Physics • Year 11 • Module 2 • Lesson 13
Momentum Synthesis
Build HSC Band 5–6 extended-response technique on multi-stage problems, collision classification, and cross-phase connections.
1. Full quantitative chain — car crash investigation (Band 5–6)
8 marks Band 5–6
Scenario. A 1400 kg car travelling at +18 m/s collides with a stationary 900 kg car. They lock together (perfectly inelastic collision). The combined wreck skids on the road surface (μk = 0.45, g = 9.8 m/s²) before coming to rest. A forensic engineer measures the skid marks and needs to verify the initial speed.
Q1. Solve the full chain and answer all parts. Show working for every step with correct significant figures.
- (a) Define the positive direction and state the total momentum of the system before the collision. Show the calculation. (1 mark)
- (b) Calculate the post-collision velocity of the combined wreck. (2 marks)
- (c) Calculate the kinetic energy of the combined wreck immediately after the collision. (1 mark)
- (d) Calculate the friction force on the combined wreck and hence find the skid distance. (2 marks)
- (e) The engineer measures skid marks of 14 m. Is the initial speed of 18 m/s consistent with this evidence? Explain any discrepancy. (2 marks)
Working:
2. Source critique — spot the physics error (Band 5)
6 marks Band 5
“When a small bullet embeds in a heavy block, most of the bullet’s momentum is lost in the collision. The block moves slowly because momentum is not conserved in perfectly inelastic collisions — only kinetic energy is. This is why the block slides only a short distance: it starts with less momentum than the bullet had, so it has less energy to dissipate against friction.”
— Student textbook summary, Year 11 Physics revision notes.
2.1 Identify two distinct physics errors in the passage. For each, state exactly what is wrong and write the correct physics. 4 marks (2 per error)
Error 1: What is wrong?
Correction:
Error 2: What is wrong?
Correction:
2.2 Despite the errors, the passage arrives at a broadly correct observation (the block slides only a short distance). Explain the correct physics reason for why the block’s slide distance is short in a bullet-block scenario. 2 marks
3. Evaluate a scenario — comparing two vehicle safety systems (Band 5–6)
7 marks Band 5–6
Scenario. Engineers are comparing two vehicle safety systems. System A (rigid bumper): the collision lasts ΔtA = 0.05 s. System B (crumple zone): the collision lasts ΔtB = 0.20 s. In both cases a 1000 kg car travelling at +15 m/s collides with a rigid wall and comes to rest.
| Measurement | System A (rigid bumper) | System B (crumple zone) |
|---|---|---|
| Collision time Δt (s) | 0.05 | 0.20 |
| Change in momentum Δp (N s) | ||
| Average force on car F (N) | ||
| Impulse J (N s) |
Illustrative data. Assume both cars start at 15 m/s and come to rest. g = 9.8 m/s².
Q3. Evaluate which safety system better protects vehicle occupants. In your response you must:
- Complete the table above and use it to compare the forces in both systems.
- Explain why the change in momentum and impulse are the same for both systems, even though the forces are different.
- Identify which system produces the smaller force, state the ratio of forces, and link this to occupant safety.
- Discuss one real-world limitation of extending Δt without bound (i.e., why crumple zones cannot be made infinitely long).
- Reach an explicit, evidence-based judgement about which system is safer and why.
Q1 — Full quantitative chain (8 marks)
(a) Positive direction: direction of initial car’s motion = positive. ptotal,before = 1400×18 + 900×0 = 25 200 N s [1].
(b) Perfectly inelastic: (1400+900)vf = 25 200. vf = 25 200/2300 = +10.96 m/s ≈ 11.0 m/s [2: 1 for equation, 1 for answer].
(c) KEf = ½×2300×10.96² = ½×2300×120.1 = 138 100 J ≈ 138 kJ [1].
(d) fk = μk(m1+m2)g = 0.45×2300×9.8 = 10 143 N. Wnet = ΔKE: −10 143×s = −138 100. s = 138 100/10 143 = 13.6 m ≈ 13.6 m [2: 1 for fk, 1 for s].
(e) Calculated skid distance 13.6 m; measured 14 m. These are consistent to within 3%, which lies within normal experimental uncertainty (surface roughness variability, measurement error in skid mark length) [1]. The initial speed of 18 m/s is consistent with the physical evidence [1]. Accept answers that correctly identify the calculated value is close to 14 m and provide a physical justification for the small discrepancy.
Q2 — Source critique (6 marks)
Error 1: The passage states that “momentum is not conserved in perfectly inelastic collisions — only kinetic energy is.” This is the exact opposite of the truth [1]. Correction: In a perfectly inelastic collision, momentum is conserved and kinetic energy is not fully conserved. The total momentum before equals the total momentum after; kinetic energy decreases because energy is converted to heat, sound and deformation [1].
Error 2: The passage states that “the block starts with less momentum than the bullet had.” This is false [1]. Correction: By conservation of momentum, the momentum of the combined block-plus-bullet immediately after the collision equals the bullet’s momentum before the collision. No momentum is lost in the collision — it is fully transferred to the combined system [1].
2.2: The block slides only a short distance because the perfectly inelastic collision converts most of the bullet’s kinetic energy (KE = ½mv²) to heat and deformation [1]. The post-collision KE is much smaller than the pre-collision KE because the combined mass is large but vf is small (since KE = p²/2m, and the momentum is the same but the mass is much larger). Only this small residual KE is available to be dissipated by friction over the slide distance [1].
Q3 — Vehicle safety evaluation (7 marks)
Table completed: Δp = mΔv = 1000×(0−15) = −15 000 N s (same for both). J = Δp = −15 000 N s (same for both). FA = Δp/ΔtA = 15 000/0.05 = 300 000 N. FB = Δp/ΔtB = 15 000/0.20 = 75 000 N.
Marking criteria (7 marks):
1. Table completed correctly: Δp and J the same for both systems; FA = 300 000 N; FB = 75 000 N. [1]
2. Correctly explains that Δp and J are the same because the car starts at the same speed and stops in both cases; the impulse required to change the momentum is fixed by the initial and final velocities, independent of the collision time. [1]
3. States System B produces the smaller force (75 000 N vs 300 000 N; ratio 4:1) because the longer collision time allows the same momentum change to be achieved with a smaller average force, as J = FΔt. [1]
4. Links smaller force to occupant safety: the 300 000 N force in System A far exceeds the human tolerance limit (∼35 000 N for a 70 kg person restrained by a seatbelt, approximately 50 g); occupants are at high risk of serious injury. System B’s 75 000 N is also large but represents a fourfold improvement. [1]
5. Real-world limitation of extending Δt: the crumple zone has a finite length; a longer zone requires a longer car, adding weight, cost and reducing cabin space. Additionally, the crumple zone must not intrude into the passenger cell. Beyond a certain depth, the structure collapses into the occupant space. [1]
6. Explicit evidence-based judgement: System B (crumple zone) is the safer design because it reduces average collision force by a factor of 4, directly reducing the risk of fatal injury to occupants. The physics of impulse-momentum (J = FΔt = Δp) dictates that extending collision time is the only way to reduce force when momentum change is fixed. [1]
7. Quality of physics reasoning: uses correct terminology (impulse, momentum, force, collision time) consistently and integrates the impulse-momentum theorem across both systems. [1]