IB Physics – YOUR EDUCATION PLUS

A. Space, time and motion

A.1 KINEMATICS

Recommended Time: 9 Hours | Standard Level and Higher Level

Guiding Questions

How can the motion of a body be described quantitatively and qualitatively?
How can the position of a body in space and time be predicted?
How can the analysis of motion in one and two dimensions be used to solve real-life problems?

1. Describing Motion: Position and Reference Direction

Kinematics is the description of motion. The starting point is a coordinate system: a chosen origin and a chosen positive direction. In one-dimensional motion, a position can therefore be represented by a positive or negative coordinate. The choice of positive direction is arbitrary, but once chosen it must be used consistently. Horizontal position is commonly written x, vertical position y, and a general one-dimensional position s.

Idea	Meaning / Implication
Reference point (origin)	The point from which position is measured. Changing the origin changes position coordinates, but it does not change the actual motion.
Positive direction	Chosen by the solver. Motion opposite to this direction is represented by negative displacement or velocity.
Position	Location relative to the origin. In one dimension its sign indicates which side of the origin the body lies on.
Change in position	The displacement: final position minus initial position. It is a vector quantity.

Distance follows the actual path; displacement connects initial and final positions directly and includes direction.

Scalar and Vector Quantities

A scalar is completely specified by magnitude. A vector requires magnitude and direction. This distinction is essential throughout A.1 because distance and speed are scalars, whereas displacement, velocity and acceleration are vectors. A negative vector component identifies direction relative to the chosen axis; it does not mean that the magnitude itself is negative.

2. Distance, Displacement, Speed and Velocity

The most common early errors in kinematics come from treating distance as if it were displacement, or speed as if it were velocity. These ideas must be kept distinct.

Quantity	Definition / Interpretation	Type	SI unit
Distance	Total length of the route travelled, irrespective of direction.	Scalar	m
Displacement	Change in position from start to finish; in one dimension Δs = s_f - s_i.	Vector	m
Speed	Rate at which distance is travelled; magnitude of velocity.	Scalar	m/s
Velocity	Rate of change of position/displacement; direction must be specified.	Vector	m/s

Average Quantities

Average values describe an entire time interval. They do not show what happens at each instant within the interval.

Quantity	Relationship	What the numerator represents
Average speed	total distance / total time	Total path length, always non-negative.
Average velocity	total displacement / total time	Net change in position, including sign/direction.
Average acceleration	change in velocity / time interval	Vector change in velocity, not merely change in speed.

Instantaneous Speed and Velocity

The instantaneous value is the value at a particular moment. On a curved distance- or displacement-time graph, the instantaneous speed or velocity is found from the gradient of a tangent at the required time. A large tangent triangle reduces the effect of reading uncertainty from the graph. For straight-line constant-velocity motion, instantaneous velocity equals average velocity over any interval.

For a curved position-time graph, the gradient of the tangent at a point gives the instantaneous velocity at that instant.

Practical Measurement of Motion

These definitions connect directly to measurement. Constant velocity may be determined by recording positions at known times, while modern motion analysis can use sensors or video. Motion can also be analysed using image/video methods and graphical representations of data. Whatever the method, the central calculation remains change in position divided by the corresponding time interval.

Worked Comparison: Average Speed vs. Average Velocity

A straight-track example shows why the two averages must be kept separate. A runner first travels in one direction, stops briefly, and then runs back in the opposite direction. The total distance is obtained by adding the lengths of both running segments, whereas the final displacement is the algebraic difference because the second segment is in the opposite direction. Dividing each by the same total elapsed time gives two different average quantities. The example is valuable because it also shows that periods of rest contribute to total time even though they add neither distance nor displacement.

Step	What to do	Why
1. Draw a one-dimensional axis	Mark the start, turning point and final position, and choose + direction.	Turns direction information into signed displacement.
2. Find total distance	Add the magnitudes of every path segment.	Distance is scalar and path-dependent.
3. Find displacement	Final position minus initial position.	Only start and finish matter.
4. Divide by total time	Use total elapsed time, including stops.	Average quantities refer to the whole interval.

3. Acceleration and the Meaning of Changing Velocity

Acceleration is the rate of change of velocity. Because velocity is a vector, acceleration can result from a change in speed, a change in direction, or both. In A.1 the main quantitative treatment is one-dimensional, so signs are used to indicate the directions of velocity and acceleration.

Velocity v	Acceleration a	Effect on speed
positive	positive	Speed increases in the positive direction.
positive	negative	Speed decreases; acceleration opposes the motion.
negative	negative	Speed increases in the negative direction.
negative	positive	Speed decreases; acceleration opposes the negative motion.

Average and Instantaneous Acceleration

Average acceleration over a finite interval is the change in velocity divided by the time taken. Instantaneous acceleration is the value at a particular instant and is obtained from the gradient of the tangent to a velocity-time graph. The SI unit m/s² means that the velocity changes by so many metres per second during each second, when acceleration is constant.

Uniform and Non-Uniform Acceleration

Uniform acceleration means constant acceleration. Equal time intervals then produce equal changes in velocity, so the velocity-time graph is a straight line. Non-uniform acceleration means the acceleration changes with time; a velocity-time graph is then curved or piecewise, and graphical analysis is often needed. Graphical analysis becomes particularly useful when acceleration is not constant.

Core graph connections in A.1: gradient moves from position to velocity to acceleration; area reverses the process from acceleration to velocity change and from velocity to displacement.

Position-Time Graphs: How Shape Describes Motion

Feature on displacement-time graph	Physical meaning
Horizontal line	Position is constant: the body is at rest.
Straight line with positive gradient	Constant positive velocity.
Straight line with negative gradient	Constant velocity in the negative direction.
Curve becoming steeper	Magnitude of velocity is increasing.
Curve becoming flatter	Magnitude of velocity is decreasing.
Turning point with horizontal tangent	Instantaneous velocity is zero at that instant.

4. Reading Velocity-Time & Acceleration-Time Graphs

Graph interpretation is not an optional add-on to A.1. The IB skills section requires students to interpret gradients, changes in gradient, intercepts, maxima and minima, and areas under graphs. In kinematics these graphical features have direct physical meanings.

Graph	Gradient gives	Area gives	Typical interpretation
displacement-time	velocity	-	Sign and steepness of slope give direction and speed.
velocity-time	acceleration	displacement	Area below the time axis contributes negative displacement.
speed-time	rate of change of speed	distance travelled	Speed is a scalar and is never negative.
acceleration-time	rate of change of acceleration	change in velocity	A horizontal line means constant acceleration.

Area Under a Velocity-Time Graph

For a small time interval, displacement is approximately velocity multiplied by time. Adding many such small intervals leads to the general result that the signed area under a velocity-time graph equals displacement. A staircase approximation can be used to motivate this result. For constant acceleration the velocity-time graph is a straight line, so the area is a trapezoid.

How to Estimate Values from Curved Graphs

Draw a tangent at the required point when an instantaneous gradient is needed.
Use a large gradient triangle on the tangent and read both axes carefully.
Estimate the area under an irregular curve by counting squares or dividing the region into simple shapes.
Check units: gradient units and area units should match the physical quantity claimed.
When a graph crosses v = 0, the direction of motion changes if the sign of velocity changes.

Graph Transformations for Constant Acceleration

For constant acceleration, v-t is linear and a-t is horizontal. The corresponding displacement-time graph is parabolic because displacement contains a term proportional to t². The concavity provides a useful check: with positive acceleration the position-time curve is concave upward; with negative acceleration it is concave downward. These linked graph shapes allow one representation to be checked against another.

Interpreting Multi-Stage Motion

Many examination graphs combine several types of motion. The safest method is to divide the graph at every point where the slope changes, where the velocity crosses zero, or where the mathematical form changes. Treat each interval separately and then combine the results. For example, on a velocity-time graph a body may accelerate positively, travel at constant velocity, decelerate to rest, and then move in the negative direction. The displacement of the whole journey is the algebraic sum of the signed areas of all four regions, whereas total distance is the sum of their magnitudes.

Graph event	What it tells you
v changes sign	The body reverses direction.
v = 0 at an isolated instant	The body is momentarily at rest; motion may continue in the opposite direction.
a = 0 over an interval	Velocity is constant during that interval.
gradient of v-t becomes steeper	Magnitude of acceleration is increasing.
area above and below axis cancel	Net displacement can be small or zero even when distance is large.

5. Uniformly Accelerated Motion and the Four Equations

When acceleration is constant, five quantities describe the one-dimensional motion over an interval: initial velocity u, final velocity v, acceleration a, displacement s (or Δs), and time t. The A.1 requirements list four equations linking these quantities.

Symbol	Meaning
u	initial velocity
v	final velocity
a	constant acceleration
s	displacement during the interval
t	time interval

Equation	Useful when the omitted quantity is...
v = u + at	displacement s
s = (u + v)t / 2	acceleration a
s = ut + ½at²	final velocity v
v² = u² + 2as	time t

Equation-selection guide: choose the relationship that excludes the one quantity you do not know.

Where the Equations Come from

For constant acceleration, acceleration is the gradient of the straight velocity-time graph, giving v = u + at. The displacement equals the trapezoidal area under that graph, giving s = (u + v)t/2. Substituting v = u + at into the area expression produces s = ut + ½at². Eliminating t between the equations gives v² = u² + 2as. Understanding this connection is more useful than treating the four equations as unrelated formulas.

Reliable Equation-Selection Method

Choose and state a positive direction. Convert all displacement, velocity and acceleration values into signed quantities.
List the five variables u, v, a, s and t. Mark the known values and the required unknown.
Select the equation containing the required unknown but not the one other quantity you do not know.
Substitute values with consistent SI units and solve algebraically.
Interpret the sign and magnitude of the answer, and confirm that constant acceleration was a justified assumption.

Common Algebraic and Physical Checks

A time obtained from a quadratic may produce two mathematical roots; only roots consistent with the physical interval are retained.
A negative displacement or velocity may be correct and simply indicates direction.
If the equation v² = u² + 2as is used, taking a square root later requires choosing the sign of v from the physical direction of motion.
Dimensional consistency is a useful check: every term in a displacement equation must have units of metres; every term in a velocity equation must have units of m/s.

6. Free-fall and One-Dimensional Vertical Motion

Free fall is motion under the influence of gravity when fluid resistance is neglected. Close to Earth, the acceleration due to gravity is treated as constant, with magnitude g about 9.8 m/s². For A.1 projectile calculations close to Earth, g is treated as constant; the same constant-acceleration treatment applies to vertical free-fall problems.

With upward chosen positive, gravitational acceleration is negative throughout the motion. At maximum height the velocity is zero momentarily, but acceleration is still -g.

Situation (upward is +)	Initial/instantaneous velocity	Acceleration	Interpretation
Dropped from rest	u = 0	a = -g	Speed increases downward.
Thrown upward	u > 0	a = -g	Velocity decreases to zero as the body rises.
At maximum height	v = 0	a = -g	The body is momentarily at rest, but gravity still acts.
Falling downward	v < 0	a = -g	Magnitude of the downward velocity increases.

Symmetry When Resistance is Negligible

For a body that returns to the same vertical level from which it was launched, the upward and downward parts are symmetric in the ideal model. The time to rise to maximum height equals the time to return from that height to the launch level. The speed on return has the same magnitude as the launch speed but the opposite direction. These statements cease to be exact when fluid resistance is significant.

Graphical View of a Vertical Throw

With upward positive, the velocity-time graph is a straight line of gradient -g. It crosses v = 0 at the maximum height. The displacement-time graph is a downward-opening parabola; its gradient decreases to zero at the top and becomes negative during descent. The acceleration-time graph is a constant horizontal line at -g.

Typical Vertical-Motion Procedure

Sketch the vertical line of motion and mark the chosen positive direction.
Identify the start and end of the interval being analysed; displacement is final position minus initial position.
Use v = 0 only at the highest point, not a = 0.
Apply one of the four constant-acceleration equations using a = ±g.
If the motion is split into stages, use the final velocity of one stage as the initial velocity of the next.

7. Projectile Motion: Two-Independent Components

A projectile is a body that has been launched and then moves under gravity, with fluid resistance either neglected or treated qualitatively. The essential modelling idea is that horizontal and vertical motions can be treated independently and then recombined. This independence is the basis of the standard component method.

Resolving the launch velocity into perpendicular components. With no fluid resistance, horizontal acceleration is zero while vertical acceleration is -g.

Horizontal motion	Vertical motion
u_x = u cos θ	u_y = u sin θ
a_x = 0	a_y = -g (if upward is positive)
v_x = u_x = constant	v_y = u_y - gt
x = u_x t	y = u_y t - ½gt²

What is Common to Both Components?

The same elapsed time t applies to the horizontal and vertical motions.
The resultant velocity is obtained by vector addition of v_x and v_y and is tangent to the trajectory.
At the highest point, only the vertical component of velocity is zero. The horizontal component remains non-zero if the projectile was launched with horizontal motion.
Gravity changes the vertical velocity only in the no-resistance model; it does not alter the horizontal component.

Why the Trajectory is Parabolic

Without resistance, horizontal displacement is proportional to time while vertical displacement contains a t² term. Eliminating time therefore produces a quadratic relation between vertical and horizontal position, which corresponds to a parabola. For A.1, know that the ideal path is parabolic; deriving the equation of the trajectory is not required.

8. Solving Projectile Problems

The same procedure works for all required launch types. The main difference is the initial vertical component and the vertical displacement to the final point.

Three required launch situations: horizontal, above the horizontal, and below the horizontal. The component method is the same in each case.

General Method

Draw x-y axes and choose signs, normally +x in the direction of launch and +y upward.
Resolve the initial velocity: u_x = u cos θ and u_y = u sin θ. If the launch is below the horizontal, u_y is negative when +y is upward.
Write separate horizontal and vertical data sets. Horizontal acceleration is zero; vertical acceleration is -g.
Use one component to determine the shared time, then substitute that time into the other component.
If final velocity is required, find v_x and v_y separately, then recombine them to obtain magnitude and direction.

Useful Same-Height Results

For a projectile launched at speed u and angle θ and returning to the same vertical height, the component method gives the following convenient results. They are consequences of the required kinematic method rather than separate principles.

