A muon 10566 MeV with a laboratory kinetic energy of 1 GeV c

A muon (105.66 MeV) with a laboratory kinetic energy of 1 GeV collides head-on with a proton (938.3 MeV) at rest, in the laboratory frame. What is the center-of-momentum velocity of this system in the laboratory frame?

Solution

Centre of momentum will have the velocity of Centre of mass

let location of muon be (x,0), that of proton be (y,0)
Mass of muon = m, Mass of proton M
Location of com be (z,0)
Velocity of muon = u, velocity of proton v

(m+M)z = mx + My
z = (mx+My)/(m+M)
z\' = (mu + Mv)/(m+M)
m = 105.658*1.783*10^-30 = 1.88*10^-28kg

to find u

E = m*c^2 /(sqroot(1-(v^2)/(c^2)))
c = speed of light = 3*10^8
v = u

HENCE U = 0.994 C (C = SPEED OF LIGHT)
v = 0
M = 1.67*10^-27Kg

so z\' = 2.73*10^8 m/s

towards proton