In the following reaction, the incident Kaon has kinetic energy of 1.63GeV.Calculate the total energy to be divided between the four recoiling particles.

\bar{K}^-\ + \ p ^+ \rightarrow\Sigma^-\ + \pi^+\ + \pi^-\ + \pi^+

Given Given mass energy of pi meson=139.6MeV, \Sigma=1197.3MeV, proton=938.3MeV and kaon=493.8 MeV

Using conservation of energy straightway, we get

493.8+938.3+1630=(3x139.6)+3(T_pi)+1197.3+(T_Sigma )
This gives, 3(T_pi)+(T_sigma)=1446

Now,it appears that we are to use conservation of momentum;but I am having difficulty in finding out the appropriate equation.Since, the particles are charged, they might move in the opposite direction.

Can anyone please help?

Is this homework you keep getting? :)

neelakash---it looks like you have the answer... The four particles have to share 1446 MeV of energy. Did you phrase the question incorrectly?

Hi Ben,I am not sure,I thought they want the amount to be sharedby each of them.Otherwise, it is a plus-minus problem.What do they really want?

I think they just want you to use conservation of energy to find how much kinetic energy the total system has after the collision. It would be easier if they told you to assume that the \Sigma was produced at rest, for example.