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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  ?
```

6 years ago

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```										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
```										Dear Raj
see the pic for solution
All the best.
AKASH GOYAL

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
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```
6 years ago

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