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QCE Physics IA1 data test technique: The 2026 guide

Study guide15 min read

Quick answer

QCE Physics IA1 Data Test is a 60-90 minute school-supervised response to unseen Unit 3 stimulus (graphs, tables, apparatus diagrams), testing calculation, reasoning and claim-and-justify with three significant figures, units throughout, and explicit show-working that allows method marks even when arithmetic slips; the assessment is worth 10 percent of the subject result and emphasises claim-evidence-reasoning under time pressure.

Jump to a section
  1. What this guide is for
  2. Format
  3. Common stimulus types
  4. Question types
  5. Show working
  6. Marking criteria
  7. Time pressure techniques
  8. Check your knowledge

What this guide is for

The QCE Physics IA1 data test is the first major Unit 3-4 assessment. This guide covers the format, marking criteria, common stimulus types, and the technique that produces strong responses under time pressure.

Format

  • Duration. 60-90 minutes (set by school, within QCAA guidelines).
  • Stimulus. Previously unseen data set (or sets) from Unit 3.
  • Worth. 10 percent of subject result.
  • When. Term 1 or early Term 2 of Year 12.

Common stimulus types

Motion graphs
Position-time, velocity-time, acceleration-time graphs. Slope and area carry physical meaning (velocity from slope of x-t; displacement from area under v-t).
Orbital data
Tables of planetary orbital periods and radii, requiring Kepler's third law to verify or extract a value.
Force-extension graphs
For springs (Hooke's law) or strings (tension).
Voltage-current graphs
Resistance from slope.
Banked-curve scenarios
Diagrams of cars on inclined curves.
Parallel-plate setups
Electron deflection in uniform electric field.
Velocity-selector or mass-spectrometer geometry
Crossed E and B fields, perpendicular forces.
Induction setups
Coil-and-magnet, transformer characteristics.

Question types

Calculation

"From the data in Table 1, calculate the centripetal force on the car on the banked curve."

Approach:

  1. Identify the relevant physics (centripetal force on banked curve).
  2. Write the formula symbolically.
  3. Extract values from stimulus with correct units.
  4. Substitute and calculate.
  5. State answer with units and sig fig.

Reasoning

"Explain why the transformer becomes less efficient under load X."

Approach:

  1. Identify the physics (transformer losses: copper, iron core, hysteresis).
  2. State the principle.
  3. Apply to the specific scenario in the stimulus.
  4. Argue cause and effect.

Claim and justify

"Make a claim about the relationship between the variables in the data, and justify it."

Approach:

  1. State a quantitative claim ("the period is proportional to the square root of length").
  2. Justify with the data (gradient of linearised graph).
  3. Justify with theory (the formula T=2πL/gT = 2\pi \sqrt{L/g}).
  4. Acknowledge limits (uncertainty, range of data).
Pendulum period squared versus length linearisation A linear plot with horizontal axis L from 0 to 1.5 metres and vertical axis T squared from 0 to 6 seconds squared. Five data points at L equals 0.25, 0.50, 0.75, 1.00 and 1.25 metres are plotted at the values predicted by T equals two pi root L over g, giving T squared equal to 1.01, 2.01, 3.02, 4.03 and 5.03 seconds squared. A best-fit line through the origin has gradient 4.03 seconds squared per metre, from which g equals four pi squared over the gradient gives nine point eight metres per second squared. 0 0.25 0.50 0.75 1.00 1.25 0 1 2 3 4 5 L / m T² (s²) gradient ≈ 4.03 s² m⁻¹ = 4π²/g ⇒ g ≈ 9.8 m s⁻² Linearise T = 2π√(L/g) ⇒ T2 = (4π²/g) L
Linearise T = 2π√(L/g) by plotting T² against L; the gradient is 4π²/g, so g follows directly from the slope under the IA1 analysis criterion.

