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Momentum and energy of rebounding balls

  1. Apr 6, 2014 #1
    Greetings,
    I was thinking about bouncing a tennis ball against a wall and how its momentum and kinetic energy would change. I asked a friend of mine and he answered that the ball would transfer more forward momentum than it had to the wall but its kinetic energy would remain constant. How is that possible? I know that the ball will strike the wall with momentum p and and bounce back with a momentum -p ignoring any forces (including gravity) that affect my ball-wall system. The energy should be conserved as the wall is assumed to be frictionless. How is it then, that the ball rebounds off the wall by transferring more momentum than it had? I assumed that this is a closed system, so therefore this should be an elastic collision. Am I wrong in doing so? And if so, how does the momentum actually reverse direction? Thanks for any answers
     
  2. jcsd
  3. Apr 6, 2014 #2

    UltrafastPED

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    The ball strikes the wall with forward momentum p_forward = mv.

    Assuming a perfect ball, it bounces backward with momentum p_backward = -mv.

    The change in momentum is (p_forward - p_backward) = (mv) - (-mv) = 2*mv.

    Hence the impulse felt by the wall is double the momentum of the ball. But if the bounce is perfect, the speed is unchanged - thus the kinetic energy of the ball is unchanged: only the direction changes.

    For imperfect balls, walls, and bounces you get something less for both.

    PS: the ball reverses direction because it is _elastic_.
     
  4. Apr 6, 2014 #3
    Thanks a lot. Makes a lot more sense
     
  5. Apr 6, 2014 #4

    BruceW

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    Also, total momentum of the earth and ball and wall is conserved. The ball has gained -2P momentum, and the earth and wall will gain 2P momentum (since the wall is firmly stuck in the earth). say the mass of earth and wall is ##M_E + M_W## this will be of the order of 10^24 kg, and the momentum 2P will be of the order of 10 kg m/s So the velocity of the earth and wall, which results from the transfer of momentum will be of the order of 10^(-24) m/s In other words, you just wouldn't notice it.
     
  6. Apr 6, 2014 #5
    I see. Very useful. Thanks
     
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