Question

A particle is moving along a circular path with a constant speed of $10\text{m/s}$. Find the magnitude of 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 $10\text{m/s}$

Figure shows the initial and final velocity of the particle, where particle has revolved around the circle through an angle ${60}^{\circ }$


Change in velocity is given by, $\Delta \overrightarrow{v}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1$
Here, ${\overrightarrow{v}}_2$ and ${\overrightarrow{v}}_1$ represents the final and initial velocity respectively.
$\Rightarrow \left|\Delta \overrightarrow{v}\right|=\sqrt{v_1^2+v_2^2+2v_1{v}_2\cos \theta }...\left(i\right)$
$\Rightarrow \left|\Delta \overrightarrow{v}\right|=\sqrt{{10}^2+{10}^2-2\times 10\times 10\times \cos {60}^{\circ }}$
$\because v_1=v_2=10\text{m/s}$
$\Rightarrow \left|\Delta \overrightarrow{v}\right|=\sqrt{100}=10\text{m/s}$
Magnitude of change in velocity is $10\text{m/s}$