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Elastic collision.

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A 43.0 marble moving at 1.90 strikes a 30.0 marble at rest. What is the speed of each marble immediately after the collision?

    (vfx)1=?

    (vfx)2=?

    2. Relevant equations

    v2f=2 m1 v1 / (m1 + m2)

    3. The attempt at a solution

    I tried using the equation above to get the first speed and is not the right answer. I did this 1.9m/s(2(.43)/.43+.30)=2.23m/s that apparently is wrong. Any help?
     
  2. jcsd
  3. Feb 16, 2012 #2

    ehild

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    Marbles usually collide elastically. You assumed completely inelastic collision, when the marbles stick together.

    ehild
     
  4. Feb 16, 2012 #3
    so what's the formula for that?
     
  5. Feb 16, 2012 #4
    Think about what an elastic collision is... When all of the energy from on marble is transferred to another - where all kinetic energy is transferred to the next marble?

    Where kinetic energy E = 1/2 mv^2

    The initial energy of 1 and the initial energy of 2 must equal the final energy...
     
  6. Feb 16, 2012 #5

    ehild

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    The energy is not transferred to one marble to the other, but conserved, just like the momentum.

    The sum of the energies of both marbles is the same before and after collision.

    ehuild
     
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