JEE PYQ: Motion in a Straight Line - Question ID 475ec9ae16ee (JEE Main 2023)

Core Formula & Concept:

In this problem, we analyze the motion of a tennis ball undergoing a collision with the floor. The key concepts involved are:

• Free-fall motion: When the ball is dropped or rebounds, it moves under gravity with acceleration $g = 10 \, \text{ms}^{-2}$.
• Velocity just before and after impact: Using the kinematic equation for free fall, the velocity $v$ of the ball when it hits the floor or leaves it can be found from the height $h$ using: $v = \sqrt{2gh}$ This gives the speed just before impact (downward) and just after rebound (upward).
• Average acceleration during contact: The ball changes velocity during the short contact time $\Delta t = 0.2 \, \text{s}$. The average acceleration $a_{\text{avg}}$ is defined as: $a_{\text{avg}} = \frac{\Delta v}{\Delta t}$ where $\Delta v$ is the change in velocity vector, accounting for direction.

Step-by-Step Derivation:

Step 1: Compute velocity just before impact (downward)

The ball is dropped from height $h_1 = 9.8 \, \text{m}$. Using the free-fall velocity formula: $v_1 = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 9.8} = \sqrt{196} = 14 \, \text{ms}^{-1}$ Since the ball is moving downward, we take $v_1 = -14 \, \text{ms}^{-1}$ (assuming upward as positive).

Step 2: Compute velocity just after rebound (upward)

The ball rebounds to height $h_2 = 5.0 \, \text{m}$. The velocity just after rebound is: $v_2 = \sqrt{2gh_2} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \, \text{ms}^{-1}$ Since the ball is now moving upward, $v_2 = +10 \, \text{ms}^{-1}$.

Step 3: Compute change in velocity $\Delta v$

The change in velocity is: $\Delta v = v_2 - v_1 = (+10) - (-14) = 10 + 14 = 24 \, \text{ms}^{-1}$ This accounts for the reversal in direction during the collision.

Step 4: Compute average acceleration during contact

The contact time is $\Delta t = 0.2 \, \text{s}$. The average acceleration is: $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{24}{0.2} = 120 \, \text{ms}^{-2}$

Common Traps & Exam Tip:

Students often make the following mistakes:

• Ignoring direction of velocity: Forgetting to assign signs (positive/negative) to velocities before and after impact leads to incorrect $\Delta v$. Always define a coordinate system (e.g., upward as positive).
• Using speed instead of velocity: Taking $\Delta v = v_2 + v_1$ (as speeds) instead of $v_2 - v_1$ (as vectors) gives $24 \, \text{ms}^{-1}$ by coincidence here, but this is conceptually wrong and may fail in other problems.
• Misapplying kinematic equations: Using $v = u + at$ during free fall without recognizing that $u = 0$ when dropped or $v = 0$ at maximum rebound height.
• Unit confusion: Not converting all units consistently (e.g., using $g = 9.8 \, \text{ms}^{-2}$ instead of the given $10 \, \text{ms}^{-2}$) can lead to numerical errors.

Exam Tip: Always draw a velocity-time sketch for collision problems. Label velocities with direction. This helps visualize $\Delta v$ and avoid sign errors.