Physics: Foundations of Physical Science

MGU Physics · Preparatory Volume

Part III — Energy and Momentum · Chapter 17

Collisions

17.1 What Counts as a Collision in Physics

A collision in physics is any brief, strong interaction between two or more objects in which the interaction forces are much larger than any external forces acting during the interaction. The short duration and large internal forces mean we can treat the system as approximately isolated during the collision itself, even if external forces such as gravity or friction act over longer timescales.

Collisions include a bat hitting a ball, two cars crashing, and a proton hitting a nucleus. They also include soft collisions such as two magnets pushing each other apart without touching, or a ball bouncing off a wall at slow speed, as long as the interaction is brief and the internal forces dominate.

What makes collision analysis tractable is that we typically know the states before and after the collision but not the details of the force during it. Conservation laws, especially momentum and, for elastic collisions, kinetic energy, connect before and after without requiring those details.

17.2 Isolated Systems in Collision Analysis

As established in Chapter 16, conservation of momentum requires that the net external force on the system is zero. In a collision, internal forces between the colliding objects are enormous but are Newton’s third law pairs. They cancel in the system sum. External forces, such as gravity or friction from the floor, are typically much smaller than the collision forces and act for too short a time to produce significant impulse.

This is why we can treat a collision as occurring in an isolated system even when external forces exist: the external impulse Fext Δt is negligible compared to the internal impulse during the brief collision time Δt. After the collision, when the objects move apart over longer times, external forces again matter and must be included.

17.3 Conservation of Momentum in Collisions

For any collision between two objects in an isolated system:

m1v1i+m2v2i=m1v1f+m2v2f

This single equation in one dimension has two unknowns if both final velocities are unknown. A second equation is needed: either the collision type provides it, as in elastic collisions or perfectly inelastic collisions, or one final velocity is measured.

Table 17.1 — Equations Available for Different Collision Types
Collision TypeEquations AvailableUnknowns Solvable
Perfectly inelasticConservation of momentum; v1f = v2fBoth final velocities, because both are the same
Elastic, 1DConservation of momentum; conservation of kinetic energyBoth final velocities
General inelasticConservation of momentum onlyOnly one final velocity if the other is measured
2D collision, any typeMomentum conserved in x and y separatelyMore complex; see Section 17.10

17.4 Elastic Collisions

A perfectly elastic collision is one in which both total momentum and total kinetic energy are conserved. No kinetic energy is converted to heat, sound, or deformation; the collision is completely bouncy.

For a one-dimensional elastic collision between m1, with initial velocity v1, and m2, initially at rest:

v1f = [(m1 − m2) / (m1 + m2)] × v1

v2f = [2m1 / (m1 + m2)] × v1

Table 17.2 — Elastic Collision Results for Common Mass Relationships
Mass Relationshipv1fv2fPhysical Meaning
m1 = m20v1, full transferBall 1 stops; ball 2 moves at the original speed of ball 1.
m1 ≫ m2Approximately v1, barely slowsApproximately 2v1 forwardHeavy object barely changes; light object flies forward at about twice the speed.
m1 ≪ m2Approximately −v1, bounces backApproximately 0, barely movesLight object bounces back; heavy object barely moves.

Worked Example 17.1 — Elastic Collision, Equal Masses

A 0.50 kg billiard ball moving at +3.0 m/s strikes an identical stationary ball. Find both final velocities.

  1. Equal masses: v1f = (0.50 − 0.50) / (0.50 + 0.50) × 3.0 = 0.
  2. v2f = 2(0.50) / (0.50 + 0.50) × 3.0 = 1.0 × 3.0 = +3.0 m/s.
  3. Result: Ball 1 stops; ball 2 moves at 3.0 m/s. This is the classic equal-mass elastic result.
  4. Momentum check: 0.50 × 3.0 = 0.50 × 0 + 0.50 × 3.0 = 1.5 kg·m/s.
  5. Kinetic energy check: KEi = ½(0.50)(3.0)2 = 2.25 J; KEf = 0 + ½(0.50)(3.0)2 = 2.25 J.