Quantity	Result	Reason
Time to maximum height	T = u sin θ / g	Set v_y = 0 in v_y = u_y - gt.
Maximum height	H = u² sin²θ / (2g)	Use v_y² = u_y² + 2a_yH with v_y = 0.
Total time of flight	2u sin θ / g	Same-height symmetry.
Range	R = u² sin(2θ) / g	Horizontal speed multiplied by total flight time.

Common Projectile Errors

Using the total launch speed u in both directions instead of resolving it into u cos θ and u sin θ.
Assuming the whole velocity is zero at maximum height; only v_y is zero.
Putting g into the horizontal equations when resistance is neglected.
Using same-height symmetry when the projectile lands at a different vertical level.
Recombining velocity components incorrectly or omitting the direction of the final velocity.

Worked Example: Launch at 20 m/s and 30°

Consider a projectile launched at 20 m/s at 30° to illustrate the component method. With resistance neglected and g taken close to 9.8 m/s², the initial components are u_x = 20 cos30° ≈ 17.3 m/s and u_y = 20 sin30° = 10.0 m/s. At maximum height v_y = 0, so the time to rise is about 10.0/9.8 = 1.02 s. The maximum height is about 5.1 m. Returning to the launch level takes twice the rise time, about 2.04 s, while the horizontal speed remains 17.3 m/s. The range is therefore about 35 m.

Stage	Horizontal component	Vertical component
Launch	17.3 m/s	+10.0 m/s
Maximum height	17.3 m/s	0
Return to launch height	17.3 m/s	-10.0 m/s

The example shows three important ideas at once: horizontal and vertical motions share the same time but have different accelerations; the projectile is still moving at the top because v_x is not zero; and the equal-magnitude upward/downward vertical speeds follow from the symmetry of the ideal model. If the landing height were different, the same-height time and range shortcuts would no longer apply.

9. Fluid Resistance and Terminal Speed

Real bodies moving through a gas or liquid experience fluid resistance. The resistance acts opposite to the instantaneous velocity and generally increases as speed increases. This makes acceleration non-uniform, so the constant-acceleration equations cannot normally be applied to the whole drag-affected motion.

Qualitative behaviour of a falling body with resistance: speed approaches a constant terminal value while acceleration falls toward zero.

Development of Terminal Speed in a Vertical Fall

Stage	Fluid resistance	Resultant acceleration / motion
Immediately after release	Very small because speed is near zero.	Acceleration is approximately g downward.
As speed increases	Resistance increases upward, opposite to the fall.	Resultant downward acceleration becomes smaller.
Terminal speed	Upward resistance balances the downward weight (and, in a more complete description, any other vertical fluid forces).	Resultant force and acceleration are zero; speed remains constant.

The corresponding graph changes are: with no resistance the speed increases linearly and acceleration remains constant; with resistance the speed curve gradually flattens toward terminal speed, acceleration decreases toward zero, and the distance-time graph becomes less strongly curved. A useful comparison is a constant-g velocity-time line versus a curve that approaches a horizontal terminal-speed line.

Effect of Resistance on a Projectile

Feature	Ideal no-resistance model	With fluid resistance
Horizontal velocity	constant	decreases during the flight
Vertical acceleration while rising	g downward	downward effect is larger because drag also acts downward
Vertical acceleration while falling	g downward	magnitude is reduced because drag acts upward
Trajectory	parabolic and, for equal heights, symmetric	not parabolic and not symmetric
Range / maximum height	larger for the same launch conditions	usually reduced in the standard drag-only model
Use of SUVAT over whole flight	valid separately in x and y	not valid because acceleration changes

Modelling Limits and What the IB Expects

For A.1, the required treatment of fluid resistance is qualitative. Simple drag models may use a resistance proportional to speed at lower speeds and proportional to speed squared at higher speeds. The important qualitative result is that acceleration decreases as resistance grows, and terminal speed is reached when the resultant force becomes zero. More detailed drag equations are not required unless a question provides the model explicitly.

This distinction matters when solving questions. If a problem merely states that air resistance is significant, you should not assume constant acceleration or invent a drag equation. Instead, explain how the resistance direction changes with the velocity, how its magnitude grows with speed, and what that implies for acceleration, trajectory and terminal behaviour. Quantitative projectile calculations in A.1 are restricted to the no-resistance model.

10. A.1 Synthesis: What You Must be able to Do

A.1 develops the ability to describe motion, predict position in space and time, and apply one- and two-dimensional analysis to real situations. The following checklist condenses the required knowledge and the most useful problem-solving habits.

Area	Minimum mastery
Describe motion	Distinguish position, distance and displacement; distinguish speed and velocity; identify scalar and vector quantities; use a consistent reference direction.
Use rates of change	Calculate average speed/velocity/acceleration and determine instantaneous velocity or acceleration from tangent gradients.
Analyse graphs	Interpret shape, slope, intercepts and turning points; use the area under v-t for displacement and under a-t for change in velocity.
Uniform acceleration	Choose and use the four kinematic equations only when acceleration is constant; connect them to the straight-line v-t graph.
Free fall	Use constant g near Earth with correct sign; know that v = 0 but a ≠ 0 at the highest point.
Projectiles	Resolve initial velocity into x and y components, use the same time in both, and recombine final vector components.
Fluid resistance	Explain why acceleration changes, why terminal speed develops, and how trajectory, range, velocity and time of flight are altered qualitatively.

Essential Equations at a Glance

Purpose	Equation
average velocity	v_avg = Δs / Δt
average acceleration	a_avg = Δv / Δt
uniform acceleration	v = u + at
uniform acceleration	s = (u + v)t / 2
uniform acceleration	s = ut + ½at²
uniform acceleration	v² = u² + 2as
projectile components	u_x = u cos θ, u_y = u sin θ
near-Earth vertical motion	a = ±g, with g ≈ 9.8 m/s²

High-Value Checks Before Finalizing an Answer

Have I identified whether the question needs distance or displacement, speed or velocity?
Have I chosen a positive direction and used it consistently?
Is acceleration constant over the interval before I use a kinematic equation?
On a graph, am I using the correct gradient or area relationship, with the correct sign?
For a projectile, have I kept horizontal and vertical quantities separate until the final recombination?
At a maximum height, have I set only the vertical velocity to zero rather than the acceleration?
If resistance is present, have I stopped treating the motion as uniformly accelerated unless the problem explicitly provides a constant-acceleration interval?

Linking Ideas Highlighted by the A.1 Requirements

A.1 is intended to connect forward to other parts of the course. Students should relate linear equations of motion to rotational motion, compare gravitational and electric-field motion, consider when conservation of energy can replace kinematic equations in projectile problems, connect kinematics to Newton’s laws, and use graphical analysis to determine other physical quantities. These links reinforce that kinematics is a modelling language used throughout physics, not an isolated collection of formulas.

Common Misconceptions and Corrections

Misconception	Correction / diagnostic check
Distance and displacement are interchangeable.	Distance follows the path; displacement is the vector change from initial to final position. Ask whether direction matters.
Negative acceleration always means slowing down.	Compare the signs of v and a. Same sign means speed is increasing; opposite signs mean speed is decreasing.
At the top of a vertical flight, acceleration is zero.	Only the vertical velocity is zero. The acceleration remains g downward.
The area under every motion graph gives distance.	Area under v-t gives signed displacement; area under speed-t gives distance; area under a-t gives change in velocity.
All projectile formulas can be used in every problem.	Special same-height results require equal launch/landing heights and negligible resistance. Otherwise use the component SUVAT method.
If air resistance acts, use a smaller constant value of g.	Resistance makes the resultant acceleration change with speed and direction. Do not replace g by a fixed reduced value over the whole motion.
At maximum height a projectile is at rest.	Only v_y is zero. Unless the launch was purely vertical, v_x remains non-zero.

A Disciplined A.1 Problem-Solving Workflow

Successful solutions follow a consistent pattern even when the wording differs. First translate the physical description into a model; then choose the mathematics that matches that model. This prevents the most common mistake in kinematics: selecting an equation before deciding what the signs, directions and assumptions mean.

Represent the situation: sketch the path or graph, mark initial and final points, and define the coordinate axes.
Classify the motion: one-dimensional or two-dimensional; constant or non-uniform acceleration; resistance negligible or significant.
Identify the target quantity and its type: scalar or vector. If it is a vector, decide how direction will be reported.
Choose the relationship: definition, graph gradient/area, constant-acceleration equation, or separate projectile components.
Calculate with consistent units, then interpret the sign and direction rather than reporting a bare number.
Check the result against the physical picture: should the body be speeding up or slowing down, above or below the start, moving left or right?
State the modelling assumption when it matters, especially constant acceleration, constant g near Earth, and negligible resistance in quantitative projectile work.

Using the IB Data Booklet and Graph Skills

The four constant-acceleration equations are provided in the Physics data booklet, so A.1 is not primarily a test of memorization. The skill is to recognize when each equation applies and to combine it with correct vector signs and graph interpretation. The IB skills framework also makes graph construction and interpretation assessable: students may need to obtain a gradient, estimate an area, identify a turning point, compare uniform with non-uniform acceleration, or infer a physical quantity from a graph. A complete A.1 revision should therefore practise both algebraic and graphical representations of the same motion.

Precise Terminology to Retain

Term	Use it to mean...
Reference point	The chosen origin relative to which position is measured.
Trajectory	The path followed by a projectile.
Range	The horizontal distance travelled by a projectile before the stated impact point.
Uniform acceleration	Acceleration that remains constant with time.
Non-uniform acceleration	Acceleration whose magnitude and/or direction changes with time.
Terminal speed	A constant speed reached when the resultant force becomes zero because resistive effects balance the driving force in the stated situation.

A.2 Forces and Momentum

Recommended Time: 10 Hours | Standard Level and Higher Level

IB Syllabus status. A.2 is prescribed for both SL and HL, with a recommended 10 teaching hours and no separate additional-HL section. It covers forces and free-body diagrams; Newton’s laws; contact and field forces; momentum and impulse; conservation of momentum in collisions and explosions; and circular motion.

Guiding Questions

How can forces acting on a system be represented both visually and algebraically?
How can Newton’s laws be modelled mathematically?
How can knowledge of forces and momentum be used to predict the behaviour of interacting bodies?

1. A.2 at a Glance

A.2 links the description of motion from A.1 to the causes of changes in motion. Its central ideas are forces, momentum and circular motion. The same physical situation can often be analysed in more than one way: by resolving forces and applying Newton’s laws, by considering impulse and momentum, or by identifying the inward resultant force needed for circular motion.

Core Idea	What You Need to Be Able to Do
Forces	Identify interactions, draw free-body diagrams and determine the resultant force in one or two dimensions.
Newton’s laws	Relate resultant force to translational equilibrium, acceleration and action–reaction force pairs.
Contact and field forces	Recognise the relevant force types and use the required A.2 relationships when they apply.
Momentum and impulse	Treat momentum as a vector, connect impulse to change in momentum and interpret force–time graphs.
Collisions and explosions	Apply momentum conservation to an isolated system and use kinetic-energy information to classify the interaction.
Circular motion	Connect angular and linear quantities and identify the real inward force or force resultant.

2. Forces as Interactions and Free-Body Diagrams

A force is an interaction between bodies. Because force is a vector quantity, both magnitude and direction matter. When several forces act on one body, their vector sum is the resultant or net force. A free-body diagram isolates one chosen body and represents every external force acting on that body with a labelled arrow.

Figure 1. A typical free-body diagram. The normal force, weight, applied force and friction all act on the same chosen body.

Free-Body Diagram Rules

Choose one body or system and mentally isolate it from its surroundings.
Draw only the forces acting on the chosen body. Do not draw forces that the body exerts on other objects.
Label each force clearly, such as the normal force, frictional force, weight or tension, and show the direction with an arrow in the diagram.
Show the correct direction of each force. In two dimensions, resolve forces into perpendicular components only after the physical picture is clear.
Add the force components vectorially to determine the resultant force. A zero resultant corresponds to translational equilibrium.

Important Distinction. Balanced forces act on the same body and can sum to zero. A Newton’s third-law pair acts on two different bodies, so those two forces never cancel on a single free-body diagram.

3. Newton’s Laws and Translational Equilibrium

Newton’s First Law

A body remains at rest or continues with constant velocity unless acted upon by a resultant external force. Therefore, if the resultant force is zero, the acceleration is zero and the body is in translational equilibrium.

Newton’s Second Law

For a body of constant mass, the resultant force vector and acceleration vector are related by Newton’s second law.

The more general momentum form of Newton’s second law is introduced later in this topic. The distinction matters because F = ma assumes constant mass.

Newton’s Third Law

When body A exerts a force on body B, body B simultaneously exerts a force of equal magnitude and opposite direction on body A. The forces are the same type of interaction, act for the same time and act on different bodies.

Situation	Third-Law Pair
Book on a table	Table pushes upward on book; book pushes downward on table.
Person walking	Foot pushes ground backward; ground pushes foot forward.
Car towing trailer	Car pulls trailer forward; trailer pulls car backward.

Worked Example 1. Resultant Force with Friction

A 5.0 kg block is pulled horizontally by a 20 N force across a horizontal surface. The coefficient of dynamic friction is 0.20. Take g = 9.8 m/s². Determine the acceleration.

No vertical acceleration occurs, thus the normal force equals the weight in magnitude: (5.0 kg)(9.8 m/s²) = 49 N.
Dynamic friction equals the coefficient of dynamic friction multiplied by the normal force: (0.20)(49 N) = 9.8 N opposite the motion.
The horizontal resultant is ΣF = 20 N − 9.8 N = 10.2 N.
Apply ΣF = ma: a = (10.2 N)/(5.0 kg) = 2.04 m/s².
Therefore, the block accelerates at 2.04 m/s² in the direction of the applied force.

4. Contact Forces

Contact forces arise when bodies or fluids interact through physical contact. A.2 requires the nature and use of the following contact forces.

