A block of mass `m_(1)` rests on a horizontal table. A string tied to the block is passed on a frictionless pulley fixed at the end of the table and to the other end of string is hung another block of mass `m_(2)`. The acceleration of the system is

Options

A. `(m_(2)g)/(m_(1)+m_(2))`
B. `(m_(1)g)/(m_(1)+m_(2))`
C. g
D. `(m_(2)g)/(m_(1))`

Explanation

To find the acceleration of the system consisting of two blocks connected by a string over a frictionless pulley, we can follow these steps:

### Step 1: Identify the Forces Acting on Each Block
- For block \( m_1 \) on the table, the only force acting on it horizontally is the tension \( T \) in the string.
- For block \( m_2 \) hanging vertically, the forces acting on it are its weight \( m_2 g \) (downward) and the tension \( T \) (upward).

### Step 2: Write the Equations of Motion
- For block \( m_1 \) (on the table):
\[ T = m_1 a \quad \text{(1)} \]
- For block \( m_2 \) (hanging):
\[ m_2 g - T = m_2 a \quad \text{(2)} \]

### Step 3: Solve the Equations Simultaneously
- From equation (1), we can express \( T \) in terms of \( a \):
\[ T = m_1 a \]
- Substitute this expression for \( T \) into equation (2):
\[ m_2 g - m_1 a = m_2 a \]

### Step 4: Rearrange the Equation
- Rearranging the equation gives:
\[ m_2 g = m_1 a + m_2 a \]
- Factor out \( a \) from the right side:
\[ m_2 g = (m_1 + m_2) a \]

### Step 5: Solve for Acceleration \( a \)
- Now, we can solve for \( a \):
\[ a = \frac{m_2 g}{m_1 + m_2} \]

### Final Result
Thus, the acceleration of the system is:
\[ a = \frac{m_2 g}{m_1 + m_2} \]

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