# Collision of 2 particles - calculating the mass and a speed after the collision

Lets say we have a particle of mass $m_1$ which has a kinetic energy $W_{k1}$. This particle collides with another same particle. How can i calculate mass $m_2$ and the speed $v_2$ of the particle which is formed out of the colliding two particles?

I know that relativistic preservation of energy $W$ and momentum $p$ must hold and i know these equations:

\begin{align} W^2 &= p^2c^2 + {W_0}^2\\ W &= W_k + W_0\\ W &= mc^2 \gamma(v)\\ p &= mv\gamma(v)\\ \end{align}

My attempt was something like this:

\begin{align} p_2 &= p_1\\ p_2 c&= p_1 c\\ {p_2}^2 c^2&= {p_1}^2 c^2\\ {W_1}^2-{W_{01}}^2 &= {W_2}^2-{W_{02}}^2 \longleftarrow \smash{\substack{\text{Doesnt it hold that $W_1 = W_2$?}\\ \text{Can i cross them out to get}\\\text{the line below?}}}\\ {W_{01}}^2 &= {W_{02}}^2\\ \left(2m_1c^2\right)^2 &= \left(m_2 c^2\right)^2\\ 4{m_1}^2c^4 &= {m_2}^2 c^4\\ m_2 &= 2m_1 \end{align}

Is this even correct? My professor said it is not... Why?

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Basic rule one for these things: don't muck about with velocity, get into four momentum form and stay there: four momentum is conserved and doesn't involve any #=!^%@+&# $\gamma$s. Basic rule two: mass is not conserved at this level, but it doesn't matter because you can find it from the four momenta. – dmckee Jun 30 '13 at 19:21
Only thing i know about four momentum form is that ${P^\mu}^T = [p_x,p_y,p_z,W/c]$. So i am kind of inexperienced here... I have only started solving these types of problems... – 71GA Jun 30 '13 at 19:59
I wonder what does it mean "kinetic energy"? In relativity, it is not absolute and depends on reference object velocity. – Val Jun 30 '13 at 20:01
@71GA That, the relations you have over the hrule and $\sum_{i \in \text{initial state particles}} \mathbf{p}_i = \sum_{i \in \text{final state particles}} \mathbf{p}_i$ (where bold face means a four vector) are all you need to know. – dmckee Jun 30 '13 at 20:28
@Val Kinetic energy is not absolute in Newtonian mechanics either. – dmckee Jun 30 '13 at 20:30