= Equal-speed elastic impact on a free rod
{title2=$|h|=\sqrt{2I/m}$}
A <point mass> $m$ strikes a free rod of equal <mass>, initially at rest, perpendicularly at distance $h$ from its <center of mass>. Assume a smooth <elastic collision> and rod <moment of inertia> $I$ about its centre. The final translational <speeds> are equal precisely when $h^2=2I/m$. Indeed, <momentum conservation> then fixes both final <velocities> to $v_0/2$; angular <impulse> gives $I\omega=mv_0h/2$, and conservation of <kinetic energy> gives $I\omega^2=mv_0^2/2$. Eliminating $\omega$ proves the formula. A uniform rod has $I=ml^2/12$, hence $|h|=l/\sqrt6$, which lies within the rod.
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