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Momentum problem with mass transfer

  1. Jul 29, 2011 #1
    1. The problem statement, all variables and given/known data
    (Kleppner & Kolenkow - Introduction to Mechanics - 3.12)
    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
    Momentum equation

    3. The attempt at a solution
    Let the mass of the freight car be m'. At some time t, the momentum of the car is given by:

    [itex]P(t)=(m'+ t \frac{dm}{dt})u [/itex]

    where u is the velocity at that time.

    At a time t+dt, the momentum would be:

    [itex]P(t+dt)=(m'+ (t+dt)\frac{dm}{dt})(u+du) [/itex]

    The momentum change:

    [itex]dP=P(t+dt)-P(t)=(m'+t\frac{dm}{dt})u+udm+(m'+t\frac{dm}{dt})du [/itex]

    Basically I'm trying to make a differential equation for u. I'm not sure if the above is correct or not, and I also need some other expression for dP to equate this to. I'm assuming that will use the fact that the sand moves at 5 m/s relative to the locomotive and car setup (since they maintain a constant distance). How should I proceed?
     
  2. jcsd
  3. Jul 30, 2011 #2

    BruceW

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    [tex] P(t)=(m'+ t \frac{dm}{dt})u [/tex]
    Yep, you got this equation right.
    To work out the rate of change of the mometum of the car, you can think of the sand as being made up of lots of small particles, and each time one of them hits the car, their momentum adds to the momentum of the car. Therefore:
    [tex] \frac{dP}{dt} = \frac{dm}{dt} (u+v) [/tex]
    (where v is the velocity of the sand relative to the locomotive).
    Now, you can differentiate your equation for momentum and equate these two equations. Then after rearranging, you will get acceleration of the car as a function of time, which you can integrate to get velocity.
     
  4. Aug 1, 2011 #3
    If I differentiate the first equation with respect to time, I get:

    [itex]\frac{dP}{dt}=(m'+t\frac{dm}{dt})\frac{du}{dt}[/itex]

    But I don't know what m' is, how will equating this expression with the one you mentioned yield a solution then?
     
  5. Aug 1, 2011 #4

    BruceW

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    That's not quite right, you have to use product rule, since the first bit in brackets contains t.
    Also, m' is just the initial mass of the car, which you are given.
     
  6. Aug 1, 2011 #5
    Oh right, forgot my own question for a bit. Anyway, the correct expression would then be:

    [itex]\frac{dP}{dt}=m'\frac{du}{dt}+u\frac{dm}{dt}+t \frac{dm}{dt}\frac{du}{dt}[/itex]

    The differential equation I got was:

    [itex]m'\frac{du}{dt}+u\frac{dm}{dt}+t \frac{dm}{dt}\frac{du}{dt}
    = u\frac{dm}{dt}+v\frac{dm}{dt}[/itex]

    Which gave the solution:

    [itex]u=v\ln(m'+t\frac{dm}{dt})[/itex]

    Does this seem reasonable?
     
  7. Aug 1, 2011 #6

    BruceW

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    Yep, that's the same as I got. Qualitatively, it means the acceleration keeps decreasing with time. Which makes sense, because the car's mass keeps increasing.
     
  8. Aug 1, 2011 #7

    gneill

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    Staff: Mentor

    How do you deal with the fact that the parameters of the ln() function appear to have units (in particular: mass)?

    What value do you get for the final velocity (at t = 100s)?
     
  9. Aug 2, 2011 #8
    I noticed that later as well, I'm not sure why that's happening. The differential equation I got was dimensionally consistent; it could be expressed as:

    [itex]\large \frac{du}{dt}=\frac{v}{m'+t \frac{dm}{dt} } \frac{dm}{dt}[/itex]

    I'm sure the integration I'm doing is right, yet why do I get the units of mass?
     
  10. Aug 2, 2011 #9

    gneill

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    Let r = dm/dt; It's a constant after all.
    [tex]\frac{du}{dt}=\frac{v}{m' + r\;t } r [/tex]
    [tex] du = \frac{r\;v}{m' + r\;t} dt [/tex]
    [tex] u(t) = \int_0^t \frac{r\;v}{m' + r\;t} dt [/tex]
    What do you get for the definite integral?
     
  11. Aug 2, 2011 #10
    [itex]\large \int_{0}^{t}\frac{rv}{m'+rt}dt= \frac{rv}{r}\small\ln{(m'+rt)}[/itex]

    The same thing?
     
  12. Aug 2, 2011 #11

    gneill

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    Staff: Mentor

    [tex]u(t) = \int_0^t \frac{r\;v}{m' + r\;t} dt[/tex]
    [tex]\large \left. u(t) = ln(m' + r\;t)v \right|_0^t [/tex]
     
  13. Aug 2, 2011 #12
    Ah right, what a silly mistake! The right answer is

    [itex]u=v\ln{(1+\frac{t}{m'}\frac{dm}{dt})}[/itex]

    Thanks!
     
  14. Aug 2, 2011 #13

    BruceW

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    Ah, yeah I made the same mistake also. thanks gneill.
     
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