Question

Two identical balls $A$ and $B$ are released from the position shown in the figure. They collide elastically with each other on the horizontal portion. The ratio of height attained by $A$ and $B$ after the collision is: (neglect friction)

• A. $1:4$
• B. $2:1$
• C. $4:13$
• D. $2:5$

Answer 1

The correct option is A

$1:4$

Step 1: Given Data

Height of ball $A,h_A=4h$

Height of ball $B,h_B=h$

The angle of inclination of $B,\theta =60°$

Let the velocity of balls $A$ and $B$ be $v_A$ and $v_B$ respectively.

Let $g$ be the acceleration due to gravity.

Let $m$ be the equal mass of both the balls.

Step 2: Formula Used

Maximum height in projectile motion,

$H=\frac{v^2{\sin }^2\theta }{2g}$

Step 3: Calculate the Velocities

According to the work-energy theorem,

Work done by gravity = Change in kinetic energy

$\therefore W=△KE=K{E}_f$

The potential energy is converted into kinetic energy when the balls start to roll.

Since it does not have any initial kinetic energy.

$$\begin{gathered} \therefore mgH=\frac{1}{2}m{v}^2 \\ \Rightarrow v=\sqrt{2gH} \\ \therefore v_A=\sqrt{2g{h}_A}=\sqrt{2g\left(4h\right)} \\ \therefore v_B=\sqrt{2g{h}_B}=\sqrt{2g\left(h\right)} \end{gathered}$$

Step 4: Calculate the Ratio

Since identical balls with equal mass collide elastically, their velocities get exchanged after the collision and move in the opposite direction.

Applying the work-energy theorem again,

$$\begin{gathered} -mgH=0-\frac{1}{2}m{v}^2 \\ \Rightarrow H=\frac{v^2}{2g} \\ \therefore \frac{h_A}{h_B}=\frac{{\displaystyle \frac{{v_B}^2}{2g}}}{\frac{{v_A}^2}{2g}} \\ =\frac{2gh}{2g.4h} \\ =1:4 \end{gathered}$$

Hence, the correct answer is option (A).