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Linear Momentum of A System of Particles

  1. Mar 1, 2010 #1
    At time t = 0, a ball is struck at ground level and sent over level ground. The figure below gives the magnitude p of the ball's momentum versus time t during the flight. (p1 = 7.0 kg·m/s and the vertical axis is marked in increments of 0.5 kg·m/s.) At what initial angle above the horizontal is the ball launched?

    http://www.webassign.net/hrw/W0170-Nalt.jpg


    I feel like this question is relatively easy, I'm just really stuck for some reason! At the highest point, I know the vertical component of momentum is 0, making the horizontal component 7. I don't know how to find the initial momentum though; I know we can probably find it using the graph (see link), and I found it to be 9.0 kg m/s, and, after I did the math, the computer told me my answer was wrong. Is there something I'm overlooking??
     
  2. jcsd
  3. Mar 1, 2010 #2

    kuruman

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    If the graph shows the magnitude of the momentum, what is the lowest possible value of this magnitude at about 2.25 s as read from the graph?
     
  4. Mar 1, 2010 #3
    It's 7 right? So, would that make the initial magnitude 5?
     
  5. Mar 1, 2010 #4

    kuruman

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    Are we talking about the same graph? The one you posted has no numbers along the momentum axis and the value at t=2.25 s is smaller than the value at t=0.
     
  6. Mar 1, 2010 #5
    Well, it's in increments of 0.5, so I just assumed the momentum axis went from 0-2.
     
  7. Mar 1, 2010 #6

    kuruman

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    Do you know for a fact that the increments are 0.5? Why not 1 or 2? Your graph does not show. Be that as it may, I think it is a fair assumption that at 2.25 s the momentum is zero. What do you think that says about the angle of projection? In other words, for what angle of projection is the momentum zero at some point of the trajectory?
     
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