Elastic and Inelastic Collisions - AP Physics 1

SECTION 1

Historical Context & Motivation

The study of collisions stands among the oldest problems in classical mechanics. Long before Newton formalized the laws of motion, natural philosophers puzzled over what happens when two objects strike each other—whether billiard balls on a table or celestial bodies in space. The concept of momentum as a conserved quantity emerged from decades of debate, experiment, and mathematical refinement in seventeenth-century Europe. Understanding how collisions redistribute momentum and energy remains central to modern physics, engineering, and safety design.

1668 • Royal Society Collision Experiments

John Wallis, Christopher Wren, and Christiaan Huygens independently submitted papers to the Royal Society analyzing collisions. Wallis treated perfectly inelastic impacts, while Huygens derived the rules for perfectly elastic collisions, recognizing that something akin to kinetic energy is conserved in those cases.

1687 • Newton's Principia

Isaac Newton published the Principia Mathematica, establishing the three laws of motion. His third law—action and reaction—provided the theoretical backbone for conservation of momentum in all collisions.

1743 • d'Alembert and the Vis Viva Debate

Jean le Rond d'Alembert helped resolve the long-standing debate between Leibniz's vis viva (living force, proportional to mv²) and the Cartesian concept of momentum (mv), clarifying that both quantities play distinct roles in collision analysis.

1960s • Particle Collider Era

High-energy particle accelerators began using conservation of momentum and energy in collisions to discover subatomic particles. These principles, first tested with billiard balls and pendulums, now probe the fundamental structure of matter.

The central question that collision theory addresses is straightforward yet powerful: given that total momentum is always conserved in an isolated system, how do we determine what happens to the kinetic energy? The answer depends on the type of collision—elastic, inelastic, or perfectly inelastic—and this classification lies at the heart of AP Physics 1 momentum problems.

SECTION 2

Core Principles & Definitions

Before classifying collisions, it is essential to recall the foundational conservation law that governs every collision in an isolated system. The law of conservation of momentum states that the total linear momentum of a system remains constant when no net external force acts on it. This law applies universally—regardless of whether the collision is elastic, inelastic, or perfectly inelastic. What distinguishes these types is the behavior of kinetic energy during the interaction.

Elastic Collision

Both momentum and kinetic energy are conserved. Objects bounce off each other with no permanent deformation or heat generation. Ideal gas molecule collisions approximate this behavior.

Inelastic Collision

Momentum is conserved, but kinetic energy is not conserved. Some kinetic energy is transformed into thermal energy, sound, or deformation. Most real-world collisions fall in this category.

Perfectly Inelastic Collision

A special case of inelastic collision where the objects stick together after impact and move as a single combined mass. This produces the maximum possible kinetic energy loss consistent with momentum conservation.

Isolated System Requirement

Momentum is conserved only when the net external force on the system is zero. In practice, we often treat collisions as occurring over such short time intervals that external impulses are negligible.

SECTION 3

Visual Explanation

Before and After: Three Types of Collisions

In the elastic row, both objects separate with individual velocities and total KE is preserved. In the inelastic row, dashed outlines indicate deformation, and KE decreases. In the perfectly inelastic row, the two masses merge into a single object moving with a common velocity, losing the maximum kinetic energy.

The diagram above illustrates the fundamental distinction among the three collision types for a one-dimensional scenario in which mass m₁ approaches a stationary mass m₂. Notice that in every case, the total momentum vector before the collision equals the total momentum vector after—the dashed center line separates the initial and final states. What changes is how kinetic energy is partitioned. In the elastic case, the total kinetic energy bar is the same height on both sides; in the perfectly inelastic case, the combined object moves slower, and a significant fraction of the original kinetic energy has been converted into internal energy of the system.

SECTION 4

Mathematical Framework

Conservation of Momentum (All Collisions)

CONSERVATION OF MOMENTUM

m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

Where m₁ and m₂ are the masses of the two objects, v₁ᵢ and v₂ᵢ are the initial velocities, and v₁f and v₂f are the final velocities. This equation holds for every type of collision in an isolated system.

Elastic Collision: Kinetic Energy Conservation

KINETIC ENERGY CONSERVATION (ELASTIC ONLY)

½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f²

This second equation is available only in elastic collisions. With two equations (momentum and KE), you can solve for two unknowns (v₁f and v₂f).

For the special case of a one-dimensional elastic collision, combining the two conservation equations yields a powerful result: the relative velocity of approach equals the relative velocity of separation. Mathematically, v₁ᵢ − v₂ᵢ = −(v₁f − v₂f). This relation is algebraically equivalent to KE conservation and is often faster to use on the AP exam than manipulating squared velocity terms.

