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Ice skaters collision with angle

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Two ice skaters crash into each other. Before they collide, one of them (50 kg) is skating in a straight line at 5 m/s, the other (40 kg) is skating at 4 m/s in a straight line at 90 degrees to the 50 kg skater’s direction. After the collision, the 50 kg skater is moving at 4 m/s at an angle of 25 degrees relative to their original direction. Calculate the velocity of the 40 kg skater after the collision. Is the total kinetic energy constant?

    2. Relevant equations
    m1v1ix+m2v2ix=m1v1fx+m2v2fx
    m1v1iy+m2v2iy=m1v1fy+m2v2fy

    3. The attempt at a solution
    I drew a diagram before collision and after collision. Really, all I need to know is where in the formula(s) would I need to include the angle and I SHOULD be able to figure it out from there.
    FullSizeRender.jpg
     
  2. jcsd
  3. Apr 23, 2016 #2

    ehild

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    You have v1fx and v1fy in the equations for components of momentum. How do you get them, knowing magnitude and angle of velocity ?
     
  4. Apr 25, 2016 #3
    Is v1fx=(4 m/s)(cos(25°)) and v1fy=(4 m/s)(sin(25°))?
     
  5. Apr 25, 2016 #4

    ehild

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    Yes.
     
  6. Apr 25, 2016 #5
    And then with that information, solve for v2fx and v2fy?
     
  7. Apr 25, 2016 #6

    ehild

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    Yes.
     
  8. Apr 25, 2016 #7
    And then once I figure out what those two velocities are, I would square them individually and take the square root of the sums? So, sqrt((v2fx)2+(v2fy)2? And that would be the answer for the first question?
     
  9. Apr 25, 2016 #8

    ehild

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    The final velocity is asked. It is a vector with magnitude and direction. I think the problem wants both the magnitude and angle.
     
  10. Apr 25, 2016 #9
    To find the angle, would it be arctan(v2fy/v2fx)?
     
  11. Apr 25, 2016 #10

    ehild

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    Yes.
     
  12. Apr 25, 2016 #11
    Alright. I can solve the second subquestion myself. Thank you again!
     
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