Question

A $U-$shaped smooth wire has a semi-circular bending between $A$ and $B$ as shown in the figure. A bead of mass ${}^{\prime }{m}^{\prime }$ 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.

• A. $0$
• B. $\frac{4m{v}^2}{\pi d}$
• C. $\frac{2m{v}^2}{\pi d}$
• D. $Noneofthese$

Answer 1

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

Choosing the positive $x-y$ axis as shown in the figure, the momentum of the bead at $A$ is $\overrightarrow{p_i}=+m\overrightarrow{v}$. The momentum of the bead at $B$ is $\overrightarrow{p_f}=-m\overrightarrow{v}$. Therefore, the magnitude of the change in momentum between $A$ and $B$ is
$\Delta \overrightarrow{p}=\overrightarrow{p_f}-\overrightarrow{p_i}=-2m\overrightarrow{v}$
i.e. $\Delta p=2mv$
The time interval taken by the bead to reach from $A$ to $B$ is
$\Delta t=\frac{\pi .d/2}{v}=\frac{\pi d}{2v}$.
From Newton's third law, the average force exerted by the bead on the wire is
$F_{av}=\frac{\Delta p}{\Delta t}=\frac{2mv}{\frac{\pi d}{2v}}=\frac{4m{v}^2}{\pi d}$
Therefore $\left(B\right)$.