3000 Solved Problems in Physics

by

Alvin Halpern, Ph.D.

Brooklyn College

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• Alvin Halpern. Ph.D .• Professor of Physics at Brooklyn College Dr. Halpern has extensive teaching experience in physics and is the chairman of the physics department at Brooklyn College. He is a member of the executive committee for the doctoral program in physics at CUNY and has written numerous research articles.

Project supervision was done by The Total Book. Index by Hugh C. Maddocks, Ph.D.

Library of Congress Cataloging-in-Publication Data Halpern, Alvin M.

Schaum's 3000 solved problems in physics. I. Physics-Problems, exercises, etc. I. Title. II. Title: Schaum's three thousand solved problems in physics.

QC32.H325 1988 530'.076 87-31075

ISBN 0-07-025636-5

14 15 16 17 18 19 VLP VLP 0 5 4 3 2

ISBN 0-07-025734-5

(Formerly published under ISBN 0-07-025636-5.)

Copyright © 1988 The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher.

McGraw-Hill

~

CONTENTS

CHAPTER SKELETONS WITH EXAMS

ix

Chapter 1

MATHEMATICAL

INTRODUCTION

1

1.1 Planar Vectors, Scientific Notation, and Units / 1.2 Three-Dimensional Vectors; Dot and Cross Products

Chapter 2

EQUILIBRIUM

OF CONCURRENT

FORCES

21

2.1 Ropes, Knots, and Frictionless Pulleys / 2.2 Friction and Inclined Planes / 2.3 Graphical and Other Problems

Chapter 3

KINEMATICS

IN ONE DIMENSION

36

3.1 Dimensions and Units; Constant-Acceleration Problems

Chapter 4

NEWfON'S

LAWS OF MOTION

51

4.1 Force, Mass, and Acceleration / 4.2 Friction; Inclined Planes; Vector Notation / 4.3 Two-Object and Other Problems

Chapter 5

MOTION

IN A PLANE

I

76

5.1 Projectile Motion / 5.2 Relative Motion

Chapter 6

MOTION

IN A PLANE

II

94

6.1 Circular Motion; Centripetal Force / 6.2 Law of Universal Gravitation; Satellite Motion / 6.3 General Motion in a Plane

Chapter 7

WORK AND ENERGY

111

7.1 Work Done by a Force / 7.2 Work, Kinetic Energy,. and Potential .Energy / 7.3 Conservation of Mechanical Energy / 7.4 Additional Problems

Chapter 8

POWER

AND SIMPLE MACHINES

136

8.1 Power / 8.2 Simple Machines

Chapter 9

IMPULSE

AND MOMENTUM

146

9.1 Elementary Problems / 9.2 Elastic Collisions / 9.3 Inelastic Collisions and Ballistic Pendulums / 9.4 Collisions in Two Dimensions / 9.5 Recoil and Reaction / 9.6 Center of Mass (see also Chap. 10)

Chapter 10

STATICS OF RIGID

BODIES

176

10.1 Equilibrium of Rigid Bodies / 10.2 Center of Mass (Center of Gravity)

Chapter 11

ROTATIONAL

MOTION

I: KINEMATICS

AND DYNAMICS

207

11.1 Angular Motion and Torque / 11.2 Rotational Kinematics / 11.3 Torque and Rotation / 11.4 Moment of Inertia / 11.5 Translational-Rotational Relationships / 11.6 Problems Involving Cords Around Cylinders, Rolling Objects, etc.

