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v4.1

a

Acknowledgments

The Princeton Review would like to give a very special thanks to Nick Owen, Jenkang Tao, Felicia Tam, for their work on revising existing practice tests and developing the three additional practice tests in the new Premium edition of this book. In addition, The Princeton Review would like to thank top notch physics teacher TJ Smolka for his content review and input on this edition.

Contents

Cover

Title Page

Copyright

Acknowledgments

Register Your Book Online!

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 1 Exam

The Structure of the AP Physics 1 Exam

A Quick Word About Equations

How AP Exams Are Used

Other Resources

Designing Your Study Plan

Part IV: Test-Taking Strategies for the AP Physics 1 Exam

Preview

1 How to Approach Multiple-Choice Questions

2 How to Approach Free-Response Questions

Reflect

Part V: Content Review for the AP Physics 1 Exam

3 Vectors

Definition

Scalar Multiplication

Vector Subtraction

Two-Dimensional Vectors

Vector Operations Using Components

Magnitude of a Vector

Direction of a Vector

Chapter 3 Review Questions

Summary

4 Kinematics

Position

Displacement

Looking at Distance vs. Time Graphs

Speed and Velocity

Acceleration

Uniformly Accelerated Motion and the Big Five

Additional Kinematic Graphical Aspects

Qualitative Graphing

Free Fall

Projectile Motion

Chapter 4 Review Questions

Summary

5 Newton’s Laws

Introduction to Forces

Newton’s First Law

Newton’s Second Law

Newton’s Third Law

Newton’s Laws: A Summary

Weight

Normal Force

An Overall Strategy

Friction

Pulleys

Inclined Planes

Chapter 5 Review Questions

Summary

6 Work, Energy, and Power

Energy: An Overview

Work

Work at an Angle

Work Done by a Variable Force

Kinetic Energy

The Work–Energy Theorem

Potential Energy

Conservation of Mechanical Energy

Conservation of Energy with Nonconservative Forces

Power

Chapter 6 Review Questions

Summary

7 Linear Momentum

Introduction to Momentum

What is Momentum?

Impulse

Conservation of Linear Momentum

Collisions

Chapter 7 Review Questions

Summary

8 Uniform Circular Motion, Newton’s Law of Gravitation, and Rotational Motion

Uniform Circular Motion

Newton’s Law of Gravitation

Banking

Rotational Motion

Center of Mass

Torque

Equilibrium

Rotational Inertia

Chapter 8 Review Questions

Summary

9 Oscillations

Simple Harmonic Motion

The Spring-Block Oscillator: Vertical Motion

Pendulums

Chapter 9 Review Questions

Summary

10 Waves

Transverse Traveling Waves

Wave Speed on a Stretched String

Superposition of Waves

Standing Waves

Sound Waves

Resonance for Sound Waves

The Doppler Effect

Chapter 10 Review Questions

Summary

11 Electric Forces and Fields

Electric Charge

Coulomb’s Law

The Electric Field

Chapter 11 Review Questions

Summary

12 Direct Current Circuits

Electric Current

Resistance

Voltage

Electric Circuits

Circuit Analysis

Chapter 12 Review Questions

Summary

13 Answers and Explanations to the Chapter Review Questions

Part VI: More Practice Tests

Practice Test 2

Practice Test 2: Answers and Explanations

Practice Test 3

Practice Test 3: Answers and Explanations

Practice Test 4

Practice Test 4: Answers and Explanations

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Part I

Using This Book to Improve Your AP Score

PREVIEW: YOUR KNOWLEDGE, YOUR EXPECTATIONS

Your route to a high score on the AP Physics 1 Exam depends a lot on how you plan to use this book. Respond to the following questions.

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 1 Exam.

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.

HOW TO BEGIN

Part II

Practice Test 1

Practice Test 1

Click here to download the PDF version.

The Exam

AP^® Physics 1 Exam

SECTION I: Multiple-Choice Questions

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

Instructions

Section I of this examination contains 50 multiple-choice questions. Fill in only the ovals for numbers 1 through 50 on your answer sheet.

CALCULATORS MAY BE USED ON BOTH SECTIONS OF THE AP PHYSICS 1 EXAM.

Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. Please note that there are two types of multiple-choice questions: single-select and multi-select questions. After you have decided which of the suggested answers is best, completely fill in the corresponding oval(s) on the answer sheet. For single-select, you must give only one answer; for multi-select you must give BOTH answers in order to earn credit. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer.

Sample Question

Chicago is a

(A) state

(B) city

(C) country

(D) continent

Sample Answer

Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all the multiple-choice questions.

About Guessing

Many candidates wonder whether or not to guess the answers to questions about which they are not certain. Multiple-choice scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers, and no points are awarded for unanswered questions. Because points are not deducted for incorrect answers, you are encouraged to answer all multiple-choice questions. On any questions you do not know the answer to, you should eliminate as many choices as you can, and then select the best answer among the remaining choices.

AP PHYSICS 1

SECTION I

Note: To simplify calculations, you may use g = 10 m/s² in all problems.

Directions: Each of the questions or incomplete statements is followed by four suggested answers or completions. Select the one that is best in each case and then fill in the corresponding circle on the answer sheet.

Questions 15–17 all refer to the below circuit diagram.

Questions 36–37 refer to the graph above.

Directions: For each of the questions 46-50, two of the suggested answers will be correct. Select the two answers that are best in each case, and then fill in both of the corresponding circles on the answer sheet.

END OF SECTION I

DO NOT CONTINUE UNTIL INSTRUCTED TO DO SO.

AP PHYSICS 1

SECTION II

Free-Response Questions

Time—90 minutes

Percent of total grade—50

General Instructions

Use a separate piece of paper to answer these questions. Show your work. Be sure to write CLEARLY and LEGIBLY. If you make an error, you may save time by crossing it out rather than trying to erase it.

AP PHYSICS 1

SECTION II

Directions: Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.

STOP

Practice Test 1: Answers and Explanations

ANSWER KEY AP PHYSICS 1 TEST 1

SECTION I

SECTION II

Part III

About the AP Physics 1 Exam

THE STRUCTURE OF THE AP PHYSICS 1 EXAM

The AP Physics 1 Exam consists of two sections: a multiple-choice section and a free-response section. The multiple-choice section consists of two question types. Single-select questions are each followed by four possible responses, only one of which is correct. Multi-select questions are a new addition to the AP Physics 1 Exam, and require two of the listed answer choices to be selected to answer the question correctly. There are five multi-select questions that always appear at the end of the multiple-choice section.