RELATIVE VELOCITY RELATION (ELASTIC, 1-D)

v₁ᵢ − v₂ᵢ = −(v₁f − v₂f)

The approach speed equals the separation speed. This replaces the quadratic KE equation with a linear one, simplifying algebra considerably.

Perfectly Inelastic Collision

PERFECTLY INELASTIC COLLISION

m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vf

Since the objects stick together, v₁f = v₂f = vf. Solving: vf = (m₁v₁ᵢ + m₂v₂ᵢ) / (m₁ + m₂). The lost kinetic energy equals ΔKE = KEᵢ − KEf, and this quantity is always positive.

ℹ

AP Exam Tip

SECTION 5

Detailed Classification & Energy Analysis

A useful way to classify collisions is by the fraction of kinetic energy retained after the impact. We can define a quantity often called the coefficient of restitution (e), which equals the ratio of relative speed of separation to relative speed of approach: e = |v₂f − v₁f| / |v₁ᵢ − v₂ᵢ|. For a perfectly elastic collision e = 1; for a perfectly inelastic collision e = 0; and for all other inelastic collisions 0 < e < 1. While the AP exam does not explicitly test this coefficient, understanding it clarifies the spectrum of collision behavior.

The top spectrum bar shows the coefficient of restitution ranging from 0 (perfectly inelastic) to 1 (elastic). The bottom energy bar charts compare outcomes for equal-mass collisions: in the elastic case, all kinetic energy transfers from m₁ to m₂; in the perfectly inelastic case, 50% of the kinetic energy is lost to internal energy (shown in red).

Property	Elastic	Inelastic	Perfectly Inelastic
Momentum conserved?	Yes	Yes	Yes
KE conserved?	Yes	No	No
Objects stick together?	No	No	Yes
Coefficient of restitution	e = 1	0 < e < 1	e = 0
# unknowns after collision	2 (use momentum + KE)	2 (need extra info)	1 (use momentum only)
Example	Ideal gas molecules, Newton's cradle	Car crash with bounce, tennis ball	Catching a ball, railroad coupling

Comparison of collision types

A level capture block hangs on parallel cords inside a protective enclosure, with an embedded pellet visible in a conceptual window. Treat capture and swing as separate stages. Horizontal momentum is conserved during the short impact; mechanical energy is conserved during the subsequent ideal swing.

The impact that becomes a pendulum swing. Treat capture and swing as separate stages. Horizontal momentum is conserved during the short impact; mechanical energy is conserved during the subsequent ideal swing. Treat capture and swing as separate stages. Horizontal momentum is conserved during the short impact; mechanical energy is conserved during the subsequent ideal swing.

SECTION 6

Worked Example

Worked example

1
Step 1 — Identify Given Values

A 4.0 kg cart moving at 6.0 m/s to the right collides with a 2.0 kg cart initially at rest on a frictionless track. The two carts lock together after the collision. Find (a) the final velocity and (b) the fraction of kinetic energy lost.
2
Step 2 — Classify the Collision

The carts lock together, so this is a perfectly inelastic collision. We use: m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vf.
3
Step 3 — Apply Conservation of Momentum

Substituting: (4.0 kg)(6.0 m/s) + (2.0 kg)(0 m/s) = (4.0 + 2.0 kg) × vf → 24 kg·m/s = 6.0 kg × vf.

vf = 4.0 m/s to the right
4
Step 4 — Calculate Initial Kinetic Energy

KEᵢ = ½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½(4.0)(6.0)² + 0 = 72 J.

KEᵢ = 72 J
5
Step 5 — Calculate Final Kinetic Energy

KEf = ½(m₁ + m₂)vf² = ½(6.0)(4.0)² = 48 J.

KEf = 48 J
6
Step 6 — Determine Fraction of KE Lost

ΔKE = 72 − 48 = 24 J. Fraction lost = ΔKE / KEᵢ = 24 / 72 = 1/3 ≈ 0.33. One-third of the kinetic energy was converted into internal energy (heat, sound, deformation). Notice that in any perfectly inelastic collision where one object is initially at rest, the fraction of KE lost equals m₂ / (m₁ + m₂), which here is 2/6 = 1/3.

