STATICS AND DYNAMICS • DYNAMICS

Elastic vs. Inelastic Collisions — Distinguish elastic vs inelastic collisions and compute post-impact velocities (intro)

Understand how momentum and energy conservation govern collision outcomes in engineering analysis.

Historical Context & Motivation

The study of collisions is among the oldest problems in classical mechanics, stretching back to the earliest attempts to formalize the laws of motion. Before Newton published his Principia, natural philosophers debated whether a quantity proportional to mass times velocity or mass times the square of velocity was conserved during impact. This debate, which pitted the Cartesian concept of momentum against Leibniz's vis viva (living force, proportional to mv²), ultimately revealed that both quantities play distinct but complementary roles in collision dynamics. Understanding when each is conserved forms the foundation for distinguishing elastic from inelastic collisions—a classification essential to modern engineering design, from crash safety systems to particle accelerator experiments.

1668
Wallis, Wren, and Huygens Submit to the Royal Society
John Wallis presented rules for perfectly inelastic collisions, while Christopher Wren and Christiaan Huygens independently formulated the laws of perfectly elastic collisions, demonstrating that both momentum and vis viva are conserved when bodies rebound without deformation.
1687
Newton's Principia and the Coefficient of Restitution
Isaac Newton formalized the three laws of motion and introduced an experimental law of restitution, defining the ratio of relative separation velocity to relative approach velocity. This coefficient of restitution (e) provided a quantitative bridge between perfectly elastic (e = 1) and perfectly inelastic (e = 0) impacts.
1743
d'Alembert's Traité de Dynamique
Jean le Rond d'Alembert synthesized the vis viva controversy by showing that both momentum and kinetic energy could be derived from the same mechanical principles, clarifying the conditions under which each quantity is or is not conserved.
1960s–Present
Computational Impact Mechanics
With the advent of finite element analysis, engineers began modeling collisions in full detail—accounting for material plasticity, fracture, and thermal dissipation. These computational tools rely on the same fundamental conservation laws, extended to deformable bodies and complex contact conditions.

The central question that collision theory addresses is straightforward yet profound: given two bodies with known masses and pre-impact velocities, what are their velocities after the collision, and how much kinetic energy is lost to deformation, heat, or sound? Answering this question requires two conservation principles—conservation of linear momentum (always valid for an isolated system) and conservation of kinetic energy (valid only for elastic collisions)—together with the coefficient of restitution that characterizes real-world impacts.

Core Principles & Definitions

Collision analysis in dynamics rests on a few foundational principles. In any collision between two bodies within an isolated system—one where no net external impulse acts during the brief collision interval—the total linear momentum is conserved. Whether kinetic energy is also conserved determines the collision classification. These principles apply to both particle models and rigid-body impacts, though the latter introduce additional considerations such as angular momentum and contact geometry that are treated in more advanced courses.

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Conservation of Linear Momentum

For an isolated system of two colliding particles, the vector sum of momenta before impact equals the vector sum after impact: m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′. This holds regardless of collision type and provides one scalar equation per spatial dimension.
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Elastic Collision

A collision in which the total kinetic energy of the system is conserved. No energy is lost to permanent deformation, heat, or sound. Perfectly elastic collisions are idealizations; atomic-scale collisions (e.g., noble gas molecules) come closest.
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Inelastic Collision

A collision in which some kinetic energy is converted to other forms—internal energy, plastic deformation, sound, or heat. Most real-world engineering collisions are inelastic. A perfectly inelastic collision is the extreme case where the two bodies coalesce and move as a single mass after impact.
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Coefficient of Restitution (e)

Defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact: e = (v₂′ − v₁′)/(v₁ − v₂). For perfectly elastic collisions e = 1; for perfectly inelastic collisions e = 0; real impacts fall between 0 and 1.
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Impulse–Momentum Theorem

The impulse (∫F dt) exerted on each body during the collision equals its change in momentum. Because the collision forces are internal and equal-and-opposite (Newton's Third Law), the net impulse on the system is zero, guaranteeing momentum conservation.
KEY TAKEAWAY
Think of a collision as a transaction at a bank. Momentum is the total balance across all accounts—it is always conserved no matter how the money is shuffled. Kinetic energy is like cash on hand: in an elastic collision, all cash remains liquid; in an inelastic collision, some cash is permanently converted into non-recoverable assets (deformation, heat). The coefficient of restitution tells you the fraction of 'liquidity' retained—it is the efficiency rating of the collision.

