Elastic Collisions. Definition: two point masses on which no external forces act collide without losing any energy.

Transcription

1 Elastc Collsons Defnton: to pont asses on hch no external forces act collde thout losng any energy v Prerequstes: θ θ collsons n one denson conservaton of oentu and energy occurs frequently n everyday applcatons Suary: The scatterng angles and the target s fnal speed are gven by here v = cos θ = ( + α ) β + α β β α Go to dervaton and cos θ = + α α = and β = v, v Why study t? llustrates conservaton las llustrates center of ass frae Go to Java applet

2 Scatterng n to densons Let s apply conservaton of oentu and energy to a non-trval proble: the elastc scatterng of one body off another (elastc eans no energy s lost n the collson) v θ θ We ll suppose body (the target), s ntally at rest Conservaton of energy v = + Conservaton of oentu parallel to v v = cos θ + cos θ Conservaton of oentu perpendcular to v sn θ = sn θ Wthout loss of generalty, e ay take and to be postve, and the scatterng angles to le beteen 0 and π You ay recall our earler treatent of a collson n one denson There, there ere to equatons n to unknons, the fnal speeds The fnal speeds ere therefore fxed, once the ntal speeds ere knon Here, e have three equatons n four unknons, the to fnal speeds and the to scatterng angles Therefore, e on t be able to solve for the all Let s regard as a varable, and solve for the other three n ters of Let s solve for θ (leavng θ and as exercses) We do ths n several steps, hch e on t reproduce here ) Elnate θ fro the to oentu equatons by solvng the for cos θ and sn θ, and then usng the dentty cos θ + sn θ = ) Elnate by solvng the energy equaton for and substtutng t nto the result of the frst step 3) Solve the resultng equaton for cos θ The result s

3 cos θ = ( + α ) β + α β, here e have ntroduced to abbrevatons: α =, hch tells the relatve szes of the to asses, and β = v, hch just gves n unts of v Here s a plot of θ for varous values of the ass rato: (target ass less than projectle ass, an uncoon stuaton) then there s a axu of the scatterng angle θ It s easy to sho that t s gven by sn θ ax = α For scatterng through angles less than ths, there are actually to values of the fnal speed for a gven value of θ The next fgure shos the oenta n a typcal case (α=/, θ =π/6) The to red lnes correspond to one soluton, and the to blue lnes correspond to the other π α> θ α= α< 0 β We see several nterestng ponts Frst, f α< Here s a ove of the process represented by the to red lnes, and another of the one represented by the to blue ones On the other hand, f α> (target ore assve than projectle) then scatterng through any angle s possble, and there s only one value of the fnal speed for a gven value of θ The follong graph shos the oenta n several cases

4 here T s the energy of the ncong body Here s a ove of the process represented by the to red lnes, and another ove of the process shon n green Lnks to oves shong the sae processes n the center of ass frae ll be provded later Another nterestng pont s that for all values of α, there s a nu value of the fnal speed of the projectle It s gven by n v = + Ths nu occurs at θ =0 hen α< and at θ =π hen α> It corresponds to a axu aount of knetc energy transferred to body : T ax = 4 ( + ) T, In cases here the ncong ass s uch less than the target ass (as often happens n fxed-targed partcle physcs experents), the factor ultplyng T on the rght-hand sde s uch less than, and not very uch of the ncdent partcle s energy can be transferred to the target Exercse: Sho that the speed and angle θ of the target after the collson are gven by v = β α and cos θ = + α v Analyss n the center of ass frae In our secton on the center of ass frae, e sho that ths frae s specal because the su of the oenta n ths frae s zero Ths allos us to splfy the analyss consderably In the follong, e ll denote oenta n the center of ass frae by pres There are only to oenta before the collson, so they are equal and opposte n the center of ass frae:

5 p P N = p P N There are only to oenta after the collson (e ll call the q P and q P ), hch are also equal and opposte to one another n the center of ass frae: q P N = q P N We no apply conservaton of energy n the center of ass frae: p P N + p P N = q P N + q P N We thus fnd that the agntudes of the oenta n the center of ass frae do not change durng the collson: q P N = p P N and q P N = p P N Snce the asses of the bodes don t change durng the collson, e therefore conclude that the speeds don t change, hen veed n the center of ass frae Only the drecton of oton changes Here are the dagras: lab frae v v c center of ass frae v c v N θ N θ θ θ N v c v N In labellng the second dagra, e have already used the facts that the speed of the target s v c, that the fnal speeds are the sae as the ntal speeds, and that the fnal veloctes are back-toback n the center of ass frae Much of our ork s already done The fnal speeds do not depend on the scatterng angle θ N n the center of ass frae The angular dependence of the fnal speeds n the lab frae coes n hen e transfor beteen the

6 to fraes: x-coponent of fnal velocty of projectle: cos θ = v N cos θ N + v c y-coponent of fnal velocty of projectle: sn θ = v N sn θ N In order to copare these processes n the to fraes, here s a ove of the process represented by the to red lnes as seen n the lab frae, and another of the sae process as seen n the center of ass frae And here s a ove of the process shon n green n the lab frae, and another n the center of ass frae We easly calculate fro ther defntons v c = v + and v N = v + We solve the to velocty transforaton equatons for cos θ N and sn θ N, and then elnate θ N usng the dentty cos θ N + sn θ N = Solvng the resultng equaton for cos θ yelds the sae result as before Here are the oenta of soe typcal scatterngs as seen n the lab frae:

