Two identical balls A and B are released from the position shown in fig. They collide elastically with each other on the horizontal portion. The ratio of heights attained by A and B after collision is (neglect friction and neglect collision of balls at corners with horizontal surface )

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$1 : 4$

$2 : 1$

$2 : 5$

$4 : 13$

answer is C.

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Detailed Solution

When the two balls collide with each other, as the mass of the two balls is equal, they exchange their velocities on colliding elastically. Let the speed of the ball B when it reaches back to the initial position be v. Then
$4 m g h = \frac{1}{2} m v^{2} + m g h \Rightarrow v = \sqrt{6 g h}$
Height reaches by particle B (from highest point on the incline) is
$H_{B} = \frac{v^{2} \sin^{2} 60^{0}}{2 g} = \frac{9 h}{4};$ total height $= h + \frac{9 h}{4} = \frac{13 h}{4}$
After collision the particle A reaches the maximum height $= h$
Ratio $= \frac{H_{A}}{H_{B}} = \frac{4}{13}$