The free-response section consists of five multi-part questions, which require you to write out your solutions, showing your work. The total amount of time for this section is 90 minutes. Unlike the multiple-choice section, which is scored by a computer, the free-response section is graded by high school and college teachers. They have guidelines for awarding partial credit, so you may still receive partial points should you not correctly respond to every part of the question. You are allowed to use a calculator on the entire AP Physics 1 Exam—including both the multiple-choice and free-response sections. Scientific or graphing calculators may be used, provided that they don’t have any unapproved features or capabilities (a list of approved graphing calculators is available on the College Board’s website). In addition, a table of equations commonly used in physics will be provided to you at the exam site. This can be found online and we’ve included it with both practice tests.

Grades on the AP Physics 1 Exam are reported as a number: either 1, 2, 3, 4, or 5. Here is a description of each of these five numerical scores plus data on how students scored on the May 2016 test:

Score (Meaning)	Percentage of Test Takers	Equivalent Grade in a first-year college course
5 (extremely qualified)	4.6%	A
4 (well qualified)	14%	A–, B+, B
3 (qualified)	21.2%	B–, C+, C
2 (possibly qualified)	30.2%	C-
1 (no recommendation)	30%	D

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 each grade? Each test is curved so scores vary from year to year, but as we see above, in May 2016 around 60% of test takers earned scores of 1 or 2, so you’ll want to study hard and prepare for this very difficult exam. We’ll get to that soon, we promise!

So, what is on the exam and how do you prepare? Take a look at the following list of the major topics covered on the AP Physics 1 Exam.

AP Physics 1: Algebra-Based Course Content

You may be using this book as a supplementary text as you take an AP Physics 1 course at your high school or you may be using it on your own. The College Board is very detailed in what they require your AP teacher to cover in his or her AP Physics 1 course. They explain that you should be familiar with the following topics:

The College Board also maps out the areas that you should study as “Big Ideas” and those are as follows:

Naturally, it’s important to be familiar with the topics—to understand the fundamentals of the theory, to define fundamental quantities, and to recognize and be able to use the equations. Then, you must practice at applying what you’ve learned to answering questions like those you will see on the exam. This book is designed to review all of the content areas covered on the exam. Also, each chapter is followed by practice multiple-choice and free-response questions, and perhaps even more important, answers and explanations 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.

A QUICK WORD ABOUT EQUATIONS

As you know, you will be given an Equation Sheet on the day of your exam and that Equation Sheet is yours for the duration of the test. You will see many equations in a different forms and it can be hard to recall what is equivalent to what, so we have made a handy chart for you. Here is a tally of a few equations that are equivalent to other equations and where you can find them in this book and on your Equation Sheet.

Equation(s)	Page	Location on AP Physics 1 Equations Sheet
Big Five #2, #3, #5	90	First three equations in top left.
Newton’s Second Law	111	Fourth equation below the first three.
Force of Friction	121	Fifth equation from top.
Power	149	P = ∆E/∆t is the equation on the sheet. Note specifically that the change in energy is the work done, so these two equations are equivalent.
Gravitational Force	176	Third equation from the bottom of the second column. Don’t be thrown by the use of absolute value bars; they mean the same 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 categories below represent the main ways in which AP Exam scores can be used.

OTHER RESOURCES

There are many resources available to help you improve your score on the AP Physics 1 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 1 and ask if the teacher will review your essays or otherwise help you with content.

Another wonderful resource is AP Central, the official site of the AP Exams. The scope of the information at this site is quite broad and includes

The AP Central home page address is: http://apcentral.collegeboard.com.

Finally, The Princeton Review offers tutoring for the AP Physics 1 Exam. 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

In Part I, you identified some areas for potential improvement. Let’s now delve further into your performance on 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:

Part IV

Test-Taking Strategies for the AP Physics 1 Exam

PREVIEW

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

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 think critically about the ways you can modify your approach.

A QUICK NOTE ABOUT NAMING CONVENTIONS

Each classroom uses different conventions for naming things. Here in this book we are using using the AP conventions set by the College Board, but please note that your AP Physics 1 course teacher may or may not decide to norm to the test the same way.

For example, v₀ and v_i are usually the same thing but not always the same thing. Some teachers and books may treat these as the same, though. These mean velocity at time zero and velocity initial, respectively.

Take a look at this example:

A car starts from rest and accelerates to 20 m/s in 5 seconds. The car holds this velocity for 10 seconds. the driver then speeds up to 28 m/s in 5 more seconds. What is the magnitude of the second acceleration?

v₀ = 0 miles per hour but v_i in this instance will be 20 m/s because the problem is looking at the second acceleration. Be careful not to confuse v₀ and v_i. This may seem silly, but it’s a common error so we wanted to address it right up front. Now, onto the good stuff!

Chapter 1

How to Approach Multiple-Choice Questions

CRACKING THE MULTIPLE-CHOICE SECTION

All the multiple-choice questions will have a similar format: Each will be followed by four answer choices. At times, it may seem that there could be more than one possible correct answer. With the exception of the five multi-select questions, there is only one! Answers resulting from common mistakes are often included in the four answer choices to trap you.

The Two-Pass System

The AP Physics 1 Exam covers a broad range of topics. There’s no way, even with our extensive review, that you will know everything about every topic in algebra-based physics. So, what should you do?

Adopt a two-pass system. The two-pass system entails going through the test and answering the easy questions first. Save the more time-consuming questions for later. (Don’t worry—you’ll have time to do them later!) First, read the question and decide if it a “now” or “later” question. If you decide this is a “now” question, answer it in the test booklet. If it is a “later” question, come back to it. Once you have finished all the “now” questions on a double page, transfer the answers to your bubble sheet. Flip the page and repeat the process.

Once you’ve finished all the “now” questions, move on to the “later” questions. Start with the easier questions first. These are the ones that require calculations or that require you to eliminate the answer choices (in essence, the correct answer does not jump out at you immediately). Transfer your answers to your bubble sheet as soon as you answer these “later” questions.

Watch Out for Those Bubbles!

Because you’re skipping problems, you need to keep careful track of the bubbles on your answer sheet. One way to accomplish this is by answering all the questions on a page and then transferring your choices to the answer sheet. If you prefer to enter them one by one, make sure you double-check the number beside the ovals before filling them in. We’d hate to see you lose points because you forgot to skip a bubble!

Process of Elimination (POE)

On most tests, you need to know your material backward and forward in order to get the right answer. In other words, if you don’t know the answer beforehand, you probably won’t answer the question correctly. This is particularly true of fill-in-the-blank and essay questions. We’re taught to think that the only way to get a question right is by knowing the answer. However, that’s not the case on Section I of the AP Physics 1 Exam. You can get a perfect score on this portion of the test without knowing a single right answer—provided you know all the wrong answers!

What are we talking about? This is perhaps the most important technique to use on the multiple-choice section of the exam. Let’s take a look at an example that probably looks quite familiar:

Now, if this were a fill-in-the-blank-style question, you might be in a heap of trouble. But let’s take a look at what we’ve got and how we can eliminate answer choices in order to reach the correct answer.

