Physics - Mechanics and Properties of Matter - How to Solve: Centre of Mass & Linear Momentum (Collisions, Coefficient of Restitution, Explosions) – NEET UG Physics Guide

How to Solve: Centre of Mass & Linear Momentum (Collisions, Coefficient of Restitution, Explosions) – NEET UG Physics Guide

Introduction

Mastering centre of mass and linear momentum unlocks 5-7 marks in NEET UG Physics—enough to push you from 150 to 160+ in the exam. These concepts explain car crashes, rocket launches, and even how bullets work—and NEET loves testing them in collisions, explosions, and coefficient of restitution problems.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Newton’s Laws of Motion (especially F = ma and action-reaction).
2. Conservation of Linear Momentum (momentum before = momentum after in isolated systems).
3. Basic Kinematics (equations of motion, velocity, acceleration).

If any of these are shaky, stop here and review them first—this topic builds on them.

KEY TERMS & FORMULAS

1. Centre of Mass (COM)

Definition: The average position of all mass in a system, weighted by mass. Formula (for discrete masses): MEMORISE THIS [ x_{COM} = \frac{m_1x_1 + m_2x_2 + \dots + m_nx_n}{m_1 + m_2 + \dots + m_n} ] - (x_{COM}) = position of centre of mass (m) - (m_1, m_2, \dots) = masses of objects (kg) - (x_1, x_2, \dots) = positions of objects (m)

For continuous objects (e.g., rods, spheres): [ x_{COM} = \frac{\int x \, dm}{\int dm} ] (Given on exam sheet, but understand the concept.)

Key Idea: - COM moves as if all mass is concentrated there. - No external force? COM velocity remains constant.

2. Linear Momentum (p)

Definition: Product of mass and velocity. Formula: MEMORISE THIS [ \vec{p} = m \vec{v} ] - (\vec{p}) = momentum (kg·m/s) - (m) = mass (kg) - (\vec{v}) = velocity (m/s)

Key Idea: - Momentum is conserved in isolated systems (no external forces). - Impulse (J) = Change in momentum = (F \Delta t = \Delta p).

3. Collisions

Types of Collisions:

Formulas for Collisions:

MEMORISE THESE
1. Conservation of Momentum (always true): [ m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2 ] - (\vec{u}_1, \vec{u}_2) = initial velocities - (\vec{v}_1, \vec{v}_2) = final velocities

1. Coefficient of Restitution (e): [ e = \frac{\text{Relative speed after collision}}{\text{Relative speed before collision}} = \frac{v_2 - v_1}{u_1 - u_2} ] (For head-on collisions only!)

2. Kinetic Energy (for elastic collisions only): [ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 ]

4. Explosions

Key Idea: - Internal forces cause explosions (e.g., gunpowder, spring release). - Momentum is conserved (initial momentum = final momentum). - COM velocity remains unchanged (no external force).

Formula: [ m_1 \vec{v}1 + m_2 \vec{v}_2 + \dots = (m_1 + m_2 + \dots) \vec{v} ] (If initially at rest, total momentum after explosion = 0.)

STEP-BY-STEP METHOD

Step 1: Identify the System & External Forces

• Isolated system? (No external forces → momentum conserved.)
• Non-isolated? (External forces → momentum not conserved, but COM still moves predictably.)

Step 2: Draw a Diagram

• Sketch before and after the event (collision/explosion).
• Label masses, velocities, directions.

Step 3: Write Down Knowns & Unknowns

• List given values (masses, velocities, e).
• Identify what you need to find.

Step 4: Apply Conservation Laws

• Momentum conserved? Use (m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2).
• Elastic collision? Use KE conservation + momentum.
• Inelastic? Use momentum only (KE not conserved).
• Explosion? Total momentum after = 0 (if initially at rest).

Step 5: Use Coefficient of Restitution (if given)

• For head-on collisions, use: [ e = \frac{v_2 - v_1}{u_1 - u_2} ]
• Solve two equations (momentum + e) for two unknowns.

Step 6: Solve for Unknowns

• Substitute values.
• Solve algebraically (avoid plugging numbers too early).
• Check units and signs (direction matters!).

Step 7: Verify Answer

• Does momentum add up?
• Does KE make sense? (Elastic → KE conserved; inelastic → KE lost.)
• Does the answer match the physical scenario?

WORKED EXAMPLES

Example 1 – Basic (Elastic Collision)

Problem: A 2 kg ball moving at 5 m/s collides head-on with a 3 kg stationary ball. If the collision is elastic, find their velocities after collision.

Solution: Step 1: System = two balls. No external forces → momentum conserved. Step 2: Draw diagram: - Before: (m_1 = 2\,kg, u_1 = 5\,m/s; m_2 = 3\,kg, u_2 = 0) - After: (v_1 = ?, v_2 = ?)

