Two blocks of masses `6 kg` and `4 kg` are attached to the two ends of a massless string passing over a smooth fixed pulley. if the system is released, the acceleration of the centre of mass of the system will be

Options

A. `g`, vertically downwards
B. `g/5`, vertically downwards
C. `g/25`, vertically downwards
D. zero

Explanation

To find the acceleration of the center of mass of the system consisting of two blocks of masses 6 kg and 4 kg, we can follow these steps:

### Step 1: Identify the masses and the gravitational acceleration
Let:
- \( m_1 = 6 \, \text{kg} \) (mass of the first block)
- \( m_2 = 4 \, \text{kg} \) (mass of the second block)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 2: Determine the net force acting on the system
When the system is released, the heavier mass (6 kg) will accelerate downwards while the lighter mass (4 kg) will accelerate upwards. The net force \( F \) acting on the system can be calculated as:
\[ F = m_1 g - m_2 g = (m_1 - m_2)g \]
Substituting the values:
\[ F = (6 \, \text{kg} - 4 \, \text{kg}) \cdot g = 2g \]

### Step 3: Calculate the total mass of the system
The total mass \( M \) of the system is:
\[ M = m_1 + m_2 = 6 \, \text{kg} + 4 \, \text{kg} = 10 \, \text{kg} \]

### Step 4: Find the acceleration of the system
Using Newton's second law, the acceleration \( a \) of the system can be calculated as:
\[ a = \frac{F}{M} = \frac{2g}{10 \, \text{kg}} = \frac{g}{5} \]

### Step 5: Determine the acceleration of the center of mass
The acceleration of the center of mass \( a_{cm} \) of the system is the same as the acceleration of the system since the two blocks are connected by a string. Thus:
\[ a_{cm} = \frac{g}{5} \]

### Step 6: Direction of acceleration
Since the 6 kg mass is moving downwards, the acceleration of the center of mass will also be directed downwards.

### Final Answer
The acceleration of the center of mass of the system is:
\[ \frac{g}{5} \text{ downwards} \]

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