Two blocks of masses 5kg and 2kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse provides a velocity of 7m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is :-

Options

A. 4 m/s
B. 5 m/s
C. 2 m/s
D. 3 m/s

Answer

To find the velocity of the center of mass of the system consisting of two blocks connected by a spring, we can follow these steps:

### Step 1: Identify the masses and their velocities
- Let \( m_1 = 5 \, \text{kg} \) (mass of the heavier block).
- Let \( m_2 = 2 \, \text{kg} \) (mass of the lighter block).
- The heavier block is given a velocity \( V_1 = 7 \, \text{m/s} \) in the positive direction (towards the lighter block).
- The lighter block is initially at rest, so \( V_2 = 0 \, \text{m/s} \).

### Step 2: Use the formula for the velocity of the center of mass
The velocity of the center of mass \( V_{cm} \) can be calculated using the formula:
\[ V_{cm} = \frac{m_1 V_1 + m_2 V_2}{m_1 + m_2} \]

### Step 3: Substitute the values into the formula
Now, substituting the values into the formula:
\[ V_{cm} = \frac{(5 \, \text{kg} \times 7 \, \text{m/s}) + (2 \, \text{kg} \times 0 \, \text{m/s})}{5 \, \text{kg} + 2 \, \text{kg}} \]

### Step 4: Calculate the numerator and denominator
Calculating the numerator:
\[ 5 \times 7 = 35 \, \text{kg m/s} \]
Calculating the denominator:
\[ 5 + 2 = 7 \, \text{kg} \]

### Step 5: Final calculation of \( V_{cm} \)
Now, substituting these values back into the equation:
\[ V_{cm} = \frac{35 \, \text{kg m/s}}{7 \, \text{kg}} = 5 \, \text{m/s} \]

### Conclusion
The velocity of the center of mass is \( 5 \, \text{m/s} \) in the direction of the lighter block.