to use this book. Please respond to the following questions.

1. Rate your level of confidence about your knowledge of the content tested by the AP Physics C Exam.

A. Very confident—I know it all

B. I’m pretty confident, but there are topics for which I could use help

C. Not confident—I need quite a bit of support

D. I’m not sure

2. If you have a goal score in mind, highlight your goal score for the AP Physics C Exam.

5 4 3 2 1 I’m not sure yet

3. What do you expect to learn from this book? Highlight all that apply to you.

A. A general overview of the test and what to expect

B. Strategies for how to approach the test

C. The content tested by this exam

D. I’m not sure yet

YOUR GUIDE TO USING THIS BOOK

This book is organized to provide as much—or as little—support as you need, so you can use this book in whatever way will be most helpful for improving your score on the AP Physics C Exam.

• The remainder of Part I will provide guidance on how to use this book and help you determine your strengths and weaknesses.

• Part II of this book contains your first practice test, along with the answers and explanations. (Bubble sheets can be downloaded here.) This is where you should begin your test preparation in order to realistically determine

your starting point right now

which question types you’re ready for and which you might need to practice

which content topics you are familiar with and which you will want to carefully review

Note that the answer key for Practice Test 1 has been specifically designed to help you self-diagnose any potential areas of weakness, so that you can best focus your test preparation and be efficient with your time.

• Part III of this book will

inform you about the structure, scoring, and content of the AP Physics C Exam

help you make a study plan

point you toward additional resources

• Part IV of this book will explore various strategies including

how to attack multiple-choice questions

how to write high scoring free-response answers

how to manage your time to maximize the number of points available to you

• Part V of this book covers the content you need to know for your exam.

• Part VI of this book contains Practice Test 2 and its answers and explanations. (Again, bubble sheets can be downloaded here) If you skipped Practice Test 1, we recommend that you take both tests (with at least a day or two between them) so that you can compare your progress between them. Additionally, this will help to identify any external issues: If you get a certain type of question wrong both times, you probably need to review it. If you only got it wrong once, you may have run out of time or been distracted by something. In either case, this will allow you to focus on the factors that caused the discrepancy and to be as prepared as possible on the day of the test.

You may choose to use some parts of this book over others, or you may work through the entire book. This will depend on your needs and how much time you have. Let’s now look at how to make this determination.

HOW TO BEGIN

1. Take Practice Test 1

Before you can decide how to use this book, you need to take a practice test. Doing so will give you insight into your strengths and weaknesses, and the test will also help you make an effective study plan. If you’re feeling test-phobic, remind yourself that a practice test is a tool for diagnosing yourself—it’s not how well you do that matters but how you use information gleaned from your performance to guide your preparation.

So, before you read further, take Practice Test 1 starting on this page of this book. Be sure to do so in one sitting, following the instructions that appear before the test.

2. Check Your Answers

Using the answer key on this page, count how many multiple-choice questions you got right and how many you missed. Don’t worry about the explanations for now, and don’t worry about why you missed questions. We’ll get to that soon.

3. Reflect on the Test

After you take your first test, respond to the following questions:

• How much time did you spend on the multiple-choice questions?

• How much time did you spend on each free-response question?

• How many multiple-choice questions did you miss?

• Do you feel you had the knowledge to address the subject matter of the free-response questions?

• Do you feel you wrote well-organized, thoughtful free responses?

4. Read Part III of this Book and Complete the Self-Evaluation

Part III will provide information on how the test is structured and scored. It will also set out areas of content that are tested.

As you read Part III, re-evaluate your answers to the questions above. At the end of Part III, you will revisit the questions on the previous page and refine your answers to them. Use the diagnostic answer key to identify the content chapters of this book in which you missed the most questions, as that may present you with a good place to begin your review. Make a study plan, based on your needs and time available, that will help you to use this book most effectively.

5. Engage with Parts IV and V as Needed

Notice the word engage. You’ll get more out of this book if you use it intentionally than if you read it passively, hoping for an improved score through osmosis.

Strategy chapters will help you think about your approach to the question types on this exam. Part IV will open with a reminder to think about how you approach questions now and close with a reflection section asking you to think about how/whether you will change your approach in the future.

Content chapters are designed to provide a review of the content tested on the AP Physics C Exam, including the level of detail you need to know and how the content is tested. You will have the opportunity to assess your mastery of the content of each chapter through test-appropriate questions and a reflection section.

6. Take Practice Test 2 and Assess Your Performance

Once you feel you have developed the strategies you need and gained the knowledge you lacked, you should take Practice Test 2. You should do so in one sitting, following the instructions at the beginning of the test.

When you are done, check your answers to the multiple-choice sections. See if a teacher will read your essays and provide feedback.

Once you have taken the test, reflect on what areas you still need to work on, and revisit the chapters in this book that address those deficiencies. Through this type of reflection and engagement, you will continue to improve.

7. Keep Working

After you have revisited certain chapters in this book, continue the process of testing, reflecting, and engaging with the next practice test in this book. Consider what additional work you need to do and how you will change your strategic approach to different parts of the test. You can continue to explore areas that can stand to be improved and engage in those areas right up to the day of the test. As we will discuss in Part III, there are other resources available to you, including a wealth of information on AP Students.

Part II

Practice Test 1

• Practice Test 1

• Practice Test 1: Answers and Explanations

Practice Test 1

Click here to download the PDF.

PHYSICS C

Physics C has two exams: Physics C (Mechanics) and Physics C (Electricity & Magnetism):

	Physics C (Mechanics)	Physics C (Electricity & Magnetism)
First 45 min.	Sec. I, Multiple Choice 35 questions	Sec. I, Multiple Choice 35 questions
Second 45 min.	Sec. II, Free Response 3 questions	Sec. II, Free Response 3 questions

You may take just Mechanics or just Electricity and Magnetism, or both. If you take both, you will receive a separate grade for each. Each section of each examination is 50 percent of the total grade; each question in a section has equal weight. Calculators are permitted on both sections of the exam. However, calculators cannot be shared with other students and calculators with typewriter-style (QWERTY) keyboards will not be permitted. On the following pages you will find the Table of Information that is provided to you during the exam.

If you are taking

— Mechanics only, please be careful to answer numbers 1–35;

— Electricity and Magnetism only, please be careful to answer numbers 36–70;

— the entire examination (Mechanics and Electricity and Magnetism), answer numbers 1–70 on your answer sheet.

PHYSICS C

SECTION I, MECHANICS

Time—45 minutes

35 Questions

Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. Select the one that is best in each case and then mark it on your answer sheet.

1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t₁ seconds. If it falls the second half of the distance in t₂ seconds, what is the value of t₂/t₁? (Ignore air resistance.)

(A) 1/(2)

(B) 1/

(D) 1 −(1/)

(E) − 1

2. A bubble starting at the bottom of a soda bottle experiences constant acceleration, a, as it rises to the top of the bottle in some time, t. How much farther does it travel in the last second of its journey than in the first second? Assume that the journey takes longer than 2 seconds.

(A) a(t + 1 s)²

(B) a(t – 1 s)²

(D) a(t + 1 s)(1 s)

(E) a(t – 1 s)(1 s)

3. An object initially at rest experiences a time-varying acceleration given by a = (2 m/s³)t for t ≥ 0. How far does the object travel in the first 3 seconds?

(A) 9 m

(B) 12 m

(D) 24 m

(E) 27 m

4. What is the fewest number of the following conditions to ensure that angular momentum is conserved?

I. Conservation of linear momentum

II. Zero net external force

III. Zero net external torque

(A) II only

(B) III only

(D) I and III only

(E) II and III only

5. In the figure shown, a tension force F_T causes a particle of mass m to move with constant angular speed ω in a horizontal circular path (in a plane perpendicular to the page) of radius R. Which of the following expressions gives the magnitude of F_T? Ignore air resistance.

(A) mω²R

(B)

(C)

(D) m(ω²R − g)

(E) m(ω²R + g)

6. An object (mass = m) above the surface of the Moon (mass = M) is dropped from an altitude h equal to the Moon’s radius (R). With what speed will the object strike the lunar surface?

(A)

(B)

(C)

(D)

(E)

7. Pretend that someone managed to dig a hole straight through the center of the Earth all the way to the other side. If an object were dropped down that hole, which of the following would best describe its motion? Assume ideal conditions (Earth is a perfect sphere, there are no dissipative forces) and that the object cannot be destroyed.

(A) It would fall to the center of the Earth and stop there.

(B) It would fall through the hole to the other side, continue past the opposite side’s opening, and fly into space.

(D) It would oscillate back and forth, but the amplitude would decrease each time, eventually settling at the center of the Earth.

(E) It would fall to the other side and stop there.

8. A uniform cylinder of mass m and radius r unrolls without slipping from two strings tied to a vertical support. If the rotational inertia of the cylinder is mr², find the acceleration of its center of mass.

(A) g

(B) g

(D) g

(E) g

9. A uniform cylinder, initially at rest on a frictionless, horizontal surface, is pulled by a constant force F from time t = 0 to time t = T. From time t = T on, this force is removed. Which of the following graphs best illustrates the speed, v, of the cylinder’s center of mass from t = 0 to t = 2T?

(A)

(B)

(C)

(D)

(E)

10. A space shuttle is launched from Earth. As it travels up, it moves at a constant velocity of 150 m/s straight up. If its engines provide 1.5 × 10⁸ W of power, what is the shuttle’s mass? You may assume that the shuttle’s mass and the acceleration due to gravity are constant.

(A) 6.7 × 10² kg

(B) 1.0 × 10⁵ kg

(D) 1.0 × 10⁶ kg

(E) 2.3 × 10⁶ kg

11. A satellite is in circular orbit around Earth. If the work required to lift the satellite to its orbit height is equal to the satellite’s kinetic energy while in this orbit, how high above the surface of Earth (radius = R) is the satellite?

(A) R

(B) R

(D) R

(E) 2R

12. The figure above shows a uniform bar of mass M resting on two supports. A block of mass M is placed on the bar twice as far from Support 2 as from Support 1. If F₁ and F₂ denote the downward forces on Support 1 and Support 2, respectively, what is the value of F₂/F₁?

(A)

(B)

(C)

(D)

(E)

13. A rubber ball (mass = 0.08 kg) is dropped from a height of 3.2 m and, after bouncing off the floor, rises almost to its original height. If the impact time with the floor is measured to be 0.04 s, what average force did the floor exert on the ball?

(A) 0.16 N

(B) 16 N

(D) 36 N

(E) 64 N

14. A disk of radius 0.1 m initially at rest undergoes an angular acceleration of 2.0 rad/s². If the disk only rotates, find the total distance traveled by a point on the rim of the disk in 4.0 s.

(A) 0.4 m

(B) 0.8 m

(D) 1.6 m

(E) 2.0 m

15. In the figure above, a small object slides down a frictionless quarter-circular slide of radius R. If the object starts from rest at a height equal to 2R above a horizontal surface, find its horizontal displacement, x, at the moment it strikes the surface.

(A) 2R

(B) R

(D) R

(E) 4R

16. The figure above shows a particle executing uniform circular motion in a circle of radius R. Light sources (not shown) cause shadows of the particle to be projected onto two mutually perpendicular screens. The positive directions for x and y along the screens are denoted by the arrows. When the shadow on Screen 1 is at position x = –(0.5)R and moving in the +x direction, what is true about the position and velocity of the shadow on Screen 2 at that same instant?

(A) y = –(0.866)R; velocity in –y direction

(B) y = –(0.866)R; velocity in +y direction

(D) y = +(0.866)R; velocity in –y direction

(E) y = +(0.866)R; velocity in +y direction

17. The figure shows a view from above of two objects attached to the end of a rigid massless rod at rest on a frictionless table. When a force F is applied as shown, the resulting rotational acceleration of the rod about its center of mass is kF/(mL). What is k?

(A)

(B)

(C)

(D)

(E)

18. A toy car and a toy truck collide. If the toy truck’s mass is double the toy car’s mass, then, compared to the acceleration of the truck, the acceleration of the car during the collision will be

(A) double the magnitude and in the same direction

(B) double the magnitude and in the opposite direction

(D) half the magnitude and in the opposite direction

(E) dependent on the type of collision

19. A homogeneous bar is lying on a flat table. Besides the gravitational and normal forces (which cancel), the bar is acted upon by exactly two other external forces, F₁ and F₂, which are parallel to the surface of the table. If the net force on the rod is zero, which one of the following is also true?

(A) The net torque on the bar must also be zero.

(B) The bar cannot accelerate translationally or rotationally.

(D) The net torque will be zero if F₁ and F₂ are applied at the same point.

(E) None of the above

20. An astronaut lands on a planet whose mass and radius are each twice that of Earth. If the astronaut weighs 800 N on Earth, how much will he weigh on this planet?

(A) 200 N

(B) 400 N

(D) 1,600 N

(E) 3,200 N

21. A particle of mass m = 1.0 kg is acted upon by a variable force, F(x), whose strength is given by the graph given above. If the particle’s speed was zero at x = 0, what is its speed at x = 4 m?

(A) 5.0 m/s

(B) 8.7 m/s

(D) 14 m/s

(E) 20 m/s

22. The radius of a collapsing spinning star (assumed to be a uniform sphere with a constant mass) decreases to of its initial value. What is the ratio of the final rotational kinetic energy to the initial rotational kinetic energy?

(A) 4

(B) 16

(D) 16³

(E) 16⁴

23. A ball is projected with an initial velocity of magnitude v₀ = 40 m/s toward a vertical wall as shown in the figure above. How long does the ball take to reach the wall?

(A) 0.25 s

(B) 0.6 s

(D) 2.0 s

(E) 3.0 s

24. If C, M, L, and T represent the dimensions of charge, mass, length, and time respectively, what are the dimensions of the permittivity of free space (ɛ₀)?

(A) T²C²/(M²L²)

(B) T²C²/(ML³)

(D) C²M/(T²L²)

(E) T²L²/(C²M)

25. The figure shown is a view from above of two clay balls moving toward each other on a frictionless surface. They collide perfectly inelastically at the indicated point and are observed to then move in the direction indicated by the post-collision velocity vector, v’. If m₁ = 2m₂, and v’ is parallel to the negative y-axis, what is v₂?

(A) v₁(sin 45°)/(2 sin 60°)

(B) v₁(cos 45°)/(2 cos 60°)

(D) v₁(2 sin 45°)/(sin 60°)

(E) v₁(cos 45°)/(2 sin 60°)

26. In the figure above, the coefficient of static friction between the two blocks is 0.80. If the blocks oscillate with a frequency of 2.0 Hz, what is the maximum amplitude of the oscillations if the small block is not to slip on the large block?

(A) 3.1 cm

(B) 5.0 cm

(D) 7.5 cm

(E) 9.4 cm

27. When two objects collide, the ratio of the relative speed after the collision to the relative speed before the collision is called the coefficient of restitution, e. If a ball is dropped from height H₁ onto a stationary floor, and the ball rebounds to height H₂, what is the coefficient of restitution of the collision?

(A) H₂/H₁

(B) H₂/H₁

(C)

(D)

(E) (H₁/H₂)²

28. The figure above shows a square metal plate of side length 40 cm and uniform density, lying flat on a table. A force F of magnitude 10 N is applied at one of the corners, as shown. Determine the torque produced by F relative to the center of rotation.

(A) 0 N•m

(B) 1.0 N•m

(D) 2.0 N•m

(E) 4.0 N•m

29. A small block of mass m = 2.0 kg is pushed from the initial point (x_i, z_i) = (0 m, 0 m) upward to the final point (x_f, z_f) = (3 m, 3 m) along the path indicated. Path 1 is a portion of the parabola z = x², and Path 2 is a quarter circle whose equation is (x – 1)² + (z – 3)² = 4. How much work is done by gravity during this displacement?

(A) −60 J

(B) −80 J

(D) −100 J

(E) −120 J

30. In the figure shown, the block (mass = m) is at rest at x = A. As it moves back toward the wall due to the force exerted by the stretched spring, it is also acted upon by a frictional force whose strength is given by the expression bx, where b is a positive constant. What is the block’s speed when it first passes through the equilibrium position (x = 0)?

(A) A

(B) A

(D) A

(E) A

31. The rod shown above can pivot about the point x = 0 and rotates in a plane perpendicular to the page. Its linear density, λ, increases with x such that λ(x) = kx, where k is a positive constant. Determine the rod’s moment of inertia in terms of its length, L, and its total mass, M.

(A) ML²

(B) ML²

(D) ML²

(E) 2ML²

32. A particle is subjected to a conservative force whose potential energy function is

U(x) = (x – 2)³ – 12x

where U is given in joules when x is measured in meters. Which of the following represents a position of stable equilibrium?

(A) x = –4

(B) x = –2

(D) x = 2

(E) x = 4

33. A light, frictionless pulley is suspended from a rigid rod attached to the roof of an elevator car. Two masses, m and M (with M > m), are suspended on either side of the pulley by a light, inextendable cord. The elevator car is descending at a constant velocity. Determine the acceleration of the masses.

(A) (M − m)g

(B) (M + m)g

(C)

(D)

(E) (M – m)(M + m)g

34. A particle’s kinetic energy is changing at a rate of –6.0 J/s when its speed is 3.0 m/s. What is the magnitude of the force on the particle at this moment?

(A) 0.5 N

(B) 2.0 N

(D) 9.0 N

(E) 18 N

35. An object of mass 2 kg is acted upon by three external forces, each of magnitude 4 N. Which of the following could NOT be the resulting acceleration of the object?

(A) 0 m/s²

(B) 2 m/s²

(D) 6 m/s²

(E) 8 m/s²

STOP

END OF SECTION I, MECHANICS

PHYSICS C

SECTION II, MECHANICS

Time—45 minutes

3 Questions

Directions: Answer all three questions. The suggested time is about 15 minutes per question for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight.

Mech 1. An ideal projectile is launched from the ground at an angle θ to the horizontal, with an initial speed of v₀. The ground is flat and level everywhere. Write all answers in terms of v₀, q, and fundamental constants.

(a) Calculate the time the object is in the air.

(b) Calculate the maximum height the object reaches.

(d) Calculate the range (horizontal displacement) of the object.

(e) What should θ be so that the projectile’s range is equal to its maximum vertical displacement?

Mech 2. A narrow tunnel is drilled through Earth (mass = M, radius = R), connecting points P and Q, as shown in the diagram on the left below. The perpendicular distance from Earth’s center, C, to the tunnel is x. A package (mass = m) is dropped from Point P into the tunnel; its distance from P is denoted y and its distance from C is denoted r. See the diagram on the right.

(a) Assuming that Earth is a homogeneous sphere, the gravitational force F on the package is due to m and the mass contained within the sphere of radius r < R. Use this fact to show that

F =

(b) Use the equation F(r) = –dU/dr to find an expression for the change in gravitational potential energy of the package as it moves from Point P to a point where its distance from Earth’s center is r. Write your answer in terms of G, M, m, R, and r.

(d) (i) At what point in the tunnel—that is, for what value of y—will the speed of the package be maximized?

(ii) What is this maximum speed? (Write your answer in terms of G, M, R, and x.)

Mech 3. The diagram below is a view from above of three sticky hockey pucks on a frictionless horizontal surface. The pucks with masses m and 2m are connected by a massless rigid rod of length L and are initially at rest. The puck of mass 3m is moving with velocity v directly toward puck m. When puck 3m strikes puck m, the collision is perfectly inelastic.

(a) Immediately after the collision,

(i) where is the center of mass of the system?

(ii) what is the speed of the center of mass? (Write your answer in terms of v.)

(iii) what is the angular speed of the system? (Write your answer in terms of v and L.)

(b) What fraction of the system’s initial kinetic energy is lost as a result of the collision?

STOP

END OF SECTION II, MECHANICS

PHYSICS C

SECTION I, ELECTRICITY AND MAGNETISM

Time—45 minutes

35 Questions

Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. Select the one that is best in each case and mark it on your answer sheet.

36. A nonconducting sphere is given a nonzero net electric charge, +Q, and then brought close to a neutral conducting sphere of the same radius. Which of the following will be true?

(A) An electric field will be induced within the conducting sphere.

(B) The conducting sphere will develop a net electric charge of –Q.

(D) The spheres will experience an electrostatic repulsion.

(E) The spheres will experience no electrostatic interaction.

37. Which set of capacitors, C₁ and C₂, would reach maximum charge most rapidly?

(A) C₁ = 2 μF and C₂ = 5 μF, arranged in parallel

(B) C₁ = 2 μF and C₂ = 5 μF, arranged in series

(D) C₁ = 3 μF and C₂ = 4 μF, arranged in series

(E) More than one of the above would be most rapid.

38. Each of the following ionized isotopes is projected with the same speed into a uniform magnetic field B such that the isotope’s initial velocity is perpendicular to B. Which combination of mass and charge would result in a circular path with the largest radius?

(A) m = 16 u, q = –5 e

(B) m = 17 u, q = –4 e

(D) m = 19 u, q = –2 e

(E) m = 20 u, q = –1 e

39. An ellipsoid-shaped conductor is negatively charged. Which one of the following diagrams best illustrates the charge distribution and electric field lines?

(A)

(B)

(C)

(D)

(E)

40. The four wires shown above are each made of aluminum. Which wire will have the greatest resistance?

(A) Wire A

(B) Wire B

(D) Wire D

(E) All the wires have the same resistance because they’re all composed of the same material.

41. Which of the following is NOT equal to one tesla?

(A) 1 J/(A•m²)

(B) 1 kg/(C•s)

(D) 1 V·s/m²

(E) 1 A·N/V

42. The figure above shows two Gaussian surfaces: a cube with side length d and a sphere with diameter d. The net electric charge enclosed within each surface is the same, +Q. If Φ_C denotes the total electric flux through the cubical surface, and Φ_S denotes the total electric flux through the spherical surface, then which of the following is true?

(A) Φ_C = (π/6)Φ_S

(B) Φ_C = (π/3)Φ_S

(D) Φ_C = (3/π)Φ_S

(E) Φ_C = (6/π)Φ_S

43. The figure above shows two large vertical conducting plates that carry equal but opposite charges. A ball of mass m and charge –q is suspended from a light string in the region between the plates. If the voltage between the plates is V, which of the following gives the angle θ?

(A) cos⁻¹ (qV/mgx)

(B) sin⁻¹ (qV/mgx)

(D) cos⁻¹ (qV/x)

(E) sin⁻¹ (qV/x)

44. An object carries a charge of –1 C. How many excess electrons does it contain?

(A) 6.25 × 10¹⁸

(B) 8.00 × 10¹⁸

(D) 3.20 × 10¹⁹

(E) 6.25 × 10¹⁹

Questions 45-46

Each of the resistors shown in the circuit below has a resistance of 200 Ω. The emf of the ideal battery is 24 V.

45. How much current is provided by the source?

(A) 30 mA

(B) 48 mA

(D) 72 mA

(E) 90 mA

46. What is the ratio of the power dissipated by R₁ to the power dissipated by R₄?

(A) 1/9

(B) 1/4

(D) 4

(E) 9

47. What is the value of the following product?

20 μF × 500 Ω

(A) 0.01 henry

(B) 0.01 ampere per coulomb

(D) 0.01 second

(E) 0.01 volt per ampere

48. A copper wire in the shape of a circle of radius 1 m, lying in the plane of the page, is immersed in a magnetic field, B, that points into the plane of the page. The strength of B varies with time, t, according to the equation

B(t) = 2t(1 − t)

where B is given in teslas when t is measured in seconds. What is the magnitude of the induced electric field in the wire at time t = 1 s?

(A) (1/π) N/C

(B) 1 N/C

(D) π N/C

(E) 2π N/C

49. In the figure above, the top half of a rectangular loop of wire, x meters by y meters, hangs vertically in a uniform magnetic field, B. Describe the magnitude and direction of the current in the loop necessary for the magnetic force to balance the weight of the mass m supported by the loop.

(A) I = mg/xB, clockwise

(B) I = mg/xB, counterclockwise

(D) I = mg/B, counterclockwise

(E) I = mg/(x+y)B, clockwise

50. A solid nonconducting cylinder of radius R and length L contains a volume charge density given by the equation ρ(r) = (+3 C/m⁴)r, where r is the radial distance from the cylinder’s central axis. This means that the total charge contained within a concentric cylinder of radius r < R and length ℓ < L is equal to 2πℓr³. Find an expression for the strength of the electric field inside this cylinder.

(A) 1/ε₀r²

(B) r/2ε₀

(D) r²/ε₀

(E) r²/2ε₀

51. The figure above shows a pair of long, straight current-carrying wires and four marked points. At which of these points is the net magnetic field zero?

(A) Point 1 only

(B) Points 1 and 2 only

(D) Points 3 and 4 only

(E) Point 3 only

52. The figure above shows two positively charged particles. The +Q charge is fixed in position, and the +q charge is brought close to +Q and released from rest. Which of the following graphs best depicts the acceleration of the +q charge as a function of its distance r from +Q?

(A)

(B)

(C)

(D)

(E)

53. Once the switch S in the figure above is closed and electrostatic equilibrium is regained, how much charge will be stored on the positive plate of the 6 μF capacitor?

(A) 9 μC

(B) 18 μC

(D) 27 μC

(E) 36 μC

54. A metal bar of length L is pulled with velocity v through a uniform magnetic field, B, as shown above. What is the voltage produced between the ends of the bar?

(A) vB, with Point X at a higher potential than Point Y

(B) vB, with Point Y at a higher potential than Point X

(D) vBL, with Point Y at a higher potential than Point X

(E) None of the above

55. An electric dipole consists of a pair of equal but opposite point charges of magnitude 4.0 nC separated by a distance of 2.0 cm. What is the electric field strength at the point midway between the charges?

(A) 0

(B) 9.0 × 10⁴ V/m

(D) 3.6 × 10⁵ V/m

(E) 7.2 × 10⁵ V/m

56. The figure above shows a cross section of two concentric spherical metal shells of radii R and 2R, respectively. Find the capacitance.

(A) 1/(8πε₀R)

(B) 1/(4πε₀R)

(D) 4πε₀R

(E) 8πε₀R

57. Traveling at an initial speed of 1.5 × 10⁶ m/s, a proton enters a region of constant magnetic field, B, of magnitude 1.0 T. If the proton’s initial velocity vector makes an angle of 30° with the direction of B, compute the proton’s speed 4 s after entering the magnetic field.

(A) 5.0 × 10⁵ m/s

(B) 7.5 × 10⁵ m/s

(D) 3.0 × 10⁶ m/s

(E) 6.0 × 10⁶ m/s

Questions 58-60

There is initially no current through any circuit element in the following diagram.

58. What is the current through r immediately after the switch S is closed?

(A) 0

(B)

(C)

(D)

(E)

59. After the switch has been kept closed for a long time, how much energy is stored in the inductor?

(A)

(B)

(C)

(D)

(E)

60. After having been closed for a long time, the switch is suddenly opened. What is the current through r immediately after S is opened?

(A) 0

(B)

(C)

(D)

(E)

61. A solid, neutral metal sphere of radius 6a contains a small cavity, a spherical hole of radius a as shown above. Within this cavity is a charge, +q. If E_X and E_Y denote the strength of the electric field at points X and Y, respectively, which of the following is true?

(A) E_Y = 4E_X

(B) E_Y = 16E_X

(D) E_Y = (11/5)E_X

(E) E_Y = (11/5)²E_X

62. Two particles of charge +Q are located on the x-axis, as shown above. Determine the work done by the electric field to move a particle of charge –Q from very far away to point P.

(A)

(B)

(C)

(D)

(E)

63. A battery is connected in a series with a switch, a resistor of resistance R, and an inductor of inductance L. Initially, there is no current in the circuit. Once the switch is closed and the circuit is completed, how long will it take for the current to reach 99% of its maximum value?

(A) (ln )RL

(B) (ln 99)RL

(D) (ln )

(E) (ln 100)

64. What is the maximum number of 40 W light bulbs that could be connected in parallel with a 120 V source? The total current cannot exceed 5 A or the circuit will blow a fuse.

(A) 3

(B) 6

(D) 12

(E) 15

65. The metal loop of wire shown above is situated in a magnetic field B pointing out of the plane of the page. If B decreases uniformly in strength, the induced electric current within the loop is

(A) clockwise and decreasing

(B) clockwise and increasing

(D) counterclockwise and constant

(E) counterclockwise and increasing

66. A dielectric of thickness is placed between the plates of a parallel-plate capacitor, as shown above. If K is the dielectric constant of the slab, what is the capacitance?

(A)

(B)

(C)

(D)

(E)

67. Consider the two source charges shown above. At how many points in the plane of the page, in a region around these charges are both the electric field and the electric potential equal to zero?

(A) 0

(B) 1

(D) 3

(E) 4

68. The figure above shows four current-carrying wires passing perpendicularly through the interior of a square whose vertices are W, X, Y, and Z. The points where the wires pierce the plane of the square (namely, R, S, T, and U) themselves form the vertices of a square each side of which has half the length of each side of WXYZ. If the currents are as labeled in the figure, what is the absolute value of

∮ B•dℓ

where the integral is taken around WXYZ?

(A) µ₀I

(B) µ₀I

(D) 2µ₀I

(E) 5µ₀I

69. A particle with charge +q is moved from point A to point B along path 1 in the picture above. Its change in potential energy is U. If a charge of –q is then moved from A to B along path 2, its change in potential energy will be

(A) −U

(B) −U

(D) U

(E) Cannot be determined without knowing the values of Q₁ and Q₂

70. Two point charges, each +Q, are fixed a distance L apart. A particle of charge –q and mass m is placed as shown in the figure above. What is this particle’s initial acceleration when released from rest?

(A)

(B)

(C)

(D)

(E)

STOP

END OF SECTION I, ELECTRICITY AND MAGNETISM

PHYSICS C

SECTION II, ELECTRICITY AND MAGNETISM

Time—45 minutes

3 Questions

E & M 1. A uniformly charged, nonconducting, circular ring of radius R carries a charge +Q. Its central axis is labeled the z-axis, and the center of the ring is z = 0. Points above the ring on the z-axis have positive z-coordinates; those below have negative z-coordinates.

(a) Calculate the electric potential at an arbitrary point on the z-axis. Write your answer in terms of Q, z, R, and fundamental constants.

(b) (i) At what point(s) on the z-axis will the potential have its greatest value?

(ii) What is this maximum potential value?

(c) Find an expression for the electric field (magnitude and direction) at an arbitrary point on the z-axis. Write your answer in terms of Q, z, R, and fundamental constants.

(d) A positive charge is released from rest at z = +2R on the z-axis. Describe the subsequent motion of the positive charge. Justify your answer.

E & M 2. In the circuit shown below, the capacitor is initially uncharged and there is no current in any circuit element.

In each of the following, k is a number greater than 1; write each of your answers in terms of ε, r, R, C, k, and fundamental constants.

(a) At t = 0, the switch S is moved to position 1.

(i) At what time t is the current through R equal to of its initial value?

(ii) At what time t is the charge on the capacitor equal to of its maximum value?

(iii) At what time t is the energy stored in the capacitor equal to of its maximum value?

(b) After the switch has been at position 1 for a very long time, it is then moved to position 2. Let this redefine t = 0 for purposes of the following questions.

(i) How long will it take for the current through R to equal of its initial value?

(ii) At what time t is the charge on the capacitor equal to of its initial value?

E & M 3. The diagram below shows two parallel conducting rails connected by a third rail of length L, raised to an angle of θ with the horizontal (supported by a pair of insulating columns). A metal strip of length L, mass m, and resistance R is free to slide without friction down the rails. The apparatus is immersed in a vertical, uniform magnetic field, B. The resistance of the stationary rails may be neglected.

(a) As the strip slides down the rails with instantaneous speed v, determine

(i) the magnitude of the induced emf. (Write your answer in terms of L, B, v, and θ.)

(ii) the direction of the induced current in the strip (X to Y, or Y to X?).

(iii) the magnitude of the induced current. (Write your answer in terms of L, B, R, v, and θ.)

(b) The strip XY will eventually slide down the rails at a constant velocity. Derive an expression for this velocity in terms of L, B, R, m, g, and θ.

(c) When the strip is sliding with the constant velocity determined in part (b), show that the power dissipation in the strip is equal to the rate at which gravity is doing work on the strip.

STOP

END OF EXAM

Practice Test 1: Answers and Explanations

PRACTICE TEST 1 ANSWER KEY

Mechanics

Question Number	Question Answer	See Chapter(s) #
1	E	Chapter 5
2	E	Chapter 5
3	A	Chapter 5
4	B	Chapter 9
5	C	Chapters 6, 9
6	A	Chapter 10
7	C	Chapters 7, 11
8	D	Chapters 6, 9
9	B	Chapters 5, 6
10	B	Chapters 6, 7
11	A	Chapters 7, 10
12	D	Chapters 6, 9
13	C	Chapters 7, 8
14	D	Chapter 9
15	C	Chapters 5, 7
16	A	Chapter 11
17	C	Chapter 9
18	B	Chapter 6
19	D	Chapters 6, 9
20	B	Chapter 10
21	C	Chapter 7
22	C	Chapter 9
23	C	Chapters 4, 5
24	B	Chapter 12
25	D	Chapters 4, 8
26	B	Chapters 6, 11
27	D	Chapters 5, 7
28	D	Chapter 9
29	A	Chapter 7
30	B	Chapters 6, 11
31	D	Chapter 9
32	E	Chapter 7
33	D	Chapter 6
34	B	Chapter 7
35	E	Chapter 6

Question Number	Question Answer	See Chapter(s) #
1(a)	*	Chapter 5
1(b)	*	Chapters 4, 5
1(c)	*	Chapters 4, 5
1(d)	*	Chapters 4, 5
1(e)	*	Chapters 4, 5
2(a)	*	Chapter 10
2(b)	*	Chapter 7
2(c)	*	Chapter 7
2(d)(i)	*	Chapter 7
2(d)(ii)	*	Chapter 7
3(a)(i)	*	Chapters 8, 9
3(a)(ii)	*	Chapters 8, 9
3(a)(iii)	*	Chapters 8, 9
3(b)	*	Chapters 7, 9

*See explanations beginning on this page.

Electricity and Magnetism

Question Number	Question Answer	See Chapter(s) #
36	C	Chapter 12
37	B	Chapter 13
38	E	Chapter 15
39	A	Chapter 12
40	B	Chapter 14
41	E	Chapter 15
42	C	Chapter 12
43	C	Chapters 12, 13
44	A	Chapter 12
45	E	Chapter 14
46	E	Chapter 14
47	D	Chapters 13, 14
48	B	Chapter 16
49	A	Chapter 16
50	D	Chapter 12
51	A	Chapter 15
52	A	Chapter 12
53	C	Chapter 13
54	E	Chapter 16
55	E	Chapter 12
56	E	Chapter 13
57	C	Chapter 15
58	B	Chapter 14
59	B	Chapter 14
60	A	Chapter 14
61	C	Chapter 12
62	B	Chapter 13
63	E	Chapter 16
64	E	Chapter 14
65	D	Chapter 16
66	D	Chapter 13
67	A	Chapters 12, 13
68	D	Chapter 15
69	B	Chapter 13
70	A	Chapters 6, 12

Question Number	Question Answer	See Chapter(s) #
1(a)	*	Chapter 13
1(b)(i)	*	Chapter 13
1(b)(ii)	*	Chapter 13
1(c)	*	Chapter 13
1(d)	*	Chapters 6, 12
2(a)(i)	*	Chapters 13, 14
2(a)(ii)	*	Chapters 13, 14
2(a)(iii)	*	Chapters 13, 14
2(b)(i)	*	Chapters 13, 14
2(b)(ii)	*	Chapters 13, 14
3(a)(i)	*	Chapter 16
3(a)(ii)	*	Chapter 16
3(a)(iii)	*	Chapter 16
3(b)	*	Chapter 16
3(c)	*	Chapter 16

*See explanations beginning on this page.

SECTION I, MECHANICS

1. E Let y denote the total distance that the rock falls and let T denote the total time of the fall. Then y = and y = gT², so

2. E A trick for solving problems of this nature is to make up numbers for a and t. Pretend that total time = 10 seconds, and acceleration = 2 m/s². Then make a table:

	Time, s	Speed, m/s
	0	0
	1	2
	9	18
	10	20

So in the first second, v_avg = 1 m/s, and d = 1 m. In the last (tenth) second, v_avg = 19 m/s, and d = 19 m. The answer to the question is 19 m – 1 m = 18 m. Which answer choice gives us 18 m?

(A) a(t + 1 s)² = (2 m/s²)(10 s + 1 s)² = (2 m/s²) (11 s)² = (2 m/s²)(121 s²) = 242 m

(B) a(t – 1 s)² = (2 m/s²)(10 s – 1 s)² = (2 m/s²)(9 s)² = (2 m/s²)(81 s²) = 162 m

(D) a(t + 1 s)(1 s) = (2 m/s²)(10 s + 1 s)(1s) = (2 m/s²)(11 s)(1 s) = 22 m

(E) a(t – 1 s)(1 s) = (2 m/s²)(10 s – 1 s)(1 s) = (2 m/s²)(9 s)(1 s) = 18 m

Therefore (E) is the correct answer.

If you really want to do it algebraically, here’s one way. Solve for the distance traveled in the first second.

d = v₀Δt + aΔt²

= (0 m/s)(1 s) + (1/2)(a)(1 s)²

= (1/2)(2 s²) a

Next, solve for the distance in the final second.

d = v₀Δt + aΔt²

= a(t – 1 s)(1 s) + (a)(1 s)²

= a(t – 1 s)(1 s) + (1 s²) (a)

The difference between the two is a(t – 1 s)(1 s) meters.

3. A Integrating a(t) gives v(t):

Since v₀ = 0, we have v(t) = t², which is never negative. This means the object never moves backward, so displacement is equal to the distance traveled. Now integrating v(t) from t = 0 to t = 3 s gives the object’s displacement during this time interval:

4. B Angular momentum is conserved when no net external torque is applied, which is (III). However, a body can experience a net nonzero torque even when the net external force is zero (which is the condition that guarantees conservation of linear momentum), so neither (I) nor (II) necessarily ensure conservation of angular momentum.

5. C The figure below shows that F_T sin θ = mg and F_T cos θ = mv²/R = mω²R:

We can eliminate θ from these equations by remembering that sin² θ + cos² θ is always equal to 1:

That’s enough information to get you to the correct answer, (C), but here’s another way to think about this problem.

