# Angular momentum, block plus bullet

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1. Dec 3, 2014

### timnswede

1. The problem statement, all variables and given/known data
A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length l and of negligible mass. The rod is pivoted at the other end. A bullet of mass m travelling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.
a.) What is the angular momentum of the bullet-block system about a vertical axis through the pivot.

2. Relevant equations
L=r x p

3. The attempt at a solution
So I got L initial, which is mvl and L final which is (m+M)vfl. What I thought the angular momentum would be would be L final - L initial, but the answer is apparently mvl down, which is just the bullet. Where did I go wrong?

2. Dec 3, 2014

### Staff: Mentor

What question are you trying to answer? (You only listed part a.) Would you expect the angular momentum to change upon collision?

3. Dec 3, 2014

### timnswede

I only put part A since I understand part B.
Is it really that easy and I am just overcomplicating it? Since angular momentum is conserved, then L initial equals L final and that is why the answer is just mvl, since it is is the same as the final angular momentum?

4. Dec 3, 2014

### Staff: Mentor

Yes, that's all there is to it. Note that the question (at least as you quoted it) does not specify before or after the collision. It doesn't matter: angular momentum is conserved.

5. Dec 3, 2014

### timnswede

OK, thank you. That was actually a really easy question then.