Worked Example 17.2 — Elastic Collision, Unequal Masses

A 3.0 kg ball at +4.0 m/s collides elastically with a 1.0 kg ball at rest.

  1. v1f = (3.0 − 1.0) / (3.0 + 1.0) × 4.0 = (2.0 / 4.0) × 4.0 = +2.0 m/s.
  2. v2f = 2(3.0) / (3.0 + 1.0) × 4.0 = (6.0 / 4.0) × 4.0 = +6.0 m/s.
  3. Momentum check: 3.0(4.0) = 3.0(2.0) + 1.0(6.0) = 6.0 + 6.0 = 12.0.
  4. Kinetic energy check: KEi = ½(3.0)(16) = 24 J; KEf = ½(3.0)(4) + ½(1.0)(36) = 6 + 18 = 24 J.

17.5 Inelastic Collisions

An inelastic collision is one in which momentum is conserved but kinetic energy is not. Some kinetic energy is converted to thermal energy, sound, or deformation during the impact. Most real-world collisions are inelastic to some degree.

The degree of inelasticity is characterized by the coefficient of restitution e:

e = (v2f − v1f) / (v1i − v2i) = relative speed after / relative speed before

For a general inelastic one-dimensional collision, momentum conservation alone gives one equation with two unknowns. A second equation, such as measured e or the final velocity of one object, is required to solve for both final velocities.

17.6 Perfectly Inelastic Collisions

A perfectly inelastic collision is one in which the colliding objects stick together and move with a common final velocity. This represents the maximum possible loss of kinetic energy consistent with conservation of momentum.

m1v1 + m2v2 = (m1 + m2)vf

vf = (m1v1 + m2v2) / (m1 + m2)

Worked Example 17.3 — Perfectly Inelastic Collision

A 1500 kg car moving at +20.0 m/s strikes a 3000 kg truck at rest. They crumple together. Find vf and the energy lost.

  1. Momentum conservation: vf = 1500(20.0) / (1500 + 3000) = 30,000 / 4500 = +6.67 m/s.
  2. Initial kinetic energy: KEi = ½(1500)(20.0)2 = 300,000 J.
  3. Final kinetic energy: KEf = ½(4500)(6.67)2 = ½(4500)(44.5) = 100,125 J.
  4. Energy lost: KEi − KEf = 300,000 − 100,125 = 199,875 J, approximately 200 kJ, converted to deformation, heat, and sound.

17.7 Kinetic Energy in Collisions

The kinetic energy analysis of collisions reveals a key insight: kinetic energy is never gained in a collision in the absence of an internal energy source. It can only be conserved in elastic collisions or lost in inelastic collisions. This is unlike momentum, which is always conserved.

ΔKE = KEf − KEi ≤ 0 for collisions without an internal energy source.

Table 17.3 — Momentum, Kinetic Energy, and Coefficient of Restitution
Collision TypeMomentumKinetic Energye Value
Perfectly elasticConservedConserved1
Partially inelasticConservedDecreases0 < e < 1
Perfectly inelasticConservedMaximum decrease0
Explosion or separationConservedIncreases because internal energy is releasede > 1, not a collision in the usual sense

Where does the kinetic energy go in an inelastic collision? It converts to thermal energy, sound, and deformation energy. Total energy is conserved; only kinetic energy decreases.

17.8 Why Momentum Is Conserved Even When Energy Is “Lost”

Students sometimes wonder: if energy can be “lost” in a collision, why can’t momentum also be lost? The answer lies in the nature of Newton’s third law and the structure of the interaction.

During a collision, the forces between the objects are internal to the system and come in Newton’s third law pairs: F12 = −F21. When summed over the system, internal forces cancel exactly. The total impulse of all internal forces on the system is zero. Therefore, the total momentum change of the system is zero. Total momentum is conserved regardless of what happens internally.

Kinetic energy, by contrast, is a scalar and does not cancel by Newton’s third law. The kinetic energy lost by one object is not necessarily gained by another. It converts to internal energy, such as deformation and heat, of the objects themselves. Momentum is transferred; kinetic energy is not necessarily conserved.

Standard result of classical mechanics.

