Two bodies of masses `m_(1)` and `m_2 (m_1 gt m_2)` respectively are tied to the ends j of a string which passes over a light frictionless pulley. The masses are intitially at rest and released. What is the acceleration of the centre of mass ?

Explanation

To find the acceleration of the center of mass of the two-body system connected by a string over a pulley, we can follow these steps:

### Step 1: Identify the forces acting on the masses
- For mass \( m_1 \) (which is greater than \( m_2 \)), the forces acting are:
- Weight: \( m_1 g \) (downward)
- Tension: \( T \) (upward)

- For mass \( m_2 \):
- Weight: \( m_2 g \) (downward)
- Tension: \( T \) (upward)

### Step 2: Write the equations of motion
- For mass \( m_1 \):
\[ m_1 g - T = m_1 a \]
(Equation 1)

- For mass \( m_2 \):
\[ T - m_2 g = m_2 a \]
(Equation 2)

### Step 3: Solve the equations simultaneously
- From Equation 1, we can express \( T \):
\[ T = m_1 g - m_1 a \]

- Substitute \( T \) in Equation 2:
\[ (m_1 g - m_1 a) - m_2 g = m_2 a \]
Simplifying this gives:
\[ m_1 g - m_1 a - m_2 g = m_2 a \]
Rearranging terms:
\[ m_1 g - m_2 g = m_1 a + m_2 a \]
\[ (m_1 - m_2) g = (m_1 + m_2) a \]

### Step 4: Solve for acceleration \( a \)
- From the above equation, we can solve for \( a \):
\[ a = \frac{(m_1 - m_2) g}{m_1 + m_2} \]

### Step 5: Find the acceleration of the center of mass
- The acceleration of the center of mass \( a_{cm} \) for a two-body system is given by the formula:
\[ a_{cm} = \frac{m_2}{m_1 + m_2} g \]

### Final Result
Thus, the acceleration of the center of mass is:
\[ a_{cm} = \frac{m_2}{m_1 + m_2} g \]

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