Elastic and Inelastic Collisions

Elastic and Inelastic Collisions

Book Subtopics

• Meaning of collision
• Elastic collision
• Inelastic collision
• One-dimensional elastic collision
• Conservation of momentum
• Conservation of kinetic energy
• Relative speed of approach and separation
• Final velocities after elastic collision
• Special collision cases

1. Learning Outcomes

After studying this topic, students should be able to:

• Define a collision.
• Differentiate between elastic and inelastic collisions.
• Explain which quantities remain conserved.
• Apply momentum conservation to collision problems.
• Apply kinetic-energy conservation to elastic collisions.
• Derive the relative-velocity relation.
• Calculate the final velocities of two objects after an elastic collision.
• Explain important special cases involving equal and unequal masses.
• Solve perfectly inelastic collision problems.
• Explain why momentum may remain conserved while kinetic energy changes.

2. Definition of Collision

Examples include:

• Two laboratory carts interacting
• Two billiard balls striking each other
• A ball interacting with a bat
• Atomic and subatomic particle interactions

3. Types of Collisions

Collisions are mainly classified as:

• Elastic collisions
• Inelastic collisions

A special type of inelastic collision is called a perfectly inelastic collision.

4. Elastic Collision

An elastic collision is one in which both total momentum and total kinetic energy remain conserved.

Momentum Conservation

Σ p⃗_i = Σ p⃗_f

Kinetic-Energy Conservation

Σ K_i = Σ K_f

Ideal interactions between atomic or subatomic particles may be treated as elastic.

5. Inelastic Collision

An inelastic collision is one in which total momentum is conserved, but total kinetic energy is not conserved.

Σ p⃗_i = Σ p⃗_f

but:

Σ Ki ≠ Σ Kf

Some initial kinetic energy changes into other forms, such as:

• Thermal energy
• Sound energy
• Internal energy
• Deformation energy

The total energy remains conserved, but kinetic energy does not remain completely kinetic.

6. Perfectly Inelastic Collision

A perfectly inelastic collision occurs when the objects remain together after collision and move with a common velocity.

If the final common velocity is v:

m₁u₁ + m₂u₂ = (m₁ + m₂)v

Therefore:

v = m₁u₁ + m₂u₂m₁ + m₂

This type of collision produces the greatest loss of kinetic energy allowed by momentum conservation for the given initial conditions.

7. Diagram Placement

Elastic and Inelastic Collisions

The definitions and equations must also remain available as searchable website text.

8. Elastic and Inelastic Collision Comparison

Elastic and Inelastic Collision Comparison

9. Important Clarification

All perfectly inelastic collisions are inelastic.

However, not every inelastic collision is perfectly inelastic.

In an ordinary inelastic collision, the objects may separate afterward.

In a perfectly inelastic collision, they remain together.

3.4.1 One-Dimensional Elastic Collision

10. Initial Conditions

Consider two objects with:

• Masses m₁ and m₂
• Initial velocities u₁ and u₂
• Final velocities v₁ and v₂

Their motion occurs along the same straight line.

Take rightward velocity as positive and leftward velocity as negative.

11. Conservation of Momentum

For an isolated system:

Initial momentum = Final momentum

Therefore:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Rearranging:

m₁u₁ − m₁v₁ = m₂v₂ − m₂u₂

Taking common factors:

m₁(u₁ − v₁) = m₂(v₂ − u₂)

12. Conservation of Kinetic Energy

For an elastic collision:

Initial kinetic energy = Final kinetic energy

Therefore:

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

Multiplying by 2:

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

Rearranging:

m₁(u₁² − v₁²) = m₂(v₂² − u₂²)

Using the difference of squares:

m₁(u₁ − v₁)(u₁ + v₁) = m₂(v₂ − u₂)(v₂ + u₂)

13. Relative Speed Relation

From momentum conservation:

m₁(u₁ − v₁) = m₂(v₂ − u₂)

From kinetic-energy conservation:

m₁(u₁ − v₁)(u₁ + v₁) = m₂(v₂ − u₂)(v₂ + u₂)

Dividing the second equation by the first:

u₁ + v₁ = v₂ + u₂

Rearranging:

u₁ − u₂ = v₂ − v₁

Therefore:

Relative speed of approach = Relative speed of separation

This relation applies to a one-dimensional elastic collision.

14. Physical Meaning of Relative Speed

Before collision, the relative speed of approach is:

u₁ − u₂

After collision, the relative speed of separation is:

v₂ − v₁

For a perfectly elastic collision:

u₁ − u₂ = v₂ − v₁

The relation uses velocities with signs, not just positive speed values.

15. Final Velocity of the First Object

The final velocity of mass m₁ is:

v₁ = (m₁ − m₂m₁ + m₂) u₁ + (2m₂/(m₁ + m₂)) u₂

It can also be written as:

v₁ = (m₁ − m₂)u₁ + 2m₂u₂m₁ + m₂

16. Final Velocity of the Second Object

The final velocity of mass m₂ is:

v₂ = (2m₁/(m₁ + m₂)) u₁ + (m₂ − m₁m₁ + m₂) u₂

It can also be written as:

v₂ = 2m₁u₁ + (m₂ − m₁)u₂m₁ + m₂

17. Symbols and Units

Symbols and Units

Special Cases of Elastic Collisions

18. Case 1: Objects Have Equal Masses

If:

m₁ = m₂

Then:

v₁ = u₂

and:

v₂ = u₁

The objects exchange their velocities.

