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A body of mass 1 kg is moving with velocity $30 m s^{- 1}$ due north. It is acted on by a force of 10 N due west for 4 seconds. Find the velocity of the body after the force ceases to act.

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50 m/s ; $\tan^{- 1} (\frac{3}{4})$

100 m/s ; $\tan^{- 1} (\frac{4}{3})$

50 m/s ; $\tan^{- 1} (\frac{3}{5})$

100 m/s ; $\tan^{- 1} (\frac{4}{5})$

answer is A.

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Detailed Solution

$m = 1 k g \Rightarrow a_{x} = \frac{F_{x}}{m} = \frac{10}{1} = 10 m^{- 2} v_{x} = u_{x} + a_{x} t = 0 + 10 (4) = 40 m s^{- 1} v_{y} = u_{y} + a_{y} t = 30 + 0 (4) = 30 m s^{- 1}$

$V = \sqrt{V_{x}^{2} + V_{y}^{2}} = \sqrt{30^{2} + 40^{2}}$

$= 50 m s^{- 1}$

$T a n θ = \frac{v_{y}}{v_{x}} = \frac{30}{40} = \frac{3}{4}$

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