Visual Explanation — Collision Types Compared

Three collision scenarios with mass m₁ (cyan) striking mass m₂ (violet) initially at rest. Top: perfectly elastic—both bodies separate with total KE conserved. Middle: general inelastic—bodies separate but some KE is lost. Bottom: perfectly inelastic—bodies coalesce into a combined mass, yielding maximum kinetic energy loss.

The diagram above illustrates the three canonical collision scenarios for a one-dimensional, central impact. In each case, the total momentum of the system is identical before and after the collision—the arrows' momentum contributions (mass × velocity) sum to the same value on both sides of the dashed dividing line. What distinguishes the three types is the kinetic energy balance. In the perfectly elastic case (top row), both bodies exchange velocity according to their mass ratio, and all kinetic energy is retained as translational kinetic energy. In the perfectly inelastic case (bottom row), the two bodies lock together and share a common final velocity, and the kinetic energy loss is maximized. The general inelastic case (middle row) lies between these extremes, characterized by a coefficient of restitution 0 < e < 1.

Mathematical Framework

For a one-dimensional collision between two particles, we have at most two unknowns—the post-impact velocities v₁′ and v₂′. We therefore need two independent equations. The first is always momentum conservation. The second depends on the collision type: kinetic energy conservation (elastic) or the restitution equation (general case).

CONSERVATION OF LINEAR MOMENTUM
m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′
m₁, m₂ = particle masses; v₁, v₂ = pre-impact velocities; v₁′, v₂′ = post-impact velocities. This equation holds for all collision types in an isolated system.
CONSERVATION OF KINETIC ENERGY (ELASTIC ONLY)
½m₁v₁² + ½m₂v₂² = ½m₁v₁′² + ½m₂v₂′²
Applicable only to perfectly elastic collisions (e = 1). Combined with momentum conservation, this yields a system of two equations in two unknowns.
COEFFICIENT OF RESTITUTION
e = (v₂′ − v₁′) / (v₁ − v₂)
e = coefficient of restitution (0 ≤ e ≤ 1). The numerator is the relative separation speed; the denominator is the relative approach speed. For elastic: e = 1; for perfectly inelastic: e = 0.

Derived Formulas for Elastic Collisions

By simultaneously solving the momentum and kinetic energy equations (or equivalently, momentum and the restitution equation with e = 1), one obtains closed-form expressions for the post-impact velocities. The key algebraic insight is to factor the energy equation as the difference of squares, which yields a second linear equation that is far easier to solve simultaneously with momentum conservation.

ELASTIC POST-IMPACT VELOCITY OF PARTICLE 1
v₁′ = ((m₁ − m₂) / (m₁ + m₂)) v₁ + (2m₂ / (m₁ + m₂)) v₂
When m₁ = m₂, this simplifies to v₁′ = v₂ — the particles exchange velocities, a result familiar from billiard-ball dynamics.
ELASTIC POST-IMPACT VELOCITY OF PARTICLE 2
v₂′ = (2m₁ / (m₁ + m₂)) v₁ + ((m₂ − m₁) / (m₁ + m₂)) v₂
Symmetric counterpart to the expression for v₁′. Note that the sum m₁v₁′ + m₂v₂′ reduces identically to m₁v₁ + m₂v₂, confirming momentum conservation.

Perfectly Inelastic Collision

COMMON FINAL VELOCITY (PERFECTLY INELASTIC)
v′ = (m₁v₁ + m₂v₂) / (m₁ + m₂)
With e = 0, the two bodies move as one after impact. This single equation from momentum conservation suffices because there is only one unknown. The kinetic energy loss is ΔKE = ½ × (m₁m₂ / (m₁ + m₂)) × (v₁ − v₂)².

Classification & Energy Analysis

One of the most powerful ways to characterize a collision is through the fractional kinetic energy loss, which quantifies how much of the initial kinetic energy is dissipated. For a head-on collision where the target is initially at rest, the fractional loss depends only on the mass ratio and the coefficient of restitution. The spectrum below and the accompanying table illustrate how the collision type maps onto the energy retention.

Collision Spectrum by Coefficient of Restitution
Perfectly Inelastic
Inelastic
Nearly Elastic
Perfectly Elastic
e = 0
e = 0.5
e = 1
Max KE lossNo KE loss
Fractional kinetic energy retained as a function of the coefficient of restitution e for three different mass ratios (target initially at rest). When m₁ = m₂ (cyan), a perfectly inelastic collision (e = 0) retains 0% of the initial KE, while a perfectly elastic collision (e = 1) transfers all KE to the target. As the mass ratio increases, the heavy striker retains more energy regardless of e, and the sensitivity to e diminishes.
Summary of collision classifications and their key properties.
PropertyPerfectly Elastic (e = 1)Inelastic (0 < e < 1)Perfectly Inelastic (e = 0)
Momentum conserved?YesYesYes
KE conserved?YesNo — partial lossNo — maximum loss
Post-impact bodiesSeparateSeparateCoalesce
Equations neededMomentum + Energy (or e = 1)Momentum + RestitutionMomentum only
Real-world exampleAtomic/molecular collisions, Newton's cradle (approx.)Sports ball impacts, car bumpers with crumple zonesBallistic pendulum, clay ball sticking to surface

