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

1 : 4

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2 : 1

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4 : 13

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2 : 11

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Solution

The correct option is C $4 : 13$
After collision balls exchange their velocities
i.e., $v_{A} = \sqrt{2 g h}$ and $v_{B} = \sqrt{2 g (4 h)} = 2 \sqrt{2 g h}$
Height attained by $A$ will be $h_{A} = \frac{v_{A}^{2}}{2 g} = h$
But path of $B$ will be first straight line and then parabolic as shown in figure.
Using energy conservation for ball $B$ ,
$\frac{1}{2} m v_{B}^{2} = \frac{1}{2} m v^{2} + m g h$ or $v = \sqrt{6 g h}$
$∴ h_{B} = h + \frac{v^{2} {sin}^{2} 60^{\circ}}{2 g} = h + \frac{9 h}{4} = \frac{13 h}{4}$
Hence, $\frac{h_{A}}{h_{B}} = \frac{4}{13}$ .

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

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