 Solution

For two particles, let $M=m_1+m_2$ and ${\mathbf{V}}_{\mathrm{cm}}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/M$. The centre-of-mass frame velocities are ${\mathbf{v}}_i^{\ast }={\mathbf{v}}_i-{\mathbf{V}}_{\mathrm{cm}}$, so $m_1{\mathbf{v}}_1^{\ast }+m_2{\mathbf{v}}_2^{\ast }=0$. Thus before and after an isolated collision the two momentum vectors are equal and opposite. For an elastic collision their common magnitude is also unchanged: only their direction rotates.

The kinetic energy decomposition about the center of mass separates unchanged bulk translation from relative motion:The reduced mass $\mu$ reduces the calculation to one relative velocity. After solving the geometry in the centre-of-mass frame, add ${\mathbf{V}}_{\mathrm{cm}}$ to each velocity to return to the laboratory. The four panels below illustrate this for the subsequent example.