Two identical balls A and B are released from | vedclass

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Question

After collision the balls exchange their velocities,

i.e.,

${\mathrm{v}_{\mathrm{A}}=\sqrt{2 \mathrm{gh}}}$

and      ${\mathrm

Vedclass · Accepted Answer

$4 : 13$

Vedclass · Answer

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,

${\mathrm{h}_{\mathrm{B}}=\frac{13}{4} \mathrm{h}}$

${\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}}=\frac{4}{13} \mathrm{h}}$

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)

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