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Why speed of the railroad hopper is the same? [please help]

  1. Aug 22, 2011 #1
    There is this problem and I do not understand how could this be:

    A railroad hopper car filled with sand is rolling with an initial speed of 15.0 m/s on straight,
    horizontal tracks. You can ignore frictional forces on the railroad car. The total mass of the car
    plus sand is 85,000 kg. The hopper door on the underside of the car is not fully closed, so sand
    leaks out the bottom of the car and falls to the ground. After 20 minutes, 13,000 kg of sand has leaked out. What is the speed of the railroad car now?

    The answer is 15 m/s. How could change in mass could not affect speed?

    initial momentum of the system where sand is about to start leaking:
    p_i = mv

    and final momentum:
    p_f = (m - 13000)v'

    and solving gives v' = 17.7 m/s, but book gives answer of 15 m/s, what am I missing here?

    Since most of the sand is on the ground now and do not have speed?
     
    Last edited: Aug 22, 2011
  2. jcsd
  3. Aug 22, 2011 #2
    Yes, the car lost momentum. Perhaps you could explain why do believe the speed would change.
     
  4. Aug 22, 2011 #3
    What about this force F=dp/dt = v*dm/dt + mdv/dt

    dm/dt is not 0 here, since the mass is changing, that would mean that a force acts on the railroad hopper?

    And the and which fall from the railroad hopper reached the ground sooner or later and it has no momentum, so how car could loose momentum?
     
  5. Aug 22, 2011 #4
    I think the problem assumes that there is zero force acting on the car. Rather, there is only an initial velocity: like if you pushed a toy car and let go.
     
  6. Aug 22, 2011 #5
    The answer definitely is NOT 15 m/s (assuming no external forces are acting on it). Perhaps it's a typo, but you are correct that the velocity will increase.
     
  7. Aug 22, 2011 #6
    Okay, so what is final and initial momentum of the whole system?
     
  8. Aug 22, 2011 #7
    Could you explain why exactly the velocity would increase?

    I believe the equations you posted in your original post are correct.
     
  9. Aug 22, 2011 #8
    That is the case, If I'm posted them correctly then the answer is 17.7 m/s and that means that the speed has increased, but the book gives answer 15 which means that speed does not change.
     
  10. Aug 22, 2011 #9

    rcgldr

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    Assume the sand falls onto a frictionless surface and continues to move at the same speed as the railroad hopper. Depending on where the sand exits the railroad hopper, there will be some displacement of the railroad hopper and sand, but once the sand stops pouring out the railroad car, both should have the same speed as they did at the beginning.
     
  11. Aug 22, 2011 #10

    A.T.

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    The sand didn't transfer it's horizontal momentum to the car, but to the ground

    The "whole system" would include the entire Earth if you want to consider conservation of momentum in this case. Or you simply assume that what happens between sand and ground has no significant effect on the car.
     
    Last edited: Aug 22, 2011
  12. Aug 22, 2011 #11

    A.T.

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    If the sand exits at the back, with a horizontal velocity component relative to the car, the car will accelerate of course.
     
  13. Aug 22, 2011 #12
    This assumption that the sand does not stop after falling makes it right, thanks everyone.
     
  14. Aug 22, 2011 #13

    PeterO

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    You have the answer now, but the key is WHEN did the sand lose its forward momentum - and the answer is "when it hit the ground", not "as it left the rail car"
     
  15. Aug 22, 2011 #14

    Ray Vickson

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    The equation F = dp/dt is *false* in non-relativistic mechanics with time-varying mass, but this is perhaps less well understood than it should be. Even as late as the 1990s, clarifications were being published and older errors uncovered. Google "variable mass dynamics", and download (free) the expository paper by Plastino and Muzzio.

    RGV
     
    Last edited: Aug 22, 2011
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