Chapter 12

ROTATIONAL

MOTION

II: KINETIC

ENERGY,

ANGULAR

IMPULSE,

ANGULAR

MOMENTUM

228

12.1 Energy and Power / 12.2 Angular Impulse; the Physical Pendulum / 12.3 Angular Momentum

Chapter 13

MATIER

IN BULK

247

13.1 Density and Specific Gravity / 13.2 Elastic Properties

iv

D

CONTENTS

Chapter 14

SIMPLE HARMONIC

MOTION

256

14.1 Oscillations of a Mass on a Spring / 14.2 SHM of Pendulums and Other Systems

Chapter 15

HYDROSTATICS

271

15.1 Pressure and Density / 15.2 Pascal's and Archimedes' Principles; Surface Tension

Chapter 16

HYDRODYNAMICS

285

16.1 Equation of Continuity, Bernoulli's Equation, Torricelli's Theorem / 16.2 Viscosity, Stokes' Law, Poiseuille's Law, Turbulence, Reynolds Number

Chapter 17

TEMPERATURE

AND THERMAL

EXPANSION

297

17.1 Temperature Scales; Linear Expansion / 17.2 Area and Volume Expansion

Chapter 18

HEAT

AND CALORIMETRY

307

18.1 Heat and Energy; Mechanical Equivalent of Heat / 18.2 Calqrimetry, Specific Heats, Heats of Fusion and Vaporization

Chapter 19

HEAT TRANSFER

316

19.1 Conduction / 19.2 Convection / 19.3 Radiation

Chapter 20

GAS LAWS AND KINETIC

THEORY

326

20.1 The Mole Concept; the Ideal Gas Law / 20.2 Kinetic Theory / 20.3 Atmospheric Properties; Specific Heats of Solids

Chapter 21

THE FIRST LAW OF THERMODYNAMICS

345

21.1 Basic Thermodynamic Concepts / 21.2 The First Law of Thermodynamics, Internal Energy, p -V Diagrams, Cyclical Systems

Chapter 22

THE SECOND

LAW OF THERMODYNAMICS

357

22.1 Heat Engines; Kelvin - Planck and Clausius Statements of the Second Law / 22.2 Entropy

Chapter 23

WAVE MOTION

366

23.1 Characteristic Properties / 23.2 Standing Waves and Resonance

Chapter 24

SOUND

379

24.1 Sound Velocity; Beats; Doppler Shift / 24.2 Power, Intensity, Reverberation Time, Shock Waves

Chapter 25

COULOMB'S

LAW AND ELECTRIC

FIELDS

387

25.1 Coulomb's Law of Electrostatic Force / 25.2 The Electric Field; Continuous Charge Distributions; Motion of Charged Particles in an Electric Field / 25.3 Electric Flux and Gauss's Law

Chapter 26

ELECTRIC

POTENTIAL

AND CAP ACIT ANCE

407

26.1 Potential Due to Point Charges or Charge Distributions / 26.2 The Potential Function and the Associated Electric Field / 26.3 Energetics; Problems with Moving Charges / 26.4 Capacitance and Field Energy / 26.5 Capacitors in Combination

Chapter 27

SIMPLE ELECTRIC

CIRCUITS

432

27.1 Ohm's Law, Current, Resistance / 27.2 Resistors in Combination / 27.3 EMF and Electrochemical Systems / 27.4 Electric Measurement / 27.5 Electric Power / 27.6 More Complex Circuits, Kirchhoff's Circuit Rules, Circuits with Capacitance

Chapter 28

THE MAGNETIC

FIELD

467

28.1 Force on a Moving Charge / 28.2 Force on an Electric Current / 28.3 Torque and Magnetic Dipole Moment / 28.4 Sources of the Magnetic Field; Law of Biot and Savart / 28.5 More Complex Geometries; Ampere's Law

CONTENTS

_{D v}

Chapter 29

MAGNETIC PROPERTIES

OF MATTER

510

29.1 The Hand MFields; Susceptibility; Relative Permeability / 29.2 Magnets; Pole Strength

Chapter 30

INDUCED EMF: GENERA TORS AND MOTORS

526

30.1 Change in Magnetic Flux, Faraday's Law, Lenz's Law / 30.2 Motional EMF; Induced Currents and Forces / 30.3 Time-Varying Magnetic and Induced Electric Fields / 30.4 Electric Generators and Motors

