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Center of Mass, Man, Woman, and Frictionless Ice

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A 58 kg woman and an 78 kg man stand 8.00 m apart on frictionless ice. How far from the woman is their CM? 4.6 m

    If each holds one end of a rope, and the man pulls on the rope so that he moves 2.6 m, how far from the woman will he be now? Use two significant figures in answer. 1.9 m

    How far will the man have moved when he collides with the woman?

    2. Relevant equations

    mass man * Δx man = mass woman * -Δx woman

    3. The attempt at a solution


    This can't be right because it's too close to letter B's answer.

    I'm assuming because the ice is frictionless that they'll just keep going until the hit each other.

    (78 kg)(x) = -(58 kg)(4.6 m)

    x = -3.42 m

    8.00 m - 2.6 m - 3.42 m = 1.98 m
     
    Last edited: Mar 7, 2012
  2. jcsd
  3. Mar 7, 2012 #2

    Doc Al

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    Staff: Mentor

    Think this way: When they finally hit each other, where will they both end up? How far from their starting points is that?
     
  4. Mar 7, 2012 #3
    DocAl:

    The man starts 8.00 m apart from the woman and I knew he was 4.6 m from her CM. 8.00m - 4.6m = 3.4m

    Thank you
     
  5. Mar 8, 2012 #4

    Doc Al

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    Staff: Mentor

    I'll rephrase: The man was initially 3.4 m from the center of mass, so when they collide he will have traveled 3.4 m.
     
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