Transcript
by
Alvin Halpern, Ph.D.
Brooklyn College
SCHAUM'S OUTLINE SERIESMcGraw-Hili
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• Alvin Halpern. Ph.D .• Professor of Physics at Brooklyn CollegeDr. Halpern has extensive teaching experience in physics and is the chairman ofthe physics department at Brooklyn College. He is a member of the executivecommittee for the doctoral program in physics at CUNY and has written numerousresearch 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 problemsin physics.QC32.H325 1988 530'.076 87-31075ISBN 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. Printedin the United States of America. Except as permitted under the United StatesCopyright Act of 1976, no part of this publication may be reproduced ordistributed in any form or by any means, or stored in a data base or retrievalsystem, without the prior written permission of the publisher.
McGraw-Hill ~A Division ofTheMcGraw·HiU Companies
CONTENTS
CHAPTER SKELETONS WITH EXAMS ix
Chapter 1 MATHEMATICAL INTRODUCTION 11.1 Planar Vectors, Scientific Notation, and Units / 1.2 Three-Dimensional Vectors; Dotand Cross Products
Chapter 2 EQUILIBRIUM OF CONCURRENT FORCES 212.1 Ropes, Knots, and Frictionless Pulleys / 2.2 Friction and Inclined Planes /2.3 Graphical and Other Problems
Chapter 3 KINEMATICS IN ONE DIMENSION 363.1 Dimensions and Units; Constant-Acceleration Problems
Chapter 4 NEWfON'S LAWS OF MOTION 514.1 Force, Mass, and Acceleration / 4.2 Friction; Inclined Planes; VectorNotation / 4.3 Two-Object and Other Problems
Chapter 5 MOTION IN A PLANE I 765.1 Projectile Motion / 5.2 Relative Motion
Chapter 6 MOTION IN A PLANE II 946.1 Circular Motion; Centripetal Force / 6.2 Law of Universal Gravitation; SatelliteMotion / 6.3 General Motion in a Plane
Chapter 7 WORK AND ENERGY 1117.1 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 1368.1 Power / 8.2 Simple Machines
Chapter 9 IMPULSE AND MOMENTUM 1469.1 Elementary Problems / 9.2 Elastic Collisions / 9.3 Inelastic Collisions and BallisticPendulums / 9.4 Collisions in Two Dimensions / 9.5 Recoil and Reaction / 9.6 Centerof Mass (see also Chap. 10)
Chapter 10 STATICS OF RIGID BODIES 17610.1 Equilibrium of Rigid Bodies / 10.2 Center of Mass (Center of Gravity)
Chapter 11 ROTATIONAL MOTION I: KINEMATICS AND DYNAMICS 20711.1 Angular Motion and Torque / 11.2 Rotational Kinematics / 11.3 Torque andRotation / 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 22812.1 Energy and Power / 12.2 Angular Impulse; the Physical Pendulum /12.3 Angular Momentum
Chapter 13 MATIER IN BULK 24713.1 Density and Specific Gravity / 13.2 Elastic Properties
iii
iv D CONTENTS
Chapter 14 SIMPLE HARMONIC MOTION 25614.1 Oscillations of a Mass on a Spring / 14.2 SHM of Pendulums and Other Systems
Chapter 15 HYDROSTATICS 27115.1 Pressure and Density / 15.2 Pascal's and Archimedes' Principles; Surface Tension
Chapter 16 HYDRODYNAMICS 28516.1 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 29717.1 Temperature Scales; Linear Expansion / 17.2 Area and Volume Expansion
Chapter 18 HEAT AND CALORIMETRY 30718.1 Heat and Energy; Mechanical Equivalent of Heat / 18.2 Calqrimetry, SpecificHeats, Heats of Fusion and Vaporization
Chapter 19 HEAT TRANSFER 31619.1 Conduction / 19.2 Convection / 19.3 Radiation
Chapter 20 GAS LAWS AND KINETIC THEORY 32620.1 The Mole Concept; the Ideal Gas Law / 20.2 Kinetic Theory / 20.3 AtmosphericProperties; Specific Heats of Solids
Chapter 21 THE FIRST LAW OF THERMODYNAMICS 34521.1 Basic Thermodynamic Concepts / 21.2 The First Law of Thermodynamics, InternalEnergy, p -V Diagrams, Cyclical Systems
Chapter 22 THE SECOND LAW OF THERMODYNAMICS 35722.1 Heat Engines; Kelvin - Planck and Clausius Statements of the SecondLaw / 22.2 Entropy
Chapter 23 WAVE MOTION 36623.1 Characteristic Properties / 23.2 Standing Waves and Resonance
Chapter 24 SOUND 37924.1 Sound Velocity; Beats; Doppler Shift / 24.2 Power, Intensity, Reverberation Time,Shock Waves
Chapter 25 COULOMB'S LAW AND ELECTRIC FIELDS 38725.1 Coulomb's Law of Electrostatic Force / 25.2 The Electric Field; Continuous ChargeDistributions; Motion of Charged Particles in an Electric Field / 25.3 Electric Flux andGauss's Law
Chapter 26 ELECTRIC POTENTIAL AND CAP ACIT ANCE 40726.1 Potential Due to Point Charges or Charge Distributions / 26.2 The PotentialFunction and the Associated Electric Field / 26.3 Energetics; Problems with MovingCharges / 26.4 Capacitance and Field Energy / 26.5 Capacitors in Combination
Chapter 27 SIMPLE ELECTRIC CIRCUITS 43227.1 Ohm's Law, Current, Resistance / 27.2 Resistors in Combination / 27.3 EMFand 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 46728.1 Force on a Moving Charge / 28.2 Force on an Electric Current / 28.3 Torque andMagnetic Dipole Moment / 28.4 Sources of the Magnetic Field; Law of Biot andSavart / 28.5 More Complex Geometries; Ampere's Law
CONTENTS D v
Chapter 29 MAGNETIC PROPERTIES OF MATTER 51029.1 The Hand M Fields; Susceptibility; Relative Permeability / 29.2 Magnets; PoleStrength
Chapter 30 INDUCED EMF: GENERA TORS AND MOTORS 52630.1 Change in Magnetic Flux, Faraday's Law, Lenz's Law / 30.2 MotionalEMF; Induced Currents and Forces / 30.3 Time-Varying Magnetic and InducedElectric Fields / 30.4 Electric Generators and Motors
Chapter 31 INDUCTANCE 55231.1 Self-Inductance / 31.2 Mutual Inductance: The Ideal Transformer
Chapter 32 ELECTRIC CIRCUITS 56632.1 R-C, R-L, L-C and R-L-C Circuits; Time Response / 32.2 AC Circuits in theSteady State / 32.3 Time Behavior of AC Circuits
Chapter 33 ELECTROMAGNETIC WAVES 59033.1 Displacement Current, Maxwell's Equations, the Speed of Light / 33.2Mathematical Description of Waves in One and Three Dimensions / 33.3 TheComponent Fields of an Electromagnetic Wave; Induced EMF / 33.4 Energy andMomentum Fluxes
Chapter 34 LIGHT AND OPTICAL PHENOMENA 60734.1 Reflection and Refraction / 34.2 Dispersion and Color / 34.3 Photometryand Illumination
Chapter 35 MIRRORS, LENSES, AND OPTICAL INSTRUMENTS 63435.1 Mirrors / 35.2 Thin Lenses / 35.3 Lensmaker's Equation; CompositeLens Systems / 35.4 Optical Instruments: Projectors, Cameras, the Eye / 35.5Optical Instruments: Microscopes and Telescopes
Chapter 36 INTERFERENCE, DIFFRACTION, AND POLARIZATION 66836.1 Interference of Light / 36.2 Diffraction and the Diffraction Grating / 36.3Polarization of Light
Chapter 37 SPECIAL RELATIVITY 68837.1 Lorentz Transformation, Length Contraction, Time Dilation, and VelocityTransformation / 37.2 Mass-Energy Relation; Relativistic Dynamics
Chapter 38 PARTICLES OF LIGHT AND WAVES OF MATTER 70838.1 Photons and the Photoelectric Effect / 38.2 Compton Scattering; X-rays; PairProduction and Annihilation / 38.3 de Broglie Waves and the UncertaintyPrinciple
Chapter 39 MODERN PHYSICS: ATOMS, NUCLEI, SOLID-STATE ELECTRONICS 72039.1 Atoms and Molecules / 39.2 Nuclei and Radioactivity / 39.3 Solid-StateElectronics
INDEX 737
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 asmall 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 arule, a problem can be solved once you have learned the ideas behind it; sometimes these very ideasare brought into sharper focus by looking at sample problems and their solutions. If you hav.edifficultywith a topic, you can select a few problems in that area, examine the solutions carefully, and thentry 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. Youshould try to gain understanding of how to approach various classes of problems, rather than memorizingparticular 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 generalphysics sequence. Ranging from the simple to the complex, they will provide you with plenty of practiceand food for thought.
