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Conservation of momentum between a moving ball and a stationary ball?

  1. Apr 14, 2009 #1
    I had a squash ball going down a ramp, which collided with another squash ball of a slightly different mass at the bottom of the ramp. They then both went off the end of the ramp into a sand pit. However they both fell in the sand pit at different lengths, and so it appears that it is not a perfectly inelastic collision as they do not stay together.

    I have the speed that it took the first ball to descend the ramp, as well as the distance the first ball would travel off the end of the ramp if there was no second ball.

    Also, the combined distances of the two balls together is greater than the distance of just one ball. Im confused. Any help? It seems like partial inelastic collision?

    The ramp is set on a table and is level at the end, and the sand pit is on the floor.
     
  2. jcsd
  3. Apr 14, 2009 #2

    jambaugh

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    Remember that the balls are rolling and being hollow they have a pretty high moment of inertia to mass ratio. This will "eat up" some of the momentum in the form of angular momentum (coupled by the contact friction forces on the balls as they roll and collide). I think that the one ball rolling will have more angular momentum per linear momentum than the two balls rolling together so if their masses are equal they should show a bit more linear momentum. I think this is a dominating effect in your experiment.

    Can you try it with toy cars with velcro to make them stick on impact?

    The relevant equations are:

    Moment of inertia for hollow sphere: [tex] I = \frac{2}{5} m r^2[/tex]
    Angular momentum: [tex] L = I \omega[/tex]
    Rolling Relationship: [tex] V = r\omega[/tex]
    Total Kinetic Energy: [tex] T = \frac{1}{2}(I\omega^2 + m V^2)[/tex]

    One can work out an effective mass:
    [tex] T= \frac{1}{2}(I V^2/r^2 + m V^2) = \frac{1}{2}\left(\frac{2}{5}m + m\right)V^2[/tex]
    [tex] m_{eff} = \frac{7}{5}m[/tex]

    Use this to figure the velocity of the first ball from the potential energy going down the ramp. I'd wager you are getting an almost elastic collision but right after the collision your squash balls are redistributing energy between linear and angular momentum. Use the effective mass again from your velocity calculations after the collision to see how much energy has been lost.

    To get the velocity figure the time to drop to the floor is:
    [tex] t = \sqrt{2h/9.8}[/tex]
    where h is height of drop in meters and t is time of flight in seconds.
    Then the distance the balls travel give:
    [tex] V=d/t[/tex]

    You may or may not be loosing some energy but you will have conservation of linear momentum at the instant of collision. But then the linear plus angular momentum of the first ball will quickly become rolling momentum as will the linear momentum of the second ball.
    There may be some energy loss in this secondary process as well depending on how much the balls skid before resuming rolling. There's also the possibility of the rotation of the first ball applying some torque to the second ball. You'll just have to see how the numbers work out.

    I know all these effects are significant enough to affect certain billiard shots where you have solid balls so they certainly should have a more dramatic effect on the hollow squash balls.
     
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