Question

A particle is moving along a circular path with a constant speed of $10{\text{ms}}^{-1}$. 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. $10\sqrt{2}\text{m/s}$
• B. $10\text{m/s}$
• C. $10\sqrt{3}\text{m/s}$
• D. $\text{Zero}$

Answer 1

The correct option is B $10\text{m/s}$


Change in velocity is given as:
$\Delta \overrightarrow{v}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1={\overrightarrow{v}}_2+(-{\overrightarrow{v}}_1)$

Here, $\theta ={120}^{\circ }$ is the angle between the vectors ${\overrightarrow{v}}_2$ and $-{\overrightarrow{v}}_1$ as represented in the figure.

Therefore, magnitude of change in velocity is given by:

${\left|\Delta \overrightarrow{v}\right|}^2=v_1^2+v_2^2+2v_1{v}_2\cos {120}^{\circ }$

$\because v_1=v_2=v=10\text{m/s}$
$\left|\Delta \overrightarrow{v}\right|=\sqrt{v^2+v^2+2v\times v\times (-\frac{1}{2})}$
$\Rightarrow \left|\Delta \overrightarrow{v}\right|=v=10\text{m/s}$