Remember that the electric field for any given particle will point toward a negative charge and away from a positive charge. Anywhere between the two charges here (regardless of x-coordinate) would have an electric field that pointed up. The negative charge would pull it up, and the positive one would push it up. This eliminates (A) and (B). Finally, the negative charge is greater magnitude. Therefore, the fields could only be in balance somewhere closer to the positive charge. This eliminates (C). So the correct answer must be (D).

We think we’ve illustrated our point: Process of Elimination is the best way to approach the multiple-choice questions. Even when you don’t know the answer right off the bat, you’ll surely know that two or three of the answer choices are not correct. What then?

Aggressive Guessing

As mentioned earlier, you are scored only on the number of questions you get right, so we know guessing can’t hurt you. But can it help you? It sure can. Let’s say you guess on four questions; odds are you’ll get one right. So you’ve already increased your score by one point. Now, let’s add POE into the equation. If you can eliminate as many as two answer choices from each question, your chances of getting them right increase, and so does your overall score. Remember, don’t leave any bubbles blank on test day!

General Advice

Answering 50 multiple-choice questions in 90 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 three wrong answers and only one correct one (and your odds are even better with the multi-select questions!), 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 under the same conditions that you will encounter during the real exam.

Reflect

Respond to the following questions:

Chapter 2

How to Approach Free-Response Questions

FREE-RESPONSE SECTION

On the free-response section, be sure to show the graders what you’re thinking. Write clearly—that is very important—and show your steps. If you make a mistake in one part and carry an incorrect result to a later part of the question, you can still earn valuable points if your method is correct. However, the graders cannot give you credit for work they can’t follow or can’t read. And, where appropriate, be sure to include units in your final answers.

Free-Response Terms Defined

You will be asked to both describe and explain natural phenomena. Both terms require the ability to demonstrate an understanding of physics principles by providing an accurate and coherent description or explanation. You will also be asked to justify a previously given answer. A justification is an argument supported by evidence. Evidence may consist of statements of physical principles, equations, calculations, data, graphs, and diagrams as appropriate. The argument, or equation used to support justifications and explanations, may in some cases refer to fundamental ideas or relations in physics, such as Newton’s laws, conservation of energy, or Coulomb’s Law. In other cases, the justification or explanation may take the form of analyzing the behavior of an equation for large or small values of a variable in the equation.

Calculate means that a student is expected to show work leading to a final answer, which may be algebraic but more often is numerical. Derive is more specific and indicates that you need to begin your solutions with one or more fundamental equations, such as those given on the AP Physics 1 Exam equation sheet. The final answer, usually algebraic, is then obtained through the appropriate use of mathematics. What is and determine are indicators that work need not necessarily be explicitly shown to obtain full credit. Showing work leading to answers, however, is a good idea, as you can earn partial credit in the case of an incorrect answer. Strict rules regarding significant digits are usually not applied to the scoring of numerical answers. However, in some cases, answers containing too many digits may be penalized. In general, two to four significant digits are acceptable. Exceptions to these guidelines usually occur when rounding makes a difference in obtaining a reasonable answer.

The words sketch and plot relate to producing graphs. “Sketch” means to draw a graph that illustrates key trends in a particular relationship, such as slope, curvature intercept(s), or asymptote(s). Numerical scaling or specific data are not required in a sketch. “Plot” means to draw the data points given in the problem on the grid provided, either using the given scale or indicating the scale and units when none are provided.

Exam questions that require the drawing of free-body diagrams or force diagrams will direct the students to draw and label the forces (not components) that act on the [object], where [object] is replaced by a reference specific to the question, such as “the car when it reaches the top of the hill.” Any components that are included in the diagram will be scored in the same way as incorrect or extraneous forces. In addition, in any subsequent part asking for a solution that would typically make use of the diagram, the following will be included: “If you need to draw anything other than what you have shown in part [x] to assist in your solutions, use the space below. Do NOT add anything to the figure in part [x].” This will give you the opportunity to construct a working diagram showing any components that are appropriate to the solution of the problem. This second diagram will not be scored.

Some questions will require students to design an experiment or outline a procedure that investigates a specific phenomenon or would answer a guiding question. You are expected to provide an orderly sequence of statements that specifies the necessary steps in the investigation needed to reasonably answer the questions or investigate the phenomenon.

CRACKING THE FREE-RESPONSE SECTION

Section II is worth 50 percent of your grade on the AP Physics 1 Exam. This section is composed of five free-response questions. You’re given a total of 90 minutes for this section. It is recommended that you spend the first 60 minutes on the three short answer questions, and the next 30 minutes on the experimental design and qualitative/quantitative translation questions.

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 1. 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.

Read 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 one free-response question that asked students to compute the weight of a newborn baby on the Moon, 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.

Don’t Forget Units!

In your free-response questions, you’ll be doing calculations right there by hand. Once you get to your final answer, DO NOT FORGET to include units or dimensions in it.

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 a comprehensive knowledge of physics and communicating that knowledge. Again, responses should be written as clearly as possible in complete sentences. You don’t need to show all the steps of a calculation, but you must explain how you got your answer and why you chose the technique you used.

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.

REFLECT

Respond to the following questions:

Part V

Content Review for the AP Physics 1 Exam

Chapter 3

Vectors

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. 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, A,

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

Displacement (which is net distance traveled including direction) is an example of a vector:

Vectors obey the Commutative Law for Addition, which states:

The vector sum A + B means the vector A followed by B. The vector sum of B + A means the vector B followed by A, and the result is an identical vector to A + B. Vectors are always added tail to end to find their sum, so A + B or B + A—both are examples of tail to end.

Two vectors are equal when they have the same magnitude and the same direction.

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:

Solution.

VECTOR SUBTRACTION

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)

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.

TWO-DIMENSIONAL VECTORS

Two-dimensional vectors are vectors that lie flat in a plane and can be written as the sum of a horizontal vector and a 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 î and ĵ.

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

The vectors B and C are called the vector components of A, and the scalar multiples of î and ĵ 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 A_xî + A_y ĵ, 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. In general, any vector in a plane can be described in this manner.

VECTOR OPERATIONS USING COMPONENTS

The use of components makes the vector operations of addition, subtraction, and scalar multiplication pretty straightforward:

Vector addition: Add the respective components.

A + B = (A_x + B_x)î + (A_y + B_y)ĵ

Vector subtraction: Subtract the respective components.

A – B = (A_x – B_x)î + (A_y – B_y)ĵ

Scalar multiplication: Multiply each component by k.

kA = (kA_x)î + (kA_y)ĵ

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

A + B = (2 – 4)î + (–3 + 2)ĵ = –2î – ĵ

A – B = [2 – (–4)]î + (–3 –2)ĵ = 6î – 5ĵ

2A = 2(2)î + 2(–3)ĵ = 4î – 6ĵ

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

MAGNITUDE OF A VECTOR

The magnitude of a vector can be computed with the Pythagorean Theorem. The magnitude of vector A can be denoted in several ways: A or |A| or ||A||. In terms of its components, the magnitude of A = A_xî + A_yĵ is given by the equation

which is the formula for the length of the hypotenuse of a right triangle with sides of lengths A_x and A_y.

