A uniform thin rod of length L lies at rest on a smooth horizontal table. It receives a sharp horizontal impulse perpendicular to its length at a…

Let the impulse $J$ act at distance $x=L/3$ from the centre. Then $v_{cm}=J/m$ and, from the angular impulse about the centre, $\omega = \dfrac{Jx}{mL^2/12}$. A point at distance $y$ from the centre on the side opposite to the impulse has velocity $v_{cm}-\omega y$, which vanishes when $y = \dfrac{v_{cm}}{\omega} = \dfrac{L^2}{12x} = \dfrac{L^2}{4L} = \dfrac{L}{4}$. This point lies on the other side of the centre from the impulse, so its distance from the struck point is $x+y = \dfrac{L}{3}+\dfrac{L}{4} = \dfrac{7L}{12}$. The value $L/4$ is only its distance from the centre.