Two particles, each with a mass of 50 g , possess opposite charges of +4.0 10⁻⁵ C and -4.0 10⁻⁵ C . They are released from rest when the separation between them is 1.0 m . Determine the speeds of the particles once their separation is reduced to 50 cm .

Step-by-step solution

By the principle of conservation of mechanical energy: U_i + K_i = U_f + K_f Since the particles are released from rest and have equal masses, they will have equal speeds v at any instant to conserve linear momentum. 1 4 ₀ q₁ q₂ r_i + 0 = 1 4 ₀ q₁ q₂ r_f + 1 2 mv^2 + 1 2 mv^2 Substituting the given values q₁ = 4.0 10⁻⁵ C , q₂ = -4.0 10⁻⁵ C , r_i = 1.0 m , r_f = 0.5 m , and m = 0.05 kg : mv^2 = 1 4 ₀ q₁ q₂ ( 1 r_i - 1 r_f ) 0.05 v^2 = (9 10^9) (4.0 10⁻⁵) (-4.0 10⁻⁵) ( 1 1.0 - 1 0.5 ) 0.05 v^2 = (9 10^9) (-16 10⁻¹⁰)