Question

A particle of mass $m$ is mode 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. 2$m{v}_0\sin \pi /6$
• B. $m{v}_0\sin \pi /6$
• C. $m{v}_0\sin \pi /3$
• D. 2$m{v}_0\sin \pi /3$

Answer 1

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

Given,

Radius of hexagon $R$

Mass of particle is $m$

Velocity of particle is $v_0$

Angle between two side is $\frac{360}{6}={60}^{\circ }$

Magnitude of impulse is resultant force of momentum

$I=\sqrt{{\left(m{v}_0\right)}^2+{\left(m{v}_0\right)}^2-2\left(m{v}_0\right)\left(m{v}_0\right)cos\left(60\right)}I=m{v}_0\sqrt{2-2cos\left(60\right)}I=m{v}_0\sqrt{2\times 2{sin}^2\left(30\right)}I=2m{v}_{0}sin\left(\frac{\pi }{6}\right)$

Hence, correct option is $A$