An open water tight railway wagon of mass $5 \times 10^{3} k g$ 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 $10^{3} k g$ of water will be:

0.5 m / s

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2 m / s

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1 m / s

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1.5 m / s

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Solution

The correct option is C $1 m / s$
Given, Initial mass of wagon, $m = 5 \times 10^{3} k g$

Initial velocity of wagon, $v = 1.2 m s^{- 1}$

After collecting rain drops the mass of wagon becomes
$m^{'} = m + 10^{3} k g = 6 \times 10^{3} k g$

Let, $v^{'}$ 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,

$m v = m^{'} v^{'}$

$v^{'} = \frac{m v}{m^{'}}$

$v^{'} = \frac{5 \times 10^{3} \times 1.2}{6 \times 10^{3}}$

$v^{'} = \frac{5 \times 1.2}{6} = 1 m s^{- 1}$

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An open water tight railway wagon of mass $5 \times 10^{3} k g$ 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 $10^{3} k g$ of water will be: