A girl of mass 50 kg stands on a railroad car of mass 75 kg moving with velocity `20 ms^(-1)` . Now , the girl begins to run with a velocity of `10ms^(-1)` with respect to the car in the same direction , as that of the car. The velocity of the car at this instant will be

Options

A. `20 ms ^(-1)`
B. `24 ms ^(-1)`
C. `16ms ^(-1)`
D. `18 ms ^(-1)`

Answer

To solve the problem, we will use the principle of conservation of momentum. The total momentum of the system (girl + car) before the girl starts running must equal the total momentum after she starts running, since no external forces are acting on the system.

### Step-by-Step Solution:

1. **Identify the masses and initial velocities:**
- Mass of the girl, \( m_g = 50 \, \text{kg} \)
- Mass of the car, \( m_c = 75 \, \text{kg} \)
- Initial velocity of the car, \( v_c = 20 \, \text{m/s} \)
- Velocity of the girl with respect to the car, \( v_{g/c} = 10 \, \text{m/s} \)

2. **Calculate the initial momentum of the system:**
The initial momentum \( p_{initial} \) is given by the sum of the momentum of the girl and the car:
\[ p_{initial} = m_g \cdot v_c + m_c \cdot v_c = 50 \cdot 20 + 75 \cdot 20 = 1000 + 1500 = 2500 \, \text{kg m/s} \]

3. **Determine the final velocity of the girl:**
The girl runs in the same direction as the car, so her final velocity \( v_g \) is:
\[ v_g = v_c + v_{g/c} = 20 + 10 = 30 \, \text{m/s} \]

4. **Set up the equation for final momentum:**
The final momentum \( p_{final} \) is given by:
\[ p_{final} = m_g \cdot v_g + m_c \cdot v_c' \]
where \( v_c' \) is the final velocity of the car that we need to find.

5. **Apply conservation of momentum:**
Since \( p_{initial} = p_{final} \):
\[ 2500 = 50 \cdot 30 + 75 \cdot v_c' \]
Simplifying this gives:
\[ 2500 = 1500 + 75 \cdot v_c' \]

6. **Solve for \( v_c' \):**
Rearranging the equation:
\[ 75 \cdot v_c' = 2500 - 1500 \]
\[ 75 \cdot v_c' = 1000 \]
\[ v_c' = \frac{1000}{75} = \frac{40}{3} \approx 13.33 \, \text{m/s} \]

7. **Final velocity of the car:**
The final velocity of the car \( v_c' \) is approximately \( 13.33 \, \text{m/s} \).