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## Homework Statement

A D

^{0}meson (with a rest mass of 1.86 GeV/c

^{2}), initially at rest, decays into a K

^{0}meson (with a rest mass of .51 GeV/c

^{2}) and a [itex]\pi[/itex]

^{0}meson (with a rest mass of .12 GeV/c

^{2}). What is the Kinetic Energy of the [itex]\pi[/itex]

^{0}meson?

## Homework Equations

E=[itex]\gamma[/itex]m

_{0}c

^{2}

p=[itex]\gamma[/itex]m

_{0}v

## The Attempt at a Solution

Subtracting final rest energy from intial energy its easy to see that the KE(K

^{0}+[itex]\pi[/itex]

^{0})=1.23 GeV

Also: [itex]\gamma[/itex]

_{K0}m

_{K0}c

^{2}+[itex]\gamma[/itex]

_{[itex]\pi[/itex]0}m

_{[itex]\pi[/itex]0}c

^{2}= 1.86 GeV/c^2 (The total energy of the two Mesons must equal the first meson)

Without going through all the horrible steps me and my friend did, eventually we got a equation that solved: V

_{K0}=f(V

_{[itex]\pi[/itex]0}) (I don't have the final equation we got on me)

When then plugged that back into: [itex]\gamma[/itex]

_{K0}m

_{K0}c

^{2}+[itex]\gamma[/itex]

_{[itex]\pi[/itex]0}m

_{[itex]\pi[/itex]0}c

^{2}= 1.86 GeV/c^2

Now at this point we had a solvable equation, the only variable was V

_{[itex]\pi[/itex]0}, but it was a hellish equation. After about 20 minutes working on trying to solve it we ran out of time and had to give up- the algebra was just to hard.

Had we had enough time, I'm completely confident we could have solved it, as it was simply a matter of foiling again and again (I think we needed to do it three levels down).

My question is: Is there a way of solving this problem without having to go in and solve for one of the variables in terms of the other, or doing that in a simpler equation?

Thank you, especially if you made it through this.