Two identical balls A and B are released from the positions shown in figure. They collide elastically on horizontal portion MN. The ratio of heights attained by A and B after collision will be (neglect friction)

A & B of same mass particles & elastic collision. Velocities are interchanged after collision .

${\text{v}}_{\text{A}}={\text{u}}_{\text{B}}$ So it reaches height ‘h’

${\text{v}}_{\text{B}}={\text{u}}_{\text{A}}$ So it reaches height ‘4h’

But there is only a height h inclination. So that at point ‘P’ the body b projects obliquely

At point ‘P’ it possess both P.E & K.E

${\left(\text{Total}\text{energy}\right)}_{\text{Q}}={\left(\text{T}\text{.E}\right)}_{\text{P}}$

$\Rightarrow 4\text{mgh}=\text{mgh}+\frac{1}{2}{\text{mv}}^{\text{2}}$

“v is velocity of B at P”

$3\text{mgh}=\frac{1}{2}{\text{mv}}^{\text{2}}$

$\text{6gh}={\text{v}}^{\text{2}}$

${\text{h}}_{\text{max}}$ from point ‘P’$\Rightarrow \frac{{\text{u}}^{\text{2}}{\sin }^2\theta }{2\text{g}}$

$=\frac{{\text{v}}^{\text{2}}{\sin }^2\theta }{\text{2g}}$

$=\frac{\text{6gh}}{2\text{g}}\times \frac{3}{4}=\frac{9}{4}\text{h}$

${\text{h}}_1=\text{h}$

${\text{h}}_2=\frac{9}{4}\text{h}+\text{h}$                                     

${\text{h}}_1:{\text{h}}_{\text{2}}=\text{h}:\frac{9}{4}\text{h}+\text{h}$

$=\text{h:}\frac{13}{4}\text{h}$

$=\text{4h}:\text{13h}$

${\text{h}}_1:{\text{h}}_{\text{2}}=4:13$