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Simple Elastic Collision

  1. Nov 14, 2009 #1

    v3r

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    1. The problem statement, all variables and given/known data
    Particle 1 of mass m moving with speed v in the +x direction has an elastic
    collision with particle 2 of mass 3m which was originally at rest. After the collision,
    particle 2 is moving in the +x direction. What is its speed?

    2. Relevant equations
    deltap = 0
    pf1 + pf2 = pi1 + pi2

    3. The attempt at a solution
    Wouldn't it just be
    pf2 = pi1
    3mvf2 = mv1
    vf2 = v1/3

    How come the answer is v1/2 instead? Help please..
     
  2. jcsd
  3. Nov 14, 2009 #2

    Doc Al

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    Staff: Mentor

    You can't assume that the first particle stops and all its momentum goes to the second particle.

    Hint: What does elastic mean?
     
  4. Nov 14, 2009 #3

    v3r

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    Elastic means no change in internal energy
    DeltaE = 0
    DeltaK + DeltaEint = 0
    DeltaK = 0
    Kf1 - Ki1 + Kf2 = 0

    0.5mvf1^2 - 0.5mvi1^2 + 0.5(3m)vf2^2 = 0
    0.5(3m)vf2^2 = 0.5mvi1^2 - 0.5mvf1^2

    There's two unknowns, vf2 and vf1
    Is this setup wrong?
     
  5. Nov 14, 2009 #4

    Doc Al

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    Staff: Mentor

    No, it's fine. You just need another equation--conservation of momentum. Then you'll have two equations and two unknowns.

    In any collision, momentum is conserved; in an elastic collision, kinetic energy is also conserved.
     
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