Two 2.4-mm-diameter beads, C and D, are 9.0 mm apart, measured between their cen

Question

Two 2.4-mm-diameter beads, C and D, are 9.0 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 1.9 nC. Bead D has mass 1.5 g and charge -1.0 nC. If the beads are released from rest, what is the speed upsilon_c of C at the instant the beads collide? Express your answer to two significant figures and include the appropriate units. What is the speed upsilon_D of D at the instant the beads collide? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

initial momentum Pi = 0

when they collide

final momentum Pf = mC*vc + mD*vD

from momentum conservation

Pf =Pi

vD = (mC*vC)/mD

vD = (1/1.5)*vC = 0.67*vC

initial potetnial emergy Ui = k*q1*q2/r1

final potential energy Uf = k*q1*q2/r2

initial kinetic energy Ki = 0

final kineitc energy of the beads Kf = (1/2)*mC*vC^2 (1/2)*mD*vD^2

from energy conservation

Ui + ki = Uf + Kf

Uf - Ui = kf - ki

k*qc*qD*(1/r2 - 1/r1) = (1/2)*mC*vC^2 (1/2)*mD*vD^2

r1 = 9 mm = 9*10^-3 m

r2 = 2.4 mm = 2.4*10^-3

-9*10^9*1.9*10^-9*1*10^-9(1/(9*10^-3) - 1/(2.4*10^-3)) = ((1/2)*10^-3*vc^2) + ((1/2)*1.5*10^-3*0.6^2*vc^2)

vc = 0.0823 m/s

======================

vD = 0.055 m/s

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