Physics - Circular Motion Question with Solution | TestHub

Let half the cord be $a=\frac{3\sqrt{3}}{2}$ m; when the load descends $y$, each sleeve sits at $x=\sqrt{a^2-y^2}$, and $x\dot{x}+y\dot{y}=0$ gives $v_{sleeve}=\frac{y}{x}u$.

Energy: $2mgy=\frac{1}{2}(2m)u^2+m{\left(\frac{yu}{x}\right)}^2$ ⇒ $u^2=\frac{2gy(a^2-y^2)}{a^2}$.

Maximum at $y=\frac{a}{\sqrt{3}}=1.5$ m - (C); $u_{max}^2=\frac{2\times 10\times 1.5\times 4.5}{6.75}=20$, so $u_{max}=2\sqrt{5}$ m/s - (B).

Sleeve speed $v_s^2=\frac{2g{y}^3}{a^2}$ rises monotonically till the sleeves collide, so $\sqrt{10}$ m/s is not their maximum - (A) wrong.

At the maximum-speed instant the load's acceleration vanishes and the sleeves' momenta cancel horizontally, so the net force on the system becomes zero once - (D)