Relativistic Kinematics - Proton-proton collision I am stuck on the following question: 
A proton of total energy $E$ collides with a proton at rest and creates a pion in addition to the two protons:
$$p+p \rightarrow p+p+ \pi^{0} $$ 
Following the collision, all three particles move at the same velocity. 
Show that $$E = m_p c^2 + \left(2+ \displaystyle \frac{m_{\pi}}{2m_p} \right) m_{\pi}c^2 $$ where $m_p$ and $m_{\pi}$ are the rest masses of the proton and pion respectively. 
I have tried conserving energy to get: 
$$E + m_p c^2 = 2m_p \gamma (v) c^2 +\gamma (v)  m_{\pi} c^2$$
The issue is, what do we do about the $\gamma$ (where $\gamma$ is the Lorentz factor)? 
Thanks - all help is appreciated. 
 A: Conservation of energy on its own is usually not sufficient for this sort of problems; you also need to work with conservation of momentum, then resort to a number of nasty substitutions and whatnot. Instead, why not go for conservation of 4-momentum?
$$p_1+p_2=q_1+q_2+q_\pi,$$
where $1$ and $2$ label the protons, and $q$ are final state momenta (for simplicity, I've dropped the "$\mu$" indices). Squaring on both sides (or, if using indices, taking the inner products):
$$p_1^2+p_2^2+p_1\cdot p_2=q_1^2+q_2^2+q_\pi^2+2q_1\cdot q_2+2q_1\cdot q_\pi+2q_2\cdot q_\pi.$$
Now remember that $p=(E,\vec{p})$ and so $p^2=E^2-|\vec{p}|^2=m^2$ :
$$2p_1\cdot p_2=2m_p^2+m_\pi^2+4m_pm_\pi.$$
Note that to simplify the RHS I've used the fact that all decay products have the same momentum. To reduce the LHS, I now use the fact that proton $1$ is at rest, i.e. $p_1=(m_p,0)$ :
$$2p_1\cdot p_2=2Em_p.$$
Plugging it in the above equation yields the desired result (up to factors of $c=1$ in natural units):
$$E=m_p+\left(2+\dfrac{m_\pi}{2m_p}\right)m_\pi.\quad\square$$
