Complete Question
Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.
A) Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two speeds reach their separations?
A) Electrons A is moving faster than electron B.
B) Electron B is moving faster than electron A.
C) Both electrons are moving at the same (nonzero) speed in the opposite direction
.D) Both electrons are moving at the same (nonzero) speed in the same direction.
E) Both electrons are momentarily stationary.
2) What is the minimum separation[tex]r_{min}[/tex] that the electrons reach?
Answer:
1
The correct option is E
2
[tex]r_{min} = \frac{kq^2}{4 mv^2}[/tex]
Explanation:
From the question we are told that
The mass of each electron is m
The charge of each electron is q
The speed of electron A is v
The speed of electron B is 3v
Generally at their point of separation the repulsion force is equal to the force that is propelling the electrons due to this the electrons are momentarily stationary
Generally the total initial kinetic energy of both electron is mathematically represented as
[tex]K_{inT} = K_A + K_B[/tex]
=> [tex]K_{inT} = \frac{1}{2}m (v)^2 + \frac{1}{2} m (3v)^2[/tex]
=> [tex]K_{inT} = \frac{1}{2} (mv^2 + 9v^2m)[/tex]
=> [tex]K_{inT} = 5mv^2 [/tex]
Generally the total final kinetic energy of both electron is mathematically represented as
[tex]K_{fT} = \frac{1}{2} *m * v^2 + \frac{1}{2} *m * v^2[/tex]
Here v is the velocity due to the repulsion force
[tex]K_{fT} = mv^2 [/tex]
Generally the final potential energy of the both electrons is
[tex]P_f = \frac{ k * q^2}{r_{min}}[/tex]
Here k is the coulombs constant
So according to energy conservation law
[tex]K_{inT} = K_{fT} + P_f[/tex]
=> [tex]5mv^2 = mv^2 + \frac{ k * q^2}{r_{min}} [/tex]
=> [tex]r_{min} = \frac{kq^2}{4 mv^2}[/tex]