Step 3: Knowns: - (m_1 = 2\,kg, u_1 = 5\,m/s) - (m_2 = 3\,kg, u_2 = 0) - (e = 1) (elastic)

Step 4: Apply momentum conservation: [ 2(5) + 3(0) = 2v_1 + 3v_2 \implies 10 = 2v_1 + 3v_2 \quad (1) ]

Step 5: Apply e = 1: [ 1 = \frac{v_2 - v_1}{5 - 0} \implies v_2 - v_1 = 5 \quad (2) ]

Step 6: Solve equations (1) and (2): From (2): (v_2 = v_1 + 5) Substitute into (1): [ 10 = 2v_1 + 3(v_1 + 5) \implies 10 = 5v_1 + 15 \implies 5v_1 = -5 \implies v_1 = -1\,m/s ] Then, (v_2 = -1 + 5 = 4\,m/s)

Step 7: Verify: - Momentum before = (10\,kg·m/s) - Momentum after = (2(-1) + 3(4) = -2 + 12 = 10\,kg·m/s) ✅ - KE before = (\frac{1}{2}(2)(5^2) = 25\,J) - KE after = (\frac{1}{2}(2)(1^2) + \frac{1}{2}(3)(4^2) = 1 + 24 = 25\,J) ✅

Answer: - (v_1 = -1\,m/s) (rebounds backward) - (v_2 = 4\,m/s) (moves forward)

What we did and why: - Used momentum conservation (always true). - Used e = 1 (elastic collision) to get a second equation. - Solved two equations for two unknowns. - Negative sign means direction reversed.

Example 2 – Medium (Inelastic Collision + COM)

Problem: A 1 kg block moving at 6 m/s collides with a 2 kg stationary block. They stick together. Find: (a) Final velocity. (b) Loss in kinetic energy. (c) Position of COM 1 second after collision (assuming no friction).

Solution: Step 1: System = two blocks. No external forces → momentum conserved. Collision is perfectly inelastic (e = 0).

Step 2: Draw diagram: - Before: (m_1 = 1\,kg, u_1 = 6\,m/s; m_2 = 2\,kg, u_2 = 0) - After: Combined mass = (3\,kg, v = ?)

Step 3: Knowns: - (m_1 = 1\,kg, u_1 = 6\,m/s) - (m_2 = 2\,kg, u_2 = 0) - (e = 0) (stick together)

Step 4: Apply momentum conservation: [ 1(6) + 2(0) = (1 + 2)v \implies 6 = 3v \implies v = 2\,m/s ]

Step 5: Calculate KE loss: - KE before = (\frac{1}{2}(1)(6^2) + \frac{1}{2}(2)(0^2) = 18\,J) - KE after = (\frac{1}{2}(3)(2^2) = 6\,J) - KE lost = (18 - 6 = 12\,J)

Step 6: Find COM position after 1 s: - COM velocity = (\frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{6 + 0}{3} = 2\,m/s) (same as final velocity). - Distance moved in 1 s = (2 \times 1 = 2\,m). - Initial COM position (let (x_1 = 0, x_2 = d)): [ x_{COM} = \frac{1(0) + 2(d)}{3} = \frac{2d}{3} ] - After 1 s: (x_{COM} = \frac{2d}{3} + 2)

Answer: (a) (v = 2\,m/s) (b) KE lost = (12\,J) (c) COM position after 1 s = (\frac{2d}{3} + 2\,m)

What we did and why: - Used momentum conservation (inelastic collision). - Calculated KE loss (not conserved in inelastic collisions). - Found COM position using velocity (no external force → COM moves at constant velocity).

Example 3 – Exam-Style (Explosion + COM)

Problem: A 5 kg bomb at rest explodes into three fragments: - 1 kg at 10 m/s east. - 2 kg at 5 m/s north. - 2 kg at v m/s southwest. Find v and the direction of the 2 kg fragment.

Solution: Step 1: System = bomb fragments. No external forces → momentum conserved. Initial momentum = 0 (bomb at rest).

Step 2: Draw diagram: - Fragment 1: (1\,kg, 10\,m/s) east → (\vec{p}_1 = 10\hat{i}) - Fragment 2: (2\,kg, 5\,m/s) north → (\vec{p}_2 = 10\hat{j}) - Fragment 3: (2\,kg, v\,m/s) southwest → (\vec{p}_3 = -2v \cos 45° \hat{i} - 2v \sin 45° \hat{j})

Step 3: Apply momentum conservation: [ \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 ] [ 10\hat{i} + 10\hat{j} + (-2v \cos 45° \hat{i} - 2v \sin 45° \hat{j}) = 0 ]

Step 4: Separate x and y components: - x-component: [ 10 - 2v \cos 45° = 0 \implies 10 = 2v \times \frac{1}{\sqrt{2}} \implies v = \frac{10 \sqrt{2}}{2} = 5\sqrt{2}\,m/s ] - y-component: [ 10 - 2v \sin 45° = 0 \implies 10 = 2v \times \frac{1}{\sqrt{2}} \implies v = 5\sqrt{2}\,m/s ] (Both give same (v), so consistent.)

Step 5: Direction is southwest (given).

Answer: - (v = 5\sqrt{2}\,m/s) (≈ 7.07 m/s) - Direction: Southwest

What we did and why: - Used vector momentum conservation (explosion). - Broke into x and y components (NEET loves this!). - Solved for magnitude and direction separately. - Southwest means 45° below negative x-axis.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for NEET momentum problems.

1. Momentum is always conserved if no external forces act. Write (m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2) first.
2. Elastic collision? Use KE conservation + momentum. Inelastic? Only momentum.
3. Coefficient of restitution (e) gives a second equation for head-on collisions: (e = \frac{v_2 - v_1}{u_1 - u_2}).
4. Explosions? Total momentum after = 0 if initially at rest. Break into x and y if 2D.
5. Centre of mass moves at constant velocity if no external force. Use (x_{COM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}).
6. Watch signs! Right = +, left = –. Negative velocity = opposite direction.
7. NEET loves 2D problems—always split into components.

You’ve got this. Now go solve those 5-7 marks like a pro."