# Relativistic collison

1. Nov 26, 2008

### captainjack2000

1. The problem statement, all variables and given/known data
kaon of ss 498 MeV/c^2 travelling through laboratory decays into two pions each of mass 137MeV/c^2. Onee of the pions is produced at rest in the lab frame. What is the energy of the kaon in the lab frame?

2. Relevant equationsI think you need to consider 4 vectors for the momentum. Since one pion is stationary its 3 momentum is 0 right?
Not really sure how to work this out!
Do you equates
P = p1 +p2 where these are 4 vectors
gamma mkaon c^2 = gamma2 mpion c^2 + 0

Any suggestions of how to continue would be appreciated!

the energy of the second pion would be it's rest mass plus the deltamass times c^2 plus a factor which equals the kaon's kinetic energy. I think....

3. The attempt at a solution

2. Nov 26, 2008

### Confundo

Just need to use conservation of energy and momentum. Then use the common expression involving energy and momentum.

Last edited: Nov 26, 2008