Two 2.3-mm-diameter beads, C and D, are8.0 mm apart, measured between their cent

Question

Two 2.3-mm-diameter beads, C and D, are8.0 mm apart, measured between their centers.Bead C has mass 1.0 g and charge 1.7 nC .Bead D has mass 2.2 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.

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.

Explanation / Answer

Momentum will be conserved

0 = 0.001vC - 0.0022vD

vD = 0.454vC .....(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 1.7 x 10^-9 x 1x10^-9 [ 1 / (2.3x10^-3) - 1 / (2.3+8 x 10^-3)] = 0.001vC^2 /2 + 0.0022 vD^2 /2

0.001vC^2 + 0.0022 vD^2 = 1.033 x 10^-5

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

0.001vC^2 + 0.0022(0.454vC)^2 =1.033 x 10^-5   

vC = 0.0719 m/s ..............Ans

vD = 0.454vC = 0.0326 m/s ..........Ans

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