Pulley Dynamics in Horizontal plane

While block $B$ has not yet hit the pulley, let $x$ be the distance of block $A$ from the pulley, and let $T$ be the tension in the string. The only force acting on $A$ is central, so the angular momentum of $A$ is conserved during this motion, so that $x^2\dot{\theta} = \ell v$ . We also have the radial equation of motion $m(\ddot{x} - x\dot{\theta}^2) \; = \; -T$ while the equation of motion of $B$ is simply $m\ddot{x} \; = \; T$ and hence $2\ddot{x} - \ell^2v^2 x^{-3} \; = \; 0$ The usual change of variable $u = x^{-1}$ yields $\dot{x} \,=\, -\ell v \frac{du}{d\theta}$ and $2\frac{d^2u}{d\theta^2} + u \; = \; 0$ and hence (since $u = \ell^{-1}$ and $\tfrac{du}{d\theta}=0$ when $\theta=0$ ) we have $u \; = \; \tfrac{1}{\ell}\cos\big(\tfrac{\theta}{\sqrt{2}}\big)$ This stage of the motion lasts until $x = 2\ell$ , which happens when $\cos\big(\tfrac{\theta}{\sqrt{2}}\big) = \tfrac12$ , so when $\theta = \tfrac{\sqrt{2}\pi}{3}$ . During this motion $\dot{\theta} \; = \; \ell v x^{-2} \; = \; \ell v u^2 \; = \; \tfrac{v}{\ell} \cos^2\big(\tfrac{\theta}{\sqrt{2}}\big)$ so that the first stage of the motion lasts for time $t_1$ , where $t_1 \; = \; \tfrac{\ell}{v}\int_0^{\frac{\sqrt{2}\pi}{3}}\sec^2\big(\tfrac{\theta}{\sqrt{2}}\big)\,d\theta \; = \; \tfrac{\ell}{v}\sqrt{6}$ For the second stage of the motion (since $(2\ell)^2\dot{\theta} = v \ell$ ), particle $A$ moves in a circular arc of radius $2\ell$ with constant angular velocity $\tfrac{v}{4\ell}$ , and so the additional time until particle $A$ hits the wall is $t_2 \; = \; \tfrac{4\ell}{v}\left(\tfrac12\pi - \tfrac{\sqrt{2}\pi}{3}\right) \; = \; \tfrac{2\pi\ell}{v}\left(1 - \tfrac{2\sqrt{2}}{3}\right)$ making the required total time $t_1 + t_2 \; = \;\tfrac{\ell}{v}\left(\sqrt{6} + 2\pi\left(1 - \tfrac{2\sqrt{2}}{3}\right)\right)$ which makes the answer $6 + 2 + 2 + 3 = \boxed{13}$ .