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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