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Speed of freight car (momentum transfer)

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data

    This is problem 3.9 from Kleppner & Kolenkow's Intro to Mech.

    A freight car of mass M contains a mass of sand of mass m. At t = 0, a force is applied in the direction of motion, and simultaneously sand is let out of the bottom of the car at a rate dm/dt. Assuming the car is at rest at t = 0, find the speed of the freight car when all the sand is gone.

    2. Relevant equations

    $$\vec{F} = \frac{\mathbb{d}\vec{p}}{dt}$$
    $$\vec{p} = m\vec{v}$$

    3. The attempt at a solution

    I attempted to form the differential of the momentum of the system, as follows:

    $$p(t) = \big(M + m - (\frac{dm}{dt})(t)\big)v$$
    $$p(t + dt) = \big(M + m - (\frac{dm}{dt})(t + dt)\big)\big(v + dv\big)$$
    $$dp = p(t + dt) - p(t) = -\frac{dm}{dt}(v dt + t dv + dv dt) + M dv + m dv$$

    Dividing through by dt and neglecting a leftover dv, I get:

    $$\frac{dp}{dt} = \big(M + m - t \frac{dm}{dt}\big)\big(\frac{dv}{dt}\big) - v \frac{dm}{dt}$$

    Unfortunately, I have very little idea of what to do from here. Furthermore, thinking about the problem more makes me realize that I have no real idea of how to utilize the force F. It seems reasonable to place it into my last equation, but that seems a bit arbitrary to me and I can't imagine how I would even progress from there. I'm sure I'm making this more complicated than it should be, but still, some help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Oct 5, 2013 #2
    You should know that whenever mass is ejected from system,it applies a force whose magnitude is equal to=(dm/dt*v)
    v=Relative velocity of ejected part will respect to system.Here v=0..So no force will apply on system when the sand is draining although mass of system will continuously decrease.
    At any instant of time t
    F(net)=(mass at time t)*a
    F=(mg-ut)*a (u=dm/dt)
    Put a=dv/dt and integrate the above equation under proper limits
     
  4. Oct 6, 2013 #3
    I'm not sure I understand this step. Should it be F = mg - ut * a instead? And even then, where does this come from?
     
  5. Oct 6, 2013 #4

    gneill

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    Staff: Mentor

    Not in this case. The sand is just falling out the bottom and neither imparts nor absorbs an impulse to the train in the direction of motion. We can assume that the sand interacting with the ground when it hits the tracks does not alter the speed of the Earth :smile:
     
  6. Oct 6, 2013 #5
    Your expression for P(t+dt) is incorrect,which leads to further errors.

    The correct way of doing it is -

    P(t) = [M+(m-(dm/dt)t]v

    P(t+dt) = [M+(m-(dm/dt)t - dm](v+dv) + dm(v+dv)

    dP = [M+(m-(dm/dt)t](dv)

    or,dP/dt = [M+(m-(dm/dt)t](dv/dt)

    Now dP/dt =Fext =F ,

    So,F = [M+(m-(dm/dt)t](dv/dt)

    Fdt/[M+(m-(dm/dt)t] = dv

    Now,integrate with proper limits.
     
  7. Oct 6, 2013 #6
    Thank you, that makes much more sense. But now I have the problem of being unable to integrate the left hand side of the equation. I'm not entirely sure that integration is even possible, unless I'm just not seeing something crucial here.

    ##F## is a constant, right? So I can pull it out of the integral?
     
    Last edited: Oct 6, 2013
  8. Oct 6, 2013 #7
    Oh, no, I realize now how to integrate that! Thank you for all the help, everyone!
     
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