CCEA GCE Physics (1210) · Unit AS 3: Practical Techniques and Data Analysis
Mini-Lesson
Practical Techniques & Data Analysis
This mini-lesson covers CCEA Unit AS 3: implementing (choosing and using apparatus correctly and safely), analysis (tables and graphs), evaluation (reliability, absolute and percentage uncertainty), refinement (attacking systematic and random error) and communication.
Why this unit matters: AS 3 is assessed on skills, not on a list of content topics. The physics context can come from anywhere in AS 1 or AS 2 — what is being tested is whether you can measure, tabulate, plot, quantify uncertainty and draw a valid conclusion.
Press Start to begin.
3.1 Implementing
Choosing the right instrument
CCEA names the apparatus you must be able to use: spring and top-pan balances (mass), rule, micrometer and calipers (length), graduated cylinder (volume), clock/stopwatch (time), thermometer and sensor (temperature), ammeter, voltmeter, multimeter, protractor (angle). Digital versions are fine.
The resolution of an instrument is the smallest change it can register. For an analogue scale, the uncertainty in a single reading is usually taken as half the smallest division; for a digital display, ±1 in the last digit.
metre rule — resolution 1 mm → uncertainty ±0.5 mm on each end, so ±1 mm on a length (two readings!)
vernier calipers — resolution 0.1 mm · micrometer screw gauge — resolution 0.01 mm
stopwatch — resolution 0.01 s, but human reaction time (~0.2 s) dominates the real uncertainty
The A-level move: pick the instrument whose resolution gives a small percentage uncertainty for your measurement. Measuring a 0.40 mm wire with a rule is hopeless (±1 mm is 250%!); with a micrometer it is ±0.01 mm, i.e. 2.5%.
3.3 Evaluation
Precision, accuracy, random and systematic error
Precise — repeated readings are close to each other (small spread).
Accurate — readings are close to the true value.
Random error — scatters readings either side of the mean (reaction time, fluctuating conditions). Reduced by repeating and averaging.
Systematic error — shifts every reading the same way (zero error, mis-calibrated scale, always reading a meniscus from above). Repeating does not help; you must find and correct the cause.
The pair that catches people out: a set of readings can be highly precise (tightly clustered) and completely inaccurate (all shifted by the same zero error). Precision says nothing about accuracy.
Quick check
Precise or accurate?
?A student measures a wire's diameter five times with a micrometer that reads +0.02 mm when fully closed. The readings are 0.42, 0.42, 0.43, 0.42, 0.42 mm. Which best describes the data?
3.3.3 Uncertainty
Absolute, fractional and percentage uncertainty
A measurement is quoted as value ± absolute uncertainty, e.g. L = 0.850 ± 0.001 m.
% uncertainty = (absolute uncertainty ÷ value) × 100fractional uncertainty = absolute ÷ value (the same thing, without the ×100)
For a set of repeated readings, the best estimate is the mean, and a good estimate of the uncertainty is half the range: (max − min) ÷ 2.
Worked example
Diameter d = 0.42 mm measured with a micrometer of resolution 0.01 mm.
% uncertainty = (0.01 ÷ 0.42) × 100 = 2.4%
Golden rule: uncertainties are quoted to 1 significant figure, and the value is then rounded to the same decimal place: 2.6137 ± 0.05 s should be written 2.61 ± 0.05 s.
Calculate
Your turn — percentage uncertainty
1A wire's diameter is measured as 0.42 mm with a micrometer of resolution 0.01 mm. Calculate the percentage uncertainty in the diameter. Give your answer to 2 significant figures.
%
Hint: (0.01 ÷ 0.42) × 100.
3.3.3 Uncertainty
Combining uncertainties
Three rules cover everything at AS:
Adding or subtracting quantities → add the absolute uncertainties. (Yes — even for a subtraction. A difference of two similar numbers has a horrible percentage uncertainty.)
Multiplying or dividing → add the percentage uncertainties.
Raising to a power n → multiply the percentage uncertainty by n.
