Question

An open water tight railway wagon of mass $5\times 10^{3}kg$ coasts at an initial velocity of $1.2\,ms^{-1}$ without friction on a railway track. Rain falls vertically downwards into the wagon. What change occurs in KE of the wagon, when it has collected $10^3\, kg$ of water?

• A. 900 J
• B. 300 J
• C. 600 J
• D. 1200 J

Answer 1

Answer: (C)

600 J

Answer 2

If $v'$ is final velocity of wagon, then applying principle of conservation of linear momentum, we get, $5 \times 10^{3} \times 1.2 =\left(5 \times 10^{3}+10^{3}\right) \times v'$ $v'=1 \,ms ^{-1}$ Change in KE $=\frac{1}{2}\left(6 \times 10^{3}\right) \times 1^{2}-\frac{1}{2}\left(5 \times 10^{3}\right)(1.2)^{2}$ $=600\, J$