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Elastric Collisions: Shortcut to Solving

  1. Jan 14, 2008 #1

    bfr

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    I given to objects of masses m1 and m2 and initial velocities v1 and v2, I know how to find out out v1' and v2', their velocities after the collision by settings up simultaneous equations. However, that takes a long time to solve, and there is a shortcut that my teacher told my class that I wrote down on a sheet of paper but then lost. I tried searching Google and haven't gotten any luck yet. I remember the shortcut was something like:

    v1'=2((m1v1+m2v2)/(m1+m2))-v1
    v2'=2((m1v1+m2v2)/(m1+m2))-v2

    But I'm not sure. Would anybody know what the shortcut is?
     
  2. jcsd
  3. Jan 14, 2008 #2
    I just remember this equation for elastic collisions:

    [tex]v_1 + v_1' = v_2 + v_2'[/tex]
    or
    [tex]v_1 - v_2 = -(v_1' - v_2')[/tex]

    Then it's fairly simple to plug this into the conservation of momentum equation to find [tex]v_1'[/tex] or [tex]v_2'[/tex]. (solve for [tex]v_1'[/tex] and plug in to get [tex]v_2'[/tex] and vice versa.)

    This doesn't look anything like what you remember though.
     
    Last edited: Jan 14, 2008
  4. Jan 14, 2008 #3

    rock.freak667

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    Homework Helper

    If I remember correctly that comes from solving:
    [itex]\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/itex]

    AND

    [itex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/itex]
     
  5. Jan 14, 2008 #4
    Yeah, energy and momentum is conserved for elastic collisions.
     
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