If you look at the image of F_T sin θ and F_T cos θ, you might recognize that this is a right triangle, to which you can apply the Pythagorean theorem. and F_T,x = = mω²R and F_T,y = mg, which means that = (mg)² + (mω²R)² → = m²(g² + ω⁴R²) → m.

6. A You can eliminate some choices right off the bat if you do some basic dimensional analysis. Choices (A), (B), and (C) all have dimensions of . Looking up the units of G, you’ll see that GM/R has units of N*m/kg = m²/s². Choices (D) and (E) have an extra factor of mass in the numerator () which doesn’t belong in an expression of speed, so they can be ruled out. Because the initial height of the object is comparable to the radius of the Moon, we cannot simply use mgh for its initial potential energy. Instead, we must use the more general expression U = –GMm/r, where r is the distance from the center of the Moon. Notice that since h = R, the object’s initial distance from the Moon’s center is h + R = R + R = 2R. Conservation of Energy then gives

K_i + U_i = k_f + U_f

0 − = mv² −

= mv²

v =

7. C This situation is just a spring–block system. It would start with some potential energy at one extreme edge, turn all of that energy into kinetic energy as it moved to the center, and continue to the other edge due to its momentum (turning the energy back into potential energy). This process would continue indefinitely in ideal conditions.

8. D Let F_T be the tension in each string. Then F_net = ma becomes mg – 2F_T = ma. Also, the total torque exerted by the tension forces on the cylinder is 2rF_T, so τ_net = Iα becomes 2rF_T = (mr²)α. Because the cylinder doesn’t slip, α = a/r, so 2F_T = ma. These equations can be combined to give mg − ma = ma, which implies mg = ma. Therefore, a = g.

9. B The cylinder slides across the surface with acceleration a = F/m until time t = T, when a drops to zero (because F becomes zero). Therefore, from time t = 0 to t = T, the velocity is steadily increasing (because the acceleration is a positive constant), but, at t = T, the velocity remains constant. This is illustrated in graph (B).

10. B P = Fv = mgv

m = P/(gv)

= (1.5 • 10⁸ W)/[(10 m/s²)(150 m/s)] = 1.0 • 10⁵ kg

11. A Let m and M denote the mass of the satellite and Earth, respectively. The work required to lift the satellite to height h above the surface of Earth is

W = ΔU = −

Earth’s gravitational pull provides the necessary centripetal force on the satellite, so F_G = F_c.

Because we’re told that W = K, we find that

12. D First note that if L is the total length of the bar, then the distance of the block from Support 1 is L and its distance from Support 2 is L. Let P₁ denote the point at which Support 1 touches the bar. With respect to P₁, the upward force exerted by Support 1, F₁, produces no torque, but the upward force exerted by Support 2, F₂, does. Since the net torque must be zero if the system is in static equilibrium, the counterclockwise torque of F₂ must balance the total clockwise torque produced by the weight of the block and of the bar (which acts at the bar’s midpoint).

Now, with respect to P₂, the point at which Support 2 touches the bar,

These results give F₂/F₁ = = .

13. C Because the ball rebounded to the same height from which it was dropped, its takeoff speed from the floor must be the same as the impact speed. Calling up the positive direction, the change in linear momentum of the ball is p_f − p_i = mv − (−mv) = 2mv, where v = (this last equation comes from the equation mv² = mgh). The impulse–momentum theorem, J = Δp, now gives

14. D Using the Big Five equation Δθ = ω₀t + αt², we find that

Δθ = (2 rad/s²)(4.0 s)² = 16 rad

Therefore, Δs = rΔθ = (0.1 m)(16 rad) = 1.6 m. If you’re worried about memorizing the Big Five equations, here’s another method. The angular acceleration is 2.0 rad/s², which means it gets 2.0 rad/s faster every second. It starts from rest. After 1 s, ω = 2 rad/s. After 2 s, ω = 4 rad/s. After 3 s, ω = 6 rad/s. After 4.0 s, ω = 8.0 rad/s. That makes the average angular velocity during these four seconds 4.0 rad/s (ω increased linearly from 0 s to 8 s). Since the disk averages 4.0 rad/s for 4 seconds, the angular distance traveled will be 16 rad, which brings you right back to ∆s = r∆θ.

One final alternative is to make a graph with ω on the vertical axis and t on the horizontal axis. The slope of an ω vs. t plot is α = 2.0 rad/s². The area under the graph from 0 to 4 is the angular distance traveled by the point in the first 4.0 seconds.

15. C The object’s initial velocity from the slide is horizontal, so v₀_y = 0, which implies that Δy = −gt². Since Δy = −R,

−R = −g t² ⇒ t =

The (horizontal) speed with which the object leaves the slide is found from the conservation of energy: mgR = mv², which gives v = . Therefore,

Δx = v_0xt = · = 2R

Since the object travels a horizontal distance of 2R from the end of the slide, the total horizontal distance from the object’s starting point is R + 2R = 3R.

16. A From the diagram below,

if R cos θ = −(0.5)R, then θ = 240°. Therefore, y = R sin θ = R sin 240° = –(0.866)R. Also, it is clear that the particle’s subsequent motion will cause the shadow on Screen 2 to continue moving in the –y direction.

17. C The center of mass of the system is at a distance of

y_cm = = L

below the mass m. With respect to this point, the clockwise torque produced by the force F has magnitude

τ = rF = F = LF

Since the rotational inertia of the system about its center of mass is

I = Σm_i = m² + (2m)² = mL²

the equation τ = Iα becomes

18. B We know that the forces during a collision are an example of an action-reaction pair. This means the force on the car will be equal in magnitude but opposite in direction compared to the force on the car. Because the forces are in opposite directions, the accelerations will be as well. Eliminate (A) and (C). Because the forces are equal in magnitude, Newton’s Second Law tells us that the acceleration of the lighter object will be double the acceleration of the heavier object. Eliminate (D). Finally, these things will be true regardless of what type of collision takes place. Therefore, the answer is (B).

19. D Since F_net = F₁ + F₂ = 0, the bar cannot accelerate translationally, so (C) is false. The net torque does not need to be zero, as the following diagram shows, eliminating (A) and (B).

However, since F₂ = −F₁, (D) is true; one possible illustration of this is given below:

20. B The value of g near the surface of a planet depends on the planet’s mass and radius:

Therefore, calling this Planet X, we find that

Since g is half as much on the planet as it is on Earth, the astronaut’s weight (mg) will be half as much as well. The astronaut would weigh half of 800 N, which is 400 N, (B). Alternatively, you can plug in some numbers, although understand that this can introduce new errors if you’re not careful with your calculations. This allows you to make some unreal assumptions with numbers that are easy to work with, in this case, that the Earth has a mass of 10 and a radius of 1. Now it’s a lot easier to solve for Earth, g = GM/R² = G*10/1² = 10G. Planet X has twice the mass and radius of Earth, so for Planet X, M = 20 and r = 2. This gives g = GM/R² = G*20/2² = 5G. Since g is half as much on the planet as it is on Earth, the astronaut’s weight will be half as much as well.

21. C The work done by the force F(x) is equal to the area under the given curve. The region under the graph can be broken into two triangles and a rectangle, and it is then easy to figure out that the total area is 50 N·m. Since work is equal to change in kinetic energy (and v_i = 0),

22. C Since no external torques act on the star as it collapses (just like a skater when she pulls in her arms), angular momentum, Iω, is conserved, and the star’s rotational speed increases. The moment of inertia of a sphere of mass m and radius r is given by the equation I = kmr² (with k = , but the actual value is irrelevant), so we have:

Therefore,

23. C The projectile’s horizontal speed is v₀ cos 60° = v₀ * 1/2 = 20 m/s, so it reaches the wall in 1.0 s.

24. B Start with the formula for electric force.

F_e = [1/(4πε₀)](q₁q₂/r²)

ε₀ = [1/(4πF_e)](q₁q₂/r²)

Next, substitute the terms with the proper units.

[ε₀] = (1/N)(C ²/m²)

= [1/(kg m/s²)](C ²/m²) = (s² C ²)/(kg m³)

Last, just substitute the letters as the problem describes, and you end up with (B).

25. D Right off the bat, you can eliminate (A) and (E), as they are equivalent to one another. Whether you noticed that or not, the diagram given with the question shows that after the clay balls collide, they move in the –y direction, which means that the horizontal components of their linear momenta canceled. In other words, p₁_x + p₂_x = 0:

m₁v_1x + m₂v_2x = 0

(m₁v₁ sin 45°) + (m₂v₂ sin 60°) = 0

v₂ =

26. B The horizontal position of the blocks can be given by the equation

x = A cos(ωt + ϕ₀)

where A is the amplitude, ω is the angular frequency, and ϕ₀ is the initial phase. Differentiating this twice gives the acceleration:

a = = −Aω² cos (ωt + ϕ₀) ⇒ a_max = Aω²

This means that the maximum force on block m is F_max = ma_max = mAω². The static friction force must be able to provide this same amount of force; otherwise, the block will slip. Therefore,

27. D The ball strikes the floor with speed v₁ = , and it leaves the floor with speed v₂ = . Therefore,

28. D The center of rotation is the center of mass of the plate, which is at the geometric center of the square because the plate is homogeneous. Since the line of action of the force coincides with one of the sides of the square, the lever arm of the force, ℓ, is simply equal to s. Therefore,

τ = ℓ F = sF = (0.40 m)(10 N) = 2.0 N·m

29. A Since gravity is a conservative force, the actual path taken is irrelevant. If the block rises a vertical distance of z = 3 m, then the work done by gravity is

W = −mgz = −(2.0 kg)(10 N/kg)(3 m) = −60 J

30. B The work done by the friction force as the block slides from x = A to x = 0 is

Now use Conservation of Energy, using U(x) = kx²:

31. D Consider an infinitesimal slice of width dx at position x; its mass is dm = λ dx = kx dx. Then, by definition of rotational inertia,

Because the total mass of the rod is

we see that I = ML².

32. E Points of equilibrium occur when dU/dx is equal to zero:

= 0 ⇒ 3(x −2)² − 12 = 0 ⇒ (x − 2)² = 4 ⇒ x = 0 or x = 4

There’s more than one way to determine which equilibrium points are stable equilibrium points. One way is to graph U(x) and see which equilibrium point is a minimum. Another way is to use the first derivative. The following explanation uses the second derivative test, but feel free to use the method you’re most comfortable with. Points of stable equilibrium occur when dU/dx = 0 and d²U/dx² is positive (because then the equilibrium point is a relative minimum). Since

= 6(x − 2)

the point x = 4 is a point of stable equilibrium (x = 0 is unstable).

33. D Since the pulley is not accelerating, it can serve as the origin of an inertial reference frame. Applying Newton’s Second Law to the masses using the following free-body diagrams,

gives

F_T − mg = ma and Mg − F_T = Ma

Adding these equations eliminates F_T, and we find that

Mg − mg = (m + M)a ⇒ a =

The acceleration of block m is (M – m)g/(M + m) upward, and the acceleration of block M is this value, (M – m)g/(M + m), downward, which is (D).

Alternatively, you can treat the two objects as one system. Tension is an internal force and cancels. The net force on the system is Mg – mg. The total mass is M + m. If you simplify this expression, you get the same answer, (D).

34. B Differentiate the equation K = mv² with respect to time:

This tells us that

35. E The maximum net force on the object occurs when all three forces act in the same direction, giving F_net = 3F = 3(4 N) = 12 N, and a resulting acceleration of a = F_net/m = (12 N)/(2 kg) = 6 m/s². These three forces could not give the object an acceleration greater than 6 m/s². Because the question looks for the acceleration that is NOT possible, the answer is 8 m/s, (E).

SECTION II, MECHANICS

1. This problem involves motion in two dimensions, so it’s important to begin by drawing a large picture of the situation. Note that our coordinate system is set up in the usual way with x pointing to the right, y pointing up, and the origin at the point where the projectile is launched.

Next, draw a second diagram that splits ₀ into vertical and horizontal components. Note that you can use trigonometric properties to derive the values of v₀_x and v₀_y.

Now we’re ready to start answering the questions.

(a) For this problem, time of flight is determined by vertical quantities. This is because the flight ends when the object hits the ground, that is, when the y-coordinate is zero again. If ∆y = 0, the projectile is either beginning or ending its flight.

You need to find t and you’re missing v_f,y, so use Big Five #3.

Δy = v_{0, y}t + a_yt²
0 = (v₀ sin θ)t + (−g)t²
(g)t² − (v₀ sin θ)t = 0
t = 0

At the beginning of the flight, t is equal to 0. However, at the end of the flight, you would have the following:

t − v₀ sin θ = 0
t = v₀ sin θ
t =

The total amount of time in the air, then, is − 0, or just .

As an alternative, you could use Big Five #2 to find the amount of time in which the projectile moves upward and then multiply that number by 2 to get the total time.

(b) As with the previous step, you are looking for a vertical quantity; because it’s t that you no longer care about and Δy that you need, use Big Five #5.

At the top of the trajectory, v_fy = 0, which means you can now solve for Δy.

0 = + 2a∆y

− = 2a∆y

= ∆y

If you plug in the known values of v₀_y and a, you get the following:

∆y =

Because y₀ = 0, that value of Δy is also the value of y_max.

(d) You’re asked for a horizontal quantity, so let’s review our horizontal components.

∆x = ?

v_x = v₀ cosθ

t =

Velocity is defined as v_x = , and this can be rewritten as ∆x = v_xt. You’ve found two of these values already, so plug them in and solve.

∆x = (v₀cosθ)

∆x =

(e) We’re told that the range equals the maximum vertical displacement, so ∆x = y_max.

Cancelling out from this equation, you’re left with the following:

2cosθ =

4 = tanθ

θ = tan⁻¹(4)

You can check this by noting that the arctan of 4 is a fairly large angle (about 76°), which indicates that ₀ points more upward than rightward.

2. (a) The mass contained in a sphere of radius r < R is

(b) Since dU = –F(r) dr,

By applying the Pythagorean theorem to the two right triangles in the figure,

we see that

so the expression given above for v can be rewritten in the form

(d) (i) From the expression derived in part (c) above, we can see that the value of y which will maximize v is

This is the midpoint of the tunnel.

(ii) The maximum value for v is

3. (a) (i) Once the 3m puck sticks to the m puck, the center of mass of the system is

above the bottom pucks on the rod.

(ii) Applying Conservation of Linear Momentum, we find that

(3m)v = (3m + m + 2m)v′_cm ⇒ v′_cm = v

(iii) To keep this as straightforward as possible, choose our origin to be the center of mass of the system. Now, applying Conservation of Angular Momentum, we find that

(b) The kinetic energy before the collision was the initial translational kinetic energy of the 3m puck:

K_i = (3m)v²

After the collision, the total kinetic energy is the sum of the system’s translational kinetic energy and its rotational kinetic energy:

Therefore, the fraction of the initial kinetic energy that is lost due to the collision is

SECTION I, ELECTRICITY AND MAGNETISM

36. C The proximity of the charged sphere will induce negative charge to move to the side of the uncharged sphere closer to the charged sphere. Since the induced negative charge is closer than the induced positive charge to the charged sphere, there will be a net electrostatic attraction between the spheres.

37. B First, find the equivalent capacitance of each pair. Remember that capacitors in parallel simply add, whereas capacitors in series add as reciprocals.

C_A = 2 μF + 5 μF = 7 μF

1/C_B = 1/(2 μF) + 1/(5 μF) = 7/(10 μF)

C_B = (10/7) μF

C_C = 3 μF + 4 μF = 7 μF

1/C_D = 1/(3 μF) + 1/(4 μF) = 7/(12 μF)

C_D = (12/7) μF

Finally, remember that the time constant is proportional to capacitance, and a lower time constant allows a circuit to charge its capacitor(s) more rapidly. Therefore, the arrangement in (B) will be quickest. Even if you don’t know about time constants, you can still narrow it down to two choices using process of elimination. Arranging the capacitances from smallest to biggest, we have C_B < C_D < C_A = C_C. Since (A) and (C) are the same, neither answer choice can be correct; if they were tied for first, the answer would be (E). Also, we’re asked for an extreme value, and (D) isn’t the largest or the smallest, so it’s wrong. Just by looking at extreme values, we’ve narrowed it down to (B) (the smallest) or (E) (the two largest).

38. E When the particle enters the magnetic field, the magnetic force provides the centripetal force to cause the particle to execute uniform circular motion:

Since v and B are the same for all the particles, the largest r is found by maximizing the ratio m/|q| Furthermore, since the ratio of u/e (atomic mass unit/magnitude of elementary charge) is a constant, the answer depends only on the ratio of “factor of u” to “factor of charge.” The ratio 20/1 is largest, so it will have the largest r.

39. A Electric field lines are always perpendicular to the surface of a conductor, and always point into a negatively charged one, eliminating (B) and (E). Excess charge on a conductor always resides on the surface, thereby eliminating (D). Finally, there is a greater density of charge at points where the radius of curvature is smaller, so between (A) and (C), pick (A).

40. B The resistance of a wire made of a material with resistivity r and with length L and cross-sectional area A is given by the equation R = ρL/A. Since Wire B has the greatest length and smallest cross-sectional area, it has the greatest resistance.

41. E From the equation F = qvB, we see that 1 tesla is equal to 1 N·s/(C·m). Since

(A) and (C) are eliminated.

Furthermore,

eliminating (B). Finally, since

(D) is also eliminated.

42. C Gauss’s law states that the total electric flux through a closed surface is equal to (1/ɛ₀) times the net charge enclosed by the surface. Both the cube and the sphere contain the same net charge, so Φ_C must be equal to Φ_S.

43. C The electric field between the plates is uniform and equal to V/x, so the magnitude of the electric force on the charge is F_E = qE = qV/x. From the following free-body diagram

we see that

F_T sin θ = F_E = qE = and F_Tcos θ = mg

Dividing the first equation by the second one, we find that

44. A The Table of Information provides the following conversion factor: 1 e = 1.6 × 10⁻¹⁹ C. Multiplying the magnitude of charge, 1 C, by this conversion factor gives the following:

There are 6.25 × 10¹⁸ electrons in 1 coulomb of negative charge.

45. E Resistors R₂, R₃, and R₄ are in parallel, so their equivalent resistance, R_2-3-4, satisfies

Since this is in series with R₁, the total resistance in the circuit is

R = R + R₂₋₃₋₄ =

The current provided by the source must therefore be

Drawing a series of equivalent circuits can make it easier to see what the resistance should be at each step.

46. E The current through R₄ is the current through R₁, and R₄ = R₁, so

47. D Note that all answers have the same number, so for this problem, don’t even worry about multiplying the numbers—just focus on the units. Because the time constant for an RC circuit is equal to the product of resistance and capacitance, τ = RC, this product has the dimensions of time. If you don’t know that off the top of your head, just be sure to carefully break the farad and ohm into their subunits, simplifying where possible:

1 F × 1 Ω = 1 C/V × 1 V/A = 1 C/A = 1 s

48. B Apply Faraday’s Law of Electromagnetic Induction:

Since r = 1 m, the value of E at t = 1 s is E = 1 N/C.

49. A The magnetic force, F_B, on the top horizontal wire of the loop must be directed upward and have magnitude mg. (The magnetic forces on the vertical portions of the wire will cancel.) By the right-hand rule, the current in the top horizontal wire must be directed to the right—because B is directed into the plane of the page—in order for F_B to be directed upward; therefore, the current in the loop must be clockwise, eliminating (B) and (D). Since F_B = IxB, we must have

IxB = mg ⇒ I =

50. D Construct a cylindrical Gaussian surface of radius r < R and length ℓ < L within the given cylinder. By symmetry, the electric field must be radial, so the electric flux through the lids of the Gaussian cylinder is zero. The lateral surface area of the Gaussian cylinder is 2πrℓ;, so by Gauss’s law,

51. A Call the top wire (the one carrying a current I to the right) Wire 1, and call the bottom wire (carrying a current 2I to the left) Wire 2. Then in the region between the wires, the individual magnetic field vectors due to the wires are both directed into the plane of the page, so they could not cancel in this region. Therefore, the total magnetic field could not be zero at either Point 2 or Point 3. This eliminates (B), (C), (D), and (E), so the answer must be (A). Since the magnetic field created by a current-carrying wire is proportional to the current and inversely proportional to the distance from the wire, the fact that Point 1 is in a region where the individual magnetic field vectors created by the wires point in opposite directions and that Point 1 is twice as far from Wire 2 as from Wire 1 implies that the total magnetic field there will be zero.

52. A If the mass of the +q charge is m, then its acceleration is

Because a should decrease as r increases, you can eliminate (C) and (E). The graph in (A) best depicts an inverse-square relationship between a and r.

53. C Once the switch is closed, the capacitors are in parallel, which means the voltage across C₁ must equal the voltage across C₂. Since V = Q/C,

This makes sense qualitatively: C₂ has twice the capacitance of C₁, so it should hold twice the charge. The total charge, 24 μC + 12 μC = 36 μC, must be redistributed so that . Therefore, we see that = 24 μC (and = 12 μC).

54. E Because v is parallel to B, the charges in the bar feel no magnetic force, so there will be no movement of charge in the bar and no motional emf.

55. E Along the line joining the two charges of an electric dipole, the individual electric field vectors point in the same direction, so they add constructively. At the point equidistant from both charges, the total electric field vector has magnitude

56. E Start with basic dimensional analysis. Choices (A) and (B) can be eliminated because those expressions do not yield units of capacitance. Imagine placing equal but opposite charges on the spheres, say, +Q on the inner sphere and –Q on the outer sphere. Then the electric field between the spheres is radial and depends only on +Q, and its strength is

E =

Therefore, the potential difference between the spheres is

Now, by definition, C = Q/V, so

57. C Since the magnetic force is always perpendicular to the object’s velocity, it does zero work on any charged particle. Zero work means zero change in kinetic energy, so the speed remains the same. Remember, the magnetic force can only change the direction of a charged particle’s velocity, not its speed.

58. B The presence of the inductor in the rightmost branch effectively removes that branch from the circuit at time t = 0 (the inductor produces a large back emf when the switch is closed and the current jumps abruptly). Initially, then, current only flows in the loop containing the resistors r and R. Since their total resistance is r + R, the initial current is ε/(r + R).

59. B The correct answer should have dimensions of energy. Using the equation U_L = LI² from the Table of Information, L should have units of J/A². Choices (A), (B), (C), and (E) all have units of J/A² × V²/Ω² = J/A² × A² = J, which is a unit of energy. Choice (D) has an extra factor of resistance in the numerator, and can be eliminated. After a long time, the current through the branch containing the inductor increases to its maximum value. With all three resistors in play, the total resistance is R_eq = r + , since is the equivalent resistance of the two resistors in parallel. The current provided by the source, , splits at the junction leading to the parallel combination. The amount which flows through the inductor is half the total.

so the energy stored in the inductor is

60. A Once the switch is opened, the resistor r is cut off from the circuit, so no current passes through it. (Current will flow around the loop containing the resistors labeled R, gradually decreasing until the energy stored in the inductor is exhausted.)

61. C If a conducting sphere contains a charge of +q within an inner cavity, a charge of –q will move to the wall of the cavity to “guard” the interior of the sphere from an electrostatic field, regardless of the size, shape, or location of the cavity. As a result, a charge of +q is left on the exterior of the sphere (and it will be uniform). So, at points outside the sphere, the sphere behaves as if this charge +q were concentrated at its center, so the electric field outside is simply kQ/r². Since points X and Y are at the same distance from the center of the sphere, the electric field strength at Y will be the same as at X.

62. B A particle of negative charge is attracted to the two positive charges in the problem, and will gain kinetic energy as it approaches. By the work-energy theorem, the electric field must be doing positive work on the particle. Alternatively, the particle is attracted to the two charges it’s moving toward, which means F_E and d are in the same direction, so the work done by E is positive. Eliminate (C) and (E).

To decide between the remaining choices, first determine the electric potential at Point P. Realize that Point P is 5 meters away from each charge because it is a 3-4-5 right triangle.

The work done by the electric field will be positive because a –Q charge would be attracted to point P by the two positive Q charges. Also the work done by the force is the negative of the change in potential energy.

63. E The current in an L–R series circuit—in which initially no current flows—increases with time t according to the equation

I(t) = I_max(1 – e^-^t^/^τ)

where I_max = ε/R and τ is the inductive time constant, L/R. The time t at which I = (99%)I_max is found as follows:

Alternatively, the solution to this question can be found by using process of elimination. Choice (B) can be eliminated because it does not incorporate the 99%. Choices (A) and (C) can be eliminated because they give negative results. Choice (D) can be eliminated because ln(100/99) is close to zero. This leaves only (E), which satisfies the given conditions.

64. E First solve for the resistance of each lightbulb as shown below:

P = V²/R

R = V²/P = (120 V)²/(40 W) = 360 Ω

Next solve for the equivalent resistance, R_eq, of the circuit as follows:

R_eq = V/I = (120 V)/(5 A) = 24 Ω

If you used a current less than 5 A (the maximum allowable before the fuse blows), then the R_eq would increase. Finally, solve for the number of 360 Ω resistors (lightbulbs) that can be placed in a parallel circuit with an equivalent resistance of 24 Ω.

1/R_eq = n/R

n = R/R_eq = (360 Ω)/(24 Ω) = 15

Thus, a maximum of 15 lightbulbs can be connected in parallel with the 120 V source.

65. D Since the magnetic flux out of the page is decreasing, the induced current will oppose this change (as always), attempting to create more magnetic flux out of the page. In order to do this, the current must circulate in the counterclockwise direction (remember the right-hand rule). Eliminate (A) and (B). As B decreases uniformly (that is, while dB/dt is negative and constant), the induced emf,

ε = =

is nonzero and constant, which implies that the induced current, I = ε/R, is also nonzero and constant.

66. D Treat the configuration as three capacitors in series, each with a distance between the plates of . The capacitance of the two vacuum capacitors will be , and the capacitance of the dielectric capacitor will be that value times K. Now solve for the total capacitance for the three capacitors in series.

Alternatively, this problem can be solved by process of elimination. First, use dimensional analysis. The Table of Information provided with the AP Test notes that capacitance should have units consistent with ε₀A/d; (B) fails this test. Second, pretend that the dielectric constant K = 1. This is the same as not having a dielectric, so plugging in K = 1 should give C = ε₀A/d. Eliminate (A) and (E). Finally, as K increases, the capacitor becomes stronger, so C should increase. Once (C) fails this test, only (D) is left.

67. A The potential is zero at the point midway between the charges, but nowhere is the electric field equal to zero (except at infinity).

68. D Ampere’s law states that

B · dℓ = μ₀I_{net through loop}

Because a total current of 2I + 4I = 6I passes through the interior of the loop in one direction, and a total current of I + 3I = 4I passes through in the opposite direction, the net current passing through the loop is 6I – 4I = 2I. Therefore, the absolute value of the integral of B around the loop WXYZ is equal to μ₀(2I).

69. B Because ΔU = q(ΔV), our answer must be negative. ΔV will be the same in both cases, and the sign on q changed, so the sign on ΔU has to change as well. Eliminate (C) and (D). Next, notice that the formula for ΔU does not depend on the distance traveled. That means the different path will not impact the magnitude of ΔU. Eliminate (A). Finally, knowing the values of Q₁ and Q₂ would be important if we wanted to calculate a numerical answer, but that knowledge is unnecessary for determining the relationship in the question.

70. A First, note that the distance between –q and each charge +Q is L. Now, refer to the following diagram, where we’ve invoked Coulomb’s law to determine the two electrostatic forces:

We find that the magnitude of F is

so dividing this by m gives the initial acceleration of the charge –q.

SECTION II, ELECTRICITY AND MAGNETISM

1. (a) No calculus is necessary, because all the charge is the same distance from P. Using the definition of electric potential, V = (1/4πε₀) Σ q_i/r_i = (1/4πε₀) Σ q_i/ = Q/4πε₀. If you want to see how to do it with calculus, the solution is below, but note that the integral is of a constant times dθ. Integrating a constant is one sign that calculus isn’t necessary.

Consider a small arc on the ring, subtended by an angle dθ. The length of the arc is ds = R dθ, so the charge it carries is dQ = (Q/2πR)(R dθ) = (Q/2π) dθ, since λ = Q/(2πR) is the linear charge density.

The distance from this arc to an arbitrary Point P on the z-axis is (z² + R²)^1/2, so the contribution to the electric potential at P due to this arc is

Integrating this from θ = 0 to θ = 2π (i.e., all the way around the ring), we find that

(b) (i) From the expression derived in part (a), we see that V will be maximized when z = 0. That is, the point on the z-axis where the potential due to the ring is greatest is at the center of the ring.

(ii) Putting z = 0 into the expression from part (a), we find that

To include the direction—and thereby find E—let’s use to denote the unit vector that defines the positive direction along the z-axis. Then

(d) The force on the positive charge is in the +z direction, so the charge will accelerate in that direction. As it moves along the z-axis, the force will decrease, so it will move with increasing speed and decreasing acceleration away from the origin.

2. (a) (i) The current decays exponentially according to the equation

I(t) = ^−t/(r+R)C

where the initial current, I(0), is equal to ε/(r + R). To find the time t at which I(t) = I(0)/k, we solve the following equation:

(ii) The charge on the capacitor increases according to the equation

Q(t) = Cε[1 − e^−t/(r+R)C]

The maximum value is Cε (when the capacitor is fully charged). To find the time t at which Q(t) = Cε/k, we solve the following equation:

(iii) The field energy stored in the capacitor can be written in the form U_E(t) = [Q(t)]²/(2C), where Q(t) is the function of time given in part (ii) above. We are asked to find the time t at which

(b) (i) Once the switch is turned to position 2, the capacitor discharges through resistor R (only). Since the capacitor is fully charged (to V = ε) when the switch is moved to position 2, the current will decrease exponentially according to the equation

I(t) = I₀^−t/RC

where I₀ = I_max = ε/R. To find the time at which I is equal to I₀/k, we solve the following equation:

(ii) Since the charge on the capacitor obeys the same equation as the current—as given in (b) (i) above, simply replacing I by Q—the time t at which the charge drops to 1/k its initial value is the same as the time t at which the current drops to 1/k times its initial value: t = (ln k)RC.

3. (a) (i) As the strip XY slides down the rails, the magnetic flux through the rectangular region bounded by the strip increases. This change in magnetic flux induces an emf. If ℓ denotes the distance the strip has slid down the rails, then the magnetic flux through the rectangular region is

Φ_B = BA cos θ = BLℓ cos θ

The rate of change of Φ_B is

where v = dℓ/dt is the speed with which the strip slides. By Faraday’s law, this is the (magnitude of the) induced emf.

(ii) As the strip slides down, the area increases, so the magnetic flux upward increases. To oppose this change (in accordance with Lenz’s law), the induced current must flow in such a way as to create some magnetic flux downward. By the right-hand rule, then, current must flow from Y to X.

(iii) The induced current depends on the speed of the sliding strip and is equal to

(b) In order for the strip to slide with constant velocity, the net force on it must be zero. The force it feels parallel to the rails and downward is F_w sin θ = mg sin θ. The following diagram

which is a side view of the apparatus looking at the X end of the strip, shows that the force the strip feels parallel to the rails and upward is

The net force on the strip is zero when these two opposing forces balance; the speed at which this occurs will be denoted v_f:

where the last equality follows from the formula P = Fv for the power produced by a constant force F acting on an object whose velocity, v, is parallel to F.

Part III

About the AP Physics C Exam

• The Structure of the AP Physics C Exam

• How the AP Physics C Exam Is Scored

• Overview of Content Topics

• How AP Exams Are Used

• Other Resources

• Designing Your Study Plan

THE STRUCTURE OF THE AP PHYSICS C EXAM

The AP Physics C Exam is actually composed of two separate exams: one in Mechanics and one in Electricity and Magnetism (E & M). You can take just the Mechanics, just the E & M, or both. They are administered at different times, and separate scores are reported for each.

Whether you are taking Mechanics or E & M, your exam will contain two sections: a multiple-choice section and a free-response section. Questions in the multiple-choice section are each followed by five possible responses (only one of which is correct), and your job, of course, is to choose the right answer. Each right answer is worth one point, and there is no penalty for a wrong answer. There are 35 total multiple-choice questions, and you are given 45 minutes to complete this section. (Note that the Mechanics and E & M tests have completely different sets of questions; the timing is the only thing they share in common.). Calculators can be used for the whole exam on both the multiple-choice and free-response sections. Scientific or graphing calculators are allowed, including the approved calculators listed on the College Board website at www.collegeboard.org/ap/calculators.

The free-response section consists of three multi-part questions that require you to write out your solutions, showing your work. The total amount of time for this section is 45 minutes, so you have an average of 15 minutes per question. Unlike the multiple-choice section, which is scored by computer, the free-response section is graded by high school and college teachers. They have guidelines for awarding partial credit, so you don’t need to correctly answer every part in order to get points. Again, you are allowed to use a calculator for the free-response sections, and a table of information (mostly equations) is provided for your use. (We’ve included a set of AP Physics C equation tables in both Practice Tests.) The two sections—multiple choice and free response—are weighted equally, so each is worth 50 percent of your grade.

HOW THE AP PHYSICS C EXAM IS SCORED

Grades on the AP Physics C Exam are reported as a number: either 1, 2, 3, 4, or 5. The descriptions for each of these five numerical scores are as follows, along with the percentage of test-takers who got each grade.

AP Exam Grade	Description	Mechanics	E & M
5	Extremely well qualified	32.3%	34.6%
4	Well qualified	27.0%	22.7%
3	Qualified	18.1%	13.2%
2	Possibly qualified	13.1%	17.9%
1	No recommendation	9.5%	11.6%

Colleges are generally looking for a 4 or 5, but some may grant credit for a 3. How well do you have to do to earn such a grade? Each test is curved, and specific cut-offs for each grade vary a little from year to year, but here’s a rough idea of how many points you must earn—as a percentage of the maximum possible raw score—to achieve each of the grades 2 through 5:

	AP Exam Grade	Percentage Needed
	5	≥ 75%
	4	≥ 60%
	3	≥ 45%
	2	≥ 35%

The percentages needed are usually a little lower for the E & M section of Physics C.

OVERVIEW OF CONTENT TOPICS

So, what exactly is on the exams and how do you prepare for them? Here’s a listing of the major topics covered on the AP Physics C Exams and an average amount of the test devoted to each subject.

Mechanics

Kinematics (18%)

Newton’s Laws of Motion (20%)

Work, Energy, and Power (14%)

Systems of Particles, Linear Momentum (12%)

Circular Motion and Rotation (18%)

Oscillations and Gravitation (18%)

Electricity and Magnetism

Electrostatics (30%)

Conductors, Capacitors, and Dielectrics (14%)

Electric Circuits (20%)

Magnetic Fields (20%)

Electromagnetism (16%)

Naturally, it’s important to be familiar with the topics—to understand the basics of the theory, to know the definitions of the fundamental quantities, and to recognize and be able to use the equations. Then, you must practice applying what you’ve learned to answering questions like you’ll see on the exam. This book is designed to review all of the content areas covered on the exam, illustrated by hundreds of examples. Also, each chapter (except the first) is followed by practice multiple-choice and free-response questions, and perhaps even more important, answers and explanations are provided for every example and question in this book. You’ll learn as much—if not more—from actively reading the solutions as you will from reading the text and examples. Also, two full-length practice tests (with solutions) are provided, one in Part II and one at the end of the book in Part VI. The difficulty level of the examples and questions is equal to or slightly above AP level, so if you have the time and motivation to attack these questions and learn from the solutions, you should feel confident that you can do your very best on the real thing.

HOW AP EXAMS ARE USED

Different colleges use AP Exams in different ways, so it is important that you go to a particular college’s website to determine how it uses AP Exams. The three items below represent the main ways in which AP Exam scores can be used:

• College Credit. Some colleges will give you college credit if you score well on an AP Exam. These credits count towards your graduation requirements, meaning that you can take fewer courses while in college. Given the cost of college, this could be quite a benefit, indeed.

• Satisfy Requirements. Some colleges will allow you to “place out” of certain requirements if you do well on an AP Exam, even if they do not give you actual college credits. For example, you might not need to take an introductory-level course, or perhaps you might not need to take a class in a certain discipline at all.

• Admissions Plus. Even if your AP Exam will not result in college credit or even allow you to place out of certain courses, most colleges will respect your decision to push yourself by taking an AP Course or even an AP Exam outside of a course. A high score on an AP Exam shows mastery of more difficult content than is taught in many high school courses, and colleges may take that into account during the admissions process.

OTHER RESOURCES

There are many resources available to help you improve your score on the AP Physics C Exam, not the least of which are your teachers. If you are taking an AP class, you may be able to get extra attention from your teacher, such as obtaining feedback on your essays. If you are not in an AP course, reach out to a teacher who teaches AP Physics C and ask if the teacher will review your essays or otherwise help you with content.

Another wonderful resource is AP Students, the College Board’s official site for all AP Exam materials and information for students. The home page is: http://apstudent.collegeboard.org. The scope of the information at this site is quite broad and includes the following:

• Course description, which includes details on what content is covered

• Free-response question prompts and multiple-choice questions from previous years

• Updates on future changes to the AP Physics C Exams

The AP Students home page address of AP Physics C: Electricity and Magnetism is: http://apstudent.collegeboard.org/apcourse/ap-physics-c-electricity-and-magnetism.

The home page address of AP Physics C: Mechanics is: http://apstudent.collegeboard.org/apcourse/ap-physics-c-mechanics.

Finally, The Princeton Review offers tutoring and small group instruction. Our expert instructors can help you refine your strategic approach and add to your content knowledge. For more information, call 1-800-2REVIEW.

DESIGNING YOUR STUDY PLAN

As part of the Introduction, you identified some areas of potential improvement. Let’s now delve further into your performance on Practice Test 1, with the goal of developing a study plan appropriate to your needs and time commitment.

Read the answers and explanations associated with the multiple-choice questions (starting at this page). After you have done so, respond to the following questions:

• Review the topics list on this page, and next to each topic, indicate your rank as follows: “1” means “I need a lot of work on this,” “2” means “I need to beef up my knowledge,” and “3” means “I know this topic well.”

• How many days/weeks/months away is your exam?

• What time of day is your best, most focused study time?

• How much time per day/week/month will you devote to preparing for your exam?

• When will you do this preparation? (Be as specific as possible: Mondays and Wednesdays from 3:00 to 4:00 P.M., for example)

• Based on the answers above, will you focus on strategy (Part IV) or content (Part V) or both?

• What are your overall goals in using this book?

