Two bodies of masses 2kg and 4 kg are moving with velocities 20 m/s and 10m/s towards each other due to mutual gravitational attraction . What is the velocity of their centre of mass ?

Options

A. `5.3 "ms"^(-1)`
B. `6.4 "ms"^(-1)`
C. Zero
D. `8.1"ms"^(-1)`

Answer

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

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

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

### Step-by-step Solution:

1. **Identify the masses and velocities**:
- Mass of body 1, \(m_1 = 2 \, \text{kg}\)
- Velocity of body 1, \(v_1 = 20 \, \text{m/s}\) (towards the right)
- Mass of body 2, \(m_2 = 4 \, \text{kg}\)
- Velocity of body 2, \(v_2 = -10 \, \text{m/s}\) (towards the left, hence negative)

2. **Substitute the values into the formula**:
\[ V_{cm} = \frac{(2 \, \text{kg} \times 20 \, \text{m/s}) + (4 \, \text{kg} \times -10 \, \text{m/s})}{2 \, \text{kg} + 4 \, \text{kg}} \]

3. **Calculate the numerator**:
\[ V_{cm} = \frac{(40 \, \text{kg m/s}) + (-40 \, \text{kg m/s})}{6 \, \text{kg}} \]
\[ = \frac{0 \, \text{kg m/s}}{6 \, \text{kg}} = 0 \, \text{m/s} \]

4. **Conclusion**:
The velocity of the center of mass of the system is \(0 \, \text{m/s}\).