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Elastic Collison Problem

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A 10-kg mass moving at a speed of V[tex]_{1}[/tex] in a positive direction collides elastically with a 2-kg mass which is initially at rest. The magnitude of the net force that acts on each object during the collision is shown in the figure. Find the initial velocity, V[tex]_{1}[/tex], of the 10-kg mass prior to collision, the final velocity, v[tex]_{3}[/tex], of the 10-kg mass, and the final velocity, v[tex]_{4}[/tex], of the 2-kg mass.

    I can't post the figure up, but it's a force with respect to time chart. It starts with nothing, then starts to spike upwards at a constant rate from time 1 millisecond to time 3 milliseconds, where it peaks at 10000N, then goes back down at the same rate, hitting 0N at 5 milliseconds. I've already figured the Impulse is 20 N*s from this.

    2. Relevant equations

    p = mv

    I = F[tex]/delta[/tex]T

    [tex]/delta[/tex]p = mv[tex]_{final}[/tex] - mv[tex]_{initial}[/tex]

    mv[tex]_{before collision}[/tex] = the some of mv[tex]_{after collision}[/tex]

    Possibly assorted others.

    3. The attempt at a solution

    So far, I know the Impulse is 20 NS.

    Impulse = Momentum, I do believe.


    20 = (10)V

    Making Velocity of the first object initially = 2 m/s, correct?

    That would answer the first question, but I'm not totally sure where to go from there.
  2. jcsd
  3. Mar 9, 2008 #2
    Here's a quick idea of what I tried to do...lemme know if it looks right.

    20 = (10)V[tex]_{3}[/tex] + 2(V[tex]_{4}[/tex])
    20 = .5(10)V[tex]_{3}[/tex][tex]^{2}[/tex] + .5(2)V[tex]_{4}[/tex][tex]^{2}[/tex]

    Those are the conservation of momentum and energy equations.

    I solved the momentum one for V[tex]_{3}[/tex] and got .55279 m/s. I can post steps if anyone wants to see.

    Then I plugged that into the energy one and got 1.7061 m/s. Sound about right?

    It looks like it MIGHT be accurate, but the sums of mv before and the sums of mv after aren't equal.

    I got 2(10) = 20 for before and 10(.55279) + 2(1.7061) = 8.94 for after. =/
    Last edited: Mar 9, 2008
  4. Mar 10, 2008 #3
    my answers:

    v1 + V3 = 10
    2V1 = 12
    V1 = 6 m/s
    V3 = 4 m/s
  5. Mar 10, 2008 #4
    How is V1 6m/s?

    Am I doing the impulse equation wrong?
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