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Two identical ball A and B are release 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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1 : 4

2 : 5

4 : 13

2 : 1

answer is C.

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Detailed Solution

Mass is equal so they exchange their velocity.

The speed of 'B' when it reaches to initial position is
$4 mgh = \frac{1}{2} {mv}^{2} + mgh \Rightarrow v = \sqrt{6 gh}$

$H = \frac{v^{2} \sin^{2} θ}{2 g} = \frac{9 h}{4}$
total height of 'B' :
$H_{B} = h + \frac{9 h}{4} = \frac{13 h}{4}$
height of 'A' is H_A = h
$\Rightarrow \frac{H_{A}}{H_{B}} = 4 : 13$

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