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Average force the floor exerted on the ball

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball of mass 200 g is released from rest at a height of 2.00 m above the floor and it rebounds straight up to a height of 0.900 m. (a) Determine the ball's change in momentum due to its contact with the floor. (b) If the contact time with the floor was 0.0950 seconds, what was the average force the floor exerted on the ball, and in what direction?



    2. Relevant equations
    Δp=m(v-vo)
    Favg=Δp/Δt
    w=mg


    3. The attempt at a solution

    The answer to a) is 2.09 kg x m/s, and the answer to b) is 24.0 N upward.

    I understood everything except for a portion of b. I used Favg=Δp/Δt=2.09 kg x m/s/.0950s to calculate 22.0 N, but you're supposed to add weight to that as well. That is where I get stuck. I don't understand why you're supposed to add w=(.200kg)(9.8 m/s) to the 22 N.

    Can someone draw a picture of this and also explain why you add g/why it's upward?

    My book says "The floor also needs to support the weight of the ball during impact."

    I just don't get that :confused:

    Thanks so much!
     
  2. jcsd
  3. Oct 21, 2008 #2

    PhanthomJay

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    When you solve for F_avg in your rate of momentum change equation, which is newton's 2nd law, don't forget that what you calculate for F_avg , 22N, is the NET force acting on the ball. The net force on the ball consists of both its weight acting down on it, and the normal average contact force of the floor acting up on it. Draw a FBD of the ball in contact with the floor and note the direction of the forces.
     
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