Particle Kinetic energy problem

[Punchlinegirl]

Particles A (of mass m and charge Q) and B (of mass m and charge 5Q) are released from rest with the distance between them equal to 0.9773 m. If Q = 8 $$\mu C$$, what is the kinetic energy of particle B at the instant when the particles are 2.9773 m apart? Answer in units of J.
I really don't know how to do this problem. I tried to set it up by using conservation of energy.
KE_A + KE_B + U_A + U_B = KE_A + KE_B + U_A + U_B
qkQ/.9773 + qk5Q/.9773 = KE_A + KE_B + qkQ/2.9773 + qk5Q/2.9773.
I really don't think this setup is right, and I have no idea how I would be able to solve for the final kinetic energy of B when I don't know the final velocity of A.
Suggestions?

 

[marlon]

1)Start out with determining what force is acting between those particles. Also, make sure that you have the correct direction of the force vector with respect to an X and Y axis.

2)Apply Newton's second Law in both the X and Y direction.

3) Solve the equations to get the particles' velocity and positions as a function of time

marlon

 

[Punchlinegirl]

Since the particles both have a positive charge, I know they will repel. Aside from that force, the normal force and gravity, those are the only forces I know of. Is that right?

 

[marlon]

Coulombic force and gravity, yes.

Though in this problem, i think you can omit gravity.

marlon

 

[Punchlinegirl]

Wouldn't the columbic force just be in the x-dir? So it would be like
qE_b- qE_a= ma ?
Sorry... I'm still confused.