Part IV

Test-Taking Strategies for the AP Physics C Exam

• Preview

1 How to Approach Multiple-Choice Questions

2 How to Approach Free-Response Questions

3 Using Time Effectively to Maximize Points

• Reflect

PREVIEW

Review your Practice Test 1 results and then respond to the following questions:

• How many multiple-choice questions did you miss even though you knew the answer?

• On how many multiple-choice questions did you guess blindly?

• How many multiple-choice questions did you miss after eliminating some answers and guessing based on the remaining answers?

• Did you find any of the free-response questions easier or harder than the others—and, if so, why?

HOW TO USE THE CHAPTERS IN THIS PART

For the following Strategy chapters, think about what you are doing now before you read the chapters. As you read and engage in the directed practice, be sure to appreciate the ways you can change your approach.

Chapter 1 How to Approach Multiple-Choice Questions

MULTIPLE-CHOICE QUESTIONS ON THE AP PHYSICS C EXAM

Use the Answer Sheet

For the multiple-choice section, you write the answers not in the test booklet but on a separate answer sheet (very similar to the ones we’ve supplied here). Five oval-shaped bubbles follow the question number, one for each possible answer. Don’t forget to fill in all your answers on the answer sheet. Marks in the test booklet will not be graded. Also, make sure that your filled-in answers correspond to the correct question numbers! Check your answer sheet after every five answers to make sure you haven’t skipped any bubbles by mistake.

Should You Guess?

Use Process of Elimination (POE) to rule out answer choices you know are wrong and increase your chances of guessing the right answer. Read all the answer choices carefully. Eliminate the ones that you know are wrong. If you only have one answer choice left, choose it, even if you’re not completely sure why it’s correct. Remember: Questions in the multiple-choice section are graded by a computer, so it doesn’t care how you arrived at the correct answer.

Even if you can’t eliminate answer choices, go ahead and guess. The AP exams no longer include a guessing penalty of a quarter of a point for each incorrect answer. You will be assessed only on the total number of correct answers, so be sure to fill in all the bubbles even if you have no idea what the correct answers are. When you get to questions that are too time-consuming, or you don’t know the answer to (and can’t eliminate any options), don’t just fill in any answer. Use what we call your “letter of the day” (LOTD). Selecting the same answer choice each time you guess will increase your odds of getting a few of those skipped questions right.

Use the Two-Pass System

Remember that you have about one and a quarter minutes per question on this section of the exam. Do not waste time by lingering too long over any single question. If you’re having trouble, move on to the next question. After you finish all the questions, you can come back to the ones you skipped.

The best strategy is to go through the multiple-choice section twice. The first time, do all the questions that you can answer fairly quickly—the ones where you feel confident about the correct answer. On this first pass, skip the questions that seem to require more thinking or the ones you need to read two or three times before you understand them. Circle the questions that you’ve skipped in the question booklet so that you can find them easily in the second pass. You must be very careful with the answer sheet by making sure the filled-in answers correspond correctly to the questions.

Once you have gone through all the questions, go back to the ones that you skipped in the first pass. But don’t linger too long on any one question even in the second pass. Spending too much time wrestling over a hard question can cause two things to happen: One, you may run out of time and miss out on answering easier questions in the later part of the exam. Two, your anxiety might start building up, and this could prevent you from thinking clearly, making it even more difficult to answer other questions. If you simply don’t know the answer, or can’t eliminate any of them, just use your LOTD and move on.

General Advice

Answering 35 multiple-choice questions in 45 minutes can be challenging. Make sure to pace yourself accordingly and remember that you do not need to answer every question correctly to do well. Exploit the multiple-choice structure of this section. There are four wrong answers and only one correct one, so even if you don’t know exactly which one is the right answer, you can eliminate some that you know for sure are wrong. Then you can make an educated guess from among the answers that are left and greatly increase your odds of getting that question correct.

Problems with graphs and diagrams are usually the fastest to solve and problems with an explanation for each answer usually take the longest to work through. Do not spend too much time on any one problem or you may not get to easier problems further into the test.

These practice exams are written to give you an idea of the format of the test, the difficulty of the questions, and to allow you to practice pacing yourself. Take them in the same circumstances as you will encounter during the real exam: 45 minutes for each of the two multiple-choice sections. And remember, you are allowed to use a calculator!

Chapter 2 How to Approach Free-Response Questions

FREE-RESPONSE QUESTIONS ON THE AP PHYSICS C EXAM

Section II is worth 50 percent of your grade on the AP Physics C Exam. You’re given a total of 45 minutes to answer three free-response questions on both the Mechanics test and the E & M test, which means you must answer a total of six questions in 90 minutes.

In the Mechanics free-response section, there is almost always one general mechanics question that involves a variety of principles: energy, momentum, Newton’s laws, and other physics concepts. This problem synthesizes a lot of concepts, but each part is usually straightforward. Another one of the questions is almost always a rotational motion problem: rolling motion, fixed-axis rotation, and/or angular momentum. Though you might encounter a second general mechanics question, one of the three problems usually comes from among the following topics: resistive forces, simple harmonic motion, potential energy functions, circular motion and orbits, and gravitational forces. One of these three questions usually involves an experimental component. This is a problem in which an experiment is explained, data or a graph is given, and you must use the information given to solve the problem. The other type of experimental problem asks you to design an experiment that would determine some value you have previously solved for. The experimental problems may make up just one part of one question, or an entire question may be centered on the experiment.

Pace Yourself

The time constraints of answering three multiple-part free-response questions in 45 minutes can be a challenge. Again, pace yourself so you get to part of each problem. Make sure to skim all the questions quickly to determine which will be the easiest to solve and start with that one. Often some specific parts of a question are easy, so even if the rest of the question is very difficult, make sure to answer these easier parts. Often students spend too long to get all the parts of one question correct and get almost nothing correct on the other two problems because they are rushed. Pace yourself so that you get to answer the easiest parts of all three questions, and leave the parts that stump you for last.

On the Electricity and Magnetism free-response section, you will almost always see one electrostatics question, one circuits question, and one magnetism question. The electrostatics question is often on Gauss’s law or several point charges distributed in a plane. The circuit question almost always involves an RC circuit and occasionally includes an inductor. The circuit will feature a switch to add different electrical components at different times. The magnetism question almost always involves induced emf, Faraday’s law, and occasionally includes Ampere’s law to determine the magnetic field of a wire, solenoid, or toroid. This portion of the test can also include an experimental component. Infrequently there are two magnetism questions and no circuits questions on the free-response section of the test.

Clearly Explain and Justify Your Answers

Remember that your answers to the free-response questions are graded by readers and not by computers. Communication is a very important part of AP Physics C. Compose your answers in precise sentences. Just getting the correct numerical answer is not enough. You should be able to explain your reasoning behind the technique that you selected and communicate your answer in the context of the problem. Even if the question does not explicitly say so, always explain and justify every step of your answer, including the final answer. Do not expect the graders to read between the lines. Explain everything as though somebody with no knowledge of physics is going to read it. Be sure to present your solution in a systematic manner using solid logic and appropriate language. And remember: Although you won’t earn points for neatness, the graders can’t give you a grade if they can’t read and understand your solution!

Use Only the Space You Need

Do not try to fill up the space provided for each question. The space given is usually more than enough. The people who design the tests realize that some students write in big letters and some students make mistakes and need extra space for corrections. So if you have a complete solution, don’t worry about the extra space. Writing more will not earn you extra credit. In fact, many students tend to go overboard and shoot themselves in the foot by making a mistake after they’ve already written the right answer.

Complete the Whole Question

Some questions might have several subparts. Try to answer them all, and don’t give up on the question if one part is giving you trouble. For example, if the answer to part (b) depends on the answer to part (a), but you think you got the answer to part (a) wrong, you should still go ahead and do part (b) using your answer to part (a) as required. Chances are that the grader will not mark you wrong twice, unless it is obvious from your answer that you should have discovered your mistake.

Use Common Sense

Always use your common sense in answering questions. For example, on a free-response question that asked students to compute the mean weight of newborn babies from given data, some students answered 70 pounds. It should have been immediately obvious that the answer was probably off by a decimal point. A 70-pound baby would be a giant! This is an important mistake that should be easy to fix. Some mistakes may not be so obvious from the answer. However, the grader will consider simple, easily recognizable errors to be very important.

Think Like a Grader

When answering questions, try to think about what kind of answer the grader is expecting. Look at past free-response questions and grading rubrics on the College Board website. These examples will give you some idea of how the answers should be phrased. The graders are told to keep in mind that there are two aspects to the scoring of free-response answers: showing physics knowledge and communicating that knowledge. Again, responses should be written as clearly as possible in complete sentences.

Think Before You Write

Abraham Lincoln once said that if he had eight hours to chop down a tree, he would spend six of them sharpening his axe. Be sure to spend some time thinking about what the question is, what answers are being asked for, what answers might make sense, and what your intuition is before starting to write. These questions aren’t meant to trick you, so all the information you need is given. If you think you don’t have the right information, you may have misunderstood the question. In some calculations, it is easy to get confused, so think about whether your answers make sense in terms of what the question is asking. If you have some idea of what the answer should look like before starting to write, then you will avoid getting sidetracked and wasting time on dead-ends.

Chapter 3 Using Time Effectively to Maximize Points

BECOMING A BETTER TEST TAKER

Very few students stop to think about how to improve their test-taking skills. Most assume that if they study hard, they will test well, and if they do not study, they will do poorly. Most students continue to believe this even after experience teaches them otherwise. Have you ever studied really hard for an exam and then blown it on test day? Have you ever aced an exam for which you thought you weren’t well prepared? Most students have had one, if not both, of these experiences. The lesson should be clear: Factors other than your level of preparation influence your final test score. This chapter will provide you with some insights that will help you perform better on the AP Physics C Exam and on other exams as well.

PACING AND TIMING

A big part of scoring well on an exam is working at a consistent pace. The worst mistake made by inexperienced or unsavvy test takers is that they come to a question that stumps them and rather than just skip it, they panic and stall. Time stands still when you’re working on a question you cannot answer, and it is not unusual for students to waste five minutes on a single question (especially a question involving a graph or the word EXCEPT) because they are too stubborn to cut their losses. It is important to be aware of how much time you have spent on a given question and on the section you are working. There are several ways to improve your pacing and timing for the test:

• Know your average pace. While you prepare for your test, try to gauge how long you take on 5, 10, or 20 questions. Knowing how long you spend on average per question will help you identify how many questions you can answer effectively and how best to pace yourself for the test.

• Have a watch or clock nearby. You are permitted to have a watch or clock nearby to help you keep track of time. However, it’s important to remember that constantly checking the clock is in itself a waste of time and can be distracting. Devise a plan. Try checking the clock every 15 or 30 questions to see if you are keeping the correct pace or whether you need to speed up. This will ensure that you’re cognizant of the time but will not permit you to fall into the trap of dwelling on it.

• Know when to move on. Since all questions are scored equally, investing appreciable amounts of time on a single question is inefficient and can potentially deprive you of the chance to answer easier ones later on. You should eliminate answer choices if you are able to, but don’t worry about picking a random answer and moving on if you cannot find the correct answer. Remember, tests are like marathons; you do best when you work through them at a steady pace. You can always come back to a question you don’t know. When you do, very often you will find that your previous mental block is gone and you will wonder why the question perplexed you the first time around (as you gleefully move on to the next question). Even if you still don’t know the answer, you will not have wasted valuable time you could have spent on easier questions.

• Be selective. You don’t have to do any of the questions in a given section in order. If you are stumped by an essay or multiple-choice question, skip it or choose a different one. Select the questions or essays that you can answer and work on them first. This will make you more efficient and give you the greatest chance of getting the most questions correct.

• Use Process of Elimination (POE) on multiple-choice questions. Many times, one or more answer choices can be eliminated. Every answer choice that can be eliminated increases the odds that you will answer the question correctly.

Remember, when all the questions on a test are of equal value, no one question is that important and your overall goal for pacing is to get the most questions correct. Finally, you should set a realistic goal for your final score. In the next section, we will break down how to achieve your desired score and ways of pacing yourself to do so.

GETTING THE SCORE YOU WANT

Depending on the score you need, it may be in your best interest not to try to work through every question. Check with the schools to which you are applying to determine your needed score.

AP exams in all subjects no longer include a “guessing penalty” of a quarter of a point for every incorrect answer. Instead, students are assessed only on the total number of correct answers. A lot of AP materials, even those you receive in your AP class, may not include this information. It’s really important to remember that if you are running out of time, you should fill in all the bubbles before the time for the multiple-choice section is up. Even if you don’t plan to spend a lot of time on every question or even if you have no idea what the correct answer is, you need to fill something in.

TEST ANXIETY

Everybody experiences anxiety before and during an exam. To a certain extent, test anxiety can be helpful. Some people find that they perform more quickly and efficiently under stress. If you’ve ever pulled an all-nighter to write a paper and ended up doing good work, you know the feeling.

However, too much stress is definitely a bad thing. Hyperventilating during the test, for example, almost always leads to a lower score. If you find that you stress out during exams, here are a few preemptive actions you can take.

• Take a reality check. Evaluate your situation before the test begins. If you have studied hard, remind yourself that you are well prepared. Remember that many others taking the test are not as well prepared, and (in your classes, at least) you are being graded against them, so you have an advantage. If you didn’t study, accept the fact that you will probably not ace the test. Make sure you get to every question you know something about. Don’t stress out or fixate on how much you don’t know. Your job is to score as high as you can by maximizing the benefits of what you do know. In either scenario, it’s best to think of a test as if it were a game. How can you get the most points in the time allotted to you? Always answer questions you can answer easily and quickly before tackling those that will take more time.

• Try to relax. Slow, deep breathing works for almost everyone. Close your eyes, take a few, slow, deep breaths, and concentrate on nothing but your inhalation and exhalation for a few seconds. This is a basic form of meditation that should help you to clear your mind of stress and, as a result, concentrate better on the test. If you have ever taken yoga classes, you probably know some other good relaxation techniques. Use them when you can (obviously, anything that requires leaving your seat and, say, assuming a handstand position won’t be allowed by any but the most free-spirited proctors).

• Eliminate as many surprises as you can. Make sure you know where the test will be given, when it starts, what type of questions are going to be asked, and how long the test will take. You don’t want to be worrying about any of these things on test day or, even worse, after the test has already begun.

The best way to avoid stress is to study both the test material and the test itself. Congratulations! By buying or reading this book, you are taking a major step toward a stress-free AP Physics C Exam.

REFLECT

Respond to the following questions:

• How much time will you spend on multiple-choice questions?

• How will you change your approach to multiple-choice questions?

• What is your multiple-choice guessing strategy?

• How much time will you spend on the free-response questions?

• What will you do before you begin writing your free-response answers?

• Will you seek further help, outside of this book (such as a teacher, tutor, or AP Students), on how to approach the questions that you will see on the AP Physics C Exam?

Part V

Content Review for the AP Physics C Exam

4 Vectors

5 Kinematics

6 Newton’s Laws

7 Work, Energy, and Power

8 Linear Momentum

9 Rotational Motion

10 Laws of Gravitation

11 Oscillations

12 Electric Forces and Fields

13 Electric Potential and Capacitance

14 Direct Current Circuits

15 Magnetic Forces and Fields

16 Electromagnetic Induction

17 Chapter Drills: Answers and Explanations

Chapter 4 Vectors

INTRODUCTION

Vectors will show up all over the place in our study of physics. Some physical quantities that are represented as vectors are: displacement, velocity, acceleration, force, momentum, and electric and magnetic fields. Since vectors play such a recurring role, it’s important to become comfortable working with them; the purpose of this chapter is to provide you with a mastery of the fundamental vector algebra we’ll use in subsequent chapters. For now, we’ll restrict our study to two-dimensional vectors (that is, ones that lie flat in a plane).

DEFINITION

A vector is a quantity that involves both magnitude and direction and obeys the commutative law for addition, which we’ll explain in a moment. A quantity that does not involve direction is a scalar. For example, the quantity 55 miles per hour is a scalar, while the quantity 55 miles per hour to the north is a vector. Other examples of scalars include: mass, work, energy, power, temperature, and electric charge.

Vectors can be denoted in several ways, including:

A, A,

In textbooks, you’ll usually see one of the first two, but when it’s handwritten, you’ll see the last one.

Displacement (which is the difference between the final and initial positions) is the prototypical example of a vector:

When we say that vectors obey the commutative law for addition, we mean that if we have two vectors of the same type, for example another displacement,

then A + B must equal B + A. The vector sum A + B means the vector A followed by B, while the vector sum B + A means the vector B followed by A. That these two sums are indeed identical is shown in the following figure:

VECTOR ADDITION (GEOMETRIC)

The figure above illustrates how vectors are added to each other geometrically. Place the tail (the initial point) of one vector at the tip of the other vector, then connect the exposed tail to the exposed tip. The vector formed is the sum of the first two. This is called the “tip-to-tail” method of vector addition.

Example 1 Add A + B using the two following vectors:

Solution. Place the tail of B at the tip of A and connect them:

SCALAR MULTIPLICATION

A vector can be multiplied by a scalar (that is, by a number), and the result is a vector. If the original vector is A and the scalar is k, then the scalar multiple kA is as follows:

magnitude of kA = |k|× (magnitude of A)

direction of kA =

Example 2 Sketch the scalar multiples 2A, A, –A, and –3A of the vector A:

Solution.

Keep in mind that when you multiply a vector times a scalar of k, the vector becomes k times longer.

VECTOR SUBTRACTION (GEOMETRIC)

To subtract one vector from another, for example, to get A – B, simply form the vector –B, which is the scalar multiple (–1)B, and add it to A:

A – B = A + (–B)

Example 3 For the two vectors A and B, find the vector A – B.

Solution. Flip B around—thereby forming –B—and add that vector to A:

It is important to know that vector subtraction is not commutative: you must perform the subtraction in the order stated in the problem.

STANDARD BASIS VECTORS

Two-dimensional vectors, that is, vectors that lie flat in a plane, can be written as the sum of a horizontal vector and a perpendicular vertical vector. For example, in the following diagram, the vector A is equal to the horizontal vector B plus the vertical vector C:

The horizontal vector is always considered a scalar multiple of what’s called the horizontal basis vector, i, and the vertical vector is a scalar multiple of the vertical basis vector, j. Both of these special vectors have a magnitude of 1, and for this reason, they’re called unit vectors.

Unit vectors are often represented by placing a hat (caret) over the vector; for example, the unit vectors i and j are sometimes denoted and , or and .

For instance, the vector A in the figure below is the sum of the horizontal vector B = 3 and the vertical vector C = 4.

The vectors B and C are called the vector components of A, and the scalar multiples of and which give A—in this case, 3 and 4—are called the scalar components of A. So vector A can be written as the sum B + C = A_x + A_y, where A_x and A_y are the scalar components of A. The component A_x is called the horizontal scalar component of A, and A_y is called the vertical scalar component of A.

VECTOR OPERATIONS USING COMPONENTS

The vector operations of addition, subtraction, and scalar multiplication can now be shown using components.

Vector addition: Add the respective components

A + B = (A_x + B_x) + (A_y + B_y)

Vector subtraction: Subtract the respective components

A – B = (A_x – B_x) + (A_y – B_y)

Scalar multiplication: Multiply each component by k

kA = (kA_x) + (kA_y)

Example 4 If A = 2 – 3 and B = –4 + 2, compute each of the following vectors: A + B, A – B, 2A, and A + 3B.

Solution. It’s very helpful that the given vectors A and B are written explicitly in terms of the standard basis vectors and :

A + B = (2 – 4) + (–3 + 2) = –2 –

A – B = [2 – (–4)] + (–3 –2) = 6 – 5

2A= 2(2) + 2(–3) = 4 – 6

A + 3B = [2 + 3(–4)] + [–3 + 3(2)] = –10 + 3

MAGNITUDE OF A VECTOR

As shown in the figure, vector A is equal to the sum of the perpendicular component vectors A_x + A_y. In terms of the components of A_x and A_y, vector A is equal to A_x + A_y.

The magnitude of any vector A, can by computed by applying the Pythagorean theorem to the scalar components of vectors A_x and A_y, A_x and A_y.

The magnitude of vector A can be denoted in several ways: A or |A| or ||A||.

DIRECTION OF A VECTOR

The direction of a vector can be specified by the angle it makes with the positive x-axis. You can sketch the vector and use its components to determine the angle. For example, if θ denotes the angle that the vector A = 3 + 4 makes with the +x axis, then tan θ = A_y/A_x = 4/3. Solving for θ using the inverse trig function, θ = tan⁻¹(4/3) = 53.1°. Make sure your calculator is in the correct mode, either radian or degree, when using trig functions.

If A makes the angle θ with the +x axis, then its x- and y-components are A cos θ and A sin θ, respectively (where A is the magnitude of A).

In general, any vector in the plane can be written in terms of two perpendicular component vectors. For example, vector W (shown below) is the sum of two component vectors whose magnitudes are W cos θ and W sin θ:

THE DOT PRODUCT

A vector can be multiplied by a scalar to yield another vector, as you saw in Example 2. For instance, we can multiply the vector A by the scalar 2 to get the scalar multiple 2A. The product is a vector that has twice the magnitude and, since the scalar is positive, the same direction as A.

Additionally, we can form a scalar by multiplying two vectors. The product of the vectors in this case is called the dot product or the scalar product, since the result is a scalar. Several physical concepts (work, electric and magnetic flux) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that’s parallel to the first. The dot product was invented specifically for this purpose.

Consider these two vectors, A and B:

In order to find the component of B that’s parallel to A, we do the following:

As this figure shows, the component of B that’s parallel to A has the magnitude B cos θ. So if we multiply the magnitude of A by the magnitude of the component of B that’s parallel to A, we would form the product A(B cos θ). This is the definition of the dot product of the vectors A and B

A • B = AB cos θ

where θ is the angle between A and B. Notice that the dot product of two vectors is a scalar.

The angle between the vectors is crucial to the value of the dot product. If θ = 0, then the vectors are already parallel to each other, so we can simply multiply their magnitudes: A • B = AB. If A and B are perpendicular, then there is no component of B that’s parallel to A (or vice versa), so the dot product should be zero. And if θ is greater than 90°,

then the component of one that’s parallel to the other is actually antiparallel (backwards), and this will give the dot product a negative value (because cos θ < 0 if 90° < θ ≤ 180°).

The value of the dot product can, of course, be figured out using the definition above (AB cos θ) if θ is known. If θ is not known, the dot product can be calculated from the components of A and B in this way:

A • B =(A_x + A_y) • (B_x + B_y) = A_xB_x + A_yB_y

This means that, to form the dot product, simply add the product of the scalar x components and the product of the scalar y components.

Example 5

(a) What is the dot product of the unit vectors and ?

(b) What is the dot product of the unit vectors and ?

Solution.

(a) Since and are perpendicular to each other (θ = 90°), their dot product must be zero, because cos 90° = 0. (In fact, unless A or B already has magnitude zero, it’s also true that two vectors are perpendicular to each other when their dot product is zero.)

(b) Because and are parallel to each other, their dot product is just the product of the magnitudes, which is 1 • 1 = 1.

Example 6 If A = –2 + 4 and B = 6 + B_y, find the value of B_y such that the vectors A and B will be perpendicular to each other.

Solution. Two vectors are perpendicular to each other if their dot product is zero. That is, A • B = 0. The scalar x component is (–2)(6) = –12. The scalar y component is (4)(B_y). Since the scalar x component + scalar y component = 0, –12 + 4B_y = 0. This means that B_y must equal 3.

THE CROSS PRODUCT

Some physical concepts (torque, angular momentum, magnetic force) require that we multiply the magnitude of one vector by the magnitude of the component of the other vector that’s perpendicular to the first. The cross product was invented for this specific purpose.

Consider these two vectors A and B:

In order to find the component of one of these vectors that’s perpendicular to the other one, we do the following:

As this figure shows, the component of B that’s perpendicular to A has magnitude B sin θ. Therefore, if we multiply the magnitude of A by the magnitude of the component of B that’s perpendicular to A, we form the product A(B sin θ). This is the magnitude of what’s called the cross product of the vectors A and B, denoted A × B:

|A×B|= AB sin θ

where θ is the angle between A and B (such that 0° ≤ θ ≤ 180°).

The previous equation gives the magnitude of the cross product. The cross product of two vectors is another vector that’s always perpendicular to both A and B, with its direction determined by a procedure known as the right-hand rule. The direction of A × B is perpendicular to the plane that contains A and B, but this leads to an ambiguity, since there are two directions perpendicular to a plane (one on either side; they point in opposite directions). The following description resolves this ambiguity.

Make sure you are using your right hand. Point your index finger in the direction of the first vector, A, then point your middle finger in the direction of the second vector, B, and your thumb now points in the direction of the cross product, A × B.

This is called the right-hand rule.

You know that the standard basis vectors and can be used to write any two-dimensional vector, but now that we have introduced the cross product we need a third unit vector to write three-dimensional vectors. This third unit vector is denoted , and, like and , it points along the direction of a coordinate axis. The vector is the unit vector that points along the x-axis, along the y, and now along the z.

The three unit vectors are mutually perpendicular:

The coordinate axes shown above define a right-handed coordinate system, because the directions of the x-, y-, and z-axes obey the right-hand rule. That is, × points in the direction of . In fact, × actually equals , since the magnitude of × is 1. As a right-handed coordinate system, × = and × = .

Unlike the dot product, the cross product is not commutative; A × B is not equal to B × A. This is because applying the right-hand rule to determine the direction of B × A would give a vector that points in the direction opposite to that of A × B. Therefore, B × A = –(A × B).

The cross product can be computed directly from the scalar components of A and B without first determining the angle θ, as follows: If A = A_x + A_y + A_z and B = B_x + B_y + B_z, then the cross product (magnitude and direction) of A and B is

A × B = (A_yB_z – A_zB_y) + (A_zB_x – A_xB_z) + (A_xB_y – A_yB_x)

This formula requires a lot of memorization. Another method for determining the cross product is to realize that the cross product is the determinant of the following 3 × 3 matrix.

By taking the determinant of each 2 × 2 matrix you will realize that you get the same formula as shown above. This may appear to be equally difficult to memorize; however, you can just memorize two facts. Each 2 × 2 matrix is missing the column associated with the basis vector that multiplies it and the terms alternate in sign.

Example 7 Calculate the cross product of the vectors A = 2 + 3 and B = – + + 4, and verify that it’s perpendicular to both A and B.

Solution. One way to calculate the cross product is by using the formula method.

A × B = [(3)(4) – (0)(1)] – [(2)(4) – (0)(–1)] + [(2)(1)– (3(–1)] = 12 –8 + 5

Using the determinant method,

Now, to verify that this vector is perpendicular to both A and B, we use the property of the dot product: Two vectors are perpendicular to each other if their dot product is zero. Extending the computation of the dot product in terms of the scalar components to three dimensions, we get:

(A × B) · A= (12 –8 + 5) · (2 + 3) = (12)(2) + (–8)(3) + (5)(0) = 0

(A × B) · B= (12 –8 + 5) · (– + + 4) = (12)(–1) + (–8)(1) + (5)(4) = 0

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 5 Kinematics

INTRODUCTION

Kinematics is the study of an object’s motion in terms of its displacement, velocity, and acceleration. Questions such as How far does this object travel? Or, How fast and in what direction does it move? Or, At what rate does its speed change? All properly belong to kinematics. In the next chapter, we will study dynamics, which delves more deeply into why objects move the way they do.

POSITION, DISTANCE, AND DISPLACEMENT

Position is an object’s relation to a coordinate axis system. Distance is a scalar that represents the total amount traveled by an object. Displacement is an object’s change in position. It’s the vector that points from the object’s initial position to its final position, regardless of the path actually taken. Since displacement means change in position, it is generically denoted ∆s, where ∆ denotes change in and s means spatial location. (The letter p is not used for position because it’s reserved for another quantity: momentum.) If it’s known that the displacement is horizontal, then it can be called ∆x; if the displacement is vertical, then it’s ∆y. The magnitude of this vector is the net distance traveled and, sometimes, the word displacement refers just to this scalar quantity. Since a distance is being measured, the correct unit for displacement is the meter.

Example 1 Traveling along a single axis, a car starts 10 m from the origin. The car then moves 8 m directly away from the origin and then turns around and moves 12 m back toward the origin. Determine the final position of the car, the distance the car traveled and the displacement of the car.

Solution. The car ends at the 6 m mark, so the final position is 6 m. The car moves a total of 8 m + 12 m = 20 m, so the distance traveled is 20 m. The displacement of the car only refers to the car’s final position minus its initial position. Because the car ended 4 m behind where it started, its displacement would be –4 m.

Example 2 An infant crawls 5 m east, then 3 m north, then 1 m east. Find the magnitude of the infant’s displacement.

Solution. Although the infant crawled a total distance of 5 m + 3 m + 1 m = 9 m, this is not the magnitude of the infant’s displacement, which is merely the net distance traveled.

Using the Pythagorean theorem, we can calculate that the magnitude of the displacement is

Example 3 In a track-and-field event, an athlete runs exactly once around an oval track, a total distance of 400 m. Find the runner’s displacement for the race.

Solution. If the runner returns to the same position from which she left, then her displacement is zero.

The total distance covered is 400 m, and the displacement is 0 m.

SPEED AND VELOCITY

When we’re in a moving car, the speedometer tells us how fast we’re going; it gives us our speed. But what does it mean to have a speed of say, 10 m/s? It means that we’re covering a distance of 10 meters every second. By definition, average speed is the ratio of the total distance traveled to the time required to cover that distance:

average velocity =

The car’s speedometer doesn’t care in what direction the car is moving (as long as the wheels are moving forward). You could be driving north, south, east, west, whatever; the speedometer would make no distinction. 55 miles per hour, north and 55 miles per hour, east register the same on the speedometer: 55 miles per hour. Speed is a scalar.

However, we will also need to include direction in our descriptions of motion. We just learned about displacement, which takes both (net) distance and direction into account. The single concept that embodies both displacement and direction is called velocity, and the definition of average velocity is:

average velocity =

(The bar over the v means average.) Because ∆s is a vector, is also a vector, and because ∆t is a positive scalar, the direction of is the same as the direction of ∆s. The magnitude of the velocity vector is called the object’s speed, and is expressed in units of meters per second (m/s).

Note the distinction between speed and velocity. In everyday language, they’re often used interchangeably. In physics, however, speed and velocity are technical terms whose definitions are not the same. The magnitude of the velocity is the speed, so velocity is speed plus direction. Because speed is a scalar, average speed is defined as the total distance traveled divided by the elapsed time. On the other hand, the magnitude of the average velocity, which is a vector, is the displacement divided by the elapsed time.

However, the car’s speedometer is not intended to tell you an average speed over a long period of time: It’s supposed to tell you the current speed at the present moment! For instantaneous speed or velocity, we need calculus. Any shift from an average quantity with deltas in the definition to an instantaneous quantity, especially if the quantity is a rate of change (that is, if Δt is in the denominator), is likely to involve changing “Δ” a difference, to “d” a derivative, and leaving everything else the same, and this is no exception:

Often, a derivative with respect to time is denoted by a dot above the quantity, so an alternative notation for the equation above is v = .

Because taking an integral is the inverse of taking a derivative, this definition also implies that displacement is given by the integral of velocity with respect to time:

s = ∫ v dt

Example 4 If the infant in Example 2 completes his journey in 20 seconds, find the magnitude of his average velocity.

Solution. Since the magnitude of the displacement is 6.7 m, the magnitude of his average velocity is

= ∆s/∆t = (6.7 m)/(20 s) = 0.34 m/s

Example 5 Assume that the runner in Example 3 completes the race in 1 minute and 18 seconds. Find her average speed and the magnitude of her average velocity.

Solution. Average speed is total distance divided by elapsed time, a scalar. Since the length of the track is 500 m, the runner’s average speed was (500 m)/(78 s) = 6.4 m/s. However, since her displacement was zero, the magnitude of her average velocity was zero also: = ∆s/∆t = (0 m)/(78 s) = 0 m/s.

Example 6 Is it possible to move with constant speed but not constant velocity? Is it possible to move with constant velocity but not constant speed?

Solution. The answer to the first question is yes. For example, if you set your car’s cruise control at 55 miles per hour but turn the steering wheel to follow a curved section of road, then the direction of your velocity changes (which means your velocity is not constant), even though your speed doesn’t change.

The answer to the second question is no. Velocity means speed and direction; if the velocity is constant, then that means both speed and direction are constant. If speed were to change, then the velocity vector’s magnitude would change (by definition), which immediately implies that the vector changes.

Example 7 An object’s position x, in meters, obeys the equation x = sin(t), where t is the time in seconds since the object began moving. How fast is the object moving at t = seconds?

Solution. This question asks for the velocity of the object at t = seconds. Since velocity is the time derivative of position, the velocity is given by v = dsin(t) = cos(t). Then plug in the value of t: v() = cos() = 0 m/s.

ACCELERATION

When you step on the gas pedal in your car, the car’s speed increases; step on the brake, and the car’s speed decreases. Turn the wheel, and the car’s direction of motion changes. In all of these cases, the velocity changes. To describe this change in velocity, we need a new term: acceleration. In the same way that velocity measures the rate-of-change of an object’s position, acceleration measures the rate-of-change of an object’s velocity. An object’s average acceleration is defined as follows:

average acceleration =

The units of acceleration are meters per second, per second: [a] = m/s². Because ∆v is a vector, is also a vector; and because ∆t is a positive scalar, the direction of is the same as the direction of ∆v.

Furthermore, if we take an object’s original direction of motion to be positive, then an increase in speed corresponds to a positive acceleration, while a decrease in speed corresponds to a negative acceleration (deceleration).

Note that an object can accelerate even if its speed doesn’t change. (Again, it’s a matter of not allowing the everyday usage of the word accelerate to interfere with its technical, physics usage.) This is because acceleration depends on ∆v, and the velocity vector v changes if (1) speed changes, or (2) direction changes, or (3) both speed and direction change. For instance, a car traveling around a circular racetrack is constantly accelerating even if the car’s speed is constant because the direction of the car’s velocity vector is constantly changing.

Just as a transition from average velocity to instantaneous velocity changes “Δ” to “d” in an equation, so too does a transition from average acceleration to instantaneous acceleration. Instantaneous acceleration is given by:

Since velocity is itself the derivative of position, this implies that acceleration is the second derivative of position:

In dot notation, a = = . The above definition of acceleration also implies that velocity is the integral of acceleration:

v = ∫ a dt

The relationships described here can be summarized in the following diagram:

Example 8 A car is traveling in a straight line along a highway at a constant speed of 80 miles per hour for 10 seconds. Find its acceleration.

Solution. Neither the direction nor the speed change, so the car is traveling at a constant velocity. If the velocity is constant, then the acceleration is zero. If there’s no change in velocity, then there’s no acceleration.

Example 9 A car is traveling along a straight highway at a speed of 20 m/s. The driver steps on the gas pedal and, 3 seconds later, the car’s speed is 32 m/s. Find its average acceleration.

Solution. Here the direction does not change but the speed does, so there is a change in velocity. To obtain the average acceleration, simply divide the change in velocity, 32 m/s – 20 m/s = 12 m/s, by the time interval during which the change occurred: = ∆v/∆t = (12 m/s)/(3 s) = 4 m/s².

Example 10 Spotting a police car ahead, the driver of the car in the previous example slows from 32 m/s to 20 m/s in 2 seconds. Find the car’s average acceleration.

Solution. Dividing the change in velocity, 20 m/s – 32 m/s = –12 m/s, by the time interval during which the change occurred, 2 s, gives us = ∆v/∆t = (–12 m/s) / (2 s) = –6 m/s². The negative sign here means that the direction of the acceleration is opposite the direction of the velocity, which describes slowing down.

Example 11 The position of an object (measured in meters from the origin, where x = 0) moving along a straight line is given as a function of time t (measured in seconds) by the equation x(t) = 4t² – 6t – 40. Find

(a) its velocity at time t

(b) its acceleration at time t

(d) the object’s velocity and acceleration at the time calculated in (c)

Solution.

(a) The direction does not change but the position does. The velocity as a function of t is the derivative of the position function:

v(t) = (t) = (4t² − 6t − 40) = 8t − 6(in m/s)

(b) The acceleration as a function of t is the derivative of the velocity function:

a(t) = (t) = (8t − 6) = 8 m/s²

Notice that the acceleration is constant (because it doesn’t depend on t).

4t² – 6t – 40 = 0

2(2t² – 3t – 20) = 0

2(2t + 5)(t – 4) = 0

t = – or 4 seconds

(d) Disregarding the negative value for t, we can say that the object passes through the origin at t = 4 s. At this time, the object’s velocity is

v(4) = v(t)|_{t= 4} = (8t − 6)|_{t= 4} = (8)(4) − 6 = 26 m/s

The object’s acceleration is a constant 8 m/s² throughout its motion, so, in particular, at t = 4 s, the acceleration is 8 m/s².

Example 12 An object is moving along the x-axis with an acceleration given by the function, a(t) = (4t + 7) m/s². At time t₀ = 0, the object is at x = 6 m, and it is moving at 2 m/s. How fast will the object be traveling at time t = 4 s? Where will the object be at time t = 4 s?

Solution. By integrating the acceleration with respect to time, we find the velocity as a function of time.

v(t) = ∫ a(t)dt = ∫(4t + 7)dt = 2t² + 7t + c

We determine the value of c, the constant of integration, by using the given initial velocity, v(0) = 2 m/s.

v(0) = 2 ⇒ (2t² + 7t + c)|_t=0 = 2 ⇒ c = 2

(Notice that c is the initial velocity, v₀). So the velocity function is given by: v(t) = 2t²+ 7t + 2 (in m/s). Therefore, we can evaluate the function at t = 4 s and determine the velocity at that moment of time.

v(4) = (2t² + 7t + 2)|_t=4 (2(4)² + 7(4) + 2) = 62 m/s

Integrating this velocity function with respect to time will give us the position function.

x(t) = ∫ v(t)dt = ∫(2t² + 7t + 2)dt = t³ + t² + 2t + c₁

Again we determine the constant of integration, c₁, by using the given initial position, x(0) = 6 m.

(Again, notice that c₁, is just the initial position, x₀). So the position function is given by:

x(t) = in meters

Therefore, we can evaluate the function at t = 4 s and determine the position at that moment of time.

