Explanation

(a) Since, linear momentum is conserved
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$mv_{0} = Mv_{CM}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
Angular momentum is also conserved
$mv_{0}\frac{L}{2} = \frac{ML^{2}}{12}\omega\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots..(ii)\ $
Where $\frac{ML^{2}}{12}$ is the moment of inertia of the rod about the axis of rotation
Since, collision is completely elastic , kinetic energy is also conserved .Thus,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}mv_{0}^{2} = \frac{1}{2}{M_{v}}_{CM} + \frac{1}{2}\left( \frac{ML^{2}}{12} \right)^{2}\omega^{2}$
From Eqs. (i)and (ii),We get
${\ \ v}_{CM} = \frac{1}{6}\omega L$
Putting this value in Eq. (iii),we get
$\ \ \ \frac{1}{2}mv_{0}^{2} = \frac{1}{2}M\left( \frac{1}{36}\omega^{2}L^{2} \right) + \frac{1}{2}M\left( \frac{1}{12}\omega^{2}L^{2} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{18}M\omega^{2}L^{2}$
OR $\frac{m}{M} = \frac{\omega^{2}L^{2}}{9V_{O}^{2}}$