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  # 9
  
  SYSTEMSOFPARTICLES
  
  EXERCISES
  Section9.1CenterofMass
  12.
  
  INTERPRETThisonedimensionalprobleminvolvesfindingthecenterofmassofasystemwithtwoobjects
  (childandfather).
  DEVELOPInonedimension,Equation9.2forthecenterofmassreducesto
  
  m x m x
  
  M
  m
  i i
  
  xcm
  
  i i
  
  Takingtheoriginofthecoordinatesystemtobeatthechild(whowedenotewithsubscript1),wehavex1=0and
  m1=28kg.Thecenteroftheseesawisthenatxcm=3.5/2m=1.75m(whereweretainanextrasignificantfigure
  becausethisisanintermediateresult).Thepositionofthefatheristheunknown,andislabeledx2.Themassofthe
  fatherism2=65kg.
  EVALUATEInsertingtheknownquantitiesintotheexpressionforcenterofmassgives
  
  xcm
  
  } 0
  x m x m
  1 1 2 2
  m1 m2
  
  x2
  
  xcm m1 m2
  m2
  
  1.75 m 28 kg 65 kg
  65 kg
  
  2.5 m
  
  fromthechild.
  ASSESSThealgebrawassimplifiedsomewhatbychoosingtheoriginofthecoordinatesystemtobeatunderthe
  childsposterior.Becausex2<3.5m,thefathercansitontheseesawandbalanceitwithhisdaughter.If m2 ? m1 ,
  thenx2=xcm,becauseitdoesnotreallymatterwherethechildsitsifthefatherweighsaton!
  13.
  
  locatedattheverticesofanequilateraltriangle.Twomassesareknownandthelocationofthecenterofmassis
  given,sowecanfindthelocationofthethirdmass.
  DEVELOPThecenterofmassofasystemofparticlesisgivenbyEquation9.2:
  
  r
  rcm
  
  i i
  
  i i
  
  m r m r
  
  M
  m
  i
  
  Weshallchoosexycoordinateswithorigin(0,0)atthemidpointofthebase.Withthisarrangement,thecenterof
  themassislocatedatxcm=0andycm=y3/2,wherey3isthepositionofthethirdmass(and,ofcourse,y1=y2=0for
  theequalmassesm1=m2=monthebase).
  EVALUATEUsingEquation9.2,theycoordinateofthecenterofmassis
  
  ycm
  
  ## m1 y1 m2 y2 m3 y3 m(0) m(0) m3 y3 m3 (2 ycm )
  
  m1 m2 m3
  m m m3
  2m m3
  91
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  92
  
  Chapter9
  
  ## Solvingform3,wehave 2m m3 2m3 , or m3 2m.
  
  ASSESSFromsymmetryconsideration,itisapparentthatxcm=0.However,wehavem+m=2matthebottom
  twoverticesofthetriangle.Because ycm y3 /2 (i.e., ycm ishalfwaytothetopvertex),weexpectthemassthereto
  be2m(SeeExample9.2).
  14.
  
  ofatwobodysystem.
  DEVELOPWiththeoriginatthecenterofthebarbell,x1=75cmandx2=75cm.UseEquation9.2tofindthe
  centerofmass.
  EVALUATE(EvaluatingEquation9.2gives
  
  xcm
  
  50 kg 75 cm 80 kg 75 cm
  50 kg 80 kg
  
  17 cm
  
  totwosignificantfigures.
  ASSESSWefindthatthecenterofmassis17cmfromthecentertowardtheheaviermass,or75cm+17cm=92
  cmfromthelightmass.ThisagreeswiththeresultofExample1.
  15.
  
  equalmasseslocatedattheverticesofanequilateraltriangleofsideL.
  DEVELOPWetakexycoordinateswiththeoriginatthecenterofonesideasshowninthefigurebelow.The
  centerofmassofasystemofparticlesisgivenbyEquation9.2:
  
  r
  rcm
  
  i i
  
  i i
  
  m r m r
  
  M
  m
  i
  
  EVALUATEFromthesymmetry(foreverymassatx,thereisanequalmassatx)wehavexcm=0.Asforycm,
  becausey=0forthetwomassesonthexaxis,and y3 L sin 60 L 3 2 forthethirdmass,Equation9.2
  gives
  
  ycm
  
  m1 y1 m2 y2 m3 y3 m(0) m(0) mL 3 2
  3
  
  L 0.289 L
  m1 m2 m3
  mmm
  6
  
  ASSESSFromsymmetryconsideration,itisapparentthatxcm=0.Ontheotherhand,wehavem+m=2matthe
  bottomtwoverticesofthetriangle,andmatthetopofthevertex.Therefore,weshouldexpectycmtobeonethird
  ofy3.Thisindeedisthecase,asycmcanberewrittenasycm=y3/3.
  16.
  
  INTERPRETThisisaonedimensionalproblemforwhichweneedtofindthecenterofmassofatwobody
  system.
  DEVELOPTaketheorigintobeatthecenteroftheEarth(seedrawingbelow),andapplyEquation9.2forthe
  centerofmass.FromAppendixE,weknowthattheEarthMoondistanceisr=3.85105m,ME=5.97Mkg,and
  MM=0.0735Mkg.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles93
  
  EVALUATEMeasuredfromthecenteroftheEarth,thecenterofmassoftheEarthMoonsystemis
  
  xcm
  
  5
  M E (0) M M r 7.35 Mkg 3.85 10 km
  
  4680 km
  ME MM
  597 Mkg 7.35 Mkg
  
  ASSESSNoticethatwearenotobligedtouseSIunits,providedallmassesareexpressedinthesameunitssothat
  theunitsofmasscancelout.
  
  Section9.2Momentum
  17.
  
  INTERPRETThisprobleminvolvesconservationoflinearmomentum(Equation9.7),whichwecanapplytofind
  thespeedofoneoutoftwoparticlesthatseparateafteranexplosion,giventhespeedoftheotherparticleandmass
  ofbothparticles.
  
  DEVELOPBeforetheexplosion,thepopcornkernelhaszeromomentum, m1 m2 v 0 .Aftertheexplosion,
  
  ## thetotalmomentumofthetwoparticlesmuststillsumtozero,sowehave m1v1 m2 v2 0 .Thus,
  
  m1 r
  v1
  m2
  
  v2
  
  sowecansolveforv2givenm1=91mg,m2=64mg,andv1=(47m/s) i .
  v
  EVALUATEInsertingthegivenquantitiesgives v2 91 mg 64 mg 47 cm/s i 67 cm/s i .
  ASSESSNoticethatthesmallerpiecemovesfasterthanthelargerpiece.Alsonoticethattotalmechanicalenergy
  2
  2
  isnotconservedbecauseK0=(m1+m2)v2=0and K m1v1 m2 v2 0 .
  18.
  
  INTERPRETTheobjectofinterestistheskater.Wewanttofindhervelocityaftershetossesasnowballina
  certaindirection.
  
  DEVELOPOnfrictionlessice,momentumwouldbeconservedintheprocess.Sincetheinitialmomentumoftheskater
  
  snowballsystemiszero,theirfinaltotalmomentummustalsobezero:
  
  r
  r
  0 m1v1 m2 v2
  
  wheresubscripts1and2refertothesnowballandskater,respectively.
  EVALUATEBymomentumconservation,thefinalvelocityoftheskateris
  
  m r
  12 kg
  r
  v2 1 v1
  53.0i 14.0 j m/s 10.6i 2.8 j m/s
  m2
  60 kg
  ASSESSAsexpected,theskatermovesintheoppositedirectionofthesnowball.Thisisaconsequenceof
  
  momentumconservation.
  19.
  
  INTERPRETThisprobleminvolvesconservationoflinearmomentum(Equation9.7),whichwecanapplytofind
  thespeedofoneoutoftwoparticlesthatseparateafteranexplosion,giventhespeedoftheotherparticleandmass
  ofbothparticles.
  
  DEVELOPBeforetheexplosion,theuraniumatomhaszeromomentum,so m1 m2 v 0 .Afterfission,the
  
  ## totalmomentumofthetwoparticlesmuststillsumtozero,sowehave m v mU 235 vU235 0 .Thus,
  
  m r
  r
  vU 235 v
  mU235
  r
  Theinitialspeedcanbeobtainedfromthekineticenergy, v 2 K m i 2 K m i ,sowecansolvefor
  vU 235 usingdatafromAppendixDforthemassesoftheparticles.
  EVALUATESolvingfor vU 235 gives
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  94
  
  Chapter9
  
  1/ 2
  
  2m K
  r
  
  vU 235
  i
  i 2.68 105 m/s i
  mU 235
  
  ## 235 u 2 1 .66 1027 kg/u
  
  2
  ASSESSBecause K 5.15 MeV << m c 3.73 Gev, relativitycanbeignored.
  
  20.
  
  INTERPRETThisprobleminvolvesusingconservationoflinearmomentumtofindthefinalspeedofamoving
  tobogganaftersomesnowdropsontoit.
  DEVELOPBecausethereisnonetexternalhorizontalforce,thetotalmomentumofthesnowtoboggansystemis
  conserved.Theinitialmomentumofthesystemis Pi mt vti . Becausethesnowandthetobogganmovetogether
  withthesamespeed v f , thefinalmomentumis Pf mt ms v f .
  EVALUATEByconservationofmomentum, Pi Pf , thefinalspeedofthetoboggansnowsystemis
  
  vf
  
  mt
  8.6 kg
  vti
  23 km/h 8.4 km/h
  mt ms
  8.6 kg 15 kg
  
  ## ASSESSToseethatourresultmakessense,letsconsiderthefollowinglimitingcases: (i) ms 0. Inthis
  
  situation,wehave v f vti , whichindicatesthatthetoboggancontinueswiththesamespeed.(ii) ms . Inthe
  situationwherealargequantityofsnowisdumpedontothetoboggan,weexpectthesystemtoslowdown
  considerably,whichisindeediswhatourequationgives(vf=0).
  
  Section9.3KineticEnergyofaSystem
  21.
  
  explodes.Wecanapplyconservationoflinearmomentumtosolvethisproblem.
  DEVELOPApplyingconservationoflinearmomentumtothebaseballgives
  
  r r
  Pi Pf
  
  ## m1 m2 vr0 m1vr1 m2 vr2
  
  2
  1
  Theinitialkineticenergyofthesystemis K i 2 m1 m2 v0 , andthetotalfinalkineticenergyis
  
  ## Kf 12 m1v12 12 m2 v22 . Therefore,thechangeinkineticenergyis
  
  1
  1
  1
  1
  1
  m1v12 m2 v22 m1 m2 v02 m1 v12 v02 m2 v22 v02
  2
  2
  2
  2
  2
  
  K K f Ki
  
  EVALUATELettheforwarddirectionbepositive.Byconservationofmomentum,thevelocityofthesecond
  piece,withmass m2 m m1 150 g 38 g 112 g, is
  
  v2
  
  m2
  
  112 g
  
  ## InSIunits v0 16.67 m/s and v1 23.6 m/s, sothedifferenceinkineticenergyis
  
  K K1 K 2
  
  1
  1
  m1 v12 v02 m2 v22 v02
  2
  2
  
  1
  1
  2
  2
  2
  38 10 3 kg 23.6 m/s 16.7 m/s 112 103 kg 14.3 m/s 16.7 m/s 02
  2
  2
  1.21 J
  ASSESSThechangeinkineticenergyforthefirstpiece(K1)ispositivebecausev1>v0,butnegativeforthe
  second(K2<0becausev2<v0).
  
  22.
  
  ## INTERPRETBeforeanexplosion,anobjecthaskineticenergy K 2 mvi . Aftertheexplosion,ithastwopieces
  
  ( m1 and m2 )eachmovingattwicetheinitialspeed, vf 2vi . Weareaskedtofindandcomparetheinternaland
  centerofmassenergiesaftertheexplosion.
  DEVELOPAftertheexplosion,thefinalkineticenergyisacombinationofthecenterofmassandinternal
  energies: K f K cm K int (Equation9.9).WeassumethattheexplosionisliketheradioactivedecayinExample
  9.6,inwhichcasemomentumisconserved.Butwecan'tassumethat m1 m2 , whichmeansthedirectionthatthe
  twopiecesflyoffrelativetotheoriginaldirectionisunknown.Butwedon'tneedtoknowthesequantitiestofind
  thefinalkineticenergy:
  1
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles95
  
  ## K f 12 m1v1f2 12 m2 v2f2 12 m1 2vi 12 m2 2vi 4
  
  2
  
  1
  2
  
  mvi2 4 K
  
  wherewehaveassumedthatthemassisconserved: m1 m2 m.
  EVALUATEBecausemomentumisconserved,thecenterofmassvelocityisconstant: vcm vi , whichimpliesthat
  thekineticenergyofthecenterofmassisthesameasthekineticenergybeforetheexplosion, K cm K . Plugging
  thisintoEquation9.9,wefind K int K f K cm 3K . Inotherwords,theinternalkineticenergyis3timesthe
  centerofmassenergy.
  ASSESSIfonedidassumethatthetwopieceshaveequalmass,thentheangle, , betweentheinitialandfinal
  velocitywouldbethesameforeachpiece,seethefigurebelow.
  
  Inthiscase,bothparticleswouldhavethesamespeedrelativetothecenterofmass:
  2
  vrel vf2 vcm
  
  2vi 2 vi 2
  
  Andtheinternalenergywouldbe
  2
  2
  K int 12 mi vi2rel 12 m1vrel
  12 m2 vrel
  3
  
  3 vi
  
  1
  2
  
  mvi2 3K
  
  Thisisexactlywhatwefoundforthegeneralcaseabove.
  
  Section9.4Collisions
  23.
  
  period.
  
  DEVELOPAnimpulseisachangeinmomentumproducedbyaforceactingonanobjectoveraspecifictimeperiod.The
  
  gravitationalforcewillbeconstantoverthetimeofthecollision,sowecanfindthegravity'simpulseoneach
  r r
  r
  spacecraftusingEquation9.10a: J Fg t p.
  EVALUATEWe'renotconcernedwiththedirectionoftheimpulse,butjustthemagnitude:
  
  ## J mg t 140 kg 8.7 m/s 2 120 103 s 146 N s
  
  Theimpulseimpartedbygravityis0.08%ofthecollisionimpulse.
  ASSESSDuringthecollision,theinfluenceofgravitycanbeneglected.That'sbecausetheaverageforcefromthe
  r r
  collisionisverylarge: F J / t 1.5 MN.
  24.
  
  INTERPRETWewanttodeterminetheaverageforceandimpulseactingonajumpingflea.
  