UNIFORMLY ACCELERATED MOTION AND THE BIG FIVE

The simplest type of motion to analyze is motion in which the acceleration is constant (possibly equal to zero). Although true uniform acceleration is rarely achieved in the real world, many common motions are governed by approximately constant acceleration. In these cases, the kinematics of uniformly accelerated motion provide an approximate description of what’s happening. Notice that if the acceleration is constant, then the average acceleration is also constant, so = a.

We can simplify our analysis by considering only motion that takes place along a straight line. In these cases, there are only two possible directions of motion. One is positive, and the other, which is in the opposite direction, is negative. Displacement, velocity, and acceleration are all vectors, which means that they include both a magnitude and a direction. With straight-line motion, direction can be specified simply by attaching a + or – sign to the magnitude of the quantity. Therefore, although we will often abandon the use of bold letters to denote the vector quantities of displacement, velocity, and acceleration, the fact that these quantities include direction will still be indicated by a positive or negative sign.

Let’s review the quantities we’ve seen so far. The fundamental quantities are displacement (Δx), velocity (v), and acceleration (a). Acceleration is a change in velocity, from an initial velocity (v_i or v₀) to a final velocity (v_f or simply v—with no subscript) during some elapsed time interval, ∆t. At t₀ = 0, Δt = t – 0 = t. We will use Δt = t to match the AP equation sheet. Therefore, we have five kinematics quantities: ∆x, v₀, v, a, and t.

These five quantities are related by a group of five equations that we call the Big Five. They work in cases where acceleration is uniform, which are the ones we’re considering.

		Variable that is missing
Big Five #1	Δx = x − x₀ = t	a
Big Five #2		∆x
Big Five #3		v
Big Five #4	x = x₀ + vt − at²	v₀
Big Five #5		t

The change in the position of an object is the displacement: Δx = x – x₀. Equations #2, #3, and #5 are given on the AP Physics C Table of Information. They are written here exactly as they will appear on the sheet, which is why we are using ∆x rather than ∆s for the displacement. Because the acceleration is constant, the average velocity is = (v₀ + v).

Each of the Big Five equations is missing one of the five kinematic quantities. To decide which equation to use when solving a problem, determine which of the kinematic quantities is missing from the problem—that is, which quantity is neither given nor asked for—and then use the equation that doesn’t contain that variable. A good strategy is to make a list of your “knowns” and your “unknowns.” For example, if the problem never mentions the final velocity—v is neither given nor asked for—the equation that will work is the one that’s missing v. That’s Big Five #3.

Big Five #1 and #2 are simply the definitions of and written in forms that don’t involve fractions. The other Big Five equations can be derived from these two definitions and the equation = (v₀ + v) by using a bit of algebra.

Example 13 An object with an initial velocity of 4 m/s moves along a straight axis under constant acceleration. Three seconds later, its velocity is 14 m/s. How far did it travel during this time?

Solution. We’re given v₀, t, and v, and we’re asked for a change in position, ∆x. So a is missing; it isn’t given and it isn’t asked for, and we use Big Five #1:

Δx = t

Δx = (14 + 4)(3) = 27 m

Example 14 A car that’s initially traveling at 10 m/s accelerates uniformly for 4 seconds at a rate of 2 m/s², in a straight line. How far does the car travel during this time?

Solution. We’re given v₀, t, and a, and we’re asked for ∆x. So, v is missing; it isn’t given and it isn’t asked for, and we use Big Five #3:

Δx = v₀t + a(t)² = (10 m/s)(4 s) + (2 m/s²)(4 s)² = 56 m

Example 15 A rock is dropped off a cliff that’s 80 m high. If it strikes the ground with an impact velocity of 40 m/s, what acceleration did it experience during its descent?

Solution. If something is dropped, then that means it has no initial velocity: v₀ = 0. So, we’re given v₀, ∆x, and v, and we’re asked for a. Since t is missing, we use Big Five #5, choosing positive as downward:

v² = + 2aΔx ⇒ v² = 2aΔx (since v₀ = 0)

a = = 10 m/s² downward

Note that since a has the same sign as ∆x, the acceleration vector points in the same direction as the displacement vector. This makes sense since the object moves downward and the acceleration it experiences is due to gravity, which also points downward.

KINEMATICS WITH GRAPHS

So far, we have dealt with kinematics problems algebraically, but you should also be able to handle kinematics questions in which information is given graphically. The two most popular graphs in kinematics are position-versus-time graphs and velocity-vs.-time graphs. For example, consider an object that’s moving along an axis in such a way that its position x as a function of time t is given by the following position-vs.-time graph:

What does this graph tell us? It says that at time t = 0, the object was at position x = 0. Then, in the next two seconds, its position changed from x = 0 to x = 10 m. Then, at time t = 2 s, it reversed direction and headed back toward its starting point, reaching x = 0 at time t = 3 s, and continued, reaching position x = –5 m at time t = 3.5 s. Then the object remained at this position, x = –5 m, at least through time t = 6 s. Notice how economically the graph embodies all this information!

We can also determine the object’s average velocity (and average speed) during particular time intervals. For example, its average velocity from time t = 0 to time t = 2 s is equal to the object’s displacement, 10 – 0 = 10 m, divided by the elapsed time, 2 s:

= = 5 m/s

Note, however, that the ratio that defines the average velocity, ∆x/∆t, also defines the slope of the x-vs.-t graph. Therefore, we know the following important fact:

The slope of a position-vs.-time graph gives the average velocity.

What was the average velocity from time t = 2 s to time t = 3.5 s? The slope of the line segment joining the point (t, x) = (2 s, 10 m) to the point (t, x) = (3.5 s, – 5m) is

= = −10 m/s

The fact that is negative tells us that the object’s displacement was negative during this time interval; that is, it moved in the negative x direction. The fact that is negative agrees with the observation that the slope of a line that falls to the right is negative. What is the object’s average velocity from time t = 3.5 s to time t = 6 s? Since the line segment from t = 3.5 s to t = 6 s is horizontal, its slope is zero, which implies that the average velocity is zero, but we can also figure this out from looking at the graph, since the object’s position did not change during that time.

Finally, let’s figure out the object’s average velocity and average speed for its entire journey (from t = 0 to t = 6 s). The average velocity is

= = −0.83 m/s

This is the slope of the imagined line segment that joins the point (t, x) = (0 s, 0 m) to the point (t, x) = (6 s, –5 m). The average speed is the total distance traveled by the object divided by the elapsed time. In this case, notice that the object traveled 10 m in the first 2 s, then 15 m (albeit backward) in the next 1.5 s; it covered no additional distance from t = 3.5 s to t = 6 s. Therefore, the total distance traveled by the object is x = 10 + 15 = 25 m, which took 6 s, so

average speed = = 4.2 m/s

Let’s next consider an object moving along a straight axis in such a way that its velocity, v, as a function of time, t, is given by the following velocity-vs.-time graph:

What does this graph tell us? It says that, at time t = 0, the object’s velocity was v = 0. Over the first two seconds, its velocity increased steadily to 10 m/s. At time t = 2 s, the velocity then began to decrease (eventually becoming v = 0, at time t = 3 s). The velocity then became negative after t = 3 s, reaching v = –5 m/s at time t = 3.5 s. From t = 3.5 s on, the velocity remained a steady –5 m/s.

What can we ask about this motion? First, the fact that the velocity changed from t = 0 to t = 2 s tells us that the object accelerated. The acceleration during this time was

a = = 5 m/s²

Note, however, that the ratio that defines the acceleration, ∆v/∆t, also defines the slope of the v-vs.-t graph. Therefore,

The slope of a velocity-vs.-time graph gives the average acceleration.

What was the acceleration from time t = 2 s to time t = 3.5 s? The slope of the line segment joining the point (t, v) = (2 s, 10 m/s) to the point (t, v) = (3.5 s, –5 m/s) is

a = = −10 m/s²

The fact that a is negative tells us that the object’s velocity change was also negative during this time interval; that is, the object accelerated in the negative direction, or slowed down. In fact, after time t = 3 s, the velocity became more negative, indicating that the direction of motion was negative with increasing speed. What is the object’s acceleration from time t = 3.5 s to time t = 6 s? Since the line segment from t = 3.5 s to t = 6 s is horizontal, its slope is zero, which indicates that the acceleration is zero, but you can also see this from looking at the graph; the object’s velocity did not change during this time interval.

Another question can be asked when a velocity-vs.-time graph is given: How far did the object travel during a particular time interval? For example, let’s figure out the displacement of the object from time t = 4 s to time t = 6 s. During this time interval, the velocity was a constant –5 m/s, so the displacement was ∆x = v∆t = (–5 m/s)(2 s) = –10 m.

Geometrically, we’ve determined the area between the graph and the horizontal axis. After all, the area of a rectangle is base × height and, for the shaded rectangle shown below, the base is ∆t, and the height is v. So, base × height equals ∆t × v, which is displacement.

Counting areas above the horizontal axis as positive and areas below the horizontal axis as negative, we can make the following claim:

Given a velocity-vs.-time graph, the area between the graph and the t-axis equals the object’s displacement.

What is the object’s displacement from time t = 0 to t = 3 s? Using the fact that displacement is the area bounded by the velocity graph, we can calculate the area of the triangle shown below:

Since the area of a triangle is × base × height, we find that

Δx = (3 s)(10 m/s) = 15 m

A Note About Calculus

The use of graphs for solving kinematics questions provides a link to some basic definitions in calculus. The slope of a curve is the geometric definition of the derivative of a function. Since the slope of a position-vs.-time graph gives the velocity, the derivative of a position function gives the velocity. That is, given an equation of the form x = x(t) for the position x of an object as a function of time t, the derivative of x(t)—with respect to t—gives the object’s velocity v(t):

Often, a derivative with respect to time is denoted by a dot above the quantity, so an alternative notation for the equation above is v(t) = (t).

Next, since the slope of a velocity-vs.-time graph gives the acceleration, the derivative of a velocity function gives the acceleration. That is, given an equation of the form v = v(t) for the velocity v of an object as a function of time t, the derivative of v(t)—with respect to t—gives the object’s acceleration a(t):

Combining this equation with the previous one, we see that the acceleration is the second derivative of the position:

Using the dot notation, these last two equations may be written as a(t) = (t) and a(t) = (t), respectively.

The area bounded by a curve and the horizontal axis is the geometric definition of the definite integral of a function. Since the area bounded by a velocity-vs.-time graph gives the displacement, the definite integral of a velocity function gives the displacement. That is, given the equation v = v(t) for the velocity v of an object as a function of time t, the integral of v(t) from time t = t₁ to time t = t₂ equals the displacement during this time interval:

displacement =

For the velocity-vs.-time graph examined above, the slope of the line segment from (t, v) = (0, 0) to (t, v) = (2, 10) can be described by the equation v(t) = 5t. Therefore, the acceleration of the object from time t = 0 to t = 2 s is

as we found above.

The line segment from (t, v) = (2, 10) to (t, v) = (3, 0) can be described by the equation v(t) = –10 + 30. The displacement of the object from time t = 0 to t = 3 s is the value of the integral from t = 0 to t = 3 s, which we can get by adding the integral of v(t) = 5t from t = 0 to t = 2 s to the integral of v(t) = –10t + 30 from t = 2 s to t = 3 s:

which also agrees with the value computed above.

Example 16 The velocity of an object as a function of time is given by the following graph:

At which point (A, B, C, D, or E) is the magnitude of the acceleration the greatest? How would you answer this same question if the graph shown were a position-vs.-time graph?

Solution. The acceleration is the slope of the velocity-vs.-time graph, or dv/dt. Although this graph is not composed of straight lines, the concept of slope still applies; at each point, the slope of the curve is the slope of the tangent line to the curve. The slope is essentially zero at Points A and D (where the curve is flat), small and positive at B, and small and negative at E. The slope at Point C is large and positive, so this is where the object’s acceleration is the greatest.

If the graph shown were a position-vs.-time graph, then the slope would be the velocity. The slope of the given graph starts at zero (around Point A), slowly increases to a small positive value at B, continues to slowly increase to a large positive value at C, then, at around Point D, this large positive slope decreases quickly to zero. Of the points designated on the graph, Point D is the location of the greatest slope change, which means that this is the point of the greatest velocity change. Therefore, this is the point at which the magnitude of the acceleration is greatest.

Average vs. Instantaneous Quantities

We have a variety of methods to determine the velocity and the acceleration of an object. We also need to distinguish between average quantities and instantaneous quantities. Average velocity is the displacement over some time interval, while instantaneous velocity is how fast the object is traveling at a specific instant of time. The speedometer on a car gives us the instantaneous speed (it does not indicate the direction, so it does not indicate the instantaneous velocity). Typically, it is easiest to solve for average quantities using the definitions mentioned before. It is easiest to solve for instantaneous quantities by using one of the Big Five equations, or taking the slope of a given graph or derivative of a given function. Make sure you carefully read the question and understand whether they are asking for an average quantity or an instantaneous quantity.

FREE FALL

The simplest real-life example of motion under relatively constant acceleration is the motion of objects in Earth’s gravitational field, ignoring any effects due to the air (mainly air resistance). Under these conditions, an object can fall freely, that is, it can fall experiencing only acceleration due to gravity. Near the surface of Earth, the gravitational acceleration has a constant magnitude of about 9.8 m/s²; this quantity is denoted g (for gravitational acceleration). On the AP Physics C Exam, you may use g = 10 m/s² as a simple approximation to g = 9.8 m/s². This is particularly helpful because, as we mentioned earlier, you will not be permitted to use a calculator on the multiple-choice section. In this book, we will always use g = 10 m/s² . And, of course, the gravitational acceleration vector, g, points downward.

Since the acceleration is constant, we can use the Big Five with a replaced by +g or –g. We will use y for displacement rather than s or x because the motion is vertical. To decide which of these two values to use for a, make a decision at the beginning of your calculations whether to call “down” the positive direction or the negative direction. If you call “down” the positive direction, then a = +g. If you call “down” the negative direction, then a = –g. We will always assume up is the positive direction unless all of the motion is downward.

In each of the following examples, we’ll ignore effects due to the air.

Example 17 A rock is dropped from an 80-meter cliff. How long does it take to reach the ground?

Solution. Since all of the rock’s motion is down, we call down the positive direction, so a = +g. We’re given v₀, ∆y, and a, and we are asked for t. So v is missing; it isn’t given and it isn’t asked for, and we use Big Five #3:

y = y₀ + v₀t + gt² ⇒ y = g t² (since y₀ = 0, v₀ = 0)

Example 18 A baseball is thrown straight upward with an initial speed of 20 m/s. How high will it go?

Solution. Since the ball travels upward, call up the positive direction. Therefore, down is the negative direction, so a = –g. The ball’s velocity drops to zero at the instant the ball reaches its highest point, so we’re given a, v₀, and v, and asked for ∆y. Since t is missing, we use Big Five #5:

Example 19 One second after being thrown straight down, an object is falling with a speed of 20 m/s. How fast will it be falling 2 seconds after it was traveling 20 m/s?

Solution. Because all the motion is downward, call down the positive direction, so a = +g and v₀ = +20 m/s. We’re given v₀, a, and t, and asked for v. Since ∆y is missing, we use Big Five #2:

v = v₀ + at = (+20 m/s) + (+10 m/s²)(2 s) = 40 m/s

Example 20 If an object is thrown straight upward with an initial speed of 8 m/s and takes 3 seconds to strike the ground, from what height was the object thrown?

Solution. The figure below shows the path of the ball. We will use up as positive because not all the motion is downward. Therefore a = –g. We’re given a, v₀, and t, and we need to find ∆y, which is y – y₀. Since v is missing, we use Big Five #3:

PROJECTILE MOTION

In general, an object that moves near the surface of Earth will not follow a straight-line path (for example, a baseball hit by a bat, a golf ball struck by a club, or a tennis ball hit from the baseline). If we launch an object at an angle other than straight upward and consider only the effect of acceleration due to gravity, then the object will travel along a parabolic trajectory.

To simplify the analysis of parabolic motion, we analyze the horizontal and vertical motions separately, using the Big Five. This is the key to doing projectile motion problems. Calling down the negative direction, we have the following:

	Horizontal motion:	Vertical motion:
	∆x = v₀_xt	∆y = v₀_yt + (–g)t²
	v_x = v₀_x (constant!)	v_y = v₀_y + (–g)t
	a_x = 0	a_y = –g
		v_y² = v₀²_y + 2 (–g)Δy

The quantity v₀_x, which is the horizontal (or x) component of the initial velocity, is equal to v₀ cos θ₀, where θ₀ is the launch angle, the angle that the initial velocity vector, v₀, makes with the horizontal. Similarly, the quantity v₀_y, the vertical (or y) component of the initial velocity, is equal to v₀ sin θ₀.

Example 21 An object is thrown horizontally off a cliff with an initial speed of 10 m/s. How far will it drop in 4 seconds assuming it does not hit the ground first?

Solution. The first step is to decide whether this is a horizontal question or a vertical question, since you must consider these motions separately. The question asks how far the object will drop. This is a vertical question, so the set of equations we will consider are those listed above under vertical motion. Next, “How far will it drop?” implies that we will use the first of the vertical-motion equations, the one that gives vertical displacement, ∆y.

Now, since the object is thrown horizontally, there is no vertical component to its initial velocity vector v₀ ; that is, v₀_y = 0. Therefore,

The fact that ∆y is negative means that the displacement is down. Also, notice that the information given about v₀_x is irrelevant to the question.

Example 22 From a height of 100 m, a ball is thrown horizontally with an initial speed of 15 m/s. How far does it travel horizontally in the first 2 seconds?

Solution. The question, “How far does it travel horizontally?” immediately tells us that we should use the first of the horizontal-motion equations:

∆x = v_0xt = (15 m/s)(2 s) = 30 m

The information that the initial vertical position is 100 m above the ground is irrelevant (except for the fact that it’s high enough that the ball doesn’t strike the ground before the two seconds have elapsed).

Example 23 A projectile is traveling in a parabolic path for a total of 6 seconds. How does its horizontal velocity 1 s after launch compare to its horizontal velocity 4 s after launch?

Solution. The only acceleration experienced by the projectile is due to gravity, which is purely vertical, so that there is no horizontal acceleration. If there’s no horizontal acceleration, then the horizontal velocity cannot change during flight, and the projectile’s horizontal velocity 1 s after it’s launched is the same as its horizontal velocity 3 s later.

Example 24 An object is projected upward with a 30° launch angle and an initial speed of 40 m/s. How long will it take for the object to reach the top of its trajectory? How high is this?

Solution. When the projectile reaches the top of its trajectory, its velocity vector is momentarily horizontal; that is, v_y = 0. We are asked how long it will take the object to reach this point with θ = 30 degrees. Using the vertical-motion equation for v_y, we can set it equal to 0 and solve for t:

At this time, the projectile’s vertical displacement is

Next, we are asked for the projectile’s vertical displacement, Δy, given t = 2 s.

Example 25 An object is projected upward with a 30° launch angle and an initial speed of 60 m/s. For how many seconds will it be in the air? How far will it travel horizontally before returning to its original height?

Solution. The total time the object spends in the air is equal to twice the time required to reach the top of the trajectory (because the parabola is symmetrical). So, as we did in the previous example, we find the time required to reach the top by setting v_y equal to 0, and now double that amount of time:

v_0y + (−g)t = 0

Therefore, the total flight time (that is, up and down) is T = 2t = 2 × (3 s) = 6 s.

Now, using the first horizontal-motion equation, we can calculate the horizontal displacement after 6 seconds:

Δx = v_0x T = (v₀ cos θ₀)T = [(60 m/s) cos 30°] (6 s) = 310 m

By the way, the full horizontal displacement of a projectile to the same height is called the projectile’s range.

A Note About Notation

We are trying to be as consistent as possible with the test by using equations identical to those on the AP Physics C Table of Information that you will be given to use during the exam. When the motion is vertical, we use y instead of x, and occasionally we use s to indicate when the motion is in two dimensions. It is important that you judge which variable best represents the values you are given and those you need to solve for.

Chapter 5 Drill

The answers and explanations can be found in Chapter 17.

Click here to download the PDF.

Section I: Multiple Choice

1. An object that’s moving with constant speed travels once around a circular path. Which of the following is/are true concerning this motion?

I. The displacement is zero.

II. The average speed is zero.

III. The acceleration is zero.

(A) I only

(B) I and II only

(D) III only

(E) II and III only

2. At time t = t₁, an object’s velocity is given by the vector v₁ shown below:

A short time later, at t = t₂, the object’s velocity is the vector v₂:

If v₂ = v₁, which one of the following vectors best illustrates the object’s average acceleration between t = t₁ and t = t₂?

(A)

(B)

(C)

(D)

(E)

3. A rock is thrown off a 30 m cliff at a 45° angle above the horizontal. Which is the following is true regarding the acceleration of the rock?

(A) The acceleration will be of magnitude g and have both horizontal and vertical components.

(B) At the peak of the rock’s path, the magnitude of acceleration will be half of what it was when the rock was initially thrown.

(D) The acceleration will be downward during the rock’s ascent and upward during the rock’s descent.

(E) The acceleration will increase throughout the rock’s flight.

4. A baseball is thrown straight upward. What is the ball’s acceleration at its highest point?

(A) 0

(B) g, downward

(D) g, upward

(E) g, upward

5. How long would it take a car, starting from rest and accelerating uniformly in a straight line at 5 m/s², to cover a distance of 200 m ?

(A) 9.0 s

(B) 10.5 s

(D) 15.5 s

(E) 20.0 s

6. A rock is dropped off a cliff and strikes the ground with an impact speed of 30 m/s. How high was the cliff?

(A) 15 m

(B) 20 m

(D) 45 m

(E) 60 m

7. A stone is thrown horizontally with an initial speed of 10 m/s from a bridge. If air resistance could be ignored, how long would it take the stone to strike the water 80 m below the bridge?

(A) 1 s

(B) 2 s

(D) 6 s

(E) 8 s

8. A soccer ball, at rest on the ground, is kicked with an initial velocity of 10 m/s at a launch angle of 30°. Calculate its total flight time, assuming that air resistance is negligible.

(A) 0.5 s

(B) 1 s

(D) 2 s

(E) 4 s

9. A stone is thrown horizontally with an initial speed of 30 m/s from a bridge. Find the stone’s total speed when it enters the water 4 seconds later. (Ignore air resistance.)

(A) 30 m/s

(B) 40 m/s

(D) 60 m/s

(E) 70 m/s

10. Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of 45° to the horizontal?

(A) The acceleration vector points opposite to the velocity vector on the way up and in the same direction as the velocity vector on the way down.

(B) The speed at the top of the trajectory is zero.

(D) The horizontal speed decreases on the way up and increases on the way down.

(E) The vertical speed decreases on the way up and increases on the way down.

11. The position of an object moving in a straight line is given by the equation x = (3t³ + 7t + 5) m, where t is in seconds. What is the object’s acceleration at time t = 5 seconds?

(A) 415 m/s²

(B) 232 m/s²

(D) 90 m/s²

(E) 35 m/s²

12. The velocity of an object moving in a straight line is graphed above. If x = 3.0 m at t = 0 s, what is the position of the particle at t = 3.0 s?

(A) 6 m

(B) 10 m

(D) 11 m

(E) –5 m

Section II: Free Response

1. This question concerns the motion of a car on a straight track; the car’s velocity as a function of time is plotted below.

(a) Describe what happened to the car at time t = 1 s.

(b) How does the car’s average velocity between time t = 0 and t = 1 s compare to its average velocity between times t = 1 s and t = 5 s?

(d) Plot the car’s acceleration during this interval as a function of time.

(e) Plot the object’s position during this interval as a function of time. Assume that the car begins at s = 0.

2. Consider a projectile moving in a parabolic trajectory under constant gravitational acceleration. Its initial velocity has magnitude v₀, and its launch angle (with the horizontal) is θ₀. Solve the following in terms of given quantities and the acceleration of gravity, g.

(a) Calculate the maximum height, H, of the projectile.

(b) Calculate the (horizontal) range, R, of the projectile.

(d) If 0 < h < H, compute the time that elapses between passing through the horizontal line of height h in both directions (ascending and descending); that is, compute the time required for the projectile to pass through the two points shown in this figure:

3. A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40° toward a castle wall 220 m away. The height of the wall is 30 m. Assume that effects due to the air are negligible. (For this problem, use g = 9.8 m/s².)

(a) How long will it take the cannonball to reach the vertical plane of the wall?

(b) Will the cannonball strike the wall? If the cannonball strikes the wall, how far below the top of the wall does it strike? If the cannonball does not strike the wall, how much does it clear the wall by?

4. A particle moves along a straight axis in such a way that its acceleration at time t is given by the equation a(t) = 6t (m/s²). If the particle’s initial velocity is 2 m/s and its initial position is x = 4 m, determine

(a) The time at which the particle’s velocity is 14 m/s

(b) The particle’s position at time t = 3 s

Summary

Definitions

Position refers to where an object is relative to a coordinate axes system.

Distance refers to the total measure of the ground traveled by an object.

Displacement is how far an object is from where it started. It can be represented as the final position minus the initial position: Δs = s_f – s_i.

average speed =

Constant Acceleration Equations

The Big 5 acceleration equations are:

Δx = x − x₀ = t

v = v₀ + at

x = x₀ + v₀t + at²

x = x₀ + vt − at²

v² = + 2a(x − x₀)

Graphs

The slope of a position-vs.-time graph is the average velocity.

The area between the t-axis and the velocity function is the displacement.

The slope of a velocity-vs.-time graph is the average acceleration.

Motion Functions

The derivate of x(t) is the velocity.

The integral of v(t) is the displacement.

The derivative of v(t) is the acceleration.

Free Fall and Projectiles

The acceleration due to gravity, g, is a constant 9.8 m/s² downward, for all objects close to the surface of Earth. Note that the test directions state that to simplify calculations you may use 10 m/s² in all problems.

Remember to analyze the vertical (constant acceleration) and horizontal (constant velocity) motions separately. You can use the constant acceleration equations with a replaced by g and x replaced by y to indicate the acceleration is in the vertical direction. The motion in the x direction has a constant velocity, so the only equation you need for that is x = v_xt.

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 6 Newton’s Laws

INTRODUCTION

In the previous chapter we studied the vocabulary and equations that describe motion. Now we will learn why things move the way they do; this is the subject of dynamics.

An interaction between two bodies—a push or a pull—is called a force. If you lift a book, you exert an upward force (created by your muscles) on it. If you pull on a rope that’s attached to a crate, you create a tension in the rope that pulls the crate. When a skydiver is falling through the air, Earth is exerting a downward pull called gravitational force, and the air exerts an upward force called air resistance. When you stand on the floor, the floor provides an upward, supporting force called the normal force. If you slide a book across a table, the table exerts a frictional force against the book, so the book slows down and then stops. Static cling provides a directly observable example of the electrostatic force. Protons and neutrons are held together in the nuclei of atoms by the strong nuclear force and radioactive nuclei decay through the action of the weak nuclear force.

The Englishman Sir Isaac Newton published a book in 1687 called Philosophiae Naturalis Principia Mathematica (The Mathematical Principles of Natural Philosophy)—referred to nowadays as simply The Principia—which began the modern study of physics as a scientific discipline. Three of the laws that Newton stated in The Principia form the basis for dynamics and are known simply as Newton’s Laws of Motion.

THE FIRST LAW

Newton’s First Law says that an object will continue in its state of motion unless compelled to change by a net force impressed upon it. That is, unless an unbalanced force acts on an object, the object’s velocity will not change: If the object is at rest, then it will stay at rest; and if it is moving, then it will continue to move at a constant speed in a straight line.

Basically, no net force means no change in velocity. This property of objects, their natural resistance to changes in their state of motion, is called inertia. In fact, the First Law is often referred to as the Law of Inertia.

THE SECOND LAW

Newton’s Second Law predicts what will happen when a net force does act on an object: The object’s velocity will change; the object will accelerate. More precisely, it says that its acceleration, a, will be directly proportional to the strength of the total—or net—force (F_net) and inversely proportional to the object’s mass, m:

The mass of an object is the quantitative measure of its inertia; intuitively, it measures how much matter is contained in an object. Two boxes, one empty and one full, have different masses; the box that is full has the greater mass. Mass is measured in kilograms, abbreviated as kg. (Note: An object whose mass is 1 kg weighs about 2.2 pounds.) It takes twice as much force to produce the same change in velocity of a 2 kg object than of a 1 kg object. Mass is a measure of an object’s inertia, its resistance to acceleration.

Forces are represented by vectors; they have magnitude and direction. If several different forces act on an object simultaneously, then the net force, F_net, is the vector sum of all these forces. (The phrase resultant force is also used to mean net force.)

Since F_net = ma, and m is a positive scalar, the direction of a always matches the direction of F_net. Finally, since F = ma, the units for F equal the units of m times the units of a:

[F] = [m][a]
= kg·m/s²

A force of 1 kg·m/s² is renamed 1 newton (abbreviated as N). A medium-size apple weighs about 1 N.

THE THIRD LAW

This is the law that’s commonly remembered as to every action, there is an equal, but opposite, reaction. More precisely, if Object 1 exerts a force on Object 2, then Object 2 exerts a force back on Object 1, equal in strength but opposite in direction. These two forces, F_1-on-2 and F_2-on-1, are called an action/reaction pair.

While the forces in an action/reaction pair are equal in magnitude, the ‘reaction’ (i.e., the resulting acceleration of the object) depends on the mass of each object. The greater the mass, the smaller the ‘reaction’ (acceleration); the smaller the mass, the greater the ‘reaction’ (acceleration).

Example 1 What net force is required to maintain a 5,000 kg object moving at a constant velocity of magnitude 7,500 m/s?

Solution. The First Law says that any object will continue in its state of motion unless a force acts on it. Therefore, no net force is required to maintain a 5,000 kg object moving at a constant velocity of magnitude 7,500 m/s. Here’s another way to look at it: Constant velocity means a = 0, so the equation F_net = ma immediately gives F_net = 0.

Example 2 How much force is required to cause an object of mass 2 kg to have an acceleration of 4 m/s²?

Solution. According to the Second Law, F_net = ma = (2 kg)(4 m/s²) = 8 N.

Example 3 An object feels two forces: one of strength 8 N pulling to the left and one of strength 20 N pulling to the right. If the object’s mass is 4 kg, what is its acceleration?

Solution. Forces are represented by vectors and can be added and subtracted. Therefore, an 8 N force to the left added to a 20 N force to the right yields a net force of 20 – 8 = 12 N to the right. Newton’s Second Law gives a = F_net/m = (12 N to the right)/(4 kg) = 3 m/s² to the right.

WEIGHT

Mass and weight are not the same thing—there is a clear distinction between them in physics—but they are often used interchangeably in everyday life. The weight of an object is the gravitational force exerted on it by a gravitational field. Mass, by contrast, is an intrinsic property of an object that measures its inertia. An object’s mass does not change with location. An astronaut, for example, has the same mass on Earth as on the Moon. The astronaut’s weight on the Moon, however, is 1/6 that on Earth because the gravitational force on the Moon is 1/6 that of Earth’s.

Since weight is a force, we can use F = ma to compute it, where the acceleration is the gravitational force imposed on an object. Therefore, setting a = g, the equation F = ma becomes

F_w = mg

This is the equation for the weight of an object of mass m. (F_g and F_w are both commonly used to represent the force of gravity. Weight is often symbolized merely by w, rather than F_w.) Notice that mass and weight are proportional but not identical. Furthermore, mass is measured in kilograms, while weight is measured in newtons.

Example 4 What is the mass of an object that weighs 500 N?

Solution. We are given a weight, so we use the equation F_w = mg, where g is the acceleration due to gravity on Earth (10 m/s²). Rearranging this equation gives

m = F_w/g = (500 N)/(10 m/s²) = 50 kg

Example 5 A person weighs 150 pounds. Given that a pound is a unit of weight equal to 4.45 N, what is this person’s mass?

Solution. This person’s weight in newtons is (150 lb)(4.45 N/lb) = 667.5 N. Again using the equation F_w = mg, where g = 10 m/s², gives

m = F_w/g = (667.5 N)/(10 m/s²) = 67 kg

Example 6 A book whose mass is 2 kg rests on a table. Find the magnitude of the force exerted by the table on the book.

Solution. The book experiences two forces: The downward pull of Earth’s gravity and the upward, supporting force exerted by the table. Since the book is at rest on the table, its acceleration is zero, so the net force on the book must be zero. Therefore the magnitude of the upward force must equal the magnitude of the book’s weight, which is F_w = mg = (2 kg)(10 m/s²) = 20 N.

Example 7 A can of paint with a mass of 6 kg hangs from a rope. If the can is to be pulled up to a rooftop with an acceleration of 1 m/s², what must the force of the tension be in the rope?

Solution. First draw a picture. Represent the object of interest (the can of paint) as a heavy dot, and draw the forces that act on the object as arrows connected to the dot. This is called a free-body (or force) diagram.

We have the tension force in the rope, F_T (sometimes symbolized merely by T), which is upward, and the downward force, F_w. Calling up the positive direction, the net force is F_T – F_w. The Second Law, F_net = ma, becomes F_T – F_w = ma, so

F_T = F_w + ma = mg + ma = m(g + a) = 6(10 + 1) = 66 N

Example 8 A can of paint with a mass of 6 kg hangs from a rope. If the can is to be pulled up to a rooftop with a constant velocity of 1 m/s, what must the tension in the rope be?

Solution. The phrase “constant velocity” automatically means a = 0 and, therefore, F_net = 0. In the diagram above, F_T would need to have the same magnitude as F_w in order to keep the can moving at a constant velocity. Thus, in this case, F_T = F_w = mg = (6)(10) = 60 N.

Example 9 How much tension must a rope have to lift a 50 N object with an acceleration of 10 m/s²?

Solution. First draw a free-body diagram:

We have the tension force, F_T, which is upward, and the weight, F_w, which is downward. Calling up the positive direction, the net force is F_T – F_w. The Second Law, F_net = ma, becomes F_T – F_w = ma, so F_T = F_w + ma. Remembering that m = F_w/g, we find that

THE NORMAL FORCE

When an object is in contact with a surface, the surface exerts a contact force on the object. The component of the contact force that’s perpendicular to the surface is called the normal force on the object. The normal force comes from electrostatic interactions among atoms. The normal force is what prevents objects from falling through tabletops or you from falling through the floor. The normal force is denoted by F_N, or simply by N. (If you use the latter notation, be careful not to confuse it with N, the abbreviation for the newton.)

Example 10 A book whose mass is 2 kg rests on a table. Find the magnitude of the normal force exerted by the table on the book.

Solution. The book experiences two forces: The downward gravitational pull and the upward, normal force exerted by the table. Since the book is at rest on the table, its acceleration is zero, so the net force on the book must be zero. Therefore, the magnitude of the normal force must equal the magnitude of the book’s weight, which is F_w = mg = (2)(10) = 20 N. This means the normal force must be 20 N as well: F_N = 20 N. (Note that this is a repeat of Example 6, except now we have a name for the “upward, supporting force exerted by the table”; it’s called the normal force.)

FRICTION

When an object is in contact with a surface, the surface exerts a contact force on the object. The component of the contact force that’s parallel to the surface is called the friction force on the object. Friction, like the normal force, arises from electrostatic interactions between the object and the surface.

We’ll look at two main categories of friction: (1) static friction and (2) kinetic (sliding) friction. If you attempt to push a heavy crate across a floor, at first you meet with resistance. Eventually, though, you can push hard enough to get the crate moving. The force that acted on the crate that cancelled the force of your initial push was static friction, and the force that acts on the crate as it slides across the floor is kinetic friction. Static friction occurs when there is no relative motion between the object and the surface (no sliding); kinetic friction occurs when there is relative motion (when there’s sliding).

The strength of the friction force depends, in general, on two things: The nature of the surfaces and the strength of the normal force. The nature of the surfaces is represented by the coefficient of friction, denoted by μ, which stands for the Greek letter mu. The greater the coefficient of friction, the stronger the friction force will be. For example, the coefficient of friction between rubber-soled shoes and a wooden floor is 0.7, but between rubber-soled shoes and ice is only 0.1. Also, since kinetic friction is generally weaker than static friction (it’s easier to keep an object sliding once it’s sliding than it is to start the object sliding in the first place), there are two coefficients of friction: one for static friction (μ_s) and one for kinetic friction (μ_k). For a given pair of surfaces, it’s virtually always true that μ_k < μ_s. The strengths of these two types of friction forces are given by the following equations:

F_{static friction, max} = μ_sF_N or F_{static friction} ≤ μ_s F_N

F_{kinetic friction} = μ_kF_N

On the equation sheet for the free-response section, this information is represented as follows:

Note that the equation for the strength of the static friction force is for the maximum or lesser value. This is because static friction can vary, precisely counteracting weaker forces that attempt to move an object. For example, suppose an object feels a normal force of F_N = 100 N and the coefficient of static friction between it and the surface it’s on is 0.5. Then, the maximum force that static friction can exert is (0.5)(100 N) = 50 N. However, if you push on the object with a force of, say, 20 N, then the static friction force will be 20 N (in the opposite direction), not 50 N; the object won’t move. The net force on a stationary object must be zero. Static friction can take on all values, up to a certain maximum; you must overcome the maximum static friction force to get the object to slide. The direction of F_{kinetic friction} = F_{f (kinetic)} is opposite to that of motion (sliding), and the direction of F_{static friction} = F_{f (static)} is opposite to that of the intended motion.

Example 11 A crate of mass 20 kg is sliding across a wooden floor. The coefficient of kinetic friction between the crate and the floor is 0.3.

(a) Determine the strength of the friction force acting on the crate.

(b) If the crate is being pulled by a force of 90 N (parallel to the floor), find the acceleration of the crate.

Solution. First draw a free-body diagram.

In part (a) F = 0, and in part (b) F = 90 N for our free-body diagram. Reminder: Separate the horizontal and vertical forces and use ΣF_x = ma_x and ΣF_y = ma_y.

(a) The normal force on the object balances the object’s weight, so F_N = mg = (20 kg)(10 m/s²) = 200 N. Therefore, F_(kinetic) = μ_kF_N = (0.3)(200 N) = 60 N.

(b) The net horizontal force that acts on the crate is F – F_f = 90 N – 60 N = 30 N, so the acceleration of the crate is a = F_net/m = (30 N)/(20 kg) = 1.5 m/s².

Example 12 A crate of mass 100 kg rests on the floor. The coefficient of static friction is 0.4. If a force of 250 N (parallel to the floor) is applied to the crate, what’s the magnitude of the force of static friction on the crate?

