Elastic collision - Wikipedia
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Two-dimensional
Elasticcollision
FromWikipedia,thefreeencyclopedia
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Collisioninwhichkineticenergyisconserved
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Aslongasblack-bodyradiation(notshown)doesn'tescapeasystem,atomsinthermalagitationundergoessentiallyelasticcollisions.Onaverage,twoatomsreboundfromeachotherwiththesamekineticenergyasbeforeacollision.Fiveatomsarecoloredredsotheirpathsofmotionareeasiertosee.
Inphysics,anelasticcollisionisanencounter(collision)betweentwobodiesinwhichthetotalkineticenergyofthetwobodiesremainsthesame.Inanideal,perfectlyelasticcollision,thereisnonetconversionofkineticenergyintootherformssuchasheat,noise,orpotentialenergy.
Duringthecollisionofsmallobjects,kineticenergyisfirstconvertedtopotentialenergyassociatedwitharepulsiveorattractiveforcebetweentheparticles(whentheparticlesmoveagainstthisforce,i.e.theanglebetweentheforceandtherelativevelocityisobtuse),thenthispotentialenergyisconvertedbacktokineticenergy(whentheparticlesmovewiththisforce,i.e.theanglebetweentheforceandtherelativevelocityisacute).
Collisionsofatomsareelastic,forexampleRutherfordbackscattering.
Ausefulspecialcaseofelasticcollisioniswhenthetwobodieshaveequalmass,inwhichcasetheywillsimplyexchangetheirmomenta.
Themolecules—asdistinctfromatoms—ofagasorliquidrarelyexperienceperfectlyelasticcollisionsbecausekineticenergyisexchangedbetweenthemolecules’translationalmotionandtheirinternaldegreesoffreedomwitheachcollision.Atanyinstant,halfthecollisionsare,toavaryingextent,inelasticcollisions(thepairpossesseslesskineticenergyintheirtranslationalmotionsafterthecollisionthanbefore),andhalfcouldbedescribedas“super-elastic”(possessingmorekineticenergyafterthecollisionthanbefore).Averagedacrosstheentiresample,molecularcollisionscanberegardedasessentiallyelasticaslongasPlanck'slawforbidsenergyfrombeingcarriedawaybyblack-bodyphotons.
Inthecaseofmacroscopicbodies,perfectlyelasticcollisionsareanidealneverfullyrealized,butapproximatedbytheinteractionsofobjectssuchasbilliardballs.
Whenconsideringenergies,possiblerotationalenergybeforeand/orafteracollisionmayalsoplayarole.
Contents
1Equations
1.1One-dimensionalNewtonian
1.1.1Examples
1.1.2Derivationofsolution
1.1.3Centerofmassframe
1.2One-dimensionalrelativistic
1.3Relativisticderivationusinghyperbolicfunctions
2Two-dimensional
2.1Two-dimensionalcollisionwithtwomovingobjects
3Seealso
4References
4.1Generalreferences
5Externallinks
Equations[edit]
One-dimensionalNewtonian[edit]
ProfessorWalterLewinexplainingone-dimensionalelasticcollisions
Inanelasticcollision,bothmomentumandkineticenergyareconserved.[1]Considerparticles1and2withmassesm1,m2,andvelocitiesu1,u2beforecollision,v1,v2aftercollision.Theconservationofthetotalmomentumbeforeandafterthecollisionisexpressedby:[1]
m
1
u
1
+
m
2
u
2
=
m
1
v
1
+
m
2
v
2
.
{\displaystylem_{1}u_{1}+m_{2}u_{2}\=\m_{1}v_{1}+m_{2}v_{2}.}
Likewise,theconservationofthetotalkineticenergyisexpressedby:[1]
1
2
m
1
u
1
2
+
1
2
m
2
u
2
2
=
1
2
m
1
v
1
2
+
1
2
m
2
v
2
2
.
{\displaystyle{\tfrac{1}{2}}m_{1}u_{1}^{2}+{\tfrac{1}{2}}m_{2}u_{2}^{2}\=\{\tfrac{1}{2}}m_{1}v_{1}^{2}+{\tfrac{1}{2}}m_{2}v_{2}^{2}.}
Theseequationsmaybesolveddirectlytofind
v
1
,
v
2
{\displaystylev_{1},v_{2}}
when
u
1
,
u
2
{\displaystyleu_{1},u_{2}}
areknown:[2]
v
1
=
m
1
−
m
2
m
1
+
m
2
u
1
+
2
m
2
m
1
+
m
2
u
2
v
2
=
2
m
1
m
1
+
m
2
u
1
+
m
2
−
m
1
m
1
+
m
2
u
2
{\displaystyle{\begin{array}{ccc}v_{1}&=&{\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}}}u_{1}+{\dfrac{2m_{2}}{m_{1}+m_{2}}}u_{2}\\[.5em]v_{2}&=&{\dfrac{2m_{1}}{m_{1}+m_{2}}}u_{1}+{\dfrac{m_{2}-m_{1}}{m_{1}+m_{2}}}u_{2}\end{array}}}
Ifbothmassesarethesame,wehaveatrivialsolution:
v
1
=
u
2
{\displaystylev_{1}=u_{2}}
v
2
=
u
1
.