  DEVELOPWe'regiventheaverageaccelerationduringthejump,sothegroundmustsupplyanaverageforceonthefleaof
  
  F ma isjustmultipliedbytheflea'smass.WecanthenuseEquation9.9a J F t p tofindtheimpulse
  
  impartedbythegroundandtheresultingmomentumchangefortheflea.
  EVALUATE(a)Theaverageforceexertedbythegroundonthefleais
  
  ## 2.16 104 N ; 220 N
  
  (b)Multiplyingtheaverageforcebythetimegivestheimpulse:
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  96
  
  Chapter9
  
  ## J F t 2.16 104 N 1.2 ms 2.6 107 N s
  
  (c)Themomentumchangeforthefleaisequaltotheimpulseprovidedbythefloor:
  
  ## p J 2.6 107 N s 2.6 107 kg m/s
  
  Noticethatwecanwritethemomentumchangeinunits kg m/s thatmightbemorefamiliarformomentum.
  ASSESSIfweassumethefleastartsitsjumpfromrest,thenattheendofitsjumpitreachesavelocityof
  v v p / m 1.2 m/s. Thatseemsreasonable.
  25.
  
  INTERPRETYouneedtodeterminehowtofirearockettoobtaintheneededimpulse.
  
  DEVELOPWe'regiventheaveragethrust,sotheneededtimecomesfromEquation9.9a: t J / F .
  EVALUATEFortherequiredimpulse,thespaceprobesrocketmustfirefor
  
  J
  5.64 N s
  
  41.8 s
  F 135 103 N
  
  ASSESSThismightseemlikealongtimeforsuchasmallimpulse.Buttherocketexertsatinyforceonthespace
  
  orientation.
  
  Section9.5TotallyInelasticCollisions
  26.
  
  INTERPRETThisprobleminvolvesconservationoftotallinearmomentum.Wearetouseittofindthefinal
  theycouple.
  DEVELOPIfweassumetheswitchyardtrackisstraightandlevel,thecollisionisonedimensional,totally
  inelastic,andEquation9.11applies,so
  
  r
  r
  r m v m2 v2 m1v1 m2 v2
  vf 1 1
  
  i
  m1 m2
  m1 m2
  
  Oncewehavefoundthefinalvelocityvf,wecaninsertitintotheexpressionforthechangeinkineticenergyto
  findthefractionofkineticenergylost.
  EVALUATE(a)Insertingthegivenquantitiesintotheexpressionforvfgives
  
  ## r 56 ton 7.0 mi /h 31 ton 2.6 mi /h
  
  vf
  i 5.4 mi /h i
  56 ton 31 ton
  
  (b)Theinitialandfinalkineticenergiesare
  
  2
  
  Kf
  
  1
  2
  
  56 31
  
  1477 T mi/h ;
  
  ## T 5.43 mi/h 1284 T mi/h
  
  2
  
  whereweretainedmoresignificantfiguresthanwarrantedbythedatabecausetheseareintermediateresults.The
  fractionofkineticenergylostis K f K i K i 13% .
  ASSESSNoticethatwedidnotneedtochangetoSIunitsforpart(b)becausewetooktheratioofinitialandfinal
  correct.
  
  27.
  
  totallyinelasticcollisionbetweentwoequalmasses.
  r
  r
  DEVELOPSupposewehavetwomassesm1andm2movingwithvelocities v1 and v2 , respectively.After
  r
  undergoingatotallyinelasticcollision,thetwomassessticktogetherandmovewithfinalvelocity v f . Although
  thecollisionistotallyinelastic,momentumconservationstillapplies,andwehave(Equation9.11):
  
  r
  r
  r
  r
  r
  r m v m2 v2
  m1v1 m2 v2 m1 m2 vf vf 1 1
  m1 m2
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles97
  
  Theinitialtotalkineticenergyofthetwoparticlesystemis K i 2 m v 2 m v , whereasthefinalkineticenergy
  2
  1
  ofthesystemaftercollisionis K f 2 m1 m2 v f .Therefore,thechangeinkineticenergyisgivenby
  1
  
  K K f Ki
  
  2
  1 1
  
  2
  2 2
  
  1
  1
  1
  1
  1
  m1 m2 vf2 m1v12 m2 v22 m1 vf2 v12 m2 vf2 v22
  
  2
  2
  2
  2
  2
  
  EVALUATEInourcase,wehavem1=m2=m,v1=v,andv2=0.Theinitialkineticenergyofthesystemis
  2
  1
  therefore K i 2 mv . Thefinalspeedis
  
  m1v1 m2 v2
  mv
  1
  
  v
  m1 m2
  mm 2
  
  vf
  
  Therefore,thechangeintotalkineticenergyis
  
  1
  1
  1 v2
  1 v2
  1
  m1 vf2 v12 m2 vf2 v22 m
  v 2 m
  0 mv 2
  2
  2
  2 4
  2
  4
  4
  
  Thus,weseethathalfofthetotalinitialkineticenergyislostinthecollisionprocess.
  ASSESSForatotallyinelasticcollision,onemayshowthatthegeneralexpressionfor Kis
  
  m1m2
  1 m1v1 m2 v2
  1
  1
  m1v12 m2 v22
  v1 v2 2
  2
  m1 m2
  2
  2
  2 m1 m2
  2
  
  Clearly,Kisalwaysnegative,anditdependsontherelativespeedbetweenm1andm2.
  28.
  
  INTERPRETThisisatwodimensionalproblemthatinvolvesconservationoflinearmomentum.Wearetofind
  theinitialvelocityofoneparticle,giventheinitialvelocityoftheother,thefinalvelocityofthecombined
  particles,andthemassesofeach.
  DEVELOPApplyconservationoflinearmomentum,Equation9.11,foratotallyinelasticcollision,andsolvefor
  thevelocityofthedeuteron.
  EVALUATEInsertingthegivenquantitiesintoEquation9.11andsolvingfor vdgives
  
  v
  v
  m v mn vn 3 u 12i 20 j 1 u 28i 17 j Mm
  r
  vd t t
  
  4i 22 j Mm/s
  md
  2u
  s
  
  totwosignificantfigures.
  ASSESSThechangeinkineticenergyinthecollisionis
  
  1
  1
  1
  1
  mt vt2 mvd2 mvn2 122 20 2 3 42 21.52 2 282 17 2 1 400 u Mm/s 2
  2
  2
  2
  2
  
  whichisnegative,indicatingthatenergyisstoredinthetritium(i.e.,isconvertedtomass).Wecanregainthis
  energybysplittingtritium,whichisthebasisofthehydrogenbomb.
  29.
  
  INTERPRETThisisatotallyinelasticcollision,sincethetrucksmovetogetherasoneaftertheyhit.Youshould
  beabletofindthemassofthesecondtruckusingconservationofmomentum.
  
  DEVELOPAccordingtoEquation9.11,conservationofmomentuminthetruckcollisionimplies
  
  r
  r
  r
  m1v1 m2 v2 m1 m2 vf
  r
  
  EVALUATEThefirsttruckisatrest v1 0 , whichmeansthefinalvelocityhastobeinthesamedirectionas v2 .
  
  ## Giventhatthefirsttruckhasamassof m1 5500 kg 3800 kg 9300 kg, wecansolveforthemassofthe
  
  second:
  
  m2 m1
  
  vf
  40 km/h
  9300 kg
  14,880 kg
  v2 vf
  65 km/h 40 km/h
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  98
  
  Chapter9
  
  wouldhavebeen38km/h.
  
  Section9.6ElasticCollisions
  30.
  
  INTERPRETThisonedimensionalprobleminvolvesanelasticcollisionbetweentwoparticles,theAunucleus
  andthealphaparticle.WearetofindthefractionofthealphaparticleskineticenergythatistransferredtotheAu
  nucleus.
  DEVELOPBecausethisproblemisonedimensional,wecanapplyEquation9.15btofindthefinalvelocityofthe
  alphaparticle.Wecanthenusethisresulttofindthefinalkineticenergyofthealphaparticleinordertocalculate
  thefractionofkineticenergylosttotheAunucleus.
  EVALUATEWithm1=4u,m2=197u,andv2i=0,wefindthat
  v2 f 2 4.00 u v1i 4.00 197 u 8.00/201 v1i . Thefractionoftheinitialenergytransferredis
  
  K1i K 2f
  
  197 u v2f2
  2
  1
  2 4.00 u v1i
  1
  2
  
  197 8.00
  
  4.00 201
  
  7.80%.
  
  31.
  
  theballafteritreboundselasticallyfromamovingcar.
  DEVELOPBothmechanicalenergyandlinearmomentumareconservedinanelasticcollision.Inthisone
  dimensionalcase,conservationoflinearmomentumgives
  
  ## m1v1i m2 v2i m1v1f m2 v2f
  
  Conservationofenergygives
  
  1
  1
  1
  1
  m1v1i2 m2 v2i2 m1v1f2 m2 v2f2
  2
  2
  2
  2
  Usingthetwoconservationequations,thefinalspeedsofm1andm2are(seeEquations9.15aand9.15b):
  
  v1f
  
  m1 m2
  2m2
  v1i
  v2i
  m1 m2
  m1 m2
  
  v2f
  
  2m1
  m m1
  v1i 2
  v2i
  m1 m2
  m1 m2
  
  EVALUATELetthesubscripts1and2beforthecarandtheball,respectively.Wechoosepositivevelocitiesin
  thedirectionofthecar.Thespeedoftheballafteritreboundsis
  
  v2f
  
  2m1
  m m1
  v1i 2
  v2i 2v1i v2i 2 14 m/s 18 m/s 46 m/s
  m1 m2
  m1 m2
  
  wherewehaveused m1 ? m2 .
  ASSESSSimilarly,thefinalspeedofthecaris
  
  v1f
  
  m1 m2
  2m2
  v1i
  v2i v1i 14 m/s
  m1 m2
  m1 m2
  
  Wedonotexpectthespeedofthecartochangemuchaftercollidingwithaball.However,theballreboundswith
  amuchgreaterspeedthanbefore.Ifthecarwerestationarywith v1i 0, thenwewouldfind v2 f v2i 18 m/s.
  32.
  
  INTERPRETThisprobleminvolvesaonedimensionalelasticcollisionbetweentwomasses,soconservationof
  giventhattheobjectshavethesamespeedaftercolliding.
  DEVELOPApplyEquations9.15aand9.19bandsolveforM,giventhatv2i=0.Wearealsotoldthattheblocks
  havethesamespeedafterthecollision,soweknowthatv2f=v1f,wherewehaveinsertedthenegativesign
  becausetheblocksmustmoveinoppositedirectionsifthisisanelasticcollision.
  EVALUATEEquations9.15aand9.15bgive
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles99
  
  mM
  v1i
  mM
  2m
  v2f
  v1i
  mM
  v1f
  
  Usingv2f=v1f,wefind
  
  mM
  2m
  v1i
  v1i
  mM
  mM
  M 3m
  ASSESSAfterthecollision,thelargerblockwillhavethreetimesthekineticenergyofthesmallerblock.
  33.
  
  Theproblemisthusonedimensionalandinvolvedconservationofmechanicalenergyandlinearmomentum.
  r
  r
  DEVELOPConsiderthegeneralsituationwheretwomassesm1andm2movingwithvelocities v1 and v2 ,
  undergoelasticcollision.Bothmomentumandenergyareconservedinthisprocess.Usingtheconservation
  equations,thefinalspeedsofm1andm2are(seeEquations9.15aand9.15b):
  
  v1f
  
  m1 m2
  2m2
  v1i
  v2i
  m1 m2
  m1 m2
  
  2m1
  m m1
  v1i 2
  v2i
  m1 m2
  m1 m2
  r
  EVALUATEWechoosepositivevelocitiestobeinthedirectionof v1 . Withm1=m2=m,andv1=v=11Mm/s,
  v2f
  
  andv2=v1=v=11Mm/s,thefinalspeedsare
  
  v1f
  
  m1 m2
  2m2
  v1i
  v2i v2i v 11 Mm/s
  m1 m2
  m1 m2
  
  v2f
  
  2m1
  m m1
  v1i 2
  v2i v1i v 11 Mm/s
  m1 m2
  m1 m2
  
  ASSESSInthiscase,theprotonssimplyexchangeplacesthefinalspeedofthefirstprotonisequaltotheinitial
  speedofthesecondproton,whilethefinalspeedofthesecondprotonisequaltotheinitialspeedofthefirstproton.
  34.
  
  INTERPRETThisonedimensionalprobleminvolvesanelasticcollision,sowecanapplyconservationof
  mechanicalenergyandlinearmomentum.
  DEVELOPApplyEquations9.15aand9.15btofindtherequisiterelationships.Wearegiventhat v1i=v21=v,
  v1f=0,andm1>m2.
  
  m1 m2
  2m2
  v
  v
  m1 m2
  m1 m2
  
  m1 m2 2m2
  m1
  3
  m2
  (b)Thefinalspeedofthelessmassiveparticleis
  
  v2 f
  
  2m1
  m m2
  6 2
  v 1
  v v 2v
  4 4
  m1 m2
  m1 m2
  
  ASSESSWeknowthattheinitialspeedsmustbeintheoppositedirectionbecausethatistheonlywaythatthey
  
  PROBLEMS
  35.
  
  INTERPRETInthisproblemwewanttofindthecenterofmassofapentagonofsideawithonetrianglular
  sectionmissing.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  910
  
  Chapter9
  
  DEVELOPWechoosecoordinatesasshowninthefigurebelow.Ifthefifthisoscelestriangle(withthesame
  uniformdensityastheothers)werepresent,thecenterofmassofthewholepentagonwouldbeattheorigin,so
  
  ## my5 4mycm y5 4 ycm
  
  5m
  5
  
  whereycmgivesthepositionofthecenterofmassofthefigurewewanttofind,andy5isthepositionofthecenter
  ofmassofthefifthtriangle.Ofcourse,themassofthefigureisfourtimesthemassofthetriangle.
  EVALUATEFromsymmetry,thexcoordinateofthecenterofmassis xcm 0. Now,tocalculateycm,wemakeuse
  oftheresultobtainedinExample9.3wherethecenterofmassofanisoscelestriangleiscalculated.Thisgivesy5=
  centerofmassis
  
  ycm
  
  1
  L a
  y5 cot 36 0.115 a
  4
  6 12
  
  ASSESSFromsymmetryargument,thecenterofmassmustliealongthelinethatbisectsthefigure.Withthemissing
  triangle,weexpectittobelocatedabove y 0, whichwouldhavebeenthecenterofmassforacompletepentagon.
  36.
  
  INTERPRETThisproblemconcernsstoppingachargingrhinowithrubberbulletsthatlosealltheirmomentum
  whentheyhittheanimal.
  