Solution. The normal force on the object balances its weight, so F_N = mg = (100 kg)(10 m/s²) = 1,000 N. Therefore, F_{static friction, max} = F_{f (static), max} = μ_sF_N = (0.4)(1,000 N) = 400 N. This is the maximum force that static friction can exert, but in this case it’s not the actual value of the static friction force. Since the applied force on the crate is only 250 N, which is less than the F_{f (static), max}, the force of static friction will be less also: F_{f (static)} = 250 N, and the crate will not slide.

PULLEYS

Pulleys are devices that change the direction of the tension force in the cords that slide over them. Here we’ll consider each pulley to be frictionless and massless, which means that their masses are so much smaller than the objects of interest in the problem that they can be ignored.

Example 13 In the diagram above, assume that the tabletop is frictionless. Determine the acceleration of the blocks once they’re released from rest.

Solution. There are two blocks, so draw two free-body diagrams: The positive directions for each block must coincide. If the block on the table travels to the right then the hanging block travels down. This is why down is positive for the hanging block.

To get the acceleration of each one, we use Newton’s Second Law, F_net = ma. Notice that while F_w = F_N for the block on the table, F_w does not equal F_T for the hanging block, because that block is in motion in the downward direction.

Note that there are two unknowns, F_T and a, but we can eliminate F_T by adding the two equations, and then we can solve for a.

Example 14 Using the same diagram as in the previous example, assume that m = 2 kg, M = 10 kg, and the coefficient of kinetic friction between the small block and the tabletop is 0.5. Compute the acceleration of the blocks.

Solution. Once again, draw a free-body diagram for each object. Note that the only difference between these diagrams and the ones in the previous example is the inclusion of the force of (kinetic) friction, F_f, that acts on the block on the table.

As before, we have two equations that contain two unknowns (a and F_T):

F_T − F_f = ma (1)
Mg − F_T = Ma (2)

Add the equations (thereby eliminating F_T) and solve for a. Note that, by definition, F_f = μF_N, and from the free-body diagram for m, we see that F_N = mg, so F_f = μmg:

Substituting in the numerical values given for m, M, and μ, we find that a = g (or 7.5 m/s²).

Example 15 In the previous example, calculate the tension in the cord.

Solution. Since the value of a has been determined, we can use either of the two original equations to calculate F_T . Using Equation (2), Mg – F_T = Ma (because it’s simpler), we find

As you can see, we would have found the same answer if Equation (1) had been used:

INCLINED PLANES

An inclined plane is basically a ramp. If an object of mass m is on the ramp, then the force of gravity on the object, F_w = mg, has two components: One that’s parallel to the ramp (mg sin θ) and one that’s normal to the ramp (mg cos θ), where θ is the incline angle. The force driving the block down the inclined plane is the component of the block’s weight that’s parallel to the ramp: mg sin θ.

When analyzing objects moving up or down inclined planes it is almost always easiest to rotate the coordinate axes such that the x-axis is parallel to the incline and the y-axis is perpendicular to the incline, as shown in the diagram. The object would accelerate in both the x- and y-directions as it moved down along the incline if you did not rotate the axis. However, with the rotated axes, the acceleration in the y-direction is zero. Now we only have to worry about the acceleration in the x-direction.

Example 16 A block slides down a frictionless inclined plane that makes a 30° angle with the horizontal. Find the acceleration of this block.

Solution. Let m denote the mass of the block, so the force that pulls the block down the incline is mg sin θ, and the block’s acceleration down the plane is

Example 17 A block slides down an inclined plane that makes a 30° angle with the horizontal. If the coefficient of kinetic friction is 0.3, find the acceleration of the block.

Solution. First, draw a free-body diagram. Notice that, in the diagram shown below, the weight of the block, F_w = mg, has been written in terms of its scalar components: F_w sin θ parallel to the ramp and F_w cos θ normal to the ramp:

The force of friction, F_f, that acts up the ramp (opposite to the direction in which the block slides) has magnitude F_f = μF_N. But the diagram shows that F_N = F_w cos θ, so F_f = μ(mg cos θ). Since the net force is the force down the ramp minus the frictional force, we have

F_w sin θ – F_f = mg sin θ − μmg cos θ = mg(sin θ − cos θ)

Setting F_net equal to ma, we solve for a:

a =
= g(sin θ − μ cos θ)
= (10 m/s²)(sin 30° − 0.3 cos 30°)
= 2.4 m/s²

UNIFORM CIRCULAR MOTION

In Chapter 5, we considered two types of motion: straight-line motion and parabolic motion. We will now look at motion that follows a circular path, such as a rock on the end of a string, a horse on a merry-go-round, and (to a good approximation) the Moon around Earth and Earth around the Sun.

If an object’s speed around its path is a constant, then the object has uniform circular motion. You should remember that although the speed may be constant, the velocity is not, because the direction of the velocity is always changing. Since the velocity is changing, there must be acceleration. This acceleration does not change the speed of the object; it only changes the direction of the velocity to keep the object on its circular path. Also, in order to produce an acceleration, there must be a force; otherwise, the object would move off in a straight line (Newton’s First Law).

The figure on the left shows an object moving counterclockwise along a circular trajectory, along with its velocity vectors at two nearby points. The vector v₁ is the object’s velocity at time t = t₁, and v₂ is the object’s velocity vector a short time later (at time t = t₂). The velocity vector is always tangential to the object’s path (whatever the shape of the trajectory). Notice that since we are assuming constant speed, the lengths of v₁ and v₂ (their magnitudes) are the same.

Notice that ∆v = v₂ – v₁ points toward the center of the circle (see the figure on the right). This means that the acceleration does also, since a = ∆v/∆t. Because the acceleration vector points toward the center of the circle, it’s called centripetal acceleration, or a_c. The centripetal acceleration is what turns the velocity vector to keep the object traveling in a circle. The magnitude of the centripetal acceleration depends on the object’s speed, v, and the radius, r, of the circular path according to the equation

When this is given on the free-response equation sheet, it is given in terms of angular velocity as well, which will be discussed later.

Example 18 An object of mass 5 kg moves at a constant speed of 6 m/s in a circular path of radius 2 m. Find the object’s acceleration and the net force responsible for its motion.

Solution. By definition, an object moving at constant speed in a circular path is undergoing uniform circular motion. Therefore, it experiences a centripetal acceleration of magnitude v²/r, always directed toward the center of the circle:

The force that produces the centripetal acceleration is given by Newton’s Second Law, coupled with the equation for centripetal acceleration:

This equation gives the magnitude of the force. As for the direction, recall that because F_net = ma, the directions of F_net and a are always the same. Since centripetal acceleration points toward the center of the circular path, so does the force that produces it. Therefore, it’s called centripetal force. Centripetal force is provided by everyday forces such as tension, friction, gravity, or normal forces. The centripetal force acting on this object has a magnitude of F_c = ma_c = (5 kg)(18 m/s²) = 90 N.

The following diagrams show examples of a ball on a string traveling in a horizontal circle and a vertical circle.

Example 19 A 10 kg mass is attached to a string that has a breaking strength of 200 N. If the mass is whirled in a horizontal circle of radius 80 cm, what maximum speed can it have? Assume the string is horizontal.

Solution. The first thing to do in problems like this is to identify what force(s) provide the centripetal force. In this example, the tension in the string (F_T) provides the centripetal force (F_c):

F_T provides F_c ⇒ F_T = m ⇒ v = ⇒ v_max =
=
= 4 m/s

Example 20 An athlete who weighs 800 N is running around a curve at a speed of 5.0 m/s in an arc whose radius of curvature, r, is 5.0 m. Assuming his weight is equal to the force of static friction, find the centripetal force acting on him. What could happen to him if r were smaller?

Solution. Using the equation for the strength of the centripetal force, we find that

In this case, static friction provides the centripetal force. Since his weight is equal to the force of static friction, this means the coefficient of static friction between his shoes and the ground is 1, so the maximum force that static friction can exert is μ_sF_N ≈ F_N = F_w = 800 N. Fortunately, 800 N is greater than 400 N. But notice that if the radius of curvature of the arc were much smaller, then F_c would become greater than what F_w, and he would slip.

Example 21 A roller-coaster car enters the circular-loop portion of the ride. At the very top of the circle (where the people in the car are upside down), the speed of the car is 25 m/s, and the acceleration points straight down. If the diameter of the loop is 50 m and the total mass of the car (plus passengers) is 1,200 kg, find the magnitude of the normal force exerted by the track on the car at this point. Also find the normal force exerted by the track on the car when it is at the bottom of the loop. Assume it is still traveling 25 m/s at that location as well.

Solution. When analyzing circular motion, consider all forces pointing toward the center to be positive and all forces pointing away to be negative. There are two forces acting on the car at its topmost point: the normal force exerted by the track and the gravitational force, both of which point downward.

The combination of these two forces, F_N + F_w, provides the centripetal force:

Now we will determine the normal force exerted by the track on the car at the bottom of the loop.

Notice that the normal force is much greater when the car is at the bottom of the track than when it is at the top. When it is at the top of the track, gravity is helping the car travel in a circle.

Example 22 In the previous example, if the net force on the car at its topmost point is straight down, why doesn’t the car fall straight down?

Solution. Remember that force tells an object how to accelerate. If the car had zero velocity at this point, then it would certainly fall straight down, but the car has a non-zero velocity (to the left) at this point. The fact that the acceleration is downward means that, at the next moment, v will point down to the left at a slight angle, ensuring that the car remains on a circular path, in contact with the track. The minimum centripital acceleration of the car at the top of the track would be equal to the acceleration of gravity, g = 9.8 m/s². If a_c were less than g then the car would fall off its circular path.

Chapter 6 Drill

The answers and explanations can be found in Chapter 17.

Click here to download the PDF.

Section I: Multiple Choice

1. Consider a box being dragged across a table at a constant velocity by a string. Which of the following would be considered an action-reaction pair?

I. The force of gravity pulling the box down and the normal force pushing the box up

II. The force of friction on the box and the tension force pulling the box

III. The force of the box pushing down on the table and the normal force pushing the box up

(A) I only

(B) I and II only

(D) I and III only

(E) III only

2. A person who weighs 800 N steps onto a scale that is on the floor of an elevator car. If the elevator accelerates upward at a rate of 5 m/s², what will the scale read?

(A) 400 N

(B) 800 N

(D) 1,200 N

(E) 1,600 N

3. A frictionless inclined plane of length 20 m has a maximum vertical height of 5 m. If an object of mass 2 kg is placed on the plane, which is the net force it feels?

(A) 5 N

(B) 10 N

(D) 20 N

(E) 30 N

4. A 20 N block is being pushed across a horizontal table by an 18 N force. If the coefficient of kinetic friction between the block and the table is 0.4, find the acceleration of the block.

(A) 0.5 m/s²

(B) 1 m/s²

(D) 7.5 m/s²

(E) 9 m/s²

5. The coefficient of static friction between a box and a ramp is 0.5. The ramp’s incline angle is 30°. If the box is placed at rest on the ramp, the box will do which of the following?

(A) Accelerate down the ramp

(B) Accelerate briefly down the ramp but then slow down and stop

(D) Not move

(E) Cannot be determined from the information given

Assuming a frictionless, massless pulley, determine the acceleration of the blocks once they are released from rest.

(A)

(B)

(C)

(D)

(E)

7. If all of the forces acting on an object balance so that the net force is zero and the object’s mass remains constant, then

(A) the object must be at rest

(B) the object’s speed will decrease

(D) the object’s direction of motion can change, but not its speed

(E) none of the above

8. An object of mass m is allowed to slide down a frictionless ramp of angle θ, and its speed at the bottom is recorded as v. If this same process was followed on a planet with twice the gravitational acceleration as Earth, what would be its final speed?

(A) 2v

(B) v

(D)

(E)

9. An engineer is designing a loop for a roller coaster. If the loop has a radius of 25 m, how fast do the cars need to be moving at the top to ensure people would be safe even if the safety bars malfunctioned?

(A) 12.7 m/s

(B) 15.8 m/s

(D) 29.3 m/s

(E) 33.3 m/s

10. An object moves at constant speed in a circular path. Which of the following statements is/are true?

I. The velocity is constant.

II. The acceleration is constant.

III. The net force on the object is zero.

(A) II only

(B) I and III only

(D) I and II only

(E) None of the above

Questions 11–12:

A 60 cm rope is tied to the handle of a bucket which is then whirled in a vertical circle. The mass of the bucket is 3 kg.

11. At the lowest point in its path, the tension in the rope is 50 N. What is the speed of the bucket?

(A) 1 m/s

(B) 2 m/s

(D) 4 m/s

(E) 5 m/s

12. What is the critical speed below which the rope would become slack when the bucket reaches the highest point in the circle?

(A) 0.6 m/s

(B) 1.8 m/s

(D) 3.2 m/s

(E) 4.8 m/s

13. An object moves at a constant speed in a circular path of radius r at a rate of 1 revolution per second. What is its acceleration?

(A) 0

(B) 2π²r

(D) 4π²r

(E) 4π²r²

Section II: Free Response

1. This question concerns the motion of a crate being pulled across a horizontal floor by a rope. In the diagram below, the mass of the crate is m, the coefficient of kinetic friction between the crate and the floor is μ, and the tension in the rope is F_T .

(a) Draw and label all of the forces acting on the crate.

(b) Compute the normal force acting on the crate in terms of m, F_T, θ, and g.

(d) Assume that the magnitude of the tension in the rope is fixed but that the angle may be varied. For what value of θ would the resulting horizontal acceleration of the crate be maximized?

2. In the diagram below, a massless string connects two blocks—of masses m₁ and m₂, respectively—on a flat, frictionless tabletop. A force F pulls on Block #2, as shown:

Solve for the following in terms of given quantities.

(a) Draw and label all of the forces acting on Block #1.

(b) Draw and label all of the forces acting on Block #2.

(d) What is the tension in the string connecting the two blocks?

(e) If the string connecting the blocks were not massless, but instead had a mass of m, find

(i) the acceleration of Block #1, and

(ii) the difference between the strength of the force that the connecting string exerts on Block #2 and the strength of the force that the connecting string exerts on Block #1.

3. In the figure shown, assume that the pulley is frictionless and massless.

Solve for the following in terms of given quantities and the acceleration of gravity, g.

(a) If the surface of the inclined plane is frictionless, determine what value(s) θ of will cause the box of mass m₁ to

(i) accelerate up the ramp

(ii) slide up the ramp at constant speed

(b) If the coefficient of kinetic friction between the surface of the inclined plane and the box of mass m₁ is μ_k, derive (but do not solve) an equation satisfied by the value of θ which will cause the box of mass m₁ to slide up the ramp at constant speed.

4. A sky diver is falling with speed v₀ through the air. At that moment (time t = 0), she opens her parachute and experiences the force of air resistance whose strength is given by the equation F = kv, where k is a proportionality constant and v is her descent speed. The total mass of the sky diver and equipment is m. Assume that g is constant throughout her descent.

(a) Draw and label all the forces acting on the sky diver after her parachute opens.

(b) Determine the sky diver’s acceleration in terms of m, v, k, and g.

(d) Sketch a graph of v as a function of time, starting at t = 0 and going until she lands, being sure to label important values on the vertical axis.

(e) Derive an expression for her descent speed, v, as a function of time t since opening her parachute in terms of m, k, and g.

5. An amusement park ride consists of a large cylinder that rotates around its central axis as the passengers stand against the inner wall of the cylinder. Once the passengers are moving at a certain speed v, the floor on which they were standing is lowered. Each passenger feels pinned against the wall of the cylinder as it rotates. Let r be the inner radius of the cylinder.

Solve for the following in terms of given quantities and the acceleration of gravity, g.

(a) Draw and label all the forces acting on a passenger of mass m as the cylinder rotates with the floor lowered.

(b) Describe what conditions must hold to keep the passengers from sliding down the wall of the cylinder.

6. A curved section of a highway has a radius of curvature of r. The coefficient of friction between standard automobile tires and the surface of the highway is μ_s.

(a) Draw and label all the forces acting on a car of mass m traveling along this curved part of the highway.

(b) Compute the maximum speed with which a car of mass m could make it around the turn without skidding in terms of μ_s, r, g, and m.

City engineers are planning on banking this curved section of highway at an angle of θ to the horizontal.

(c) Draw and label all of the forces acting on a car of mass m traveling along this banked turn. Do not include friction.

(d) The engineers want to be sure that a car of mass m traveling at a constant speed v (the posted speed limit) could make it safely around the banked turn even if the road were covered with ice (that is, essentially frictionless). Compute this banking angle θ in terms of r, v, g, and m.

Summary

Newton’s Laws

Newton’s First Law (Law of Inertia) states that objects will continue in their state of motion unless acted upon by an unbalanced force.

Newton’s Second Law: F_net = ma

Newton’s Third Law states that whenever two objects interact, the force that the first object exerts on the second object is equal to, but in the opposite direction of, the force the second object exerts on the first object.

Weight

The weight of an object is given by F_w = mg.

The Normal Force

The normal force is the component of the contact force exerted on an object in contact with a surface and is perpendicular to the surface.

The normal force can be represented by F_N or N.

Friction

Friction is the component of the contact force exerted on an object in contact with a surface and is parallel to the surface.

Static friction occurs when there is no relative motion between the object and the surface. Its strength is given by the following equation: F_{static friction, max} = µ_sF_N

Kinetic friction occurs when there is relative motion between the two surfaces. Its strength is given by the following equation: F_{kinetic friction} = µ_kF_N

Inclined Planes

There are two components to the force of gravity on an object on an inclined plane: the force parallel to the ramp (mg sin θ) and the force normal to the ramp (mg cos θ).

To simplify analysis of an object moving up or down a ramp, rotate the coordinate axes so that the x-axis is parallel to the incline and the y-axis is perpendicular to the incline.

Uniform Circular Motion

The velocity vector is tangent to the circle.

The centripetal acceleration points toward the center of the circle, and therefore the centripetal force must also point to the center of the circle.

Any force, or component of a force, that points toward the center of the circle is positive and any force, or component of a force, that points away from the center is negative when using the following equation:

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 7 Work, Energy, and Power

INTRODUCTION

It wasn’t until more than one hundred years after Newton that energy became incorporated into physics, but today its inclusion permeates every branch of the subject.

There are different forms of energy because there are different kinds of forces. There’s gravitational energy (a meteor crashing into Earth), elastic energy (a stretched rubber band), thermal energy (an oven), radiant energy (sunlight), electrical energy (a lamp plugged into a wall socket), nuclear energy (nuclear power plants), and mass energy (the heart of Einstein’s equation E = mc²). Energy can come into a system or leave it through various interactions that produce changes. While force is the agent of change, energy is often defined as the ability to do work, and work is one way of transferring energy from one system to another. One of the most important laws in physics (the Law of Conservation of Energy, also known as the First Law of Thermodynamics) says that if you account for all its various forms, the total amount of energy in a given process will stay constant; that is, it will be conserved. For example, electrical energy can be converted into light and heat (this is how a light bulb works), but the amount of electrical energy going into the light bulb equals the total amount of light and heat given off.

WORK

When you lift a book from the floor you exert a force on it over a distance, and when you push a crate across a floor, you also exert a force on it over a distance. When you hold a book in your hand, you exert a force on the book (normal force), but since the book is at rest, its displacement = 0, so you do no work. If F is the force, and dr is an infinitesimal amount of displacement, then the work done is:

W = ∫F · dr

If the force is constant, it can be removed from of the integral, and since F · r = (F cosθ)r for any angle θ, then W = (F cos θ)r. The unit of work is the newton-meter (N·m), which is also called a joule (J).

Notice that work depends on two vectors (F and r), but work itself is not a vector. Work is a scalar quantity. This is a result of the dot product, whereby two vectors are multiplied to produce a scalar. Another result of the dot product is that only the component of the force that is parallel (or antiparallel) to the displacement does any work. Any force or component of a force is perpendicular to the direction that an object actually moves cannot do work because the displacement in that direction is zero.

Also, since an integral is the area under a curve, if a graph of force as a function of position or displacement is given, the work done by the force is the area under the curve whose boundaries correspond to the displacement, Δx.

Example 1 You slowly lift a book of mass 2 kg at constant velocity through a distance of 3 m. How much work did you do on the book?

Solution. In this case, the force you exert must balance the weight of the book (other-wise the velocity of the book wouldn’t be constant), so F = mg = (2 kg)(10 m/s²) = 20 N. Since this force is straight upward and the displacement, d, of the book is also straight upward, F and d are parallel, so the work done by your lifting force is W = Fd = (20 N)(3 m) = 60 N·m = 60 J.

Example 2 A 15 kg crate is moved along a horizontal floor by a warehouse worker who’s pulling on it with a rope that makes a 30° angle with the horizontal. The tension in the rope is 200 N and the crate slides a distance of 10 m. How much work is done on the crate by the rope?

Solution. The figure below shows that F_T and d are not parallel. It’s only the component of the force acting along the direction of motion, F_T cos θ, that does work.

Therefore,

W = (F_T cos θ)d = (200 N · cos 30°)(10 m) = 1,730 J

Example 3 In the previous example, assume that the coefficient of kinetic friction between the crate and the floor is 0.4.

(a) How much work is done by the normal force?

(b) How much work is done by the friction force?

Solution.

(a) Since the angle between F_N and d is 90° (by definition of normal) and cos 90° = 0, the normal force does zero work.

(b) The friction force, F_f, is also not parallel to the motion; it’s antiparallel. That is, the angle between F_f and d is 180°. Since cos 180° = –1, and since the strength of the normal force is F_N = F_T,y – F_W = F_T sin θ – mg = 200 N*sin 30° – (15 kg)(10 m/s²) = 100 N – 150 N = 50 N, the work done by the friction force is:

W = –F_f d = –μ
μ_kF_N d = –(0.4)(50 N)(10 m) = –200 J

The two previous examples show that work, which is a scalar quantity, may be positive, negative, or zero. If the angle between F and d (θ) is less than 90°, then the work is positive (because cos θ is positive in this case); if θ = 90°, the work is zero (because cos 90° = 0); and if θ > 90°, then the work is negative (because cos θ is negative). Intuitively, if a force helps the motion, the work done by the force is positive, but if the force opposes the motion, then the work done by the force is negative.

Example 4 A box slides down an inclined plane (incline angle = 37°). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, μ_k, is 0.3, and the length of the ramp, d, is 8 m.

(a) How much work is done by gravity?

(b) How much work is done by the normal force?

(d) What is the total work done?

Solution.

(a) Recall that the force that’s directly responsible for pulling the box down the plane is the component of the gravitational force that’s parallel to the ramp so the angle between F and d is 0°: W = (F_w|| cos (θ) d = F_w|| mg sin θ) (where θ is the incline angle). This component is parallel to the motion, so the work done by gravity is

W_{by gravity} = (mg sin θ)d = (35 kg)(10 N/kg)(sin 37°)(8 m) = 1,680 J

Note that the work done by gravity is positive, as we would expect it to be, since gravity is helping the motion. Also, be careful with the angle θ. The general definition of work reads W = (F cos θ)d, where θ is the angle between F and d. However, the angle between F_w and d is not 37° here, it was 0°, so the work done by gravity is not (mg cos 37°)d. The angle θ used in the calculation above is the incline angle.

(b) Since the normal force is perpendicular to the motion, the work done by this force is zero.

(c) The strength of the normal force is F_w cos θ (where θ is the incline angle), so the strength of the friction force is F_f = μ_kF_N = μ_kF_w cos θ = μ_kmg cos θ. Since F_f is antiparallel to d, the cosine of the angle between these vectors (180°) is –1, so the work done by friction is

W_{by friction} = –F_f d = –(μ_kmg cos θ)(d) = –(0.3)(35 kg)(10 N/kg)(cos 37°)(8 m) = –672 J

Note that the work done by friction is negative, as we expect it to be, since friction is opposing the motion.

(d) The total work done is found simply by adding the values of the work done by each of the forces acting on the box:

W_total = ΣW = W_{by gravity} + W_{by normal force} + W_{by friction} = 1,680 + 0 + (–672) = 1,008 J

Example 5 The force exerted by a spring when it’s displaced by x from its natural length is given by the equation F(x) = –kx, where k is a positive constant. This equation is known as Hooke’s law. What is the work done by a spring as it pushes out from x = –x₂ to x = –x₁ (where x₂ > x₁)?

Solution. Since the force is variable, we calculate the following definite integral:

Another solution would involve sketching a graph of F(x) = –kx and calculating the area under the graph from x = –x₂ to x = –x₁.

Here, the region is a trapezoid with area A = (base₁ + base₂) × height, so

W = A = (kx₂ + kx₁) (x₂ − x₁)

= k(x₂ + x₁) (x₂ − x₁)

=

KINETIC ENERGY

Consider an object at rest (v₀ = 0), and imagine that a steady force is exerted on it, causing it to accelerate. Let’s be more specific; let the object’s mass be m, and let F be the net force acting on the object, pushing it in a straight line. The object’s acceleration is a = F_net/m, so after the object has traveled a distance ∆x under the action of this force, its final speed, v, is given by Big Five #5:

But the quantity F_net∆x is the total work done by the force, so W_T = mv². The work done on the object has transferred energy to it, in the amount mv². The energy an object possesses by virtue of its motion is therefore defined as mv² and is called kinetic energy:

K = mv²

THE WORK–ENERGY THEOREM

Kinetic energy is expressed in joules, just like work. The derivation above can be extended to an object with a non-zero initial speed, and the same analysis will show that the total work done on an object—or, equivalently, the work done by the net force—will equal its change in kinetic energy; this is known as the work–energy theorem:

W_total = ∆K

Note that kinetic energy, like work, is a scalar quantity.

Example 6 What is the kinetic energy of a ball (mass = 0.10 kg) moving with a speed of 30 m/s?

Solution. From the definition,

K = m v² = (0.10 kg)(30 m/s)² = 45 J

Example 7 A tennis ball (mass = 0.06 kg) is hit straight upward with an initial speed of 50 m/s. How high would it go if air resistance were negligible?

Solution. This could be done using the Big Five, but let’s try to solve it using the concepts of work and energy. As the ball travels upward, gravity acts on it by doing negative work. [The work is negative because gravity is opposing the upward motion. F_w and d are in opposite directions, so θ = 180°, which tells us that W = (F_w cos θ)d = −F_wd.] At the moment the ball reaches its highest point, its speed is 0, so its kinetic energy is also 0. The work–energy theorem says

Example 8 Consider the box sliding down the inclined plane in Example 4. If it starts from rest at the top of the ramp, with what speed does it reach the bottom?

Solution. It was calculated in Example 4 that W_total = 1,008 J. According to the work–energy theorem,

Example 9 A pool cue striking a stationary billiard ball (mass = 0.25 kg) gives the ball a speed of 2 m/s. If the average force of the cue on the ball was 200 N, over what distance did this force act?

Solution. The kinetic energy of the ball as it leaves the cue is

K = mv² = (0.25 kg)(2 m/s)² = 0.50 J

The work W done by the cue gave the ball this kinetic energy, so

POTENTIAL ENERGY

Kinetic energy is the energy an object has by virtue of its motion. Potential energy is independent of motion; it arises from the object’s position (or the system’s configuration). For example, a ball at the edge of a tabletop has energy that could be transformed into kinetic energy if it falls from the table. An arrow in an archer’s pulled-back bow has energy that could be transformed into kinetic energy if the archer releases the arrow. Both of these are examples of potential energy, the energy an object or system has by virtue of its position or configuration. In each case, work was done on the object to put it in the given configuration (the ball was lifted to the tabletop, the bowstring was pulled back), and since work is the means of transferring energy, these things have stored energy that can be converted into kinetic energy. This is potential energy, denoted by U.

Because there are different types of forces, there are different types of potential energy. The ball at the edge of the tabletop provides an example of gravitational potential energy, U_grav, which is the energy stored by virtue of an object’s position in a gravitational field. This energy would be converted to kinetic energy as gravity pulled the ball down to the floor. For now, let’s concentrate on gravitational potential energy. Some textbooks label this U_grav = PE_grav .

Assume the ball has a mass m of 2 kg, and that the tabletop is h = 1.5 m above the floor. How much work did gravity do as the ball was lifted from the floor to the table? The strength of the gravitational force on the ball is F_w = mg = (2 kg) (10 N/kg) = 20 N. The force F_w points downward, and the ball’s motion was upward, so the work done by gravity during the ball’s ascent was

W_{by gravity} = −F_wh = −mgh = −(20 N)(1.5 m) = –30 J

So, someone performed +30 J of work to raise the ball from the floor to the tabletop. That energy is now stored and, if the ball was given a push to send it over the edge, by the time the ball reached the floor it would release 30 J of kinetic energy. We therefore say that the change in the ball’s gravitational potential energy in moving from the floor to the table was +30 J. That is,

∆U_g = –W_{by gravity}

In general, if an object of mass m is raised a height h (which is small enough that g stays essentially constant over this altitude change), then the increase in the object’s gravitational potential energy is

∆U_g = mgh

In this equation the work done by gravity as the object is raised does not depend on the path taken by the object. The ball could be lifted straight upward, or in some curvy path; it would make no difference. Gravity is said to be a conservative force because of this property.

If we decide on a reference level whereby h = 0, then we can say that the gravitational potential energy of an object of mass m at a height h is U_grav = mgh. Let’s consider a passenger in an airplane reading a book. If the book is 1 m above the floor of the plane (our reference point), then, to the passenger, the gravitational potential energy of the book is mgh, where h = 1 m. However, to someone on the ground looking up, the floor of the plane may be 9,000 m above the ground. To this person, the gravitational potential energy of the book is mgH, where H = 9,001 m. Differences, or changes, in potential energy are unambiguous, but calculated values of potential energy are relative.

Example 10 A stuntwoman (mass = 60 kg) scales a 40-meter-tall rock face. What is her gravitational potential energy (relative to the ground)?

Solution. Calling the ground h = 0, we find

U_grav = mgh = (60 kg)(10 m/s2)(40 m) = 24,000 J

Example 11 If the stuntwoman in the previous example were to jump off the cliff, what would be her final speed as she landed on a large, air-filled cushion lying on the ground?

Solution. The gravitational potential energy would be transformed into kinetic energy. So

CONSERVATION OF MECHANICAL ENERGY

We have seen energy in its two basic forms: Kinetic energy (K) and potential energy (U). The sum of an object’s kinetic and potential energies is called its mechanical energy, E:

E = K + U

Assuming that no nonconservative forces (friction, for example) act on an object or system while it undergoes some change, then mechanical energy is conserved. That is, the initial mechanical energy, E_i, is equal to the final mechanical energy, E_f, or

K_i + U_i = K_f + U_f

This is the simplest form of the Law of Conservation of Total Energy, which we mentioned at the beginning of this section.

Example 12 A ball of mass 2 kg is dropped from a height of 5.0 m above the floor. Find the speed of the ball as it strikes the floor.

Solution. Ignoring the friction due to the air, we can apply Conservation of Mechanical Energy. Calling the floor our h = 0 reference level, we write

Note that the ball’s potential energy decreased, while its kinetic energy increased. This is the basic idea behind conservation of mechanical energy: One form of energy decreases while the other increases.

Example 13 A box is projected up a long ramp (incline angle with the horizontal = 37°) with an initial speed of 10 m/s. If the surface of the ramp is very smooth (essentially frictionless), how high up the ramp will the box go? What distance along the ramp will it slide?

Solution. Because friction is negligible, we can apply Conservation of Mechanical Energy. Calling the bottom of the ramp our h = 0 reference level, we write

Since the incline angle is θ = 37°, the distance d it slides up the ramp is found in this way:

Example 14 A skydiver jumps from a hovering helicopter that’s 3,000 m above the ground. If air resistance can be ignored, how fast will he be falling when his altitude is 2,000 m?

Solution. Ignoring air resistance, we can apply Conservation of Mechanical Energy. Calling the ground our h = 0 reference level, we write

That’s over 300 mph! The terminal velocity of a human falling is about 100 mph, which shows that air resistance does play a role, even before the parachute is opened.

The equation K_i + U_i = K_f + U_f holds if no nonconservative forces are doing work. However, if work is done by such forces during the process under investigation, then the equation needs to be modified to account for this work as follows:

K_i + U_i + W_other = K_f + U_f

Example 15 A crash test dummy (mass = 40 kg) falls off a 50-meter-high cliff. On the way down, the force of air resistance has an average strength of 100 N. Find the speed with which he crashes into the ground.

Solution. The force of air resistance opposes the downward motion, so it does negative work on the dummy as he falls: W_r = –F_rh. Calling the ground h = 0, we find that

Example 16 A skier starts from rest at the top of a 20° incline and skis in a straight line to the bottom of the slope, a distance d (measured along the slope) of 400 m. If the coefficient of kinetic friction between the skis and the snow is 0.2, calculate the skier’s speed at the bottom of the run.

Solution. The strength of the friction force on the skier is F_f = μ_kF_N = μ_k(mg cos θ), so the work done by friction is −F_f d = μ_k(mg cos θ) d. The vertical height of the slope above the bottom of the run (which we designate the h = 0 level) is h = d sin θ. Therefore, Conservation of Mechanical Energy (including the negative work done by friction) gives

POTENTIAL ENERGY CURVES

The behavior of a system can be analyzed if we are given a graph of its potential energy, U(x), and its mechanical energy, E. Since K + U = E, we have mv² + U(x) = E which can be solved for v, the velocity at position x:

For example, consider the following potential energy curve:

The graph shows how the potential energy, U, varies with position, x. A particular value of the total energy, E = E₀, is also shown. Motion of an object whose potential energy is given by U(x) and which has a mechanical energy of E₀ is confined to the region –x₀ ≤ x ≤ x₀, because only in this range is E₀ ≥ U(x). At each position x in this range, the kinetic energy K = E₀ – U(x) is positive. However, if x > x₀ (or if x < –x₀), then U(x) > E₀, which is physically impossible because the difference, E₀ – U(x), which should give K, is negative.

This particular energy curve with U(x) = kx², describes one of the most important physical systems: a simple harmonic oscillator. The force felt by the oscillator can be recovered from the potential energy curve. Recall that, in the case of gravitational potential energy we defined ∆U_grav = –W_{by grav}. In general, ∆U = –W. If we account for a variable force of the form F = F(x), which does the work W, then over a small displacement ∆x, we have ∆U(x) = –W = –F(x) ∆x, so F(x) = –∆U(x)/∆x. In the limit as ∆x → 0, this last equation becomes

Therefore, in this case, we find F(x) = −(d/dx)(kx²) = −kx, which specifies a linear restoring force, a prerequisite for simple harmonic motion. This equation, F(x) = –kx, is called Hooke’s law and is obeyed by ideal springs (see Example 5).

With this result, we can appreciate the oscillatory nature of the system whose energy curve is sketched above. If x is positive (and not greater than x₀), then U(x) is increasing, so dU/dx is positive, which tells us that F is negative. So the oscillator feels a force—and an acceleration—in the negative direction, which pulls it back through the origin (x = 0). If x is negative (and not less than –x₀), then U(x) is decreasing, so dU/dx is negative, which tells us that F is positive. So the oscillator feels a force—and an acceleration—in the positive direction, which pushes it back through the origin (x = 0).

Furthermore, the difference between E₀ and U, which is K, decreases as x approaches x₀ (or as x approaches –x₀), dropping to zero at these points. The fact that K decreases to zero at ±x₀ tells us that the oscillator’s speed decreases to zero as it approaches these endpoints. It then changes direction and heads back toward the origin—where its kinetic energy and speed are maximized—for another oscillation. By looking at the energy curve with these observations in mind, you can almost see the oscillator moving back and forth between the barriers at x = ±x₀.

The origin is a point at which U(x) has a minimum, so the tangent line to the curve at this point is horizontal; the slope is zero. Since F = –dU/dx, the force F is 0 at this point [which we also know from the equation F(x) = –kx]; this means that this is a point of equilibrium. If the oscillator is pushed from this equilibrium point in either direction, the force F(x) will attempt to restore it to x = 0, so this is a point of stable equilibrium. However, a point where the U(x) curve has a maximum is also a point of equilibrium, but it’s an unstable one, because if the system were moved from this point in either direction, the force would accelerate it away from the equilibrium position.

Consider the following potential energy curve:

Point C is a position of stable equilibrium and E is a point of unstable equilibrium. Since U(x) is decreasing at points A and F, F(x) is positive, accelerating the system in the positive x-direction. Points B and D mark the barriers of oscillation if the system has a mechanical energy E₀ of 0. To imagine an object moving in an area with this U(x), imagine an object sliding on a frictionless hill of this shape. If released from rest at point B, it would accelerate toward point C, then slow down, stop at point D, and then return to point B. It would oscillate between these two points.

Example 17 An object of mass m = 4 kg has a potential energy function

U(x) = (x – 2) – (2x – 3)³

Where x is measured in meters and U in joules. The following graph is a sketch of the potential energy function.

(a) Determine the positions of points A and B, the equilibrium points.

(b) If the object is released from rest at the point B, can it reach point A or C? Explain.

Solution.

(a) The points A and B are a local minimum and maximum, respectively, so the derivative of U(x) will be zero at these locations.

(b) The object has a negative total amount of mechanical energy at point B because all of its energy is potential energy. It will not be able to reach point C, because that position has a potential energy well above zero. However the object would be able to reach point A because the potential energy at A is less than the energy the object started with at point B. As the object moved from B to A its potential energy would decrease (become more negative) and its kinetic energy would increase.

(c) First, we need to find how much potential energy the object has at point C, and this will define the total mechanical energy of the object. Then determine the potential energy at point A, x = 1.3 m, and use conservation of energy to determine the speed at point A.

