Maths for IGCSE Physics 0625

This page brings the main information for Maths for IGCSE Physics 0625 into one place. Check the current syllabus cycle, then use the related practice to test what you can reproduce independently.

• they do not understand the Physics concept
• they understand the concept but cannot express it mathematically

The treatment should differ. This guide isolates the main mathematical skills used in Cambridge IGCSE Physics 0625 and then reconnects them to Physics questions.

Start with this diagnostic

Complete the questions without notes. Record the type of every error.

Units and standard form

1. Convert $72\text{ km/h}$ to m/s.
2. Convert $350\text{ cm}^3$ to m³.
3. Write $0.000\,004\,8$ in standard form.
4. Calculate $(6.0 \times 10^7)(2.0 \times 10^{-3})$.

Rearranging equations

5. Rearrange $v = s/t$ to make $t$ the subject.
6. Rearrange $R = V/I$ to make $I$ the subject.
7. Rearrange $E_k = \frac{1}{2}mv^2$ to make $v$ the subject.
8. Rearrange $p_1V_1 = p_2V_2$ to make $V_2$ the subject.

Ratios, percentages and graphs

9. A device receives 500 J and transfers 320 J usefully. Calculate its efficiency. State the appropriate tier for the calculation.
10. A graph rises from $(2.0\text{ s}, 5.0\text{ m/s})$ to $(8.0\text{ s}, 17.0\text{ m/s})$. Calculate the gradient.
11. A model diagram uses a scale of 1 cm to 5 N. What length represents 35 N?
12. A right-angled vector triangle has perpendicular sides 6.0 N and 8.0 N. Calculate the resultant magnitude.

Answers and diagnosis appear later on this page.

Skill 1: Units and prefixes

A correct calculation with inconsistent units can still produce a wrong answer.

Common prefixes

Check the quantity as well as the prefix.

• $1\text{ cm} = 10^{-2}\text{ m}$
• $1\text{ cm}^2 = 10^{-4}\text{ m}^2$
• $1\text{ cm}^3 = 10^{-6}\text{ m}^3$

A volume conversion is cubed. This is why converting 350 cm³ by dividing by 100 gives the wrong result.

Speed conversion

To convert km/h to m/s:

$\text{m/s} = \text{km/h} \times \frac{1000}{3600}$

Therefore:

$72\text{ km/h} = 72 \times \frac{1000}{3600} = 20\text{ m/s}$

To reverse the conversion, multiply m/s by 3.6.

A pre-substitution unit check

Before substituting, write:

mass = 250 g = 0.250 kg
area = 40 cm² = 0.0040 m²
current = 35 mA = 0.035 A

Do the conversion before inserting values into the equation. This makes the working easier to mark and debug.

Skill 2: Standard form

Standard form is written as:

$a \times 10^n$

where $1 \leq a < 10$.

Examples:

• $48\,000 = 4.8 \times 10^4$
• $0.000\,004\,8 = 4.8 \times 10^{-6}$

Multiplication

Multiply the leading numbers and add the powers:

$(6.0 \times 10^7)(2.0 \times 10^{-3}) = 12.0 \times 10^4 = 1.2 \times 10^5$

Division

Divide the leading numbers and subtract the powers:

$\frac{8.0 \times 10^9}{2.0 \times 10^3} = 4.0 \times 10^6$

Calculator entry

Use the calculator’s exponent key consistently, often labelled EXP, EE or ×10ˣ. Do not enter an extra multiplication by 10 if the exponent key already represents it.

After calculating, check the order of magnitude. Dividing by $10^{-18}$ should produce an extremely large value, not a small one.

Skill 3: Rearranging equations

Rearrangement should preserve equality by performing the same operation on both sides.

One-step examples

From:

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

multiply both sides by $t$:

$vt = s$

then divide by $v$:

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

From:

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

multiply by $I$:

$RI = V$

then divide by $R$:

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

Equations containing a square

From:

$E_k = \frac{1}{2}mv^2$

multiply by 2:

$2E_k = mv^2$

divide by $m$:

$\frac{2E_k}{m} = v^2$

take the square root:

$v = \sqrt{\frac{2E_k}{m}}$

A simple formula triangle cannot handle this safely.

Products on both sides

From:

$p_1V_1 = p_2V_2$

make $V_2$ the subject by dividing by $p_2$:

$V_2 = \frac{p_1V_1}{p_2}$

Keep paired units consistent.

Skill 4: Ratios and proportionality

Physics often asks how one quantity changes when another changes.

Direct proportion

If $y \propto x$, doubling $x$ doubles $y$.

Examples include resistance and wire length under the required conditions:

$R \propto l$

Inverse proportion

If $y \propto 1/x$, doubling $x$ halves $y$.

For a fixed mass of gas at constant temperature:

$p \propto \frac{1}{V}$

because $pV = \text{constant}$.

Square relationships

For kinetic energy:

$E_k \propto v^2$

Doubling speed makes kinetic energy four times as large when mass is constant.

For cable heating loss:

$P_{\text{loss}} \propto I^2$

Halving current reduces the loss to one quarter when resistance is constant.

