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A normal impulse $j$ on a free planar rigid body of mass $M$ and moment of inertia $I$, at perpendicular lever arm $b$, changes the contact point's normal velocity by $j(1/M+b^2/I)$. Its effective mass in that direction is therefore $(1/M+b^2/I{)}^{-1}$. This follows by adding the translational change $j/M$ to the rotational contribution $b(jb/I)$.

For an incoming point mass $m$, the inverse reduced contact mass is $1/m+1/M+b^2/I$. A smooth elastic collision with initial relative approach speed $u$ has impulse $2u/(1/m+1/M+b^2/I)$, since the relative normal velocity reverses. This separates effective inertia at a contact from the body's total mass and explains the lever-arm dependence in an equal-speed elastic impact on a free rod.

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, $I=m{l}^2/12$, 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 giveBoth 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 henceEqual 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,Thus $I{\omega }^2=m{v}_0^2/2$. Since $I\omega =mVh=m{v}_{0}h/2$, elimination of $\omega$ givesEither side of the center of mass works; the nearer end is a distance $l(1/2-1/\sqrt{6})$ away. The final angular speed is $|\omega |=\sqrt{6}{v}_0/l$. 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.