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?

Question

Arun · Answer

Dear Student Conservation of momentu p i = p f 5 * 10 3 * (1.2) = (5*10 3 + 10 3 ) * V V = 1 m/sec now you can calculate the change in K.E. Regards Arun (askIITians forum expert)

Deepak Kumar · Answer

Mass of railway wagon = M r = 5*10 3 Kg. Initial speed of railway wagon = 1.2m/s. Mass of collected water = M w = 10 3 kg. There is no external force is working in velocity direction So, we can apply Momentum conservation. Initial momentum = Final momentum M r *Initial velocity = (M r + M w )*Final velocity 5*10 3 *1.2 = (5*10 3 +10 3 )*V So, V= 1m/s. Hence, Final speed of wagon and water = 1m/s.

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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?

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?

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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?

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?

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