How Much Kinetic Energy is Lost in a Bullet-Block Collision?

[roeb]

Homework Statement

A wooden block of mass M resting on a frictionless horizontal surface is attached to a rigid rod of length l and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

What fraction of the original kinetic energy is list in the collision?

Homework Equations

KE = 1/2 m v^2
L = cross(r,p)

The Attempt at a Solution

The only way I can think of doing this problem is using linear momentum conservation.
Say if pi = m*v, pf = (m+M)v'
v' = v * m / (M+m)

final / initial Kinetic Energy = m / (M+m)

Unfortunately, the answer is M/(M+m).

I believe that I should be using conservation of angular momentum. According to part a of this problem (not stated) the angular momentum of the bullet+block = mvl.

If I were to use cons. of angular momentum, I don't really know to formulate it.
Initial Li = 0
Final Lf = mvl

Does anyone have any hints?

 

[davieddy]

fraction lost = (init - final)/init
=1 - m/(m+M)
=M/(m+M)

Read the question!

The rod is an irrelevance, since it's massless.
If it wasn't, you would indeed conserve angular momentum
to solve it.