C. Energy and Momentum · Momentum and CollisionsElastic and Inelastic Collisions
C2.1 C2.6 C3.3About 65 minutes
Learning goals- I can distinguish between elastic, inelastic and perfectly inelastic collisions.
- I can classify a collision by comparing the total kinetic energy before and after.
- I can calculate the kinetic energy lost in a collision and explain where it goes.
- I can use the coefficient of restitution to describe a bounce.
Words to know
elastic collision
A collision in which the total kinetic energy of the system is conserved.
inelastic collision
A collision in which some kinetic energy is converted to other forms, so the total kinetic energy decreases.
perfectly (completely) inelastic collision
An inelastic collision in which the objects stick together and move with a common velocity.
coefficient of restitution (e)
Separation speed ÷ approach speed; e = 1 for elastic and e = 0 for perfectly inelastic collisions.
deformation
A change in an object's shape; permanent deformation uses up kinetic energy.
Where did the energy go?
In the last two lessons, momentum was conserved in every collision. But look back at Challenge 3 in the lesson on conservation of momentum: three carts stuck together one after another, and the kinetic energy fell from 18 J to 9 J to 6 J while the momentum stayed at 6.0 kg·m/s. Kinetic energy can be lost in a collision even when momentum is not. Physicists sort collisions into types by asking one question: what happens to the total kinetic energy?
Types of collision
In every collision of an isolated system, momentum is conserved. Kinetic energy is a different story:
- Elastic collision: the total kinetic energy after equals the total kinetic energy before. Nothing is lost to heat, sound or permanent deformation. Collisions between gas molecules are treated as elastic in the kinetic theory of gases; collisions between billiard balls, steel ball bearings, or carts with magnetic or spring bumpers are nearly elastic.
- Inelastic collision: the total kinetic energy after is less than before. Some kinetic energy becomes thermal energy, sound, and the energy of permanent deformation (dents, crushed metal). Most everyday collisions are inelastic.
- Perfectly (completely) inelastic collision: the objects stick together and move with one common velocity. This loses the most kinetic energy that can be lost while still conserving momentum. A car crash where the vehicles lock together, a dart hitting a board, and Velcro carts are examples.
A collision can also gain kinetic energy if stored energy is released during it (for example, a spring that is latched and then released on contact, or a chemical reaction). This is really an explosion combined with a collision; some textbooks call it superelastic.
Total energy is always conserved. "Losing kinetic energy" means kinetic energy turns into other forms; it does not vanish. The law of conservation of energy and the law of conservation of momentum are both obeyed in every collision. What differs is how the energy is shared among its forms.
How to classify a collision
- Check that momentum is conserved (it must be, for an isolated system).
- Calculate the total kinetic energy before and after.
- Equal → elastic. Less after → inelastic. Objects stuck together → perfectly inelastic. More after → stored energy was released.
Do not decide from appearance alone. A collision that looks bouncy can still lose a lot of kinetic energy.
Worked exampleExample 1: Carts that stick
In the conservation of momentum lesson, a 0.50 kg cart at 2.0 m/s hit a 1.5 kg cart at rest and they stuck, moving at 0.50 m/s. Classify the collision and find the percentage of kinetic energy lost.
- Ek before = ½(0.50)(2.0)² = 1.0 J.
- Ek after = ½(2.0)(0.50)² = 0.25 J.
- Lost: 1.0 − 0.25 = 0.75 J, which is 0.75 ÷ 1.0 = 75%.
Perfectly inelastic: the carts stuck together and 75% of the kinetic energy became thermal energy, sound and deformation.
Worked exampleExample 2: Pucks that separate
A 0.20 kg puck at 5.0 m/s hit a 0.30 kg puck at rest. Afterwards the pucks moved at 1.0 m/s and 2.67 m/s in the original direction. Classify the collision.
- Momentum check: before (0.20)(5.0) = 1.0; after (0.20)(1.0) + (0.30)(2.667) = 0.20 + 0.80 = 1.0 kg·m/s. ✓
- Ek before = ½(0.20)(5.0)² = 2.5 J.
- Ek after = ½(0.20)(1.0)² + ½(0.30)(2.667)² = 0.10 + 1.067 = 1.167 J.
- Ek after ÷ Ek before = 1.167 ÷ 2.5 ≈ 0.47, so about 53% was lost.
