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Conservation of momentum in x and y directions

  1. Dec 14, 2008 #1
    1. The problem statement, all variables and given/known data
    An m2 = 1.7 kg can of soup is thrown upward (north) with a velocity of v2 = 6 m/s. It
    is immediately struck from the side by an m1 = 0.5 kg rock traveling at v1 = 7.8 m/s. (west)
    The rock ricochets off at an angle of 46 degrees with a velocity of v3 = 5.6 m/s.
    What is the angle of the can’s motion after the collision? Answer in units of degrees.

    can someone look over my work? I'm not getting the right answer even after 2 attempts...

    1. I broke up momentum into x and y components:
    m1v1x + m2v2x = m1v1xf+ m2v2xf
    m1v1x= m1v1xf + m2v2xf
    I did 5.6cos46 to find V1xf (rock)
    so plugging in m1= .5kg m2= 1.7kg and V1xf= 3.890086875m/s
    i got V2xf of can= 1.149974449m/s

    m1v1y + m2v2y = m1v1yf + m2v2yf
    m2v2y= m1v1yf + m2v2yf
    I did 5.6sin46 to find V1yf rock)
    so plugging in m1= .5kg m2= 1.7kg and v1yf= 4.028302882 m/s
    i got v2yf of can= 4.815205035 m/s

    2. tantheta= V2yf/V2xf
    so plugging in above numbers i got theta= 13.4318702 degrees
  2. jcsd
  3. Dec 14, 2008 #2


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    Homework Helper

    I get 13.39° angle with the West axis. (The rock is traveling west)

    What is the angle reference is it looking for?
  4. Dec 14, 2008 #3
    it doesnt say...and how did you get 13.9? did you round any of the numbers?
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