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Conceptual understanding of moment transfer for bouncing ball

  1. Oct 29, 2014 #1
    Hi,

    I am trying to wrap my head around what happens when a ball of mass m bounces against a hard floor. I assume a system that includes the ball and the floor (and eventually the entire planet) and an elastic collision. Before the ball hits the floor it has a momentum of mv. After the bounce the ball momentum change to -mv (that is changes by -2mv) and the floor momentum increases by +2mv. This way the momentum in the system is conserved.

    At the instant just before the impact, t_a, the boll momentum is +mv and floor momentum is 0. At time t_a the ball touches the floor. According to Newton's 3rd law, the ball exerts a force F on the floor and the floor responds with a corresponding force -F (a force of equal magnitude, but in opposite direction). The ball starts compressing due to the force from the floor, loosing velocity and converting its kinetic energy into elastic (potential) energy. At the instant t_0 the ball velocity is 0 and maximum compression was achieved. Between t_a and t_0 an amount of +mv of momentum is transfered to the floor due to the force from the ball and -mv momentum is transfered from the floor to the ball due to the force from the floor. Thus, the floor momentum has increased by mv and the ball momentum is 0. After t_0, the ball's elastic energi begins to convert back to kinetic energi while the ball regains form. This results in a force F exerted by the ball on the floor and corresponding force -F from the floor that accelerates the ball upward. At instant t_b, when the ball looses contact with the floor, the floor has gained another +mv momentum and the ball lost the same amount. Consequently, the floor has now +2mv momentum and the ball has -mv momentum. Is this an accurate (albeit idealistic) explanation?

    Also, am I correct in assuming that the force F decreases to 0 between t_a and t_0 and increases from 0 to F between t_0 and t_b? I am bit conflicted about what happens at t_0. At that instant I would think the ball still exerts a force F=mg on the floor and a corresponding force is exerted by the floor on the ball. In that case, F never decreases below mg... Hopefully somebody can help with a good explanation :-)
     
  2. jcsd
  3. Oct 29, 2014 #2

    A.T.

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    Science Advisor
    Gold Member

    No, the other way around

    No, more than mg

    It does, around t_a and t_b.
     
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