Do not apply a linear rule to a squared relationship.

Skill 5: Percentages and efficiency

Percentage change is:

$\text{percentage change} = \frac{\text{change}}{\text{original}} \times 100\%$

Supplement efficiency calculations use:

$\text{efficiency} = \frac{\text{useful output}}{\text{total input}} \times 100\%$

For a 500 J input and 320 J useful output:

$\text{efficiency} = \frac{320}{500} \times 100\% = 64\%$

The useful output belongs on top. An efficiency greater than 100% indicates an error in this model.

Skill 6: Graphs and gradients

A gradient is:

$\text{gradient} = \frac{\text{change in vertical quantity}}{\text{change in horizontal quantity}}$

For the points $(2.0,5.0)$ and $(8.0,17.0)$:

$\text{gradient} = \frac{17.0 - 5.0}{8.0 - 2.0} = \frac{12.0}{6.0} = 2.0\text{ m/s}^2$

Use a large triangle on a best-fit line where possible. The two points used for the gradient do not have to be original data points if they lie accurately on the drawn line.

Interpreting a gradient

The meaning depends on the axes:

• distance-time gradient gives speed
• speed-time gradient gives acceleration
• current-voltage gradient is not automatically resistance unless the axes and relationship are considered correctly

Always state the gradient unit from the vertical unit divided by the horizontal unit.

Area under a graph

The meaning also depends on the axes. For example, the area under a speed-time graph gives distance travelled for the specified situations.

Count units and dimensions before assuming that an area represents a physical quantity.

Skill 7: Scale drawings, geometry and trigonometry

Scale drawings

If 1 cm represents 5 N, then 35 N is represented by:

$\frac{35}{5} = 7\text{ cm}$

Label the scale and measure from the correct point.

Pythagoras

For perpendicular components 6.0 N and 8.0 N:

$R = \sqrt{6.0^2 + 8.0^2} = 10.0\text{ N}$

This applies only to a right-angled triangle.

Trigonometry

For a right-angled triangle:

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

Check whether the calculator is in degree mode for IGCSE angle work.

Refractive-index equations use sine directly:

$n = \frac{\sin i}{\sin r}$

Do not replace the sine values with the angles themselves.

Skill 8: Significant figures and decimal places

Use the precision of the data and the instructions in the question.

General habits:

• keep extra digits during intermediate calculation
• round once at the end
• include trailing zeros when they show required precision
• match repeated raw measurements sensibly in a table
• do not write an unjustifiably long calculator display

Examples:

• 3.246 to 3 significant figures is 3.25
• 0.004856 to 2 significant figures is 0.0049
• 12.00 has four significant figures

In practical work, decimal places can be linked to instrument resolution. Follow the context and the paper instructions.

Skill 9: Estimation and sense checks

Before accepting an answer, ask:

• Is the sign sensible?
• Is the order of magnitude sensible?
• Is the unit correct?
• Is an efficiency between 0% and 100%?
• Is a combined parallel resistance smaller than either branch resistance?
• Does a step-down transformer have fewer secondary turns?
• Does doubling speed affect a squared quantity correctly?

A ten-second check catches many calculator and algebra errors.

Diagnostic answers

1. $20\text{ m/s}$
2. $3.50 \times 10^{-4}\text{ m}^3$
3. $4.8 \times 10^{-6}$
4. $1.2 \times 10^5$
5. $t = s/v$
6. $I = V/R$
7. $v = \sqrt{2E_k/m}$
8. $V_2 = p_1V_1/p_2$
9. $64\%$; the calculation equation is Supplement
10. $2.0\text{ m/s}^2$
11. $7\text{ cm}$
12. $10\text{ N}$

What does the result mean?

• Errors in Questions 1 to 4 suggest unit or standard-form work.
• Errors in Questions 5 to 8 suggest algebraic rearrangement.
• Errors in Question 9 suggest ratio or tier confusion.
• Errors in Question 10 suggest gradient work.
• Errors in Questions 11 and 12 suggest scale or geometry work.

Do not convert the twelve questions into a predicted Physics grade. Use them to choose the next practice set.

A four-week Mathematics repair plan

Week 1: Units and standard form

• ten conversions per day
• three standard-form calculations per session
• Physics examples from density, electricity and space

Week 2: Rearrangement

• begin with one-step equations
• progress to fractions and squares
• state the required subject before rearranging

Week 3: Ratios, percentages and proportionality

• efficiency
• transformer ratios
• direct and inverse relationships
• squared relationships

Week 4: Graphs and mixed calculations

• gradients and units
• areas where required
• multi-stage questions
• full method with sense check

Retest the original diagnostic at the end using new numbers rather than memorising the answers.

A calculation checklist

Before submitting a numerical answer, check:

• I identified the required quantity.
• I selected an equation from the current syllabus.
• I converted units before substitution.
• I rearranged correctly.
• I entered brackets and powers correctly.
• I rounded appropriately.
• I wrote the unit.
• I checked whether the answer is physically sensible.

Mathematics should become a visible, repeatable method rather than an invisible source of anxiety. Use the topic question bank to apply each skill inside original Physics questions.