# Proton moving from infinity

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1. Feb 19, 2016

### kuokius

1. The problem statement, all variables and given/known data
A proton is moving at speed v from infinity toward a second stationary proton, as shown below. Determine the minimal distance between them.

http://s27.postimg.org/lmw3d21j7/Untitled.png

2. Relevant equations
$$W = \frac{kq_1q_2}{r}$$
$$E_k = \frac{mv^2}{2}$$
3. The attempt at a solution

Let's say that the minimal distance between two protons is x and at that moment their speeds are v_1. Then the initial energy would be E_1 and the final E_2:
$$E_1 = \frac{mv^2}{2}$$
$$E_2 = \frac{ke^2}{x} + mv_1^2$$
According to the law of conservation of energy: $$E_1 = E_2$$
And now I just don't know how to find the speed v_1.

2. Feb 19, 2016

### vela

Staff Emeritus
Can you think of any other conserved quantities?

3. Feb 19, 2016

### kuokius

Maybe an electric field or potential will remain constant at some point?

4. Feb 19, 2016

### alw34

5. Feb 19, 2016

### kuokius

So, I see that I can use conservation law of linear momentum. But how to include a given distance r?

6. Feb 19, 2016

### alw34

good

that was my problem too....
been away from solving any of these for, well 50 plus years,

amazing I can remember my own name!!

https://en.wikipedia.org/wiki/Momentum#Conservation

"If the velocities of the particles are u1 and u2 before the interaction, and afterwards they arev1 and v2, then

"

Think about whether you can use these as a second set of equations.....m's are all the same, only proton is moving initially, right?? two equations, two velocity unknowns

the stationary proton accelerates, reaches some velocity which you'll know.....
so you'll need another equation relating velocity to minimal distance....unsure what that is

7. Feb 19, 2016

### vela

Staff Emeritus
Is the second proton free to move or is its location fixed? I think you're supposed to assume the latter.

8. Feb 19, 2016

### alw34

Of course....duh!!
thank you

9. Feb 20, 2016

### kuokius

Yes, the second proton is fixed. Then I suppose I could use conservation law of angular momentum.

10. Feb 20, 2016

### haruspex

Yes, but only if you choose the axis carefully.
Linear momentum is not conserved because the second proton is being held in place by some external force. How do you avoid that being a problem for angular momentum?

11. Feb 20, 2016

### kuokius

Let's say I choose an axis going through the second proton which is fixed. Vector of electrical force creating external forces momentum goes through the axis, so the momentum of external forces equals zero. Then the initial and final angular moments would be:

$$L_i = mvr$$
$$L_f = mv_1x$$
$$L_i = L_f$$

According to the law of conservation of energy:

$$\frac{mv^2}{2} = \frac{mv_1^2}{2} + \frac{ke^2}{x}$$

Am I right?

12. Feb 20, 2016

### haruspex

Looks right.

13. Feb 20, 2016

### vela

Staff Emeritus
This doesn't really make sense. I know what you're trying to say, but what you've written is nonsense.