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 (and an inverse trig function) to determine the angle. For example, if θ denotes the angle that the vector A = 3î + 4ĵ makes with the +x-axis, then tan θ = 4/3, so θ = tan⁻¹(4/3) = 53.1°.

In general, the axis that θ is made to is known as the adjacent axis. The adjacent component is always going to get the cos θ. For example, 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:

Chapter 3 Review Questions

Answers and Explanations can be found in Chapter 13.

Section I: Multiple-Choice

Section II: Free-Response

Summary

Chapter 4

Kinematics

 

And yet it moves.

 

—Galileo Galilei

Galileo, the “father of modern physics,” stated a distinction between the cause of motion and the description of motion. Kinematics is that modern description of motion that answers questions such as:

How far does this object travel?

How fast and in what direction does it move?

At what rate does its speed change?

Kinematics are the mathematical tools for describing motion in terms of displacement, velocity, and acceleration.

POSITION

An object’s position is its location in a certain space. Since it is difficult to describe an object’s location, in mathematics, we typically use a coordinate system to show where an object is located. Using a coordinate system with an origin also helps to determine positive and negative positions. Typically we set the object in question as the origin and relate its surroundings to the object.

DISPLACEMENT

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 displacement is only the distance from an object’s final position and initial position. The net distance takes into account the total path taken (meaning if an object went backward and then forward, those distances are included in the total distance). Since a distance is being measured, the SI unit for displacement is the meter [∆s] = m.

Solution. Displacement only refers to the object’s initial position and final position, not the details of its journey. Since the rock started on the edge of the cliff and ended up on the ground 30 m below, its displacement is 30 m, downward. Its distance traveled is 50 m; 10 m on the way up and 40 m on the way down.

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

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

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

The total distance covered is 500 m, but the net distance—the displacement—is 0.

A Note About Notation

∆s is a more general term that works in any direction in space. The term x or “∆x = x_f – x_i” has a specific meaning that is defined in the x direction. However, to be consistent with AP notation and to avoid confusion between spacial location and speed, from this point on we will use x in our development of the concepts of speed, velocity, and acceleration. x is just a reference and can refer to any direction in terms of the object’s motion (for example straight down or straight up).

LOOKING AT DISTANCE VS. TIME GRAPHS

You should also be able to handle kinematics questions in which information is given graphically. One of the popular graphs in kinematics is the position-versus-time graph. 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-versus-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 first second, its position changed from x = 0 to x = 10 m. Then, at time t = 1 s to 3 s, it stopped (that is, it stayed 10 m away from wherever it started). From t = 3 s to t = 6 s, it reversed direction, reaching x = 0 at time t = 5 s, and continued, reaching position x = –5 m at time t = 6 s.

Solution. Displacement is just the final position subtracted from the initial position. In this case, that’s –5 – 0 = –5 m (or 5 meters to the left, or negative direction).

What is the total distance traveled?

10 m  +  15 m  = 25 meters

Segment #1  Segment #3

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:

The car’s speedometer doesn’t care what direction the car is moving. 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. Remember that speed is scalar, so it does not take into account the direction.

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

Note that the bar over the v means average. Because ∆x is a vector, is also a vector, and because ∆t is a positive scalar, the direction of is the same as the direction of ∆x. 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. Velocity is speed plus direction. An important note: The magnitude of the velocity is the speed. The magnitude of the average velocity is not called the average speed (as we will see in these next two examples).

Solution. Average speed is total distance divided by elapsed time. 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, her average velocity was zero also: = ∆x/∆t = (0 m)/(78 s) = 0 m/s.

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.

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:

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.

Solution. Since the car is traveling at a constant velocity, its acceleration is zero. If there’s no change in velocity, then there’s no acceleration.

Solution. Assuming that the direction of the velocity doesn’t change, it’s simply a matter of dividing the change in velocity, v_f − v_i, 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.

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, give 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.

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-versus-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.

Segment #1: 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

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

Segment #2: 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

Between time t = 2 s to time t = 3.5 s, the object experienced a negative acceleration. Between time t = 2 s to time t = 3 s, the object’s velocity was slowed down till it reached 0 m/s. Between time t = 3 s to time t = 3.5 s, the object began increasing its velocity in the opposite direction.

Segment #3: After time t = 3.5 s, the slope of the graph is zero meaning the object experienced zero acceleration. This, however, does not mean the object did not move since its velocity was constant and in the negative direction.

Let’s look at Example 9 on this page again: 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.

We say signed area because regions below the horizontal axis are negative quantities (since the object’s velocity is negative, its displacement is negative). Thus, by counting areas above the horizontal axis as positive and areas below the horizontal axis as negative, we can make the following claim:

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 figure out 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.

Solution.

Let’s consider the relationship between acceleration and velocity in a velocity versus time graph.

If we take an object’s original direction of motion to be positive, then an increase in speed corresponds to positive acceleration. This is indicative of Segment #1 and Segment #3.

A decrease in velocity corresponds to negative acceleration (as indicated in top part of Segment #5).

However, if an object’s original direction is negative, then an increase in speed corresponds to negative acceleration, indicated by the bottom part of Segment #5. Whatever is below the x-axis shows us that you are speeding up backward.

When the velocity and acceleration are in opposite directions, the object slows down, as is indicated in Segment #7. Note that Segments #2, #4, and #6 indicate no acceleration.

UNIFORMLY ACCELERATED MOTION AND THE BIG FIVE

The simplest type of motion to analyze is motion in which the acceleration is constant.

Another restriction that will make our analysis easier is to consider 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 opposite direction is negative. Most of the quantities we’ve been dealing with—displacement, velocity, and acceleration—are 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 position (x), 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). And, finally, the motion takes place during some elapsed time interval, ∆t. Also, if we agree to start our clocks at time 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 these equations a little easier to write down. 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 cases we’re considering.

Big Five #1 is the definition of velocity (this is the area under a velocity versus time graph, which will be covered a little further in this chapter). Big Five #2 is the definition of acceleration (this is the slope at any given moment of a velocity versus time graph).

Big Five #1 and #2 are simply the definitions of and ā 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), using a bit of algebra.

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

x = t = (v₀ + v)t = (4 m/s + 14 m/s)(3 s) = 27 m

It’s okay to leave off the units in the middle of the calculation as long as you remember to include them in your final answer. Leaving units off your final answer will cost you points on the AP Exam.