Fraction lost = 1/3 (about 33%)

Worked example

1
Step 1 — Identify Given Values

A 3.0 kg ball moving at 5.0 m/s collides head-on and elastically with another 3.0 kg ball at rest. Find the final velocities of both balls.
2
Step 2 — Apply Momentum Conservation

(3.0)(5.0) + (3.0)(0) = 3.0 v₁f + 3.0 v₂f → 15 = 3v₁f + 3v₂f → v₁f + v₂f = 5.0 m/s.
3
Step 3 — Apply Relative Velocity Relation

For elastic collisions: v₁ᵢ − v₂ᵢ = −(v₁f − v₂f). Substituting: 5.0 − 0 = −(v₁f − v₂f) → v₂f − v₁f = 5.0 m/s.
4
Step 4 — Solve the System

Adding the two equations: v₁f + v₂f = 5.0 and v₂f − v₁f = 5.0 → 2v₂f = 10 → v₂f = 5.0 m/s, v₁f = 0 m/s. The moving ball stops completely, and the stationary ball takes on its entire velocity.

v₁f = 0 m/s, v₂f = 5.0 m/s

AP Physics 1: a 4 kg cart at 6 m/s strikes a stationary 2 kg cart. An elastic collision gives velocities 2 and 8 m/s; sticking gives both 4 m/s. Momentum remains 24 kg·m/s. The C companion uses different masses.

SECTION 7

Strengths, Limitations & Common Misconceptions

Aspect	Strengths	Limitations / Pitfalls
Momentum conservation	Universal—applies to all collision types regardless of internal forces.	Only valid in isolated systems; external forces (friction, gravity during long collisions) violate the assumption.
KE conservation (elastic)	Provides a second equation, making two-body elastic problems fully solvable.	Perfectly elastic collisions are idealizations—real macroscopic collisions always lose some KE.
Perfectly inelastic model	Simplest to solve—only one unknown (vf). Directly applicable to ballistic pendulums.	Students often incorrectly assume KE is conserved or forget to account for the combined mass.
1-D vs. 2-D	1-D problems are algebraically clean with scalar equations.	AP Physics 1 may include 2-D glancing collisions; you must apply momentum conservation independently in x and y.

Strengths and limitations of collision models

ℹ

Common Misconception

SECTION 8

Connections to Advanced Theory

The collision framework you learn in AP Physics 1 extends naturally into more advanced domains. In AP Physics C, you will encounter collisions analyzed through calculus-based impulse integrals, where the force-time profile during impact is modeled explicitly. At the university level, relativistic collisions in special relativity require modifications to the momentum expression (p = γmv) and use the invariant mass-energy relation E² = (pc)² + (mc²)². In nuclear and particle physics, collisions are the primary experimental tool—particles are smashed together at near-light speeds, and conservation of four-momentum determines what new particles can be created.

Feature	AP Physics 1 Treatment	Advanced Treatment
Momentum	p = mv (classical)	p = γmv (relativistic); four-momentum in spacetime
Collision analysis	Before/after snapshots; algebra	Impulse integrals ∫F dt; Lagrangian/Hamiltonian formulations
Energy accounting	KE conserved or lost	Mass-energy equivalence; particle creation/annihilation
Dimensions	Primarily 1-D; some 2-D	Full 3-D with center-of-mass reference frames

AP Physics 1 vs. Advanced collision physics

One particularly elegant extension is the center-of-mass reference frame, in which the total momentum of the system is zero by definition. In this frame, elastic collisions simply reverse the velocities of the two objects, and perfectly inelastic collisions bring both objects to rest. Mastering the AP-level framework gives you the conceptual foundation to work in any reference frame and to appreciate why particle physicists build ever-larger colliders to reach higher center-of-mass energies.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Two ice skaters push off from each other on a frictionless rink. Skater A has twice the mass of Skater B. Which of the following statements is true about the push-off?

A. Skater A gains twice as much momentum as Skater B.
B. Both skaters gain equal magnitudes of momentum but in opposite directions.
C. Both skaters gain equal kinetic energy.
D. Skater A gains more kinetic energy than Skater B because A has greater mass.

Show answer (B)

B is correct. By Newton's third law, the force on each skater during the push-off is equal in magnitude and opposite in direction. Since both forces act over the same time interval, the impulse on each skater is equal in magnitude and opposite in direction. Therefore, both skaters gain equal magnitudes of momentum but in opposite directions. A is wrong because Newton's third law guarantees that both skaters experience the same magnitude of impulse, so they gain equal magnitudes of momentum regardless of their masses. C is wrong because kinetic energy depends on mass: since both skaters gain the same magnitude of momentum p, and KE = p²/(2m), the lighter Skater B gains more kinetic energy than the heavier Skater A. D is wrong for the same reason—Skater A has greater mass and therefore a smaller speed, resulting in less kinetic energy than Skater B, not more.

Question 2: BASIC CALCULATION

A 5.0 kg ball moving at 3.0 m/s to the right collides with and sticks to a 10.0 kg ball at rest. What is the velocity of the combined mass after the collision?

