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What is the resulting increase in the speed of the flatcar?

  1. Feb 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

    2. Relevant equations


    3. The attempt at a solution

    v=13.05 m/s

    Well this is the wrong answer and I have no clue why. I know that the fact that the weights are given and not mass is not an issue since W=ma and a would cancel out on both sides so I cannot figure out why this answer is wrong. Anything else that I can try???
  2. jcsd
  3. Feb 27, 2007 #2
    my physics teacher always says when you are dealing with speed or velocity, you always use energy.
    So, since the net vertical change is 0, change in potential energy is negligible.
    So, start with this equation
    PEi + KEi = PEf + KEf
    You can rearrange to get -(deltaPE) = deltaKE. We know that deltaPE = 0, so we have this formula
    KEf - KEi = 0
    KEf = summation of final kinetic energy
    KEi = summartion of initial kinetic energy

    you can find m because w = mg and you are given w and g is known.
    Once you find m, use the above KE equations and solve for v final of flatcar, and that should give you the right answer.
  4. Feb 27, 2007 #3
    why do you have (v+18)?
  5. Feb 27, 2007 #4
    pakmingki- be careful here. Total mechanical energy is not conserved here.
  6. Feb 27, 2007 #5
    Why wouldn't it be? The only force that could possibly act against the system is friction, which is negligible.
  7. Feb 27, 2007 #6
    I have v+18 because its asking for the resulting increase in speed. do you have a suggestion of doing it another way???
  8. Feb 27, 2007 #7
    KE is not involved in this problem, only conservation of momentum.
  9. Feb 27, 2007 #8
    Looks inelastic to me....

    but it is late cause I dont see anything wrong with the work shown above. :zzz:
  10. Feb 27, 2007 #9
    using KE could be a different approach to the problem, and if there is mass and velocity involved, it could very well be a KE problem as much as a P problem.
  11. Feb 27, 2007 #10
    KE is not conserved in inelastic collisions.
  12. Feb 27, 2007 #11
    maybe so but in the section of this book it says not to use KE as KE is conserved. Thus, using KE for this problem would be ridiculous.
  13. Feb 27, 2007 #12
    if its an inelastic collision, you can use the inelastic collision equation.
    Vfinal = sum P / sum m
    where Vfinal is the velocity of the system, which is the man and the train.
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