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Energy loss of bouncing ball

  1. Jul 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball bounces down a uniform flight of stairs of height H , rising after each
    bounce to to the level h of the previous stair .
    the velocity of the ball before the impact is equal in all the impacts .

    what is the average power of energy loss ?

    2. Relevant equations


    3. The attempt at a solution
    2e22ssx.jpg
    first attempt :
    at points A ,B, the ball has just a potential energy, ball’s horizontal velocity seems to be
    not relevant (cancel out ).

    in A - E = mg(h+H)
    in B - E = mgh

    EA = EB + Wothers
    Wothers = mg(h+H) - mgh = mgh
    ....
    So the energy loss is actually a constant ....
    what i'm missing ?
     
  2. jcsd
  3. Jul 2, 2015 #2

    Nathanael

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    mg(h+H) - mgh = mgH

    The energy lost per cycle is constant. So you found the energy lost after one cycle but the question asked about the average rate that energy is lost, so you're not quite finished.


    Side note: no calculations are necessary to find the energy lost per cycle:
    For the velocity to be the same before each bounce, it must lose as much energy in the collision as it gains from falling down the next step. Therefore the energy lost is mgH.
     
  4. Jul 2, 2015 #3
    thanks :)
    ...mgH of course:)
    Q : So what is the average rate that energy is lost ?
    some scrawling ...
    if [itex]mgH [/itex]is the energy lost per cycle ...and if at some point it was [itex]E_0 [/itex]then after ...
    say [itex] 5 [/itex] steps it become [itex]E_0 - 5 (mgH)[/itex] ...

    so they ask me to find [itex] \frac{E_0-5(mgH)}{5} [/itex] or [itex]\frac{E_0-12(mgH)}{12}[/itex]...?
    don't really understand the question .
     
  5. Jul 3, 2015 #4

    haruspex

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    You need to use H and h to find the duration of each cycle.
     
  6. Jul 3, 2015 #5
    How ?
    the ball bounce at some parabolic curve .
    the time of flight depends on the initial vertical velocity , [itex] v_{0y} [/itex] .
    i know that [itex] |\vec v_0| =\sqrt{2gh} [/itex] ,and that it .[itex] |v_{0y}| = ?[/itex] .
     
  7. Jul 3, 2015 #6

    Nathanael

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    Not quite.
    Think about the energy between the moment after the collision and the moment when it has risen a height h. What is the change in kinetic energy?
     
  8. Jul 3, 2015 #7
    Obviously im missing something :)
    29uwuva.png
     
  9. Jul 3, 2015 #8

    Nathanael

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    What is the kinetic energy at the top of the path? Your equation assumes it's zero.
     
  10. Jul 3, 2015 #9

    haruspex

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    That is not relevant, and indeed unknown. The horizontal speed is constant, so is not involved in power loss.
    Don't worry about the velocity. A ball is thrown up a height h. How long does it take o reach its highest point? (You need a SUVAT equation that involves t, a, s and final speed of 0.). Then it falls h+H.
     
  11. Jul 3, 2015 #10

    Nathanael

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    My point was that the horizontal component of velocity cancels out in the energy equation... Therefore you can find the vertical component of velocity in terms of h (which the OP wanted to use to solve for the time).
     
  12. Jul 3, 2015 #11

    haruspex

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    Ok.
     
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