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A fixed U-shaped smooth wire has a semi-...

A fixed `U`-shaped smooth wire has a semi-circular bending between `A` and `B` as shown in the figure. A bead of mass `'m'` moving with uniform speed `v` through the wire enters the semicircular bend at `A` and leaves at `B`. The magnitude of average force exerted by the bead on the part `AB` of the wire is

A

`0`

B

`(4mv^(2))/(pid)`

C

`(2mv^(2))/(pid)`

D

None of these

Text Solution

Verified by Experts

Choosing the positive `X-Y` axis as shown in the figure, the momentum of the bead at `A` is `vecp_(i)=+mvecv`. The momentum of the bead at `B` is `p_(f)=-mvecv`.
Therefore, the magnitude of the change in momentum between `A` and `B` is
`Delta vecp=vecp_(f)-vecp_(i)= -2mvecv`
The time interval taken by the bead to reach from `A` to `B` is
`Deltat=(pi*d//2)/(v)=(pid)/(2v)`.
Therefore, the average force exerted by the bend on the wire is
`F_(av)=(Delta p)/(Delta t)`
`=(2mv//(pid)/(2v))=(4mv^(2))/(pid)`
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