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Revision Notes

Forces and Motion: Revision Notes

Condensed recall notes on Newton’s laws, F = ma, friction, terminal velocity and stopping distance for Cambridge O Level Physics 5054.

Subject
Physics
Level
O LEVELS
Topic
Motion, forces and energy
Updated

Aligned to Cambridge O Level Physics (5054), 2026-2028. Official specification .

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Condensed for the final weeks. For the full explanation, use the Forces and Motion study guide, and for exam-style practice with mark-scheme answers see the Forces and Motion practice questions.

Types of force

The syllabus names nine forces you should be able to recognise: weight, friction, drag, air resistance, tension, electrostatic force, magnetic force, thrust, and contact force. A free-body diagram shows only the object of interest, with one arrow per force acting on it — arrow length roughly shows size, arrow direction shows direction. Draw one before calculating anything; it is the fastest way to spot the resultant.

Newton’s three laws

Law Statement What it means in practice
First An object stays at rest or at constant velocity unless acted on by a resultant force Constant velocity ⇒ zero resultant force
Second Resultant force produces acceleration in its direction F = m a
Third If object A exerts a force on object B, then object B exerts an equal and opposite force on object A (“action and reaction”) The pair acts on two different objects

The equation

F = m a        F in newtons, m in kg, a in m/s2

weight:  W = m g       (g = 9.8 or 10 N/kg)

Mass is the amount of matter in kg and never changes. Weight is a force in newtons and changes with gravitational field strength.

Worked example. A resultant force of 15 N acts on a 3 kg object.

a = F / m = 15 / 3 = 5 m/s2

Resultant force

Forces along one line add algebraically. Direction matters — forces are vectors.

Driving force 4000 N forward, drag 1500 N backward
Resultant = 4000 - 1500 = 2500 N forward

The driving force is not the resultant — the resultant is what remains after drag and friction. This is the single most common error in the topic.

Circular motion (qualitative)

A resultant force acting perpendicular to an object’s motion changes its direction without (necessarily) changing its speed — this is how circular motion works. With everything else held constant:

  • Speed increases if force increases.
  • Radius decreases if force increases — a tighter circle needs a bigger force at the same speed.
  • A bigger mass needs a bigger force to keep speed and radius the same.

F = mv2/r is not required at this level — give the qualitative relationships only, never the equation.

Terminal velocity

  1. Object falls; only weight acts → maximum acceleration.
  2. Speed rises → drag increases.
  3. Drag grows until drag = weight → resultant = 0.
  4. Acceleration = 0 → constant terminal velocity.

On a velocity–time graph the line flattens — the object does not slow down.

Stopping distance

stopping distance = thinking distance + braking distance
Increased by
Thinking distance Speed (∝ v), tiredness, alcohol, drugs, distraction
Braking distance Speed (∝ v²), wet or icy roads, worn tyres, worn brakes, heavy load

Doubling speed doubles thinking distance but quadruples braking distance.

Third-law pairs — how to check

A genuine pair: same type of force · equal size · opposite direction · acting on two different bodies. Same type of force is the check most students skip — a weight force can only pair with another weight force, never with a contact force, even if the sizes happen to match.

A book on a table: weight (Earth on book) and normal contact force (table on book) both act on the book, so they are a first-law balance, not a third-law pair.

Exam traps

  • Confusing mass with weight, or their units.
  • Treating the driving force as the resultant.
  • Saying an object at terminal velocity “has no forces on it” — it has no resultant force.
  • Naming a third-law pair that acts on the same object.
  • Forgetting braking distance scales with v², not v.
  • Trying to use F = mv2/r — it is explicitly excluded from this syllabus; answer circular-motion questions with the qualitative trends only.
  • Skipping the free-body diagram and guessing the resultant — draw the arrows first.

Self-test

  1. A 1500 kg car accelerates at 2 m/s². Find the resultant force.
  2. If drag on that car is 600 N, what is the driving force?
  3. Explain terminal velocity in three steps.
  4. Why is the weight of a book and the table’s push on it not a third-law pair?
  5. A car doubles its speed. What happens to thinking and braking distance?
  6. Name the nine force types the syllabus expects you to recognise.
  7. A ball on a string is swung in a horizontal circle at constant speed. If the string is shortened while the speed stays the same, what happens to the force needed, and why?

Answers: 1. F = 1500 × 2 = 3000 N. 2. 3000 + 600 = 3600 N. 3. Weight causes acceleration; drag increases with speed; when drag equals weight the resultant is zero and velocity becomes constant. 4. Both forces act on the same object (the book); a third-law pair must act on two different bodies. 5. Thinking distance doubles; braking distance quadruples. 6. Weight, friction, drag, air resistance, tension, electrostatic force, magnetic force, thrust, contact force. 7. The force needed increases — with speed and mass unchanged, a smaller radius needs a bigger centre-seeking force to keep the ball on its (now tighter) circular path.

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