Two small bodies of masses 10 kg and 20 kg are initially held at a distance of 1.0 m apart and then released. Assuming that only mutual gravitational forces act, determine the speeds of the particles when the separation decreases to 0.5 m .

Step-by-step solution

By the principle of conservation of linear momentum, the initial momentum of the system is zero, so the final momentum is also zero. m₁ v₁ = m₂ v₂ 10 v₁ = 20 v₂ v₁ = 2 v₂ By the principle of conservation of mechanical energy: K_i + U_i = K_f + U_f 0 - G m₁ m₂ r_i = 1 2 m₁ v₁^2 + 1 2 m₂ v₂^2 - G m₁ m₂ r_f 1 2 m₁ v₁^2 + 1 2 m₂ v₂^2 = G m₁ m₂ ( 1 r_f - 1 r_i ) Substituting m₁ = 10 kg , m₂ = 20 kg , r_i = 1.0 m , and r_f = 0.5 m : 1 2 (10) (2 v₂)^2 + 1 2 (20) v₂^2 = G (10)(20) ( 1 0.5 - 1 1.0 ) 5 (4 v₂^2) + 10 v₂^2 = 2