In Activity 5-1-3 you learned that momentum is conserved whenever two objects interact in an isolated system. That rule holds for every collision. What changes from one collision to the next is the kinetic energy, and that is how collisions get sorted:
Type
Momentum
Kinetic energy
Do they stick?
e
Examples
Elastic
Conserved
Conserved
No, they bounce apart
1
Magnets repelling, billiard balls, atoms in a gas
Inelastic
Conserved
Some is lost
No, but they don’t bounce fully
between 0 and 1
A tennis ball, most real bounces, a car bumper tap
Perfectly inelastic
Conserved
The most is lost
Yes, they move together
0
Clay, Velcro, train couplings, most car crashes
💡Truly elastic collisions only happen between atoms and molecules. Everyday “bounces” always lose a little kinetic energy to sound and heat, so we call very bouncy collisions nearly elastic.
BMath
Counting the Kinetic Energy
To classify a collision, compare the kinetic energy of everything before with everything after:
% lost = (KE_before − KE_after) ÷ KE_before × 100
Example 1: two identical cars
Given
A 1,000 kg car at +20 m/s hits an identical car at rest
They lock together
Solve
v_f = (1,000 × 20) ÷ 2,000 = +10 m/s
KE_before = ½ (1,000)(20²) = 200,000 J
KE_after = ½ (2,000)(10²) = 100,000 J
% lost = 100,000 ÷ 200,000 × 100 = 50%
✓ Half of the kinetic energy disappears from the motion.
Example 2: a heavier car hits a lighter one
Given
A 2,000 kg car at +15 m/s hits a 1,000 kg car at rest
They lock together
Solve
v_f = (2,000 × 15) ÷ 3,000 = +10 m/s
KE_before = ½ (2,000)(15²) = 225,000 J
KE_after = ½ (3,000)(10²) = 150,000 J
% lost = 75,000 ÷ 225,000 × 100 = 33%
✓ A third of the kinetic energy is lost, which is less than in Example 1.
🔑If object A hits object B at rest and they stick, the fraction of kinetic energy lost is always m_B ÷ (m_A + m_B). Equal masses lose half. A big A hitting a tiny B loses very little. A tiny A hitting a huge B loses almost all of it.
CMath
Solving an Elastic Collision
In an elastic collision both momentum and kinetic energy are conserved. That gives two equations for the two unknown final velocities. A little algebra turns them into something simple. Write each in terms of the changes:
Momentum: m₁(v₁ − v₁′) = m₂(v₂′ − v₂)
Energy: m₁(v₁² − v₁′²) = m₂(v₂′² − v₂²)
Divide the second by the first, and use a² − b² = (a − b)(a + b):
v₁ + v₁′ = v₂′ + v₂
v₁ − v₂ = −(v₁′ − v₂′)
In words, in an elastic collision the relative velocity reverses: the objects separate at exactly the speed they approached. Combine that with conservation of momentum and you can solve for both final velocities. When object 2 starts at rest, the results are:
v₁′ = ((m₁ − m₂) ÷ (m₁ + m₂)) v₁
v₂′ = (2m₁ ÷ (m₁ + m₂)) v₁
Example 1: equal masses
Given
1.0 kg at +2.0 m/s hits an identical 1.0 kg cart at rest, elastic
Solve
v₁′ = ((1.0 − 1.0) ÷ 2.0)(2.0) = 0
v₂′ = (2(1.0) ÷ 2.0)(2.0) = +2.0 m/s
✓ The carts swap velocities: the first stops dead and the second leaves at 2.0 m/s.
Example 2: a heavy cart hits a light one
Given
0.50 kg at +1.0 m/s hits a 0.25 kg cart at rest, elastic
✓ The light cart bounces back at 0.333 m/s while the heavy cart moves forward at 0.667 m/s.
DExample
Collision Explorer
Start with e = 1 and two equal masses, with cart 2 at rest. Then slide e down to 0. Try unequal masses, and try sending the carts toward each other.
