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v0.1.1039
NotesPhysicsTopic 6.1
Unit 6 · Experimental skills · Topic 6.1

IB Physics — Experimental skills

Topic 6.1 of IB Physics covers Experimental skills, which is part of Unit 6: Experimental skills. Students explore key concepts including Measurement technique & choosing instruments, Uncertainties & error propagation, Graphing: plotting, best-fit lines & gradients, and more. A strong understanding of experimental skills is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Experimental skills

Key Idea: This topic is the experimental-skills toolkit — how you take a measurement, attach an uncertainty to it, push that uncertainty through a calculation, and turn a table of readings into a straight-line graph you can read a physics quantity off. These skills are examined as a whole paper of their own — Paper 1B (data analysis) — built around one experiment. Expect: pick the right instrument and justify it; quote and combine uncertainties; draw a best-fit line and read its gradient (with the steepest/shallowest lines for the gradient's uncertainty); linearize a law and decide whether the data support it; and evaluate the method (random vs systematic error, repeats, anomalies). The same habits — units, significant figures, ± uncertainties — also score easy marks all through Paper 2.

📋 Key rules & formulas

The multiply/divide and power propagation rules carry the data-booklet badge (look for it). The add/subtract rule, the gradient and the gradient-uncertainty are not printed — you reproduce those from the definitions.

ΔxxandΔxx×100%\frac{\Delta x}{x}\quad\text{and}\quad \frac{\Delta x}{x}\times 100\%xΔx​andxΔx​×100%
The three forms of an uncertainty carry the SAME information: absolute Δx (a ± in the unit), fractional = absolute ÷ value, percentage = fractional × 100%.
Δx\Delta xΔx
the absolute uncertainty — a ± value in the SAME unit as x
xxx
the measured value
Δxx\tfrac{\Delta x}{x}xΔx​
the fractional uncertainty — a plain number, no unit
Δxx×100%\tfrac{\Delta x}{x}\times 100\%xΔx​×100%
the percentage uncertainty
if y=a±b  ⇒  Δy=Δa+Δb\text{if } y = a \pm b \;\Rightarrow\; \Delta y = \Delta a + \Delta bif y=a±b⇒Δy=Δa+Δb
Add or subtract → add the ABSOLUTE uncertainties. (Not in the booklet, but you must use it.)
yyy
a result found by adding or subtracting measurements
a,ba, ba,b
the measured quantities added or subtracted
Δa,Δb\Delta a, \Delta bΔa,Δb
their absolute uncertainties (same unit as the quantity)
Δy\Delta yΔy
the absolute uncertainty in the result (same unit)
y=abc  ⇒  Δyy=Δaa+Δbb+Δccy = \frac{ab}{c} \;\Rightarrow\; \frac{\Delta y}{y} = \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}y=cab​⇒yΔy​=aΔa​+bΔb​+cΔc​
Multiply or divide → add the FRACTIONAL (or percentage) uncertainties. Given in the data booklet (Uncertainties page).
yyy
the calculated result (e.g. a density = m ÷ V)
a,b,ca, b, ca,b,c
the measured quantities multiplied or divided to get y
Δaa\tfrac{\Delta a}{a}aΔa​
the fractional uncertainty in a (no unit)
Δyy\tfrac{\Delta y}{y}yΔy​
the fractional uncertainty in the result y
y=an  ⇒  Δyy=∣n∣Δaay = a^{n} \;\Rightarrow\; \frac{\Delta y}{y} = \left| n \right|\frac{\Delta a}{a}y=an⇒yΔy​=∣n∣aΔa​
Power → multiply the fractional uncertainty by the SIZE of the power |n| (½ for a square root, 2 for a square, 3 for a cube). Given in the data booklet.
y=any = a^{n}y=an
the result is a raised to a power n (e.g. area = πr², so r²)
nnn
the power (2 for a square, 3 for a cube, ½ for a square root)
Δaa\tfrac{\Delta a}{a}aΔa​
the fractional uncertainty in a (no unit)
Δyy\tfrac{\Delta y}{y}yΔy​
the fractional uncertainty in the result (no unit)
m=ΔyΔxm = \frac{\Delta y}{\Delta x}m=ΔxΔy​
Gradient of a best-fit line = rise ÷ run, read off TWO well-separated points ON THE LINE (never the raw data points).
mmm
gradient (slope) of the best-fit line — usually a physics quantity
Δy\Delta yΔy
rise — change in the y-value between two points ON THE LINE
Δx\Delta xΔx
run — change in the x-value over the same interval
Δm=mmax−mmin2\Delta m = \frac{m_{max} - m_{min}}{2}Δm=2mmax​−mmin​​
Uncertainty in the gradient = half the spread between the steepest and shallowest lines that still pass through all the error bars.
Δm\Delta mΔm
uncertainty in the gradient
mmaxm_{max}mmax​
gradient of the STEEPEST line that still passes through all the error bars
mminm_{min}mmin​
gradient of the SHALLOWEST such line
Y=mX+cY = mX + cY=mX+c
Straight-line form. Rearrange the law so your two plotted quantities match Y and X; the gradient m and intercept c are then physics quantities. c = 0 means 'directly proportional'.
YYY
the quantity plotted UP the vertical axis (chosen so the graph is straight)
XXX
the quantity plotted ACROSS the horizontal axis
mmm
the gradient — equals a physics constant you are trying to find
ccc
the vertical intercept — 0 for a 'directly proportional' law

