Solution Manual For College Physics: A Strategic Approach, 4th Edition

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College Physics:A Strategic ApproachFourthEditionRandall D. KnightBrian JonesStuart FieldInstructor’s Solutions ManualforCollegePhysics: AStrategicApproachChris Porter

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I-1PptI.1.Reason:The orbital speed depends on the mass of the central body. If there were no dark matter then thestars would orbit around the Milky Way slower so the period would be longer. The answer is A.Assess:Ifthe sun’s mass were smaller then the earth would take more than 365 days to orbit.PptI.2.Reason:Assumecircular orbits. The gravitational force is centripetally directed. If Saturn is10fartherthan the earth then Kepler’s third law says its periodTis1000thatof the earth. Use the formula for centripetalacceleration.2222SSSeeSSSe222eeeeeeeee(2)(2 101000)10100101100010001010100(2)(2)vR /TR /TaRRRavR /TR /TRRR======The answer is B.Assess:Anotherway is to say the gravitational field is100weakerat that distance, so the acceleration will be100less(fromnet)Fma=PptI.3.Reason:Ifthe orbital speeds are the same, then we simply compare the centripetal accelerations.22SeSSe22eeeee10110vvaRRavvRR===The answer is A.Assess:Theacceleration is greater with the dark matter than without.PptI.4.Reason:Newton’slawforthegravitationalforcedependsonthemass.Moremassmeansmoregravitational force. The answer is A.Assess:Weexpect more massive things to pull harder on each other.PptI.5.Reason:Solveav/t= fort 220 m/s3 2 s6 0 m/svta ===The correct answer is C.Assess:Itwould take a greyhound less time to reach top speed, but 3.2 s seems reasonable for a horse.FORCE ANDMOTIONP A R TI

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I-2PartIPptI.6.Reason:Solve22fi2vva x=+forx The equation is simple becausei0v=22f2(20 m/s)33 m22(6 0 m/s )vxa===The correct choice is B.Assess:Itwould take the horse 33 m to reach top speed; this seems reasonable.PptI.7.Reason:Solve2c/avr=forr222c(15 m/s)31 7 m7 1 m/svra===The correct answer is B.Assess:Fromthe photograph it appears 32 m is in the ballpark for the radius of the turn.PptI.8.Reason:Usenetby block on foot,FFma==where212 m/sa=(twice that of a horse).2by block on foot(70 kg)(12 m/s )840 NF==The answer is B.Assess:Sincethe acceleration is moderately close to,gwe expect the answer to be moderately close to the runner’sweight.PptI.9.Reason:BecausenetFma=the net force on the dog must be in the same direction as the dog’sacceleration. In uniform circular motion the acceleration is toward the center of the circle, so the answer is D.Assess:Theforce is provided by friction of the ground acting on the dog’s paws.PptI.10.Reason:Letus call the direction of motion.x+Solvenet,xxFma=(wherenet,xFisjustthe drag force6)xDrv= −forxa.221166(0 0010 N s/m )(25m)(0 25 mm/s)1 812 m/s6 510kgxxDrvamm−−−==== −  The magnitude of this is0 2 ,gso the correct choice is A.Assess:Thisis a reasonable acceleration through a viscous fluid.PptI.11.Reason:Theacceleration is constant so we can use kinematics.Solve22fi2vva x=+forx Theequation is simple becausef0v=228i2(0 25 mm/s)1 7210m0 02m22( 1 812 m/s )vxa−−−===− The answer is A.Assess:Theparamecium comes to rest in a distance much less than its own length.PptI.12.Reason:Thedrag force6Drv=is directly proportional to the speed, so doubling the speed woulddouble the drag force. The answer is C.Assess:Thisis straight-forward given the model of the drag force as proportional to speed.PptI.13.Reason:At terminal speed the acceleration is zero, so the net force is also zero. The answer is B.Assess:At terminal speed the magnitude of the drag force is the same as the magnitude of the gravitational force.

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Forceand MotionI-3PptI.14.Reason:BecausenetFma=the net force on the falcon must be in the same direction as the falcon’sacceleration. In uniform circular motion the acceleration is toward the center of the circle, so the answer is C.Assess:Thiscentripetally-directed acceleration is provided by the lift.PptI.15.Reason:Themass of the falcon is(8 0 N)0 816 kg/g=22netliftlift()(0 816 kg)(15 m/s9 8 m/s )20 NFFmgmaFm ag=−==+=+ =The correct answer is D.Assess:Thegravitational force cannot be neglected in this problem.PptI.16.Reason:Combinethe kinematic equation with the centripetal acceleration equation.222c22(9 8 m/s )(30 m)11 76 m/s50 mva xarr====This value of acceleration is greater than the acceleration during free-fall2(9 8 m/s ),g=so the answer is B.Assess:Thecentripetal acceleration is still less than the stated maximum of215 m/s ,so this is reasonable.PptI.17.Reason:Fanddare directly proportional, so if the force (due to the weight of the person) is only half,then the deflection will also be only half. The answer is B.Assess:Thismakes intuitive sense since the variables are proportional.PptI.18.Reason:Sincedeflection is proportional to the force, then the force is three times as great when thedeflection is. So when the deflection is 12 cm the force must be233(70 kg)(9 8 m/s )2058 N2100 Nmg==Thecorrect answer is C.Assess:Thespringboard must be able to withstand forces much greater than the weight of the person.PptI.19.Reason:With all other variables held constant, the deflection is proportional to3L334dFLYwt=If the length is decreased to half, then the deflection will only be3(1 2)1 8//=as much.1 (4 0 cm)0 50 cm8d ==The correct choice is A.Assess:This makes sense since the variables are proportional.PptI.20.Reason:UseNewton’s second law in both directions. Position thex-axis down the slide.40=J0 45=

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I-4PartIycos0cosFnmgnmg=−==xsinFmgnma=−=Substitute in forn,cancelm,and solve fora.sincosmgmgma−=(sincos )ag=−Assume the acceleration is constant so we can use the kinematic equations withi0v=21()2xat=2222(3 0 m)1 4 s(sincos)(9 8 m/s )(sin 400 45cos40 )xxtag ==== − −Assess:Theslide is short, so it doesn’t take very long to slide down.PptI.21.Reason:(a)For uniform circular motion there must be a net force toward the center of the circle.(b)The new reading is,nand the amount the reading is reduced ismgn−The radius of the earth is66 3710 mThe mass of the person is2(800 N)/(9 8 m/s )81 6 kg=1d86400 s=22222rvtmgnFmammmrrrt−= =====262(81 6 kg)(6 3710m)2 8 N86400 s=Assess:Thereading is only 2.8 N less than 800 N.PptI.22.Reason:Wewill need the speed of the dolphin as it enters the water. We compute this from the 7.0 m freefall withi0v=22ff222(9 8 m/s )(7 0 m)11 71 m/svaxvax====(a)Usingother kinematic equations for the time in the water:222(1 5 m)0 2561s0 26 s11 71 m/s() /2vvvxattavvx= =====

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Forceand MotionI-5(b)Whenthe dolphin is in the water, there is an upward buoyant force due to the water that mostly cancels thedownward force of gravity on the dolphin; that is why the dolphin can generally float while underwater withoutaccelerating toward the bottom. Thus we will ignore both the buoyant and weight forces of the dolphin while it isunderwater.Then we can use Newton’s second law to write,aF/m=whereFis the force of the water on the dolphin. We findafrom the kinematic equation22fi2vva y=+Here,i11 7 m/sv=is the speed just before the dolphin hits the water,f0 m/s,v=andyis the stopping distance ofthe dolphin. Solving forawe get222i(11 7 m/s)45 6 m/s22(1 5 m)vay===so that2(210 kg)(45 6 m/s )9.6 kNFma===Assess:It takes a lot of force to stop a massive object in a short distance.

