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Five hoboes jumping off a cart

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Five hoboes, each with mass m, stand on a railway flatcar of mass M. They jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite direction without friction.
    (a) What is the final velocity of the flatcar if all the men jump at the same time?
    Data: m = 78 kg; M = 280 kg; u = 3.3 m/s.

    (b) What is the final velocity of the flatcar if they jump off one at a time? Does case (a) or case (b) yield the largest final velocity of the flat car? Can you give a simple physical explanation for your answer?


    2. Relevant equations
    Pi=Pf


    3. The attempt at a solution
    The answer for part a) is 4.596m/s

    and for part b) i just use consveration of momentum starting with

    0=-mu+(M+4m)v
    v=(mu)/(M+4m)
    then continuing on:
    v=0.435 m/s
    then the next one jumps:
    m(0.435 m/s)=-mu+(M+3m)v
    and just continuing this on until I get to where there is only one person left to jump and calculate the v, and they are the same, and the question seems to imply that they should not be the same. ideas? thanks.
     
  2. jcsd
  3. Jun 20, 2011 #2

    Dick

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    Homework Helper

    Sure, the physical reason is conservation of momentum. So it doesn't matter if they jump all at once or one at a time. I think you have reached the correct conclusion.
     
  4. Jun 20, 2011 #3
    Right, that's what I thought. It seems there is no reason why the v's should differ.
     
  5. Jun 20, 2011 #4

    gneill

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    Staff: Mentor

    Careful! In the first case, where all five hobos jump at once, they are all pushing off against only the inertial resistance of the flatcar. Effectively, they don't have to accelerate each other nor share any of their momentum with them; each contributes an equal chunk of momentum to the flatcar equal to m*u.

    In the second case, each hobo but the last who jumps is giving momentum to the flatcar and the friends he left behind. The net delta-V that he produces is smaller than he produced when they all jumped together (except for the last jumper who has the same effect as before).

    If you look at it another way, in the first case when all five jump at once, all five end up with a momentum (in the rest frame) of -m*u, for a total of -5*m*u because all five end up with speed u in the rest frame. By conservation the flatcar must take up equal and opposite momentum. In the second case, each successive hobo who jumps does so from an already moving platform, and since his launch speed is with respect to that platform, his final speed in the rest frame must be less, too, so his momentum in the rest frame will be less. This should mean a lower overall contribution of momentum to the flatcar.
     
  6. Jun 20, 2011 #5

    Dick

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    Right. I ignored the 'relative to the car' part. That was stupid. Sorry.
     
  7. Jun 21, 2011 #6
    Ok, makes sense, but I am a little confused as to how to account for this in my equations?
     
  8. Jun 21, 2011 #7

    gneill

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    Staff: Mentor

    One way would be to determine the individual flatcar Δv's that take place for each hobo that jumps, and then sum them. Each hobo that jumps does so with relative velocity -u. Each hobo has mass m. So he has a fixed Δp with respect to the flatcar plus its remaining contents. So the flatcar plus remaining contents gains momentum Δp while the jumping hobo has -Δp, leaving the net change zero. You can thus determine the Δv of the flatcar plus remaining contents. You should be able to work out an expression for the Δv for the i'th jumper, for i = 1 to 5.
     
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