A man of 80 kg attempts to jump from the small boat of mass 40 kg on to the shore. He can generate a relative velocity of 6 m/s between him and boat. His velocity towards shore is

Options

A. (a)`4 ms^(-1)`
B. (b)`8 ms^(-1)`
C. (c)`2 ms^(-1)`
D. (d)`3 ms^(-1)`

Answer

To solve the problem step by step, we will use the principle of conservation of momentum and the concept of relative velocity.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the man, \( m_m = 80 \, \text{kg} \)
- Mass of the boat, \( m_b = 40 \, \text{kg} \)
- Relative velocity of the man with respect to the boat, \( V_{mb} = 6 \, \text{m/s} \)

2. **Define the Variables:**
- Let \( V_m \) be the velocity of the man towards the shore.
- Let \( V_b \) be the velocity of the boat in the opposite direction (away from the shore).

3. **Use the Relative Velocity Equation:**
The relative velocity of the man with respect to the boat is given by:
\[ V_{mb} = V_m - V_b \]
Rearranging this gives:
\[ V_m = V_{mb} + V_b \]

4. **Apply Conservation of Momentum:**
Since there are no external forces acting on the system, the total momentum before and after the jump must be conserved. Initially, both the man and the boat are at rest, so the initial momentum is zero:
\[ 0 = m_m V_m + m_b (-V_b) \]
Here, we take the velocity of the boat as negative because it moves in the opposite direction to the man.

5. **Substituting Values:**
Rearranging the momentum equation gives:
\[ m_m V_m = m_b V_b \]
Substituting the masses:
\[ 80 V_m = 40 (-V_b) \]
Simplifying this gives:
\[ 2 V_m = -V_b \quad \text{(1)} \]

6. **Substituting into the Relative Velocity Equation:**
From equation (1), we can express \( V_b \) in terms of \( V_m \):
\[ V_b = -2 V_m \]
Now substitute this into the relative velocity equation:
\[ V_m = 6 + (-2 V_m) \]
Rearranging gives:
\[ V_m + 2 V_m = 6 \]
\[ 3 V_m = 6 \]
Therefore:
\[ V_m = \frac{6}{3} = 2 \, \text{m/s} \]

7. **Conclusion:**
The velocity of the man towards the shore is \( V_m = 2 \, \text{m/s} \).

### Final Answer:
The velocity of the man towards the shore is **2 m/s**.