Water Flowing through U-tubes (Velocity Given Cross-Sectional Areas)

[SignaturePF]

Homework Statement

http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf
See question 23

A' = 1/2A

Homework Equations

Av = A'v'

The Attempt at a Solution

I thought this was just a simple Av = A'v' problem
which would lead to v' = 2v. But there is apparently more to it, as the answer is √2v.
It could have to do with the momentum because F_net = 0 so it is conserved.
In that case,
mv = mv but I'm confused about how to use that to solve the problem.

 

[SignaturePF]

Any thoughts?

 

[Chestermiller]

Use conservation of momentum. In terms of the fluid density, the area, and the velocity v, what is the rate of change of momentum of the fluid passing through the U tube on the left? Same question for the U tube on the right.

 

[SignaturePF]

momentum = mv
dp/dt = m/t(v)
dp/dt = (ρAv)v
dp/dt = ρAv^2 - This is the left

Right:
dp/dt = pA'v'^2
A' = .5A
dp/dt = ρAv'^2/2

The two rate changes are equal ( net forces are zero).
v^2 = v'^2/2
2v^2 = v'^2
v' = √2v

This is the right answer. Thanks.

Just a question; from where did you get the motivation to use rate change of momentum?

 

[Chestermiller]

momentum = mv
dp/dt = m/t(v)
dp/dt = (ρAv)v
dp/dt = ρAv^2 - This is the left

Right:
dp/dt = pA'v'^2
A' = .5A
dp/dt = ρAv'^2/2

The two rate changes are equal ( net forces are zero).
v^2 = v'^2/2
2v^2 = v'^2
v' = √2v

This is the right answer. Thanks.

Just a question; from where did you get the motivation to use rate change of momentum?

The problem said that the assembly of u tubes is in equilibrium. That means that the forces exerted by the fluids on the u tubes must be equal. The force exerted by each fluid on each u tube is equal to the force exerted by the u tube on the fluid. The latter is equal to the rate of change of momentum of the fluid.

 

[SignaturePF]

Ok I see where you were going.
Essentially, the conceptual link is:
Equlibrium → ƩF = 0 → F_L = F_R
F = dp/dt
dp/dt = mv/dt
Given, that v is constant, m is changing
dp/dt = dm/dt*v
dm/dt = ρAv
dp/dt = ρAv^2
And that's it right.

I'm just making sure so I don't end up memorizing scenarios.

 

[Chestermiller]

Ok I see where you were going.
Essentially, the conceptual link is:
Equlibrium → ƩF = 0 → F_L = F_R
F = dp/dt
dp/dt = mv/dt
Given, that v is constant, m is changing
dp/dt = dm/dt*v
dm/dt = ρAv
dp/dt = ρAv^2
And that's it right.

I'm just making sure so I don't end up memorizing scenarios.

Yes. Well done. I hold in very high regard a student like you who focuses on fundamentals. You can look forward to a bright future.

Chet