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Homework Help: Momentum, man running on flatbed

  1. Jun 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Man who weighs 915 N stands on a flatcar that weighs 2640 N as it rolls at 16.0 m/s in positive x direction. No friction. Man runs in negative x direction at 53 m/s relative to the flatcar. What is the resulting increase in speed of flatcar?

    2. Relevant equations

    Velocity(man, flatcar) = Velocity(man, ground) + velocity(ground, flatcar)
    F(g) = ma(g)
    m(1i)v(1i) + m(2i)v(2i) = m(1f)v(1f) + m(2f)v(2f)

    3. The attempt at a solution

    I divided weight by 9.81 to arrive at mass(man) = 93.3 kg and mass(flatcar) = 269 kg. I then found the velocity of the man relative to the ground to be -37 m/s. Using conservation of momentum, I found that the final velocity of the flatbed to be 34.4 m/s. This led to an increase of 18.4 m/s for the flatbed.

    93.3(16) + 269(16) = 93.3(-37) + 269v(flatbed, final)
  2. jcsd
  3. Jun 6, 2009 #2


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    More simply can't you just consider the motion of the center of mass of the system?

    Won't it remain in motion undisturbed by an external force, so won't the ratio of the masses need to be inversely proportional to their velocities to maintain the constraint on the center of mass?

    915/2640 = v/53 or v = 18.4 as you found.

    Since he is running counter to the 16 m/s, the car must be moving faster by the 18.4.
  4. Jun 6, 2009 #3
    Submitted that solution online, but feedback stated that I was incorrect. Huh?
  5. Jun 6, 2009 #4
    A man (weighing 915 N) stands on a long railroad flatcar (weighing 2640 N) as it rolls at 16.0 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 53.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

    Here's a cut-n-paste of the problem from the website where it's posted.
  6. Jun 6, 2009 #5


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    Does your program want more significant digits perhaps?
  7. Jun 6, 2009 #6
    I did the calculations without any rounding and came up with an answer of 18.369 or something like that. I have until 11pm tonight to figure this out. I can't see where I went wrong...
  8. Jun 6, 2009 #7


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    I get the same thing, so it seems like you did the problem correctly. Maybe they want only 2 significant digits, because of the 53 m/s? Did you make sure to submit the answer with the proper units? Or in the proper units?

    I tend to think these online homework things are pretty stupid, because they're so inflexible - you not only have to come up with the right answer, but guess the "right" way to type it in to get credit.
  9. Jun 6, 2009 #8


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    As soon as he stops of course it goes back to 16 m/s. And it has to be a pretty long car at that speed before he must stop.

    His speed is given with 2 place after the decimal, so maybe they want 18.37 m/s as the increase.

    Edit: As diazona has noted as well.
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