 Solution

Write the incoming particle speed as $u>0$, its final signed speed as $v$, and the rod's final centre-of-mass velocity along the impact direction as $V$. Let $b$ be the signed impact distance from the centre, and choose the sign of the angular velocity $\Omega$ so that the contact point's velocity component is $V+b\Omega$. The rod's central moment of inertia is $I=M{L}^2/12$.

The impulse delivered to the rod is $MV$. Conservation of linear and angular momentum givesUsing these in conservation of kinetic energy,This is the equal-mass instance of the effective contact mass calculation. For $u>0$, both final speeds are nonnegative, so equal speeds require $v=V=u/2$. Therefore $M{b}^2/I=2$ andBoth symmetric impact points lie on the rod, since $L/\sqrt{6}<L/2$.

More generally, the relative contact velocity immediately afterwards isThus its magnitude equals the incoming relative speed $u$ and its direction reverses. At the specified impact point, $V+b\Omega =3u/2$ and $v=u/2$, which makes the reversal particularly explicit.