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Conservation of Momentum Question.

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data

    An 800 kg airplane travelling at a velocity of 120 m/s [30o N of E @ 20o above the horizon] collides with a stationary 1200 kg helicopter. If the velocity of the airplane after the collision is 65 m/s [40o N of E @ 60o above the horizon] what is the final velocity of the helicopter?

    2. Relevant equations



    3. The attempt at a solution

    I dont understand what to do with the 20 degrees above the horizon part in both the parts.
     
  2. jcsd
  3. Jan 20, 2013 #2

    SammyS

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    First, choose a coordinate system.

    The 20° angle should allow you to find a vertical and a horizontal component of the velocity of the airplane. Then break the horizontal component into an eastward component and a northward component.

    ...
     
  4. Jan 20, 2013 #3

    tms

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    The way I read it, that is the plane's attitude. That is, it is not in level flight, but has a 20 degree angle up before the collision. The collision appears to take place in two dimensions, [STRIKE]in a vertical plane.[/STRIKE]
     
    Last edited: Jan 21, 2013
  5. Jan 20, 2013 #4
    Isnt, 20 degree above the horizon same as 20 degree N of E? 20 degrees from 0 right
     
  6. Jan 20, 2013 #5

    tms

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    Nope. Angles east or west of north are in a horizontal plane. Angles above or below the horizon (itself essentially a horizontal plane) are in a vertical plane.
     
  7. Jan 21, 2013 #6

    SammyS

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    The plane of the collision may be found from the airplane's velocity vector before the collision and it's velocity vector after the collision. This plane is not vertical.
     
  8. Jan 21, 2013 #7

    tms

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    You're right, of course, and neither is it horizontal.
     
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