# Momentum and Mech. Energy

1. Mar 11, 2009

### PnotConserved

1. The problem statement, all variables and given/known data
A bullet of mass (m) is fired into a block of mass (M). The block with the embedded bullet slides across a frictionless table and collides with a horizontal spring whose constant is (k). The springs maximum compression (d) is measured.
Find an expression for the bullets initial speed (Vb) in terms of m, M, k, d.

2. Relevant equations
mv+Mv=(m+M) V

1/2mv^2 + 1/2 kd^2 = 1/2 mv^2 +1/2 kd^2

3. The attempt at a solution

I attempted to add the equations together which got me to:
Vb = (m+M)V
1/2mV`^2 = 1/2 kd^2

....I'm lost

2. Mar 12, 2009

### windsupernova

Well, I think the spring starts in equilibrium so its potential energy is cero.

The block I suppose starts at rest.

I think that you can go from there tbqh

3. Mar 12, 2009

### LowlyPion

You have an inelastic collision. So hopes of simply equating Kinetic energy of the bullet directly to Potential energy in the spring can't be used.

But ... The kinetic Energy of the block can be used.

That depends on its V' which you know from conservation of momentum is:

V' = Vb*m/(M + m)

So yes you do equate just as you have done.

Instead of being lost, you should merely have substituted V' with V' = Vb*m/(M + m) of your first equation and you'd have slept soundly satisfied that you had the right answer.