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Impulse and Momentum problem on skaters

  1. Dec 13, 2008 #1
    [SOLVED] Impulse and Momentum problem on skaters

    1. The problem statement, all variables and given/known data

    A 50.0-kg skater is traveling due east at a speed of 3.00 m/s. A 70.0-kg skater is moving due south at a speed of 7.00 m/s. They collide and hold on to each other after the collision, managing to move off at an angle theta south of east, with a speed of v(f). Find (a) the angle theta and (b) the speed v(f), assuming that friction can be ignored.

    2. Relevant equations

    Tangent of
    Momentum = mass * volume

    3. The attempt at a solution

    I first decided to find the angle theta. I drew a picture on a separate sheet of paper (sorry, I don't have a way of posting an image here, but it should be easy to reproduce) to make the problem look easier. I used tan theta=(opposite/adjacent) but I got 66.8 degrees as my answer with 7 as the opposite and 3 as the adjacent, whereas my answer booklet says 73.0 degrees. I thought velocities could be treated as vectors.

    Also, for part b, the answer is 4.28 m/s, but I also don't know how to get this answer. I thought I could use the pythagorean theorem, but that doesn't work. It seems like I'm missing some vital concepts here. Any help would be appreciated.
    Last edited: Dec 14, 2008
  2. jcsd
  3. Dec 13, 2008 #2
    The "vital concept here" is simply conservation of momentum. we have 2 objects each moving in one direction then they collide and begin moving in a 2 dimensional path. So find the initial momentum south m*v and set it equal to the new mass of the skaters together. This will give us the first component of our final velocity vector. Then do the same with the eastward momentum. then once you have a final east velocity and a south velocity you can add them using the Pythagorean theorem.
  4. Dec 13, 2008 #3


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    Homework Helper

    Hi DMOC! :smile:
    You're finding the relative velocity! :rolleyes:

    Hint: momentums can be treated as vectors also. :wink:
  5. Dec 14, 2008 #4
    Thank you! I found both answers (well, I got 4.27 m/s for part b but that's just 0.01 off) correctly!
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