A particle of mass m, initially at rest, is acted upon by a variable force F for a brief interval of time T. It begins to move with a velocity u after the force stops acting. F is shown in the graph as a function of time. The curve is a semicircle.

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$u = \frac{π F_{0}^{2}}{2 m}$

$u = \frac{π T^{2}}{8 m}$

$u = \frac{π F_{0} T}{4 m}$

$u = \frac{F_{0} T}{2 m}$

answer is C.

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Detailed Solution

Initially particle was at rest. By the application of force its momentum increases.
Final momentum of the particle = Area of F - t graph
$\Rightarrow m u =$ Area of semi circle
$m u = \frac{π r^{2}}{2} = \frac{π r_{1} r_{2}}{2} = \frac{π (F_{0}) (T / 2)}{2} \Rightarrow u = \frac{π F_{0} T}{4 m}$

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