\(m v=M V \quad \ldots(1)\)
By conservation of angular momentum
\(m v(L / 2)=\left(\frac{M L^{2}}{12}\right) \omega \quad \ldots(2)\)
As the collision is elastic, we have
\(\frac{1}{2} m v^{2}=\frac{1}{2} M V^{2}+\frac{1}{2} I \omega^{3}\) \(...(3)\)
Substituting the values, we get \(m=M / 4\)