📐 Choosing the instrument (resolution)

Choose the instrument whose smallest division is small compared with what you are measuring. A 0.5 mm wire on a mm ruler is hopeless (±0.5 mm is the whole thing) but easy on a micrometer (±0.005 mm). Reading uncertainty from a single instrument = ± half its smallest division.

⚖️ The three propagation rules side by side

🎯 Random vs systematic error

A straight line alone only means the law is linear (Y = mX + c). It is directly proportional only if the line also passes through the origin (c = 0). Equivalently, the ratio Y/X is constant across every row of the table.

✏️ Worked exam-style questions

IB-style question — measure a multiple to shrink the uncertainty

A student measures the thickness of one sheet of paper. A micrometer (resolution 0.01 mm, so ±0.005 mm) on a single sheet would give a fractional uncertainty of nearly 7%. Instead they measure a stack of 80 sheets and get 8.40 mm with the same ±0.005 mm. (a) Find the thickness of one sheet. (b) Find the absolute and percentage uncertainty in the thickness of one sheet, and say why this method is better.

Solution:

  1. (a) The stack thickness divided by the number of sheets:

    t=8.4080=0.105 mmt = \frac{8.40}{80} = 0.105\ \text{mm}t=808.40​=0.105 mm
  2. (b) Dividing by an EXACT count (80) divides the absolute uncertainty by the same number:

    Δt=0.00580=6.25×10−5 mm\Delta t = \frac{0.005}{80} = 6.25\times10^{-5}\ \text{mm}Δt=800.005​=6.25×10−5 mm
  3. (b) Percentage uncertainty = (Δt ÷ t) × 100%:

    Δtt×100%=6.25×10−50.105×100%≈0.06%\frac{\Delta t}{t}\times100\% = \frac{6.25\times10^{-5}}{0.105}\times100\% \approx 0.06\%tΔt​×100%=0.1056.25×10−5​×100%≈0.06%
Final answer:

(a) t = 0.105 mm. (b) Δt ≈ 6 × 10⁻⁵ mm, about 0.06% — far better than the ≈7% a single sheet would give. Measuring a MULTIPLE (a stack, or 10 swings) and dividing by an exact count keeps the same small absolute reading uncertainty while making the measured value large.

IB-style question — propagate uncertainty through a density

A small cube has side L = 2.00 ± 0.01 cm and mass m = 64.0 ± 0.5 g. The density is ρ = m ÷ V, where V = L³. (a) Find the density. (b) Find its percentage uncertainty. (c) Quote ρ with its absolute uncertainty.