Contact Force	Direction / Meaning	Required Relationship
Normal force, F_N	Component of the contact force perpendicular to the surface. It is not automatically equal to weight.	Determined from the force balance or acceleration perpendicular to the surface.
Surface friction, F_f	Parallel to the contact surface and opposes relative motion or the tendency to slip.	Static: F_f ≤ μ_sF_N. Dynamic: F_f = μ_dF_N.
Tension, T	Pulling force transmitted by a taut string, cable or similar connector.	No single universal A.2 formula; obtain it from the free-body diagram and Newton’s laws.
Elastic restoring force, F_H	Acts toward the equilibrium length, opposing extension or compression.	F_H = −kx.
Viscous drag, F_d	Opposes the motion of a small sphere through a fluid.	F_d = 6πηrv.
Buoyancy, F_b	Upward force due to displacement of fluid.	F_b = ρVg.

Static and Dynamic Friction

Static friction adjusts to the value needed to prevent sliding, up to a maximum. For a stationary body, the actual frictional force can be any value up to the maximum value μ_s times the normal force. Equality applies only at limiting equilibrium, just before slipping begins. Once sliding begins, the dynamic-friction model uses the coefficient of dynamic friction multiplied by the normal force.

Figure 2. Static friction increases as required until its limiting value; after motion begins, the dynamic-friction model applies.

Elastic Restoring Force

F_H = −kx

The negative sign shows that the restoring force acts opposite to the displacement from equilibrium.

Viscous Drag and Buoyancy

F_d = 6πηrv
F_b = ρVg

In A.2, the drag expression above applies to the specific small-sphere model required by the syllabus.

5. Field Forces

Field forces act without direct contact. A.2 requires recognition of gravitational, electric and magnetic forces. The detailed laws and fields for electric and magnetic interactions are developed later in the course.

Field Force	A.2 Treatment
Gravitational force, F_g	Near Earth, weight has magnitude mg and acts vertically downward.
Electric force, F_e	Recognise it as the force on a charged body in an electric field.
Magnetic force, F_m	Recognise it as the force associated with magnetic interactions.

Modelling check. A force should be named according to the physical interaction that produces it. “Centripetal force” is not an additional type of force; it is the name given to the inward resultant of real forces during circular motion.

6. Momentum, Impulse, and Force–Time Graphs

Linear momentum measures motion using both mass and velocity. Because velocity is a vector, momentum is also a vector. A sign convention or vector-component method is therefore essential whenever direction matters.

Momentum relation: p = mv. The SI unit is kg m/s, and the direction is the same as the velocity direction.

Impulse and Change in Momentum

An external resultant force acting for a time interval produces an impulse. Using the average resultant force over that interval, the impulse equals the change in momentum of the system.

Impulse relation: J = FΔt = Δp and F = Δp/Δt.

Force–Time Graphs

When force varies during an interaction, the area under a resultant force–time graph is the impulse and therefore the change in momentum. This graphical interpretation is especially useful for collisions and safety problems.

Figure 3. The signed area under a resultant force–time graph gives impulse J and hence the change in momentum Δp.

Worked Example 2. Impulse During a Rebound

A 0.20 kg ball moves toward a wall at +12 m/s and rebounds at −8.0 m/s. The contact time is 0.050 s. Determine the impulse on the ball and the average resultant force.

Choose the initial direction as positive: u = +12 m/s and v = −8.0 m/s.
The initial momentum is (0.20 kg)(12 m/s) = +2.4 kg m/s.
The final momentum is (0.20 kg)(−8.0 m/s) = −1.6 kg m/s.
The change in momentum is final momentum minus initial momentum: (−1.6 kg m/s) − (+2.4 kg m/s) = −4.0 kg m/s.
Impulse J = Δp = −4.0 N s.
Average resultant force F = Δp/Δt = (−4.0 N s)/(0.050 s) = −80 N.

7. Conservation of Momentum

For a system with no resultant external force, total linear momentum remains constant. Internal forces can redistribute momentum between parts of the system, but they do not change the total momentum of the isolated system.

Conservation statement: total momentum before the interaction equals total momentum after the interaction.

Choosing the System

Choose the system boundary.
Identify which forces are internal.
Determine whether a significant external impulse acts during the interval.
If the external impulse is negligible, total momentum is conserved.

One-Dimensional Momentum Method

Choose a positive direction before writing momentum terms.
Calculate each initial momentum using signed velocity and add them to obtain the total initial momentum.
Write the final momenta using the same sign convention.
Set total momentum before equal to total momentum after and solve algebraically.

8. Collisions and Explosions

Collisions and explosions are interactions in which momentum is transferred between parts of a system. In an isolated system, total momentum is conserved in all such interactions. Kinetic energy, however, may or may not be conserved.

Interaction	Momentum	Kinetic Energy
Elastic collision	Conserved for an isolated system.	Total kinetic energy is conserved.
Inelastic collision	Conserved for an isolated system.	Total kinetic energy decreases.
Perfectly inelastic collision	Conserved for an isolated system.	Bodies move together and kinetic energy decreases.
Explosion	Conserved for an isolated system.	May increase because stored internal energy becomes kinetic energy.

IB Scope. The use of simultaneous equations involving conservation of momentum and conservation of energy is not required. SL quantitative collisions and explosions are one-dimensional; HL can also be quantitative in two dimensions.

Worked Example 3. Perfectly Inelastic Collision

A 2.0 kg cart moving at 4.0 m/s collides with a stationary 1.0 kg cart. They stick together. Determine their common speed and the change in total kinetic energy.

Initial momentum = (2.0 kg)(4.0 m/s) + (1.0 kg)(0 m/s) = 8.0 kg m/s.
After collision, total mass = 3.0 kg. Momentum conservation gives 8.0 kg m/s = (3.0 kg)v, so v = 2.67 m/s.
Initial kinetic energy = ½(2.0 kg)(4.0 m/s)² = 16 J.
Final kinetic energy = ½(3.0 kg)(2.67 m/s)² ≈ 10.7 J.
Change in kinetic energy = 10.7 J − 16 J = −5.3 J.

Two-Dimensional Momentum at HL

For HL quantitative collisions or explosions in two dimensions, momentum conservation is applied independently along two perpendicular directions. Choose axes, resolve every momentum vector, conserve momentum in each direction and then recombine components where necessary.

Worked Example 4. HL Two-Dimensional Momentum

A 1.0 kg ball moving east at 5.0 m/s collides with an identical stationary ball. After the collision, the first ball moves at 3.0 m/s at 30° north of east. Determine the velocity of the second ball.

The initial momentum components are 5.0 kg m/s in the x-direction and 0 kg m/s in the y-direction.
For ball 1 after the collision, the momentum components are (1.0 kg)(3.0 m/s)cos30° = 2.60 kg m/s in x and (1.0 kg)(3.0 m/s)sin30° = 1.50 kg m/s in y.
Conservation in x gives the second ball a momentum of 5.0 kg m/s − 2.60 kg m/s = 2.40 kg m/s in the x-direction.
Conservation in y gives the second ball a momentum of 0 kg m/s − 1.50 kg m/s = −1.50 kg m/s in the y-direction.
Because the second mass is 1.0 kg, its velocity components are 2.40 m/s in x and −1.50 m/s in y.
Speed = √[(2.40 m/s)² + (1.50 m/s)²] = 2.83 m/s.
Direction = tan⁻¹(1.50/2.40) ≈ 32° south of east.

9. Circular Motion

An object can accelerate even when its speed is constant, because velocity changes whenever direction changes. For motion along a circular trajectory, the inward component of acceleration is centripetal acceleration and points radially toward the centre of the circle.

Angular Velocity and Linear Speed

For circular motion of radius r, linear speed and angular velocity are related by v = 2πr/T = ωr.

Centripetal Acceleration and Force

The centripetal acceleration is given by v²/r, which is also equal to ω²r and 4π²r/T². Newton’s second law then requires an inward resultant force equal to mass times centripetal acceleration, or equivalently mv²/r.

The real interaction providing this inward resultant depends on the situation: gravity can supply it in an orbit, tension can supply it in a string, and friction can supply it for a vehicle turning on a flat road.

Uniform and Non-Uniform Circular Motion

In uniform circular motion, speed is constant and acceleration is purely radial. In non-uniform circular motion, the speed also changes, so there is a tangential acceleration in addition to the inward radial acceleration.

Worked Example 5. Centripetal Acceleration and Inward Force

An 800 kg car travels at 15 m/s around a level circular curve of radius 40 m. Determine the centripetal acceleration and the inward resultant force.

The centripetal acceleration is (15 m/s)²/(40 m) = 225 m²/s² ÷ 40 m = 5.625 m/s².
The inward resultant force is mass times centripetal acceleration: (800 kg)(5.625 m/s²) = 4500 N.
Therefore, the car has a centripetal acceleration of 5.625 m/s² and an inward resultant force of 4500 N.

10. A.2 Synthesis: What You Must Be Able to Do

Skill	Exam-Ready Outcome
Represent forces	Construct and interpret one- and two-dimensional free-body diagrams.
Use Newton’s laws	Recognise equilibrium, calculate acceleration and identify correct third-law pairs.
Model contact forces	Apply normal force, friction, Hooke’s law, viscous drag and buoyancy appropriately.
Use momentum and impulse	Apply p = mv, J = FΔt = Δp and force–time graph areas.
Conserve momentum	Define a suitable system and solve collisions and explosions with correct direction conventions.
Analyse circular motion	Connect angular and linear quantities and identify the real inward force resultant.

Essential Equations at a Glance

Purpose	Equation
Static friction	Static friction does not exceed μ_s × normal force
Dynamic friction	Dynamic friction = coefficient of dynamic friction × normal force
Elastic restoring force	Restoring force = −kx
Viscous drag	Viscous drag = 6πηrv
Buoyancy	Buoyancy = ρVg
Weight	Weight = mg
Momentum	Momentum = mass × velocity
Impulse	Impulse = force × time interval = change in momentum
Newton’s second law	Resultant force = mass × acceleration = change in momentum / time
Centripetal acceleration	Centripetal acceleration = v²/r = ω²r = 4π²r/T²
Angular and linear speed	v = 2πr/T = ωr

Equation-Selection Map

High-Value Checks Before Finalising an Answer

Have I drawn the free-body diagram for the correct body or system?
Have I included only real forces acting on that body and labelled them correctly?
If I assumed static friction equals its maximum value, is the body actually at limiting equilibrium? If not, determine the required static friction from the force balance and simply check that it does not exceed the maximum possible value.
Have I distinguished balanced forces on one body from a Newton’s third-law pair on two bodies?
Have I chosen a positive direction before calculating momentum or impulse?
For circular motion, have I identified the real force or resultant that points inward rather than drawing an extra “centripetal force”?

Common Misconceptions and Corrections

Misconception	Correction / Diagnostic Check
A moving body needs a resultant force to keep moving.	A zero resultant force gives constant velocity.
The normal force always equals mg.	Only in particular situations. Determine the normal force from the force balance.
Static friction always equals its maximum possible value.	The quantity μ_s times the normal force is only the maximum possible static friction.
Momentum is conserved only in elastic collisions.	Momentum is conserved in every isolated collision or explosion.
“Centripetal force” is an extra force.	It is the inward resultant of real forces.

A Disciplined A.2 Problem-Solving Workflow

Define the body or system. Decide exactly what is inside the system boundary.
Sketch the geometry and choose axes. For momentum problems, establish a positive direction before writing velocities.
Draw the free-body diagram if forces are involved. Name every interaction and resolve forces only after the physical picture is clear.
Classify the model: equilibrium, constant-mass Newton’s second law, impulse/change of momentum, collision/explosion, or circular motion.
Choose the equation from the physical model rather than from the list of known numerical values.
Carry vector signs or components through the calculation. Do not discard direction information too early.
Check assumptions: static versus sliding friction, isolated system, negligible external impulse, elastic versus inelastic interaction, and whether the circular-motion scope is within the required IB treatment.
Check units, magnitude and physical direction. A correct calculation should agree with the free-body diagram and the predicted motion.

Precise Terminology to Retain

Term	Use It to Mean...
Resultant / net force	The vector sum of all external forces acting on the chosen body or system.
Free-body diagram	A diagram of one isolated body showing all external forces acting on it.
Translational equilibrium	A state in which the resultant force is zero and the acceleration is zero.
Static friction	Friction that prevents relative sliding; it adjusts up to a limiting maximum.
Dynamic friction	Frictional force during sliding, modelled by the coefficient of dynamic friction times the normal force.
Restoring force	Force directed toward an equilibrium position, such as the spring relation −kx.
Momentum	Vector quantity p = mv.
Impulse	Change in momentum caused by a resultant external force acting over time.
Isolated system	System for which no resultant external impulse changes total momentum over the interaction considered.
Elastic collision	Collision in which total kinetic energy as well as momentum is conserved.
Inelastic collision	Collision in which momentum is conserved for an isolated system but total kinetic energy is not.
Centripetal acceleration	Radially inward acceleration associated with motion along a circular path.
Angular velocity	Rate of angular motion, related to linear speed by v = ωr.

A.3 Work, Energy and Power

Recommended Time: 8 Hours | Standard Level and Higher Level

Contents

Guiding Questions

How are concepts of work, energy and power used to predict changes within a system?
How can a consideration of energetics be used as a method to solve problems in kinematics?
How can transfer of energy be used to do work?

1. A.3 at a Glance

A.3 provides another way of analysing motion. In A.1, motion was described using displacement, velocity and acceleration. In A.2, forces were used to explain why motion changes. In A.3, the same physical situations can often be analysed by tracking energy transfers.

The central principle is that energy is conserved. Energy can move between stores or forms and can be transferred between a system and its surroundings, but it is not created or destroyed. Mechanical situations can therefore often be solved without explicitly calculating acceleration or time.

Core Idea	What You Need to Be Able to Do
Work	Relate work done by a force to the energy transferred and use W = Fs cos θ.
Kinetic energy	Relate mass, speed and momentum to translational kinetic energy.
Gravitational potential energy	Calculate changes in gravitational potential energy close to Earth’s surface.
Elastic potential energy	Calculate energy stored in a spring obeying Hooke’s law.
Mechanical-energy conservation	Transfer energy between kinetic, gravitational and elastic forms when resistive forces are absent.
Dissipation	Explain changes in mechanical energy when friction or other non-conservative forces act.
Sankey diagrams	Represent useful and dissipated energy transfers quantitatively.
Power	Calculate the rate at which work is done or energy is transferred.
Efficiency	Compare useful energy or power output with total input.
Energy density	Compare fuel sources using energy available per unit volume.

