Two 1.5-mm-diameter beads, C and D, are 11 mm apart, measured between their cent

ID: 1332911 • Letter: T

Question

Two 1.5-mm-diameter beads, C and D, are

11 mm apart, measured between their centers. Bead C has mass 1.0 g and charge 1.9 nC . Bead D has mass 2.4 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.

here the distance between the center of charges d = 1.5 + 1.5 = 3 mm

so net force acting on the system is zero , hence momentum is conserved

so use the law of conswervation of momentum as m1VC =m2VD

Vc = 2.4/1 = 2.4 VD

as also eenrgyis conserved

0.5 m1Vc^2 initial + 0.5 m2Vd^2 initial + Kq1q2/r^2 = 0.5 m1Vc^2 final + 0.5 m2Vd^2 final + Kq1q2/r^2

0 + 0 + (9e9 * 1.9 e -9 * 1 e -9)/(11 e -3 * 11 e-3 ) = 0.5 * 1 e-3* C^2 + 0.5 * 2.4 e -3 * D^2 + (9e9 * 1.9 e -9 * 1 e -9/(3 e-3* 3e-3)

1.41 *10^-4 = 0.5 *10^-3 *2.4 * 2.4 D^2) + 1.2 e -3* D^2 - 0.0019

so VD = 1.0429 m/s---------------Answer to part 2:

-----------------------------

so Vc = 2.4 * 1.0429 = 2.502 m/s ------------<<<<<<<<<<<<<Answer to part A

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