$A$ uniform rod of length $L$ and mass $M$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ moving with velocity $v$ strikes the rod at one end perpendicular to the rod. After the collision,the particle comes to rest. What is the angular velocity of the rod about its centre of mass after the collision?

(D) Let the rod have length $L$ and mass $M$. The particle has mass $m$ and velocity $v$.
$1$. Conservation of Linear Momentum:
Since the surface is smooth,there is no external horizontal force.
$mv = MV_{cm} + m(0) \implies V_{cm} = \frac{mv}{M}$
$2$. Conservation of Angular Momentum about the centre of mass $(CM)$ of the rod:
Initial angular momentum $L_i = m v (L/2)$
Final angular momentum $L_f = I_{cm} \omega + M V_{cm} (0) = (\frac{ML^2}{12}) \omega$
Equating $L_i = L_f$:
$m v \frac{L}{2} = \frac{ML^2}{12} \omega$
$3$. Solving for $\omega$:
$\omega = \frac{mvL/2}{ML^2/12} = \frac{6mv}{ML}$