Newton's Laws of Motion

Newton’s three laws of motion form the foundation of classical mechanics. They describe the relationship between forces acting on an object and the motion of that object. Every mechanics problem you encounter at A-Level can ultimately be traced back to these three laws.

Newton’s First Law (The Law of Inertia)

An object remains at rest or continues to move at constant velocity unless acted upon by a resultant (net) force.

This law tells us two things:

A stationary object stays stationary unless a net force acts on it.
A moving object continues at the same speed in the same direction unless a net force acts on it.

The key word is resultant. An object can have many forces acting on it, but if they all cancel out (resultant = 0), the object behaves as if no force is acting. A book on a table has weight pushing it down and a normal contact force pushing it up — these balance, so the book stays still.

Inertia is the tendency of an object to resist changes in its motion. Mass is a measure of inertia — a more massive object is harder to accelerate or decelerate.

Newton’s Second Law

The net force acting on an object is equal to the rate of change of its momentum. For constant mass, this simplifies to F = ma.

F = ma (where F is the resultant force in N, m is mass in kg, and a is acceleration in m/s²)

This is the most important equation in mechanics. It tells us:

The acceleration is in the same direction as the resultant force.
The acceleration is proportional to the net force (double the force, double the acceleration).
The acceleration is inversely proportional to the mass (double the mass, halve the acceleration).

Worked Example 1

A 1200 kg car has a driving force of 4000 N and experiences a total resistive force of 1000 N. Find the acceleration.

Net force = 4000 - 1000 = 3000 N a = F/m = 3000/1200 = 2.5 m/s²

Worked Example 2

A 70 kg sprinter accelerates from rest to 10 m/s in 1.5 s. What net force acts on the sprinter?

a = (v - u)/t = (10 - 0)/1.5 = 6.67 m/s² F = ma = 70 x 6.67 = 467 N

Free-Body Diagrams

A free-body diagram shows all the forces acting on a single object. Each force is drawn as an arrow starting from the object, with the length and direction representing the magnitude and direction of the force.

flowchart TD
    A["Drawing a Free-Body Diagram"] --> B["1. Isolate the object\n(draw as box or dot)"]
    B --> C["2. Identify ALL forces\nacting ON the object"]
    C --> D["3. Draw each force as\na labelled arrow"]
    D --> E["4. Check: Weight? Normal?\nFriction? Tension? Drag?"]
    E --> F["5. Resolve into components\nif forces act at angles"]

Rules for drawing free-body diagrams:

Isolate the object — draw it as a simple shape (box or dot).
Identify every force acting ON the object (not forces the object exerts on other things).
Draw each force as a labelled arrow from the object.
Include: weight (always downward), normal reaction (perpendicular to contact surface), friction (opposing motion along the surface), tension (along rope/string away from object), air resistance/drag (opposing motion).

Common Forces Checklist

Force	Symbol	Direction	Always Present?
Weight	W or mg	Vertically downward	Yes (near Earth)
Normal reaction	N or R	Perpendicular to surface	Only if in contact with surface
Friction	F or f	Opposing motion/tendency to move	Only if surfaces in contact and tendency to slide
Tension	T	Along string, away from object	Only if attached to taut string/rope
Drag/air resistance	D	Opposing velocity	Only if moving through fluid
Applied/push force	P or F	Direction of push/pull	Only if explicitly applied

Newton’s Third Law

When object A exerts a force on object B, object B exerts an equal and opposite force on object A.

The crucial features of Newton’s third law pairs:

Equal in magnitude
Opposite in direction
Same type of force (both gravitational, both electromagnetic, etc.)
Act on different objects

Identifying Third Law Pairs

A book rests on a table. The weight of the book (gravitational pull of Earth on book) and the normal contact force (table pushing up on book) are NOT a Newton’s third law pair. They are both forces on the same object (the book) and are different types of force.

The third law pair for the weight of the book is: the gravitational pull of the book on the Earth (same type, same magnitude, opposite direction, different objects).

The third law pair for the normal contact force of the table on the book is: the contact force of the book on the table (pushing down on the table surface).

Quick Test for Third Law Pairs

To check if two forces form a Newton’s third law pair, fill in this sentence:

"Object A exerts a [type] force on object B, and object B exerts an equal and opposite [same type] force on object A."

If both forces are the same type and act on different objects, they are a third law pair.

Apparent Weight in a Lift

When you stand on bathroom scales in a lift, the reading shows the normal contact force, not your true weight. This is your apparent weight.

Consider a person of mass 70 kg (weight = 70 x 9.81 = 687 N):

Lift Motion	Acceleration	Apparent Weight N	Feel
Stationary or constant velocity	a = 0	mg = 687 N	Normal
Accelerating upward	a = +2.0 m/s²	m(g + a) = 827 N	Heavier
Accelerating downward	a = +2.0 m/s² downward	m(g - a) = 547 N	Lighter
Free fall	a = g	0 N	Weightless

Worked Example 3

A 55 kg person stands on scales in a lift. The scales read 440 N. Find the acceleration of the lift and state its direction.

