Electric potential energy - how is it conserved???

[vaxopy]

obviously the answer is yes , its conserved.. but something bugs me about it.

One particle has a mass of 3.00×10–3 kg and a charge of +8.00 mC. A second particle has a mass of 6.00×10–3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00×10–3-kg particle is 125 m/s. Find the initial separation between the particles.

Am i doing this correctly?
Ebefore = Eafter
kq1q2/r = kq1q2/(r+0.1) + m1v1^2/2

is this correct? if so, does that mean (according the the equation) all the energy is in the 1st particle.. then the other one should be stationary? obviously not.. then why is my equation wrong? what am i missing?

should it be
kq1q2/r = kq1q2/(r+0.1) + m1v1^2/2 + m2v2^2/2 ?
but now how can u solve for 2 unknowns? (r and v2) would you use m1v1 = m2v2 to solve for v2? why (if u can) can u use this?

[StatusX]

Energy is only conserved in an inertial reference frame. That is to say, energy is only conserved if momentum is too. So you need to pick a frame where the total momentum of the system (or the velocity of the center of mass) is the same before and after, then apply conservation of energy. The conservation of momentum condition will give you a relation between v1 and v2 and then the conservation of energy condition allows you to solve for them.

[vaxopy]

im so confused :'(

is what im doing correct?

[Curious3141]

obviously the answer is yes , its conserved.. but something bugs me about it.

One particle has a mass of 3.00×10–3 kg and a charge of +8.00 mC. A second particle has a mass of 6.00×10–3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00×10–3-kg particle is 125 m/s. Find the initial separation between the particles.

Am i doing this correctly?
Ebefore = Eafter
kq1q2/r = kq1q2/(r+0.1) + m1v1^2/2

is this correct? if so, does that mean (according the the equation) all the energy is in the 1st particle.. then the other one should be stationary? obviously not.. then why is my equation wrong? what am i missing?

should it be
kq1q2/r = kq1q2/(r+0.1) + m1v1^2/2 + m2v2^2/2 ?
but now how can u solve for 2 unknowns? (r and v2) would you use m1v1 = m2v2 to solve for v2? why (if u can) can u use this?

Why is there an (r + 0.1) in one of the denominators ? Just use "0.1" as the final distance of separation.

You're only dealing with conservation of energy as applied to the less massive particle. The work done on that particle by the repulsive electrostatic force is equal to the change in its kinetic energy. So you don't have to "worry" about the other particle at all.

[StatusX]

You just have to make sure that whatever v1 and v2 are, momentum is conserved, as you mentioned at the end of your post. The reason for this is that conservation of energy only holds if conservation of momentum is holding as well.