Kinematics — Describing Motion

Kinematics is the mathematical description of motion without asking what causes it. Before we can understand forces, we need a precise language for displacement, velocity, and acceleration.

Displacement, Velocity, and Acceleration

Displacement (s) is the change in position — a vector quantity with both magnitude and direction. It is not the same as distance: a runner who completes one full lap of a track has zero displacement but non-zero distance.

Velocity (v) is the rate of change of displacement — how fast position is changing and in which direction. Average velocity = total displacement ÷ time elapsed. Instantaneous velocity is the limit of this as the time interval approaches zero.

Acceleration (a) is the rate of change of velocity. An object accelerates whenever its speed changes, its direction changes, or both. Crucially, an object moving at constant speed around a circle is always accelerating — its direction is changing continuously.

v̄ = Δs / Δt Average velocity equals change in displacement divided by change in time. Both v̄ and Δs are vectors.
ā = Δv / Δt Average acceleration equals change in velocity divided by time. Acceleration points in the direction of the velocity change, not necessarily the velocity itself.
Key Insight

Speed is the magnitude of velocity. Velocity is a vector; speed is a scalar. A car braking at 30 m/s to 20 m/s has the same speed direction but a negative acceleration (deceleration). A car turning a corner at constant speed has zero speed change but nonzero acceleration — it's changing direction.

The SUVAT Equations (Uniform Acceleration)

When acceleration is constant, five variables describe the motion completely: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). Four equations connect them — each one missing one variable.

Equation Missing Variable When to Use
v = u + ats (displacement)When you know u, a, t and need v
s = ut + ½at²v (final velocity)When you know u, a, t and need s
v² = u² + 2ast (time)When time is unknown or not needed
s = ½(u + v)ta (acceleration)When acceleration is unknown
s = vt − ½at²u (initial velocity)When initial velocity is unknown
Worked Example
Braking Car
A car travelling at 25 m/s brakes uniformly and stops in 5 seconds. Find the deceleration and stopping distance.
1
List known variables: u = 25 m/s, v = 0 m/s (stops), t = 5 s. Need: a and s.
2
Find acceleration using v = u + at: 0 = 25 + a(5) → a = −5 m/s²
3
Find stopping distance using v² = u² + 2as: 0 = 625 + 2(−5)s → s = 625/10 = 62.5 m
✓ Deceleration = 5 m/s²  |  Stopping distance = 62.5 m

Projectile Motion

A projectile moves under gravity alone — no air resistance. The key insight is that horizontal and vertical motions are completely independent. Horizontal: constant velocity (no horizontal force). Vertical: constant downward acceleration g = 9.8 m/s².

↔️

Horizontal Motion

No horizontal force, so no horizontal acceleration. The horizontal velocity never changes throughout the flight.

x = v₀cosθ · t
↕️

Vertical Motion

Constant downward acceleration g. Vertical velocity starts at v₀sinθ and decreases to zero at maximum height, then increases downward.

y = v₀sinθ · t − ½gt²
📐

Maximum Range

Range is maximised at 45° launch angle — this is where horizontal and vertical velocity components are equal, optimally sharing the total initial speed.

R = v₀²sin2θ / g
⬆️

Maximum Height

At maximum height, the vertical velocity is exactly zero. Only horizontal motion remains at the apex of the trajectory.

H = v₀²sin²θ / 2g
Interactive Simulation — Projectile Motion
Range: 81.6 m  |  Max H: 20.4 m
Worked Example
Football Kick
A football is kicked at 20 m/s at 30° above horizontal on flat ground. Find the range, maximum height, and time of flight. (g = 9.8 m/s²)
1
Resolve initial velocity: horizontal component u_x = 20cos30° = 17.32 m/s. Vertical component u_y = 20sin30° = 10 m/s.
2
Time of flight: ball lands when vertical displacement = 0. Using s = u_y·t − ½gt²: 0 = 10t − 4.9t² → t(10 − 4.9t) = 0. So t = 0 (launch) or t = 10/4.9 = 2.04 s
3
Range: x = u_x · t = 17.32 × 2.04 = 35.3 m
4
Max height (at t = 1.02 s, half flight time): H = 10(1.02) − 4.9(1.02)² = 10.2 − 5.1 = 5.1 m
✓ Range = 35.3 m  |  Max height = 5.1 m  |  Time of flight = 2.04 s

Newton's Three Laws of Motion

Published in the Principia Mathematica in 1687, Newton's three laws remain the foundation of classical mechanics — and of every branch of engineering built upon it. Understanding them deeply, not just memorising them, is essential.

