Mechanics, conservation of momentum.

In summary, the problem involves a tank of water (mass m) on a frictionless surface, with a pipe (small compared to the tank) connected to a tap at a distance L from the center of mass of the tank. When the tap is opened, the water flows out either vertically or horizontally. If the water flows out horizontally, the tank will move to the left due to an equal and opposite force from the water. If the water flows out vertically, gravity and the normal force will also contribute to the tank's motion. The distance traveled by the tank can be calculated using the equation d = v1f(t), but this only applies if the water comes out at a constant speed in one direction. The exact solution would
  • #1
sg001
134
0

Homework Statement



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

Homework Equations


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 don't 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?
 
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  • #3
why does this question seem so intimidating?
 
  • #4
sg001 said:
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.

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?)
 
  • #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!
 
  • #6
sg001 said:
I agree Bruce, the water looks like it leaves the pipe horizontally.

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.
 
  • #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 I am 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.
 
  • #8
sg001 said:
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).
Exactly.
sg001 said:
Therefore...
m1 (v1f) + m2 (v2f) = 0
Hence, v1f = -(m2/m1) (v2f)
(the negative sign indicating the tank moving to the left)

...So I am guessing d = v1f(t)
to find an expression for distance traveled d= v1f(t)
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?
 
  • #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!
 
  • #10
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?
 
  • #11
well...
now I am thinking that v=gt
so, (m1+m2)(vi)= (m1-Δm2)(vi-v(water) + Δm2(vi + v(water)).
?
 
  • #12
and distance
d = (v-Δv)(Δt)
 
  • #13
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?
 
  • #14
BruceW said:
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?

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.
 
  • #15
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).
 
  • #16
Well so I assume then I will just approach it as a conservation of energy problem.
so...
ke i+gpei = kef + gpef ------> (they don't 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??
 
  • #17
are you calculating the speed of the water when it reaches the ground? what do m1, m2, Δm2 mean?
 
  • #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)?
 
  • #19
hold on it should just be vf =√(2*m1*g)...?

as in the kef = 1/2* m2* v^2
 
  • #20
Your equation:
sg001 said:
0 + (m1+m2)g = 1/2 (m1 -Δm2)v^2 +0
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.
 
  • #21
OK I think I finally have it..

vf= √(2((m+M)g(L) - (MgL))/m) m/s

??

and is my assumption correct about it traveling endlessly towards the right because of the frictionless surface?

Thanks!
 
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  • #22
Sorry not towards the left but towards the right!
 
  • #23
I thought we were assuming that the water came out vertically? That was the whole idea of the equation - the water loses some GPE, but gains some kinetic energy instead. In fact, you could imagine that there was no tank at all, and just see what happens to a cube of water which is falling.

But are you saying that you want to assume the water comes out horizontally instead?
 
  • #24
Ok
so the equation is correct but the tank does not move?
 
  • #25
The equation is basically saying that there is a cube of water which starts from rest, and falls under gravity. (As long as m is the mass of the water, M is the mass of the empty tank, and L is the height through which the water has fallen). So vf gives the vertical speed of the water, and you're correct, the tank is not moving.

The equation is correct for this physical model. But it is another thing to ask if the physical model will accurately reflect what will happen. If the 'tap' is really wide (e.g. the bottom of the tank is effectively just a trap door), then the physical model is an accurate reflection of what happens. But In the question, the tap is actually said to be much smaller than the area of the tank. So the water will come out gradually.

Anyway, you've now worked out that the tank does not move. And that's what the question was asking for. You could also work out what is the flow rate and velocity of the water, if you wanted the practise.
 
  • #26
BruceW said:
The equation is basically saying that there is a cube of water which starts from rest, and falls under gravity. (As long as m is the mass of the water, M is the mass of the empty tank, and L is the height through which the water has fallen). So vf gives the vertical speed of the water, and you're correct, the tank is not moving.The equation is correct for this physical model. But it is another thing to ask if the physical model will accurately reflect what will happen. If the 'tap' is really wide (e.g. the bottom of the tank is effectively just a trap door), then the physical model is an accurate reflection of what happens. But In the question, the tap is actually said to be much smaller than the area of the tank. So the water will come out gradually.

Anyway, you've now worked out that the tank does not move. And that's what the question was asking for. You could also work out what is the flow rate and velocity of the water, if you wanted the practise.
But from the diagram L is a distance not a height.

BruceW said:
Anyway, you've now worked out that the tank does not move. And that's what the question was asking for. You could also work out what is the flow rate and velocity of the water, if you wanted the practise.
Also, how would I work out flow rate?
 
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  • #27
Just to refresh
the objective was to find...
- expression for the distance traveled = 0
- expression for the final velocity = vf equation
- at least a few sentences explaining the answer for the velocity.
 
  • #28
sg001 said:
But from the diagram L is a distance not a height.
Yes, you shouldn't have used L in the equation, you should have used h, where h is the vertical distance that the centre of mass of the water has moved. In fact, I think they put that bit in about L just as a 'red herring'. If you think about inviscid, incompressible fluid flow, then do you see that the horizontal section of pipe L doesn't affect the outcome?

sg001 said:
Also, how would I work out flow rate?
Usually in these kinds of situations, we should assume inviscid, incompressible fluid flow. What equations do you know for this type of flow? (Or, we could say its a more general type of fluid flow, but then the equations get more complicated).

sg001 said:
Just to refresh
the objective was to find...
- expression for the distance traveled = 0
- expression for the final velocity = vf equation
- at least a few sentences explaining the answer for the velocity.
Yes, distance traveled is zero, since we assume the water comes out vertically. And for final velocity, your vf equation is one possible model for the speed of the fluid. But I think the question is asking for the speed of the tank.
 
  • #29
Yes the speed of the tank is what I have been trying to work out
so what changes do I have to make to this equation

vf= √(2((m+M)g(h) - (Mgh))/m)

= √(2gh)

So that it satisfies the speed of the tank?
 
  • #30
I think this is right for the speed of the tank
vf =√((2mgh)/M))
But one thing that confuses me is how can this be the speed of the tank if we found out the tank does not move??
Or was that caused by the previous eqn for vf of water.
 
  • #31
I have thought about this problem a bit and it seems to me that the tank should move to -x direction with a constant velocity(conservation of momentum) until the water hits the vertical part of the tap after which the tank would stop moving.
 
  • #32
sg001 said:
But one thing that confuses me is how can this be the speed of the tank if we found out the tank does not move??

Exactly. The tank does not move, so this tells you that the speed of the tank is zero.
 
  • #33
i checked with my physics instructor and the answer is not zero the tank does have a final velocity!
now I am really confused.
help!
 

Related to Mechanics, conservation of momentum.

1. What is mechanics?

Mechanics is the branch of physics that deals with the study of motion and the forces that cause it.

2. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant, regardless of any external forces acting on the system.

3. How is momentum conserved in a closed system?

In a closed system, the total momentum remains constant because any change in momentum of one object is balanced by an equal and opposite change in momentum of another object within the system.

4. What are some real-life examples of conservation of momentum?

Some examples of conservation of momentum include a billiard ball breaking a rack, a rocket launching into space, and a car crash.

5. How is conservation of momentum related to Newton's Third Law?

Conservation of momentum is related to Newton's Third Law, which states that for every action, there is an equal and opposite reaction. This means that when two objects interact, the total momentum of the system remains constant because the forces acting on each object are equal and opposite.

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