Frictionless Slide Propulsion

The velocity of mass $\displaystyle m$ when it is about to strike the rod is given by,

$v = \sqrt{2gh}$

Now, it is very common to commit a mistake here. We may think that we should apply Conservation of Linear Momentum .
But, note that the Conservation of Linear Momentum is valid only when there is no external force on the system. And here, the external force is that of the reactionary force at the hinge of the rod.

Hence, we cannot apply Linear momentum conservation here. But, if we consider the net torque about the hinge of the rod, then we can see that there is no external torque on the system. Hence, it is valid to apply Conservation of Angular Momentum . We use this to relate the angular velocity before and after the collision:

$L_i = L_f$

$m\sqrt{2gh}d = I_{hinge} w = \left(md^2+\frac{Md^2}{3}\right)w$

$\Rightarrow w = \frac{3m\sqrt{2gh}}{(M+3m)d}$

Now, to find the angular displacement of the system, we can use Conservation of Energy, from right after the collision to the peak of the swing:

$K_i + U_i = K_f + U_f$

$\frac{1}{2}Iw^2 = \Delta U_{rod} + \Delta U_{m}$

$\frac{3m^2gh}{M+3m} = M \frac{d}{2}(1-\cos\theta)g + md(1-\cos\theta)g$

$\frac{3m^2gh}{M+3m} = (1-\cos\theta) d \left(\frac{M+2m}{2}\right)$

$\Rightarrow \cos\theta = 1 - \frac{6m^2h}{(M+3m)(M+2m)d}$

$\Rightarrow \theta = \cos^{-1}\left(1 - \frac{6m^2h}{(M+3m)(M+2m)d}\right)$

Hence, we get,

$\alpha+\beta+\gamma+\nu = \boxed{12}$

Similar solution to Anish,but his is way better.I used calculus where it was unnecessary .

$\large{\Rightarrow \omega =\frac {3m\sqrt { 2gh } }{ (M+3m)d }}$

$\large{\Rightarrow \int _{ 0 }^{ \theta }{ \tau d\theta } =\frac { 1 }{ 2 } I{ \omega }^{2 }}$

$\large{\Rightarrow \int _{ 0 }^{ \theta }{ { X }_{ cm }(M+m)d\sin { \theta }.d\theta } =\frac { 1 }{ 2 } I{ \omega }^{ 2 }}$

$\large{\Rightarrow -\cos {\theta} + 1 = \frac{-6m^2 h}{(M+3m)(M+2m)d} }$

$\large{\Rightarrow \alpha +\beta +\gamma+v=1+2+3+6=12}$