17.9 Explosion and Fragmentation Events

An explosion or fragmentation event is the reverse scenario from a perfectly inelastic collision: a single object or system separates into multiple pieces, with internal potential energy converting to kinetic energy. Momentum is conserved; kinetic energy increases.

If the system starts from rest, total initial momentum is zero. After the explosion, the momenta of all fragments must vector-sum to zero.

Worked Example 17.4 — Fragmentation into Three Pieces

A 6.0 kg object at rest explodes into three equal 2.0 kg pieces. Two pieces fly off at right angles: piece 1 at +5.0 m/s along x; piece 2 at +4.0 m/s along y. Find the velocity of piece 3.

  1. x-momentum: 0 = 2.0(5.0) + 2.0(0) + 2.0v3x, so v3x = −5.0 m/s.
  2. y-momentum: 0 = 2.0(0) + 2.0(4.0) + 2.0v3y, so v3y = −4.0 m/s.
  3. Speed of piece 3: v3 = √[(-5.0)2 + (-4.0)2] = √(25 + 16) = √41 = 6.40 m/s.
  4. Direction: arctan(4.0 / 5.0) = 38.7° south of west.

17.10 Two-Dimensional Collision Concepts

In two-dimensional collisions, momentum is conserved separately in the x and y directions. This gives two scalar equations, which can solve for two unknowns, typically two velocity components or two angles.

x: m1v1ix + m2v2ix = m1v1fx + m2v2fx

y: m1v1iy + m2v2iy = m1v1fy + m2v2fy

For two-dimensional elastic collisions, kinetic energy conservation provides a third equation, allowing solution for three unknowns. For two-dimensional inelastic collisions, the angle of deflection must be measured experimentally to fully specify the final state.

Worked Example 17.5 — 2D Perfectly Inelastic Collision

A 2.0 kg ball moving at 4.0 m/s east collides with a 3.0 kg ball moving at 3.0 m/s north. They stick together. Find the final velocity.

  1. x-momentum: px = 2.0(4.0) + 3.0(0) = 8.0 kg·m/s.
  2. y-momentum: py = 2.0(0) + 3.0(3.0) = 9.0 kg·m/s.
  3. Final mass: mf = 5.0 kg.
  4. Final velocity components: vfx = 8.0 / 5.0 = 1.6 m/s east; vfy = 9.0 / 5.0 = 1.8 m/s north.
  5. Magnitude: vf = √(1.62 + 1.82) = √(2.56 + 3.24) = √5.80 = 2.41 m/s.
  6. Direction: arctan(1.8 / 1.6) = 48.4° north of east.

17.11 Collision Analysis Strategy

A systematic approach prevents errors in all collision problems.

  1. Classify the collision: elastic, inelastic, or perfectly inelastic.
  2. Define a positive direction; assign signed velocities to all objects before and after.
  3. Apply conservation of momentum. This is always valid for isolated collisions.
  4. Apply kinetic energy conservation only if the collision is specified as elastic.
  5. For perfectly inelastic collisions, set v1f = v2f = vf.
  6. Solve the resulting equations algebraically; then substitute numbers.
  7. Check momentum balance. If elastic, check kinetic energy balance. Verify that signs make sense.

17.12 Common Mistakes in Collision Problems

Mistake 1: Confusing Elastic and Inelastic Collisions

Elastic means both momentum and kinetic energy are conserved. Inelastic means momentum is conserved but kinetic energy decreases. If a problem says the objects bounce off each other without saying elastically, you cannot assume it is elastic. Bouncing is not sufficient.

Mistake 2: Using Kinetic Energy Conservation for Inelastic Collisions

Never apply kinetic energy conservation to an inelastic collision. It gives wrong answers. Momentum conservation is the only universal tool; kinetic energy conservation is a special property of elastic collisions only.

Mistake 3: Forgetting Vector Signs in 1D

Objects moving in opposite directions have momenta of opposite signs. Assign signs before calculating, carry them through algebra, and check that the final momentum matches the initial momentum algebraically, not just in magnitude.

Mistake 4: Using the Wrong Mass in a Perfectly Inelastic Collision

After a perfectly inelastic collision, the combined object has mass m1 + m2, not m1 or m2 alone. Write the final momentum as (m1 + m2)vf.

