Where Will a Proton Stop Before Turning Around in a Collision?

[loganblacke]

Homework Statement

A proton with a speed of 1.23x10⁴ m/s is moving from ∞ directly towards another proton. Assuming that the second proton is fixed in place, find the position where the moving proton stops momentarily before turning around.

Homework Equations

Potential Energy - U/q = V

The Attempt at a Solution

Clueless.

 

[Delphi51]

The proton is moving initially so it has kinetic energy.
You must think about what energy change takes place as it is repelled by the other proton. The calculation itself will be easy.

 

[cmb]

Assuming that the second proton is fixed in place

...in a little proton vice??

 

[loganblacke]

The proton is moving initially so it has kinetic energy.
You must think about what energy change takes place as it is repelled by the other proton. The calculation itself will be easy.

Okay so conservation of energy, I get that.

E_k = .5(1.67x10^-27kg)(1.23x10⁴m/s)²

When the proton stops, E_k = 0, I'm not understanding how this relates to the position of the proton when it stops.

 

[Delphi51]

It lost kinetic energy. But energy is conserved; it must have changed into some other kind of energy. It is the energy due to electric charges being near each other and is called electric potential energy. It depends on the distance between charges.

Write "initial kinetic energy = electric potential energy when stopped"
Put in the detailed formulas for both types of energy. Wikipedia has the potential energy one here: http://en.wikipedia.org/wiki/Electric_potential_energy
Then you can solve for the distance between charges.

 

[loganblacke]

It lost kinetic energy. But energy is conserved; it must have changed into some other kind of energy. It is the energy due to electric charges being near each other and is called electric potential energy. It depends on the distance between charges.

Write "initial kinetic energy = electric potential energy when stopped"
Put in the detailed formulas for both types of energy. Wikipedia has the potential energy one here: http://en.wikipedia.org/wiki/Electric_potential_energy
Then you can solve for the distance between charges.

I really really appreciate your help with this but it appears as if there are 50 equations for electric potential energy. I'm assuming the initial kinetic energy would be the 1/2mv², however it isn't obvious to me which equation to use for the electric potential energy..

 

[loganblacke]

I tried using 1/2mv² = q*(1/4∏ε₀)(Q/r)... and then solving for r but the answer I ended up with was wrong.

 

[Delphi51]

It is the first equation in the Wikipedia article (link in previous post).
Your choice of the form with k or the form with 1/(4∏ε0).

 

[loganblacke]

I tried using both equations and I'm still getting the wrong answer. Here is what I got..

(1/(4∏*(8.854*10^-12)))*((Q₁*Q₂)/r)=.5(1.67*10^-27)(1.23*10⁴)²

Q₁ and Q₂ would be the charge of each proton which I think is 1.6*10^-19

If i solve for r, I get 4.92*10¹⁸.

Another website said that the charge of a proton is e, so I tired that as well and got 1.89*10^-30.

Both answers are not correct.

 

[Delphi51]

Your calc looks okay but I don't get the same answer.
The elementary charge IS 1.6 x 10^-19, so you should not have got a different answer that way!
You know, it is much easier to solve the equation before putting the numbers in - just easier to manipulate letters than lengthy numbers.
k*e²/R = ½m⋅v²
R = 2k*e²/(m⋅v²)
Putting in the numbers at this stage, I get an answer in nanometers.

 

[loganblacke]

Your calc looks okay but I don't get the same answer.
The elementary charge IS 1.6 x 10^-19, so you should not have got a different answer that way!
You know, it is much easier to solve the equation before putting the numbers in - just easier to manipulate letters than lengthy numbers.
k*e²/R = ½m⋅v²
R = 2k*e²/(m⋅v²)
Putting in the numbers at this stage, I get an answer in nanometers.

Finally got it right, 1.821 x 10^-9

Apparently e and e on the graphing calculator are two very different things!

 

[Delphi51]

Looks good! Yes, the e on the calculator is 2.718, the base of the natural logarithm.