Two bodies of masses 4 kg and 6 kg connected by means of a light string are lying on a smooth horizontal surface. A horizontal pulling force is applied on the lighter body. Two seconds later the string connecting the two masses is cut. After two more seconds if the velocity of the heavier mass is `2 ms^(-1)`, the force initially applied is

Options

A. `10 N `
B. `25 N `
C. `20 N`
D. `40 N`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the System
We have two bodies with masses \( m_1 = 4 \, \text{kg} \) and \( m_2 = 6 \, \text{kg} \) connected by a light string on a smooth horizontal surface. A horizontal force \( F \) is applied to the lighter mass \( m_1 \).

### Step 2: Determine the Acceleration
Since both masses are connected, they will accelerate together when the force \( F \) is applied. The total mass of the system is:
\[ m_{\text{total}} = m_1 + m_2 = 4 \, \text{kg} + 6 \, \text{kg} = 10 \, \text{kg} \]
The acceleration \( a \) of the system can be expressed as:
\[ a = \frac{F}{m_{\text{total}}} = \frac{F}{10} \]

### Step 3: Calculate the Velocity at \( t = 2 \, \text{s} \)
Using the formula for velocity:
\[ v = u + at \]
where \( u \) is the initial velocity (which is 0), we can find the velocity of the heavier mass \( m_2 \) at \( t = 2 \, \text{s} \):
\[ v_{m_2}(2) = 0 + \left(\frac{F}{10}\right) \cdot 2 = \frac{2F}{10} = \frac{F}{5} \, \text{m/s} \]

### Step 4: Analyze the Situation After Cutting the String
At \( t = 2 \, \text{s} \), the string is cut. After this point, the heavier mass \( m_2 \) will continue to move with the velocity it had at \( t = 2 \, \text{s} \) since there is no force acting on it:
\[ v_{m_2}(4) = v_{m_2}(2) = \frac{F}{5} \, \text{m/s} \]

### Step 5: Use the Given Information
We know that after two more seconds (at \( t = 4 \, \text{s} \)), the velocity of the heavier mass \( m_2 \) is given as \( 2 \, \text{m/s} \):
\[ \frac{F}{5} = 2 \]

### Step 6: Solve for the Force \( F \)
To find \( F \), we can rearrange the equation:
\[ F = 2 \cdot 5 = 10 \, \text{N} \]

### Final Answer
The force initially applied is \( F = 10 \, \text{N} \).

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