Mistake 5: Applying the Elastic Collision Formula When the Second Object Is Not at Rest

The elastic collision formulas for v1f and v2f assume m2 is initially at rest. If m2 is moving, derive the result from scratch by applying both momentum and kinetic energy conservation simultaneously.

Mistake 6: Ignoring Direction in 2D Collisions

In two dimensions, momentum is conserved in each dimension independently. Do not add the magnitudes of momenta in different directions. Write separate equations for x and y, solve independently, and then reconstruct the magnitude and direction of the final velocity.

Chapter Summary

Collisions are brief, strong interactions treated as isolated events. Momentum is conserved in all isolated collisions; kinetic energy is conserved only in elastic collisions.

  • Elastic collision: both momentum and kinetic energy are conserved; e = 1.
  • Inelastic collision: momentum is conserved; kinetic energy decreases; 0 < e < 1.
  • Perfectly inelastic collision: momentum is conserved; objects stick together; maximum kinetic energy loss; e = 0.
  • Equal-mass elastic collisions exchange velocities.
  • When a heavy object hits a light object elastically, the light object can move at approximately 2v.
  • When a light object hits a heavy object elastically, the light object bounces back at approximately −v.
  • Kinetic energy lost in inelastic collisions converts to thermal, sound, and deformation energy. It is not destroyed.
  • Momentum is conserved because internal Newton’s third-law force pairs cancel. Kinetic energy is not always conserved because it is scalar and has no cancellation mechanism.
  • Explosions conserve momentum while kinetic energy increases from internal energy.
  • Two-dimensional collisions require momentum conservation in x and y independently.
  • A reliable strategy is: classify type, assign signs, apply momentum, apply kinetic energy only for elastic collisions, solve, and check.

Key Terms

collision
A brief, strong interaction between objects in which internal forces dominate over external forces during the interaction time.
elastic collision
A collision in which both total momentum and total kinetic energy are conserved; objects bounce without energy loss; coefficient of restitution e = 1.
inelastic collision
A collision in which momentum is conserved but kinetic energy decreases, converting to other forms; 0 < e < 1.
perfectly inelastic collision
An inelastic collision in which the objects stick together and share a common final velocity; maximum kinetic energy loss consistent with momentum conservation; e = 0.
coefficient of restitution (e)
The ratio of relative separation speed to relative approach speed: e = |v2f − v1f| / |v1i − v2i|; ranges from 0 to 1 for ordinary collisions.
fragmentation
The separation of an object into multiple pieces by an internal explosion or release of stored energy; momentum is conserved while kinetic energy increases.

Concept Review Questions

  1. What is the difference between an elastic and an inelastic collision? Is momentum conserved in each? Is kinetic energy?
  2. Two billiard balls of equal mass collide head-on elastically. One is moving at 5.0 m/s and the other is at rest. What are the velocities after the collision?
  3. A perfectly inelastic collision produces the maximum loss of kinetic energy. Why does it not violate conservation of momentum?
  4. Where does the lost kinetic energy go in an inelastic collision?
  5. Explain why momentum is always conserved in a collision even when kinetic energy is not.
  6. A very heavy ball collides elastically with a very light ball at rest. Qualitatively describe the final motion of each ball.
  7. What is the coefficient of restitution? What values does it take for elastic, inelastic, and perfectly inelastic collisions?
  8. In a two-dimensional collision, how do you apply conservation of momentum? Write the two equations.

Practice Problems

Elastic Collision Problems

  1. A 2.0 kg ball at +5.0 m/s collides elastically with a 2.0 kg ball at rest. Find both final velocities.
  2. A 4.0 kg ball at +6.0 m/s collides elastically with a 2.0 kg ball at rest. Find both final velocities.
  3. A 1.0 kg ball at +8.0 m/s collides elastically with a 3.0 kg ball at rest. Find both final velocities and verify kinetic energy conservation.
  4. A 5.0 kg ball at +3.0 m/s collides elastically with a 5.0 kg ball at −1.0 m/s. Find both final velocities. Use momentum and kinetic energy conservation simultaneously, not the rest-frame formulas.
  5. Ball A, with mass 0.30 kg and velocity +4.0 m/s, collides elastically with Ball B, with mass 0.50 kg and velocity 0. Find both final velocities, kinetic energy before, kinetic energy after, and verify that the collision is elastic.

