Instructor's Solutions Manual to Physics

1580 Pages·2012·27.535 MB·English

by  John D. Cutnell, Kenneth W. Johnson

INSTRUCTOR’S SOLUTIONS MANUAL

to accompany

Physics

Ninth Edition

John D. Cutnell
Kenneth W. Johnson

Southern Illinois University at Carbondale

Volume I

Chapters 1  17
PREFACE for Volume I

This volume contains the answers to the Focus on Concepts questions and the
complete solutions to the problems for chapters 1 through 17. These chapters
include all of mechanics (including fluids), thermal physics and wave motion. The
solutions for chapters 18 through 32 (electricity and magnetism, light and optics,
and modern physics) are contained in Volume 2.

Each chapter is organized so that the answers to the Focus on Concepts questions
appear first, followed by the solutions to the problems.

An electronic version of this manual is available on the Instructor’s Companion
Website. The files are available in three formats: Microsoft Word
97-2003, Microsoft Word 2007, and PDF files of each individual solution.

The icon SSM at the beginning of some of the problems indicates that the
solution is also available in the Student Solutions Manual.

Note to adopters regarding The Instructor’s Solutions Manual to
accompany Physics, 9e by Cutnell & Johnson

Thank you for adopting Physics, 9e. We are pleased to be able to provide you
with a variety of support material to help you in the teaching of your course. All
of this materialincluding the Instructor’s Solutions Manualis copyrighted by
John Wiley & Sons, Inc. and is explicitly intended for use only at your institution.

Please note that our providing these solutions does not carry with it permission to
distribute them beyond your institution. Before putting any of the solutions on a
Web site, we ask that you request formal permission from us to do so. Please
write to: Permissions Department, John Wiley & Sons, Inc. 111 River St.
Hoboken, NJ 07030. In most cases we will grant such permission PROVIDED
THAT THE WEB SITE IS PASSWORD PROTECTED.

Our goal is to prevent students from other campuses from being able to access
your solutions. We trust that you can understand how that might undermine the
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We appreciate your support and understanding in this matter.
CONTENTS

CHAPTER 1 Answers to Focus on 1 CHAPTER 10 Answers to Focus on 476
Concepts Questions Concepts Questions

Problems 3 Problems 479
CHAPTER 2 Answers to Focus on 42 CHAPTER 11 Answers to Focus on 539
Concepts Questions Concepts Questions
Problems 45 Problems 542
CHAPTER 3 Answers to Focus on 97 CHAPTER 12 Answers to Focus on 601
Concepts Questions Concepts Questions
Problems 99 Problems 603
CHAPTER 4 Answers to Focus on 158 CHAPTER 13 Answers to Focus on 658
Concepts Questions Concepts Questions
Problems 161 Problems 660
CHAPTER 5 Answers to Focus on 236 CHAPTER 14 Answers to Focus on 689
Concepts Questions Concepts Questions
Problems 238 Problems 691
CHAPTER 6 Answers to Focus on 273 CHAPTER 15 Answers to Focus on 729
Concepts Questions Concepts Questions
Problems 276 Problems 731
CHAPTER 7 Answers to Focus on 332 CHAPTER 16 Answers to Focus on 792
Concepts Questions Concepts Questions
Problems 334 Problems 794
CHAPTER 8 Answers to Focus on 378 CHAPTER 17 Answers to Focus on 856
Concepts Questions Concepts Questions
Problems 380 Problems 858
CHAPTER 9 Answers to Focus on 419
Concepts Questions

Problems 421
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

1. (d) The resultant vector R is drawn from the tail of the first vector to the head of the last
vector.

2. (c) Note from the drawing that the magnitude R of the resultant vector R is equal to the
shortest distance between the tail of A and the head of B. Thus, R is less than the magnitude
(length) of A plus the magnitude of B.

3. (a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides are
known, so the Pythagorean theorem can be used to determine the length R of the
hypotenuse.
4.0 km
4. (b) The angle is found by using the inverse tangent function,  tan1 53.
 
3.0 km
5. (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are not
reversed.

6. (c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector A
is not reversed.

7. (e) These vectors form a closed four-sided polygon, with the head of the fourth vector
exactly meeting the tail of the first vector. Thus, the resultant vector is zero.

8. (c) When the two vector components A and A are added by the tail-to-head method, the
x y
sum equals the vector A. Therefore, these vector components are the correct ones.