Chapter 31

INDUCTANCE

552

31.1 Self-Inductance / 31.2 Mutual Inductance: The Ideal Transformer

Chapter 32

ELECTRIC CIRCUITS

566

32.1 R-C, R-L, L-C and R-L-C Circuits; Time Response / 32.2 AC Circuits in the Steady State / 32.3 Time Behavior of AC Circuits

Chapter 33

ELECTROMAGNETIC

WAVES

590

33.1 Displacement Current, Maxwell's Equations, the Speed of Light / 33.2 Mathematical Description of Waves in One and Three Dimensions / 33.3 The Component Fields of an Electromagnetic Wave; Induced EMF / 33.4 Energy and Momentum Fluxes

Chapter 34

LIGHT AND OPTICAL PHENOMENA

607

34.1 Reflection and Refraction / 34.2 Dispersion and Color / 34.3 Photometry and Illumination

Chapter 35

MIRRORS, LENSES, AND OPTICAL INSTRUMENTS

634

35.1 Mirrors / 35.2 Thin Lenses / 35.3 Lensmaker's Equation; Composite Lens Systems / 35.4 Optical Instruments: Projectors, Cameras, the Eye / 35.5 Optical Instruments: Microscopes and Telescopes

Chapter 36

INTERFERENCE,

DIFFRACTION,

AND POLARIZATION

668

36.1 Interference of Light / 36.2 Diffraction and the Diffraction Grating / 36.3 Polarization of Light

Chapter 37

SPECIAL RELATIVITY

688

37.1 Lorentz Transformation, Length Contraction, Time Dilation, and Velocity Transformation / 37.2 Mass-Energy Relation; Relativistic Dynamics

Chapter 38

PARTICLES

OF LIGHT AND WAVES OF MATTER

708

38.1 Photons and the Photoelectric Effect / 38.2 Compton Scattering; X-rays; Pair Production and Annihilation / 38.3 de Broglie Waves and the Uncertainty

Principle

Chapter 39

MODERN PHYSICS: ATOMS, NUCLEI, SOLID-STATE ELECTRONICS

720

39.1 Atoms and Molecules / 39.2 Nuclei and Radioactivity / 39.3 Solid-State Electronics

TO THE STUDENT

This book is intended for use by students of general physics, either in calculus- or

noncalculus-based courses. Problems requiring real calculus (not merely calculus notation) are marked with a

small superscript c.

The only way to master general physics is to gain ability and sophistication in problem-solving.

This book is meant to make you a master of the art -

and should do so if used properly. As a

rule, a problem can be solved once you have learned the ideas behind it; sometimes these very ideas

are brought into sharper focus by looking at sample problems and their solutions. If you hav.e difficulty

with a topic, you can select a few problems in that area, examine the solutions carefully, and then

try to solve related problems before looking at the printed solutions.

There are numerous ways of posing a problem and, frequently, numerous ways of solving one. You

should try to gain understanding of how to approach various classes of problems, rather than memorizing

particular solutions. Understanding is better than memory for success in physics.

The problems in this book cover every important topic in a typical two- or three-semester general

physics sequence. Ranging from the simple to the complex, they will provide you with plenty of practice

and food for thought.

The Chapter Skeletons with Exams, beginning on the next page, was devised to help students with

limited time gain maximum benefit from this book. It is hoped that the use of this feature is

self-evident; still, the following remarks may help:

•

The Chapter Skeletons

divide the problems

in this book into three categories: SCAN,

HOMEWORK and EXAMS. (Turn to page ix to see an example.)

•

To gain a quick overview of the basic ideas in a chapter, review the SCAN problems and

study their printed solutions.

•

HOMEWORK problems are for practicing your problem-solving skills; cover the solution with

an index card as you read, and try to solve, the problem. Do both sets if your course is

calculus based.

•

No problem from SCANor HOMEWORKis duplicated in EXAMS,and no two Exams overlap.

Calculus-based students are urged also to take the Hard Exam. Exams run about 60 minutes,

unless otherwise indicated.

•

Still further problems constitute the two groups of Final Exams. Stay in your category(ies), and

good luck.

xxviii

_/

CHAPTER SKELETONS WITH EXAMS

Final Exams for Chapters 23-39 (160 - 180 min.)

