Module 1 · Classical Mechanics

Newton's
Laws of Motion

Three deceptively simple statements that unified terrestrial and celestial mechanics, held unchallenged for two centuries, and still govern every engineering system you will ever encounter.

Prerequisites Kinematics, Vectors
Core equations F = ma, p = mv, J = Δp
Sections 7
§ 01

Foundations: Force, Mass & Inertia

Before Newton's laws can be stated precisely, we need exact meanings for the terms they use. Imprecise vocabulary produces imprecise thinking.

Force

A force is a push or a pull — any interaction that, when unopposed, produces a change in an object's state of motion. Forces are vectors: they have both magnitude and direction. The SI unit is the newton (N), defined as the force that gives a 1 kg mass an acceleration of 1 m/s².

Definition — Newton
$$1\,\text{N} = 1\,\text{kg} \cdot \text{m}\,\text{s}^{-2}$$
A modest force: the weight of a 100 g apple is approximately 0.98 N.

Mass and Inertia

Mass quantifies the amount of matter in an object and, more importantly for dynamics, its inertia — its resistance to changes in motion. A more massive object requires more force to achieve the same acceleration.

Mass is a scalar — it has no direction. Crucially, mass is an intrinsic property of an object; it does not change based on location.

⚠ Common Confusion — Weight ≠ Mass Weight $W = mg$ is a force (vector, measured in newtons), caused by gravity acting on mass. Mass is measured in kilograms and is location-independent. An astronaut has the same mass on the Moon as on Earth; their weight is approximately one-sixth, because $g_\text{Moon} \approx 1.62\ \text{m s}^{-2}$.

Symbol Reference

SymbolQuantityTypeSI UnitNotes
$\mathbf{F}$ForceVectorN = kg m s⁻²Always has direction
$m$MassScalarkgIntrinsic, location-independent
$\mathbf{a}$AccelerationVectorm s⁻²Rate of change of velocity
$\mathbf{v}$VelocityVectorm s⁻¹Speed with direction
$\mathbf{p}$MomentumVectorkg m s⁻¹$\mathbf{p} = m\mathbf{v}$
$W$WeightVectorN$W = mg$, always downward
🍎

Historical context

Newton published the Principia Mathematica in 1687, synthesising Galileo's kinematics with Kepler's planetary laws. The apple story — whether apocryphal or not — captures the essential insight: the same force governing falling apples also governs the Moon's orbit. Gravity, for the first time, was understood as a universal force.


§ 02

The Three Laws

Each law is a distinct claim about nature. Together they form a closed, self-consistent system sufficient to solve all of classical mechanics.

First Law
The Law of Inertia
An object at rest remains at rest, and an object in uniform motion remains in uniform motion in a straight line, unless acted upon by a net external force.
$\mathbf{F}_\text{net}=0 \Rightarrow \mathbf{a}=0$
Second Law
The Law of Acceleration
The net force acting on an object equals the rate of change of its linear momentum. For constant mass, this reduces to the familiar force–acceleration relation.
$\mathbf{F}=\dfrac{d\mathbf{p}}{dt}=m\mathbf{a}$
Third Law
Action & Reaction
For every force exerted by object A on object B, there exists an equal in magnitude, opposite in direction force exerted by object B on object A. The forces act on different objects.
$\mathbf{F}_{AB}=-\mathbf{F}_{BA}$

The First Law — deeper reading

Superficially, the first law looks like a special case of the second (when $F_\text{net}=0$, then $a=0$). But Newton stated it separately for an important reason: it defines the class of reference frames in which the second law is valid — the inertial frames. In a non-inertial frame (like an accelerating car), objects accelerate even with no net force. The first law is the criterion that tells you whether you're in a valid frame for applying $F=ma$.

📐 Inertial vs Non-Inertial Frames An inertial frame is one in which the first law holds: objects with no net force either stay still or move at constant velocity. The surface of the Earth is approximately inertial (it rotates, so technically it isn't). A rotating turntable is emphatically non-inertial — objects on it experience a centrifugal pseudo-force. This will be central to Module 3 (Circular Dynamics).

The Third Law — the critical subtlety

The third law is the most consistently misapplied of the three. The key word is different objects.

✗ Common Error — Equilibrium from Third Law "The book sits still on the table because the book pushes down on the table, and the table pushes back equally — third law!" This is wrong. Third-law pairs act on different objects; they can never cancel each other.

