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Rod of m and length l is free to rotate ...

Rod of m and length l is free to rotate about point U in vertical plane. A particle having same mass m moving horizontally with velocity `v_(0)` his the rod perpendicular at distance `(l)/(4)` from the top end 'O' and stops. Find impulse due to hinge on the rod due to collision.

A

`(mv_(0))/(19)`

B

`(-10mv_(0))/(19)`

C

`mv_(0)`

D

`(5mv_(0))/(8)`

Text Solution

Verified by Experts

`mv_(0)xx(l)/(4)=(ml^(2))/(3)omega`
implse `=mv_(0)-m(3v_(0))/(4l)xx(l)/(2)`
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