The Chapter Skeletons with Exams, beginning on the next page, was devised to help students withlimited 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 andstudy their printed solutions.
• HOMEWORK problems are for practicing your problem-solving skills; cover the solution withan index card as you read, and try to solve, the problem. Do both sets if your course iscalculus 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), andgood luck.
vii
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 simplenumbers, 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 avector velocity of 40 km/h southward.
A vector quantity can be represented by an arrow drawn to scale. The length of the arrow is proportionalto the magnitude of the vector quantity (40 km/h in the above example). The direction of the arrowrepresents 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 wouldhave 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 andd~awing (to scale) each vector arrow in turn. They may be taken in any order of succession. The tail end ofeach 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 ofthe 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) 6.7 X 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 X 10-3)(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 107 m. Find its diameter in (a) millimeters,(b) megameters, (c) miles.
, (a) (1.27 x 107 m)(l000 mm/1 m) = 1.27 x 1010mm. (b) Multiply meters by 1 Mm/1Q6m to obtain 12.7 Mm.(c) Then use (1 km/1000 m)(l mi/1.61 km); the diameter is 7.89 x 103 mi.
1.11 A 100-m race is run on a 200-m-circumference circular track. The runners run eastward at the start and bendsouth. 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 onediameter = 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. Forexample, the p-component of a displacement is the distance along the p axis corresponding to the givendisplacement. It is a scalar quantity, being positive or negative as it is positively or negatively directed alongthe axis in question. In Fig. 1-4, Ap is positive. (One sometimes defines a vector component as a vectorpointing along the axis and having the size of the scalar component. If the scalar component is negativethe vector component points in the negative direction along the axis.) It is customary, and useful, to resolve avector 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 asnegative. The x component of the resultant, Rx, is the algebraic sum of all the x components. The y and zcomponents 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); thedirection 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 toendpoint 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 betaken 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-hatchsymbol on the original vector to show that it can be replaced by the sum of its vector components.) Theresultant 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 howfar 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-kmnorth displacement vectors to get a net 2.0-km south displacement vector. Similarly the 5.0-km east and2.0-km west vectors add to a 3-km east displacement vector. Because the east displacement contributes nocomponent 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 canconstruct 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 = 18N. ThenR =_31_.6_Nand tan e = 18/-26, so e =_14_5°.
• 3 kg weighs about 30 N. Since the pulleys are frictionless and with negligible mass, the tension T in thecord is the same everywhere. T holds up the weight, so T = 30 N. The forces on the leg and foot from thedevice are caused by the tensions in the cord. The horizontal or stretching force is T + T cos 30° = _56_N_,whilethe 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 tosupport 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. Theother vertical force on the man is the tension in the rope attached to the pulley above the man's head, whichmust be 2T for the pulley in equilibrium. The net vertical force is 3T, which is balanced by his weight of600 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 equalweights w. Find the angle e.
f Since the tension in the cord is w, the condition for vertical equilibrium of the mobile pulley is2w sin e = w, or sin e = t or e = 30°.
7.1 WORK DONE BY A FORCE7.1 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 thedisplacement, 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 of37° to the horizontal. (a) How much work does the child do? (b) Would you expect more or less work to bedone for the same displacement if the child pulled upward at the same angle to the horizontal?
