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Grade: 12th pass
        
  1. 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

Answers : (2)

Arun
24739 Points
							
Dear Student
 
Conservation of momentu
 
pi = pf
5 * 103 * (1.2) = (5*103 + 103) * V
V = 1 m/sec
 
now you can calculate the change in K.E.
 
Regards
Arun (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)*V
So, V= 1m/s.
Hence, Final speed of wagon and water = 1m/s.
2 years ago
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