# Homework Help: Ballistic Pendulum

1. Apr 23, 2010

### Cfem

1. The problem statement, all variables and given/known data
A 2.3 kg wood block hangs from the bottom of a 1.3 kg, 1.3 m long rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A 12 g bullet is fired into the block, where it sticks, causing the pendulum to swing out to a 35 degrees.

2. Relevant equations
Conservation of Energy.
Conservation of Momentum

3. The attempt at a solution

(1/2)(m)(v^2) = mgh

Drawing a triangle, h is 1.3 * cos 35 = 1.06m

So (1/2)(v^2) = (9.8)(1.06)
v = 4.568 m/s

Conservation of momentum:

(mB)(vB) = (mB + mP)(v)
(.012)vB = (2.3+.012)(4.568)
vB = 10.561 / .012 = 880.21 m/s

Not sure where I went wrong.

2. Apr 23, 2010

### ideasrule

"h" has to be the amount by which the system's center of mass increased in height. It's not equal to 1.3*cos35 because the pendulum itself isn't massless.

3. Apr 23, 2010

### ideasrule

Note: actually, h isn't equal to 1.3*cos35 even if the pendulum were massless.

4. Apr 24, 2010

### Cfem

h would have been L - Lcos35 if it were massless. That's my slip up.

But back to the question at hand, though, I'm lost now. Would I have to use the center of mass of the pendulum and the block? That sounds complicated.

Is there another way I could do it?