{\displaystylev_{2}=u_{1}.}
Thissimplycorrespondstothebodiesexchangingtheirinitialvelocitiestoeachother.[2]
Ascanbeexpected,thesolutionisinvariantunderaddingaconstanttoallvelocities(Galileanrelativity),whichislikeusingaframeofreferencewithconstanttranslationalvelocity.Indeed,toderivetheequations,onemayfirstchangetheframeofreferencesothatoneoftheknownvelocitiesiszero,determinetheunknownvelocitiesinthenewframeofreference,andconvertbacktotheoriginalframeofreference.
Examples[edit]
Ball1:mass=3kg,velocity=4m/s
Ball2:mass=5kg,velocity=−6m/s
Aftercollision:
Ball1:velocity=−8.5m/s
Ball2:velocity=1.5m/s
Anothersituation:
Elasticcollisionofunequalmasses.
Thefollowingillustratethecaseofequalmass,
m
1
=
m
2
{\displaystylem_{1}=m_{2}}
.
Elasticcollisionofequalmasses
Elasticcollisionofmassesinasystemwithamovingframeofreference
Inthelimitingcasewhere
m
1
{\displaystylem_{1}}
ismuchlargerthan
m
2
{\displaystylem_{2}}
,suchasaping-pongpaddlehittingaping-pongballoranSUVhittingatrashcan,theheaviermasshardlychangesvelocity,whilethelightermassbouncesoff,reversingitsvelocityplusapproximatelytwicethatoftheheavyone.[3]
Inthecaseofalarge
u
1
{\displaystyleu_{1}}
,thevalueof
v
1
{\displaystylev_{1}}
issmallifthemassesareapproximatelythesame:hittingamuchlighterparticledoesnotchangethevelocitymuch,hittingamuchheavierparticlecausesthefastparticletobouncebackwithhighspeed.Thisiswhyaneutronmoderator(amediumwhichslowsdownfastneutrons,therebyturningthemintothermalneutronscapableofsustainingachainreaction)isamaterialfullofatomswithlightnucleiwhichdonoteasilyabsorbneutrons:thelightestnucleihaveaboutthesamemassasaneutron.
Derivationofsolution[edit]
Toderivetheaboveequationsfor
v
1
,
v
2
{\displaystylev_{1},v_{2}}
,rearrangethekineticenergyandmomentumequations:
m
1
(
v
1
2
−
u
1
2
)
=
m
2
(
u
2
2
−
v
2
2
)
{\displaystylem_{1}(v_{1}^{2}-u_{1}^{2})=m_{2}(u_{2}^{2}-v_{2}^{2})}
m
1
(
v
1
−
u
1
)
=
m
2
(
u
2
−
v
2
)
{\displaystylem_{1}(v_{1}-u_{1})=m_{2}(u_{2}-v_{2})}
Dividingeachsideofthetopequationbyeachsideofthebottomequation,andusing
a
2
−
b
2
(
a
−
b
)
=
a
+
b
{\displaystyle{\tfrac{a^{2}-b^{2}}{(a-b)}}=a+b}
,gives:
v
1
+
u
1
=
u
2
+
v
2
⇒
v
1
−
v
2
=
u
2
−
u
1
{\displaystylev_{1}+u_{1}=u_{2}+v_{2}\quad\Rightarrow\quadv_{1}-v_{2}=u_{2}-u_{1}}
.
Thatis,therelativevelocityofoneparticlewithrespecttotheotherisreversedbythecollision.
Nowtheaboveformulasfollowfromsolvingasystemoflinearequationsfor
v
1
,
v
2
{\displaystylev_{1},v_{2}}
,regarding
m
1
,
m
2
,
u
1
,
u
2
{\displaystylem_{1},m_{2},u_{1},u_{2}}
asconstants:
{
v
1
−
v
2
=
u
2
−
u
1
m
1
v
1
+
m
2
v
2
=
m
1
u
1
+
m
2
u
2
.
{\displaystyle\left\{{\begin{array}{rcrcc}v_{1}&-&v_{2}&=&u_{2}-u_{1}\\m_{1}v_{1}&+&m_{2}v_{2}&=&m_{1}u_{1}+m_{2}u_{2}.\end{array}}\right.}
Once
v
1
{\displaystylev_{1}}
isdetermined,
v
2
{\displaystylev_{2}}
canbefoundbysymmetry.
Centerofmassframe[edit]
Withrespecttothecenterofmass,bothvelocitiesarereversedbythecollision:aheavyparticlemovesslowlytowardthecenterofmass,andbouncesbackwiththesamelowspeed,andalightparticlemovesfasttowardthecenterofmass,andbouncesbackwiththesamehighspeed.
Thevelocityofthecenterofmassdoesnotchangebythecollision.Toseethis,considerthecenterofmassattime
t
{\displaystylet}
beforecollisionandtime
t
′
{\displaystylet'}
aftercollision:
x
¯
(
t
)
=
m
1
x
1
(
t
)
+
m
2
x
2
(
t
)
m
1
+
m
2
{\displaystyle{\bar{x}}(t)={\frac{m_{1}x_{1}(t)+m_{2}x_{2}(t)}{m_{1}+m_{2}}}}
x
¯
(
t
′
)
=
m
1
x
1
(
t
′
)
+
m
2
x
2
(
t
′
)
m
1
+
m
2
.
