A uniform rod of mass 0.60 kg and length 1.2 m lies at rest on a smooth horizontal table…

Step 1. Three equations. Let $v$ be the puck's velocity after, $V$ that of the rod's centre, and $\omega$ the rod's angular velocity, all positive in the sense the puck pushes.

• Linear momentum: $mu = mv + MV$.
• Angular momentum about the rod's centre (the puck hits at $L/2$): $mu\dfrac{L}{2} = mv\dfrac{L}{2} + I\omega$.
• Elastic impact: the struck end, moving at $V + \omega\dfrac{L}{2}$, separates from the puck at $u$: $V + \omega\dfrac{L}{2} - v = u$.

Step 2. Solve. The first two give $V = \dfrac{m}{M}(u - v) = 0.333\,(u - v)$ and $\omega = \dfrac{mL}{2I}(u - v) = 1.667\ \text{m}^{-1} \times (u - v)$. Put them into the third: $(u - v)(0.333 + 1.0) = u + v$, which gives $v = \dfrac{u}{7} = 0.857\ \text{m/s}$. Then

The puck carries on forwards, slowly.

Step 3. The point at rest. A point at distance $s$ from the centre, on the struck side, moves at $V + \omega s$. This is zero at $s = -\dfrac{V}{\omega} = -0.20\ \text{m}$: $0.20\ \text{m}$ from the centre on the far side, that is $0.40\ \text{m}$ from the far end.

Step 4. Energy. The puck had $\tfrac12 \times 0.20\ \text{kg} \times (6.0\ \text{m/s})^2 = 3.60\ \text{J}$. After: puck $0.0735\ \text{J}$, rod moving along $0.8816\ \text{J}$, rod turning $\tfrac12 I\omega^2 = 2.6449\ \text{J}$. The rotation takes $\dfrac{2.6449}{3.60} = 73.5\%$.