IGCSE Physics: Forces and Motion — The Complete Revision Guide

Forces and motion is the backbone of IGCSE Physics, and it all starts with three fundamental quantities: speed, velocity, and acceleration. Understanding these — and being able to work with the graphs that represent them — is essential for every paper you will sit.

Speed

is the distance travelled per unit time. The formula is s = d / t, where s is speed, d is distance, and t is time. Speed is a scalar quantity, which means it has magnitude only and no direction. Velocity, on the other hand, is a vector: it is speed in a given direction. The distinction matters on the exam. If a question asks for velocity and you give speed without direction, you will lose the mark.

Acceleration

is the rate of change of velocity. The formula is a = (v - u) / t, where v is the final velocity, u is the initial velocity, and t is the time taken. Acceleration is measured in m/s². If an object is slowing down, the acceleration is negative — this is called deceleration. Cambridge examiners love testing whether you can correctly identify deceleration from a velocity-time graph.

Now, graphs. A distance-time graph shows how far an object has travelled over time. The gradient of the line equals the speed. A flat horizontal line means the object is stationary. A straight diagonal line means constant speed. A curve that gets steeper means the object is accelerating. A velocity-time graph is even more information-rich. The gradient gives you acceleration, and the area under the graph gives you the distance travelled. A flat horizontal line on a velocity-time graph means constant velocity — not that the object is stationary. This is one of the most common misinterpretations I see in student work. When calculating the area under a velocity-time graph, break it into rectangles and triangles. For a triangle, area = 0.5 x base x height. For a trapezium, area = 0.5 x (a + b) x h. These are given on the formula sheet, but you must know when to apply them.

Finally, make sure you are comfortable converting units. Speed might be given in km/h but the answer required in m/s. To convert km/h to m/s, divide by 3.6. To convert m/s to km/h, multiply by 3.6. Examiners frequently set traps with mixed units, and students who rush straight into calculations without checking units lose easy marks.

Newton's laws are the theoretical heart of the forces and motion topic. You need to be able to state each law precisely, explain it with examples, and — critically — apply it to exam scenarios. Vague descriptions will not earn marks; Cambridge wants precision.

Newton's First Law

states that an object will remain at rest or continue to move at a constant velocity in a straight line unless acted on by a resultant force. The key phrase is "resultant force." An object can have multiple forces acting on it and still obey the first law, provided those forces are balanced and the resultant is zero. A common exam question shows a car moving at constant velocity and asks students to explain why. The answer is that the driving force equals the resistive forces (friction and air resistance), so the resultant force is zero, so by Newton's First Law, the velocity does not change.

Newton's Second Law

states that the resultant force acting on an object is equal to the product of its mass and acceleration: F = ma. Force is measured in newtons (N), mass in kilograms (kg), and acceleration in m/s². This is the workhorse equation of IGCSE Physics. You will use it in calculation questions more than almost any other formula. A typical exam question might give you the mass of a car and the resultant force, and ask you to find the acceleration. Or it might give you mass and acceleration and ask for the force. Make sure you can rearrange the formula confidently in all three directions: F = ma, m = F/a, a = F/m. One important subtlety: the "F" in F = ma is the resultant force, not just any single force. If a question tells you the engine force is 2000 N and friction is 500 N, you must first calculate the resultant force (1500 N) before using F = ma.

Newton's Third Law

states that for every action, there is an equal and opposite reaction. More precisely: when object A exerts a force on object B, object B exerts an equal force in the opposite direction on object A. The two forces act on different objects — this is the detail students most commonly get wrong. If you push a wall, the wall pushes you back with the same force. Your weight pushes down on the floor; the floor pushes up on you with a normal contact force. These are Newton's Third Law pairs. But weight and normal contact force acting on the same object are NOT a Third Law pair — they are a balanced pair under the First Law. Examiners specifically test whether you understand this distinction, and it appears regularly on Paper 4.

Now that you understand Newton's laws, let us look at the specific forces you need to know for IGCSE Physics and how they interact in real scenarios — particularly free fall and terminal velocity.

