The shape of the lesson
By the end of the lesson, learners can
- Before equals after: Set the masses and velocities of two trolleys and choose elastic or inelastic.
- Two kinds of collision: A perfectly inelastic collision is the extreme case where the objects stick together and move with one common velocity v = (m₁u₁ + m₂u₂) / (m₁ + m₂).
- Momentum is a vector.
- Conserved always: total momentum, when no external force acts.
- Vector: choose a positive direction; opposite motion is negative.
- Kinetic energy: conserved only in an elastic collision.
One picture every idea returns to
The principle of conservation of momentum: for a system with no external resultant force, the total momentum stays constant. In a collision, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Momentum is a vector, so directions and signs matter, and in two dimensions it is conserved separately along each axis. Kinetic energy is conserved only in an elastic collision; in an elastic collision the relative speed of approach equals the relative speed of separation.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, What collisions keep.. In any interaction with no external resultant force, the total momentum afterwards equals the total before. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Before equals after | Set the masses and velocities of two trolleys and choose elastic or inelastic. Project tab 3 of the lab (Conservation of momentum) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two kinds of collision | A perfectly inelastic collision is the extreme case where the objects stick together and move with one common velocity v = (m₁u₁ + m₂u₂) / (m₁ + m₂). | Pairs |
| 21 to 30 min | Teach and check: Momentum is a vector | Momentum is a vector. | Mix-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Momentum is a vector, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Two traps recur. | First, momentum is a vector: a head-on collision needs opposite signs for the two velocities, or you will get a wrong total. Second, do not assume kinetic energy is conserved. Momentum is conserved in every collision with no external force, but kinetic energy is conserved only in an elastic one; in a real crash some kinetic energy becomes heat, sound and deformation. Stating that "energy is destroyed" is also wrong: total energy is always conserved, it just changes form. |
Conservation of momentum: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
State the principle of conservation of momentum, including the condition under which it applies.
A 1.5 kg trolley moving at 4.0 m s⁻¹ collides with and sticks to a stationary 2.5 kg trolley. Find their common velocity afterward.
A 800 kg cannon, initially at rest, fires a 5.0 kg shell horizontally at 240 m s⁻¹. Find the recoil velocity of the cannon.
Two trolleys are held together against a compressed spring and released from rest. A 1.0 kg trolley moves off to the left at 6.0 m s⁻¹. Find the velocity of the 2.0 kg trolley.
A 2.0 kg ball moving at 6.0 m s⁻¹ makes a head-on elastic collision with a stationary 2.0 kg ball. Use the elastic relative-speed result to find the final velocities.
A 0.50 kg lump of clay moving at 8.0 m s⁻¹ strikes and sticks to a stationary 1.5 kg block. Find the common velocity and the kinetic energy lost in the collision.
Answer key and marking notes
- The total momentum of a system stays constant. ✓
- This holds provided there is no external resultant force acting on the system. ✓
- Momentum before = 1.5 × 4.0 + 0 = 6.0 kg m s⁻¹. ✓
- Combined mass = 1.5 + 2.5 = 4.0 kg. ✓
- v = 6.0 / 4.0 = 1.5 m s⁻¹ in the original direction. ✓
- Total momentum before = 0, so total after = 0. ✓
- 800 × V + 5.0 × 240 = 0, so 800V = −1200. ✓
- V = −1.5 m s⁻¹: the cannon recoils at 1.5 m s⁻¹ in the opposite direction. ✓
- Total momentum before = 0 (both at rest). ✓
- Taking right as positive: 1.0 × (−6.0) + 2.0 × v = 0, so 2.0v = 6.0. ✓
- v = 3.0 m s⁻¹ to the right (opposite to the 1.0 kg trolley). ✓
- Momentum: 2.0 × 6.0 = 2.0 v₁ + 2.0 v₂, so v₁ + v₂ = 6.0. ✓
- Elastic: relative speed of separation equals approach, so v₂ − v₁ = 6.0. ✓
- Solving: v₁ = 0 and v₂ = 6.0 m s⁻¹. For equal masses the balls exchange velocities. ✓
- Common velocity: v = (0.50 × 8.0) / (0.50 + 1.5) = 4.0 / 2.0 = 2.0 m s⁻¹. ✓
- KE before = ½ × 0.50 × 8.0² = 16 J; KE after = ½ × 2.0 × 2.0² = 4.0 J. ✓
- KE lost = 16 − 4.0 = 12 J, transferred to heat, sound and deformation. ✓
17 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The total momentum of a system is conserved provided that:
- Athe collision is elastic
- Bno external resultant force acts on it
- Ckinetic energy is conserved
- Dthe objects have equal mass
A 4.0 kg trolley at 3.0 m s⁻¹ collides with and sticks to a stationary 2.0 kg trolley. Their common velocity is:
- A1.5 m s⁻¹
- B3.0 m s⁻¹
- C6.0 m s⁻¹
- D2.0 m s⁻¹
In an elastic collision, which quantity is conserved that is not conserved in an inelastic collision?
