# Momentum conservation (ballistic pendulum)

## Homework Statement

A .01kg bullet is fired into a 1.2kg block hanging from a 1m wire. The bullet exits the block with a speed of 200m/s and the block swings to a height of .2 meters. What is the original velocity of the bullet? What percentage of the original energy of the bullet is no longer in mechanical forms of energy?

## Homework Equations

I know that initial momentum = final momentum but I can't seem to find the velocity of the block after the collision.

## The Attempt at a Solution

I had this on a test and tried energy conservation. Obviously this isn't correct but I cant find the momentum of the block. I've been working this for about 5 hours now so i'm clearly not getting something so a step by step explanation would be great.
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GPE of block = (mgh) = (1.2 x 9.8 x 0.2m) = 2.353 Joules. This has resulted from KE of the bullet having swung the block.
The bullet exited the block. Its KE after exit = 1/2 (m*v^2) = 1/2 (0.01*200^2) = 200 Joules.
The total energy has come from the bullet, = (200 + 2.353) = 202.353 Joules.
Original bullet V = sqrt.(2KE/m) = sqrt.(404.706/0.01) = 201.2 m/sec.

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GPE of block = (mgh) = (1.2 x 9.8 x 0.2m) = 2.353 Joules.
That's the energy of the block after the bullet has passed through it. Use this to determine the speed of the block.

Ok, so this is what I ended up with:

mgh = 1/2mv2
v=sqrt(2gh) = 1.98m/s

Pi = Pf
mpvpi = mbvb + mpvpf
vpi = (mbvb + mpvpf) / mp
vpi = 437.6 m/s

Thanks for the help!