Easy A

23.52, 24,9, 26.57, 28.55, 30.20, 32.44, 32.45, 32.46, 34.92, 36.14, 38.13, 39.2

Easy B

24.10, 25.22, 27.44, 29.52, 31.9, 32.51, 33.72, 35.63, 37.9, 39.7

Hard A

23.23, 24.17, 26.38, 27.108, 28.48, 30.109, 32.63, 34.51, 36.23, 38.45

HardB

24.30, 25.47, 27.39, 28.97, 29.53, 31.47, 33.47, 35.98, 37.29, 39.31

Calc. A

23.17, 26.88, 28.120, 30.72, 30.73, 32.19, 34.48, 38.60, 39.71

Calc. B

23.50, 25.40, 27.132, 28.125, 29.56, 31.55, 33.43, 35.103, 36.16, 37.33

1.1 PLANAR VECTORS, SCIENTIFIC NOTATION,

AND UNITS

1.1

What is a scalar quantity?

• A scalar quantity has only magnitude; it is a pure number, positive or negative. Scalars, being simple numbers, are added, subtracted, etc., in the usual way. It may have a unit after it, e.g. mass=3 kg.

1.2

What is a vector quantity?

•

A vector quantity has both magnitude and direction. For example, a car moving south at 40 km/h has a

vector velocity

of 40 km/h southward.

A vector quantity can be represented by an arrow drawn to scale. The length of the arrow is proportional to the magnitude of the vector quantity (40 km/h in the above example). The direction of the arrow represents the direction of the vector quantity.

1.3

What is the 'resultant' vector?

• The resultant of a number of similar vectors, force vectors, for example, is that single vector which would have the same effect as all the original vectors taken together.

1.4

Describe the graphical addition of vectors.

• The method for finding the resultant of several vectors consists in beginning at any convenient point and d~awing (to scale) each vector arrow in turn. They may be taken in any order of succession. The tail end of each arrow is attached to the tip end of the preceding one.

The resultant is represented by an arrow with its tail end at the starting point and its tip end at the tip of the last vector added.

1.5

Describe the parallelogram method of addition of two vectors.

• The resultant of two vectors acting at any angle may be represented by the diagonal of a parallelogram. The two vectors are drawn as the sides of the parallelogram and the resultant is its diagonal, as shown in Fig. 1-1. The direction of the resultant is away from the origin of the two vectors.

2

0

CHAPTER 1

1.8 Express each of the following in scientific notation: (a) 627.4, (b) 0.000365, (c) 20001, (d) 1.0067, (e)0.0067. , (a) 6.274 x Uf. (b) 3.65 x 10-4• (c) 2.001 x 1if. (d) 1.0067 x 10°.(e)6.7X 10-3•

1.9 Express each of the following as simple numbers xlOo: (a) 31.65 x 10-3 (b) 0.415 x 106(c) 1/(2.05 X10-3)

(d)1/(43 x 1W).

, (a) 0.03165. (b) 415,000. (c) 488.(d)0.0000233.

1.10 The diameter of the earth is about 1.27 x 107m. Find its diameter in (a) millimeters,

(b) megameters, (c) miles.

, (a) (1.27 x 107m)(l000 mm/1 m)

=

1.27 x 1010mm. (b) Multiply meters by 1 Mm/1Q6 m to obtain 12.7 Mm.

(c) Then use (1 km/1000 m)(l mi/1.61 km); the diameter is 7.89 x 103mi.

1.11 A 100-m race is run on a 200-m-circumference circular track. The runners run eastward at the start and bend south. What is the displacement of the endpoint of the race from the starting point?

, The runners move as shown in Fig. 1-3. The race is halfway around the track so the displacement is one diameter

=

2oo/:rc

=

6_3_.7_m_due south.

