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Momentum, Ball being thrown on Ice

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a ball of mass 0.400kg that is traveling horizontally at 10.5 m/s . Your mass is 65.0 kg .
    If you catch the ball, with what speed do you and the ball move afterwards?


    I set the momentum before equal to the momentum after, then solved for final velocity.
    v2= (MaVa1+ MbVb1)/(Ma+Mb) and since the body isn't moving at all than the MbVb1 just is zero. you also want the 10.5 m/s into cm which is 1050 cm/s


    I keep getting 16.1 and the answer is 6.42...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 30, 2012 #2
    This seems to be a relatively simple problem, though I am getting 15.571, rather than 16.1. Given the fact that this will be an inelastic collision with the person catching the ball, I change

    m1iv1i + m2iv2i = m1fv1f + m2fv2f

    to

    m1v1 + m2v2 = m3v3

    Then,

    [STRIKE](65.0 kg)(0 m/s)[/STRIKE] + (.400 kg)(10.5 m/s) = (65.4 kg)v3
    v3 = 15.571

    Also, I may have missed something, but I do not understand why you wished to change m/s to cm/s.
     
  4. Sep 30, 2012 #3
    Im sorry, I was unclear, the problem asks us to change it to cm/s. I'm having problems because the webwork is saying that the answer is not 15 or 16...
     
  5. Sep 30, 2012 #4
    That's strange, especially considering that our answers would give 1500 and 1600 cm/s. The only way that I can think of that much of a change would be an error in units somewhere, seeing that the problem doesn't seem to want you to take energy loss into account.
     
  6. Sep 30, 2012 #5
    I'm not sure how the answer is 15.571 since I've calculated it to be 0.0642
     
  7. Sep 30, 2012 #6
    I just looked back over my work, and I divided by 4.2 instead of 65.4, and just ended up with the reciprocal. I'm not really sure how I overlooked that, but at least I figured it out. I'm guessing he did something similar, as he came out with a similar answer to what I had.
     
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