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Elastic Collision question

  1. Mar 26, 2006 #1
    Q: A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the pundulum bob will barely swing through a complete vertical circle?

    I'm having alot of trouble with this question! I just can't seem to get my head around where to start, i've implemented the conservation of momentum equation;
    mv(i) + MV(i) = mv(f) + MV(f)
    as
    mv = m(v/2) + MV
    but i really have no idea how they want the question answered, or how to go about anwering it... :uhh: anyone have any insight?
    Thanks for the time
     
  2. jcsd
  3. Mar 26, 2006 #2

    Doc Al

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    Staff: Mentor

    So far, so good! Conservation of momentum does apply during the collision of bullet with pendulum. So this allows you to relate bullet speed (pre-collision) with pendulum speed (post-collision).

    Now you need to know what minimum speed must the pendulum have to just make it swing up in a complete circle. Hint: What's conserved as the pendulum swings?

    (By the way, the collision of bullet with pendulum is not an elastic collision.)
     
  4. Mar 26, 2006 #3
    Heh, sorry, 'twas at the end of the chapter on elastic, i didn't think about it!
    Thanks alot for your help... would i be right, then, in thinking that the pendulum needs only enough E to reach just past the top of it's rotation, after which, gravity will act to complete the swing.
    In that case, all of the kinetic E transfered to the bob will have been converted to potential E at the top of the swing...
    U=mgh=K transfered=.5Mv^2
    mg2l=.5M(v/2)^2
    ...
    ...
    2*(mgl/M)^-2
    is this along the right track?
     
  5. Mar 26, 2006 #4

    Doc Al

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    Staff: Mentor

    You're almost there. The KE of the pendulum at the bottom position (just after the collision) must be enough to get it to the top position: use energy conservation. But what's the KE of the bob in terms of the bullet's speed? (In other words, what is [itex]0.5 M V^2[/itex]?)
     
  6. Mar 26, 2006 #5
    [2*(mgl/M)^1/2 not 2!!! is what i ment :P]

    Aaahhh! now i get it, Thanks very much for your help Doc, really appreciated.
     
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