What is the Momentum of Particle A in a 3-Particle Collision?

[PhMichael]

Homework Statement

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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

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

 

[vela]

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

 

[PhMichael]

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

 

[vela]

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

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.

 

[PhMichael]

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.

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} ?

 

[vela]

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.

 

[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?

 

[vela]

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.