# Distance Of Closest Approach

## Homework Statement

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?

## The Attempt at a Solution

Doc Al
Mentor
Show what you've done so far.

Hint: What's conserved?

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

ok, will do

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?

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

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

Doc Al
Mentor
You need to set total KE equal to PE.

I thought that is what I did

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

I am not sure what you are looking for

Doc Al
Mentor
I am not sure what you are looking for
The basic expression for the KE of each proton, like you used in post #5.

1/2mVo^2

Doc Al
Mentor
1/2mVo^2
Good. That's the KE of one proton. So what's the total KE of both protons?

so,

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

?

Doc Al
Mentor
Good.

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