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Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0.5 m.

Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to 0.5 m.

Grade:11

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
7 years ago
Sol. The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is also zero. So (10 kg)v base 1 = (20 kg) v base 2 Or v base 1 = v base 2 …(1) Since P.E. is conserved Initial P.E. = -6.67 * 10^-11 * 10 *20/1 = -13.34*10^-19 J When separation is 0.5 m, -13.34 * 10^-9 + 0 = -13.34 * 10^-9/(1/2) + (1/2) * 10 v base^2 + (1/2) * 20 v base 2^2 …(2) ⇒ - 13.34 * 10^-19 = -26.68 * 10^-9 + 5v base 1^2 + 10 v base 2^2 ⇒ - 13.34 * 10^-19 = -26.68 * 10^-9 + 30 v base 2^2 ⇒ v base 2^2 = 13.34 * 10^-9/30 = 4.44 * 10^-10 ⇒ v base 2 = 2.1 * 10^-15 m/s. So, v base 1 = 4.2 × 10^–5 m/s.

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