U(0.5) = (0.5 − 2) − (2 · (0.5) − 3)³ = 6.5 J
U(1.3) = −0.636 J
E_C = E_A
6.5 = −0.636 + K_A
7.136 = (4)v_A²
v_A = 1.89 m/s

POWER

Simply put, power is the rate at which work gets done (or energy gets transferred, which is the same thing). Suppose you and I each do 1,000 J of work, but I do the work in 2 minutes while you do it in 1 minute. We both did the same amount of work, but you did it more quickly; you were more powerful. Here’s the definition of power:

Since power is a rate, it can be represented as a derivative, too. The calculus (rather than algebra) equation for power is:

The unit of power is the joule per second (J/s), which is called the watt, and symbolized W (not to be confused with the symbol for work, W). One watt is 1 joule per second: 1 W = 1 J/s. Here in the United States, which still uses older English units like inches, feet, yards, miles, ounces, pounds, and so forth, you still hear of power ratings expressed in horsepower (particularly of engines). One horsepower is defined as the power output of a large horse. Horses can pull a 150-pound weight at a speed of 2 mph for quite a while. If the equation for work done by a constant force, namely W = F · r, is inserted into the equation for power, where r is the displacement, the equation for power becomes . Since r/t (displacement over time) is velocity, another equation for power is:

P = F · v

Since the theoretical horse mentioned above is pulling in the direction of its velocity,

P = Fv ⇒ 1 horsepower (hp) = (150 lb)(2 mph)

Using unit conversions to convert horsepower to watts:

Therefore,

1 hp = (667.5 N)(1.117 m/s) = 746 W

By contrast, a human in good physical condition can do work at a steady rate of about 75 W (about 1/10 that of a horse!) but can attain power levels as much as twice this for short periods of time.

Example 18 A mover pushes a large crate (mass m = 75 kg) from the inside of the truck to the back end (a distance of 6 m), exerting a steady push of 300 N. If he moves the crate this distance in 20 s, what is his power output during this time?

Solution. The work done on the crate by the mover is W = Fd = (300 N)(6 m) = 1,800 J. If this much work is done in 20 s, then the power delivered is P = W/t = (1,800 J)/(20 s) = 90 W.

Example 19 What must be the power output of a rocket engine, which moves a 1,000 kg rocket at a constant speed of 8.0 m/s?

Solution. The equation P = Fv, with F = mg, yields

P = mgv = (1,000 kg)(10 N/kg)(8.0 m/s) = 80,000 W = 80 kW

Chapter 7 Drill

The answers and explanations can be found in Chapter 17.

Click here to download the PDF.

Section I: Multiple Choice

1. A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displacement, the work it does is equal to

(A) 0 J

(B) 60 J

(D) 600 J

(E) 2,400 J

2. Under the influence of a force, an object of mass 4 kg accelerates from 3 m/s to 6 m/s in 8 s. How much work was done on the object during this time?

(A) 27 J

(B) 54 J

(D) 96 J

(E) Cannot be determined from the information given

3. A box of mass m slides down a frictionless inclined plane of length L and vertical height h. What is the change in its gravitational potential energy?

(A) –mgL

(B) –mgh

(D) –mgh/L

(E) –mghL

4. A 4 kg box is pulled up a ramp of angle θ = 30 degrees and height = 5 m at a constant velocity. How much work is done by the normal force?

(A) 160 J

(B) 80 J

(D) 20 J

(E) 0 J

5. A man stands in an elevator as it begins to ascend. Does the normal force from the floor do work on the man?

(A) Yes, and the work done will be positive.

(B) Yes, and the work done will be negative.

(D) No, the normal force will do no work in any situation.

(E) No, the normal force will do no work in this situation, but it can in others.

6. A block of mass 3.5 kg slides down a frictionless inclined plane of length 6.4 m that makes an angle of 30° with the horizontal. If the block is released from rest at the top of the incline, what is its speed at the bottom?

(A) 5.0 m/s

(B) 5.7 m/s

(D) 8.0 m/s

(E) 10 m/s

7. A block of mass m slides from rest down an inclined plane of length s and height h. If F is the magnitude of the force of kinetic friction acting on the block as it slides, then the kinetic energy of the block when it reaches the bottom of the incline will be equal to

(A) mgh

(B) mgh – Fh

(D) mgh – Fs

(E) mgs – Fs

8. If a 1,000 kg car is accelerating at a rate of 4 m/s² and experiencing 300 N of drag force, how much force do the engines have to produce? Ignore any frictional effects with the road.

(A) 16,300 N

(B) 15,700 N

(D) 4,000 N

(E) 3,700 N

9. An astronaut drops a rock from the top of a crater on the Moon. When the rock is halfway down to the bottom of the crater, its speed is what fraction of its final impact speed?

(A)

(B)

(C)

(D)

(E)

10. A force of 200 N is required to keep an object sliding at a constant speed of 2 m/s across a rough floor. How much power is being expended to maintain this motion?

(A) 50 W

(B) 100 W

(D) 400 W

(E) Cannot be determined from the information given

Section II: Free Response

1. A box of mass m is released from rest at point A, the top of a long, frictionless slide. Point A is at height H above the level of points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The coefficient of kinetic friction between the box and this surface is μ_k, and the horizontal distance between point B and C is x.

Solve for the following in terms of given quantities and the acceleration of gravity, g.

(a) Find the speed of the box when its height above point B is H.

(b) Find the speed of the box when it reaches point B.

(d) Now assume that points B and C were not on the same horizontal level. In particular, assume that the surface from B to C had a uniform upward slope so that point C was still at a horizontal distance of x from B but now at a vertical height of y above B. Answer the question posed in part (c).

2. The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins at rest from point A and moves down the frictionless track from A to B where it then enters the vertical loop (also frictionless), traveling once around the circle from B to C to D to E and back to B, after which it travels along the flat portion of the track from B to F (which is not frictionless).

Solve for the following in terms of given quantities and the acceleration of gravity, g.

(a) Find the centripetal acceleration of the car when it is at point C.

(b) What is the minimum cut-off speed v_c that the car must have at point D to make it around the loop?

(d) If H = 6r and the coefficient of friction between the car and the flat portion of the track from B to F is 0.5, how far along this flat portion of the track will the car travel before coming to rest at point F?

3. A ball m = 3 kg has the potential energy function

U(x) = 3(x – 1) – (x – 3)³

where x is measured in meters and U in joules. The following graph is a sketch of this potential energy function.

The energies indicated on the vertical axis are evenly spaced; that is, E₃ − E₂ = E₂ − E₁. The energy E₁ is equal to U(x₁), and the energy E₃ is equal to U(x₃).

(a) Determine the numerical values of x₁ and x₃.

(b) Describe the motion of the particle if its total energy is E₂.

(d) Sketch the graph of the particle’s acceleration as a function of x. Be sure to indicate x₁ and x₃ on your graph.

(e) The particle is released from rest at x = x₁. Find its speed as it passes through x = x₁.

4. The force on a 6 kg object is given by the equation: F(x) = 3x + 5, in newtons. The object is moving 2 m/s at the origin.

(a) Determine the work done on the object by the force when it is moved 4 m from the origin in the x direction.

(b) Determine the speed of the object when it has moved 4 m.

Summary

Work

Work is the dot product of force and displacement: W = F ⋅ x

When force varies with x, the work is given by the following equation: W = ∫F(x)dx

Work is positive when the force and displacement are parallel.

Work is negative when the force and displacement are antiparallel.

Work–Energy Theorem: W = ∆KE = –∆U

Energy

Kinetic energy is energy associated with motion and is given by the equation K = mv².

Potential energy is stored energy. The potential energy due to gravity is given by the equation U_g = mgh. The potential energy due to springs is given by the equation U_s = kx².

Work done by a conservative force only depends on the initial and final positions, and not on the path taken. Gravity and springs are examples of conservative forces.

Work done by a non-conservative force depends on the path taken, and mechanical energy is lost by heat, sound, and so on, when these forces act on a system. Friction and air resistance are examples of non-conservative forces.

Conservation of Mechanical Energy states that the total mechanical energy of a system is constant when there are no non-conservative forces acting on the system. It is usually written as E_i = E_f or K_i + U_i = K_f + U_f.

Potential Energy Diagrams

The potential energy can be given as U(x). Then F = −.

If = 0, then F = 0, and it is an equilibrium point.

Stable equilibrium occurs when the force restores the object back toward the equilibrium point after it is disturbed.

Unstable equilibrium occurs when the force moves the object further away from the equilibrium point after it is disturbed.

Power

Power is the rate at which work is done.

P = Fv

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 8 Linear Momentum

INTRODUCTION

When Newton first expressed his Second Law, he didn’t write F_net = ma. Instead, he expressed the law in the words, The alteration of motion is…proportional to the…force impressed…. By “motion,” he meant the product of mass and velocity, a vector quantity known as linear momentum, which is denoted by p:

So Newton’s original formulation of the Second Law read ∆p ∝ F, or, equivalently, F ∝ ∆p. But a large force that acts for a short period of time can produce the same change in linear momentum as a small force acting for a greater period of time. Knowing this, if we take the average force, , that acts over the time interval ∆t, we can turn the proportion above into an equation:

This equation becomes F = ma, since ∆p/∆t = ∆(mv)/∆t = m (∆v/∆t) = ma (assuming that m remains constant). If we take the limit as ∆t → 0, then the equation above takes the form:

Example 1 A golfer strikes a golf ball of mass 0.05 kg, and the time of impact between the golf club and the ball is 1 ms. If the ball acquires a velocity of magnitude 70 m/s, calculate the average force exerted on the ball.

Solution. Using Newton’s Second Law, we find

IMPULSE

The product of force and the time during which it acts is known as impulse; it’s a vector quantity that’s denoted by J:

J = Δt

In terms of impulse, Newton’s Second Law can be written in yet another form:

J = ∆p

Sometimes this is referred to as the impulse–momentum theorem, but it’s just another way of writing Newton’s Second Law. If F varies with time over the interval during which it acts, then the impulse delivered by the force F = F(t) from time t = t₁ to t = t₂ is given by the following definite integral:

On the equation sheet for the free-response section, this information will be represented as follows:

If a graph of force-versus-time is given, then the impulse of force F as it acts from t₁ to t₂ is equal to the area bounded by the graph of F, the t-axis, and the vertical lines associated with t₁ and t₂ as shown in the following graph.

Example 2 A football team’s kicker punts the ball (mass = 0.4 kg) and gives it a launch speed of 30 m/s. Find the impulse delivered to the football by the kicker’s foot and the average force exerted by the kicker on the ball, given that the impact time is 8 ms.

Solution. Impulse is equal to change in linear momentum, so

J = ∆p = p_f – p_i = p_f = mv = (0.4 kg)(30 m/s) = 12 kg·m/s

Using the equation F = J/Δt, we find that the average force exerted by the kicker is

F = J/Δt = (12 kg · m/s)(8 × 10⁻³s) = 1,500 N [≈ 340 lb]

Example 3 An 80 kg stuntman jumps out of a window that’s 45 m above the ground.

(a) How fast is he falling when he reaches ground level?

(b) He lands on a large, air-filled target, coming to rest in 1.5 s. What average force does he feel while coming to rest?

Solution.

(a) His gravitational potential energy turns into kinetic energy: mgh= mv², so

(You could also have answered this question using Big Five #5.)

(b) Using = Δp/Δt and p = mv, we find that

This force is equivalent to about 27 tons (!), which is more than enough to break bones and cause fatal brain damage. Notice how crucial impact time is: Increasing the slowing-down time reduces the acceleration and the force, ideally enough to prevent injury. This is the purpose of air bags in cars, for example.

Example 4 A small block of mass m = 0.07 kg, initially at rest, is struck by an impulsive force F of duration 10 ms whose strength varies with time according to the following graph:

What is the resulting speed of the block?

Solution. The impulse delivered to the block is equal to the area under the F vs. t graph. The region is a trapezoid, so its area, (base₁ + base₂) × height, can be calculated as follows:

J = (10 + 4) × 20 = 0.14 N . s

Now, by the impulse–momentum theorem,

CONSERVATION OF LINEAR MOMENTUM

Newton’s Third Law says that when one object exerts a force on a second object, the second object exerts an equal but opposite force on the first. Since Newton’s Second Law says that the impulse delivered to an object is equal to the resulting change in its linear momentum, J = ∆p, the two interacting objects experience equal but opposite momentum changes (assuming that there are no external forces), which implies that the total linear momentum of the system remains constant. In fact, given any number of interacting objects, each pair that comes in contact will undergo equal but opposite momentum changes, so the result described for two interacting objects will actually hold for any number of objects, given that the only forces they feel are from each other. This means that, in an isolated system, the total linear momentum will remain constant. This is the Law of Conservation of Linear Momentum. In equation form, for two objects colliding, we have .

Example 5 An astronaut is floating in space near her shuttle when she realizes that the cord that’s supposed to attach her to the ship has become disconnected. Her total mass (body + suit + equipment) is 89 kg. She reaches into her pocket, finds a 1 kg metal tool, and throws it out into space with a velocity of 9 m/s directly away from the ship. If the ship is 10 m away, how long will it take her to reach it?

Solution. Here, the astronaut + tool are the system. Because of Conservation of Linear Momentum,

Using distance = average speed × time, we find

COLLISIONS

Conservation of Linear Momentum is routinely used to analyze collisions. The objects whose collision we will analyze form the system, and although the objects exert forces on each other during the impact, these forces are only internal (they occur within the system). The system’s total linear momentum is conserved if there is no net external force on the system.

Collisions are classified into two major categories: (1) elastic and (2) inelastic. A collision is said to be elastic if kinetic energy is conserved. Ordinary macroscopic collisions are never truly elastic, because there is always a change in energy due to energy transferred as heat, deformation of the objects, and the sound of the impact. However, if the objects do not deform very much (for example, two billiard balls or a hard glass marble bouncing off a steel plate), then the loss of initial kinetic energy is small enough to be ignored, and the collision can be treated as virtually elastic. Inelastic collisions, then, are ones in which the total kinetic energy is different after the collision. An extreme example of inelasticism is completely (or perfectly or totally) inelastic. In this case, the objects stick together after the collision and move as one afterward. In all cases of isolated collisions (elastic or not), Conservation of Linear Momentum states that

total p_{before collision} = total p_{after collision}

Example 6 Two balls roll toward each other, one red and the other green. The red ball has a mass of 0.5 kg and a speed of 4 m/s just before impact. The green ball has a mass of 0.3 kg and a speed of 2 m/s. After the head-on collision, the red ball continues forward with a speed of 1.7 m/s. Find the speed of the green ball after the collision. Was the collision elastic?

Solution. First remember that momentum is a vector quantity, so the direction of the velocity is crucial. Since the balls roll toward each other, one ball has a positive velocity while the other has a negative velocity. Let’s call the red ball’s velocity before the collision positive; then v_red = +4 m/s, and v_green = –2 m/s. Using a prime to denote after the collision, Conservation of Linear Momentum gives us the following:

Notice that the green ball’s velocity was reversed as a result of the collision; this typically happens when a lighter object collides with a heavier object. To see whether the collision was elastic, we need to compare the total kinetic energies before and after the collision. In this case, however, an explicit calculation is not needed since both objects experienced a decrease in speed as a result of the collision. Kinetic energy was lost, so the collision was inelastic; this is usually the case with macroscopic collisions. Most of the lost energy was transferred as heat; the two objects are both slightly warmer as a result of the collision.

Example 7 The balls in Example 6 roll toward each other. The red ball has a mass of 0.5 kg and a speed of 4 m/s just before impact. The green ball has a mass of 0.3 kg and a speed of 2 m/s. If the collision is completely inelastic, determine the velocity of the composite object after the collision.

Solution. If the collision is completely inelastic, then, by definition, the masses stick together after impact, moving with a velocity, v′. Applying Conservation of Linear Momentum, we find

Example 8 An object of mass m₁ is moving with velocity v₁ toward a target object of mass m₂ which is stationary (v₂ = 0). The objects collide head-on, and the collision is elastic. Show that the relative velocity before the collision, v₂ – v₁, has the same magnitude as − , the relative velocity after the collision.

Solution. Since the collision is elastic, both total linear momentum and kinetic energy are conserved. Therefore,

Now for some algebra. Cancel the ’s in the second equation and factor to get the following pair of equations:

Next, dividing the second equation by the first gives v₁ + = , so we can write

Adding Equations (1″) and (2″) gives

Substituting this result into Equation (2″) gives

Now we have calculated the final velocities, v₁ and v₂. To verify the claim made in the statement of the question, we notice that

so the relative velocity after the collision, − , is equal (in magnitude) but opposite (in direction) to v₂ – v₁, the relative velocity before the collision. This is a general property that characterizes elastic collisions.

Example 9 An object of mass m moves with velocity v toward a stationary object of mass 2m. After impact, the objects move off in the directions shown in the following diagram:

(a) Determine the magnitudes of the velocities after the collision (in terms of v).

(b) Is the collision elastic? Explain your answer.

Solution.

(a) Conservation of Linear Momentum is a principle that establishes the equality of two vectors: p_total before the collision and p_total after the collision. Writing this single vector equation as two equations, one for the x component and one for the y, we have

Adding these equations eliminates v₂, because cos 45° = sin 45°.

mv = m(cos 30° + sin 30°)

and lets us determine v₁:

Substituting this result into Equation (2) gives us

(b) The collision is elastic only if kinetic energy is conserved. The total kinetic energy after the collision, K′, is calculated as follows:

However, the kinetic energy before the collision is just K = mv², so the fact that

tells us that K′ is less than K, so some kinetic energy is lost; the collision is inelastic.

CENTER OF MASS

The center of mass is the point where all of the mass of an object can be considered to be concentrated; it’s the dot that represents the object of interest in a free-body diagram.

For a homogeneous body (that is, one for which the density is uniform throughout), the center of mass is where you intuitively expect it to be: at the geometric center. Thus, the center of mass of a uniform sphere or cube or box is at its geometric center.

If we have a collection of discrete particles, the center of mass of the system can be determined mathematically as follows. First consider the case where the particles all lie on a straight line. Call this the x-axis. Select some point to be the origin (x = 0) and determine the positions of each particle on the axis. Multiply each position value by the mass of the particle at that location, and get the sum for all the particles. Divide this sum by the total mass, and the resulting x-value is the center of mass:

On the AP Physics C Table of Information, this information will be represented as follows:

The system of particles behaves in many respects as if all its mass, M = m₁ + m₂ +···+ m_n, were concentrated at a single location, x_cm.

If the system consists of objects that are not confined to the same straight line, use the equation above to find the x-coordinate of their center of mass, and the corresponding equation,

to find the y-coordinate of their center of mass (and one more equation to calculate the z-coordinate, if they are not confined to a single plane).

From the equation

we can derive

Mv_cm = m₁v₁ + m₂v₂ + ··· + m_nv_n

So, the total linear momentum of all the particles in the system (m₁v₁ + m₂v₂ +…+ m_nv_n) is the same as Mv_cm, the linear momentum of a single particle (whose mass is equal to the system’s total mass) moving with the velocity of the center of mass.

We can also differentiate again and establish the following:

F_net = Ma_cm

This says that the net (external) force acting on the system causes the center of mass to accelerate according to Newton’s Second Law. In particular, if the net external force on the system is zero, then the center of mass will not accelerate.

Example 10 Two objects, one of mass m and one of mass 2m, hang from light threads from the ends of a uniform bar of length 3L and mass 3m. The masses m and 2m are at distances L and 2L, respectively, below the bar. Find the center of mass of this system.

Solution. The center of mass of the bar alone is at its midpoint (because it is uniform), so we may treat the total mass of the bar as being concentrated at its midpoint. Constructing a coordinate system with this point as the origin, we now have three objects: one of mass m at (–3L/2, –L), one of mass 2m at (3L/2, –2L), and one of mass 3m at (0, 0):

We figure out the x- and y-coordinates of the center of mass separately:

Therefore, the center of mass is at

(x_cm, y_cm) = (L/4, –5L/6)

relative to the midpoint of the bar.

Example 11 A man of mass m is standing at one end of a stationary, floating barge of mass 3m. He then walks to the other end of the barge, a distance of L meters. Ignore any frictional effects between the barge and the water.

(a) How far will the barge move?

(b) If the man walks at an average velocity of v, what is the average velocity of the barge?

Solution.

(a) Since there are no external forces acting on the man + barge system, the center of mass of the system cannot accelerate. In particular, since the system is originally at rest, the center of mass cannot move. Letting x = 0 denote the midpoint of the barge (which is its own center of mass, assuming it is uniform), we figure out the center of mass of the man + barge system:

So, the center of mass is a distance of L/8 from the midpoint of the barge, and since the mass is originally at the left end, the center of mass is a distance of L/8 to the left of the barge’s midpoint.

When the man reaches the other end of the barge, the center of mass will, by symmetry, be L/8 to the right of the midpoint of the barge. But, since the position of the center of mass cannot move, this means the barge itself must have moved a distance of

L/8 + L/8 = 2L/8 = L/4

to the left.

(b) Let the time it takes the man to walk across the barge be denoted by t; then t = L/v. In this amount of time, the barge moves a distance of L/4 in the opposite direction, so the velocity of the barge is

So far we have dealt with objects that can be considered point masses, or masses with uniform density. Now we will learn how to find the center of mass of objects with non-uniform density.

Take the example of a bar that becomes denser along its length. Here we will deal with the linear density λ as a function of x, λ(x). Each small segment of the bar, ∆x, has a different mass, ∆m. We treat each ∆m as a point mass and then take the limit as ∆x approaches zero. Using the formula for calculating the center of mass of point masses, and replacing ∆m with dm, we get the integral shown below.

where M is the total mass, x is the distance to each dm, and you can substitute for dm in terms of x. Linear density is mass per length, so the equation is λ = , therefore M = ∫dm = ∫λdx.

Example 12 A bar with a length of 30 cm has a linear density λ = 10 + 6x, where x is in meters and λ is in kg/m. Determine the mass of the bar and the center of mass of this bar.

Solution. We can determine the mass of the bar by using the definition of linear density, λ = . Therefore,

To calculate the center of mass, we will use this equation:

This answer makes sense because the center of mass of the bar is beyond the midpoint.

Chapter 8 Drill

Click here to download the PDF.

The answers and explanations can be found in Chapter 17.

Section I: Multiple Choice

1. An object of mass 2 kg has a linear momentum of magnitude 6 kg·m/s. What is this object’s kinetic energy?

(A) 3 J

(B) 6 J

(D) 12 J

(E) 18 J

2. The graph below shows the force on an object over time.

If the object has a mass of 8 kg and is moving in a straight line, what is its change in speed?

(A) 16 m/s

(B) 14 m/s

(D) 10 m/s

(E) 8 m/s

3. A box with a mass of 2 kg accelerates in a straight line from 4 m/s to 8 m/s due to the application of a force whose duration is 0.5 s. Find the average strength of this force.

(A) 2 N

(B) 4 N

(D) 12 N

(E) 16 N

4. A ball of mass m traveling horizontally with velocity v strikes a massive vertical wall and rebounds back along its original direction with no change in speed. What is the magnitude of the impulse delivered by the wall to the ball?

(A) 0

(B) mv

(D) 2mv

(E) 4mv

5. Two objects, one of mass 3 kg and moving with a speed of 2 m/s and the other of mass 5 kg and speed 2 m/s, move toward each other and collide head-on. If the collision is perfectly inelastic, find the speed of the objects after the collision.

(A) 0.25 m/s

(B) 0.5 m/s

(D) 1 m/s

(E) 2 m/s

6. A student is trying to balance a meter stick on its midpoint. Given that m₁ = 6 kg and m₂ = 2 kg, how far from the left edge should the student hang a third mass, m₃ = 10 kg, to balance the meter stick?

(A) 40 cm

(B) 50 cm

(D) 70 cm

(E) 80 cm

7. Two objects move toward each other, collide, and separate. If there was no net external force acting on the objects, but some kinetic energy was lost, then

(A) the collision was elastic and total linear momentum was conserved

(B) the collision was elastic and total linear momentum was not conserved

(D) the collision was not elastic and total linear momentum was not conserved

(E) None of the above

8. Three thin, uniform rods each of length L are arranged in the shape of an inverted U:

The two rods on the arms of the U each have mass m; the third rod has mass 2m. How far below the midpoint of the horizontal rod is the center of mass of this assembly?

(A) L/8

(B) L/4

(D) L/2

(E) 3L/4

9. A car of mass 1,000 kg collides head-on with a truck of mass 2,000 kg. Both vehicles are moving at a speed of 21 m/s, and the collision is perfectly inelastic. After the crash, the two vehicles skid to a halt. Assuming friction is the only force acting on the vehicles after the collision, how much work is done by friction after the crash?

(A) 73,500 J

(B) –73,500 J

(D) –147,000 J

(E) 220,500 J

10. Which of the following best describes a perfectly inelastic collision free of external forces?

(A) Total linear momentum is never conserved.

(B) Total linear momentum is sometimes conserved.

(D) Kinetic energy is sometimes conserved.

(E) Kinetic energy is always conserved.

Section II: Free Response

1. A steel ball of mass m is fastened to a light cord of length L and released when the cord is horizontal. At the bottom of its path, the ball strikes a hard plastic block of mass M = 4m, initially at rest on a frictionless surface. The collision is elastic.

(a) Find the tension in the cord when the ball’s height above its lowest position is L. Write your answer in terms of m and g.

(b) Find the speed of the block immediately after the collision.

2. A ballistic pendulum is a device that may be used to measure the muzzle speed of a bullet. It is composed of a wooden block suspended from a horizontal support by cords attached at each end. A bullet is shot into the block, and as a result of the perfectly inelastic impact, the block swings upward. Consider a bullet (mass m) with velocity v as it enters the block (mass M). The length of the cords supporting the block each have length L. The maximum height to which the block swings upward after impact is denoted by y, and the maximum horizontal displacement is denoted by x.

(a) In terms of m, M, g, and y, determine the speed v of the bullet.

(b) What fraction of the bullet’s original kinetic energy is lost as a result of the collision? What happens to the lost kinetic energy?

(d) Once the block begins to swing, does the momentum of the block remain constant? Why or why not?

3. An object of mass m moves with velocity v toward a stationary object of the same mass. After their impact, the objects move off in the directions shown in the following diagram:

Assume that the collision is elastic.

(a) If K₁ denotes the kinetic energy of Object 1 before the collision, what is the kinetic energy of this object after the collision? Write your answer in terms of K₁ and θ₁.

(b) What is the kinetic energy of Object 2 after the collision? Write your answer in terms of K₁ and θ₁.

Summary

Momentum

Linear momentum is given by the equation: p = mv.

Linear momentum is conserved when no external force acts on a system. This is known as the Law of Conservation of Linear Momentum. It can be written as:

total p_{before collision} = total p_{after collision}

Elastic collisions conserve kinetic energy (in general, every collision conserves energy but not necessarily kinetic energy).

Inelastic collisions do not conserve kinetic energy.

When the objects stick together, the collision is known as perfectly inelastic.

Anytime you are given a problem that involves a collision or separation, first consider whether you can use the Law of Conservation of Linear Momentum.

Impulse

Impulse is given by the equation:

J = FΔt

The Impulse–Momentum Theorem states that the impulse on an object is equal to the change in momentum of the object. The equation is: J = FΔt = Δp.

Center of Mass

Usually the motion of an object is describing the motion of the center of mass. When you use Newton’s Second Law, F = ma, the acceleration you calculate is the acceleration of the center of mass.

For point masses, r_cm = , where r is used for the position of each mass.

For distributed mass (for example, a bar with non-uniform density) the equation is: r_cm = ∫ r dm.

You usually use linear density, λ dr = dm, for dm and then integrate to solve for the center of mass.

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 9 Rotational Motion

INTRODUCTION

So far we’ve studied only translational motion: objects sliding, falling, or rising, but in none of our examples have we considered spinning objects. We will now look at rotation, which will complete our study of motion. All motion is some combination of translation and rotation, which are illustrated in the figures below. Consider any two points in the object under study (on the left) and imagine connecting them by a straight line. If this line does not turn while the object moves, then the object is translating only. However, if this line does not always remain parallel to itself while the object moves, then the object is rotating.

ROTATIONAL KINEMATICS

Mark several dots along a radius on a disk, and call this radius the reference line. If the disk rotates about its center, we can use the movement of these dots to talk about angular displacement, angular velocity, and angular acceleration.

If the disk rotates as a rigid body, then all three dots shown have the same angular displacement, ∆θ. In fact, this is the definition of a rigid body: All points along a radial line always have the same angular displacement.

Just as the time rate-of-change of displacement gives velocity, the time rate-of-change of angular displacement gives angular velocity, denoted by ω (omega). The definition of the average angular velocity is:

Note that if we let the time interval ∆t approach 0, then the equation above leads to the definition of the instantaneous angular velocity:

And, finally, just as the time rate-of-change of velocity gives acceleration, the time rate-of-change of angular velocity gives angular acceleration, or α (alpha). The definition of the average angular acceleration is:

If we let the time interval ∆t approach 0, then the equation above leads to the definition of the instantaneous angular velocity:

On the rotating disk illustrated on the previous page, we said that all points undergo the same angular displacement in any given time interval; this means that all points on the disk have the same angular velocity, ω, but not all points have the same linear velocity, v. This follows from the definition of radian measure. Expressed in radians, the angular displacement, ∆θ, is related to the arc length, ∆s, by the equation

Rearranging this equation and dividing by ∆t, we find that

Or, using the equations v = ds/dt and ω = dθ/dt,

Therefore, the greater the value of r, the greater the value of v. Points on the rotating body farther from the rotation axis move more quickly than those closer to the rotation axis.

From the equation v = rω, we can derive the relationship that connects angular acceleration and linear acceleration. Differentiating both sides with respect to t (holding r constant), gives us

(It’s important to realize that the acceleration a in this equation is not centripetal acceleration; it’s tangential acceleration, which arises from a change in speed caused by an angular acceleration. By contrast, centripetal acceleration does not produce a change in speed.) Often, tangential acceleration is written as a_t to distinguish it from centripetal acceleration (a_c).

Example 1 A rotating, rigid body makes one complete revolution in 2 s. What is its average angular velocity?

Solution. One complete revolution is equal to an angular displacement of 2π radians (that is, there are 2π radians in 360°), so the body’s average angular velocity is

Example 2 The angular velocity of a rotating disk increases from 2 rad/s to 5 rad/s in 0.5 s. What’s the disk’s average angular acceleration?

Solution. By definition,

Example 3 A disk of radius 20 cm rotates at a constant angular velocity of 6 rad/s. How fast does a point on the rim of this disk travel (in m/s)?

Solution. The linear speed, v, is related to the angular speed, ω, by the equation v = rω. Therefore,

v = rω = (0.20 m)(6 rad/s) = 1.2 m/s

Note that although we typically write the abbreviation rad when writing angular measurements, the radian is actually a dimensionless quantity, since, by definition, θ = s/r. So ∆θ = 6 means the same thing as ∆θ = 6 rad.

Example 4 The angular velocity of a rotating disk of radius 50 cm increases from 2 rad/s to 5 rad/s in 0.5 s. What is the linear tangential acceleration of a point on the rim of the disk during this time interval?

Solution. The linear acceleration a is related to the angular acceleration α by the equation a = rα. Since α = 6 rad/s² (as calculated in Example 2), we find that

a = rα = (0.50 m)(6 rad/s²) = 3 m/s²

Example 5 Derive an expression for centripetal acceleration in terms of angular speed.

Solution. For an object revolving with linear speed v at a distance r from the center of rotation, the centripetal acceleration is given by the equation a_c = v²/r. Using the fundamental equation v = rω, we find that

On the equation sheet for the free-response section, this information will be represented as follows:

THE BIG FIVE FOR ROTATIONAL MOTION

The simplest type of rotational motion to analyze is motion in which the angular acceleration is constant (possibly equal to zero). Another restriction that will make our analysis easier (and which doesn’t significantly diminish the power and applicability of our results) is to consider rotational motion around a fixed axis of rotation. In this case, there are only two possible directions for motion. One direction, counterclockwise, is called positive (+), and the opposite direction, clockwise, is called negative (–).

Let’s review the quantities we’ve seen so far. The fundamental quantities for rotational motion are angular displacement (∆θ), angular velocity (ω), and angular acceleration (α). Because we’re dealing with angular acceleration, we know about changes in angular velocity, from initial velocity (ω_i or ω₀) to final velocity (ω_f or simply ω—with no subscript). And, finally, the motion takes place during some elapsed time interval, ∆t. Therefore, we have five kinematics quantities: ∆θ, ω₀, ω, α, and ∆t.

These five quantities are interrelated by a group of five equations which we call the Big Five. They work in cases in which the angular acceleration is uniform. These equations are identical to the Big Five we studied in Chapter 5 but, in these cases, the translational variables (s, v, or a) are replaced by the corresponding rotational variables (θ, ω, or α, respectively).

In Big Five #1, because angular acceleration is constant, the average angular velocity is simply the average of the initial angular velocity and the final angular velocity: = (ω₀ + ω). Also, if we set t_i = 0, then ∆t = t_f – t_i = t – 0 = t, so we can just write “t” instead of “∆t” in the first four equations. This simplification in notation makes the equations a little easier to memorize.

		Variable that’s missing
Big Five #1	Δθ = t	α
Big Five #2		∆θ
Big Five #3		ω
Big Five #4	Δθ = ωt − α(t)²	ω₀
Big Five #5	ω² = + 2αΔθ	t

Each of the Big Five equations is missing exactly one of the five kinematics quantities and, as with the other Big Five you learned, the way you decide which equation to use is to determine which of the kinematics quantities is missing from the problem, and use the equation that’s also missing that quantity. For example, if the problem never mentions the final angular velocity—ω is neither given nor asked for—then the equation that will work is the one that’s missing ω; that’s Big Five #3.

Example 6 An object with an initial angular velocity of 1 rad/s rotates with constant angular acceleration. Three seconds later, its angular velocity is 5 rad/s. Calculate its angular displacement during this time interval.

Solution. We’re given ω₀, t, and ω, and asked for ∆θ. So α is missing, and we use Big Five #1:

Δθ = Δt = (ω₀ + ω)Δt = (1 rad/s + 5 rad/s) (3 s) = 9 rad

Example 7 Starting with zero initial angular velocity, a sphere begins to spin with constant angular acceleration about an axis through its center, achieving an angular velocity of 10 rad/s when its angular displacement is 20 rad. What is the value of the sphere’s angular acceleration?

Solution. We’re given ω₀, ∆θ, and ω, and asked for α. Since t is missing, we use Big Five #5:

To summarize, here’s a comparison of the fundamental quantities of translational and rotational motion and of the Big Five (assuming constant acceleration and a fixed axis of rotation):

	Translational	Rotational	Connection
displacement:	∆x	∆θ	∆x = r∆θ
velocity:	v	ω	v = rω
acceleration:	a	α	a = rα
Big Five #1:	Δx = x − x₀ = t	Δθ = t
Big Five #2:	v = v₀ + at	ω = ω₀ + αt
Big Five #3:	x = x₀ + v₀t + at²	Δθ = ω₀t + α(t)²
Big Five #4:	x = x₀ + v₀t − at²	Δθ = ωt − α(t)²
Big Five #5:	v₂ = + 2a(x – x₀)	ω² = + 2αΔθ

ROTATIONAL DYNAMICS

The dynamics of translational motion involve describing the acceleration of an object in terms of its mass (inertia) and the forces that act on it; F_net = ma. By analogy, the dynamics of rotational motion involve describing the angular (rotational) acceleration of an object in terms of its rotational inertia and the torques that act on it.

Torque

Intuitively, torque describes the effectiveness of a force in producing rotational acceleration. Consider a uniform rod that pivots around one of its ends, which is fixed. For simplicity, let’s assume that the rod is at rest. What effect, if any, would each of the four forces in the figure below have on the potential rotation of the rod?

Our intuition tells us that F₁, F₂, and F₃ would not cause the rod to rotate, but F₄ would. What’s different about F₄? It has torque. Clearly, torque has something to do with rotation. Just like a force is a vector quantity that produces linear acceleration, a torque is a vector quantity that produces angular acceleration. Note that, just like for linear acceleration, an angular acceleration is something that either changes the direction of the angular velocity or changes the angular speed. A torque can be thought of as being positive if it produces counterclockwise rotation or negative if it produces clockwise rotation.

The torque of a force can be defined as follows. Let r be the distance from the pivot (axis of rotation) to the point of application of the force F, and let θ be the angle between vectors r and F.

Then the torque of F, denoted by τ (tau), is defined as:

τ = rF sin θ

Imagine sliding r over so that its initial point is the same as that of F. The angle between two vectors is the angle between them when they start at the same point. However, the supplementary angle θ′ can be used in place of θ in the definition of torque. This is because torque depends on sin θ, and the sine of an angle and the sine of its supplement are always equal. Therefore, when figuring out torque, use whichever of these angles is more convenient.

We will now see if this mathematical definition of torque supports our intuition about forces F₁, F₂, F₃, and F₄.

The angle between r and F₁ is 0, and θ = 0 implies sin θ = 0, so by the definition of torque, τ = 0 as well. The angle between r and F₂ is 180°, and θ = 180° gives us sin θ = 0, so τ = 0. For F₃, r = 0 (because F₃ acts at the pivot, so the distance from the pivot to the point of application of F₃ is zero); since r = 0, the torque is 0 as well. However, for F₄, neither r nor sin θ is zero, so F₄ has a nonzero torque. Of the four forces shown in that figure, only F₄ has torque and would produce rotational acceleration.

There’s another way to determine the value of the torque. Of course, it gives the same result as the method given above, but this method is often easier to use. Look at the same object and force:

Instead of determining the distance from the pivot point to the point of application of the force, we will now determine the perpendicular distance from the pivot point to what’s called the line of action of the force. This distance is the lever arm (or moment arm) of the force F relative to the pivot, and is denoted by l.

The torque of F is defined as the product

τ = lF

Just as the lever arm is often called the moment arm, the torque is called the moment of the force. That these two definitions of torque, τ = rF sin θ and τ = lF are equivalent follows immediately from the fact that l = r sin θ:

Since l is the component of r that’s perpendicular to F, it is also denoted by r_⊥ (“r perp”). So the definition of torque can be written as τ = r_⊥F.

These two equivalent definitions of torque make it clear that only the component of F that’s perpendicular to r produces torque. The component of F that’s parallel to r does not produce torque. Notice that τ = rF sin θ = rF_⊥, where F_⊥ (“F perp”) is the component of F that’s perpendicular to r:

So the definition of torque can also be written as τ = rF_⊥.

Also, since only the component of F perpendicular to r produces torque, the torque can be written as the cross product of r and F, and this is how it appears on the free-response equation sheet:

Example 8 A student pulls down with a force of 40 N on a rope that winds around a pulley of radius 5 cm.

What’s the torque of this force?

Solution. Since the tension force, F_T, is tangent to the pulley, it is perpendicular to the radius vector r at the point of contact:

Therefore, the torque produced by this tension force is simply

τ = rF_T = (0.05 m)(40 N) = 2 N·m

Example 9 What is the net torque on the cylinder shown below? The cylinder is pinned at its center.

