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NTA Abhyas JEE Main2020PhysicsGravitationPractice

Two particles of masses m and 2 m are kept at a distance a . Find the speed of particles and their relative velocity of approach when separation becomes a / 2 .

Options

  1. A2 3 a 2 G m
  2. Ba 2 G m
  3. C2 2 G m 3 a
  4. D6 G m a

Correct answer

D. 6 G m a

Step-by-step solution

In arrangements 1 and 2 , Momentum conservation, 0 = m v 1 - 2 m v 2 ...(i) Energy conservation - G m ⋅ 2 m a = 1 2 m v 1 2 + 1 2 2 m v 2 2 - G m ⋅ 2 m a / 2 ...(ii) From Equation (i), we get v 1 = 2 v 2 Putting this value v 1 = 2 v 2 in Equation (ii), we get - 2 G m 2 a = 1 2 m 2 v 2 2 + 1 2 2 m v 2 2 - 4 G m 2 a 3 m v 2 2 = 2 G m 2 a v 2 = 2 G m 3 a Relative velocity v 1 , 2 = v 1 + v 2 = 3 2 G m 3 a = 6 G m a

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