Two plastic beads, Bead C and D, are spaced 10-mm apart, measured between their

Question

Two plastic beads, Bead C and D, are spaced 10-mm apart, measured between their centers. Each has a diameter of 2.0-mm. Bead C has a mass of 1.0-g and is charged to 2.0 nC. Bead D has a mass of 2.0 g and is charged to -1.0 nC. If the beads are released from rest what will be the final speeds of each of them before impact? For full credit Draw a before-and-after pictorial representation, identifying all important symbols.

Please explain process if possible, I would like to understand it and not just get the answer!

Explanation / Answer

Given that,

distance between the beads = r = 10 mm = 10 x 10-3 m

mass of C = Mc = 1 g = 10-3 kg; charge of C = Qc = 2 nC = 2 x 10-9 C

mass of D = Md = 2 g = 2 x 10-3 kg; charge of D = Qd =- 1 nC = - 1 x 10-9 C

Intially both are at rest so, intial speed must be zero and hence the intial momentum is also zero.

The intial energy of the system is electric potential energy given by:

Ei = PE = k Qc Qd / r (1)

And the final energy of the system is all kinetic So,

Ef = KE = 1/2 Mc Vc2 + 1/2 Md Vd2 (2)

From energy conservation we have:

Ei = Ef

k Qc Qd / r =  1/2 Mc Vc2 + 1/2 Md Vd2 (3)

From conservation of momentum , Pi = Pf, but we know that Pi = 0,

So Finally, Mc x Vc = Md x Vd => Vd = Mc x Vc / Md

Putting this value in (3)

K Qc Qd / r =  1/2 Mc Vc2 + 1/2 Md Vd2 = + 1/2 Mc Vc2 + 1/2 Md x (Mc x Vc / Md)2

Solvinf for Vc we get:

Vc = sqrt [ 2 k Qc Qd Md / Mc x r x (Mc + Md) ]

Vc = sqrt [ 2 x 9 x 109 x 2 x 10-9 C x 1 x 10-9 C x 2 x 10-3 / 1 x 10-3 x 10 x 10-3 x ( 1 + 2 ) x 10-3 ]

Vc = sqrt [ 2.4 x 10-3] = 0.045 m/s

Hence, Vc = 0.045 m/s.

The speed of D will be:

Vd = Mc x Vc / Md = 1 x 10-3 x 0.045 / 2 x 10-3 = 0.023 m/s

Hence, the speed of D is = Vd = 0.023 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.