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two small bodies of masses 10kg and 2okg are kept a distance 1.0m apart and released.Assuming that only muutual gravitational forces are acting,find the speeds of the particles when the sepration decreases to 0.5m.Asked by shubham 4 months ago
Answers (1)

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Conserve the net gravitational potential energy and the kinetic energy of the two body system initially, when they are released and at the instant when they are 0.5 m apart.
Initial potential energy = -G(10)(20)/1 = -Gm1m2/r
Initial kinetic energy = 0.
Final potential energy = -G(10)(20)/0.5
Final kinetic energy = 1/2 (10)(v12) + 1/2(20)(v22)
Now equate net initial energy to the net final energy.
You can get another equation by conserving the linear momentum of the two mass system (as no external force acts on it).
Initial linear momentum = 0.
Final linear momentum = 10(v1) + 20(-v2)
Solve these equations (energy and momentum conservation) to get v1 and v2.