Newton's Laws of Motion Revision Notes | A Level AQA Physics

1. Why Newton's Laws Are the Engine of Mechanics

Picture a book sitting on a table. Nothing is moving. Now push it — it slides, slows, and stops. Why does it behave exactly that way, every single time? The answer lies in three laws that Isaac Newton published in 1687, and they remain the foundation of every mechanics problem you will face at A Level.

These laws are not just definitions to memorise. They are a framework for reasoning about any situation involving forces and motion. Master them and you can analyse a skydiver, a rocket launch, or a collision — all with the same toolkit.

2. Newton's First Law — The Law of Inertia

Think about a hockey puck sliding across frictionless ice. Nothing is pushing it forward, and nothing is slowing it down. It keeps going at the same speed in the same direction — indefinitely. That behaviour is Newton's First Law in action.

The key phrase here is resultant force. If all the forces on an object cancel out — meaning the resultant is zero — the object behaves as though no forces exist at all. This state is called equilibrium. A car travelling at a steady 30 m/s on a motorway is in equilibrium: the driving force from the engine exactly balances air resistance and friction.

The property that causes objects to resist changes in their motion is called inertia. More massive objects have greater inertia — they are harder to start moving and harder to stop. This is why a lorry takes much longer to brake than a bicycle travelling at the same speed.

A common place this law appears in exam questions is in the analysis of objects moving at constant velocity. If speed and direction are both constant, the resultant force must be zero — even if several individual forces are acting. That conclusion alone can earn you marks.

Newton's First Law establishes the concept of equilibrium and inertia — the bedrock on which the second and third laws are built. With that foundation in place, the natural next question is: what happens when the resultant force is not zero?

3. Newton's Second Law — Force, Mass, and Acceleration

When a resultant force does act on an object, the object accelerates. The second law tells you precisely how much. Push a trolley with twice the force and it accelerates twice as fast. Double the mass of the trolley and the same force produces half the acceleration. These relationships are captured in one of the most important equations in physics.

Notice that $F$ here is the resultant force — the vector sum of all forces acting. If a 1200 kg car has a driving force of 4000 N and a friction force of 1000 N acting against it, the resultant force is 3000 N forwards, and the acceleration is $a = F/m = 3000/1200 = 2.5$ m s$^{-2}$. You must always find the resultant before applying $F = ma$.

At A Level, the second law is also expressed in its more general form: $F = \Delta p / \Delta t$, where $p$ is momentum ($p = mv$). For constant mass, $\Delta p / \Delta t = m \Delta v / \Delta t = ma$, recovering the familiar equation. The momentum form becomes essential when mass changes — for example in rocket propulsion — but for most mechanics problems at this level, $F = ma$ applies directly.

Worked example: Applying Newton's Second Law with Multiple Forces

A 900 kg car accelerates from rest. The engine provides a driving force of 5400 N. A resistive force of 1800 N acts against the motion. Calculate the acceleration of the car. [3 marks]

1. Driving force (forwards): $F_d = 5400$ N Resistive force (backwards): $F_r = 1800$ N — Always list every force with its direction before calculating. This prevents sign errors. [M1 — identifying forces correctly]
2. $F_{\text{resultant}} = 5400 - 1800 = 3600$ N (forwards) — Subtract the opposing force from the driving force. The resultant acts in the direction of motion here.
3. $F = ma \Rightarrow a = \dfrac{F}{m} = \dfrac{3600}{900}$ — [M1 — correct substitution into $F = ma$]
4. $a = \dfrac{3600}{900} = 4.0$ m s$^{-2}$ — Verify: $3600 \div 900 = 4.0$. Units check: N/kg = kg m s$^{-2}$ / kg = m s$^{-2}$. [A1 — correct answer with units]

Answer: $a = 4.0$ m s$^{-2}$ (forwards)

Newton's Second Law links force, mass, and acceleration quantitatively — it is the law you will use most often in calculation questions. Now consider what happens at the boundary between the first and second laws: when the resultant force is zero, acceleration is zero, confirming the first law as a special case of the second.

4. Newton's Third Law — Action and Reaction

Here is where students most often lose marks. Newton's Third Law is frequently stated correctly but applied incorrectly. The law is about interactions between two separate objects — not about forces on a single object.

Picture a person standing on the ground. The Earth pulls the person downwards with a gravitational force (their weight, $W$). By Newton's Third Law, the person pulls the Earth upwards with an equal gravitational force. These are the Newton's Third Law pair. The normal contact force from the ground pushing the person upwards is a separate force — it is the reaction to the person pushing down on the ground, and it acts on the person, not on the Earth.

To identify a Newton's Third Law pair correctly, check four criteria: the forces must be equal in magnitude, opposite in direction, act on different objects, and be the same type of force (both gravitational, both contact, both electromagnetic, and so on). If any criterion fails, the pair is not a valid Newton's Third Law pair.

