Question

A particle is moving along a circular path with a constant speed of $10m/s$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of ${60}^{\circ }$ around the centre of the circle?

• A. $10\sqrt{3}m/s$
• B. $10m/s$
• C. $10\sqrt{2}m/s$
• D. Zero

Answer 1

The correct option is B $10m/s$

Given, $\left|{\overrightarrow{v}}_1\right|=\left|{\overrightarrow{v}}_2\right|=v=10m/s$
And angle between $v_1$ and $v_2$ is ${60}^{\circ }$
Magnitude of the change in velocity,
$|\Delta \overrightarrow{v}|=\sqrt{v_1^2+v_2^2+2v_1{v}_2\cos (\pi -\theta )}$
$|\Delta \overrightarrow{v}|=\sqrt{v^2+v^2+2v^2\cos \theta }$
$|\Delta \overrightarrow{v}|=2v\sin \frac{\theta }{2}$
$|\Delta \overrightarrow{v}|=(2\times 10)\sin \left({30}^{\circ }\right)$
$|\Delta \overrightarrow{v}|=10m/s$