Inelastic (but not perfectly inelastic, since the pucks separated). About 47% of the kinetic energy remains.
Worked exampleExample 3: A truly elastic collision
A 0.20 kg cart moving at 3.0 m/s with a magnetic bumper approaches a 0.10 kg cart at rest. Afterwards the 0.20 kg cart moves at 1.0 m/s and the 0.10 kg cart at 4.0 m/s, both forward. Classify the collision.
- Momentum: before (0.20)(3.0) = 0.60; after (0.20)(1.0) + (0.10)(4.0) = 0.20 + 0.40 = 0.60 kg·m/s. ✓
- Ek before = ½(0.20)(3.0)² = 0.90 J.
- Ek after = ½(0.20)(1.0)² + ½(0.10)(4.0)² = 0.10 + 0.80 = 0.90 J.
Kinetic energy is unchanged, so the collision is elastic. The magnets pushed the carts apart without the carts ever touching, so nothing was crushed or heated.
Total kinetic energy after each collision in Examples 1 to 3, as a percentage of the kinetic energy before. Momentum is conserved in all three; only kinetic energy tells them apart.Worked exampleExample 4: How much energy can a sticking collision lose?
An object of mass m₁ moving at v hits an object of mass m₂ at rest, and they stick. Show that the fraction of kinetic energy lost is m₂ ÷ (m₁ + m₂). Then apply it to a 1500 kg car that hits a stopped 1000 kg car and locks bumpers.
- Momentum: m₁v = (m₁ + m₂)v′, so v′ = m₁v ÷ (m₁ + m₂).
- Ek before = ½m₁v². Ek after = ½(m₁ + m₂)v′² = ½(m₁ + m₂) × m₁²v² ÷ (m₁ + m₂)² = ½m₁v² × m₁ ÷ (m₁ + m₂).
- So Ek after ÷ Ek before = m₁ ÷ (m₁ + m₂), and the fraction lost is 1 − m₁ ÷ (m₁ + m₂) = m₂ ÷ (m₁ + m₂).
- Cars: 1000 ÷ (1500 + 1000) = 0.40.
40% of the kinetic energy is lost. The more massive the target compared with the moving object, the bigger the fraction lost: a clay ball hitting a heavy block loses nearly all of its kinetic energy.
Measuring "bounciness": the coefficient of restitution
Engineers and sports scientists describe how elastic a collision is with the coefficient of restitution, e:
e = (speed at which the objects separate) ÷ (speed at which they approach)
- e = 1 for an elastic collision.
- e = 0 for a perfectly inelastic collision (no separation at all).
- Real collisions are in between.
For a ball dropped on a hard floor (which does not move), the approach speed is the impact speed and the separation speed is the rebound speed. Since v = √(2gh), the ratio of speeds is √(h_bounce ÷ h_drop), so
e = √(h_bounce ÷ h_drop), and the fraction of kinetic energy kept in the bounce is h_bounce ÷ h_drop = e².
This is not a new law, just a useful way to describe how much kinetic energy each bounce keeps.
Worked exampleExample 5: A bouncing ball
A ball is dropped from 1.20 m and bounces back up to 0.75 m. Find the fraction of kinetic energy it keeps in the bounce, the coefficient of restitution, and its speeds just before and just after the bounce.
- Fraction kept = h_bounce ÷ h_drop = 0.75 ÷ 1.20 = 0.625, so 62.5%.
- e = √0.625 ≈ 0.79.
- Speed just before: v = √(2gh) = √(2 × 9.8 × 1.20) ≈ 4.85 m/s.
- Speed just after: v = √(2 × 9.8 × 0.75) ≈ 3.83 m/s. Check: 3.83 ÷ 4.85 ≈ 0.79 = e. ✓
The ball keeps 62.5% of its kinetic energy; e ≈ 0.79; it hits at about 4.8 m/s and leaves at about 3.8 m/s.
Just for funSquash players hit the ball around for a few minutes before a match because a cold squash ball has a low coefficient of restitution and barely bounces. It may be the only sport where the ball needs a warm-up as much as the players do.
Hands-on activityThe great bounce test
You need: 3 or 4 different balls (for example a tennis ball, a ping-pong ball, a rubber bouncy ball and a basketball), a metre stick or tape measure, masking tape, a hard floor, and a phone that can record slow-motion video. Optional: a freezer.
