Exam

JEE Main

Year

2019

Date

11 January

Shift

Morning Shift

Subject

Physics

Chapter

Motion in Two Dimensions

JEE Main 2019 · 11 January · Morning Shift · Physics · Motion in Two Dimensions

A particle is moving along a circular path with a constant speed of...

A particle is moving along a circular path with a constant speed of $10 \mathrm{~ms}^{-1}$. 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?

Step-by-step solution

Change in velocity,
\\$$\begin{array}{l}\\ |\Delta \overline{\mathrm{v}}|=\sqrt{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+2 \mathrm{v}_{1} \mathrm{v}_{2} \cos (\pi-\theta)}\\ \\ =2 \mathrm{v} \sin \frac{\theta}{2} \quad\left(\because\left|\overrightarrow{\mathrm{v}}_{1}\right|=\left|\overrightarrow{\mathrm{v}}_{2}\right|\right)=\mathrm{v}\\ \\ =(2 \times10) \times \sin \left(30^{{}^{\circ}}\right)=2 \times10 \times \frac{1}{2}\\ \\ =10 \mathrm{~m}/ \mathrm{s}\\ \end{array}$$\\