2. Work and Energy Transfer

Work is the transfer of energy that occurs when a force acts through a displacement.

For a constant force,

W = Fs cos θ

where:

W = work done, in joules, J
F = |F| = magnitude of the force, in N
s = |s| = magnitude of the displacement, in m
θ = angle between F and s

The joule is therefore equivalent to:

1 J = 1 N m

The equation uses only the component of the force parallel to the displacement.

Positive, Negative and Zero Work

Angle Between Force and Displacement	Sign of Work	Interpretation
0°	Positive	Force acts in the direction of displacement.
Between 0° and 90°	Positive	Force has a component in the direction of displacement.
90°	Zero	Force is perpendicular to displacement.
Between 90° and 180°	Negative	Force has a component opposite the displacement.
180°	Negative	Force acts directly opposite the displacement.

This explains an important circular-motion result. In uniform circular motion the inward force is perpendicular to the instantaneous displacement, so that force does no work on the body. The direction of v changes, but its magnitude and therefore its kinetic energy remain constant.

Worked Example 1. Work Done by an Angled Force

A constant force of 50 N pulls an object through 6.0 m. The force makes an angle of 30° with the displacement.

W = Fs cos θ

W = (50 N)(6.0 m) cos 30°

W = 260 J

Therefore, approximately 260 J of energy is transferred by the force.

Work Done by the Resultant Force

The syllabus requires the connection between resultant work and the change in the energy of the system. In translational mechanics, the familiar work–kinetic-energy result is:

W_net = ΔE_k

A positive resultant work increases kinetic energy; negative resultant work decreases it.

Force–Displacement Graphs

When force changes with displacement, the work done can be interpreted from the area under a force–displacement graph. This is particularly useful for springs because the force changes as the spring is stretched or compressed.

3. Mechanical Energy

The total mechanical energy of a system is the sum of:

E_mech = E_k + E_p + E_H

where:

E_k is kinetic energy
E_p is gravitational potential energy
E_H is elastic potential energy.

Kinetic Energy

For translational motion:

E_k = ½mv² = p²/(2m)

Kinetic energy depends on the square of speed. Doubling speed therefore increases kinetic energy by a factor of four.

Because E_k contains v², kinetic energy is a scalar and does not indicate the direction of motion.

Gravitational Potential Energy

Close to the Earth’s surface:

ΔE_p = mgΔh

where Δh is the change in vertical height.

Only the change in height matters. The route taken between the two positions does not alter ΔE_p.

A reference position is chosen for zero gravitational potential energy. It is usually the difference in potential energy that has physical significance in A.3.

Elastic Potential Energy

For a spring obeying Hooke’s law:

E_H = ½k(Δx)²

where:

k = spring constant in N/m
Δx = extension or compression from the natural length.

The factor of ½ occurs because the spring force increases from zero to kΔx during extension. The energy stored is the triangular area underneath the force–extension graph.

Worked Example 2. Elastic Potential Energy

A spring has spring constant 250 N/m and is compressed by 0.080 m.

E_H = ½k(Δx)²

E_H = ½(250 N/m)(0.080 m)²

E_H = 0.80 J

The spring stores 0.80 J of elastic potential energy.

4. Conservation of Mechanical Energy

The general conservation principle states:

For an ideal mechanical system in which no frictional or other resistive forces remove mechanical energy:

E_{mech, initial} = E_{mech, final}

or:

E_k,i + E_p,i + E_H,i = E_k,f + E_p,f + E_H,f

This provides a powerful method for solving motion problems. Unlike the constant-acceleration equations from A.1, an energy method does not require the acceleration to be uniform.

Typical Energy Transformations

Situation	Energy Transformation
Falling object	E_p → E_k
Object moving upward	E_k → E_p
Compressed spring launching an object	E_H → E_k
Object moving upward while compressing a spring	E_k → E_p + E_H

Worked Example 3. Falling Using Conservation of Energy

A 2.0 kg object is released from rest 5.0 m above the ground. Ignore air resistance.

Initially:

E_k,i = 0

E_p,i = (2.0 kg)(9.8 m/s²)(5.0 m) = 98 J

Immediately before reaching the ground:

E_p,f = 0

Conservation of mechanical energy gives:

98 J = ½(2.0 kg)v²

98 J = (1.0 kg)v²

v² = 98 m²/s²

v = 9.9 m/s

The speed immediately before impact is approximately 9.9 m/s.

Energy Method Versus Kinematics

For a vertical fall without resistance,

mgΔh = ½mv² − ½mu²

The mass cancels, which is why bodies falling through the same vertical height under the same ideal conditions undergo the same change in speed regardless of mass.

5. Non-Conservative Forces and Energy Dissipation

Mechanical energy is not necessarily conserved when friction, drag or another non-conservative force acts.

The change in total mechanical energy can be interpreted in terms of work done by non-conservative forces:

W_{non-conservative} = ΔE_mech

with the sign interpreted according to whether the force transfers energy into or out of the mechanical-energy store of the system.

Friction does not destroy energy. Instead, mechanical energy is transferred into other forms, commonly internal energy of the objects and surroundings.

initial energy = final useful mechanical energy + energy dissipated

Energy that spreads into the surroundings and becomes difficult to recover for useful work is often described as dissipated or degraded energy.

Worked Example 4. Energy Dissipated by Friction

A 5.0 kg block descends through a vertical height of 2.0 m from rest. Its speed at the bottom is 5.0 m/s.

Initial gravitational potential energy:

E_p,i = (5.0 kg)(9.8 m/s²)(2.0 m) = 98 J

Final kinetic energy:

E_k,f = ½(5.0 kg)(5.0 m/s)² = 62.5 J

Energy dissipated:

E_dissipated = 98 J − 62.5 J = 35.5 J

Therefore, 35.5 J has been transferred away from the mechanical-energy store, principally through frictional processes.

6. Energy Transfers and Sankey Diagrams

A Sankey diagram represents the transfer of energy or power through a system.

The width of each arrow is proportional to the quantity of energy or power represented. Useful output normally continues in the main direction, while dissipated transfers branch away.

For example:

Reading a Sankey Diagram

The total width entering the system represents total input.
Useful output and dissipated output are identified correctly.
The widths of all outputs add to the input.
The diagram represents either energy or power consistently.
Efficiency can be inferred from the ratio of useful output to total input.

Sankey diagrams are particularly useful for comparing energy-transfer systems and visualising where energy becomes dissipated.

7. Power

Power measures how quickly work is done or energy is transferred.

P = ΔW/Δt

The SI unit is the watt:

1 W = 1 J/s

A high-power device does not necessarily transfer more total energy. It transfers energy at a greater rate.

Two machines may perform the same 10 000 J of work, but the one that performs it in less time has the greater power.

The Relationship P = Fv

For a force acting along the direction of motion:

P = Fv

This follows from:

P = ΔW/Δt

ΔW = FΔs

P = FΔs/Δt = F(Δs/Δt) = Fv

This relationship is especially useful for vehicles moving at constant speed against resistive forces.

Worked Example 5. Power at Constant Speed

A car travels at a constant speed of 20 m/s. The total resistive force is 600 N.

At constant velocity, the engine must provide a forward force of 600 N.

P = Fv

P = (600 N)(20 m/s)

P = 12 000 W = 12 kW

The useful mechanical power required is 12 kW.

8. Efficiency

Real energy-transfer systems normally produce both useful output and dissipated output.

Efficiency measures the fraction of total input that becomes useful output.

η = E_{useful output}/E_{total input}

or:

η = P_{useful output}/P_{total input}

Efficiency has no unit. It may be written either as a decimal or percentage:

η = 0.75 = 75%

Worked Example 6. Efficiency

A motor receives 600 J of electrical energy and performs 450 J of useful mechanical work.

η = 450 J / 600 J

η = 0.75 = 75%

The remaining energy is:

600 J − 450 J = 150 J

so 150 J is transferred into non-useful forms. A corresponding Sankey diagram would therefore contain a 600 J input, a 450 J useful branch and a 150 J dissipated branch.

9. Energy Density of Fuel Sources

The syllabus requires understanding of the energy density of fuel sources.

Energy density is the amount of energy that can be transferred from a unit volume of fuel:

energy density = energy available / volume

SI unit:

J/m³

Fuel sources with greater energy density contain more transferable energy in a given volume.

The approved texts emphasise that energy densities vary enormously. Nuclear fuels have exceptionally large energy densities compared with conventional chemical fuels.

Energy Density Versus Specific Energy

Quantity	Definition	Unit
Energy density	energy / volume	J/m³
Specific energy	energy / mass	J/kg

The IB A.3 syllabus specifically names energy density of fuel sources.

Worked Example 7. Energy Density

Suppose a fuel sample releases 2.4 × 10⁹ J from a volume of 0.10 m³.

energy density = (2.4 × 10⁹ J)/(0.10 m³)

energy density = 2.4 × 10¹⁰ J/m³

10. A.3 Synthesis: What You Must Be Able to Do

Essential Equations at a Glance

Concept	Required Relationship
Work	W = Fs cos θ
Kinetic energy	E_k = ½mv² = p²/(2m)
Change in gravitational potential energy	ΔE_p = mgΔh
Elastic potential energy	E_H = ½k(Δx)²
Mechanical energy	E_mech = E_k + E_p + E_H
Power	P = ΔW/Δt = Fv
Efficiency using energy	η = E_{useful output}/E_{total input}
Efficiency using power	η = P_{useful output}/P_{total input}
Energy density	energy / volume

The principal A.3 equations appear together on the Space, Time and Motion page of the IB Physics data booklet.

Equation-Selection Map

If the Question Gives You...	Think...	Main Relationship
Force, displacement and angle	Work done	W = Fs cos θ
Mass and speed	Kinetic energy	E_k = ½mv²
Momentum and mass	Kinetic energy	E_k = p²/(2m)
Mass and vertical-height change	Gravitational potential energy	ΔE_p = mgΔh
Spring constant and extension	Elastic potential energy	E_H = ½k(Δx)²
Initial and final mechanical states with no resistance	Conservation of mechanical energy	E_mech,i = E_mech,f
Friction or drag plus an energy change	Non-conservative work	Compare initial and final mechanical energy
Work or energy and time	Power	P = ΔW/Δt
Force and speed	Mechanical power	P = Fv
Useful and total energy or power	Efficiency	output ÷ input
Energy and fuel volume	Energy density	energy ÷ volume

Equation-Selection Infographic

WORK

Force + displacement + angle

W = Fs cos θ

MECHANICAL ENERGY

Mass/speed, height or spring extension

E_k, ΔE_p, E_H

CONSERVATION

Initial and final states; negligible resistance

E_mech,i = E_mech,f

POWER & EFFICIENCY

Energy/work with time, or useful versus total

P = ΔW/Δt = Fv; η = output/input

High-Value Checks Before Finalising an Answer

Define the system. Decide what objects and energy stores are inside the system boundary.
Identify the initial and final states. Do not start substituting numbers before deciding which energies exist at each state.
Check whether mechanical energy is actually conserved. Friction or resistance normally means it is not.
Use vertical height for gravitational potential energy. Path length along a slope is not Δh.
Use extension or compression from the spring’s natural length in E_H.
Square speed and spring extension carefully.
Check the angle in the work equation. It is the angle between force and displacement.
Do not confuse energy with power. Energy is measured in J; power in W.
Do not confuse energy density with specific energy.
Show units through every numerical calculation, not only in the final answer.
Check the energy balance. Inputs must equal all outputs once every relevant energy transfer has been included.

Common Misconceptions and Corrections

Misconception	Correction / Diagnostic Check
Energy is “used up”.	Energy is transferred. Total energy is conserved.
Mechanical energy is always conserved.	Only when no frictional or other relevant non-conservative forces remove mechanical energy.
A force always does work if it acts on a moving body.	Work depends on the component of force along displacement. A perpendicular force does zero work.
W = Fs can always be used directly.	Only when force and displacement are parallel. Otherwise use W = Fs cos θ.
Kinetic energy has direction because velocity has direction.	Kinetic energy is a scalar.
Gravitational potential energy depends on the distance travelled.	Near Earth’s surface, the change depends on the change in vertical height.
A stretched spring stores kx².	The correct elastic potential energy is ½k(Δx)².
Power and energy mean the same thing.	Power is the rate of energy transfer.
Efficiency can exceed 100%.	Useful output cannot exceed total input.
Dissipated energy has disappeared.	It has been transferred to less useful forms, commonly internal energy of the surroundings.
Energy density means energy per kilogram.	In A.3, energy density means energy per unit volume, J/m³.

A Disciplined A.3 Problem-Solving Workflow

1. Define the system and the two states. Identify exactly what is being compared: initial and final positions, speeds, spring extensions and surroundings.

2. List the energy present initially. E_k,i, E_p,i, E_H,i.

3. List the energy present finally. E_k,f, E_p,f, E_H,f.

4. Decide whether non-conservative forces transfer energy. If no: conserve mechanical energy. If yes: include the corresponding mechanical-energy change or dissipated energy.

5. Choose the physical principle before choosing an equation. Do not select an equation simply because its symbols match the numbers in the question.

6. Write the energy equation symbolically. For example: mgh = ½mv² + E_dissipated. Only then substitute values.

7. Keep units throughout the calculation.

8. Check the result physically. A calculated speed, energy, power or efficiency should agree with the expected direction of the energy transfer.

Using the IB Data Booklet

The A.3 section of the data booklet supplies the central relationships for W, E_k, ΔE_p, E_H, P and η, so examination success depends less on memorising isolated formulas and more on recognising the energy model represented by the problem.

Precise Terminology to Retain

Term	Use It to Mean...
Work	Energy transferred when a force acts through a displacement.
Energy	A conserved physical quantity that can be transferred between forms or systems.
Kinetic energy	Mechanical energy associated with translational motion.
Gravitational potential energy	Energy associated with position in a gravitational interaction; in A.3 use ΔE_p = mgΔh close to Earth’s surface.
Elastic potential energy	Energy stored through elastic deformation, such as a stretched or compressed spring.
Mechanical energy	Sum of kinetic, gravitational potential and elastic potential energies.
Conservative force	A force for which mechanical-energy transfer does not depend on the path between two positions.
Non-conservative force	A force such as friction that changes the mechanical energy of the system through dissipative transfer.
Dissipated energy	Energy transferred into forms that spread into the surroundings and become less available for useful work.
Power	Rate of doing work or transferring energy.
Efficiency	Ratio of useful output energy or power to total input energy or power.
Sankey diagram	Diagram in which arrow widths represent amounts of transferred energy or power.
Energy density	Energy available from a unit volume of a fuel source, in J/m³.
Specific energy	Energy available per unit mass, in J/kg; not the same as energy density.