NEWTON'S FIRST LAW — The Law of Inertia

An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a net external force.

This overthrew Aristotle's 2,000-year-old idea that objects naturally come to rest. The natural state is constant velocity — including zero. Inertia is the tendency of matter to resist changes to its state of motion. Mass is the measure of inertia.

Common Misconception

"Moving objects need a force to keep moving." — False. A hockey puck sliding on frictionless ice continues at constant velocity indefinitely. Friction is a real force that decelerates objects on Earth — remove it and no driving force is needed to maintain constant speed.

NEWTON'S SECOND LAW — The Law of Acceleration

The net force on an object equals its mass times its acceleration: F = ma

This is the equation of motion — the most practically powerful of the three laws. Net force and acceleration are always vectors pointing in the same direction. Mass is the proportionality constant: doubling mass halves acceleration for the same force.

F_net = ma Net force (N) = mass (kg) × acceleration (m/s²). F_net is the vector sum of all forces. Always draw a free-body diagram first.
Worked Example
Two-Block System
Blocks of mass 4 kg and 6 kg are connected by a light string on a frictionless horizontal surface. A force of 20 N pulls the 6 kg block. Find: (a) the acceleration of the system, (b) the tension in the string.
1
Treat both blocks as one system. Total mass = 4 + 6 = 10 kg. Net force = 20 N.
2
System acceleration: a = F/m = 20/10 = 2 m/s²
3
Isolate the 4 kg block. The only horizontal force on it is tension T. Apply Newton's 2nd: T = 4 × 2 = 8 N
✓ Acceleration = 2 m/s²  |  Tension = 8 N
NEWTON'S THIRD LAW — Action & Reaction

For every action there is an equal and opposite reaction — forces always come in pairs acting on different objects.

The critical phrase is "on different objects." Action-reaction pairs can never cancel each other because they act on separate bodies. A rocket pushes exhaust backward; the exhaust pushes the rocket forward. The rocket accelerates because only the forward force acts on it.

🚀

Rocket Propulsion

Exhaust gases pushed backward → reaction force pushes rocket forward. The paired forces act on different objects (rocket vs. gases).

🚶

Walking

Foot pushes backward on ground → ground pushes foot (and you) forward. Without this reaction force you couldn't move forward.

🏊

Swimming

Hand pushes water backward → water pushes swimmer forward. The same principle governs fish fins, bird wings, and boat propellers.

Energy, Work, and Power

Energy is the capacity to do work. It cannot be created or destroyed — only converted between forms. The work-energy theorem connects forces directly to changes in kinetic energy, making it one of the most powerful tools in mechanics.

Work and the Work-Energy Theorem

Work is done when a force causes displacement in the direction of the force. If a force F acts at angle θ to the direction of motion over displacement s, the work done is W = Fs·cosθ. Work is a scalar — it has no direction, only magnitude and sign.

W = Fs·cosθ Work (J) = Force (N) × displacement (m) × cosine of angle between them. W is positive when force and displacement are in the same direction.

The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy: W_net = ΔKE = ½mv² − ½mu². This is one of the most powerful shortcuts in mechanics — it bypasses the need to know forces at every instant.

Kinetic Energy

Energy of motion. Depends on mass and the square of velocity — doubling speed quadruples kinetic energy.

KE = ½mv²
🏔️

Gravitational PE

Energy stored by an object due to its height above a reference level. Depends on mass, gravity, and height.

