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Theoretical Expression for %loss in Kinetic Energy

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm trying to derive an equation which should look like this in the end:
    %loss in Kinetic energy= M/m+M*100%, I'm just not sure how to substitute everything, and cancel things out to get the expression. Can someone show me the steps?


    2. Relevant equations
    mVi=(M+m)Vf
    %loss in Kinetic Energy= 1/2mVi^2-1/2(M+m)Vf^2/ 1/2mVi^2 *100%


    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 1, 2009 #2

    rl.bhat

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    Pi = mVi
    Pi^2 = (mVi)^2 = 2m*(1/2*m*Vi^2) = 2mEi.
    So Ei = Pi^2/2m
    Similarly find Ef
    Then percent change = (Ei - Ef)/Ei*100
     
  4. Nov 1, 2009 #3
    I'm sorry i really don't see how that's going to work out to M/M+m
     
  5. Nov 1, 2009 #4

    rl.bhat

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    Substitute the values of Ei and Ef. While simplification substitute the value of Vf in terms of Vi using the first equation in relevant equations.
     
  6. Nov 1, 2009 #5
    yea...that means nothing to me sorry..
     
  7. Nov 2, 2009 #6

    rl.bhat

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    (Ei - Ef)/Ei = [1/2*m*vi^2 - 1/2*(m+M)*vf^2]/1/2*m*vi^2.
    Cancel 1/2m
    = vi^2 - [(m+M)/m*vf^2]/vi^2
    Put vi^2 = [(m + M)/m*vf]^2 and simplify.
     
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