Solution:

  1. (a) Volume first (a cube, so L³), then density = mass ÷ volume:

    V=(2.00)3=8.00 cm3,ρ=64.08.00=8.00 g cm−3V = (2.00)^{3} = 8.00\ \text{cm}^{3},\quad \rho = \frac{64.0}{8.00} = 8.00\ \text{g cm}^{-3}V=(2.00)3=8.00 cm3,ρ=8.0064.0​=8.00 g cm−3
  2. (b) ρ = m ÷ L³ is a quotient with a power, so ADD fractional uncertainties — the side counts THREE times (|n| = 3):

    Δρρ=Δmm+3 ΔLL=0.564.0+3×0.012.00\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 3\,\frac{\Delta L}{L} = \frac{0.5}{64.0} + 3\times\frac{0.01}{2.00}ρΔρ​=mΔm​+3LΔL​=64.00.5​+3×2.000.01​
  3. (b) Evaluate each term, then add, then ×100%:

    Δρρ=0.0078+0.0150=0.0228≈2.3%\frac{\Delta\rho}{\rho} = 0.0078 + 0.0150 = 0.0228 \approx 2.3\%ρΔρ​=0.0078+0.0150=0.0228≈2.3%
  4. (c) Absolute uncertainty = fractional × value, rounded to 1 s.f.:

    Δρ=0.0228×8.00≈0.2 g cm−3\Delta\rho = 0.0228 \times 8.00 \approx 0.2\ \text{g cm}^{-3}Δρ=0.0228×8.00≈0.2 g cm−3
Final answer:

(a) ρ = 8.00 g cm⁻³. (b) ≈ 2.3%. (c) ρ = 8.0 ± 0.2 g cm⁻³. The L³ makes the side's fractional uncertainty count THREE times — that power term dominates, which is why you measure the side most carefully.

IB-style question — gradient and its uncertainty from a graph

On a force-versus-extension graph the best-fit line passes through (0, 0). Two points read off THE LINE are (0.020 m, 2.4 N) and (0.080 m, 9.6 N). The steepest and shallowest lines through the error bars have gradients 124 N m⁻¹ and 116 N m⁻¹. (a) Find the spring constant from the best-fit gradient. (b) Find the uncertainty in the gradient and quote k properly.

Solution:

  1. (a) Gradient = rise ÷ run, using the two points ON THE LINE:

    m=ΔyΔx=9.6−2.40.080−0.020=7.20.060m = \frac{\Delta y}{\Delta x} = \frac{9.6 - 2.4}{0.080 - 0.020} = \frac{7.2}{0.060}m=ΔxΔy​=0.080−0.0209.6−2.4​=0.0607.2​
  2. (a) Evaluate — and the gradient of a force-extension line IS the spring constant k:

    k=120 N m−1k = 120\ \text{N m}^{-1}k=120 N m−1
  3. (b) Uncertainty = half the spread of the steepest and shallowest gradients:

    Δm=mmax−mmin2=124−1162=4 N m−1\Delta m = \frac{m_{max} - m_{min}}{2} = \frac{124 - 116}{2} = 4\ \text{N m}^{-1}Δm=2mmax​−mmin​​=2124−116​=4 N m−1
Final answer:

(a) k = 120 N m⁻¹. (b) k = 120 ± 4 N m⁻¹. Always read the gradient off the LINE (not the data points), and get its uncertainty from the steepest/shallowest lines through the error bars — never guess it.

IB-style question — linearize a law and test it

Theory predicts the depth d a marker sinks is d = k√P, where P is the water pressure. (a) State what to plot on each axis to get a straight line through the origin, and what the gradient represents. (b) A classmate instead claims d is directly proportional to P. Using the rows (P = 4.0 kPa, d = 3.1 cm) and (P = 9.0 kPa, d = 4.6 cm), show that this claim is wrong.

