Two masses m and m' are tied with a thread passing over a pulley, m' is on a frictionless horizontal surface and m is hanging freely. If acceleration due to gravity is g, the acceleration of m' in this arrangment will be `:`

Options

A. g
B. `mg //( m + m ')`
C. `mg //m'`
D. `mg // ( m - m')`

Answer

To solve the problem, we need to analyze the forces acting on both masses and apply Newton's second law of motion. Here's a step-by-step solution:

### Step 1: Identify the system
We have two masses:
- Mass \( m' \) is on a frictionless horizontal surface.
- Mass \( m \) is hanging freely.

### Step 2: Draw Free Body Diagrams (FBD)
1. For the hanging mass \( m \):
- The force acting downward is the weight \( mg \).
- The force acting upward is the tension \( T \) in the thread.

The net force acting on mass \( m \) is:
\[ F_{net} = mg - T \]

2. For the mass \( m' \) on the surface:
- The only horizontal force acting on it is the tension \( T \).

The net force acting on mass \( m' \) is:
\[ F_{net} = T \]

### Step 3: Apply Newton's Second Law
Using Newton's second law, we can write the equations for both masses:

1. For mass \( m \):
\[ mg - T = ma \quad \text{(Equation 1)} \]

2. For mass \( m' \):
\[ T = m'a \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 2, we can express tension \( T \) in terms of \( m' \) and \( a \):
\[ T = m'a \]

Substituting this expression for \( T \) into Equation 1:
\[ mg - m'a = ma \]

Rearranging gives:
\[ mg = ma + m'a \]

Factoring out \( a \):
\[ mg = (m + m')a \]

### Step 5: Solve for acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{mg}{m + m'} \]

### Conclusion
The acceleration of mass \( m' \) on the frictionless surface is:
\[ a = \frac{mg}{m + m'} \]