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Bouncing Ball Dissipative Interaction

  1. Sep 30, 2014 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    A 0.62-kg basketball dropped on a hardwood floor rises back up to 64% of its original height. If the basketball is dropped from a height of 1.6m , how much energy is dissipated in the first bounce? How much energy is dissipated in the fourth bounce?

    2. Relevant equations
    U(x) = mgh

    3. The attempt at a solution
    I answered the first part correctly, and found that the change in Energy, or dissipated energy, after the first bounce is 3.5 J. I am having trouble answering the second part however, I was able to convince myself that U(x) = 0.62-kg*9.8*(1.6*0.64^n) where n=the number of bounces by thinking about it for a while. So I found the initial potential energy to be 9.7216 J and when I calculated the final potential energy after four bounces I got ~1.63 J. Subtracting final from initial, I got an answer of ~8.1 J as the change in energy, but that is wrong. I don't know how, can anyone steer me in the right direction (preferably in the next 20 minutes (I promise I didn't procrastinate :oops:))
     
  2. jcsd
  3. Oct 1, 2014 #2

    olivermsun

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    You got the right change in energy, but that was for the total change from initial to after the 4th bounce.

    What about the change in the 4th bounce (difference of 3rd to 4th)?
     
  4. Oct 1, 2014 #3

    B3NR4Y

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    Hm, so the energy after the 3rd bounce is U(x) = ~2.54 and then after the forth bounce it would be ~1.63 so I subtract the two and get ~0.92 J
     
  5. Oct 1, 2014 #4

    olivermsun

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    Looks right.
     
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