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Inelastic Collision and average force between 2 trains

  1. Apr 26, 2013 #1
    I solved the question , but i need to check it with you guys ... I believe there is something wrong with part b .
    1. The problem statement, all variables and given/known data

    The 15000kg train A is running at 1.5 m/s on the horizontal tack (to the right) when it encounters a 12000kg train B running at 0.75m/s toward it (to the left) . If the trains meet and
    couple together, determine (a) the speed of both cars just after the couple, and (b) the
    average force between them if the coupling takes place in 0.8sec.

    2. Relevant equations
    m1v1 + m2v2 = (m1+m2)v3
    momentum 1 + F(avg)t = momentum 2

    3. The attempt at a solution
    a-
    v3 = (m1v1 + m2v2)/m1+m2
    v3 = 0.5 m/s to the right

    b-
    momentum 1 + F (avg) t = momentum 2
    (m1+m2)v3 + F(avg) 0.8 = 0
    F(avg) = -16875 N
     
  2. jcsd
  3. Apr 26, 2013 #2

    Doc Al

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    Staff: Mentor

    Good.

    To find the average force, just look at one of the cars. It doesn't matter which one, since the force on either will be the same magnitude.
     
  4. Apr 26, 2013 #3
    Ok.....
    then :
    -m2v2 + F(avg) t = (m1+m2) v3
    F(avg) = 28125 N
     
  5. Apr 26, 2013 #4

    Doc Al

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    Staff: Mentor

    Still not right. Stick to a single mass, all the way through. How does the velocity, and thus the momentum, of mass m2 change?
     
  6. Apr 26, 2013 #5
    Aha ... I guess I was confused ...
    for train a which is going to the right
    m1v1 - F(avg from b) t = m2v3
    F(avg from b) = (m1v1 - m2v3 ) / t
    F(avg from b) = 18750 N to the left
     
  7. Apr 26, 2013 #6

    Doc Al

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    Staff: Mentor

    I believe you have a typo. The only mass that should appear is m1.

    Good!
     
  8. Apr 26, 2013 #7
    Oh yes ... it was a typo
    Thanks very much for helping me through the problem
     
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