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A Student and A Bowling Ball

  1. Feb 5, 2014 #1
    Ok so when one drops a bowling ball hanging from a rope, it won't come back and smack him in the face. This I can understand, due to potential energy of gravity and conservation of energy. However when someone pushes the bowling ball, how can you show that it will rise higher than the original height and hit the person in the face? Is there a Work external or is it the Kinetic Energy starting out the Work-Energy Theorem?
  2. jcsd
  3. Feb 5, 2014 #2
    Law of Conservation of Energy

    If you push a bowling ball of mass m from some initial height, hi, so that it starts out with some initial velocity, v, then, applying the Law of Conservation of Energy, we have: PEi + KEi = PEf + KEf, or equivalently: mghi + (1/2)mvi2 = mghf + (1/2)mvf2 + E"lost". If we neglect air drag, we can discard the E"lost" term.

    We want to know what's happening when the bowling ball returns to you at hi, so our initial and final states are both taken to be at height hi, the height from which it was pushed.

    mghi + (1/2)mvi2 = mghi + (1/2)mvf2

    Subtracting the gravitational potential energy from both states reveals that the kinetic energy of the bowling ball as it returns to you is the same as that when it leaves your hands, so it will hit you at the same speed you pushed it.

    You may wonder how high the bowling ball would swing if you managed to get out of its way. To solve this, acknowledge that the bowling ball's maximum height, hmax, is attained when all of its energy is in the form of gravitational potential energy, with no kinetic energy remaining. So, mghi + (1/2)mvi2 = mghmax. Dividing both sides by m, we have ghi + (1/2)vi2 = ghmax. To get a nice expression for hmax, divide both sides by g, and then multiply both the top and bottom of the resulting expression by 2, yielding hmax = (2ghi + vi2)/(2g).
  4. Jun 8, 2014 #3
    Thank you so much, this was a great help.
  5. Jun 9, 2014 #4

    Simon Bridge

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    Science Advisor
    Homework Helper

    Welcome to PF;
    Looks a lot like homework to me - so you are lucky that someone was prepared to do all that work for you.
    The short answer is that the energy added from the push has to be higher than the energy shortfall from what is needed to bounce back to the right height.
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