Proton collision problem

1. Oct 23, 2011

loganblacke

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Potential Energy - U/q = V

3. The attempt at a solution
Clueless.

2. Oct 23, 2011

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.

3. Oct 23, 2011

cmb

:surprised ...in a little proton vice??

4. Oct 23, 2011

loganblacke

Okay so conservation of energy, I get that.

Ek = .5(1.67x10-27kg)(1.23x104m/s)2

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

5. Oct 23, 2011

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.

6. Oct 23, 2011

loganblacke

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..

7. Oct 23, 2011

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.

8. Oct 23, 2011

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).

9. Oct 24, 2011

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.

10. Oct 24, 2011

Delphi51

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.

11. Oct 24, 2011

loganblacke

Finally got it right, 1.821 x 10-9

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

12. Oct 24, 2011

Delphi51

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