The small marble is projected with a velocity of 10 m s^-1 in a direction 45^o from the horizontal y-direction on the smooth inclined plane. Calculate the magnitude v of its velocity after 2 s.

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$10 \sqrt{2} {ms}^{- 1}$

$5 {ms}^{- 1}$

$10 {ms}^{- 1}$

$5 \sqrt{2} {ms}^{- 1}$

answer is C.

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Detailed Solution

$Given (V = 10 {ms}^{- 1})$

$After 2s: V_{x} = \frac{10}{\sqrt{2}} - \frac{10}{\sqrt{2}} \times 2 \Rightarrow V_{x} = - \frac{10}{\sqrt{2}} {ms}^{- 1}$

$\begin{array}{l} V_{x} = - \frac{10}{\sqrt{2}} {ms}^{- 1} and V_{y} = - \frac{10}{\sqrt{2}} {ms}^{- 1} \\ V = \sqrt{\frac{100}{2} + \frac{100}{2}} = 10 \sqrt{\frac{1}{2} + \frac{1}{2}} = 10 {ms}^{- 1} \end{array}$

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