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 :

Solution

Physics sir · 4 steps

For the centre of mass to stay put, the total mass-times-shift must add up to zero.

1. Condition. The centre of mass is x_(cm) = (m₁ x₁ + m₂ x₂)/(m₁ + m₂). It stays fixed only if the shifts balance, so m₁ Δ x₁ + m₂ Δ x₂ = 0.

   10 × (+6) + 30 × Δ x₂ = 0

2. Signs. Take +x along the line from the 10 kg block towards the 30 kg block. The 10 kg block moves towards the other, so Δ x₁ = +6 cm.
3. Solve. Δ x₂ = -60/30 = -2 cm. The minus sign means the 30 kg block moves in the -x direction, back towards the 10 kg block.
4. So. The 30 kg block must move 2 cm towards the 10 kg block, which is option (c).

The idea to keep. To keep the centre of mass fixed, use Σ mᵢ Δ xᵢ = 0. A heavier body moves a smaller distance, and if one body moves forward the other must move back.

Answer from the NTA final answer key. Solution by PrepWiser.