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Angular momentum, block plus bullet

  1. Dec 3, 2014 #1
    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. jcsd
  3. Dec 3, 2014 #2

    Doc Al

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    Staff: Mentor

    What question are you trying to answer? (You only listed part a.) Would you expect the angular momentum to change upon collision?
     
  4. Dec 3, 2014 #3
    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?
     
  5. Dec 3, 2014 #4

    Doc Al

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    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.
     
  6. Dec 3, 2014 #5
    OK, thank you. That was actually a really easy question then.
     
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