Proton-Proton Interaction at Minimum Separation

AI Thread Summary
The discussion focuses on calculating the kinetic energy required for one proton to approach another proton at rest, ensuring their minimum separation is 1 femtometer (fm). The key approach involves using the center of mass system, where both protons have equal velocities as they approach each other. The total kinetic energy is zero at the closest approach, as both protons stop momentarily before repelling each other due to their like charges. The confusion initially stemmed from misidentifying the particles as electrons instead of protons, but the mass cancellation in calculations remains valid. Ultimately, the correct kinetic energy needed for the approaching proton is determined to be 0.46 picojoules (pJ).
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I don't even know why I am having trouble with this problem.

Homework Statement


Two protons are initially an infinite distance apart. One of the protons is initially at rest, and the other is approaching the other at a certain speed. What kinetic energy must the other proton have so that their minimum separation is again 1 fm?

Homework Equations


F=kq1q2/r^2 \hat{r}
E=kq/r^2\hat{r}
Ei=Ef (No external force)
W=\DeltaU=∫F*dr=kq1q2/r

The Attempt at a Solution


Assuming that at r=infinity, U=0:
Ki=Uf+2Ke
since the electron will try to repel the other electron, the minimum separation happens when both electrons move at the same speed.
ki=2.304*10^-18J+2Ke

I don't know where to go from here.
 
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The answer is 0.46pJ but I don't know why it is that.
 
f25274 said:
I don't even know why I am having trouble with this problem.

Homework Statement


Two protons are initially an infinite distance apart. One of the protons is initially at rest, and the other is approaching the other at a certain speed. What kinetic energy must the other proton have so that their minimum separation is again 1 fm?

Homework Equations


F=kq1q2/r^2 \hat{r}
E=kq/r^2\hat{r}
Ei=Ef (No external force)
W=\DeltaU=∫F*dr=kq1q2/r

The Attempt at a Solution


Assuming that at r=infinity, U=0:
Ki=Uf+2Ke
since the electron will try to repel the other electron, the minimum separation happens when both electrons move at the same speed.
ki=2.304*10^-18J+2Ke

I don't know where to go from here.

Work in the center of mass system. If the initial velocity of the proton at infinity is v with zero potential energy then in the center of mass system you have two protons approaching each other with velocity v/2 also at infinity and final kinetic energy at closest approach of both protons is zero.
 
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.. and note that the question says protons, not electrons
 
haruspex said:
.. and note that the question says protons, not electrons

Good point. But as far as kinetic energy goes it doesn't really matter what they are except for charge and that they are equal mass. The mass should cancel.
 
Haha. Yeah, I forgot that it was a proton. The center of mass way worked! Thanks.
Though I don't get how both their final kinetic energies would be zero. The other proton would start moving as the moving one came closer, right?
 
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f25274 said:
Haha. Yeah, I forgot that it was a proton. The center of mass way worked! Thanks.
Though I don't get how both their final kinetic energies would be zero. The other electron would start moving as the moving one came closer, right?

In the center of mass system they are approaching each other with equal velocities. When they are 1 fm apart both will stop and then move apart. At that point total kinetic energy is zero. Makes it easy.
 
Oh, they aren't moving in opposite directions since the two protons are the same charge.
One is moving in a direction where the other proton is but the other is initially at rest, not moving towards the other proton.
 
f25274 said:
Oh, they aren't moving in opposite directions since the two protons are the same charge.
One is moving in a direction where the other proton is but the other is initially at rest, not moving towards the other proton.
Right, but if you take a frame of reference which is the mass centre of the system at all times then it will move with constant velocity, so constitutes an inertial frame. (It follows from conservation of momentum.) In this frame, the protons will approach each other with equal speed.
 
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f25274 said:
Oh, they aren't moving in opposite directions since the two protons are the same charge.
One is moving in a direction where the other proton is but the other is initially at rest, not moving towards the other proton.

That's the point to using center of mass coordinates! There one proton has velocity v/2 and the other has velocity -v/2. You can transform back to rest coordinates if you want to but you don't have to.
 
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