7 Moentu and conservaton las The concept of oentu n three densons s not aterally dfferent fro that n one denson The oentu vector of a body s ts ass tes ts velocty vector: p P = v P Suppose e have a any-body syste In such systes, the concept of oentu becoes partcularly useful Let s see hy In general, to knds of forces can act on the bodes n the syste There are nternal forces beteen the bodes, and external forces actng on ndvdual bodes In hat follos, e ll ake a specfc assupton about the nature of the forces actng beteen bodes We ll assue that they satsfy Neton s thrd la : The force of body j on body s equal n agntude and opposte n drecton to the force of body on body j (Here, and j stand for nubers hch label the bodes) Ths eans that f the subsyste contanng only those to bodes ere placed n solaton (no external forces actng), then no net force ould act on t Although ths ay see obvous, t does not hold for all types of force Neton s thrd la s not a la at all; t s just a descrpton of a nce knd of force We ll sybolze the force on body due to body j by F P j Then the thrd la looks lke ths n pctures: P 3 F P = 3 F P F 3 and lke ths n sybols: F P j = F P j We ll assue that a body exerts no force on tself, so F P = 0 (Ths s contaned n the above equaton) Let s also let F P be the external force actng on body We ake no assuptons about ts nature Then Neton s second la appled to body reads 4 3

8 a P = F P + 3 j F P j The rght-hand sde s the su of the external force and all the nternal forces actng on body No let s consder the total oentu of the syste; t s the su of the ndvdual oenta: p P tot = 3 v P Ho does the total oentu change n te? d dt 3 v P = 3 a P = 3 F P + 3 F P j, j The second ter on the rght-hand sde s zero because the forces, taken n pars, cancel out by Neton s thrd la Hence, e fnd d p P tot dt = F P ext, here the su of the external forces actng on the syste s F P ext = 3 F P The nternal forces have no effect on the rate of change of the total oentu No let s suppose that the total force actng on the syste s zero Then e have a conservaton la: dp P tot = 0 dt Ths s not as trval as t ght see, because t s true even hen the nternal forces are nonzero The only thng that can change the total oentu s a nonzero total external force on the syste The center of ass frae Suppose you have an object ade up of several saller bodes connected together soeho You toss ths coposte body nto the ar, possbly gvng t soe spn and gvng the constuents soe nontrval nternal otons The resultng oton can be rather coplcated As s often done n physcs, e ould lke to fnd soe aspect of the net oton hch s sple You have probably already guessed hat ths aspect s, especally after havng studed the center of ass n one denson The hole thng carres over very easly nto three densons, the only change beng that any quanttes becoe vectors We return to our equaton for the rate of change of the total oentu of the syste Wrtten n

9 ters of the poston vectors, t reads d dt 3 P r = F P ext We can ake ths look lke Neton s la for a sngle body hose total ass s the su of all the asses n the syste, by ultplyng and dvdng by the total ass M: M d P r c = F P dt ext here P r c = M 3 P r The quantty P r c s called the center of ass It s the average of the postons of all the bodes, eghted by ther asses The above to equatons say that the oton of the center of ass of the syste s the sae as the oton of a sngle body hose ass s equal to the total ass of the syste; hch s located at the center of ass of the syste; and hch s acted on by a force equal to the su of the external forces actng on the bodes of the syste In general, the center of ass oves as te goes on The velocty of the center of ass s just the te dervatve of r P c, hch s v P c = M 3 v P No, let s do a trck We ll ve the hole syste fro a frae of reference hch s ovng along th velocty equal to the velocty of the center of ass Let s say that a body s ovng at 0 s - n the orgnal reference frae, and that the center of ass s ovng n the sae drecton at 6 s - Then the velocty of the body hen veed n the center of ass frae s 4 s - To get the velocty n the center of ass frae, you just subtract v P c We ll sybolze quanttes n the center of ass frae by puttng pres on the Then v P N = v P v P c The oentu of a body n the center of ass frae s just p P N = v P N = ( v P v P c ) No for the punch lne We can sho that the su of the oenta n the center of ass frae s zero:

10 3 p N = 3 ( v P v P c ) P = 3 v P M v P c = 0 The last lne follos drectly fro the defnton of the velocty of the center of ass Energy and the center of ass frae An portant property of the center of ass frae s that the total knetc energy of a syste s equal to the knetc energy of the center of ass plus the su of the knetc energes of the bodes of the syte, as calculated n the center of ass frae The proof s just a short calculaton Usng the velocty transforaton la v P = v P c + v N, P 3 v P = 3 ã ä v P c + v P c A v P N + v P N ë í = ä å 3 ë ì v P c + v P c A 3 v P N + 3 ã í v = M v P c + 3 v P N P N The frst ter on the last lne s the knetc energy of the center of ass Another ay to rte the last equaton s T = T c + TN, here T stands for knetc energy and the fact that the su of the oenta n the center of ass frae s zero: e fnd 3 v P N = 0,