Physics • Year 11 • Module 2 • Lesson 13
Momentum Synthesis
Lock in the vocabulary, formula set and stage-identification skills before tackling multi-step problems.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: elastic collision, inelastic collision, perfectly inelastic collision, explosion from rest, impulse-momentum theorem, conservation of momentum, vector protocol, work-energy theorem, kinetic energy, Phase 2 bridge. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A collision in which both kinetic energy and momentum are conserved; objects bounce apart. | |
| 1.2 | A collision in which momentum is conserved but total kinetic energy decreases. | |
| 1.3 | A collision in which objects stick together and move as one; maximum possible kinetic energy is lost. | |
| 1.4 | An interaction that starts from rest; total momentum is zero both before and after, so the two parts move in opposite directions with equal and opposite momenta. | |
| 1.5 | The principle that the net impulse on an object equals its change in momentum: J = FΔt = Δp. | |
| 1.6 | The principle that in a closed system with no net external force, total momentum does not change. | |
| 1.7 | The procedure of choosing and stating a positive direction before writing any momentum equation; then assigning signs consistently to all velocities. | |
| 1.8 | The principle that net work done on an object equals its change in kinetic energy: Wnet = ΔKE. | |
| 1.9 | The energy a moving object possesses due to its motion: KE = ½mv². | |
| 1.10 | The technique of using the post-collision velocity from momentum conservation to calculate a subsequent sliding distance or height using Wnet = ΔKE. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 In a perfectly inelastic collision, momentum is conserved and kinetic energy is also fully conserved. T / F
2.2 When a rifle is fired from rest, the total momentum of the rifle-bullet system after firing is zero. T / F
2.3 In a multi-stage problem, the velocity calculated at the end of Stage 1 is not needed for Stage 2; each stage is solved independently from scratch. T / F
2.4 The work-energy theorem Wnet = ΔKE belongs to Phase 2 (dynamics/energy) and is used to find the sliding distance after a collision. T / F
2.5 A heavier object always has greater momentum than a lighter object. T / F
2.6 Impulse has the same units as momentum (kg m/s), so the impulse-momentum theorem is dimensionally consistent. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 9 marks (1 per blank)
Word bank:
chain · conserved · explosion · friction · impulse · inelastic · momentum · positive · velocity
In a multi-step momentum problem, the output of each stage becomes the input for the next, forming a calculation ___________. The first step is always to define a ___________ direction so that every ___________ is assigned the correct sign. When a system starts from rest, such as a rifle firing a bullet, the interaction is called an ___________ from rest; total ___________ remains zero both before and after. When the bullet strikes a stationary block and embeds, the collision is perfectly ___________; momentum is still ___________ but kinetic energy is not. After the collision, the block slides to rest against ___________; this stage uses the Phase 2 work-energy theorem because ___________ (force × time) is no longer the relevant tool.
4. Function recall
Answer each question in 1–2 sentences using precise physics terms. 8 marks (2 each)
4.1 What is the function of the “vector protocol” in multi-step momentum problems?
4.2 Why does the bullet-block collision destroy most of the system’s kinetic energy even though momentum is conserved?
4.3 What connects the output of Stage 2 (perfectly inelastic collision) to Stage 3 (sliding block) in the Phase 2–3 bridge?
4.4 In an explosion from rest, why must the two pieces always travel in opposite directions?
5. Connect the formula chain
Draw labelled arrows between the six terms below to show how they connect in a multi-step momentum problem. Each arrow must carry a linking phrase (e.g. “feeds into”, “applies when”, “equals”, “gives the input for”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: explosion from rest · conservation of momentum · perfectly inelastic collision · post-collision velocity · Wnet = ΔKE · slide distance.
Q1 — Term–definition match
1.1 elastic collision • 1.2 inelastic collision • 1.3 perfectly inelastic collision • 1.4 explosion from rest • 1.5 impulse-momentum theorem • 1.6 conservation of momentum • 1.7 vector protocol • 1.8 work-energy theorem • 1.9 kinetic energy • 1.10 Phase 2 bridge.
Q2 — True / false with correction
2.1 False. In a perfectly inelastic collision, momentum is conserved but kinetic energy is not conserved — kinetic energy decreases, with the lost energy converted to heat, sound and deformation.
2.2 True. Conservation of momentum: total p before = 0 (at rest) = total p after. Rifle and bullet carry equal and opposite momenta so they sum to zero.
2.3 False. The output of each stage (especially the velocity) is the input for the next stage. This is the essential feature of multi-stage chain problems.
2.4 True. The work-energy theorem belongs to Phase 2 dynamics and is the correct tool for the sliding stage after a collision.
2.5 False. Momentum depends on both mass and velocity (p = mv). A lighter, faster object can have greater momentum than a heavier, slower one.
2.6 True. Units of impulse: N s = kg m/s² × s = kg m/s. Units of momentum: kg m/s. Dimensionally identical, confirming J = Δp is valid.
Q3 — Cloze paragraph
In order: chain / positive / velocity / explosion / momentum / inelastic / conserved / friction / impulse.
Q4.1 — Function of the vector protocol
The vector protocol requires the student to choose and state a positive direction before writing any equation. This ensures all velocities are assigned the correct sign, preventing sign errors that invalidate momentum calculations (especially in explosions and collisions where objects move in opposite directions).
Q4.2 — Why the bullet-block collision destroys most KE
In a perfectly inelastic collision, the bullet (small mass, very high speed) embeds in the block (large mass, stationary). The final combined mass is much larger, so it moves at a much lower speed (from conservation of momentum). KE = ½mv², so the large drop in v² means most of the original KE is converted to heat and deformation during the collision, even though momentum is fully conserved.
Q4.3 — Phase 2–3 bridge connection
The post-collision velocity vf from Stage 2 (perfectly inelastic) is used to calculate the kinetic energy of the combined mass: KEf = ½(m1+m2)vf². This KE is then dissipated by friction in Stage 3 using Wnet = ΔKE: −fk × s = 0 − KEf. Velocity is the bridge quantity that passes information from the momentum world (Phase 3) to the energy world (Phase 2).
Q4.4 — Why explosion pieces travel in opposite directions
The system starts from rest, so total p = 0. After the explosion, conservation requires that the two momenta sum to zero: m1v1′ + m2v2′ = 0. Therefore v1′ = −(m2/m1)v2′. The negative sign means the two velocities are always in opposite directions.
Q5 — Sample concept map
Correct maps should include arrows such as:
- explosion from rest — is a special case of → conservation of momentum
- perfectly inelastic collision — applies → conservation of momentum
- conservation of momentum — gives → post-collision velocity
- post-collision velocity — feeds into → Wnet = ΔKE
- Wnet = ΔKE — gives → slide distance
- perfectly inelastic collision — outputs → post-collision velocity
Award 1 mark per valid labelled arrow (minimum 6, maximum 6 marked).