  DEVELOPTheimpulseimpartedontherhinobyonebulletisequaltotherhino'schangeinmomentum(Equation9.10a:
  
  r
  r
  J b pr ).Butwedon'tknowthemassoftherhino,soitiseasiertodealwiththebullets.Theirchangein
  momentumisequalandoppositetothechangeinmomentumoftherhino: pr pb . Notethatwe'vedropped
  thevectornotation,sincethemomentaarecollinear,butwe'llassumethatthebulletsareinitiallymovinginthe
  positivedirection.We'regiventheinitialvelocityofthebullets,andweknowthattheyfallstraighttotheground
  afterimpact,sotheirfinalvelocitymustbezero.Puttingallthistogether,wehave:
  
  J b pr pb 0 mb vb0 mb vb0
  
  Wecanfindthemassoftherhinobycalculatingthetotalimpulsesuppliedbyallthebulletsfiredattherhino:
  J tot N b J b . Thistotalimpulseiswhatsupposedlybringstherhinotorest: J tot pr,tot 0 mr vr0 mr vr0 . The
  minussignisnotaproblem,sincetherhino'sinitialvelocityisnegativecomparedtothepositivevelocityofthe
  bullets.
  EVALUATE(a)Fromtheexpressionabove,theimpulseimpartedbyonebulletis
  
  ## J b mb vb0 20 g 73 m/s 1.46 kg m/s ; 1.5 N
  
  s
  
  (b)Tofindthemassoftherhino,wefirstneedtocalculatethenumberofbullets,whichistheratethegunisfired
  multipliedbythetime:
  
  ## N b rt 15 bullets/s 34 s 510 bullets
  
  Themasscanthenbefoundfromthetotalimpulsefromallthesebullets:
  
  mr
  
  pr,tot
  vr0
  
  ## N b J b 510 1.46 kg m/s
  
  920 kg
  vr0
  0.81 m/s
  
  37.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles911
  
  r
  r
  DEVELOPThepositionofthecenterofmassisgivenbyEquation9.2: rcm mi ri / M . Inthiscase,themassesareall
  equal, mi m, sothetotalmassis M 3m. Oncewefindthecenterofmassposition,thevelocityand
  accelerationcanbefoundthroughdifferentiating.
  EVALUATEThemasstermdividesout,sothecenterofmasspositionisthesumofthethreegivenvectors:
  
  r
  rcm
  
  1
  3
  
  r a i b j t
  r
  i
  
  1
  3
  
  103 t 73 i
  
  2
  3
  
  t 83 j
  
  Thecenterofmassvelocityisthefirstderivative:
  
  r
  dr
  r
  vcm cm 2t 103 i
  dt
  
  j
  2
  3
  
  Thecenterofmassaccelerationisthesecondderivative:
  
  r
  dv
  r
  acm cm 2 i
  dt
  
  ASSESSTheaccelerationisconstantandinthexdirection.Thisisduetothe t 2 terminthepositionofthe
  
  particle1.Theremustbeaforcethatacceleratesthisparticle,andcorrespondinglyacceleratesthecenterofmass.
  38.
  
  massofthesystem.
  DEVELOPThisproblemissimilartoExample9.4.Takethexaxistobehorizontalfrombowtostern,withthe
  originatthecenterofmass(CM)oftheboatandpeople.Intheabsenceofexternalhorizontalforceslikefriction,
  theCMremainsstationary.Thus
  
  ## 0 mp xpi mB xBi mp xpf mB xBf
  
  where xBi istheinitialpositionoftheCMoftheboat, xBf isitsfinalposition,and xpi and xpf aretheinitialand
  finalpositionofthepeople.Notethat xpi 0, xBi 0, xpf 0, and xBf 0. Thisequationcanberewrittenas
  
  ## since xBi xBf isthedistancetheboatmovesrelativetothefixedCM.Thedistancesarerelatedtothedimensions
  
  oftheboat,sincethelengthoftheboatisequalto
  
  ## | xpi | xBi | xBf | xpf xpf xpi xBi xBf 6.5 m
  
  EVALUATESubstitutingintothefirstequation,onefinds
  
  Thus,wefind
  
  ## xBi xBf 6.5 m
  
  mp
  mp mB
  
  6.5 m
  
  1500 kg
  72 cm
  1500 kg 12, 000 kg
  
  totwosignificantfigures,whichistheprecisionofthedata.NotethatwedidnothavetoassumethattheCMof
  theboatwasatthecenteroftheboat.
  
  ASSESSTheboatsdisplacementof72cmislessthanthedistancethepeoplewalked.Thismakessensebecause
  theboatismuchmoremassivethanthepeople.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  912
  
  39.
  
  Chapter9
  
  INTERPRETThisprobleminvolvesthecenterofmassofatwobodysystem,whichremainsstationaryinthe
  absenceofexternalhorizontalforces.
  DEVELOPWhenthemousestartsattherim,thecenterofmassofthemousebowlsystemhas xcomponent:
  
  xcm mb xb mm xm mb mm mm R mb mm
  
  sinceinitiallyxb=0andxm=R.Becausethereisnoexternalhorizontalforce(nofriction),xcmremainsconstantas
  themousedescends.Whenitreachesthecenterofthebowl,thecenterofmassofthesystemis
  
  xcm mb xb mm xm mb mm mb d 10 mm d 10
  
  mb mm
  
  Becausethecenterofmassdoesnotmove,wecanequatethesetwoexpressionsforthecenterofmasstofindthe
  ratioofmbtomm.
  
  EVALUATEUsingthefactthat2R=d,wefind
  
  mm R mb mm mb R 5 mm R 5
  
  mb mm
  
  mb 4mm
  ASSESSThebowlis4timesmoremassivethanthemouse,whichmakessensebecausethebowlhasbeen
  horizontallydisplaced.
  40.
  
  INTERPRETThisprobleminvolvestheimpulseexertedonaneedleshotintothebodyinordertoobtainasample
  ofinternalorgans.
  
  DEVELOPTheneedlestartsatrestandisacceleratedbytheforceofthespring.Wedon'tknowthisforceorhowlongitis
  
  applied,butweknowthatthemomentumgainfromthespring, p, islostwhentheneedleisstoppedbytheskin.
  Tosayitanotherway,theimpulseimpartedbythespringisequalbutoppositetotheimpulseimpartedbythe
  skin'sstoppingforce.Andwehavetheinformationneededtocalculatetheimpulsefromtheskin: J skin F t.
  Forpart(b),todeterminethepenetrationdistance,wetaketheaccelerationoftheneedleastheforceofthetissue
  actsonit: an F / mn . Sincethisaccelerationisconstant,wecanusetheformalismfromChapter2.
  EVALUATE(a)Asexplainedabove,theimpulseimpartedbythespringhasthesamemagnitudeastheimpulse
  impartedbytheskin:
  
  ## J spring J skin F t 41 mN 90 ms 3.7 mN s
  
  (b)Asfaraswecantell,theforceandcorrespondingaccelerationareconstant.Theinitialspeedoftheneedle(just
  beforeenteringtheskin)musthavebeen v0 an t , andthedistancetravelledthroughthebodyis(Equation2.10)
  
  x v0 t 12 an t 2
  
  41 mN 90 ms 2 20 cm
  F
  t 2
  
  2mn
  2 8.3 mg
  
  41.
  
  DEVELOPChoosethezaxisalongtheaxisofthecone,withtheoriginatthecenterofthebase(seefigure
  atposition(x,y,z)thereisanequalmasselementatposition(x,y,z),sotheintegraloverxandygiveszero].
  Thus,weonlyneedtofindthezcoordinateofthecenterofmass,soEquation9.4reducesto
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles913
  
  zcm
  
  zdm
  M
  
  2
  dm r 2 dz R 2 1 z h dz whereisthedensityofthecone,and M 13 R 2 h isthetotalmassofthe
  cone.
  
  EVALUATEForthezcoordinateofthecenterofmass,theintegralabovegives
  
  zcm
  
  1
  M
  
  z dm
  0
  
  3
  2
  z R 2 1 z h dz
  2
  R h 0
  
  h
  3 h 2 2h 2 h 2
  3
  2 z 2 z 3
  1
  z
  2 dz
  
  h
  h 0
  h
  h
  3 4
  4
  h 2
  
  sothecompletecenterofmasscoordinateis(0,0,h/4).
  ASSESSTheresultmakessensebecauseweexpectzcmtobeclosertothebottomoftheconebecausemoremass
  isdistributedinthisregion.
  42.
  
  INTERPRETThisprobleminvolvesconservationoflinearmomentum,whichwecanusetofindthemassandthe
  directionofmotionofthesecondfirecrackerfragment.
  DEVELOPApplyconservationoflinearmomentum.Becausethefirecrackerisinitiallyatrest,theinitial
  momentumofthesystemiszero.Aftertheexplosion(ignoringairresistanceandrelativisticeffects),thetotal
  linearmomentumisstillzero,andisexpressedas
  
  r
  r
  0 m1v1 m2 v2
  
  ## wherem1=14g, v1 48 m/s i ,and v2 32 m/s .
  
  r
  r
  EVALUATEBecausemassisalwayspositive,weknowthatthedirectionof v2 isoppositeto v1 ,so
  r
  r
  v2 32 m/s i ,sothedirectionofmotionis i ,oroppositetothedirectionof v1 .Evaluatingtheexpression
  fortotallinearmomentumgivesm2=m1v1/v2=(14g)(48m/s)/(32m/s)=21g.
  ASSESSWefindthattheslowermovingmassisgreaterthanthefastermovingmass,asexpected.
  43.
  
  DEVELOPSincethetotalmomentumofthesystemisconservedintheprocess,wehave
  
  Pi Pf
  
  ## m1v1 (m1 m2 )vf
  
  Thepotentialenergyofthespringatmaximumcompressionequalsthekineticenergyofthetwocarsystemprior
  2
  2
  1
  1
  tocontactwiththespring: 2 kxmax 2 (m1 m2 )vf .
  For(b),wenotethatwhenthecarsrebound,theyarecoupledtogetherandbothhavethesamevelocity.Sincethe
  2
  , istransformedbackintokineticenergyof
  springisideal(byassumption),itsmaximumpotentialenergy, 12 k xmax
  thecars.
  EVALUATE(a)Thesecondcarisinitiallyatrestso v2 0. Bymomentumconservation,thespeedofthecars
  aftercollisionis
  
  vf
  
  m1v1
  (9, 400 kg)(8.5 m/s)
  
  3.92 m/s
  m1 m2 11, 000 kg 9, 400 kg
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  914
  
  Chapter9
  
  xmax vf
  
  m1 m2
  11, 000 kg 9, 400 kg
  (3.92 m/s)
  0.99 m
  k
  0.32 106 N/m
  
  (b)Thespring'spotentialenergyisconvertedbackintothekineticenergyofthecars,sothereboundspeedshould
  bethesame(onlyintheoppositedirection)asthespeedpriortothespringbeingcompressed:
  
  ## vreb vf 3.9 m/s
  
  whereweonlykeepthesignificantfigures.
  ASSESSDuringthecollisioninthefirstpartofthemotion,themomentumisconservedbutenergyisnot.
  However,duringthespringcompressionandreleaseinthesecondpart,energyisconserved.Therefore,thecars
  reboundwiththesamespeedasthatbeforecomingintocontactwiththespring.
  44.
  
  INTERPRETThisonedimensionalprobleminvolvesaninelasticcollisiononafrictionlesssurface,sokinetic
  energyisnotconserved,buttotallinearmomentumisconserved.Wecanusethistofindthespeedofthe
  threevehiclewreckage.
  momentum,wecanequatethetotallinearmomentumbeforeandafterthecollision.Beforethecollision,thetotal
  momentumofthethreevehiclesisp=m1v1+m2v2+m3v3.Aftertheaccident,wehavep=(m1+m2+m3)v.
  EVALUATEEquatingthetwoexpressionsfortotallinearmomentum,wefind
  
  m1v1 m2 v2 m3 v3
  m1 m2 m3
  
  ## 1200 kg 50 km/h 4400 kg 35 km/h 1500 kg 65 km/h
  
  1200 kg 4400 kg 1500 kg
  
  44 km/h
  
  ASSESSNoticethatthetruckhasincreaseditsspeed,whereasthecarshavereducedtheirspeed,asexpected.
  45.
  
  INTERPRETThisprobleminvolvestheNewtonssecondlawintheformofEquation9.6.Wecanusethistofind
  anexpressionfortheinitialaccelerationofthecarduetothewaterjetthatbouncesoffitsrearwindow,andtofind
  thefinalspeedofthecar.
  rest.FromEquation9.6,weknowthattheforceexertedonthecarbythewateristhenegativeoftherateofchangeof
  momentumofthewater:
  
  r
  r
  d pw
  Fc
  
  dt
  
  rest,thisspeedisthesamebeforeandafterreflectingoffthecarsrearwindow.Inthiscase,
  
  r
  dp
  d
  d
  dm
  d pw
  
  pw j pw i w j i mv0 j i v0
  j i
  dt
  
  dt
  dt
  dt
  dt
  
  Thus,theinitialforceexertedonthecarbythewaterjetis
  
  r
  r
  dm
  d pw
  Fc
  v0
  ij
  
  dt
  dt
  r
  r
  r
  r
  IfweapplyNewtonssecondlaw Fnet dpc dt tothecar,itreducesto Fnet Marc ,becausethecarsmassdoes
  notchange.Thenetforceactingonthecarissimplythehorizontalcomponentof Fc ,becauseitsvertical
  
  r
  r
  dm
  r
  Fnet Fc i v0
  i Mac
  dt
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles915
  
  EVALUATE(a)SolvingNewtonssecondlawfortheaccelerationacofthecargives
  
  v dm
  r
  ac 0
  i
  M dt
  (b)Whenthecarstartsmoving,thechangeinthewatersmomentumisreducedbecausethespeedvofthewater
  thusbecomes
  
  v dm v0 v c dm
  r
  
  ac
  i
  
  M dt
  M
  dt
  Thus,whenvc=v0,thecarwillnolongeraccelerate,sothefinalvelocityofthecarisv0.
  ASSESSAsexpected,theaccelerationofthecarincreaseswiththewaterspeed v0andthewatermassratedm/dt,
  butdecreaseswithM,themassofthecar.
  46.
  
  INTERPRETThistwodimensionalprobleminvolvestheprincipleofconservationoflinearmomentuminan
  inelastictwobodycollision.Wecanapplythisprincipletofindthefinalvelocityofthetwobodysystemafterthe
  collision.
  DEVELOPIftherearenoexternalhorizontalforcesactingonthecarwagonsystem,momentum(inthe xyplane)
  isconserved,so
  
  r
  r
  pi pf
  v
  v
  v
  m1v1 m2 v2 m1 m2 v .
  
  whichwecansolveforthefinalvelocity v .
  EVALUATEInsertingthegivenquantitiesintotheexpressionabovegives
  
  ## v 950 kg 32i 17 j m/s 450 kg 12i 14 j m/s
  
  v
  26i 16j m/s
  950 450 kg
  
  2
  2
  ASSESSThemagnitudeofthisvelocityis v vx v y 31 m/s ,anditsdirectionis atan v y vx 32
  withrespecttothexaxis.
  
