Practice, explain, then check
Six original scenarios, 24 short tasks. Work on paper first, then reveal the reasoning and mark the tasks you have reviewed.
These are independent practice exercises, not official IB questions or a predicted paper. Self-review counts are not examination marks. Progress stays in this tab and resets on reload.
Paper 1B toolkit · Reasoning workbook
0 of 24 tasks reviewed.
A small-amplitude pendulum is used to test T² ∝ L. Four timings of 12 complete oscillations are 18.6, 18.8, 18.7 and 18.5 s. The stated timing uncertainty floor is 0.2 s per total timing.
Mean total = 18.65 s. Half-range = (18.8−18.5)/2 = 0.15 s, so use 0.2 s. T = 18.65/12 = 1.55417 s; ΔT = 0.2/12 = 0.01667 s. To one uncertainty significant figure: T = (1.55 ± 0.02) s.
A similar absolute start/stop uncertainty is divided by the number of complete oscillations. Its percentage contribution is smaller for a longer total timing. Counting errors or drift would limit the benefit.
g = 4π²/4.08 = 9.67608 m/s². Relative uncertainty ≈ 0.08/4.08 = 0.01961; Δg = 0.18973 m/s². Report g = (9.7 ± 0.2) m/s².
Keep the starting angle small and approximately constant. Larger amplitudes increase the period relative to the small-angle model and could confound a comparison across lengths.
Check calculations: Repeated measurements & timing · Physical constants from gradients
Three measurements have negligible x uncertainty: (x,y,Δy) = (0,1,0.2), (1,3,0.2), (2,5,0.2). Both axes use SI quantities; x is time in seconds and y is displacement in metres.
m = (5−1)/(2−0) = 2 m/s. c = 1 m. The gradient represents velocity in this displacement–time example.
Minimum gradient joins (0,1.2) to (2,4.8): 1.8 m/s. Maximum joins (0,0.8) to (2,5.2): 2.2 m/s. Both give y = 3 at x = 1, so they intersect the middle bar. Δm = (2.2−1.8)/2 = 0.2 m/s.
No. Any straight line through both endpoint intervals has its midpoint y between 2.8 and 3.2. The new middle interval is 3.6 to 4.0. No single line intersects all three bars. Check the measurement and the model; do not silently remove the point.
No. It may represent a physical initial displacement, an offset, an omitted term or a fitting artefact. Interpret it using the model and uncertainty before naming a cause.
Check calculations: Error bars & acceptable lines · Gradient uncertainty
An idealized supplied dataset is x = 1, 2, 4, 8 m and y = 5, 20, 80, 320 J. A power law y = A xⁿ is proposed.
Plot ln(y / 1 J) against ln(x / 1 m). The model is ln(y / 1 J) = n ln(x / 1 m) + a constant. The arguments of the logs are dimensionless.
Doubling x multiplies y by 4, so 2ⁿ = 4 and n = 2. A = 5 J/m². On the stated log plot, the intercept is ln 5 ≈ 1.60944.
x² = 4.00 m². Relative uncertainty ≈ 2 × 0.04/2.00 = 0.04; absolute uncertainty ≈ 0.16 m². This is a first-order estimate.
1 cm = 10⁻² m, so 1 cm² = 10⁻⁴ m². Therefore 2.5 cm² = 2.5 × 10⁻⁴ m², not 0.025 m².
Check calculations: Linearization & log plots · SI units & dimensional checks
A supplied simulation gives a decaying signal after background subtraction. Values at t = 0, 2, 4, 6 s are 80, 40, 20, 10 arbitrary units. No apparatus is needed for this exercise.
The signal halves every 2 s. The gradient is ln(1/2)/2 = −0.346574 s⁻¹. A positive reference changes the intercept, not this slope.
τ = −1/m = 2/ln 2 = 2.88539 s. The time constant is longer than the half-life; t_half = τ ln 2.
At late times the background becomes a larger fraction of the reading. The log plot tends to flatten, so a single fitted decay slope can be biased towards zero. A straight-looking section alone does not justify ignoring the background.
Logarithms change the residual scale and usually the variance pattern. An unweighted fit in log space is not equivalent to an unweighted fit in the original signal. State the assumed error model.
Check calculations: Linearization & log plots · Physical constants from gradients
A velocity–time graph is represented by straight segments through (t/s, v/(m/s)) = (0,0), (2,4), (4,0), (6,−2).
The first segment has slope (4−0)/(2−0) = 2 m/s². Interpolation gives v = 0 + 2 × 1.5 = 3 m/s.
Acceleration is the gradient of velocity against time: (0−4)/(4−2) = −2 m/s². A negative acceleration does not by itself imply negative velocity.
Trapezoidal areas are 4 m, 4 m and −2 m. Total displacement = 6 m. The total distance is 4 + 4 + 2 = 10 m because the final segment is below zero velocity.
It is outside the measured range. Extending the last line gives −5 m/s, but that assumes the acceleration remains −1 m/s² after 6 s. The data alone do not establish this.
Check calculations: Graph gradient, area & prediction · SI units & dimensional checks
A student compares two independent determinations: a = (1.84 ± 0.06) cm and b = (1.95 ± 0.03) cm. Separately, a scale reads +0.2 mm when it should read zero.
a spans 1.78–1.90 cm and b spans 1.92–1.98 cm. They do not overlap: the separation 0.11 cm exceeds the sum of uncertainties 0.09 cm. This is an interval comparison, not a p-value.
Corrected value = observed − offset = 8.7 − 0.2 = 8.5 mm. Any uncertainty in the offset correction remains relevant.
Repeats can reduce the influence of random fluctuations on the mean. A shared calibration bias shifts all readings in the same way, so averaging does not remove it. Compare with a standard or assess calibration.
Select an instrument with a suitable range and smaller justified uncertainty, check its zero, and take readings at several positions if thickness varies. Name how each change addresses the identified limitation rather than merely saying “use better equipment”.
Check calculations: Agreement within uncertainty · Instrument readings & zero error