Physics Mechanics - How to Solve: Elastic and Inelastic Collisions (Coefficient of Restitution) – IIT JEE Guide

How to Solve: Elastic and Inelastic Collisions (Coefficient of Restitution) – IIT JEE Guide

Introduction

Mastering collisions unlocks 5-7 marks in IIT JEE—enough to push you from a 90th to a 99th percentile rank. Whether it’s a bullet embedding in a block or two cars crashing, this topic appears in every major exam, including JEE Main, Advanced, and even Olympiad papers.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Conservation of Linear Momentum – Total momentum before collision = Total momentum after collision (if no external force acts).
2. Kinetic Energy (KE) – KE = ½mv². In elastic collisions, KE is conserved; in inelastic collisions, it is not.
3. Relative Velocity – The velocity of one object as seen from another (e.g., v₁ – v₂).

KEY TERMS & FORMULAS

Key Terms

1. Collision – Interaction between two bodies where forces act for a short time.
2. Elastic Collision – Both momentum and kinetic energy are conserved.
3. Inelastic Collision – Momentum is conserved, but kinetic energy is not (some is lost as heat, sound, etc.).
4. Perfectly Inelastic Collision – The two bodies stick together after collision (maximum KE loss).
5. Coefficient of Restitution (e) – Measures "bounciness" of a collision.
6. e = 1 → Perfectly elastic collision.
7. 0 < e < 1 → Inelastic collision.
8. e = 0 → Perfectly inelastic collision.

Formulas

1. Coefficient of Restitution (e)

Formula: [ e = \frac{\text{Relative speed after collision}}{\text{Relative speed before collision}} = \frac{v_2' - v_1'}{v_1 - v_2} ] - ( v_1, v_2 ) → Velocities before collision. - ( v_1', v_2' ) → Velocities after collision. - MEMORISE THIS – This is the most important formula for collisions.

2. Conservation of Momentum

Formula: [ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' ] - Given on exam sheet (but you must know how to apply it).

3. Kinetic Energy in Elastic Collisions

Formula: [ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2 ] - MEMORISE THIS – Only applies to elastic collisions (e = 1).

4. Final Velocities in Elastic Collision (Special Case: One Body at Rest)

If ( m_1 ) collides with ( m_2 ) at rest (( v_2 = 0 )): [ v_1' = \frac{(m_1 - m_2)v_1}{m_1 + m_2} ] [ v_2' = \frac{2m_1v_1}{m_1 + m_2} ] - MEMORISE THIS – Saves time in exams.

5. Final Velocity in Perfectly Inelastic Collision

[ v' = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} ] - MEMORISE THIS – Both bodies move together after collision.

STEP-BY-STEP METHOD

Step 1: Identify the Type of Collision

• Elastic (e = 1) → Use momentum + KE conservation.
• Inelastic (0 < e < 1) → Use momentum + coefficient of restitution.
• Perfectly Inelastic (e = 0) → Use momentum + final velocity formula.

Step 2: Draw a Diagram

• Sketch before and after collision.
• Label masses (( m_1, m_2 )), initial velocities (( v_1, v_2 )), and final velocities (( v_1', v_2' )).

Step 3: Write Down Known Quantities

• List all given values (masses, velocities, e).
• If any velocity is unknown, assign a variable (e.g., ( v_1' )).

Step 4: Apply Conservation of Momentum

• Write the momentum equation: [ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' ]
• Solve for one unknown if possible.

Step 5: Apply Coefficient of Restitution (e)

• Write the restitution equation: [ e = \frac{v_2' - v_1'}{v_1 - v_2} ]
• If e = 1, you can also use KE conservation for a second equation.

Step 6: Solve the Equations

• If two unknowns (( v_1', v_2' )), solve the two equations (momentum + restitution) simultaneously.
• If e = 0, use the perfectly inelastic formula directly.

Step 7: Check Units and Signs

• Ensure all velocities are in same direction (positive/negative).
• If a body reverses direction, its velocity sign changes.

Step 8: Calculate Final Quantities (if asked)

• Find KE before/after, percentage loss, etc.

WORKED EXAMPLES

Example 1 – Basic (Elastic Collision)

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

Solution:
1. Identify collision type: Elastic (e = 1).
2. Diagram: - Before: ( m_1 = 2 ) kg, ( v_1 = 5 ) m/s; ( m_2 = 3 ) kg, ( v_2 = 0 ). - After: ( v_1' = ? ), ( v_2' = ? ).
3. Momentum conservation: [ 2(5) + 3(0) = 2v_1' + 3v_2' ] [ 10 = 2v_1' + 3v_2' \quad \text{(Equation 1)} ]
4. Restitution (e = 1): [ 1 = \frac{v_2' - v_1'}{5 - 0} ] [ v_2' - v_1' = 5 \quad \text{(Equation 2)} ]
5. Solve Equations 1 & 2: - From Equation 2: ( v_2' = v_1' + 5 ). - Substitute into Equation 1: [ 10 = 2v_1' + 3(v_1' + 5) ] [ 10 = 5v_1' + 15 ] [ 5v_1' = -5 ] [ v_1' = -1 \text{ m/s} ] - Then, ( v_2' = -1 + 5 = 4 \text{ m/s} ).
6. Final Answer: - ( v_1' = -1 ) m/s (reverses direction). - ( v_2' = 4 ) m/s.

