Question

One particle has a mass of 3.84 x 10-3 kg and a charge of +8.90 μC. A second particle has a mass of 8.33 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.158 m, the speed of the 3.84 x 10-3 kg-particle is 190 m/s. Find the initial separation between the particles.

Answer 1

Given Data

m₁ = 3.84 x 10^-3 kg,

q = + 8.90 uC = +8.90 x 10^-6 C,

m₂ = 8.33 x 10^-3 kg,

d = 0.158 m,

v₁ = 190 m/s.

To Find the initial separation d₀

Solution:-

initial energy = kq²/d₀

final energy = m₁v₁²/2 + m₂v₂²/2 + kq²/d

momentum conservation: 0 = m₁v₁ + m₂v₂, so v₂ = -m₁v₁/m₂ = -(3.84x 10^-3 kg) * 190 / (8.33 x 10^-3 kg)

                                                                        = - 87.58 m/s

kq²/d₀ = m₁v₁²/2 + m₂v₂²/2 + kq²/d

d₀ = [1/d + (m₁v₁² + m₂v₂²)/(2kq²)]^-1

    = [1/0.158 + { (3.84 x 10^-3 kg)*190² + (8.33x 10^-3 kg)*87.58²) / { (2*(9*10⁹)*(8.90 x 10^-6)² }]^-1

    = 0.00673 m