= Solution
For two particles, let $M=m_1+m_2$ and $\mathbf V_{\rm cm}=(m_1\mathbf v_1+m_2\mathbf v_2)/M$. The <centre-of-mass frame> velocities are $\mathbf v_i^*=\mathbf v_i-\mathbf V_{\rm cm}$, so $m_1\mathbf v_1^*+m_2\mathbf v_2^*=0$. 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:
$$
K=\frac12M V_{\rm cm}^2+\frac12\mu|\mathbf v_1-\mathbf v_2|^2,
\qquad \mu=\frac{m_1m_2}{M}.
$$
The <reduced mass> $\mu$ reduces the calculation to one relative <velocity>. After solving the geometry in the <centre-of-mass frame>, add $\mathbf V_{\rm cm}$ to each <velocity> to return to the laboratory. The four panels below illustrate this for the subsequent example.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017-resit-collision-frames.png]
{title=Velocity vectors before and after an elastic collision in laboratory and centre-of-mass frames}
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