Question

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

• A. $50m/s;{\tan }^{-1}\left(\frac{3}{4}\right)$
• B. $100m/s;{\tan }^{-1}\left(\frac{4}{3}\right)$
• C. $50m/s;{\tan }^{-1}\left(\frac{3}{5}\right)$
• D. $100m/s;{\tan }^{-1}\left(\frac{4}{5}\right)$

Answer 1

The correct option is A $50m/s;{\tan }^{-1}\left(\frac{3}{4}\right)$

Given,

Force act in direction of west $F=10N$

Velocity toward north Is $V_n=30m{s}^{-1}$

Initial velocity in west direction is $u=0m/s$

Acceleration in west direction is $a=\frac{F}{M}=\frac{10}{1}=10m/s^2$

Final velocity after time 4 sec, in west direction is $V_s=u+at=0+10\times 4=40m/s$

Resultant velocity is $V_R=\sqrt{{V_s}^2+{V_n}^2}=\sqrt{{30}^2+{40}^2}=50m{s}^{-1}$

Direction of resultant velocity makes angle $\theta$ with the direction of force act.

$\tan \theta =\frac{V_n}{V_s}=\frac{30}{40}=\frac{3}{4}$

$\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$

Hence the velocity of the body after the force ceases will be $50m/s$ todards an angle $\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$ with the direction of force.