Practice Makes Perfect Physics

By Connie J. Wells

Don't be perplexed by physics. Master this science with practice, practice, practice!

Practice Makes Perfect: Physics is a comprehensive guide and workbook that covers all the basics of physics that you need to understand this subject. Each chapter focuses on one major topic, with thorough explanations and many illustrative examples, so you can learn at your own pace and really absorb the information. You get to apply your knowledge and practice what you've learned through a variety of exercises, with an answer key for instant feedback. Offering a winning formula for getting a handle on science right away, Practice Makes Perfect: Physics is your ultimate resource for building a solid understanding of physics fundamentals.

• Language: English
• Publisher: McGraw-Hill Education
• Release date: Jun 10, 2011
• ISBN: 9780071746960

Book preview

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Preface

For many students, difficulty with a course in physics stems from difficulty in using mathematical techniques effectively to solve problems. The understanding of physics concepts is built in many ways: by careful reading of a well-written textbook, class discussions and presentations, hands-on laboratory work, practice in solving related problems, and observations of how physics works in everyday applications. This algebra-based problem-solving guide is designed to supplement your work in all other areas of your physics education—including the development of an understanding of concepts.

The sample problems here are designed to help strengthen you, the student, as you are shown methods of solution for problems in different topic areas. You are then given conceptual questions and practice problems to test your understanding. Practicing with these problems and checking the solutions may not make you perfect, but the practice should improve your understanding of physics applications and your ability to solve related problems.

The biggest mistake many students (and teachers) of physics make when faced with a physics problem is to jump in and start manipulating numerical values to find an answer—what I’ll call step 5 in problem solving. They leave out the preliminary steps and follow-up steps that will help develop an understanding of the physics and make the next problem more understandable. After all, finding an answer means nothing by itself. How many times in your daily life will you apply an answer such as 22 meters per second directly to a situation? You are likely, however, to use a relationship, such as doubling the speed of my car as I turn a corner requires a centripetal force that is four times as great. You will understand and learn from problem solving if you follow—every time—a few simple steps:

Read the problem and try to develop a mental picture or understanding of what’s really happening.

When possible, make a diagram to help you picture the situation.

List the quantities given (or knowns) and define what is unknown—the quantity you want to find. This can sometimes be done directly on the diagram.

Decide what equations you will use to define the mathematical relationships among the known and unknown quantities.

Now substitute the known quantities into the equation(s), with appropriate units. (Here’s where you’ll catch any unit conversions necessary to work the problem correctly.)

Solve the equations for the unknown quantity, and express your answer to the correct number of significant digits with the appropriate units.

Consider your answer. Does the answer make sense in the context of the problem?

Think about what the problem means conceptually, and how this answer might vary if one or more of the given quantities changes.

You should see now that just putting the numbers together until you get an answer will keep you from understanding the physics—and not thinking about the physics will keep you from getting the answer! Better to use all the steps—and all you know about the physics—when working on problems. This book is full of examples that will show you how to apply these problem-solving techniques.

Making the most of this book

Each unit starts with a brief set of definitions and descriptions of concepts and equations, called Terms and equations. While a textbook is necessary for a full understanding of physics concepts, the descriptions in this book will clarify the notations that will be used in equations.

Sample Problems with solutions are provided as instruction to get you started on problem solving. Read through these carefully—step by step—to make sure you understand. You can also cover the solution with a card, solve the problem yourself, and then check your solution by reading through the one given.

Conceptual questions are given in each unit to make sure you understand the basic concepts. These are followed by Practice problems that enable you to put to use the techniques shown by the sample problems. An answer key is provided at the back of the book; but try working the practice problems yourself, if possible, before referring to the answers provided. And the appendix contains some reference data that you should find useful as you make your way through the book.

I hope the resources in this book are helpful to you in your study of physics. Good luck!

PRACTICE MAKES PERFECT™

Physics

1 A mathematical review

Most of the skills reviewed in this section should be familiar from previous studies in algebra. These concepts are used throughout the study of physics, so this section should be used to build and test skills prior to starting physics and can be used as reference throughout.

