A rod AB of length L and mass M is free to move on a frictionless horizontal surface is moving with a velocity v as shown in fig. End B of rod AB strikes end of wall. Assuming elastic impact, angular velocity of rod AB just after impact is 

Let the impulse from the wall on the rod be J,

After this impulse the velocity of centre of mass will be $\mathrm{v}\text{'}=\frac{\mathrm{mv}-\mathrm{J}}{\mathrm{m}}$....(1)

And the angular velocity of the rod about centre of mass be $\mathrm{\omega }$, this is given by $\mathrm{\omega }=\frac{\mathrm{J}.{\displaystyle \frac{\mathrm{L}}{2}}}{{\displaystyle \frac{{\mathrm{mL}}^2}{12}}}$.....(2)

So the velocity of the point B will be given by ${\mathrm{v}}_{\mathrm{B}}=\mathrm{v}\text{'}-\frac{\mathrm{L}}{2}\mathrm{\omega }$, ......(3)

Since the collision is elastic ${\mathrm{v}}_{\mathrm{B}}=-\mathrm{v}$

using the equations, 

$\frac{\mathrm{mv}-\mathrm{J}}{\mathrm{m}}-\frac{\mathrm{L}}{2}\left(\frac{\mathrm{J}.{\displaystyle \frac{\mathrm{L}}{2}}}{{\displaystyle \frac{{\mathrm{mL}}^2}{12}}}\right)=-\mathrm{v}$

$\Rightarrow \mathrm{J}=\frac{\mathrm{mv}}{2}$

Using the value of J in (2)

$\mathrm{\omega }=\frac{{\displaystyle \frac{\mathrm{mv}}{2}\frac{\mathrm{L}}{2}}}{{\displaystyle \frac{{\mathrm{mL}}^2}{12}}}=\frac{3\mathrm{v}}{\mathrm{L}}$