A bob of mass $m$, suspended by a string of length $\ell_1$, is given the minimum velocity required to complete a full circle in the vertical plane. At the
highest point, it collides elastically with another bob of mass $m$ suspended by a string of length $\ell_2$ which is initially at rest. Both strings are
mass-less and inextensible. If the second bob, after collision, acquires the minimum speed required to complete a full circle in the vertical
minimum speed required to complete a full circle in the vertical plane, find the ratio $\ell_1/\ell_2$.
Proposed by the user "phalit" on the website dailyintegral.com.
In order to find such minimum speed, it is convenient to think of the two most important points of the motion of the first ball, namely the starting point
where it will have a speed $v_0$, and the apogee of its trajectory, whose speed will be named $v_1$.
At the last scenario, there are two forces acting upon the ball which constrain its movement to a circular one: gravity and the tension of the string, i.e., if $a_\perp$ is the centripetal acceleration at that point, it follows that $a_\perp = g + T/m$, and so $$ \frac{v_1^2}{\ell_1} = a_\perp = g + \frac{T}{m} \implies v_1 = \sqrt{\ell_1 \left(g + \frac{T}{m}\right)}. $$ We can now invoke the law of conservation of energy to find $v_0$, setting the origin of our frame of reference at the rest position of the first bob: $$ \begin{align*} K_1 + U_1 &= \frac{1}{2} m \ell_1 \left(g + \frac{T}{m}\right) + 2mg\ell_1 = \frac{1}{2}mv_0^2 = K_0 + U_0\\ \implies v_0 &= \sqrt{\ell_1 \left(5g + \frac{T}{m} \right)}, \end{align*} $$so since $m, \ell_1, g$ and $m$ are all constant, we conclude that the minimum speed is achieved when $T=0$, and thus $$ v_0 = \sqrt{5g\ell_1},\qquad v_1 = \sqrt{g\ell_1}. $$ None of the past assumptions are intrinsic to the first ball only, the same reasoning applies to the second one: $$ v_0' = \sqrt{5g\ell_2},\qquad v_1' = \sqrt{g\ell_2}. $$
Now consider the collision happening at the top; because of Newton's third law and the fact that the collision is elastic, the momenta of the system is preserved. From this relation we can finally solve for $\ell_1/\ell_2$: $$ mv_1 = mv_0' \implies \sqrt{g\ell_1} = \sqrt{5g\ell_2} \implies \color{green}\boxed{\frac{\ell_1}{\ell_2} = 5.} $$
At the last scenario, there are two forces acting upon the ball which constrain its movement to a circular one: gravity and the tension of the string, i.e., if $a_\perp$ is the centripetal acceleration at that point, it follows that $a_\perp = g + T/m$, and so $$ \frac{v_1^2}{\ell_1} = a_\perp = g + \frac{T}{m} \implies v_1 = \sqrt{\ell_1 \left(g + \frac{T}{m}\right)}. $$ We can now invoke the law of conservation of energy to find $v_0$, setting the origin of our frame of reference at the rest position of the first bob: $$ \begin{align*} K_1 + U_1 &= \frac{1}{2} m \ell_1 \left(g + \frac{T}{m}\right) + 2mg\ell_1 = \frac{1}{2}mv_0^2 = K_0 + U_0\\ \implies v_0 &= \sqrt{\ell_1 \left(5g + \frac{T}{m} \right)}, \end{align*} $$so since $m, \ell_1, g$ and $m$ are all constant, we conclude that the minimum speed is achieved when $T=0$, and thus $$ v_0 = \sqrt{5g\ell_1},\qquad v_1 = \sqrt{g\ell_1}. $$ None of the past assumptions are intrinsic to the first ball only, the same reasoning applies to the second one: $$ v_0' = \sqrt{5g\ell_2},\qquad v_1' = \sqrt{g\ell_2}. $$
Now consider the collision happening at the top; because of Newton's third law and the fact that the collision is elastic, the momenta of the system is preserved. From this relation we can finally solve for $\ell_1/\ell_2$: $$ mv_1 = mv_0' \implies \sqrt{g\ell_1} = \sqrt{5g\ell_2} \implies \color{green}\boxed{\frac{\ell_1}{\ell_2} = 5.} $$