- Tape the metre stick upright against a wall, with 0 at the floor.
- Hold each ball so its bottom is at 1.00 m. Drop it (do not throw it) and film the bounce in slow motion beside the metre stick.
- Step through the video to find the highest point the bottom of the ball reaches after the first bounce. Do three trials per ball and average.
- For each ball calculate the fraction of kinetic energy kept (h_bounce ÷ h_drop) and e = √(h_bounce ÷ h_drop).
- Optional: put a tennis ball in the freezer for 30 minutes and repeat. Then try a carpet or a lawn instead of the hard floor.
What to look for: bouncy rubber balls often keep around 70% to 90% of their kinetic energy; a tennis ball on a hard floor typically keeps a little over half. Many balls bounce lower when cold, and every ball bounces lower on carpet or grass, because the surface itself deforms and absorbs energy. None reaches 1.00 m: no real collision is perfectly elastic.
Your turn
1A 2.0 kg cart moving at 3.0 m/s hits a 1.0 kg cart at rest. Afterwards they move forward at 1.0 m/s and 4.0 m/s. Classify the collision.
Momentum: 6.0 = 2.0 + 4.0 ✓. Ek before = ½(2.0)(9.0) = 9.0 J; after = ½(2.0)(1.0) + ½(1.0)(16) = 1.0 + 8.0 = 9.0 J. Elastic.
2A 2.0 kg cart moving at 3.0 m/s hits a 1.0 kg cart at rest and they stick, moving at 2.0 m/s. Find the kinetic energy lost and classify the collision.
Ek before = 9.0 J; after = ½(3.0)(2.0)² = 6.0 J. 3.0 J lost (33%). Perfectly inelastic (they stuck).
3A 0.50 kg puck at 4.0 m/s hits an identical puck at rest. Afterwards they move forward at 1.0 m/s and 3.0 m/s. Classify the collision.
Momentum: 2.0 = 0.50 + 1.5 ✓. Ek before = ½(0.50)(16) = 4.0 J; after = ½(0.50)(1.0) + ½(0.50)(9.0) = 0.25 + 2.25 = 2.5 J. Inelastic; 1.5 J (37.5%) lost.
4A 0.40 kg ball of clay hits a 1.6 kg block at rest and sticks. What fraction of the kinetic energy is lost?
m₂ ÷ (m₁ + m₂) = 1.6 ÷ 2.0 = 0.80 (80%).
5A ball dropped from 2.0 m bounces to 1.5 m. Find the fraction of kinetic energy kept and the coefficient of restitution.
Fraction kept = 1.5 ÷ 2.0 = 0.75 (75%). e = √0.75 ≈ 0.87.
6In an inelastic collision between two cars, where does the "lost" kinetic energy go?
Into thermal energy (the metal heats up), sound, and the energy of permanent deformation (bending and crushing metal and plastic).
7Can a collision in which the objects stick together ever be elastic? Explain.
No. Two separate objects with different velocities have more kinetic energy than the same total momentum carried by one combined object (except in the trivial case where they already had the same velocity, so there is no collision). Sticking always turns some kinetic energy into other forms.
More practice: warm-up
1What is an elastic collision?
A collision in which the total kinetic energy is conserved (as well as momentum).
2What is a perfectly inelastic collision?
A collision in which the objects stick together and move with a common velocity afterward.
3Is momentum conserved in an inelastic collision of an isolated system?
Yes. Momentum is conserved in every collision of an isolated system.
4Give an example of a nearly elastic collision.
Billiard balls colliding, steel ball bearings colliding, or carts with magnetic bumpers. (Collisions between gas molecules are treated as elastic.)
5What is the coefficient of restitution for a perfectly inelastic collision?
e = 0, since the objects do not separate.
More practice: on your own
1In the hockey collision from the conservation of momentum lesson, an 85 kg player at 6.0 m/s [E] and a 95 kg player at 4.0 m/s [W] held on and moved at 0.722 m/s [E]. How much kinetic energy was lost, and what percentage is that?
Before: ½(85)(6.0)² + ½(95)(4.0)² = 1530 + 760 = 2290 J. After: ½(180)(0.7222)² ≈ 46.9 J. Lost ≈ 2243 J, about 98%. Head-on sticking collisions lose almost all their kinetic energy.
