Question

A particle of mass $m$ is made to move with uniform speed $v$ along the perimeter of a regular hexagon. The magnitude of impulse applied at each corner is:

• A. $mv$
• B. $mv\sqrt{3}$
• C. $\frac{mv}{2}$
• D. $\frac{mv}{\sqrt{3}}$

Answer 1

The correct option is A $mv$

Let,
Initial momentum of particle is $p_i$,
Final momentum of particle is $p_f$
Now,
$p_i=mv\overrightarrow{i}$

$p_f=mv(\frac{1}{2}\overrightarrow{i}+\frac{1}{2}\overrightarrow{j})$

Impulse is change in momentum hence,
Impulse $=p_f-p_i$

Impulse $=mv(\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j})-mv\hat{i}$

Impulse $=mv(\frac{\sqrt{3}}{2}\hat{j}-\frac{1}{2}\hat{i})$

Hence, the magnitude is,

Impulse $=(mv{\left(\frac{\sqrt{3}}{2}\right)}^2+mv{\left(\frac{1}{2}\right)}^2)$

Impulse $=mv\sqrt{\frac{4}{4}}$

Impulse$=mv$