Worked Example — Two-Car Rear-End Collision

Consider a rear-end collision between two vehicles on a level, frictionless surface during the very short collision interval. Car A (m₁ = 1500 kg) travels at v₁ = 20 m/s and strikes Car B (m₂ = 1000 kg), which is moving in the same direction at v₂ = 5 m/s. The coefficient of restitution is measured as e = 0.3. Determine the post-impact velocities of both vehicles and the kinetic energy lost during the collision.

Rear-End Collision with e = 0.3
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Step 1 — Identify Given Values and Unknownsm₁ = 1500 kg, v₁ = 20 m/s, m₂ = 1000 kg, v₂ = 5 m/s, e = 0.3. Unknowns: v₁′ and v₂′. We need two equations: momentum conservation and the restitution equation.
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Step 2 — Apply Conservation of Linear Momentumm₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′ → (1500)(20) + (1000)(5) = 1500 v₁′ + 1000 v₂′ → 30000 + 5000 = 1500 v₁′ + 1000 v₂′.
1500 v₁′ + 1000 v₂′ = 35000 …(1)
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Step 3 — Apply the Restitution Equatione = (v₂′ − v₁′) / (v₁ − v₂) → 0.3 = (v₂′ − v₁′) / (20 − 5) → v₂′ − v₁′ = 0.3 × 15 = 4.5 m/s.
v₂′ − v₁′ = 4.5 …(2)
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Step 4 — Solve the System of EquationsFrom (2): v₂′ = v₁′ + 4.5. Substitute into (1): 1500 v₁′ + 1000(v₁′ + 4.5) = 35000 → 2500 v₁′ + 4500 = 35000 → 2500 v₁′ = 30500 → v₁′ = 12.2 m/s. Then v₂′ = 12.2 + 4.5 = 16.7 m/s.
v₁′ = 12.2 m/s, v₂′ = 16.7 m/s
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Step 5 — Verify Momentum ConservationCheck: 1500(12.2) + 1000(16.7) = 18300 + 16700 = 35000 kg·m/s. This matches the initial momentum of 35000 kg·m/s. ✓
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Step 6 — Compute Kinetic Energy LossKE₀ = ½(1500)(20²) + ½(1000)(5²) = 300000 + 12500 = 312500 J. KE_f = ½(1500)(12.2²) + ½(1000)(16.7²) = ½(1500)(148.84) + ½(1000)(278.89) = 111630 + 139445 = 251075 J. ΔKE = 312500 − 251075 = 61425 J.
ΔKE ≈ 61.4 kJ lost (≈ 19.7% of initial KE)
🔍 Sanity Check
The post-impact velocity of Car A (12.2 m/s) is less than its pre-impact velocity (20 m/s), while Car B's post-impact velocity (16.7 m/s) exceeds its initial 5 m/s—both consistent with physical intuition. The energy loss of ~19.7% is reasonable for e = 0.3, lying between the perfectly elastic limit (0% loss) and the perfectly inelastic limit (maximum loss).

Strengths & Limitations of the Particle Collision Model

The one-dimensional particle collision model taught in introductory dynamics is remarkably powerful, but it is also built upon simplifying assumptions. Understanding where these assumptions hold and where they break down is essential for engineering judgment. The table below summarizes the strengths and limitations of the approach presented in this lesson.

Strengths and limitations of the 1-D particle collision model.
StrengthsLimitations
Closed-form solutions for post-impact velocities enable rapid preliminary design calculations (e.g., bumper design, impact testing).Assumes the collision is instantaneous and that external forces (gravity, friction) are negligible during impact—not always valid for prolonged contact.
Conservation of momentum is exact for isolated systems, providing a robust constraint regardless of internal complexity.The coefficient of restitution is treated as a constant, but in reality e depends on impact speed, temperature, and material condition.
The coefficient of restitution provides a single parameter to characterize collision inelasticity, simplifying experimental correlation.The model treats bodies as particles—rotational effects, oblique impacts, and deformable-body mechanics are not captured.
Directly applicable to ballistic pendulum experiments, Newton's cradle analysis, and vehicle crash reconstruction.Energy dissipation mechanisms (plastic work, sound, fragmentation) are lumped into a single parameter; detailed energy budgets require FEA or experimental instrumentation.
⚙️ ENGINEERING PERSPECTIVE
In practice, the particle collision model serves as the first-pass sizing tool for impact problems—analogous to how beam theory gives preliminary stresses before a full finite element analysis. A crash engineer, for example, might use these equations with an experimentally determined e to estimate peak deceleration and energy absorption before committing to a detailed computational simulation. Knowing the model's boundaries prevents misapplication and guides the decision of when to invest in higher-fidelity analysis.

