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 for the given system of two masses connected by a string over a pulley, we can follow these steps:

### Step 1: Identify the forces acting on each mass
- For the 6 kg mass (m1):
- Weight (downward force) = m1 * g = 6 kg * 10 m/s² = 60 N
- Tension in the string (upward force) = T

- For the 4 kg mass (m2):
- Weight (downward force) = m2 * g = 4 kg * 10 m/s² = 40 N
- Tension in the string (upward force) = T

### Step 2: Write the equations of motion for each mass
- For the 6 kg mass:
\[ m_1 g - T = m_1 a \quad \text{(1)} \]
Substituting m1 = 6 kg and g = 10 m/s²:
\[ 60 - T = 6a \quad \text{(1)} \]

- For the 4 kg mass:
\[ T - m_2 g = m_2 a \quad \text{(2)} \]
Substituting m2 = 4 kg and g = 10 m/s²:
\[ T - 40 = 4a \quad \text{(2)} \]

### Step 3: Solve the equations simultaneously
- From equation (1):
\[ T = 60 - 6a \quad \text{(3)} \]

- Substitute equation (3) into equation (2):
\[ (60 - 6a) - 40 = 4a \]
Simplifying this gives:
\[ 20 - 6a = 4a \]
\[ 20 = 10a \]
\[ a = 2 \, \text{m/s}^2 \]

### Step 4: Calculate the acceleration of the center of mass
The acceleration of the center of mass (a_cm) can be calculated using the formula:
\[ a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} \]
Where:
- \(a_1 = a\) (downward for m1, so we take it as positive)
- \(a_2 = -a\) (upward for m2, so we take it as negative)

Substituting the values:
\[ a_{cm} = \frac{6 \cdot 2 + 4 \cdot (-2)}{6 + 4} \]
\[ = \frac{12 - 8}{10} \]
\[ = \frac{4}{10} = 0.4 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the center of mass is \(0.4 \, \text{m/s}^2\).

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