Q.45 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 45 Solution

Topic: Laws of Motion · Official NTA Booklet Code M1

Question

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly:

✓

Verified Against NTA Final Answer Key

Option (2) : 4.2 kg m/s

📚 Laws of Motion Easy

📋 Given Parameters & Problem Setup

Mass of Ball (m) 0.15 kg

Drop Height (h) 10 m

Rebound Height (h) 10 m (same height)

Acceleration Due to Gravity (g) 10 m/s^2

⚡

The 30-Second Exam Hall Shortcut: ⚡ Rebound Impulse: J = 2 m √(2gh)

⏱️ Save ~20 seconds

Impact speed $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 10\sqrt{2}\text{ m/s}$ . Since it rebounds to the same height, velocity reverses ( $+v$ upward, $-v$ downward). Magnitude of impulse $J = |\Delta p| = 2mv = 2 \times 0.15 \times 10\sqrt{2} = 3\sqrt{2} = 3 \times 1.414 = 4.24\text{ kg m/s} \approx 4.2\text{ kg m/s}$ ! Match Option (2)!

1 Downward velocity just before impact:

v_1 = -\sqrt{2gh} = -\sqrt{2(10)(10)} = -10\sqrt{2}\text{ m/s}

2 Upward velocity after rebound:

v_2 = +\sqrt{2gh} = +10\sqrt{2}\text{ m/s}

3 Impulse is change in linear momentum:

J = m (v_2 - v_1) = m (10\sqrt{2} - (-10\sqrt{2})) = 2 m (10\sqrt{2})

4 Substitute

m = 0.15\text{ kg}

J = 2(0.15)(10\sqrt{2}) = 3\sqrt{2}\text{ kg m/s}

3 \times 1.414 = 4.242\text{ kg m/s} \approx 4.2\text{ kg m/s}

6 Match Option (2).

🛡️

Why You Can 100% Trust This Shortcut: Canonical impulse-momentum theorem and kinematics of free fall.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating Velocity Just Before Striking the Ground

For a body falling freely from rest from height $h = 10\text{ m}$ under gravity $g = 10\text{ m/s}^2$ , taking upward as positive:

v_1 = -\sqrt{2gh} = -\sqrt{2 \times 10\text{ m/s}^2 \times 10\text{ m}} = -\sqrt{200} = -10\sqrt{2}\text{ m/s}

Step 2

Calculating Velocity Just After Rebound

The ball rebounds back to the same height $h = 10\text{ m}$ . To reach height $h$ , its initial upward velocity must be:

v_2 = +\sqrt{2gh} = +10\sqrt{2}\text{ m/s}

Step 3

Applying the Impulse-Momentum Theorem

Impulse imparted by the ground to the ball is defined as the change in its linear momentum vector:

J = \Delta p = m v_2 - m v_1 = m \left(v_2 - v_1\right) = m \left[10\sqrt{2} - (-10\sqrt{2})\right] = 2 m (10\sqrt{2})

Step 4

Numerical Evaluation

Substitute $m = 0.15\text{ kg}$ and $\sqrt{2} \approx 1.414$ :

J = 2 \times 0.15\text{ kg} \times 10\sqrt{2}\text{ m/s} = 3\sqrt{2} = 3 \times 1.414 = 4.242\text{ kg m/s} \approx 4.2\text{ kg m/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 1

✗ Incorrect

Verdict & Scientific Reason:

Assumes scalar momentum difference ( $mv - mv = 0$ ), completely ignoring that velocity reverses direction.

⚠️

Student Fallacy (Why candidates pick this trap): Subtracting momentum magnitudes instead of vectors (

v - v = 0

2 2

✓ 100% Correct

Verdict & Scientific Reason:

Exact magnitude of impulse: $J = 2mv = 3\sqrt{2} \approx 4.2\text{ kg m/s}$ .

💡

Why this is airtight: None; textbook impulse-momentum theorem.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Calculates only single momentum $mv = 0.15 \times 14.14 \approx 2.1\text{ kg m/s}$ , forgetting to double for rebound.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming impulse equals initial momentum without rebound reflection.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Arithmetic approximation error.

⚠️

Student Fallacy (Why candidates pick this trap): Dividing by 3 or miscalculating

\sqrt{2}

⚠️

NTA Trap Alert: Vector Reversal vs. Scalar Cancellation Trap

Impulse is a vector quantity: $\vec{J} = m\vec{v}_f - m\vec{v}_i$ . Because $\vec{v}_f$ is upward and $\vec{v}_i$ is downward, the two terms ADD together: $m(v - (-v)) = 2mv$ . Treating velocity as a scalar yields $mv - mv = 0$ (Option 1).

🛡️ How to avoid the -1 mark: Whenever an object rebounds, change in momentum is

2mv

, NOT zero!

📖

Core NEET Syllabus Standard

NCERT Physics · Laws of Motion Standard Benchmark

Core Scientific Principle Tested:

"Impulse is the change in linear momentum: $\vec{J} = \Delta\vec{p} = m\vec{v}_2 - m\vec{v}_1$ . For complete rebound, $|J| = 2mv$ ."

Syllabus Focus: Impulse and Momentum

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 45 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Rebound Impulse: J = 2 m √(2gh)

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating Velocity Just Before Striking the Ground

Calculating Velocity Just After Rebound

Applying the Impulse-Momentum Theorem

Numerical Evaluation

360° Option Elimination Engine Key Feature

NTA Trap Alert: Vector Reversal vs. Scalar Cancellation Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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