Generated ERQ

✓ passed A.3 Work, energy and power × B.5 Current and circuits 12 marks SL 3 passes 106.2s $0.7005

## ERQ · 12 marks · Topics: A.3 Work, energy and power + B.5 Current and circuits · Archetype: data_response **Integration:** primary=A.3 Work, energy and power, secondary=B.5 Current and circuits (strength: supporting) **Stem.** A student investigates the performance of a small electric motor used to lift a load vertically. The motor is connected to a 12.0 V battery and lifts a mass of 2.0 kg via a light inextensible string wound on a spindle. For each trial, the student measures the time *t* taken to lift the mass through a fixed height of *h* = 1.50 m at approximately constant speed, and records the average current *I* drawn from the battery during the lift. The resistance of the leads and the internal resistance of the battery may be neglected. Take *g* = 9.81 m s⁻². The recorded data are shown below. | Trial | Current *I* / A (±0.05) | Time *t* / s (±0.2) | |:-:|:-:|:-:| | 1 | 0.85 | 4.8 | | 2 | 1.10 | 3.9 | | 3 | 1.45 | 3.2 | | 4 | 1.95 | 2.6 | | 5 | 2.60 | 2.1 | The manufacturer claims the motor operates at an efficiency of at least 75% across this range of loads. ### Part (a) Define [2 marks] · AO1 · Topic: A.3 State what is meant by (i) the *work done* by a force, and (ii) the *efficiency* of an energy transfer. ### Part (b)(i) Calculate [2 marks] · AO2 · Topic: A.3+B.5 For Trial 3, calculate the electrical energy supplied by the battery to the motor during the lift. ### Part (b)(ii) Calculate [2 marks] · AO2 · Topic: A.3 For Trial 3, calculate the gravitational potential energy gained by the 2.0 kg mass. ### Part (c) Determine [3 marks] · AO3 · Topic: A.3+B.5 Using the data table, determine how the efficiency of the motor varies as the current drawn increases. Support your answer with at least two calculated efficiency values. ### Part (d) Evaluate [3 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: A.3 Evaluate the manufacturer's claim that the motor operates at an efficiency of at least 75% across this range of loads. --- ## Mark Scheme ### Part (a) [2 marks] — Define - M1: work done = force × displacement in the direction of the force (OR energy transferred when a force moves its point of application through a displacement) [no ECF] - M2: efficiency = useful energy (or power) output / total energy (or power) input [no ECF] ### Part (b)(i) [2 marks] — Calculate - M1: substitution *E* = *VIt* = 12.0 × 1.45 × 3.2 [no ECF] - M2: *E* = 55.7 J (accept 55–56 J) [no ECF] ### Part (b)(ii) [2 marks] — Calculate - M1: substitution *E*_p = *mgh* = 2.0 × 9.81 × 1.50 [no ECF] - M2: *E*_p = 29.4 J (accept 29 J) [no ECF] ### Part (c) [3 marks] — Determine (data-based trend with support) - M1: efficiency calculated for at least two trials using η = *mgh*/(*VIt*); e.g., Trial 1: η = 29.4/(12.0 × 0.85 × 4.8) = 0.601 (60%); Trial 3: η = 29.4/55.7 = 0.528 (53%); Trial 5: η = 29.4/(12.0 × 2.60 × 2.1) = 0.449 (45%) — award for any two correct values [ECF from (b)(i) and (b)(ii)] - M2: identifies trend — efficiency **decreases** as current increases [ECF from M1] - M3: links trend to physics — at higher currents the *I*²*R* (resistive/heat) dissipation in the motor windings grows faster than the useful mechanical output, so a smaller fraction of *VIt* appears as *mgh* [ECF from M1] ### Part (d) [3 marks] — Evaluate (position + support + limit, per §4.4.1) - M1: position — the manufacturer's claim is **not supported** by the data; the maximum efficiency obtained is approximately 60% (Trial 1), which is below 75% [ECF from (c)] - M2: supporting consideration — efficiency falls further as current increases, reaching only ~45% in Trial 5, so the claim of ≥75% fails across the *entire* range, not just at high load [ECF from (c)] - M3: limiting consideration — the stated uncertainties (±0.05 A in *I*, ±0.2 s in *t*) give a relative uncertainty of roughly ±6–10% in each efficiency value; even at the upper bound, Trial 1's efficiency (≈65%) remains clearly below 75%, so the conclusion is robust within experimental uncertainty ### Marker notes - Alternative method accepted for (c): comparing power ratios *P*_out/*P*_in = *mgh*/(*VIt* ) trial-by-trial, or plotting η vs *I* and citing the negative gradient. - (d) accept equivalent uncertainty-based limits: e.g., propagating ±0.05/0.85 ≈ 6% and ±0.2/4.8 ≈ 4% in Trial 1 gives η = 60% ± ~6%, upper bound ≈ 66% < 75%. - (d) also accept as a valid limiting consideration: only five data points sampled over a limited load range; the claim might hold outside this range, but cannot be supported within it. - Do NOT award (d) M3 for re-defining the system (e.g., "losses might be in the leads") since leads are stated to have negligible resistance.

