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An open water tight railway wagon of mass $5 \times 10^3 kg$ coasts at an initial velocity $1.2 m/s$ without friction on a railway track. Rain drops fall vertically downwards into the wagon. The velocity of the wagon after it has collected $10^3 kg$ of water will be .............. $\mathrm{m}/ \mathrm{s}$
$0.5$
$2$
$1$
$1.5$
Solution
Given, Initial mass of wagon, $m=5 \times 10^{3} k g$
Initial velocity of wagon, $v=1.2 \mathrm{ms}^{-1}$
After collecting rain drops the mass of wagon becomes
$m^{\prime}=m+10^{3} k g=6 \times 10^{3} k g$
Let, $v^{\prime}$ be the velocity of the wagon after collecting rain drops. Wagon is moving without friction on a railway track, thus momentum will remain conserved, hence,
$m v=m^{\prime} v^{\prime}$
$v^{\prime}=\frac{m v}{m^{\prime}}$
$v^{\prime}=\frac{5 \times 10^{3} \times 1.2}{6 \times 10^{3}}$
$v^{\prime}=\frac{5 \times 1.2}{6}=1 m s^{-1}$