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II-1PptII.1.Reason:The water molecules that become the snowflake become more ordered, so the entropy of thewater decreases. The answer is A.Assess:Butthe entropy of the entire universe increases. The entropy of the air increases by more than the entropy ofthe water decreases.PptII.2.Reason:Theoverall entropy always increases, so the answer is B.Assess:Itis possible for the entropy of an isolated system to decrease, but it is fantastically improbable for areasonably sized system.PptII.3.Reason:Whenenergy is transferred between hot and cold (as in convection cells) the entropy increases,so the process is not reversible. The answer is B.Assess:Theentropy of the universe never decreases.PptII.4.Reason:In an isolated system the energy is conserved, but the entropy never decreases, so the answeris C.Assess:Theother options are easily eliminated.PptII.5.Reason:Therubber bands slow down the rider over a longer period of time so the force is reduced.netFtp = shows this. The answer is D.Assess:Thisis the same reason it doesn’t hurt to land on a trampoline from a large height.PptII.6.Reason:Theheight the rider will go is proportional to the maximum spring energy in the bands.212.kxmgh=But the energy in the bands is proportional to the square of the stretch.21s2.Ukx=So half the stretchwill produce a height212as high.14(2 0 m)0 5 m.=The answer isD.Assess:Halfthe stretch gives less than half the energy, so also less than half the height.PptII.7.Reason:Solve212kxmgh=for,hso the height the rider will go is inversely proportional to the mass(given that that spring energy in the numerater is the same)212/,hkxmg=so the rider will go twice as high with halfthe mass. The answer is C.Assess:Fourmeters is dangerously high.CONSERVATIONLAWSP A R TII

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II-2Part IIPptII.8.Reason:By Newton’s third law the force the pogo stick exerts on the ground is the same as the force theground exerts on the pogo stick. The work-kinetic energy theorem says the net force on the stick times the distancethe force acts(1 4 m)y= is212mvmgh=wherehis the 1.6 m free fall. The net force on the stick is the normalforce up (which we seek) minus the gravitiational force down:netF.nmg=−Put this all together.net()mghWFynmgymghnmgy==−=−=22(80 kg)(9 8 m/s )(1 6 m)(80 kg)(9 8 m/s )4000 N0 4 mmghnmgy=+=+=The answer is A.Assess:We can’t forget the gravitational force during the slowing phase.PptII.9.Reason:SinceFkx=reducing the spring constant will reduce the force on the rider. And since212kxmgh=reducing the spring constant will also reduce the final height (all else remaining equal). So the answer isA.Assess:It makes sense that a smaller force would produce a lower jump height.PptII.10.Reason:Theelastic potential energy is what makes the ball spring back up. There is some thermalenergy, but there must be more elastic energy than thermal because the ball bounces back up to more than half itsoriginal height. The answer is D.Assess:Theother choices are eliminated because0h=and0.v=PptII.11.Reason:Theforce is what makes the ball change momentum and bounce back up, so if it doesn’t bounceas high it must be because the force wasn’t as great. The answer is C.Assess:Theforce must also be less to slow the ball down over a longer time.PptII.12.Reason:Assumethe ball is in free fall.(a)22ff222( 9 8 m/s )( 2 5 m)7 0 m/svayvay===− − =(b)22ii222( 9 8 m/s )(1 4 m)5 2 m/svayvay= −=−=−− =(c)The “lost” energy is transformed into thermal energy.(d)if()(0 057 kg)(5 2 m/s7 0 m/s)120 N6 0 mspm vvFtt−+ ====Assess:Thesenumbers all seem reasonable.PptII.13.Reason:Useconservation of momentum, with the initial momentum of the system zero.wssswwss(0 30 kg)(10 m/s)0 75 m/s4 0 kgm vm vm vvm−−= −=== − The squid’s speed is thus 0.75 m/s, so the answer is D.Assess:Becausethe water’s mass is less than the squid’s, we expect the squid’s speed to be less than the water’s.PptII.14.Reason:20 75 m/s7 5 m/s0 20 svat===22(7 5 m/s ) (9 8 m/s )0 75,a g=so the answer is D.Assess:Thefirst two choices seem toobig, so we are happy with an answer in the range of1 .g

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Conservation LawsII-3PptII.15.Reason:Computethe average force fromnet(0 30 kg)(10 m/s)30 N0 10 spmvFtt====So the answer is B.Assess:This force is not very big.PptII.16.Reason:Define“what you get” as the kinetic energy of the squid and “what you had to pay” as the sumof the kinetic energies of the squid and the water.221ss2222211ssww22(4 0 kg)(0 75 m/s)7 0%(4 0 kg)(0 75 m/s)(0 30 kg)(10 m/s)m vem vm v===++Assess:In this case speed must be more important evolutionarily than efficiency.PptII.17.Reason:(0 046 kg)(63 m/s)2.9 kg m/spm v===The answer is C.Assess:Momentumis conserved for the club-ball system.PptII.18.Reason:Useconservation of momentum of the club-ball system.cciccfbbf()()()mvmvmv=+ccibbfcfc()()(0 30 kg)(40 m/s)(0 046 kg)(63 m/s)()30.3 m/s30 m/s0 30 kgmvmvvm−−===The answer is A.Assess:We expect the club to be slower after than right before.PptII.19.Reason:Seeif the kinetic energy is the same before and after.22icci11()(0 30 kg)(40 m/s)240 J22Kmv===2222fccfbbf1111()()(0 30 kg)(30.3 m/s)(0 046 kg)(63 m/s)229 J2222Kmvmv=+=+=Kinetic energy isnot conserved, so the answer is B.Assess:We would be worried iffi.KKPptII.20.Reason:221bbf2221cci2()(0 046 kg)(63 m/s)0 38()(0 30 kg)(40 m/s)mvemv===The anwer is C.Assess:Thisprocess is fairly efficient.PptII.21.Reason:221122(80 kg)(11 m/s)1180 W1200 W4 1 smvEPtt====The answer is D.Assess:It is hard to select between the choices based on ballpark expectations because they are all in a reasonablerange.