Solution:

  1. (a) Match d = k√P to Y = mX + c: Y is d, X is √P, and there is no '+ c'.

    plot d (up) against P (across);m=k\text{plot } d \text{ (up) against } \sqrt{P}\text{ (across)};\quad m = kplot d (up) against P​ (across);m=k
  2. (b) If d ∝ P then the ratio d ÷ P must be CONSTANT. First row:

    dP=3.14.0=0.78 cm kPa−1\frac{d}{P} = \frac{3.1}{4.0} = 0.78\ \text{cm kPa}^{-1}Pd​=4.03.1​=0.78 cm kPa−1
  3. (b) Second row:

    dP=4.69.0=0.51 cm kPa−1\frac{d}{P} = \frac{4.6}{9.0} = 0.51\ \text{cm kPa}^{-1}Pd​=9.04.6​=0.51 cm kPa−1
  4. (b) The two ratios differ, so d ÷ P is NOT constant:

    0.78≠0.510.78 \neq 0.510.78=0.51
Final answer:

(a) Plot d (up) against √P (across); the gradient equals the constant k. (b) d ÷ P is not constant (0.78 ≠ 0.51 cm kPa⁻¹), so d is NOT directly proportional to P — it fits d ∝ √P. 'Directly proportional' always means a straight line through the origin / a constant ratio.


🧠 Quick self-check

Tap each card to reveal the answer.


🎯 Exam tips

Exam Tips

  • Match the instrument's RESOLUTION to the quantity, and reading uncertainty = ± half the smallest division. To shrink it, measure a MULTIPLE (a stack of sheets, 10 swings) and divide by the exact count.
  • Pick the right propagation rule by the operation: + or − → add ABSOLUTE uncertainties; × or ÷ → add FRACTIONAL/percentage uncertainties; a power aⁿ → multiply the fractional uncertainty by |n|. The last two are in the data booklet.
  • In a quotient with a power (like ρ = m ÷ L³), the side's fractional uncertainty counts |n| times — here three times — so that term usually dominates. Measure the powered quantity most carefully.
  • Round the absolute uncertainty to 1 significant figure, then match the value to the same decimal place: ρ = 8.0 ± 0.2 g cm⁻³, never 8.00 ± 0.23.
  • Read a gradient off TWO well-separated points ON the best-fit line, not the data points. State what the gradient represents and give its units — it is always a physics quantity.
  • For the gradient's uncertainty, draw the steepest and shallowest lines that still pass through all the error bars, then Δm = (mₘₐₓ − mₘᵢₙ) ÷ 2.
  • Linearize by rearranging the law to Y = mX + c, then plot Y against X. 'Directly proportional' demands a straight line THROUGH THE ORIGIN (or a constant ratio across rows) — a straight line alone is only 'linear'.
  • Random error scatters readings → cut it by repeating and averaging. Systematic error shifts them all the same way → averaging won't help; fix the instrument or method. Always discard a clear anomaly before averaging.
  • Dimensional analysis: balance the fundamental SI units (kg, m, s, A) on both sides to find an unknown exponent, or to state the units of a constant read off a gradient (y-axis units ÷ x-axis units).

What you'll learn in Topic 6.1

  • 6.1.1 Measurement technique & choosing instruments
  • 6.1.2 Uncertainties & error propagation
  • 6.1.3 Graphing: plotting, best-fit lines & gradients
  • 6.1.4 Linearizing relationships & testing a law
  • 6.1.5 Evaluating method & dimensional analysis
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 6.1 Experimental skills

6.1.1

Measurement technique & choosing instruments

Notes
6.1.2

Uncertainties & error propagation

Notes
6.1.3

Graphing: plotting, best-fit lines & gradients

Notes
6.1.4

Linearizing relationships & testing a law

Notes
6.1.5

Evaluating method & dimensional analysis

Notes

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Topic 6.1 Experimental skills forms a core part of Unit 6: Experimental skills in IB Physics. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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