4.4 Elastic and Inelastic Collisions

Unit 4: Linear Momentum. Unit overview · Games · Practice set

Six little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last. Tap a round play button to hear a step read aloud. After every three stories there is one quick check.

1. Bouncy bumper cars

Read the story as text

You watch two kinds of crashes on the same evening. At a pool table, the cue ball hits a still ball dead centre: the cue ball stops, and the other ball rolls off at almost the same speed. Out in the parking lot, a slow car bumps into the back of a parked car and the bumpers lock: both cars roll forward together, slowly, with a crunch.

In both crashes the total momentum is the same before and after. So what is different? Where did the "motion energy" go in the parking lot, and how can we tell which kind of collision we are looking at?

2. Clay sticks to a cart

3. A head-on bump

Quick check. One question before the next stories.

Two balls of clay collide and stick together (no outside forces). Which is true?

4. Heavy hits light

5. Light hits heavy

6. Catching up and locking together

Quick check. One question before the next stories.

A light ball hits a much heavier ball at rest in an elastic collision. What does the light ball do?

7. Check yourself

Think of your answer first, then tap to see it.

a) A collision is called elastic. Which two things are the same before and after it?

Show answer

The total momentum and the total kinetic energy.

b) Two objects collide and stick together. Is momentum conserved? Is kinetic energy conserved?

Show answer

Momentum: yes (no outside push during the crash). Kinetic energy: no, some turns into heat, sound and bent material. This is a perfectly inelastic collision.

c) A 2 kg cart at 3 m/s hits a 1 kg cart at rest, and they stick. How fast do they move? How much kinetic energy is lost?

Show answer

p = 2 × 3 = 6 kg·m/s, shared by 3 kg: v = 6 ÷ 3 = 2 m/s. KE before = ½ × 2 × 3² = 9 J, after = ½ × 3 × 2² = 6 J, so 3 J is lost.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

In which collisions is the total momentum kept (no outside forces)?

What makes a collision elastic?

A 2 kg cart at 3 m/s hits a 1 kg cart at rest, and they stick. How fast do they move together?

The idea

Picture two carts rolling into each other on a track.

With springy bumpers, they bounce apart. With clay on the front, they stick and roll on together. Two real cars do something in between: they bounce a little and dent.

Before A B 4 m/s → rest Elastic A B A 0, B 4 m/s Stick both 2 m/s

In every collision between objects of an isolated system (no outside push during the crash), the total momentum is conserved. That never changes, whatever the objects are made of.

What can change is the total kinetic energy. That splits collisions into kinds:

Always: m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′

Elastic only: ½m₁v₁² + ½m₂v₂² = ½m₁v₁′² + ½m₂v₂′²

Stick together: m₁v₁ + m₂v₂ = (m₁ + m₂)v′

Two handy facts for head-on elastic collisions:

Trap: saying the moving object always stops and the target takes its speed in any elastic collision. Instead: that swap happens only for equal masses with one at rest; in Example 2 the 0.50 kg cart bounces back at −2.0 m/s and the 1.5 kg cart leaves at +2.0 m/s.

Even in an elastic collision, kinetic energy is not constant during the contact. While the bumpers squeeze, some KE is stored as spring energy, then it comes back. "Elastic" only compares just before with just after.

The same idea with numbers

Read the steps as text

Cart A (1.0 kg) moves right at 4.0 m/s toward cart B (1.0 kg) at rest. Right is positive. Compare an elastic collision with a stick-together collision.

  1. Momentum before: p = (1.0)(4.0) + (1.0)(0) = 4.0 kg·m/s. KE before: ½(1.0)(4.0)² = 8.0 J. Why: we always need both totals to classify a collision.
  2. Elastic, equal masses: A stops, B moves at 4.0 m/s. Check momentum: 0 + (1.0)(4.0) = 4.0 kg·m/s. Check KE: 0 + ½(1.0)(4.0)² = 8.0 J. Why: both totals match, so this is elastic.
  3. Stick together: (1.0 + 1.0)v′ = 4.0 → v′ = 2.0 m/s. Why: momentum is conserved and they share one velocity.
  4. KE after sticking: ½(2.0)(2.0)² = 4.0 J. Lost: 8.0 − 4.0 = 4.0 J, which is 50 %. Why: the missing 4.0 J became sound, heat and squashing. Momentum did not go anywhere; energy changed form.

Lab: the collision lab

Set masses, velocities and the bounciness e (e = 1 is elastic, e = 0 sticks together). Predict which way cart 1 goes after, then play. Watch four graphs: momentum (the dashed total stays flat), velocity, the contact force (equal and opposite on the two carts; the shaded area is the impulse), and kinetic energy (the total dips during contact, then comes back fully only when e = 1).