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. Therefore, we use Big Five #3:

x = x₀ + v₀Δt + a(Δt)² = (10 m/s)(4s) + (2 m/s²)(4s)² = 56 m

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:

Note that since a has the same sign as (x – x₀), the acceleration vector points in the same direction as the displacement vector. This makes sense here, since the object moves downward, and the acceleration it experiences is due to gravity, which also points downward.

ADDITIONAL KINEMATIC GRAPHICAL ASPECTS

Not all graphs have nice straight lines as shown so far. Straight line segments represent constant slopes and therefore constant velocities, or accelerations, depending on the type of graph. What happens as an object changes its velocity in a position versus time graph? The “lines” become “curves.” Let’s look at a typical question that might be asked about such a curve.

Solution. This is familiar territory. To find the average velocity, use

a)

b)

c)

The instantaneous velocity is the velocity at a given moment in time. When you drive in a car and look down at the speedometer, you see the magnitude of your instantaneous velocity at that time.

To find the instantaneous velocity at 10 seconds, we will need to approximate. The velocity from 9–10 seconds is close to the velocity at 10 seconds, but it is still a bit too slow. The velocity from 10–11 seconds is also close to the velocity at 10 seconds, but it is still a bit too fast. You can find the middle ground between these two ideas, or the slope of the line that connects the point before and the point after 10 seconds. This is very close to the instantaneous velocity. A true tangent line touches the curve at only one point, but this line is close enough for our purposes.

Solution. Draw a tangent line. Find the slope of the tangent by picking any two points on the tangent line. It usually helps if the points are kind of far apart and it also helps if points can be found at “easy spots” such as (0, 5) and (15, 1,000) and not (6.27, 113) and (14.7, 983)

QUALITATIVE GRAPHING

Beyond all the math, you are at a clear advantage when you start to recognize that position-versus-time and velocity-versus-time graphs have a few basic shapes, and that all the graphs you will see will be some form of these basic shapes. Having a feel for these building blocks will go a long way toward understanding kinematics graphs in physics.

Either of the following two graphs represents something that is not moving.

Either of the following two graphs represents an object moving at a constant velocity in the positive direction.

Either of the following two graphs represents an object moving at a constant velocity in the negative direction.

Either of the following two graphs represents an object speeding up in the positive direction.

Either of the following two graphs represents an object slowing down in the positive direction.

Either of the following two graphs represents an object slowing down in the negative direction.

Either of the following two graphs represents an object speeding up in the negative direction.

Solution. Part A is a constant speed moving away from the origin, part B is at rest, part C is speeding up moving away from the origin, part D is slowing down still moving away from the origin, part E is speeding up moving back toward the origin, and part F is slowing down moving back toward the origin.

The area under an acceleration-versus-time graph gives the average velocity.

Solution. The acceleration is the slope of the velocity-versus-time graph. 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.

FREE FALL

The simplest real-life example of motion under pretty constant acceleration is the motion of objects in Earth’s gravitational field, near the surface of the Earth and ignoring any effects due to the air (mainly air resistance). With these effects ignored, an object can fall freely. That is, it can fall experiencing only acceleration due to gravity. Near the surface of the 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 1 Exam, you may use g = 10 m/s² as a simple approximation to g = 9.8 m/s². Even though you can use the exact value on your calculator, this approximation helps to save time, so in this book, we will always use g = 10 m/s². And, of course, the gravitational acceleration vector, g, points downward.

Since gravity points downward, for the sake of keeping things consistent throughout the rest of this chapter, we will set gravity as g = –10 m/s².

Solution. We are given v_o, and asked for t. Unless you specifically see words to the contrary (such as, “you are on the moon where the acceleration due to gravity is…”), assume you are also given a = –10 m/s². Because v is missing, and it isn’t asked for, we can use Big Five equation #3.

y = y₀ + v₀t + at² ⇒ y = at²

If we set the origin y₀ = 0 at the base of the cliff and v₀ = 0, we get:

Note: The negative in front of the 80 is inserted because the rock fell in the down direction.

Solution. We are given v₀, a = –10 m/s² is implied, and we are asked for y. Now, neither t nor v is expressly given; however, we know the vertical velocity at the top is 0 (otherwise the baseball would still rise). Consequently, we use Big Five equation #5.

v² = v₀² + 2a(y − y₀) ⇒ −2ay = v₀²

We set y₀ = 0 and we know that v = 0, so that leaves us with:

The time it takes for an object to be thrown straight up is the same as the time it takes for an object to fall down from the peak back into your hand.

Solution. We’re given v₀, a, and t, and asked for v. Since x is missing, we use Big Five #2:

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

The negative sign in front of the 40 simply indicates that the object is traveling in the down direction.

Solution. We’re given a, v₀, and t, and we need to find y₀. Because v is missing, we use Big Five #3:

PROJECTILE MOTION

In general, an object that moves near the surface of the 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.

The quantity v_0x, 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 v_0y, the vertical (or y) component of the initial velocity, is equal to v₀ sin θ₀.

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 How far did it drop? is a vertical question, so the set of equations we will consider are those listed above under vertical motion. Next, How far…? 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, v_0y = 0. Therefore,

The fact that ∆y is negative means that the displacement is down. Also, notice that the information given about v_0x is irrelevant to the question.

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).

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.

Solution. When the projectile reaches the top of its trajectory, its velocity vector is momentarily horizontal; that is, v_y = 0. 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

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:

Therefore, the total flight time (that is, up and down) is t_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_0xt_t = (v₀ cos θ)t_t = [(60 m/s)cos 30°](6 s) = 312 m

By the way, assuming it lands back at its original height, the full horizontal displacement of a projectile is called the projectile’s range.

Chapter 4 Review Questions

Answers and Explanations can be found in Chapter 13.

Section I: Multiple-Choice

Section II: Free-Response

Summary

Graphs are very useful tools to help visualize the motion of an object. They can also help you solve problems once you learn how to translate from one graph to another. As you work with graphs, keep the following things in mind:

Chapter 5

Newton’s Laws

 

“Nature is pleased with simplicity. And nature is no dummy.”

 

—Sir Isaac Newton

The Englishman Sir Isaac Newton published a book in 1687 called Philosphiae Naturalis Principa 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.

In kinematics we discovered the nature of how objects move, but there still existed the question of why objects move the way they do? For this fundamental task of understanding the cause of motion, we must turn our attention to dynamics.

INTRODUCTION TO FORCES

An interaction between two bodies—a push or a pull—is force. If an apple falls from a tree, it falls to the ground. If you pull on a door handle, it opens a door. If you push on a crate, you move it. In all these cases, some force is required for these actions to happen. In the first case of an apple falling from a tree, the Earth is exerting a downward pull called gravitational force. When you stand on the floor, the floor provides an upward force called the normal force. When you slide the crate across the floor, the floor exerts a frictional force against the crate.

