The following link directs to the picture of the problem:

https://imgur.com/gallery/1a8cX?s=wa

Questions:

A) Are momentum, angular momentum and energy conserved? Under which circumstances?

B) Angular velocity of the ball immediately before the collision?

C) M/m rate so as to make the system (rod+ball) stay at rest after collision.

D) Velocity of the block after the collision.

E) Obtain an equation for D taking into account there is friction (Mu coefficient) ground-block along distance d and height h.

These are my answers:

A) - Linear momentum is conserved before/after the collision, due to the lack of external forces. However, there is a force exerted by the pivot on the system during the collision. Therefore, linear momentum of the system (ball+rod) IS NOT CONSERVED.

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- Angular momentum is conserved due to the fact that torque exerted on the rod by the ball is equal and opposite to the torque exerted on the ball by the rod. (Meaning that AM of the rod is equal and opposite to AM of the ball; therefore AM of the system IS CONSERVED).
- Energy IS CONSERVED IF FRICTION of any kind IS NEGLECTED. If we considered friction ground-block along distance d, energy would not be conserved after the elastic collision.
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B) First, I obtain system's center of mass: I consider the rod a puntual mass and remembering both rod and ball masses are equal to m: Xcm= (m(l/2)+ml)/2m= 3/4l

As energy is conserved before the collision and system start and ends at rest, we can assure gravitational potential energy transforms to kinetic energy(note omega equals to angular velocity):

2mg(3/4l)=1/2(2m(3/4l)^2)omega^2 Therefore: omega= sqroot(8g/3l)

C) First we obtain CoM's velocity before the collision:

V=(omega)radius; therefore Vcm:sqroot(3gl/2)

Now to obtain the velocity of the block after the collision we use momentum conservation principle:

2mVcm=MVblock; therefore: Vblock=2m/Msqroot(3gl/2)

Finally, to relate both M and m we use the principle of conservation of kinetic energy at an elastic collision:

1/22mVcm=1/2MVblock; therefore: M/m=2

D) Vblock= sqroot(3gl/2)

E) Energy is not conserved along d, however we now deltaE=muMgd. Therefore, Vblock= (mu)gd.

We have a projectile trajectory. Block has an initial velocity just at its horizontal component. Therefore:

D= d + Vblock(t) and y= h - 1/2g(t^2) Solving for t at second equation: t=sqroot(2h/g)

Finally, D= d(1 + mu(g)sqroot(2h/g))

Please if you could check my answers I would be greatful. If there's anything wrong or that can be improved let me know.

Thanks