Critic verdicts

Iteration 0 — failed

ao_ladder pass

The ERQ follows a strict cognitive ladder: (a) AO1 (Define), (b)(i)–(ii) AO2 (Calculate), (c) AO3 (Determine), (d) AO3 (Evaluate). Each tier uses appropriate command terms and increases in cognitive demand. The progression from definition → calculation → analysis → critical evaluation satisfies IB's cognitive load hierarchy.
command_structure fail

Part (c) mark scheme contains a critical error: M2 identifies Trial 1 as highest efficiency, but the subsequent note and M3 contradict this by stating the correct answer is Trial 5 (≈22%). This violates the fundamental requirement that a mark scheme must have a single, unambiguous correct answer. Additionally, part (d) lacks explicit ECF annotations on M2 and M3 as required by §4.6 for method marks depending on earlier calculations.

FIX: Part (c): Delete the incorrect sentence 'identifies Trial 1 as highest efficiency ✓' from M2. Replace M2 entirely with: 'M2: identifies Trial 5 as highest efficiency ✓ [ECF from M1]'. Remove or clarify the confusing parenthetical note so it unambiguously states that Trial 5 (η ≈ 21.5%) is the answer. Part (d): Add ECF annotations to M2 and M3 as '[ECF from (b)(i), (b)(ii), (c)]' since the limiting considerations in M3 reference the system efficiency calculated in (c). Alternatively, restructure M1–M3 to explicitly show: M1 = position statement; M2 = supporting consideration + justification; M3 = limiting consideration + justification, each with appropriate ECF flags.
cross_theme pass

Both A.3 (Work, energy and power) and B.5 (Current and circuits) are genuinely integrated throughout. The question requires students to calculate electrical power using P = IV (B.5), mechanical power using P = mgv (A.3), and then blend them to find efficiency η = P_useful/P_input in part (c), where both concepts are essential to the solution. Part (d) further demands critical evaluation of assumptions linking electrical input to mechanical output, reinforcing the authentic integration.
discriminator pass

Part (d) explicitly asks the student to evaluate a claim about motor efficiency by reference to assumptions made in the analysis. The mark scheme requires students to identify that the calculated value represents system efficiency (not motor efficiency alone) and to discuss at least one limiting factor or unaccounted loss—e.g., friction, air resistance, wire heating, or internal resistance. This is genuine model critique, not a calculation.
rule_of_one fail

Part (c) claims 3 marks but the mark scheme lists only 2 distinct award criteria (M2 identifies the trial; M3 states the percentage). M1 requires 'calculation of at least two trials' but this is a procedural step, not a separate criterion. The part should either be 2 marks or the scheme needs explicit separation of the calculation step from identification and statement of the result.

FIX: Either: (1) reduce part (c) to [2 marks] with M1 = identification of Trial 5 + efficiency percentage stated (combined); or (2) expand to [3 marks] and explicitly separate: M1 = formula setup (η = mgv/IV); M2 = calculation of at least two trials (numerical work); M3 = identification of Trial 5 and statement of ≈22%. The current scheme conflates procedural work into M1 without treating it as a distinct mark.
topic pass

The ERQ authentically integrates both topics. A.3 (Work, energy and power) is primary: parts (a), (b)(ii), and (c) require efficiency definition, mechanical power calculation (P = mgv), and efficiency determination. B.5 (Current and circuits) is genuinely necessary: parts (b)(i) and (c) require electrical power calculation (P = IV) from battery voltage and measured current, making current a non-superficial input. The scenario demands both domains—students cannot solve this without understanding power in mechanical and electrical forms and how to compare them.

