Two masses 6 kg and 4 kg are connected by massless flexible and inextensible string passing over massless and frictionless pulley.The acceleration of centre of mass is (g=10ms^2)

Options

A. `0.4ms^-2`
B. `2ms^-2`
C. `50ms^-2`
D. Zero

Answer

To find the acceleration of the center of mass of the system consisting of two masses (6 kg and 4 kg) connected by a massless string over a frictionless pulley, we can follow these steps:

### Step 1: Identify the masses and gravitational acceleration
Let:
- \( m_1 = 6 \, \text{kg} \) (mass 1)
- \( m_2 = 4 \, \text{kg} \) (mass 2)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 2: Calculate the net force acting on the system
The net force (\( F \)) acting on the system can be calculated as the difference in weight of the two masses:
\[ F = m_1 g - m_2 g = (6 \, \text{kg} \cdot 10 \, \text{m/s}^2) - (4 \, \text{kg} \cdot 10 \, \text{m/s}^2) \]
\[ F = 60 \, \text{N} - 40 \, \text{N} = 20 \, \text{N} \]

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

### Step 4: Calculate the acceleration of the system
Using Newton's second law (\( F = M a \)), we can find the acceleration (\( a \)) of the system:
\[ a = \frac{F}{M} = \frac{20 \, \text{N}}{10 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Step 5: Calculate the acceleration of the center of mass
The acceleration of the center of mass (\( a_{cm} \)) can be calculated using the formula:
\[ a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} \]
Since both masses have the same acceleration (equal to the acceleration of the system), we can use \( a_1 = a_2 = a \):
\[ a_{cm} = \frac{m_1 a + m_2 a}{m_1 + m_2} = \frac{(m_1 + m_2) a}{m_1 + m_2} = a \]
Thus, the acceleration of the center of mass is:
\[ a_{cm} = 2 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the center of mass is \( 2 \, \text{m/s}^2 \).

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