9. (b) The three vectors form a right triangle, so the magnitude of A is given by the
2 2
Pythagorean theorem as A A  A . If A and A double in size, then the magnitude of
x y
x y
 2  2
2 2 2 2
A doubles: 2A  2A  4A 4A  2 A  A  2A.
x y x y x y
 A 
1 y
10. (a) The angle  is determined by the inverse tangent function,  tan  . If Ax and
 A 
x
A both become twice as large, the ratio does not change, and remains the same.
y

11. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar
component along the x axis (A = 0 m) and its scalar component along the y axis is negative.
x
2 INTRODUCTION AND MATHEMATICAL CONCEPTS

12. (e) The scalar components are given by A = (450 m) sin 35.0 = 258 m and
x
A = (450 m) cos 35.0 = 369 m.
y

13. (d) The distance (magnitude) traveled by each runner is the same, but the directions are
different. Therefore, the two displacement vectors are not equal.

14. (c) A and B point in opposite directions, and A and B point in the same direction.
x x y y

15. (d)

16. A = 3.4 m, B = 3.4 m
y y

17. R = 0 m, R = 6.8 m
x y

18. R = 7.9 m, degrees
Chapter 1 Problems 3
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS

PROBLEMS
____________________________________________________________________________________________

1. REASONING We use the fact that 1 m = 3.28 ft to form the following conversion factor:
(1 m)/(3.28 ft) = 1.

2 2
SOLUTION To convert ft into m , we apply the conversion factor twice:

  1 m  1 m 
Area = 1330 ft2  124 m2
  
3.28 ft 3.28 ft 
_____________________________________________________________________________

2. REASONING
a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we need
to convert miles to kilometers. This conversion is achieved by using the relation 1.609 km =
1 mi (see the page facing the inside of the front cover of the text).

b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must
convert miles to meters and hours to seconds. This is accomplished by using the conversions
1 mi = 1609 m and 1 h = 3600 s.

SOLUTION a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi)
= 1, we find the speed of the bicyclists is

 mi  mi 1.609km km
Speed = 34.0 134.0   54.7
 h   h  1mi  h

b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and
(1 h)/(3600 s) = 1, the speed of the bicyclists is

 mi  mi 1609m 1h  m
Speed = 34.0 1134.0    15.2
 h   h  1mi 3600s s
______________________________________________________________________________

3. SSM REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and 1 yd = 3 ft.
With these facts we construct three conversion factors: (5280 ft)/(1 mi) = 1, (1 m)/(3.281 ft)
= 1, and (3 ft)/(1 yd) = 1.
4 INTRODUCTION AND MATHEMATICAL CONCEPTS
SOLUTION By multiplying by the given distance d of the fall by the appropriate
conversion factors we find that

5280 ft  1 m   3 ft  1 m 
   
d  6 mi    551 yd    10 159 m
 1 mi 3.281 ft  1 yd 3.281 ft 

4. REASONING The word “per” indicates a ratio, so “0.35 mm per day” means 0.35 mm/d,
which is to be expressed as a rate in ft/century. These units differ from the given units in
both length and time dimensions, so both must be converted. For length, 1 m = 103 mm, and
1 ft = 0.3048 m. For time, 1 year = 365.24 days, and 1 century = 100 years. Multiplying the
resulting growth rate by one century gives an estimate of the total length of hair a long-lived
adult could grow over his lifetime.

SOLUTION Multiply the given growth rate by the length and time conversion factors,
making sure units cancel properly:

 mm  1 m  1 ft 365.24d  100y 
Growth rate0.35 d 103mm 0.3048m  1y century 42 ft/century

______________________________________________________________________________

5. REASONING In order to calculate d, the units of a and b must be, respectively, cubed and
squared along with their numerical values, then combined algebraically with each other and
the units of c. Ignoring the values and working first with the units alone, we have

a3 m3 m32 m2
d   = =
cb2 m/ss2 m/ss21 s

Therefore, the units of d are m2/s.

