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## ERQ · 12 marks · Topics: B.1 Thermal energy transfers + A.2 Forces and momentum · Archetype: theory_application
**Integration:** primary=B.1 Thermal energy transfers, secondary=A.2 Forces and momentum (strength: supporting)
**Stem.** A railway buffer car at the end of a siding uses a hydraulic spring-damper system to bring runaway freight wagons to rest. A wagon of mass 4.2 × 10⁴ kg rolls into the stationary buffer car at 1.8 m s⁻¹ along a level track. During the collision the spring compresses by a maximum of 0.70 m, and the wagon and buffer momentarily move together before the spring partially rebounds. The damping fluid inside the cylinder has a total mass of 25 kg and a specific heat capacity of 2100 J kg⁻¹ K⁻¹. Engineers model the collision by assuming that all of the wagon's kinetic energy that is not stored elastically is transferred as internal energy to the damping fluid, and that no thermal energy is lost to the cylinder casing or surroundings during the 0.78 s contact time.
### Part (a) Define [2 marks] · AO1 · Topic: B.1
Define *specific heat capacity* and state its SI unit.
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: A.2
The buffer car has a mass of 8.0 × 10³ kg and is initially at rest. Using conservation of momentum, calculate the common speed of the wagon and buffer car at the instant of maximum spring compression.
### Part (b)(ii) Show that [3 marks] · AO2 · Topic: B.1
Using your answer to (b)(i), show that the temperature rise of the damping fluid is about 0.21 K, assuming the engineers' model holds.
### Part (c) Suggest [2 marks] · AO3 · Topic: B.1+A.2
Measurements show that after the collision the buffer car rebounds with the wagon at 0.25 m s⁻¹. Suggest, with reference to energy partitioning, what fraction of the wagon's initial kinetic energy was actually converted to internal energy in the damping fluid.
### Part (d) Evaluate [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
Evaluate the engineers' assumption that no thermal energy is lost to the cylinder casing or surroundings during the collision.
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## Mark Scheme
### Part (a) [2 marks] — Define
- M1: energy required to raise the temperature of unit mass (of a substance) by one kelvin (or 1 °C) [no ECF]
- M2: SI unit J kg⁻¹ K⁻¹ [no ECF]
### Part (b)(i) [3 marks] — Calculate
- M1: applies conservation of momentum: (4.2 × 10⁴)(1.8) = (4.2 × 10⁴ + 8.0 × 10³)v [no ECF]
- M2: total mass = 5.0 × 10⁴ kg AND total initial momentum = 7.56 × 10⁴ kg m s⁻¹ [no ECF]
- M3: v = 1.51 m s⁻¹ (accept 1.5 m s⁻¹) [no ECF]
### Part (b)(ii) [3 marks] — Show that
- M1: energy converted to internal energy = ½(4.2 × 10⁴)(1.8)² − ½(5.0 × 10⁴)(1.51)² [ECF from (b)(i)]
- M2: ΔE = 6.80 × 10⁴ − 5.70 × 10⁴ ≈ 1.10 × 10⁴ J [ECF from (b)(i)]
- M3: ΔT = Q/(mc) = 1.10 × 10⁴ / (25 × 2100) = 0.209 K ≈ 0.21 K [ECF from (b)(i)]
### Part (c) [2 marks] — Suggest (proposal + physics-based justification per §4.4.1)
- M1: proposal with mechanism — most (≈ 86%) of the initial KE was converted to internal energy in the damping fluid, since after rebound the system retains only ½(5.0 × 10⁴)(0.25)² ≈ 1.6 × 10³ J of the initial ½(4.2 × 10⁴)(1.8)² ≈ 6.8 × 10⁴ J of KE
- M2: justification invoking work–energy theorem — the "missing" KE (≈ 6.6 × 10⁴ J, or ≈ 97% of initial KE) equals the net work done by the (non-conservative) damping force during compression and rebound, and this work appears as internal energy of the fluid
### Part (d) [2 marks] — Evaluate (position + supporting + limiting per §4.4.1)
- M1: position + supporting consideration — the assumption is reasonable over the brief 0.78 s contact time AND because the contact time is much shorter than the timescale for significant conductive heat transfer through the steel cylinder casing, so most of the thermal energy remains in the fluid during the collision
- M2: limiting consideration — however, some energy is inevitably dissipated as sound, vibration of the rails, or plastic deformation of couplings, AND some heat will conduct into the casing especially given the large surface area of the cylinder, so ΔT is likely an over-estimate
### Marker notes
- Alternative method accepted for (b)(ii): compute elastic + kinetic energy at maximum compression as ½(m₁+m₂)v² and subtract from initial KE; both routes give ≈ 1.1 × 10⁴ J
- (c) accept any clearly reasoned fraction between 0.95 and 0.98 with supporting numerical calculation
- (d) accept any of: conduction to casing, radiation losses, sound emission, vibration of rails/track bed, plastic deformation of couplings, heating of brake pads
- Show-that target in (b)(ii): 0.209 K given to 3 sig figs; student-derived 0.20–0.22 K acceptable
- ECF chain: error in (b)(i) propagates to (b)(ii) and (c); award method marks throughout
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