JEE PYQ: Motion in a Straight Line - Question ID ab77f17a1ea6 (JEE Main 2009)

This problem involves motion under constant acceleration (gravity) and elastic collisions. The key concepts and formulas are:

• Free-fall motion: For an object dropped from rest at height $h$, the velocity $v(t)$ and height $y(t)$ as functions of time are: $v(t) = -gt \quad \text{(downward direction taken as negative)}$ $y(t) = h - \frac{1}{2}gt^2$ The time to fall from height $h$ to the ground is: $t_{\text{fall}} = \sqrt{\frac{2h}{g}}$ The velocity just before impact is: $v_{\text{impact}} = -\sqrt{2gh}$
• Elastic collision with a stationary plate: In a perfectly elastic collision with a stationary horizontal surface, the velocity reverses direction but retains its magnitude. So, if the ball hits the plate with velocity $v$ downward, it rebounds with velocity $v$ upward.
• Periodic motion: After rebounding, the ball moves upward, decelerates under gravity, momentarily stops, then falls again. This creates a periodic motion with symmetric velocity and height profiles over time.

Given $h = 4.9\,m$ and $g = 9.8\,m/s^2$, we can compute the time of fall: $t_{\text{fall}} = \sqrt{\frac{2 \times 4.9}{9.8}} = \sqrt{1} = 1\,s$ So the ball hits the plate at $t = 1\,s$, rebounds, and repeats the motion every $2\,s$ (up and down).

Step 1: Define coordinate system and initial conditions

• Let $y = 0$ be the position of the plate (ground level).
• Positive $y$-direction is upward.
• Ball is released from $y = h = 4.9\,m$ at $t = 0$, with initial velocity $v = 0$.

Step 2: Motion before first impact ($0 \leq t < 1\,s$)

• Acceleration: $a = -g = -9.8\,m/s^2$
• Velocity: $v(t) = -gt = -9.8t$
• Height: $y(t) = h - \frac{1}{2}gt^2 = 4.9 - 4.9t^2$
• At $t = 1\,s$: $v = -9.8\,m/s$, $y = 0$

Step 3: Elastic collision at $t = 1\,s$

• Velocity reverses: $v \to +9.8\,m/s$ (upward)
• Duration of collision is negligible, so position remains $y = 0$.

Step 4: Motion after first rebound ($1\,s < t < 3\,s$)

• Ball moves upward with initial velocity $+9.8\,m/s$, decelerates at $-9.8\,m/s^2$.
• Velocity: $v(t) = 9.8 - g(t - 1) = 9.8 - 9.8(t - 1) = 9.8(2 - t)$
• Height: $y(t) = 0 + 9.8(t - 1) - \frac{1}{2}g(t - 1)^2 = 9.8(t - 1) - 4.9(t - 1)^2$
• At $t = 2\,s$: $v = 0$, $y = 4.9\,m$ (ball reaches max height again)
• At $t = 3\,s$: $v = -9.8\,m/s$, $y = 0$ (ball hits plate again)

Step 5: Generalize for all $t \geq 0$

The motion is periodic with period $T = 2\,s$. We can express $v(t)$ and $y(t)$ using the sawtooth and parabolic patterns:

• Velocity $v(t)$: - Decreases linearly from 0 to $-9.8\,m/s$ in $0 \leq t < 1\,s$ - Jumps to $+9.8\,m/s$ at $t = 1\,s$ - Decreases linearly to 0 at $t = 2\,s$ - Repeats every $2\,s$
  This creates a sawtooth wave with slope $-g$, jumping from $-v_{\text{max}}$ to $+v_{\text{max}}$ at each impact.
• Height $y(t)$: - Parabolic descent from $4.9\,m$ to 0 in $0 \leq t \leq 1\,s$ - Parabolic ascent from 0 to $4.9\,m$ in $1 \leq t \leq 2\,s$ - Repeats every $2\,s$
  This creates a symmetric parabolic arch every $2\,s$, peaking at $t = 2n\,s$, hitting zero at $t = 2n + 1\,s$.

Step 6: Match with given options

- Option A: Shows velocity as a smooth sine-like wave — incorrect. - Option B: Shows velocity as a sawtooth wave with correct slope and jumps; height as symmetric parabolic arches — correct. - Option C: Velocity has incorrect slope and jumps; height is asymmetric — incorrect. - Option D: Velocity is piecewise constant — incorrect.

Students often make these mistakes:

• Ignoring velocity reversal in elastic collision: Some assume the ball stops or loses speed, leading to incorrect velocity profiles.
• Miscounting time intervals: The time to fall is $1\,s$, not $0.7\,s$ or $1.4\,s$. Always compute $t_{\text{fall}} = \sqrt{2h/g}$.
• Confusing velocity and height graphs: Velocity is linear (sawtooth), height is parabolic (smooth arches). Mixing them up leads to wrong option selection.
• Forgetting periodicity: The motion repeats every $2\,s$. The graphs should reflect this symmetry.

Exam Tip: Always sketch the motion: downward parabola, elastic rebound, upward parabola. Then match the shape of $v(t)$ and $y(t)$ to the options. The sawtooth velocity and symmetric parabolic height are hallmarks of elastic bouncing under gravity.