Calculating Minimal Distance between Two Protons in Motion

AI Thread Summary
The discussion centers on calculating the minimal distance between two protons, one moving towards a stationary proton. Key equations involve conservation of energy and momentum, with initial and final energy states expressed in terms of kinetic and potential energy. Participants explore the implications of the stationary proton being fixed and how this affects momentum conservation. The use of angular momentum conservation is debated, particularly regarding the choice of axis for calculations. The conversation highlights the complexities of applying these conservation laws to the problem at hand.
kuokius
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Homework Statement


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

Homework Equations


W = \frac{kq_1q_2}{r}
E_k = \frac{mv^2}{2}

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.
 
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Can you think of any other conserved quantities?
 
Maybe an electric field or potential will remain constant at some point?
 
So, I see that I can use conservation law of linear momentum. But how to include a given distance r?
 
kuokius said:
I see that I can use conservation law of linear momentum.
good

kuokius said:
But how to include a given distance r?

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

dccfa0969fdbd08e06981e85f2813970.png
"

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

Of course...duh!
thank you
 
vela said:
Is the second proton free to move or is its location fixed? I think you're supposed to assume the latter.
Yes, the second proton is fixed. Then I suppose I could use conservation law of angular momentum.
 
  • #10
kuokius said:
Yes, the second proton is fixed. Then I suppose I could use conservation law of angular momentum.
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
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
kuokius said:
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?
Looks right.
 
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  • #13
kuokius said:
Vector of electrical force creating external forces momentum goes through the axis, so the momentum of external forces equals zero.
This doesn't really make sense. I know what you're trying to say, but what you've written is nonsense.
 
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