19. Case 2: Equal Masses and Target Initially at Rest

If:

m₁ = m₂

and:

u₂ = 0

Then:

v₁ = 0

and:

v₂ = u₁

The first object stops, while the second object moves with the first object’s initial velocity.

This behaviour can be approximately observed with identical laboratory carts or ideal billiard balls.

20. Case 3: Light Object Strikes a Very Massive Stationary Object

If:

m₂ ≫ m₁

and:

u₂ = 0

Then approximately:

v₁ ≈ −u₁

and:

v₂ ≈ 0

The light object rebounds with almost the same speed in the opposite direction.

The massive object experiences a negligible change in velocity.

21. Case 4: Very Massive Object Strikes a Light Stationary Object

If:

m₁ ≫ m₂

and:

u₂ = 0

Then approximately:

v₁ ≈ u₁

and:

v₂ ≈ 2u₁

The massive object continues with nearly its original velocity.

The light object may move with nearly twice the massive object’s initial velocity.

These are limiting approximations for a perfectly elastic interaction.

22. Book-Based Solved Example: Elastic Collision

Question: A 5 kg object moving at 10 m s⁻¹ interacts elastically with a 10 kg object moving in the same direction at 5 m s⁻¹. Find their final velocities.

Given

m₁ = 5 kg u₁ = 10 m s⁻¹ m₂ = 10 kg u₂ = 5 m s⁻¹

Final Velocity of the First Object

v₁ = (m₁ − m₂m₁ + m₂) u₁ + (2m₂/(m₁ + m₂)) u₂ v₁ = (5 − 105 + 10)(10) + (2105 + 10)(5) v₁ = (−515)(10) + (2015)(5) v₁ = −5015 + 10015 v₁ = 5015 v₁ = 3.33 m s⁻¹

Final Velocity of the Second Object

v₂ = (2m₁/(m₁ + m₂)) u₁ + (m₂ − m₁m₁ + m₂) u₂ v₂ = (255 + 10)(10) + (10 − 55 + 10)(5) v₂ = (1015)(10) + (515)(5) v₂ = 10015 + 2515 v₂ = 12515 v₂ = 8.33 m s⁻¹

Final Answers

v₁ = 3.33 m s⁻¹ v₂ = 8.33 m s⁻¹

23. Verification of Momentum

Initial Momentum

p_i = m₁u₁ + m₂u₂ p_i = (5)(10) + (10)(5) p_i = 50 + 50 p_i = 100 kg m s⁻¹

Final Momentum

Using unrounded final velocities:

p_f = m₁v₁ + m₂v₂ p_f = 100 kg m s⁻¹

Therefore:

p_i = p_f

Small differences may appear if rounded velocities are substituted too early.

24. Solved Example: Perfectly Inelastic Collision

Question: A 3 kg cart moving right at 8 m s⁻¹ interacts with a stationary 5 kg cart. They remain together. Find their common velocity.

Given

m₁ = 3 kg u₁ = 8 m s⁻¹ m₂ = 5 kg u₂ = 0

Formula

m₁u₁ + m₂u₂ = (m₁ + m₂)v

Solution

(3)(8) + (5)(0) = (3 + 5)v 24 = 8v v = 3 m s⁻¹

Final Answer

v = 3 m s⁻¹ rightward

25. Kinetic-Energy Check

Initial Kinetic Energy

K_i = ½ m₁u₁² K_i = ½(3)(8)² K_i = ½(3)(64) K_i = 96 J

Final Kinetic Energy

K_f = ½(m₁ + m₂)v² K_f = ½(8)(3)² K_f = 4(9) K_f = 36 J

Change in Kinetic Energy

ΔK = K_f − K_i ΔK = 36 − 96 ΔK = −60 J

The negative sign shows that 60 J of kinetic energy changed into other forms.

Total energy remains conserved.

26. Key Logic

In Every Isolated Collision

Momentum is conserved

Only in an Elastic Collision

Kinetic energy is also conserved

In a Perfectly Inelastic Collision

Objects move together afterward

Therefore, begin every collision problem with momentum conservation. Apply kinetic-energy conservation only when the collision is stated to be elastic.

27. Coefficient of Restitution

The coefficient of restitution is sometimes used to classify collisions.

However, according to the learning scope of this chapter, detailed knowledge or calculation of the coefficient of restitution is not required.

Students should focus on:

• Momentum conservation
• Kinetic-energy conservation
• Relative speed of approach
• Relative speed of separation

28. Real-Life Applications

Collision principles are useful in:

• Laboratory-cart experiments
• Billiard and snooker-ball motion
• Ball-and-bat interactions
• Vehicle-safety analysis
• Particle-physics experiments
• Packaging and protective materials
• Sports-equipment design

29. Important Facts

• Momentum is conserved in every isolated collision.
• Kinetic energy is conserved only in an elastic collision.
• Total energy is always conserved.
• A perfectly inelastic collision has a common final velocity.
• Equal masses exchange velocities in a one-dimensional elastic collision.
• A velocity’s negative sign indicates direction.
• Objects do not need to remain together in every inelastic collision.
• Relative speed of approach equals relative speed of separation only for an elastic collision.

