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

Options

A. `6ms^-1`
B. `8ms^-1`
C. `4.8ms^-1`
D. `4.5ms^-1`

Explanation

To solve the problem of finding the final speed of the coal wagon after dropping out some coal, we can use the principle of conservation of momentum. Here’s a step-by-step solution:

### Step 1: Identify the initial conditions
- Mass of the coal wagon (M_initial) = 20,000 kg
- Initial velocity of the wagon (V_initial) = 6 m/s
- Mass of coal dropped (M_dropped) = 5,000 kg

### Step 2: Calculate the initial momentum
The initial momentum (P_initial) of the system (wagon + coal) can be calculated using the formula:
\[ P_{\text{initial}} = M_{\text{initial}} \times V_{\text{initial}} \]
Substituting the values:
\[ P_{\text{initial}} = 20,000 \, \text{kg} \times 6 \, \text{m/s} = 120,000 \, \text{kg m/s} \]

### Step 3: Determine the mass of the wagon after dropping coal
After the coal is dropped, the mass of the wagon (M_final) becomes:
\[ M_{\text{final}} = M_{\text{initial}} - M_{\text{dropped}} = 20,000 \, \text{kg} - 5,000 \, \text{kg} = 15,000 \, \text{kg} \]

### Step 4: Calculate the momentum of the dropped coal
The momentum of the coal that was dropped can be calculated as:
\[ P_{\text{dropped}} = M_{\text{dropped}} \times V_{\text{initial}} = 5,000 \, \text{kg} \times 6 \, \text{m/s} = 30,000 \, \text{kg m/s} \]

### Step 5: Apply the conservation of momentum
According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
The final momentum consists of the momentum of the remaining wagon and the dropped coal:
\[ P_{\text{final}} = M_{\text{final}} \times V_{\text{final}} + P_{\text{dropped}} \]
Substituting the known values:
\[ 120,000 \, \text{kg m/s} = 15,000 \, \text{kg} \times V_{\text{final}} + 30,000 \, \text{kg m/s} \]

### Step 6: Solve for the final velocity of the wagon
Rearranging the equation to solve for \( V_{\text{final}} \):
\[ 120,000 \, \text{kg m/s} - 30,000 \, \text{kg m/s} = 15,000 \, \text{kg} \times V_{\text{final}} \]
\[ 90,000 \, \text{kg m/s} = 15,000 \, \text{kg} \times V_{\text{final}} \]
Dividing both sides by 15,000 kg:
\[ V_{\text{final}} = \frac{90,000 \, \text{kg m/s}}{15,000 \, \text{kg}} = 6 \, \text{m/s} \]

### Conclusion
The final speed of the wagon after dropping the coal remains **6 m/s**.

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