What we did and why: - Used momentum + restitution for elastic collisions. - Solved two equations for two unknowns. - Negative sign indicates direction change.

Example 2 – Medium (Inelastic Collision)

Problem: A 4 kg block moving at 6 m/s collides with a 2 kg block moving at 3 m/s in the same direction. If ( e = 0.5 ), find their velocities after collision.

Solution:
1. Identify collision type: Inelastic (( e = 0.5 )).
2. Diagram: - Before: ( m_1 = 4 ) kg, ( v_1 = 6 ) m/s; ( m_2 = 2 ) kg, ( v_2 = 3 ) m/s. - After: ( v_1' = ? ), ( v_2' = ? ).
3. Momentum conservation: [ 4(6) + 2(3) = 4v_1' + 2v_2' ] [ 30 = 4v_1' + 2v_2' ] [ 15 = 2v_1' + v_2' \quad \text{(Equation 1)} ]
4. Restitution (e = 0.5): [ 0.5 = \frac{v_2' - v_1'}{6 - 3} ] [ v_2' - v_1' = 1.5 \quad \text{(Equation 2)} ]
5. Solve Equations 1 & 2: - From Equation 2: ( v_2' = v_1' + 1.5 ). - Substitute into Equation 1: [ 15 = 2v_1' + (v_1' + 1.5) ] [ 15 = 3v_1' + 1.5 ] [ 3v_1' = 13.5 ] [ v_1' = 4.5 \text{ m/s} ] - Then, ( v_2' = 4.5 + 1.5 = 6 \text{ m/s} ).
6. Final Answer: - ( v_1' = 4.5 ) m/s. - ( v_2' = 6 ) m/s.

What we did and why: - Used momentum + restitution for inelastic collisions. - Solved two equations for two unknowns. - Both bodies continue in the same direction (no sign change).

Example 3 – Exam-Style (Perfectly Inelastic + KE Loss)

Problem: A 3 kg mass moving at 8 m/s collides with a 5 kg mass at rest. They stick together. Find: (a) Final velocity. (b) Percentage loss in kinetic energy.

Solution:
1. Identify collision type: Perfectly inelastic (( e = 0 )).
2. Diagram: - Before: ( m_1 = 3 ) kg, ( v_1 = 8 ) m/s; ( m_2 = 5 ) kg, ( v_2 = 0 ). - After: Combined mass ( m = 8 ) kg, ( v' = ? ).
3. (a) Final velocity: [ v' = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{3(8) + 5(0)}{8} = \frac{24}{8} = 3 \text{ m/s} ]
4. (b) KE before collision: [ KE_{\text{initial}} = \frac{1}{2}(3)(8)^2 + \frac{1}{2}(5)(0)^2 = 96 \text{ J} ]
5. KE after collision: [ KE_{\text{final}} = \frac{1}{2}(8)(3)^2 = 36 \text{ J} ]
6. Percentage loss: [ \text{Loss} = \frac{96 - 36}{96} \times 100 = 62.5\% ]

Final Answer: (a) ( v' = 3 ) m/s. (b) 62.5% KE loss.

What we did and why: - Used perfectly inelastic formula for final velocity. - Calculated KE before/after to find loss. - Exam trap: They may ask for percentage loss, not just final KE.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is all you need to remember for collisions in JEE:

1. Momentum is always conserved—write ( m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' ) first.
2. Coefficient of restitution (e) tells you the collision type:
3. e = 1? Elastic → Use momentum + KE conservation.
4. 0 < e < 1? Inelastic → Use momentum + restitution formula.
5. e = 0? Perfectly inelastic → They stick together → Use ( v' = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} ).
6. For elastic collisions with one body at rest, memorize these shortcuts: [ v_1' = \frac{(m_1 - m_2)v_1}{m_1 + m_2} ] [ v_2' = \frac{2m_1v_1}{m_1 + m_2} ]
7. Always draw a diagram—label directions, masses, and velocities.
8. Check signs! If a body reverses, its velocity becomes negative.
9. If they ask for KE loss, calculate initial and final KE and find the difference.

That’s it. Now go crush those 5-7 marks in your exam!