A rod AC of length l and mass m is kept on a horizontal smooth plane. It is free to rotate and move. A particle of same mass m moving on the plane with velocity v strikes the rod at point B making angle 37° with the rod. The collision is elastic. After collision

Let the linear velocity of the rod be u and angular velocity be $\mathrm{\omega }$ after collision.

After the collision the velocity of the ball will be unchanged in the direction parallel to the rod.

Since the collision is elastic, velocity of separation = velocity of approach 

$\frac{3\mathrm{v}}{5}=\left(\frac{\mathrm{\omega l}}{4}+\mathrm{u}\right)+{\mathrm{v}}^{\text{'}}$...(1), where v' is the velocity of mass m in the perpendicular direction.

On conserving linear momentum of the system in the direction perpendicular to the rod

$\mathrm{mv}\frac{3}{5}=\mathrm{mu}-{\mathrm{mv}}^{\text{'}}\Rightarrow \frac{3\mathrm{v}}{5}=\mathrm{u}-{\mathrm{v}}^{\text{'}}$....(2)

On conserving the total angular momentum about the point of collision, we get

$0=0+\left[\mathrm{mu}\frac{\mathrm{l}}{4}-\frac{{\mathrm{ml}}^2}{12}\mathrm{\omega }\right]\Rightarrow \frac{{\mathrm{ml}}^2}{12}\mathrm{\omega }=\frac{\mathrm{mul}}{4^2}\Rightarrow \mathrm{u}=\frac{\mathrm{\omega l}}{3}$....(3)

So from (3) we can say that by the time the rod makes half rotation the center of rod would have displaced by $\frac{\mathrm{\pi l}}{3}$

On adding (1) and (2), we get

$\frac{6\mathrm{v}}{5}=2\mathrm{u}+\mathrm{\omega l}$

using (3)

$\frac{6\mathrm{v}}{5}=2\mathrm{u}+\frac{3\mathrm{u}}{4}\Rightarrow \mathrm{u}=\frac{24}{55}\mathrm{v}$

Impulse imparted will be $\mathrm{mu}=\frac{24}{55}\mathrm{mv}$

So $\mathrm{\omega }=\frac{3\mathrm{u}}{\mathrm{l}}=\frac{72\mathrm{v}}{55\mathrm{l}}$