OxfordAQA International GCSE Physics: Resultant forces and Newton's laws (9203) – Revision Notes

These notes follow section 3.1.3, Resultant forces (statements a to h), of the 9203 OxfordAQA International GCSE Physics specification for exams from May/June 2018. It is an untiered qualification. If a point below is unclear, the study guide explains it step by step.

Also see the practice questions, the OxfordAQA IGCSE Physics hub, the printable checklist, the free diagnostics and the wider forces notes for the course.

Key facts at a glance

Idea	Spec	Remember
Newton’s Third Law	3.1.3 a	Pair: equal size, opposite direction, same type, two different objects
Resultant force	3.1.3 b	One force with the same effect as all the forces together
Resultant and acceleration	3.1.3 c	Non-zero resultant means acceleration
Acceleration	3.1.3 d	a = Δv / t; deceleration is negative acceleration
Gradient of v-t graph	3.1.3 e	Acceleration
Area under v-t graph	3.1.3 f	How far the object moves
Newton’s First Law	3.1.3 g	Zero resultant: constant velocity or at rest
Newton’s Second Law	3.1.3 h	F = ma, in the direction of the resultant

Equations and units

Quantity	Equation	Units
Acceleration	a = Δv / t	m/s²
Change in velocity	Δv = final - initial	m/s
Resultant force	F = ma	N
Mass	m = F / a	kg
Acceleration from force	a = F / m	m/s²

Rearrange before you substitute, and convert grams to kilograms and minutes to seconds first.

Newton’s Third Law

• The two forces of a pair sit on different objects, so neither is part of the other’s resultant.
• Weight (the Earth pulling the object) and the push of a surface on it are not a pair: one object, two different types of force.
• Pair for weight: the object pulling the Earth upwards, same size.
• Pair for a push of a wall on your hand: your hand pushing the wall.

Method for naming a pair: write “A exerts a force on B” and swap to “B exerts the same type of force on A”.

Finding a resultant

Forces in a line:

1. Choose a positive direction.
2. Add forces that point that way, subtract the others.
3. State size and direction of the answer.

Scale drawing for two coplanar forces:

1. Pick a scale (for example 1 cm to 1 N).
2. Draw force 1 with correct length and direction.
3. From its end, draw force 2 at the correct angle, head to tail.
4. Join start to finish: this is the resultant.
5. Measure length (convert with the scale) and measure the angle with a protractor.
6. State the angle relative to a named force.

Sense check: the resultant lies between the difference and the sum of the two force sizes.

Worked reminder: forces of 10 N and 6 N with 60° between them give a resultant of about 14 N, roughly 22° from the 10 N force.

Acceleration

• Acceleration is the rate of change of velocity.
• Constant speed with changing direction is still acceleration, because velocity changes.
• Deceleration: negative acceleration, so Δv is negative.
• Circular path at steady speed: the resultant force points to the centre.

Worked reminder: velocity 5 m/s to 17 m/s in 4 s.

a = (17 - 5) / 4 = 3.0 m/s²

Velocity-time graphs

Feature of the line	Meaning
Straight line sloping up	Constant positive acceleration
Horizontal line	Constant velocity, zero resultant force
Straight line sloping down	Constant deceleration
Steeper line	Larger acceleration
Area under the line	Distance travelled
Curve getting steeper	Increasing acceleration

Method for area: split the graph into rectangles and triangles, work out each, then add. Use half x base x height for a triangle.

Worked reminder: velocity rises from 0 to 9 m/s in 3 s, then stays at 9 m/s for 6 s.

Distance = 1/2 x 3 x 9 + 6 x 9 = 13.5 + 54 = 67.5 m

Do not confuse the graph types: gradient of a distance-time graph is speed, gradient of a velocity-time graph is acceleration.

Newton’s First Law

• Resultant zero, object moving: same speed, same direction.
• Resultant zero, object at rest: stays at rest.
• A steady speed in a straight line needs balanced forces, not a resultant force.
• When a vehicle brakes, loose passengers carry on forwards because nothing slows them until a seat belt or seat does.

Newton’s Second Law

Method in steps:

1. Draw and label all forces on the object.
2. Find the resultant (driving minus resistive, or net of all forces).
3. Rearrange F = ma for the unknown.
4. Substitute in base units and state the unit.