R = V / I → %R = %V + %IA = πr² → %A = 2 × %r (the power rule)
Worked example
V = 6.0 ± 0.1 V → 1.7% · I = 0.50 ± 0.02 A → 4.0%
% uncertainty in R = 1.7 + 4.0 = 5.7%, and R = 12 Ω, so R = 12 ± 0.7 Ω.
Where to attack: the term with the biggest percentage uncertainty (or the biggest power) dominates. In the example above, improving the ammeter reading helps far more than improving the voltmeter reading.
Calculate
Your turn — combining uncertainties
2A resistance is found from R = V / I with V = 6.0 ± 0.1 V and I = 0.50 ± 0.02 A. Calculate the percentage uncertainty in R.
3A circle's radius is 2.5 ± 0.1 cm. Its area is A = πr². Calculate the percentage uncertainty in the area.
%
Hint: % uncertainty in r = (0.1/2.5) × 100 = 4%. A depends on r², so multiply by the power: 2 × 4%.
Sort it
Random, systematic, or good practice?
Tap a statement, then tap the box it belongs in.
🎲 Random error
📉 Systematic error
✅ Good practice
3.2 Analysis
Graphs, gradients and error bars
Rearrange your relationship into the form y = mx + c so that the graph is a straight line, then get the physics from the gradient (and sometimes the intercept).
s = ½at² → plot s against t² → gradient = ½aV = E − Ir → plot V against I → gradient = −r, intercept = E
Good graph practice: label axes with quantity and unit; choose a scale that uses more than half the paper; plot points as small crosses; draw a single smooth best-fit line (not dot-to-dot); circle any anomaly, ignore it when drawing the line, and say you would repeat that reading.
Error bars show the absolute uncertainty in each point. Draw the best-fit line and then the line of worst fit — the steepest (or shallowest) line that still passes through all the error bars. Then:
uncertainty in gradient = |mbest − mworst|% uncertainty in the gradient = (that difference ÷ mbest) × 100
Calculate
Your turn — from a gradient
4A student plots displacement s against t² for an object released from rest, and the graph is a straight line through the origin passing through (0.20 s², 0.98 m) and (1.00 s², 4.90 m). Since s = ½at², calculate the acceleration.
m s⁻²
Hint: gradient = (4.90 − 0.98) ÷ (1.00 − 0.20) = 3.92 ÷ 0.80 = 4.90. The gradient equals ½a, so a = 2 × gradient.
Calculate
Your turn — timing many oscillations
5A pendulum is timed over 10 complete oscillations: the stopwatch reads 14.2 s, with an uncertainty of ±0.2 s from reaction time. Calculate the percentage uncertainty in the period T.
%
Hint: The ±0.2 s applies to the whole 14.2 s reading: (0.2 ÷ 14.2) × 100. Dividing by 10 to get T divides the value AND the absolute uncertainty by 10, so the percentage is unchanged.
Quick check
Why time 20 swings?
?Why does timing 20 oscillations and dividing by 20 give a better value for the period than timing a single oscillation?
Quick check
Dealing with an anomaly
?One point on a student's graph lies far from an otherwise convincing straight line. What is the correct thing to do?
Quick check
Estimating the gradient uncertainty
?A student has plotted points with error bars. How should they estimate the uncertainty in the gradient?
Quick check
Refining the experiment
?A student is measuring the extension of a spring, and the percentage uncertainty in their extension is unacceptably large. Which change would most directly reduce it?
Match it
Match the uncertainty rule to the situation
Tap an item on the left, then its partner on the right.
Rule
When you use it
Recap
Unit AS 3 — the big ideas
Instruments: analogue → ± half a division; digital → ±1 in the last digit; micrometer 0.01 mm
Precision ≠ accuracy: precise = tight cluster; accurate = close to the true value
Random error: repeat and average. Systematic error: find and correct the cause
Uncertainty: % = (absolute ÷ value) × 100; repeats → half the range
Combining: add absolutes (+ / −) · add percentages (× / ÷) · × the power (rⁿ)
Graphs: rearrange to y = mx + c; gradient carries the physics; worst-fit line gives the gradient uncertainty
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