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., $v_A=\sqrt{2gh}$ 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 $B$,
$\frac{1}{2}m{v}_B^2=\frac{1}{2}m{v}^2+mgh$ or $v=\sqrt{6gh}$
$\therefore h_B=h+\frac{v^{2}si{n}^{2}60^{\circ }}{2g}$
$=h+\frac{9h}{4}=\frac{13h}{4}$
Hence, $\frac{h_A}{h_B}=\frac{4}{13}$