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Impulse-Momentum Theory

  1. Oct 23, 2009 #1
    1. A 0.280 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.650 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?



    2. Impluse-Momentum Theory: Sum of Forces * time inerval = m Vf - mVo
    Momentum = p= mV


    3. I found the final velocity of the ball when it has been dropped (4.85 m/s, I found this out using the equation y= (Vf^2 -Vo^2)/2a ), but when I tried to plug it into the IMT, I got an answer of 1.36 kg*m/s, which is apparently wrong. Where did I go wrong? I assumed the initial velocity to be zero since it was dropped from rest. Is there something I was supposed to do with the rebounded part?
     
  2. jcsd
  3. Oct 23, 2009 #2
    "Initial velocity" for the equation is not zero; it is the velocity of the ball upon hitting the ground. In this case we are interested in the change in momentum of the ball due to the net force acting upwards on it when it hits the ground, so initial momentum = instantaneous momentum of the ball as it hits the ground, and final momentum = instantaneous momentum of the ball as it rebounds from the ground.
     
  4. Oct 23, 2009 #3

    Doc Al

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    Staff: Mentor

    In this context, initial and final velocity mean the velocities immediately before and immediately after the collision. Vo = 4.85 m/s (not zero!) downward.
    Yes. Use the given data to find the final (post-collision) velocity. (Just like you found the initial (pre-collision) velocity.)
     
  5. Oct 23, 2009 #4
    Ok, so I got a final (post-collision) velocity of 3.57 m/s. Does this seem right? What would I do next? I notice we still don't have a number for the time interval, how would I go about figuring that out?
     
  6. Oct 23, 2009 #5
    The question asked for the impulse, not average force, so you don't need the time interval. (not enough information is provided to compute it in any case)
     
  7. Oct 23, 2009 #6
    So I would just solve the equation mVf -mVi?

    Which would be (.28kg)(3.57 m/s) - (.28kg)(-4.85 m/s) = 2.3576 kg*m/s?
     
  8. Oct 23, 2009 #7
    Looks correct to me :)
     
  9. Oct 23, 2009 #8
    Thank you very much, I'm not too good at this stuff...
     
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