A uniform rod of mass 5m and length L is placed over a smooth horizontal surface. A particle of mass m moving horizontally with velocity  $v_0$  hits the rod at one end perpendicularly as shown in figure and comes to rest. Then 
 

a) Velocity of centre of rod after collision is  $\frac{v_0}{5}$

b) Angular velocity of rod after collision is  $\frac{3v_0}{5L}$

c) Coefficient of restitution is  $\frac{4}{5}$

d) Kinetic energy of rod after collision is  $\frac{4}{5}m{v}_0^2$

Ans: 1, 3
Let  $V_C$  and  $\omega$  be the velocity of centre of rod and its angular velocity just after collision

As no external force on the system
So using C.O.L.M
 $\overrightarrow{P_{in}}=\overrightarrow{P_{inal}}$
 $\Rightarrow m{v}_0=5m{v}_c\Rightarrow v_c=\frac{v_0}{5}$
As no external torque about fixed point near to centre of rod (A) of system. So using C.O.A
 $$\begin{gathered} \overrightarrow{L_{in}}={\overrightarrow{L}}_{inal} \\ m{v}_0\frac{L}{2}=\frac{5m{L}^2}{12}\omega \\ \Rightarrow \omega =\frac{6v_0}{5L} \end{gathered}$$
Coefficient of restitution
 $$\begin{gathered} e=\frac{velocityofseparation}{velocityofapproach} \\ e=\frac{v_c+\frac{\omega L}{2}}{v_0} \end{gathered}$$
 $e=\frac{\frac{v_0}{5}+\frac{6v_0}{5L}.\frac{L}{2}}{v_0}=\frac{4}{5}$
K.E. of rod  $=\frac{1}{2}5m{v}_c^2+\frac{1}{2}{I}_c{\omega }^2$
 $=\frac{1}{2}5m{\left(\frac{v_0}{5}\right)}^2+\frac{1}{2}\frac{\left(5m\right)L^2}{12}{\left(\frac{6v_0}{5L}\right)}^2=\frac{2m{v}_0^2}{5}$