One block of mass m moves at velocity $v_{0}$ over a horizontal table of height H without friction. When it reaches the edge of the table, the block collides elastically against a rod of mass M and length L which turns about an axis (perpendicular to the paper) located at the opposed extreme of the collision (look at the image). Mass ratio: M/m=3/2.

Calculate :

a) Velocity of the block and angular velocity of the rod after the collision

b) Maximum height that the CM of the rod reaches after the collision (initially located as shown at the image)

c) Horizontal distance that the block reaches after the fall.

What I tried:

Note the energy immediately before the collision is the right side of the equation and immediately after the collision the left side (in the image is written the other way around and is wrong).

At a) I obtained an equation of the velocity of the block after the collision in function of the angular velocity of the rod after the collision. I need another equation so as to obtain both values. I think we have to use conservation of angular momentum here. I thought about this equation:

$$\omega_{f} = \frac{mL^2\omega_{o}}{\frac{ML^2}{3}+mL^2}$$

$$\omega_{o}=\frac{v_{o}}{L}$$

$$v_{f}^2=v_{o}-\frac{L^2\omega^2_{f}}{2}$$

Then I could obtain equations for final velocity and angular velocity in function of $v_{o}$ and $L$

At b) the solution is said to be $z=\frac{8v_{0}^2}{27g}$ But I do not get it.

a) Yes you need to use conservation of angular momentum. You also need to use conservation of kinetic energy because the collision is perfectly elastic. The easiest way to do the latter is through Coefficient of Restitution $e=1$ : relative velocity of separation = relative velocity of approach.

b) KE of rod immediately after collision = gravitational PE of rod at maximum swing.