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Homework Help: Conservation of momentum railroad car collision

  1. Jun 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A railroad car of mass 27000 kg moving at 2.00 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 1.20 m/s. (a) What is the speed of the three coupled cars after the collision?


    2. Relevant equations
    mv + mv = mv + mv


    3. The attempt at a solution
    Mrc = 27000 kg
    Vrc = 2.00 m/s
    M2rc = 54000 kg
    V2rc = 1.20 m/s

    (Mrc)(Vrc)i + (M2rc)(V2rc)i = (Mrc)(Vrc)f + (M2rc)(V2rc)f
    (27000 kg)(2.00 m/s) + (54000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (54000 kg)(V2rc)f
    V2rcf=2.20 m/s
     
  2. jcsd
  3. Jun 28, 2010 #2

    rock.freak667

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    The left side of your equation is correct. Remember, the three cars couple together, so your new mass would be the sum of the masses which are all moving at the same final velocity.
     
  4. Jun 28, 2010 #3
    (27000 kg)(2.00 m/s) + (81000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (81000 kg)(V2rc)f
    V2rcf= 1.87 m/s

    or

    (27000 kg)(2.00 m/s) + (54000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (81000 kg)(V2rc)f
    V2rcf= 1.47m/s
     
  5. Jun 28, 2010 #4

    rock.freak667

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    Remember that conservation of momentum is basically that momentum before = momentum after

    The terms on the left side didn't happen before.

    This is the correct one.
     
  6. Jun 28, 2010 #5
    (b) How much kinetic energy is lost in the collision?
    KE = 1/2 mv2
    KE = 1/2 (81000 kg)(1.47 m/s)2
    KE = 87516.45 J

    what is the other KE that I need to calculate to determine the difference?
     
  7. Jun 28, 2010 #6

    rock.freak667

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    You calculated the final KE. Initially, the car is moving at 2 m/s and the couple at 1.20 m/s. So find the total KE then and find the difference.
     
  8. Jun 28, 2010 #7
    KE= 1/2 (27000 kg)(2.0 m/s)^2
    KE = 54000 J

    KE = KE= 1/2 (54000 kg)(1.2 m/s)^2
    KE = 38880 J

    KEi = 92880 J
    KEf = 87516.45 J

    KElost = 5363.55
     
  9. Jun 28, 2010 #8

    rock.freak667

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    That should be correct.
     
  10. Jun 28, 2010 #9
    it says this:
    Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.
    I reworked it and I keep getting the same exact answer, but I only have one more submission left before I will get it wrong.
     
  11. Jun 28, 2010 #10

    rock.freak667

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    From this equation, use the exact value of V2rcf and see what you get.
     
  12. Jun 28, 2010 #11
    okay that worked because the value was 1.4666666667 and I rounded it to 1.47. Thanks
     
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