NEWTON’S FIRST LAW

Newton’s First Law says that an object will continue in its state of motion unless compelled to change by a 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; if it is moving, then it will continue to move at a constant speed in a straight line.

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.

NEWTON’S SECOND LAW

Newton’s Second Law predicts what will happen when an unbalanced 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:

This is the most important equation in mechanics!

Two identical boxes, one empty and one full, have different masses. The box that’s full has the greater mass, because it contains more stuff; more stuff, more mass. Mass (m) is measured in kilograms, abbreviated as kg. (Note: An object whose mass is 1 kg weighs about 2.2 pounds.) If a given force produces some change in velocity for a 2 kg object, then a 1 kg object would experience twice that change. Note that mass is a proxy for the extent of inertia inherent in an object and thus inertia is a reflection of an object’s mass.

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.

NEWTON’S 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.

Solution. The First Law says that any object will continue in its state of motion unless an unbalanced 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.

You might be asking, “If no net force is needed to keep a car moving at a constant speed, why does the driver need to press down on the gas pedal in order to maintain a constant speed?” There is a big difference between force and net force. As the car moves forward, there is a frictional force (more on that shortly) opposite the direction of motion that would be slowing the car down. The gas supplies energy to the engine to spin the tires to exert a forward force on the car to counteract friction and make the net force zero, which maintains a constant speed.

Here’s another way to look at it: Constant velocity means a = 0, so the equation F_net = ma immediately gives F_net = 0.

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

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. Then Newton’s Second Law gives a = F_net/m = (12 N to the right)/(4 kg) = 3 m/s² to the right.

NEWTON’S LAWS: A SUMMARY

These three laws will appear time and time again in later concepts. Let’s sum up those three laws in easier terms.

Newton’s First Law

Newton’s Second Law

Newton’s Third Law

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 the Earth (or by whatever planet it happens to be on). Mass, by contrast, is a measure of the quantity of matter that comprises an object. An object’s mass does not change with location. Weight changes depending on location. For example, you weigh less on the Moon than you do on Earth.

Since weight is a force, we can use F = ma to compute it. What acceleration would the gravitational force impose on an object? The gravitational acceleration, of course! Therefore, setting a = g, the equation F = ma becomes

This is the equation for the weight of an object of mass m. (Weight is often symbolized as F_g, 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.

Solution. Since weight is m multiplied by g, mass is F_w (weight) divided by g. Therefore,

m = F_w/g = (500 N)/(10 m/s²) = 50 kg

Solution. This person’s weight in newtons is (150 lb)(4.45 N/lb) = 667.5 N, so his mass is

m = F_w/g = (667.5 N)/(10 m/s²) = 66.75 kg

Solution. The book experiences two forces: the downward pull of the 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 support force must equal the magnitude of the book’s weight, which is F_w = mg = (2 kg)(10 m/s²) = 20 N.

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. (In physics, the word normal means perpendicular.) 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.)

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 support 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.)

AN OVERALL STRATEGY

The previous examples are the lowest level of understanding Newton’s laws. They are pretty straightforward thinking sometimes referred to as “plug and chug.” Most of physics is not that simple. Frequently there is more than one force acting on an object, and many times angles are involved. Following the below strategy can greatly increase your chance of success for all but the most trivial of Newton’s Second Law problems.

As you go though the following examples, notice how this strategy is used.

Solution. Notice that gravity always points down (even on ramps). Also the normal force is perpendicular to the surface (even on ramps). Do not always put the normal opposite the direction of gravity, because the normal is relative to the surface, which may be tilted. Finally, friction is always parallel to the surface (or perpendicular to the normal) and tends to point in the opposite direction from motion.

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 (also symbolized merely by 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 = mg + ma = m(g + a) = 6(10 + 1) = 66 N

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.

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

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 electrical interactions between atoms of which the object is composed and those of which the surface is composed.

We’ll look at two main categories of friction: (1) static friction and (2) kinetic (sliding) friction. Static friction results from the weak electrostatic bonds formed between the surfaces when the object is at rest. These bonds need to be broken before the object will slide. As a result, the static friction force is higher than the kinetic friction force. When the object begins to slide, the weak electronegative bonds cannot form fast enough, so the kinetic friction is lower. It is easier to keep an object moving than it is to start an object moving because the Kinetic μ is lower than the static μ. 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, which is denoted by μ (mu) and has no units. The greater this number is, 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, it’s 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:

Note that the equation for the strength of the static friction force is for the maximum value only. This is because static friction can vary, precisely counteracting weaker forces that attempt to move an object. For example, suppose an object experiences 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, and 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 usually, but not always, opposite to that of the intended motion.

Solution. First draw a free-body diagram:

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. Pulley systems multiply the force by however many strings are pulling on the object

Solution. There are two blocks, so we draw two free-body diagrams:

To get the acceleration of each one, we use Newton’s Second Law, F_net = ma.

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.

Newton’s Second Law is defined as F = ma, whereas the Force required to accelerate an object is equal to the mass of the object multiplied by the magnitude of the desired acceleration. If the net force in a system is not zero, then the object in the system is accelerating. Newtons second law can be used to calculate the acceleration of objects with unbalanced forces. Once the net force is determined, substitute this force and mass into the F = ma (F_net = ma) equation to determine the acceleration of the object.

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)

F_T − Mg = M(−a) (2)

Subtract 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:

   Mg − F_f = ma + Ma

  Mg − μmg = a(m + M)

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

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), F_T – Mg = M(−a) (because it’s simpler), we find

F_T = Mg − Ma = Mg − M · g = Mg = (10)(10) = 25 N

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 you look at the forces acting on a block that sits on a ramp using a standard coordinate system, it initially looks straightforward. However, part of the normal force acts in the x direction, part acts in the y direction, and the block has acceleration in both the x and y directions. If friction is present, it also has components in both the x and y directions. The math has the potential to be quite cumbersome.

This is a non-rotated coordinate system—notice F_f, F_N, and a will each have to be broken into x- and y-components.

This is a rotated coordinate system—F_N acts only in the y direction, F_f and the acceleration act only in the x direction, and only F_w must be broken into components. 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.

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

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). Therefore, the net force down the ramp is

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

Then, setting F_net equal to ma, we solve for a:

Chapter 5 Review Questions

Answers and Explanations can be found in Chapter 13.

Section I: Multiple-Choice

Section II: Free-Response

Summary

Chapter 6

Work, Energy, and Power

 

“Energy cannot be created or destroyed, it can only be changed from one form to another.”

 

—Albert Einstein

Kinematics and dynamics are about change. Simple observations of our environment show us that change is occurring all around us. But what is needed to make an object change, and where did that change go to? It wasn’t until more than one hundred years after Newton that the idea of energy became incorporated into physics, but today it permeates every branch of the subject.

