Question

Two 2.0-mm diameter beads, c and d, are 12mm mm apart, measured between their centers. bead c has mass 1.0 g and charge 2.5nc nc . bead d has mass 1.7g g and charge -1.0 nc. if the beads are released from rest.
a. what is the speed vc at the instant the beads collide?
b. what is the speed vd at the instant the beads collide?

Answer 1

The speed of Vc at the instant bead collide is 1.38 m/s and the instant speed of Vd at the instant bead collide is 0.106 m/s.

The force of attraction on the beads are due to both gravitational force and electrostatic force

As gravitational force is very weak force as compared to electrostatic force the net force will be due to electrostatic force only.

Net Force = Electrostatic force

Net Force = K q1 q2 / r²

Net Force = (9 × 10⁹) (2.5 × 10⁻⁹) (1 × 10⁻⁹) / (14 × 10⁻³) (as distance between there centre of mass is 14 mm)

Net Force = (9 × 2.5 × 10⁻⁶) / 14

Net Force = 1.6 × 10⁻⁶

For bead c

Force = Mc × Ac = 1.6 × 10⁻⁶

(1 × 10⁻³) × Ac = 1.6 × 10⁻⁶

Ac = 1.6 × 10⁻³

Acceleration of bead C is 1.6 × 10⁻³ m/s²

For bead d

Force = Md × Ad = 1.6 × 10⁻⁶

(1.7 × 10⁻³) × Ad = 1.6 × 10⁻⁶

Ad = 0.94 × 10⁻³

Acceleration of bead D is 0.94 × 10⁻³ m/s²

Let us suppose bead C travels y distance before collision

So bead D should travel 12-y distance before collision

So speed of both the beads before the collision can be find out by using newton third law of motion

For bead C

Vc² - Uc² = 2 Ac s

Vc² = 2 × 1.6 × 10⁻³ × y

Vc² = 3.2 × 10⁻³ × y

Vc = √(3.2 × 10⁻³ × y)

For bead D

Vd² - Ud² = 2 Ad s

Vd² = 2 × 0.94 × 10⁻³ × (12-y)

Vd² = 1.88 × 10⁻³ × (12-y)

Vd = √(1.88 × 10⁻³ × (12-y))

Applying energy conservation

initial = final

0 + 0 = (1/2)McVc² - (1/2)MdVd²

(1/2)McVc² = (1/2)MdVd²

McVc² = MdVd²

1 × 10⁻³ × 3.2 × 10⁻³ × y = 1.7 × 10⁻³ × 1.88 × 10⁻³ × (12-y)

3.2 × y = 3.2 × (12-y)

y = 12-y

2y = 12

y = 6

Putting the value of y in Vc and Vd

Vc = √(3.2 × 10⁻³ × y)

Vc = √(3.2 × 10⁻³ × 6)

Vc = √(19.2 × 10⁻³)

Vc = √0.0192

Vc = 1.38 m/s   (approximately)

Vd = √(1.88 × 10⁻³ × (12-y))

Vd = √(1.88 × 10⁻³ × (12-6))

Vd = √(1.88 × 10⁻³ × 6)

Vd = √(11.28 × 10⁻³)

Vd = √0.01128

Vd = 0.106 m/s   (approximately)

So the speed of Vc at the instant beads collide will be approximately 1.38 m/s and Vd at the instant speed the beads collide will be approximately 0.106 m/s.

Learn more about Fundamental Forces here:

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