SOLUTION With the units known, the numerical value may be calculated:

9.73
d  m2/s 0.75 m2/s
694.22

6. REASONING The dimensions of the variables v, x, and t are known, and the numerical
factor 3 is dimensionless. Therefore, we can solve the equation for z and then substitute the
known dimensions. The dimensions L and T can be treated as algebraic quantities to
determine the dimensions of the variable z.
Chapter 1 Problems 5
3v
SOLUTION Since v 1zxt2, it follows that z  . We know the following dimensions:
3 xt2
vL/T, xL, and t T. Since the factor 3 is dimensionless, z has the dimensions
of
v L/T 1
 
xt2 LT2 T3
____________________________________________________________________________________________

7. SSM REASONING This problem involves using unit conversions to determine the
number of magnums in one jeroboam. The necessary relationships are

1.0 magnum1.5 liters
1.0 jeroboam0.792 U. S. gallons
1.00 U. S. gallon = 3.78510–3 m3 3.785 liters

These relationships may be used to construct the appropriate conversion factors.

SOLUTION By multiplying one jeroboam by the appropriate conversion factors we can
determine the number of magnums in a jeroboam as shown below:

0.792 gallons 3.785 liters 1.0 magnum
 
1.0 jeroboam 1.0 jeroboam  1.0 gallon  1.5 liters  2.0 magnums
____________________________________________________________________________________________

8. REASONING By multiplying the quantity 1.78  103 kg/sm by the appropriate
conversions factors, we can convert the quantity to units of poise (P). These conversion
factors are obtainable from the following relationships between the various units:

1 kg1.00  103 g
1 m1.00  102 cm
1 P1 g/scm

SOLUTION The conversion from the unit kg/sm to the unit P proceeds as follows:

 kg 1.00  103 g  1 m  1 P 
1.78  103     1.78  102 P
 s  m  1 kg 1.00  102 cm 1 g/s  cm
_____________________________________________________________________________________________
6 INTRODUCTION AND MATHEMATICAL CONCEPTS
9. REASONING Multiplying an equation by a factor of 1 does not alter the equation; this is
the basis of our solution. We will use factors of 1 in the following forms:

1 gal
1, since 1 gal = 128 oz
128 oz

3.785103 m3
1, since 3.785  103 m3 = 1 gal
1 gal

1 mL
1, since 1 mL = 106 m3
106m3

SOLUTION The starting point for our solution is the fact that

Volume = 1 oz

Multiplying this equation on the right by factors of 1 does not alter the equation, so it
follows that
 1 gal 3.785103 m3  1 mL 
Volume 1 oz1111 oz    29.6 mL
128 oz  1 gal 106 m3 

Note that all the units on the right, except one, are eliminated algebraically, leaving only the
desired units of milliliters (mL).

10. REASONING To convert from gallons to cubic meters, use the equivalence
1 U.S. gal = 3.785×10−3 m3. To find the thickness of the painted layer, we use the fact that
the paint’s volume is the same, whether in the can or painted on the wall. The layer of paint
on the wall can be thought of as a very thin “box” with a volume given by the product of the
surface area (the “box top”) and the thickness of the layer. Therefore, its thickness is the
ratio of the volume to the painted surface area: Thickness = Volume/Area. That is, the larger
the area it’s spread over, the thinner the layer of paint.

SOLUTION
a. The conversion is
3.785103 m3
0.67 U.S. gallons  2.5103 m3
 U.S. gallons 
 

b. The thickness is the volume found in (a) divided by the area,

Volume 2.5103 m3
Thickness   1.9104 m
Area 13 m2
Chapter 1 Problems 7
11. SSM REASONING The dimension of the spring constant k can be determined by first
solving the equation T 2 m/k for k in terms of the time T and the mass m. Then, the
dimensions of T and m can be substituted into this expression to yield the dimension of k.

2 2
SOLUTION Algebraically solving the expression above for k gives k 4 m/T . The
2
term 4 is a numerical factor that does not have a dimension, so it can be ignored in this
analysis. Since the dimension for mass is [M] and that for time is [T], the dimension of k is

M
Dimension of k 
T2

12. REASONING AND SOLUTION The following figure (not drawn to scale) shows the
geometry of the situation, when the observer is a distance r from the base of the arch.
The angle  is related to r and h by tanh/r.
Solving for r, we find
h = 192 m

h 192 m
r   5.5103 m = 5.5 km
tan tan 2.0 r
______________________________________________________________________________

13. SSM REASONING The shortest distance between the two towns is along the line that
joins them. This distance, h, is the hypotenuse of a right triangle whose other sides are
h = 35.0 km and h = 72.0 km, as shown in the figure below.
o a

SOLUTION The angle  is given by tan h /h
o a
so that

135.0 km
tan  25.9 S of W
72.0 km

We can then use the Pythagorean theorem to find h.

h h2 h2  (35.0 km)2 (72.0 km)2  80.1 km
o a