Physics

7.3 - Processing Data & Calculations

7 min

note 1 of 4

What a calculation question looks like

Most numerical questions give you data to use with an equation from the relationships sheet.

Example: a student measures a potential difference of 6.0 V across a resistor and a current of 0.25 A through it - calculate the resistance.

The three parts of a calculation

Write down the relationship you're using: $R = \frac{V}{I}$
Substitute the values into it: $R = \frac{6.0}{0.25}$
Give the answer with its unit: $R = 24$ Ω

Why write it out in full

Write the relationship down even if you get stuck afterwards - It earns credit on its own.
If you substitute an incorrect earlier value but calculate correctly, subsequent steps can still gain marks if your working is shown.
The relationships sheet is given in the exam, so you don't need to memorise the equations - you just have to choose the right one and set it out clearly.

A typical question

A student measures a potential difference of 6.0 V across a resistor and a current of 0.25 A through it. Calculate the resistance of the resistor.

note 2 of 4

Setting out a calculation

Write every calculation in the same three lines so each step can be credited separately:

$R = \frac{V}{I}$

$R = \frac{6.0}{0.25}$

$R = 24 Ω$

Line one is the relationship, written out as an equation.
Line two is the substitution - The same equation with numbers in place of the symbols, converted into the required units beforehand.
Line three is the answer with its unit. A number without a unit is incomplete.

Rules for line one

A magic triangle is not a relationship. Drawing a formula triangle doesn't count on its own. You must write $V = I R$ or $R = \frac{V}{I}$ as an equation.
Choose one relationship, don't list several. Writing multiple equations without choosing one to substitute into earns nothing.

Recurring general calculations

Calculation	How to do it	Worked example
Mean	Add the repeats and divide by how many there are.	Resistances of 22, 24, and 26 Ω: $\frac{22 + 24 + 26}{3} = 24$ Ω
Percentage	The part over the whole, times 100.	18 of 25 bulbs working: $\frac{18}{25} \times 100 = 72 %$
Percentage change	The change over the original value, times 100.	Resistance rises from 20 Ω to 26 Ω: $\frac{26 - 20}{20} \times 100 = 30 %$ increase

note 3 of 4

Significant figures, units, and prefixes

How many significant figures

Round your final answer to about the same number of significant figures as the starting data.
SQA allows an answer with up to two figures more, or one figure fewer, than the piece of data with the fewest significant figures.
For $R = \frac{6.0}{0.25}$ , both values have two significant figures, so 24 Ω is ideal (acceptable range: 20 Ω to 24.00 Ω).

Significant figures vs decimal places:

Significant figures are counted from the first non-zero digit - so 47.316 to three significant figures is 47.3.
Decimal places are counted after the decimal point - so 6.028 to one decimal place is 6.0, but to one significant figure it's 6.
Don't round intermediate values during a calculation - carry the full figures and round only the final answer.
Don't use a recurring dot notation, as it implies infinite precision.

Convert units before substituting

Convert quantities into standard units before line two. For a distance of 4 km over a time of 2 minutes:

$v = \frac{d}{t}$

$v = \frac{4,000}{120}$

$v = 33 {m s}^{- 1}$

Prefixes and abbreviations

Use the correct case for unit prefixes (e.g. 5 ms is 5 milliseconds, while 5 Ms is 5 megaseconds). Standard abbreviations must be used:

Write this	Not this
second, s	sec, secs
metres per second, m/s or m s^-1	mps, m/s^-1
metres per second squared, m/s² or m s^-2	m/s/s, mpsps

note 4 of 4

Common exam questions

Question 1

A calculation is written out in three lines. What are they?

The relationship you are using, written as an equation; the substitution, with your values in place of the symbols; and the final answer with its unit.

Question 2

Why should you write down the relationship even if you cannot finish the calculation?

Because writing the correct relationship is looked at in its own right, so it stands even when the rest of the working goes wrong or is never completed.

Question 3

A student draws a triangle with $V$ on top and $I$ and $R$ below it, then writes the answer. Why is the triangle not enough?

Because a magic triangle on its own does not count as stating the relationship. It has to be written as an equation, such as $V = I R$ or $R = \frac{V}{I}$ .

Question 4

A current of 0.40 A flows through a 15 Ω resistor. Calculate the potential difference across it.

$V = I R$ , so $V = 0.40 \times 15$ , giving $V = 6.0$ V.

Question 5

A car travels 4 km in 2 minutes. Calculate its average speed in metres per second.

$v = \frac{d}{t}$ , so $v = \frac{4,000}{120}$ , giving $v = 33$ m s^-1. The distance has to be converted to metres and the time to seconds before substituting.

Question 6

A student calculates a speed of 33.33333 m s^-1 from data given to two significant figures. What is wrong with the answer as written?

It carries far more significant figures than the data supports. Rounded to two or three significant figures, 33 or 33.3 m s^-1, it is within the accepted range.

Question 7

A student finishes a calculation with "= 24". What is missing?

The unit. A final answer in physics needs both a value and a unit, so it should read 24 Ω.

Question 8

The resistance of a wire rises from 20 Ω to 26 Ω. Calculate the percentage increase.

$\frac{26 - 20}{20} \times 100 = 30 %$ increase. The change of 6 is divided by the original value, 20.

key terms

TermDefinition

Significant figuresThe number of digits in a value, counted from the first non-zero digit, which shows how precisely it is given.