The book is in equilibrium because two forces acting on the book cancel: gravity (Earth pulling book down) and the normal force (table pushing book up). The book pushing down on the table is its third-law partner of the normal force, but that force acts on the table, not the book.

How the Laws Relate

Momentum conservation — one of the deepest principles in physics — emerges directly from the third law. If two objects interact and there is no external force:

Momentum conservation from Newton III
$$\mathbf{F}_{AB} + \mathbf{F}_{BA} = 0 \implies \frac{d\mathbf{p}_A}{dt} + \frac{d\mathbf{p}_B}{dt} = 0 \implies \frac{d}{dt}(\mathbf{p}_A + \mathbf{p}_B) = 0$$
Total momentum is constant — it doesn't change over time. Newton III is momentum conservation.

§ 03

$F = ma$ — Full Derivation

The most famous equation in mechanics is a special case of Newton's deeper momentum statement. Understanding the derivation reveals when it applies and when it doesn't.

From the momentum definition

Newton's second law in its most general form involves momentum $\mathbf{p} = m\mathbf{v}$:

Second Law — General Form (Product Rule)
$$\mathbf{F}_\text{net} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt} = m\frac{d\mathbf{v}}{dt} + \mathbf{v}\frac{dm}{dt}$$
The second term $\mathbf{v}\,\frac{dm}{dt}$ is non-zero in variable-mass systems such as rockets (which expel mass as exhaust). For constant mass, this term vanishes.

Derivation for constant mass

$\mathbf{F}_\text{net} = \dfrac{d(m\mathbf{v})}{dt}$
Newton's Second Law — momentum form. This is the foundational statement.
$\mathbf{F}_\text{net} = m\,\dfrac{d\mathbf{v}}{dt} + \mathbf{v}\,\dfrac{dm}{dt}$
Product rule of differentiation: $\frac{d}{dt}(fg) = f\frac{dg}{dt} + g\frac{df}{dt}$
$\dfrac{dm}{dt} = 0$
Assumption: mass is constant (not ejected or accreted). Valid for most everyday objects.
$\mathbf{F}_\text{net} = m\,\dfrac{d\mathbf{v}}{dt}$
The $\mathbf{v}\,\frac{dm}{dt}$ term drops out.
$$\boxed{\mathbf{F}_\text{net} = m\mathbf{a}}$$
Since $\dfrac{d\mathbf{v}}{dt} \equiv \mathbf{a}$ by definition. This is a vector equation valid in all directions simultaneously.

Component form

Being a vector equation, $\mathbf{F}=m\mathbf{a}$ separates into independent scalar equations along each coordinate axis. This is the key to solving 2D and 3D problems:

Cartesian Component Form
x-direction
$$\sum F_x = ma_x$$
y-direction
$$\sum F_y = ma_y$$
z-direction
$$\sum F_z = ma_z$$
Component independence: a force in the $x$-direction produces only $x$-acceleration. This is why a projectile's horizontal and vertical motions are solved separately.

Rearrangements

Find acceleration
$$a = \frac{F_\text{net}}{m}$$
More force → more acceleration.
More mass → less acceleration.
Find mass
$$m = \frac{F_\text{net}}{a}$$
Used to measure mass experimentally — apply known force, measure resulting acceleration.
Find force
$$F_\text{net} = ma$$
Given motion data ($a$) and mass, compute the net force that caused it.
Worked Example 1 — Single Force, Horizontal
F = ma

A 5 kg box rests on a frictionless horizontal surface. A horizontal force of 20 N is applied. Find the acceleration.

m = 5 kg F = 20 N surface: frictionless find: a
  1. Identify all forces. Vertical: weight $W=mg=(5)(9.8)=49\ \text{N}$ down; normal $N=49\ \text{N}$ up. These cancel. No friction.
  2. Net force is the 20 N applied force only: $F_\text{net} = 20\ \text{N}$.
  3. Apply $F_\text{net}=ma$: $\quad a = \dfrac{F_\text{net}}{m} = \dfrac{20}{5}$
Answer: a = 4.0 m/s² in the direction of the applied force
Worked Example 2 — Multiple Forces
Vector sum

A 3 kg block is pulled right with 18 N and experiences a 6 N friction force opposing its motion. Find acceleration.

m = 3 kg F_applied = 18 N → f = 6 N ← find: a
  1. Choose positive direction as rightward (+).
  2. Net force: $F_\text{net} = +18 - 6 = +12\ \text{N}$
  3. $a = \dfrac{F_\text{net}}{m} = \dfrac{12}{3} = 4.0\ \text{m s}^{-2}$
Answer: a = 4.0 m/s² rightward
Worked Example 3 — Object on an Incline
Components

