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Standard XII

Physics

Second Law of Motion

A train start...

Question

A train starts from rest with a constant acceleration $a = 2 m s^{- 2}$ . After 5 s, a stone is dropped from the window of the train. If the window is at a height of 2 m from the ground, what is the magnitude of the velocity of the stone 0.2 s after it was dropped? Take $g = 10 m s^{- 2}$ .

$4 \sqrt{6} m s^{- 1}$

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$10 m s^{- 1}$

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$2 \sqrt{26} m s^{- 1}$

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$12 m s^{- 1}$

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Solution

The correct option is C
$2 \sqrt{26} m s^{- 1}$

At t = 5 s, velocity of the train, $v = u + a t = 0 + 2 \times 5 = 10 m s^{- 1}$
As the stone is dropped from the moving train, its initial velocity will be equal to the velocity of the train. However, its acceleration will be equal to acceleration due to gravity in the downward direction.
Now, $v_{x} = u_{x} + a_{x} t = 10 + 0 \times 0.2 = 10 m / s$
$v_{y} = u_{y} + a_{y} t = 0 + 10 \times 0.2 = 2 m / s$
Magnitude of velocity at t = 0.2 s, $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(10)^{2} + (2)^{2}} = \sqrt{104} = 2 \sqrt{26} m s^{- 1}$
Hence, the correct choice is (c).

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A train starts from rest with a constant acceleration a=2 ms−2. After 5 s, a stone is dropped from the window of the train. If the window is at a height of 2 m from the ground, what is the magnitude of the velocity of the stone 0.2 s after it was dropped? Take g=10 ms−2.

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