Question

A U-shaped wire has a semicircular bending between A and B as shown in figure. A bead of mass m moving with uniform speed v through a wire enters the semicircular bend at A and leaves at B with velocity v/2 after time T. The average force exerted by the bead on the part AB of the wire is:

• A. 0
• B. $\frac{3mv}{2T}$
• C. $\frac{3mv}{T}$
• D. none of these

Answer 1

Answer: (B)

$\frac{3mv}{2T}$

Velocity of bead at A and B is $v$ and $\frac{v}{2}$ respectively.

$\therefore$ Change in momentum $\Delta P=\frac{mv}{2}-(-mv)=\frac{3mv}{2}$

Thus the average force $F=\frac{\Delta P}{T}=\frac{3mv}{2T}$