JEE PYQ: Motion in a Straight Line - Question ID 9417e5e48f1b (JEE Main 2021)

Core Formula & Concept:

To find the average speed of the rubber ball over its entire motion, we use the fundamental definition:

$\text{Average Speed} = \frac{\text{Total Distance Traveled}}{\text{Total Time Taken}}$

Key physics concepts involved:

• Free-fall motion: The ball is released from rest and falls under gravity. The time to fall from height $h$ is given by: $t_{\text{fall}} = \sqrt{\frac{2h}{g}}$ where $g = 10 \, \text{ms}^{-2}$.
• Bounce dynamics: After each bounce, the ball rises to $\frac{81}{100}$ of the previous maximum height. This forms a geometric progression of heights.
• Time symmetry: The time to rise to a height $h$ is the same as the time to fall from that height, due to symmetry in uniformly accelerated motion.

Step-by-Step Derivation:

Step 1: Compute the time to fall from initial height

Initial height, $h_0 = 5 \, \text{m}$. Time to fall from $h_0$: $t_0 = \sqrt{\frac{2h_0}{g}} = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1 \, \text{s}$

Step 2: Model the bouncing heights and times

After the first bounce, the ball rises to $h_1 = \frac{81}{100} h_0 = 0.81 \times 5 = 4.05 \, \text{m}$. Time to rise to $h_1$ and fall back down is: $t_1 = 2 \times \sqrt{\frac{2h_1}{g}} = 2 \times \sqrt{\frac{2 \times 4.05}{10}} = 2 \times \sqrt{0.81} = 2 \times 0.9 = 1.8 \, \text{s}$ Similarly, the next bounce reaches $h_2 = 0.81 \times h_1 = 0.81^2 \times 5$, and so on.

Step 3: Express total distance and total time as infinite series

Total distance $D$:

• Downward: $h_0 = 5 \, \text{m}$
• Upward: $h_1 + h_2 + h_3 + \cdots = 5 \times 0.81 + 5 \times 0.81^2 + \cdots$
• Downward after first bounce: $h_1 + h_2 + \cdots = 5 \times 0.81 + 5 \times 0.81^2 + \cdots$

So, $D = h_0 + 2(h_1 + h_2 + \cdots) = 5 + 2 \times 5 \times (0.81 + 0.81^2 + \cdots)$ This is a geometric series with first term $a = 0.81$, ratio $r = 0.81$: $S = \frac{a}{1 - r} = \frac{0.81}{1 - 0.81} = \frac{0.81}{0.19} = \frac{81}{19}$ Thus, $D = 5 + 2 \times 5 \times \frac{81}{19} = 5 + \frac{810}{19} = \frac{95 + 810}{19} = \frac{905}{19} \approx 47.63 \, \text{m}$

Total time $T$:

• Initial fall: $t_0 = 1 \, \text{s}$
• Subsequent up-down cycles: $t_1 + t_2 + \cdots$, where $t_n = 2 \sqrt{\frac{2h_n}{g}} = 2 \sqrt{\frac{2 \times 5 \times 0.81^n}{10}} = 2 \times \sqrt{0.81^n} = 2 \times (0.9)^n$

So, $T = t_0 + t_1 + t_2 + \cdots = 1 + 2(0.9 + 0.9^2 + 0.9^3 + \cdots)$ This is a geometric series with $a = 0.9$, $r = 0.9$: $S = \frac{a}{1 - r} = \frac{0.9}{0.1} = 9$ Thus, $T = 1 + 2 \times 9 = 1 + 18 = 19 \, \text{s}$

Step 4: Compute average speed

$\text{Average Speed} = \frac{D}{T} = \frac{905/19}{19} = \frac{905}{361} \approx 2.507 \, \text{ms}^{-1}$ Rounding to two decimal places gives $2.50 \, \text{ms}^{-1}$, which matches option A.

Common Traps & Exam Tip:

Common mistakes students make:

• Confusing average speed with average velocity: Average velocity would be zero over a complete cycle, but average speed is non-zero.
• Incorrectly summing the series: Forgetting that after the first fall, each bounce involves both an upward and downward journey, doubling the distance and time contributions.
• Arithmetic errors in geometric series: Misapplying the formula $S = \frac{a}{1 - r}$, especially when $r$ is a fraction like 0.81 or 0.9.
• Ignoring the initial fall time: Some students only sum the bounce times and forget the first 1-second fall.

Exam Tip: Always write down the first few terms of the series explicitly to visualize the pattern. Use symmetry: the time to rise equals the time to fall for each bounce. Double-check series convergence and arithmetic.