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Two identical balls $A$ and $B$ are released from the positions shown in figure. They collide elastically on horizontal portion $MN$. The ratio of the heights attained by $A$ and $B$ after collision will be (neglect friction)
$1 : 4$
$2 : 1$
$4 : 13$
$2 : 5$
Solution
After collision the balls exchange their velocities,
i.e.,
${\mathrm{v}_{\mathrm{A}}=\sqrt{2 \mathrm{gh}}}$
and ${\mathrm{v}_{\mathrm{B}}=\sqrt{2 \mathrm{g}(4 \mathrm{h})}=2 \sqrt{2 \mathrm{gh}}}$
Height attained by $A$ will be,
$\mathrm{h}_{\mathrm{A}}=\frac{\mathrm{v}_{\mathrm{A}}^{2}}{2 \mathrm{g}}=\mathrm{h}$
But path of $\mathrm{B}$ will be first straight line and then parabolic as shows in the figure.
After calculation we can show that,
${\mathrm{h}_{\mathrm{B}}=\frac{13}{4} \mathrm{h}}$
${\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}}=\frac{4}{13} \mathrm{h}}$
