# E-P invariant on relativistic mecahnics problem

1. Apr 9, 2012

### Froskoy

1. The problem statement, all variables and given/known data
A particle of mass $m$ moving at speed $\frac{3}{5}c$ collides with an identical particle at rest, and forms a new particle of mass $M$ which moves off at speed $v$. Find v.

2. Relevant equations
E-P invariant: $E_1^2-p_1c^2=E_2^2-p_2^2c^2=\mathrm{const.}$
Momentum: $p^2=m^2c^2\left({\gamma_v^2-1}\right)$
Energy: $E=\sum_i\gamma_im_ic^2$

3. The attempt at a solution
There is a single particle after the collision, so $E_2^2-p_2^2c^2=M^2c^4$
where $E_2=\gamma_vMc^2$
and $p_2^2=M^2c^2\left({\gamma_v^2-1}\right)$

so

$$\gamma_v^2M^2c^4-M^2c^4\left({\gamma_v^2-1}\right)^2=M^2c^4\\ \gamma_v^2-\left({\gamma_v^4-2\gamma_v^2+1}\right)=1\\ \left({\gamma_v^2-1}\right)\left({\gamma_v^2-2}\right)=0$$

Reject the $\gamma_v^2=1$ solution, since this would mean $v=0$.

The $\gamma_v^2=2$ solution gives $v=\frac{\sqrt{2}}{2}c$, which is incorrect - the answer is $v=\frac{1}{3}c$.

I expect there's something fundamentally wrong with my method, but not sure what? Thank you!

2. Apr 10, 2012

### vela

Staff Emeritus
You shouldn't have squared the factor $(\gamma_v^2-1)$ in the first line. With this approach, all you'll end up with is M=M, so it's not helpful.
You haven't used any information from before the collision. Use conservation of energy and momentum.