Question

The centre of mass, or centre of momentum, (CMS) frame is that in which the sum of the momenta (i.e., the total momentum) of all particles is zero. An electron has a kinetic energy K of 1 MeV and is incident on a proton at rest in the laboratory. The goal of this exercise is to calculate the speed of the CMS frame moving relative to the laboratory. (a) Express the initial energies Ee, Ep and initial momenta pe, pp of the electron and proton respectively (with c, me, mp, K). (b) What are the total energy Etot and the total momentum ptot before collision? (c) Express the transformed momentum p 0 tot in the CMS frame and obtain the speed v of the CMS frame with respect to the laboratory frame. The rest mass of the electron is 0.511 MeV/c2 . The rest mass of a proton is 938 MeV/c2

Answer 1

its relative mass is a) When the object moves reduced by a factor To so, the election relative man is Ime: Een kinetic energy & Mech Ep = Kinetic energy & imp. c? since the proton is rest, KE=0 → Ep = 0+ mph K (kinetic energy = h (mot → 2km = (mus 2km 2 p² (P=my) P- v2km Pp =0 (ask:o) Pe = /2kTimes by The total energy, Etot = Ep the Etat = kt med Ange? Priz. Pe tpp. = /2 kt me to z Vakitme i Noma = Total momentum. Total mass Pe metme

The velocity of electron and proton swarot CMS is, Vecms = Vems - Ve. Vp cos = Vems - OVP=0 The total initial momentum (Perry) wart CMS is , Pems = Pp the amplems + the (Verms - Ve) cab initial momentum a final moment after collision - Pems = mp voms + me (Vorms - Ve).