A 4 kg block slides down a frictionless 30° incline. Find the acceleration down the slope.

m = 4 kg θ = 30° frictionless find: a along slope
  1. Set axes: positive $x$ down the slope, positive $y$ perpendicular to slope (away from surface).
  2. Weight components: along slope $= mg\sin\theta$; perpendicular $= mg\cos\theta$.
  3. Along slope ($x$): $F_x = mg\sin30° = (4)(9.8)(0.5) = 19.6\ \text{N}$
  4. Perpendicular ($y$): $N - mg\cos30° = 0 \Rightarrow N = 33.9\ \text{N}$ (block doesn't sink into slope).
  5. $a_x = \dfrac{F_x}{m} = \dfrac{19.6}{4} = 4.9\ \text{m s}^{-2}$
  6. Note: $a = g\sin\theta$, independent of mass — confirmed by Galileo's incline experiments.
Answer: a = g sin 30° = 4.9 m/s² down the slope

§ 04

Free Body Diagrams

A free body diagram (FBD) is a sketch of a single isolated object with every force acting on it drawn as a labelled vector. It is the essential first step in any dynamics problem.

Step 1: Draw the object alone — nothing it is touching, no surroundings.
Step 2: Identify every force acting on it (not forces it exerts on other things).
Step 3: Draw each force as an arrow from the contact point (surface forces) or centre of mass (body forces).
Step 4: Label magnitude and direction. The net force determines the acceleration: $\sum\mathbf{F} = m\mathbf{a}$.

Common forces

Weight
$W = mg$
Gravitational force. Always acts vertically downward from the centre of mass. Never omit this one.
Normal Force
$N$ (perpendicular to surface)
A contact force perpendicular to the surface. Arises because surfaces resist compression. Equals $mg\cos\theta$ on a slope.
Friction
$f \leq \mu_s N$; $f_k = \mu_k N$
Parallel to the surface, opposing relative motion or tendency of motion. Static and kinetic variants differ.
Applied Force
$\mathbf{F}_\text{app}$
Any external push or pull. May be at any angle; resolve into components before applying Newton II.
🔗
Tension
$T$ (along string/rope)
Force in a taut rope or string. Always directed along the rope, pulling toward the attachment. In a massless rope, $T$ is constant throughout.
💨
Drag / Air Resistance
$F_D = \tfrac{1}{2}\rho v^2 C_D A$
Opposes motion through a fluid. Proportional to $v^2$ at high speed. Neglected in basic problems unless stated.

Diagram examples

Equilibrium — flat surface
Incline — accelerating down slope
Hanging mass — Atwood machine (one side)
Applied force at angle θ above horizontal
📏 FBD Rules of Thumb 1. Draw the FBD before writing any equations. Every force in your equations must appear on the diagram.
2. Choose a coordinate system aligned with the motion (along slope, for example).
3. The net force arrow (sum of all force arrows) points in the same direction as the acceleration.

§ 05

Linear Momentum

Momentum is the quantity that truly underlies Newton's second law. It is also — in some sense — more fundamental than force: it is conserved even when forces are not well-defined, such as in quantum mechanics.

Momentum — Definition
$$\mathbf{p} = m\mathbf{v}$$
Momentum is a vector with the same direction as velocity. Its magnitude is mass × speed. Unit: kg m s⁻¹ = N s.

Newton's actual statement of his second law was not "$F = ma$" but "force equals rate of change of momentum." This is more general because it handles variable-mass systems (rockets, raindrop collecting water) where the first form fails.

Conservation of momentum

When no external force acts on a system of particles, the total momentum is conserved — it remains constant in magnitude and direction. This follows from Newton III, as shown in §2:

Conservation Law
$$\mathbf{p}_\text{total} = \sum_i m_i\mathbf{v}_i = \text{constant} \quad (\text{when } \mathbf{F}_\text{ext} = 0)$$

Collision types

Momentum is conserved in all collisions (assuming no external forces during the brief collision time). What varies is whether kinetic energy is also conserved:

TypeMomentumKinetic EnergyTypical Example
Elastic Conserved Conserved Billiard balls, atomic/subatomic collisions
Inelastic Conserved Lost (heat, sound, deformation) Car crash, ball of clay on wall
Perfectly inelastic Conserved Maximum loss possible Objects stick together after collision