1.12 What is a component of a vector?

,

A component of a vector is its "shadow" (perpendicular drop) on an axis in a given direction. For example, the p-component of a displacement is the distance along the p axis corresponding to the given displacement. It is a scalar quantity, being positive or negative as it is positively or negatively directed along the axis in question. In Fig. 1-4,Ap is positive. (One sometimes defines a vector component as a vector pointing along the axis and having the size of the scalar component. If the scalar component is negative the vector component points in the negative direction along the axis.) It is customary, and useful, to resolve a vector into components along mutually perpendicular directions (rectangular components).

1.13 What is the component method for adding vectors?

, Each vector is resolved into its x, y, and z components, with negatively directed components taken as negative. The x component of the resultant, Rx, is the algebraic sum of all the x components. The y and z components of the resultant are found in a similar way.

1.14 Define the multiplication of a vector by a scalar.

, The quantity bF is a vector having magnitude Ibl F (the absolute value of b times the magnitude of F); the direction of bF is that of For -F, depending on whether b is positive or negative.

1.15 Using the graphical method, find the resultant of the following two displacements: 2 mat 40° and 4 mat 127°,

the angles being taken relative to the +x axis.

,

Choose x, y axes as shown in Fig. 1-5 and layout the displacements to scale tip to tail from the origin. Note that all angles are measured from the +x axis. The resultant vector, R, points from starting point to endpoint as shown. Measure its length on the scale diagram to find its magnitude, 4.6 m. Using a protractor, measure its angle

e

to be 101°, The resultant displacement is therefore 4.6 mat 101°.

1.16 Find the x and y components of a 25-m displacement at an angle of 210°.

I

The vector displacement and its components are shown in Fig. 1-6. The components are

x component = -25 cos 30°= -21.7 m y component = -25 sin 30°=-12.5 m

Note in particular that each component points in the negative coordinate direction and must therefore be taken as negative.

1.17 Solve Prob. 1.15 by use of rectangular components.

I

Resolve each vector into rectangular components as shown in Fig. 1-7(a) and (b). (Place a cross-hatch symbol on the original vector to show that it can be replaced by the sum of its vector components.) The resultant has the scalar components

1.20 (a) Let F have a magnitude of 300 N and make angle

e

= 30° with the positive x direction. Find F'x and Fy.

(b) Suppose that F = 300 Nand

e

= 145° (F is here in the second quadrant). Find F'x and Fy.

I

(a) F'x = 300 cos 30° = 2_5_9._8_N,Fy = 300 sin 30° = _15_0_N.(b) F'x = 300 cos 145° = (300)( -0.8192) = -245.75 N

(in the negative direction of X), Fy = 300 sin 145° = (300)( +0.5736) = 172.07 N

1.21 A car goes 5.0 km east, 3.0 km south, 2.0 km west, and 1.0 km north. (a) Determine how far north and how far east it has been displaced. (b) Find the displacement vector both graphically and algebraically.

I

(a) Recalling that vectors can be added in any order we can immediately add the 3.0-km south and 1.0-km north displacement vectors to get a net 2.0-km south displacement vector. Similarly the 5.0-km east and 2.0-km west vectors add to a 3-km east displacement vector. Because the east displacement contributes no component along the north-south line and the south displacement has no component along the east-west line, the car is -2.0 km north and 3.0 km east of its starting point. (b) Using the head-to-tail method, we easily can construct the resultant displacement D as shown in Fig. 1-11. Algebraically we note that

1.22 Find the x and y components of a 400-N force at an angle of 125° to the x axis.

I

Formal method (uses angle above positive x axis):

F'x = (400 N) cos 125° = - 229 N F;. = (400 N) sin 125° = 327 N

Visual method (uses only acute angles above or below positive or negative x axis):

1F'x1

= F cos ¢ = 400 cos 55° = 229 N

IFyI

= F sin ¢ = 400 sin 55° = 327 N By inspection of Fig. 1-12, F'x =

-1F'x1

= _-_22_9_N;F; =

IF;· I

= _32_7_N.

