question_answer

A U-shaped wire has a rough semicircular 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 \[B\] with velocity \[v/2\]after time \[T\]. The average force, exerted by the bead on the part \[A\]\[B\] of the wire is

A) \[\frac{mv}{2T}\]                     

B) \[\frac{3mv}{2T}\]

C) \[\frac{3mv}{T}\]         

D) None of these

Correct Answer: B

Solution :

[b] The momentum of the bead at A is \[{{\vec{P}}_{i}}=-mv\hat{i}\] The momentum of the bead at B \[{{\vec{P}}_{f}}=(mv/2)i\] Therefore, the magnitude of the change in momentum between A and B is \[\Delta p=\left| {{{\vec{p}}}_{f}}-{{{\vec{p}}}_{i}} \right|=(3/2)mv\] Average force exerted by the bead on the wire is \[{{F}_{av}}=\frac{\Delta p}{\Delta t}=\frac{3mv}{2T}\]