Question

A particle of mass $m$ is made to move with uniform speed $v_0$ along the perimeter of a regular hexagon, inscribed in a circle of radius $R$. The magnitude of impulse applied at each corner of the hexagon is

• A. $2m{v}_0\sin \frac{\pi }{6}$
• B. $m{v}_0\sin \frac{\pi }{6}$
• C. $m{v}_0\sin \frac{\pi }{3}$
• D. $2m{v}_0\sin \frac{\pi }{3}$

Answer 1

The correct option is A $2m{v}_0\sin \frac{\pi }{6}$

Impulse = change in momentum

Final velocity is given by :

$v=v\cos 60\hat{i}-v\sin 60\hat{j}$

$=\frac{v}{2}\hat{i}-\frac{\sqrt{3}}{2}v\hat{j}$

Change in velocity

$\Delta v=v-u$

$=v_0\hat{i}-(\frac{v_0}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j})$

$=\frac{v_0}{2}(\hat{i}+\sqrt{3}\hat{j})$

Change in momentum $\begin{align}

$=m\Delta v$

$=m\frac{v_0}{2}(\hat{i}+\sqrt{3}\hat{j})$

Since, impulse = change in momentum

So,

$impulseJ=m\frac{v_0}{2}(\hat{i}+\sqrt{3}\hat{j})$

$\left|J\right|=m{v}_0$

$or$

$\left|J\right|=2m{v}_{0}sin\frac{\pi }{6}$