• (a) Work = Fs cos (J = 6(4)(0.80) = _19_.2_J.(b) Less work; since the normal force on the block is less, thefriction force will be less and the needed F will be smaller.
7.4 Figure 7-1 shows the top view of two horizontal forces pulling a box along the floor: (a) How much workdoes each force do as the box is displaced 70cm along the broken line? (b) What is the total work done bythe 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 byeach 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 frictionbetween 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 isthe 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 therope 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 doesthe 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 isparallel 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 pullingforce 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 is xF cos 37°, where F cos 3r = f = J.tFN' In this case FN = mg - F sin 37°,
111
, (a) When the applied force F acts through a distance ~s = 2:rcR, the upper gear turns through 2:rcrad.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 cmare 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 adownward force F on the chain to lift the load w. (a) Determine the IMA. (b) What is the efficiency of themachine 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 toturn 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 isshortened 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 thisinstant is v dt. The increment of momentum imparted to the table by this length in coming to rest is m(v dt)v.Thus, the rate at which momentum is transferred to the table is
dp = mv2 = (2my)gdt
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 weightof a length 3y of rope.
9.14 An astronaut is doing maintenance work outside a space station. He is coasting along the station at a speed of1.00 m/s. He wishes to change his direction of motion by 90° and to increase his speed to 2.00 m/s. His totalmass is 100 kg, including his spacesuit and rocket belt, which provides a thrust of 50 N. (a) Find themagnitude and direction of the impulse needed to accomplish the desired change in motion. (b) What is theshortest 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. Theinitial 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_°withrespect to Vi' (b) Since 1= L F 6.t the shortest possible time occurs for F II I and is given by T = I IF =(I00VS kg' m/s)/(50 N) = 4_._47_s.In order to accomplish the desired change in this minimum firing time, therocket'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 exhaustby 90°, and then accelerating up to the desired final speed. How long would this take? How much rocket fuelis used, compared to the minimum?
• The deceleration to rest requires a time tl = (mv;! F); the subsequent acceleration to velocity vf requires atime t2 = (mvfl F). The total required time T' = tl + t2 = m(vi + vf)1 F. 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 theminimum, 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.
• If U\ and Uz are the initial velocities, and VI and Vz are the final velocities of objects 1 and 2, thenmomentum conservation gives mlul + mzuz = mlv1 + mzvz, or ml(ul - VI) = mz(vz - uz). Energyconservation 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 = fit gives the recoil distance as x = (3.2 fi/s)(0.25 s) = _0._8_0_fi.
9.93 A 0.25-kg ball moving in the +x direction at 13 mls is hit by a bat. Its final velocity is 19 mls in the -xdirection. 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= 13 mls 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)(13 m/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-gbulletparallel 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 remainsat x = 0, (500 g)x + (10 g)( -5 m) = 0, or x = _1O_c_m.
9.95 While coasting along a street at a constant velocity of 0.50 mis, a 20-kg girl in a 5-kg wagon sees a vicious dogin front of her. She has with her only a 3.0-kg bag of sugar which she is bringing from the grocery, and shethrows it at the dog with a forward velocity of 4.0 mls relative to her original motion. How fast is she movingafier 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 = 0.020 m/s.
9.96 A 6O-kg man dives from the stern of a 9O-kg boat with a horizontal component of velocity of 3.0 mls 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 2.0 mls 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 theboxcar be M. He throws a ball of mass m with velocity Votoward the other end, where it collides elasticallywith the wall and travels back down the length (L) of the car, striking the opposite side inelastically andcoming 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
9.US Consider the system composed of a I-kg body and a 2-kg body initially at rest at a center-to-center distance of1 m. All numerical values quoted in this exercise are to be considered exact, and the 2-kg body is to the rightof the I-kg body. (a) How far is the system's center of mass from the center of the I-kg body? (b) Beginningat t = 0 s, a net rightward force of 2 N acts on the 2-kg body. What is its resultant acceleration? (c) How fardoes the 2-kg body move between t = 0 sand t = 1 s? (d) How far is the center of mass from the I-kg body att = 1 s? (e) How far did the center of mass move between t = 0 sand t = 1 s? if) What is the acceleration ofthe center of mass, beginning at t = 0 s? (g) Suppose that all the mass in both objects were concentrated atthe center of mass, and the net rightward force of 2 N acted on this concentrated mass. What would itsacceleration be? (h) State the general theorem which is illustrated by the results of parts if) and (g) .
