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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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