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Raj Deol Grade:
        

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  ?

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

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

6 years ago
AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

Dear Raj


see the pic for solution


All the best.


AKASH GOYAL


AskiitiansExpert-IITD


 


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1279_23499_Photo-0104.jpg

6 years ago
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