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II-4Part IIPptII.22.Reason:21 kg1609 m1 h(0 10)(1200 lb)(9 8 m/s )(3 0 mi/h)717 W720 W2 2 lb1 mi3600 sPFv===1 hp720 W720 W0 96 hp746 W==Assess:We expect a horse to be able to exert about a horsepower of power.PptII.23.Reason:Momentumis conserved in this inelastic collision.11i22i12f()()()m vmvmmv+=+11i22if12()()(100 kg)(6 0 m/s E)(130 kg)(5 0 m/s W)(0 22 m/s W)230 kgm vmvvmm++===+Assess:Sincethey are about equal in momentum the final speed is small.PptII.24.Reason:(a)Thekinetic energy of the hand needs to be sufficient to bend the board 1.2 cm, sowe set thekinetic energy of the hand equal to the spring potential energy of the board at the breaking point. Thespring constantk(800 N) (1 2 cm)6667 N/m.==221166667 N/m (0 012 m)4.38 m/s4 4 m/s220 50 kgkmvkxvxm−===(b)()224.38 m/s(0.50 kg)400 N22(0.012 m)vFmamx====Another way to look at this is when the had just touches the board its force on the board is 0 N; just before the boardbreaks the force is 800 N. Thus the average force is 400 N.Assess:That’sa fast speed to swing a hand, but possible in karate.PptII.25.Reason:(a)Use conservation of momentum.LLiLLiLsffLs()(30 kg)(4 0 m/s)()()3 0 m/s40 kgmvmvmmvvmm=+===+(b)(10 kg)(3 0 m/s)120 N0 25 spFt===(c)Use212.Kmv=22if11(30 kg)(4 0 m/s)240 J(40 kg)(3 0 m/s)180 J22KK====fi180 K240 J60 JKKK=−=−= −This energy was dissipated as thermal energy.Assess:Inpart(c)it seems reasonable to “lose” a quarter of the energy.

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III-1PptIII.1.Reason:The scaling law for specific metabolic rate vs.body mass is0 25.M− Because the wolf has amass 16 times as great as the jackrabbit’s we expect the specific metabolic rate of the jackrabbit to be0 25(16)2=times the wolf’s. The answer is A.Assess:Remember, thespecificmetabolic rate is the power used per kilogram of tissue.PptIII.2.Reason:Thewolf weighs 16 times more than the jackrabbit, but the jackrabbit’s specific metabolic rateis twice as high as the wolf’s, so the total energy used by the wolf per day would be16/ 28=times as much as thejackrabbit. The answer is C.Assess:The energy used per day is the specific metabolic rate times the mass.PptIII.3.Reason:Thespecific metabolic rate for a rat is5 W/kgso the metabolic rate is(5W/kg)(0 20 kg)1W1J/s.==In a day the rat would use1 cal1 Cal24 h3600 s(1 J/s)(1 d)20 Cal4 19 J1000 cal1 d1 h The answer is B.Assess:Thisis about1/100what a human would use; this makes sense.PptIII.4.Reason:Thepassage talks about heat transfer and that larger animals have a smaller surface-to-volumeratio, so a large animal loses less heat to the cold environment. The answer is D.Assess:Choice D is also the one that best addresses the issue.PptIII.5.Reason:Jump height is the answer that can be explained by scaling laws. We know that all animals can’tjump exactly the same height, but the variation in jump height is tiny compared to the ratio of the masses of theanimals. The scaling laws must introduce offsetting factors that leave the jump height approximately the same. Theanswer is D.Assess:Theother choices are true but don’t reflect scaling laws.PptIII.6.Reason:Assumeno friction or air resistance, and use conservation of energy.15 km/h4 17 m/s=22212221(4 17 m/s)1 m222(9 8 m/s )mvvmghmvhmgg====The answer is D.Assess:Theother options seem too big.PROPERTIES OFMATTERP A R TIII

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III-2Part IIIPptIII.7.Reason:If25% of the 480W is converted to mechanical energy of motion, then the other 75% isconverted to thermal energy in his body. So (0.75)(480 W) = 360 W. The answer is B.Assess:Ittakes some effort to stay cool on a strenuous bike ride.PptIII.8.Reason:Theheat of vaporizationvLof water is522 610 J/kg.UsevQML=to calculate the amountof water one would need to evaporate to get rid of 360 J of thermal energy every second for two hours.5v(360 J/s)(2 h)3600 s1 1 kg22.610 J/kg1 hQML===Because one liter ofwaterhas a mass of one kilogram, the answer is C.Assess:Theseems to be in the reasonable range for the amount of water a cyclist would drink in two hours.PptIII.9.Reason:If the cyclist is unable to get rid of this thermal energy thenQMc T=will tell us how muchthe temperature will go up. For mammalian bodies3400 J(kg K).c=(360 J/s)(10 min)60 s0 9 K0 9C(68 kg)(3400 J/(kg K))1 minQTMc====The answer is C.Assess:Lessthan one degree doesn’t seem like much, but that is a dangerous temperature rise for humans.PptIII.10.Reason:Thebuoyant force needs to be about the same magnitude as the weight force, so we computethe weight force on the balloon by assuming its mass is about the mass of the displaced air.332b(1 2 kg/m )(4 m )(9.8 m/s ) = 47 N50 NFmgVg===The answer is A.Assess:If the buoyant force is a little bigger than the weight force, then the balloon accelerates upward.PptIII.11.Reason:Thebuoyant force needs to be about the same magnitude as the weight force, so we computethe weight force on the balloon by assuming its mass is about the same as the mass of the displaced air. Read fromthe figure that the density of air at 10km altitude is0.43kg/m .332b(0 4 kg/m )(12 m )(9.8 m/s )47 N50 NFmgVg====The answer is A.Assess:Theweight of the balloon has hardly changed at all (being a little bit farther from the earth), so thebuoyant force doesn’t need to either.PptIII.12.Reason:Ifthe temperature were unchanged, then as the pressure halved the volume would double to38.0 m .But because the temperature in the balloon drops, this will somewhat shrink the gas volume, leading to avolume less than38.0 m .Or we can use an equation.Since there is no heat transfer(= 0),Qthen the process isadiabatic. The air is diatomic, so we use the adiabatic equation with1.4.=11.43331.4iffiifif2(4.0 m )2(4.0 m )8.0 mpp VpVVVp====Either way, the answer is C.Assess:Wedo not need to actually compute11.42to know it is less than 2.