  47.
  
  ## sowehave v2f v1f 2v.
  
  EVALUATEPlugging v2f 2v v1f intotheonedimensionalconservationofmomentumequation(Equation
  9.12a):
  
  m1 m2 v m1v1f m2 2v v1f
  
  m1 m2 v m1 m2 v1f
  
  v1f v
  
  ## And v2f v, usingthepreviousrelation.Sothefinalspeedofeachobjectisequal(butoppositetotheinitial
  
  speed).
  ASSESSDotheminussignsmakessenseinthederivation?Let'sassumethefirstobjectapproachesfromtheleft
  withpositivevelocity,anditbouncesofftotheleftinthenegativedirection.Thesecondobjectdoestheexact
  opposite,approachingfromtherightwithnegativevelocityandbouncingbackwithpositivevelocity.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  916
  
  48.
  
  Chapter9
  
  INTERPRETThisisatwodimensionalproblemthatinvolvesconservationoflinearmomentum.Thequantityof
  interestistherecoilvelocityofthethoriumnucleus,producedfromthedecayofthe 238Unucleus.
  DEVELOPBecausenoexternalforcesareactingonthesystem(ignoringgravity),linearmomentumisconserved,
  asinExample9.6.Equatingtheinitialandfinalmomentagives
  
  r
  r
  r
  mU vU mHe vHe mTh vTh
  
  Intermsofthexandycomponents,thisvectorequationgivesthefollowingtwoscalarequations:
  
  mU vU mHe vHe, x mTh vTh, x mHe vHe cos mTh vTh cos
  
  ( x component)
  
  0 mHe vHe, y mTh vTh, y mHe vHe sin mTh vTh sin
  r
  Theseequationscanbeusedtosolveforthemagnitudeanddirectionof vTH .
  
  ( y component)
  
  EVALUATESolvingthetwoequations,weobtain
  
  5
  7
  mU vU mHe vHe cos 238 u 5 10 m/s 4 u 1.4 10 m/s cos 22
  
  2. 9 105 m/s
  mTh
  234 u
  
  and
  
  mHe vHe sin
  
  ## 9.0 104 m/s
  
  mTh
  234 u
  
  totwosignificantfigures.Thustherecoilvelocityofthethoriumatomis
  
  r
  2
  2
  5
  vTh 2.9 105 m/s i 9.0 10 4 m/s j ,or vTh vTh
  andthedirectionis
  , x vTh, y 3.0 10 m/
  
  ## atan vTh, y vTh, x
  
  17.
  
  ASSESSThefactthat isnegativetellsusthatthevelocityofthethoriumatomisdownward,asexpectedto
  compensatefortheupwardvelocityofthealphaparticle.
  49.
  
  INTERPRETThisprobleminvolvesfindingthecenterofmassofanobjectcomposedofseveralparts(walls,
  base,andsilage).
  DEVELOPHereitsconvenienttofindthecentersofmassofsubpartsandthentreatthesepartsaspointparticles
  itsaninfinitelythindiskwiththesamediameterasthesilo.Useacoordinatesystemwiththeoriginatthecenterof
  thecylindersbaseandthezaxisrunningalongthecenterofthesilocylinder.Becausethesystemissymmetric
  EVALUATE(a)Tofindthecenterofmasswhenthesiloisempty,findthecenterofmassofthecylindricalwall
  andbaseseparately,andthentreattheproblemasifthemassofeachobjectwereconcentratedattheirrespective
  centerofmasspoints.Bysymmetry,thezcoordinateofthecenterofmassofthewallmustbeathalftheheight,or
  wall
  base
  zcm
  15 m .Thezcoordinateofthecenterofmassofthebaseisat zcm
  0 m becauseitisinfinitelythin.
  InsertingtheseresultsintoEquation9.2tofindthecenterofmassoftheemptysilogives
  
  silo
  cm
  
  wall
  base
  3.8 104 kg 15 m 6 103 kg 0
  mwall zcm
  mbase zcm
  
  13 m
  M
  3.8 104 kg 6 103 kg
  
  sothecompletecoordinatesofthecenterofmassare(0,0,13m),totwosignificantfigures.
  (b)Treattheemptysiloandsilageasiftheirentiremasswereconcentratedattheirrespectivecenterofmass
  points.Wefoundthecenterofmassoftheemptysiloinpart(a).Thezcoordinateofthecenterofmassofthe
  silage
  silageishalfwayupits20meterheight,so zcm 10 m andthesilagesmassis
  
  ## msilage silageVsilage silage r 2 h 800 kg/m3 4 m 2 20 m 2.01 105 kg
  
  InsertingtheseresultsintoEquation9.2gives
  silage
  zcm
  
  silo
  silage
  msilo zcm
  msilage zcm
  
  4.4 10
  
  kg 13 m (2.01 105 kg 10 m
  
  4.4 10 kg 2.01 10 kg
  4
  
  11 m
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles917
  
  totwosignificantfigures.Thus,thecompletecoordinatesofthecenterofmassare(0,0,11m).
  bottomofthesilo.
  50.
  
  hasexplodedintothreepieces.Withthemassandvelocityoftwopiecesgiven,wecanuseconservationoflinear
  momentumtofindthevelocityofthethirdpiece.
  totalmomentumofthethreebodysystem(i.e.,thethreefirecrackerfragments)isstillzero.Expressed
  mathematically,thisis
  
  r
  r
  r
  r
  Ptot 0 m1v1 m2 v2 m3 v3
  
  whichwecansolvetofind v3 .
  EVALUATEThemassofthethirdpieceis m3 m m1 m2 42 g 12 g 21 g 9 g. Itsvelocityis
  
  r
  r
  12 g 35 m/s i 21 g 29 m/s j
  m v m2 v2
  r
  v3 1 1
  
  47 m/s i 68 m/s j
  m3
  9g
  
  ASSESSSincetheinitialmomentumofthefirecrackeriszero,weexpectthemomentumofthethirdpieceto
  
  ## completelycancelthemomentumofthefirsttwopieces.Thus, v3 hascomponentsthatareoppositeto v1 and v2 .
  
  Sincem3issmallerthanm1andm2,weexpectthemagnitudeof v3 tobegreaterthanthemagnitudesof v1 and v2 .
  51.
  
  INTERPRETNoexternalforcesactonthethreebodysystem,sototallinearmomentumisconserved.Wecanuse
  DEVELOPIntherestframeoftheastronaut(i.e.,intheinertialframeofreferenceinwhichtheastronautisat
  momentummuststillbezero,so
  
  r
  r
  r
  m1v1 m2 v2 m3v3 0
  
  wherethesubscripts1,2,and3refertotheastronaut,theaircanister,andthecamera,respectively.Decomposing
  thisvectorequationintotwoscalarequationsgives
  
  m v sin 200 m v
  
  sin 0 m v
  
  ## m1v1 cos 200o m2 v2 cos 0o m3v3, x 0
  
  o
  
  r
  whichwecansolvefor v3 .
  
  1 1
  
  2 2
  
  3 3, y
  
  EVALUATESolvingfirstforthexcomponentofthecamerasvelocity,wefind
  
  v3 x
  
  ## 60 kg 0.85 m/s cos 200 14 kg 1.6 m/s
  
  5.8 kg
  
  4.4 m/s
  
  Similarly,theycomponentis
  
  ## 60 kg 0.85 m/s sin 200
  
  3.0 m/s
  5.8 kg
  r
  Sothevelocityofthecamerais v3 4.4 m/s i 3.0 m/s j
  v3 y
  
  ASSESSAlternatively,wecanexpresstheresultintermsofthemagnitudeanddirectionofthevelocity.This
  gives v3 4.4 2 3.0 2 m/s 5.3 m/s and3=atan(3.0ms/4.4ms)=34counterclockwisefromthexaxis.
  52.
  
  ## INTERPRETBeforeanexplosion,anobjecthaskineticenergy K 2 mvi . Aftertheexplosion,ithastwopieces:
  
  1
  eachwithmassof 2 m, andeachmovingattwicetheinitialspeed, vf 2vi . Weareaskedtofindtheanglesat
  whichtwopiecesofanobjectflyofffromtheexplosion.
  1
  
  DEVELOPLet'sassumethattheoriginalobjectwasmovinginthexdirection,withnomomentumintheydirection.After
  
  theexplosion,conservationofmomentumimpliesthattheymomentumofthetwopiecessumstozero:
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  918
  
  Chapter9
  
  y component: 0
  
  m 2v sin m 2v sin
  1
  2
  
  1
  2
  
  ForTHEycomponentstocancel,theanglesthateachpiecemakeswiththexaxisareequalandopposite:
  
  1 2 .
  
  EVALUATETofindthevalueoftheseangles,weconsiderthemomentuminthexdirection:
  
  x component: mvi
  
  m 2v cos m 2v cos
  1
  2
  
  1
  2
  
  ## Thisreducesto cos 1 2 , whichmeanstheanglesare60and60.
  
  ASSESSInthiscase,thespeedrelativetothecenterofmassisjusttheycomponentoftheirvelocity:
  vrel 2vi sin 60o 3vi . Thisimpliesthattheinternalenergywouldbe
  1
  
  K int 12 mi vi2rel
  
  1
  2
  
  m
  1
  2
  
  3vi
  
  12
  
  m
  1
  2
  
  3vi
  
  1
  2
  
  mvi2 3K
  
  ThisAGREESwiththeresultinProblem9.22.
  53.
  
  INTERPRETThisonedimensionalprobleminvolvesconservationoflinearmomentumandrelativemotion.We
  canusetheformertofindthespeedofthesprinterwithrespecttothecartandthelattertofindherspeedrelativeto
  theground.
  DEVELOPWechooseacoordinatesysteminwhichthecartmovesinthe i direction,andthesprinterrunsin
  the i direction.Theinitialmomentumofthesystemis
  
  p ms mc vcm
  Thefinalmomentumofthesystemis
  
  p ms vs mc vc mc vc
  becauseshehaszerovelocitywithrespecttotheground(vs=0).Equatingthesetwoexpressionsfortotallinear
  momentum(byconservationoftotallinearmomentum),wehave
  
  ms mc vcm ms vs mc vc mc vc
  
  vs vrel vc
  vrel vc
  becausevs=0.Wearegivenvcm=7.6m/s,mc=240kg,andms=55kg.
  EVALUATESolvingtheequationsaboveforvrel,wefind
  
  ms mc vcm mc vc mc vrel
  m mc vcm 55 kg 240 kg 7.6 m/s
  v s
  rel
  
  mc
  
  240 kg
  
  9.3 m/s
  
  ASSESSThefactthatthesprinteracceleratesbypushingagainstthecartacceleratesthecartfrom7.6m/sto
  9.3m/s,whichisreasonable.
  54.
  
  INTERPRETThisprobleminvolvesexertingaforceonaconveyorbelttocompensateforthechangein
  DEVELOPIftheconveyorbeltishorizontalandmovingwithspeedv=50cm/sandthemoundsofdoughfall
  vertically,thenthechangeinthehorizontalmomentumduetoeachmoundofmass mis p (m)v. The
  averagehorizontalforceneededisequaltotherateatwhichmoundsaredropped(anumberNintime t , orN/t)
  timesthechangeinmomentumduetoasinglemound.Thus,forthisproblemEquation9.6takestheform
  
  r
  N
  r N
  Fav p
  t
  t
  
  m v
  
  EVALUATEInsertingthevaluesgivenintheproblemstatement,wefindthattheaverageforcetheconveyorbelt
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles919
  
  N
  1
  Fav m v
  t
  2 s
  
  ## 0.012 kg 0.50 m/s 3.0 103 N
  
  r
  
  ASSESSTheaverageforceisjustthetotalchangeinmomentum, P N p N m v , dividedbythetime,
  
  t.Thegreateristhechangeinmomentumoveragiventimeinterval,thegreateristheaverageforce.
  55.
  
  DEVELOPThecollisionisonedimensional,soEquations9.15(a)and9.15(b)arerelevant.Theinformationthatwe'regiven
  
  ## isthat m1 m, v1i 2v, m2 4m, and v2i v. Noticethatbothobjectsareinitiallymovinginthesame
  
  (positive)direction.
  EVALUATEPluggingtheparametersintoEquations9.15,
  
  v1f
  
  2 4m
  m 4m
  6 8
  2
  2v
  v v v
  m 4m
  m 4m
  5 5
  5
  
  v2f
  
  2m
  4m m
  4 3
  7
  2v
  v v v
  m 4m
  m 4m
  5 5
  5
  
  ## "push"fromthecollision v2i v2f . Noticethat v1f v2f , otherwiseitwouldn'tmakesensehowthefirstobject
  
  56.
  
  ofenergy.
  
  DEVELOPTheconservationofenergyinacollisionisexpressedinEquation9.13.
  EVALUATETheinitialkineticenergyis:
  1
  2
  
  ## m1v1i2 12 m2 v2i2 12 m 2v 12 4m v 2 4mv 2
  
  2
  
  Usingtheresultsfromthepreviousproblemwehaveforthefinalkineticenergy:
  1
  2
  
  m1v1f2 12 m2 v2f2 12 m
  
  v
  2
  5
  
  12 4m
  
  v
  2
  
  7
  5
  
  4
  50
  
  196
  mv 2 4mv 2
  50
  
  Soyes,theenergyisconservedinthiscollision.
  ASSESSAlthoughbothobjectshavethesamekineticenergyinitially,thesecondparticleleavesthecollisionwith
  mostofthekineticenergy.
  57.
  
  INTERPRETThisprobleminvolvesconservationofmomentumappliedtoatwobodysystem.Thecenterofmass
  ofthissystemdoesnotmovebecauseofconservationofmomentum.Wewillapplytheseprinciplestofindthe
  initialangleatwhichwethrewtherockandthespeedatwhichyoumustbemoving.
  DEVELOPWechooseacoordinatesysteminwhichyourinitialpositionisattheorigin(seefigurebelow).Apply
  Equation3.15tofindtheangleatwhichyouthrowtherock,
  
  x1
  
  v02
  sin 2
  g
  
  withx1=15.2mx2andv0=12.0m/s.Wecanfindx2becauseweknowthecenterofmassofthetwobody
  systemdoesnotchangesincetherearenohorizontalforcesactingonit.Thus,
  
  xcm 0
  x2
  
  m1 x1 m2 x2
  m1 m2
  
  m1 x1
  m2
  
  so
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  920
  
  Chapter9
  
  x1 15.2 m
  x1
  
  m1 x1
  m2
  
  15.2 m
  1 m1 m2
  
  whichallowsustosolveEquation3.15fortheangle.Tofindthespeedatwhichyoumoveafterthrowingthe
  rock,applyconservationoflinearmomentum.Yourinitialhorizontalmomentumiszero,soyourfinalmomentum
  mustalsobezero,or
  
  m1v1x m2 v2 x 0
  where v1x v0 cos ,withbeingtheanglewithrespecttothehorizontalatwhichyouthrowtherock.Solvethis
  equationforv2x,whichisthespeedatwhichyoumoveasaresultofthrowingtherock.
  
