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Bouncing ball

  1. Feb 24, 2016 #1
    1. What forces and impulse impact on a bouncing basketball?

    I understand that it is the gravitational force, when it falls, then the normal force when it is on the way back. Are there any additional powers?

    The momentum, I came to the potential energy = kinetic energy, where I got the speed. The impulse I got through F = 0.81 * 9.82 and t = 0,34s = 2,70kgm / s. I have observed that the ball bounces about 43% of the height at which it is released from. How can I link to such impulse and power?

    But what shall I write in the discussion? Why the result as it gets, aiming for an A, it must have the nuance and detail, but gets nowhere ..


    2. mv^2=mgh
    p = mv
    Imp = F*t
    F = mg

    3. I figured out the momentum via mv^2=mgh , where I got the speed. I got the impulse through F = 0.81 * 9.82 and t = 0,34s = 2,70kgm / s. I have observed that the ball bounces about 43% of the height at which it is released from. I must discuss why the result is as it is? I truly have no idea.
     
    Last edited by a moderator: Feb 24, 2016
  2. jcsd
  3. Feb 24, 2016 #2

    jbriggs444

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    Consider splitting up the process into four parts, not just two. Instead of "falling down" and "bouncing back", what about "falling down, not yet touching the ground", "smooshing into the ground", "unsmooshing from the ground" and "rising back up, after leaving the ground".

    In an ideal world, would smooshing and unsmooshing be mirror images of one another? What about in the real world?
     
  4. Feb 24, 2016 #3
    I can divide it into four parts, but why the force and impulse impact in the way they do still remains a mystery for me.

    Vaguely put, it will be pullen down due to gravity, when it touches the ground, it will continue to be pulled downwards hence compressing, the energy of compression will thereafter convert to potentiall energy and is greater than gravitional force, thus the ball will shoot backwards (v^2 = 2gd, there d is the compressiong upon the ball) because for instance on a bowling ball is very small, so is the energy - umm, does this make any sense or am I completely wrong) and last stage it's flying up.

    But again, I don't know what to write in the discussion in my lab, this is the sole result. So any help?
     
  5. Feb 24, 2016 #4

    jbriggs444

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    If you drop a bowling ball, a properly inflated basketball and an under-inflated basketball on the ground, which one smooshes most? Are you accounting for that in your analysis?
     
  6. Feb 24, 2016 #5

    haruspex

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    It would help if you were to explain what these numbers represent. It looks suspiciously as though 0.81 kg is the mass of the ball and 9.81 m/s2 is g. If so, you have calculated the impulse on the ball from the ground while it rests on the ground for 0.34 s. What about the impact velocity?
    Again, although gravity will add to the compressing, most of it comes from the impact momentum.

    Since it is imperfectly elastic, you need to think about the compression and decompression in more detail, as jbriggs suggests.
    This link might help: https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/. See section 4.
     
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