A particle of mass m is made to move with uniform sped v along the perimeter of a regular polygon of $2n$ sides. The magnitude of impulse applied at each corner of the polygon is:
$\mathbb {Impulse} = \vec{p_f}-\vec{p_i} $
I assumed the particle to be moving on the top edge of a regular hexagon. Here's my diagram:
Now, when the particle moves to the other edge, it still has speed v but velocity vector is split into two components : $v\cos\theta \hat{i}- v\sin\theta\hat{j}$
which gives $\Delta vec{p} = mv\cos\theta \hat{i}- mv\sin\theta\hat{j} – mv\hat{i}$
$||\vec{\Delta p}||= \sqrt{m^2v^2\cos^2\theta +m^2v^2 – 2mv^2\cos^2\theta+m^2v^2\sin^2\theta} \\ = mv(\sqrt{2(1-cos\theta)}) $
Now, it's clear that $\theta = \dfrac{\pi}{2}- \dfrac{\pi}{2n} \\ \implies ||\vec{\Delta p}||= mv\sqrt{(2(1- \sin \dfrac{\pi}{2n}))}$
But answer given is: $2mv\sin\dfrac{\pi}{2n}$
Where have I gone wrong? WolframAlpha doesn't simply my result too.
Best Answer
Use the fact that exterior angles of any polygon sum up to $2\pi$. Then each exterior angle of this $2n$-gon will be $\dfrac{\pi}{n}$. Then the change in momentum vector will have an angle $\theta=\pi-\frac{\pi}{n}$. This is the angle between blue and red vectors shown below:
Resolve along and perpendicular the angular bisector of this angle (grey line). Components along grey line add up, while components along black line cancel. So the answer is simply
$$\begin{align}\Delta p &= 2 mv \cos\left(\frac{\theta}{2}\right)\\ &= 2 mv \cos\left(\frac{\pi-\tfrac{\pi}{n}}{2}\right)\\ &= 2 mv \sin\left(\frac{\pi}{2n}\right)\end{align}$$