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(C) Let the masses of the balls be $m$. The velocity of ball $A$ at $M$ before collision is $v_A = \sqrt{2g(4h)} = 2\sqrt{2gh}$. The velocity of ball

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$4 : 13$

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(C) Let the masses of the balls be $m$. The velocity of ball $A$ at $M$ before collision is $v_A = \sqrt{2g(4h)} = 2\sqrt{2gh}$. The velocity of ball $B$ at $N$ before collision is $v_B = \sqrt{2gh}$.Since the balls are identical and the collision is elastic,they exchange their velocities. After the collision,ball $A$ moves towards the left with velocity $v_A' = \sqrt{2gh}$ and ball $B$ moves towards the right with velocity $v_B' = 2\sqrt{2gh}$.For ball $A$,the maximum height attained is $H_A = \frac{(v_A')^2}{2g} = \frac{2gh}{2g} = h$.For ball $B$,it moves up the incline at $N$ with velocity $v_B' = 2\sqrt{2gh}$. It reaches a height $h$ with velocity $v_P$ given by $v_P^2 = (v_B')^2 - 2gh = 8gh - 2gh = 6gh$. At this point,it leaves the incline as a projectile with an angle of $60^{\circ}$ to the horizontal. The additional height attained is $h' = \frac{(v_P \sin 60^{\circ})^2}{2g} = \frac{6gh \cdot (3/4)}{2g} = \frac{9h}{4}$.The total maximum height attained by $B$ is $H_B = h + h' = h + \frac{9h}{4} = \frac{13h}{4}$.The ratio of heights is $\frac{H_A}{H_B} = \frac{h}{13h/4} = \frac{4}{13}$.

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

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