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Maximum force and Impulse

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A 41 rubber ball is dropped from a height of 1.8 and rebounds to two-thirds of its initial height.
    http://session.masteringphysics.com/problemAsset/1070447/4/09.P29.jpg
    The figure (Part C figure) shows the impulse received from the floor. What maximum force does the floor exert on the ball?
    t=5ms

    2. Relevant equations

    change in p = F*change in time
    F=ma

    3. The attempt at a solution

    I tried and got Fmax = .176 by plugging into the 1st equation for impulse
     
  2. jcsd
  3. Mar 1, 2009 #2

    LowlyPion

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    Homework Helper

    If you solved the other problem you posted you should know how to solve this so long as you calculated your Impulse correctly from the info given.
     
  4. Mar 1, 2009 #3
    So I tried to do F*change in t = change in m*v but it didn't work. i calculated velocity by doing change in d/ change in t. i used 2.5 for t because that is where fmax occurs and i also used d as 1.8 because that is the distance the ball travels in 2.5 seconds

    so i got v=.72 and then mv=.029
    so 2.5F = .029
    F = .0116

    and that was wrong. so i dont really know what to do?
     
  5. Mar 1, 2009 #4

    LowlyPion

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    Where did you get v = .72 from? It fell 1.8m

    V2 = 2*9.8*1.8m

    How fast does it bounce back?

    Isn't that going to give you the ΔV that you want to use?

    The 2.5 ms* Fmax looks ok though.
     
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