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Q.

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

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a

$\frac{3 m v}{T}$

b

$\frac{m v}{T}$

c

$0$

d

$\frac{3 m v}{2 T}$

answer is B.

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Detailed Solution

The momentum of the bead at A is $p_{i} = m v \hat{i}$
The momentum of the bead at B ${\vec{p}}_{f} = (- \frac{m v}{2}) \hat{i}$
Therefore, the magnitude of the change in momentum between A and B is
$Δ p = | {\vec{p}}_{f} - {\vec{p}}_{i} | = (\frac{3}{2}) m v$
Average force exerted by the bead on the wire is
$F_{a v} = \frac{Δ p}{Δ t} = \frac{3 m v}{2 T}$

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