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Conservation of Momentum and Dissipation of Kinetic Energy - AP Problem

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data

    An open top railroad car (initially empty and of mass M.) rolls with negigible friction along a straight horizontal track and passes under the sprout of a sand conveyor. When the car is under the conveyor, sand is dispensed from the conveyer in a narrow stream at a steady rate delta-M/delta-t=C and falss vertically from an average height h above the florr of the railroad car. The car has an initial speed v. and sand is filling it from time t = 0 to t = T. Express your answers to the following in terms of the given quatities and g.

    a) Determine the mass M of the car plus the sand that it catches as a function of time t for 0 < t < T.

    b) Determine the speed v of the car as a function of time t for for 0 < t < T.

    c) i.) Determine the initial kinetic energy of the empty car.
    ii.) Determine the final kinetic energy K final of the car and its load
    iii.) Is kinetic energy conserved? Explain.

    d.) Determine the expressions for the normal force exerted on the car by the tracks at the following times
    i.) before t=0
    ii.) for for 0 < t < T.
    iii.) after t=T

    2. Relevant equations

    Conservation of Momentum: m1(v1i) +m2(v2i) = m1(v1f) + m2(v2f)

    3. The attempt at a solution

    I got part A by using Calculus: CT + Mo

    But I have a question about part b. I was going through the homework archives on this website and I found a previous poster said that:

    v(t) = momentum / the mass as a function of time:
    v(t) = p/m(t) = m0v0/(m0 + Ct).

    My question is:

    Why does v(t) = momentum / the mass as a function of time?

    And is the reason why can you use m0v0 (the initial momentum) as the momentum for V(t) for the entire time from o<t<T because momentum is conserved throughout the entire time interval?

    Thanks!! :)
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2

    Doc Al

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    Staff: Mentor

    If you consider the "car + enclosed sand" as a system, what is its momentum? Does it change?
  4. Dec 8, 2007 #3
    The car + the sand as a system makes the momentum= rate of mass (velocity)

    Ahhhhhh I see! A little math makes velocity=momentum/rate of mass, thanks! :)
  5. Dec 8, 2007 #4
    For part c iii I got that when I set the inital and final kinetic energies equal to each other:

    0 =CT

    So I'm guessing that means Kinetic Energy is not conserved because CT must be a nonzero since its mass of sand falling into the railroad car.

    On part d, Fn=g(mass)

    i) Mog ii) g(Ct+Mo) mass as a function of time times acceleration of gravity iii) g(CT=Mo) since the railroad car is no longer filling up.

    Is this work correct? Thank You!! :)
  6. Dec 8, 2007 #5

    Doc Al

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    OK. But rather than set initial and final KE equal, just compare them. KE is not conserved. (The sand collides with the moving car--essentially an inelastic collision.)

    Careful here. Don't neglect the height from which the sand falls.
  7. Dec 8, 2007 #6
    I'm confused. Why does the height matter is we care about the Normal Force?
  8. Dec 8, 2007 #7

    Doc Al

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    I'll ask you this: If you step gently on a bathroom scale, what does it read? But what if you jumped down onto it from some height, would it read any different?

    Hint: Consider the change in vertical momentum.
  9. Dec 8, 2007 #8
    I'm still confused. I don't know how to apply this...although now I understand that the normal force would be greater, I don't know the physics to change this..
  10. Dec 8, 2007 #9

    Doc Al

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    What force must the car exert on the falling sand to stop its vertical motion?
  11. Dec 8, 2007 #10
    So would it be

    ii) g(Ct+Mo) + Fg of the sand falling? Whats the mass of the falling sand? Is it Ct+Mo?

    g(Ct+Mo) + g(Ct+Mo)

    so 2g(Ct+Mo) for ii

    and iii:


    :) ?
  12. Dec 8, 2007 #11

    Doc Al

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    Force = rate of change of momentum. What's the rate of change of the vertical component of the sand's momentum as it hits the car?
  13. Dec 8, 2007 #12
    I found something on the internet that said:


    I don't understand how they came to that conclusion. Any explanation would be great! =)
  14. Dec 8, 2007 #13

    Doc Al

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    Which step in that chain do you not understand? The first step is key: Force equals the rate of change of momentum.
  15. Dec 8, 2007 #14
    Okay, I think I see what they did. Thank you!
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