Potential Energy between two protons?

[MCATPhys]

Say you have two protons 10nm apart (at rest). If they are released, they naturally tend to accelerate away from each other. But what's the velocity after they are very far apart?

Basically, all the initial potential energy gets converted to kinetic energy. So I equal them to each other like this:

PE = qV = kq^2/.000000010 = 0.5mv^2

q is the charge of the proton, k is the constant, m is the mass (1.66*10^-27), v is the final velocity

When I solved it, the answer I get is around 5.3*10^3 m/s... but the right answer is supposed to be 3.8*10^3

Can someone please tell me what I am doing wrong... just the theory.

 

[kuruman]

When they are very away both protons move with speed v each (this conserves momentum). What is an expression for the kinetic energy of two protons each moving with speed v?

 

[MCATPhys]

I tried equaling the initial potential energy to mv^2 (with the mass being that of a proton) - and the answer is still wrong

 

[kuruman]

Using mv² for the kinetic energy is the correct way to go. Show what number you got and how and maybe I will be able to find out what went wrong.

 

[MCATPhys]

(9*10^9)(1.6*10^-19)^2/.000000010 = (1.66*10^-27)v^2
v = 3725 = which is pretty close to 3.8*10^3 i guess

Is that the answer you are talking about?

But I don't understand why I would use the combined mass of both protons in the kinetic energy 1/2mv^2.

The potential energy should equal the kinetic energy of each of the two protons:
PE = 1/2mv^2 + 1/2mv^2 (m is the mass on of one proton)
PE = mv^2

But why do we use the combined mass for m?

 

[kuruman]

(9*10^9)(1.6*10^-19)^2/.000000010 = (1.66*10^-27)v^2
v = 3725 = which is pretty close to 3.8*10^3 i guess

Is that the answer you are talking about?

I got 3795 m/s - the difference is probably due to rounding off.

But I don't understand why I would use the combined mass of both protons in the kinetic energy 1/2mv^2.

Anything that moves has kinetic energy. Here we have two protons moving at the same speed. Proton 1 has kinetic energy (1/2)mv² and proton2 also has kinetic energy (1/2)mv². Their total kinetic energy is (1/2)mv² +(1/2)mv² = 2x(1/2)mv² = mv²

The potential energy should equal the kinetic energy of each of the two protons:
PE = 1/2mv^2 + 1/2mv^2 (m is the mass on of one proton)
PE = mv^2

You are confused. The potential energy should equal the sum of the kinetic energies of the two protons. That's what
PE = 1/2mv^2 + 1/2mv^2
is saying.

But why do we use the combined mass for m?

That's because the masses are the same and the speeds of the protons are the same because of momentum conservation. If the speeds and masses were not the same, you would have to write

PE = (1/2)m₁v₁² + (1/2)m₂v₂².

 

[MCATPhys]

I understand what you are saying... the initial potential energy should equal the sum of the kinetic energies of each of the two protons... which means..

PE = 1/2mv^2 + 1/2 mv^2 (but each represents one proton, therefore the m should be for one proton... why do we use the mass of both protons for m?)

btw... you are being very helpful... and thanks so much :)

 

[MCATPhys]

nevermind nevermind - bit of a confusion in the book... now i get it... we were using the single mass only

 

[MCATPhys]

btw... did you round off to get 3795 m/s - because i didn't do any rounding off

 

[kuruman]

PE = 1/2mv^2 + 1/2 mv^2 (but each represents one proton, therefore the m should be for one proton... why do we use the mass of both protons for m?)

I am not sure what you are asking here. In the equation, the symbol "m" stands for "the mass of one proton" and can be replaced with 1.67x10^-27 kg.

 

[kuruman]

btw... did you round off to get 3795 m/s - because i didn't do any rounding off

I did the calculation on a spreadsheet and got 3794.733192. I calculated

$$v=\sqrt{\frac{kq^2}{md}}$$

 

[MCATPhys]

I did the calculation on a spreadsheet and got 3794.733192. I calculated

$$v=\sqrt{\frac{kq^2}{md}}$$

yeah - that looks right... but I'm guessing you meant q^2 in the above equation

 

[MCATPhys]

haha - u got it