GPE = mgh
🌀

Elastic PE

Energy stored in a compressed or stretched spring. Depends on the spring constant and deformation squared.

EPE = ½kx²

Power

Rate of doing work or transferring energy. Measured in watts (W = J/s). Also equals force × velocity for constant motion.

P = W/t = Fv

Conservation of Energy

In an isolated system with no non-conservative forces (like friction), total mechanical energy is conserved: KE + PE = constant. When friction acts, mechanical energy is not conserved — but total energy (including heat) still is. This is always true — it's one of the deepest laws in physics.

Conservation of Energy

The total energy of an isolated system never changes. Energy converts between kinetic, potential, thermal, and other forms — but the total is constant. This law has never been found to be violated in any experiment.

Worked Example
Roller Coaster Drop
A roller coaster (mass 800 kg) starts from rest at the top of a 40 m drop. Assuming no friction, find the speed at the bottom.
1
Apply conservation of energy. Initial state: all PE, no KE. Final state: all KE, no PE (taking bottom as reference).
2
mgh = ½mv² — the mass cancels: gh = ½v²
3
v² = 2gh = 2 × 9.8 × 40 = 784 → v = √784 = 28 m/s
✓ Speed at bottom = 28 m/s (≈ 101 km/h)

Momentum and Collisions

Momentum is the "quantity of motion" — a vector that combines both mass and velocity. Its conservation is one of the most powerful principles in physics, applying even when energy is not conserved.

p = mv Momentum (kg·m/s) = mass (kg) × velocity (m/s). A vector in the same direction as velocity. Total momentum of a closed system is always conserved.
J = Δp = FΔt Impulse equals change in momentum equals force multiplied by time. This explains why extending collision time (airbags, crumple zones) reduces the force.
🎱

Elastic Collision

Both momentum AND kinetic energy are conserved. Ideal billiard balls approximate this. Particles bounce off each other without energy loss.

🚗

Inelastic Collision

Momentum is conserved but kinetic energy is not — some converts to heat or deformation. Real-world collisions are inelastic.

💥

Perfectly Inelastic

Objects stick together after collision — maximum kinetic energy loss while momentum is still conserved. Car crashes, clay impacts.

Worked Example
Perfectly Inelastic Collision
A 1500 kg car moving at 15 m/s east collides and sticks with a stationary 2500 kg truck. Find the velocity of the combined wreckage after the collision.
1
Conservation of momentum: p_before = p_after. m₁u₁ + m₂u₂ = (m₁ + m₂)v
2
(1500)(15) + (2500)(0) = (1500 + 2500)v → 22500 = 4000v
3
v = 22500 / 4000 = 5.625 m/s east
✓ Combined velocity = 5.625 m/s east

Circular Motion

An object moving in a circle at constant speed is constantly accelerating — because its direction is constantly changing. This centripetal acceleration always points toward the centre of the circle and requires a centripetal force to maintain.

a_c = v²/r Centripetal acceleration points toward the centre of the circle. v is the speed, r is the radius. For a given radius, faster speed requires larger centripetal force.
F_c = mv²/r = mω²r Centripetal force required to maintain circular motion. This is not a new type of force — it is provided by gravity, tension, friction, or normal force depending on the context.
Important

"Centrifugal force" is not a real force — it is a fictitious force felt in a rotating reference frame. In an inertial frame, there is only centripetal force pulling inward. When a car turns a corner and you feel pushed outward, your body's inertia is resisting the inward centripetal force — you are not being pushed outward by any real force.

🌍

Planetary Orbits

Gravity provides the centripetal force for orbital motion. For a circular orbit: GMm/r² = mv²/r, giving orbital speed v = √(GM/r).

🎡

Vertical Circles

At the top of a loop, gravity and normal force both point down (toward centre). At the bottom, normal force points up and gravity down — the difference provides centripetal force.

🏎️

Banked Curves

Roads banked at angle θ allow cars to navigate curves without friction — the horizontal component of the normal force provides centripetal force: tanθ = v²/rg.

