Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of `14 m//s` to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

Options

A. `30 m//s`
B. `20m//s`
C. `10 m//s`
D. `5m//s`

Explanation

To solve the problem, we need to find the velocity of the center of mass of the system consisting of two blocks of masses 10 kg and 4 kg, after an impulse is applied to the 10 kg block.

### Step-by-step Solution:

1. **Identify the Masses and Initial Velocities:**
- Mass of block 1 (heavier block, m1) = 10 kg
- Mass of block 2 (lighter block, m2) = 4 kg
- Initial velocity of block 1 (v1) = 14 m/s (after impulse)
- Initial velocity of block 2 (v2) = 0 m/s (at rest)

2. **Calculate the Total Mass of the System:**
\[ \text{Total mass} (M) = m1 + m2 = 10 \, \text{kg} + 4 \, \text{kg} = 14 \, \text{kg} \]

3. **Use the Formula for the Velocity of the Center of Mass (V_cm):**
The velocity of the center of mass (V_cm) can be calculated using the formula:
\[ V_{cm} = \frac{m1 \cdot v1 + m2 \cdot v2}{M} \]

4. **Substitute the Values into the Formula:**
\[ V_{cm} = \frac{(10 \, \text{kg} \cdot 14 \, \text{m/s}) + (4 \, \text{kg} \cdot 0 \, \text{m/s})}{14 \, \text{kg}} \]
\[ V_{cm} = \frac{140 \, \text{kg m/s}}{14 \, \text{kg}} \]

5. **Calculate the Velocity of the Center of Mass:**
\[ V_{cm} = 10 \, \text{m/s} \]

### Conclusion:
The velocity of the center of mass of the system is **10 m/s** towards the direction of the 4 kg mass.

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