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I can't get started can someone help. Kinetic Energy in term of momentum

  1. Sep 16, 2008 #1
    1. The problem statement, all variables and given/known data
    The question is this:
    Expressing the kinetic energy in terms of momentum (K=1/2mv2 = p2/2m),
    prove using symbols, not numbers, that the fractional loss during the collision is
    equal to M/(m+M)


    2. Relevant equations



    3. The attempt at a solution
    I'm not sure where to even begin as I don't understand it
    Thanks,
    Kevin
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2008 #2
    "During the collision"

    You have to be more specific, what kind of collision occured?
     
  4. Sep 16, 2008 #3
    This question was asked from a lab for a ballistic pendulum experiment
    Thanks,
    Kevin
     
  5. Sep 16, 2008 #4
    Hi Kevin..
    Thats relatively simple.
    This is a case of a Ballistic Pendulum, therefore I take it as a complete;y inelastic collision, i.e. the bullet gets embedded.
    Conserve the momentum:
    mv=(M+m)(v1)
    Let v1 b the velocity of the pendulum when the bullet gets embedded.
    v1=mv/(M+m)
    now find the change in kinetic energy:
    [tex]\frac{1}{2}mv^{2}- \frac{1}{2}\frac{(M+m)m^{2}v^{2}}{(M+m)^{2}}[/tex]

    Simplify this to get:

    Loss in KE = [tex]\frac{mMv^{2}}{2(M+m)}[/tex]

    Divide this by initial Kinetic energy to get the fraction loss:

    =M/(M+m)
     
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