Combining Uncertainties | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Combining Uncertainties

Most results are calculated from two or more measurements. Each measurement has its own uncertainty, so the result has a combined uncertainty.

The 9702 worst-case method does not let uncertainties cancel. Add either absolute or percentage uncertainties according to the operation in the equation.

The three rules

When the formula has …	… combine like this
a sum or a difference (+, −)	add the ABSOLUTE uncertainties
a product or a quotient (×, ÷)	add the PERCENTAGE uncertainties
a power xⁿ	multiply the percentage uncertainty by n

Why add even for a difference: the two readings can be wrong in opposite directions at the same time, so the worst case for the result is the two uncertainty ranges added together.
The power rule is the product rule repeated: $x^2 = x \times x$ adds the same percentage twice.

Always add, never subtract, even for a difference. Absolute for $+,\,-$ ; percentage for $\times,\,\div$ ; times $n$ for a power.

Switch the operation below to see which rule applies, and which measurement adds the most uncertainty:

Uncertainty Combiner

Pick an operation and enter each quantity as value ± uncertainty. Watch the rule switch between adding absolute and adding percentage uncertainties, and see which measurement dominates.

Quantity A2.5%

Quantity B5.0%

Product or quotient → add the PERCENTAGE uncertainties.

% from A

2.5%

% from B

5.0%

% combined

7.5%

Result

8.0±0.6

≈ 7.5% uncertainty

Using the rules, one change at a time

Worked example

Smallest case: a quotient

$V = 6.0 \pm 0.2 \text{ V}$ , $I = 2.4 \pm 0.1 \text{ A}$ . Find $R = V/I$ with its uncertainty.

Percentages first: $\tfrac{0.2}{6.0} = 3.3\%$ and $\tfrac{0.1}{2.4} = 4.2\%$ .
A quotient, so add: $3.3\% + 4.2\% = 7.5\%$ .
Value: $R = 2.5\ \Omega$ . Absolute uncertainty: $7.5\% \times 2.5 = 0.2\ \Omega$ .

Answer

R = 2.5 \pm 0.2\ \Omega

Worked example

One change: a difference

A thermometer reads to ±0.5 °C. It measures a rise from 40 °C to 100 °C. Find the percentage uncertainty in the rise.

The rise: $100 - 40 = 60 \text{ °C}$ .
A difference, so add the absolutes: $0.5 + 0.5 = 1.0 \text{ °C}$ .
$\tfrac{1.0}{60} \times 100\% = 1.7\%$ .

Answer

1.7\%

(9702/11/M/J/24 Q6)

Worked example

One change: a power

A ball is dropped from rest and $g = 2s/t^2$ . The distance uncertainty is negligible; the time uncertainty is 4%. Find the percentage uncertainty in $g$ .

The formula uses $t^2$ , so multiply the time uncertainty by 2.
$2 \times 4\% = 8\%$ .

Answer

8\%

(9702/12/O/N/25 Q4)

Worked example

The full workflow: g from a pendulum

$g = \dfrac{4\pi^2 l}{T^2}$ , with $l = 1.50 \text{ m} \pm 2\%$ and $T = 2.48 \text{ s} \pm 3\%$ . Find $g$ with its absolute uncertainty.

Value first: $g = \dfrac{4\pi^2 (1.50)}{2.48^2} = 9.63 \text{ m s}^{-2}$ .
$l$ contributes 2%. $T^2$ contributes $2 \times 3\% = 6\%$ . The constant $4\pi^2$ is exact, so it adds no uncertainty.
A quotient, so add: $2\% + 6\% = 8\%$ .
Absolute: $8\% \times 9.63 = 0.77 \approx 0.8 \text{ m s}^{-2}$ .

Answer

g = 9.6 \pm 0.8 \text{ m s}^{-2}

(9702/22/F/M/20 Q1(b))

Your turn9702/11/O/N/24 Q3 · 1 mark · adapted

A block has mass (25.0 ± 0.1) g and dimensions (5.00 ± 0.01) cm, (2.00 ± 0.01) cm and (1.00 ± 0.01) cm. Its density is 2.50 g cm⁻³. Using density = mass ÷ (length × width × height), find the absolute uncertainty in the density.

Show worked answer

Density = mass ÷ (length × width × height): all products and quotients, so add all four percentages.
$\tfrac{0.1}{25.0} = 0.4\%$ , $\tfrac{0.01}{5.00} = 0.2\%$ , $\tfrac{0.01}{2.00} = 0.5\%$ , $\tfrac{0.01}{1.00} = 1.0\%$ .
Total: $2.1\%$ . Absolute: $2.1\% \times 2.50 \approx 0.05 \text{ g cm}^{-3}$ .

