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
- 4 cm towards the 10 kg block
- 2 cm away from the 10 kg block
- 2 cm towards the 10 kg block
- 4 cm away from the 10 kg block
The Correct Option is C
Solution and Explanation
For COM to remain unchanged,
\(m_1x_1 = m_2m_2\)
\(⇒ 10 × 6 = 30 × x_2\)
\(⇒ x_2 = 2\) \(cm\) towards \(10\) \(kg\) block.
Therefore, the correct option is (C): \(2\) \(cm\) towards the \(10\) \(kg\) block
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Concepts Used:
Center of Mass
The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.
The formula for the Centre of Mass:
Center of Gravity
The imaginary point through which on an object or a system, the force of Gravity is acted upon is known as the Centre of Gravity of that system. Usually, it is assumed while doing mechanical problems that the gravitational field is uniform which means that the Centre of Gravity and the Centre of Mass is at the same position.