A man of mass 62 kg is standing on a stationary boat of mass 238 kg. The man is carrying a sphere of mass 0.5 kg in his hands. If the man throws the sphere horizontally with a velocity of 12 `ms^(-1)`, find the velocity with which the boat will move (in magnitude)

Options

A. `0.02 ms^(-1)`
B. `0.5 ms^(-1)`
C. `0.04 ms^(-1)`
D. `0.06 ms^(-1)`

Answer

To solve the problem, we will use the principle of conservation of linear momentum. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a man, a boat, and a sphere. Initially, all are at rest, so the total initial momentum of the system is zero.

### Step 2: Define the Variables
- Mass of the man (m₁) = 62 kg
- Mass of the boat (m₂) = 238 kg
- Mass of the sphere (m₃) = 0.5 kg
- Velocity of the sphere when thrown (v₃) = 12 m/s
- Velocity of the boat (v₂) = ? (to be determined)

### Step 3: Apply Conservation of Momentum
According to the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]

Initially, the momentum is:
\[ P_i = 0 \]

After the man throws the sphere, the final momentum can be expressed as:
\[ P_f = m₃ \cdot v₃ + m₂ \cdot v₂ + m₁ \cdot v₁ \]

Since the man is not moving relative to the boat, we can assume that the velocity of the man (v₁) is equal to the velocity of the boat (v₂). Thus:
\[ P_f = m₃ \cdot v₃ + (m₂ + m₁) \cdot v₂ \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ 0 = (0.5 \text{ kg} \cdot 12 \text{ m/s}) + (238 \text{ kg} + 62 \text{ kg}) \cdot v₂ \]

Calculating the momentum of the sphere:
\[ 0 = 6 \text{ kg m/s} + 300 \text{ kg} \cdot v₂ \]

### Step 5: Solve for v₂
Rearranging the equation to solve for v₂:
\[ 300 \text{ kg} \cdot v₂ = -6 \text{ kg m/s} \]
\[ v₂ = \frac{-6 \text{ kg m/s}}{300 \text{ kg}} \]
\[ v₂ = -0.02 \text{ m/s} \]

### Step 6: Determine the Magnitude
The magnitude of the velocity of the boat is:
\[ |v₂| = 0.02 \text{ m/s} \]

### Conclusion
The velocity with which the boat will move (in magnitude) is **0.02 m/s**.

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