Two 2.3-mm-diameter beads, C and D, are 9.0 mm apart, measured between their cen

Question

Two 2.3-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 2.5 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.

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

A)
Let

q1 = 2.1 nC

m1 = 1 g = 1*10^-3 kg

q2 = -1 nc

m2 = 2.5 g = 2.5*10^-3 kg

Let v1 and v2 are the speed of m1 and m2 when they collide.

Apply conservation of momentum

initial momentum = final momentum

0 = m1*v1 + m2*v2

v2 = -v1*m1/m2

= -v1*1/2.5

= -v1/2.5 ----(1)

Apply conservation of energy

initial mechanicalenergy = final mechsncal energy

Ui + Ki = Uf + Kf

k*q1*q2/d1 + 0 = k*q1*q2/d2 + 0.5*m1*v1^2 + 0.5*m2*v2^2

k*q1*q2*(1/d1 - 1/d2) = 0.5*m1*v1^2 + 0.5*m2*(-v1/2.5)^2

9*10^9*2.1*10^-9*(-1)*10^-9*(1/0.009 - 1/0.0023) = 0.5*1*10^-3*v1^2 + 0.5*2.5*10^-3*v1^2/6.25

6.12*10^-6 = 5*10^-4*v1^2 + 2*10^-4*v1^2

6.12*10^-6 = 7*10^-4*v1^2

v1 = sqrt(6.12*10^-2/7)

= 0.094 m/s <<<<<<---------Answer

B)

from equation 1

v2 = -v1/2.5

= -0.094/2.5

= -0.038 m/s

speed, |v2| = 0.038 m/s <<<<<<---------Answer

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