{\displaystyle{\bar{x}}(t')={\frac{m_{1}x_{1}(t')+m_{2}x_{2}(t')}{m_{1}+m_{2}}}.}
Hence,thevelocitiesofthecenterofmassbeforeandaftercollisionare:
v
x
¯
=
m
1
u
1
+
m
2
u
2
m
1
+
m
2
{\displaystylev_{\bar{x}}={\frac{m_{1}u_{1}+m_{2}u_{2}}{m_{1}+m_{2}}}}
v
x
¯
′
=
m
1
v
1
+
m
2
v
2
m
1
+
m
2
.
{\displaystylev_{\bar{x}}'={\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}}.}
Thenumeratorsof
v
x
¯
{\displaystylev_{\bar{x}}}
and
v
x
¯
′
{\displaystylev_{\bar{x}}'}
arethetotalmomentabeforeandaftercollision.Sincemomentumisconserved,wehave
v
x
¯
=
v
x
¯
′
{\displaystylev_{\bar{x}}=v_{\bar{x}}'}
.
One-dimensionalrelativistic[edit]
Accordingtospecialrelativity,
p
=
m
v
1
−
v
2
c
2
{\displaystylep={\frac{mv}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}}
wherepdenotesmomentumofanyparticlewithmass,vdenotesvelocity,andcisthespeedoflight.
Inthecenterofmomentumframewherethetotalmomentumequalszero,
p
1
=
−
p
2
{\displaystylep_{1}=-p_{2}}
p
1
2
=
p
2
2
{\displaystylep_{1}^{2}=p_{2}^{2}}
m
1
2
c
4
+
p
1
2
c
2
+
m
2
2
c
4
+
p
2
2
c
2
=
E
{\displaystyle{\sqrt{m_{1}^{2}c^{4}+p_{1}^{2}c^{2}}}+{\sqrt{m_{2}^{2}c^{4}+p_{2}^{2}c^{2}}}=E}
p
1
=
±
E
4
−
2
E
2
m
1
2
c
4
−
2
E
2
m
2
2
c
4
+
m
1
4
c
8
−
2
m
1
2
m
2
2
c
8
+
m
2
4
c
8
2
c
E
{\displaystylep_{1}=\pm{\frac{\sqrt{E^{4}-2E^{2}m_{1}^{2}c^{4}-2E^{2}m_{2}^{2}c^{4}+m_{1}^{4}c^{8}-2m_{1}^{2}m_{2}^{2}c^{8}+m_{2}^{4}c^{8}}}{2cE}}}
u
1
=
−
v
1
.
{\displaystyleu_{1}=-v_{1}.}
Here
m
1
,
m
2
{\displaystylem_{1},m_{2}}
representtherestmassesofthetwocollidingbodies,
u
1
,
u
2
{\displaystyleu_{1},u_{2}}
representtheirvelocitiesbeforecollision,
v
1
,
v
2
{\displaystylev_{1},v_{2}}
theirvelocitiesaftercollision,
p
1
,
p
2
{\displaystylep_{1},p_{2}}
theirmomenta,
c
{\displaystylec}
isthespeedoflightinvacuum,and
E
{\displaystyleE}
denotesthetotalenergy,thesumofrestmassesandkineticenergiesofthetwobodies.
Sincethetotalenergyandmomentumofthesystemareconservedandtheirrestmassesdonotchange,itisshownthatthemomentumofthecollidingbodyisdecidedbytherestmassesofthecollidingbodies,totalenergyandthetotalmomentum.Relativetothecenterofmomentumframe,themomentumofeachcollidingbodydoesnotchangemagnitudeaftercollision,butreversesitsdirectionofmovement.
