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. `2/5`
B. `2/3`
C. `8/5`
D. `8/3`

Answer

To solve the problem, we will use the principle of conservation of momentum. Here are the steps to find the final velocity of the boat:

### Step 1: Identify the masses and velocities
- Mass of the man (m1) = 50 kg
- Mass of the boat (m2) = 200 kg
- Length of the boat = 25 m
- Velocity of the man with respect to the boat (v_m) = 2 m/s

### Step 2: Define the velocities
Let:
- v_b = final velocity of the boat with respect to water
- v_m = velocity of the man with respect to water when he reaches the other end of the boat

Since the man runs from one end of the boat to the other, we can express the velocity of the man with respect to water as:
\[ v_m = v_b + 2 \, \text{m/s} \]

### Step 3: Apply the conservation of momentum
Initially, the total momentum of the system (man + boat) is zero because both are at rest:
\[ \text{Initial momentum} = 0 \]

When the man runs, the momentum of the system must still equal zero:
\[ m_1 \cdot v_m + m_2 \cdot v_b = 0 \]

Substituting the values:
\[ 50 \cdot (v_b + 2) + 200 \cdot v_b = 0 \]

### Step 4: Expand and simplify the equation
Expanding the equation gives:
\[ 50v_b + 100 + 200v_b = 0 \]

Combine like terms:
\[ 250v_b + 100 = 0 \]

### Step 5: Solve for the final velocity of the boat
Rearranging the equation:
\[ 250v_b = -100 \]
\[ v_b = -\frac{100}{250} \]
\[ v_b = -\frac{2}{5} \, \text{m/s} \]

### Step 6: Interpret the result
The negative sign indicates that the boat moves in the opposite direction to the man's running direction. Thus, the magnitude of the final velocity of the boat is:
\[ |v_b| = \frac{2}{5} \, \text{m/s} \]

### Final Answer
The final velocity of the boat is \( \frac{2}{5} \, \text{m/s} \) in the direction opposite to the man's running direction.

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