Question

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

• A. $t=\frac{l}{\sqrt{3gh}}$
• B. $t=\frac{3l}{\sqrt{2gh}}$
• C. $t=\frac{l}{\sqrt{2gh}}$
• D. $t=\frac{l}{\sqrt{gh}}$

Answer 1

The correct option is C $t=\frac{l}{\sqrt{2gh}}$

Before collision, the velocity of each ball is $v=\sqrt{2gh}$ in the same direction.

After collision, velocity of each ball is $v=\sqrt{2gh}$ in opposite direction.
Since, collision is elastic,
the velocity of approach $=$ velocity of seperation
$\Rightarrow v_a=v_s=2\sqrt{2gh}$

Hence, the two balls will collide after a time, $t=\frac{l}{2\sqrt{2gh}}$

Also, the string becomes taut after, $t=\frac{l}{2\sqrt{2gh}}$

Total time $=\frac{l}{2\sqrt{2gh}}+\frac{l}{2\sqrt{2gh}}=\frac{l}{\sqrt{2gh}}$