A rocket in space provides another vivid example. The rocket expels gas backwards at high speed. By Newton's Third Law, the gas exerts an equal and opposite force on the rocket — forwards. There is no ground, no air, no external surface needed. The interaction is entirely between the rocket and the expelled gas.

Newton's Third Law explains why isolated objects cannot accelerate themselves — every force they exert on another object results in an equal force back on themselves. External forces are always required to change the momentum of a system, which connects directly back to the first and second laws.

5. Connecting the Three Laws — Worked Analysis

The three laws work together in every real situation. A single scenario can require you to apply all three. The worked example below shows how to move between them systematically — the kind of multi-step reasoning that distinguishes a top-band response from a mid-band one.

Worked example: Multi-Law Analysis: Lift Accelerating Upwards

A lift of mass 500 kg accelerates upwards at 2.0 m s$^{-2}$. Calculate the tension in the lift cable. Then state the Newton's Third Law pair to the tension force. Use $g = 9.8$ m s$^{-2}$. [4 marks]

1. Tension $T$ acts upwards. Weight $W = mg$ acts downwards. — Draw a free-body diagram mentally: two forces on the lift, both vertical. The lift is accelerating upwards, so the resultant must be upwards.
2. $W = mg = 500 \times 9.8 = 4900$ N (downwards) — Verify: $500 \times 9.8 = 4900$. This is the gravitational force on the lift. [B1]
3. $F_{\text{resultant}} = T - W = ma$ $T - 4900 = 500 \times 2.0$ $T - 4900 = 1000$ — The resultant upward force equals $ma$. Since the lift accelerates upwards, $T > W$. [M1 — correct application of $F = ma$ with direction]
4. $T = 4900 + 1000 = 5900$ N — Verify: $4900 + 1000 = 5900$. The tension exceeds the weight, consistent with upward acceleration. [A1]
5. The cable pulls the lift upwards with 5900 N. By Newton's Third Law: the lift pulls the cable downwards with 5900 N. — The pair acts on the cable (not the lift), is equal in magnitude (5900 N), opposite in direction, and is the same type of force (tension/contact). [B1]

Answer: Tension in the cable = 5900 N. The Newton's Third Law pair is the 5900 N downward pull that the lift exerts on the cable.

6. Common Misconceptions and Exam Technique

The three laws generate predictable errors in exams. Knowing where students typically drop marks — and how to avoid those traps — is as important as knowing the physics itself.

Exam questions on Newton's laws frequently use command words that require different response structures. 'State' requires a brief, precise answer — quote the law directly. 'Explain' requires cause and effect linked by 'because' or 'therefore'. 'Calculate' requires formula, substitution, and answer with units, each earning a separate mark.

For extended-response questions asking you to analyse a scenario using Newton's laws, structure your answer by identifying all forces, finding the resultant, applying the appropriate law, and then interpreting the result in context. This four-step approach maps directly onto the mark scheme structure that AQA examiners use.

7. Consolidation and Exam-Style Practice

You have now worked through all three laws in sequence. Before moving on, it is worth consolidating them side by side — the table below shows how each law applies to a different type of situation, which is exactly the kind of comparison an examiner might ask you to make.

Use the signal phrases in the right-hand column to identify which law an exam question is targeting. Examiners embed these phrases deliberately — recognising them quickly saves time and directs your response to the correct law.

Worked example: Exam-Style: Two-Object System on a Surface

Two blocks, A (mass 3 kg) and B (mass 2 kg), are in contact on a frictionless surface. A horizontal force of 20 N is applied to block A, pushing both blocks together. Calculate: (i) the acceleration of the system, and (ii) the contact force that block A exerts on block B. [4 marks]

1. $m_{\text{total}} = 3 + 2 = 5$ kg — When two objects move together with the same acceleration, treat them as one system for the first calculation.
2. $a = \dfrac{F}{m_{\text{total}}} = \dfrac{20}{5} = 4.0$ m s$^{-2}$ — Verify: $20 \div 5 = 4.0$. The surface is frictionless so the only horizontal force is the applied 20 N. [M1, A1]
3. Apply $F = ma$ to block B alone: $F_{\text{contact}} = m_B \times a = 2 \times 4.0 = 8.0$ N — Verify: $2 \times 4.0 = 8.0$. The only horizontal force on block B is the contact force from A. This is the force A exerts on B. [M1, A1]
4. Block B exerts an equal and opposite contact force of 8.0 N on block A (backwards). — This is not asked here, but noting it demonstrates full understanding and would earn marks in an extended question.

Answer: (i) Acceleration = 4.0 m s$^{-2}$. (ii) Contact force on block B = 8.0 N (in the direction of motion).

All three of Newton's laws — equilibrium and inertia, quantitative force-acceleration relationships, and interaction force pairs — are now covered at the depth required by the AQA A Level specification. Every calculation, explanation, and evaluation question on this topic draws on these three principles in combination.