Solution. Each of the two forces produces a torque, but these torques oppose each other. The torque of F₁ is counterclockwise, and the torque of F₂ is clockwise. This can be visualized either by imagining the effect of each force, assuming that the other was absent, or by using the vector definition of torque, τ = r × F. In the case of F₁, we have, by the right-hand rule,

τ₁ = r₁ × F₁ points out of the plane of the page: ⊙, while

τ₂ = r₂ × F₂ points into the plane of the page: ⊗

These symbols are easy to remember if you think of a dart: ⊙ is the point of the dart coming at you and ⊗ is the pattern of feathers at the back of the dart you see as it flies away from you toward the dartboard.

If the torque vector points out of the plane of the page, this indicates a tendency to produce counterclockwise rotation and, if it points into the plane of the page, this indicates a tendency to produce clockwise rotation.

The net torque is the sum of all the torques. Counting a counterclockwise torque as positive and a clockwise torque as negative, we have

τ₁ = +r₁F₁ = +(0.12 m)(100 N) = +12 N·m

and

τ₂ = –r₂F₂ = –(0.08 m)(80 N) = –6.4 N·m

τ_net = Στ = τ₁ + τ₂ = (+12 N·m) + (–6.4 N·m) = +5.6 N·m

Rotational Inertia

Our goal is to develop a rotational analog of Newton’s Second Law, F_net = ma. We’re almost there; torque is the rotational analog of force and, therefore, τ_net is the rotational analog of F_net. The rotational analog of translational acceleration, a, is rotational (or angular) acceleration, α. We will now look at the rotational analog of inertial mass, m.

Consider a small point mass m at a distance r from the axis of rotation, being acted upon by a tangential force F.

From Newton’s Second Law, we have F = ma. Substituting a = rα, this equation becomes F = mrα. Now multiply both sides of this last equation by r to yield

rF = mr²α

or, since rF = τ,

τ = mr²α

In the equation F = ma, the quantity m is multiplied by the acceleration produced by the force F, while in the equation τ = mr²α, the quantity mr² is multiplied by the rotational acceleration produced by the torque τ.

So for a point mass at a distance r from the axis of rotation, its rotational inertia (also called moment of inertia) is defined as mr². If we now take into account all the point masses that comprise the object under study, we can get the total rotational inertia, I, of the body by adding them:

I = Σm_ir_i²

For a continuous solid body, this sum becomes the integral

This formula can be used to calculate expressions for the rotational inertia of cylinders (disks), spheres, slender rods, and hoops. Notice that the rotational inertia depends not only on m, but also on r; both the mass and how it’s distributed about the axis of rotation determine I. By summing up all point masses and all external torques, the equation τ = mr²α becomes

or τ_net = Iα

On the equation sheet for the free-response section, this information will be represented as follows:

We’ve reached our goal:

Translational motion	Rotational motion
force, F	torque, τ
acceleration, a	rotational acceleration, α
mass, m	rotational inertia, I
F_net = ma	τ_net = Iα

Example 10 Three beads, each of mass m, are arranged along a rod of negligible mass and length L. Figure out the rotational inertia of the assembly when the axis of rotation is through the center bead and when the axis of rotation is through one of the beads on the ends.

(a)

(b)

Solution.

(a) In the first case, both the left bead and the right bead are at a distance of L/2 from the axis of rotation, while the center bead is at distance zero from the axis of rotation. Therefore,

(b) In the second case, the left bead is at distance zero from the rotation axis, the center bead is at distance L/2, and the right bead is at distance L. Therefore,

Note that, although both assemblies have the same mass (namely, 3m), their rotational inertias are different, because of the different distribution of mass relative to the axis of rotation.

If the rotational inertia of a body is known relative to an axis that passes through the body’s center of mass, then the rotational inertia, I′, relative to any other rotation axis (parallel to the first one) can be calculated as follows. Let I_cm be the rotational inertia of a body relative to a rotation axis that passes through the body’s center of mass, let the mass of the body be M, and let x be the distance from the axis through the center of mass to the rotation axis. Then

I = I_cm + mx²

This is called the parallel–axis theorem. Let’s use this result to calculate the rotational inertia of the three-bead assembly in part (b) from the value obtained in part (a), which is I_cm. Since M = 3m and x = L/2, we have

which agrees with the value calculated above.

Example 11 Show that the rotational inertia of a homogeneous cylinder of radius R and mass M, rotating about its central axis, is given by the equation I = MR^2.

Solution. A solid is homogeneous if its density is constant throughout. Let ρ be the density of the cylinder. In order to compute I using the formula I = ∫ r² dm we choose our mass element to be an infinitesimally thin cylindrical ring of radius r:

The mass of this ring is equal to its volume, 2πr dr × H, times the density; that is,

dm = ρ × 2πrH dr

Therefore,

To eliminate ρ, use the fact that the total mass of the cylinder is

M = ρV = ρ · π R²H

Putting this into the expression derived for I, we find that

I = πρHR⁴ = (ρ · πR²H)R² = MR²

The height of the cylinder (the dimension parallel to the axis of rotation) is irrelevant. Therefore, the formula I = MR² gives the rotational inertia of any homogeneous solid cylinder revolving around its central axis. This includes a disk (which is just a really short cylinder).

Example 12 Again, consider the system of Example 9:

Assume that the mass of the cylinder is 50 kg. Given that the rotational inertia of a cylinder of radius R and mass M rotating about its central axis is given by the equation I = MR², determine the rotational acceleration produced by the two forces shown.

Solution. In Example 9, we figured out that τ_net = +5.6 N·m. The rotational inertia of the cylinder is

I = MR² = (50 kg)(0.12 m)² = 0.36 kg·m²

Therefore, from the equation τ_net = Iα, we find that

This angular acceleration will be counterclockwise, because τ_net is counterclockwise.

Example 13 A block of mass m is hung from a pulley of radius R and mass M and allowed to fall. What is the acceleration of the block?

Solution. Remember that earlier, we treated pulleys as if they were massless, and no force was required to make them rotate. Now, however, we know how to take the mass of a pulley into account, by including its rotational inertia in our analysis. The pulley is a disk, so its rotational inertia is given by the formula I = MR².

First we draw a free-body diagram for the falling block:

and apply Newton’s Second Law:

mg – F_T = ma (1)

Now the tension F_T in the cord produces a torque, τ = RF_T, on the pulley:

Since this is the only torque on the pulley, the equation τ_net = Iα becomes RF_T = Iα. But I = MR² and α = a/R. Therefore,

which tells us that

F_T = Ma (2)

Substituting Equation (2) into Equation (1), we find that

KINETIC ENERGY OF ROTATION

A rotating object has rotational kinetic energy, just as a translating object has translational kinetic energy. The formula for kinetic energy is, of course, K = mv², but this can’t be directly used to calculate the kinetic energy of rotation because each point mass that makes up the body can have a different v. For this reason, we need a definition of K_rotational that involves ω instead of v.

In short, then:

Note that this expression for rotational kinetic energy follows the general pattern displayed by our previous results: I is the rotational analog of m and ω is the rotational analog of v. Therefore, the rotational analog of mv² should be Iω².

Rolling Motion

One of the main types of motion associated with rotational motion is rolling motion. We will primarily deal with rolling motion without slipping.

Consider a disk rolling down an incline—without slipping:

The point of contact of the object with the surface P is instantaneously at rest. If this were not the case, then the disk would be slipping down the incline, so the contact point must not be moving relative to the surface. In this case, the velocity of the center of mass of the disk is equal to the radius times the angular velocity of the disk.

You can take the torque around any point to determine the acceleration of the disk. It is often easiest to take it around the contact point P because then only gravity provides a torque about this point, and if you know the mass of the object, then you know the force of gravity. Make sure to use the parallel–axis theorem in this case since you are considering the rotational inertia about point P, not the center of mass. This will allow you to calculate the acceleration of the disk. Once you know the acceleration, you can calculate the necessary coefficient of friction to produce rolling without slipping using Newton’s Second Law.

The total motion for an object that is rolling without slipping is the combined motion of the entire object translating with the velocity of the center of mass, and the object rotating about its center of mass, as shown below. This shows the object is instantaneously rotating about the contact point P.

For rolling motion the total kinetic energy is the translational kinetic energy and the rotational kinetic energy.

Example 14 A cylinder of mass M and radius R rolls (without slipping) down an inclined plane whose incline angle with the horizontal is θ. Determine the acceleration of the cylinder’s center of mass, and the minimum coefficient of friction that will allow the cylinder to roll without slipping on this incline.

Solution. First, we draw a free-body diagram for the cylinder:

We know that the cylinder rolls without slipping, so the force of friction is not kinetic friction. Since the speed of the point on the cylinder in contact with the ramp is zero with respect to the ramp, static friction supplies the torque that allows the cylinder to roll smoothly.

Take the torque about the contact point to solve for the acceleration because the frictional force will not be part of the equation and we do not know it yet.

Now we will use Newton’s Second Law to solve for µ_s since we know the acceleration.

Example 15 A cylinder of mass M and radius R rolls (without slipping) down an inclined plane (of height h and length L) whose incline angle with the horizontal is θ. Determine the linear speed of the cylinder’s center of mass when it reaches the bottom of the incline (assuming that it started from rest at the top).

Solution. We will attack this problem using Conservation of Mechanical Energy. As the cylinder rolls down the ramp, its initial gravitational potential energy is converted into kinetic energy, which is a combination of translational kinetic energy (since the cylinder’s center of mass is translating down the ramp) and rotational kinetic energy:

Since I = MR² and ω= v_cm/R, this equation becomes

Therefore,

We can verify this result using the result of the previous example. There we found that the acceleration of the cylinder’s center of mass as it rolled down the ramp was a = gsin θ. Applying Big Five #5 gives us:

WORK AND POWER

Consider a small point mass m at distance r from the axis of rotation, acted upon by a tangential force F.

As it rotates through an angular displacement of ∆θ, the force does work on the point mass: W = F∆s, where ∆s = r∆θ. Therefore,

W = Fr∆ θ = τ ∆θ

If the force is not purely tangential to the object’s path, then only the tangential component of the force does work; the radial component does not (since it’s perpendicular to the object’s displacement). Therefore, for a general constant force F, the equation above would read W = F_tr∆θ = τ ∆θ, where F_t denotes the tangential component of F.

If we want to allow for a varying F—and a varying τ—then the work done is equal to the definite integral:

Again, notice the analogy between this equation and the one that defines work by F:

The work–energy theorem (W = ∆K) also holds in the rotational case, where W is the work done by net torque and ∆K is the resulting change in the rotational kinetic energy.

The rate at which work is done, or the power (P), is defined by the equation

Over an infinitesimal angular displacement dθ, the torque τ does an amount of work dW given by dW = τ dθ. This implies that

Once again, notice the parallel to P = Fv, the translational version of this equation.

Example 16 A block of mass m = 5 kg is hung from a pulley of radius R = 15 cm and mass M = 8 kg and then released from rest.

(a) What is the speed of the block as it strikes the floor, 2 m below its initial position?

(b) What is the rotational kinetic energy of the pulley just before the block strikes the floor?

Solution.

(a) Apply Conservation of Mechanical Energy. The initial gravitational potential energy of the block is transformed into the purely rotational kinetic energy of the pulley and translational kinetic energy of the falling block:

Substituting in the given numerical values, we get

(b) The rotational kinetic energy of the pulley as the block strikes the floor is

(c) The rate at which work is done on the pulley is the power produced by the torque. One way to compute this is to first use the work–energy theorem to determine the work done by the torque and then divide this by the time during which the block fell. So,

W = ΔK = K_f − K_i = K_f = 44 J

The time during which this work was done is the time required for the block to drop to the ground. Using Big Five #1 and the result of part (a), we find that

Therefore,

ANGULAR MOMENTUM

So far we’ve developed rotational analogs for displacement, velocity, acceleration, force, mass, and kinetic energy. We will finish by developing a rotational analog for linear momentum; it’s called angular momentum.

Consider a small point mass m at distance r from the axis of rotation, moving with velocity v and acted upon by a tangential force F.

Then, by Newton’s Second Law,

If we multiply both sides of this equation by r and notice that rF = τ, we get

Therefore, to form the analog of the law F = ∆p/∆t (force equals the rate-of-change of linear momentum), we say that torque equals the rate-of-change of angular momentum, and the angular momentum (denoted by L) of the point mass m is defined by the equation

L = rmv

If we now take into account all the point masses that comprise the object under study, we can get the angular momentum of the body by adding up all the individual contributions. This gives

L = I ω

Note that this expression for angular momentum follows the general pattern we saw previously: I is the rotational analog of m, and ω is the rotational analog of v. Therefore, the rotational analog of mv should be Iω.

If the point mass m does not move in a circular path, we can still define its angular momentum relative to any reference point.

If r is the vector from the reference point to the mass, then the angular momentum is

L = rmv_⊥

where v_⊥ is the component of the velocity that’s perpendicular to r. This is the perpendicular component of v relative to r that’s important for figuring out angular momentum, and this fact leads to the general vector definition with the cross product. The equation L = (r)(mv_⊥) = (r)(p_⊥) becomes

Example 17 A solid uniform sphere of mass M = 8 kg and radius R = 50 cm is revolving around an axis through its center at an angular speed of 10 rad/s. Given that the rotational inertia of the sphere is equal to MR², what is the spinning sphere’s angular momentum?

Solution. Apply the definition:

L = I ω = MR²ω = (8 kg)(0.50 m)²(10 rad/s) = 8 kg · m²/s

If you want to specify the direction of the angular momentum vector, L, use the right-hand rule. Let the fingers of your right hand curl in the direction of rotation of the body. Your thumb gives the direction of L, pointing along the rotation axis:

CONSERVATION OF ANGULAR MOMENTUM

Newton’s Second Law says that

so if F_net = 0, then p is constant. This is Conservation of Linear Momentum.

The rotational analog of this is

So if τ_net = 0, then L is constant. This is Conservation of Angular Momentum. Basically, this says that if the torques on a body balance so that the net torque is zero, then the body’s angular momentum can’t change.

An often cited example of this phenomenon is the spinning of a figure skater. As the skater pulls his or her arms inward, he or she moves more of his or her mass closer to the rotation axis and decreases his or her rotational inertia. Since the external torque on the skater is negligible, his or her angular momentum must be conserved. Since L = Iω, a decrease in I causes an increase in ω, and the skater spins faster.

Example 18 A child of mass m = 30 kg stands at the edge of a small merry-go-round that’s rotating at a rate of 1 rad/s. The merry-go-round is a disk of radius R = 2.5 m and mass M = 100 kg. If the child walks in toward the center of the disk and stops 0.5 m from the center, what will happen to the angular velocity of the merry-go-round (if friction can be ignored)?

Solution. The child walking toward the center of the merry-go-round does not provide an external torque to the child + disk system, so angular momentum is conserved. Let’s denote the child as a point mass, and consider the following two views of the merry-go-round (looking down from above):

In the first picture, the total rotational inertia, I_i, is equal to the sum of the rotational inertia of the merry-go-round (MGR) and the child:

I_i = I_MGR + I_child = MR² + m = MR² + mR² = (M + m)R²

In the second picture, the total rotational inertia has decreased to

I_f = I_MGR + = MR² + m

So, by Conservation of Angular Momentum, we have

and substituting the given numerical values gives us

Notice that ω increased as I decreased, just as Conservation of Angular Momentum predicts.

EQUILIBRIUM

An object is said to be in translational equilibrium if the sum of the forces acting on it is zero; that is, if F_net = 0. Similarly, an object is said to be in rotational equilibrium if the sum of the torques acting on it is zero; that is, if τ_net = 0. The term equilibrium by itself means both translational and rotational equilibrium. A body in equilibrium may be in motion; F_net = 0 does not mean that the velocity is zero; it only means that the velocity is constant. Similarly, τ_net = 0 does not mean that the angular velocity is zero; it only means that it’s constant. If an object is at rest, then it is said to be in static equilibrium.

Example 19 A uniform bar of mass m and length L extends horizontally from a wall. A supporting wire connects the wall to the bar’s midpoint, making an angle of 55° with the bar. A sign of mass M hangs from the end of the bar.

If the system is in static equilibrium, determine the tension in the wire and the strength of the force exerted on the bar by the wall if m = 8 kg and M = 12 kg.

Solution. Let F_C denote the (contact) force exerted by the wall on the bar. In order to simplify our work, we can write F_C in terms of its horizontal component, F_C_x, and its vertical component, F_C_y. Also, if F_T is the tension in the wire, then F_T_x = F_T cos 55° and F_T_y = F_T sin 55° are its components. This gives us the following force diagram:

The first condition for equilibrium requires that the sum of the horizontal forces is zero and the sum of the vertical forces is zero:

ΣF_x = 0: F_Cx − F_T cos 55° = 0 (1)

ΣF_y = 0: F_Cy + F_T cos 55° − mg – Mg = 0 (2)

We notice immediately that we have more unknowns (F_C_x, F_C_y, F_T) than equations, so this system cannot be solved as is. The second condition for equilibrium requires that the sum of the torques about any point is equal to zero. Choosing the contact point between the bar and the wall as our pivot, only three of the forces in the diagram above produce torque. F_T_y produces a counterclockwise torque, and both mg and Mg produce clockwise torques, which must balance. From the definition τ = lF, and taking counterclockwise torque as positive and clockwise torque as negative, we have

Στ = 0: (L/2)F_Ty – (L/2)(mg) – LMg = 0 (3)

This equation contains only one unknown and can be solved immediately:

F_Ty = mg + Lmg
F_Ty = mg + 2Mg = (m + 2M)g

Since F_T_y = F_T sin 55°, we can find that

Substituting this result into Equation (1) gives us F_C_x:

And finally, from Equation (2), we get

The fact that F_C_y turned out to be negative simply means that in our original force diagram, the vector F_C_y points in the direction opposite to how we drew it. That is, F_C_y points downward. Therefore, the magnitude of the total force exerted by the wall on the bar is

Chapter 9 Drill

The answers and explanations can be found in Chapter 17.

Click here to download the PDF.

Section I: Multiple Choice

1. A compact disc has a radius of 6 cm. If the disc rotates about its central axis at an angular speed of 5 rev/s, what is the linear speed of a point on the rim of the disc?

(A) 0.3 m/s

(B) 1.9 m/s

(D) 52 m/s

(E) 83 m/s

2. A compact disc has a radius of 6 cm. If the disc rotates about its central axis at a constant angular speed of 5 rev/s, what is the total distance traveled by a point on the rim of the disc in 40 min?

(A) 180 m

(B) 360 m

(D) 720 m

(E) 4.5 km

3. A disc starts at rest and experiences constant angular acceleration of 4 rad/s². When the disc has gone through a total angular displacement of 50 rad, what is its angular speed?

(A) 12 rad/s

(B) 15 rad/s

(D) 20 rad/s

(E) 23 rad/s

4. An object, originally at rest, begins spinning under uniform angular acceleration. In 10 s, it completes an angular displacement of 60 rad. What is the numerical value of the angular acceleration?

(A) 0.3 rad/s²

(B) 0.6 rad/s²

(D) 2.4 rad/s²

(E) 3.6 rad/s²

5. Joe and Alice are sitting on opposite ends of a seesaw. The fulcrum is beneath the center of the seesaw, which has a length of 4 m. Given that Joe’s mass is 60 kg and Alice’s mass is 30 kg, what will be the magnitude of the net torque on the seesaw when it is perfectly level?

(A) 300 N•m

(B) 600 N•m

(D) 1,200 N•m

(E) 1,500 N•m

6. In the figure above, what is the torque about the pendulum’s suspension point produced by the weight of the bob, given that the length of the pendulum, L, is 80 cm and m = 0.50 kg?

(A) 0.5 N•m

(B) 1.0 N•m

(D) 2.0 N•m

(E) 3.4 N•m

7. A uniform meter stick of mass 1 kg is hanging from a thread attached at the stick’s midpoint. One block of mass m = 3 kg hangs from the left end of the stick, and another block, of unknown mass M, hangs below the 80 cm mark on the meter stick. If the stick remains at rest in the horizontal position shown above, what is M?

(A) 4 kg

(B) 5 kg

(D) 8 kg

(E) 9 kg

8. What is the rotational inertia of the following body about the indicated rotation axis? (The masses of the connecting rods are negligible.)

(A) 4mL²

(B) mL²

(D) mL²

(E) mL²

9. The moment of inertia of a solid uniform sphere of mass M and radius R is given by the equation I = MR ². Such a sphere is released from rest at the top of an inclined plane of height h, length L, and incline angle θ. If the sphere rolls without slipping, find its speed at the bottom of the incline.

(A)

(B)

(C)

(D)

(E)

10. When a cylinder rolls down an inclined plane without slipping, which force is responsible for providing the torque that causes rotation?

(A) The force of gravity parallel to the plane

(B) The force of gravity perpendicular to the plane

(D) The force of static friction

(E) The force of kinetic friction

Section II: Free Response

1. In the figure below, the pulley is a solid disk of mass M and radius R, with rotational inertia MR²/2. Two blocks, one of mass m₁ and one of mass m₂, hang from either side of the pulley by a light cord. Initially the system is at rest, with Block 1 on the floor and Block 2 held at height h above the floor. Block 2 is then released and allowed to fall. Give your answers in terms of m₁, m₂, M, R, h, and g.

(a) What is the speed of Block 2 just before it strikes the ground?

(b) What is the angular speed of the pulley at this moment?

(d) How long does it take for Block 2 to fall to the floor?

2. The diagram below shows a solid uniform cylinder of radius R and mass M rolling (without slipping) down an inclined plane of incline angle θ. A thread wraps around the cylinder as it rolls down the plane and pulls upward on a block of mass m. Ignore the rotational inertia of the pulley.

(a) Show that “rolling without slipping” means that the speed of the cylinder’s center of mass, v_cm, is equal to Rω, where ω is its angular speed.

(b) Show that, relative to P (the point of contact of the cylinder with the ramp), the speed of the top of the cylinder is 2v_cm.

(c) What is the relationship between the magnitude of the acceleration of the block and the linear acceleration of the cylinder?

(d) What is the acceleration of the cylinder?

(e) What is the acceleration of the block?

3. Two slender uniform bars, each of mass M and length 2L, meet at right angles at their midpoints to form a rigid assembly that’s able to rotate freely about an axis through the intersection point, perpendicular to the page. Attached to each end of each rod is a solid ball of clay of mass m. A bullet of mass m_b is shot with velocity v as shown in the figure (which is a view from above of the assembly) and becomes embedded in the targeted clay ball.

(a) Show that the moment of inertia of each slender rod about the given rotation axis, not including the clay balls, is ML²/3.

(b) Determine the angular velocity of the assembly after the bullet has become lodged in the targeted clay ball.

(d) Determine the ratio of the final kinetic energy of the assembly to the kinetic energy of the bullet before impact.

Summary

Relating Linear and Angular Quantities

s = rθ

v = rω

a_tan = rα

Linear Equations and Angular Equivalents

Basic Rotation Information

Rotational inertia is the rotational analog of inertia, essentially a measure of how difficult it is to change an object’s rotational motion.

For point masses, I = mr²

For distributed mass, I = ∫ r² dm

Parallel Axis Theorem: I = I_cm + mx²

Torque is a force’s ability to cause an object to rotate. The equation for torque is τ = r × F.

Newton’s Second Law for rotation: ∑ τ = Iα

Rotating objects have rotational kinetic energy although the object does not necessarily translate.

The equation for rotational kinetic energy is k_rotation = Iω².

If the object is rolling, the rotational kinetic energy is

k_rolling = K_rotation + k_translation =

I_cmω + mv_cm²

Angular Momentum

Angular momentum for a point particle is given by the equation L = Iω.

Angular momentum for a rigid object is given by the equation .

Angular momentum is conserved unless a net torque acts on the object. This is expressed by the equation Στ = dL/dt.

For an object to be in static equilibrium, the net force and the net torque must be zero: ΣF = 0, Στ = 0.

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 10 Laws of Gravitation

KEPLER’S LAWS

Johannes Kepler spent years of exhaustive study distilling volumes of data collected by his mentor, Tycho Brahe, into three simple laws that describe the motion of planets.

Kepler’s First Law

Every planet moves in an elliptical orbit, with the Sun at one focus.

Kepler’s Second Law

As a planet moves in its orbit, a line drawn from the Sun to the planet sweeps out equal areas in equal time intervals.

Kepler’s Third Law

If T is the period, the time required to make one revolution, and a is the length of the semimajor axis of a planet’s orbit, then the ratio T²/a³ is the same for all planets orbiting the same star.

NEWTON’S LAW OF GRAVITATION

Newton eventually proved that Kepler’s first two laws imply a law of gravitation: Any two objects in the universe exert an attractive force on each other—called the gravitational force—whose strength is proportional to the product of the objects’ masses and inversely proportional to the square of the center-to-center distance between them. If we let G be the universal gravitational constant, then the strength of the gravitational force is given by the equation:

Recall that is the unit vector from r¹ to r². Consider a mass, m₁, close to the surface of Earth. We will use Newton’s Law of Gravitation and Newton’s Second Law to show that the gravitational acceleration of m₁ is independent of the mass of the object, as shown below.

The forces F_1-on-2 and F_2-on-1 act along the line that joins the bodies and form an action/reaction pair.

The first reasonably accurate numerical value for G was determined by Cavendish more than one hundred years after Newton’s Law was published. To three decimal places, the currently accepted value of G is

G = 6.67 × 10⁻¹¹ N • m²/kg²

Kepler’s Third Law then follows from Newton’s Law of Gravitation. We’ll show how this works for the case of a circular orbit of radius R (which can be considered an elliptical orbit with eccentricity zero). If the orbit is circular, then Kepler’s Second Law says that the planet’s orbit speed, v, must be constant. Therefore, the planet executes uniform circular motion, and centripetal force is provided by the gravitational attraction of the Sun. If we let M be the mass of the Sun and m be the mass of the planet, then this last statement can be expressed mathematically as:

The period of a planet’s orbit is the time it requires to make one revolution around the Sun, so dividing the distance covered, 2πR, by the planet’s orbit speed, v, we have

Equation (1) implies that v² = GM/R. Squaring both sides of Equation (2) and then substituting v² = GM/R, we find that

Therefore,

which is Kepler’s Third Law for a circular orbit of radius R.

Acceleration of Gravity Due to Large Bodies

We have been using the acceleration due to gravity g = 10 m/s² for all objects falling near the surface of Earth. We have assumed that the mass does not affect the acceleration of gravity and now we will show why.

Assume a small mass m is located near a large body (i.e., a planet or star) of mass M. The gravitational force on the object near the surface will equal the mass of the object times the acceleration of gravity a_g. The equation below shows how the mass of the object cancels out, and the acceleration of gravity is independent of that mass of the object.

From this expression we can see that the acceleration due to gravity for any object on a planet would be related to the mass and radius of the planet, but not the mass of the object.

THE GRAVITATIONAL ATTRACTION DUE TO AN EXTENDED BODY

Newton’s Law of Gravitation is really a statement about the force between two point particles: objects that are very small in comparison to the distance between them. Newton also proved that a uniform sphere attracts another body as if all of the sphere’s mass were concentrated at its center.

For this reason, we can apply Newton’s Law of Gravitation to extended bodies, that is, to objects that are not small relative to the distance between them.

Additionally, a uniform shell of mass does not exert a gravitational force on a particle inside it. This means that if a spherical planet is uniform, then as we descend into it, only the mass of the sphere underneath us exerts a gravitational force; the shell above exerts no force because we’re inside it.

Example 1 What is the gravitational force on a particle of mass m at a distance x from the center of a spherically symmetric planet of uniform density ρ, total mass M, and radius R for

(a) x ≥ R

(b) x < R

Solution.

(a) If x ≥ R, then the planet can be treated as a point particle with all its mass concentrated at its center, and

(b) However, if x < R, then only the mass within the sphere of radius x exerts a gravitational force on the particle. Since the volume of such a sphere is (4/3)πx³, its mass is (4/3)πx³ρ; we’ll denote this by M_{within x}. Since the mass of the entire planet is (4/3)πR³ρ, we see that

Therefore, the force that this much mass exerts on the particle of mass m is

In summary then,

Example 2 Given that the radius of Earth is 6.37 × 106 m, determine the mass of Earth.

Solution. Consider a small object of mass m near the surface of Earth (mass M). Its weight is mg, but its weight is just the gravitational force it feels due to Earth, which is GMm/R². Therefore,

Since we know that g = 10 m/s² and G = 6.67 × 10⁻¹¹ N·m²/kg², we can substitute to find

Example 3 We can derive the expression GM/R² by equating mg and GMm/R² (as we did in the previous example), and this gives the magnitude of the absolute gravitational acceleration, a quantity that’s sometimes denoted g₀. The notation g is acceleration, but with the spinning of Earth taken into account. Show that if an object is at the equator, its measured weight (the weight that a scale would measure), mg, is less than its true weight, mg₀, and compute the weight difference for a person of mass m = 60 kg.

Solution. Imagine looking down at Earth from above the North Pole.

The net force toward the center of Earth is F₀ – F_N, which provides the centripetal force on the object. Therefore,

Since v = 2πR/T, where T is Earth’s rotation period we have

or, since F₀ = mg₀ and F_N = mg,

Since the quantity 4π²mR/T² is positive, mg must be less than mg₀. The difference between mg₀ and mg, for a person of mass m = 60 kg, is only:

and the difference between g₀ and g is

Example 4 Communications satellites are often parked in geosynchronous orbits above Earth’s surface. Such satellites have orbit periods that are equal to Earth’s rotation period, so they remain above the same position on Earth’s surface. Determine the altitude and the speed that a satellite must have to be in a geosynchronous orbit above a fixed point on Earth’s equator. (The mass of Earth is 5.98 × 1024 kg.)

Solution. Let m be the mass of the satellite, M be the mass of Earth, and R be the distance from the center of Earth to the position of the satellite. The gravitational pull of Earth provides the centripetal force on the satellite, so

The orbit speed of the satellite is 2πR/T, so

which implies that

Now, the key feature of a geosynchronous orbit is that its period matches Earth’s rotation period, T = 24 hr. Substituting the numerical values of G, M, and T into this expression, we find that

Therefore, if r_E is the radius of Earth, then the satellite’s altitude above Earth’s surface must be

h = R – r_E = (4.23 × 10⁷ m) – (6.37 × 10⁶ m) = 3.59 × 10⁷ m

which is equal to 5.6r_E. The speed of the satellite in this orbit is, from the first equation in our calculations,

regardless of the mass of the satellite (as long as m < < M).

Example 5 A uniform, slender bar of mass M has length L. Determine the gravitational force it exerts on the point particle of mass m shown below:

Solution. Since the bar is an extended body (and not spherically symmetric), we must calculate F using an integral. Select an arbitrary segment of length dx and mass dM in the bar, at a distance x from its left-hand end.

Then, since the bar is uniform, dM = (M/L)dx, so the gravitational force between m and dM is

Now, by adding (that is, by integrating) all of the contributions dF, we get the total gravitational force, F:

GRAVITATIONAL POTENTIAL ENERGY

When we developed the equation U = mgh for the gravitational potential energy of an object of mass m at height h above the surface of Earth, we took the surface of Earth to be our U = 0 reference level and assumed that the height, h, was small compared to Earth’s radius. In that case, the variation in g was negligible, so g was treated as constant. The work done by gravity as an object was raised to height h was then simply –F_grav × ∆s = –mgh, so U_grav, which by definition equals –W_{by grav}, was mgh.

But now we’ll take variations in g into account and develop a general equation for gravitational potential energy, one that isn’t restricted to small altitude changes.

Consider an object of mass m at a distance r₁ from the center of Earth (or any spherical body) moving by some means to a position r₂:

How much work did the gravitational force perform during this displacement? The answer is given by the equation:

Therefore, since ∆U_grav = –W_{by grav}, we get

Let’s choose our U = 0 reference at infinity. That is, we decide to allow U₂ → 0 as r₂ → ∞. Then this equation becomes

Notice that, according to this equation (and our choice of U = 0 when r = ∞), the gravitational potential energy is always negative. This just means that energy has to be added to bring an object (mass m) bound to the gravitational field of M to a point very far from M, at which U = 0.

PROVING THE EQUATION

Because the gravitational force is not constant over the displacement, the work done by this force must be calculated using a definite integral:

Displacement can be broken into a series of infinitesimal steps of two types: Those that are at a constant distance from Earth’s center and those that are along a radial line going away from Earth’s center. The result is that the displacement is equivalent to the sum of two curves C₁ + C₂:

The gravitational force along C₁ does no work, because it’s always perpendicular to the displacement. Along C₂, the inward gravitational force is in the opposite direction from the outward displacement, so along this vector,

Therefore,

Example 6 With what minimum speed must an object of mass m be launched in order to escape Earth’s gravitational field? (This is called escape speed, v_esc.)

Solution. When launched, the object is at the surface of Earth (r_i = r_E) and has an upward, initial velocity of magnitude v_i. To get it far away from Earth, we want to bring its gravitational potential energy to zero, but to find the minimum launch speed, we want the object’s final speed to be zero by the time it gets to this distant location. So, by Conservation of Energy,

which gives

Substituting the known numerical values for G, M_E, and r_E gives us:

Example 7 A satellite of mass m is in a circular orbit of radius R around Earth (radius r_E, mass M).

(a) What is its total mechanical energy (where U_grav is considered zero as R approaches infinity)?

(b) How much work would be required to move the satellite into a new orbit, with radius 2R?

Solution.

(a) The mechanical energy, E, is the sum of the kinetic energy, K, and potential energy, U. You can calculate the kinetic energy since you know that the centripetal force on the satellite is provided by the gravitational attraction of Earth:

Therefore,

(b) From the equation K_i + U_i + W = K_f + U_f, we see that

W = (K_f + U_f − (K_i + U_i)
= E_f − E_i

Therefore, the amount of work necessary to effect the change in the satellite’s orbit radius from R to 2R is

A Note on Elliptical Orbits

The expression for the total energy of a satellite in a circular orbit of radius R [derived in Example 7(a)] is:

E = − (circular orbit)

And this also holds for a satellite traveling in an elliptical orbit if the radius R is replaced by a, (the length of the semimajor axis):

E = − (elliptical orbit)

Example 8 An asteroid of mass m is in an elliptical orbit around the Sun (mass M). Assume that m << M.

(a) What is the total energy of the asteroid?

(b) What is the ratio of v₁ (the asteroid’s speed at aphelion) to v₂ (the asteroid’s speed at perihelion)? Write your answer in terms of r₁ and r₂.

Solution.

(a) The total energy of the asteroid is equal to –GMm/2a, where a is the semimajor axis. However, notice in the figure above that 2a = r₁ + r₂. Therefore, the total energy of the asteroid is

(b) One way to answer this question is to invoke Conservation of Angular Momentum. When the asteroid is at aphelion, its angular momentum (with respect to the center of the Sun) is L₁ = r₁mv₁. When the asteroid is at perihelion, its angular momentum is L₂ = r₂mv₂. Therefore,

This tells us that the asteroid’s speed at aphelion is less than its speed at perihelion (because r₂/r₁ < 1), as implied by Kepler’s Second Law (a line drawn from the Sun to the asteroid must sweep out equal areas in equal time intervals). The closer the asteroid is to the Sun, the faster it has to travel to make this true.

(c) As you know, the time necessary for the asteroid to make a complete orbit around the Sun is the orbit period, T. Using Kepler’s Third Law, with a = (r₁ + r₂), we find

ORBITS OF THE PLANETS

Kepler’s First Law states that the planets’ orbits are ellipses, but the ellipses that the planets in our solar system travel are nearly circular. The deviation of an ellipse from a perfect circle is measured by a parameter called its eccentricity. The eccentricity, e, is the ratio of c (the distance between the center and either focus) to a, the length of the semimajor axis. For every point on the ellipse, the sum of the distances to the foci (plural of focus) is a constant (and is equal to 2a in the figure below).

Kepler’s First Law also states that one of the foci of a planet’s elliptical orbit is located at the position of the Sun. Actually, the focus is at the center of mass of the Sun-planet system, because when one body orbits another, both bodies orbit around their center of mass, a point called the barycenter.

For most of the planets, which are much less massive than the Sun, this correction to Kepler’s First Law has little significance, because the center of mass of the Sun and the planet system is close enough to the Sun’s center. For example, let’s figure out the center of mass of the Sun-Earth system. The mass of Earth is m = 5.98 × 10²⁴ kg, the mass of the Sun is M = 1.99 × 10³⁰ kg, and the Sun-Earth distance averages R = 1.496 × 10¹¹ m. Therefore, letting x = 0 be at the Sun’s center, we have

So the center of mass of the Sun-Earth system is only 450 km from the center of the Sun, a distance of less than 0.1% of the Sun’s radius.

Example 9 Derive a corrected version of Kepler’s Third Law for the following orbiting system. Both bodies have orbit period T.

Solution. The centripetal force on each body is provided by the gravitational pull of the other body, so

which imply

But, since both bodies have the same orbit period, T, we have

Substituting these results into the preceding pair of equations gives us

which simplify to

Adding this last pair of equations gives us the desired result:

Note that this final equation is a general version of Kepler’s Third Law for a circular orbit derived earlier, T²/R³ = 4π²/GM, where it was assumed that the planet orbited at a distance R from the center of the Sun.

Example 10 An artificial satellite of mass m travels at a constant speed in a circular orbit of radius R around Earth (mass M). What is the speed of the satellite?

Solution. The centripetal force on the satellite is provided by Earth’s gravitational pull. Therefore,

Solving this equation for v yields

Notice that the satellite’s speed doesn’t depend on its mass; even if it were a baseball, if its orbit radius were R, then its orbit speed would still be .

Chapter 10 Drill

The answers and explanations can be found in Chapter 17.

Click here to download the PDF.

Section I: Multiple Choice

1. An object has an altitude of 2 times the Earth’s radius, r_E, and experiences some force of gravity, F_g,0. If the object’s altitude is doubled, then the new force of gravity will be

(A) 4F_g,0

(B) F_g,0

(D) F_g,0

(E) F_g,0

2. At the surface of Earth, an object of mass m has weight w. If this object is transported to a height above the surface that’s twice the radius of Earth, then, at the new location,

(A) its mass is m/2 and its weight is w/2

(B) its mass is m and its weight is w/2

(D) its mass is m and its weight is w/4

(E) its mass is m and its weight is w/9

3. A moon of mass m orbits a planet of mass 100m. Let the strength of the gravitational force exerted by the planet on the moon be denoted by F₁, and let the strength of the gravitational force exerted by the moon on the planet be F₂. Which of the following is true?