ENERGY: AN OVERVIEW

So what is energy? How do we determine the energy of a system? These are not easy questions. It is difficult to give a precise definition of energy; there are different forms of energy as a result of different kinds of forces. Energy can come as a result of gravitational force, the speed of an object, stored energy in springs, heat loss to nuclear energy. But one truth remains the same for all of them: the Law of Conservation of Energy. Energy can not just appear out of nowhere nor can it dissapear in a closed system; it must always take on another form. Force is the agent for this change, energy is the measure of that change, and work is the method of transferring energy from one system to another.

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. The application of force over a distance and the resulting change in energy of the system that the force acted on, give rise to the concept of work.

Definition. If a constant force F acts over a distance d and F is parallel to d, then the work done by F is the product of force and distance:

Work Units

The unit for work, the newton-meter (N·m), is renamed a joule and abbreviated as J.

Named after English physicist James Prescott Joule, one joule is the work required to produce one watt of power for one second.

Notice that, although work depends on two vectors (F and d), work itself is not a vector. Work is a scalar quantity. However, even being a scalar quantity there exists positive, negative, and zero work.

Solution. In this case, the force you exert must balance the weight of the book (otherwise 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 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, or 60 J.

WORK AT AN ANGLE

The previous formula works only when work is done completely parallel to the intended distance of travel. What happens when the force is done at an angle? The formula becomes:

Let’s compare the work done in a few instances with a force being applied between certain angles:

Note that when we do positive work we increase the speed of an object, whereas when we do negative work we slow an object down. This will be important to note when we relate work with kinetic energy and potential energy.

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 = (69 N ∙ cos 30°)(10 m) = 600 J

Solution.

Solution.

W_total = ΣW = W_{by gravity} + W_{by normal force} + W_{by friction} = 1,690 + 0 + (−671) = 1,019 J

WORK DONE BY A VARIABLE FORCE

If a force remains constant over the distance through which it acts, then the work done by the force is simply the product of force and distance. However, if the force does not remain constant, then the work done by the force is given by the area under the curve of a force versus displacement graph. In physics language, the term “under the curve” really means between the line itself and zero.

Solution. The area under the curve will be equal to the work done. In this case, we have some choices. You may recognize this shape as a trapezoid (it might help to momentarily rotate your head, or this book, 90 degrees) to see this.

A = (b₁ + b₂)h

A = (40N + 80N)(0.20m)

  = 12 N • m or 12 J

Similar to the previous chapter, units for work and energy should be confined to kg, m, and s, which is why we converted here.

An alternative choice is to recognize this shape as a triangle sitting on top of a rectangle. The total area is simply the area of the rectangle plus the area of the triangle.

A_total = A_rectangle + A_triangle

= bh₁ + (bh₂)

    = (0.20m)(40N − 0N) + (0.20m)(80N − 40N)

   = 8 N • m + 4 N • m = 12 J

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 force acting on the object, pushing it in a straight line. The object’s acceleration is a = F/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(x–x₀) is the work done by the force, so W = 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:

THE WORK–ENERGY THEOREM

In the previous section, we derived kinetic energy from Big Five #5. Let’s solve it a different way this time:

	v2 = vo2 + 2ad	Recall: a = F/m
	2(F/m)d = v2 − vo2	Recall: W = Fd
	W = (1/2)mv2 − (1/2)mvo2
	W = Kfinal − Kinitial

One of the questions posed at the beginning of this chapter was “How does a system gain or lose energy?” The Work–Energy Theorem begins to answer that question by stating that a system gains or loses kinetic energy by transferring it through work between the environment (forces being introduced into the system) and the system. In basic terms, doing positive work means increasing kinetic energy.

Solution. From the definition,

K = mv² = (0.10 kg)(30 m/s)² = 45 J

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

Solution. It was calculated in Example 4 that W_total = 1,019 J. According to the Work–Energy Theorem,

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 off. 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 examples illustrate the concept 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 retrieved as kinetic energy. Also remember that potential energy can be found in multiple sources such as chemical sources, mechanical sources, and objects in gravitational fields. 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_g, 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.

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 were given a push to send it over the edge, by the time the ball reached the floor it would acquire a kinetic energy of 30 J. 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,

Note that potential energy, like work (and kinetic energy), is expressed in joules.

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

An important fact that makes the above equation possible is that 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. Conversely, any work done by a nonconservative force is path-dependent. Such is the case for friction and air resistance, as different paths taken may require more or less work to get from the initial to final positions.

If we decide on a reference level to call h = 0, then we can say that the gravitational potential energy of an object of mass m at a height h is U_g = mgh. In order to use this last equation, it’s essential that we choose a reference level for height. For example, consider a passenger in an airplane reading a book. If the book is 1 m above the floor of the plane, 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, say, 9,000 m above the ground. So, to this person, the gravitational potential energy of the book is mgH, where H = 9,001 m. What both would agree on, though, is that the difference in potential energy between the floor of the plane and the position of the book is mg × (1 m), since the airplane passenger would calculate the difference as mg × (1 m – 0 m), while the person on the ground would calculate it as mg × (9,001 m – 9,000 m). Differences, or changes, in potential energy are unambiguous, but values of potential energy are relative.

Solution. Calling the ground h = 0, we find

U_g = mgh = (60 kg)(10 m/s2)(40 m) = 24,000 J

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 total mechanical energy, E.

(Note that because U is relative, so is E.) Assuming that no nonconservative forces (friction, for example) act on an object or system while it undergoes some change, then mechanical energy is conserved. Mechanical energy, that is the sum of potential and kinetic energies, is dissipated or converted into other energy forms such as heat, by nonconservative forces. So, if there are no nonconservative forces acting on the system, the initial mechanical energy, E_i, is equal to the final mechanical energy, E_f, or

This is the simplest form of the Law of Conservation of Total Energy.

Let’s evaluate a situation in which an object, initially at rest, is falling from a building and calculate the total mechanical energy in each situation:

What can you say about the calculated total mechanical energy for each situation? They all are equal. An independent system obeys the Law of Conservation of Energy. Since the total mechanical energy remains the same throughout its travel, sometimes using energy to solve problems is a lot simpler than using kinematics.

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.

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:

h = d sinθ    

    

Solution. Ignoring air resistance, we can apply Conservation of Mechanical Energy. Calling the ground our h = 0 reference level, we write

CONSERVATION OF ENERGY WITH NONCONSERVATIVE FORCES

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

Solution. The force of air resistance opposes the downward motion, so it does negative work on the coyote as he falls: W_r = –F_rh. Calling the ground h = 0, we find that

k_i + U_i + W_r = K_f + U_f

0 + mgh + (−f_rh) = mv² + 0

v = = 27 m/s

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

So far, any of the problems we have solved in this chapter could have been solved using the kinematics equations and Newton’s laws. The truly powerful thing about energy is that in a closed system, changes in energy are independent of the path you take. This allows you to solve many problems you would not otherwise be able to solve. With many energy problems you do not need to measure time with a stopwatch, you do not need to know the mass of the object, you do not need a constant acceleration (remember that is required for our Big Five equations from kinematics), and you do not need to know the path the object takes.