  EVALUATE(a)InsertingtheknownquantitiesintoEquation3.15andsolvingfor gives
  
  15.2 m g
  15.2 m 9.8 m/s 2
  x1 g
  1
  1
  1
  asin 2 asin
  asin
  37.7o
  2
  2
  1 4.50 kg 65.0 kg 12.0 m/s
  2
  2
  2
  v0
  1 m1 m2 v0
  
  (b)Thespeedatwhichyourecoilafterthrowingtherockis
  
  v2 x
  
  ## 4.50 kg 12.0 m/s cos 37.7o
  
  m1v1x
  m v cos
  1 0
  
  65.8 cm /s
  m2
  m2
  65.0 kg
  
  ASSESSThehorizontalspeedoftherockisv0cos()=9.50m/s.Thus,yourrecoilspeedismuchlessthanthe
  rockshorizontalspeed,asexpected.
  58.
  
  isnotconservedinthisprocess,butmomentumis.
  frictionbeequaltothechangeofthekineticenergyofbothcars,orWnc=K(seeEquation7.5).Since
  
  Wnc f k x k nx k (m1 m2 ) gx
  and
  
  K 12 m1 m2 v 2
  
  1
  k gx v 2
  2
  Theequationcanbeusedtosolveforv.Oncevisknown,wecanapplymomentumconservationtofindtheinitial
  speedofthedrunkdriver.
  EVALUATEFromtheabove,wefindthespeedofthecars(wreckage)justafterthecollisionis
  
  v 2 k gx
  Momentumisconservedtheinstantofthecollision,soifv1isthespeedofthedrunkdriverscarjustbeforethe
  collision(andv2=0fortheparkedcar),then m1v1 m1 m2 v or
  
  v1
  
  m1 m2
  m m2
  v 1
  m1
  m1
  
  2 k gx
  
  1600 kg 1300 kg
  2 0.77 9.8 m/s 2 25 m 35 m/s
  1600 kg
  
  skiddingdistancex.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles921
  
  59.
  
  INTERPRETThistwobodyprobleminvolveskinematics(Chapters2and3)andmotionofthecenterofmass.
  Therocketsexplodesintotwoequalmassfragmentsatitspeakheight,whichwecancalculatefromthekinematic
  equationsofChapter2.Wearegiventhetimeittakesfromtheexplosionforonefragmenttohitthegroundandare
  DEVELOPAtthepeakoftherocketstrajectory(justbeforetheexplosion),itscenterofmass yvelocityiszero,
  or vcm 0 .Themotionofthecenterofmassisunaffectedbytheexplosion,sojustaftertheexplosion,the
  
  0 .Expressingthecenterofmassvelocityintermsofthevelocitiesv1andv2offragments1and2
  velocity vcm
  gives
  
  vcm
  0
  v1 v2
  
  mv1 mv2
  mm
  
  wheretheequalmassfragmentseachhavemassm.Wecannowusethekinematicequationstofindv1.Theheight
  hatwhichtherocketexplodesmaybefoundusingEquation2.11,whichgives(withaslightchangeinnotation)
  
  } 0
  vcm v02 2 gh
  h
  
  v02
  2g
  
  wherevisthevelocityatthepeakofthetrajectoryandv0=40m/s.Knowingtheheightandthetimet1for
  fragment1tohittheground,wecanfinditsinitialvelocityfromEquation3.13.Thisgives
  h
  }
  1
  y y0 v1t1 gt12
  2
  2
  h gt1 2 v02 2 g gt12 2 v02 gt1
  v1
  
  t1
  t1
  2 gt1 2
  
  Knowingv1(andthusv2),wecanfindthetimeforfragment2tohitthegroundbyusingthesameEquation(i.e.,
  
  h v2t2
  t2
  
  1 2
  gt2
  2
  
  v2 v22 2 gh
  g
  
  v1 v12 2 gh
  
  EVALUATEEvaluatingfirstv1,wefind
  
  v1
  
  v1 v12 v02
  g
  
  2
  9.81 m/s 2 2.87 s
  40 m/s
  v02 gt1
  
  14.38 m/s
  2 gt1 2
  2
  2 9.81 m/s 2 2.87 s
  
  wherewehaveretainedoneextrasignificantfigurebecausethisisanintermediateresult.Insertingthisresultinto
  theexpressionfort2gives
  
  t2
  
  v1 v12 v02
  g
  
  14.38 m/s
  
  ## 14.38 m/s 2 40 m/s 2
  
  9.81 m/s
  2
  
  5.80 s, 2.87 s
  
  Thephysicallysignificantresultist2=5.80s.
  ASSESSThetimeforfragment2toreachthegroundwillincreasewithincreasing|v1|,whichisreasonable
  becauseiffragment1hasalargerdownwardvelocityinitially,thenfragment2hasalargerinitialupwardvelocity.
  Also,notethatifv1,v20,thent2=t1=4.08s,whichisintermediatebetween2.87sand5.80s,asexpected.
  60.
  
  INTERPRETInthisproblem,atotallyinelasticcollisionresultsinhalfofthekineticenergybeinglost.Weare
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  922
  
  Chapter9
  
  DEVELOPTheparticlescomeateachotherwithequalbutoppositevelocities v1 v2 . Inordertoobeyconservationof
  
  momentum(Equation9.11),thefinalvelocityhastobeparalleltotheinitialvelocities.Inotherwords,theproblem
  isonedimensional:
  
  m m2
  r
  r
  r
  r
  m1v1 m2 v1 m1 m2 v1 m1 m2 vf vf 1
  v
  m1 m2
  EVALUATEWe'retoldtheinitialkineticenergyistwicethefinalkineticenergy:
  1
  2
  
  m1v12 12 m2 v22
  
  1
  2
  
  m1 m2 v 2 2 12 m1 m2 vf2
  
  vf
  
  1
  2
  
  We'llchoosethepositiveroot,whichpluggedintotheaboverelationgives:
  
  m1
  
  m2
  
  2 1
  2 1
  
  2 1 ; 5.8
  
  m1 / m2 1/
  
  61.
  
  boosterafterthegivenimpulse.Wecanapplyconservationofmomentumbecausetherearenoexternalforces
  actingonthesystem.Finally,thiswillbeaonedimensionalproblembecausewearedealingwithonlytwobodies,
  sotheirrelativemotionmustbelineartosatisfyconservationofmomentum.
  DEVELOPByNewtonsthirdlaw,theexplosionappliesaforceofequalmagnitudetoeachbody(satelliteand
  r
  r
  booster),butintheoppositedirection.Wethereforehave J s J b ,wherethesubscriptssandbrefertothe
  r
  J J 350 N s .AsexplainedforEquation9.10b,animpulse J is
  satelliteandbooster,respectively,and
  r sr b r
  equaltothechangeofmomentum: J p mv , whichallowsustofindthespeedofthesatelliteandthe
  r r
  boosterinthe(stationary)centerofmassframe.Therelativespeedofseparationis vrel vs vb .
  EVALUATEInitiallyboththesatelliteandtheboosterareatrest.Afterexplosion,theirvelocitiesare
  
  r
  r
  r ps J s
  vs
  
  ms
  ms
  
  r
  r
  r
  pb J b J s
  r
  vb
  
  mb
  mb
  mb
  
  Thus,therelativespeedofseparationis
  
  r
  r
  1
  Js Js
  
  1
  1
  1
  r r
  | vs vb |
  
  Js
  (350 N s)
  
  0.92 m/s
  ms mb
  950 kg 640 kg
  
  ms mb
  
  ASSESSTherelativespeedisshowntodependonJs,themagnitudeoftheimpulse.Thegreatertheimpulse,the
  fasterthesatelliteandtheboosterseparatefromeachother.
  62.
  
  INTERPRETYouhavetocalculatetheimpulseimpartedbyaforcethatisnotconstant.
  
  DEVELOPTheimpulseofavariableforcerequiresintegration(Equation9.10b): J F t dt.
  EVALUATETherocket'sthrustisonedimensional,sowecandropthevectornotation.Integratingthegivenforce
  
  equationovertheburntimegives
  t
  
  J at t t dt 13 at 3 12 at 2 t
  Plugginginthegivenvalues:
  
  16 at 3
  
  ## J 16 4.6 N/s 2 2.8 s 17 N s
  
  3
  
  Yes,therocketmeetsitsspecs.
  1
  1
  2
  ASSESSThethruststartsatzero,thenrisestoapeakat t 2 t where F 4 at (recallaisnegative),before
  fallingbacktozeroat t t. Wecouldgetaroughestimateoftheimpulsebyassumingthattheaverageforceis
  1
  2
  approximatelyequaltohalfofthepeakvalue F : 8 a t . Multiplyingbythetimetheforceisappliedgives:
  1
  3
  J : 8 a t , whichisn'ttoofarofffromthepreciseresultabove.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles923
  
  63.
  
  wanttoshowthatthespeedofoneofthecarsexceeded25km/h.Energyisnotconservedinthisprocess,but
  momentumis.Furthermore,becauseworkisdonebyfriction,thisprobleminvolvestheworkenergytheorem.
  frictionbeequaltothechangeofthekineticenergyofbothcars.Wnc=K(seeEquation7.5).Because
  Wnc f k x k nx k m1 m2 gx, and K K f Ki 0 12 m1 m2 v 2 , wherevisthespeedofthe
  wreckageimmediatelyaftercollision,weareledto
  
  1
  k gx v 2
  2
  Therefore,thespeedofthewreckagejustafterthecollisionis v 2 k gx . Next,momentumconservation
  requiresthattheinitialandfinalmomentumarethesame,so
  
  r
  r
  r
  m1v1 m2 v2 m1 m2 v
  r
  r
  r m1v1 m2 v2
  v
  m1 m2
  
  where v istheinitialvelocityofthewreckage.Tofindthechangeinkineticenergy,weneedtocalculatethe
  r r
  scalarproduct v v :
  
  r
  r
  r
  r
  r r m v m2 v2 m1v1 m2 v2
  v 2 v v 1 1
  
  m1 m2
  m1 m2
  
  0
  }
  r r
  2 2
  2 2
  m v m2 v2 2m1m2 v1 v2
  1 1
  
  m1 m2 2 m1 m2 2
  
  ## m12 v12 m22 v22
  
  m1 m2 2
  
  r r
  
  wherewehaveusedthefactthatthescalarproduct v1 v2 0 becausetheinitialvelocitiesareperpendicularto
  eachother.Inthenextstep,weinsertthemaximumspeedforonecartofindtheminimumspeedfortheothercar.
  EVALUATEInserting v 2 k gx intotheaboveexpressionfortheinitialvelocityofthewreckageleadsto
  
  v2
  
  ## m12 v12 m22 v22
  
  m1 m2 2
  
  2 k gx
  
  Solvingfor v1 gives
  
  2 k gx m1 m2 m22 v22
  2
  
  v1
  
  m12
  
  wherewehavetakenthepositivesquareroot.Considernowthefollowingsituations:
  Letsubscript1correspondtotheToyota2totheBuick.IfthespeedoftheBuickisv2=25km/h=6.94m/s,then
  thespeedoftheToyotawouldbe
  
  2 k gx m1 m2 m22 v22
  2
  
  v1
  
  m12
  
  2
  
  6.94 m/s 2
  
  (1200 kg)2
  
  ## 55 m/s 200 km/h
  
  Thus,weconcludethatthespeedoftheToyotaexceeded25km/h.
  (2)Here,wereversetheassignmentofthesubscripts1and2.IfthespeedoftheToyotaisv2=25km/h=6.94m/s,
  thenthespeedoftheBuickwouldbe
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  924
  
  Chapter9
  
  2 k gx m1 m2 m22 v22
  2
  
  v1
  
  m12
  
  2
  
  6.94 m/s 2
  
  2200 kg 2
  
  ## 30 m/s 110 km/h
  
  Thus,weconcludethatthespeedoftheBuickexceeded25km/h.
  Fromtheanalysisabove,weconcludethatifonecarisgoingat25km/h,thentheotheronemusthavebeen
  speeding,soatleastonecarmusthavebeenspeeding.
  ASSESSIfweknewthedirectionofthewreckagevelocity,wecouldeasilyfindthecarthatwasspeeding.
  64.
  
  INTERPRETThisonedimensionalproblemconsidersanexplosiononahorizontalfrictionlesssurface,so
  momentumisconservedbecausetherearenohorizontalforcesactingonthepopcorn.Kineticenergyisnot
  conservedbecausetheexplosiondoesworkonthepopcornfragments,sotheworkenergytheoremtellsusthatthe
  kineticenergymustchange(seeEquation7.5).
  DEVELOPByconservationofmomentum,wecanequatetheinitialandfinalmomenta,whichgives
  
  ## 2mv0 mv1 mv2 m v1 v2
  
  } 0
  2v0 v1 v2
  v2 2v0
  wherem=200g,andwehavearbitrarilychosenfragment1tobetheonewithzerovelocityaftertheexplosion.
  Knowingtheinitialandfinalspeeds,wecanfindtheinitialandfinalkineticenergiesandsocalculatethechange
  inkineticenergyK=KfKi:
  
  } 0
  
  1
  1 2 1 2
  2
  
  K 2m v0 mv1 mv2
  2
  2
  2
  
  1
  K mv02 mv22
  2
  EVALUATEInsertingthev2=2v0intotheexpressionforKandevaluatingtheresultgives
  1
  2
  2
  K mv02 m 2v0 mv02 200 106 kg 0.082 m/s 1.3 J
  2
  
  ASSESSWefindthatthekineticenergyofthesystemdecreases,asexpected,becausetheexplosiondoesworkon
  thesystembypushingthefragmentsapart,andtheworkenergytheoremtellsusthatthisworkmustcomeatthe
  expenseofthesystemskineticenergy.
  65.
  
