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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

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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.

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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

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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

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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:

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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

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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

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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

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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.

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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.

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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

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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

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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.

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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:

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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

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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.

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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

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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.

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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
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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

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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
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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:
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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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