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Homework Help: Conservation of Energy problem

  1. Mar 8, 2008 #1
    [SOLVED] Conservation of Energy problem

    Assuming there is no air friction, A 1.4 kg ball is dropped from a height of 2.68 m. It hits the ground, losing 1/12 of its energy in the collision. How high will the ball bounce upward before it comes momentarily to rest?


    E = K sys + U sys
    Ef = Ei
    1/2mv^2 + mgh = 1/2mv^2 + mgh

    I've attempted this problem but I seem to be missing some piece of information and I have run out of ideas hopefully someone can point me in the right direction.

    Since there are no non-conservative forces then W non conservative = 0 and there are no external forces so that Work is equal to zero.

    change in mechanical E = Ef = Ei

    Ef = Ei
    1/2 m vf^2 + mghf = 1/2 m vi^2 + mghi

    I assumed that the final hieght is zero and the initial height is 2.68m. I also assumed that the initial velocity is 0.

    1/2 mvf^2 + mg0 = 1/2 m 0 + mghi
    1/2 mvf^2 = 1/2 mghi
    1/2 (1.4) vf^2 = (1.4)(9.81)(2.68)

    i found vf^2 using kinematic equations = 52.5816 m^2/s^2

    36.80712 = 36.80712

    Since the final energy is 36.80712 and when it collides with the ground the energy disspates by 1/12.
    So :
    36.80712/12 = 3.06726

    So I made the the inital energy of the first bounce.

    Ei = Ef
    3.06726 = 1/2mv^f + mgh
    3.06726 = 0 + mgh
    3.06726 = mgh
    3.067/mg = h
    .22333 = h

    This is what I calculated the height to be but this is wrong.

    I'm not sure what I'm missing or if I should be looking at it in a different way.
     
  2. jcsd
  3. Mar 8, 2008 #2
    PART 1
    Potential Energy
    E=mgh
    E=(1.4)(9.8)(2.68)
    E=36.77N
    Energy Lost= 36.77/12=3.06N
    Total Energy Left= 36.77-3.06=33.71N
    The amount of 33.71N is being carried on to the next step.

    PART 2
    E=mgh
    33.71=(1.4)(9.8)(h)
    33.71=13.72h
    h=2.45m

    I am not sure that my work is correct.
     
  4. Mar 8, 2008 #3

    Dick

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    Science Advisor
    Homework Helper

    If it loses 1/12 of it's energy KE in the collision, then it's energy after the collision is (11/12)*KE, isn't it? And while I think you are doing this essentially correctly, you could do it a lot more simply. (2.68m)*m*g=(1/2)*m*v^2=KE. After the bounce the KE is (11/12)*KE. So (x)*m*g=(11/12)*KE, where x is the rebound height. So x*mg=(11/12)*(2.68m)*mg. You don't need to know m or g.
     
  5. Mar 8, 2008 #4

    Dick

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    Science Advisor
    Homework Helper

    Dead right. (11/12)*2.68~2.45. Now simplify the calculation.
     
  6. Mar 9, 2008 #5
    Thank you. I did not subtract the dissipated energy from the original. Thanks to all that helped.
     
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