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Problem involving two Ice Skaters' Positions

  1. Aug 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Two skaters on a frictionless pond push apart from one another. One skater has mass M much greater than the mass m of the second skater. After some time the two skaters are a distance d apart. How far has the lighter moved from original position.

    The answer is somehow d((M)/(M+m))


    2. Relevant equations

    conservation of momentum, delta x = v*time


    3. The attempt at a solution

    I understand the two skaters have equal but opposite momentum after they push each other.

    so M*v1 = m*v2

    and the position of is given by x2 = v2*t = M/m*v1

    d = x2 - x1

    .... whenever I try to substitute for v2*t, I get an answer of d...


    I just can't seem to get the solution they get...
     
    Last edited: Aug 18, 2010
  2. jcsd
  3. Aug 18, 2010 #2

    cepheid

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    Welcome to PF

    You can use the conservation of momentum to find the RATIO of the velocity of the heavier skater to the velocity of the lighter skater. The former will be moving more slowly than the latter by this ratio. Therefore, in the same amount of time, the distances travelled by each skater will also be related by this ratio.

    You also know that the sum of these two distances has to add up to d, which means that you have two equations in two unknowns and can solve.
     
  4. Aug 18, 2010 #3
    Thank you so much! Ratios make everything easier. I was getting bogged down by too many unknowns. I am glad I joined PF.
     
  5. Aug 18, 2010 #4

    cepheid

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    Umm, well that's basically what you were doing in your attempt anyway (you even wrote that v2 = M/m*v1, so you had the ratio, you just didn't add it to your post until after I had already replied). So I don't think I told you anything new, but if I somehow helped, then I am glad.
     
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