1.23 Add the following two coplanar forces: 30 N at 37° and 50 N at 180°.

I

Split each into components and find the resultant: Rx = 24 - 50 = - 26 N, R, = 18 + 0 = 18 N. Then

• 3 kg weighs about 30 N. Since the pulleys are frictionless and with negligible mass, the tension T in the cord is the same everywhere. T holds up the weight, so T=30 N. The forces on the leg and foot from the device are caused by the tensions in the cord. The horizontal or stretching force is T

+

T cos 30°=_56_N_,while the upward force is T

+

T sin 30°=_45_N_.

2.25 For the situation shown in Fig. 2-19, with what force must the 6OO-Nman pull downward on the rope to support himself free from the floor? Assume the pulleys have negligible friction and weight .

• Call T the tension in the rope the man is holding; T is the same throughout the one piece of rope. The other vertical force on the man is the tension in the rope attached to the pulley above the man's head, which must be 2T for the pulley in equilibrium. The net vertical force is 3T, which is balanced by his weight of 600 N. Therefore the man exerts a downward pull of _200_N_.

2.26 In the setup of Fig. 2-20, the mobile pulley and the fixed pulley, both frictionless, are associated with equal weights w. Find the angle

e.

f

Since the tension in the cord is w, the condition for vertical equilibrium of the mobile pulley is

7.1

WORK DONE BY A FORCE

7.1

A force of 3 N acts through a distance of 12 m in the direction of the force. Find the work done. • Force and displacement are in the same direction, so W= Fs

=

(3 N)(12 m)=3_6_J.

7.2

A horizontal force of 25 N pulls a box along a table. How much work does it do in pulling the box 80 cm? • Work is force times displacement through which the force acts. Here, force is in the same direction as the displacement, so W=(25 N)(0.80 m)=2_0_J.

7.3

A child pushes a toy box 4.0 m along the floor by means of a force of 6 N directed downward at an angle of 37° to the horizontal. (a) How much work does the child do? (b) Would you expect more or less work to be done for the same displacement if the child pulled upward at the same angle to the horizontal?

• (a) Work

=

Fscos(J

=

6(4)(0.80)

=

_19_.2_J.(b) Less work; since the normal force on the block is less, the friction force will be less and the needed Fwill be smaller.

7.4

Figure 7-1 shows the top view of two horizontal forces pulling a box along the floor: (a) How much work does each force do as the box is displaced 70cm along the broken line? (b) What is the total work done by the two forces in pulling the box this distance?

• (a)In each case take the component of the force in the direction of the displacement:

(85 cos 300N)(0.70 m)

=

51.5 J, (60 cos 45° N)(0.70 m)=2_9_.7_J.(b) Work is a scalar, so add the work done by each force to give _81_.2_J.

7.5

A horizontal force F pulls a 20-kg carton across the floor at constant speed. If the coefficient of sliding friction between carton and floor is 0.60, how much work does F do in moving the carton 3.0 m?

• Because horizontal speed is constant, the carton is in horizontal equilibrium: F

=

f

=

J.tFN'Normal force is the weight, 20(9.8)

=

196 N. Therefore W

=

Fx

=

0.60(196)(3.0)

=

3_53_J.

7.6

A box is dragged across a floor by a rope which makes an angle of 000 with the horizontal. The tension in the rope is 100 N while the box is dragged 15 m. How much work is done?

• Only the horizontal component of the tension, T"=100 cos 60°, does work. Thus, W

=

T"x =

(100 cos 000)(15)

=

7_50_J.

7.7

An object is pulled along the ground by a 75-N force directed 28° above the horizontal. How much work does the force do in pulling the object 8 m?

• The work done is equal to the product of the displacement, 8 m, and the component of the force that is parallel to the displacement, (75 N) cos 28°.

work =[(75 N) cos 28°](8 m)

=

5_30_J.

7.8

The coefficient of kinetic friction between a 20-kg box and the floor is 0.40. How much work does a pulling force do on the box in pulling it 8.0 m across the floor at constant speed? The pulling force is directed 37° above the horizontal.

