Homework Help: physics

Posted by sara on Wednesday, March 4, 2009 at 10:36pm.

One particle has a mass of 3.38 x 10-3 kg and a charge of +7.76 錀. A second particle has a mass of 8.46 x 10-3 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 10-3 kg-particle 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 * 10-3 / 2 (1 + 3.38 / 8.46) * 1402 / (2 * 9 * 10E9 * 7.762 * 10E-12) + 1 / .159

1 / Ri = 52.6 and Ri = .019 m

physics - drwls, Wednesday, March 4, 2009 at 11:34pm
The change is PE is kQ1*Q2(1/Ri - 1/Rf)

What you computed is potential, not PE

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