• (a) The initial configuration is as shown in Fig. 9-31. The center of mass of the system is located betweenbody 1 and body 2, at a distance dl from body 1, where mIdI = m2(D - dl). With D = 1 m, ml = 1 kg, andm2= 2 kg, we find that dl =~. (b) Applying Newton's second law, we obtain a2 = F;/ m2 = (2 N) -;-(2 kg) =1.00 m/s2 rightward. (c) The constant-acceleration kinematic equations give 52 = !a2t2 = O.5(1. 00)(1. oif) =O_._50_m_.(d) As before, we write mld~ = m2(D' - dD, so that d~ = 2D' /3 = [2(1.5 m)]!3 = _1._00_m_.(e) Therightward displacement is 1 - ~ = _l m_.if) The x coordinate Xc of the center of mass is given byXc = (mlxl + m2x2)/(ml + m2) = (Xl+ 2x2)/3. Therefore ac = (al + 2a2)/3. Since al = 0, ac = 2a2/3 =~ m/s2 rightward. (g)ln this case a = F /(ml + m2) = 2 N/3 kg = ~ m/s2 rightward. (h) The center of mass ofany system moves with an acceleration Be = F/M, where F == E Fext is the resultant of all external forces actingon the system, and
244 D CHAPTER 12
12.67 A student volunteer is sitting stationary on a piano stool with her feet off the floor. The stool can turn freelyon its axle.
(a) The volunteer is handed a nonrotating bicycle wheel which has handles on the axle. Holding the axlevertically with one hand, she grasps the rim of the wheel with the other and spins the wheel clockwise (asseen from above). What happens to the volunteer as she does this? (b) She now grasps the ends of thevertical axle and turns the wheel until the axle is horizontal. What happens? (c) Next she gives the rotatingwheel to the instructor, who turns the axle until it is vertical with the wheel rotating clockwise, as seen fromabove. The instructor now hands the wheel back to the volunteer. What happens? (d) The volunteer graspsthe ends of the axle and turns the axle until it is horizontal. What happens now? (e) She continues turning theaxle until it is vertical but with the wheel rotating counterclockwise as viewed from above. What is the result?
• Since the axle of the piano stool is frictionless, there are no vertical torques exerted on the stool-volunteersystem, so the vertical component of angular momentum is conserved. (There are horizontal torques; theseresult from forces that the floor exerts on the base of the stool.)
(a) The initial angular momentum is zero, so the final angular momentum must also be zero. Therefore, thevolunteer spins counterclockwise. (b) The angular momentum of the wheel is now horizontal. The volunteer'svertical component of angular momentum must now be zero, so she stops spinning. (c) When the wheel ishanded back to the volunteer, the system of wheel and volunteer has a downward vertical angularmomentum, all contributed by the wheel. The volunteer remains stationary. (d) Since the vertical componentof the total angular momentum must not change, the volunteer must rotate clockwise. (e) The wheel's angularmomentum is now upward. The volunteer must therefore have a downward vertical angular momentum tokeep the total angular momentum pointing down. She must therefore spin clockwise. (Her spin rate is twiceas fast as in part (d).)
12.68 A top consists of a uniform disk of mass mo and radius '0 rigidly attached to an axial rod of negligible mass.The top is placed on a smooth table and set spinning about its axis of symmetry with angular speed w•. Howmuch work must be done in setting the top spinning? Evaluate your result for mo = 0.050 kg, '0 = 2.0 cm, andw. = 200.1rrad/s (or 6000 rotations per minute).
• The moment of inertia 10 of the top is given by 10 = !mo'~' The work required to set the top spinning withangular speed w. is equal to the spin kinetic energy !/ow;. For the given numerical values, we find10 = (0.50)(0.050)(2.0 X 10-2)2= 10-5 kg· m2. The work required is (0.5)(1O-5)(200.1r)2= _1._97_J.