• The work done by the force isxF cos 37°, where F cos

3r

=

f

=

J.tFN'In this caseFN

=

mg - F sin 37°,

, (a) When the applied force F acts through a distance

~s

=2:rcR, the upper gear turns through 2:rc rad.

Therefore the lower gear turns through 2:rc/ N rad and the object of mass M is elevated by a distance

~h

=

2:rcr/ N. Since Mg ~h

=

F ~s, the mechanical advantage Mg / F

=

~s / ~h

=

2:rcR + (2:rcr/ N)

=

NR / r.

(b) With N

=

3.0, R

=

40 cm, and r

=

5.0 cm, we find NR/r

=

(3.0)(40)/(5.0)

=

24.

8.31 A differential pulley (chain hoist) is shown in Fig. 8-5. Two toothed pulleys of radii r

=

10 cm and R

=

11 cm are fastened together and turn on the same axle. A continuous chain passes over the smaller (lO-cm) pulley, then around the movable pulley at the bottom, and finally around the 11-cm pulley. The operator exerts a downward force F on the chain to lift the load w. (a) Determine the IMA. (b) What is the efficiency of the machine if an applied force of 50 lb is required to lift a load of 700 lb?

, (a) Suppose that the force F moves down a distance sufficient to cause the upper rigid system of pulleys to turn one revolution. Then the smaller upper pulley unwinds a length of chain equal to its circumference, 2:rcr, while the larger upper pulley winds a length 2:rcR. As a result, the chain supporting the lower pulley is shortened by a length 2:rcR - 2:rcr. The load w is lifted half this distance, !(2:rcR - 2:rcr)=:rc(R - r) when the

descended a distance y. The length of the rope which lands on the table during an interval dt following this instant isv dt. The increment of momentum imparted to the table by this length in coming to rest ism(v dt)v.

Thus, the rate at which momentum is transferred to the table is

dp

=

mv2=(2my)g dt

and this is the force arising from stopping the downward fall of the rope. Since a length of rope y, of weight

(my)g, already lies on the tabletop, the total force on the tabletop is(2my)g

+

(my)g

=

(3my)g, or the weight of a length 3yof rope.

9.14 An astronaut is doing maintenance work outside a space station. He is coasting along the station at a speed of

1.00 m/s. He wishes to change his direction of motion by 90° and to increase his speed to 2.00m/s. His total mass is 100 kg, including his spacesuit and rocket belt, which provides a thrust of 50 N. (a)Find the

magnitude and direction of the impulse needed to accomplish the desired change in motion. (b) What is the shortest time in which the astronaut can complete the change in motion? How must the rocket be pointed?

• (a)We let i be along the initial direction of motion and y be along the desired final direction. The initial momentum mVi

=

(100 kg)(1.00 m/s)i

=

(100 kg' m/s)i. The desired final momentum mVf =

(100 kg)(2.00 m/s)y

=

(200 kg· m/s)y. The required impulse 1= mVf - mVi

=

(-100 N . s)i

+

(200 N .s)y.

The magnitude 1= 100 VS

=

224 N . s; the direction is at an angle of arccos (-I00/100VS)

=

_11_6_.6_°with respect to Vi' (b) Since 1=

L

F6.t the shortest possible time occurs for F

II

I and is given by T =I

I

F=

(I00VS kg' m/s)/(50 N)

=

4_._47_s.In order to accomplish the desired change in this minimum firing time, the rocket's exhaust must be pointed opposite to I, or at an angle of _-_63_.4_°with respect to Vi'

9.15 Suppose that the astronaut of Prob. 9.14 makes the change by decelerating to rest, turning the rocket exhaust by 90°, and then accelerating up to the desired final speed. How long would this take? How much rocket fuel is used, compared to the minimum?

• The deceleration to rest requires a time tl

=

(mv;! F); the subsequent acceleration to velocity vfrequires a time t2

=

(mvfl F). The total required time T' =tl

+

t2=m(vi

+

vf)1F. Using the given numerical values,

T'

=

(100)(1.00

+

2.00)/(50)

=

_6._00_s.Since the firing time is (6.00 - 4.47)/4.47 = 34 percent longer than the minimum, the fuel consumption is 34 percent more than the minimum.