Connection to Advanced Impact Theory

The introductory framework presented here forms the foundation for several advanced topics that you will encounter in later courses on dynamics, vibrations, and computational mechanics. The table below maps the concepts from this lesson to their more sophisticated counterparts, giving you a roadmap for deeper study.

Mapping introductory collision concepts to advanced impact mechanics.
This Lesson (Intro)Advanced Extension
1-D central impact between particlesOblique impact in 2-D/3-D with tangential friction (Coulomb friction at contact)
Constant coefficient of restitution eVelocity-dependent e, energetic COR, and Stronge's hypothesis for energy-consistent restitution
Particle (point-mass) modelRigid-body impact with angular momentum and eccentric impact points (Whittaker's theory)
Instantaneous collision assumptionContact-force models (Hertzian contact, Hunt–Crossley, nonlinear spring-dashpot) resolving force vs. time during impact
ΔKE as a scalar lossExplicit energy dissipation via elastoplastic material models, fracture mechanics, and thermal effects in FEA simulations

As you progress through your engineering curriculum—particularly in courses on vibrations, mechanical design, and computational mechanics—you will revisit impact problems with increasingly sophisticated tools. The transition from a scalar e to a contact-force-time history, for instance, mirrors the broader engineering theme of moving from lumped-parameter models to distributed-parameter (continuum) models as design fidelity requirements increase.

Practice Problems

PROBLEM 1CONCEPTUAL
A 2-kg ball traveling at 5 m/s collides head-on with a 2-kg ball at rest. After the collision, the first ball is stationary and the second ball moves at 5 m/s. Is this collision elastic, inelastic, or perfectly inelastic? Justify your answer by considering both momentum and kinetic energy.
PROBLEM 2BASIC CALCULATION
A 4-kg cart moving at 6 m/s to the right collides perfectly inelastically with a 2-kg cart initially at rest. Compute the common post-impact velocity and the kinetic energy lost in the collision.
PROBLEM 3INTERMEDIATE
A 1200-kg vehicle traveling at 15 m/s strikes the rear of an 800-kg vehicle moving at 8 m/s in the same direction. After the collision, the lighter vehicle moves at 17 m/s. (a) Find the post-impact velocity of the heavier vehicle. (b) Determine the coefficient of restitution. (c) Calculate the percentage of kinetic energy lost.
PROBLEM 4APPLIED
A ballistic pendulum consists of a 10-g bullet fired horizontally into a 2.5-kg wooden block suspended by cords. After the bullet embeds in the block, the block-bullet system swings upward to a maximum height of 0.12 m. Determine the bullet's impact speed. (Use g = 9.81 m/s².)
PROBLEM 5CRITICAL THINKING
Two particles of masses m and 3m undergo a head-on collision. Particle 1 (mass m) moves to the right with speed v₀ and particle 2 (mass 3m) is at rest. (a) Derive general expressions for v₁′ and v₂′ in terms of v₀ and the coefficient of restitution e. (b) Show that the fractional kinetic energy lost is ΔKE/KE₀ = ¾(1 − e²). (c) For what value of e is the kinetic energy loss exactly 50%? Discuss whether this is achievable with common engineering materials.

Summary

Collisions between two bodies are classified by the coefficient of restitution e, which measures the ratio of relative separation speed to relative approach speed. Perfectly elastic collisions (e = 1) conserve both linear momentum and kinetic energy, yielding two independent equations that fully determine the two post-impact velocities. Perfectly inelastic collisions (e = 0) result in the bodies coalescing, maximizing kinetic energy loss while still conserving momentum. General inelastic collisions (0 < e < 1) are solved by combining momentum conservation with the restitution equation.

The key equations are: m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′ (momentum) and e = (v₂′ − v₁′)/(v₁ − v₂) (restitution). For elastic collisions, closed-form velocity expressions exist in terms of mass ratios. The fractional kinetic energy loss depends on both the mass ratio and e², providing a direct metric for collision severity. This particle-level framework serves as the foundation for more advanced impact models involving oblique contact, rigid-body rotation, and deformable-body mechanics.

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