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Find Coefficient of Restitution with just height of the ball after one bounce

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A ball is dropped from a height of 9.02 m. After hitting the ground, the ball then rebounds to a height of 1.78 m. What is the coefficient of restitution associated with the ball and ground impact?

    2. Relevant equations
    e=(V'-v')/(V-v)

    3. The attempt at a solution
    I thought maybe this was a trick question and you just do 9.02-1.78=7.24 and that was the answer but I am wrong... I need a bit of guidance on how to proceed with this question.

    EDIT:
    I just realized I should be using e=v'/v instead since there is only one ball... And I think I have realized that v' is -9.81? Not completely sure though because I also think that that might only be acceleration... CONFUSED lol
    I thought that maybe I could use acceleration to find velocity, but I don't have the time....
     
  2. jcsd
  3. Feb 17, 2016 #2

    haruspex

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    Yes, that's the acceleration. What SUVAT equation do you know that relates acceleration, distance and speeds?
     
  4. Feb 17, 2016 #3
    vf2=vi2+2ad perhaps?
    Thanks for the reply, I always seem to forget about the SUVAT equations...

    vf2=vi2+2ad
    vf2-vi2=2(9.81)(9.02)

    I used 9.02m and not 1.78m because it travelled down 9.02m when gravity was acting on it and I don't know what it's acceleration upwards is. Hopefully this is correct!

    vf2-vi2=2(9.81)(9.02)
    vf2-vi2=176.9724
    vf-vi=13.30
    Δv=13.30m/s

    Is the above something I can do? So now I have the change in velocity of the ball...
    But only on the way down correct? or is this all together?
     
  5. Feb 17, 2016 #4

    haruspex

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    You have found the landing speed, yes. A few doubtful steps along the way, though.
    You should be careful with signs. If up is positive, the acceleration is -9.81m/s2, and the distance is -9.02m. That way you get a positive Δ(v2).
    Secondly, you cannot go straight from vf2-vi2 to vf-vi by square rooting. You need to plug in the value for vi (0) first.

    Now you need to find the rebound speed by the same method.
     
  6. Feb 17, 2016 #5
    How do I get the acceleration for the rebound speed though? Since it's travelling upwards it's not gravity right? And I don't have enough information to solve for acceleration with Δv/Δt.
     
  7. Feb 17, 2016 #6

    haruspex

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    why not? Does gravity stop acting on a body just because it is moving upwards?
     
  8. Feb 17, 2016 #7
    No, I guess gravity is causing it to slow down as it approaches its peak height! Okay I understand that part now, thanks!

    I solved for the rebound speed and got 5.91m/s.

    e=v'/v
    e=5.91/13.30
    e=0.44 which comes up as correct!

    This turned out to be pretty simple, thanks for helping me understand how to go about it haruspex! :)
     
  9. Feb 17, 2016 #8

    haruspex

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    Well done.

    By the way, if you work it symbolically instead of plugging in numbers you will find the relationship between e and the height ratio. You don't actually need to calculate the velocities.
     
  10. Dec 15, 2017 at 2:15 PM #9

    haruspex

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    How about you try what I suggested, solving the problem but entirely symbolically, i.e. not plugging in any numerical values.
     
  11. Dec 15, 2017 at 3:20 PM #10

    haruspex

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    There was no sarcasm, none at all. It was a serious suggestion for what would be far more useful to you than my merely providing an equation.
    Indeed, that is how these forums work - we provide hints, corrections, advice, but require the student to show effort. You had effectively asked me to violate forum rules.
     
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