# Two masses colliding relativistically

Tags:
1. Jan 22, 2016

### TristanRE

1. The problem statement, all variables and given/known data
A mass m moving at speed 4c/5 collides with another mass m at rest. The collision produces a photon with energy E traveling perpendicular to the original direction, and a mass M traveling in another direction. In terms of E and m, what is M? What is the largest value of E (in terms of m) for which this setup is possible?

2. Relevant equations
p=Ymv
E=Ymc^2
(E^2-P^2c^2)_before = (E^2-P^2c^2)_after

3. The attempt at a solution
I tried using momentum only, and I get the calculation of p_x=4/3mc, p_y=-E/c (I used four momenta) so now this doesn't give me much to go on alone since I'm looking for M first. I tried using equation 3 but it becomes so convoluted I'm not sure how that could help me with the answer at all.

Any suggestions?

2. Jan 22, 2016

### TSny

OK. These are the x and y components of the momentum of the particle of mass M.

Try conservation of energy. For the particle of mass M, use the relativistic expression for energy in terms of mass and momentum. That way, you can use your results for px and py.

3. Jan 23, 2016

### TristanRE

I'm not sure this is what you meant, but I ended up doing the following.

P1 = incoming mass 4 momentum = (E1),p1,0,0)
P2 = created particle = (E2,p1,-E,0)
P3 = photon = (E,0,E,0)

I then equate: P2=P1 - P3
I squared it and rearranged, and plugged in for E1 = 5/3m, which yielded M = (m2-10/3mE)1/2

Then for the second part it seems easy (assuming I'm right up to this point). I just solved for how large E can be without the the radical being negative: Emax=3m/10

How does this look? I found an example comparable to this in my textbook, thought its not exactly the same, I think this is right.

4. Jan 23, 2016

### TSny

What about the 4-momentum of the particle that was initially at rest?

5. Jan 24, 2016

### TristanRE

It completely slipped my mind! Thank you.

I redid did it the same way except replacing P1 with PInitial=(8/3m,,p1,0,0)

Thereby obtaining M = (m2-16/3mE)1/2,

and Emax = 3m/16.

6. Jan 24, 2016

### TSny

OK.

I'm getting the 16/3 in a different place: $M = \sqrt{\frac{16}{3}\left(m^2-mE\right)}$.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted