Question

A small ball of mass $m$ is connected by an inextensible massless string of length $l$ with another ball of mass $M=4\text{m}$. They are released with zero tension in the string from a height $h$ as shown in the figure. The time after which string becomes taut for the first time after the release(after the mass $M$ collides with the ground ) is $\frac{l}{\sqrt{ngh}},$ where $n=$
(Assume all collisions to be elastic)

Answer 1

Before collision, the velocity of each ball is $V=\sqrt{2gh}$ in the same direction.
After collision between M and ground,
the velocity of $M$ is $V=\sqrt{2gh}$ in the opposite direction while velocity of $m$ remains same.
After collision, the velocity of approach is $\sqrt{2gh}-(-\sqrt{2gh})=2\sqrt{2gh}$

$\therefore$ The two balls will collide after a time$=\frac{\text{Distance between them}}{V_{app}}=\frac{l}{2\sqrt{2gh}}$
After collision between M and m,

their velocities can be witten as,
$V_{Mf}=\frac{M-m}{M+m}\sqrt{2gh}-\frac{2m}{M+m}\sqrt{2gh}$
$V_{mf}=\frac{2M}{M+m}\sqrt{2gh}-\frac{m-M}{M+m}\sqrt{2gh}$
Given that $M=4m$
$\therefore$
$V_{Mf}=\frac{4m-m}{4m+m}\sqrt{2gh}-\frac{2m}{4m+m}\sqrt{2gh}=\frac{\sqrt{2gh}}{5}$
$V_{mf}=\frac{8m}{4m+m}\sqrt{2gh}-\frac{m-4m}{4m+m}\sqrt{2gh}=\frac{11\sqrt{2gh}}{5}$
After collision, the velocity of seperation is $\frac{11\sqrt{2gh}}{5}-\left(\frac{\sqrt{2gh}}{5}\right)=2\sqrt{2gh}$

$\therefore$ The string joining them will be taut after time $t_2=\frac{\text{length of string}}{V_{sep}}=\frac{l}{2\sqrt{2gh}}$
$\therefore$ Total time is $t_1+t_2=\frac{l}{\sqrt{2gh}}$
$\therefore n=2$