An open water tight railway wagon of mass `5xx10^(3)` kg coasts at initial velocity of `1.2 m//s` without friction on a railway track. Rain falls vertically downwards into the wagon. What change then occurred in the kinetic energy of the wagon, when it has collected `10^(3)` kg of water

Options

A. 1200 J
B. 300 J
C. 600 J
D. 900 J

Answer

To solve the problem, we need to calculate the change in kinetic energy of the railway wagon after it collects rainwater. We will follow these steps:

### Step 1: Calculate the initial momentum of the wagon
The initial momentum \( P_i \) of the wagon can be calculated using the formula:
\[ P_i = m \cdot v \]
where:
- \( m = 5 \times 10^3 \, \text{kg} \) (mass of the wagon)
- \( v = 1.2 \, \text{m/s} \) (initial velocity of the wagon)

Substituting the values:
\[ P_i = 5 \times 10^3 \, \text{kg} \times 1.2 \, \text{m/s} = 6000 \, \text{kg m/s} \]

### Step 2: Determine the final mass of the wagon after collecting rainwater
The final mass \( m_f \) of the wagon after collecting rainwater is the sum of the mass of the wagon and the mass of the rainwater:
\[ m_f = m + m_{\text{water}} = 5 \times 10^3 \, \text{kg} + 10^3 \, \text{kg} = 6 \times 10^3 \, \text{kg} \]

### Step 3: Apply the conservation of momentum to find the final velocity
Since no external horizontal forces are acting on the system, we can use the conservation of momentum:
\[ P_i = P_f \]
where \( P_f \) is the final momentum:
\[ P_f = m_f \cdot v_f \]
Setting the initial momentum equal to the final momentum:
\[ 6000 \, \text{kg m/s} = 6 \times 10^3 \, \text{kg} \cdot v_f \]
Solving for \( v_f \):
\[ v_f = \frac{6000 \, \text{kg m/s}}{6 \times 10^3 \, \text{kg}} = 1 \, \text{m/s} \]

### Step 4: Calculate the initial kinetic energy of the wagon
The initial kinetic energy \( KE_i \) can be calculated using the formula:
\[ KE_i = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE_i = \frac{1}{2} \times 5 \times 10^3 \, \text{kg} \times (1.2 \, \text{m/s})^2 \]
Calculating:
\[ KE_i = \frac{1}{2} \times 5 \times 10^3 \times 1.44 = 3600 \, \text{J} \]

### Step 5: Calculate the final kinetic energy of the wagon
The final kinetic energy \( KE_f \) can be calculated similarly:
\[ KE_f = \frac{1}{2} m_f v_f^2 \]
Substituting the values:
\[ KE_f = \frac{1}{2} \times 6 \times 10^3 \, \text{kg} \times (1 \, \text{m/s})^2 \]
Calculating:
\[ KE_f = \frac{1}{2} \times 6 \times 10^3 \times 1 = 3000 \, \text{J} \]

### Step 6: Calculate the change in kinetic energy
The change in kinetic energy \( \Delta KE \) is given by:
\[ \Delta KE = KE_f - KE_i \]
Substituting the values:
\[ \Delta KE = 3000 \, \text{J} - 3600 \, \text{J} = -600 \, \text{J} \]

### Conclusion
The change in kinetic energy of the wagon after collecting \( 10^3 \, \text{kg} \) of water is \( -600 \, \text{J} \).

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