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3-2.Motion in Plane
medium
A particle is moving on a circular path of radius $r$ with uniform speed $v$. The magnitude of change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^o)$
A$2v\, cos\, 40^o$
B$2v\, sin\, 40^o$
C$2v\, sin\, 20^o$
D$2v\, cos\, 20^o$
Solution
$\vec{v}_{i}=v \cos \theta \hat{i}+v \sin \theta \hat{j}$$\vec{v}_{f}=v \hat{i}$
$ \therefore \Delta \vec{v}=\vec{v}_{f}-\vec{v}_{i}$
$=v(1-\cos \theta) \hat{i}+v \sin \theta \hat{j}$
or $\quad \Delta \mathrm{v}=\mathrm{v} \sqrt{(1-\cos \theta)^{2}+(\sin \theta)^{2}}$
${=v \sqrt{1-2 \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta}}$
${=v \sqrt{2(1-\cos \theta)}}$
${=v \sqrt{2\left[1-\left(1-2 \sin ^{2} \frac{\theta}{2}\right)\right]}}$
${v=\sqrt{2 \times 2 \sin ^{2} \frac{\theta}{2}}}$
${=2 v \sin \frac{\theta}{2}}$
$\Rightarrow \Delta \mathrm{v}=2 \mathrm{v} \sin 20^{\circ} \quad\left(\because \quad \theta=40^{\circ}\right)$
Standard 11
Physics
