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Time for ball to stop bouncing

  1. Jan 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello!
    A ball is dropped and falls to the floor (no horizontal velocity). It hits the floor and bounces with inelastic collisions. The velocity after each bounce is [itex] \mu [/itex] times the velocity of the previous bounce (here [itex]\mu [/itex] is the constant of restitution). The velocity of the first bounce is just [itex] v_0[/itex]. Find the time it takes for the ball to stop bouncing.

    2. Relevant equations
    Newtons Laws

    3. The attempt at a solution

    Well:
    I know this will turn into a convergent geometric series. Im just trying to find what that series will look like.

    using the formula [itex]h=x_0+v_0t+1/2at^2 [/itex] its easy to see that the time it takes for the ball to reach the ground is:

    [itex] h=1/2gt^2[/itex] so [itex] t=\sqrt{2h/g}[/itex].
    Using energy I also have: [itex] mgh=1/2mv_0^2[/itex] so [itex]gh=1/2v_0^2 [/itex]

    Time for the next bounce: well, the ball now has an upward velocity of [itex] \mu v_0[/itex] and the height of the first bounce is [itex] h'=\mu v_0t-1/2gt^2[/itex].

    I realize this is a simple problem but for some reason i'm not seeing it. If I solve this equation for time, (using quadratic formula) the resulting series for the times [itex]t=t_1+t_2+... [/itex] isn't geometric and actually quite complicated. Is my approach right?
     
  2. jcsd
  3. Jan 25, 2016 #2

    BvU

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    Hi,
    You don't want the height of the next bounce, but the time for the ball to reach the ground again. Easier to solve, too!
     
  4. Jan 25, 2016 #3
    I solved the quadratic for the time. Is this not the right approach?
     
  5. Jan 25, 2016 #4

    BvU

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    Should lead to the same answer - with a lot more work.
    What did you use for ##h'## ?
     
  6. Jan 25, 2016 #5
    Ahh, I see what you mean. So after the first bounce, I have:

    Time for ball to reach ground again: [itex]0= \mu v_0 -1/2gt^2 [/itex] solvig for t yields: [itex] 2 \mu v_0/g[/itex] so

    [itex]t_1 =2 \mu v_0/g[/itex]

    Time for ball the reach the ground the third time:

    [itex]t_2= 2 (\mu)^2 v_0/g[/itex]
    and so on. Is this the right direction?
     
  7. Jan 25, 2016 #6

    BvU

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    And there you have your geometric sequence !
    Make sure you have the right summation: the first t is only 'half a bounce'
     
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