= 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 $I=ML^2/12$.
The <impulse> delivered to the rod is $MV$. Conservation of linear and angular <momentum> gives
$$
u=v+V,\qquad I\Omega=MbV.
$$
Using these in conservation of <kinetic energy>,
$$
u^2=v^2+V^2+\frac IM\Omega^2
\quad\Longrightarrow\quad
V=\frac{2u}{2+Mb^2/I},\qquad
v=\frac{(Mb^2/I)u}{2+Mb^2/I}.
$$
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 $Mb^2/I=2$ and
$$
\boxed{|b|=\frac L{\sqrt6}.}
$$
Both symmetric impact points lie on the rod, since $L/\sqrt6<L/2$.
More generally, the relative contact <velocity> immediately afterwards is
$$
v-(V+b\Omega)=u-(2+Mb^2/I)V=-u.
$$
Thus 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.
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