Two 1.9-mm-diameter beads, C and D, are12 mm apart, measured between their cente

Question

Two 1.9-mm-diameter beads, C and D, are12 mm apart, measured between their centers.Bead C has mass 1.0 g and charge 2.5 nC .Bead D has mass 2.5 g and charge -1.0 nC.

Part A:

If the beads are released from rest, what is the speed vC of C at the instant the beads collide?

Express your answer to two significant figures and include the appropriate units.

Incorrect answers: 0.052 m/s, 1.6*10^-3 m/s

Part B:

What is the speed vD of D at the instant the beads collide?

Express your answer to two significant figures and include the appropriate units.

Incorrect answers: none guessed

Explanation / Answer

Momentum will be conserved

0 = 0.001vC - 0.0025vD

vD = 0.4vC .....(i)

From energy conservation electric potential energy converts into KE.

electric potential energy between 2 charge system = kq1q2 / d

so, 9 x 10^9 x 2.5 x 10^-9 x 1x10^-9 [ 1 / (1.9x10^-3) - 1 / (12 x 10^-3)] = 0.001vC^2 /2 + 0.0022 vD^2 /2

0.001vC^2 + 0.0025 vD^2 = 9.967 x 10^-6

putting value of vD in terms of vC from eqtion(i)

0.001vC^2 + 0.0025(0.4vC)^2 =9.967 x 10^-6   

vC = 0.0706m/s ..............Ans

vD = 0.4vC = 0.0282 m/s ..........An

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