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Homework Help: Varying mass/momentum transfer question

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    This is not a homework problem, just a question from an old exam I'm looking at. You have an elephant (M1) hanging on a sling, connected by a string and two pulleys to a hanging bucket of water (M2). The string doesn't stretch, so if you pull down on either the elephant or the bucket the other will move up. The elephant drinks water from the bucket at a known rate, so it's constantly gaining mass while the bucket is constantly losing mass. The problem is to find the velocity as a function of time.

    I have a copy of the solution, but I don't fully understand the first step. They write:
    d/dt(P1) = T - M1g - 2vd/dt(M2)
    d/dt(P2) = T - M2g + 2vd/dt(M2)

    I don't fully understand where they got the last term in each equation. Obviously it's due to the momentum transfer of the water as it's drunk. But I would have expected that to come in the following form:
    T-M1g = d/dt(P1) = M1dv/dt + vd/dt(M1)
    I.e. the second term is the momentum change due to varying mass. That's how I've always seen it done in the standard rocket problem, for example. But the solutions say that IN ADDITION to that term there is a force due to the mass transfer.

    Can anyone explain where the momentum transfer comes from? Is there a general equation that derives this? Is dp/dt = sum of forces still correct in this case?

  2. jcsd
  3. Feb 28, 2010 #2


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    Hi Chaosmarch! :smile:

    If this was a rocket in "zero gravity", there would be no T and no mg, and the fuel would be the water (with a change in momentum of 2v).

    Consider the system comprising the rocket and a small amount of fuel (and ignore all the fuel that's already gone) … then that is a closed system with no external forces, so you can use F = ∑ dp/dt

    Similarly, the F = ∑ dp/dt equation for the elephant and a small amount of water (ignroing all the water remaining in the bucket) would simply be 0 = d/dt(P1) + 2vd/dt(M2) …

    add the external forces (the T and the mg), and you get the equation given. :wink:

    (your equation, T-M1g = d/dt(P1) = M1dv/dt + vd/dt(M1) only gives the change of momentum of what's already there, not the change of momentum of the water arriving: it's not a closed system)
  4. Feb 28, 2010 #3
    I think I understand your reasoning, but something still isn't adding up. I found another problem (with solution) in another test, as follows. An elephant on a cliff is attempting to pull up a bag of peanuts (M1) vertically with a constant force (F0). Squirrels are attempting to prevent him from doing this by jumping on the bag at a constant rate (thus increasing its mass with time). Again, we're asked for v(t).

    The solution gives the equation as follows:

    M1dv/dt + vd/dt(M1) = F0 - M1g

    From what you said, however, it seems like another term should be added to take into account not just the increasing bag mass, but also the squirrel momentum added to the 'bag system', something like:

    M1dv/dt + vd/dt(M1) = F0 - M1g - ud/dt(M2)

    Where M2 is the total mass of the squirrel population, and u is the velocity of the squirrels in the lab frame.

    So why in the first problem is there a force term due to the mass transfer, but in the second there isn't one?
  5. Mar 1, 2010 #4


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    Dunno … the book's solution takes no account of how the squirrels jump, which is obviously wrong.

    Perhaps the book means the squirrels jump up so that their speed just matches that of the penanuts?

    Your formula is more general (but it shouldn't have u, the squirrel-velocity, it should have u - v).
  6. Mar 1, 2010 #5
    Ah, that makes more sense. Thanks!
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