Solution.

K_i + U_i = K_f + U_f

or

Canceling the mass, setting v_i = 0 m/s, and rearranging terms, we get

POWER

Simply put, power is the rate at which work gets done (or energy gets transferred, which is the same thing). Suppose Scott and Jean each do 1,000 J of work, but Scott does the work in 2 minutes, while Jean does it in 1 minute. They both did the same amount of work, but Jean did it more quickly; thus Jean was more powerful. Here’s the definition of power:

The unit of power is the joule per second (J/s), which is renamed 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 units like inches, feet, yards, miles, ounces, pounds, and so forth, you still hear of power ratings (particularly of engines) expressed in horsepower. One horsepower is defined as 1 hp = 746 W.

Note that this conversion will be provided on the test.

	P = W/t	Recall: W = Fd
	P = Fd/t	Recall: v = d/t

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.

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 6 Review Questions

Answers and Explanations can be found in Chapter 13.

Section I: Multiple-Choice

Section II: Free-Response

Summary

Chapter 7

Linear Momentum

 

“Nothing will happen until something moves.”

 

—Albert Einstein

Previously we discovered the nature of objects, the reason why objects move the way they do, and the energy required to make objects move. Now we will predict the nature of objects when they interact with other objects and the resulting outcome.

INTRODUCTION TO MOMENTUM

A pool stick hitting the cue ball, a car collision, the Death Star exploding—physics is about the interaction of objects. A collision is a complex interaction between two objects. But what happens after two objects interact with each other? If we use Newton’s Second Law, it will prove to be quite a tedious challenge. The forces occurring during a car collision are unimaginably complex. And not all collisions end with the same results: sometimes they move away from each other, sometimes they move in the same direction, and sometimes the objects stick together. If we were to measure their speeds before the collision and their speeds after the collision, we would also get different results. Using our observations and a general understanding of Newton’s Second Law, we could predict that the smaller car in a collision (head on and both cars traveling with the same speed) would end up having the greater speed in the end. How can such a complex interaction give rise to a simple outcome?

WHAT IS MOMENTUM?

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 and 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, we can turn the proportion above into an equation, if we take the average force that acts over the time interval ∆t:

This equation becomes F = ma, since ∆p/∆t = ∆(mv)/∆t = m (∆v/∆t) = ma.

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:

Impulse is equal to change in linear momentum. In terms of impulse, Newton’s Second Law can be written in yet another form:

Sometimes this is referred to as the Impulse-Momentum Theorem, but it’s just another way of writing Newton’s Second Law. The impulse delivered to an object may be found by taking the area under a force-versus-time graph.

The Impulse-Momentum Theorem basically states that an impulse that is delivered on an object changes its momentum. The momentum “after” the collision is equal to the momentum “before” the collision added in with the impulse required to get your final outcome. In equation form, this states:

p_final = p_initial + J

Impulse eliminates the need to use F = ma and simplifies the intricate forces delivered into mass, initial, and final velocities.

Solution. As we know, 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 × 10⁻³ s) = 1,500 N

Solution.

The negative signs in the vector answers to (b) and (c) simply tell you that the forces are acting in the opposite direction of motion and will cause the object to slow down. This force is equivalent to about 27 tons (!), 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 instance.

Solution. The impulse delivered to the block is equal to the area under the F-versus-t graph. The region is a trapezoid, so its area, (base₁ + base₂) × height, can be calculated as follows:

J = [(10 ms − 0) + (6 ms − 2 ms)] × (20 N) = 0.14 N · s

Now, by the Impulse-Momentum Theorem,

CONSERVATION OF LINEAR MOMENTUM

Newton’s Third Law states that when one object exerts a force on a second object, the second object exerts an equal but opposite force on the first. Newton’s Third Law combines with Newton’s Second Law when two objects interact with each other.

In the previous section, we redefined Newton’s Second Law as Impulse-Momentum Theorem, J = Δp. If we combine the laws and interpret them in terms of momentum, it states that two interacting objects experience equal but opposite momentum changes (assuming we have an isolated system, meaning no external forces).

The momentum “before” equals the momentum “after.” This is the Law of Conservation of Momentum, which states:

Solution. Here, the astronaut + tool are the system. Because of Conservation of Linear Momentum,

Using distance = rate × 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), and the system’s total linear momentum is conserved.

Let’s break down the collision types:

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:

total p_before = total p_after 

m_redv_red + m_greenv_green = m_redv′_red + m_greenv′_green

(0.5)(+4) + (0.2)(−2) = (0.5)(+2) + (0.2)v′_green

v′_green = +3.0 m/s

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.

Initially

K_t = K₁ + K₂

    = m₁v_i1² + m₂v_i2²

    = (0.5)(+4)² + (0.2)(−2)²

= 4.4 J

There are 4.4 joules at the beginning. At the end,

K_t = K₁ + K₂

    = m₁v_f1² + m₂v_f2²

    = (0.5)(+2)² + (0.2)(3)²

= 1.9 J

So, there is less kinetic energy at the end compared to the beginning. Kinetic energy was lost (so the collision was inelastic). Most of the lost energy was transferred as heat; the two objects are both slightly warmer as a result of 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

total p_before = total p_after

m_redv_red + m_greenv_green = (m_red +m_green)v′ 

(0.5)(4)+(0.3)(−2) = (0.5+0.3)v′ 

  v′ = +1.8 m/s

Solution. This problem illustrates the vector nature of numbers. First, look only at the x (east-west) direction. There is only one car moving west and its momentum is given by:

p = mv ⇒(500 kg)(30 m/s) = 15,000 kg · m/s west

Next, look only at the y (north-south) direction. There is only one car moving north and its momentum is given by:

p = mv ⇒(500 kg)(20 m/s) = 10,000 kg · m/s north

The total final momentum is the resultant of these two vectors. Use the Pythagorean Theorem:

(10,000)² + (15,000)² = p_f²

p = 18,027 kg · m/s

To find the total velocity you need to solve for v using

p_f = mv ⇒ 18,027 kg · m/s = (500 kg + 500 kg)(v)

   v = 18 m/s

However, we are not done yet because velocity has both a magnitude (which we now know is 18 m/s) and a direction. The direction can be expressed using tan θ = 10,000/15,000 so θ = tan⁻¹ (10,000/15,000), or 33.7 degrees north of west. Again, most of the AP Physics 1 Exam is in degrees, not radians.

Solution.

Chapter 7 Review Questions

Answers and Explanations can be found in Chapter 13.

Section I: Multiple-Choice

Section II: Free-Response

Summary