# Relativity (3 particles collision)

1. Aug 17, 2010

### PhMichael

1. The problem statement, all variables and given/known data
http://img687.imageshack.us/img687/1364/13082447.jpg [Broken]

We have 3 particles: Particle A with an unknown mass moves towards two particles, B and C, each having a mass of m and is at rest. A collides with B and both move as one body and hit C. As a result of the second collision, two new particles are formed with masses: $$m_{1}=3m$$ and $$m_{2}=5m$$ . The kinetic energy of the particle $$m_{1}=3m$$ in the center of mass frame is $$K_{1cm}=2mc^{2}$$.
What is the momentum of particle A in the lab frame?

2. The attempt at a solution

http://img204.imageshack.us/img204/4766/89310897.jpg [Broken]

and of course, this can't be true .... What have I done wrong?

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2. Aug 18, 2010

### vela

Staff Emeritus
How do you know the energy of particle A is 10mc2?

3. Aug 18, 2010

### PhMichael

I really didn't mention that but it's a given quantity.

4. Aug 18, 2010

### vela

Staff Emeritus
Ah, that helps. I couldn't see how to solve it without a little more info.

Try working it backwards. In the CM frame, you can find the total energy E1 of m1 and then its momentum. From this, you should be able to find the energy E2 of m2. The quantity E1+E2 is equal to an invariant.

5. Aug 18, 2010

### PhMichael

Before i'll try that, I want to ask about something which i think is essential to obtain a valid solution:

When I consider the first collision, i.e. between A and B , need I take into account also particle C[/t] that is, temporarily, not involved in that collision? In other words, need I add its rest energy to the energy expression of the first collision so that it will be $$12mc^{2}$$ instead of $$11mc^{2}$$ ?

6. Aug 18, 2010

### vela

Staff Emeritus
It depends what you're trying to calculate. If you want to compare to the final state with m1 and m2, then yes, because C contributed to it. But if you're just trying to see what happens when A and B collide before it hits C, then no.

7. Aug 18, 2010

### PhMichael

ummm, i didn't get it ... let's say, if I want to follow the method I used in the solution above, need I consider particle C in the first collision?

8. Aug 18, 2010

### vela

Staff Emeritus
I don't see why you would. You're calculating what happens when A and B collide and form the composite AB particle, so there's no reason to include C.