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Simple question on Conservation of Momentum in 2 directions

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A firework at rest with a mass of 0.90kg explodes into 3 pieces. One piece has a mass of 0.25kg and moves horizontally at 36m/s [N]. Another has a mas of 0.35kg and moves horizontally at 32m/s [S50°W]
    Find the velocity of the third peice



    2. Relevant equations
    SOH CAH TOA
    Conservation of Momentum
    Pythagorean Theorem

    3. The attempt at a solution
    Let N and E be positive

    I first split up both the x and y components
    v1y=36m/s
    v1x=0m/s
    v2y-32cos50=-20.57m/s
    v2x=-32sin50=-24.51m/s

    Now I subtracted the 2 masses from .90kg which gave me a mass of .30kg.
    Then I used the conservation of momentum for both the x and y components, which must equal zero as the object was at rest initially
    PT=0
    0=m1v1fx-m2v2fx+m3v3fx
    0=(0.25kg)(0m/s)-(0.35kg)(24.51m/s)+(0.30kg)v3fx
    v3fx=28.59m/s

    0=m1v1fy-m2v2fy+m3v3fy
    0=(0.25)(36m/s)-(0.35)(20.57m/s)+(0.30kg)v3fy
    v3fy=6.0m/s

    To find the final velocity with the x and y components
    a2+b2=c2
    62+28.592=820
    √820=28.63
    v3f=28.63m/s

    Now we find the angle of the magnitude
    tan-1=(6/28.59)
    =11.85°

    However, this is where I'm stuck. How do I know if it is north or south or west or east? I looked at the book and my calculations are correct, but they say it is S11.9W. How do I determine this?
     
  2. jcsd
  3. Apr 22, 2015 #2

    jedishrfu

    Staff: Mentor

    Try drawing a diagram of the event and trajectories and see if things make sense from it.
     
  4. Apr 22, 2015 #3
    Thanks for the suggestion, I think there's a typo in the text as on the diagram I drew it makes sense that it will be E11.85S. Will point it out to my teacher.
     
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