Cart 1 runs into cart 2. The elasticity slider goes from 0 (they stick) to 1 (a perfect bounce). Watch the total momentum bar stay the same while the kinetic energy bar changes.
Momentum is conserved: +8.0 before and +8.0 after.
EMath
The Coefficient of Restitution
The coefficient of restitution, e, scores how bouncy a collision was. It compares how fast the objects move apart with how fast they came together:
e = (v₂′ − v₁′) ÷ (v₁ − v₂)
For an elastic collision, e = 1. If the objects stick, they separate at zero speed, so e = 0. Everything in between is inelastic. Here are some rough typical values. They depend on the surfaces and the speed:
Collision
Typical e
Steel on steel
about 0.9
Superball on concrete
about 0.85
Basketball on a wood floor
about 0.8
Tennis ball on a hard court
about 0.7
Wooden block on wood
about 0.5
Clay or putty
about 0
When a ball hits a fixed floor, the floor’s velocity is zero before and after, so e is just the ball’s rebound speed divided by its impact speed. Since speeds come from heights, v = √(2gh):
e = v_after ÷ v_before = √(h₂ ÷ h₁)
From the bounce lab in Activity 5-1-2
Given
A ball is dropped from h₁ = 1.0 m
It rebounds to 0.62 m on a hard floor and to 0.30 m on a foam pad
Solve
Floor: e = √(0.62 ÷ 1.0) = 0.79
Foam: e = √(0.30 ÷ 1.0) = 0.55
Clay: h₂ = 0, so e = 0
✓ The harder surface is closer to elastic. Clay is perfectly inelastic.
FConcept
Where the Lost Energy Goes
Energy is never destroyed. When kinetic energy disappears from the motion in an inelastic collision, it becomes other forms: thermal energy as materials heat up, sound, and the work done bending, breaking, and crumpling. That last one is the key in a crash. From the work-energy theorem in Unit 4, the energy absorbed by crumpling is the average crushing force times the distance crushed:
KE lost ≈ F_crush × d_crush
Reading a crash
Given
The two-car crash from Example 1 lost 100,000 J of kinetic energy
The two cars crumpled by a total of 1.0 m
Solve
F_crush ≈ KE lost ÷ d_crush
F_crush ≈ 100,000 ÷ 1.0
F_crush ≈ 100,000 N
✓ The average crushing force was about 100,000 N. Bigger damage means more energy was lost, which means the cars were going faster.
🔑This is the idea behind crash reconstruction in Problem 5-3-1. Momentum tells you what the cars did during the collision. The damage tells you how much kinetic energy was lost. Together, they let an investigator work backward to the speeds.
GWatch Out
Three Common Mix-Ups
⚠️“Momentum isn’t conserved in an inelastic collision.” It is. Momentum is conserved in every collision in an isolated system. It’s kinetic energy that goes missing.
⚠️“Perfectly inelastic means everything stops.” It means the objects stick and move off together. They only stop if the total momentum was zero to begin with.
⚠️“The lost kinetic energy is destroyed.” It isn’t. It becomes thermal energy, sound, and the work of deforming the objects. Total energy is always conserved.
HConcept
Quick Check
Four questions. They are not graded. Answer them to find out what still needs work.
Q1Which quantity is conserved in every collision in an isolated system, elastic or not?
Q2A 1.0 kg cart at +2.0 m/s hits an identical 1.0 kg cart at rest in an elastic collision. What happens?
Q3A ball dropped from 1.0 m rebounds to 0.64 m. What is its coefficient of restitution?
Q4A 1,000 kg car at 20 m/s hits an identical car at rest, and they lock together. What fraction of the kinetic energy is lost?
Score: 0 of 4
IConcept
Looking Ahead
You can now sort collisions by their energy. Next you’ll use momentum conservation on full two-object problems, and then measure it yourself.