A man of `50 kg` is standing at one end on a boat of length `25m` and mass `200kg`.If he starts running and when he reaches the other end, has a velocity `2ms^(-1)` with respect to the boat.The final velocity of the boat is

Options

A. `2/3ms^(-1)`
B. `2/5ms^(-1)`
C. `8/5ms^(-1)`
D. `8/3ms^(-1)`

Explanation

To solve the problem, we can use the principle of conservation of momentum. Let's break down the steps:

### Step 1: Understand the system
We have a man with a mass of \( m_1 = 50 \, \text{kg} \) and a boat with a mass of \( m_2 = 200 \, \text{kg} \). The length of the boat is \( 25 \, \text{m} \). The man runs from one end of the boat to the other and reaches a velocity of \( v_{m, \text{boat}} = 2 \, \text{m/s} \) with respect to the boat.

### Step 2: Define the velocities
Let:
- \( v_b \) = final velocity of the boat with respect to the ground.
- \( v_m \) = final velocity of the man with respect to the ground.

The velocity of the man with respect to the ground can be expressed as:
\[ v_m = v_{m, \text{boat}} + v_b \]
Substituting the known value:
\[ v_m = 2 \, \text{m/s} + v_b \]

### Step 3: Apply conservation of momentum
Initially, both the man and the boat are at rest, so the total initial momentum is zero:
\[ \text{Initial momentum} = 0 \]

When the man runs, the total momentum must still equal zero:
\[ m_1 v_m + m_2 v_b = 0 \]
Substituting \( m_1 \) and \( m_2 \):
\[ 50 \, \text{kg} \cdot (2 \, \text{m/s} + v_b) + 200 \, \text{kg} \cdot v_b = 0 \]

### Step 4: Solve for \( v_b \)
Expanding the equation:
\[ 100 \, \text{kg m/s} + 50 \, \text{kg} \cdot v_b + 200 \, \text{kg} \cdot v_b = 0 \]
Combining the terms:
\[ 100 + 250 v_b = 0 \]
Solving for \( v_b \):
\[ 250 v_b = -100 \]
\[ v_b = -\frac{100}{250} = -0.4 \, \text{m/s} \]

### Step 5: Conclusion
The final velocity of the boat is \( -0.4 \, \text{m/s} \). The negative sign indicates that the boat moves in the opposite direction to the man's movement.