Worked reminder: a 2000 kg van has a driving force of 5000 N and resistive forces of 1400 N.

Resultant = 5000 - 1400 = 3600 N
a = 3600 / 2000 = 1.8 m/s²

Proportionality: more resultant force means more acceleration; more mass means less acceleration for the same force.

Must-know distinctions

• Speed has size only; velocity has size and direction.
• Balanced forces (zero resultant) differ from no forces.
• Resultant force differs from any single force acting.
• Mass (kg) differs from weight (N).
• Deceleration is acceleration with a negative value, not a separate quantity.
• a Third Law pair involves two objects; balanced forces involve one.

One extended reminder

A 900 kg car speeds up from 4 m/s to 24 m/s in 10 s, with resistive forces of 500 N. Find the acceleration, the resultant force, the driving force and the distance travelled, assuming constant acceleration.

a = (24 - 4) / 10 = 2.0 m/s²
Resultant force F = ma = 900 x 2.0 = 1800 N
Driving force = resultant + resistive = 1800 + 500 = 2300 N
Distance = area under the line = (4 + 24) / 2 x 10 = 140 m

The order is always the same: acceleration from the graph or from Δv / t, then resultant force from F = ma, then the individual force from the resultant.

Command words and answer shape

• State: a short fact, such as the wording of a law.
• Calculate: show the equation, substitution and answer with units.
• Explain: give a reason that links cause and effect, such as “the resultant force is zero, so the velocity does not change”.
• Describe: say what happens, using the graph features (gradient, horizontal section, area).
• Determine from a graph: show where the gradient or area comes from, with numbers read from the axes.

For a scale drawing, give the scale, the measured length, the converted force and the angle with its reference line.

Quick self-test

1. State two features of a pair of forces in Newton’s Third Law.
2. A box has 90 N pushing it right, 35 N and 20 N pushing it left. Find the resultant.
3. A boat’s speed stays constant as it moves round a circular course. Is it accelerating? Explain.
4. A cyclist speeds up from 3 m/s to 15 m/s in 8 s. Calculate the acceleration.
5. A tram slows from 16 m/s to rest in 5.0 s. Give its acceleration.
6. Which quantity is found from the area under a velocity-time graph?
7. An object’s velocity rises steadily from 0 to 16 m/s over 5.0 s. How far does it travel?
8. A car moves at constant velocity on a level road. What can you say about the forces on it?
9. A resultant force of 600 N acts on a 150 kg object. Find its acceleration.
10. A 0.25 kg trolley accelerates at 6.0 m/s². What resultant force acts on it?
11. A skater of mass 60 kg is given an acceleration of 0.50 m/s². Find the resultant force.
12. A force of 24 N and a force of 7 N act at right angles. Find the resultant size.

Answers

1. Equal in size, opposite in direction (also: same type of force, acting on different objects).
2. 90 - 35 - 20 = 35 N to the right.
3. Yes. Its direction is changing, so its velocity is changing, even though its speed is constant.
4. a = (15 - 3) / 8 = 1.5 m/s².
5. a = (0 - 16) / 5.0 = -3.2 m/s² (deceleration of 3.2 m/s²).
6. The distance travelled.
7. Triangle: 1/2 x 5.0 x 16 = 40 m.
8. The forces are balanced: the resultant force is zero.
9. a = 600 / 150 = 4.0 m/s².
10. F = 0.25 x 6.0 = 1.5 N.
11. F = 60 x 0.50 = 30 N.
12. Pythagoras: √(24² + 7²) = √625 = 25 N.

Where marks are usually lost

• Writing “the forces cancel” for a Third Law pair instead of saying they act on different objects.
• Subtracting forces in the wrong order and giving a negative resultant without a direction.
• Putting only the driving force into F = ma.
• Not subtracting the starting velocity when finding Δv.
• Missing the minus sign on a deceleration.
• Forgetting that motion in a circle at constant speed involves acceleration.
• Calculating the area of a triangle without the half.
• Giving a scale-drawing angle without saying which force it is measured from.
• Leaving out units, or leaving mass in grams.

Official syllabus

Source document: OxfordAQA International GCSE Physics (9203), version 5.2, exams May/June 2018 onwards. Scope here: section 3.1.3, a to h.