(A) F₁ = 100F₂

(B) F₁ = 10F₂

(D) F₂ = 10F₁

(E) F₂ = 100F₁

4. Humans cannot survive for long periods under gravity more than 4 times what we experience on Earth. If a planet were discovered with the same mass as Earth, what is the smallest radius it could have (in terms of Earth’s radius, r_E) without being dangerous to humans?

(A) r_E

(B) r_E

(D) 2 r_E

(E) 4 r_E

5. Humans cannot survive for long periods under gravity more than 4 times what we experience on Earth. If a planet were discovered with the same mass as Earth, what is the smallest volume it could have (in terms of Earth’s volume, V_E), without being dangerous to humans? Assume that the planet is a sphere.

(A) V_E

(B) V_E

(D) 2 V_E

(E) 4 V_E

6. A moon of Jupiter has a nearly circular orbit of radius R and an orbit period of T. Which of the following expressions gives the mass of Jupiter?

(A) 2πR/T

(B) 4π²R/T ²

(D) 4πR ²/(GT ²)

(E) 4π²R ³/(GT ²)

7. Two large bodies, Body A of mass m and Body B of mass 4m, are separated by a distance R. At what distance from Body A, along the line joining the bodies, would the gravitational force on an object be equal to zero? (Ignore the presence of any other bodies.)

(A) R/16

(B) R/8

(D) R/4

(E) R/3

8. The mean distance from Saturn to the Sun is 9 times greater than the mean distance from Earth to the Sun. How long is a Saturn year?

(A) 18 Earth years

(B) 27 Earth years

(D) 243 Earth years

(E) 729 Earth years

9. The Moon has mass M and radius R. A small object is dropped from a distance of 3R from the Moon’s center. The object’s impact speed when it strikes the surface of the Moon is equal to for k =

(A)

(B)

(C)

(D)

(E)

10. Two satellites, A and B, orbit a planet in circular orbits having radii R_A and R_B, respectively, as shown above. If R_B = 3R_A, the velocities v_A and v_B of the two satellites are related by which of the following?

(A) v_B = v_A

(B) v_B = 3v_A

(D) v_B = v_A

(E) v_B =

Section II: Free Response

1. Consider two uniform spherical bodies in deep space. Sphere 1 has mass m₁ and Sphere 2 has mass m₂. Starting from rest from a distance R apart, they are gravitationally attracted to each other.

(a) Compute the acceleration of Sphere 1 when the spheres are a distance R/2 apart.

(b) Compute the acceleration of Sphere 2 when the spheres are a distance R/2 apart.

(d) Compute the speed of Sphere 2 when the spheres are a distance R/2 apart.

Now assume that these spheres orbit their center of mass with the same orbit period, T.

(e) Determine the radii of their orbits. Write your answer in terms of m₁, m₂, T, and fundamental constants.

2. A satellite of mass m is in the elliptical orbit shown below around Earth (radius r_E, mass M). Assume that m << M.

(a) Determine v₁, the speed of the satellite at perigee (the point of the orbit closest to Earth). Write your answer in terms of r₁, r₂, M, and G.

(b) Determine v₂, the speed of the satellite at apogee (the point of the orbit farthest from Earth). Write your answer in terms of r₁, r₂, M, and G.

(d) What is the satellite’s angular momentum (with respect to Earth’s center) when it’s at apogee?

(e) Determine the speed of the satellite when it’s at the point marked X in the figure.

(f) Determine the period of the satellite’s orbit. Write your answer in terms of r₁, r₂, M, and fundamental constants.

Summary

Newton’s Law of Gravitation

Newton’s Law of Gravitation gives the force between any two point masses, regardless of their mass or location. This uniform circular motion can also be described by an angular velocity and a centripetal acceleration.

Circular Orbits

F_g = F_c

= v²

v =

General Orbits

The gravitational potential energy is represented by the following formula:

Both mechanical energy and angular momentum are conserved for orbits.

E_total = 0

Systems are bound when the total mechanical energy is less than zero. This means that the gravitational potential energy is greater than the kinetic energy of the system.

For an object to escape, its kinetic energy must be greater than or equal to its gravitational potential energy. So

to escape the pull of M₂.

Gravity of Spheres and Shells

The gravity due to a spherical shell of mass, M, and radius, R, where the mass is at a distance, r, away from the center of the shell is represented by the equations below.

Outside the shell is and the gravity inside the shell is F_g = 0

The gravity due to a uniform, solid sphere where a small mass, m, is at a distance, r, away from the center of the sphere is represented by the equations below.

Outside the sphere is and the gravity inside the shell is .

The gravity of a spherically symmetric sphere of mass, M, and radius, R (that is, ρ(r)), is represented by the equation

where dM = ρ(r)dV

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 11 Oscillations

INTRODUCTION

In this chapter, we’ll concentrate on a kind of periodic motion that’s straight-forward and that, fortunately, actually describes many real-life systems. This type of motion is called simple harmonic motion. The prototypical example of simple harmonic motion is a block that’s oscillating on the end of a spring. What we learn about this simple system can be applied to many other oscillating systems.

SIMPLE HARMONIC MOTION (SHM): THE SPRING–BLOCK OSCILLATOR

When a spring is compressed or stretched from its natural length, a force is created. If the spring is displaced by x from its natural length, the force it exerts in response is given by the equation

This is known as Hooke’s law. The proportionality constant, k, is a positive number called the spring (or force) constant that indicates how stiff the spring is. The stiffer the spring, the greater the value of k. The minus sign in Hooke’s law tells us that F_S and x always point in opposite directions. For example, referring to the figure on the next page, when the spring is stretched (x is to the right), the spring pulls back (F is to the left); when the spring is compressed (x is to the left), the spring pushes outward (F is to the right). In all cases, the spring wants to return to its original length. As a result, the spring tries to restore the attached block to the equilibrium position, which is the position at which the net force on the block is zero. For this reason, we say that the spring provides a restoring force.

Example 1 A 12 cm-long spring has a force constant (k) of 400 N/m. How much force is required to stretch the spring to a length of 14 cm?

Solution. The displacement of the spring has a magnitude of 14 – 12 = 2 cm = 0.02 m so, according to Hooke’s law, the spring exerts a force of magnitude F = –kx = –(400 N/m)(0.02 m) = –8 N. This is the restoring force, so we’d have to exert a force of 8 N to keep the spring in this stretched state.

Springs that obey Hooke’s law (called ideal or linear springs) provide an ideal mechanism for defining the most important kind of vibrational motion: simple harmonic motion (SHM).

Consider a spring with force constant k, attached to a vertical wall, with a block of mass m on a frictionless table attached to the other end.

Grab the block, pull it some distance from its original position, and release it. The restoring force will pull the block back toward equilibrium. Of course, because of its inertia, the block will pass through the equilibrium position and compress the spring. At some point, the block will stop, and the compressed spring will push the block back. In other words, the block will oscillate.

During the oscillation, the force on the block is zero when the block is at equilibrium (the point we designate as x = 0). This is because Hooke’s law says that the strength of the spring’s restoring force is given by the equation F_S = –kx, so F_S = 0 at equilibrium. The acceleration of the block is also equal to zero at x = 0, since F_S = 0 at x = 0 and a = F_S/m. At the endpoints of the oscillation region, where the block’s displacement, x, has the greatest magnitude, the restoring force and the magnitude of the acceleration are both at their maximum. Only the direction of the restoring force is different.

Simple Harmonic Motion in Terms of Energy

Another way to describe the block’s motion is in terms of energy transfers. A stretched or compressed spring stores elastic potential energy, which is transformed into kinetic energy (and back again). This shuttling between potential and kinetic energy causes the oscillations. For a spring with spring constant k, the elastic potential energy it possesses—relative to its equilibrium position—is given by the equation

In terms of energy transfers, we can describe the block’s oscillations as follows: When you initially pull the block out, you increase the elastic potential energy of the system. Upon releasing the block, this potential energy turns into kinetic energy, and the block moves. As it passes through its equilibrium position, U_S = 0. Then, as the block continues through that position, the spring is compressed, and the kinetic energy is transformed back into elastic potential energy.

By Conservation of Mechanical Energy, the sum K + U_S is a constant. Therefore, when the block reaches the maximum displacement (that is, when x = ±x_max), U_S is maximized, so K must be minimized; in fact, K = 0 at the endpoints of the oscillation region. As the block is passing through equilibrium, x = 0, so U_S = 0 and K is maximized.

The maximum displacement from equilibrium is called the amplitude of oscillation and is denoted by A. So instead of writing x = x_max, we write x = A (and x = –x_max will be written as x = –A).

Example 2 A block of mass m = 0.05 kg oscillates on a spring whose force constant k is 500 N/m. The amplitude of the oscillations is 4.0 cm. Calculate the maximum speed of the block.

Solution. First, let’s get an expression for the maximum elastic potential energy of the system:

U_s = kx² ⇒ U_{S, max} = kx²_max = kA²

When all this energy has been transformed into kinetic energy—which occurs at the equilibrium position—the block will have maximum kinetic energy and maximum speed

Example 3 Show why U_s = kx² if F_S = −kx.

Solution. By definition, ∆U_S = –W_S, and, since F is not constant, the work done by F must be calculated using a definite integral. As the end of the spring moves from position x = x₁ to x = x₂, the work it does is equal to

Therefore,

ΔU_S = U_S2 − U_S1 = −W_S = k − k

So if we designate U_S₁ = 0 at x₁ = 0, the equation above yields U_s = kx², as desired. (Even if a spring does not obey Hooke’s law, this method can still be used to find the work done and the potential energy stored.)

Example 4 A block of mass m = 2.0 kg is attached to an ideal spring of force constant k = 500 N/m. The amplitude of the resulting oscillations is 8.0 cm. Determine the total energy of the oscillator and the speed of the block when it’s 4.0 cm from equilibrium.

Solution. The total energy of the oscillator is the sum of its kinetic and potential energies. By Conservation of Mechanical Energy, the sum K + U_S is a constant. When the block is at its amplitude position, x = 8 cm, and its kinetic energy is zero:

E = K + U_S = 0 + kA² = (500 N/m)(0.08 m)2 = 1.6 J

At any position x, we have

Therefore, at x = 4.0 cm,

Example 5 A block of mass m = 3.0 kg is attached to an ideal spring of force constant k = 500 N/m. The block is at rest at its equilibrium position. An impulsive force acts on the block, giving it an initial speed of 2.0 m/s. Find the amplitude of the resulting oscillations.

Solution. When the impulsive force acts on the block at its equilibrium position, U_i = 0. Then, when all of the kinetic energy has been transformed into potential energy, the block is at its maximum displacement from equilibrium and it’s at one of its amplitude positions. At this point K_f = 0, and

THE KINEMATICS OF SIMPLE HARMONIC MOTION

Now that we’ve explored the dynamics of the block’s oscillations in terms of force and energy, let’s talk about motion—or kinematics. As you watch the block oscillate, you should notice that it repeats each cycle of oscillation in the same amount of time. A cycle is a round-trip: for example, from position x = A over to x = –A and back again to x = A. The amount of time it takes to complete a cycle is called the period of the oscillations, or T. If T is short, the block is oscillating rapidly, and if T is long, the block is oscillating slowly.

Another way of indicating the rapidity of the oscillations is to count the number of cycles that can be completed in a given time interval: the greater the number of completed cycles, the more rapid the oscillations. The number of cycles that can be completed per unit time is called the frequency of the oscillations, or f, and is expressed in cycles per second. One cycle per second is one hertz (abbreviated Hz).

One of the most basic equations of oscillatory motion expresses the inverse relationship between period and frequency:

Therefore:

Example 6 A block oscillating on the end of a spring moves from its position of maximum spring stretch to maximum spring compression in 0.25 s. Determine the period and frequency of this motion.

Solution. The period is defined as the time required for one full cycle. Moving from one end of the oscillation region to the other is only half a cycle. Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s. Because frequency is the reciprocal of period, the frequency of the oscillations is f = 1/T = 1/(0.5 s) = 2 Hz.

Example 7 A student observing an oscillating block counts 45.5 cycles of oscillation in one minute. Determine its frequency (in hertz) and period (in seconds).

Solution. The frequency of the oscillations in hertz (which is the number of cycles per second) is

Therefore,

One of the defining properties of the spring–block oscillator is that the frequency and period can be determined from the mass of the block and the force constant of the spring. The equations are as follows:

Let’s analyze these equations. Suppose we had a small mass on a very stiff spring. Intuitively, we would expect that this strong spring would make the small mass oscillate rapidly, with a high frequency and a short period. Both of these predictions are substantiated by the equations above, because if m is small and k is large, then the ratio k/m is large (high frequency) and the ratio m/k is small (short period).

Example 8 A block of mass m = 2.0 kg is attached to a spring whose force constant, k, is 300 N/m. Calculate the frequency and period of the oscillations of this spring–block system.

Solution. According to the equations above,

Example 9 A block is attached to a spring and set into oscillatory motion, and its frequency is measured. If this block were removed and replaced by a second block with 1/4 the mass of the first block, how would the frequency of the oscillations compare to that of the first block?

Solution. Since the same spring is used, k remains the same. According to the equation given above, f is inversely proportional to the square root of the mass of the block: f ∝ 1/. Therefore, if m decreases by a factor of 4, then f increases by a factor of = 2.

The equations we saw above for the frequency and period of the spring–block oscillator do not contain A, the amplitude of the motion. In simple harmonic motion, both the frequency and the period are independent of the amplitude. The reason for this is that the strength of the restoring force is proportional to x, the displacement from equilibrium, as given by Hooke’s law: F_S = –kx.

Example 10 A student performs an experiment with a spring–block simple harmonic oscillator. In the first trial, the amplitude of the oscillations is 3.0 cm, while in the second trial (using the same spring and block), the amplitude of the oscillations is 6.0 cm. Compare the values of the period, frequency, and maximum speed of the block between these two trials.

Solution. If the system exhibits simple harmonic motion, then the period and frequency for a trial using the same spring and block are independent of amplitude. However, the maximum speed of the block will be greater in the second trial than in the first. Since the amplitude is greater in the second trial, the system possesses more total energy (E = kA²). So when the block is passing through equilibrium (its position of greatest speed), the second system has more kinetic energy, meaning that the block will have a greater speed. In fact, from Example 2, we know that v_max = so A is twice as great in the second trial than in the first, v_max will be twice as great in the second trial than in the first.

Example 11 For each of the following arrangements of two springs, determine the effective spring constant, k_eff. This is the force constant of a single spring that would produce the same force on the block as the pair of springs shown in each case.

(a)

(b)

(c)

(d) Determine k_eff in each of these cases if k₁ = k₂ = k.

Solution.

(a) Imagine that the block was displaced a distance x to the right of its equilibrium position. Then the force exerted by the first spring would be F₁ = –k₁x and the force exerted by the second spring would be F₂ = –k₂x. The net force exerted by the springs would be

F₁ + F₂ = –k₁x + –k₂x = –(k₁ + k₂)x

Since F_eff = −(k₁ + k₂)x, we see that k_eff = k₁ + k₂.

(b) Imagine that the block was displaced a distance x to the right of its equilibrium position. Then the force exerted by the first spring would be F₁ = −k₁x and the force exerted by the second spring would be F₂ = –k₂x. The net force exerted by the springs would be

F₁ + F₂ = –k₁x + −k₂x = −(k₁ + k₂)x

As in part (a), we see that since F_eff = −(k₁ + k₂)x, we get k_eff = k₁ + k₂.

(c) Imagine that the block was displaced a distance x to the right of its equilibrium position. Let x₁ be the distance that the first spring is stretched, and let x₂ be the distance that the second spring is stretched. Then x = x₁ + x₂. But x₁ = –F/k₁ and x₂ = –F/k₂, so

Therefore,

(d) If the two springs have the same force constant, that is, if k₁ = k₂ = k, then in the first two cases the pairs of springs are equivalent to one spring that has twice their force constant: k_eff = k₁ + k₂ = k + k = 2k. In (c), the pair of springs is equivalent to a single spring with half their force constant:

THE SPRING–BLOCK OSCILLATOR: VERTICAL MOTION

So far we’ve looked at a block sliding back and forth on a horizontal table, but the block could also oscillate vertically. In vertical oscillation, gravity causes the block to move downward to an equilibrium position where the spring is not at its natural length.

Consider a spring of negligible mass hanging from a stationary support. A block of mass m is attached to its end and allowed to come to rest, stretching the spring a distance d. At this point, the block is in equilibrium; the upward force of the spring is balanced by the downward force of gravity. Therefore,

Next, imagine that the block is pulled down a distance A and released. The restoring force is greater than the block’s weight, and, as a result, the block accelerates upward. As the block’s momentum carries it up through the equilibrium position, F_S is less than the block’s weight. As a result, the block decelerates, stops, and accelerates downward again, and the up-and-down motion repeats.

When the block is at a distance y below its equilibrium position, the spring is stretched a total distance of d + y, so the upward spring force is equal to k(d + y). The downward force stays the same, mg. The net force on the block is

F = k(d + y) – mg

but this equation becomes F = ky, because kd = mg (as we saw above). Since the resulting force on the block, F = ky, has the form of Hooke’s law, we know that the vertical simple harmonic oscillations of the block have the same characteristics as the horizontal oscillations. The equilibrium position, y = 0, is not at the spring’s natural length, but at the point where the hanging block is in equilibium.

Example 12 A block of mass m = 1.5 kg is attached to the end of a vertical spring of force constant k = 300 N/m. After the block comes to rest, it is pulled down a distance of 2.0 cm and released.

(a) What is the frequency of the resulting oscillations?

(b) What are the minimum and maximum amounts of stretch of the spring during the oscillations of the block?

Solution.

(a) The frequency is given by

(b) Before the block is pulled down, to begin the oscillations, it stretches the spring by a distance

Since the amplitude of the motion is 2 cm, the spring is stretched a maximum of 5 cm + 2 cm = 7 cm when the block is at the lowest position in its cycle, and a minimum of 5 cm – 2 cm = 3 cm when the block is at its highest position.

THE SINUSOIDAL DESCRIPTION OF SIMPLE HARMONIC MOTION

The position of the block during its oscillation can be written as a function of time. Take a look at the experimental setup below.

A small pen is attached to the oscillating block, and it makes a mark on the paper as the paper is pulled along by the roller on the right. Clearly, the simple harmonic motion of the block is sinusoidal.

The basic mathematical equation for describing simple harmonic motion is

y = A sin (ωt)

where y is the position of the oscillator, A is the amplitude, ω is the angular frequency (defined as 2πf, where f is the frequency of the oscillations), and t is time. Since sin(ωt) oscillates between –1 and +1, the quantity A sin(ωt) oscillates between –A and +A; this describes the oscillation region.

If t = 0, then the quantity A sin (ωt) is also equal to zero. This means that y = 0 at time t = 0. However, what if y ≠ 0 at time t = 0? For example, if the oscillator is pulled to one of its amplitude positions, say y = A, and released at time t = 0, then y = A at t = 0. To account for the fact that the oscillator can begin anywhere in the oscillation region, the basic equation for the position of the oscillator given above is generalized as follows:

y = A sin (ωt + ϕ₀)

where ϕ₀ is called the initial phase. The argument of the sine function, ωt + ϕ₀, is called the phase (or phase angle). By carefully choosing ϕ₀, we can be sure that the equation correctly specifies the oscillator’s position no matter where it may have been at time t = 0. The value of ϕ₀ can be calculated from the equation

Example 13 A simple harmonic oscillator has an amplitude of 3.0 cm and a frequency of 4.0 Hz. At time t = 0, its position is y = 3.0 cm. Where is it at time t = 0.3 s?

Solution. First, A = 3 cm and ω = 2πf = 2π(4.0 s⁻¹) = 8π s⁻¹. The value of the initial phase is

Therefore, the position of the oscillator at any time t is given by the equation

y = (3 cm) · sin [(8 πs⁻¹]t + π

So, at time t = 0.3 s, we find that y = (3 cm) · sin [(8 πs⁻¹) (0.3 s) + π] = 0.93. (Make sure your calculator is in radian mode when you evaluate this expression!)

Example 14 The position of a simple harmonic oscillator is given by the equation

y = (4 cm)· sin [(6 π s⁻¹)t − π]

(a) Where is the oscillator at time t = 0?

(b) What is the amplitude of the motion?

(d) What is the period?

Solution.

(a) To determine y at time t = 0, we simply substitute t = 0 into the given equation and evaluate:

γ_{at t=0} = (4 cm) · sin(−π) = −4 cm

(b) The amplitude of the motion, A, is the coefficient in front of the sin(ω t + ϕ₀) expression. In this case, we read directly from the given equation that A = 4 cm.

(c) The coefficient of t is the angular frequency, ω. In this case, we see that ω = 6π s⁻¹. Since ω = 2πf by definition, we have f = ω/(2π) = 3 Hz.

(d) Since T = 1/f, we calculate that T = 1/f = 1/(3 Hz) = 0.33 s.

Instantaneous Velocity and Acceleration

If the position of a simple harmonic oscillator is given by the equation y = A sin (ω t + ϕ₀), its velocity and acceleration can be found by differentiation:

v(t) = (t) = [Asin(ωt + ϕ₀)] = Aωcos(ωt + ϕ₀)

and

a(t) = (t) = [Aωcos(ωt + ϕ₀)] = −Aω²sin (ωt + ϕ₀)

Note that both the velocity and acceleration vary with time.

Differential Equation for Simple Harmonic Motion

The differential equation, = −ω²y has a solution of y(t) = A sin (ωt + ϕ₀). To check that this function, y(t), is a solution, substitute y(t) and the second derivative of y(t), which is a(t), into the equation, and see if both sides of the equation are the same.

= −ω²y

−Aω² sin(ωt + ϕ₀) = −ω²(A sin (ωt + ϕ₀))

check

−Aω² sin (ωt + ϕ₀) = −ω²(A sin(ωt + ϕ₀))

As you can see, the two sides of the equation are the same so y(t) = A sin (ωt + ϕ₀) is a solution to the differential equation = −ω²y. Therefore, if we can derive a differential equation that takes this form we will know the solution to the equation and that the object is undergoing SHM.

Let’s do this for a spring/mass system sliding along a frictionless, horizontal floor, as shown above.

This is the same basic form of the differential equation from above, and for the spring mass system:

ω = and x(t) = A sin(ωt)

Example 15 The position of a simple harmonic oscillator is given by the equation

y = 4 sin (6πt − π)

where y is in cm and t is in seconds. What are the maximum speed and maximum acceleration of the oscillator?

Solution. Since the maximum value of cos (ωt + ϕ₀) is 1, the maximum value of v is v_max = Aω ; similarly, since the minimum value of (ωt + ϕ₀) is –1, the maximum value of a is a_max = Aω². From the equation for y, we see that A = 4 cm = 0.04 m and ω = 6π s⁻¹. Therefore, v_max = Aω = (0.04 m)⋅(6π s⁻¹) = 0.75 m/s and a_max = Aω² = (0.04 m)(6π s⁻¹)² = 14 m/s².

PENDULA

A simple pendulum consists of a weight of mass m attached to a massless rod that swings, without friction, about the vertical equilibrium position. The restoring force is provided by gravity and, as the figure below shows, the magnitude of the restoring force when the bob is θ to an angle to the vertical is given by the equation:

F_restoring = mg sin θ

Although the displacement of the pendulum is measured by the angle that it makes with the vertical, rather than by its linear distance from the equilibrium position (as was the case for the spring–block oscillator), the simple pendulum shares many of the important features of the spring–block oscillator. For example,

• Displacement is zero at the equilibrium position.

• At the endpoints of the oscillation region (where θ = ±θ_max), the restoring force and the tangential acceleration (a_t) have their greatest magnitudes, the speed of the pendulum is zero, and the potential energy is maximized.

• As the pendulum passes through the equilibrium position, its kinetic energy and speed are maximized.

Despite these similarities, there is one important difference. Simple harmonic motion results from a restoring force that has a strength that’s proportional to the displacement. The magnitude of the restoring force on a pendulum is mg sin θ, which is not proportional to the displacement θ. Strictly speaking, then, the motion of a simple pendulum is not really simple harmonic. However, if θ is small, then sin θ ≈ θ (measured in radians). So, in this case, the magnitude of the restoring force is approximately mgθ, which is proportional to θ. So if θ_max is small, the motion can be treated as simple harmonic.

If the restoring force is given by mgθ, rather than mg sin θ, then the frequency and period of the oscillations depend only on the length of the pendulum and the value of the gravitational acceleration, according to the following equations:

Differential Equation for a Pendulum

We have derived a differential equation for a spring-mass system that has a solution of y(t) = A sin (ωt + ϕ_o). Now we will do the same for a pendulum with small amplitude oscillations, which allows us to approximate sin θ = θ. We will use s for the arc length displacement.

Consider a pendulum of length L and mass m undergoing small amplitude oscillations.

This is the same basic form of the differential equation for the spring-mass system.

Therefore, for a pendulum: ω = and θ(t) = θ_max sin(ωt + ϕ₀).

Note that neither frequency nor period depends on the amplitude (the maximum angular displacement, θ_max); this is a characteristic feature of simple harmonic motion. Also notice that neither depends on the mass of the weight.

Example 16 A simple pendulum has a period of 1 s on Earth. What would its period be on the Moon (where g is one-sixth of its value here)?

Solution. The equation T = 2π shows that T is inversely proportional to , so if g decreases by a factor of 6, then T increases by a factor of . That is,

T_{on Moon} = ×T_{on Earth} = (1 s) = 2.4 s

Chapter 11 Drill

The answers and explanations can be found in Chapter 17.

Click here to download the PDF.

Section I: Multiple Choice

1. Which of the following is/are characteristics of simple harmonic motion?

I. The acceleration is constant.

II. The restoring force is proportional to the displacement.

III. The frequency is independent of the amplitude.

(A) II only

(B) I and II only

(D) II and III only

(E) I, II, and III

2. What combination of values will result in a spring–block system with the greatest frequency?

(A) A = 50 cm; k = 80 N/m; m = 2 kg

(B) A = 50 cm; k = 100 N/m; m = 2 kg

(D) A = 75 cm; k = 80 N/m; m = 2 kg

(E) A = 75 cm; k = 100 N/m; m = 4 kg

3. A block attached to an ideal spring undergoes simple harmonic motion about its equilibrium position (x = 0) with amplitude A. What fraction of the total energy is in the form of kinetic energy when the block is at position x = A ?

(A)

(B)

(C)

(D)

(E)

4. A student measures the maximum speed of a block undergoing simple harmonic oscillations of amplitude A on the end of an ideal spring. If the block is replaced by one with twice the mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will

(A) decrease by a factor of 4

(B) decrease by a factor of 2

(D) remain the same

(E) increase by a factor of 2

5. A spring with a natural length of 40 cm and a spring constant of 400 N/m is hung vertically with a 10 kg mass attached to the end. Assuming the spring’s mass is negligible, what will be the final length of the spring when it reaches equilibrium?

(A) 25 cm

(B) 35 cm

(D) 50 cm

(E) 65 cm

6. A linear spring of force constant k is used in a physics lab experiment. A block of mass m is attached to the spring and the resulting frequency, f, of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If f² is plotted versus 1/m, the graph will be a straight line with slope

(A) 4π²/k²

(B) 4π²/k

(D) k/4π²

(E) k²/4π²

7. A block of mass m = 4 kg on a frictionless, horizontal table is attached to one end of a spring of force constant k = 400 N/m and undergoes simple harmonic oscillations about its equilibrium position (x = 0) with amplitude A = 6 cm. If the block is at x = 6 cm at time t = 0, then which of the following equations (with x in centimeters and t in seconds) gives the block’s position as a function of time?

(A) x = 6 sin(10t + π)

(B) x = 6 sin(10πt + π)

(D) x = 6 sin(10t)

(E) x = 6 sin(10t – π)

8. A block attached to an ideal spring undergoes simple harmonic motion about its equilibrium position with amplitude A and angular frequency ω. What is the maximum magnitude of the block’s velocity?

(A) Aω

(B) A²ω

(D) A/ω

(E) A/ω²

9. A simple pendulum swings about the vertical equilibrium position with a maximum angular displacement of 5° and period T. If the same pendulum is given a maximum angular displacement of 10°, then which of the following best gives the period of the oscillations?

(A) T/2

(B) T/

(D) T

(E) 2T

10. Which of the following best describes the relationship between the tension force, F_T, in the string of a pendulum and the component of gravity that pulls antiparallel to the tension, F_G? Assume that the pendulum is only displaced by a small amount.

(A) F_T > F_G

(B) F_T ≥ F_G

(D) F_T ≤ F_G

(E) F_T < F_G

Section II: Free Response

1. The figure below shows a block of mass m (Block 1) that’s attached to one end of an ideal spring of force constant k and natural length L. The block is pushed so that it compresses the spring to 3/4 of its natural length and then released from rest. Just as the spring has extended to its natural length L, the attached block collides with another block (also of mass m) at rest on the edge of the frictionless table. When Block 1 collides with Block 2, half of its kinetic energy is lost to heat; the other half of Block 1’s kinetic energy at impact is divided between Block 1 and Block 2. The collision sends Block 2 over the edge of the table, where it falls a vertical distance H, landing at a horizontal distance R from the edge.

(a) What is the acceleration of Block 1 at the moment it’s released from rest from its initial position? Write your answer in terms of k, L, and m.

(b) If v₁ is the velocity of Block 1 just before impact, show that the velocity of Block 1 just after impact is v₁.

(c) Determine the amplitude of the oscillations of Block 1 after Block 2 has left the table. Write your answer in terms of L only.

(d) Determine the period of the oscillations of Block 1 after the collision, writing your answer in terms of T₀, the period of the oscillations that Block 1 would have had if it did not collide with Block 2.

(e) Find an expression for R in terms of H, k, L, m, and g.

2. A bullet of mass m is fired horizontally with speed v into a block of mass M initially at rest, at the end of an ideal spring on a frictionless table. At the moment the bullet hits, the spring is at its natural length, L. The bullet becomes embedded in the block, and simple harmonic oscillations result.

(a) Determine the speed of the block immediately after the impact by the bullet.

(b) Determine the amplitude of the resulting oscillations of the block.

(d) Derive an equation which gives the position of the block as a function of time (relative to x = 0 at time t = 0).

3. A block of mass M oscillates with amplitude A on a frictionless horizontal table, connected to an ideal spring of force constant k. The period of its oscillations is T. At the moment when the block is at position x = A and moving to the right, a ball of clay of mass m dropped from above lands on the block.

(a) What is the velocity of the block just before the clay hits?

(b) What is the velocity of the block just after the clay hits?

(d) What is the new amplitude of the oscillations? Write your answer in terms of A, k, M, and m.

(e) Would the answer to part (c) be different if the clay had landed on the block when it was at a different position? Support your answer briefly.

(f) Would the answer to part (d) be different if the clay had landed on the block when it was at a different position? Support your answer briefly.

4. An object of total mass M is allowed to swing around a fixed suspension point P. The object’s moment of inertia with respect to the rotation axis perpendicular to the page through P is denoted by I. The distance between P and the object’s center of mass C, is d.

(a) Compute the torque τ produced by the weight of the object when the line PC makes an angle θ with the vertical. (Take the counterclockwise direction as positive for both θ and τ.)

(b) If θ is small, so that sin θ may be replaced by θ, write the restoring torque τ computed in part (a) in the form τ = –kθ.

A simple harmonic oscillator whose displacement from equilibrium, z, satisfies an equation of the form has a period of oscillation given by the formula T = .

(c) Setting z equal to θ in the equation above, use the result of part (b) to derive an expression for the period of small oscillations of the object shown above.

(d) Answer the question posed in part (c) if the object were a uniform bar of mass M and length L (whose moment of inertia about one of its ends is given by the equation I = ML².)

Summary

Simple Harmonic Motion

Simple harmonic motion will occur when there is a restoring force on an object that is proportional to the displacement from equilibrium.

The mathematical equation of an object undergoing SHM is y(t) = A sin(ωt + ϕ).

This equation is a solution to the differential equation for SHM: = −ω²y.

The mathematical equation for angular frequency is ω = 2πf.

The period is the length of time it takes the object to complete one cycle:

The period can be expressed as an equation proportional to the angular frequency:

The frequency is the number of cycles the object completes in one unit of time:

An object undergoing SHM typically transforms potential energy to kinetic energy and back to potential energy in a repeating cycle.

The two examples of SHM that we have studied thus far are a mass oscillating on a spring and a simple pendulum.

REFLECT

Respond to the following questions:

• For what content topics discussed in this chapter do you feel you have achieved sufficient mastery to answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you have achieved sufficient mastery to discuss effectively in a free response?

• For which content topics discussed in this chapter do you feel you need more work before you can answer multiple-choice questions correctly?

• For which content topics discussed in this chapter do you feel you need more work before you can discuss effectively in a free response?

• What parts of this chapter are you going to re-review?

• Will you seek further help, outside this book (such as a teacher, tutor, or AP Students), on any of the content in this chapter—and if so, on what content?

Chapter 12 Electric Forces and Fields

INTRODUCTION

The basic components of atoms are protons, neutrons, and electrons. Protons and neutrons form the nucleus (and are referred to collectively as nucleons), while the electrons occupy regions of space outside of the nucleus. Most of an atom consists of empty space. In fact, if a nucleus were the size of the period at the end of this sentence, then the nearest electrons would be 5 meters away. So what holds such an apparently tenuous structure together? One of the most powerful forces in nature: the electromagnetic force. Protons and electrons are characterized by their electric charge that gives them an attractive force. Electric charge can be either positive, negative, or neutral. A positive particle always attracts a negative particle, and particles of the same charge always repel each other. Protons are positively charged, and electrons are negatively charged.

Protons and electrons are intrinsically charged, but bulk matter is not. Since neutral atoms contain an equal number of protons and electrons, their overall electric charge is 0, because the negative charges cancel out the positive charges. Therefore, in order for matter to be charged, an imbalance between the numbers of protons and electrons must exist. This can be accomplished by either the removal or addition of electrons (that is, by the ionization of some of the object’s atoms). If you remove electrons, then the object becomes positively charged, while if you add electrons, then it becomes negatively charged. Furthermore, charge is conserved. For example, if you rub a glass rod with a piece of silk, then the silk will acquire a negative charge and the glass will be left with an equal positive charge. Net charge cannot be created or destroyed. (Charge can be created or destroyed—it happens all the time—but net charge cannot.)

The magnitude of charge on an electron (and therefore on a proton) is denoted by e. This stands for elementary charge because it’s the basic unit of electric charge. The charge of an ionized atom must be a whole number times e because charge can be added or subtracted only in units of e. For this reason we say that charge is quantized. To remind us of the quantized nature of electric charge, the charge of a particle (or object) is denoted by the letter q. In the SI system of units, charge is expressed in coulombs (abbreviated C). One coulomb is a tremendous amount of charge; the value of e is about 1.6 × 10⁻¹⁹ C.

COULOMB’S LAW

The electric force between two charged particles obeys a law that is very similar to that describing the gravitational force between two masses: they are both inverse-square laws. The electric force between two particles with charges of q₁ and q₂, separated by a distance r, is given by the equation

This is Coulomb’s law. We interpret a negative F_E as an attraction between the charges (q₁ and q₂ have opposite signs) and a positive F_E as a repulsion. The value of the proportionality constant, k, called Coulomb’s constant, is approximately 9 × 10⁹ N · m²/C². For reasons that will become clear later in this chapter, k is usually written in terms of a fundamental constant known as the permittivity of free space, denoted ε₀, whose numerical value is approximately 8.85 × 10⁻¹² C²/N · m². The equation that gives k in terms of ε₀ is:

Coulomb’s law for the force between two point charges is then written as

Let’s compare the electric force to the gravitational force between two protons that are a distance r apart.

Cracking the AP Physics C Exam 2018

ACKNOWLEDGMENTS

Contents

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Part I

Using This Book to Improve Your AP Score

PREVIEW: YOUR KNOWLEDGE, YOUR EXPECTATIONS

YOUR GUIDE TO USING THIS BOOK

HOW TO BEGIN

Part II

Practice Test 1

Practice Test 1

Practice Test 1: Answers and Explanations

PRACTICE TEST 1 ANSWER KEY

Mechanics

Electricity and Magnetism

SECTION I, MECHANICS

SECTION II, MECHANICS

SECTION I, ELECTRICITY AND MAGNETISM

SECTION II, ELECTRICITY AND MAGNETISM

Part III

About the AP Physics C Exam

THE STRUCTURE OF THE AP PHYSICS C EXAM

HOW THE AP PHYSICS C EXAM IS SCORED

OVERVIEW OF CONTENT TOPICS

HOW AP EXAMS ARE USED

OTHER RESOURCES

DESIGNING YOUR STUDY PLAN

Part IV

Test-Taking Strategies for the AP Physics C Exam

PREVIEW

HOW TO USE THE CHAPTERS IN THIS PART

Chapter 1

How to Approach Multiple-Choice Questions

MULTIPLE-CHOICE QUESTIONS ON THE AP PHYSICS C EXAM

Use the Answer Sheet

Should You Guess?

Use the Two-Pass System

General Advice

Chapter 2

How to Approach Free-Response Questions

FREE-RESPONSE QUESTIONS ON THE AP PHYSICS C EXAM

Clearly Explain and Justify Your Answers

Use Only the Space You Need

Complete the Whole Question

Use Common Sense

Think Like a Grader

Think Before You Write

Chapter 3

Using Time Effectively to Maximize Points

BECOMING A BETTER TEST TAKER

PACING AND TIMING

GETTING THE SCORE YOU WANT

TEST ANXIETY

REFLECT

Part V

Content Review for the AP Physics C Exam

Chapter 4

Vectors

INTRODUCTION

DEFINITION

VECTOR ADDITION (GEOMETRIC)

SCALAR MULTIPLICATION

VECTOR SUBTRACTION (GEOMETRIC)

STANDARD BASIS VECTORS

VECTOR OPERATIONS USING COMPONENTS

MAGNITUDE OF A VECTOR

DIRECTION OF A VECTOR

THE DOT PRODUCT

THE CROSS PRODUCT

REFLECT

Chapter 5

Kinematics

INTRODUCTION

POSITION, DISTANCE, AND DISPLACEMENT

SPEED AND VELOCITY

ACCELERATION

UNIFORMLY ACCELERATED MOTION AND THE BIG FIVE

KINEMATICS WITH GRAPHS