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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

Two particles of masses 2   kg and 3   kg are projected horizontally in opposite directions from the top of a tower of height 39 . 2   m with velocities 5   m   s - 1 and 10   m   s - 1 respectively. The horizontal range of the centre of mass of two particles is [ g = 9 . 8   m   s - 2 ]

Options

  1. A8 2 m in the direction of 2 kg
  2. B8 2 m in the direction of 3 kg
  3. C8 m in the direction of 2 kg
  4. D8 m in the direction of 3 kg

Correct answer

B. 8 2 m in the direction of 3 kg

Step-by-step solution

Range of C.M = V c m 2 h g V c m = m 1 v 1 + m 2 v 2 m 1 + m 2 V c o m = − 2 × 5 + 3 × 10 2 + 3 = 4 m / s Range = 4 2 × 39.2 9.8 = 8 2

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