Try this: (1) Equal masses, v₂ = 0, e = 1: do they swap velocities? (2) Same masses, e = 0: what fraction of KE is lost? (It should be 50 %.) (3) Make cart 1 light (0.5 kg) and cart 2 heavy (5 kg) at rest with e = 1: cart 1 bounces back. (4) Change only the contact time: the peak force changes but the shaded area (impulse) and the final velocities do not.

Examples

Example 1: bumpers lock basic

A 1200 kg car moving at 15 m/s runs into the back of an 800 kg car at rest. The bumpers lock and the cars move together. Find their speed just after, and the kinetic energy lost.

Show solution
Read the steps as text
  1. Momentum before: (1200)(15) + (800)(0) = 18 000 kg·m/s. Why: the crash is quick, so friction from the road barely changes momentum during it.
  2. (1200 + 800)v′ = 18 000 → v′ = 9.0 m/s. Why: stuck together means one shared velocity.
  3. KE before: ½(1200)(15)² = 135 000 J. KE after: ½(2000)(9.0)² = 81 000 J. Why: compare totals to see the loss.
  4. Lost: 54 000 J (40 %), mostly to crumpling metal, heat and sound. Why: perfectly inelastic, but not 100 % lost, because the pair still moves.
Trap: saying the crash "lost momentum" because the cars slowed down. Instead: momentum is conserved (18 000 kg·m/s before and after); only kinetic energy (54 000 J) was lost.

Example 2: is it elastic? medium

A 0.50 kg cart at +4.0 m/s hits a 1.5 kg cart at rest. Afterwards the 0.50 kg cart moves at −2.0 m/s and the 1.5 kg cart at +2.0 m/s. Check momentum, then decide if the collision is elastic.

Show solution
Read the steps as text
  1. Momentum before: (0.50)(4.0) = 2.0 kg·m/s. After: (0.50)(−2.0) + (1.5)(2.0) = −1.0 + 3.0 = 2.0 kg·m/s. Why: the data must pass this test first; if not, something outside pushed.
  2. KE before: ½(0.50)(4.0)² = 4.0 J. Why: only the moving cart has KE.
  3. KE after: ½(0.50)(2.0)² + ½(1.5)(2.0)² = 1.0 + 3.0 = 4.0 J. Why: KE is a scalar; the minus sign disappears when squared.
  4. Same KE, so elastic. Quick check: approach speed 4.0 m/s, separation speed 2.0 − (−2.0) = 4.0 m/s. Why: equal approach and separation speeds is the head-on elastic test.

Example 3: ballistic pendulum AP

A 0.020 kg bullet hits and sticks in a 2.0 kg wooden block hanging on strings. The block and bullet swing up 0.10 m. Find the bullet's speed and how much KE was lost in the collision.

Show solution
Read the steps as text
  1. Split into two stages: (1) the collision, (2) the swing. Why: each stage conserves a different quantity. The collision conserves momentum (not KE). The swing conserves mechanical energy (not momentum, because gravity and the string push on it).
  2. Swing: ½Mv′² = Mgh → v′ = √(2gh) = √(2 × 9.8 × 0.10) = √1.96 = 1.4 m/s. Why: KE just after the collision becomes gravitational PE at the top.
  3. Collision: (0.020)v = (2.02)(1.4) → v = 2.828 / 0.020 = 141 m/s, about 140 m/s. Why: stick together, momentum conserved.
  4. KE before: ½(0.020)(141.4)² ≈ 200 J. KE after: ½(2.02)(1.4)² ≈ 1.98 J. About 198 J (99 %) is lost to heat and splintering wood. Why: this is why you cannot use energy conservation across the collision.
Trap: setting the bullet's KE equal to the block-plus-bullet KE for the hit. Instead: when objects stick, use momentum for the collision (99% of the KE is lost here); use energy only for the swing afterwards.

Example 4: a glancing hit in 2D AP

On an air table, puck A (2.0 kg) moves east at 3.0 m/s and hits puck B (1.0 kg) at rest. Afterwards A moves with components 2.0 m/s east and 0.50 m/s north. Find B's velocity and decide if the collision is elastic.

