Calculating Velocity in a Charge/Field Interaction: A Homework Challenge

[Callix]

Homework Statement

Homework Equations

F=kqQ/D^2=Ma
E=kQ/D^2
D=vt?

The Attempt at a Solution

So far, I've been able to label the forces on each ship through superposition. But that is all I currently able to understand. I know that my end solution is a speed/velocity, but I am not sure how to get there with the given (Q, t, D, and M)

Any help and direction would be greatly appreciated!

 

[gneill]

Have you considered a conservation law approach?

 

[Callix]

I have not because we haven't discussed it. How might I use said approach?

 

[gneill]

I have not because we haven't discussed it. How might I use said approach?

Usually one considers the trade off between potential and kinetic energy. What form of potential energy is involved here?

 

[Callix]

Usually one considers the trade off between potential and kinetic energy. What form of potential energy is involved here?

There is definitely potential energy between each ship.

 

[gneill]

There is definitely potential energy between each ship.

Yes... what type and how much?

 

[Callix]

Yes... what type and how much?

Electrical Potential Energy
U12=kq1q2/D
U23=kq2q3/D
U13=kq1q3/D

 

[gneill]

What's the total PE of the system in terms of the variables that you were given?

 

[Callix]

U(total) = 3* kq^2/D, right?

 

[gneill]

U(total) = 3* kq^2/D, right?

Looks good.

So how will that change as the ships all tend towards infinite separation? Where will it go?

 

[Callix]

Looks good.

So how will that change as the ships all tend towards infinite separation? Where will it go?

All the potential will simply turn into kinetic energy right?

 

[gneill]

All the potential will simply turn into kinetic energy right?

That's right.

 

[Callix]

That's right.

Alright, so then 3* kq^2/D=1/2Mv^2 and just solve for v

 

[gneill]

Alright, so then 3* kq^2/D=1/2Mv^2 and just solve for v

Remember that the energy will be split three ways.

 

[Callix]

Remember that the energy will be split three ways.

So is the multiplying by 3 completely necessary if the charges and D are the same? Couldn't I just solve it as kq^2/D=1/2Mv^2

 

[gneill]

So is the multiplying by 3 completely necessary if the charges and D are the same? Couldn't I just solve it as kq^2/D=1/2Mv^2

Sure. In your presented solution be sure to state why you are justified in doing it.

 

[Callix]

Sure. In your presented solution be sure to state why you are justified in doing it.

Ah okay, I understand now! Thank you very much! :)