A.4 Rigid Body Mechanics

Recommended Time: 7 Hours | Higher Level Only

Contents

Guiding Questions

How can the understanding of linear motion be applied to rotational motion?
How is the understanding of the torques acting on a system used to predict changes in rotational motion?
How does the distribution of mass within a body affect its rotational motion?

1. A.4 at a Glance

Earlier mechanics often treated an object as though all of its mass were concentrated at one point. That model is excellent for describing the translation of an object’s centre of mass, but it cannot describe what happens when a force acts away from that centre and the object begins to rotate. A.4 extends the mechanics of A.1–A.3 from point objects to extended rigid bodies.

A rigid body is modelled as an extended object whose shape does not change significantly while forces act on it. The position of its centre of mass can still describe translational motion, while its orientation about an axis is described using angular quantities.

Fig. 1

A resultant force through the centre of mass can change translational motion without producing a turning effect. An off-centre resultant force can change both translation and rotation.

Linear Motion

Position, velocity and acceleration describe the motion of the centre of mass.

resultant force → linear acceleration

Rotational Motion

Angular displacement, angular speed and angular acceleration describe rotation about an axis.

resultant torque → angular acceleration

2. Torque and the Turning Effect of a Force

Torque measures the turning effect of a force about an axis. Its magnitude is:

τ = Fr sin θ

where F is the magnitude of the force, r is the distance from the axis to the point where the force acts, and θ is the angle between the direction of F and the direction of r.

Equivalently, torque can be calculated as:

τ = Fd_⊥

where d_⊥ is the perpendicular distance from the axis to the line of action of the force.

Fig. 2

The lever arm d⊥ is the perpendicular distance from the axis to the force’s line of action. The angle θ is the angle between the directions of r and F.

Maximum and Zero Torque

When the force is perpendicular to the radius, θ = 90°, so sin θ = 1 and the torque is maximum: τ = Fr.
When the line of action passes through the axis, the perpendicular distance is zero and τ = 0.
Increasing either F or the perpendicular lever arm increases the turning effect.

Worked Example 1. Torque from an Angled Force

A 40 N force acts at the end of a 0.50 m spanner. The angle between the spanner and the force is 60°. Determine the torque about the bolt.

τ = Fr sin θ

τ = (40 N)(0.50 m) sin 60°

τ = 17.3 N m

The magnitude of the torque is 17 N m to two significant figures. Its sign depends on whether the force tends to turn the spanner clockwise or counter-clockwise.

Rotational Equilibrium

A body in rotational equilibrium has zero resultant torque:

Στ = 0

This means that clockwise and counter-clockwise turning effects balance. Rotational equilibrium does not by itself guarantee translational equilibrium; a body in complete static equilibrium must satisfy both zero resultant force and zero resultant torque.

Fig. 3

For rotational equilibrium, the sum of clockwise torques equals the sum of counter-clockwise torques.

Worked Example 2. Balancing a Beam

A 300 N load acts 0.80 m to the left of a pivot. A second downward force acts 1.20 m to the right. Determine the force required for rotational equilibrium.

Counter-clockwise torque = (300 N)(0.80 m) = 240 N m.

For equilibrium, clockwise torque = 240 N m.

F(1.20 m) = 240 N m.

F = 200 N.

3. Angular Kinematics

The rotational equivalents of displacement, velocity and acceleration are angular displacement, angular speed and angular acceleration. Although the term angular velocity is used, a formal vector treatment is not required in this topic.

Linear Quantity	Rotational Analogue	Meaning
displacement	angular displacement Δθ	change in angular position, measured in rad
speed	angular speed ω	rate of change of angular displacement, in rad/s
acceleration	angular acceleration α	rate of change of angular speed, in rad/s²

Fig. 4

Because every point turns through the same angle in the same time, the whole rigid body has one angular speed. A point farther from the axis has a larger linear speed because v = rω.

Uniform Angular Acceleration

When angular acceleration is constant, the familiar constant-acceleration equations from linear kinematics have rotational counterparts:

Δθ = ((ω_f + ω_i)/2)t

ω_f = ω_i + αt

Δθ = ω_it + ½αt²

ω_f² = ω_i² + 2αΔθ

Use + and − to include direction. A chosen positive rotational sense is essential. For example, if counter-clockwise is positive, clockwise angular displacement, angular speed and angular acceleration are negative.

Worked Example 3. Uniform Angular Acceleration

A wheel starts from rest and has a constant angular acceleration of 3.0 rad/s² for 4.0 s. Determine its final angular speed and angular displacement.

ω_f = ω_i + αt = 0 + (3.0 rad/s²)(4.0 s) = 12 rad/s.

Δθ = ω_it + ½αt² = 0 + ½(3.0 rad/s²)(4.0 s)² = 24 rad.

4. Moment of Inertia

Moment of inertia, I, describes how strongly a body resists angular acceleration about a particular axis. It depends not only on the total mass, but also on how that mass is distributed relative to the axis.

For a system of point masses:

I = Σmr²

The squared-distance term is crucial: moving the same mass farther from the axis can greatly increase the moment of inertia.

Fig. 5

The two systems can have the same total mass but very different moments of inertia because the distances from the axis are different.

Worked Example 4. Moment of Inertia of Point Masses

Two 1.5 kg masses are attached to a light rigid rod at distances 0.20 m and 0.60 m from the axis. Calculate the system’s moment of inertia.

I = Σmr²

I = (1.5 kg)(0.20 m)² + (1.5 kg)(0.60 m)²

I = 0.060 kg m² + 0.540 kg m² = 0.600 kg m²

5. Newton’s Second Law for Rotation

The rotational form of Newton’s second law is:

τ = Iα

where τ is the resultant or average torque, I is the moment of inertia about the chosen axis, and α is the angular acceleration.

Fig. 6

For the same resultant torque, increasing the moment of inertia reduces the angular acceleration: α = τ/I.

Worked Example 5. Torque to Angular Acceleration

A rigid system has moment of inertia 0.60 kg m². A resultant torque of 3.0 N m acts on it. Determine the angular acceleration.

α = τ/I = (3.0 N m)/(0.60 kg m²) = 5.0 rad/s².

6. Angular Momentum and Angular Impulse

An extended rigid body rotating with angular speed ω has angular momentum:

L = Iω

This is the rotational analogue of linear momentum. A resultant torque changes angular momentum. For a constant or average torque acting over a time interval:

ΔL = τΔt = Δ(Iω)

The product τΔt is the angular impulse.

Fig. 7

For a constant torque, the area under a torque–time graph equals the angular impulse and therefore the change in angular momentum.

7. Conservation of Angular Momentum

Angular momentum remains constant unless a resultant external torque acts on the system. Therefore:

L_initial = L_final

For a single rotating body whose moment of inertia changes:

I_iω_i = I_fω_f

If I decreases while external torque is negligible, ω must increase. If I increases, ω must decrease.

Fig. 8

The skater changes the distribution of mass. With negligible external torque, angular momentum is conserved, so a reduction in moment of inertia produces an increase in angular speed.

Coupled Rotating Bodies

A.4 also includes situations in which two bodies become rotationally coupled. If the total external torque on the chosen system is negligible, the total angular momentum before coupling equals the total angular momentum after coupling.

Fig. 9

When an initially separate body joins a rotating body, the total moment of inertia can increase and the common angular speed can decrease while total angular momentum remains constant.

Worked Example 6. Coupled Disks

A disk with moment of inertia 0.050 kg m² rotates at 8.0 rad/s. A second disk, initially at rest, is dropped concentrically onto it. Their combined moment of inertia is 0.080 kg m². Neglect external torque. Determine their common final angular speed.

Initial angular momentum: L_i = (0.050 kg m²)(8.0 rad/s) = 0.40 kg m²/s.

Conservation: L_i = L_f.

0.40 kg m²/s = (0.080 kg m²)ω_f.

ω_f = 5.0 rad/s.

8. Rotational Kinetic Energy

A rotating rigid body has kinetic energy because every mass element is moving. The total rotational kinetic energy is:

E_k = ½Iω² = L²/(2I)

The first form is most useful when I and ω are known. The second form is particularly revealing in angular-momentum-conservation situations.

Important distinction. Conservation of angular momentum does not automatically mean conservation of rotational kinetic energy. In a skater pulling in their arms, internal biochemical energy can be transferred into rotational kinetic energy. In two bodies coupling through friction, some rotational kinetic energy can be transferred to internal energy even though total angular momentum remains conserved.

Worked Example 7. Rotational Kinetic Energy

A body has moment of inertia 0.30 kg m² and rotates at 6.0 rad/s.

E_k = ½Iω² = ½(0.30 kg m²)(6.0 rad/s)² = 5.4 J.

9. Rolling Without Slipping

Simultaneous translational and rotational motion in A.4 is restricted to rolling without slipping. The centre of mass translates while the object rotates about its centre.

For rolling without slipping:

v = rω

The total kinetic energy is the sum of translational and rotational kinetic energies:

E_k,total = ½mv² + ½Iω²

Fig. 10

For pure rolling, the point of contact is instantaneously at rest relative to the surface, the centre moves at speed v, and the top point moves at speed 2v relative to the surface.

Energy in Rolling Motion

If a rolling body moves down through a vertical height and dissipative losses are negligible, gravitational potential energy can be transferred into both translational and rotational kinetic energy:

mgΔh = ½mv² + ½Iω²

Using v = rω links the two kinetic-energy terms. This makes the distribution of mass—through I—important to how quickly different rolling bodies accelerate.

10. A.4 Synthesis: What You Must Be Able to Do

A.4 is easiest when treated as a rotational extension of mechanics you already know. The central pattern is:

Fig. 11

The close mathematical analogy between linear and rotational mechanics is a powerful organizing idea throughout A.4.

Essential Equations at a Glance

Purpose	Relationship	Use When...
Torque	τ = Fr sin θ	A force acts at a known distance and angle from an axis.
Rotational equilibrium	Στ = 0	Clockwise and counter-clockwise turning effects balance.
Uniform angular acceleration	Δθ = ((ω_f + ω_i)/2)t	Average angular speed is useful.
Uniform angular acceleration	ω_f = ω_i + αt	Angular displacement is not needed.
Uniform angular acceleration	Δθ = ω_it + ½αt²	Final angular speed is not needed.
Uniform angular acceleration	ω_f² = ω_i² + 2αΔθ	Time is not needed.
Point-mass moment of inertia	I = Σmr²	The body is modelled as discrete point masses.
Rotational Newton II	τ = Iα	Resultant torque causes angular acceleration.
Angular momentum	L = Iω	Moment of inertia and angular speed are known.
Angular impulse	ΔL = τΔt = Δ(Iω)	A resultant torque acts for a time interval.
Rotational kinetic energy	E_k = ½Iω² = L²/(2I)	Energy of rotational motion is required.
Rolling without slipping	v = rω	Translation and rotation occur together without slipping.

Equation-Selection Infographic

FORCE + DISTANCE + ANGLE

Find a turning effect.

τ = Fr sin θ

TORQUES BALANCE

No angular acceleration.

Στ = 0

TORQUE + MOMENT OF INERTIA

Find angular acceleration.

τ = Iα

CONSTANT α

Choose the angular kinematics equation that omits the unneeded variable.

ω, α, Δθ, t

NO EXTERNAL RESULTANT TORQUE

Use angular-momentum conservation.

ΣL_i = ΣL_f

TORQUE ACTS FOR A TIME

Find the change in angular momentum.

ΔL = τΔt

ROTATIONAL ENERGY

Use moment of inertia with angular speed, or angular momentum.

E_k = ½Iω² = L²/(2I)

ROLLING WITHOUT SLIPPING

Link translation and rotation.

v = rω

Common Misconceptions and Corrections

Misconception	Correction
Any force on an extended object causes rotation.	A force whose line of action passes through the axis or centre of mass can have zero torque about that axis.
Torque is always simply Fr.	Only when force and radius are perpendicular. In general use τ = Fr sin θ or Fd_⊥.
Zero resultant force means zero resultant torque.	Not necessarily. Translational and rotational equilibrium are separate conditions.
All points on a rotating rigid body have the same linear speed.	They have the same angular speed. Linear speed depends on radius: v = rω.
Moment of inertia depends only on total mass.	It also depends strongly on how mass is distributed relative to the chosen axis.
Angular momentum conservation means rotational kinetic energy is conserved.	Angular momentum can be conserved while rotational kinetic energy changes.
Moment of inertia formulas for every shape must be memorized.	The equation for a required specific mass distribution will be provided when necessary; know I = Σmr² for point masses and understand mass distribution qualitatively.

A Disciplined A.4 Problem-Solving Workflow

Choose the system and the axis. Torque and moment of inertia are always defined about an axis.
Draw the forces clearly. Identify each line of action and the perpendicular distance to the axis.
Choose a rotational sign convention. State which sense is positive.
Decide the physics regime. Equilibrium? Constant angular acceleration? Resultant torque? Angular-momentum conservation? Rolling?
Find I before using rotational dynamics. If the mass distribution changes, calculate the initial and final moments of inertia separately.
Check whether external torque is negligible before conserving L. Internal friction can redistribute angular momentum within the system without changing the total.
Do not assume energy conservation just because L is conserved. Check whether internal energy changes or work is done internally.
For rolling without slipping, link v and ω immediately. Use v = rω before combining translational and rotational energy.
Check units. Torque: N m; I: kg m²; α: rad/s²; L: kg m²/s; rotational energy: J.

Final Retrieval Check. You have covered A.4 when you can move confidently from forces to torques, from torque to angular acceleration, from mass distribution to moment of inertia, from changing moment of inertia to angular-momentum conservation, and from pure rotation to rolling without slipping.

