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`        A particle of mass m is made to move with uniform speed v along the perimeter of a regular hexagon the magnitude of the impulse applied at each corner of the hexagon is`
one year ago

Arun
23345 Points
```							Dear Rahul  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 =viat 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/2change in velocity is = final velocity - initial velocity                             =(v/2-v)i + sqrt3.vj/2                             =-vi/2 + sqrt3.vj/2impulse = mdv=mv/2 (-i + sqrt3j)           I=  mv in magnitude RegardsArun
```
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions