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Ballistic Pendulum

  1. Apr 23, 2010 #1
    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. jcsd
  3. Apr 23, 2010 #2

    ideasrule

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    "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.
     
  4. Apr 23, 2010 #3

    ideasrule

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    Note: actually, h isn't equal to 1.3*cos35 even if the pendulum were massless.
     
  5. Apr 24, 2010 #4
    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?
     
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