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
- A4   cm towards the 10   kg block
- B2   cm away from the 10   kg block
- C2   cm towards the 10   kg block
- 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