Universal Gravitation

Newton's Law of Universal Gravitation describes the attractive force between any two masses — from falling apples to orbiting planets. It was the first unified physical law, explaining both terrestrial and celestial mechanics with one equation.

F = GMm / r² Gravitational force between masses M and m separated by distance r. G = 6.674 × 10⁻¹¹ N·m²/kg². Force is always attractive and acts along the line joining the centres of mass.

The inverse-square relationship (1/r²) means that doubling the distance reduces the gravitational force by a factor of 4. This law applies at any scale — from a person on Earth to galaxies attracting each other across millions of light-years.

Gravitational Field Strength

The gravitational field strength g at any point is the force per unit mass experienced by a test mass at that point: g = GM/r². At Earth's surface (r = 6.37 × 10⁶ m), this gives g ≈ 9.8 N/kg = 9.8 m/s². The field decreases with altitude — astronauts on the ISS (400 km altitude) still experience about 89% of surface gravity, which is why they fall freely around Earth rather than floating weightlessly in zero gravity.

v_esc = √(2GM/r) Escape velocity — the minimum speed needed to escape a body's gravity from distance r. For Earth: v_esc ≈ 11.2 km/s (independent of mass of escaping object).
Worked Example
Satellite Orbital Speed
Calculate the orbital speed of a satellite in a circular orbit 400 km above Earth's surface. (M_Earth = 5.97×10²⁴ kg, R_Earth = 6.37×10⁶ m, G = 6.67×10⁻¹¹ N·m²/kg²)
1
Orbital radius: r = R_Earth + altitude = 6.37×10⁶ + 4×10⁵ = 6.77×10⁶ m
2
Set gravitational force = centripetal force: GMm/r² = mv²/r → v = √(GM/r)
3
v = √(6.67×10⁻¹¹ × 5.97×10²⁴ / 6.77×10⁶) = √(5.88×10⁷) ≈ 7,668 m/s
✓ Orbital speed ≈ 7,668 m/s ≈ 7.7 km/s (ISS orbits at ~7.66 km/s ✓)

Rotational Dynamics

Rotational dynamics is the angular equivalent of Newton's laws — every concept from linear mechanics has a rotational counterpart. Understanding the parallels makes both systems far easier to master.

Linear ConceptLinear SymbolRotational AnalogueRotational Symbol
DisplacementsAngular displacementθ (rad)
VelocityvAngular velocityω (rad/s)
AccelerationaAngular accelerationα (rad/s²)
MassmMoment of inertiaI (kg·m²)
ForceFTorqueτ (N·m)
F = maτ = Iα
Momentum p = mvAngular momentum L = Iω
KE = ½mv²KE_rot = ½Iω²
τ = Iα = r × F Torque (the rotational equivalent of force) equals moment of inertia × angular acceleration. Also equals the cross product of position vector and force — only the perpendicular component of force creates torque.
Conservation of Angular Momentum

In the absence of external torques, the angular momentum L = Iω of a system is conserved. This is why a spinning figure skater spins faster when pulling their arms in — reducing r reduces I, and since L = Iω must be constant, ω must increase.

Classical Mechanics Formula Sheet

Every essential equation for classical mechanics, organized by topic. Bookmark this page as your go-to reference.

Kinematics

v = u + atSUVAT (no s)
s = ut + ½at²SUVAT (no v)
v² = u² + 2asSUVAT (no t)
s = ½(u+v)tSUVAT (no a)
R = v₀²sin2θ/gProjectile Range
H = v₀²sin²θ/2gMax Height

Newton's Laws & Forces

F = maNewton's 2nd Law
W = mgWeight
f = μNFriction Force
F_c = mv²/rCentripetal Force
a_c = v²/rCentripetal Accel.
F = GMm/r²Gravitation

Energy & Momentum

KE = ½mv²Kinetic Energy
GPE = mghGravitational PE
EPE = ½kx²Elastic PE
W = Fs·cosθWork Done
P = W/t = FvPower
p = mvMomentum
J = FΔt = ΔpImpulse

Rotational Dynamics

τ = IαNewton's 2nd (rot.)
L = IωAngular Momentum
KE = ½Iω²Rotational KE
τ = rF·sinθTorque
v = ωrLinear ↔ Angular

Practice Problems

Test your understanding. Try each problem before revealing the solution — the struggle before the answer is where real learning happens.

