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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 $\mathrm{P}$ to $\mathrm{Q}$ is :$\left(\angle \mathrm{P~POQ}=40^{\circ}\right)$
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Verified Answer
The correct answer is:
$2 v \sin 20^{\circ}$
Let the velocity at point $P$
$\overrightarrow{v_1}=v \hat{1}$
So the velocity at point $\mathrm{Q}$ will be
$\overrightarrow{v_2}=v \cos 40^{\circ} \hat{i}-v \sin 40^{\circ} \hat{j}$
Change in velocity
$\overrightarrow{\Delta v}=\overrightarrow{v_2}-\overrightarrow{v_1}=\left(v \cos 40^{\circ} \hat{i}-v \sin 40^{\circ} \hat{j}\right)-v \hat{i}$
Magnitude of change of velocity
$|\Delta \mathrm{v}|=\sqrt{\left(\mathrm{v} \cos 40^{\circ}-\mathrm{v}\right)^2+\left(-\mathrm{v} \sin 40^{\circ}\right)^2}$
On solving we get
$|\Delta \mathrm{v}|=2 \mathrm{v} \sin 20^{\circ}$
$\overrightarrow{v_1}=v \hat{1}$
So the velocity at point $\mathrm{Q}$ will be
$\overrightarrow{v_2}=v \cos 40^{\circ} \hat{i}-v \sin 40^{\circ} \hat{j}$
Change in velocity
$\overrightarrow{\Delta v}=\overrightarrow{v_2}-\overrightarrow{v_1}=\left(v \cos 40^{\circ} \hat{i}-v \sin 40^{\circ} \hat{j}\right)-v \hat{i}$
Magnitude of change of velocity
$|\Delta \mathrm{v}|=\sqrt{\left(\mathrm{v} \cos 40^{\circ}-\mathrm{v}\right)^2+\left(-\mathrm{v} \sin 40^{\circ}\right)^2}$
On solving we get
$|\Delta \mathrm{v}|=2 \mathrm{v} \sin 20^{\circ}$
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