From conservation of linear momentum we have,
                    
                      $\text{v}={\text{v}}_1+{\text{v}}_2$                                     ...(1)
From conservation of angular momentum about centre
of rod we have,
                  $\text{mv}a={\text{mv}}_{2}a+\frac{\text{m}{a}^2}{3}·\mathrm{\omega }$
or                    $\text{v}={\text{v}}_2+\frac{a\mathrm{\omega }}{3}$                                   ...(2)
Further from the definition of coefficient of restitution
(e = 1) at point of impact.
Relative speed of approach = relative speed of separation
∴                    $\text{v}={\text{v}}_1+a\mathrm{\omega }-{\text{v}}_2$                           ...(3)
solving these three Eqs. (1), (2) and (3) we get,
                      ${\text{v}}_1=\frac{2}{5}\text{v}\text{and}\mathrm{\omega }=\frac{6\text{v}}{5a}$
∴      Kinetic energy of rod,
                        $\text{K}=\frac{1}{2}\times \text{m}\times {\left(\frac{2}{5}\text{v}\right)}^2+\frac{1}{2}\times \frac{\text{m}{a}^2}{3}\times {\left(\frac{6\text{v}}{5a}\right)}^2$
                           $=\frac{8}{25}\text{m}{\text{v}}^2$