Angular momentum, block plus bullet

[timnswede]

Homework Statement

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 traveling 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.

Homework Equations

L=r x p

The Attempt at a Solution

So I got L initial, which is mvl and L final which is (m+M)v_fl. 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?

 

[Doc Al]

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?

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

 

[timnswede]

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

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?

 

[Doc Al]

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?

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.

 

[timnswede]

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.

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