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Momentum of five-car train

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Three identical train cars, coupled together, are rolling east at 3.97 m/s. A fourth car travelling east at 4.59 m/s catches up with the three and couples to make a four-car train. A moment later, the train cars hit a fifth car that was at rest on the tracks, and it couples to make a five-car train. What is the speed of the five-car train?

    2. Relevant equations

    p=mv
    Momentum is conserved


    3. The attempt at a solution

    Given- V(1,2,3)= 3.97 m/s
    v(initial of 4)= 4.59

    Find- v(1,2,3,4,5)
    p at the instant 4 hits 1,2,3
    p at the instant 5 hits 1,2,3,4

    Cart 4- P=mv
    P= 4.59m (initially)

    Cart 1,2,3
    P=mv
    P=3m(3.97)=11.91m

    When 4 collides with 1,2,3
    p(of 4)=p(of 1,2,3)
    4.59m=11.91m
    Change in momentum=11.91m-4.59m=7.32m

    When 1,2,3,4 collides with 5
    P(1,2,3,4)=7.32m
    P(1,2,3,4,5)=7.32m-mv

    Really stuck after this.:confused: ..is anything i did above incorrect?:grumpy:
    any help will be appreciated greatly :!!)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 12, 2007 #2

    Dick

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    Homework Helper

    Huh? How can 4.59m=11.91m? You are making this much too complicated. What is the total momentum of the five cars before any collisions? How should this relate to the total momentum of the 5 car train after all of the collisions? So write down an expression for the final momentum in terms of the unknown final velocity and solve for it.
     
  4. May 15, 2009 #3
    hi i was trying to do this question and i was wondering since all the masses are the same for the carts do we just ignore the masses?? then wouldnt the final velocity of the 5cars be the same as the 4cars?? can you please explain how one would go about doing this question??
     
  5. May 15, 2009 #4

    Dick

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    Homework Helper

    No, it won't be the same. Use conservation of momentum. The momentum of the 4 cars is the same as the momentum of the 5 cars, but the velocity isn't because the mass has changed. In the future, try to post a new question instead of adding onto an old thread. You'll probably get a much quicker response.
     
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