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An open water tight railway wagon of mass 5 ×10^3 kg coasts at an initial velocity of 1.2 m/s without friction on a railway track. Rain falls vertically downwards into the wagon. What change occurs in kinetic energy of the wagon, when it has collected 10^3 kg of water? [AFMC] (a) 900 J (b) 300 J (c) 1560 J (d) 1200 J?
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An open water tight railway wagon of mass 5 ×10^3 kg coasts at an init...
Calculation of Kinetic Energy

To solve this problem, we need to calculate the initial kinetic energy of the wagon and then calculate the kinetic energy of the wagon after it has collected 1000 kg of water. The change in kinetic energy will be the difference between the two values.

Initial Kinetic Energy

The initial kinetic energy of the wagon can be calculated using the formula:

KE = 1/2 mv^2

where KE is the kinetic energy, m is the mass of the wagon, and v is the initial velocity of the wagon.

Substituting the given values, we get:

KE = 1/2 x 5000 kg x (1.2 m/s)^2 = 3600 J

Kinetic Energy After Collecting Water

The kinetic energy of the wagon after it has collected 1000 kg of water can be calculated using the same formula as before, but with a slightly different value for the mass of the wagon. We need to add the mass of the water collected to the mass of the wagon.

The mass of the wagon and water collected is:

m = 5000 kg + 1000 kg = 6000 kg

The initial velocity of the wagon remains the same, so we can calculate the kinetic energy as follows:

KE = 1/2 x 6000 kg x (1.2 m/s)^2 = 4320 J

Change in Kinetic Energy

The change in kinetic energy is the difference between the kinetic energy of the wagon after it has collected water and its initial kinetic energy. Therefore:

Change in KE = 4320 J - 3600 J = 720 J

Therefore, the change in kinetic energy when the wagon has collected 1000 kg of water is 720 J.

Answer: (b) 300 J
Community Answer
An open water tight railway wagon of mass 5 ×10^3 kg coasts at an init...
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.
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An open water tight railway wagon of mass 5 ×10^3 kg coasts at an initial velocity of 1.2 m/s without friction on a railway track. Rain falls vertically downwards into the wagon. What change occurs in kinetic energy of the wagon, when it has collected 10^3 kg of water? [AFMC] (a) 900 J (b) 300 J (c) 1560 J (d) 1200 J?
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An open water tight railway wagon of mass 5 ×10^3 kg coasts at an initial velocity of 1.2 m/s without friction on a railway track. Rain falls vertically downwards into the wagon. What change occurs in kinetic energy of the wagon, when it has collected 10^3 kg of water? [AFMC] (a) 900 J (b) 300 J (c) 1560 J (d) 1200 J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An open water tight railway wagon of mass 5 ×10^3 kg coasts at an initial velocity of 1.2 m/s without friction on a railway track. Rain falls vertically downwards into the wagon. What change occurs in kinetic energy of the wagon, when it has collected 10^3 kg of water? [AFMC] (a) 900 J (b) 300 J (c) 1560 J (d) 1200 J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An open water tight railway wagon of mass 5 ×10^3 kg coasts at an initial velocity of 1.2 m/s without friction on a railway track. Rain falls vertically downwards into the wagon. What change occurs in kinetic energy of the wagon, when it has collected 10^3 kg of water? [AFMC] (a) 900 J (b) 300 J (c) 1560 J (d) 1200 J?.
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