Weight

is the force of gravity acting on an object. It is calculated using W = mg, where m is the mass in kg and g is the gravitational field strength. On Earth, g = 9.8 N/kg (often rounded to 10 N/kg in IGCSE calculations). Weight acts downward, toward the centre of the Earth. Do not confuse weight with mass. Mass is a measure of the amount of matter in an object and does not change with location. Weight depends on the gravitational field strength, so the same object weighs less on the Moon (where g is approximately 1.6 N/kg) than on Earth.

Friction

is a contact force that opposes motion between two surfaces. It acts in the opposite direction to the movement. Friction can be useful (it allows us to walk without slipping, it lets car tyres grip the road) or wasteful (it generates unwanted heat in engines). On the exam, you may be asked to identify friction on a free-body diagram and show its direction.

Air resistance

(or drag) is a type of friction that acts on objects moving through air. It increases as the speed of the object increases. This is the key to understanding terminal velocity. When a skydiver jumps from a plane, initially the only significant force is weight acting downward, so the skydiver accelerates downward. As speed increases, air resistance increases. The resultant downward force decreases, so acceleration decreases (the skydiver is still speeding up, but more slowly). Eventually, air resistance equals weight, the resultant force becomes zero, and by Newton's First Law the skydiver moves at constant velocity — this is terminal velocity. When the parachute opens, air resistance suddenly increases dramatically, becoming much greater than weight. The resultant force is now upward, so the skydiver decelerates. As speed decreases, air resistance decreases again until a new, much lower terminal velocity is reached.

Free-body diagrams

are essential for this topic. A free-body diagram shows all the forces acting on a single object, drawn as arrows from the centre of the object. The length of each arrow represents the magnitude of the force, and the direction shows which way the force acts. On the exam, label each arrow clearly (e.g., "weight," "air resistance," "normal contact force," "friction"). Examiners award marks for correct direction, correct labelling, and correct relative sizes of the arrows.

Momentum is a topic that appears reliably on Paper 4, and the calculation questions tend to follow predictable patterns. If you practise the method, you can pick up full marks consistently.

Momentum

is defined as the product of an object's mass and velocity: p = mv. The unit is kg m/s (or N s). Momentum is a vector quantity — it has both magnitude and direction. This means that if two objects are moving in opposite directions, one will have positive momentum and the other negative momentum (depending on which direction you define as positive).

The principle of conservation of momentum states that in a closed system (where no external forces act), the total momentum before an event equals the total momentum after the event. This applies to collisions and explosions. For a collision between two objects: m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2, where u represents initial velocities and v represents final velocities.

Let me walk you through a typical exam calculation. A trolley of mass 2 kg moving at 3 m/s collides with a stationary trolley of mass 1 kg. After the collision, they stick together. Find the velocity after the collision. Step 1: Calculate total momentum before the collision. Momentum of trolley 1 = 2 x 3 = 6 kg m/s. Momentum of trolley 2 = 1 x 0 = 0 kg m/s. Total momentum before = 6 kg m/s. Step 2: Apply conservation of momentum. Total momentum after = 6 kg m/s. Step 3: Calculate the velocity. Combined mass = 2 + 1 = 3 kg. Velocity = 6 / 3 = 2 m/s. That is the complete method, and you must show every step to earn full marks.

Impulse

is the change in momentum of an object. It equals the force multiplied by the time for which the force acts: impulse = F x t = change in momentum = mv - mu. This relationship explains why car safety features work. A crumple zone increases the time over which the car decelerates during a collision. Since impulse (the change in momentum) is the same regardless of the time, increasing the time reduces the force experienced by the passengers. The same principle applies to airbags, seatbelts, and crash helmets — they all increase the time of deceleration to reduce the maximum force.

A common exam trap involves collisions where objects move in opposite directions. If object A moves to the right at 4 m/s and object B moves to the left at 2 m/s, you must assign one direction as positive. If right is positive, then object A has momentum +m1 x 4 and object B has momentum -m2 x 2. Forgetting the negative sign is one of the most frequent errors in momentum calculations.

Stopping distance questions bridge physics and real-world application, and they appear frequently on IGCSE papers. The concept is straightforward, but the detail in the factors is where marks are won and lost.

Stopping distance = thinking distance + braking distance.

Thinking distance is the distance the car travels during the driver's reaction time — the time between seeing a hazard and pressing the brake. Braking distance is the distance the car travels from the moment the brakes are applied until the car comes to a complete stop.

