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JEE Main202227 Jun 2022Morning ShiftPhysicsCenter of Mass, Momentum and CollisionActual

Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates 0 , 0 cm and x , 0 cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is

Options

  1. A4   cm towards the 10   kg block
  2. B2   cm away from the 10   kg block
  3. C2   cm towards the 10   kg block
  4. D4   cm away from the 10   kg block

Correct answer

C. 2   cm towards the 10   kg block

Step-by-step solution

Initial position of centre of mass: x C M = 10 0 + 30 × x 40 = 3 x 4 Final position of the centre of mass: x ' C M = 10 6 + 30 × x - y 40 = 6 + 3 x - y 4 As position of centre of mass remains same, x C M = x ' C M   ⇒ 3 x 4 = 6 + 3 x - y 4 ⇒ 3 x = 6 + 3 x - 3 y ⇒ y = 2   cm Alternate method, ∆ x COM = m 1 ∆ x 1 + m 2 ∆ x 2 m 1 + m 2 ⇒ 0 = 10 × 6 + 30 y ⇒ y = - 2   cm

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