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Conservation of linear momentum at the skating rink

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    a) At the skating rink, a girl of 35 kgs slides at a velocity of 3.0 m/s against her father. Her father, who is standing with his skates in the direction of motion, has a mass of 90 kgs. He takes hold of her and they slide together across the ice. Friction is ignored. What is their common speed?

    b) While they're gliding along at this speed, the father shoves his daughter, so that they move in opposite directions. The father still moves with the same velocity as his daughter, but in the other direction. What is their speed now.



    2. Relevant equations

    p=p0

    3. The attempt at a solution

    a) I have calculated their speed while moving as one body : [itex]p=p_0\Leftrightarrow (m+M)v=mv_0+0\Leftrightarrow v=(mv_0)/(m+M)=\frac{35 kg\cdot 3.0 m/s}{35 kg+90 kg}=0.84 m/s[/itex]. This complies with the answer key of my book.

    b) I don't manage to calculate the right answer here, the answer key of my book says it shall be approximately 1.9 m/s. Could you give me some directions?
     
  2. jcsd
  3. Jan 17, 2012 #2

    Doc Al

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    Staff: Mentor

    Momentum is still conserved. See if you can express the given conditions mathematically.
     
  4. Jan 17, 2012 #3
    I see

    I must solve:
    [itex](m+M)v=Mv_2 +m(-v_2)[/itex] for v_2

    When I posted the original question I had tried to "conserve" kinetical energy and solve it for velocity. :lol:
     
  5. Jan 17, 2012 #4
    Thanks a lot!
     
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