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Bowling Ball Return

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A bowling ball encounters a h = 0.662 m vertical rise on the way back to the ball rack, as the figure below illustrates.

    Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3.70 m/s at the bottom of the rise. Find the translational speed at the top.

    2. Relevant equations

    KE= 1/2mv^2

    U=mgh

    Conservation of energy and momentum?


    3. The attempt at a solution
    I tried basic conservation of energy but that didn't really work out. I'm not sure where to start on this one since it doesn't give the mass of the ball. Though that probably cancels out anyway. When I set KE = U the answer is incorrect. I'm not sure on what else I can do.
     
  2. jcsd
  3. Apr 8, 2009 #2

    minger

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    The initial kinetic energy is not equal to the change in potential energy. The initial energy is equal to the change in potential energy plus the final kinetic energy.
    (KE + U)_1 = (KE + U)_2
     
  4. Apr 8, 2009 #3
    So:

    1/2mvi2=mgh + 1/2mvf2?


    When I plug in the values for this the answer is incorrect as well.
     
  5. Apr 8, 2009 #4

    minger

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    The answer is not 0.837 m/s?
     
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