Question

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)

• A. $1 : 4$
• B. $2 : 1$
• C. $4 : 13$
• D. $2 : 11$

Answer 1

Answer: (C)

After collision balls exchange their velocities
i.e., $v_{A} = \sqrt{2\,gh}$ and $v_{B} = \sqrt{2g\left(4h\right)} = 2\sqrt{2gh}$
Height attained by $A$ will be $h_{A} = \frac{v^{2}A}{2g} = h$

But path of $B$ will be first straight line and then parabolic as shown in figure.
Using energy conservation for ball $B$,
$\frac{1}{2}mv^{2}_{B} = \frac{1}{2}mv^{2} + mgh$ or $v = \sqrt{6\,gh}$
$\therefore h_{B} = h+\frac{v^{2}\,sin^{2}\,60^{\circ}}{2\,g}$
$= h + \frac{9h}{4} = \frac{13h}{4}$
Hence, $\frac{h_{A}}{h_{B}} = \frac{4}{13}$