Inelastic Collision Problems

  1. A 3.0 kg lump of clay moving at +4.0 m/s strikes a stationary 2.0 kg lump and sticks. Find final velocity, kinetic energy before, kinetic energy after, and kinetic energy lost.
  2. A 1200 kg car at +15.0 m/s collides with a 2000 kg truck at rest and they lock together. Find final velocity, kinetic energy lost, and percent of kinetic energy lost.
  3. A 0.050 kg bullet traveling at 300 m/s embeds in a 2.0 kg stationary block. Find the final velocity of the block and bullet, kinetic energy before and after, and percent of kinetic energy lost.
  4. A 60.0 kg person running at 4.0 m/s jumps onto a stationary 20.0 kg skateboard. Find the speed of the person and board and the kinetic energy lost.
  5. Two balls collide: m1 = 0.40 kg at +5.0 m/s; m2 = 0.60 kg at −2.0 m/s. After the collision, m1 has velocity +1.0 m/s. Find the velocity of m2 and check momentum conservation.

Momentum Conservation Practice

  1. Verify conservation of momentum in each case: (a) 3.0 kg at +4.0 m/s strikes 2.0 kg at rest; result: 3.0 kg at +1.6 m/s, 2.0 kg at +3.6 m/s. (b) 1.0 kg at +5.0 m/s and 2.0 kg at −1.0 m/s stick together; result: 3.0 kg at +1.0 m/s.
  2. A 5.0 kg cart at +3.0 m/s collides with a 3.0 kg cart at +1.0 m/s in the same direction. After collision, the 5.0 kg cart moves at +2.25 m/s. Find the velocity of the 3.0 kg cart.
  3. A 0.010 kg bullet at +500 m/s passes through a 2.0 kg stationary block and exits at +200 m/s. Find the velocity of the block after.
  4. A 4.0 kg ball moving east at 6.0 m/s and a 3.0 kg ball moving west at 4.0 m/s collide and stick. Find the final velocity, including magnitude and direction.
  5. An ice skater, 60.0 kg, at rest pushes off a 5.0 kg box and sends it at 3.0 m/s to the east. Find the skater’s recoil velocity.

Mixed Collision Analysis

  1. A ballistic pendulum: a 0.010 kg bullet traveling at 400 m/s embeds in a 1.0 kg pendulum bob and swings it upward. Find the speed of the bob and bullet immediately after impact. Then, using energy conservation, find the height the pendulum rises.
  2. Two identical carts, 1.0 kg each, are on a frictionless track. Cart A moves at +2.0 m/s; Cart B is at rest. They collide. After the collision, Cart A moves at +0.50 m/s. Determine whether this collision is elastic or inelastic, find Cart B’s final velocity, and find the kinetic energy lost.
  3. A 2.0 kg object moving at +3.0 m/s north collides and sticks with a 3.0 kg object moving at +4.0 m/s east. Find the magnitude and direction of the final velocity.
  4. A 1.0 kg ball with velocity +4.0 m/s collides elastically with a 2.0 kg ball moving at −1.0 m/s. Find both final velocities, verify momentum, and verify kinetic energy. Use simultaneous equations.

Chapter 17 Checkpoint

Chapter 17 Checkpoint
I can…Section
Classify a collision as elastic, inelastic, or perfectly inelastic.17.4–17.6
Apply conservation of momentum to any one-dimensional collision.17.3
Apply the elastic collision formulas for m2 at rest.17.4
Solve perfectly inelastic collisions using (m1 + m2)vf.17.6
Calculate kinetic energy before and after to find kinetic energy lost.17.7
Explain why momentum is conserved even when kinetic energy is not.17.8
Solve explosion and fragmentation problems using ptotal = 0.17.9
Apply x and y momentum equations for two-dimensional collisions.17.10
Check momentum conservation after every collision calculation.17.11
Avoid the six most common collision analysis mistakes.17.12