Time to fall: Using s = ½gt² with s = 45 m: t = √(2×45/10) = √9 = 3 s.

Horizontal distance: x = v_x × t = 12 × 3 = 36 m from the base.
Normal force: N = mg = 2 × 9.8 = 19.6 N
Friction force: f = μN = 0.3 × 19.6 = 5.88 N
Net force: F_net = 10 − 5.88 = 4.12 N
Acceleration: a = F/m = 4.12/2 = 2.06 m/s²
Conservation of momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v
(60)(4) + (20)(0) = (80)v
240 = 80v → v = 3 m/s
Centripetal acceleration: a_c = v²/r = 25/0.8 = 31.25 m/s²
Tension (provides centripetal force): T = ma_c = 0.3 × 31.25 = 9.375 N
Energy conservation: EPE → KE: ½kx² = ½mv²
½(400)(0.15)² = ½(0.1)v²
4.5 = 0.05v² → v² = 90 → v = √90 ≈ 9.49 m/s

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Frequently Asked Questions

Classical mechanics is the branch of physics that describes the motion of macroscopic objects — from thrown balls to orbiting planets — using Newton's laws of motion and the principles of energy and momentum conservation. It applies at everyday speeds (well below the speed of light) and everyday scales (much larger than atoms). At very high speeds, Special Relativity takes over; at atomic scales, Quantum Mechanics applies. For everyday engineering and physics, classical mechanics is extraordinarily accurate.
First Law: An object remains at rest or moves at constant velocity unless acted upon by a net external force (inertia).

Second Law: F = ma — the net force on an object equals its mass times its acceleration. The most useful law for solving mechanics problems.

Third Law: Every action has an equal and opposite reaction — forces come in pairs acting on different objects. A book on a table pushes the table down; the table pushes the book up with equal force.
Velocity is the rate of change of position — how fast and in which direction an object is moving. Acceleration is the rate of change of velocity — how quickly the velocity itself is changing. An object can move at high velocity with zero acceleration (constant speed straight line). It can also be momentarily stationary with high acceleration (ball thrown upward at its peak). Acceleration always points in the direction the velocity is changing, not necessarily the direction of motion.
A heavier object experiences a larger gravitational force (F = mg), but it also has more inertia (more mass resists acceleration). These two effects cancel exactly. Applying Newton's 2nd Law: a = F/m = mg/m = g. The mass cancels, leaving the same acceleration g for all objects regardless of mass. This was famously demonstrated by Galileo dropping objects from the Leaning Tower of Pisa, and dramatically confirmed on the Moon by astronaut David Scott dropping a hammer and feather simultaneously — they hit the ground together.
In an elastic collision, both momentum AND kinetic energy are conserved — objects bounce off each other with no energy lost to heat or deformation. Ideal billiard balls approximate this. In an inelastic collision, momentum is conserved but some kinetic energy converts to heat, sound, or deformation. Almost all real-world collisions are inelastic. In a perfectly inelastic collision, objects stick together — this is the maximum possible kinetic energy loss while still conserving momentum.
Follow these steps every time: (1) Resolve the initial velocity into horizontal (v₀cosθ) and vertical (v₀sinθ) components. (2) Treat horizontal and vertical motion completely separately. (3) Horizontal: no acceleration, constant velocity. Use x = v_x·t. (4) Vertical: constant downward acceleration g. Use SUVAT equations. (5) The time of flight connects both — find it from the vertical motion, then use it in the horizontal. (6) At maximum height, vertical velocity = 0. At landing (on level ground), vertical displacement = 0.

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