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Force exerted by a wall

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Above you had a ball travelling horizontally at 25.0 m/s bouncing off a brick wall and rebounding at 20.0 m/s. The ball was in contact with the wall for 3.50 ms, and you found the magnitude of the average acceleration during the time the ball was in contact with the wall. Now, what was the average force that the wall exerted on the ball during this collision, if the mass of the ball is 180

    2. Relevant equations

    Where it says above all I solved a question asking for acceleration which was 1428.57 m/s^2

    3. The attempt at a solution
    Not sure.. my only guess is that it goes with a free fall equation or an x not known since the ball was throw horizontally
     
  2. jcsd
  3. Mar 13, 2013 #2

    Simon Bridge

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    Hint: Newton's laws.
     
  4. Mar 14, 2013 #3
    [tex]\vec{F}=\frac{d\vec{p}}{dt}[/tex], this is related to impulse.

    Impulse is the change in momentum.
     
    Last edited: Mar 14, 2013
  5. Mar 14, 2013 #4

    haruspex

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    But that's the wrong answer for the scenario in the OP.
    Presumably you arrived at that by dividing the change in velocity by the time taken, Δv/Δt.
    Average force is defined similarly: change in momentum divided by time taken.
     
  6. Mar 14, 2013 #5
    I probably should have posted my optional answers:

    2.32 * 10^3 N

    1.61 * 10^3 N

    952 N

    4.06 * 10^4 N


    I'll post my attempt later tonight when I am home.
     
  7. Mar 15, 2013 #6

    haruspex

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    But you need to get the average acceleration right first. it's nearly ten times the figure you posted.
     
  8. Mar 17, 2013 #7
    1.43 * 10^3 was the answer for acceleration
     
  9. Mar 17, 2013 #8

    haruspex

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    Are you saying that's the answer you got or that it is the correct answer? And are you saying it is the average acceleration for this scenario?:
    (20+25)/.0035 = 1.29*104.
    You appear to have calculated (25*2)/0.035.
     
  10. Mar 17, 2013 #9
    Ah thanks for the correction. So.. this makes F = Ma easier to use correct?

    1.29*10^4*180 = 2.32*10^3 N since mass and acceleration are known
     
  11. Mar 17, 2013 #10

    haruspex

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    That's it.
     
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