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(A) Let the mass of each ball be $m$. By conservation of energy,the velocity of ball $A$ just before collision is $v_A = \sqrt{2g(4h)} = 2\sqrt{2gh}$

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$1 : 4$

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(A) Let the mass of each ball be $m$. By conservation of energy,the velocity of ball $A$ just before collision is $v_A = \sqrt{2g(4h)} = 2\sqrt{2gh}$ and the velocity of ball $B$ is $v_B = \sqrt{2gh}$.Since the collision is elastic and the balls are identical,they exchange their velocities upon collision.After the collision,ball $A$ moves with velocity $v_A' = v_B = \sqrt{2gh}$ towards the left incline. The maximum height attained by $A$ is $h_A = \frac{(v_A')^2}{2g} = h$.Ball $B$ moves with velocity $v_B' = v_A = 2\sqrt{2gh}$ towards the right incline at an angle of $60^\circ$ with the horizontal.Ball $B$ will travel along the incline and then become a projectile. The maximum height $h_B$ reached by ball $B$ is given by the formula for projectile motion: $h_B = h_{incline} + \frac{(v_y)^2}{2g}$,where $v_y = v_B' \sin 60^\circ = 2\sqrt{2gh} \cdot \frac{\sqrt{3}}{2} = \sqrt{6gh}$.Thus,$h_B = h + \frac{(\sqrt{6gh})^2}{2g} = h + 3h = 4h$.Wait,re-evaluating the projectile motion: The ball $B$ leaves the incline at height $h$. The vertical component of its velocity is $v_y = 2\sqrt{2gh} \sin 60^\circ = \sqrt{6gh}$. The additional height gained is $\frac{v_y^2}{2g} = \frac{6gh}{2g} = 3h$. Total height $h_B = h + 3h = 4h$.Therefore,the ratio $\frac{h_A}{h_B} = \frac{h}{4h} = \frac{1}{4}$.

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

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