What is the original speed of a bullet fired into a ballistic pendulum?

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Homework Statement


A bullet of mass 10.0 g is fired into a ballistic pendulum of mass 3.00 kg. The pendulum swings back and rises 12.0 cm. Find the original speed of the bullet.

Homework Equations


p1 + p2 = p1' + p2'
We're learning momentum right now, and I think that's what I need.

The Attempt at a Solution


I'm too confused to even write much down. I've only thought about it. If the string's length isn't given, how can I solve this? If a string was 50000 meters long and someone hit it, wouldn't it swing less than a string 10 meters long? It would travel a different distance along the circumference with differing string lengths, right?I wrote this down and can't go any further: .5 * .01 * v^2 = 9.8 * .12 * 3.01(wrong answer for v) The only way I could write anything down was using energy, but that's not what this chapter is about :(. I don't even mind if you don't solve it all the way; a step in the right direction would be appreciated.

I should have posted this in introductory physics. Sorry, I was in a rush.
 
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I figured it out. SOLVED Thanks for the... forum space? lol

I'll explain how I did it just in case someone finds this post in search of the answer i sought.

(m1 + m2) * g * h = 1/2 * (m1 + m2) * v^2

the potential energy gained by the masses together(since they stuck) equals the kinetic energy of both masses RIGHT when the bullet struck.

Momentum in this system is conserved so p1 + p2 = p1' + p2'

m1 * v + m2 * 0 = (m1 + m2) * (the v we found up there)

solve for v
 
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