In a system two particles of masses m₁=3 kg and m₂=2 kg are placed at certain distance from each other. The particle of mass m₁ is moved towards the center of mass of the system through a distance 2 cm. In order to keep the center of mass of the system at the original position, the particle of mass m₂ should move towards the center of mass by the distance _________ cm.

Solution

Physics sir · 4 steps

Both particles move towards the CM, so their moments must balance: m₁ d₁ = m₂ d₂.

1. Set up the condition. Take the original centre of mass as the origin. For it to stay put, the mass-weighted shifts must cancel: m₁ Δ x₁ + m₂ Δ x₂ = 0.
2. Directions. The two particles sit on opposite sides of the centre of mass. So moving each one towards it means moving them in opposite directions, and their shifts can cancel.

   m₁ d₁ = m₂ d₂

3. Put in numbers. 3 × 2 = 2 × d₂, so d₂ = 6/2 = 3 cm.
4. So. The particle of mass m₂ must move 3 cm towards the centre of mass.

The common slip: 1.33. Writing the ratio upside down, d₂ = m₂/m₁ d₁, gives about 1.33 cm, not 3. The lighter particle must move farther, not less.

The idea to keep. To keep the centre of mass fixed, each mass times its shift must balance. The lighter particle has to move farther, in the ratio d₂/d₁ = m₁/m₂.

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