Show solution
Read the steps as text
  1. East (x): (2.0)(3.0) = (2.0)(2.0) + (1.0)vBx → vBx = 6.0 − 4.0 = 2.0 m/s. Why: momentum is conserved in each direction separately.
  2. North (y): 0 = (2.0)(0.50) + (1.0)vBy → vBy = −1.0 m/s (south). Why: there was no y momentum before, so the y momenta after must cancel.
  3. B: speed √(2.0² + 1.0²) = 2.24 m/s at tan⁻¹(1.0/2.0) = 26.6° south of east. Why: combine perpendicular components.
  4. KE before: ½(2.0)(3.0)² = 9.0 J. After: ½(2.0)(2.0² + 0.50²) + ½(1.0)(2.0² + 1.0²) = 4.25 + 2.50 = 6.75 J. Why: KE uses speed squared = sum of squared components.
  5. 2.25 J (25 %) lost, so the collision is inelastic, even though the pucks bounced apart. Why: bouncing apart does not prove elastic; only equal KE does.
Trap: calling a collision elastic because the objects bounced apart. Instead: test the kinetic energy: 9.0 J before and 6.75 J after, so this one is inelastic.

Practice (AP style)

1. Which statement is true for every collision between two objects that form an isolated system?

Show answer

Isolated system means no net outside impulse, so total momentum is conserved. KE is conserved only in elastic collisions.

2. A 3.0 kg cart moving at 4.0 m/s hits a 1.0 kg cart at rest, and they stick together. What is their speed afterwards?

Show answer

(3.0)(4.0) = (4.0)v′ → v′ = 12 / 4.0 = 3.0 m/s. (12 is the momentum, not the speed.)

3. Steel ball A moves at 5.0 m/s and hits an identical steel ball B at rest, head-on. The collision is elastic. What happens?

Show answer

Equal masses, elastic, one at rest: they swap velocities. Check: momentum 5m before and after; KE ½m(25) before and after. Choice A conserves momentum but loses half the KE (that is the stick-together answer). Choice D breaks momentum.

4. A moving lump of clay hits an identical lump at rest and they stick. What fraction of the original kinetic energy is lost?

Show answer

v′ = v/2. KE after = ½(2m)(v/2)² = ¼mv², which is half of ½mv². So half is lost.

5. Two carts with spring bumpers have an elastic collision. At the instant the springs are squeezed the most, how does the total kinetic energy of the carts compare with its value before the collision?

Show answer

At maximum squeeze both carts move at the same velocity (vcm), so KE is at its lowest; the rest is spring energy. It all comes back as KE after, which is why the collision is elastic. The carts do not both stop unless total momentum is zero. Watch the KE graph in the lab.

6. Two 1.0 kg carts collide on a track. Velocity data (right is +): cart 1 goes from +4.0 m/s to +1.0 m/s; cart 2 goes from 0 to +3.0 m/s. Which describes the collision?

Show answer

Momentum: 4.0 before, 1.0 + 3.0 = 4.0 after. KE: 8.0 J before, 0.5 + 4.5 = 5.0 J after, so 3.0 J lost. Not perfectly inelastic, because the final velocities differ. Also approach speed 4.0, separation speed 2.0: not elastic.

7. A tennis ball hits a wall head-on at 20 m/s and bounces straight back at 20 m/s. Which is true about the ball alone?

Show answer

Same speed means same KE. But the velocity reversed, so Δp = m(−20 − 20) = −40m. The ball alone is not an isolated system: the wall pushed on it. (The ball + Earth system does conserve momentum; Earth's change in velocity is far too small to see.)

8. (Free response, mathematical routine) A ball of mass m moving at speed v₀ hits a ball of mass 2m at rest. They stick together.
(a) Derive the final speed in terms of v₀.
(b) Derive the fraction of the original kinetic energy that is lost.
(c) Explain in words why momentum is conserved here but kinetic energy is not.

Show answer
  1. (a) mv₀ = (3m)v′ → v′ = v₀/3.
  2. (b) Kbefore = ½mv₀². Kafter = ½(3m)(v₀/3)² = mv₀²/6 = ⅓Kbefore. Fraction lost = 1 − ⅓ = 2/3.
  3. (c) During the collision the only big forces are the balls pushing on each other. By Newton's third law these are equal and opposite for the same time, so the impulses cancel and total momentum stays the same. Nothing like that protects KE: the forces do negative work deforming the balls, turning KE into internal (thermal) energy and sound.

9. (Free response, experimental design) You have two carts on a low-friction track, a balance, two motion sensors, and extra masses. Describe a procedure to test whether collisions between the carts with magnetic bumpers are elastic. Say what you measure, how you use it, and what result would support "elastic".

Show answer
  1. Measure each cart's mass with the balance.
  2. Put a motion sensor at each end of the track. Push cart 1 toward cart 2 (at rest). Record both velocities just before and just after the collision (take the flat parts of each v-t graph next to the collision).
  3. Compute total momentum before and after (to check the system was nearly isolated) and total KE = ½m₁v₁² + ½m₂v₂² before and after.
  4. Repeat several times with different speeds and with extra mass added to one cart.
  5. If KE after equals KE before within measurement uncertainty (for example within a few percent) in every trial, the collisions are elastic. If KE after is consistently lower, they are inelastic. A graph of Kafter vs Kbefore with slope near 1 supports elastic.

