= Solution
{c}
For an <equal-speed elastic impact on a free rod>, let $h$ be the signed distance of the impact point from the rod's <center of mass>, $I=ml^2/12$, and $\mathcal J$ the horizontal <impulse> delivered to the rod. For a smooth impact the <impulse> is perpendicular to the rod. If the ball's signed final <velocity> is $u$, the rod's <center of mass> <velocity> is $V$ and its <angular velocity> is $\omega$, <momentum> balance and angular <impulse> give
$$
m(v_0-u)=\mathcal J=mV,\qquad I\omega=\mathcal J h.
$$
Both objects have positive final translational <velocities> for a nonzero impact away from the centre. More explicitly, restitution in an <elastic collision> gives $V+\omega h-u=v_0$, and hence
$$
V=\frac{2v_0}{2+mh^2/I},\qquad
u=\frac{(mh^2/I)v_0}{2+mh^2/I}\ge0.
$$
Equal final <speeds> thus require $u=V=v_0/2$. One can also see that the alternative $u=-V$ would violate $u+V=v_0$.
By conservation of <kinetic energy> in the <elastic collision>,
$$
\frac12mv_0^2=\frac12m\left(\frac{v_0}{2}\right)^2
+\frac12m\left(\frac{v_0}{2}\right)^2+\frac12I\omega^2.
$$
Thus $I\omega^2=mv_0^2/2$. Since $I\omega=mVh=mv_0h/2$, elimination of $\omega$ gives
$$
\boxed{h^2=\frac{2I}{m}=\frac{l^2}{6},\qquad |h|=\frac l{\sqrt6}.}
$$
Either side of the <center of mass> works; the nearer end is a distance $l(1/2-1/\sqrt6)$ away. The final <angular speed> is $|\omega|=\sqrt6\,v_0/l$. These values also satisfy the restitution condition because the struck point's <velocity> is $V+\omega h=3v_0/2$, reversing the initial relative contact <velocity>.
Back to article page