N = m(g - a) for downward acceleration, or N = m(g + a) for upward acceleration.

440 = 55(9.81 - a) [assuming downward acceleration] 440/55 = 9.81 - a 8.0 = 9.81 - a a = 9.81 - 8.0 = 1.81 m/s² downward

Check: The reading is less than the true weight (55 x 9.81 = 540 N), confirming the lift accelerates downward (or decelerates while going up).

Terminal Velocity

When an object falls through a fluid (air or liquid), it experiences:

Weight (downward, constant for a given object near Earth’s surface)
Drag / air resistance (upward, increasing with speed)

When first released, drag is zero (speed is zero), so the net force is the full weight and the object accelerates at g. As speed increases, drag increases. The net downward force decreases, so acceleration decreases. Eventually, drag equals weight, the net force is zero, and the object moves at constant velocity — this is terminal velocity.

At terminal velocity: drag = weight, net force = 0, acceleration = 0.

flowchart TD
    A["Object released from rest"] --> B["Speed = 0, Drag = 0\nNet force = Weight\nAcceleration = g"]
    B --> C["Speed increases\nDrag increases\nNet force decreases"]
    C --> D["Acceleration decreases\n(but still positive)"]
    D --> E{"Drag = Weight?"}
    E -->|No| C
    E -->|Yes| F["Net force = 0\nAcceleration = 0\nTERMINAL VELOCITY"]

Worked Example 4

A 75 kg skydiver reaches terminal velocity. What is the drag force at this point?

At terminal velocity, drag = weight = mg = 75 x 9.81 = 736 N

When the skydiver opens their parachute, the drag force suddenly increases (much larger surface area). Drag now exceeds weight, so there is a net upward force and the skydiver decelerates. As speed decreases, drag decreases until a new, lower terminal velocity is reached.

Worked Example 5: Ball bearing in oil

A steel ball bearing of mass 8.0 g falls through oil. At terminal velocity, the drag force is 47 mN and the upthrust is 12 mN. Is the ball at terminal velocity?

Weight = mg = 0.0080 x 9.81 = 0.0785 N = 78.5 mN Upward forces = drag + upthrust = 47 + 12 = 59 mN Downward force = weight = 78.5 mN

Since 78.5 mN > 59 mN, the ball is NOT at terminal velocity — it is still accelerating downward. Net force = 78.5 - 59 = 19.5 mN downward.

Applying Newton’s Second Law: Method

Draw a free-body diagram showing all forces on the object.
Choose a positive direction.
Resolve forces into components if necessary.
Write F_net = ma in each direction.
Solve for the unknown.

Common Mistakes and Misconceptions

“Force causes motion”: Force causes acceleration, not motion. An object moving at constant velocity has zero net force.
“Heavier objects fall faster”: In a vacuum, all objects fall at the same rate regardless of mass. In air, drag effects make the situation more complex, but this is not because gravity pulls harder relative to mass.
Newton’s third law confusion: The weight of a book and the normal reaction from the table are NOT a third law pair. They happen to be equal only when the book is in equilibrium.
“The force of motion”: There is no such thing. If an object moves at constant velocity, no net force is needed to maintain that motion.
Forgetting to include all forces: Students often forget drag, or the component of weight on a slope.

Exam Tips for Edexcel Physics

When asked to "explain using Newton’s laws," name the specific law you are using (first, second, or third).
For Newton’s third law questions, always identify both forces, both objects, and state that the forces are the same type.
In free-body diagram questions, only draw forces on the specified object — do not include forces the object exerts on other things.
For terminal velocity questions, describe the process step by step: initial state, what changes, what stays constant, and the final equilibrium.
When using F = ma, always identify the resultant force first, not just one of the individual forces.

Key Points

First law: no net force means no change in velocity (constant velocity or rest).
Second law: F = ma — acceleration is proportional to net force and inversely proportional to mass.
Third law: forces come in equal and opposite pairs acting on different objects of the same force type.
Free-body diagrams are essential for identifying all forces on an object.
Apparent weight changes in an accelerating lift: N = m(g ± a).
Terminal velocity occurs when drag equals weight (plus upthrust in liquids).

A-Level Deep Dive: Newton's Laws of Motion

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.3 (Forces) requires candidates to use Newton's first, second and third laws of motion to analyse situations involving constant velocity, acceleration produced by a resultant force, and interaction pairs (refer to the official Pearson Edexcel specification document for exact wording). Although the three laws are stated here, they recur structurally across Topic 4 (Materials) when analysing tensions in stretched wires and equilibrium in suspended-mass problems, Topic 6 (Further mechanics) where the second law is rewritten in its momentum form $F = \dfrac{\Delta p}{\Delta t}$ to handle variable-mass and impulsive collisions, Topic 7 (Electric and magnetic fields) for charged-particle acceleration in uniform fields, and Topic 8 (Nuclear and particle physics) for momentum conservation in decays — itself a third-law consequence. The Edexcel data, formulae and relationships booklet supplies $F = ma$ and $F = \dfrac{\Delta(mv)}{\Delta t}$ , but interaction-pair reasoning and free-body-diagram conventions must be deployed unaided.