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Properties of MatterIII-3PptIII.13.Reason:If the temperature were unchanged, then as the pressure halved the volume would double to38.0 m .But because the temperature in the balloon drops, this will somewhat shrink the gas volume, leading to avolume less than38.0 m .The answer is C.Assess:Wedo not need to actually compute11.42to know it is less than 2.PptIII.14.Reason:Examine4termmgAv=to see thattermvdecreases asincreases if,m,gandAremain thesame. The answer is C.Assess:This is in contradiction to the statement in the book “once an object has reached terminal speed, it willcontinue falling at that speed until it hits the ground” because that statement assumesisn’t changing.PptIII.15.Reason:Assumethe helium is an ideal gas. IffiTT=thenffii,p VpV=so reducing the pressure by afactor of three must triple the volume. The answer is D.Assess:Theratios confirm our intuition.PptIII.16.Reason:Assumethe air is an ideal gas. We are told the volume and the temperature of the air areconstant, butncan change.12221121pRTppnnnVnp===Since/m V=withVconstant, the density is proportional tomwhich is proportional to the number ofmolecules. Since211 3pp= thenn(and therefore the density) also decreases by a factor of 3. So the answer isD.Assess:It is intuitive that as the balloon rises the density of the air decreases.PptIII.17.Reason:Thebuoyancy force is in the opposite direction from the weight, so it is up. The drag force isin the opposite direction from the motion, so it is up too. The answer is A.Assess:Sinceit is descending at a constant rate the sum of the three forces is zero.PptIII.18.Reason:To get the air through a smaller cross section requires it to move faster. So the air movesslower in the trachea than in the nostrils.Assess:If this weren’t the case then air would pile up at the nostrils.PptIII.19.Reason:Estimatetheareaofthediaphragmtobe215 cm30 cm0 045 m .=Sincepressureisforce/area then2(7 0 kPa)(0 045 m )315 N300 NFPA===Assess:Theestimate is probably good to only one significant figure.PptIII.20.Reason:Theforcethe block exerts on the bottom is the same size as the normal force the tank exerts onthe block. In a free body diagram of the block we have the weight force down, and the buoyant force and the normalforce up. The buoyant force is the weight of the water displaced.bw,Fm g=but because the density of aluminum is32 7 kg/mthe buoyant force is1bAl2 7.Fm g=2AlbAlAlAl111(20 kg)(9 8 m/s )(0 6296)120 N2 72 7nm gFm gm gm g=−=−=−==Assess:Thisis less than the weight of the aluminum, which we expected.

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III-4Part IIIPptIII.21.Reason:(a)Assumethe air in the bladder is an ideal gas. We solve the ideal gas equation for.nThepressure at a depthdis0.ppgd=+80 ft24 384 m.=15 C288 K.=3(0 070)(7 0 L)0 00049 m .V==3230()()[101 3 kPa(1030 kg/m )(9 8 m/s )(24 384 m)](0 00049 m )0.07113 mol(8 31 J/(mol K))(288 K)pVpgdVnRTRT++====Which we report as0.071 molto two significant figures.(b)50 ft15 24 m.=34332(0.07113 mol)(8 310 J/(mol K))(288 K)0 000667 m6 710m101 3 kPa(1030 kg/m )(9 8 m/s )(15 24 m)nRTVp−===+(c)Findthe new number of moles and then subtract.323[101 3 kPa(1030 kg/m )(9 8 m/s )(15 24 m)](0 00049 m )0 05226 mol(8 31 J/(mol K))(288 K)pVnRT+===So0 07113 mol0 05226 mol0 01887 mol0 019 mol− =need to be removed.Assess:Weexpect the needed moles at 50ft to be less than at 80 ft.

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IV-1PptIV.1.Reason:Nothing goes faster than light in a vacuum, so it is first. Waves through the solid earth are nextfastest, followed by sound in air. According to the passage, tsunamis travel at hundreds of kilometers per hour, (about200 m/s),but not quite as fast as the speed of sound in air.lightearthquakesoundtsunamivvvvAssess:This ranking matches expectations.PptIV.2.Reason:As the front of the wave nears shore, its speed decreases, and the back of the wave moves fasterthan the front, so the water piles up.The answer is C.Assess:This is as the passage explained it.PptIV.3.Reason:Use1/Tf=in.vf=150 000 m750 s12 5 min15 min200 m sTv====The answer is D.Assess:This is a long period, but it makes sense given the long wavelength.PptIV.4.Reason:The frequency is unaffected. Since the speed decreases, so does the wavelength.The answer is C.Assess:The peak-to-peak distance gets smaller.PptIV.5.Reason:A tsunami is not a standing wave, but it does reflect as shown in the simulation. The reflectedwave can interfere with the primary wave to make extra large crests.The answer is B.Assess:A standing wave would need a continuous source of new pulses.PptIV.6.Reason:Use1/Tf=invf=and the given equation forvin deep water.275 m6 9 s7 s/2(9.8 m/s )(75 m)/((2)Tvg====The answer is C.Assess:Imagining oneself on a boat, this sounds reasonable.OSCILLATIONS ANDWAVESP A R TIV

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IV-2Part IVPptIV.7.Reason:Use1/Tf=invf=and the given equation forvin deep water. The effective speed withwhich the waves pass the ship is the speed of the waves plus the speed of the ship toward the waves.()2ship75 m4.9 s5 s(9.8 m/s ) 75 m4.5 m/s22Tvgv====++The answer is B.Assess:We expect the waves to reach the ship more frequently if the ship is sailing toward them.PptIV.8.Reason:Frequency and wavelength are inversely related, so higher frequency means shorter wavelength.Therefore, the answer is C.Assess:The concept here is similar to the dispersion of light in glass where the speed of blue (short wavelength)light is slower than the speed of red (long wavelength) light.PptIV.9.Reason:We need to know how far 40 wavelengths is, so we find.In Table15.1 find the speed ofultrasound in human tissue:1540m/s.v=1540 m/s0 00154 m1 0 MHzvf===Now find the number of wavelengths in 12cm.()1 wavelength0.00154 m12 cm12 cm78 wavelengths.==This is about twice the 40-wavelength half-distance, so the intensity is halved twice. In other words, the new intensityis 2502W/m .The answer is C.Assess:The penetration depth depends on the frequency, but this is reasonable for 1MHz ultrasound.PptIV.10.Reason:Doubling the frequency halves the wavelength; so twice as many wavelengths fit in the samedistance. So instead of 40 wavelengths, there would be 80; this means the intensity is halved twice, so the finalintensity is2250 W/m .The answer is C.Assess:Putting more wavelengths in the same distance means the intensity would have to be less, so this eliminateschoices A and B.PptIV.11.Reason:Higher frequency provides less penetration, as we saw in the previous problem, but withsmaller wavelength it provides better resolution. See Example 15.7. The correct answer is B.Assess:There are tradeoffs in deciding which frequency to use.PptIV.12.Reason:The beat frequency is the difference betweenthe originalfrequencyandthereflectedfrequency, but this is justo02.vvff= So increasing0fwill also increase the beat frequency. The answer is A.Assess:Increasing0fwould allow the physician to detect slower movements of the heart.PptIV.13.Reason:There are three full oscillations in0.004 sso the period is0 00133 s.The frequency is theinverse of this.13750Hz0.004 sfT===The answer is C.Assess:All of the choices are plausible answers, so we must have faith in the math.

