Question

A fixed U-shaped smooth wire has a semi-circular bending between A and B as shown in 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 average force exerted by the bead on the part AB of the wire is

• A. $0$
• B. $\frac{4m{v}^2}{\pi d}$
• C. $\frac{2m{v}^2}{\pi d}$
• D. None of these

Answer 1

The correct option is B $\frac{4m{v}^2}{\pi d}$

$\begin{array}{c}AsthebeadtravelsfromAtoB,\\ Changeofmomentum=2mv\\ Timeoftravel=\frac{s}{v}=\frac{\left(\frac{\pi d}{2}\right)}{v}=\frac{\pi d}{2v}\\ \therefore Averageforce=\frac{changeinmomentum}{timeoftravel}=\frac{2mv}{\left(\frac{\pi d}{2v}\right)}=\frac{4m{v}^2}{\pi d}\end{array}$

Hence

option $\left(B\right)$ is correct answer.