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Momentum and Isolated Systems

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A 68.6 kg archer stands at rest on frictionless ice and fires a 0.46 kg arrow horizontally. He fires and slides back -0.37m/s. He fires another arrow while he is still sliding that is going 76m/s. What will be his velocity after he fires the second arrow?


    2. Relevant equations
    m1v1f + m2v2f = -0.37

    3. The attempt at a solution
    I solved for v1f and plugged in the numbers to get

    v1f = -.37/68.6 - .46(76)/68.6 = -.515
     
  2. jcsd
  3. Apr 10, 2009 #2

    LowlyPion

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    That doesn't look quite right. Just check your units. You have an equation in which momentum = velocity?

    Maybe figure the speed at which the initial arrow was shot? And then the total momentum of both arrows will yield the reaction momentum of the archer and hence his velocity?
     
  4. Apr 11, 2009 #3
    since the archer is moving -0.37m/s should I multiply that by the mass of the archer so I get momentum on the right side of the equation? The velocity of the first arrow is 54.8 m/s
     
  5. Apr 11, 2009 #4

    LowlyPion

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    I think it amounts to the same thing. Whether you start with no momentum and then figure the sum of the 2 momenta as one, or figure them serially.

    There is the more subtle treatment of the mass of the arrows as to whether he was holding the second when he shot the first and hence his mass is reduced with the second shot. But I wouldn't put that fine a point on it.
     
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