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Oscillations and WavesIV-3PptIV.14.Reason:The second batch of wiggles (after reflection from the closed end) is not inverted; but the thirdbatch is, after reflecting from the open end, and then the closed end again. The answer is C.Assess:This is similar to a wave pulse on a string inverting upon reflection if the end is tied down, but not if the endis free.PptIV.15.Reason:The wave pulse travels down the tube and back in0.008 sdistspeedtime2(343m/s)(0 008s)1 4mLL=== The answer is C.Assess:This is a reasonable length for a tube for this measurement.PptIV.16.Reason:In the fundamental or lowest resonance mode only half a wavelength can fit in the tube. Inother words,14,L=so4 .L=See Figure16.19. The answer is D.Assess:An open-closed organ pipe sounds an octave lower than one of the same length open at both ends.PptIV.17.Reason:It will take half a period to return to the starting spot.25.0 m2.2 s29.8 m/sTLg===The answer is C.Assess:This seems about right for a5.0 mlong swing.PptIV.18.Reason:The period is independent of the mass of a simple pendulum, which is our model for thisproblem. So the time is still2.2 s.The answer is C.Assess:Try this on a long swing by timing the period for different swingers.PptIV.19.Reason:At the farthest point you are1.0 mhigher than at the bottom. See figure.We now use conservation of energy to find the speed at the bottom.22122(9 8 m/s )(1 0 m)4 4 m/s2mvmghvgh====The correct answer is C.Assess:This seems like a reasonable speed for a big swing.

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IV-4Part IVPptIV.20.Reason:(a)The instant the kangaroo touches ground the spring is still unstretched. When the kangarooleaves the ground the spring is unstretched again. This is the bottom half of a typical mass-on-spring motion (weignore the kangaroo’s trajectory through the air), so the full period is twice the time the kangaroo is in contact withthe ground. That is,0 20s.T=(b)For a mass on a spring,2.mkT=222244(70 kg)69 kN/m(0 20 s)mkT===(c)The amplitude would have to increase, but the contact time of each bounce would remain the same since theperiod does not depend on the amplitude.Assess:It would take a largekto make such a short period.PptIV.21.Reason:It is a logarithmic scale.2 700(dB)10 log27dB10500IIII=−==Assess:The answer may be surprising to those who don’t realize the decibel scale is logarithmic.PptIV.22.Reason:The air chambers allow the sound to reflect back and forth. With each reflection a bit isabsorbed, so the intensity decreases.Assess:This is a reasonable explanation of how to get a series of clicks with decreasing intensity from one loudinitial click.

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V-1PptV.1.Reason:Certainly a diverging lens diverges parallel rays, but so does a converging lens after the focalpoint. The answer is C.Assess:Ina converging lens parallel rays initially converge, but then after the focus point they diverge again.PptV.2.Reason:The light would be more spread out, making its intensity lower. Further, the image would beblurry because regions near to the focus would also contribute light that could get through the aperture. So the answeris C.Assess:Itis important to have high-quality lenses.PptV.3.Reason:Smallerwavelengths can probe smaller features, just as in a regular microscope. The answer is A.Assess:Forthis reason more information fits on a Blu-ray disk than a CD.PptV.4.Reason:Sincethe laser source is small and then gets diverged, after the light leaves the object thediverging needs to be undone to return to a small hole in the screen. So the two distances are the same through thesame optics. The answer is B.Assess:Symmetryleads us to this conclusion.PptV.5.Reason:A nearsighted person has an eyeball that is too long, but can still focus on near objects. So wherethe distance from the cornea to the retina in the horse’s eye is the longest is the answer. The answer is A.Assess:Close objects focus farther away.PptV.6.Reason:A farsighted person has an eyeball that is too short, but can still focus on distant objects. Sowhere the distance from the cornea to the retina in the horses’ eye is the shortest is the answer. The answer is B.Assess:One would expect that in general close objects are at the bottom of the field of view and distant objects moreoften in the upper part of the field of view.PptV.7.Reason:Closethings that are at the bottom of the field of vision are in focus, so the feet will be in focus;but in the upper part of the field of vision distant objects are in focus, so the close head of the person will be out offocus. The answer is B.Assess:Seethe previous two problems.PptV.8.Reason:Ifall the distances are increased then near objects would be focused at the bottom of the retina(the top of the field of view), but distant objects could not be focused at all, so the horse would be nearsighted. Theanswer is A.Assess:Havingan eyeball that is too long is a cause of nearsightedness in humans as well.OPTICSP A R TV

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V-2Part VPptV.9.Reason:Recall that the refraction at an interface depends on the indices of refraction according to Snell’slaw:()()1122sinsin.nn=If the indices of refraction of the lens and the surrounding water are very similar, then()()1221sin1.sinnn=Thus, the angles will be very similar. The correct answer is B. If the indices of refraction are verysimilar, the refraction at the lens-water interface will be minimal.Assess:Thecurvature of the lens is important for focusing, but not for the basic refraction that occurs at theinterface.PptV.10.Reason:Recallthat there is a phase change due to reflection when light is incident from a medium oflower index of refraction onto one of higher index of refraction. So when the light is in the cytoplasm (1.34n=) andreflects off the guanine (1.83n=) there will be a phase change. However, at the back of the guanine crystals, wherethe reflection occurs off the interface with cytoplasm, there is not a phase change. So the correct answer is B. Thereis one phase change.Assess:Thisis analogous to waves on strings flipping over when they reflect off an interface with a more massiverope.PptV.11.Reason:Inorderfor there to be a strong reflection, a full number of wavelengthsNplus one half awavelength/ 2must fit into the path length difference (which is twice the thickness of the crystal). Thus2/ 2.tN=+We want the largest possible,which corresponds to the smallest possible.NWith0,N=wehave()44 80 nm320 nm.t===This is the wavelength in guanine (since that is where the extra path length is). Inair, this light would have a wavelength:()() ()guanairguanair1.83320 nm586 nm1.0nn===The correct answer isC, sincethis is closest to 600 nm.Assess:Thismeans the scallops can see at least part of the spectrum that is visible to us, but not all.PptV.12.Reason:A spherical mirrorfocuses light down to make a real image. Light rays cross the axis during thefocusing process, leaving the image inverted. The correct answer is B: the image is real and inverted.Assess:The fact that the image is inverted is not important to the scallop. The brain can still flip the image inprocessing.PptV.13.Reason:Thef numberis the ratio of the focal length to the aperture of an optical system. For humans,the focal length is close to the 2.5 cm diameter of the eye, and the aperture is given as 3.0 mm. The ratio is thus about8.3. The closest answer is D.Assess:Thisis significantly higher than the f number for scallops.PptV.14.Reason:Theincident power depends on the intensity (which we will assume is constant once thecontraction begins) and the area of the pupil:2.PIAIr==The area depends on (half) the diameter squared. Sohalving the diameter means reducing the power by a factor of four. The correct answer is B.Assess:Ofcourse, your pupil contracts because the intensity increased. In this problem we are only dealing with theeffect of the pupil’s contraction once the intensity has already changed.PptV.15.Reason:Spherical aberration occurs when light rays strike the edge of an optical device and don’t quiteform the image in the same place as the other rays. Since contracting the pupil removes these aberrant rays formerlyat the edge of the pupil, the contraction should reduce the spherical aberration. The correct option is C.Assess:The eye often has a positive spherical aberration.PptV.16.Reason:TheRayleigh criterion is1.22D =for a circular aperture. If the diameter of the pupildecreases, the minimum angular separation that can be resolved goes up. The correct answer is A.Assess:Thiscan be semantically confusing. The “resolution” refers to the minimum angular spread that can beresolved. So when that goes up, it means it is more difficult for you to resolve nearby objects.