Iteration 1 — failed

ao_ladder pass

The ERQ obeys the cognitive ladder correctly. Part (a) is AO1 with State command terms (definitions). Parts (b)(i)–(b)(ii) are AO2 with Calculate. Part (c) is AO3 with Determine (requires selection and justification). Part (d) is AO3 with Evaluate (requires judgment, supporting evidence, and limiting considerations). The progression from lower to higher order thinking is strict and appropriate for a 12-mark SL question.
command_structure fail

Part (d) fails the §4.4.1 AO3 'Evaluate' structural requirement. While M1 provides a position and M2 provides a supporting consideration, M3 is framed as a 'limiting consideration' but does not constitute a genuine limitation on the student's conclusion. Instead, it introduces speculative alternative mechanisms (motor-vs-system losses) that shift the burden of proof away from evaluating the data against the claim. A true limiting consideration must acknowledge a constraint on the analysis itself (e.g., uncertainty in measurements, incomplete sampling, assumptions made) or concede a valid counter-argument—not redefine the boundary of the system. The current M3 is a post-hoc escape clause rather than a critical evaluation limitation.

FIX: Rewrite Part (d) M3 to provide a genuine limiting consideration. Example: 'M3: limiting consideration — the measurement uncertainties (±0.05 A in current, ±0.2 s in time) introduce approximately ±5% relative error in the calculated efficiencies; within this uncertainty range, Trial 5's efficiency of 18% ± ~1% remains far below 75%, so the conclusion is robust.' Alternatively: 'M3: limiting consideration — the analysis assumes all electrical power drawn is converted within the motor system; if significant power is dissipated in battery internal resistance or leads (though stated to be neglected), the true motor input would be lower and efficiency estimates may be underestimated, but the claim would still not be supported.' This preserves the evaluate structure (position + support + genuine limit) without illegitimately widening the system boundary.
cross_theme pass

Both A.3 (Work, energy and power) and B.5 (Current and circuits) are genuinely integrated throughout the question. B.5 concepts (electrical power P=IV, current measurement) are essential to parts (b)(i), (c), and (d), not merely name-checked. The efficiency calculation fundamentally requires both electrical power input (B.5) and mechanical power output (A.3), making neither topic dispensable.
discriminator pass

Part (d) explicitly asks students to evaluate whether the data support a manufacturer's claim by comparing measured system efficiencies to a stated motor efficiency, and requires consideration of limitations (losses outside the motor itself). This is a genuine model-critique task that goes beyond calculation.
rule_of_one pass

Each part's mark count matches the number of distinct award criteria. Parts (a), (b)(i), and (b)(ii) each have 2 marks with 2 bullets; part (c) has 3 marks with 3 bullets (method, calculation, identification); part (d) has 3 marks with 3 bullets (position, support, limitation). All allocations are consistent with the stated mark values.
topic pass

The ERQ genuinely integrates both topics. Topic A.3 (work, energy, power) is the primary focus: parts (a), (b)(ii), and (c)–(d) require calculation and evaluation of mechanical power output, useful work, and efficiency. Topic B.5 (current and circuits) is substantively invoked in part (b)(i) and throughout (c)–(d): students must use electrical power P = IV (requiring current measurement and voltage) to compute system efficiency as the ratio of mechanical to electrical power. Neither topic is merely nominal; both are required to solve the problem coherently.

Iteration 2 — failed

ao_ladder pass

The ERQ correctly implements the cognitive ladder. Part (a) uses AO1 (Define); parts (b)(i) and (b)(ii) use AO2 (Calculate); part (c) uses AO3 (Determine with justification); part (d) uses AO3 (Evaluate with position, support, and limitation). Each tier is appropriately sequenced and command terms match cognitive demand.
command_structure fail

Part (c) fails the AO3 command-term structure requirement. 'Determine' with data-based support requires an explicit trend identification linked to physics mechanism, but M3 uses language ('I²R dissipation in motor windings grows faster than useful mechanical output') that attempts causal explanation without the required 'X therefore Y' explicit chain. More critically, part (d) 'Evaluate' violates §4.4.1: while M1 (position) and M2 (supporting consideration) are present, M3 (limiting consideration) frames uncertainty as a robustness check rather than as a genuine limit to the conclusion's validity. A proper limiting consideration should acknowledge what weakens or constrains the evaluation (e.g., 'however, we cannot test conditions outside this current range' or 'systematic errors in calibration could shift all values uniformly'), not argue that uncertainty does not undermine the conclusion.