A.5 Galilean and Special Relativity

Recommended Time: 8 Hours | Higher Level Only

Guiding Questions

How do observers in different reference frames describe events in terms of space and time?
How does special relativity change our understanding of motion compared with Galilean relativity?
How are spacetime diagrams used to represent relativistic motion?

1. Reference Frames and Events

Relativity begins with a simple fact: measurements of position and velocity must be made relative to a chosen reference frame. An event is something that occurs at a particular position and a particular time. In one-dimensional relativity, an event is assigned coordinates (x, t).

Reference Frame

A coordinate system, together with synchronized clocks, used to assign positions and times to events.

Inertial Reference Frame

A non-accelerating frame. Newton's first law holds in an inertial frame. Any two inertial frames move at constant velocity relative to one another.

Fig. 1. The same event may be assigned different spatial coordinates by two inertial frames.

2. Galilean Relativity

Galilean relativity is the classical description appropriate when relative speeds are much smaller than the speed of light. Suppose frame S′ moves at constant speed v in the +x direction relative to S, with origins coincident when t = 0.

x′ = x − vtt′ = tu′ = u − v

Here u, u′ and v are signed one-dimensional velocities along the x-axis. Galilean relativity assumes absolute time: all inertial observers agree on a time interval.

Fig. 2. Classical velocity addition works for everyday motion because ordinary speeds are tiny compared with c.

Worked Example 1: Galilean Velocity Addition

A train travels at 20.0 m/s relative to the ground. A student walks forward at 1.50 m/s relative to the train. Determine the student's velocity relative to the ground.

Step 1: u = u′ + v

Step 2: u = 1.50 m/s + 20.0 m/s

Step 3: u = 21.5 m/s

Answer: 21.5 m/s in the forward direction relative to the ground.

Worked Example 2: Galilean Transformation of Position

A cyclist travels 350 m along a road in 50.0 s. A car moves in the same direction at 12.0 m/s. Determine the cyclist's displacement during the same interval in the reference frame of the car.

Step 1: Δx′ = Δx − vΔt

Step 2: Δx′ = 350 m − (12.0 m/s)(50.0 s)

Step 3: Δx′ = 350 m − 600 m = −250 m

Answer: −250 m. The negative sign means the cyclist moves backward relative to the car.

3. Special Relativity

Galilean relativity predicts that ordinary velocities simply add. Applied to light, this would imply that an observer moving toward a light beam should measure a speed greater than c, while an observer chasing the beam should measure less than c. Special relativity replaces this classical assumption.

Postulate 1

The laws of physics are identical in all inertial reference frames. No inertial frame is physically privileged.

Postulate 2

The speed of light in vacuum has the same value c for every inertial observer, independent of the motion of the source or observer.

Fig. 3. All inertial observers measure the same vacuum light speed c.

4. Lorentz Factor and Lorentz Transformations

The strength of relativistic effects is controlled by the Lorentz factor:

γ = 1 / √(1 − v²/c²)

For ordinary speeds γ is almost exactly 1. As v approaches c, γ increases rapidly.

Fig. 4. Relativistic effects grow rapidly only when the relative speed becomes a substantial fraction of c.

Worked Example 3: Calculating the Lorentz Factor

Calculate γ for a spacecraft moving at 0.800c.

Step 1: γ = 1 / √(1 − v²/c²)

Step 2: γ = 1 / √(1 − 0.800²) = 1 / √0.360

Step 3: γ = 1 / 0.600 = 1.67

Answer: γ = 1.67. The Lorentz factor has no units.

The Lorentz transformations replace the Galilean transformations:

x′ = γ(x − vt)t′ = γ(t − vx/c²)

The transformed time coordinate depends on both t and x. Space and time are therefore linked.

Worked Example 4: Transforming an Event

An event occurs in S at x = 1200 m and t = 6.00 × 10⁻⁶ s. Frame S′ moves at 0.600c in the +x direction. Find x′ and t′.

Step 1: γ = 1 / √(1 − 0.600²) = 1.25

Step 2: v = 0.600(3.00 × 10⁸ m/s) = 1.80 × 10⁸ m/s

Step 3: vt = (1.80 × 10⁸ m/s)(6.00 × 10⁻⁶ s) = 1.08 × 10³ m
x′ = 1.25(1200 m − 1080 m) = 150 m

Step 4: vx/c² = [(1.80 × 10⁸ m/s)(1200 m)]/(9.00 × 10¹⁶ m²/s²) = 2.40 × 10⁻⁶ s
t′ = 1.25[(6.00 × 10⁻⁶ s) − (2.40 × 10⁻⁶ s)] = 4.50 × 10⁻⁶ s

Answer: x′ = 150 m and t′ = 4.50 × 10⁻⁶ s.

5. Relativistic Velocity Addition

u′ = (u − v) / (1 − uv/c²)

An equivalent rearrangement is:

u = (u′ + v) / (1 + u′v/c²)

Critical point. The denominator prevents ordinary sub-light velocities from combining to give a result greater than c.

Fig. 5. Ordinary addition can predict a speed above c. Relativistic addition approaches c without crossing it.

Worked Example 5: Spacecraft Launches a Probe

A spacecraft moves at 0.800c relative to Earth. It launches a probe forward at 0.700c relative to the spacecraft. Determine the probe speed relative to Earth.

Step 1: u = (u′ + v)/(1 + u′v/c²)

Step 2: u = (0.700c + 0.800c)/[1 + (0.700c)(0.800c)/c²]

Step 3: u = 1.500c/1.560 = 0.962c

Step 4: u = 0.962(3.00 × 10⁸ m/s) = 2.89 × 10⁸ m/s

Answer: 0.962c = 2.89 × 10⁸ m/s.

6. Proper Time and Time Dilation

The proper time interval Δt₀ is measured in the frame in which the two events occur at the same position.

Δt = γΔt₀

Because γ ≥ 1, an observer who sees the clock moving measures a longer time interval.

Fig. 6. A light clock makes time dilation visual: the external observer sees a longer light path while still measuring the same light speed c.

Worked Example 6: Muon Time Dilation

A muon has a proper mean lifetime of 2.20 × 10⁻⁶ s and moves at 0.980c relative to Earth. Calculate its mean lifetime measured from Earth and its mean Earth-frame travel distance.

Step 1: γ = 1/√(1 − 0.980²) = 5.03

Step 2: Δt = (5.03)(2.20 × 10⁻⁶ s) = 1.11 × 10⁻⁵ s

Step 3: v = 0.980(3.00 × 10⁸ m/s) = 2.94 × 10⁸ m/s

Step 4: d = vΔt = (2.94 × 10⁸ m/s)(1.11 × 10⁻⁵ s) = 3.26 × 10³ m

Answer: 1.11 × 10⁻⁵ s and approximately 3.26 km.

7. Proper Length and Length Contraction

The proper length L₀ is measured in the frame where the object is at rest. A moving observer measures contraction only in the direction parallel to relative motion.

L = L₀/γ

Fig. 7. The moving spacecraft is measured shorter only along its direction of motion.

Worked Example 7: Length Contraction

A spacecraft has proper length 120 m and passes Earth at 0.800c. Determine its Earth-frame length.

Step 1: γ = 1.67

Step 2: L = 120 m/1.67

Step 3: L = 71.9 m

Answer: approximately 72.0 m.

Fig. 8. Length contraction becomes significant only at relativistic speeds.

8. Relativity of Simultaneity

Spatially separated events that are simultaneous in one inertial frame are not generally simultaneous in another frame moving relative to the first. This disagreement is built into spacetime itself and is not merely caused by signal travel time.

Fig. 9. Simultaneity is frame-dependent for events separated in space.

Δt′ = γ(Δt − vΔx/c²)

Worked Example 8: Simultaneous Lightning Strikes

In the Earth frame, two lightning strikes occur simultaneously and are separated by Δx = +600 m. A rocket moves at 0.800c in the +x direction. Determine Δt′.

Step 1: Δt = 0 s and γ = 1.67.

Step 2: Δt′ = −γvΔx/c²

Step 3: vΔx/c² = [(2.40 × 10⁸ m/s)(600 m)]/(9.00 × 10¹⁶ m²/s²) = 1.60 × 10⁻⁶ s

Step 4: Δt′ = −(1.67)(1.60 × 10⁻⁶ s) = −2.67 × 10⁻⁶ s

Answer: Δt′ = −2.67 μs. With the stated sign convention, the event at the larger x-coordinate occurs earlier in S′.

9. Invariant Spacetime Interval

(Δs)² = (cΔt)² − (Δx)²

Observers can disagree about Δx and Δt separately while agreeing on the spacetime interval.

Useful Interpretation — Extension / Enrichment. The formal A.5 requirement is to use the invariant spacetime interval. The time-like, light-like and space-like labels below are useful for interpreting its sign, but they are not listed as a separate required understanding in the syllabus statement.

Interval Type	Condition	Meaning
Time-like	(Δs)² > 0	A frame can exist in which the two events occur at the same position.
Light-like	(Δs)² = 0	The events can be connected by a light signal.
Space-like	(Δs)² < 0	A frame can exist in which the events are simultaneous.

Worked Example 9: Classifying an Interval

Two events are separated by Δx = 600 m and Δt = 1.00 × 10⁻⁶ s.

Step 1: cΔt = (3.00 × 10⁸ m/s)(1.00 × 10⁻⁶ s) = 300 m

Step 2: (Δs)² = (300 m)² − (600 m)²

Step 3: (Δs)² = 9.00 × 10⁴ m² − 3.60 × 10⁵ m² = −2.70 × 10⁵ m²

Answer: −2.70 × 10⁵ m²; space-like interval.

10. Spacetime Diagrams and Worldlines

A spacetime diagram places x on the horizontal axis and ct on the vertical axis so that both coordinates have dimensions of length. The path of an object through spacetime is its worldline.

Vertical worldline: object at rest in that frame.
Straight tilted worldline: constant velocity.
Photon worldline: 45° when the x and ct axes use the same scale.
For a constant-velocity particle, tan θ = v/c, where θ is measured from the ct-axis.
The moving-frame axes x′ and ct′ are tilted relative to x and ct.
The x/x′ and ct/ct′ scales are not identical; lines of constant spacetime interval define the scale.

Fig. 10. Massive-particle worldlines remain inside the light boundaries.

Worked Example 10: Speed from a Worldline Angle

A constant-velocity worldline makes an angle of 31.0° with the ct-axis. Determine the speed.

Step 1: tan θ = v/c

Step 2: v = c tan 31.0°

Step 3: v = (3.00 × 10⁸ m/s)(0.601) = 1.80 × 10⁸ m/s

Answer: 1.80 × 10⁸ m/s ≈ 0.600c.

Fig. 11. The moving-frame axes are tilted and require their own calibration.

11. Muon Decay as Experimental Evidence

Muon-decay measurements give direct experimental evidence for relativistic time dilation and length contraction. Muons formed high in Earth's atmosphere have short proper lifetimes, yet many reach low altitude.

Earth frame: the moving muon's lifetime is dilated, so it travels farther before decaying.
Muon frame: the atmosphere moves toward the muon and the atmospheric distance is length-contracted.

Fig. 12. The Earth-frame and muon-frame explanations are different descriptions of the same experiment.

Worked Example 11: Finding Speed from Lifetime Data

A particle has proper lifetime 2.60 × 10⁻⁸ s. In the Earth frame its mean lifetime is 1.00 × 10⁻⁷ s. Determine its speed.

Step 1: γ = Δt/Δt₀ = (1.00 × 10⁻⁷ s)/(2.60 × 10⁻⁸ s) = 3.85

Step 2: v/c = √(1 − 1/γ²)

Step 3: v/c = √[1 − 1/(3.85)²] = 0.966

Step 4: v = 0.966(3.00 × 10⁸ m/s) = 2.90 × 10⁸ m/s

Answer: v ≈ 0.966c = 2.90 × 10⁸ m/s.

12. Equation-Selection Guide

1. Identify the FramesState S, S′ and the sign/direction of v.

2. Identify the QuantityCoordinates? Velocity? Clock interval? Length? Event separation? Worldline angle?

3. Identify the Proper QuantitySame place → proper time. Object at rest → proper length.

4. Select and CheckKeep units throughout; require γ ≥ 1 and v < c for massive particles.

Problem Type	Best Starting Relationship	Key Check
Low-speed classical frames	x′ = x − vt; t′ = t; u′ = u − v	Relativistic effects negligible.
Coordinates of one event	x′ = γ(x − vt), t′ = γ(t − vx/c²)	Use consistent signs and synchronized origins.
Relativistic velocity	u′ = (u − v)/(1 − uv/c²)	Massive-particle result stays below c.
Clock, decay or lifetime	Δt = γΔt₀	Proper time: same-place events.
Moving length	L = L₀/γ	Proper length: object at rest.
Two-event comparison	(Δs)² = (cΔt)² − (Δx)²	Interval is invariant.
Worldline angle	tan θ = v/c	Measure θ from the ct-axis.

Data-Booklet Equations at a Glance

Relationship	Use
x′ = x − vt	Galilean position transformation.
t′ = t	Absolute-time assumption in Galilean relativity.
u′ = u − v	Galilean velocity addition.
x′ = γ(x − vt)	Lorentz position transformation.
t′ = γ(t − vx/c²)	Lorentz time transformation.
γ = 1/√(1 − v²/c²)	Lorentz factor.
u′ = (u − v)/(1 − uv/c²)	Relativistic velocity addition.
(Δs)² = (cΔt)² − (Δx)²	Invariant spacetime interval.
Δt = γΔt₀	Time dilation.
L = L₀/γ	Length contraction.
tan θ = v/c	Worldline angle and speed.

13. Common Errors and Exam Strategy

Common Error	Correction
Using ordinary velocity addition near c.	Use the relativistic velocity-addition equation.
Calling any measured time the proper time.	Proper time belongs to the frame where both events occur at the same place.
Calling a moving length L₀.	L₀ is measured only in the object's rest frame.
Thinking a moving observer notices their own clock running slowly.	Each local clock behaves normally in its own frame.
Assuming simultaneous events remain simultaneous.	Spatially separated simultaneity is frame-dependent.
Treating spacetime geometry as ordinary Euclidean geometry.	Use the invariant interval and spacetime rules.
Reading equal drawn distances on x and x′ as equal coordinate intervals.	The moving-frame axes require their own scale.
Measuring θ from the x-axis.	For tan θ = v/c, θ is measured from the ct-axis.

