A body of mass 1 kg is moving with velocity 30 ms^–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(3/5)

100 m/s ; tan^-1(4/5)

100 m/s ; tan^-1(4/3)

50 m/s ; tan^-1(3/4)

answer is A.

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Detailed Solution

$\large \vec v_i = 30\vec j\;m/s, \vec F=-10\vec i\;N$
$\large \therefore m\vec v_f-m\vec v_i=\vec F\times t$
$\fn_phv \large \Rightarrow \vec v_f-\vec V_i=\frac {\vec F. t}{m}$
$\large =30\hat j-{(\frac {10}{1}\times 4)}\hat i$
$\large =(30\hat j-40\hat i)m/s$
$\large \therefore tan\theta=\frac {30}{40}=\frac 34\Rightarrow \theta=tan^{-1}(\frac 34)$
$\large |\vec v_f|=\sqrt {30^2+40^2}=50m/s$
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