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Bullet, Collision, Block

  1. Jan 31, 2009 #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 traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

    What fraction of the original kinetic energy is list in the collision?

    2. Relevant equations

    KE = 1/2 m v^2
    L = cross(r,p)

    3. The attempt at a solution

    The only way I can think of doing this problem is using linear momentum conservation.
    Say if pi = m*v, pf = (m+M)v'
    v' = v * m / (M+m)

    final / initial Kinetic Energy = m / (M+m)

    Unfortunately, the answer is M/(M+m).

    I believe that I should be using conservation of angular momentum. According to part a of this problem (not stated) the angular momentum of the bullet+block = mvl.

    If I were to use cons. of angular momentum, I don't really know to formulate it.
    Initial Li = 0
    Final Lf = mvl

    Does anyone have any hints?
     
  2. jcsd
  3. Jan 31, 2009 #2
    fraction lost = (init - final)/init
    =1 - m/(m+M)
    =M/(m+M)

    Read the question!

    The rod is an irrelevance, since it's massless.
    If it wasn't, you would indeed conserve angular momentum
    to solve it.
     
    Last edited: Jan 31, 2009
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