Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speed of `2 m//s` and 6 m/s respectively on a smooth horizontal surface. The speed of centre of mass of the system is

Options

A. `(10)/(3) m//s`
B. `(10)/(7) m//s`
C. `(11)/(2) m//s`
D. `(12)/(3) m//s`

Explanation

To find the speed of the center of mass of the system consisting of two particles, we can use the formula for the speed of the center of mass (V_cm) given by:

\[ V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]

where:
- \(m_1\) is the mass of the first particle,
- \(v_1\) is the speed of the first particle,
- \(m_2\) is the mass of the second particle,
- \(v_2\) is the speed of the second particle.

### Step-by-step Solution:

1. **Identify the masses and velocities:**
- Mass of the first particle, \(m_1 = 1 \, \text{kg}\)
- Velocity of the first particle, \(v_1 = 2 \, \text{m/s}\)
- Mass of the second particle, \(m_2 = 0.5 \, \text{kg}\)
- Velocity of the second particle, \(v_2 = 6 \, \text{m/s}\)

2. **Substitute the values into the formula:**
\[ V_{cm} = \frac{(1 \, \text{kg} \cdot 2 \, \text{m/s}) + (0.5 \, \text{kg} \cdot 6 \, \text{m/s})}{1 \, \text{kg} + 0.5 \, \text{kg}} \]

3. **Calculate the numerator:**
- For the first particle: \(1 \cdot 2 = 2 \, \text{kg m/s}\)
- For the second particle: \(0.5 \cdot 6 = 3 \, \text{kg m/s}\)
- Total: \(2 + 3 = 5 \, \text{kg m/s}\)

4. **Calculate the denominator:**
\[ 1 + 0.5 = 1.5 \, \text{kg} \]

5. **Calculate the speed of the center of mass:**
\[ V_{cm} = \frac{5 \, \text{kg m/s}}{1.5 \, \text{kg}} = \frac{10}{3} \, \text{m/s} \approx 3.33 \, \text{m/s} \]

### Final Answer:
The speed of the center of mass of the system is \(\frac{10}{3} \, \text{m/s}\) or approximately \(3.33 \, \text{m/s}\).

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