Two indentical ball A and B are release from the positions shown in figure. They collied elastically on horizontal portionMN. The ratio of the heights attained by A and B after collision will be: (Neglect friction)

The velocities of the balls fluctuate after collision.

$v_A=\sqrt{2gh} \& v_B=\sqrt{2g\left(4h\right)}=2\sqrt{2gh}$

Height attained by A,

$h_A=\frac{{v_A}^2}{2g}=h$

But as depicted in the image, the course of B will initially be linear and subsequently parabolic.
Using energy-saving techniques for ball B,

$$\begin{gathered} \frac{1}{2}m{v_B}^2=\frac{1}{2}m{v}^2+mgh \\ v=\sqrt{6gh} \\ {h}_B=h+\frac{v^2{\sin }^2{60}^{°}}{2g}=h+\frac{9h}{4}=\frac{13h}{4} \\ \frac{h_A}{h_B}=\frac{4}{13} \end{gathered}$$

Hence the correct answer is 4:13.