= Solution
Use nonrelativistic <mechanics> and let $\mathbf V=(m_1\mathbf v_1+m_2\mathbf v_2)/(m_1+m_2)$ be the constant <centre-of-mass velocity>. In the <centre-of-mass frame>, <conservation of momentum> gives $m_1\mathbf u_1+m_2\mathbf u_2=0$ before and after the <elastic collision>. Thus $\mathbf u_2=-(m_1/m_2)\mathbf u_1$, and the <kinetic energy> is
$$
T_*=\frac12m_1u_1^2+\frac12m_2u_2^2=\frac12m_1\left(1+\frac{m_1}{m_2}\right)u_1^2.
$$
The decomposition $T_{\rm lab}=T_*+\tfrac12(m_1+m_2)V^2$ and constant <centre-of-mass velocity> show that the <elastic collision> preserves $T_*$, so the positive constant multiplying $u_1^2$ forces that <speed> to remain unchanged. The same <momentum> relation fixes the second <speed>:
$$
\boxed{|\mathbf u'_1|=|\mathbf u_1|,\qquad |\mathbf u'_2|=|\mathbf u_2|.}
$$
Only their directions can change; the two <velocities> remain antiparallel in this frame. This derivation does not require a head-on <collision>.
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