Q. A uniform rod of mass M and length a lies on a smooth horizontal plane. A partical of mass m moving at a speed v perpendicular to the length of rod strikes it at a distance a/4 from the centre and stops after the collision. Find (a) the vel. of centre of the rod and (b) the angular velocity of the rod about its centre just after the collision.

The situation is shown in figure
Let the velocity of centre of mass and angular velocity of centre of be V and w respectively.
Given velocity of particle is v and mass is m.
As there is no external force in horizontal direction so applying conservation of linear momentum, we get
mv = MV or V =(m/M)v 
 now, applying conservation of angular momentum about point A,
$$\begin{gathered} mv\times \frac{a}{4}=I\omega =\frac{1}{12}M{a}^2\omega \\ it gives \omega =\frac{3mv}{Ma} \\ Thus linear velocity of rod is \frac{m}{M}v and angular velocity is \frac{3mv}{Ma} \end{gathered}$$