A particle is moving on a circular path with a constant speed $v$. Its change of velocity as it moves from $A$ to $B$ in the figure is

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Aicharya · Accepted Answer

The given situation is shown below

From above figure we can observe that velocity vector (tangential to circle) can be resolved into horizontal $\left(v \cos \frac{\theta}{2}\right)$ and vertical $\left(v \sin \frac{\theta}{2}\right)$ components. As, particle is executing uniform circular motion, magnitudes of components remains same but direction of vertical component is reversed. So, Initial velocity $ v_1=v \cos \frac{\theta}{2} \hat{i}+v \sin \frac{\theta}{2} \hat{j} \text { and } $ final velocity, $v_2=v \cos \frac{\theta}{2} \hat{i}-v \sin \frac{\theta}{2} \hat{j}$ Change in velocity $ \begin{aligned} & =\text { final velocity }- \text { Initial velocity } \\ & =v_2-v_1 \\ & =\left(v \cos \frac{\theta}{2} \hat{i}-v \sin \frac{\theta}{2} \hat{j}\right)-\left(v \cos \frac{\theta}{2} \hat{i}+v \sin \frac{\theta}{2} \hat{j}\right) \\ & =-2 v \sin \frac{\theta}{2} \hat{j} \end{aligned} $ Magnitude of change of velocity will be $ \begin{aligned} |\Delta \mathbf{v}| & =\left|v_2-v_1\right| \\ & =2 v \sin \frac{\theta}{2} \end{aligned} $

A particle is moving on a circular path with a constant speed $v$. Its change of velocity as it moves from…

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