Inelastic Collision?

Question

Here is the question:
A 77kg man holding a .575 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 10.9 m/s (relative to the ground) and then catches the all after it rebounds from the wall. Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. How fast is he moving after he catches the ball? Answer in m/s.

I used conservation of momentum as follows.

I said that Vman = (Mball*Vball) / (Mman + Mball). I got .080793 as my answer but this is wrong. What am I not seeing correctly?

Part B of this question asks how many times does the man have to go through this process before his speed reaches at least 2.7 m/s relative to the ground?

Here would you just take the velocity of part 1 of the question and divide 2.7 by that number?

Answer 1

In an elastic collision between 2 bodies both momentum & kinetic energy of the system is conserved.
Whereas in an inelastic or plastic collision only momentum is conserved but kinetic energy is not conserved

Answer 2

You have to apply conservation of momentum twice: first when he throws the ball and next when he catches it.

From the throw, Mball*Vball = -Mman*Vman (momentum sums to zero)

After the catch (Mman+Mball)*Vboth = Mman*Vman+Mball*Vball, where the velocities on the right side are the ones from the previous equation (except that Vball is now of opposite sign). If you do this you will see that you are missing a factor of two in your equation.

You know Vball, Mman, Mball; unknown Vboth and Vman. You have two equations for the two unknowns.

You have to repeat this with the new velocities, but remember if the man throws the ball while he is moving at a velocity V, the ball velocity will be Vball - V.

EDIT:
I now see that the problem states that the ball's velocity is relative to the ground. If we assume that means the ball's velocity is Vball even when the man is moving, the general result for the man's velocity after the nth throw is

Vn = n*V1,

where V1 is the velocity of part 1. So your final statement is correct.

You can find the details worked out here:
http://img120.imageshack.us/img120/9148/gencasemanballsr5.png

Answer 3

You have to apply conservation of momentum twice: first when he throws the ball and next when he catches it.

From the throw, Mball*Vball = Mman*Vman (momentum sums to zero)

After the catch (Mman+Mball)*Vboth = Mman*Vman+Mball*Vball, where the velocities on the right side are the ones from the previous equation.

You know Vball, Mman, Mball; unknown Vboth and Vman. You have two equations for the two unknowns.