# Distance Of Closest Approach

1. Jul 10, 2008

### kdrobey

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

Two Protons are moving directly toward one another. When they are very far apart, their initial speeds are 2.1 x 10^6 m/s. What is the distance of closest approach?

2. Relevant equations

3. The attempt at a solution

2. Jul 10, 2008

### Staff: Mentor

Show what you've done so far.

Hint: What's conserved?

3. Jul 10, 2008

### nicksauce

Your are expected to show an attempt at the problem. What are your thoughts on the problem?

4. Jul 10, 2008

### kdrobey

ok, will do

5. Sep 28, 2008

### nckaytee

Okay I have the same problem except my initial speed is 1.2*10^6

So, I used the equation 1/2mVo^2 = Kq^2/r

I am solving for r and I keep getting 1.92 * 10^-13, but it is wrong. What am I doing wrong?

6. Sep 29, 2008

### Staff: Mentor

Realize that both protons are moving and thus have kinetic energy.

7. Sep 29, 2008

### nckaytee

So, I am using the wrong formula? Not getting it...

8. Sep 29, 2008

### Staff: Mentor

You need to set total KE equal to PE.

9. Sep 29, 2008

### nckaytee

I thought that is what I did

10. Sep 29, 2008

### Staff: Mentor

What's the KE of each proton? (Symbolically--no need for numbers yet.)

11. Sep 29, 2008

### nckaytee

I am not sure what you are looking for

12. Sep 29, 2008

### Staff: Mentor

The basic expression for the KE of each proton, like you used in post #5.

13. Sep 29, 2008

### nckaytee

1/2mVo^2

14. Sep 29, 2008

### Staff: Mentor

Good. That's the KE of one proton. So what's the total KE of both protons?

15. Sep 29, 2008

### nckaytee

so,

1/2mVo^2 + 1/2mVo^2= Kq^2/r

?

16. Sep 29, 2008

### Staff: Mentor

Good.

17. Sep 29, 2008

### nckaytee

Okay I got the correct answer 9.6e-14 .... finally :-)