Algebraic manipulations

SAMPLE PROBLEM 1

Solve for v

SOLUTION

(Add −xto each side of the equation)

Geometry

Area of rectangle: A = bh

Area of triangle: A

Area of parallelogram: A = bh

Volume of parallelepiped: V = bwh

Circumference of circle: C = 2πr

Area of circle: A = πr²

Surface area of sphere: A = 4πr²

Volume of sphere: V

Volume of cylinder: V = πr²h

Surface area of cylinder: A = πrh + 2πr²

Right-triangle trigonometry

a² + b² = c²

a = c sin θ

b = c cos θ

Useful trigonometric relationships

sin²θ + cos² θ = 1

Special triangles

SAMPLE PROBLEM 2

Suppose point B is 6 m directly west of point A, and point C is 8 m directly north of point B. (a) How far is point C from point A? (b) What direction is point C from point A?

SOLUTION

A sketch of the situation shows that it is a right triangle with legs of length 6 m and 8 m, which are in the ratio 3:4.

We recognize this as a special 3:4:5 right triangle with angles of 37° and 53°. Using this ratio, the hypotenuse, or distance from point A to point C, is 10 m. The angle at point A has to be the 53° angle, since the larger angle is opposite the longer leg. Thus, point C is in a direction of 53° north of west.

Vectors and vector components

Quantities that are described only in terms of magnitude are called scalars. These include measured quantities such as energy, time, distance, temperature, speed, charge, electric potential, pressure, and volume.

Quantities that are described in terms of both their magnitude (or size) and direction are called vectors. These include measured quantities such as force, acceleration, velocity, displacement, torque, gravitational field, electric field, and magnetic field.

The directions of vectors are expressed relative to a frame of reference, such as a cardinal direction (east, west, north, south) or a defined coordinate axis (for example, the x-axis). An example of a vector with its direction is a force of 3.0 newtons at an angle of 30° north of east (shown here with its coordinate frame of reference).

It’s easy enough to combine vectors when they are in line with each other (or in one dimension). However, when you are working with vectors in two dimensions (starting in Unit 3), vectors need to be divided into components in each reference direction so that the components can be added separately. The following sample problems show how to do this.

SAMPLE PROBLEM 3

Find the horizontal and vertical components of a force vector that has a magnitude of 40 newtons and is directed at an angle of 30° above horizontal.

SOLUTION

Construct the vector and its components, as shown.

Note that the component opposite the angle is a sine function and the component next to the angle is a cosine function.

Horizontal component: Fx = 40 cos 30° = 34.6 N

Vertical component: Fy = 40 sin 30° = 20.0 N

SAMPLE PROBLEM 4

Show diagrammatically the resultant, by addition, of vector X and vector Y.

SOLUTION

The resultant, labeled R, is found by placing the vectors tip to tail and constructing a new vector from the beginning of the first vector to the end of the last vector.

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Note: To subtract vector Y from vector X, simply add the opposite of vector Y to vector X.

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SAMPLE PROBLEM 5

Calculate the resultant of the addition of the following vector quantities:

SOLUTION

Step 1

Diagram the two vectors and find the east-west component and the north-south component of each vector. We’ll call east the positive direction and west the negative direction in that dimension, and we’ll call north positive and south negative in the second dimension.

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Note: We could also work this using the x and y coordinates, with east as +x and north as +y.

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Step 2

Add the east-west components to find the east-west component of the resultant (Rx), and add the north-south components to find the north-south component of the resultant (Ry). Then apply the Pythagorean theorem to these two perpendicular components to find the resultant, R.

Rx = 2.0 cos 40° + 4.0 cos 25° = 5.2 N(east)

Ry = 2.0 sin 40° + 4.0 sin 25° = 3.0 N(north)

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Since both components are positive, the resultant must be directed east and north.

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Step 3

Diagram the addition of the two components to help in determining the direction of the resultant.

Step 4

Find the direction of the resultant by solving for θ.

Step 5

Express the final vector with magnitude and direction:

R = 6.0 N at 30° north of east.

SAMPLE PROBLEM 6

Diagram the resultant vector for the subtraction of vector A from vector B.

SOLUTION

To subtract A from B, add the opposite of vector A to vector B. The opposite, or negative, of a vector is the same size but in the opposite direction. Then add the vectors tip to tail.

