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Average force on the walls

  1. May 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A ball of mass m bounces between two parallel walls, such that its velocity v is perpendicular to the walls.The collisions with the walls are totally elastic. What is the average force exerted by the ball on the walls, if the distance between the walls is d?


    2. Relevant equations



    3. The attempt at a solution
    When the ball collides with one of the walls and bounces back, the total change in momentum is ##2mv##. The average force can be calculated by
    [tex]F_{avg}=\frac{2mv}{\Delta t}[/tex]
    What can I substitute for ##\Delta t## here? :confused:

    Thanks!
     
  2. jcsd
  3. May 9, 2013 #2

    Doc Al

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    Staff: Mentor

    What's the time between collisions?
     
  4. May 9, 2013 #3
    2l/v? But why do I need the time between collisions? ##\Delta t## is the time of contact between the wall and the ball during the collision. :confused:
     
  5. May 9, 2013 #4

    Doc Al

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    I think they mean overall average, not average just during each collision.
     
  6. May 9, 2013 #5
    In my previous post, I meant to say l/v. Is that correct?
     
  7. May 9, 2013 #6

    Doc Al

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    What is l? If you mean d, then no.
     
  8. May 9, 2013 #7
    Yes, I mean d. Sorry.

    Why is it wrong? Do you mean the time taken between the collisions on the same wall?
     
  9. May 9, 2013 #8

    Doc Al

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    Yes, exactly.
     
  10. May 9, 2013 #9
    Its 2d/v then. Hence the average force is mv^2/d. Thank you Doc Al! :smile:
     
  11. May 9, 2013 #10

    Doc Al

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    Good! :approve:
     
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