10. (Short free response: translation between representations) Cart 1 (1.0 kg) moves at +6.0 m/s toward cart 2 (2.0 kg) at rest. A velocity-time graph of the collision shows: cart 1's line drops from +6.0 m/s to −2.0 m/s and cart 2's line rises from 0 to +4.0 m/s during the short contact. (a) Check that momentum is conserved. (b) Decide whether the collision is elastic. (c) Add the center of mass velocity to the graph: describe its line. (d) At the moment of greatest squeeze the two lines cross. At what velocity?

Show answer
  1. (a) Before: 1.0 × 6.0 = 6.0 kg·m/s. After: 1.0 × (−2.0) + 2.0 × 4.0 = −2.0 + 8.0 = 6.0 kg·m/s. Conserved.
  2. (b) KE before: ½(1.0)(6.0)² = 18 J. After: ½(1.0)(2.0)² + ½(2.0)(4.0)² = 2.0 + 16 = 18 J. Elastic. (Also: approach speed 6.0 m/s, separation speed 4.0 − (−2.0) = 6.0 m/s.)
  3. (c) vcm = 6.0 / 3.0 = 2.0 m/s: a horizontal line at +2.0 m/s for the whole graph.
  4. (d) At greatest squeeze both carts move together, at vcm = 2.0 m/s, so the lines cross on the center of mass line.

Point guide (4 points):

  • 1 point: momentum check, 6.0 = 6.0.
  • 1 point: KE 18 J both, so elastic.
  • 1 point: vcm = 2.0 m/s as a flat line.
  • 1 point: crossing at 2.0 m/s.

11. (Short free response: qualitative/quantitative translation) Cart 1 (mass m1) moves at speed v toward cart 2 (mass m2) at rest. They stick together. (a) Without numbers: if m2 is much bigger than m1, is most of the KE kept or lost? (b) Derive the fraction of the KE that is lost, in terms of m1 and m2. (c) Check it with m1 = 1.0 kg, m2 = 3.0 kg, v = 4.0 m/s by working out the KE before and after.

Show answer
  1. (a) Mostly lost: a light cart hitting a heavy one hardly moves the pair, so little KE is left.
  2. (b) Momentum: m1v = (m1 + m2)v′, so v′ = m1v / (m1 + m2). KE after / KE before = (m1 + m2)v′² / (m1v²) = m1 / (m1 + m2). Fraction lost = m2 / (m1 + m2), which is close to 1 when m2 is big, matching (a).
  3. (c) v′ = 1.0 × 4.0 / 4.0 = 1.0 m/s. KE before: ½(1.0)(4.0)² = 8.0 J. After: ½(4.0)(1.0)² = 2.0 J. Lost 6.0 J = 75%. Formula: 3.0 / 4.0 = 75%. Agrees.

Point guide (4 points):

  • 1 point: mostly lost, with a reason.
  • 1 point: uses momentum (not KE) to find v′.
  • 1 point: fraction lost m2/(m1 + m2).
  • 1 point: 8.0 J, 2.0 J, 75% check.

Common mistakes

The mistake: "momentum is lost in an inelastic collision."

Why it is wrong: in an isolated system momentum is always conserved. It is kinetic energy that is lost (turned into other forms).

How to spot it: if your momentum totals before and after differ, look for an outside force or an arithmetic sign error, not for "inelastic".

The mistake: using KE conservation for objects that stick together (or for a ballistic pendulum collision).

Why it is wrong: sticking always loses KE (unless nothing was moving relative to each other). Use momentum for the collision, and energy only for the parts before or after it.

How to spot it: do the words say "sticks", "embeds", "couples", or "move together"? Then only momentum works across the collision.

The mistake: "they bounced apart, so it was elastic."

Why it is wrong: many bouncing collisions lose some KE (Example 4 lost 25 %). Elastic is a statement about energy, not about bouncing.

How to spot it: always compute KE before and after, or compare approach and separation speeds for a head-on hit.

The mistake: dropping the minus sign on a velocity that reverses.

Why it is wrong: momentum is a vector; a rebound has the opposite sign. (KE has no sign, so it hides the error.)

How to spot it: state the positive direction at the top; any object moving the other way gets a minus sign.

The mistake: thinking a perfectly inelastic collision loses all the kinetic energy.

Why it is wrong: the stuck objects still move with vcm = ptotal/M, so they keep ½Mvcm². All KE is lost only when the total momentum is zero.

How to spot it: compute the total momentum. If it is not zero, the pair keeps moving and keeps some KE.