physics
posted by sara on .
One particle has a mass of 3.38 x 103 kg and a charge of +7.76 ìC. A second particle has a mass of 8.46 x 103 kg and the same charge. The two particles are initially held in place and then released. The partcles fly apart, and when the separation between them is 0.159 m, the speed of the 3.38 x 103 kgparticle is 140 m/s. Find the initial separation between the particles.
This is what I tried. Sorry if it confusing. The answer I found was .019m and I am told it is wrong. PLease help!
Change in PE = Change in KE
KQ2 / Ri  KQ2 / Rf = 1/2 m v2 + 1/2 M V2 initial KE = 0
by conservation of momentum mv = MV and V = m v / M
KQ2 (1 / Ri  1 / Rf) = 1/2 m v2 + 1/2 m2 / M v2 = m / 2 (1 + m / M) v2
1 / Ri = [m / 2 (1 + m / M) v2 ] / (2 K Q2 ) + 1 / Rf
1 / Ri = (3.38 * 103 / 2 (1 + 3.38 / 8.46) * 1402 / (2 * 9 * 10E9 * 7.762 * 10E12) + 1 / .159
1 / Ri = 52.6 and Ri = .019 m

The change is PE is kQ1*Q2(1/Ri  1/Rf)
What you computed is potential, not PE