Two 2.2-mm-diameter beads. C and D, are 12 mm apart, measured between their cent

Question

Two 2.2-mm-diameter beads. C and D, are 12 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 2.2 nC. Bead D has mass 2.5 g and charge -1.0 nC. If the beads are released from rest, what is the speed v_C of C at the instant the beads collide? Express your answer to two significant figures and include the appropriate units. Submit Part B What is the speed v_D of D at the instant the beads collide? Express your answer to two significant figures and include the appropriate units. Submit

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 at seperation r2 = D = 2.2 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 = (1/2.5)vC = 0.4vC

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*(0.4*vc)^2

from energy conservation

Ef = Ei

k*q2*q1/r2 + 0.5*mc*vc^2 + 0.5*mD*(0.4*vc)^2 = k*q1*q2/r1

0.5*mc*vc^2 + 0.5*mD*(0.4*vc)^2 = k*q1*q2(1/r1 - 1/r2)

(0.5*0.001*vc^2)+(0.5*0.0025*0.4^2*vc^2)=-9*10^9*2.2*10^-9*1*10^-9*((1/0.012-1/0.0022)

a)vc = 0.10246 m/s   <<---answer

b) vD = 0.4*0.095 = 0.038 m/s <<---answer

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