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. 5m/s
C. 2m/s
D. 3m/s

Answer

To find the velocity of the center of mass of the two blocks, we can follow these steps:

### Step 1: Identify the masses and initial conditions
- Mass of block 1 (heavier block, \( m_1 \)) = 5 kg
- Mass of block 2 (lighter block, \( m_2 \)) = 2 kg
- Initial velocity of block 1 (\( u_1 \)) = 7 m/s (after the impulse)
- Initial velocity of block 2 (\( u_2 \)) = 0 m/s (at rest)

### Step 2: Use the formula for the velocity of the center of mass
The velocity of the center of mass (\( V_{cm} \)) of a system of particles is given by the formula:

\[ V_{cm} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \]

### Step 3: Substitute the known values into the formula
Substituting the values we have:

\[ V_{cm} = \frac{(5 \, \text{kg} \cdot 7 \, \text{m/s}) + (2 \, \text{kg} \cdot 0 \, \text{m/s})}{5 \, \text{kg} + 2 \, \text{kg}} \]

### Step 4: Calculate the numerator and denominator
Calculating the numerator:

\[ 5 \cdot 7 + 2 \cdot 0 = 35 + 0 = 35 \]

Calculating the denominator:

\[ 5 + 2 = 7 \]

### Step 5: Calculate the velocity of the center of mass
Now substituting back into the equation:

\[ V_{cm} = \frac{35}{7} = 5 \, \text{m/s} \]

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

---