30. Tips and Tricks

• Select the positive direction before writing velocities.
• Convert all masses into kilograms.
• Use negative values for motion opposite to the positive direction.
• Apply momentum conservation first.
• Use kinetic-energy conservation only for elastic collisions.
• Use a common final velocity only if the objects remain together.
• Retain calculator values until the final answer.
• Verify elastic-collision answers using both conservation laws.
• Interpret negative final velocity as reversed direction.

31. Common Mistakes

• Assuming kinetic energy is conserved in every collision.
• Assuming objects always remain together in an inelastic collision.
• Ignoring velocity directions.
• Adding speeds instead of signed velocities.
• Confusing total energy with kinetic energy.
• Applying elastic final-velocity formulas to an inelastic collision.
• Forgetting the square of velocity in kinetic energy.
• Rounding final velocities too early.
• Using a common velocity when objects separate.
• Treating momentum as a scalar.

32. Remember Box

33. Exam Facts

• Collision forces are internal forces within the selected system.
• Elastic collision conserves momentum and kinetic energy.
• Inelastic collision conserves momentum but not kinetic energy.
• Perfectly inelastic objects remain together.
• Equal masses exchange velocities during a one-dimensional elastic collision.
• Relative approach and separation speeds are equal in an elastic collision.
• A negative final velocity means reversed direction.
• Final-velocity formulas apply to one-dimensional elastic collisions.

34. Short-Question Preparation

Short-Question Preparation

35. Long-Question Preparation

Derive the relative-speed relation for an elastic collision.

Consider two objects moving along one straight line. Their masses are m₁ and m₂. Their initial velocities are u₁ and u₂. Their final velocities are v₁ and v₂. Momentum conservation gives: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Rearranging: m₁(u₁ − v₁) = m₂(v₂ − u₂) Kinetic-energy conservation gives: ½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂² Multiplying by two and rearranging: m₁(u₁² − v₁²) = m₂(v₂² − u₂²) Factoring the differences of squares: m₁(u₁ − v₁)(u₁ + v₁) = m₂(v₂ − u₂)(v₂ + u₂) Divide this equation by the rearranged momentum equation. The common terms cancel: u₁ + v₁ = v₂ + u₂ Rearranging gives: u₁ − u₂ = v₂ − v₁ The left side is the relative speed of approach. The right side is the relative speed of separation. Hence, both relative speeds are equal in an elastic collision.

36. MCQs

37. Practice Questions

Beginner

• Define collision.
• Define an elastic collision.
• Define an inelastic collision.
• What is a perfectly inelastic collision?
• Which quantity is conserved in every isolated collision?

Intermediate

• Two equal-mass carts interact elastically. One is initially stationary. Explain their final motion.
• A 2 kg cart moving at 6 m s⁻¹ remains attached to a stationary 4 kg cart. Find their common velocity.
• Calculate the initial and final kinetic energies for Question 2.
• State the relative-speed relation for an elastic collision.
• Explain why momentum conservation does not guarantee kinetic-energy conservation.

Advanced

• Derive the final-velocity formulas for a one-dimensional elastic collision.
• Derive the relative-speed relation.
• Explain all four special cases described in the textbook.
• Verify momentum and kinetic-energy conservation for an elastic-collision result.
• Compare elastic, inelastic and perfectly inelastic collisions.

38. Practice Answer Key

For the 2 kg cart moving at 6 m s⁻¹ and the stationary 4 kg cart: m₁u₁ + m₂u₂ = (m₁ + m₂)v (2)(6) + (4)(0) = (2 + 4)v 12 = 6v v = 2 m s⁻¹ Initial kinetic energy: Ki = ½(2)(6)² Ki = 36 J Final kinetic energy: Kf = ½(6)(2)² Kf = 12 J Kinetic-energy change: ΔK = 12 − 36 ΔK = −24 J Momentum is conserved, while 24 J of kinetic energy changes into other forms.

39. Sticky Notes

Sticky Notes

• Elastic: momentum and kinetic energy conserved.
• Inelastic: momentum conserved, kinetic energy not conserved.
• Perfectly inelastic: objects remain together.
• Equal elastic masses exchange velocities.
• Velocity signs represent directions.
• Begin collision problems with momentum conservation.
• Use kinetic-energy conservation only when the collision is elastic.
• Avoid early rounding.

40. 60-Second Revision

Momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Elastic-collision kinetic energy: ½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂² Relative-speed relation: u₁ − u₂ = v₂ − v₁ Perfectly inelastic common velocity: v = m₁u₁ + m₂u₂m₁ + m₂

41. Topic Summary

In every isolated collision, total momentum remains conserved. An elastic collision also conserves total kinetic energy. In an inelastic collision, some kinetic energy changes into other forms. During a perfectly inelastic collision, objects remain together and move with a common velocity.