Answer

\pm 0.05\ \text{g cm}^{-3}

Common mistake

Three common errors: subtracting uncertainties on a difference (always add), forgetting the ×2 on a squared quantity, and adding absolute uncertainties for a product (use percentages there).

Combining Uncertainties

Most results are calculated from two or more measurements. Each measurement has its own uncertainty, so the result has a combined uncertainty.

The 9702 worst-case method does not let uncertainties cancel. Add either absolute or percentage uncertainties according to the operation in the equation.

The three rules

When the formula has …	… combine like this
a sum or a difference (+, −)	add the ABSOLUTE uncertainties
a product or a quotient (×, ÷)	add the PERCENTAGE uncertainties
a power xⁿ	multiply the percentage uncertainty by n

Why add even for a difference: the two readings can be wrong in opposite directions at the same time, so the worst case for the result is the two uncertainty ranges added together.
The power rule is the product rule repeated: $x^2 = x \times x$ adds the same percentage twice.

Always add, never subtract, even for a difference. Absolute for $+,\,-$ ; percentage for $\times,\,\div$ ; times $n$ for a power.

Switch the operation below to see which rule applies, and which measurement adds the most uncertainty:

Uncertainty Combiner

Pick an operation and enter each quantity as value ± uncertainty. Watch the rule switch between adding absolute and adding percentage uncertainties, and see which measurement dominates.

Quantity A2.5%

Quantity B5.0%

Product or quotient → add the PERCENTAGE uncertainties.

% from A

2.5%

% from B

5.0%

% combined

7.5%

Result

8.0±0.6

≈ 7.5% uncertainty

Using the rules, one change at a time

Worked example

Smallest case: a quotient

$V = 6.0 \pm 0.2 \text{ V}$ , $I = 2.4 \pm 0.1 \text{ A}$ . Find $R = V/I$ with its uncertainty.

Percentages first: $\tfrac{0.2}{6.0} = 3.3\%$ and $\tfrac{0.1}{2.4} = 4.2\%$ .
A quotient, so add: $3.3\% + 4.2\% = 7.5\%$ .
Value: $R = 2.5\ \Omega$ . Absolute uncertainty: $7.5\% \times 2.5 = 0.2\ \Omega$ .

Answer

R = 2.5 \pm 0.2\ \Omega

Worked example

One change: a difference

A thermometer reads to ±0.5 °C. It measures a rise from 40 °C to 100 °C. Find the percentage uncertainty in the rise.

The rise: $100 - 40 = 60 \text{ °C}$ .
A difference, so add the absolutes: $0.5 + 0.5 = 1.0 \text{ °C}$ .
$\tfrac{1.0}{60} \times 100\% = 1.7\%$ .

Answer

1.7\%

(9702/11/M/J/24 Q6)

Worked example

One change: a power

A ball is dropped from rest and $g = 2s/t^2$ . The distance uncertainty is negligible; the time uncertainty is 4%. Find the percentage uncertainty in $g$ .

The formula uses $t^2$ , so multiply the time uncertainty by 2.
$2 \times 4\% = 8\%$ .

Answer

8\%

(9702/12/O/N/25 Q4)

Worked example

The full workflow: g from a pendulum

$g = \dfrac{4\pi^2 l}{T^2}$ , with $l = 1.50 \text{ m} \pm 2\%$ and $T = 2.48 \text{ s} \pm 3\%$ . Find $g$ with its absolute uncertainty.

Value first: $g = \dfrac{4\pi^2 (1.50)}{2.48^2} = 9.63 \text{ m s}^{-2}$ .
$l$ contributes 2%. $T^2$ contributes $2 \times 3\% = 6\%$ . The constant $4\pi^2$ is exact, so it adds no uncertainty.
A quotient, so add: $2\% + 6\% = 8\%$ .
Absolute: $8\% \times 9.63 = 0.77 \approx 0.8 \text{ m s}^{-2}$ .

Answer

g = 9.6 \pm 0.8 \text{ m s}^{-2}

(9702/22/F/M/20 Q1(b))

Your turn9702/11/O/N/24 Q3 · 1 mark · adapted

Show worked answer

Density = mass ÷ (length × width × height): all products and quotients, so add all four percentages.
$\tfrac{0.1}{25.0} = 0.4\%$ , $\tfrac{0.01}{5.00} = 0.2\%$ , $\tfrac{0.01}{2.00} = 0.5\%$ , $\tfrac{0.01}{1.00} = 1.0\%$ .
Total: $2.1\%$ . Absolute: $2.1\% \times 2.50 \approx 0.05 \text{ g cm}^{-3}$ .

Answer

\pm 0.05\ \text{g cm}^{-3}

Common mistake

Three common errors: subtracting uncertainties on a difference (always add), forgetting the ×2 on a squared quantity, and adding absolute uncertainties for a product (use percentages there).