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Mechanics, conservation of momentum.

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Tank (mass M) contains water
    (mass m). L is distance from
    centre of mass of container to
    pipe.(x) is noted as the direction to the right. Pipe is small compared to
    tank. Tank is sitting on a frictionless surface
    Open the tap, water flows out.
    After, where is the tank and
    which way is it moving?
    Explain your answer in terms of
    forces on the tank.

    see link two posts down for picture...

    I need to find
    expression for the distance travelled
    - expression for the final velocity



    2. Relevant equations



    3. The attempt at a solution

    I assume as the water flows out, the tank would move to the left as the water is flowing out of the tank to the right so there would be an equal but opposite in direction force acting on the tank. Hence the tank moves to the left.

    But I dont know how to work out where the tank is, given There is no variables to work with.

    Please correct anything that is wrong with my logic above?
     
    Last edited: Mar 2, 2012
  2. jcsd
  3. Mar 2, 2012 #2
  4. Mar 2, 2012 #3
    why does this question seem so intimidating???
     
  5. Mar 2, 2012 #4

    BruceW

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    From the picture, it looks like the water would go straight downwards when it leaves the pipe... If the water was coming out horizontally I think you would need to use Bernoulli's equation to solve it. Have you learned that stuff yet? (Or, at least Torricelli's law?)
     
  6. Mar 2, 2012 #5
    I agree Bruce, the water looks like it leaves the pipe horizontally.
    Apart from gravity which always acts downwards are there any other forces that could influence the system that would need to considered, ie, pressure?
    Also I would assume that as mentioned in the picture the pipe connected to the tap is quiet thin compared to the tank, hence having an affect on the amount of water leaving the tank at a time, ie, the pressure would be greatest when the tank is full so this would force more water out at a given time... as the water level decreases so would the pressure, and ultimately the amount of water leaving the tank at a time.
    But would this cause a notable force on the tank?
    Do I need to consider Torricelli's or Bernoulli's law if the water is leaving VERTICALLY?
    And also when considering an expression for final velocity and distance would I need to include a time (t) in these expressions?

    Thanks for the help!
     
  7. Mar 2, 2012 #6

    BruceW

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    You mean vertically, right? If the water really does leave vertically, then the answer is very simple. (almost a trick question, really). Think about the equation for horizontal momentum in this case.
     
  8. Mar 2, 2012 #7
    Think about the equation for horizontal momentum....
    Ok so I think I have it...

    So considering gravity and the normal force are vertical and perpindicular to the motion of the system we can think of the system being closed.
    Since the tap is initially off, the total horizontal momentum of the system is zero as there is no motion in the horizontal direction (ignoring of course the jitters of the excited water particles). Hence the total momentum in the horizontal position after the water is let out must also be zero. (conservation). take the innitial system as 1 and the final system (ie water on floor 2)....
    Therefore...
    m1 (v1f) + m2 (v2f) = 0
    Hence, v1f = -(m2/m1) (v2f)
    (the negative sign indicating the tank moving to the left)

    ...So im guessing d = v1f(t)
    to find an expression for distance traveled d= v1f(t)

    Please correct me as i have a feeling I am wrong.
     
  9. Mar 3, 2012 #8

    BruceW

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    Exactly.
    This would be correct for the horizontal velocities, if the water all came out at one speed. As you said in an earlier post, the pressure of the tank changes with time, so the speed of the water leaving the tank will also change. In other words, you're going to need to use Torricelli's law and some calculus if we are assuming the water leaves horizontally. (The system is a lot like a rocket, if you think about it, where the water is the propellant).

    But if the water leaves vertically, you pretty much have the answer already. Your equations are using the horizontal velocity of the water and tank, and if the water leaves vertically, what will be its horizontal velocity?
     
  10. Mar 3, 2012 #9
    I think it should be the same
    But just swap the signs of the horizontal velocity for vertical.
    Is this assumption correct??
    How about the distance traveled, does that expression seem to fit?
    Thanks in advance!
     
  11. Mar 3, 2012 #10

    BruceW

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    Its definitely not the same for horizontal and vertical motion. There is also gravity and the normal force in the vertical direction (as you said in post 7). So if the water comes out vertically, what would happen?

    About the distance travelled, d=v1f(t) Would be correct if all the water came out at the same speed, all at once, all in the horizontal direction. (I'm assuming the (t) is just multiplying the velocity of water, right?)

    Anyway, looking at the pdf, I think it is just asking for a qualitative description of what will happen (not an exact analytical equation). So what way will the water come out, vertically or horizontally? And what will happen to the tank due to this?
     
  12. Mar 3, 2012 #11
    well....
    now I am thinking that v=gt
    so, (m1+m2)(vi)= (m1-Δm2)(vi-v(water) + Δm2(vi + v(water)).
    ????
     
  13. Mar 3, 2012 #12
    and distance
    d = (v-Δv)(Δt)
     
  14. Mar 3, 2012 #13

    BruceW

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    They just want an explanation, not an equation. Talk it through. What direction will the water come out? And so what direction is the force on the tank?
     
  15. Mar 3, 2012 #14
    No, it says in the description
    - expression for the distance travelled
    - expression for the final velocity
    - at least a few sentences explaining the answer for the velocity.
     
  16. Mar 3, 2012 #15

    BruceW

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    hmm. It says 'where is the tank and which way is it moving'. And since the pdf is an introduction to the course, I was guessing that they just want to provoke some thought. But anyway, it sounds like you want to answer the question as fully as possible.

    So, we're assuming the water comes out vertically? In this case, the answer is nice and simple (assuming the water in the tank is not being kept at high pressure).
     
  17. Mar 3, 2012 #16
    Well so I assume then I will just approach it as a conservation of energy problem.
    so...
    ke i+gpei = kef + gpef ------> (they dont mention an element of height in the prob.)...
    0 + (m1+m2)g = 1/2 (m1 -Δm2)v^2 +0
    Hence,
    √((2(m1+m2)g)/(m1-Δm2)) = vf.

    Is this the right approach??
     
  18. Mar 3, 2012 #17

    BruceW

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    are you calculating the speed of the water when it reaches the ground? what do m1, m2, Δm2 mean?
     
  19. Mar 3, 2012 #18
    m1 meaning the mass of the tank without water
    m2 meaning the total mass of the water ie when tank is full
    Δm2 meaning the changes in mass of the water.
    or should I just assume the water at a time 1 and later at a time 2.
    ie time1 - tap off (initial conditions)
    time 2- tap on all the water on the floor.
    so it would then be (m1 -m2) instead 0f (m1-Δm2)?
     
  20. Mar 3, 2012 #19
    hold on it should just be vf =√(2*m1*g)...?

    as in the kef = 1/2* m2* v^2
     
  21. Mar 4, 2012 #20

    BruceW

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    Your equation:
    So on the left hand side, initial KE is zero. That's right. And you've got m1+m2, so that's the mass of the tank with water in it, good. But you need to multiply by the centre of mass of the combined tank and water, to get the initial PE.
    On the right hand side, you've got the PE=0, but the tank itself doesn't squish itself down to the floor. So there should be the PE of the tank without water in it. And about the KE term: think about what is moving. What will be moving, the water or the tank? And then the mass you need to use is just the mass of the thing that is moving.
     
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