Precise Terminology to Retain

Term	Use It to Mean...
Event	An occurrence assigned a definite position and time in a reference frame.
Inertial reference frame	A non-accelerating reference frame.
Galilean relativity	The classical framework in which Newton's laws have the same form in all inertial frames and time is absolute.
Lorentz factor	The dimensionless factor γ controlling relativistic effects.
Proper time interval	The time between two events measured where those events occur at the same position.
Proper length	The length of an object measured in its rest frame.
Spacetime interval	The invariant event separation (Δs)² = (cΔt)² − (Δx)².
Worldline	The path of an object through spacetime.
Relativity of simultaneity	Spatially separated events simultaneous in one inertial frame need not be simultaneous in another.

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B. The particulate nature of matter

B.1 THERMAL ENERGY TRANSFERS

Recommended Time: 6 Hours | Standard Level and Higher Level

Guiding Questions

How do macroscopic observations provide a model of the microscopic properties of a substance?
How is energy transferred within and between systems?
How can observations of one physical quantity be used to determine the other properties of a system?

1. B.1 at a Glance

B.1 connects observable thermal behaviour to the particle model of matter. The section moves from the arrangement and motion of particles, through temperature and internal energy, to the three mechanisms of thermal energy transfer and finally to radiation from black bodies and stars.

Core Idea	What You Need to Be Able to Do
Molecular model	Relate the arrangements and motion of particles to the properties of solids, liquids and gases; calculate density.
Temperature and internal energy	Use Celsius and Kelvin scales correctly, connect kelvin temperature to average molecular kinetic energy, and distinguish temperature from internal energy.
Heating and phase change	Use Q = mcΔT for temperature changes and Q = mL for changes of state at constant temperature.
Conduction and convection	Explain conduction using particle kinetic energy, calculate conduction rate, and describe convection through fluid density differences.
Thermal radiation	Use black-body luminosity, apparent brightness and Wien’s displacement law, and interpret black-body spectra.

2. Molecular Model and Density

The molecular model treats matter as being made of extremely small particles. In solids the particles are close together and vibrate about approximately fixed positions. In liquids the particles remain close together but can move past one another. In gases the particles are much farther apart and move freely and randomly.

Figure 1. A particle model for solids, liquids and gases. The visual emphasizes relative arrangement and freedom of motion.

ρ = m / V

Density is the mass per unit volume. In SI units, density is measured in kg/m³.

Worked Example 1. Density

A sample has mass 125 g and volume 137 cm³. Determine its density in SI units.

1. Convert the mass: 125 g = 0.125 kg.

2. Convert the volume: 137 cm³ = 137 × 10⁻⁶ m³ = 1.37 × 10⁻⁴ m³.

3. Substitute into ρ = m/V.

ρ = (0.125 kg) / (1.37 × 10⁻⁴ m³) = 9.12 × 10² kg/m³

Answer. The density is approximately 9.1 × 10² kg/m³.

3. Temperature, Kelvin Scale and Internal Energy

Temperature can be expressed on the Celsius or Kelvin scale. A temperature difference has the same numerical value in kelvin and degrees Celsius, but absolute temperature used in molecular-energy and radiation equations must be expressed in kelvin.

T / K = θ / °C + 273.15

Kelvin temperature is a measure of average molecular kinetic energy. For the molecular model used in B.1:

E_k = 3/2 k_BT

Internal energy is not the same as temperature. It is the sum of the total random kinetic energy of the particles and the total intermolecular potential energy associated with the forces between them.

Figure 2. Temperature, internal energy and phase change are related but are not the same concept.

Worked Example 2. Average Molecular Kinetic Energy

Calculate the average kinetic energy of a particle at 300 K. Use k_B = 1.38 × 10⁻²³ J/K.

E_k = 3/2 × (1.38 × 10⁻²³ J/K) × (300 K)

E_k = 6.21 × 10⁻²¹ J

Answer. 6.21 × 10⁻²¹ J.

4. Thermal Equilibrium and Direction of Energy Transfer

A temperature difference determines the direction of the resultant thermal energy transfer. Energy is transferred from the body at higher temperature to the body at lower temperature until thermal equilibrium is reached.

Figure 3. Left: thermal energy transfer proceeds from higher to lower temperature until equilibrium. Right: a thermogram-style visual shows how temperature can vary across a real object.

5. Phase Changes, Specific Heat Capacity and Specific Latent Heat

A phase change represents a change in particle behaviour caused by energy transfer at constant temperature. The terms melting, freezing, boiling, condensing and evaporation should be familiar.

Specific Heat Capacity

Q = mcΔT

Use this relationship when thermal energy transfer produces a temperature change but no phase change.

Figure 4. Experimental arrangements for determining specific heat capacity of a metal block and of water.

Worked Example 3. Specific Heat Capacity

A 1.50 kg metal block is heated for 300 s using an 18 W heater. Its temperature rises from 18.0 °C to 27.5 °C. Assume no energy is transferred to the surroundings. Determine the specific heat capacity.

1. Energy supplied: Q = Pt = (18 J/s)(300 s) = 5400 J.

2. Temperature change: ΔT = 27.5 °C − 18.0 °C = 9.5 K.

3. Rearrange Q = mcΔT:

c = Q/(mΔT) = (5400 J)/[(1.50 kg)(9.5 K)] = 3.79 × 10² J kg⁻¹ K⁻¹

Answer. 3.8 × 10² J kg⁻¹ K⁻¹.

Specific Latent Heat

Q = mL

Use this relationship when a substance changes state at constant temperature.

Figure 5. A method-of-mixtures experiment for specific latent heat of fusion. The energy bookkeeping must include both melting the ice and warming the meltwater.

Worked Example 4. Heating and Then Melting

A 0.080 kg sample of ice at 0 °C is completely melted. Take L_f = 3.34 × 10⁵ J/kg.

Q = mL_f = (0.080 kg)(3.34 × 10⁵ J/kg) = 2.672 × 10⁴ J

Answer. 2.67 × 10⁴ J.

Worked Example 5. A Multi-Stage Thermal Calculation

Calculate the total energy needed to heat 0.25 kg of water from 20 °C to 100 °C and then vaporize it completely. Use c = 4200 J kg⁻¹ K⁻¹ and L_v = 2.26 × 10⁶ J/kg.

1. Heating stage:

Q₁ = mcΔT = (0.25 kg)(4200 J kg⁻¹ K⁻¹)(80 K) = 8.40 × 10⁴ J

2. Vaporization stage:

Q₂ = mL_v = (0.25 kg)(2.26 × 10⁶ J/kg) = 5.65 × 10⁵ J

3. Total:

Q_total = 8.40 × 10⁴ J + 5.65 × 10⁵ J = 6.49 × 10⁵ J

Answer. 6.49 × 10⁵ J.

6. Conduction

Conduction transfers thermal energy because particles with greater kinetic energy interact with neighbouring particles. In solids this transfer occurs through particle interactions; in metals, mobile electrons make conduction especially effective.

ΔQ/Δt = kA(ΔT/Δx)

The rate increases with thermal conductivity k, cross-sectional area A and temperature difference ΔT, but decreases as the distance Δx through which the energy must be conducted increases.

Figure 6. Conduction through a slab, showing the variables in the thermal-conductivity equation and the effect of placing identical slabs in series.

Worked Example 6. Rate of Thermal Conduction

A wall has area 12.0 m², thickness 0.20 m and thermal conductivity 0.80 W m⁻¹ K⁻¹. Its two faces are at 22 °C and 5 °C.

1. Temperature difference: ΔT = 22 °C − 5 °C = 17 K.

2. Substitute:

ΔQ/Δt = (0.80 W m⁻¹ K⁻¹)(12.0 m²)(17 K)/(0.20 m)

ΔQ/Δt = 816 W

Answer. The conduction rate is 816 W.

7. Convection

Convection is thermal energy transfer by the bulk movement of a fluid. Heating usually lowers the density of part of a fluid. The warmer, less dense fluid rises while cooler, denser fluid moves downward to replace it, establishing a circulation.

Figure 7. Convection in water and air. Density differences drive the circulating motion of the fluid.

8. Thermal Radiation

Thermal radiation transfers energy by electromagnetic waves. It does not require matter, so it can transfer energy across a vacuum. Every body emits thermal radiation; the amount and wavelength distribution depend strongly on temperature.

9. Black-Body Radiation, Luminosity and Apparent Brightness

A black body is an idealized perfect absorber and emitter. The power emitted by a black body is its luminosity.

L = σAT⁴

For an isotropic source at distance d, apparent brightness is:

b = L / (4πd²)

The peak wavelength of black-body radiation is related to temperature by:

λ_maxT = 2.9 × 10⁻³ m K

Figure 8. Black-body spectra at different temperatures. Increasing temperature raises the total emitted power and shifts the peak toward shorter wavelengths.

Worked Example 7. Wien’s Displacement Law

A star has surface temperature 5800 K. Determine its peak wavelength.

λ_max = (2.9 × 10⁻³ m K)/(5800 K) = 5.0 × 10⁻⁷ m

Answer. 5.0 × 10⁻⁷ m, or approximately 500 nm.

Worked Example 8. Black-Body Luminosity

A spherical object has emitting area 0.40 m² and temperature 600 K. Treat it as a black body. Use σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

L = (5.67 × 10⁻⁸ W m⁻² K⁻⁴)(0.40 m²)(600 K)⁴

L = 2.94 × 10³ W

Answer. 2.9 × 10³ W.

Figure 9. Apparent brightness decreases with the square of the distance because the same luminosity is spread over a larger spherical area.

Worked Example 9. Apparent Brightness

A source has luminosity 3.6 × 10²⁶ W and is at distance 2.0 × 10¹¹ m. Determine the apparent brightness.

b = (3.6 × 10²⁶ W)/[4π(2.0 × 10¹¹ m)²]

b = 7.16 × 10² W/m²

Answer. 7.2 × 10² W/m².

10. B.1 Synthesis: What You Must Be Able to Do

B.1 requires both microscopic explanations and quantitative calculations. The following table condenses the minimum mastery expected across the section.

Area	Minimum Mastery
Molecular model	Describe solids, liquids and gases using particle arrangement and motion; calculate density from mass and volume.
Temperature	Convert between °C and K and know when absolute temperature is required.
Internal energy	Separate random kinetic and intermolecular potential energy contributions.
Heating and phase change	Choose correctly between Q = mcΔT and Q = mL, including multi-stage calculations.
Conduction	Explain the mechanism and calculate transfer rate using thermal conductivity and temperature gradient.
Convection	Explain circulation through fluid density differences.
Radiation	Use L = σAT⁴, b = L/(4πd²) and Wien’s law, and interpret black-body spectra.

Essential Equations at a Glance

Purpose	Equation
density	ρ = m/V
temperature conversion	T/K = θ/°C + 273.15
average molecular kinetic energy	E_k = 3/2 k_BT
temperature change	Q = mcΔT
phase change	Q = mL
conduction rate	ΔQ/Δt = kAΔT/Δx
black-body luminosity	L = σAT⁴
apparent brightness	b = L/(4πd²)
black-body peak wavelength	λ_maxT = 2.9 × 10⁻³ m K

Equation-Selection Map

Figure 10. Equation-selection map for B.1. Identify the physical process first; then choose the relationship that matches it.

High-Value Checks Before Finalising an Answer

Have I decided whether the process is a temperature change, a phase change, conduction or radiation?
Have I converted mass to kilograms and volume to m³ where required?
Am I using kelvin rather than Celsius in molecular-energy, Stefan–Boltzmann and Wien calculations?
If the temperature changes, have I used ΔT rather than the final temperature?
If a phase change occurs, have I kept the temperature constant during that stage?
For a multi-stage problem, have I calculated and then added each energy contribution separately?
For conduction, have I used the thickness in metres and the area perpendicular to the energy flow?
For apparent brightness, have I squared the distance?

Linking Ideas Highlighted by the B.1 Requirements

B.1 connects strongly to later work on the greenhouse effect, ideal gases, thermodynamics and stellar physics. The same ideas of energy transfer and radiation reappear when studying the Earth’s energy balance, gas behaviour, heat engines and the temperature and luminosity of stars.

Common Misconceptions and Corrections

Misconception	Correction
A hotter body must have more internal energy.	Temperature concerns average particle kinetic energy; internal energy also depends on the amount of matter and intermolecular potential energy.
Temperature always rises when energy is supplied.	During a phase change, energy can be supplied while temperature remains constant.
Convection occurs in solids.	Convection requires bulk motion of a fluid, so it occurs in liquids and gases.
Radiation needs a material medium.	Electromagnetic radiation can transfer energy through a vacuum.
Use Celsius in every thermal equation.	Temperature differences may be in K or °C, but absolute temperature in molecular-energy and radiation equations must be in kelvin.
Doubling distance halves apparent brightness.	Apparent brightness follows an inverse-square relationship, so doubling distance reduces it by a factor of four.

A Disciplined B.1 Problem-Solving Workflow

Identify the physical process: temperature change, phase change, conduction, convection or radiation.
Write down the known quantities with units and convert them to the required SI units.
Decide whether the temperature in the equation is a change in temperature or an absolute temperature.
Choose the equation that matches the process, not merely the variables that happen to be present.
For multi-stage heating or cooling, separate the process into stages and calculate each energy transfer independently.
Carry units through every line of numerical working.
Check the result physically: direction of transfer, sign, order of magnitude and whether a phase change should have occurred.

Using the IB Data Booklet and Experimental Skills

The key B.1 relationships are provided in the Physics data booklet, so the main challenge is selecting and applying them correctly. Experimental work in this section also requires careful measurements of temperature, mass, time and electrical energy input. Thermal experiments are especially sensitive to unwanted energy transfer to or from the surroundings, so insulation, stirring, steady-state conditions and measurement uncertainty must be considered when evaluating results.

