Question

A particle of mass m is made to move with speed v along the perimeter of a regular polygon of side 2n.The magnitude of the impulse applied at each corner of the polygon is?

Answer 1

Impulse=pf→−pi→Impulse=pf→−pi→

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 : vcosθi^−vsinθj^vcos⁡θi^−vsin⁡θj^

which gives Δvecp=mvcosθi^−mvsinθj^−mvi^Δvecp=mvcos⁡θi^−mvsin⁡θj^−mvi^

||Δp→||=m2v2cos2θ+m2v2−2mv2cos2θ+m2v2sin2θ−−−−−−−−−−−−−−−−−−−−−−−−−−√=mv(2(1−cosθ)−−−−−−−−−√)||Δp→||=m2v2cos2⁡θ+m2v2−2mv2cos2⁡θ+m2v2sin2⁡θ=mv(2(1−cosθ))

Now, it's clear that θ=π2−π2n⟹||Δp→||=mv(2(1−sinπ2n))−−−−−−−−−−−−√θ=π2−π2n⟹||Δp→||=mv(2(1−sin⁡π2n))

But answer given is: 2mvsinπ2n

Answer 2

Impulse is ∆p.
Now, ∆p in original direction is -mv2.
∆p in new direction is √3mv/2. 
Original direction means the direction of the particle along the side it's travelling along, and new direction is the direction it acquires after it turns to travel along the other side, located at 120°. 
Now, Magnitude of impulse is difference between the momenta along the original direction and new direction. Hence, magnitude of \begin{lgathered}\frac {-mv}{2}I + \frac {\sqrt{3}mv}{2} \\ \implies \sqrt{{\frac{mv}{2}}^{2} + {\frac{\sqrt{3}mv}{2}}^{2}} \\ =mv\end{lgathered}2−mv​I+23​mv​⟹2mv​2+23​mv​2​=mv​ 
So it will be MV.