Question

please answer #1 all parts

1) Imagine you have the following hypothetical "collision(s)". A massive particle (A), with mass, MA, is at rest. It then decays into three particles, B (of mass, mg). C and D (both massless, with the same magnitude of momentum, 110, and energy, E.). a) (5 pnts, Write the general conservation of 4-momentum relation for this situation in terms of 4-vectors). For parts b-c), assume that particle B is produced at rest: Before Ater: b) (10 pnts Derive II, as a function of the masses of the particles. c) (5 pntsIf mA = 106 MeV/c and mp = 6.0 MeV/c", evaluate E. (the total energy of either one of the massless particles, C or D.) For parts d-f), assume that particles C and D go off in the same (say + direction): d) (10 pnts) Repeat b) for this new configuration. e) (5 pnts) Repeat c) for this new configuration. f) (5 pnts) What is the magnitude of the velocity of particle B after the decay. (Hint: you might want to keep at least 4 significant digits in your calculation.)

Answer 1

Ans: since a mass mA is decaying into 3 particle in which 2 Particles are massless and third one has mass mo as shown - ac mezo mBoB, VB=0 VA= moso (a) since four momentum is also include space - time. component of momentum. Hence total 4-momentum of a particle if its total energy E, is given. as - (Ę, Px, Py, z) = ( 4 ) Hence momentum conservation of above reaction is EA, O = (Eso) + ( Et ) + . Ĉ and b particles are massless Hence the for total momentum s can be written as E shows the direction of momentum in right side . since the momentum of mass less particles are To then from above reaction - . EA = mach [. kinetic energy is zero] (6) and EB = MB c2 E MAC² = math + T. + to MBC + MAC 2. To

= (mA-molc To The total energy of either one mass less partick in E = Toc = (mama) c² - MAC² - mach . MAC2 = 106 Mev Mac? = &. = 6 mey 106 - 6 mev - 50 meva (d) Now Particle is decaying in following way B Now from y-momentum conservation principle moet ( hate) + + + 21 E B = P22 Ma c' Potch m et mc - Po C2 (Est, Ps) = mec moc - mis c + 2 ( mm) Hence momentum still remains same.