Comparingwithclassicalmechanics,whichgivesaccurateresultsdealingwithmacroscopicobjectsmovingmuchslowerthanthespeedoflight,totalmomentumofthetwocollidingbodiesisframe-dependent.Inthecenterofmomentumframe,accordingtoclassicalmechanics,
m
1
u
1
+
m
2
u
2
=
m
1
v
1
+
m
2
v
2
=
0
{\displaystylem_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}={0}\,\!}
m
1
u
1
2
+
m
2
u
2
2
=
m
1
v
1
2
+
m
2
v
2
2
{\displaystylem_{1}u_{1}^{2}+m_{2}u_{2}^{2}=m_{1}v_{1}^{2}+m_{2}v_{2}^{2}\,\!}
(
m
2
u
2
)
2
2
m
1
+
(
m
2
u
2
)
2
2
m
2
=
(
m
2
v
2
)
2
2
m
1
+
(
m
2
v
2
)
2
2
m
2
{\displaystyle{\frac{(m_{2}u_{2})^{2}}{2m_{1}}}+{\frac{(m_{2}u_{2})^{2}}{2m_{2}}}={\frac{(m_{2}v_{2})^{2}}{2m_{1}}}+{\frac{(m_{2}v_{2})^{2}}{2m_{2}}}\,\!}
(
m
1
+
m
2
)
(
m
2
u
2
)
2
=
(
m
1
+
m
2
)
(
m
2
v
2
)
2
{\displaystyle(m_{1}+m_{2})(m_{2}u_{2})^{2}=(m_{1}+m_{2})(m_{2}v_{2})^{2}\,\!}
u
2
=
−
v
2
{\displaystyleu_{2}=-v_{2}\,\!}
(
m
1
u
1
)
2
2
m
1
+
(
m
1
u
1
)
2
2
m
2
=
(
m
1
v
1
)
2
2
m
1
+
(
m
1
v
1
)
2
2
m
2
{\displaystyle{\frac{(m_{1}u_{1})^{2}}{2m_{1}}}+{\frac{(m_{1}u_{1})^{2}}{2m_{2}}}={\frac{(m_{1}v_{1})^{2}}{2m_{1}}}+{\frac{(m_{1}v_{1})^{2}}{2m_{2}}}\,\!}
(
m
1
+
m
2
)
(
m
1
u
1
)
2
=
(
m
1
+
m
2
)
(
m
1
v
1
)
2
{\displaystyle(m_{1}+m_{2})(m_{1}u_{1})^{2}=(m_{1}+m_{2})(m_{1}v_{1})^{2}\,\!}
u
1
=
−
v
1
{\displaystyleu_{1}=-v_{1}\,\!}
Thisagreeswiththerelativisticcalculation
u
1
=
−
v
1
{\displaystyleu_{1}=-v_{1}}
,despiteotherdifferences.
OneofthepostulatesinSpecialRelativitystatesthatthelawsofphysics,suchasconservationofmomentum,shouldbeinvariantinallinertialframesofreference.Inageneralinertialframewherethetotalmomentumcouldbearbitrary,
m
1
u
1
1
−
u
1
2
/
c
2
+
m
2
u
2
1
−
u
2
2
/
c
2
=
m
1
v
1
1
−
v
1
2
/
c
2
+
m
2
v
2
1
−
v
2
2
/
c
2
=
p
T
{\displaystyle{\frac{m_{1}\;u_{1}}{\sqrt{1-u_{1}^{2}/c^{2}}}}+{\frac{m_{2}\;u_{2}}{\sqrt{1-u_{2}^{2}/c^{2}}}}={\frac{m_{1}\;v_{1}}{\sqrt{1-v_{1}^{2}/c^{2}}}}+{\frac{m_{2}\;v_{2}}{\sqrt{1-v_{2}^{2}/c^{2}}}}=p_{T}}
m
1
c
2
1
−
u
1
2
/
c
2
+
m
2
c
2
1
−
u
2
2
/
c
2
=
m
1
c
2
1
−
v
1
2
/
c
2
+
m
2
c
2
1
−
v
2
2
/
c
2
=
E
{\displaystyle{\frac{m_{1}c^{2}}{\sqrt{1-u_{1}^{2}/c^{2}}}}+{\frac{m_{2}c^{2}}{\sqrt{1-u_{2}^{2}/c^{2}}}}={\frac{m_{1}c^{2}}{\sqrt{1-v_{1}^{2}/c^{2}}}}+{\frac{m_{2}c^{2}}{\sqrt{1-v_{2}^{2}/c^{2}}}}=E}
Wecanlookatthetwomovingbodiesasonesystemofwhichthetotalmomentumis
p
T
{\displaystylep_{T}}
,thetotalenergyis
E
{\displaystyleE}
anditsvelocity
v
c
{\displaystylev_{c}}
isthevelocityofitscenterofmass.Relativetothecenterofmomentumframethetotalmomentumequalszero.Itcanbeshownthat
v
c
{\displaystylev_{c}}
isgivenby:
v
c
=
p
T
c
2
E
{\displaystylev_{c}={\frac{p_{T}c^{2}}{E}}}
Nowthevelocitiesbeforethecollisioninthecenterofmomentumframe
u
1
′
{\displaystyleu_{1}'}
and
u
2
′
{\displaystyleu_{2}'}
are:
u
1
′
=
u
1
−
v
c
1
−
u
1
v
c
c
2
