# Velocity in relation to Electric charges.

• ejezisek
For part (c), you can use conservation of momentum to find the final velocity of particle m1. For part (d), you can use conservation of energy to find the final velocity of particle m2.In summary, you are trying to find the speed at the instant of closest approach, the distance of closest approach, and the velocities of the two particles after the interaction. This involves using the equations for force, acceleration, conservation of energy, and conservation of momentum. By simplifying the equation for velocity at the center of mass, you can find the speed at the instant of closest approach. Using conservation of energy, you can find the distance of closest approach. And for the final velocities of the particles, you can use conservation of momentum

## Homework Statement

From a large distance away, a particle of mass m1 and positive charge q1 is fired at speed v in the positive x direction straight toward a second particle, originally stationary but free to move, with mass m2 and positive charge q2. Both particles are constrained to move only along the x axis. (Use any variable or symbol stated above along with the following as necessary: ke for 1
4πε0
.)
(a) At the instant of closest approach, both particles will be moving at the same speed. Find this speed.

(b) Find the distance of closest approach.

After the interaction, the particles will move far apart again. At this time, find the velocity of the following.
(c) the particle of mass m1

(d) the particle of mass m2

## Homework Equations

F=ma
F=q1*q2/r^2
There are probably more but, I'm not entirely sure what they are.

## The Attempt at a Solution

For part a. I got the correct answer v-(v/(m1*(1/m1+1/m2))

For part b I tried to use the antiderivative of Part a. I also tried extracting r^2 from the formulae in part a. ie. vi-Δv1=Δv2
However I made a few errors in this part and do not know how to calculate Δv1. I am not entirely sure why part a is correct because I, instead of plugging in values for Δv I used values for acceleration.

For part C I tried using the acceleration formula I used in Part A. F/m=a and I've been trying to figure out the Δv. However I forgot/never knew how to do it.

I have not yet started part D. However I figure it to be about the same as part c.

Some updates I've made regarding the work to part 3. Due to conservation of energy, the velocity of particle 1 - change in velocity = velocity of particle 2 + change in velocity of particle 1 * m2/m1

ejezisek said:

## Homework Statement

From a large distance away, a particle of mass m1 and positive charge q1 is fired at speed v in the positive x direction straight toward a second particle, originally stationary but free to move, with mass m2 and positive charge q2. Both particles are constrained to move only along the x axis. (Use any variable or symbol stated above along with the following as necessary: ke for 1
4πε0
.)
(a) At the instant of closest approach, both particles will be moving at the same speed. Find this speed.

(b) Find the distance of closest approach.

After the interaction, the particles will move far apart again. At this time, find the velocity of the following.
(c) the particle of mass m1

(d) the particle of mass m2

## Homework Equations

F=ma
F=q1*q2/r^2
There are probably more but, I'm not entirely sure what they are.

## The Attempt at a Solution

For part a. I got the correct answer v-(v/(m1*(1/m1+1/m2))

For part b I tried to use the antiderivative of Part a. I also tried extracting r^2 from the formulae in part a. ie. vi-Δv1=Δv2
However I made a few errors in this part and do not know how to calculate Δv1. I am not entirely sure why part a is correct because I, instead of plugging in values for Δv I used values for acceleration.

For part C I tried using the acceleration formula I used in Part A. F/m=a and I've been trying to figure out the Δv. However I forgot/never knew how to do it.

I have not yet started part D. However I figure it to be about the same as part c.