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Five train cars and their final speed

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Three identical train cars, coupled together, are rolling east at 1.13 m/s. A fourth car travelling east at 6.43 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. All 5 cars are identical. What is the speed of the five-car train?


    2. Relevant equations

    pinitial = pfinal
    m1v1+m2v2 = mtotalvfinal
    3. The attempt at a solution

    so this is what i did

    3mV+ mV= 4mV [the masses cancel]
    (1.13)+(6.43)= V
    7.56 m/s = v

    4mV+ mv = 5mv
    Vtotal+0 = vfinal
    7.56 m/s = vfinal

    can someone pls tell me what i'm doin wrong thanx :))
     
  2. jcsd
  3. May 16, 2009 #2
    Looks like this exact same problem was recently discussed here:

    https://www.physicsforums.com/showthread.php?t=155891

    I haven't read it, but it might help quite a bit. I also recommend you label your velocities like v1, v2, etc.

    Edit: Seems you noticed it already, I see your post in there now.
     
  4. May 16, 2009 #3
    You do not add the velocities. You use the conservation of momentum equation. So for the first part you would have...

    3m*(1.13 m/s) + m*(6.43 m/s) = 4m*v_final

    Then you solve for v_final. You do the same for the last part of the problem.
     
  5. May 16, 2009 #4
    thank u sooo much :)) i feel so foolish that was really straighforward..thanx again
     
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