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OpticsV-3PptV.17.Reason:Inbright light the pupil is small and so many off-axis rays do not enter the pupil. Aberrationsare reduced. The focus is better with a smaller aperture (pupil).Assess:This is the same in cameras. The smaller the aperture, the greater the depth of field (objects at differentdistances are all in focus).PptV.18.Reason:Thisis a simple ratio problem. The pupil acts like a pinhole.8 8 mm (24 mm)35m6 1 mhhhhssss====Assess:Theangle between the outermost rays is the same on both sides of the pupil.PptV.19.Reason:Thefocal length is six times longer so the image will be six times taller, or 7.2mm. The totallight hitting the detector is the same, but is now spread out over62times the area, so the intensity is reduced by 36;the new intensity is20 069W/m .Assess:Asthe text says, “iffis large, the image is large and its light is spread out and dim. Iffis small, theimage is small and the light is concentrated and bright.”PptV.20.Reason:Weneed to compute the wavelength of1200 Hzsound at20 C.The speed of sound in air atthat temperature is343m/s.343 m/s0 286 m1200 Hzvf===Use the formula for the positions of dark fringes in the single-slit diffraction pattern with1.p=(1)(0 286 m)(12 m)3 4 m1 0 mppLya===Assess:Thisseems to be a reasonable diffraction pattern width given the dimensions in the problem.

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VI-1PptVI.1.Reason:The intensity of sunlight decreases with the square of the distance from the sun. If Mars is1.5 times farther away then the sunlight is21 1 5 times as intense.22Mars21400 W/m620 W/m1 5I==The correct choice is B.Assess:Thisis one reason it is quite cold on Mars. The lack of a greenhouse effect is another.PptVI.2.Reason:If the earth received appreciably more or less energy than it emitted then the temperature wouldincrease or decrease. The overall average temperature of the earth is fairly stable over the long haul (global warmingnotwithstanding). So energy in must equal energy out. The correct choice is C.Assess:Thegreenhouse effect prevents the earth from radiating as much energy as it would have otherwise, so theequilibrium temperature is higher than without the greenhouse effect. But whatever the temperature is, if it is stablethen energy in equals energy out.PptVI.3.Reason:Asblackbodiesgethotterthepeakoftheiremissionspectrumshiftstowardshorterwavelengths, so the correct choice is C.Assess:Thepeak of the sun’s radiation curve is at a much shorter wavelength than the earth’s because the surface ofthe sun is so much hotter.PptVI.4.Reason:Theenergy of a photon is158(4 1410eV s)(3 010 m/s)0 12 eV10mhcEhf−====The correct choice is D.Assess:Theunits work out correctly.PptVI.5.Reason:A water molecule has no net charge, so the net force on it due to an electric field is zero.However, because of the molecule’s electric dipole moment, an electric field does exert a torque on the molecule.The correct answer is B.Assess:Thealternating net torque on the dipole causes the water molecule to rotate back and forth. This causes themolecules to jostle each other and the temperature rises.PptVI.6.Reason:Electronslike to go against the field, so the field is to the left since the electrons are acceleratingto the right. The correct choice is A.Assess:Electronsroll up the potential hill.ELECTRICITY ANDMAGNETISMP A R TVI

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VI-2Part VIPptVI.7.Reason:UseNewton’s second law.1918231(1 610C)(70kV/1 0 cm)1 210m/s9 11 10kgVdqFqEammm−−===== The correct choice is A.Assess:Thisis a surprisingly large acceleration, but so were the other choices, so we trust the math.PptVI.8.Reason:Sincecurrent is charge per time, thenmA sis equal to mC. The correct choice is C.Assess:Thisgoes back to the fundamental definitions of current and charge in the SI system.PptVI.9.Reason:Weare given10 mAI=andV70 kV.=(10 mA)(70 kV)700 WPIV= ==The correct choice is B.Assess:Thisis a decent amount of power, but it isn’t on for very long.PptVI.10.Reason:Multiplythe usualenergy = powertime equation by 1%.0 01(700 W)(0 75 s)(0 01)5 JEP t=  ==The correct choice is C.Assess:Thisseems like a reasonable amount of energy for the x-ray pulse.PptVI.11.Reason:Assumethe maximum energy of the x-ray photon is the kinetic energy of the electron justbefore it strikes the target electrode.2311821411(2)(9 11 10kg)(1 210m/s )(0 010 m)1 1 10J22Emvma xma x−−=====  The correct choice is C.Assess:Fora photon this is fairly energetic(68 keV),as we expect for an x-ray.PptVI.12.Reason:Use.PIV=130 MJ5910 s100 min(100 A)(220 V)EEtPIV ====The correct choice is A.Assess:Thisseems like a long time at such a large current and voltage, but the point is that gasoline really stores alot of energy.PptVI.13.Reason:Use.PIV=270 kW675 A400 VPIV===To two significant digits, this would round to 680 A.Thecorrectchoice is C.Assess:Thisis a lot of current, but what is needed to produce good car performance.PptVI.14.Reason:Assumethat the transformer is ideal:12.PP=1212220 V 100 A55 A400 VVIIV===The correct choice is C.Assess:Thisis a step up transformer, so the current in the secondary is less than the current in the primary.

