Physics • Year 11 • Module 2: Dynamics • Lesson 14
Phase 3 Consolidation
Build HSC Band 5–6 extended-response technique on two-stage momentum–energy problems, source critique, and the evaluation of real-world safety applications.
1. Two-stage problem: car crash → skid (Band 5–6)
8 marks Band 5–6
Scenario. A 900 kg car travelling at +24 m/s collides with a stationary 1600 kg truck. They lock together and skid to rest. A 65 kg passenger was travelling at the same speed as the car. With the seatbelt, the passenger stops over 0.30 s. Without the seatbelt, the estimated stopping time is 0.025 s. Take g = 9.8 m/s².
Q1. Work through the following steps, showing all working and defining your positive direction at the start.
- (a) Find the velocity of the combined car+truck immediately after the collision using conservation of momentum. (2 marks)
- (b) Calculate the kinetic energy of the combined mass just after the collision. Calculate the kinetic energy lost in the collision. (2 marks)
- (c) Using Wnet = ΔKE, find the coefficient of kinetic friction (μk) between the tyres and the road if the vehicles skid 12.5 m to rest. (2 marks)
- (d) Calculate the average force on the passenger with and without a seatbelt. Express both in kN, and explain the safety significance using J = FΔt = Δp. (2 marks)
2. Source critique — a journalist’s report on a train collision (Band 5–6)
7 marks Band 5–6
Source. “In this morning’s train collision in Sydney, a 320-tonne locomotive travelling at 40 km/h struck a stationary 80-tonne goods wagon. Because the momentum of the locomotive was so large and the wagon was small, momentum was clearly not conserved — the locomotive kept most of its momentum while the wagon was destroyed. The impact force on the wagon was smaller than the impact force on the locomotive because the locomotive is much heavier and harder to stop.”
Fictional media article for assessment purposes. Illustrative values.
Q2. This passage contains two factual physics errors. Identify each error, explain the correct physics using precise terminology, and, where possible, use a calculation to illustrate the correct result. In your response you must:
- Identify Error 1 and explain what the correct principle states.
- Identify Error 2 and explain what Newton’s Third Law actually implies about the forces during the collision.
- Use numbers from the scenario (convert km/h to m/s first) to show that momentum is conserved and to illustrate the equal-and-opposite force relationship.
- Explain one way a journalist could correctly report on what actually happened.
Q1 — Sample Band 6 response (8 marks), annotated
Positive direction: direction of the car’s initial motion (+).
(a) Post-collision velocity [2 marks]
Conservation of momentum in a closed system (negligible external forces during short collision):
Σpbefore = 900×(+24) + 1600×0 = 21 600 N s
(900 + 1600)vf = 21 600 → 2500vf = 21 600 → vf = +8.64 m/s
[1 mark: correct conservation equation and setup; 1 mark: correct vf = 8.64 m/s]
(b) Kinetic energy before, after, and lost [2 marks]
KEbefore = ½×900×24² = ½×900×576 = 259 200 J
KEafter = ½×2500×8.64² = ½×2500×74.65 = 93 312 J
KElost = 259 200 − 93 312 = 165 888 J ≈ 166 kJ (64% of initial KE lost)
[1 mark: both KE values correct; 1 mark: correct KElost]
(c) Coefficient of kinetic friction [2 marks]
Wnet = ΔKE: −fk×12.5 = 0 − 93 312 → fk = 93 312/12.5 = 7464.96 N
μk = fk/(mg) = 7465/(2500×9.8) = 7465/24 500 = 0.305 ≈ 0.30
[1 mark: correct Wnet = ΔKE equation with signs; 1 mark: correct μk]
(d) Force on passenger with and without seatbelt [2 marks]
Δppassenger = 65×(0 − 8.64) = −561.6 N s (magnitude 561.6 N s)
With seatbelt: F = 561.6/0.30 = 1872 N ≈ 1.9 kN
Without seatbelt: F = 561.6/0.025 = 22 464 N ≈ 22.5 kN
The impulse (Δp = 561.6 N s) is identical in both cases — the seatbelt does not change the total momentum change required to stop the passenger. From J = FΔt, the seatbelt increases Δt by a factor of 12 (from 25 ms to 300 ms), which reduces the peak force by the same factor. The 22.5 kN force without a seatbelt is equivalent to roughly 2.3 tonnes pressing on the passenger — far exceeding the tolerances of the human body and likely fatal.
[1 mark: correct forces for both scenarios; 1 mark: clear qualitative explanation using J = FΔt = Δp with specific values]
Marking criteria summary (8 marks): 1 = correct conservation setup (closed system condition stated); 1 = vf = 8.64 m/s; 1 = both KE values correct; 1 = KElost correct; 1 = Wnet = ΔKE with correct signs; 1 = μk ≈ 0.30; 1 = both force values correct; 1 = safety explanation using J = FΔt = Δp with quantitative comparison.
Q2 — Sample Band 6 response (7 marks), annotated
Error 1: “Momentum was not conserved.”
This is incorrect. The Law of Conservation of Momentum states that in a closed system (no net external force), the total momentum before any interaction equals the total momentum after, regardless of the relative masses of the objects. The collision between the locomotive and wagon is exactly the situation where conservation applies — during the very short collision, external forces (friction, air resistance) produce negligible impulse compared to the collision force. [1 mark — identifies Error 1 correctly]
Numerical illustration: Convert 40 km/h = 40/3.6 = 11.1 m/s. pbefore = 320 000×11.1 + 0 = 3 552 000 kg m/s. After the collision (if they stick): vf = 3 552 000/(320 000 + 80 000) = 8.88 m/s. pafter = 400 000×8.88 = 3 552 000 kg m/s. Momentum is conserved. [1 mark — uses numbers to demonstrate conservation; 1 mark — states and applies closed-system condition]
Error 2: “The impact force on the wagon was smaller than on the locomotive.”
This violates Newton’s Third Law, which states that for every action there is an equal and opposite reaction. The force that the locomotive exerts on the wagon is equal in magnitude and opposite in direction to the force the wagon exerts on the locomotive. Both forces act for the same contact time (Δt), so both objects receive equal and opposite impulses. [1 mark — identifies Error 2; 1 mark — correctly states Newton’s Third Law as the relevant principle]
Why the outcomes differ: The forces are equal, but the effects differ because the masses differ. The wagon (80 000 kg) experiences the same force as the locomotive (320 000 kg) but has four times less mass, so it undergoes four times the acceleration (F = ma). This is why the wagon “appears” to be affected more — not because the force is different, but because the same force produces much greater acceleration in the lighter object. [1 mark — explains why acceleration differs despite equal forces using F = ma]
Correct journalistic framing: “The locomotive and wagon exerted equal and opposite forces on each other during the collision. Total momentum was conserved: the locomotive slowed significantly while the wagon accelerated rapidly because the same force acted on a much smaller mass. The collision was inelastic — significant kinetic energy was converted to heat, sound, and structural deformation.” [1 mark — provides a physically accurate restatement]
Marking criteria summary (7 marks): 1 = identifies Error 1 (momentum was conserved); 1 = numerical demonstration of momentum conservation; 1 = states closed-system condition correctly; 1 = identifies Error 2 (forces are equal, Newton 3); 1 = correctly applies Newton’s Third Law; 1 = explains why wagon accelerates more using F = ma; 1 = provides a correct journalistic restatement using precise physical terminology.