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Momentum problem

  1. Jul 17, 2006 #1
    Two balls A and B having different but unkown masses collide. A is initially at rest and B has a speed v moving North. After collision B has a speed v/2 and moves perpendicular (East) to the original motion. Find the direction of motion of A after collision. Is it possible to determine the speed of A from the information given? Explain.

    What I did was I drew out a grid with x and y axis, and I figured that the y-axis would have to cancel out while momentum is conserved in the x-axis, so the motion is going to be somewhere in hte lower left quadrant. But after that, I'm stuck.
     
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  3. Jul 17, 2006 #2

    Andrew Mason

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    From conservation of momentum:

    [tex]m_a\vec{v_{af}} + m_b\vec{v_{bf}} = m_b\vec{v_{bi}}[/tex]

    So:

    (1) [tex]m_av_{af}\cos\theta = -m_bv/2[/tex]

    (2) [tex]m_av_{af}\sin\theta = m_bv[/tex]

    Dividing 2 by 1:

    [tex]\tan\theta = -2[/tex]

    So you know the angle.

    If you also assume that there is conservation of energy, you can determine the speed of a in terms of the relative masses:

    [tex]\frac{1}{2}m_bv^2 = \frac{1}{2}m_bv^2/4 + \frac{1}{2}m_av_a^2[/tex]

    [tex]m_av_a^2 = \frac{3}{4}m_bv^2[/tex]

    which gives us:

    [tex]v_a = \sqrt{\frac{3m_b}{4m_a}}v[/tex]

    But I think that is as far as you can go unless you know the relative masses.

    AM

    Edit: tan = -2
     
    Last edited: Jul 18, 2006
  4. Jul 17, 2006 #3

    Office_Shredder

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    You've got it backwards there. Assuming north is positive y-axis, momentum must be conserved vertically, and horizontally it must cancel out. So the new momentum must be in the upper left quadrant.

    Does that help at all?
     
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