{\displaystyleu_{1}'={\frac{u_{1}-v_{c}}{1-{\frac{u_{1}v_{c}}{c^{2}}}}}}
u
2
′
=
u
2
−
v
c
1
−
u
2
v
c
c
2
{\displaystyleu_{2}'={\frac{u_{2}-v_{c}}{1-{\frac{u_{2}v_{c}}{c^{2}}}}}}
v
1
′
=
−
u
1
′
{\displaystylev_{1}'=-u_{1}'}
v
2
′
=
−
u
2
′
{\displaystylev_{2}'=-u_{2}'}
v
1
=
v
1
′
+
v
c
1
+
v
1
′
v
c
c
2
{\displaystylev_{1}={\frac{v_{1}'+v_{c}}{1+{\frac{v_{1}'v_{c}}{c^{2}}}}}}
v
2
=
v
2
′
+
v
c
1
+
v
2
′
v
c
c
2
{\displaystylev_{2}={\frac{v_{2}'+v_{c}}{1+{\frac{v_{2}'v_{c}}{c^{2}}}}}}
When
u
1
≪
c
{\displaystyleu_{1}\llc}
and
u
2
≪
c
{\displaystyleu_{2}\llc}
,
p
T
{\displaystylep_{T}}
≈
m
1
u
1
+
m
2
u
2
{\displaystylem_{1}u_{1}+m_{2}u_{2}}
v
c
{\displaystylev_{c}}
≈
m
1
u
1
+
m
2
u
2
m
1
+
m
2
{\displaystyle{\frac{m_{1}u_{1}+m_{2}u_{2}}{m_{1}+m_{2}}}}
u
1
′
{\displaystyleu_{1}'}
≈
u
1
−
v
c
{\displaystyleu_{1}-v_{c}}
≈
m
1
u
1
+
m
2
u
1
−
m
1
u
1
−
m
2
u
2
m
1
+
m
2
=
m
2
(
u
1
−
u
2
)
m
1
+
m
2
{\displaystyle{\frac{m_{1}u_{1}+m_{2}u_{1}-m_{1}u_{1}-m_{2}u_{2}}{m_{1}+m_{2}}}={\frac{m_{2}(u_{1}-u_{2})}{m_{1}+m_{2}}}}
u
2
′
{\displaystyleu_{2}'}
≈
m
1
(
u
2
−
u
1
)
m
1
+
m
2
{\displaystyle{\frac{m_{1}(u_{2}-u_{1})}{m_{1}+m_{2}}}}
v
1
′
{\displaystylev_{1}'}
≈
m
2
(
u
2
−
u
1
)
m
1
+
m
2
{\displaystyle{\frac{m_{2}(u_{2}-u_{1})}{m_{1}+m_{2}}}}
v
2
′
{\displaystylev_{2}'}
≈
m
1
(
u
1
−
u
2
)
m
1
+
m
2
{\displaystyle{\frac{m_{1}(u_{1}-u_{2})}{m_{1}+m_{2}}}}
v
1
{\displaystylev_{1}}
≈
v
1
′
+
v
c
{\displaystylev_{1}'+v_{c}}
≈
m
2
u
2
−
m
2
u
1
+
m
1
u
1
+
m
2
u
2
m
1
+
m
2
=
u
1
(
m
1
−
m
2
)
+
2
m
2
u
2
m
1
+
m
2
{\displaystyle{\frac{m_{2}u_{2}-m_{2}u_{1}+m_{1}u_{1}+m_{2}u_{2}}{m_{1}+m_{2}}}={\frac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2}}{m_{1}+m_{2}}}}
v
2
{\displaystylev_{2}}
≈
u
2
(
m
2
−
m
1
)
+
2
m
1
u
1
m
1
+
m
2
{\displaystyle{\frac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}}}
Therefore,theclassicalcalculationholdstruewhenthespeedofbothcollidingbodiesismuchlowerthanthespeedoflight(~300millionm/s).
Relativisticderivationusinghyperbolicfunctions[edit]
Weusetheso-calledparameterofvelocity
s
{\displaystyles}
(usuallycalledtherapidity)toget :
v
/
c
=
tanh
(
s
)
{\displaystylev/c=\tanh(s)}
henceweget
1
−
v
2
c
2
=
sech
(
s
)
{\displaystyle{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}=\operatorname{sech}(s)}
Relativisticenergyandmomentumareexpressedasfollows:
E
=
m
c
2
1
−
v
2
c
2
=
m
c
2
cosh
(
s
)
{\displaystyleE={\frac{mc^{2}}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}=mc^{2}\cosh(s)}
p
=
m
v
1
−
v
2
c
2
=
m
c
sinh
(
s
)
{\displaystylep={\frac{mv}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}=mc\sinh(s)}
Equationssumofenergyandmomentumcollidingmasses
m
1
{\displaystylem_{1}}
and
m
2
{\displaystylem_{2}}
,(velocities
v
1
{\displaystylev_{1}}
,
v
2
{\displaystylev_{2}}
,
u
1
{\displaystyleu_{1}}
,
u
2
{\displaystyleu_{2}}
