Question

4. Consider two particles of masses my and m2 and positions rı and r2 respectively. Suppose m2 exerts a force F on mı. Suppose further that the two particles are in a uniform gravitational field g. (a) Write down the equations of motion for the two particles. (b) Show that the equations of motion can be written as MR = Mg ur = F where M is the total mass, R is the centre of mass, he is the reduced mass and r is the relative position of the two masses. Show further that m2 ri = R + M mi r2 M R r. (c) Consider an elastic collision between two particles on a perfectly smooth horizontal surface. Show that in the centre of mass frame the momenta of the particles are equal in magnitude and opposite in direction. (d) Suppose that in the lab frame the second particle is initially at rest. Suppose further that the collision is head-on (that is, the scattering angle is a). Find the kinetic energy transferred to the second particle (in the lab frame). (e) Find an expression relating the total kinetic energy of the system in the lab frame and in the centre of mass frame.

Answer 1

Two PARTIES WITH MASS M, AND M₂ and & Audri M2 EXERTS FORCE Ě ON mo TWO PARTICLES, a) EQUATION OF MOTION mais USING is AND Mr Newtow's THIRD LAW. CENTRE OP MASS DE FINED AS, Ř = M, + m₂ iņam mitm2 AND you CENTRE OF MASS VELOCITY 3 cm m. v, + mv, 19, If we А As, THE sys TEM GENPRE ONE SYSTEM MASS MOTION, THEN ОР CONSIDER

MOTION OF CENTRE, Macm Feat AND R (+) VCMt. TH EXTERNAL FORCE ON THE SYSTEM. HERE Text = Mg HENCE M = M Now - min - À AND M22 in = m2 And mi PM, M₂ r M, "z M₂ HAVE, SUBTRACTING WE ш, F Z Now M, M2 (AB min = ² AND mat - m, man m, F AND m, ma l SUBTRACTING WE HAVE, CRT) m. - (-m, (M 2 tm , ) E mm Ñ 2 个个 mimiz M₂tm F.

OR, un un = Til m, mz u = REDUCED MASS, AND P--P) 4, +42 RELATIVE VECTOR Ř - mint mind mit m₂ → R (M, + 2) = mir & meht man man ŔM (m, rm2) + m₂ (m 2 ) Mhaman RM + m₂ = R + M2 E qa6 95 M 1 en en

E 9 do martmarz + m, re- mi re micha) + M 2 tm, ) 27 MŘE miña ME 1 su 981 po MR-mi = Ř - will un ELASTIC COLLISION IN COM FRAME Vem = mo, & "₂" mitm2 -61 v-am ų,- mw, tm. " M₂ (20,-₂) Mytma m, tmz $42 Vorel m,th2 q = V₂-eaan wi (4₂") Vrel м, mgtM2 11 ry, the For momentum, P m, wiz m, the Vrel - Pr m, uz maze Vorel =-u vrel m, th2 TOTAL MOMENTUM incom IS ZERO

NO COLLISION K.E (CING CONSERVATION 2 2 B2 Porta Pii 2017 P2f 2m2 2 m2 2_2 22 uveri Mures of u Terreif 2m2 2 m, 242 As, Now were i Irel, f SOLUTION Two IT ITAS 50 = ere, f Verei, EQUAL AL AND OPPOSITE TRANSFER OF MOMENTOM, E MOMENTAS ARE oppos I TE AND EQUAL

को IN LAB FRAMP FOR FLASIT COLLISION au lit M₂M₂i = a vif + M2 Va f AND, temini + 1 m, V2 = ₂ mm + 1 m. v, 4 auf W CAN GET FROM HERE, 292 Vai If m, 12 m, thiz Wii m, tm2 2u 2 м, mem, Uzi Vzt m, the mith Now Vergi =0 bo m, - m2 Vii Vuf m,+ m2 2m ܙܛܢ U2f m, t2 SECOND OF KINETIC ENERGY TO TRANS FER PARTICLB 10- tmavý ] 2 m, mz ve 2 I mom.x2 mtu2 m, the

e) IN LAB FRAME, collisiON, TOTAL MOMENTOM FOR ELAS TIC is CON SERVED 2 2 IN LAB FRAME, Žmovil £m, vier + 1 m, v 24 222 HOWEVER IN COM FRAME, ш,V, Tч, , Vem = 2 К. Е = 1/2 VELOCITELES MVCM re > HERE ARE , V = Vii-Vcm 2 vi-Vcm. Vi