Work, Power and Energy Part 2

Concept:Horizontal component of velocity remains constant for both beads since there is no horizontal force.Explanation:For bead Q on horizontal string AB, length is $2R$.No gravity along motion, so speed is constant $v$.Time taken: $t_Q = \frac{2R}{v}$.For bead P on semicircular path ACB, at $t=0$ at point S, horizontal velocity is $v$.As it slides, horizontal component stays $v$.Horizontal displacement from S to B is $S'O + OB$.$S'O = R\cos45^\circ = \frac{R}{\sqrt{2}}$, $OB = R$.Total horizontal distance = $\frac{R}{\sqrt{2}} + R = R\left(1 + \frac{1}{\sqrt{2}}\right)$.Time taken: $t_P = \frac{R\left(1 + \frac{1}{\sqrt{2}}\right)}{v}$.Compare times: $\frac{t_Q}{t_P} = \frac{2R/v}{R(1+1/\sqrt{2})/v} = \frac{2}{1+1/\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}+1}$.Simplify: $= 4 - 2\sqrt{2} \approx 1.172 > 1$.Thus $t_Q > t_P$, i.e., $t_P < t_Q$.Answer:$t_P < t_Q$, option A.