Worked example with full mark scheme

Question (8 marks):

A wooden block of mass 4.00 kg is released from rest on a rough plane inclined at 25.0° to the horizontal. The coefficient of friction between the block and the plane is $\mu = 0.180$ . Take $g = 9.81$ m s⁻².

(a) Draw a labelled free-body diagram showing the four forces acting on the block while it slides. (2)

(b) Using Newton's second law along the incline, calculate the acceleration of the block down the slope. (5)

Solution with mark scheme:

(a) The free-body diagram shows: weight $W = mg$ acting vertically downward; normal contact force $N$ acting perpendicular to the slope (away from the surface); friction $f = \mu N$ acting along the slope, opposing the direction of motion (i.e. up the slope while sliding down); and — strictly — the absence of any applied force, since the block is released from rest.

B1 — all four forces drawn with correct directions (weight vertical, normal perpendicular to slope, friction up the slope, applied force absent).

B1 — labels correct, including the angle 25.0° between the slope and the horizontal (or equivalently between the normal and the vertical).

(b) Step 1 — resolve the weight. Choose axes parallel and perpendicular to the slope. The weight $mg$ resolves into a component $mg\sin\theta$ down the slope and a component $mg\cos\theta$ into the slope.

M1 — correct resolution of weight along and perpendicular to the slope.

Step 2 — apply N2 perpendicular to the slope. There is no acceleration perpendicular to the slope (the block stays on the surface), so the normal contact force balances the perpendicular component of weight:

$N = mg\cos\theta = 4.00 \times 9.81 \times \cos 25.0° = 35.6 \text{ N}$

M1 — equilibrium statement perpendicular to the slope used to find $N$ .

Step 3 — friction. The kinetic friction magnitude is $f = \mu N$ :

$f = 0.180 \times 35.6 = 6.40 \text{ N}$

A1 — correct friction value (allow ecf from $N$ ).

Step 4 — apply N2 along the slope. Taking down-the-slope as positive, the resultant force is the down-slope weight component minus friction:

$F_{\text{net}} = mg\sin\theta - f = 4.00 \times 9.81 \times \sin 25.0° - 6.40 = 16.59 - 6.40 = 10.2 \text{ N}$

$a = \dfrac{F_{\text{net}}}{m} = \dfrac{10.2}{4.00} = 2.55 \text{ m s}^{-2}$

M1 — Newton's second law applied along the slope with friction subtracted from the down-slope weight component.

A1 — $a = 2.55$ m s⁻² (3 sf), down the slope.

$v^2 = 2 \times 2.55 \times 1.50 = 7.65 \implies v = 2.77 \text{ m s}^{-1}$

B1 — $v = 2.77$ m s⁻¹ (3 sf).

Total: 8 marks (B2 M3 A2 B1). A characteristic 9PH0 stress-test of N2 — three layers (resolve, normal force, friction, axial N2) compressed into a single five-mark part.

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A horse pulls a cart of mass 320 kg along a level road at constant velocity. The horse exerts a forward force of 480 N on the cart through the harness.

(a) State, with reference to Newton's third law, the magnitude and direction of the force exerted by the cart on the horse. (2)

(b) Explain, with reference to Newton's first law, why the cart can move at constant velocity even though the horse pulls it forward. (2)

(c) The horse increases its pull on the cart to 600 N while resistive forces remain unchanged. Calculate the cart's acceleration. (2)

Mark scheme decomposition by AO:

Part	AO weighting	Earned by
(a) B1	AO1.1a	States magnitude 480 N.
(a) B1	AO1.2	States direction "backwards on the horse, along the harness" — explicitly identifying that the N3 partner force acts on a different body (the horse, not the cart) and is the same type (tension/contact).
(b) M1	AO2.1	Identifies that constant velocity means zero resultant force on the cart.
(b) A1	AO3.1a	Concludes friction and air resistance on the cart together amount to 480 N backwards, exactly cancelling the harness pull.
(c) M1	AO1.2	Resultant force = 600 − 480 = 120 N (using the previous resistive-force value).
(c) A1	AO1.1b	$a = F/m = 120 / 320 = 0.375$ m s⁻².

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. The signature feature is part (a): a third-law question that examiners use to filter candidates who confuse N3 pairs with N1 equilibrium. The N3 partner of "horse pulls cart forward" is "cart pulls horse backward" — not "ground pushes cart backward via friction".