Decimal placesThe number of digits after the decimal point, which shows how precisely a value is given.

end of notes

Key points to remember

7.4 - Plotting Graphs & Bar ChartsNext up

Physics

7.3 - Processing Data & Calculations

7 min

note 1 of 4

What a calculation question looks like

Most numerical questions give you data to use with an equation from the relationships sheet.

Example: a student measures a potential difference of 6.0 V across a resistor and a current of 0.25 A through it - calculate the resistance.

The three parts of a calculation

Write down the relationship you're using: $R = \frac{V}{I}$
Substitute the values into it: $R = \frac{6.0}{0.25}$
Give the answer with its unit: $R = 24$ Ω

Why write it out in full

Write the relationship down even if you get stuck afterwards - It earns credit on its own.
If you substitute an incorrect earlier value but calculate correctly, subsequent steps can still gain marks if your working is shown.
The relationships sheet is given in the exam, so you don't need to memorise the equations - you just have to choose the right one and set it out clearly.

A typical question

A student measures a potential difference of 6.0 V across a resistor and a current of 0.25 A through it. Calculate the resistance of the resistor.

note 2 of 4

Setting out a calculation

Write every calculation in the same three lines so each step can be credited separately:

$R = \frac{V}{I}$

$R = \frac{6.0}{0.25}$

$R = 24 Ω$

Line one is the relationship, written out as an equation.
Line two is the substitution - The same equation with numbers in place of the symbols, converted into the required units beforehand.
Line three is the answer with its unit. A number without a unit is incomplete.

Rules for line one

A magic triangle is not a relationship. Drawing a formula triangle doesn't count on its own. You must write $V = I R$ or $R = \frac{V}{I}$ as an equation.
Choose one relationship, don't list several. Writing multiple equations without choosing one to substitute into earns nothing.

Recurring general calculations

Calculation	How to do it	Worked example
Mean	Add the repeats and divide by how many there are.	Resistances of 22, 24, and 26 Ω: $\frac{22 + 24 + 26}{3} = 24$ Ω
Percentage	The part over the whole, times 100.	18 of 25 bulbs working: $\frac{18}{25} \times 100 = 72 %$
Percentage change	The change over the original value, times 100.	Resistance rises from 20 Ω to 26 Ω: $\frac{26 - 20}{20} \times 100 = 30 %$ increase

note 3 of 4

Significant figures, units, and prefixes

How many significant figures

Round your final answer to about the same number of significant figures as the starting data.
SQA allows an answer with up to two figures more, or one figure fewer, than the piece of data with the fewest significant figures.
For $R = \frac{6.0}{0.25}$ , both values have two significant figures, so 24 Ω is ideal (acceptable range: 20 Ω to 24.00 Ω).

Significant figures vs decimal places:

Significant figures are counted from the first non-zero digit - so 47.316 to three significant figures is 47.3.
Decimal places are counted after the decimal point - so 6.028 to one decimal place is 6.0, but to one significant figure it's 6.
Don't round intermediate values during a calculation - carry the full figures and round only the final answer.
Don't use a recurring dot notation, as it implies infinite precision.

Convert units before substituting

Convert quantities into standard units before line two. For a distance of 4 km over a time of 2 minutes:

$v = \frac{d}{t}$

$v = \frac{4,000}{120}$

$v = 33 {m s}^{- 1}$

Prefixes and abbreviations

Use the correct case for unit prefixes (e.g. 5 ms is 5 milliseconds, while 5 Ms is 5 megaseconds). Standard abbreviations must be used:

Write this	Not this
second, s	sec, secs
metres per second, m/s or m s^-1	mps, m/s^-1
metres per second squared, m/s² or m s^-2	m/s/s, mpsps

note 4 of 4

Common exam questions

Question 1

A calculation is written out in three lines. What are they?

The relationship you are using, written as an equation; the substitution, with your values in place of the symbols; and the final answer with its unit.

Question 2

Why should you write down the relationship even if you cannot finish the calculation?

Because writing the correct relationship is looked at in its own right, so it stands even when the rest of the working goes wrong or is never completed.

Question 3

A student draws a triangle with $V$ on top and $I$ and $R$ below it, then writes the answer. Why is the triangle not enough?

Because a magic triangle on its own does not count as stating the relationship. It has to be written as an equation, such as $V = I R$ or $R = \frac{V}{I}$ .

Question 4

A current of 0.40 A flows through a 15 Ω resistor. Calculate the potential difference across it.

$V = I R$ , so $V = 0.40 \times 15$ , giving $V = 6.0$ V.

Question 5

A car travels 4 km in 2 minutes. Calculate its average speed in metres per second.

$v = \frac{d}{t}$ , so $v = \frac{4,000}{120}$ , giving $v = 33$ m s^-1. The distance has to be converted to metres and the time to seconds before substituting.

Question 6

A student calculates a speed of 33.33333 m s^-1 from data given to two significant figures. What is wrong with the answer as written?

It carries far more significant figures than the data supports. Rounded to two or three significant figures, 33 or 33.3 m s^-1, it is within the accepted range.

Question 7

A student finishes a calculation with "= 24". What is missing?

The unit. A final answer in physics needs both a value and a unit, so it should read 24 Ω.

Question 8

The resistance of a wire rises from 20 Ω to 26 Ω. Calculate the percentage increase.

$\frac{26 - 20}{20} \times 100 = 30 %$ increase. The change of 6 is divided by the original value, 20.

key terms

TermDefinition

Significant figuresThe number of digits in a value, counted from the first non-zero digit, which shows how precisely it is given.

Decimal placesThe number of digits after the decimal point, which shows how precisely a value is given.

end of notes

Key points to remember

7.4 - Plotting Graphs & Bar ChartsNext up