Am I on the right track with this electromagnetic and fluid qs?

[MD2000]

I have two questions that I want to make sure I'm doing correctly and understanding.

The first question is how close will two protons get to each other if they are traveling at a velocity v towards each other.

I figured you can set up 1/2mv^2 = kq1q2/r and solve for r..is that correct?

The second q I'm not really sure about..I actually drew a picture which was given to me..

Basically you have the following pipe where the area on the top is twice the area on the bottom..The question asks what should h equal in order for the pressure in both pipes to be the same..

I'm guessing that the area on top will be .5V1 (using A1V1=A2V2)..then do we use Bernoilli's principle to solve for h?

 

[MD2000]

Anyone?

 

[learningphysics]

For the first question... I'm wondering about a couple of things... when the protons begin moving... is one moving at v, and the other one at rest, or are both moving at v(one moving at v in one direction, other moving at v in the opposite direction).

Also, do they begin at in infinite distance away from each other?

Assuming they begin at an infinite distance from each other and one is moving at v in one direction, and the other moving at v in the opposite direction... we know by symmetry that they'll both come to rest eventually...

yes, you can use conservation of energy... initially all the energy is kinetic energy... potential energy = 0 (since r is infinite kq1q2/r=0)... and finally it's all potential energy and no kinetic...

So initial energy = 2*(1/2)mv^2 (since you have two protons)
final energy = kq1q2/r as you said...

set them equal and solve for r.

 

[learningphysics]

Yes problem 2 looks right to me. I'm guessing you meant to say the velocity at the top is half the velocity at the bottom? using that and equal pressures you can use bernoulli to solve for h.

 

[MD2000]

Ok yea..they are moving towards each other both at v...would it really make a difference if both or one were moving? Is that 2 x .5mv^2 really necessary?

 

[learningphysics]

Ok yea..they are moving towards each other both at v...would it really make a difference if both or one were moving? Is that 2 x .5mv^2 really necessary?

Yes, because both particles have kinetic energy. Each one has kinetic energy of (1/2)mv^2. So you need to use 2*(1/2)mv^2.

If one was moving and the other was at rest, then total energy at the beginning is just (1/2)mv^2 instead of 2*(1/2)mv^2. And also, I don't think kinetic energy will ever become zero in the case where one is moving and the other starts at rest... the first proton will slow down, while the second one accelerates... so that's a tougher problem to calculate... at the moment when the two are closest together, there will still be kinetic energy... we can't assume it's 0 at that point...

 

[MD2000]

Ahh..i see..thanks..

can anyone help me with the other q?

 

[learningphysics]

Ahh..i see..thanks..

can anyone help me with the other q?

Did you use Bernouilli's principle? Can you show what you did?

 

[MD2000]

I figured on top V2=.5V1 (because it has twice the area)

Then...

P1 + DGH + .5DV^2 = P2 + DGH + .5DV^2

I figured P would need to be equal so that cancels and then .5V1 could be plugged in for V2 and solve for H? Honestly I'm not sure

 

[learningphysics]

I figured on top V2=.5V1 (because it has twice the area)

Then...

P1 + DGH + .5DV^2 = P2 + DGH + .5DV^2

I figured P would need to be equal so that cancels and then .5V1 could be plugged in for V2 and solve for H? Honestly I'm not sure

Yes, is the tank open at the top and bottom? Then both P1 and P2 are the same (equal to the atmorpheric pressure) and they cancel. Take the height for the bottom as 0 and height for the top as H. Solve for H. Is V1 given?