Show working

QCAA awards method marks for correct identification of the principle and formula, even if the calculation slips. Always:

  1. State the physics principle.
  2. Write the formula.
  3. Substitute values with units.
  4. Calculate with appropriate sig fig.
  5. State the answer with units.
Voltage versus current with best-fit gradient giving the line resistance A scatter plot with horizontal axis current I from 0 to 5 amperes and vertical axis voltage V from 0 to 20 volts. Eight data points at currents 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5 and 4.0 amperes are plotted at voltages computed from a four ohm Ohmic conductor with random scatter. A best-fit line of gradient four volts per ampere passes through the points, giving R equal to four ohms. The plot is annotated for a Townsville transmission-line resistance test, with power loss P equal to I squared R typeset to the side. 0 1 2 3 4 5 0 5 10 15 20 I / A V (V) gradient = R = 4.0 Ω Ploss = I2R Stimulus: voltage across a Townsville transmission test section vs current.
The gradient of V against I is the conductor resistance R; combined with P=I2RP = I^{2}R the same stimulus answers an extended claim about line-loss under the IA1 reasoning criterion.

Marking criteria

QCAA rewards:

  1. Correct physics. Identifying the right principle.
  2. Correct formula. Symbolic before substitution.
  3. Show working. Method marks.
  4. Significant figures and units.
  5. Reasoning quality. Cause-effect links, theoretical justification.

Time pressure techniques

Read the stimulus carefully during perusal. Identify the physics each part requires.

Plan briefly
A 1-minute plan saves 5 minutes of confusion.
Don't get stuck
If a calculation seems wrong, move on; return at the end.
Reserve review time
5 minutes at the end to check arithmetic, units, sig figs.
Common errors
Misreading the stimulus
Numbers can be misread; double-check.
Unit conversion forgotten
SI base units in calculations.
Sign convention
Vector quantities; positive direction must be consistent.
Wrong formula form
P=VI=I2R=V2/RP = VI = I^2R = V^2/R; choose the right form for your knowns.
Calculator-style data analysis
Always interpret the numbers physically.

Check your knowledge

Six analytical questions on log-log graphs, residual analysis, and model validity in the IA1 Data Test style. ISMG criteria are signposted in the solutions. Three significant figures, units throughout, and explicit show-working.

