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Momentum in two directions(q35)

  1. Nov 7, 2007 #1
    Hi, for this question

    This is what I have done:
    in x direction
    5*60=5*v1*cos30 + 2*v2*cos30 --(1)
    In y direction
    0=5*v1sin30 + 2*v2sin30 --(2)

    From (2), v1=-(2/5)*v2 subs into (1)

    And what I get is 300=-2*v2*cos30 + 2*v2*cos30

    What is the problem?
  2. jcsd
  3. Nov 7, 2007 #2


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    Homework Helper

    you should use:

    0=5*v1sin30 - 2*v2sin30 --(2)

    the two particles have opposite directions for y-components.
  4. Nov 7, 2007 #3
    Got it, answer is b:)

    But does it mentioned in the question...or it just common sense?
  5. Nov 7, 2007 #4
    >>But does it mentioned in the question
    What I mean is "the two particles have opposite directions for y-components."
  6. Nov 7, 2007 #5

    Doc Al

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    Staff: Mentor

    You are told that each particle has a velocity that is 30 degrees from the original direction (along the x-axis). Is it possible (considering conservation of momentum) that the two particles both move to the same side of the x-axis?

    Hint: What's the y-component of total momentum?
  7. Nov 7, 2007 #6
    I see...so it is base on observation:)
  8. Nov 7, 2007 #7

    Doc Al

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    Staff: Mentor

    It is based on conservation of momentum.
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