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Bouncing ball exercise with time of contact with the surface given

  1. Feb 26, 2013 #1
    Hello. I have an exercise (not exactly homework, it's a physics contest that we are allowed to solve at home, using all possible help we can find) that goes like this:
    A ball with mass m = 0,10 kg falls vertically on a horizontal non-movable solid surface. Speed of the ball just before the contact with the surface was v0=14m/s. The time of contact of the ball and the surface during the bounce was τ = 4.0x10-3.
    I am to calculate the speed of the ball after the bounce. These are all information given (the only other instruction is to consider the surface, which is some kind of board, to be much more massive than the ball, and to consider the ball a perfectly elastic object).
    I'm not asking for a complete solution, but I would really use some help on what to do with the time of contact with surface, as I've never seen an exercise with this given (and Internet searc didn't reveal any usable equations for this particular problem). What is the theory behind this and what equation should I use? Thanks for any help.
     
  2. jcsd
  3. Feb 26, 2013 #2
    Think about the energy of the ball. What does "perfectly elastic" mean?
     
  4. Feb 27, 2013 #3
    Yeah, I've been thinking about that. So does that mean that the ball will bounce with the same speed in opposite direction, and the time of the contact that is given has no effect on this?
     
  5. Feb 27, 2013 #4
    If the time of contact has any effect, then the velocity must be different, which means an energy change. Is this possible in the situation given?
     
  6. Feb 27, 2013 #5
    I'm not sure. It shouldn't be if the ball is perfectly elastic. But would that mean that the time of contact is given just to confuse us? :)

    EDIT: Just thinking - it says that the ball is perfectly elastic, but doesn't say anything about the surface in this regard. Could that mean that the bounce doesn't have to be perfectly elastic, although the ball is?
     
    Last edited: Feb 27, 2013
  7. Feb 27, 2013 #6
    Well, if the surface is not completely elastic, they will have to specify exactly how inelastic it is - otherwise you cannot compute anything.

    As given, the description specifies a conservative system, where total energy is conserved.
     
  8. Feb 27, 2013 #7
    Thanks. So do I get it right that the ball bounces back up with the same speed, i.e., 14 m/s, and the time of contact has nothing to do with it?
     
  9. Feb 27, 2013 #8
    Correct.
     
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