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`        	An open watertight railway wagon of mass 5 x 10^3 kg is moving with an initial velocity of 1.2 m/s without friction on a railway track. Rain falls vertically downwards into the wagon. What change will occur in the K.E of wagon, when it has collected 10^3 kg of water?`
2 years ago

Arun
24739 Points
```							Dear Student Conservation of momentu pi = pf5 * 103 * (1.2) = (5*103 + 103) * VV = 1 m/sec now you can calculate the change in K.E. RegardsArun (askIITians forum expert)
```
2 years ago
Deepak Kumar
31 Points
```							Mass of railway wagon = Mr = 5*103Kg.Initial speed of railway wagon = 1.2m/s.Mass of collected water = Mw = 103kg.There is no external force is working in velocity direction So, we can apply Momentum conservation. Initial momentum = Final momentum Mr*Initial velocity = (Mr + Mw)*Final velocity 5*103*1.2 = (5*103 +103)*VSo, V= 1m/s.Hence, Final speed of wagon and water = 1m/s.
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions