Two point charges connected by a massless rope

[KiNGGeexD]

Question:

Two positive charges q(a) and q(b) and masses m(a) and m(b) are at rest, held together by a massless string of length d. Now the string is cut, and the particles fly off in opposite directions. How fast are each going when they are far apart.My attempt:

From this the first thing I done was make a formulae for the forces and apart from that my only ideas where to use the fact that when they are far apart the not energy which would need to be considered would be that of kinetic and the conservation of momentum would hold for this problem??

Any more ideas or how to develop my own is what I'm really looking for here, thanks a lot

 

[haruspex]

Yes, you need to consider conservation of energy. What energy is lost as KE is gained?
Conservation of momentum will simply tell you that the mass centre of the system does not move.

 

[KiNGGeexD]

The initial potential energy is lost as the kinetic energy is gained

 

[KiNGGeexD]

Would I use the fact that the potential energy is the difference in work done of a conservative force, I.e coulombs force?

 

[haruspex]

The initial potential energy is lost as the kinetic energy is gained

Right. How much PE did the system start with? How much PE does it have when the particles are infinitely far apart? How much PE has been lost?

 

[KiNGGeexD]

All the potential energy has been lost I would have though? As they are "far" apart

 

[haruspex]

All the potential energy has been lost I would have though? As they are "far" apart

Right.

 

[KiNGGeexD]

So my expression for this would be of the form total kinetic energy gained is equal to total initial potential energy

 

[haruspex]

So my expression for this would be of the form total kinetic energy gained is equal to total initial potential energy

Yes.

 

[KiNGGeexD]

Brilliant thanks for all your help

 

[KiNGGeexD]

Would I then have 1/2m(a)* v(a)^2 + 1/2m(b)* v(b)^2= k q(a)q(b)/ dSorry this format is messy, I could post a photograph if it's easier

 

[haruspex]

Would I then have


1/2m(a)* v(a)^2 + 1/2m(b)* v(b)^2= k q(a)q(b)/ d


Sorry this format is messy, I could post a photograph if it's easier

That's right. For your second equation see post #2.

 

[KiNGGeexD]

 

[haruspex]

View attachment 68932

Looks right.

 

[KiNGGeexD]

Excellent thanks for all you help