Generated ERQ

✓ passed C.1 Simple harmonic motion × A.1 Kinematics 10 marks SL 3 passes 85.48s $0.6256
``` ## ERQ · 10 marks · Topics: C.1 Simple harmonic motion + A.1 Kinematics · Archetype: theory_application **Integration:** primary=C.1 Simple harmonic motion, secondary=A.1 Kinematics (strength: supporting) **Stem.** A student investigates the motion of a 0.250 kg trolley attached to a horizontal spring on a frictionless air-track. The trolley is pulled 8.00 cm to the right of its equilibrium position and released from rest at time *t* = 0. A motion sensor records the position *x* of the trolley as a function of time. The recorded data show that the trolley completes 5 full oscillations in 2.50 s. The student then tries to apply the kinematic (SUVAT) equations to predict the trolley's maximum speed, using the measured initial acceleration as if it were constant throughout the motion. ### Part (a) State [1 mark] · AO1 · Topic: C.1 State the period of oscillation of the trolley. ### Part (b)(i) Show that [2 marks] · AO2 · Topic: C.1 Show that the angular frequency of the oscillation is about 12.6 rad s⁻¹. ### Part (b)(ii) Calculate [2 marks] · AO2 · Topic: C.1 Calculate the maximum speed of the trolley. ### Part (c) Determine [2 marks] · AO2 · Topic: C.1+A.1 Determine the magnitude of the trolley's acceleration at the instant of release. ### Part (d) Discuss [3 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: C.1+A.1 The student argues: "Since I now know the initial acceleration from (c), I can simply use the SUVAT equation *v*² = *u*² + 2*as* with *s* = 0.0800 m and *u* = 0 to find the maximum speed at the equilibrium position." Discuss whether this approach is valid. --- ## Mark Scheme ### Part (a) [1 mark] - M1: *T* = 2.50/5 = 0.500 s [no ECF] ### Part (b)(i) [2 marks] — Show that - M1: Uses ω = 2π/*T* with *T* = 0.500 s, i.e. ω = 2π/0.500 [ECF from (a)] - M2: ω = 12.566… rad s⁻¹ ≈ 12.6 rad s⁻¹ (shown to ≥3 s.f.) [ECF from (a)] ### Part (b)(ii) [2 marks] — Calculate - M1: Uses *v*_max = ω*x*₀ with *x*₀ = 0.0800 m and ω = 12.6 rad s⁻¹ [ECF from (b)(i)] - M2: *v*_max = 1.01 m s⁻¹ (accept 1.00–1.02 m s⁻¹) [ECF from (b)(i)] ### Part (c) [2 marks] — Determine - M1: Uses *a* = ω²*x*₀ (or *a*_max at extreme of motion) with substitution [ECF from (b)(i)] - M2: *a* = (12.6)² × 0.0800 = 12.7 m s⁻² (accept 12.6–12.7 m s⁻²) [ECF from (b)(i)] ### Part (d) [3 marks] — Discuss (balanced trade-off per §4.4.1) - M1 (perspective 1): SUVAT is superficially attractive — it is computationally simple and directly reuses the kinematics framework with the measured initial acceleration from (c), giving *v* = √(2 × 12.7 × 0.0800) ≈ 1.43 m s⁻¹ - M2 (perspective 2 — contrast): However, SUVAT assumes *uniform* acceleration, whereas in SHM the restoring force (and hence acceleration) varies continuously with displacement, decreasing to zero at the equilibrium position — so the assumption is fundamentally violated - M3 (judgement resolving trade-off): The SUVAT estimate (≈1.43 m s⁻¹) overestimates the true *v*_max (≈1.01 m s⁻¹ from (b)(ii)) by ~40%, so physical validity must outweigh computational simplicity and only the SHM model is correct here [ECF from (b)(ii), (c)] ### Marker notes - Alternative method for (b)(ii): energy conservation ½*kx*₀² = ½*mv*² with *k* = *m*ω² gives same result - Alternative method for (c): *F* = *kx* with *k* = *m*ω² = 39.6 N m⁻¹, then *a* = *F*/*m* - Show-that target in (b)(i): ω = 12.566 rad s⁻¹ given to 3 s.f.; student-derived 12.5–12.6 rad s⁻¹ acceptable - (d) accept equivalent reasoning: e.g. noting that average acceleration over the quarter-cycle is *a*_max(2/π) ≈ 0.637·*a*_max, which when used in SUVAT recovers the SHM result; this counts as a valid second perspective - ECF in (d) M3: if student's (b)(ii) and (c) values differ, accept their numerical comparison provided the SUVAT estimate exceeds the SHM estimate ```