Velocity in relation to Electric charges.

  • Thread starter ejezisek
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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.

Thank you for your help.
 

Answers and Replies

  • #2
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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
 
  • #3
SammyS
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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.

Thank you for your help.
For superscripts & subscripts: Use the X2 and X2 buttons in the toolbar in the second line at the top of the message box. (as you type in your message.)

It might be helpful to use the idea of "center of mass". You already found the velocity, Vcm, of the center of mass: Vcm = v·(m1/(m1+m2)) which is the result of simplifying v‒(v/(m1·(1/m1+1/m2)).

For (b): Use conservation of energy to find the distance of closest approach. What's the potential energy of the system when the two particles are separated by distance r ?

For parts (c) & (d): It's an elastic collision. KEinitial = KEfinal .
 

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