Physics collisions

Question

A bullet with a mass m is fired into a wooden block that lies on the top of a frictionless horizontal table. The bullet becomes embedded after the collision with the block and the block slides off the table, landing at a distance x from the edge of the table. The height of the table is h.
 
Diagram: https://imgur.com/a/t9aiM

a. Find the velocity of the bullet-block system after the collision.
b. Find the initial velocity of the bullet before the collision.
c. Find the ratio between the initial kinetic energy of the bullet and the kinetic energy of the bullet-block system.

In the second experiment the bullet is fired with the same velocity and the block is placed at a distance L from the right end of the table. The friction is present in this trial
e. Find the minimum coefficient of kinetic friction so that the block doesn’t leave the table.

Answer 1

a. Momentum is conserved: mv₀+0=(M+m)v where v is the bullet-block velocity.

So v=mv₀/(M+m).

b. The initial velocity of the bullet is v₀ according to the diagram.

c. KE of bullet is ½mv₀²; after collision the KE is ½(M+m)v²=½(M+m)m²v₀²/(M+m)²=½m²v₀²/(M+m); the ratio of the two KEs is m/(M+m):1=(KE of bullet-block):(KE of bullet).

e. Opposing frictional force F=-µ(M+m)g=(M+m)a, so acceleration a=-µg (deceleration).

Using v²-u²=2as where v (final velocity)=0 and u=mv₀/(M+m), s=L,

-u²=-2µgL, µ=u²/2gL=m²v₀²/(2gL(M+m)²).