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Change in momentum, impulse force calculations

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A 178.0g ball is dropped from a height of 2.99m, bounces on a hard floor, and rebounds to a height of 1.36m. The impulse received from the floor is shown below.

    (Is a picture of a graph, Y-axis is force, X-axis is time. No numbers are labelled and is simply a line starting at 0 and goes to a peak and back down to zero like a triangle)

    What maximum force does the floor exert on the ball if it is exerted for 2.00ms

    2. Relevant equations

    Impulse = Change in Momentum

    Momentum = mv
    Ui = mgy
    KE = 1/2 mv^2

    3. The attempt at a solution

    Ui = mgy
    =5.2157
    1/2 mv^2 = 5.2157
    v=7.655m/s

    (to find velocity just as ball hits the ground)

    Pi = (0.178)(7.655)
    =1.3626

    (Initial Momentum)

    Finding velocity just as it bounces back from floor
    Uf = KEi
    v=5.1629m/s

    Pf = 0.919

    (final momentum)

    Change in momentum = 0.433

    DeltaP(change in momentum) = 1/2 Fmax (0.002)

    Fmax = 443N (not correct answer) where did I go wrong?
     
  2. jcsd
  3. Nov 8, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Momentum is a vector quantity. You haven't taken into account the change in direction at the bounce.
     
  4. Nov 8, 2015 #3

    TSny

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    Homework Helper
    Gold Member

    Did you take into account that momentum is a vector quantity, so it has direction? How does the final direction of momentum compare to the initial direction of momentum?

    [EDIT: oops, my post is redundant to gneill's.]
     
  5. Nov 8, 2015 #4
    Ouch, thanks a lot that negative sign fixed things up!

    DeltaP =2.2816 got me the right answer!
     
  6. Nov 8, 2015 #5
    Yeah that fixed things up :)
     
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