Calculating Closest Approach Distance for Two Moving Protons

[kdrobey]

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?

Homework Equations

The Attempt at a Solution

 

[Doc Al]

Show what you've done so far.

Hint: What's conserved?

 

[nicksauce]

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

 

[kdrobey]

ok, will do

 

[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?

 

[Doc Al]

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

 

[nckaytee]

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

 

[Doc Al]

You need to set total KE equal to PE.

 

[nckaytee]

I thought that is what I did

 

[Doc Al]

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

 

[nckaytee]

I am not sure what you are looking for

 

[Doc Al]

I am not sure what you are looking for

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

 

[nckaytee]

1/2mVo^2

 

[Doc Al]

1/2mVo^2

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

 

[nckaytee]

so,

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

?

 

[Doc Al]

Good.

 

[nckaytee]

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