correspondtothevelocityparameters
s
1
{\displaystyles_{1}}
,
s
2
{\displaystyles_{2}}
,
s
3
{\displaystyles_{3}}
,
s
4
{\displaystyles_{4}}
),afterdividingbyadequatepower
c
{\displaystylec}
areasfollows:
m
1
cosh
(
s
1
)
+
m
2
cosh
(
s
2
)
=
m
1
cosh
(
s
3
)
+
m
2
cosh
(
s
4
)
{\displaystylem_{1}\cosh(s_{1})+m_{2}\cosh(s_{2})=m_{1}\cosh(s_{3})+m_{2}\cosh(s_{4})}
m
1
sinh
(
s
1
)
+
m
2
sinh
(
s
2
)
=
m
1
sinh
(
s
3
)
+
m
2
sinh
(
s
4
)
{\displaystylem_{1}\sinh(s_{1})+m_{2}\sinh(s_{2})=m_{1}\sinh(s_{3})+m_{2}\sinh(s_{4})}
anddependentequation,thesumofaboveequations:
m
1
e
s
1
+
m
2
e
s
2
=
m
1
e
s
3
+
m
2
e
s
4
{\displaystylem_{1}e^{s_{1}}+m_{2}e^{s_{2}}=m_{1}e^{s_{3}}+m_{2}e^{s_{4}}}
subtractsquaresbothsidesequations"momentum"from"energy"andusetheidentity
cosh
2
(
s
)
−
sinh
2
(
s
)
=
1
{\displaystyle\cosh^{2}(s)-\sinh^{2}(s)=1}
,aftersimplicityweget:
2
m
1
m
2
(
cosh
(
s
1
)
cosh
(
s
2
)
−
sinh
(
s
2
)
sinh
(
s
1
)
)
=
2
m
1
m
2
(
cosh
(
s
3
)
cosh
(
s
4
)
−
sinh
(
s
4
)
sinh
(
s
3
)
)
{\displaystyle2m_{1}m_{2}(\cosh(s_{1})\cosh(s_{2})-\sinh(s_{2})\sinh(s_{1}))=2m_{1}m_{2}(\cosh(s_{3})\cosh(s_{4})-\sinh(s_{4})\sinh(s_{3}))}
fornon-zeromass,usingthehyperbolictrigonometricidentitycosh(a−b)=cosh(a)cosh(b)−sinh(b)sinh(a),weget:
cosh
(
s
1
−
s
2
)
=
cosh
(
s
3
−
s
4
)
{\displaystyle\cosh(s_{1}-s_{2})=\cosh(s_{3}-s_{4})}
asfunctions
cosh
(
s
)
{\displaystyle\cosh(s)}
isevenwegettwosolutions:
s
1
−
s
2
=
s
3
−
s
4
{\displaystyles_{1}-s_{2}=s_{3}-s_{4}}
s
1
−
s
2
=
−
s
3
+
s
4
{\displaystyles_{1}-s_{2}=-s_{3}+s_{4}}
fromthelastequation,leadingtoanon-trivialsolution,wesolve
s
2
{\displaystyles_{2}}
andsubstituteintothedependentequation,weobtain
e
s
1
{\displaystylee^{s_{1}}}
andthen
e
s
2
{\displaystylee^{s_{2}}}
,wehave:
e
s
1
=
e
s
4
m
1
e
s
3
+
m
2
e
s
4
m
1
e
s
4
+
m
2
e
s
3
{\displaystylee^{s_{1}}=e^{s_{4}}{\frac{m_{1}e^{s_{3}}+m_{2}e^{s_{4}}}{m_{1}e^{s_{4}}+m_{2}e^{s_{3}}}}}
e
s
2
=
e
s
3
m
1
e
s
3
+
m
2
e
s
4
m
1
e
s
4
+
m
2
e
s
3
{\displaystylee^{s_{2}}=e^{s_{3}}{\frac{m_{1}e^{s_{3}}+m_{2}e^{s_{4}}}{m_{1}e^{s_{4}}+m_{2}e^{s_{3}}}}}
Itisasolutiontotheproblem,butexpressedbytheparametersofvelocity.Returnsubstitutiontogetthesolutionforvelocitiesis:
v
1
/
c
=
tanh
(
s
1
)
=
e
s
1
−
e
−
s
1
e
s
1
+
e
−
s
1
{\displaystylev_{1}/c=\tanh(s_{1})={\frac{e^{s_{1}}-e^{-s_{1}}}{e^{s_{1}}+e^{-s_{1}}}}}
v
2
/
c
=
tanh
(
s
2
)
=
e
s
2
−
e
−
s
2
e
s
2
+
e
−
s
2
{\displaystylev_{2}/c=\tanh(s_{2})={\frac{e^{s_{2}}-e^{-s_{2}}}{e^{s_{2}}+e^{-s_{2}}}}}
Substitutetheprevioussolutionsandreplace:
e
s
3
=
c
+
u
1
c
−
u
1
{\displaystylee^{s_{3}}={\sqrt{\frac{c+u_{1}}{c-u_{1}}}}}
and
e
s
4
=
c
+
u
2
c
−
u
2
{\displaystylee^{s_{4}}={\sqrt{\frac{c+u_{2}}{c-u_{2}}}}}
,afterlongtransformation,withsubstituting:
Z
=
(
1
−
u
1
2
/
c
2
)
(
1
−
u
2
2
/
c
2
)
{\textstyleZ={\sqrt{\left(1-u_{1}^{2}/c^{2}\right)\left(1-u_{2}^{2}/c^{2}\right)}}}
