Question

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

• A. 4 cm towards the 10 kg block
• B. 2 cm away from the 10 kg block
• C. 2 cm towards the 10 kg block
• D. 4 cm away from the 10 kg block

Answer 1

The Correct Option is C

To solve this problem, we need to ensure that the center of mass of the system remains unchanged when the 10 kg block is moved. We will use the concept of the center of mass and its formula.

The center of mass (x_{cm}) of a system of two blocks is given by:

x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}

where m_1 and m_2 are the masses of the blocks, and x_1 and x_2 are their respective initial positions.

Initially:

• m_1 = 10 \, \text{kg}, x_1 = 0 \, \text{cm}
• m_2 = 30 \, \text{kg}, x_2 = x \, \text{cm}

The initial center of mass is:

x_{cm} = \frac{10 \cdot 0 + 30 \cdot x}{10 + 30} = \frac{30x}{40} = \frac{3x}{4}

Now, the 10 kg block is moved 6 cm towards the 30 kg block, so its new position is 6 cm. Let the 30 kg block be moved d cm towards the 10 kg block to maintain the same center of mass position.

New positions:

• x_1' = 6 \, \text{cm}
• x_2' = x - d \, \text{cm}

The new center of mass should be equal to the initial center of mass:

\frac{10 \cdot 6 + 30 \cdot (x - d)}{40} = \frac{3x}{4}

Solving for d:

60 + 30x - 30d = 30x \quad \Rightarrow \quad 60 = 30d \quad \Rightarrow \quad d = 2 \, \text{cm}

Thus, to keep the center of mass of the system unchanged, the 30 kg block must be moved 2 cm towards the 10 kg block.

The correct answer is: 2 cm towards the 10 kg block