  INTERPRETThistwodimensionalprobleminvolvesatotallyinelasticcollision,somomentumisconservedbut
  kineticenergyisnotconserved.Wecanuseconservationofmomentumtofindtheanglebetweentheinitial
  velocitiesbeforeathecollision.
  DEVELOPThecollisionbetweenthetwomassesistotallyinelastic.Conservationofmomentumtellsusthat
  
  r
  r
  r
  m1v1 m2 v2 m1 m2 vf
  r
  r
  r m v m2 v2
  vf 1 1
  m1 m2
  
  Weknownthemagnitudeofallthevelocitiesinvolved,butnottheirrelativeorientation.Wecanfindthisby
  takingthescalarproductofthefinalvelocitywithitself:
  
  r
  r
  r
  r
  m1v1 m2 v2 m1v1 m2 v2
  r r
  2
  vf vf vf
  
  m1 m2
  m1 m2
  
  r r
  m12 v12 m22 v22 2m1m2 v1 v2
  
  m1 m2 2
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles925
  
  r r
  r
  r
  Usingthedefinitionofascalarproduct, v1 v2 v1v2 cos , theanglebetween v1 and v2 canbefound.
  EVALUATEWithm1=m2=mandv1=v2=v=2vf,theaboveequationcanbesimplifiedto
  
  v 2 m 2 v 2 m 2 v 2 2m 2 v 2 cos 1 2
  
  v 1 cos
  4
  2
  4m 2
  Therefore,theanglebetweenthetwoinitialvelocitiesis
  
  1
  120
  2
  
  ac os
  r
  
  ## ASSESSToseethattheresultmakessense,suppose v1 makesanangle 60 withxand v2 makesanangle
  
  60 withx.Theycomponentofthetotalmomentumcancels.Butforthexcomponent,wehave
  
  mv cos 60 mv cos 60 m m v f
  Solvingforvf,wegetvf=v/2,whichconfirmstheresultobtainedabove.
  66.
  
  INTERPRETThisisaonedimensionalproblemthatinvolvesanelasticcollisionbetweentwoparticles,so
  conservationofmomentumandoftotalmechanicalenergyapply.Wecanusetheseprinciplestofindthemassand
  finalvelocityofthenonprotonparticle.
  DEVELOPBecausethisisanelasticcollisioninonedimension,wecanapplyEquation9.15aand9.15btofind
  themassandvelocityofthesecondparticle.Forthisproblem,theprecollisionvelocitiesarev1i=6.90Mm/sand
  v2i=2.80Mm/s,andthepostcollisionvelocityoftheprotonis v1f 8.62 Mm/s .Theknownmassism1=1u.
  WecansolveEquation9.15aforthem2,whichgives
  
  m2
  
  m1 v1i v1f
  v1f v1i 2v2i
  
  WecantheninserttheresultofthiscalculationintoEquation9.15btofindv2f.
  EVALUATEInsertingthegivenquantitiesintotheexpressionaboveforthemassoftheunknownparticlegives
  
  ## 8.62 Mm/s+6.90 Mm/s 2 2.80 Mm/s
  
  m2
  
  1 u 4 u
  
  InsertingthisresultintoEquation9.15bgives
  
  v2f
  
  2
  3
  2
  3
  v1i v2i 6.90 Mm/s 2.80 Mm/s 1.08 Mm/s
  5
  5
  5
  5
  
  ASSESSBecausethesecondparticleismoremassive,theprotongainsmomentuminthecollision.Thealpha
  particle,however,losesmomentum.
  67.
  
  INTERPRETTheonedimensionalcollisioninthisproblemiselastic,sobothmomentumandenergyare
  DEVELOPMomentumisconservedinthisprocess.Inthisonedimensionalcase,wemaywrite
  
  ## m1v1i m2 v2i m1v1f m2 v2f
  
  Sincethecollisioniscompletelyelastic,energyisconserved:
  0
  1
  1 }2 1
  1
  2
  m1v1i m2 v2i m1v1f2 m2 v2f2
  2
  2
  2
  2
  
  Usingthetwoconservationequations,thefinalspeedsofm1andm2are(seeEquations9.15aand9.15b):
  
  v1f
  
  m1 m2
  2m2
  2m1
  m m1
  v1i
  v2i and v2f
  v1i 2
  v2i
  m1 m2
  m1 m2
  m1 m2
  m1 m2
  
  Giventhatv2i=0,theaboveexpressionsmaybesimplifiedto
  
  v1f
  
  m1 m2
  v1i
  m1 m2
  
  v2f
  
  2m1
  v1i
  m1 m2
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  926
  
  Chapter9
  
  Now,ifhalfofthekineticenergyofthefirstobjectistransferredtothesecond,then
  
  K 2f
  
  1
  K1i
  2
  
  1 2m1v1i
  m2
  2 m1 m2
  
  1
  m1v1i2
  4
  
  EVALUATETheaboveequationcanbefurthersimplifiedto
  
  8m1 m2 m1 m2
  
  m m1
  8 1
  1
  m2 m2
  
  2
  2
  Theresultingquadraticequation, m1 6m1 m2 m2 0 hastwosolutions:
  
  5.83m2
  
  m1 3 8 m2
  
  5.83 m2
  1
  
  interchanged.Thus,oneobjectis5.83timesmoremassivethantheother.
  Usingm1=5.83m2,wefind
  
  K 2f
  
  1
  1 2m1v1i
  m2 v2f2 m2
  2
  2 m1 m2
  
  2m1
  1 m1
  
  v1i2
  2 5.83
  m1 m1 /5.83
  
  1 m1
  2
  1
  1
  v1i2 m1v1i2 K1i
  
  2 5.83 1 1/5.83
  4
  2
  
  asexpected.
  68.
  
  INTERPRETThisisaonedimensional,threebodyproblemthatinvolveselasticcollisions,sobothconservation
  ofmomentumandenergyapply.Weneedtofindthefinalvelocityofeachblockafterthecollisions.
  DEVELOPWecananalyzeseparatelythetwocollisionsinthisproblem,andapplyEquations9.15toeach
  collision.Forthefirstcollision,betweenblocksAandB,wefind(withvafv)
  
  vAf
  
  0
  mA mB
  2mB }
  m mB
  vAi
  vBi A
  vAi
  mA mB
  mA mB
  mA mB
  
  vBf,int
  
  0
  mB mA }
  2mA
  2mA
  vBi
  vAi
  vAi
  mA mB
  mA mB
  mA mB
  
  wherevBf,intistheintermediatefinalvelocityofblockB.BlockBthenproceedstocollidewithblockC,andthe
  finalvelocitiesfromthatcollisionare
  
  vBf
  vCf
  
  0
  mB mC
  2mC }
  m mC
  vBf,int
  vCi B
  vBf,int
  mB mC
  mB mC
  mB mC
  
  0
  mC mB }
  2mB
  2mB
  
  vCi
  vBf,int
  vBf,int
  mC mB
  mA mB
  mA mB
  
  EVALUATEInsertingthemassesoftheblocksmA=m,mB=2m,andmC=m,andrecallingthatvAfv,wefind
  
  vAf
  
  m 2m
  1
  vAi v
  m 2m
  3
  
  vBf,int
  vBf
  
  2m
  2
  vAi v
  m 2m
  3
  
  2m m
  2
  1 2
  vBf,int vAi v
  3 3
  2m m
  9
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles927
  
  vCf
  
  2mB
  4m 2
  8
  vBf,int
  vAi v
  mA mB
  m 2m 3
  9
  
  ASSESSWecanverifythatmomentumisconservedinthisprocess:
  
  2
  8
  3 4 8
  1
  
  ## mvAi mvAi 2mvAi mvAi mvAi mvAi
  
  3
  
  9
  9
  9 9 9
  69.
  
  DEVELOPUseconservationofmomentum,
  
  r
  r
  r
  r
  m1v1i m2 v2i m1v1f m2 v2f .
  
  Becausethisisanelasticcollision,kineticenergyisalsoconserved,so
  
  1
  1
  1
  1
  m1v1i2 m2 v2i2 m1v1f2 m2 v2f2 .
  2
  2
  2
  2
  EVALUATEFirstsolveEquation9.14forv1ftoget v1f v2f v2i v1i . Whenwesubstitutethisresultinto
  Equation9.12,usingthesignofvtodenotethedirection,weobtain m1v1i m2 v2i m1 v2f v2i v1i m2 v2f .
  Solvingthisforv2fgives
  
  ## m1v1i m2 v2i m1v2i m1v1i m1 m2 v2f
  
  v2f
  
  2m1
  m m1
  v1i 2
  v2i
  m1 m2
  m1 m2
  
  ASSESSOurresultagreeswithEquation9.15b,asexpected.
  70.
  
  INTERPRETThistwodimensional,twobodyprobleminvolvesconservationofmomentumandenergybecause
  perpendicular.
  DEVELOPWithoneofthebodiesinitiallyatrest,conservationofenergygives
  
  r
  r
  r
  p1i p1f p2f
  
  andconservationofenergygives[usingK=p2/(2m)]
  
  ## p1i2 2m1 p1f2 2m1 p2f2 2m2
  
  p1i2 p1f2 p2f2
  2
  wherethesecondlineusesthefactthattheobjectshaveequalmass.Thesetwoexpressionsfor p1i canbeequated
  
  tofindtheanglebetweentheresultingmomentumvectors.
  2
  EVALUATEEquatingthetwoexpressionsabovefor p1i gives
  
  r
  r
  p1f2 p2f2 | p1f p2f |2
  r r
  p1f2 p2f2 p1f2 2 p1f p2f p2f2
  r r
  0 2 p1f p2f
  r r
  Recallingthatthescalarproductisdefinedas p1f p2f p1f p2f cos ,weseethattheangle=90,unlessp1f=0,
  ASSESSYoucanverifythisresultonapooltable.
  71.
  
  INTERPRETThetwodimensionalprobleminvolvesanelasticcollisionbetweenaprotonandaninitially
  stationarydeuteron.Giventheanglebetweentheirfinalvelocities,wearetofindthefractionofkineticenergy
  transferredfromtheprotontothedeuteronintheprocess.
  DEVELOPUsingthecoordinatesystemshowninthesketchbelow(thedeuteronsrecoilangle 2fisnegative),
  thecomponentsoftheconservationofmomentumequationsfortheelasticcollisionbecome
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  928
  
  Chapter9
  
  ## mp vpi mp vpf cos 1f md vdf cos df
  
  0 mp vpf sin pf md vdf sin df
  
  1
  1
  1
  mp vpi2 mp vpf2 md vdf2
  2
  2
  2
  sothefractionofinitialkineticenergytransferredtothedeuteronis
  
  md vdf
  
  m p vpi
  
  K pf
  K df
  1
  K pi
  K pi
  
  v pf
  1
  vpi
  
  EVALUATEWithmd=2mp,theconservationequationsbecome
  
  ## vpi vpf cos pf 2vpf cos pf
  
  0 vpf sin pf 2vdf sin df
  v vpf2 2vdf2 .
  2
  pi
  
  Tofindthefinalvelocities,eliminate2ffromthefirstandsecondequationsandv2ffromthethirdtoget
  
  vpi2 2vpi vpf pf vpf2 4vdf2 sin 2 df cos 2 df 4vdf2 2vpi2 2vpf2
  Thisresultsinaquadraticequationforvpf: 3v 2vpi vpf cos pf v 0, withpositivesolution
  2
  pf
  
  2
  pi
  
  1
  vpf vpi cos pf cos 2 pf 3 0.902vpi
  3
  wherewehaveused1f=37.Fromthekineticenergyequation,wehave vdf
  
  1
  2
  
  ## (vpi2 vpf2 ) 0.305vpi , and
  
  fromthetransversemomentumequation,wehave
  
  0.902vpi sin 37
  vpf sin 37
  1
  
  sin
  
  62.7
  
  2vdf
  
  2 0.305vpi
  
  df sin 1
  
  Fromeithervpforvdf,thefractionoftransferredkineticenergyisfoundtobe
  
  vpf
  K df
  K
  1 df 1
  K pi
  K pi
  vpi
  
  1 0.902 18.6%
  2
  
  ASSESSThefractionofenergytransfercanalsobeobtainedas
  
  K df md vdf
  
  K pi mp vpi
  
  ## 2 0.305 0.186 18.6%
  
  2
  
  includingthedeuteronrecoilangle,isinstructive.
  72.
  
  INTERPRETThistwodimensionalprobleminvolvesathreebody,totallyelasticcollision.Wecantherefore
  applyconservationoftotallinearmomentumandconservationofenergytofindthevelocitiesofthethreeballs
  afterthecollision.
  DEVELOPBecausetheballsarethesamesize,thedirectionoftheimpactforceisat30withrespecttothe
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles929
  
  ## componentsandthemagnitudeoftheirvelocitiesmustbeequivalent.Thus,vC=vBand vC vB, x i vB, y j .
  
  Applyingconservationofmomentuminthe i directionthereforegives
  
  ## mv0 mvA mvB cos 30 mvC cos 30
  
  v0 vA vB 3
  ApplyingconservationofenergyandusingtheresultfromconservationofmomentumandtheresultthatvC=vB
  gives
  
  1 2 1 2 1 2 1 2
  mv0 mvA mvB mvC
  2
  2
  2
  2
  2
  2
  2
  v0 vA 2vB
  v0 v A
  
  v02 vA2 2
  
  whichwecansolveforvA.OnceweknowvA,wecanusetheexpressionabovefromconservationofmomentumto
  findvB(andvC).
  
  EVALUATESolvingtheexpressionaboveforvAgives
  
  2
  v0 vA 2
  3
  2
  v0 vA v0 vA 3 v0 vA 2
  v
  vA 0
  5
  r
  v
  
  v
  5
  0 i .UsingtheresultforvAtofindvBgives
  BecausevAhasonlyahorizontalcomponent,wehave
  v02 vA2
  
  v02
  2vB2
  25
  2v
  vB 0 3
  5
  v02
  
  soincomponentformwehave
  
  r
  vB vB cos i vB sin j
  2 3
  2 3 1
  3
  j
  v0 i
  v0
  5
  2
  5
  2
  
  3 3
  v i
  v0 j
  5 0
  
  Fromthesymmetryargumentsabove,wenowhave
  
  3
  r 3
  vC v0 i
  v0 j
  5
  5
  ASSESSWecancheckthatmomentumandenergyareconserved.Formomentum,wehave
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  930
  
  Chapter9
  
  r
  r
  r
  r
  mv0 mvA mvB mvC
  1
  3
  3 3
  3
  v0i v0 i v0i
  v0 j v0i
  v0 j
  5
  5
  5
  5
  5
  v0i
  andforenergy,wehave
  
  1 2 1 2 1 2 1 2
  mv0 mvA mvB mvC
  2
  2
  2
  2
  2
  v
  9
  3
  9 3 2
  
  v02 0
  v02
  v0
  
  25
  25 25
  25 25
  v02
  73.
  
  ## use sin x dx cos x.
  
  EVALUATETheforceandtheimpulseareonedimensional,sowewillneglectthevectorformalism.Evaluating
  
  theintegraloverthegiventimeinterval:
  
  /a
  
  /a
  
  F
  F0 sin at dt 0 cos at
  a
  0
  
  F0
  2F
  cos cos 0 0
  
  a
  a
  
  ASSESSTheunitsare N s, astheyshouldbefortheimpulse.
  74.
  