9.2

ELASTIC

COLLISIONS

9.16 Prove that relative velocity is reversed by a head-on elastic collision.

• IfU\and Uz are the initial velocities, and VI and Vz are the final velocities of objects 1 and 2, then

momentum conservation gives mlul

+

mzuz = mlv1

+

mzvz, or ml(ul - VI) = mz(vz - uz). Energy

conservation gives !mlui

+

!mzu~

=

!m1vi

+

!mzv~. or ml(ui - vi) =mz(v~ - uD. or ml(ul - VI)(UI

+

vJ = m2(vZ - uz)(vz

+

uz). By division of equations. UI

+

VI

=

Uz

+

Vz• or Uz - UI

=

-(vz - VI). the desired result.

168

D

CHAPTER 9

• The loss in momentum during recoil is due to the impulse exerted on the gun by the 400-lb resisting force. Therefore, choosing the direction of recoil as positive,

COO

)

impulse

=

mVf - mvo (-400 Ib)t

=

0 - -32 slug (6.4 fils) from which t

=

0.25 s.

Since the resisting force is constant, the gun's recoil is uniformly decelerated. We may therefore write

v

=

1(0

+

6.4) fils

=

3.2 fils. Then x

=

fitgives the recoil distance asx

=

(3.2 fi/s)(0.25 s)

=

_0._8_0_fi. 9.93 A 0.25-kg ball moving in the +x direction at 13mls is hit by a bat. Its final velocity is 19mls in the -x

direction. The bat acts on the ball for 0.010 s. Find the average force F exerted on the ball by the bat. • We have Vo

=

13mls and vf

=

-19 m/s. The impulse equation then gives

Ft=mvf -mvo F(O.01 s)

=

(0.25 kg)( -19 m/s) - (0.25 kg)(13m/s)

from which F

=

_-_800_N_.

9.94 A 500-g pistol lies at rest on an essentially frictionless table. It accidentally discharges and shoots a 1O-gbullet parallel to the table. How far has the pistol moved by the time the bullet hits a wall 5 m away?

• Take the recoil direction as the positive x direction. Then, since the center of mass of the system remains atx

=

0, (500g)x

+

(10 g)( -5 m)

=

0, or x

=

_1O_c_m.

9.95 While coasting along a street at a constant velocity of 0.50mis, a 20-kg girl in a 5-kg wagon sees a vicious dog in front of her. She has with her only a 3.0-kg bag of sugar which she is bringing from the grocery, and she throws it at the dog with a forward velocity of 4.0mls relative to her original motion. How fast is she moving afier she throws the bag of sugar?

• Momentum conservation for the system of girl, wagon, and sugar is (20

+

5.0

+

3.0)(0.50)

=

(20

+

5.0)v +

3.0(4.5); she is now moving at v

=

0.020m/s.

9.96 A 6O-kg man dives from the stern of a 9O-kg boat with a horizontal component of velocity of 3.0mls north. Initially the boat was at rest. Find the magnitude and direction of the velocity acquired by the boat. • Let 1 refer to the man and 2 to the boat. Before diving Ut

=

U2

=

0, then mtVt

+

m2V2

=

0 and (60 kg)(3.0m/s)

+

(90kg)V2

=

O.

V2

=

-2.0m/s, or 2.0mls south.

9.97 Suppose that a boy stands at one end of a boxcar sitting on a railroad track. Let the mass of the boy and the boxcar be M. He throws a ball of mass m with velocity Votoward the other end, where it collides elastically with the wall and travels back down the length (L) of the car, striking the opposite side inelastically and coming to rest. If there is no friction in the wheels of the boxcar, describe the motion of the boxcar; • All forces are internal (Fig. 9-20). Therefore, if V and v are the velocities of the boxcar plus boy and the