A U-shaped smooth wire has a semi-circular bending between A and B as shown in fig. A bead of mass m moving with uniform speed v through the wire enters the semiculcular bent at A and leaves at B. Find the average force exerted by the bead on the part AB of the wire.

Options

A. 0
B. `(4mv^(2))/(pi d)`
C. `(2mv^(2))/(pi d)`
D. None of these

Answer

B

Explanation

Choosing the position x-y axis as shown in the figure, the momentum of the bead at A is `vec(P)_(i)= +mvec(v)`.
The momentum of the bead at B is `vec(P)_(f)=-mvec(v)`.
Therefore, the magnitude of the change in momentum between A and B is ` Delta vec(P)=vec(P)_(f) -vec(P)_(i)=-2mvec(v), i.e., Delta(p)=2mv` along the positive x axis.
The time taken by the bead to reach from A and B is
`Deltat=(pi.d//2)/(v)=(pi d)/(2v)`
Therefore, the average force exerted by the bead on the wire is
`F_(av) =(Delta p)/(Deltap) = (2mv//(pid)/(2v))=(4mv^(2))/(pi d)`.