Energy of Kaon in Lab Frame: Relativistic Collision

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Homework Statement


kaon of ss 498 MeV/c^2 traveling 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?


Homework Equations

I 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...



The Attempt at a Solution

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