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Momentum question

  1. Dec 13, 2006 #1
    1. The problem statement, all variables and given/known data
    "A white pool ball (m1 = .3kg) moving at a speed of Vo1 = 3m/s collisdes head-on with a red pool ball (m2 = .4kg) initially moving at a speed of Vo2 = -2m/s. Neglecting friction and assuming the collision is perfectly elastic, what is the velocity of each ball after the collision?"

    Underneath the question are two diagrams, a diagram labeled before (shows a white and red ball moving towards one another) and a diagram labeled after (shows the collision and the balls moving in opposite directions).

    2. Relevant equations

    p before = p after

    3. The attempt at a solution

    p before = p after
    p1 + p2 = p1' + p2'
    m1v1 + m2v2 = m1v1' + m2v2'
    (.3kg)(3m/s) + (.4kg)(-2m/s) = (.3kg)v1' + (.4kg)v2'
    .1 Ns = .3v1' + .4v2'

    This is where i'm getting stuck...Since the balls have different masses, the velocities v1' and v2' will be different, but how can you make the two variables into one, and solve for the speeds?

    Any help will be appreciated.
  2. jcsd
  3. Dec 13, 2006 #2
    Perfectly elastic collisions also conserve kinetic energy.
  4. Dec 13, 2006 #3
    i'm still kind of confused...how exactly do you relate kinetic energy and momentum. I know that KE = 1/2mv^2...so does that mean you can solve v1' and v2' independly using just KE?
  5. Dec 13, 2006 #4


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    Staff Emeritus
    Science Advisor

    No, the conservation of kinetic energy equation will provide you with a second equation for v1' and v2'. Consider it in the same way in which you have considered the conservation of momentum, i.e total KE before = total KE after. This and the conservation of momentum equation will enable you to solve for v1' and v2'.
  6. Dec 13, 2006 #5


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    Staff Emeritus
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    Gold Member

    You can use one equation to eliminate one variable by substituting into the other equation.
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