Insulating Spheres in Electric Fields

[faceoclock]

Hi, I'd like to ask the good people of this forum for some help.
Here's a problem I've been working on for a while, and I'm seriously at my wit's end. I guess there's something I'm missing here...

Homework Statement

Two insulating spheres have radii r₁ and r₂, masses m₁ and m₂, and uniformly distributed charges -q₁ and q₂. They are released from rest when their centers are separated by a distance d. How fast is each moving when they collide? Suggestion: Consider conservation of energy and of linear momentum.

Homework Equations

I thought these were relevant:
Momentum=mv
Kinetic energy = 1/2(mv^2)
\DeltaU = -q\intE dr

The Attempt at a Solution

First I solved for the potential energy that this system gains when the two spheres are moved apart:
\DeltaU = q1\int^{d}_{d-r1-r2}E dr = k(q1)(q2)(\frac{1}{d-r1-r2} - 1/d)

I figured this is the amount of energy the spheres would have when they collide, so...
\DeltaU = \frac{1}{2}(m1)v^{2}_{1} + \frac{1}{2}(m2)v^{2}_{2}

From conservation of momentum, v2 = (m1/m2)v1 so subbing that into the above equation I got:
\DeltaU = \frac{1}{2}m1v^{2}_{1} + \frac{1}{2}\frac{m^{2}_{1}}{m_{2}}v^{2}_{1}

So then I solved for v1 to get:

v1 = \sqrt{\frac{2kq_{1}q_{2}((1/(d-r1-r2)-(1/d))}{m_{1}+\frac{m^{2}_{1}}{m_{2}}}}

And v2 can be figured out the same way. However, I know for a fact this isn't the right answer.


In closing
I'm don't really know what I did wrong, but I suspect it's because I treated the two spheres as point charges, and I'm not sure if I'm justified in doing that.

 

[nasu]

They move from a distance d to a distance r1+r2.
What is d-r1-r2?
Look at your potential energy again. You don't need any integral. Just take the difference between the final energy and initial energy. I would treat them as point charges.

 

[faceoclock]

Yup lol, that would do it. Thanks a lot for your help Nasu!