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

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In summary, the problem involves a bullet of mass 10.0 g being fired into a ballistic pendulum of mass 3.00 kg, causing the pendulum to swing back and rise 12.0 cm. The original speed of the bullet needs to be found. The equations used are p1 + p2 = p1' + p2' and (m1 + m2) * g * h = 1/2 * (m1 + m2) * v^2, with momentum and energy being conserved. The final solution is v = 2.86 m/s.
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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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