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Grade 11Mechanics

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.

Profile image of Radhika Batra
12 Years agoGrade 11
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1 Answer

Profile image of Navjyot Kalra
12 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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