15.6: Two Dimensional Elastic Collisions - Physics LibreTexts

Consider the elastic collision between two particles in which we neglect any external forces on the system consisting of the two particles. Skiptomaincontent Two-dimensionalElasticCollisioninLaboratoryReferenceFrame Two-dimensionalElasticCollisioninLaboratoryReferenceFrame Considertheelasticcollisionbetweentwoparticlesinwhichweneglectanyexternalforcesonthesystemconsistingofthetwoparticles.Particle1ofmass\(m_{1}\)isinitiallymovingwithvelocity\(\overrightarrow{\mathbf{V}}_{1,i}\)andcollideselasticallywithaparticle2ofmassthatis\(m_{2}\)initiallyatrest.Weshallrefertothereferenceframeinwhichoneparticleisatrest,‘thetarget’,asthelaboratoryreferenceframe.Afterthecollisionparticle1moveswithvelocity\(\overrightarrow{\mathbf{V}}_{1,f}\),andparticle2moveswithvelocity\(\overrightarrow{\mathbf{V}}_{2,f}\),(Figure15.9).Theangles\(\theta_{1,f}\)and\(\theta_{2,f}\)thattheparticlesmakewiththepositiveforwarddirectionofparticle1arecalledthelaboratoryscatteringangles. Figure15.9Two-dimensionalcollisioninlaboratoryreferenceframe Generallytheinitialvelocity\(\overrightarrow{\mathbf{v}}_{1,i}\)ofparticle1isknownandwewouldliketodeterminethefinalvelocities\(\overrightarrow{\mathbf{V}}_{1,f}\)and\(\overrightarrow{\mathbf{V}}_{2,f}\),whichrequiresfindingthemagnitudesanddirectionsofeachofthesevectors,\(v_{1,f},v_{2,f},\theta_{1,f}\)and\(\theta_{2,f}\)Thesequantitiesarerelatedbythetwoequationsdescribingtheconstancyofmomentum,andtheoneequationdescribingconstancyofthekineticenergy.Thereforethereisonedegreeoffreedomthatwemustspecifyinordertodeterminetheoutcomeofthecollision.Inwhatfollowsweshallexpressourresultsfor\(v_{1,f},v_{2,f},\text{and}\theta_{2,f}\)intermsof\(v_{1,i}\text{and}\theta_{1,f}\). Thecomponentsofthetotalmomentum\[\overrightarrow{\mathbf{p}}_{i}^{\mathrm{sys}}=m_{1}\overrightarrow{\mathbf{v}}_{1,i}+m_{2}\overrightarrow{\mathbf{v}}_{2,i}\]\[\begin{array}{l} p_{x,i}^{\mathrm{sys}}=m_{1}v_{1,i}\\ p_{y,i}^{\mathrm{sys}}=0 \end{array}\]Thecomponentsofthemomentum\[\overrightarrow{\mathbf{p}}_{f}^{\mathrm{sys}}=m_{1}\overrightarrow{\mathbf{v}}_{1,f}+m_{2}\overrightarrow{\mathbf{v}}_{2,f}\]inthefinalstatearegivenby\[\begin{array}{l} p_{x,f}^{\mathrm{sys}}=m_{1}v_{1,f}\cos\theta_{1,f}+m_{2}v_{2,f}\cos\theta_{2,f}\\ p_{y,f}^{\mathrm{sys}}=m_{1}v_{1,f}\sin\theta_{1,f}-m_{2}v_{2,f}\sin\theta_{2,f} \end{array}\]Therearenoanyexternalforcesactingonthesystem,soeachcomponentofthetotalmomentumremainsconstantduringthecollision,\[p_{x,i}^{\mathrm{sss}}=p_{x,f}^{\mathrm{ss}}\]\[p_{y,i}^{\mathrm{ss}}=p_{y,f}^{\mathrm{sys}}\]Equations(15.6.3)and(15.6.4)become\[m_{1}v_{1,i}=m_{1}v_{1,f}\cos\theta_{1,f}+m_{2}v_{2,f}\cos\theta_{2,f}\],\[0=m_{1}v_{1,f}\sin\theta_{1,f}-m_{2}v_{2,f}\sin\theta_{2,f}\].Thecollisioniselasticandthereforethesystemkineticenergyofisconstant\[K_{i}^{\mathrm{sys}}=K_{f}^{\mathrm{sys}}\]Usingthegiveninformation,Equation(15.6.7)becomes\[\frac{1}{2}m_{1}v_{1,i}^{2}=\frac{1}{2}m_{1}v_{1,f}^{2}+\frac{1}{2}m_{2}v_{2,f}^{2}\]RewritetheexpressionsinEquations(15.6.5)and(15.6.6)as\[m_{2}v_{2,f}\cos\theta_{2,f}=m_{1}\left(v_{1,i}-v_{1,f}\cos\theta_{1,f}\right)\]\[m_{2}v_{2,f}\sin\theta_{2,f}=m_{1}v_{1,f}\sin\theta_{1,f}\]SquareeachoftheexpressionsinEquations(15.6.9)and(15.6.10),addthemtogetherandusetheidentity\[\cos^{2}\theta+\sin^{2}\theta=1\]yielding\[v_{2,f}^{2}=\frac{m_{1}^{2}}{m_{2}^{2}}\left(v_{1,i}^{2}-2v_{1,i}v_{1,f}\cos\theta_{1,f}+v_{1,f}^{2}\right)\]SubstitutingEquation(15.6.11)intoEquation(15.6.8)yields\[\frac{1}{2}m_{1,i}=\frac{1}{2}m_{1}v_{1,f}^{2}+\frac{1}{2}\frac{m_{1}^{2}}{m_{2}}\left(v_{1,i}^{2}-2v_{1,i}v_{1,f}\cos\theta_{1,f}+v_{1,f}^{2}\right)\]Equation(15.6.12)simplifiesto\[0=\left(1+\frac{m_{1}}{m_{2}}\right)v_{1,f}^{2}-\frac{m_{1}}{m_{2}}2v_{1,i}v_{1,f}\cos\theta_{1,f}-\left(1-\frac{m_{1}}{m_{2}}\right)v_{1,i}^{2}\]Let\(\alpha=m_{1}/m_{2}\)thenEquation(15.6.13)canbewrittenas\[0=(1+\alpha)v_{1,f}^{2}-2\alphav_{1,i}v_{1,f}\cos\theta_{1,f}-(1-\alpha)v_{1,i}^{2}\]Thesolutiontothisquadraticequationisgivenby\[v_{1,f}=\frac{\alphav_{1,i}\cos\theta_{1,f}\pm\left(\alpha^{2}v_{1,i}^{2}\cos^{2}\theta_{1,f}+(1-\alpha)v_{1,i}^{2}\right)^{1/2}}{(1+\alpha)}\]DividetheexpressionsinEquation(15.6.9),yielding\[\frac{v_{2,f}\sin\theta_{2,f}}{v_{2,f}\cos\theta_{2,f}}=\frac{v_{1,f}\sin\theta_{1,f}}{v_{1,i}-v_{1,f}\cos\theta_{1,f}}\]Equation(15.6.16)simplifiesto\[\tan\theta_{2,f}=\frac{v_{1,f}\sin\theta_{1,f}}{v_{1,i}-v_{1,f}\cos\theta_{1,f}}\] TherelationshipbetweenthescatteringanglesinEquation(15.6.17)isindependentofthemassesofthecollidingparticles.Thusthescatteringangleforparticle2is\[\theta_{2,f}=\tan^{-1}\left(\frac{v_{1,f}\sin\theta_{1,f}}{v_{1,i}-v_{1,f}\cos\theta_{1,f}}\right)\]WecannowuseEquation(15.6.10)tofindanexpressionforthefinalvelocityofparticle1\[v_{2,f}=\frac{v_{1,f}\sin\theta_{1,f}}{\alpha\sin\theta_{2,f}}\] Example15.5ElasticTwo-dimensionalcollisionofidenticalparticles Figure15.10Momentumflowdiagramfortwo-dimensionalelasticcollision Object1withmassisinitiallymovingwithaspeed=3.0\(\mathrm{m}\cdot\mathrm{s}^{-1}\)andcollideselasticallywithobject2thathasthesamemass,\(m_{2}=m_{1}\),andisinitiallyatrest.Afterthecollision,object1moveswithanunknownspeed\(v_{1,f}\)atanangle\(\theta_{1,f}\),withrespecttoitsinitialdirectionofmotionandobject2moveswithanunknownspeed\(v_{2,f}\),atanunknownangle\(\theta_{2,f}\),f(asshownintheFigure15.10).Findthefinalspeedsofeachoftheobjectsandtheangle\(\theta_{2,f}\). Solution