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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the initial parameters
- Mass of the wagon (m) = \(5 \times 10^3\) kg
- Initial velocity of the wagon (u) = 1.2 m/s
- Mass of water collected (m_water) = \(10^3\) kg

### Step 2: Calculate the total mass after collecting water
The total mass of the wagon after collecting the rainwater will be:
\[ M = m + m_{\text{water}} = 5 \times 10^3 \, \text{kg} + 10^3 \, \text{kg} = 6 \times 10^3 \, \text{kg} \]

### Step 3: Apply the conservation of momentum
Since there are no external forces acting on the wagon in the horizontal direction, the momentum before and after collecting the rainwater must be conserved.

Initial momentum (p_initial):
\[ p_{\text{initial}} = m \cdot u = (5 \times 10^3 \, \text{kg}) \cdot (1.2 \, \text{m/s}) = 6 \times 10^3 \, \text{kg m/s} \]

Let the final velocity after collecting the water be \(v\). The final momentum (p_final) will be:
\[ p_{\text{final}} = M \cdot v = (6 \times 10^3 \, \text{kg}) \cdot v \]

Setting the initial momentum equal to the final momentum:
\[ 6 \times 10^3 = (6 \times 10^3) \cdot v \]

### Step 4: Solve for the final velocity (v)
Dividing both sides by \(6 \times 10^3\):
\[ 1 = v \implies v = 1 \, \text{m/s} \]

### Step 5: Calculate the change in kinetic energy
The initial kinetic energy (KE_initial) is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 = \frac{1}{2} (5 \times 10^3) (1.2)^2 \]
Calculating this:
\[ KE_{\text{initial}} = \frac{1}{2} (5 \times 10^3) (1.44) = 3600 \, \text{J} \]

The final kinetic energy (KE_final) is given by:
\[ KE_{\text{final}} = \frac{1}{2} M v^2 = \frac{1}{2} (6 \times 10^3) (1)^2 \]
Calculating this:
\[ KE_{\text{final}} = \frac{1}{2} (6 \times 10^3) (1) = 3000 \, \text{J} \]

### Step 6: Calculate the change in kinetic energy
The change in kinetic energy (ΔKE) is:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 3000 \, \text{J} - 3600 \, \text{J} = -600 \, \text{J} \]

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

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