A wagon of `200 kg` is moving on a smooth track with velocity of `2 m//sec`. A man of `80 kg` also runs in the wagon with a velocity such that speed of the centre of mass of the system is zero. Find the velocity of man relative to the wagon (in m//s).

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To solve the problem, we need to find the velocity of the man relative to the wagon, given that the velocity of the center of mass of the system (wagon + man) is zero.

### Step-by-Step Solution:

1. **Identify the masses and velocities**:
- Mass of the wagon, \( m_w = 200 \, \text{kg} \)
- Velocity of the wagon, \( v_w = 2 \, \text{m/s} \)
- Mass of the man, \( m_m = 80 \, \text{kg} \)
- Velocity of the man, \( v_m \) (unknown)

2. **Use the formula for the velocity of the center of mass (COM)**:
The velocity of the center of mass \( v_{cm} \) of the system is given by:
\[ v_{cm} = \frac{m_w \cdot v_w + m_m \cdot v_m}{m_w + m_m} \]
We know that \( v_{cm} = 0 \) (as given in the problem).

3. **Set up the equation**:
Substituting the known values into the equation:
\[ 0 = \frac{200 \cdot 2 + 80 \cdot v_m}{200 + 80} \]
This simplifies to:
\[ 0 = \frac{400 + 80 v_m}{280} \]
Multiplying both sides by 280 gives:
\[ 0 = 400 + 80 v_m \]

4. **Solve for \( v_m \)**:
Rearranging the equation:
\[ 80 v_m = -400 \]
Dividing both sides by 80:
\[ v_m = -5 \, \text{m/s} \]

5. **Find the velocity of the man relative to the wagon**:
The velocity of the man relative to the wagon \( v_{m, \text{relative}} \) is given by:
\[ v_{m, \text{relative}} = v_m - v_w \]
Substituting the values we found:
\[ v_{m, \text{relative}} = -5 - 2 = -7 \, \text{m/s} \]

### Final Answer:
The velocity of the man relative to the wagon is \( -7 \, \text{m/s} \).

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