Factors affecting thinking distance

relate to the driver: tiredness, alcohol or drugs, distractions (using a mobile phone), age, and illness all increase reaction time. If the car is travelling faster, the thinking distance is also greater because the car covers more ground during the same reaction time. Notice that thinking distance is proportional to speed: if speed doubles, thinking distance doubles.

Factors affecting braking distance

relate to the car and the road conditions: worn brakes, worn tyres, wet or icy road surfaces, loose gravel, and the mass of the car. Heavier cars require a greater braking force to decelerate at the same rate, and poor road conditions reduce the friction between the tyres and the road. Critically, braking distance is proportional to the square of the speed. If speed doubles, braking distance quadruples. This is because the kinetic energy of the car is proportional to v², and the brakes must do work equal to the kinetic energy to bring the car to rest. The work done by the brakes = braking force x braking distance. Since work = kinetic energy = 0.5 x m x v², if v doubles, the energy quadruples, and so the braking distance quadruples (assuming the braking force stays the same).

On the exam, you may be given a table of stopping distances at different speeds and asked to explain the pattern. You might also be asked to calculate thinking distance given speed and reaction time (thinking distance = speed x reaction time), or to explain why stopping distance increases on a wet road. For the wet road question, the answer is that water acts as a lubricant between the tyres and the road, reducing friction, which reduces the braking force. Since work = force x distance, a smaller force requires a greater distance to do the same amount of work (dissipate the same kinetic energy). Always structure your answer as a chain of cause and effect — this is how examiners award marks.

Another important calculation involves the relationship between kinetic energy and braking. If a car of mass 1200 kg is travelling at 20 m/s, its kinetic energy is 0.5 x 1200 x 20² = 240,000 J. If the braking force is 8000 N, the braking distance is 240,000 / 8000 = 30 m. This type of question combines energy and forces, and it is a favourite on extended papers.

Knowing the physics is only half the battle on the IGCSE exam. The other half is presenting your knowledge in the way that earns maximum marks. After years of marking and tutoring, here are the techniques that separate A* students from the rest in forces and motion questions.

Show every step of your working.

This is the single most important piece of advice I can give. Even if you can do the calculation in your head, write out: (1) the formula you are using, (2) the substitution with numbers, (3) the final answer with the correct unit. Cambridge awards method marks at each stage. If you make an arithmetic error but your method is correct, you can still earn two out of three marks. If you only write the final answer and it is wrong, you get zero. I have seen students lose dozens of marks across a paper simply because they skipped working.

Always include units in your final answer.

If the question asks for force and you write "150" instead of "150 N," you will lose the mark. Common units you need: speed (m/s), acceleration (m/s²), force (N), momentum (kg m/s), energy (J), weight (N), mass (kg), distance (m), time (s). When calculating, check that your units are consistent — if mass is given in grams, convert to kilograms before using F = ma.

Read the command word carefully.

"State" means give a brief factual answer — one sentence is enough. "Explain" means give a reason using physics principles — you need a because-chain. "Calculate" means show numerical working. "Describe" means say what happens in sequence. "Compare" means identify similarities AND differences. Using the wrong level of detail for the command word either wastes your time or costs you marks.

Draw diagrams even when not asked.

For forces questions, a quick free-body diagram helps you identify all the forces before you start calculating. For motion questions, a quick sketch of the velocity-time graph can help you visualise what is happening. These diagrams take 30 seconds and prevent errors that cost 3-4 marks.

Common mark-losing mistakes in forces and motion:

(1) Forgetting to calculate the resultant force before using F = ma. (2) Using weight instead of mass in F = ma (mass is in kg, not newtons). (3) Confusing mass and weight in written explanations. (4) Not converting units (e.g., using grams instead of kilograms, or km/h instead of m/s). (5) On graph questions, reading values inaccurately — use a ruler to draw lines to the axes. (6) In momentum questions, forgetting to assign negative values when objects move in opposite directions. (7) In terminal velocity explanations, saying "air resistance equals gravity" instead of "air resistance equals weight" — gravity is a field, weight is a force.

Time management:

Forces and motion questions on Paper 4 typically carry 6-10 marks and should take about 10-15 minutes. If you are spending more than 15 minutes on a single question, move on and come back. The marks you can earn on easier questions elsewhere are worth more than the final mark on a question you are struggling with.