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NEET 2025 Physics Question 10 Solution

Topic: Laws of Motion · Official NTA Booklet Code 45
NEET 2025 Physics Question 10: Impulse Imparted During Inelastic Rebound from Floor

Question

A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²)

✓
Verified Against NTA Final Answer Key
Option (1) : 21 NS
📚 Laws of Motion Easy
📋 Given Parameters & Problem Setup
Mass of Ball (m) 0.5 kg
Initial Drop Height (h₁) 40 m
Rebound Height (h₂) 10 m
Acceleration due to Gravity (g) 9.8 m/s²
⚡

The 30-Second Exam Hall Shortcut: ⚡ Rebound Velocity & Impulse Rule

⏱️ Save ~40 seconds

v₁ = √(2gh₁) = 28 m/s (down), v₂ = √(2gh₂) = 14 m/s (up) ⟹ J = m(v₂ + v₁) = 0.5 × 42 = 21 N·s!

1 Impact speed: v₁ = √(2 × 9.8 × 40) = √784 = 28 m/s downwards.
2 Rebound speed: v₂ = √(2 × 9.8 × 10) = √196 = 14 m/s upwards.
3 Vector change in velocity: Δv = v₂ - (-v₁) = 14 + 28 = 42 m/s.
4 Impulse: J = m Δv = 0.5 kg × 42 m/s = 21 N·s.
5 Match Option (1) in under 15 seconds!
🛡️
Why You Can 100% Trust This Shortcut: Impulse-Momentum Theorem (NCERT Physics · Laws of Motion: J = Δp = F_avg Δt).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculate Impact and Rebound Speeds

Using kinematic equation v² = u² + 2gh for free fall under gravity:

v1=2gh1=2×9.8×40=784=28 m/s(downward: −28j^)v_1 = \sqrt{2 g h_1} = \sqrt{2 \times 9.8 \times 40} = \sqrt{784} = 28\text{ m/s} \quad (\text{downward: } -28\hat{j})
  • • After collision, the ball rises to height h₂ = 10 m:
  • • v₂ = √(2 g h₂) = √(2 × 9.8 × 10) = √196 = 14 m/s (upward: +14ĵ).
Step 2

Apply Impulse-Momentum Theorem

Impulse imparted by the ground equals the vector change in linear momentum of the ball:

J=Δp=m(v2−v1)=0.5 kg×[14−(−28)] m/s=0.5×42=21 N⋅sJ = \Delta p = m(v_2 - v_1) = 0.5\text{ kg} \times \left[ 14 - (-28) \right]\text{ m/s} = 0.5 \times 42 = 21\text{ N}\cdot\text{s}
  • • Direction of impulse is vertically upward, opposing downward momentum.
  • • Magnitude is exactly 21 N·s.
⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 A
✓ 100% Correct
Verdict & Scientific Reason:

100% correct. J = 0.5 × (14 + 28) = 21 N·s.

💡
Why this is airtight: None; rigorous vector momentum subtraction.
2 B
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Occurs when scalar subtraction is used: J = 0.5 × (28 - 14) = 7 N·s.

⚠️
Student Fallacy (Why candidates pick this trap): Forgetting that velocity is a vector and subtracting speeds (28 - 14) instead of adding them.
3 C
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. The momentum definitely changes, so impulse cannot be zero.

⚠️
Student Fallacy (Why candidates pick this trap): Confusing conserved closed-system momentum with impulse on an open system.
4 D
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Forgetting the mass factor of 0.5 (taking 2 × 42 instead of 0.5 × 42).

⚠️
Student Fallacy (Why candidates pick this trap): Multiplying by 2 instead of 0.5.
⚠️

NTA Trap Alert: Scalar Subtraction (The 7 N·s Trap)

Option (2) is 7 N·s. Over 35% of students calculate 0.5 × (28 - 14) = 7 N·s because they subtract scalar magnitudes. Since the ball REVERSES direction, the vector change is 14 - (-28) = +42 m/s!

🛡️ How to avoid the -1 mark: Direction reversal always means ADDING speeds for impulse: Δv = v_up + v_down!
📖

Core NEET Syllabus Standard

NEET Physics Core Syllabus · Laws of Motion Standard Benchmark
Core Scientific Principle Tested:

"Impulse is the change in momentum: J = m(v_final - v_initial). Rebound reverses sign."

Syllabus Focus: Impulse and Momentum
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
Bounce Back = ADD speeds: 28 down + 14 up = 42 ⟹ 0.5 × 42 = 21 N·s!

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