A loaded `20,000kg` coal wagon is moving on a level track at `6ms^-1`. Suddenly `5000kg` of coal is dropped out of the wagon. The final speed of the wagon is a) 6 m/s b) 8 m/s c) 4.8 m/s d) 4.5 m/s

Options

A. (a) `6ms^-1`
B. (b) `8ms^-1`
C. (c) `4.8ms^-1`
D. (d) `4.5ms^-1`

Explanation

To solve the problem of the final speed of the coal wagon after dropping 5000 kg of coal, we can use the principle of conservation of linear momentum.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Mass of the coal wagon (M) = 20,000 kg
- Initial speed of the wagon (u) = 6 m/s
- Mass of coal dropped (m) = 5,000 kg

2. **Calculate Initial Momentum**:
- The initial momentum (p_initial) of the system (wagon + coal) can be calculated using the formula:
\[ p_{\text{initial}} = M \cdot u = 20,000 \, \text{kg} \cdot 6 \, \text{m/s} = 120,000 \, \text{kg m/s} \]

3. **Determine Final Mass**:
- After dropping the coal, the mass of the wagon (M_final) becomes:
\[ M_{\text{final}} = M - m = 20,000 \, \text{kg} - 5,000 \, \text{kg} = 15,000 \, \text{kg} \]

4. **Apply Conservation of Momentum**:
- According to the conservation of momentum, the total momentum before the drop must equal the total momentum after the drop:
\[ p_{\text{initial}} = p_{\text{final}} \]
- The final momentum (p_final) can be expressed as:
\[ p_{\text{final}} = M_{\text{final}} \cdot v \]
- Where \( v \) is the final speed of the wagon after the coal is dropped.

5. **Set Up the Equation**:
- Equating the initial and final momentum, we have:
\[ 120,000 \, \text{kg m/s} = 15,000 \, \text{kg} \cdot v \]

6. **Solve for Final Speed (v)**:
- Rearranging the equation to solve for \( v \):
\[ v = \frac{120,000 \, \text{kg m/s}}{15,000 \, \text{kg}} = 8 \, \text{m/s} \]

7. **Conclusion**:
- The final speed of the wagon after the coal is dropped is \( 8 \, \text{m/s} \).

### Final Answer:
The final speed of the wagon is **8 m/s** (Option b).