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Locomotive throwing sand in a car

  1. Aug 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A sand-spraying locomotive sprays sand horizontally into a freight car situated ahead of it. The locomotive and freight car are not attached. The engineer in the locomotive maintains his speed so that the distance to the freight car is constant. The sand is transferred at a rate dm/dt = 10 kg/s with a velocity of 5 m/s relative to the locomotive. The car starts from rest with an initial mass of 2000 kg. Find its speed after 100 s.


    2. Relevant equations



    3. The attempt at a solution
    I am actually not sure how to start because of the given constraint that distance between the locomotive and car stays constant.

    Do I have to begin with evaluating P(t) and P(t+dt)? If so, what should be my system? Car and the incoming sand?

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Aug 22, 2013 #2
  4. Aug 22, 2013 #3
    This basically means the mass rate and the velocity of the incoming sand are constant with regard to the car, and you can forget about the locomotive.
     
  5. Aug 22, 2013 #4
    I initially thought that I have to find the distance between the locomotive and car as a function of time, differentiate it and set the derivative to zero as the distance stays constant. Is this a wrong approach?

    I still don't understand how to make the equations. Let the mass of car (without sand) be M. At time t, let the mass be M(t) and velocity of car be v(t).
    ##P(t)=M(t)v(t)+\text{momentum of incoming sand}##
    But what do I write in place of "momentum of incoming sand"? :confused:
     
  6. Aug 22, 2013 #5
    Consider what happens during some brief time ## \Delta t ##.
     
  7. Aug 22, 2013 #6
    A mass of ##\lambda \Delta t## (##\lambda=dm/dt##) is added to the car.
     
  8. Aug 22, 2013 #7
    What happens with the car's momentum, it being the product of mass with velocity?
     
  9. Aug 22, 2013 #8
    Let the velocity of car change to ##v(t+\Delta t)##. The momentum is ##P(t+\Delta t)=(M(t)+\lambda \Delta t)v(t+\Delta t)##. Correct?
     
    Last edited: Aug 22, 2013
  10. Aug 22, 2013 #9
    Correct, but not too useful. How about expressing that in terms of ## \Delta M ## and ## \Delta v ##?
     
  11. Aug 22, 2013 #10
    ##P(t+\Delta t)=(M(t)+\Delta m)v(t+\Delta t)##
    I don't see how to introduce ##\Delta v## here.
     
  12. Aug 22, 2013 #11
    ## v(t + \Delta t) = v + \Delta v ##, no?
     
  13. Aug 22, 2013 #12
    I recently posted a thread (Sand flowing out of the car) where I had to expand v(t+dt) using Taylor series. What you wrote doesn't look like it results from Taylor series. Using Taylor series, ##v(t+\Delta t)=v(t)+v'(t)\Delta t##.
     
  14. Aug 22, 2013 #13
    Well, you are just using the more complicated approach. In the limit ## \Delta X /\Delta t = X' ## by definition. Anyway, express the momentum of the car at ## t + \Delta ## and think what its difference over ## \Delta t ## should be equal to.
     
  15. Aug 22, 2013 #14
    I am still confused, I am writing down the equations blindly. :(

    ##P(t+\Delta t)=(M(t)+\Delta m)(v(t)+\Delta v)=M(t)v(t)+M(t)\Delta v+\Delta m v(t)+\Delta m\Delta v##
    I think I can drop the last term. Do I have to calculate ##P(t+\Delta t)-P(t)##? But I still don't have an expression for P(t). Sorry if I am missing something obvious.
     
  16. Aug 22, 2013 #15
    You have the expression for ## P ##. It is ## Mv ##. What you really want to know is the change of the momentum (## \Delta P ##) during ## \Delta t ##. Then you need to figure out how that change is related to the sand flowing in.
     
  17. Aug 22, 2013 #16
    Why P is Mv at time t? I mean, there is an influx of sand at time t also and you have not taken that into consideration. :confused:

    I hope you understand what I am trying to say.
     
  18. Aug 22, 2013 #17
    Momentum is ## mv ## by definition. The influx of the sand changes the momentum of the car over time, but the instantaneous momentum is given by that simple formula.
     
  19. Aug 22, 2013 #18
    $$P(t+\Delta t)-P(t)=\Delta P=M(t)\Delta v+\Delta mv$$
    Dividing both the sides by ##\Delta t##
    $$\frac{\Delta P}{\Delta t}=M(t)\frac{\Delta v}{\Delta t}+\frac{\Delta m}{\Delta t}v$$
    Is ##\Delta P/\Delta t=0##?
     
  20. Aug 22, 2013 #19
    Read #15 and #17 carefully.
     
  21. Aug 22, 2013 #20

    haruspex

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    Of course not. The incoming sand is increasing the momentum. What momentum does the incoming sand add in time Δt? Be careful to be consistent by using an inertial frame for this.
     
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