Item 10 Part A Two 2.0-mm-diameter beads, C and D, are 9.0 mm apart, measured be

Question

Item 10 Part A Two 2.0-mm-diameter beads, C and D, are 9.0 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 2.1 nC. Bead D has mass 2.4 g and charge -1.0 nC. 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. 2.13-10 vc = 2.13-10-311 m Submit My Answers Give Up Incorrect; Try Again:4 attempts remaining 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

when the beads are at seperation r1 = 12 mm

potential energy = U1 = k*q1*q2/r1

KE1 = 0

total energy Ei = k*q1*q2/r1

when the beads are in contact the seperation between their centers is equal ro the diameter

when the beads are at seperation r2 = D = 2 mm

potentail energy U2 = k*q1*Q2/r2

kinetic energy = 0.5*mC*vc^2 + 0.5*mD*vD^2

here the electtric force is internal force

momentum is conserved

mC*vC = mD*vD

vD = (mc/mD)*vc = (1/2.4)vC = 0.41vC

total energy Ef = k*q1*q2/r2 + 0.5*mC*vc^2 + 0.5*mD*vD^2

Ef = k*q2*q1/r2 + 0.5*mc*vc^2 + 0.5*mD*(0.42*vc)^2

from energy conservation

Ef = Ei

k*q2*q1/r2 + 0.5*mc*vc^2 + 0.5*mD*(0.42*vc)^2 = k*q1*q2/r1

0.5*mc*vc^2 + 0.5*mD*(0.42*vc)^2 = k*q1*q2(1/r1 - 1/r2)

(0.5*0.001*vc^2)+(0.5*0.0024*0.42^2*vc^2)=-9*10^9*2.1*10^-9*1*10^-9*((1/0.009-1/0.002)

vc = 0.101 m/s = 10.1*10^-2   <<---answer

vD = 0.42*0.101 = 0.0424 m/s = 42.4*10^-2 m/s <<---answer

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