Metric units and conversions

Physics uses exclusively metric units, but information provided in problems may need to be converted to appropriate units to solve the problems. Metric prefixes are provided for reference in the Appendix in the back of this book. SI (Système International d’Unités, the International System of Units) metric units most commonly used are the kilogram (kg) for mass, the meter (m) for length, and the second (s) for time.

Unit conversions may be necessary to find an answer with metric units or to match units when solving a problem. The unit cancellation method shown here can be used to convert units or to check that units match in an equation.

SAMPLE PROBLEM 7

Convert 30 miles per hour to meters per second.

SOLUTION

SAMPLE PROBLEM 8

SOLUTION

For the force to be calculated in newtons, the mass must be in kilograms, the velocity must be in meters per second, and the radius must be in meters—all of which are SI units of metric measurement.

Significant digits

It is important to express the results of calculations using a reasonable number of digits. Students often write down all digits showing on their calculator screen, which is unreasonable. For most situations, the following rules for multiplication, division, addition, and subtraction will allow rounding to within one digit of the scientifically significant number.

Multiplication and division

Use the numerical value that has the lowest number of significant digits to determine the number of significant digits in the answer.

SAMPLE PROBLEM 9

Multiply the following measurements and round the answer to the correct number of significant digits: 262.89 cm × 15.57 cm × 12.3 cm.

SOLUTION

When I multiply these numbers, my calculator reads 50346.32679. I round to only three digits, based on the 12.3 cm measurement. The answer I provide is 50,300 or 5.03 × 10⁴ cm. Notice that in scientific notation, the number of significant digits is obvious.

SAMPLE PROBLEM 10

Perform the following calculation and round the answer to the correct number of significant digits:

SOLUTION

When this calculation is performed on the calculator, the screen reads 3.010634885. This is obviously too many digits to record reasonably. The answer should be rounded to two significant digits: 3.0 s. The value 9.8 m/s² has the fewest significant digits—noting that the 2 is exact (no limit on significant digits) and π is known to thousands (perhaps millions) of significant digits.

Addition and subtraction

The rule is quite different for addition and subtraction. To add or subtract, line up the numbers at the decimal point and round the answer to the highest power of ten or by least significant place. (See the example below.)

SAMPLE PROBLEM 11

Add the following numbers and round the answer to the correct number of significant digits: 27.34 + 24.001 + 780.

SOLUTION

The answer is expressed as 831, which is rounded to the ones place—dictated by the number 780, which is expressed only to the ones place. Notice that it is not the number of significant digits, but the place, that determines the rounding.

SAMPLE PROBLEM 12

Perform the following operations and round the answer to the correct number of significant digits: 98.6 + 75.281 − 0.023.

SOLUTION

The answer should be expressed as 173.9, since the value 98.6 is expressed to the tenths place. Notice here that the 0.023 doesn’t affect the answer; it’s basically so small that it is insignificant in this case.

SAMPLE PROBLEM 13

In an experiment, students acquire the following times: 1.23 s, 1.21 s, 1.19 s, and 1.2 s. What is the average value the students should report?

SOLUTION

Adding these four values and dividing by 4 to average them produces 1.2075 s, but this has to be rounded to 1.2 s. The measurement 1.2 s reported by one student has only two significant digits, so the answer can only be significant to the tenths place. The division by a whole number to calculate the average does not affect the number of significant digits in the answer, since the number is exact.

This is why it is important to record every significant digit in a measurement. If the last student’s time was exactly 1.20 s, it should have been recorded that way.

There are varying rules regarding the rounding of numbers when the last digit is exactly 5. Often, the half is rounded up if the previous digit is an odd number and rounded down (or dropped) if the previous digit is even. Loosely, this rounds up or down an equal number of times. For example, if the number 3.45 needs to be rounded to two significant digits, then you would round it to 3.4. If, however, the number 7.75 needs to be rounded to two significant digits, you would round it to 7.8. Be careful about rounding subsequently. If you were given 7.746 to round to two significant digits, you would not round up the 7.746 to make it 7.75 and then round the 7.75 to 7.8.

Graphing and graphical analysis

SAMPLE PROBLEM 14

Graph the following set of data from an experiment to examine