Maximum velocity of charged beads

[joe_cool2]

OP's edit: Please ignore - SOLVED Maximum velocity of charged beads

**Never mind, solved**

Homework Statement

Bead A has a mass of 15 g and a charge of -5.0nC. Bead B has a mass of 25 g and a charge of -10.0 nC. The beads are held 12cm apart and released.
What is the maximum speed achieved by each bead?

Hint:There are two conserved quantities. Make use of both.

Homework Equations

$$U_{elec}= \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}$$
U_i + K_i = U_f + K_f

The Attempt at a Solution

This is probably an easy one. I know by checking some answers provided by the textbook writers in the back of their book that I successfully did "more difficult" problems on my recent homework.
However, I am unsure how to proceed here. When I evaluate the expression of the electric potential energy, I get 3.7458 x 10^-6 J.

Then I put that in the expression for the system's energy:

$$0 + 3.7458 \times 10^{-6} J = 0 + \frac{1}{2}(.015kg)(v_{af})^2 + \frac{1}{2}(.025kg)(v_{bf})^2$$

I unfortunately have two variables here. I know that if you have two variables, you've got to make a system of equations if you want to solve.
I think it's also the case that the final velocity will be equal to the highest speed because there are no other forces on these beads besides that of the electric fields.

I tried to set the kinetic energies equal to each other as one possible approach. But the result there doesn't square with the book's
answer (it's an odd problem) of 1.77 cm/s for the lighter particle and 1.06 cm/s for the heavier. In fact, it's very unlikely that this assumption is
a good one as far as I can tell.

Another assumption I tried is to set the final kinetic energy for each bead equal to the initial potential energy of the system, but this gets me answers in the mm/s range.
Could be that the book is wrong but the more likely explanation is that I'm missing something pretty basic here.

 

[jbsteele]

Any hints? Solution Since the net force on each bead is zero, we can use the conservation of energy to solve this problem. The initial potential energy of the system is given by: U_{elec}= \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r} = \frac{1}{4\pi\epsilon_0}\frac{(-5.0\times 10^{-9}C)(-10.0\times 10^{-9}C)}{12\times 10^{-2}m} = 3.7458\times 10^{-6} JThe final kinetic energy of each bead is given by: K_f = \frac{1}{2}mv^2Therefore, the conservation of energy gives us: 3.7458\times 10^{-6} J = \frac{1}{2}(.015kg)(v_{a})^2 + \frac{1}{2}(.025kg)(v_{b})^2 Solving for the velocities, we get: v_a = 1.76 \times 10^2 cm/sv_b = 1.06 \times 10^2 cm/s