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Collisions and conservation of momentum

  1. Oct 20, 2008 #1
    1. A railroad car of mass 28400kg moving at 2.33 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.06 m/s. What is the speed of the three coupled cars after the collision? Answer in units of m/s



    2. m1v1i +m2v2i = (m1 +m2) Vf



    3. I thought that since this is an inelastic collision, once the three rail road cars stuck together, their final velocity would be the same. So using the equation above, and solving for Vf :
    28400 kg (2.33 m/s) + 2x 28400 kg (1.06 m/s) = ( 85200 kg) Vf
    66172 + 60208 = 85200 Vf
    Vf= 0.7767 m/s
    But this isn't correct = ( Any help please?
     
  2. jcsd
  3. Oct 20, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks good.
    Redo that last calculation.
     
  4. Oct 20, 2008 #3
    28400 kg (2.33 m/s) + 2x 28400 kg (1.06 m/s) = ( 85200 kg) Vf
    66172 + 60208 = 85200 Vf
    Okay! now I got 1.483 m/s, which is correct! Thank you!

    There is a second part though which has also stumped me.
    It asks, how much KE is lost in the collision, answer in units of J?

    KEf-KEi = change in KE

    KEi = (.5) m1v1^2 + (.5)m2v2^2
    KEi = (.5) 28400kg (2.33m/s ^2) + (.5) 2x 28400kg (1.06 m/s^2)
    KEi = 77090.38 + 31910.24 = 10900.62 J


    KEf = (.5) (m1 +m2)Vf
    KEf = (.5) (85200kg) 1.483 m/s = 63190 J

    Therefore chanGE IN KE = 63190 J - 10900.62 J
    But I seem to be missing something...
     
  5. Oct 20, 2008 #4

    Doc Al

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    Staff: Mentor

    Check that last step. (Decimal point!)
    Don't forget to square the speed.

    Also, how much is "lost" will be KEi - KEf.
     
  6. Oct 20, 2008 #5
    Thanks! I was confusing "change in KE" with how much is "lost" I squared the speed and subtracted KEf from KEi and got 15310.92 J!
     
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