Question

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

Answer 1

Consider a hexagon placed in x-y plane where bottom side is on x-axis.

Let us imagine the figure,

Let its two vertex which are on x-axis are A,B & particle is moving from A to B a long +ve x-axis.

Initial velocity of particle is $u=vi$ at point B its velocity charger in direction its velocity maker as angle of 60 degree with x-axis.

Final velocity is vf $=v\cos 60+v\sin 60$

$=(i+\sqrt{3})\frac{v}{2}$

Change in velocit is $=$ final velocity $-$ initial velocity

$=(\frac{v}{2}-v)i+\frac{\sqrt{3v}j}{2}$

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

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

$\Rightarrow i=mv$ in magnitude

Hence, solve.