A. 0.5 m/s to the right
B. 1.0 m/s to the right
C. 1.5 m/s to the right
D. 3.0 m/s to the right

Show answer (B)

B is correct. This is a perfectly inelastic collision. Applying conservation of momentum: (5.0)(3.0) + (10.0)(0) = (5.0 + 10.0)vf → 15 = 15vf → vf = 1.0 m/s to the right. A (0.5 m/s) would result from an error like dividing by 30 instead of 15. C (1.5 m/s) comes from incorrectly dividing 15 by 10 instead of by the total mass 15. D (3.0 m/s) incorrectly assumes the combined mass moves at the initial velocity of m₁, ignoring the momentum sharing with m₂.

Question 3: INTERMEDIATE

A 2.0 kg object moving at 8.0 m/s to the right collides head-on with a 6.0 kg object moving at 2.0 m/s to the left. If the collision is perfectly inelastic, what is the velocity of the combined object after the collision?

A. 0.50 m/s to the right
B. 0.50 m/s to the left
C. 2.0 m/s to the left
D. 1.0 m/s to the right

Show answer (A)

A is correct. Define rightward as positive. The total initial momentum is p_total = m₁v₁ + m₂v₂ = (2.0 kg)(+8.0 m/s) + (6.0 kg)(−2.0 m/s) = +16 − 12 = +4.0 kg·m/s. In a perfectly inelastic collision the objects stick together, so the combined mass is 2.0 + 6.0 = 8.0 kg. By conservation of momentum, vf = p_total / m_total = 4.0 / 8.0 = +0.50 m/s. The positive sign indicates the combined object moves to the right at 0.50 m/s. B is wrong because the net momentum is positive (rightward), not negative (leftward). C is wrong because 2.0 m/s to the left would require a net momentum of −16 kg·m/s. D is wrong because 1.0 m/s to the right would require a net momentum of 8.0 kg·m/s.

Question 4: APPLIED

A ballistic pendulum consists of a 0.010 kg bullet fired horizontally into a 2.0 kg wooden block suspended from strings. The block (with the bullet embedded) swings upward to a maximum height of 0.20 m. Design an experimental procedure to determine the bullet's initial speed using only a ruler and a scale, and calculate the bullet's speed from the given data. (Use g = 10 m/s².)

Show answer

Solution: (1 pt) The bullet embeds in the block, making this a perfectly inelastic collision where momentum is conserved:

m_b

v_b

= (

m_b

+ M) × V. (1 pt) After collision, energy is conserved as the system swings: ½(

m_b

+ M)V² = (

m_b

+ M)gΔh, giving V = √(2gΔh). (1 pt) V = √(2 × 10 × 0.20) = √4.0 = 2.0 m/s. (1 pt)

v_b

= (

m_b

+ M)V /

m_b

= (2.01)(2.0)/(0.010) ≈ 400 m/s. (1 pt) Procedure: Use the scale to measure masses of bullet and block. Suspend the block from strings, fire the bullet into it, and use the ruler to measure the maximum height the block-bullet system reaches. Apply the two-step analysis (momentum for collision, energy for swing) to compute bullet speed.

Question 5: CRITICAL THINKING

Two objects of masses m and 3m undergo a head-on collision. Object 1 (mass m) moves to the right at speed v₀ and Object 2 (mass 3m) is at rest. (a) Derive expressions for the final velocities of both objects if the collision is perfectly elastic. (b) Show that kinetic energy is conserved using your expressions. (c) Determine what fraction of the initial kinetic energy is transferred to Object 2. (d) Explain physically why the lighter object reverses direction.

Show answer

Solution: (1 pt) Momentum: mv₀ = mv₁f + 3mv₂f, simplifying to v₀ = v₁f + 3v₂f. (1 pt) Elastic relative velocity: v₀ = v₂f − v₁f. (1 pt) Adding: 2v₀ = 4v₂f → v₂f = v₀/2. Subtracting: v₁f = v₂f − v₀ = −v₀/2. (1 pt) KE check: KEf = ½m(v₀/2)² + ½(3m)(v₀/2)² = mv₀²/8 + 3mv₀²/8 = ½mv₀² = KEᵢ. ✓ (1 pt) Fraction to Object 2: (3mv₀²/8)/(mv₀²/2) = 3/4 = 75%. (1 pt) Object 1 reverses because during contact, the normal force from the heavier object acts as a large impulse pointing leftward on Object 1. Since m < 3m, Object 1's velocity swings from +v₀ to −v₀/2 while Object 2 moves forward at only v₀/2 due to its greater inertia.