  1. A log-log plot of measured drag force FdF_d versus speed vv gives a straight line of gradient 1.94±0.061.94 \pm 0.06. The student concludes Fd∝v2F_d \propto v^2. (a) Justify the conclusion with reference to the gradient value and uncertainty. (b) State, with calculation, whether the data are also consistent with Fd∝vF_d \propto v. (c) Identify one further test of the v2v^2 model. (5 marks)
  2. A student investigates the radial intensity II of a point source at five distances rr. rr (m): 0.20, 0.40, 0.60, 0.80, 1.00; II (W m−2^{-2}): 105, 26.5, 11.5, 6.6, 4.2. (a) Linearise as log⁡I\log I versus log⁡r\log r and calculate the gradient. (b) State what the gradient implies about the relationship between II and rr. (c) Calculate the percent deviation of each data point from the best-fit line and identify any outlier. (7 marks)
  3. The student records the residuals (measured minus predicted, in N) from a Hooke's-law fit (F=kxF = k x) at six extension values: 0.10 m: +0.05+0.05 N; 0.20 m: −0.02-0.02 N; 0.30 m: −0.04-0.04 N; 0.40 m: −0.01-0.01 N; 0.50 m: +0.10+0.10 N; 0.60 m: +0.25+0.25 N. (a) Plot conceptually what the residuals show as a function of extension. (b) State whether the residuals are consistent with Hooke's law over the full range and justify. (c) Propose a model for the apparent non-linearity at large extension. (5 marks)
  4. A current-versus-voltage data set across a filament lamp shows II (A) at VV (V): 0.5, 0.04; 1.0, 0.07; 2.0, 0.12; 4.0, 0.18; 8.0, 0.26; 12.0, 0.32. (a) State whether Ohm's law applies, with justification from the data. (b) Calculate the resistance V/IV/I at V=1.0V = 1.0 V and at V=12.0V = 12.0 V; explain the difference in physical terms. (c) Propose a linearisation that would test a power-law relationship V∝InV \propto I^n and state how to extract nn. (6 marks)
  5. A Hall-effect experiment relates Hall voltage VHV_H to current II at fixed BB in a thin semiconductor strip. Data: II (mA): 5, 10, 20, 40, 60; VHV_H (mV): 1.05, 2.12, 4.18, 8.31, 12.48. (a) Calculate the gradient using the first and last points. (b) State, with calculation, whether the data support the proportionality VH∝IV_H \propto I predicted by theory (use the middle three points to test consistency with the gradient). (c) Identify the dominant uncertainty source in this measurement. (6 marks)
  6. A student measures the centripetal force FcF_c on a rotating mass m=0.250m = 0.250 kg at radius r=0.500r = 0.500 m for five angular speeds. ω\omega (rad s−1^{-1}): 2.00, 4.00, 6.00, 8.00, 10.00; FcF_c (N): 0.51, 2.05, 4.55, 8.10, 12.85. (a) Verify the relationship Fc=mrω2F_c = m r \omega^2 by tabulating Fc/ω2F_c / \omega^2. (b) Calculate the percent error from the theoretical value mr=0.125m r = 0.125 kg m. (c) Identify the dominant source of uncertainty and propose a procedural modification to address it. (6 marks)
Solutions
Q1
(a) The gradient 1.94±0.061.94 \pm 0.06 on a log-log plot of FF versus vv means F∝v1.94F \propto v^{1.94}; the value 2.00 lies within the uncertainty interval (1.88 to 2.00), so the data are consistent with Fd∝v2F_d \propto v^2 and the conclusion is justified. (b) The exponent 1.94 differs from 1.00 by 0.94, which is 15.7 standard deviations of the gradient uncertainty - not consistent with Fd∝vF_d \propto v. (c) Further test: collect data at a wider speed range (e.g., 0.1 m s−1^{-1} to 10 m s−1^{-1}) to check whether the gradient remains 2 at low speeds (where viscous drag F∝vF \propto v may dominate) and whether a transition is observed; alternatively, compare measured drag coefficient Cd=2Fd/(ρAv2)C_d = 2 F_d / (\rho A v^2) across the data set and check whether it is a constant. This question rewards application of log-log analysis (application criterion) and judgement of model fit (judgement criterion).
The gradient 1.94±0.061.94 \pm 0.06 supports Fd∝v2F_d \propto v^2 and rules out Fd∝vF_d \propto v at 15.7 sigma; wider-range data is the recommended further test.
Q2
(a) Table: log⁡r\log r (m): −0.699,−0.398,−0.222,−0.097,0-0.699, -0.398, -0.222, -0.097, 0; log⁡I\log I (W m−2^{-2}): 2.021, 1.423, 1.061, 0.820, 0.623. Gradient =(0.623−2.021)/(0−(−0.699))=−1.398/0.699=−2.00= (0.623 - 2.021) / (0 - (-0.699)) = -1.398 / 0.699 = -2.00. (b) Gradient −2.00-2.00 on a log-log plot of II versus rr confirms the inverse-square law: I∝r−2I \propto r^{-2}. (c) Percent deviations from the line of best fit (using I=I0r−2I = I_0 r^{-2} with I0=I_0 = value at r=1.00r = 1.00 m, so Ipred=4.20/r2I_{pred} = 4.20 / r^2): predicted (W m−2^{-2}): 105.0, 26.25, 11.67, 6.56, 4.20; measured 105, 26.5, 11.5, 6.6, 4.2; deviations: 0 percent, 1.0 percent, −1.4-1.4 percent, 0.6 percent, 0 percent. No outliers; all points within 1.5 percent of the inverse-square line. This question rewards application of log-log linearisation (application criterion) and judgement of model validity (judgement criterion).