Becausethemassesareequal,\(\alpha=1\).Wearegiventhat\(v_{1,i}=3.0\mathrm{m}\cdot\mathrm{s}^{-1}\).Wearegiventhat\(v_{1,i}=3.0\mathrm{m}\cdot\mathrm{s}^{-1}\)and\(\theta_{1,f}=30^{\circ}\).HenceEquation(15.5.14)reducesto\[v_{1,f}=v_{1,i}\cos\theta_{1,f}=\left(3.0\mathrm{m}\cdot\mathrm{s}^{-1}\right)\cos30^{\circ}=2.6\mathrm{m}\cdot\mathrm{s}^{-1}\] SubstitutingEquation(15.6.20)inEquation(15.6.17)yields\[\begin{aligned} \theta_{2,f}&=\tan^{-1}\left(\frac{v_{1,f}\sin\theta_{1,f}}{v_{1,i}-v_{1,f}\cos\theta_{1,f}}\right)\\ \theta_{2,f}&=\tan^{-1}\left(\frac{\left(2.6\mathrm{m}\cdot\mathrm{s}^{-1}\right)\sin\left(30^{\circ}\right)}{3.0\mathrm{m}\cdot\mathrm{s}^{-1}-\left(2.6\mathrm{m}\cdot\mathrm{s}^{-1}\right)\cos\left(30^{\circ}\right)}\right)\\ &=60^{\circ} \end{aligned}\]Theaboveresultsfor\(v_{1,f}\)and\(\theta_{2,f}\)maybesubstitutedintoeitheroftheexpressionsinEquation(15.6.9),orEquation(15.6.11),tofind\(v_{2,f}=1.5\mathrm{m}\cdot\mathrm{s}^{-1}\).Equation(15.6.11)alsohasthesolution\(v_{2,f}=0\),whichwouldcorrespondtotheincidentparticlemissingthetargetcompletely. Beforegoingon,thefactthat\(\theta_{1,f}+\theta_{2,f}=90^{\circ}\)thatis,theobjectsmoveawayfromthecollisionpointatrightangles,isnotacoincidence.AvectorderivationispresentedinExample15.6.Wecanseethisresultalgebraicallyfromtheaboveresult.SubstitutingEquation(15.6.20)\(v_{1,f}=v_{1,i}\cos\theta_{1,f}\)inEquation(15.6.17)yields\[\tan\theta_{2,f}=\frac{\cos\theta_{1,f}\sin\theta_{1,f}}{1-\cos\theta_{1,f}^{2}}=\cot\theta_{1,f}=\tan\left(90^{\circ}-\theta_{1,f}\right)\]showingthat\(\theta_{1,f}+\theta_{2,f}=90^{\circ}\),theangles\(\theta_{1,f}\)and\(\theta_{2,f}\)arecomplements. Example15.6Two-dimensionalelasticcollisionbetweenparticlesofequalmass Showthattheequalmassparticlesemergefromatwo-dimensionalelasticcollisionatrightanglesbymakingexplicituseofthefactthatmomentumisavectorquantity. Figure15.11Elasticscatteringofidenticalparticles Solution Chooseareferenceframeinwhichparticle2isinitiallyatrest(Figure15.11).Therearenoexternalforcesactingonthetwoobjectsduringthecollision(thecollisionforcesareallinternal),thereforemomentumisconstant\[\overrightarrow{\mathbf{p}}_{i}^{\mathrm{sys}}=\overrightarrow{\mathbf{p}}_{f}^{\mathrm{sys}}\]whichbecomes\[m_{1}\overrightarrow{\mathbf{v}}_{1,i}=m_{1}\overrightarrow{\mathbf{v}}_{1,f}+m_{1}\overrightarrow{\mathbf{v}}_{2,f}\]Equation(15.6.24)simplifiesto\[\overrightarrow{\mathbf{v}}_{1,i}=\overrightarrow{\mathbf{v}}_{1,f}+\overrightarrow{\mathbf{v}}_{2,f}\]Recallthevectoridentitythatthesquareofthespeedisgivenbythedotproduct\(\overrightarrow{\mathbf{v}}\cdot\overrightarrow{\mathbf{v}}=v^{2}\).Withthisidentityinmind,wetakethedotproductofeachsideofEquation(15.6.25)withitself,\[\begin{aligned} \overrightarrow{\mathbf{v}}_{1,i}\cdot\overrightarrow{\mathbf{v}}_{1,i}&=\left(\overrightarrow{\mathbf{v}}_{1,f}+\overrightarrow{\mathbf{v}}_{2,f}\right)\cdot\left(\overrightarrow{\mathbf{v}}_{1,f}+\overrightarrow{\mathbf{v}}_{2,f}\right)\\ &=\overrightarrow{\mathbf{v}}_{1,f}\cdot\overrightarrow{\mathbf{v}}_{1,f}+2\overrightarrow{\mathbf{v}}_{1,f}\cdot\overrightarrow{\mathbf{v}}_{2,f}+\overrightarrow{\mathbf{v}}_{2,f}\cdot\overrightarrow{\mathbf{v}}_{2,f} \end{aligned}\]Thisbecomes\[v_{1,i}^{2}=v_{1,f}^{2}+2\overrightarrow{\mathbf{v}}_{1,f}\cdot\overrightarrow{\mathbf{v}}_{2,f}+v_{2,f}^{2}\]Recallthatkineticenergyisthesamebeforeandafteranelasticcollision,andthemassesofthetwoobjectsareequal,soconstancyofenergy,(Equation(15.4.2))simplifiesto\[v_{1,i}^{2}=v_{1,f}^{2}+v_{2,f}^{2}\]ComparingEquation(15.6.27)toEquation(15.6.28),weseethat\[\overrightarrow{\mathbf{v}}_{1,f}\cdot\overrightarrow{\mathbf{v}}_{2,f}=0\]Thedotproductoftwononzerovectorsiszerowhenthetwovectorsareatrightanglestoeachotherjustifyingourclaimthatthecollisionparticlesemergeatrightanglestoeachother. Example15.7Two-dimensionalcollisionbetweenparticlesofunequalmass Particle1ofmass\(m_{1}\),initiallymovinginthepositivex-direction(totherightinthefigurebelow)withspeed\(v_{1,i}\)collideswithparticle2ofmass\(m_{2}=m_{1}/3\)whichisinitiallymovingintheoppositedirection(Figure15.12)withanunknownspeed\(v_{2,i}\).Assumethatthetotalexternalforceactingontheparticlesiszero.Donotassumethecollisioniselastic.Afterthecollision,particle1moveswithspeed\(v_{1,f}=v_{1,i}/2\)inthenegativey-direction.Afterthecollision,particle2moveswithanunknownspeed\(v_{2,f}\)atanangle\(\theta_{2,f}=45^{\circ}\)withrespecttothepositivex-direction.(i)Determinetheinitialspeed\(v_{2i}\)ofparticle2andthefinalspeed\(v_{2,f}\)ofparticle2intermsof\(\mathcal{V}_{1,i}\).(ii)Isthecollisionelastic? Figure15.12Two-dimensionalcollisionbetweenparticlesofunequalmass Solution Wechooseasoursystemthetwoparticles.Wearegiventhat\(v_{1,f}=v_{1,i}/2\).Weapplythetwomomentumconditions,\[m_{1}v_{1,i}-\left(m_{1}/3\right)v_{2,i}=\left(m_{1}/3\right)v_{2,f}(\sqrt{2}/2)\]\[0=m_{1}v_{1,f}-\left(m_{1}/3\right)v_{2,f}(\sqrt{2}/2)\]SolveEquation(15.5.31)for\(v_{2,f}\):\[v_{2,f}=3\sqrt{2}v_{1,f}=\frac{3\sqrt{2}}{2}v_{1,i}\]SubstituteEquation(15.6.32)intoEquation(15.6.30)andsolvefor\(v_{2,i}\)\[v_{2,i}=(3/2)v_{1,i}\]Theinitialkineticenergyisthen\[K_{i}=\frac{1}{2}m_{1}v_{1,i}^{2}+\frac{1}{2}\left(m_{1}/3\right)v_{2,i}^{2}=\frac{7}{8}m_{1}v_{1,i}^{2}\]Thefinalkineticenergyis\[K_{f}=\frac{1}{2}m_{1}v_{1,f}^{2}+\frac{1}{2}m_{2}v_{2,f}^{2}=\frac{1}{8}m_{1}v_{1,i}^{2}+\frac{3}{4}m_{1}v_{1,i}^{2}=\frac{7}{8}m_{1}v_{1,i}^{2}\]Comparingourresults,weseethatkineticenergyisconstantsothecollisioniselastic.