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Electricityand MagnetismVI-3PptVI.15.Reason:Theenergy stored in a capacitor is212() .CVHalving the voltage cuts the energy stored to 1/4the original value. The correct choice is D.Assess:Thisis a challenge for a capacitor-powered car.PptVI.16.Reason:Usethe right hand rule and curl your fingers in the clockwise direction as viewed from thesecondary coil. Your thumb points to the left. The correct choice is B.Assess:A compass needle at the center of the primary coil would verify this.PptVI.17.Reason:Theinduced current in the secondary will be in such a direction as to oppose the increase ofthe field from the primary, so the induced current will create a field to the left. The correct choice is B.Assess:Ifthe right-pointing field from the primary were decreasing, the induced current would be in the otherdirection, such as to oppose the decrease.PptVI.18.Reason:Thefrequency of the secondary is supposed to match the frequency in the primary. To doubletheresonantfrequencyinthesecondaryrequiresadecreaseinthecapacitancebyafactorof4because21.fLC= The correct choice is D.Assess:Variablecapacitors are used to tune circuits like this.PptVI.19.Reason:Use.PIV=rmsrms60 W0 50 A120 VPIV===The peak current is2more than this. The correct choice is C.Assess:We know that the peak current has to be more than the rms current; this eliminates all other choices.PptVI.20.Reason:We need to pull in lots of ideas for this integrated problem.1 0 Lof water has a mass ofwater1 0 kg.4190 J (kgC).c=Use2()VEtRP==and.Ecm T=Combine and solve for.T22()(120 V) (60 s)10C(20)(4190 J/(kgC))(1 0 kg)VtTRcm===So the final temperature isfi20 C10 C30 C.TTT=+ =+=Assess:Thepower dissipated by the resistor is 720 W, so it seems reasonable the temperature would go up by10 C.PptVI.21.Reason:Eachside of the wire loop has a mass of 50 g. To hold the loop steady we must have0.=Compute the torques around the pivot axis. The downward torque due to the weight of the two sides of the loopperpendicular to the axis plus the torque due to the weight of the side parallel to the axis is22grav2(50 g)(9 8 m/s )(5 0 cm)(50 g)(9 8 m/s )(10 cm)0 098 N m=+=The upward torque due to the current in the magnetic field must counterbalance that torque. No torque is generated inthe two sides perpendicular to the axis becausesin0,=so all the torque must be generated on the side parallel tothe axis. The current will need to be clockwise (looking from above) to producethe upward torque, and in theparallel sidesin1.=2magmag220 098 N m()19.6 A20 A(0 10 m) (0 50 T)r FL ILBIL BIL B⊥======Assess:It seems reasonable that a current of this magnitude could hold up a 200 g loop of wire.

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VII-1PptVII.1.Reason:At low speeds the quantum nature becomes more important, and the waves of the atoms canoverlap and create new forms of matter. The answer is C.Assess:Thesenew states of matter have only recently been observed, but they are very exciting.PptVII.2.Reason:Themomentum of a photon is directly related to its energy,which is directly related to itsfrequency. The ultraviolet photon has the highest frequency and therefore the greatest momentum, so it would givethe highest recoil speed to the atom due to conservation of momentum. The answer is D.Assess:We can assume that the answer has to be one of the extreme answers, not the blue or red photons which arebetween the others in energy and momentum.PptVII.3.Reason:Dueto the Doppler effect the laser photons appear to be blue-shifted which is greater energy.The answer is A.Assess:Thegreater energy pushed the atoms back to the equilibrium position.PptVII.4.Reason:Thefaster moving atoms see the photons as Doppler shifted so the photons no longer match theabsorption energy as well. The answer is B.Assess:Thewarming of the gas does not change the atomic energy levels.PptVII.5.Reason:Temperatureis related to the kinetic energy of the atoms, which is proportional to the speedsquared. Halving the speed reduces the temperature to2(1 2)1 4= the previous temperature. The answer is D.Assess:Thesenanokelvin temperatures are quite impressive.PptVII.6.Reason:Usethe equation for the de Broglie wavelength.34627h6 6310J s4 610m5000 nmmv(87)(1 6710kg)(0 0010 m/s)−−−===The answer is D.Assess:Thisis not the wavelength of a photon, it is the wavelength of the rubidium atoms themselves. With thislong wavelength they can overlap to form the Bose-Einstein condensates.MODERNPHYSICSP A R TVII

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VII-2Part VIIPptVII.7.Reason:Thesame temperature means the same average kinetic energy.22p()2 m2 mhTK==This shows that for the same temperature a more massive atom has a shorter wavelength. Sodium, with less mass, hasa longer wavelength. The answer is A.Assess:Redoingthe previous problem for sodium would allow a direct comparison.PptVII.8.Reason:Addup themass numbersof the daughter nuclei and subtract from themass numberof theoriginal uraniumand the incident neutron.(2351)(87147)2+−+=The answer is B.Assess:Becausesome neutrons escape and don’t cause further fissions, theaveragenumber of released neutronsmust be greater than one to sustain a chain reaction.PptVII.9.Reason:Thetwo fragments fly off at high speeds, and their kinetic energy is what is eventuallyconverted to electric energy by a turbine and generator. The answer is D.Assess:The released neutrons also carry away some energy, but not as much as the two large fragments.PptVII.10.Reason:Thelaw of conservation of momentum applies here, and since we are told to ignore themomentum of the neutrons, the two fragments must have momenta that are equal in magnitude and opposite indirection. The answer is C.Assess:Thespeeds will be different because the masses are different, butmvwill be the same for each.PptVII.11.Reason:Becauseof conservation of momentum (see previous problem) we know the Br nucleus isgoing faster than the La nucleus (because the Ba is lighter). In the formula for kinetic energy the speed is squared, sothe nucleus with the higher speed has the greatest kinetic energy when the momenta have the same magnitude. Theanswer is A.Assess:Wesaw this idea back in the chapter on conservation of momentum.PptVII.12.Reason:Theenergy of an electron (or a positron) is just barely over half an MeV, so a pair would beabout one MeV. So 200 MeV can create about 200 pairs. The answer is C.Assess:Thisis quite a bit of energy!PptVII.13.Reason:Inbeta decay it is like a neutron decaying into a proton and an electron. This reduces theneutron number while increasing the proton number and thus makes a more stable isotope. The answer is B.Assess:Gammadecay doesn’t change the neutron number nor the proton number. Alpha decay carries away just asmany protons as neutrons.PptVII.14.Reason:Themagnetic field will bendthe path of charged particles.Because those that go inDetector 4 are not bent very much, those particles must havea relatively high mass.The alpha particles have a massof 4 u and are positively charged. So the answer is A.Assess:Particlesof opposite charges will be bent in different directions.PptVII.15.Reason:Theradiation detected by Detector 5 was not bent by the magnetic field; this means it isn’tcharged. Gamma radiation is electromagnetic radiation (photons) and not charged. The correct choice is D.Assess:Chargedparticles are deviated by a magnetic field, but gamma rays (and all other electromagnetic radiationis not.PptVII.16.Reason:Themagnetic field will bend the path of charged particles.Because those that go inDetector 10 are bent quite a bit, those particles must have a relatively low mass.The beta-minus particles have asmall mass and are negatively charged. So the answer is B.Assess:Particlesof opposite charges will be bent in different directions.

