Q. a particle of mass m is made to move with uniform speed v along the perimeter of a regular hexagon.What is the magnitude of the impulse applied at each corner of the hexagon ?

consider a hexagon placed in X-Y plane whose bottom side is on x axis...

first u draw this figure ...

let its two vertex which are on x axis are A,B &  particle is moving from A to B along +ve x axis...

initial velocity of particle is U =vi

at point B its velocity changes in direction...its velocity makes an angel of 60 degree with x axis ..

final velocity is Vf= vcos60i + vsin60j=(i + sqrt3.j)v/2

change in velocity is = final velocity - initial velocity

                             =(v/2-v)i + sqrt3.vj/2

                             =-vi/2 + sqrt3.vj/2

impulse = mdv=mv/2 (-i + sqrt3j)

           I=  mv in magnitude

Dear Raj

see the pic for solution

All the best.

AKASH GOYAL

AskiitiansExpert-IITD

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