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New impulse and average force

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball of mass 1.5 kg is traveling downward at a speed of 10 m/s. It comes to rest in your hand in .25 s. What is the net impulse exerted on the ball? What is the average force exerted by the hand on the ball (Don't forget the gravitational force.)

    2. Relevant equations
    J = mvf-mvi
    J = F[tex]\Delta[/tex]t

    3. The attempt at a solution
    I figured out that the impulse is 15 kg m/s from the first equation. I thought that I had figured out the force exerted by the hand on the ball to be 60 N but now I'm not sure what it means by gravitational force. How does this affect the force exerted by the hand on the ball?
     
  2. jcsd
  3. Oct 23, 2008 #2

    LowlyPion

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    Over that time period that you used you also had gravity in the form of m*g acting on the hand as well.

    For instance if you had merely arrested its acceleration at the point that it contacted your hand and it traveled still at 10m/s with your hand then your hand would have been exerting a force of m*g on it to keep it from accelerating further. Not only did you do that, but you also arrested its motion all together which is the impulse force you calculated.
     
  4. Oct 23, 2008 #3
    I have to calculate another impulse for the gravitational force? So it would be something like J = (1.5 kg)(9.8 m/s2)(.25s).
     
  5. Oct 23, 2008 #4

    LowlyPion

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    No. It's just that the gravitational force was constant throughout the impulse and hence is added to the average force.
     
  6. Oct 24, 2008 #5
    Oh I see. Thank you.
     
  7. Oct 24, 2008 #6

    LowlyPion

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    No problem.

    Cheers.
     
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