weget:
v
1
=
2
m
1
m
2
c
2
u
2
Z
+
2
m
2
2
c
2
u
2
−
(
m
1
2
+
m
2
2
)
u
1
u
2
2
+
(
m
1
2
−
m
2
2
)
c
2
u
1
2
m
1
m
2
c
2
Z
−
2
m
2
2
u
1
u
2
−
(
m
1
2
−
m
2
2
)
u
2
2
+
(
m
1
2
+
m
2
2
)
c
2
{\displaystylev_{1}={\frac{2m_{1}m_{2}c^{2}u_{2}Z+2m_{2}^{2}c^{2}u_{2}-(m_{1}^{2}+m_{2}^{2})u_{1}u_{2}^{2}+(m_{1}^{2}-m_{2}^{2})c^{2}u_{1}}{2m_{1}m_{2}c^{2}Z-2m_{2}^{2}u_{1}u_{2}-(m_{1}^{2}-m_{2}^{2})u_{2}^{2}+(m_{1}^{2}+m_{2}^{2})c^{2}}}}
v
2
=
2
m
1
m
2
c
2
u
1
Z
+
2
m
1
2
c
2
u
1
−
(
m
1
2
+
m
2
2
)
u
1
2
u
2
+
(
m
2
2
−
m
1
2
)
c
2
u
2
2
m
1
m
2
c
2
Z
−
2
m
1
2
u
1
u
2
−
(
m
2
2
−
m
1
2
)
u
1
2
+
(
m
1
2
+
m
2
2
)
c
2
{\displaystylev_{2}={\frac{2m_{1}m_{2}c^{2}u_{1}Z+2m_{1}^{2}c^{2}u_{1}-(m_{1}^{2}+m_{2}^{2})u_{1}^{2}u_{2}+(m_{2}^{2}-m_{1}^{2})c^{2}u_{2}}{2m_{1}m_{2}c^{2}Z-2m_{1}^{2}u_{1}u_{2}-(m_{2}^{2}-m_{1}^{2})u_{1}^{2}+(m_{1}^{2}+m_{2}^{2})c^{2}}}}
.
Two-dimensional[edit]
Forthecaseoftwonon-spinningcollidingbodiesintwodimensions,themotionofthebodiesisdeterminedbythethreeconservationlawsofmomentum,kineticenergyandangularmomentum.Theoverallvelocityofeachbodymustbesplitintotwoperpendicularvelocities:onetangenttothecommonnormalsurfacesofthecollidingbodiesatthepointofcontact,theotheralongthelineofcollision.Sincethecollisiononlyimpartsforcealongthelineofcollision,thevelocitiesthataretangenttothepointofcollisiondonotchange.Thevelocitiesalongthelineofcollisioncanthenbeusedinthesameequationsasaone-dimensionalcollision.Thefinalvelocitiescanthenbecalculatedfromthetwonewcomponentvelocitiesandwilldependonthepointofcollision.Studiesoftwo-dimensionalcollisionsareconductedformanybodiesintheframeworkofatwo-dimensionalgas.
Two-dimensionalelasticcollision
Inacenterofmomentumframeatanytimethevelocitiesofthetwobodiesareinoppositedirections,withmagnitudesinverselyproportionaltothemasses.Inanelasticcollisionthesemagnitudesdonotchange.Thedirectionsmaychangedependingontheshapesofthebodiesandthepointofimpact.Forexample,inthecaseofspherestheangledependsonthedistancebetweenthe(parallel)pathsofthecentersofthetwobodies.Anynon-zerochangeofdirectionispossible:ifthisdistanceiszerothevelocitiesarereversedinthecollision;ifitisclosetothesumoftheradiiofthespheresthetwobodiesareonlyslightlydeflected.
Assumingthatthesecondparticleisatrestbeforethecollision,theanglesofdeflectionofthetwoparticles,
θ
1
{\displaystyle\theta_{1}}
and
θ
2
{\displaystyle\theta_{2}}
,arerelatedtotheangleofdeflection
θ
{\displaystyle\theta}
inthesystemofthecenterofmassby[4]
tan
θ
1
=
m
2
sin
θ
m
1
+
m
2
cos
θ
,
θ
2
=
π
−
θ
2
.
{\displaystyle\tan\theta_{1}={\frac{m_{2}\sin\theta}{m_{1}+m_{2}\cos\theta}},\qquad\theta_{2}={\frac{{\pi}-{\theta}}{2}}.}
Themagnitudesofthevelocitiesoftheparticlesafterthecollisionare:
v
1
′
=
v
1
m
1
2
+
m
2
2
+
2
m
1
m
2
cos
θ
m
1
+
m
2
,
v
2
′
=
v
1
2
m
1
m
1
+
m
2
sin
θ
2
.
