(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A particle of rest mass M and total energy E collides with a particle of rest m at rest. Show that the sum E' of the total energies of the two particles in the frame at which their centre of mass is at rest is given by

(E')^2=(M^2+m^2)c^4+2Emc^2

3. The attempt at solution

I'll denote particle of rest mass M as having velocity v and particle of rest mass m as having velocity w and the relevant gammas as G(v), G(w)

E=Mc^2

Conservation of four momentum implies:

E'=G(v)Mc^2+G(w)mc^2

0=G(v)Mv+G(w)mw (where v and w are vectors)

G(v)Mv=-G(w)mw

(G(v)Mv)^2=(G(w)mw)^2 (taking modulus squared of both sides)

M^2(G(v)^2-1)=m^2(G(w)^2-1) (as G(u)^2=c^2(G(u)^2-1)

I can't see how to eliminate both G(v) and G(w) and get an answer for E', I need another equation.

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# Homework Help: Special Relativity Collision

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