JEE PYQ: Motion in a Straight Line - Question ID 1ceeae2a9f19 (JEE Main 2020)

The problem involves the motion of a tennis ball under gravity, first falling freely from height \( h \) and then rebounding to height \( \frac{h}{2} \). The key physics concepts and formulas used are:

• Kinematic Equations for Free Fall: For an object in free fall (ignoring air resistance), the velocity \( v \) at any height \( y \) can be related to the initial height using energy conservation or kinematic equations. The velocity of the ball at height \( y \) during descent is given by: $v = \sqrt{2g(h - y)}$ where \( g \) is the acceleration due to gravity.
• Velocity at Impact: When the ball hits the floor (\( y = 0 \)), its velocity just before impact is: $v_{\text{impact}} = \sqrt{2gh}$
• Rebound Velocity: After rebounding, the ball reaches height \( \frac{h}{2} \). Using energy conservation, the velocity just after rebound (\( v_{\text{rebound}} \)) is: $v_{\text{rebound}} = \sqrt{2g \cdot \frac{h}{2}} = \sqrt{gh}$ This shows that the rebound velocity is less than the impact velocity due to energy loss (e.g., deformation, sound, heat).
• Velocity vs. Height Graph: The graph of velocity \( v \) versus height \( y \) should reflect:
  • During descent: \( v \) increases as \( y \) decreases (negative slope, since velocity is downward).
  • At impact (\( y = 0 \)): Velocity abruptly changes direction (from downward to upward) but with reduced magnitude.
  • During ascent: \( v \) decreases as \( y \) increases (negative slope, since velocity is upward but decreasing).
  The graph must show a discontinuity at \( y = 0 \) due to the sudden change in velocity direction and magnitude.

We analyze the motion in two phases: descent and ascent.

1. Descent Phase (Falling from \( h \) to \( 0 \)):
   • The velocity at any height \( y \) during descent is: $v = -\sqrt{2g(h - y)}$ (The negative sign indicates downward direction.)
   • At \( y = h \), \( v = 0 \) (initial release).
   • At \( y = 0 \), \( v = -\sqrt{2gh} \) (maximum downward velocity before impact).
   • The \( v \) vs. \( y \) graph for descent is a curve starting at \( (h, 0) \) and ending at \( (0, -\sqrt{2gh}) \), with a negative slope (since \( \frac{dv}{dy} = -\sqrt{\frac{g}{2(h - y)}} < 0 \)).
2. Ascent Phase (Rebounding from \( 0 \) to \( \frac{h}{2} \)):
   • The velocity at any height \( y \) during ascent is: $v = +\sqrt{2g \left( \frac{h}{2} - y \right)}$ (The positive sign indicates upward direction.)
   • At \( y = 0 \), \( v = +\sqrt{gh} \) (rebound velocity, less than \( \sqrt{2gh} \)).
   • At \( y = \frac{h}{2} \), \( v = 0 \) (maximum height after rebound).
   • The \( v \) vs. \( y \) graph for ascent is a curve starting at \( (0, +\sqrt{gh}) \) and ending at \( \left( \frac{h}{2}, 0 \right) \), with a negative slope (since \( \frac{dv}{dy} = -\sqrt{\frac{g}{2 \left( \frac{h}{2} - y \right)}} < 0 \)).
3. Graph Characteristics:
   • The graph must show two distinct curves:
     • One for descent (negative \( v \), decreasing \( y \)).
     • One for ascent (positive \( v \), increasing \( y \)).
   • At \( y = 0 \), there is a discontinuity in \( v \) (from \( -\sqrt{2gh} \) to \( +\sqrt{gh} \)).
   • The slope \( \frac{dv}{dy} \) is negative for both curves (since \( v \) decreases as \( y \) increases in both phases).
   • The graph should not pass through the origin or show symmetry about \( y = 0 \), as the rebound velocity is smaller than the impact velocity.
4. Matching with Options:
   • Option A: Incorrect. Shows a single straight line, implying constant acceleration (which is true) but no discontinuity at \( y = 0 \) and no distinction between descent/ascent.
   • Option B: Incorrect. Shows a symmetric graph about \( y = 0 \), implying equal impact and rebound velocities (not the case here).
   • Option C: Correct. Shows:
     • Two separate curves for descent and ascent.
     • A discontinuity at \( y = 0 \).
     • Negative slope for both curves.
     • Rebound velocity smaller than impact velocity.
   • Option D: Incorrect. Shows a continuous curve with no discontinuity at \( y = 0 \), implying no rebound.

Students often make the following mistakes:

• Ignoring the Discontinuity: Many assume the velocity changes smoothly at \( y = 0 \), leading them to choose Option D. However, the rebound is instantaneous, causing a sudden change in velocity direction and magnitude.
• Assuming Symmetric Velocities: Some students assume the rebound velocity is equal in magnitude to the impact velocity (Option B), ignoring energy loss during the collision.
• Misinterpreting the Slope: The slope \( \frac{dv}{dy} \) is negative for both descent and ascent. A positive slope (as in some incorrect options) would imply increasing velocity with height, which is unphysical.
• Energy Conservation Misapplication: Students may forget that the rebound height is \( \frac{h}{2} \), not \( h \), leading to incorrect velocity calculations.

Exam Tip: Always sketch the motion phases separately (descent and ascent) and mark key points (initial/final heights, impact/rebound velocities). This helps visualize the \( v \) vs. \( y \) graph and avoid common pitfalls.