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Impulse and Max Force

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A 200 g ball is dropped from a height of 1.9 m, bounces on a hard floor, and rebounds to a height of 1.5 m. The figure shows the impulse received from the floor.
    What maximum force does the floor exert on the ball?

    09.P29.jpg

    2. Relevant equations

    Kinetics:
    x-xo = vot + 0.5at2
    v2 = v02 + 2a(x - xo)
    v = vo + at

    Momentum and Force:
    SF*dt = dP = m*dv

    3. The attempt at a solution

    What I did was calculate the velocity of the ball just as it hit the ground (-6.10m/s) and just as it rebounded (5.42m/s).
    Then, using these I found the dv (5.42 + 6.10 = 11.52m/s) and plugged that into the momentum equation:
    dP = m * dv
    dP = 0.2 * 11.52 = 2.304 = SFdt

    From here I don't know how to calculate the maximum force, as I know virtually nothing about integrals. Also, my work so far is probably wrong somewhere so some help would be greatly appreciated! :)
     
  2. jcsd
  3. Nov 21, 2009 #2
    I'm in mechanics and i have not seen a question about max force. But i'll try this. From the graph we can see the Impulse which is the area under the curve is (1/2)(0.005s)F_max
    This is equal to the change in momentum which we can calculate.

    So F_max = 2*(p2-p1)/0.005

    Where p2-p1 = m(v2-v1)
     
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