{\displaystylev'_{1}=v_{1}{\frac{\sqrt{m_{1}^{2}+m_{2}^{2}+2m_{1}m_{2}\cos\theta}}{m_{1}+m_{2}}},\qquadv'_{2}=v_{1}{\frac{2m_{1}}{m_{1}+m_{2}}}\sin{\frac{\theta}{2}}.}
Two-dimensionalcollisionwithtwomovingobjects[edit]
Thefinalxandyvelocitiescomponentsofthefirstballcanbecalculatedas:[5]
v
1
x
′
=
v
1
cos
(
θ
1
−
φ
)
(
m
1
−
m
2
)
+
2
m
2
v
2
cos
(
θ
2
−
φ
)
m
1
+
m
2
cos
(
φ
)
+
v
1
sin
(
θ
1
−
φ
)
cos
(
φ
+
π
2
)
v
1
y
′
=
v
1
cos
(
θ
1
−
φ
)
(
m
1
−
m
2
)
+
2
m
2
v
2
cos
(
θ
2
−
φ
)
m
1
+
m
2
sin
(
φ
)
+
v
1
sin
(
θ
1
−
φ
)
sin
(
φ
+
π
2
)
{\displaystyle{\begin{aligned}v'_{1x}&={\frac{v_{1}\cos(\theta_{1}-\varphi)(m_{1}-m_{2})+2m_{2}v_{2}\cos(\theta_{2}-\varphi)}{m_{1}+m_{2}}}\cos(\varphi)+v_{1}\sin(\theta_{1}-\varphi)\cos(\varphi+{\tfrac{\pi}{2}})\\[0.8em]v'_{1y}&={\frac{v_{1}\cos(\theta_{1}-\varphi)(m_{1}-m_{2})+2m_{2}v_{2}\cos(\theta_{2}-\varphi)}{m_{1}+m_{2}}}\sin(\varphi)+v_{1}\sin(\theta_{1}-\varphi)\sin(\varphi+{\tfrac{\pi}{2}})\end{aligned}}}
wherev1andv2arethescalarsizesofthetwooriginalspeedsoftheobjects,m1andm2aretheirmasses,θ1andθ2aretheirmovementangles,thatis,
v
1
x
=
v
1
cos
θ
1
,
v
1
y
=
v
1
sin
θ
1
{\displaystylev_{1x}=v_{1}\cos\theta_{1},\;v_{1y}=v_{1}\sin\theta_{1}}
(meaningmovingdirectlydowntotherightiseithera−45°angle,ora315°angle),andlowercasephi(φ)isthecontactangle.(Togetthexandyvelocitiesofthesecondball,oneneedstoswapallthe'1'subscriptswith'2'subscripts.)
Thisequationisderivedfromthefactthattheinteractionbetweenthetwobodiesiseasilycalculatedalongthecontactangle,meaningthevelocitiesoftheobjectscanbecalculatedinonedimensionbyrotatingthexandyaxistobeparallelwiththecontactangleoftheobjects,andthenrotatedbacktotheoriginalorientationtogetthetruexandycomponentsofthevelocities[6][7][8][9][10][11]
Inanangle-freerepresentation,thechangedvelocitiesarecomputedusingthecentersx1andx2atthetimeofcontactas
v
1
′
=
v
1
−
2
m
2
m
1
+
m
2
⟨
v
1
−
v
2
,
x
1
−
x
2
⟩
‖
x
1
−
x
2
‖
2
(
x
1
−
x
2
)
,
v
2
′
=
v
2
−
2
m
1
m
1
+
m
2
⟨
v
2
−
v
1
,
x
2
−
x
1
⟩
‖
x
2
−
x
1
‖
2
(
x
2
−
x
1
)
{\displaystyle{\begin{aligned}\mathbf{v}'_{1}&=\mathbf{v}_{1}-{\frac{2m_{2}}{m_{1}+m_{2}}}\{\frac{\langle\mathbf{v}_{1}-\mathbf{v}_{2},\,\mathbf{x}_{1}-\mathbf{x}_{2}\rangle}{\|\mathbf{x}_{1}-\mathbf{x}_{2}\|^{2}}}\(\mathbf{x}_{1}-\mathbf{x}_{2}),\\\mathbf{v}'_{2}&=\mathbf{v}_{2}-{\frac{2m_{1}}{m_{1}+m_{2}}}\{\frac{\langle\mathbf{v}_{2}-\mathbf{v}_{1},\,\mathbf{x}_{2}-\mathbf{x}_{1}\rangle}{\|\mathbf{x}_{2}-\mathbf{x}_{1}\|^{2}}}\(\mathbf{x}_{2}-\mathbf{x}_{1})\end{aligned}}}
wheretheanglebracketsindicatetheinnerproduct(ordotproduct)oftwovectors.
Seealso[edit]
Collision
Inelasticcollision
Coefficientofrestitution
References[edit]
^abcSerway,RaymondA.(5March2013).Physicsforscientistsandengineerswithmodernphysics.Jewett,JohnW.,Peroomian,Vahé.(Ninth ed.).Boston,MA.p. 257.ISBN 978-1-133-95405-7.OCLC 802321453.
^abSerway,RaymondA.(5March2013).Physicsforscientistsandengineerswithmodernphysics.Jewett,JohnW.,Peroomian,Vahé.(Ninth ed.).Boston,MA.p. 258.ISBN 978-1-133-95405-7.OCLC 802321453.
^Serway,RaymondA.(5March2013).Physicsforscientistsandengineerswithmodernphysics.Jewett,JohnW.,Peroomian,Vahé.(Ninth ed.).Boston,MA.p. 258-9.ISBN 978-1-133-95405-7.OCLC 802321453.
^Landau,L.D.;Lifshitz,E.M.(1976).Mechanics(3rd ed.).PergamonPress.p. 46.ISBN 0-08-021022-8.
^Craver,WilliamE."ElasticCollisions."Williamecraver.wix.com.Wix.com,13Aug.2013.Web.13Aug.2013.
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