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Superball between two walls

  1. Oct 21, 2009 #1

    It is the last problem, 4.29.

    4.29 A "superball" of mass m bounces back and forth between two surfaces with speed Vo. Gravity is neglected a nd the collisions are perfectly elastic.
    a. Find the average force F on each wall.
    b. If one surface is slowly moved toward the other with speed V «v,
    the bounce rate will increase due to the shorter distance between colli-
    sions, and because the ball's speed increases when it bounces from the
    moving surface. Find F in terms of the separation of the surfaces, x.
    (Hint: Find the average rate at which the ball's speed increases as the
    surface moves.)

    Here is the link to the answer : http://physics141.uchicago.edu/2002/hw4.pdf

    This is a problem from Kleppner and Kolenkow. I have a problem with the working. It is claimed that after every collision, the velocity changes by 2V.
    My point is that initial velocity before striking the wall was Vo towards the left and after the collision, the velocity is Vo + 2V towards the right, thus making the change of velocity 2(Vo + V) and not just 2V. Can someone justify how it is that one can solve the problem as done above?
  2. jcsd
  3. Oct 21, 2009 #2


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    For the moving wall, if the ball speed is Vb and the wall speed is Vw then the collision speed is (Vb + Vw) and the rebound speed is - (Vb + 2 Vw) (using a fixed frame of reference).

    In the pdf file, V is used for the wall speed, and v is used for the ball speed, and it's noted that with each bounce cycle, the ball speed increases by 2 V, which seems to be the same as what you're asking. The distance between the walls is defined as x = L - Vt. (I swapped uppercase L for the lower case l in the pdf file). I think once you understand the variable names, then the rest of the problem solution makes sense.
    Last edited: Oct 21, 2009
  4. Oct 21, 2009 #3
    You mean velocity. Collision velocity, i.e. velocity of ball before impact, is only Vb, not Vb + Vw.
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