Two bodies of masses `3kg` and `2kg` are connected by a along string and the string is made to pass over a smooth fixed pulley Initially the bodies are held at the saem level and released from rest. The velocity of the `3kg` body after one second is `(g =10m//s^(2))` .

Options

A. `2m//s`
B. `1m//s`
C. `0.4m//s`
D. `4m//s`

Explanation

To solve the problem of finding the velocity of the 3 kg body after one second, we can follow these steps:

### Step 1: Identify the masses and the gravitational acceleration
We have two masses:
- \( m_1 = 3 \, \text{kg} \) (the heavier mass)
- \( m_2 = 2 \, \text{kg} \) (the lighter mass)

The acceleration due to gravity is given as:
- \( g = 10 \, \text{m/s}^2 \)

### Step 2: Determine the net force acting on the system
The net force acting on the system can be calculated using the difference in weights of the two masses. The force due to gravity on each mass is:
- Weight of \( m_1 = m_1 \cdot g = 3 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 30 \, \text{N} \)
- Weight of \( m_2 = m_2 \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \)

The net force \( F \) acting on the system is:
\[ F = m_1 \cdot g - m_2 \cdot g = 30 \, \text{N} - 20 \, \text{N} = 10 \, \text{N} \]

### Step 3: Calculate the total mass of the system
The total mass \( M \) of the system is the sum of the two masses:
\[ M = m_1 + m_2 = 3 \, \text{kg} + 2 \, \text{kg} = 5 \, \text{kg} \]

### Step 4: Calculate the acceleration of the system
Using Newton's second law, the acceleration \( a \) of the system can be calculated as:
\[ a = \frac{F}{M} = \frac{10 \, \text{N}}{5 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Step 5: Use the kinematic equation to find the velocity
Since the bodies are released from rest, the initial velocity \( u = 0 \). We can use the kinematic equation:
\[ v = u + at \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (0 in this case),
- \( a \) is the acceleration (2 m/s²),
- \( t \) is the time (1 second).

Substituting the values:
\[ v = 0 + (2 \, \text{m/s}^2)(1 \, \text{s}) = 2 \, \text{m/s} \]

### Final Answer
The velocity of the 3 kg body after one second is \( 2 \, \text{m/s} \).

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