movement of particles in electric field I am confused about a homework problem. Let's assume we have two electrically charged particles of which we know the charge and mass respectively. Let's say that at first they are fixed at some distance $r_1$ and then released simultaneously. I want to find their velocities at distance $r_2$.
Due to conservation of energy, we should have the equation $$ \frac{m_1 v_{1}^{2}}{2} + \frac{m_1 v_{1}^{2}}{2} = \int_{r_1}^{r_2} F \; dr $$
Where $$F = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1 Q_2}{r^2}$$.
Now I obviously need another equation. I was thinking that by the law of conservation of momentum, I'd get (as the momentum equals 0 when both particles are still fixed in position)
$$m_1 v_1 + m_2 v_2 = 0$$
But this is where I get confused: Consider the case where one particle remains fixed and  we let go of the other one. Wouldn't we get $m_1 v_1 = 0$ by conservation of momentum and something not equal to 0 by conservation of energy in the same way I obtained the first equation above?
 A: If one particle is fixed, some force is keeping it fixed, and in the presence of an external force, conservation of momentum doesn't apply. Your second equation is then $v_1 = 0$ (assuming particle #1 is the one that is fixed), not $m_1 v_1 + m_2 v_2 = 0$.
A: In the first case,momentum is conserved because force is applied to each charge from WITHIN the system.So,the center of mass of the system is constant.In the second case,in order for just one charge to move,it has to be put in an EXTERNAL electric field.So,you can see that momentum within the system which consists only of one charge can not be conserved.If you include the source of the electric field,then yes you can do this with m2 being the source of the field.
A: This equation: $$m_1 v_1 + m_2 v_2 = 0$$ will not work that way. There are multiple ways of looking at a system of moving particles:


*

*You can look at it in reference to a random origin, for example one corner of your room. This is what we usually do. And this is where you are trying to apply your equation, but it doesn't work.

*You can set your origin to the center of mass. In this case your equation would work, but it won't help much for your original problem. 
Let's dig into that equation a little: You already implied correctly that one half of it is 0, as the particles are not moving at the beginning, thus $0 *m_1 +0* m_2=0$, therefore that part of the equation is 0. The reference point, or rather the center of the whole system, your "0", stays the same at all times. It is the center of mass of your two particles.
Here an explanation: You probably imagine one particle stuck at its place while the other is moving away from is, which is the case if you choose a random origin. Let's give out some coordinates, particle 1 (p1) is at (0|0), and particle 2 (p2) is at (0|1), and your center of mass (CM) is at (0|0.5) if both particles are of the same weight. Now you start moving p2 away, giving p2 a coordinate of (0|3) for now. With that you also shifted the CM to (0|1.5), so you moved it in relation to your origin, which makes the equation wrong as the reference point stays the same ( 0 doesn't change in your equation).
What you'd have to do is look at the center of mass as your reference point.
Instead of doing what I described earlier, you must maintain the center of mass at the same point. So when you are moving away one particle, the other one actually moves away as well in relation the the center of mass. It is kind of hard to imagine as you would automatically use your desk or whatever as a reference, but you have to get away from that idea.
Imagine a seesaw: 
You are pulling the right mass away, represented by the continuous arrow. If the seesaw has an amazing feature where it could move one of the masses, it would have to move the other mass as well (represented by the dashed line) to keep the center of mass in the middle (which again is the "0" in your equation).
Conclusion: You can not use the equation thinking that one of the particles will remain at it's place. If you move one away, and take the center of mass as your reference, the other one will always move in the opposite direction with the same velocity (if $m_1=m_2$) but negative. Thus: $(-v_1)=v_2$ and thus ... well I'm not supposed to tell you. But hopefully this helps you out a bit.