FIX: For Part (c) M3: Rewrite to make the causal chain explicit: 'As current increases, resistive losses *I²R* in the motor coil increase proportionally to *I²*; since useful mechanical output *mgh* is constant, the fraction of input electrical energy *VIt* converted to useful work decreases, therefore efficiency η = *mgh*/*VIt* falls.' For Part (d) M3: Replace the uncertainty-robustness argument with a genuine limiting consideration such as: 'However, we have measured data only over a current range 0.85–2.60 A; the manufacturer's claim refers to the entire operating range, which may extend beyond these trials, so the claim cannot be fully evaluated outside the measured interval.' Alternatively: 'A limiting factor is that we assume all losses are dissipated as heat within the motor; if some energy is lost in the transmission system (spindle friction, string compliance), the true mechanical efficiency could be even lower, strengthening the conclusion, but unknown non-electrical losses introduce ambiguity.'
cross_theme pass

The question genuinely integrates both A.3 (Work, energy and power) and B.5 (Current and circuits). Part (b)(i) requires calculating electrical energy using V, I, and t from circuit theory; parts (b)(ii), (c), and (d) then use that electrical energy to compute efficiency as a ratio of gravitational potential energy gained to electrical energy supplied. The motor's efficiency emerges as the core physics concept linking mechanical work (lifting mass against gravity) to electrical power input (VIt), with the trend analysis in (c) and evaluation in (d) demanding understanding of both resistive losses in circuits (I²R dissipation) and energy conservation. Neither topic is name-checked as filler; both are essential to solving the problem.
discriminator pass

Part (d) explicitly asks the student to 'Evaluate the manufacturer's claim,' which requires critique of the model/claim against experimental data. The mark scheme confirms this is a model-evaluation question: students must assess whether the claim (≥75% efficiency) is supported, identify where it fails, and address experimental uncertainty as a limiting consideration. This is unambiguously a critique dimension, not just a calculation.
rule_of_one fail

Part (c) is marked out of 3 but lists 3 award criteria (M1: two efficiency calculations, M2: trend identification, M3: physics link). However, M1 is composite—it requires calculation of efficiencies for "at least two trials" using the formula η = mgh/(VIt). This is fundamentally a single operation (apply formula to multiple data points), not two distinct operations. The mark scheme conflates "number of trials calculated" with "distinct conceptual/procedural steps." A 3-mark Calculate part should have three distinct operations (formula, substitution, answer), not one operation applied twice plus two conceptual steps. Part (c) is labeled "Determine" (AO3), not "Calculate," yet M1 is structured as a calculation. The architecture is misaligned with the dimension's strict requirement.

FIX: Restructure part (c) to separate calculation from analysis: either (1) split into two parts—(c) as a 2-mark Calculate (formula + two substitutions → two η values), and (c)(ii) as a 1-mark Explain (trend + physics); OR (2) reframe (c) as a 3-mark Determine with M1 = identify formula for efficiency, M2 = calculate at least two values (single award), M3 = state trend and explain physics. Current scheme incorrectly bundles two calculations into one mark (M1).
topic pass

The ERQ successfully integrates both topics. A.3 (Work, energy and power) is central throughout: parts (a), (b)(ii), (c), and (d) all require explicit use of gravitational PE, work, and efficiency concepts. B.5 (Current and circuits) is genuinely required in parts (b)(i) and (c): calculating electrical energy supplied demands E = VIt (voltage, current, time), and efficiency analysis requires understanding power dissipation in motor windings (I²R losses) as current varies. The scenario is physically coherent—a motor lifting a mass requires both electrical energy input (B.5) and mechanical energy output (A.3) to be quantified and compared. Neither topic is superficial; both are operationally essential to solve the problem.

Planner output (stage 1)

Archetype: real-world device
Scenario: A small electric motor is used to lift a 2.0 kg mass vertically. The motor is powered by a 12 V battery and draws a current that varies as the load increases. Students measure the time taken to raise the mass through different heights and record the current drawn at each stage. The scenario explores the relationship between electrical power input, mechanical work output, and energy efficiency in a practical system.
Cross-pollination part: b(i)
Discriminator part: d

Part	Marks	AO	Command
a	2	AO1	Define
b(i)	2	AO2	Calculate
b(ii)	2	AO2	Calculate
c	3	AO3	Determine
d	3	AO3	Evaluate

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