- Akinetic energy
- Bmomentum
- Cmass
- Dtotal energy
In a perfectly inelastic collision, the colliding objects:
- Abounce off with no energy loss
- Balways come to rest
- Cstick together and move with one common velocity
- Dexchange velocities
A 60 kg skater, initially at rest, throws a 2.0 kg ball horizontally at 9.0 m s⁻¹. The skater's recoil speed is:
- A9.0 m s⁻¹
- B0.30 m s⁻¹
- C18 m s⁻¹
- D0.033 m s⁻¹
Momentum is a vector quantity, which means that in a one-dimensional collision you must:
- Aadd all the speeds
- Bignore direction
- Cuse only kinetic energy
- Dgive velocities in opposite directions opposite signs
An object at rest explodes into two fragments. The total momentum of the fragments is:
- Azero
- Bequal to the mass times the speed
- Calways to the right
- Dimpossible to determine
For a one-dimensional elastic collision, the relative speed of approach equals the:
- Atotal momentum
- Btotal kinetic energy
- Crelative speed of separation
- Dsum of the masses
In a one-dimensional elastic collision between two equal masses, one moving and one at rest, the result is that they:
- Astick together
- Bexchange velocities
- Cboth stop
- Dboth reverse
A 2.0 kg trolley moving right at 5.0 m s⁻¹ collides with a 3.0 kg trolley moving left at 2.0 m s⁻¹, and they stick together. Their common velocity is:
- A1.4 m s⁻¹ to the right
- B0.80 m s⁻¹ to the left
- C3.0 m s⁻¹ to the right
- D0.80 m s⁻¹ to the right
A 0.20 kg ball hits a wall at 5.0 m s⁻¹ and rebounds elastically at 5.0 m s⁻¹. The magnitude of its change in momentum is:
- A2.0 kg m s⁻¹
- B0 kg m s⁻¹
- C1.0 kg m s⁻¹
- D5.0 kg m s⁻¹
A moving trolley collides with and sticks to an identical stationary trolley. Their common speed is:
- Athe same as the original speed
- Btwice the original speed
- Chalf the original speed
- Dzero
In a real car crash where the cars crumple, the total momentum is conserved but the kinetic energy:
- Aincreases
- Bdecreases, becoming heat, sound and deformation
- Cis also conserved
- Dbecomes zero
An astronaut floating at rest in space throws a spanner away from herself. Afterward she:
- Astays exactly at rest
- Bmoves in the same direction as the spanner
- Cspins without moving
- Ddrifts in the opposite direction to the spanner
In a two-dimensional (glancing) collision with no external force, momentum is conserved:
- Aseparately along each of two perpendicular directions
- Bonly along the direction of the faster object
- Conly as total kinetic energy
- Donly if the masses are equal
A 0.10 kg ball travelling at 20 m s⁻¹ strikes a wall head-on and rebounds at 20 m s⁻¹. The magnitude of the change in momentum is:
- A2.0 kg m s⁻¹
- B0 kg m s⁻¹
- C4.0 kg m s⁻¹
- D40 kg m s⁻¹
A 1.0 kg trolley at 6.0 m s⁻¹ strikes a stationary 2.0 kg trolley and they couple together. The kinetic energy lost is:
- A0 J
- B12 J
- C18 J
- D6.0 J
Quiz key
Momentum is conserved when there is no external resultant force, regardless of whether the collision is elastic.
p = 4.0 × 3.0 = 12 kg m s⁻¹; combined mass 6.0 kg; v = 12 / 6.0 = 2.0 m s⁻¹.
Both conserve momentum and total energy. Only the elastic collision also conserves kinetic energy.
A perfectly inelastic collision is the extreme case where the bodies move off together with a single velocity.
60 × V = 2.0 × 9.0 = 18, so V = 0.30 m s⁻¹ in the opposite direction.
Because momentum has direction, opposite velocities take opposite signs when you add them.
The momentum before is zero, so the total after is zero; the fragments carry equal and opposite momenta.
In an elastic collision the relative speed of separation equals the relative speed of approach.
For equal masses in an elastic collision, the moving one stops and the stationary one moves off at the original speed: they exchange velocities.
Right positive: p = 2.0(+5.0) + 3.0(−2.0) = 10 − 6.0 = 4.0 kg m s⁻¹; v = 4.0 / 5.0 = 0.80 m s⁻¹ right.
Δp = m(v − u) = 0.20(5.0 − (−5.0)) = 0.20 × 10 = 2.0 kg m s⁻¹.
Momentum is unchanged but the mass doubles, so the speed halves.
An inelastic collision conserves momentum but loses kinetic energy to heat, sound and permanent deformation.
Total momentum stays zero, so she recoils in the opposite direction to the thrown spanner.
Momentum is a vector, so it is conserved independently along each of two perpendicular axes.
Δp = m(v − u) = 0.10(20 − (−20)) = 0.10 × 40 = 4.0 kg m s⁻¹.
v = 6.0/3.0 = 2.0 m s⁻¹. KE before = ½×1.0×36 = 18 J; after = ½×3.0×4 = 6.0 J; lost = 12 J.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.