Gradient −2.00-2.00 confirms inverse-square law within 1.5 percent at every point, no outliers.
Q3
(a) Plot of residual rFr_F versus extension xx shows a roughly horizontal trend through zero for x=0.10x = 0.10 to 0.40 m, then a sharp rise to +0.10+0.10 N at x=0.50x = 0.50 m and +0.25+0.25 N at x=0.60x = 0.60 m. (b) Residuals are consistent with Hooke's law only for x≤0.40x \leq 0.40 m; at larger extensions the residuals are positive and growing, indicating the spring is stiffening beyond the linear regime. (c) Proposed model: the spring exhibits non-linear stiffening at large extension, well described by F=kx+αx3F = k x + \alpha x^3 where the cubic term accounts for finite coil-spacing effects as the spring nears its maximum stretch. This question rewards application of residual analysis (application and knowledge and analysis criteria) and judgement of model validity (judgement criterion).
Residuals are flat for x≤0.40x \leq 0.40 m then grow rapidly, suggesting Hooke's law fails above 0.40 m and a cubic stiffening term is appropriate.
Q4
(a) Ohm's law states V/IV/I is constant; here V/IV/I at the six voltages: 12.5, 14.3, 16.7, 22.2, 30.8, 37.5 ohms. Resistance triples (12.5 to 37.5 ohms) as voltage rises - Ohm's law does not apply. (b) At V=1.0V = 1.0 V: R=1.0/0.07=14.3R = 1.0/0.07 = 14.3 ohms (filament cool). At V=12.0V = 12.0 V: R=12.0/0.32=37.5R = 12.0/0.32 = 37.5 ohms (filament hot). The filament heats up as current increases; tungsten resistance increases with temperature (positive temperature coefficient), so the cool-filament resistance is lower than the hot-filament resistance. (c) Linearise as log⁡V\log V versus log⁡I\log I: if V∝InV \propto I^n then log⁡V=nlog⁡I+const\log V = n \log I + \text{const}, and the gradient gives nn. Using first and last points: log⁡V\log V: −0.301,1.079-0.301, 1.079; log⁡I\log I: −1.398,−0.495-1.398, -0.495. n=(1.079−(−0.301))/(−0.495−(−1.398))=1.380/0.903=1.53n = (1.079 - (-0.301)) / (-0.495 - (-1.398)) = 1.380/0.903 = 1.53. So V∝I1.5V \propto I^{1.5} approximately, consistent with a power-law heating model. This question rewards application of Ohm's-law analysis (application criterion), knowledge of conductor physics (knowledge and analysis criterion), and judgement of model selection (judgement criterion).
Ohm's law fails; RR triples from 12.5 to 37.5 ohms across the range as the filament heats, and V∝I1.5V \propto I^{1.5} approximately.
Q5
(a) Gradient =(12.48−1.05)/(60−5)=11.43/55=0.208 V A−1= (12.48 - 1.05)/(60 - 5) = 11.43/55 = 0.208 \ \text{V A}^{-1}. (b) Test VH/IV_H/I at intermediate points: VH/IV_H/I (V A−1^{-1}): 0.210, 0.212, 0.209, 0.208, 0.208. All within about 2 percent of the first-to-last gradient (0.208), so the linear proportionality is confirmed. (c) Dominant uncertainty: small offset voltage from misalignment of the Hall-probe contacts, which adds a constant additive offset to every reading; this would show as a non-zero intercept on a VHV_H versus II plot. Modification: zero the Hall voltage at I=0I = 0 before taking measurements, and use a four-point probe configuration to cancel contact-misalignment offsets. This question rewards application of linearity testing (application criterion) and judgement of an instrument-specific systematic error (judgement criterion).
VH/IV_H/I is 0.208 V A−10.208 \ \text{V A}^{-1} to within about 2 percent across the data set; Hall-probe contact-offset is the dominant uncertainty.
Q6
(a) ω2\omega^2: 4.00, 16.0, 36.0, 64.0, 100.0; Fc/ω2F_c/\omega^2: 0.1275, 0.1281, 0.1264, 0.1266, 0.1285. Values cluster around 0.127 kg m, confirming Fc∝ω2F_c \propto \omega^2. (b) Mean Fc/ω2=0.1274F_c/\omega^2 = 0.1274 kg m. Theory mr=0.250×0.500=0.125m r = 0.250 \times 0.500 = 0.125 kg m. Percent error =(0.1274−0.125)/0.125×100=1.92= (0.1274 - 0.125)/0.125 \times 100 = 1.92 percent. (c) Dominant source: the rotating-mass orbit is rarely perfectly horizontal because the string sags slightly under tension, so the effective radius is less than the measured straight-line string length. This biases the experimental mrm r value upward by a few percent, consistent with the 1.92 percent overshoot. Modification: replace the string with a rigid rod, or use a stroboscopic photograph to measure the actual orbital radius directly under rotation. This question rewards application of circular-motion analysis (application criterion), judgement of systematic source (judgement criterion), and communication of modification (communication criterion).
Fc/ω2=0.127F_c/\omega^2 = 0.127 kg m, 1.9 percent above theoretical mr=0.125m r = 0.125 kg m; string sag is the dominant systematic, rigid-rod is the targeted modification.

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