Question

Two particles, each having a mass of 3.0 mg and having equal but opposite charges of magnitude of 6.0 nC, are released simultaneously from rest when they are a very large distance apart. What distance separates the two at the instant when each has a speed of 5.0 m/s?

Answer 1

To calculate the distance that separates the two particles at the instant when each has a speed of 5.0 m/s, we can equate the initial potential energy to the final kinetic energy. By plugging in the mass of each particle, acceleration due to gravity, and their speeds, we can solve for the distance. The distance is approximately 0.51 meters.

Step-by-step explanation:

To find the distance that separates the two particles at the instant when each has a speed of 5.0 m/s, we can use the principle of conservation of energy. Initially, when the particles are at rest and a very large distance apart, their potential energy is maximum and their kinetic energy is zero. As they move closer to each other, the potential energy decreases and the kinetic energy increases until they both have a speed of 5.0 m/s.

Since the particles have equal but opposite charges, they repel each other. This means that the potential energy decreases as the particles move closer. At the instant when each particle has a speed of 5.0 m/s, the potential energy has decreased to zero. This means that all the initial potential energy has been converted into kinetic energy.

To calculate the distance that separates the two particles at this instant, we can equate the initial potential energy to the final kinetic energy:mgh = (1/2)mv^2

Where m is the mass of each particle, g is the acceleration due to gravity (which we can assume to be approximately 9.8 m/s²), and h is the distance that separates the two particles at this instant.

Plugging in the values: m = 3.0 mg = 3.0 x 10^-6 kg, g = 9.8 m/s², and v = 5.0 m/s, we can solve for h:

3.0 x 10^-6 kg x 9.8 m/s² x h = (1/2) x 3.0 x 10^-6 kg x (5.0 m/s)^2

Simplifying the equation:

3.0 x 10^-6 kg x 9.8 m/s² x h = 1.5 x 10^-5 kg m²/s²

h = (1.5 x 10^-5 kg m²/s²) / (3.0 x 10^-6 kg x 9.8 m/s²)

h = 0.51 m

Therefore, the distance that separates the two particles at the instant when each has a speed of 5.0 m/s is approximately 0.51 meters.

Answer 2

The distance that separates the two particles is 7.42 cm.

Step-by-step explanation:

Given;

the mass of each particle, m = 3 mg = 3 x 10⁻⁶ kg

the magnitude of charge of each particle, q = 6.0 nC

speed of each particle, v = 5.0 m/s

`F = ma = (kq^2)/(r^2)`

a = v/t

where;

a is the acceleration of the two particles

v is the final velocity

t is time

v = u + gt

5 = 0 + 9.8t

5 = 9.8t

t = 5/9.8

t = 0.51 s

a = v/t

a = 5/0.51 = 9.8 m/s²

Total force on the two particles = (2m)a = (2* 3 x 10⁻⁶)9.8

F = 5.88 x 10⁻⁵ N

Substitute in the value of F in the above equation and calculate r

`F = (kq^2)/(r^2)`

where;

k is coulomb's constant = 8.99 x 10⁹ Nm²/c²

r is the distance of separation between the two particles

`F = (kq^2)/(r^2) \\\\r^2 = (kq^2)/(F)\\\\r = \sqrt{ (kq^2)/(F)} = \sqrt{ (8.99*10^9*(6*10^(-9))^2)/(5.88*10^(-5))} = 0.0742 \ m`

Therefore, the distance that separates the two particles at the instant when each has a speed of 5.0 m/s, is 7.42 cm.