Elastic collision equations (1D)

Post-collision velocities — elastic, 1D
Final velocity of A
$$v_A' = \frac{m_A - m_B}{m_A + m_B}v_A + \frac{2m_B}{m_A + m_B}v_B$$
Final velocity of B
$$v_B' = \frac{2m_A}{m_A + m_B}v_A + \frac{m_B - m_A}{m_A + m_B}v_B$$
Special case — equal masses ($m_A = m_B$): $v_A' = v_B$ and $v_B' = v_A$. The objects swap velocities completely — a result familiar from billiards.
Worked Example 4 — Perfectly Inelastic Collision
Conservation of Momentum

A 2 kg trolley moving at 4 m/s collides and sticks to a stationary 3 kg trolley. Find the combined velocity after collision and the kinetic energy lost.

m_A = 2 kg, v_A = 4 m/s m_B = 3 kg, v_B = 0 perfectly inelastic
  1. Momentum before: $p_i = m_Av_A + m_Bv_B = (2)(4) + (3)(0) = 8\ \text{kg m s}^{-1}$
  2. After collision, combined mass $= 5\ \text{kg}$. By conservation: $(m_A+m_B)v' = p_i$
  3. $v' = \dfrac{8}{5} = 1.6\ \text{m s}^{-1}$
  4. KE before: $\tfrac{1}{2}(2)(4^2) = 16\ \text{J}$
  5. KE after: $\tfrac{1}{2}(5)(1.6^2) = 6.4\ \text{J}$
  6. KE lost: $16 - 6.4 = 9.6\ \text{J}$ (converted to heat and sound)
v' = 1.6 m/s  |  ΔKE = −9.6 J

§ 06

Impulse and the Impulse–Momentum Theorem

If Newton's second law involves varying forces, integrating over time gives impulse — a concept as practical as it is elegant.

Starting from Newton II:

$\mathbf{F}_\text{net} = \dfrac{d\mathbf{p}}{dt}$
Newton's second law in momentum form
$\mathbf{F}_\text{net}\,dt = d\mathbf{p}$
Rearrange: multiply both sides by $dt$
$\displaystyle\int_{t_i}^{t_f}\mathbf{F}_\text{net}\,dt = \int_{\mathbf{p}_i}^{\mathbf{p}_f}d\mathbf{p}$
Integrate both sides over the time interval of interest
$$\mathbf{J} = \Delta\mathbf{p} = m\mathbf{v}_f - m\mathbf{v}_i$$
The impulse $\mathbf{J}$ (left side integral) equals the change in momentum. This is the Impulse–Momentum Theorem.

For a constant force, this simplifies to:

Impulse — Constant Force
$$\mathbf{J} = \mathbf{F}_\text{avg}\,\Delta t = \Delta\mathbf{p}$$
The area under an $F$-$t$ graph equals the impulse, which equals the change in momentum. This is why variable force problems are solved graphically or by integration.
🏃 Why This Matters — Soft Landings and Airbags For a given change in momentum (e.g., stopping a moving person), impulse $J = F\Delta t$ is fixed. Increasing $\Delta t$ (the time of the collision) decreases the average force $F$. This is the physics of:
• Airbags and crumple zones (extend $\Delta t$, reduce $F$ on occupants)
• Bending knees when landing (extend $\Delta t$ to reduce joint forces)
• Catching a ball with a moving hand (increase $\Delta t$, stings less)
Worked Example 5 — Impulse from a Variable Force
Impulse Theorem

A 0.5 kg ball travelling at 10 m/s is brought to rest by a net braking force that varies. The average force is 50 N. How long does the braking take?

m = 0.5 kg v_i = 10 m/s v_f = 0 F_avg = 50 N
  1. Change in momentum: $\Delta p = m(v_f - v_i) = 0.5(0 - 10) = -5\ \text{kg m s}^{-1}$
  2. Magnitude: $|\Delta p| = 5\ \text{N s}$
  3. Impulse theorem: $F_\text{avg}\,\Delta t = |\Delta p|$
  4. $\Delta t = \dfrac{5}{50} = 0.10\ \text{s}$
Braking time = 0.10 s = 100 ms

§ 07

Interactive Simulation

All readouts computed from $F = ma$ each frame. Force vectors are drawn at their actual magnitudes. Adjust the sliders and observe Newton's laws in real time.

Force & Motion Lab
v — t graph
F — t graph