  INTERPRETThistwodimensionaltwobodyprobleminvolvesaninelasticcollision,sowecanapply
  conservationoflinearmomentum,butnotconservationofenergy.Wecanapplyconservationofmomentumto
  findthevelocityoftheozone.
  DEVELOPForthistotallyinelasticcollision,momentumconservationgives
  
  r
  r
  r
  m1v1 m2 v2 m1 m2 vf
  r
  r
  r m v m2 v2
  vf 1 1
  m1 m2
  
  Thevelocitiesoftheoxygenmolecule(denotedwithsubscript1)andoxygenatom(subscript2)are
  
  r
  v1 580 m/s i
  r
  v2 870 m/s cos 27 i sin 27 j 775 m/s i 395 m/s j
  r
  r
  EVALUATESubstitutingtheaboveexpressionsfor v1 and v2 intothefirstequation,weobtain
  r
  r
  r
  r
  r m v m2 v2 32 u v1 16 u v2 2 r 1 r
  vf 1 1
  
  v1 v2
  m1 m2
  32 u 16 u
  3
  3
  
  2
  1
  580 m/s i 775 m/s i 395 m/s j
  3
  3
  
  ## 645 m/s i 132 m/s j
  
  r
  ASSESSThemagnitudeanddirectionof v f are v 658 m/s and x 11.5. Theresultisreasonablesincewe
  expecttheangletobebetween0and 27.
  
  75.
  
  INTERPRETThisonedimensionaltwobodyprobleminvolvesaninelasticcollisionsowecanapply
  conservationofmomentumbutnotconservationofenergy.Wewillalsoneedtoapplysomekinematicstofindthe
  maximumheightandthespeedwithwhichthecombinationhittheground.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles931
  
  DEVELOPByconservationofmomentum,wecanequatethemomentumofthetwobodysystembeforeandafter
  theFrisbeemudcollision.Thisgives
  
  mm vi mF mm vf
  UsingEquation2.11,wefindthevelocityvm,iwithwhichthemudhitstheFrisbeetobe
  2
  vm,i vm,0
  2 g y y0
  
  ## 17.7 m/s 2 2 9.8 m/s 2 7.65 m 1.23 m
  
  13.69 m/s
  
  Therefore,theinitialvelocityofthemudFrisbeecombinationis
  
  vf
  
  mm vi
  0.240 kg 13.7 m/s
  
  9.288 m/s
  mF mm
  0.240 kg 0.114 kg
  
  upward.UsethisresultinthekinematicEquation2.11tofindthemaximumheightandthespeeduponhittingthe
  groundforthemudFrisbeecombination.
  EVALUATE(a)Themaximumheightreachedis
  
  v 2 vf2
  0.00 m/s 9.28 m/s
  7.65 m
  8.10 m
  2 g
  2 9.8 m/s 2
  2
  
  y y0
  
  (b)Anobjectfallingfromthisheight,unimpededbyairresistanceorotherobstacles,wouldattainaspeedof
  
  ## v 2 gy 2 9.8 m/s 2 8.10 m 12.6 m/s
  
  whenitreachestheground(wherewehaveretainedthenegativesigntoindicatethevelocityisdownward).
  ASSESSNoticethatweretainedextrasignificantfiguresfortheintermediateresults.
  76.
  
  INTERPRETThisonedimensionaltwobodyprobleminvolvesanelasticcollisionandkinematics.Wecanapply
  conservationofenergyandmomentumtofindtheheightthesmallballreboundsafterbeingdroppedtogetherwith
  alargerballandreboundingfromtheground.
  DEVELOPTheballsreachtheground,afteraverticalfallthroughaheighth,withspeed v0 2 gh (see
  groundwithinitialvelocityv2i=v0(positiveupward),andthesmallballstillfallingdownwardwithinitialvelocity
  v1i=v0.Equation9.15agivesthefinalvelocityofthesmallballas
  
  m M
  vf
  v0
  m M
  
  2M
  3M m
  
  v
  v 3v0
  m M 0 m M 0
  
  kinematicEquation2.11).
  2
  EVALUATEConservationoftotalmechanicalenergyrequiresthat mv f 2 mghf , so
  
  v2
  v 2 3v0
  hf f
  9 0 9h
  2g
  2g
  2g
  2
  
  ASSESSThisdemonstration,sometimescalledaMinskicannon,isstriking.Tryitwithanewtennisballand
  77.
  
  INTERPRETThisonedimensional,twobodyprobleminvolvesacollisionthatisneitherelasticnorinelastic,so
  wecanapplyconservationofmomentum.Giventhatwearetoldtheamountofkineticenergythatislostinthe
  collision,wecanapplyconservationofenergyaswell.Wecanusetheseprinciplestofindthevelocityofthe
  wreckage.
  DEVELOPLettheinitialvelocitycar1bev,andthatofcarsafterthewreckagebev1andv2.Conservationof
  momentumrequiresthat
  
  mv mv1 mv2
  v v1 v2
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  932
  
  Chapter9
  
  andconservationofenergygives
  
  1 2 1 2 1 2 5 1 2
  mv mv1 mv2 mv1
  2
  2
  2
  18 2
  13 2
  v v12 v22
  18
  Solvethesetwoequationsforthefinalvelocities.
  EVALUATESolvingtheexpressionfromconservationofenergyfor v1,fgives
  
  v1
  
  13 2
  v v22
  18
  
  Takingthepositivesquareroot(thenegativesolutionsimplycorrespondstothecarsmovingintheopposite
  direction)andusingtheresultfromconservationofmomentumeliminatev2,fgives
  
  13 2
  2
  v v v1
  18
  13
  v12 v 2 v 2 2vv1 v12
  18
  v1
  
  5 2
  v 0
  18
  1
  
  5
  2 v1 v v1 v 0
  
  6
  6
  v12 2vv1
  
  whichhasthesolutionsv1=5v/6orv/6.Becausev1<v2(becausecar1isbehindcar2afterthewreck),wechosev1
  ASSESSWecaneasilyverifythatconservationofmomentumandenergy(includingtheconvertedkinetic
  energy)isrespectedwiththissolution.
  78.
  
  INTERPRETThisprobleminvolveskinematics,conservationofmomentum,andconservationofenergy.Asmall
  blockslidesdownafrictionlessinclineandcollidesonahorizontalfrictionlesssurfacewithasecond,largerblock
  inanelasticcollision.Thesmallerblockreboundsandtravelsbackuptheinclineacertaindistance,thenslides
  elapsesbeforethetwoblockscollideforthesecondtime.
  DEVELOPFromconservationofenergy,andassumingasmoothtransitionfrominclinetohorizontalsurface,the
  smallblockhasspeed v1i 2 gh whenthefirstcollisionoccurs.UseEquation9.15a,withm2=4m1andv2i=0to
  findthespeedsoftheblocksimmediatelyafterthefirstcollision(att=0):
  
  m m2
  3
  1 4
  v1f 1
  v1i
  v1i v1i
  
  1 4
  5
  m1 m2
  2m1
  2
  v2f
  v1i v1i
  
  5
  m1 m2
  Thelargerblockmoveswithconstantspeedofv2f=2v1i/5totheright;itsposition,relativetothebottomofthe
  incline,is
  
  2
  x2,f t x2,i v2f t 1.4 m v1,i t
  5
  1
  1
  Thesmallerblocktakestime t1 1.4 m 3v1,i 5 togetbacktotheincline,and t2 2 3v0 5 a togoupand
  downtheincline,where a g sin 30 g/2. (UseEquation2.7,withinitialspeed 3v1,i 5 uptheinclineand
  
  ## finalspeed 3v1,i 5 downtheincline,tocalculatet2.)Thesmallblockthenproceedswithconstantspeedin
  
  pursuitofthelargerblock,itspositionbeing
  
  ## x1,f t 3 v1,i t t1 t2 for t t1 t2
  
  5
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles933
  
  Theblockscollideforthesecondtimewhen x1 t x2 t .
  EVALUATEThecondition x1 t x2 t implies
  
  3
  2
  v1,i t t1 t2 1.4 m v1,i t
  5
  5
  Solvingfort,wefind
  
  5 1.4 m 3v1,i 5 t1 t2
  t
  7 m v1,i 3 t1 t2
  v1,i
  Numerically,wehave v0
  
  t1
  t2
  
  1.4 m
  1.05 s
  3 2.21 m/s 5
  6v1,i
  5a
  
  t 7 m
  
  12v1,i
  5g
  
  12 2.21 m/s
  
  5 9.8 m/s 2
  
  0.542 s
  
  ## 2.21 m/s 3 1.05 s 0.542 s 7.95 s
  
  ASSESSThisproblemisratherinvolved.However,thevalidityofourresultcanbecheckedbysubstitutingthe
  numericalvaluesfort1andt2backtotheequationsintheintermediatesteps.
  79.
  
  INTERPRETThistwodimensionaltwobodyprobleminvolvesaninelasticcollision.Intheverticaldirection,the
  motionisgovernedbytheforceofgravity,andinthehorizontaldirection,wecanuseconservationofmomentum
  tofindthehorizontalvelocityofthecombinedbodiesafterthecollision.Wewillneedtousekinematicequations
  tofindthevelocitiesandpositionsatthevariouspointsalongthetrajectories.
  DEVELOPThepeakoftheprojectile1isathalfitsrange,whichisgivenbyEquation3.15.Thus,thetwo
  projectilescollideatahorizontalposition,measuredfromthelaunchpointofprojectile1,of
  
  1 2
  x
  v0 sin 2
  2g
  2 9.8 m/s 2
  2
  
  ## Justbeforethecollision,thehorizontalvelocityofprojectile1is v1, x v1 cos 380 m/s cos 55 218 m/s .
  
  Fromconservationoflinearmomentum,weknowthatthehorizontalvelocityofthecombinedprojectileis
  o
  
  m1v1, x m2 v2, x m1 m2 vx
  m2 m1
  
  vx v1, x
  v2, x vx
  
  Wealsoknowthatthecombinedprojectiletravelsto9.6kmfromthelaunchpointbeforehittingtheground,so
  
  x x0 vx t
  vx
  
  x x0 9.6 km 6.92 km
  
  t
  t
  
  Finally,wecanfindtandthussolveform2usingthekinematicEquation2.7(v=v0+at,witha=gandv=0).
  Thisgives
  
  0 v0 gt
  t
  EVALUATEEvaluatingtheexpressionforvxgives
  
  v0
  g
  
  ## 9.6 km 6.92 km 9.8 m/s 2
  
  9.6 km 6.92 km
  vx
  g
  84.3 m/s
  v0
  380 m/s
  Insertingthisresultintotheexpressionform2gives
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  934
  
  Chapter9
  
  m2 m1
  
  vx v1, x
  v2, x vx
  
  14 kg
  
  ## 84.3 m/s 218 m/s
  
  8.3 kg
  140 m/s 84.3 m/s
  
  ASSESSThetimerequiredforthetwotofalltothegroundist=v0/g=(380m/s)/(9.8m/s)=31.8s.Wefindthat
  m2<m1,whichmakessensebecausethecombinedprojectilecontinuestotravelinthedirectionatwhichm1was
  initiallytraveling.
  80.
  
  INTERPRETWearetoldtheforcethatbringsacartorestinacollision.Fromthiswewillcalculatetheimpulse
  andtheaverageforce.
  DEVELOPTheforceexertedbythebarrierwallonthecarisonedimensional,sotheimpulseofthecollisionis
  J F t dt , wheretheintegralevaluatedoverthecollisionduration, t. Oncewehavetheimpulse,theaverage
  forceissimply F J / t , fromEquation9.10a.Andfinally,themasscanbefoundthroughNewton'ssecondlaw
  m F / a , withtheaccelerationdeterminedfromthecar'sinitialspeed.
  EVALUATE(a)Evaluatingtheintegraloverthegiventimeinterval:
  t
  
  J at 4 bt 3 ct 2 dt dt 15 a t 5 14 bt 4 13 ct 3 12 d t 2
  0
  
  8.86 GN
  
  3.27 GN
  4
  MN
  362
  5
  s4
  s3
  s2
  0.2s
  
  0.2s
  
  5
  4
  3
  
  12.5 MN
  
  s
  0.2s 2 25.6 kN s
  2
  
  0.2s 3
  
  (b)Asarguedabove,theaverageforceisjusttheimpulsedividedbythetimeinterval:
  
  J
  25.6 kN s
  
  128 kN
  t
  0.2 s
  
  (c)We'retoldthecarwasoriginallytravellingat50km/hbeforecomingtoarestin200ms,sothemagnitudeof
  theaverageaccelerationis
  
  v 50 km/h 1 m/s
  2
  
  69 m/s
  t
  0.2 s 3.6 km/h
  
  Fromthis,themassmustbe
  
  F 128 kN
  
  1900 kg
  a 69 m/s 2
  
  81.
  
  DEVELOPLet'ssimplifytheforceequationslightlybysubstitutingin x t / t :
  
  ## Thisfunctionequalszeroat x 0 and x 1, whichcorrespondstothebeginningandendofthecar'scollisionwith
  
  thebarrier.Theforceispositivebetweenthesetwopoints,sothereisapeakintheforcesomewherebetween
  x 0 and x 1, justaswe'dexpect.
  EVALUATEAtthepeak,thederivativeshouldbezero:
  
  dF
  56.8 x3 78.6 x 2 29.0 x 2.50 0
  dx
  wherewehavedroppedtheunitsforthetimebeing.Dividingthroughbythex3coefficientgivesafunctionwe'll
  call f x :
  
  ## Solvingacubiclikethisisratherinvolved,butweknowthatthereisarootbetween x 0 and x 1, sowecan
  
  usetrialanderrortofindwherethecubiccrossesthexaxis.Wefirstnotethat f x isnegativeat x 0 and
  positiveat x 1. Ifwetry x 0.5, wegetanegativeresult,sothecrossingpointmustbetotherightof x 0.5.
  Sowecheckandseethat f 0.75 0, but f 0.85 0. Thereforetherootisinbetweenthesepoints.Thiscanbe
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles935
  
  ## doneseveraltimestohoneinon x 0.825, whichisperhapssufficientlyaccurateforourpurposes.Intermsofthe
  
  originalvariables,thepeakoccursapproximatelyat t 165 ms withavalueof F 327 kN.
  ASSESSInthepreviousproblem,theaverageforcewasfoundtobe128kN.Thepeakvaluewefoundaboveis
  2.5timestheaverage,whichseemsreasonable.
  82.
  
  thesecondblockinthecourseofacollision.Thefractionisrelatedtothemassratio.
  
  ## DEVELOPWith v2i 0, Equations9.15aand9.15bbecome
  
  m m2
  v1f 1
  v1i
  m1 m2
  
  2m1
  v2f
  v1i
  m1 m
  2
  
  Thefractionoftheinitialkineticenergytransferredto m2 is
  
  K 2f 12 m2 v2f2
  1
  2
  K1i
  2 m1v1i
  
  ## EVALUATESubstitutingtheexpressionfor v1f intotheequationabove,weobtain
  
  K 2f m2 2m1
  
  K1i
  m1 m1 m2
  
  4m1m2
  
  m1 m2
  
  4 m1 /m2
  
  1 m1/m2 2
  
  Weplotthisratiointhefigurebelow.
  
