Let v and ω be the linear and angular speeds of the rod
after applying an impulse J at B. Then from :
                  impulse = change in momentum

we have

       $\text{mv}=\text{J or}\text{v}=\frac{\text{J}}{\text{m}}$                                    ...(1)

                   $\text{I}\mathrm{\omega }=\text{J}·\frac{\mathit{\text{l}}}{2}$

or      $\frac{\text{m}{\mathit{\text{l}}}^2}{12}\mathrm{\omega }=\text{J}·\frac{\mathit{\text{l}}}{2}\text{or}\mathrm{\omega }=\frac{6\text{J}}{\text{m}\mathit{\text{l}}}$                ...(2)

After the given time $\text{t}=\frac{\mathrm{\pi }\text{m}\mathit{\text{l}}}{12\text{J}}\text{,}$ the rod will rotate an angle

                     $\mathrm{\theta }=\mathrm{\omega }\text{t}=\left(\frac{6\text{J}}{\text{m}\mathit{\text{l}}}\right)\left(\frac{\mathrm{\pi }\text{m}\mathit{\text{l}}}{12\text{J}}\right)=\frac{\mathrm{\pi }}{2}$

               $\frac{\mathit{\text{l}}}{6}\text{.}\mathrm{\omega }=\left(\frac{\mathit{\text{l}}}{6}\right)\left(\frac{6\text{J}}{\text{m}\mathit{\text{l}}}\right)=\frac{\text{J}}{\text{m}}=\text{v}$

∴               $\left|{\text{v}}_{\text{P}}^{\overrightarrow{}}\right|=\sqrt{2}\text{v}=\sqrt{2}\frac{\text{J}}{\text{m}}$