Question

A small ball of mass '$m$' is connected by an inextensible massless string of length '$l$' with another balls of mass $M=4m$. They are released with zero tension in the string from a length '$h$' as shown. The string becomes taut for the first time after the mass '$m$' collides with the ground after a time $\frac{nl}{\sqrt{4gh}}$. Then $n^2$ = .
(Assume all collision to be elastic) (Answer upto 2 decimal digits)

Answer 1

The velocities of both the balls just after the bigger ball hits the ground will be $\sqrt{2gh}$ in opposite directions. Then both the balls move towards each other and cover $l$ meters to collide again. Since accelerations are same for both the balls or relative acceleration is zero, the time taken for collision is:
$t_1=\frac{L}{2\sqrt{2gh}}s$
At the time of collision of two balls relative velocity of approach = $2\sqrt{2gh}m/s$
Since the collision is elastic, velocity of separation = $2\sqrt{2gh}m/s$
$\therefore$ The string becomes taut after time $t_2=\frac{l}{2\sqrt{2gh}}s$
$\therefore$ Total time is $\frac{l}{\sqrt{2gh}}s$.
Hence $\frac{l}{\sqrt{2gh}}=\frac{nl}{\sqrt{4gh}}\Rightarrow \frac{1}{\sqrt{2}}=\frac{n}{2}\Rightarrow n=\sqrt{2}.$