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 a system consisting of two particles, we can use the formula for the velocity of the center of mass (V_cm):

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

Where:
- \( m_1 \) and \( m_2 \) are the masses of the two particles,
- \( v_1 \) and \( v_2 \) are the velocities of the two particles.

### Step 1: Identify the given values
- 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, \( v_2 = 6 \, \text{m/s} \)

### Step 2: Substitute the values into the formula
Now, we substitute the values into the center of mass velocity formula:

\[ V_{cm} = \frac{(1 \, \text{kg} \times 2 \, \text{m/s}) + (0.5 \, \text{kg} \times 6 \, \text{m/s})}{1 \, \text{kg} + 0.5 \, \text{kg}} \]

### Step 3: Calculate the numerator
Calculate the contributions from each mass:

\[ 1 \, \text{kg} \times 2 \, \text{m/s} = 2 \, \text{kg m/s} \]
\[ 0.5 \, \text{kg} \times 6 \, \text{m/s} = 3 \, \text{kg m/s} \]

Adding these together gives:

\[ 2 \, \text{kg m/s} + 3 \, \text{kg m/s} = 5 \, \text{kg m/s} \]

### Step 4: Calculate the denominator
Calculate the total mass:

\[ 1 \, \text{kg} + 0.5 \, \text{kg} = 1.5 \, \text{kg} \]

### Step 5: Compute the speed of the center of mass
Now, substitute the values back into the formula:

\[ V_{cm} = \frac{5 \, \text{kg m/s}}{1.5 \, \text{kg}} = \frac{5}{1.5} \, \text{m/s} \]

To simplify this fraction:

\[ \frac{5}{1.5} = \frac{10}{3} \, \text{m/s} \]

### Final Answer
The speed of the center of mass of the system is:

\[ \boxed{\frac{10}{3} \, \text{m/s}} \]