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Solution Manual For College Physics: A Strategic Approach, 4th Edition - Page 28 preview image

Modern PhysicsVII-3PptVII.17.Reason:(a)Theequationforrelativistickineticenergyis0(1).KE=−Solveforv.Themassofanelectronis20 511 MeV.c222111vcKmc+==−222719 keV511 keV11110 2658 010m/s11Kmcvccc=−=−==++(b)19881 decay/s1 610J(15 MBq)(19 keV decay)4 5610W4 610W1 Bq1 eV−−−=(c)The activity decays exponentially.1 251201115 MBq11 MBq22t tyyRR===Assess:The11 MBq in the last part seems reasonable after less than half of a half-life.PptVII.18.Reason:1% of 700 W is 7.0 W. The energy of a 0.030 nm photon is34815(6 6310J s)(3 010 m/s)6 6310J(0 030 nm)hc−−==(a)Thenumber of photons emitted per second is1515157 0 W1 05610photons s1 110photons s6 6310J photonN−== (b)The dose in grays is the energy absorbed (7.0 W for one second is 7.0 J) per kilogram of tissue.80%, or 0.80, ofthe x-ray energy is absorbed.7.0 Jdose(0.80)0.075 Gy75 kg==The dose equivalent is the dose multiplied by the RBE which is 1.0 for x-rays.dose equivalent0 075 Gy1 075 mSv==Assess:Thisdose equivalent is quite high relative to the values in Table30.5.PptVII.19.Reason:Firstexpress the speed of the ship in m/s.87(0 12)(3 010 m/s)3 610 m/s.=(a)Fora non-relativistic calculation we use26722111 (1 710 kg)(3 610 m/s)1 1 10J22Kmv===  

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Solution Manual For College Physics: A Strategic Approach, 4th Edition - Page 29 preview image

VII-4Part VII(b)Thenumber of reactions to produce twice this kinetic energy (since only half of the fusion energy ends up askinetic energy of the ship) is21323219(2)(1 110J)1 eV7 6510reactions7 710reactions18 MeV1 610JN− ==(c)Weneed327 6510of each kind of nucleus. The mass of a21Hnucleus is2727(2)(1 6710kg)3 3410kg−−=and the mass of a32Henucleus is2727(3)(1 6710kg)5 0110kg.−−=32276213227632(7 6510)(3 3410kg)2 610kg ofH(7 6510)(5 0110kg)3 810kg ofH−−==Assess:Wewould need to haul around (or collect from space) a few million kilograms of hydrogen and helium.PptVII.20.Reason:Start with0(1)KE=−where2211vc= −and solve forv.()()00222201111111KKEEKvvcEc+==−=−++Call the later speed,vwhich corresponds to4 0 GeV.K =The ratio of the new speed to the original speed is (aftercancelingc)()()()()224 0 GeV105 MeV0226 0 GeV105 MeV011111111110 99982111KEKEvv++++−− ===−−The fraction by which the speed decreases is110 9998210 00018vv−=−=Assess:Eventhough the muon lost a third of its kinetic energy its speed is only a tiny fraction slower.PptVII.21.Reason:Ifwe follow the discussion in the text and assume the muons are traveling at a speed close tothespeedoflightthenthetimeneededtoreachthegroundintheearth’sreferenceframeis8(120 km) (3 010 m/s)400s.t ==Next we findfor the 10 GeV muons.010 GeV1196 24105 MeVKE=+=+=In the muons’ reference frame the time interval (proper time) is4004 16s96 24ts===This is(4 16s) (1 5 s)2 77=half-lives, so the number reaching the ground is2 7701150002NN==Assess:Manymore muons reach the ground than expected without time dilation. From the other reference frame wecould say the atmosphere is length contracted and arrive at the same answer.

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1-1Q1.1.Reason:The softball player starts with an initial velocity but as he slides he moves slower and slower untilcoming to rest at the base. The distance he travels in successive times will become smaller and smaller until he comesto a stop. See the figure below.Assess:Compare to Figure 1.10 in the text.Q1.2.Reason:Assess:The dots are equally spaced until the brakes are applied to the car. Equidistant dots indicate constant averagespeed. On braking, the dots get closer as the average speed decreases.Q1.3.Reason:REPRESENTINGMOTION1

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1-2Chapter1Assess:The spacing between dots is initially large, since the initial speed with which the bush baby leaves theground is large. As the bush baby rises and gravity slows the ascent, the speed decreases and therefore the spacingbetween adjacent dots in the motion diagram decreases.Q1.4.Reason:As the ball drops from the tall building t02.he ball will go faster and faster the farther it falls underthe pull of gravity. The motion diagram should show the displacements for later times to be getting larger and larger.The successive displacements in the diagram given in the text get smaller and smaller. So the diagram given in theproblem is incorrect. The correct diagram is below.Assess:Compare to Figure 1.5 in the text, which shows a motion diagram for two objects falling under the influenceof gravity. The displacements increase during the fall of the object as we reasoned.Q1.5.Reason:Position refers to a location in a coordinate frame. A displacement is the difference between twopositions. In general, displacement is a vector and requires a direction. But in this one-dimensional case, we ignorethat subtlety. The four miles Mark and Sofia walked definitely refer to a difference between their starting and endingpositions. There is no information given about a reference frame. So it is more reasonable to associate the four mileswith a displacement (magnitude) than with an absolute position.Assess:Mark and Sofia’s position could be specified as (for example) 100 m west of a particular intersection. Butwhat is described is a difference between two positions. This is more like a displacement magnitude.Note that if theirstarting point had been the origin of a coordinate system, then both Mark and Sofia would be correct.Q1.6.Reason:The distance you travel will be recorded on the odometer. As you travel, the distance you travelaccumulates, is recorded by the odometer, and is independent of the direction of travel. Your displacement is thedifference between your final position and your initial position. If you travel around a 440 m track and end up whereyou started, you have traveled 440 m; however, since you ended up where you started, your change in position andhence displacement is zero.Assess:If you watch a track meet, you will observe the 440-m race. As you watch the race, it is obvious that therunners travel a distance of 440 m (assuming they complete the race). Yet since they end up where they started, theirfinal position is the same as their initial position and hence their displacement is zero.Q1.7.Reason:Both speed and velocity are ratios with a time interval in the denominator, but speed is a scalarbecause it is the ratio of the scalar distance over the time interval while velocity is a vector because it is the ratio ofthe vector displacement over the time interval. Speed and velocity have the same SI units, but one must specify thedirection when giving a velocity.An example of speed would be that your hair grows (the end of a strand of hair moves relative to your scalp) at aspeed of about 0.75 in/month.An example of velocity (where direction matters) would be when you spring off a diving board. Your velocity couldinitially be 2.0 m/s up, while later it could be 2.0 m/s down.

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