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## Homework Statement

A 2.0kg (M) block hangs from the bottom of a 1kg (m), 1m (L) rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A .01kg bullet (m') is fired into the block, where it sticks, causing the pendulum to swing out to a 30 degree angle. What was the speed of the bullet (v)? You can treat the wood block as a particle.

## Homework Equations

Using conservation of energy: .5m'v[tex]^{2}[/tex]=(M+m+m')gh, with h=L-Lcos(30degrees)

## The Attempt at a Solution

v=Square root[2(.01kg + 1kg + 2kg)(9.8m/sec[tex]^{2}[/tex])(1m-cos(30degrees))/.01kg)]=28.11m/s

However, the given answer is 3.9 x10[tex]^{2}[/tex]m/s...

I've been working on this problem for north of six hours now. I will greatly appreciate any insight anyone can give me.