# Relativistic elastic head-on collision: The momentum just reverses direction

Consider a relativistic elastic head-on collision between a projectile, with mass $m_a$ and velocity $\mathbf{v}_a$, and a stationary target of mass $m_b$. Here "head-on" means that the two particles emerge from the collision along the line of the incident velocity $\mathbf{v}_a$.

(Picture taken from John Taylor's "Classical Mechanics") By definition, the CM-frame is the frame in which $$\mathbf{p}_a^{in}+\mathbf{p}_b^{in}=0$$ where $\mathbf{p}$ denotes the three-momentum $\mathbf{p}=\gamma m\mathbf{v}$. In this frame, then, we have by conservation of momentum $$\mathbf{p}_a^{in}=-\mathbf{p}_b^{in},\quad\text{and}\quad \mathbf{p}_a^{fin}=-\mathbf{p}_b^{fin}.$$ Question: How to show that in the CM-frame, $$\mathbf{p}_a^{in}=-\mathbf{p}_a^{fin}?$$ That is, the momentum just reverses direction.

Clearly, it would be sufficient to show that $$|\mathbf{p}_a^{in}|=|\mathbf{p}_a^{fin}|,$$ but how? In think I must use conservation of energy, but all my attempts failed.

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In the COM frame the mass-b is not stationary. –  DWin Jan 5 '14 at 2:11
@DWin Indeed. Can you explain the point of your comment? –  Spenser Jan 5 '14 at 2:17
You should be able to derive the kinetic energy of both objects in the COM frame from the velocity of the incoming particle in the observer frame since you know their relative masses. –  DWin Jan 5 '14 at 2:27