12.69 Refer to Prob. 12.68. The center of the disk is a distance d from the top's point of contact with the table. Thetop is observed to precess steadily about the vertical axis with angular speed wp- Assuming that wp «w••write wp in terms of '0. d, w•• and g. Evaluate wp for d = 3.0 cm and g = 9.80 m/s2, with the other quantitiesas given in Prob. 12.68. Is your result consistent with the assumption wp «w. ?
• For wp « w. the angular speed of steady precession is approximated by
16.1 EQUATION OF CONTINUITY, BERNOULLI'S EQUATION, TORRICELLI'S THEOREM
16.1 With regard to fluid flow, define streamline, stream tube, steady flow, turbulent flow, incompressible flow, andirrotational flow.
, A streamline is an imaginary line in a fluid, taken at an instant of time, such that the velocity vector ateach point of the line is tangential to it. A stream tube is a tube whose surface is made up of streamlines,across which there is no transport of fluid (Fig. 16-1).
In steady flow, the fluid velocity at a given location is independent of time. (However, the velocity will ingeneral vary from point to point.) Streamlines and stream tubes are fixed in steady flow, and individualparticles flow along the streamlines and within the stream tubes. Steady flow is sometimes called laminar flow.
In turbulent flow the fluid velocity changes from moment to moment at any given location. Streamlines thusno longer characterize the paths of fluid particles. Turbulent flow is often characterized by constantly changingswirls or eddies of fluid.
A flow is incompressible if the fluid density p is constant; it is irrotational if there is no swirling or circularflow of fluid.
16.2 What is the equation of continuity?
, The conservation of mass requires that the net rate of flow of mass inward across any closed surface equalthe rate of increase of the mass within the surface, assuming that there are no sources or sinks of matterwithin the surface. Applying this to a stream tube in steady flow (Fig. 16-1), we obtain the continuity equationin the form PtAtvt = P2A2V2, where P is the density, assumed uniform over the cross section of area A, and vis the average velocity over the cross section (and normal to it).
If, besides being steady, the flow is incompressible, the continuity equation reduces to Al VI = A2v2•
• (a) The hydrostatic pressure on the inside surface of the disk is given by Pi =Patm+ pgh. The air pressureof the outside of the disk is Po =Paw Since the disk has area a, the net outward force is (Pi - po)a = eE!!:E..(b) Once the disk is removed, the fluid quickly attains a (relatively) steady flow. Torricelli's theorem impliesthat the exit speed is v =V2iii. Since friction is being ignored, this is the exit speed across the entire opening.Therefore the mass flux cP = pav = pa"l/2ih and the flux of rightward momentum is (pav)v = 2pgha. If thewater loses its entire rightward momentum as it strikes the disk, the disk must absorb momentum at the rate2pgha. That is, it will experience a rightward force 2pgha, which is twice the hydrostatic force found in parta. Note that the force due to the atmosphere cancels on the left and right of the disk.
16.28 A flat plate moves normally toward a discharging jet of water at the rate of 3 m/s. The jet discharges waterat the rate of 0.1 m3/s and at a speed of 18 m/s. (a) Find the force on the plate due to the jet and(b) compare it with that if the plate were stationary .
• We do part b first. With no other information we assume the plate stops the forward motion, but there isno bounce back, i.e., the water splashes along the plate at right angles to the original motion. Then the forcenormal to the plate equals the time rate of change of momentum along the direction of the water jet, orF = (pav)v, where the term in parentheses is the mass/time hitting the plate and a is the cross-sectional areaof the jet. We are given v = 18 m/s and av = 0.1 m3/s. p = 1000 kg/m3
• F = (1000 kg/m3)(0.1 m3/s)(18 m/s) =_18_oo_N_.
In part a the plate is moving toward the stream at 3 m/s. Two things effect a change in the momentumchange/time. First if the liquid again splashes at right angles to the plate it has picked up a velocity of 3 m/sopposite to the jet's direction. The total change in forward velocity is therefore not v = 18 m/s but =18+ 3 = 21 m/s. Second, the mass of water hitting the plate per second increases from av to a(v + 3). Noting
21.1 Describe the convention for work associated with a thermodynamic system.
, In physics it is usual to define W as the work done by a system on the environment. Thus W is positivewhen a chemical system expands against the environment and hence the system transfers energy to thesurroundings; W is negative when the system contracts and the system absorbs energy from the surroundings.
Sometimes, especially in chemistry, the opposite convention is adopted.
21.2 What is meant by the internal energy of a system?
, The internal energy (U) of a system is the total energy content of the system. It is the sum of the kinetic,potential, chemical, electric, nuclear, and all other forms of energy possessed by the atoms and molecules ofthe system. A form of energy may be categorized as organized or disorganized. Organized energy is associatedwith concerted behavior of the particles composing the system, e.g., macroscopic motion of the system. Also,chemical potential energy, where a definite amount of energy is to be released for each molecule formed,represents organized energy. Disorganized energy is associated with the random interactions ("collisions") ofthe particles. Thus, the temperature of an ideal gas measures its disorganized kinetic energy. Also calledthermal energy, this category is of central interest in thermodynamics.
21.3 Give the first law of thermodynamics.
, The first law states the conservation of energy: If an amount of heat energy AQ flows into a system, thenthis energy must appear as increased internal energy AU for the system andlor work AW done by the systemon its surroundings. As an equation, the first law is AQ = AU + AW.
21.4 What is the relation between the specific heats (or heat capacities) at constant pressure and constant volume?
, When a gas is heated at constant volume, the entire heat energy supplied goes to increase the internalenergy of the gas molecules. But when a gas is heated at constant pressure, the heat supplied not onlyincreases the internal energy of the molecules but also does mechanical work in expanding the gas against theopposing constant pressure. Hence the specific heat of a gas at constant pressure, cp' is greater than itsspecific heat at constant volume, C, .. It can be shown that for an ideal gas of molecular weight M,cp - c" = RI M (ideal gas), where R is the universal gas constant.
The ratio of specific heats at constant pressure and constant volume is important for various applications,especially those involving adiabatic processes, and is often given its own symbol y (y = cplc,.). As discussedabove, this ratio is greater than unity for a gas. The kinetic theory of gases indicates that for monatomic gases(such as He, Ne, Ar), y = 1.67. For diatomic gases (such as Oz, Nz), y = 1.40 at ordinary temperatures.
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23.53 A sounding tuning fork whose frequency is 256 Hz is held over an empty measuring cylinder. See Fig. 23-10.The sound is faint, but if just the right amount of water is poured into the cylinder, it becomes loud. If theoptimal amount of water produces an air column of length 0.31 m, what is the speed of sound in air to a firstapproximation?
, The loudest sound will be heard at resonance, when the frequency of vibration of the air column in thecylinder is the same as that of the tuning fork. Since the air column is open at one end and closed at theother, we conclude that the wavelength of the vibration is four times the length of the column:A:: 4L = (4)(0.31 m) = 1.24 m. Here we have assumed that the observed resonant oscillation of the air columnis its fundamental oscillation. Since the frequency v = 256 Hz, the sound speed v = VA = 317 m/s. This is anunderestimate since the displacement antinode (or the pressure node) which is located a distance of
• Figure 24-3(a) shows a set of six wave fronts in a stationary medium. The waves have been emitted by asource which is traveling toward the right with a speed Iv,1 equal to twice the wave speed Ivl. The wave frontsare numbered and the positions of the source at the times of emission of the waves are indicated.
We observe that (by construction) the center of each numbered circular wave is located at a distance fromthe current position (location 7 at the instant depicted) of the source that is (Iv,I/lvl) times the radius of thewave. The tangent lines drawn from location 7 to the numbered wave fronts all lie at the Mach angle (1' with-i, where sin (1' = Ivl/lv,l. The tangent lines are shown in Fig. 24-3(b). Considering that the situation is asshown in every plane containing the path of the source, we have established that at each instant the wavefronts are all tangent to a right circular cone whose apex is at the current location of the source, whose axispasses through the prior locations of the source, and whose apex angle 2(1' is determined by the equation(1' = sin-1(lvl/lv,l).
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