2A student reports that a 1.0 kg cart at 4.0 m/s hit an identical cart at rest, and afterwards the carts moved at 1.0 m/s backward and 5.0 m/s forward. Is this possible without any stored energy? Explain.
Momentum checks: 4.0 = −1.0 + 5.0 ✓. But Ek before = ½(1.0)(16) = 8.0 J and after = 0.5 + 12.5 = 13 J. Kinetic energy cannot increase without a source of stored energy, so this is impossible (the data must be wrong, or a spring or explosive was involved).
3A 3000 kg truck hits a 1000 kg car at rest and they lock together. What fraction of the kinetic energy is lost? What if the 1000 kg car had hit the stopped truck instead?
Truck hitting car: 1000 ÷ 4000 = 25%. Car hitting truck: 3000 ÷ 4000 = 75%. Hitting a heavier object at rest loses a larger fraction of the kinetic energy.
4A 1.7 kg cart at 2.0 m/s hits a 1.5 kg cart at rest. Afterwards the carts move forward at 0.50 m/s and 1.7 m/s. Find the percentage of kinetic energy kept and the coefficient of restitution.
Momentum: 3.4 = 0.85 + 2.55 ✓. Ek before = ½(1.7)(4.0) = 3.4 J; after = ½(1.7)(0.25) + ½(1.5)(2.89) = 0.2125 + 2.1675 = 2.38 J, so 70% kept. e = (1.7 − 0.50) ÷ (2.0 − 0) = 0.60.
5A rubber ball dropped from 1.5 m bounces to 1.2 m. Find the fraction of kinetic energy kept, e, and the speed just after the bounce.
Kept = 1.2 ÷ 1.5 = 0.80. e = √0.80 ≈ 0.89. Speed after = √(2 × 9.8 × 1.2) ≈ 4.8 m/s.
Challenge
1Two identical carts move toward each other at equal speeds and stick together. Show that all their kinetic energy is lost. Why is this the largest possible loss?
Their momenta are equal and opposite, so the total momentum is zero, and after sticking the pair is at rest: Ek after = 0, so 100% is lost. Kinetic energy can only be lost down to the amount that momentum conservation requires the system to keep. Here the total momentum is zero, so nothing has to remain in motion, and every joule can be converted.
2A ball dropped from 1.0 m keeps 64% of its kinetic energy in every bounce. Find e, and the total distance it travels before coming to rest (assume it bounces forever with shrinking heights).
e = √0.64 = 0.80. Heights after bounces: 0.64, 0.64², ... m. Each is travelled up and down. Total = 1.0 + 2(0.64 + 0.64² + ...) = 1.0 + 2 × 0.64 ÷ (1 − 0.64) = 1.0 + 3.56 ≈ 4.6 m (infinite geometric series, from MCR3U).
3Show that, for a ball bouncing on a fixed floor, the fraction of kinetic energy kept equals e².
Ek ∝ v². Fraction kept = v_after² ÷ v_before² = (v_after ÷ v_before)² = e², since e = v_after ÷ v_before for a fixed floor.
Quiz
1. In an elastic collision between two objects in an isolated system:
2. Two objects collide and stick together. This collision is:
3. Which quantity is conserved in every collision of an isolated system?
4. A 1.0 kg cart at 2.0 m/s hits an identical cart at rest and they stick. What fraction of the kinetic energy is lost?
5. A ball dropped from 1.00 m bounces to 0.81 m. What is its coefficient of restitution?
6. After a collision, the total kinetic energy is greater than before. Which statement is correct?
Big idea (C2.1, C2.6, C3.3): collisions are classified by what happens to kinetic energy. Elastic: kinetic energy is conserved. Inelastic: some kinetic energy becomes thermal energy, sound and deformation. Perfectly inelastic: objects stick, losing the most kinetic energy possible. Momentum is conserved in all of them.
How to help: Do the bounce test together. It is quick, and it turns the coefficient of restitution into something your teen can see. Ask: "Where did the missing energy go?" (warming the ball and floor, sound).
Common difficulties:
- Thinking momentum is lost in inelastic collisions. Only kinetic energy is "lost" (converted); momentum is conserved.
- Adding kinetic energies with signs. Kinetic energy is never negative; square the velocity.
- Classifying by appearance. Always calculate the kinetic energy before and after.
- Mixing up h ratio and e. The height ratio is the fraction of energy kept (e²); e is its square root.