  ASSESSThefractionofenergytransferreachesamaximumofunitywhenthemassratioequalsone.This
  
  correspondstoCase2inSection9.6where m1 m2 . Thefirstobjectstopscompletelyandtransfersallitsenergy
  tothesecondobject,whichmovesonwiththeinitialspeed v2f v1i .
  83.
  
  INTERPRETThisproblemislikethepreviousproblemwherewelookedatthetransferofkineticenergyfroma
  movingblocktoaninitiallystationaryblock.Here,weshowthatthefractionofenergytransferredisindependent
  ofwhichblockisthemovingoneandwhichisthestationaryone.
  
  DEVELOPWecanusetheresultfromthepreviousproblemtowritethefractionoftheinitialenergyinthemovingblock
  
  thatistransferredtotheinitiallystationaryblock:
  
  4 m1 /m2
  K stat
  
  K mov 1 m1 /m2 2
  
  whereweassumeinthiscasethatthemovingblockhasmass m1 andthestationaryblockhasmass m2 .
  
  4 m2 /m1
  4 m2 /m1 m1 /m2
  4 m1 /m2
  K stat
  
  2
  2
  2
  K mov 1 m2 /m1
  1 m2 /m1 m1/m2 1 m1/m2 2
  2
  
  Butthisisthesameasbefore,sotheenergytransferisindependentofwhichmassisinitiallystationary.
  ASSESSAsanexampleofthisindependence,onecanimaginealightblockcollidingwithaheavystationary
  block(likethepingpongballandbowlingballcollisioninSection9.6).Afterthecollision,theheavyblockbarely
  moves,whereasthelightblockricochetsbackwardwithessentiallythesamespeed.Inotherwords,verylittle
  energyistransferred K stat / K mov 0 . Thenimaginetheblocksswitchplaces,withthelightblockinitiallyat
  rest.Inthiscase,theheavyblockplowsintothelightblockandknocksitforward.Buttheheavyblockkeeps
  movingwithprettymuchitsinitialspeed,soagainnotmuchenergyistransferred K stat / K mov 0 .
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  936
  
  84.
  
  Chapter9
  
  weseethatthewireissymmetriclefttoright,sothecenterofmassisalongthecenterline.Weneedtofindthe
  distanceofthecenterofmassabovethecenterofthesemicircle.Wewouldexpectthatitssomewhereover
  halfwayup.
  DEVELOPUsethecoordinatesystemdefinedinthefigurebelow.Theequationforcenterofmassis(Equation
  9.4)
  
  1 r
  r
  rcm
  rdm
  M
  
  ## Becausethesystemisleftrightsymmetric,weneedonlytheverticalcomponentyof rcm ,sotheequationfor
  
  centerofmassreducesto
  
  1
  ydm
  M
  1
  M 2 R 2 M R ,sodm=Rdandy=Rsin.
  
  ycm
  Themassperunitlengthofthewireis M C 2
  
  EVALUATEWeintegrateovertheentirewire,from=0to=
  
  ycm
  
  1
  M
  R
  
  ydm
  0
  
  cos
  M
  R
  
  M
  0.637 R
  
  1
  R2
  R sin Rd
  sin d
  
  M 0
  M 0
  
  2R
  
  ASSESSThisisabitshortof2/3ofthewayup,whichiswhatweexpected.
  85.
  
  itsalongthecenteroftheslice,andclosertothecrustthantothetip.
  
  r
  rdm, whichintwodimensionscanbewrittenoutas:
  xi yj dm 1 xdm i 1 ydm j xcm i ycm j
  M
  M
  
  DEVELOPTheequationforcenterofmassis rcm
  
  r
  rcm 1
  M
  
  1
  M
  
  ycm 0. Asfor xcm , weuse x r cos andintegrateover r from0toR,andover from / 2 to / 2, where
  
  ## Theinfinitesimalmasselement, dm, isequalto dA rdrd , where isthemassperunitarea.Sincetheslice
  
  isuniform,thedensityisconstant:
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles937
  
  M
  M
  2M
  A / 2 R 2 R 2
  Onecancheckthesevaluesbyverifyingthat M dA.
  EVALUATEEvaluatingtheintegralforthexcomponentofthecenterofmassgives
  /2
  R
  xcm 1 r cos 2 M2 rdrd
  R
  M / 2 0
  R
  /2
  2 1 3
  4R
  r
  
  sin / 2 3 sin / 2
  R2 3 0
  
  ## xcm 0.65R, whichmatchesouroriginalpredictionthatthecenterofmasswillbeclosertotheoutercrustthanto
  
  thetip.Ifinstead 2 (thefullpizza),thenthecenterofmassisat xcm 0, aswewouldexpect.
  86.
  
  INTERPRETWeuseconservationofmomentumandofenergytocalculatethekineticenergyofapelletthatgoes
  energyofthependulum,sowecanfindthefinalkineticenergyofthepellet.
  DEVELOPWefindtheinitialspeedofthepelletfromitsinitialkineticenergy Ki=3.25Jandmassm=052
  103kg.Wethenworkbackwardfromthemaximumpotentialenergyofthependulumtofindthespeedofthe
  pendulum.KnowingthispotentialenergyisU=Mgh,wherethependulummassisM=0.400kgandtheheightis
  2
  h=5104m,wecanfindtheinitialkineticenergy K Mvp 2 ofthependulum.Thisgivesusthespeedofthe
  pendulumjustafterthecollision.Next,weuseconservationofmomentumatthecollisionwiththependulumto
  2
  findthepelletsspeedafterthecollision.Finally,weanswerthequestionusing K f mvf 2 .
  EVALUATETheinitialspeedofthepelletis
  
  1
  mvi 2
  2
  2 Ki
  vi
  m
  Ki
  
  Thespeedofthependulumafterthecollisionisfoundby
  
  U M gh
  
  M vp2
  
  1
  2
  
  vp 2 gh
  Weuseconservationofmomentumatthecollisiontofindthevelocityoftheexitingpellet:
  
  m
  
  2 Ki
  mvf M 2 gh
  m
  vf
  
  2Ki M
  
  m
  m
  
  2 gh
  
  2 3.25 J
  0.40 kg
  
  ## 2 9.8 m/s 2 50 103 m
  
  4
  5.2 10 kg 5.2 104 kg
  
  35.7 m/s
  Sothefinalkineticenergyis
  
  Kf
  
  1 2 1
  2
  mvf 5.2 10 4 kg 35.7 m/s 0.33 J
  2
  2
  
  ASSESSNotethatwecantjustuseconservationofenergy!Compare:
  
  Ei 3.25 J
  1
  Ef U K f Mgh mvf2 0.332 J
  2
  Mostofthekineticenergyislostinthecollision,becausethisisaninelasticcollision.
  
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  938
  
  87.
  
  Chapter9
  
  INTERPRETWeuseconservationofmomentumtofindthespeedofanastronautaftershethrowshertoolbox
  away,andusethisspeedandthegivendistancetodeterminewhethershereachessafetybeforeheroxygenruns
  out.
  DEVELOPThemassoftheastronautisma=80kg.Themassofthetoolboxismt=10kg.Theinitialspeedof
  bothiszero,sothefinalspeedofthetoolboxisvta=8m/srelativetotheastronaut.Wecanuseconservationof
  momentumtofindthespeedoftheastronaut:0=mtvt+mava.Oncewehavethisspeed,wecalculatehowlongit
  EVALUATEFirstwefindthespeedofthetoolboxrelativetotherestframe:vt=va+vta.Nextweplugthisinto
  theconservationofmomentumequation: 0 mt va vta ma va andsolvefortheastronautsspeed:
  
  0 mt va vta ma va
  
  va mt ma mt vta
  
  mt
  va vta
  0.89 m/s
  mt ma
  
  Thetimeittakesis
  
  x
  200 m
  
  ## 225 s 3.75 min
  
  va 0.89 m/s
  
  ASSESSShemakesitwith15secondstospare.
  88.
  
  INTERPRETTheSunwillrotatearoundthecenterofmassofthesolarsystem,sothisproblemisessentially
  Jupiterandneglectthemasscontributionfromtherestoftheplanets.
  
  DEVELOPWecanthinkoftheSunandJupiterastwoendsofabarbell,likethatinExample9.1.Thecenterofmasslieson
  
  ## thelinebetweenthetwoobjects,sowecandropthevectornotationfromEquation9.2: rcm M mi ri . Let'stake
  
  thecenteroftheSunasourorigin,sothattheSuncontributioniszero rS 0 . Theonlyterminthesumisthat
  forJupiter:
  1
  
  rcm
  
  mJ
  1
  mJ rJ mS rS
  r.
  
  m
  m
  S J
  S mJ J
  
  27
  EVALUATEFromAppendixE,themassofJupiteris 1.90 10 kg anditsorbitalradiusis 7.78 1011 m. Themass
  30
  oftheSunis 1.99 10 kg. Pluggingthesevaluesintheaboveexpression,
  
  rcm
  
  1.90 10 27 kg
  1.99 1030 kg 1.90 1027 kg
  
  7.78 10 m 7.42 10 m
  11
  
  8
  ASSESSThisisonlyslightlylargerthantheradiusoftheSun RS 6.96 10 m . Theactualdistancebetween
  
  theSunandthecenterofmassofthesolarsystemisconstantlychangingsincetheplanetsareconstantlymoving
  relativetoeachotherduetotheirorbitalmotion.
  89.
  
  Thedensityincreasesfromzeroatoneendtoamaximumattheotherend,sowe'dexpectthecenterofmasstobe
  closertothedenserend.
  
  ## DEVELOPWewillhavetointegratetofindthetotalmass: M tot dm dx. Thelimitsofintegrationarebetween
  
  x 0 and x L. Sincethemassisdistributedalongonedimension,thecenterofmassintegraltakesasimilar
  1
  1
  from: rcm M tot xdm M tot x dx.
  EVALUATE(a)Themassintegralgives
  L
  
  M tot dx
  
  Mx a
  M x1 a
  dx 1 a
  1 a
  L
  L 1 a
  
  M
  1 a
  
  (b)Usingtheaboveresult,thecenterofmassis
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  SystemsofParticles939
  
  rcm
  
  1
  M tot
  
  x dx
  
  1 a L Mx1 a
  1 a x2 a
  dx 1 a
  1
  
  M 0 L
  L 2 a
  
  1 a
  L
  2 a
  
  ## (c)If a 0, thedensityisconstant: M / L. ThetotalmassisM,andthecenterofmassoccursat 2 L, justas
  
  wewouldexpectforarodwithuniformdensity.
  2
  3
  ASSESSFor a 1, thecenterofmassoccursat 3 L, whilefor a 2, itoccursat 4 L. Thisagreeswithour
  premonitionthathavingthedensitygetlargertoward x L willmeanthatthecenterofmasswillbeclosertothatend.
  1
  
  90.
  
  DEVELOPThepictureshowsthattheballbouncesuptoalowerheightaftereachbounce.
  EVALUATEIfthecollisionswiththefloorweretotallyinelastic,thenwe'dexpecttheballtosticktothefloorafterthe
  
  firstbounce,whichitdoesnot.Toanalyzebetterwhatdoeshappen,let'sdivideupthevelocityintoitsxandy
  r
  components: v vx i v y j. Intermsofsolelytheverticalmotion,theballhasaheadoncollisionwiththeground.If
  ithitthegroundwith: v yf v yi . ThisfollowsfromCase1inSection9.6,whereasmallmass(theball)collideswitha
  muchlargermass(thefloor).Iftheballhasthesameverticalspeedafterthecollision,thenbyconservationofenergythe
  2
  ballshouldreturntoroughlythesameheight h v y aftereachbounce,whichitdoesnot.
  ASSESSWewouldsaythecollisionwiththegroundisinelastic(justnot"totally"),sincesomeofthekinetic
  energyislosttointernalenergy(heat)oftheballandtheground.
  
  91.
  
  2
  2
  DEVELOPRightbeforethesecondcollision,theballhaskineticenergy K i 2 mvxi 2 mv yi , whileafterthecollision,it
  1
  
  has K f mv mv . Wearguedinthepreviousproblemthatbecausetheballdoesn'treboundtothesame
  height,theverticalspeedatgroundlevelmustbegettingsmalleraftereachcollision v y 2 gh . Ifwejust
  considerthemotionintheverticaldirection,thefractionofenergylostis:
  1
  2
  
  2
  xf
  
  1
  2
  
  2
  yf
  
  K
  K
  i
  
  hf hi
  hi
  
  EVALUATEWithourfingersorwithasmallruler,wecancheckthatthepeakheightafterthesecondcollisionis
  
  energyintheverticaldirection.Assumingthelossinhorizontaldirectionwasn'tmorethanthat,thefractionofthe
  totalenergylostisalittlelessthanhalf.
  1
  1
  2
  2
  ASSESSWe'vetreatedthecomponentsofkineticenergyseparately: K x 2 mvx and K y 2 mv y . Itshouldbe
  notedthatthetwoarenotcompletelyseparate.Ifthegroundwereflatoriftheballwerespinning,acollisioncould
  transferenergyintheverticaldirectiontoenergyinthehorizontaldirection,orviceversa.
  92.
  
  DEVELOPTheverticalcomponentofthevelocityafteracollisioncanbeestimatedbytheheightthattheballreachesatthe
  
  topofthebounce: v y 2 gh . Sincetherearenohorizontalforcesactingontheballwhileit'sintheair,the
  horizontalcomponentofthevelocitybetweencollisionsisconstant.Itisequaltothedistance, x, theballtravels
  horizontallydividedbythetime,t,thatitremainsairbornebetweencollisionswiththefloor:
  
  vx
  
  x
  x
  xg
  g
  
  x
  t
  8
  h
  2 2 gh
  2v y / g
  
  EVALUATEInthepreviousproblem,weestimatedthattheheightafteracollisionis0.6timestheheightbefore
  
  distancebeforethecollision.Therefore,thehorizontalcomponentofthevelocitydecreasesby:
  be reproduced, in any form or by any means, without permission in writing from the publisher.
  
  940
  
  Chapter9
  
  vx vxf vxi xf / hf
  0.8
  
  1
  1 0
  vxi
  vxi
  xi / hi
  0.6
  Therefore,theverticalcomponentismoreaffected.
  ASSESSWemighthaveexpectedthatthehorizontalvelocityremainsroughlyconstantthroughthecollision.The
  nonconservativeforcesthatremoveenergyfromtheballarelikelytopointintheverticaldirectionwherethe
  velocitygoesthroughthebiggestchange.
  93.
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