A particle of mass m collides with the end of a spinning rod of mass 2m and length ‘l’, at  the end of the rod. If the coefficient of restitution of collision is $e=\frac{2}{3}$, then

Using conservation of linear momentum,

$m{v}_o=m{v}_1+2m{v}_2$.................(1)

Using conservation of angular momentum about center of rod,

$m{v}_o\frac{l}{2}-2m\frac{l^2}{12}\frac{v_o}{l}=m{v}_1\frac{l}{2}+\frac{2m{l}^2}{12}\omega$............(2)

Coefficient of restitution is $e=\frac{2}{3}$,

$\frac{2}{3}=\frac{\left(v_2+{\displaystyle \frac{\omega l}{2}}\right)-v_1}{v_0-\left(-{\displaystyle \frac{v_0}{2}}\right)}$........................(3)

After solving we get,

$v_1=\frac{v_0}{6}$

$v_2=\frac{5v_0}{12}$

$\omega =\frac{3v_o}{2l}$