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

Explanation

To solve the problem, we need to find the velocity of the center of mass of the system consisting of two blocks with masses 10 kg and 4 kg after an impulse is applied to the heavier block. Here’s a step-by-step solution:

### Step 1: Identify the masses and initial conditions
- Let \( m_1 = 10 \, \text{kg} \) (mass of the heavier block)
- Let \( m_2 = 4 \, \text{kg} \) (mass of the lighter block)
- The initial velocity of the heavier block \( v_1 = 14 \, \text{m/s} \) (after the impulse)
- The initial velocity of the lighter block \( v_2 = 0 \, \text{m/s} \) (it is 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:
\[ V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]

### Step 3: Substitute the known values into the formula
Substituting the values we have:
\[ V_{cm} = \frac{(10 \, \text{kg} \cdot 14 \, \text{m/s}) + (4 \, \text{kg} \cdot 0 \, \text{m/s})}{10 \, \text{kg} + 4 \, \text{kg}} \]

### Step 4: Calculate the numerator and denominator
Calculating the numerator:
\[ 10 \cdot 14 + 4 \cdot 0 = 140 + 0 = 140 \]
Calculating the denominator:
\[ 10 + 4 = 14 \]

### Step 5: Calculate the velocity of the center of mass
Now, substituting back into the equation:
\[ V_{cm} = \frac{140}{14} = 10 \, \text{m/s} \]

### Final Answer
The velocity of the center of mass is \( 10 \, \text{m/s} \).

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