Question

A body of mass 1 $kg$ travelling due East with $3m/s$ turns due North to travel with the same velocity. Find the net force exerted on the body for it to turn if it takes $0.01$ sec to turn

• A. $3\sqrt{2}\times {10}^{2}N$
• B. $5\sqrt{2}\times {10}^{2}N$
• C. $5\times {10}^{2}N$
• D. $3\times {10}^{2}N$

Answer 1

The correct option is A $3\sqrt{2}\times {10}^{2}N$

From newton's second law,
$F=m\overrightarrow{a}=m\frac{(\overrightarrow{v}-\overrightarrow{u})}{t}$
Where,
$\overrightarrow{v}=\text{Final velocity}=3\hat{j}$
$\overrightarrow{u}=\text{Initial velocity}=3\hat{i}$
Hence,
$\overrightarrow{F}=m\frac{(\overrightarrow{v}-\overrightarrow{u})}{t}=\frac{1(3\hat{j}-3\hat{i})}{0.01}$
Hence, magnitude of force $F$$=3\sqrt{2}\times {10}^{2}N$