Experiment	High-Value Evaluation Points
Specific heat capacity	Reduce energy loss to surroundings; ensure good thermal contact; stir liquids; account for thermometer and container effects where relevant.
Specific latent heat by mixtures	Dry the ice, know its initial temperature, minimize energy exchange with surroundings and stir to reach a uniform final temperature.
Thermal conductivity	Wait for steady state, insulate side surfaces, measure thickness and area accurately and monitor the temperature difference across the sample.

Precise Terminology to Retain

Term	Use It to Mean...
Density	Mass per unit volume.
Temperature	A quantity related to the average random kinetic energy of particles.
Internal energy	Total random kinetic energy plus total intermolecular potential energy of the particles in a system.
Thermal equilibrium	Condition in which bodies in thermal contact are at the same temperature and there is no resultant thermal energy transfer between them.
Specific heat capacity	Energy required per kilogram per kelvin of temperature change.
Specific latent heat	Energy required per kilogram for a change of state at constant temperature.
Thermal conductivity	Material constant that measures the ability of a substance to conduct thermal energy.
Convection	Thermal energy transfer by bulk movement of a fluid caused by density differences.
Black body	An idealized perfect absorber and emitter of thermal radiation.
Luminosity	Total power emitted by a body.
Apparent brightness	Power received per unit area from a source at a given distance.
Peak wavelength	Wavelength at which a black-body emission spectrum has its maximum intensity.

B.2 Greenhouse Effect

Recommended Time: 6 Hours | Standard Level and Higher Level

Contents

Guiding Questions

How does the greenhouse effect help to maintain life on Earth and how does human activity enhance this effect?
How is the atmosphere as a system modelled to quantify the Earth–atmosphere energy balance?

1. B.2 at a Glance

The greenhouse effect is fundamentally an energy-accounting problem. A planet receives electromagnetic radiation from its star, reflects some of it, absorbs the remainder, and emits thermal radiation. If the rate at which the planet gains energy is equal to the rate at which it loses energy, its mean temperature is steady. If the two rates are unequal, the mean temperature changes until a new balance is established.

The Earth is not a simple black sphere in empty space. Its atmosphere absorbs selected wavelengths, clouds and surface materials reflect different fractions of incoming radiation, water transports energy through evaporation and condensation, and greenhouse gases exchange infrared radiation with the surface. B.2 develops progressively more realistic models of this system.

Core Conservation Principle. In equilibrium, average incoming power to the Earth system equals average outgoing power from the Earth system. Every diagram and equilibrium-temperature calculation in this topic is an application of conservation of energy.

2. Radiation from the Sun and the Solar Constant

The Sun is treated approximately as a black-body radiator. Its surface temperature is about 5780 K. Using the Stefan–Boltzmann law, the Sun’s luminosity can be estimated from its surface area and temperature:

L_⊙ = σA_⊙T_⊙⁴

Worked Example 1. Estimating the Sun’s Luminosity and the Solar Constant

Take σ = 5.67 × 10⁻⁸ W/m² K⁴, A_⊙ = 6.05 × 10¹⁸ m² and T_⊙ = 5780 K.

L_⊙ = (5.67 × 10⁻⁸ W/m² K⁴)(6.05 × 10¹⁸ m²)(5780 K)⁴

L_⊙ = 3.83 × 10²⁶ W

At Earth’s mean distance d = 1.50 × 10¹¹ m, this luminosity is spread over a sphere of area 4πd²:

S = L_⊙/(4πd²)

S = (3.83 × 10²⁶ W) / [4π(1.50 × 10¹¹ m)²]

S = 1.35 × 10³ W/m² ≈ 1.36 × 10³ W/m²

This intensity just above Earth’s atmosphere is called the solar constant, S. It is not perfectly constant: solar output varies slightly, but the accepted mean value is approximately 1.36 × 10³ W/m².

Why the Average Incoming Intensity Is S/4

Sunlight arriving at Earth is essentially parallel. The planet intercepts sunlight according to its projected area, a disk of area πR². However, the received energy is averaged over the whole spherical surface area 4πR².

I_mean,in = SπR²/(4πR²) = S/4

For Earth:

S/4 = (1.36 × 10³ W/m²)/4 = 340 W/m²

The incoming rays are intercepted through the projected disk πR², whereas the global mean is calculated over the complete surface area 4πR². This geometrical argument produces the factor of 4.

High-Value Exam Point. The factor 4 must be justified geometrically. It is not an arbitrary averaging factor and it does not mean that the solar constant has become four times smaller in space.

Radiation Through the Atmosphere

The radiation received from the Sun contains ultraviolet, visible and infrared wavelengths. The atmosphere interacts differently with different wavelengths:

some ultraviolet is absorbed high in the atmosphere, including by oxygen and ozone;
much of the visible spectrum passes through the atmosphere;
selected infrared wavelengths are absorbed by gases such as H₂O, CO₂, CH₄ and N₂O;
the spectrum measured at the surface therefore contains absorption bands and is not identical to the spectrum at the top of the atmosphere.

A schematic comparison of the relatively smooth solar spectrum at the top of the atmosphere with the spectrum that reaches Earth’s surface after wavelength-selective absorption.

3. Interaction of Radiation with Surfaces

When radiation reaches a material, it may be reflected or scattered, absorbed, or transmitted. For an opaque planetary surface, transmission through the entire planet is not relevant, so the competition is mainly between reflection and absorption. Absorbed radiation increases the internal energy of the material and tends to raise its temperature; the warmed material then emits thermal radiation.

Albedo

Albedo measures the average fraction of incident power that is scattered or reflected by a macroscopic system:

albedo = total scattered power / total incident power

Albedo has no unit. A value close to 1 means strong reflection; a value close to 0 means strong absorption. Earth’s albedo is not fixed. It changes with cloud formation, surface conditions, season and latitude. Typical values used in the textbooks are shown below.

Surface	Approximate Albedo
Ocean water	0.1
Forest	0.1
Road surface	0.1
Soil	0.2
Grass	0.3
Desert sand	0.4
Clouds	about 0.5, highly variable
Ocean ice	about 0.6
Snow	about 0.8

Worked Example 2. Albedo and Absorbed Power

A surface receives radiation of intensity 610 W/m² and reflects 180 W/m².

α = (180 W/m²)/(610 W/m²)

α = 0.295 ≈ 0.30

Absorbed intensity = (1 − 0.295)(610 W/m²)

Absorbed intensity = 430 W/m²

Emissivity

Emissivity, e, compares the power radiated per unit area by a real surface with the power that an ideal black surface would radiate at the same absolute temperature:

e = (power radiated per unit area)/(σT⁴)

Therefore, for a real surface:

P = eσAT⁴

A black body has e = 1. Real surfaces have 0 ≤ e ≤ 1. High albedo and high emissivity are not opposites; they refer to different aspects of the interaction with radiation.

Surface	Typical Emissivity
Ocean water	0.99
Forest	0.97
Road surface	0.96
Soil	0.95
Grass	0.91
Desert sand	0.90
Clouds	about 0.55, variable
Ocean ice	0.97
Snow	0.94

Albedo concerns reflected incident radiation; emissivity concerns emitted thermal radiation. They must not be treated as the same property.

Worked Example 3. Determining a Planet’s Emissivity

A planet has surface temperature −63 °C, surface area 1.4 × 10¹⁴ m² and radiates 1.3 × 10¹⁶ W.

T = 273 K − 63 K = 210 K

P/A = (1.3 × 10¹⁶ W)/(1.4 × 10¹⁴ m²) = 92.9 W/m²

σT⁴ = (5.67 × 10⁻⁸ W/m² K⁴)(210 K)⁴ = 110.3 W/m²

e = (92.9 W/m²)/(110.3 W/m²) = 0.84

Worked Example 4. Photovoltaic Panel Irradiation

580 W of solar power is incident perpendicularly on a rectangular panel 1.86 m × 2.12 m. The panel albedo is 0.21.

Area = (1.86 m)(2.12 m) = 3.9432 m²

Intensity = (580 W)/(3.9432 m²) = 147 W/m²

Absorbed power = (1 − 0.21)(580 W) = 458 W

For solar collection, a low albedo increases absorption and placing the panel close to perpendicular to the rays maximizes the projected area receiving radiation.

4. Planetary Energy-Balance Models

For a planet in radiative equilibrium, the average power absorbed from its star is equal to the average power emitted to space. A useful first model is:

(1 − α)S/4 = eσT⁴

This can also be written using total powers:

SπR²(1 − α) = eσ(4πR²)T⁴

The radius cancels, which is why the simplest equilibrium temperature depends on the solar constant, albedo and emissivity rather than directly on planetary radius.

Worked Example 5. Earth as a Black Body with No Greenhouse Effect

Use S = 1400 W/m², α = 0.30 and e = 1.00.

Absorbed intensity = (1 − 0.30)(1400 W/m²)/4

Absorbed intensity = 245 W/m²

At equilibrium: 245 W/m² = (5.67 × 10⁻⁸ W/m² K⁴)T⁴

T⁴ = (245 W/m²)/(5.67 × 10⁻⁸ W/m² K⁴)

T = 256 K ≈ −17 °C

The actual mean surface temperature is about 288 K, so the no-greenhouse model is approximately 32 K too cold. This is strong evidence that atmospheric processes are essential to the energy balance.

Worked Example 6. A Simple Earth–Atmosphere Effective Model

Use S = 1.36 × 10³ W/m², α = 0.315 and effective emissivity e = 0.61.

Absorbed intensity = (1 − 0.315)(1.36 × 10³ W/m²)/4

Absorbed intensity = (0.685)(340 W/m²) = 232.9 W/m²

At equilibrium: 232.9 W/m² = (0.61)(5.67 × 10⁻⁸ W/m² K⁴)T⁴

T = [232.9 W/m² ÷ ((0.61)(5.67 × 10⁻⁸ W/m² K⁴))]^1/4

T = 286 K ≈ 13 °C

This simple effective model comes close to the measured global mean, but it should not be mistaken for a complete climate model. It compresses many different surface–atmosphere processes into average values of albedo and emissivity.

Worked Example 7. Mercury and the Inverse-Square Law

Mercury is 0.40 AU from the Sun and has no substantial atmosphere. Take Earth’s solar constant as 1.36 × 10³ W/m² and Mercury’s albedo as 0.11.

S_Mercury = (1.36 × 10³ W/m²)/(0.40)² = 8.5 × 10³ W/m²

Mean incident intensity = (8.5 × 10³ W/m²)/4 = 2.1 × 10³ W/m²

Mean absorbed intensity = (1 − 0.11)(2.1 × 10³ W/m²) = 1.9 × 10³ W/m²

Useful Dependence. Since S ∝ 1/d² and T⁴ ∝ S in the simplest model, T ∝ d^−1/2. A planet farther from the same star has a lower equilibrium temperature, all else being equal.

5. The Natural Greenhouse Effect

The natural greenhouse effect raises Earth’s mean surface temperature because the atmosphere is relatively transparent to much of the incoming short-wave radiation but absorbs important parts of the long-wave infrared spectrum emitted by the cooler surface.

A greenhouse can help students visualize the difference between incoming shorter-wavelength radiation and outgoing longer-wavelength radiation, but the analogy must not be taken too literally because a glass greenhouse also suppresses convection.

The sequence is:

Short-wave solar radiation enters the atmosphere; some is reflected and some is absorbed.
A substantial fraction reaches and warms the surface.
The warmed surface emits long-wave infrared radiation.
Greenhouse-gas molecules absorb selected infrared wavelengths.
Excited molecules later emit infrared radiation in all directions.
Some of this radiation escapes to space; some is directed downward and adds to the energy received by the surface.

The atmosphere does not simply act as a mirror. Greenhouse gases absorb infrared at selected wavelengths and re-emit radiation in all directions, including downward toward the surface.

Main Greenhouse Gases and Their Sources

The syllabus specifies four principal greenhouse gases: methane CH₄, water vapour H₂O, carbon dioxide CO₂ and nitrous oxide N₂O. Each has natural as well as anthropogenic origins.

Gas	Natural Origins	Human-Related Origins / Changes
H₂O	Evaporation from oceans, rivers and lakes	Mainly changed indirectly through temperature-driven changes in evaporation and atmospheric water content
CO₂	Respiration, volcanic activity, forest fires and ocean exchange	Burning fossil fuels, cement production and deforestation
CH₄	Wetlands, oceans, lakes, rivers and termites	Livestock, rice cultivation, landfill and fossil-fuel extraction or leakage
N₂O	Soils, forests and oceans	Agriculture, fertiliser use and some industrial processes

Sources increase atmospheric concentration; sinks reduce it. For example, CO₂ is removed from the atmosphere by photosynthesis and by dissolution in the oceans. Concentration therefore depends on the competition between sources and sinks.

Why Greenhouse Gases Absorb Infrared

Molecules have discrete energy states. Infrared photons can be absorbed when the photon energy matches an allowed difference between molecular energy levels. The same behaviour may also be described using resonance: the electromagnetic radiation drives a molecular vibration strongly when the radiation frequency matches a natural vibrational frequency of the molecule.

In carbon dioxide, important vibrational motions include symmetric stretching, asymmetric stretching and bending. The symmetric stretching mode of linear CO₂ does not produce the required changing dipole and is not infrared-active in the same way; asymmetric stretching and bending modes do absorb infrared. Water is a bent molecule and has strong infrared-active vibrational modes.

Greenhouse-gas absorption can be described using discrete molecular energy levels or a resonance model. Absorption is wavelength-selective, and emission after excitation occurs in random directions.

Limitation of the Resonance Model. The resonance picture is a useful analogy for matching frequencies, but real molecular absorption is quantized and constrained by molecular structure and selection rules. The molecular-energy-level description is therefore the more complete microscopic picture.

6. Earth–Atmosphere Energy-Flow Models

Energy-flow diagrams make conservation of energy visible. Different textbooks use different representative global averages because the diagrams are models and because observational datasets are updated. The numerical values therefore differ slightly, but every valid model must conserve energy within the stated approximations.

A deliberately simple global model: of 350 W/m² incident on average, about 105 W/m² is reflected and about 245 W/m² must eventually leave the Earth system for equilibrium.

Surface and Atmosphere as Coupled Subsystems

A more useful model separates the surface and atmosphere. The surface receives absorbed sunlight and downward atmospheric radiation; it loses energy by infrared emission plus non-radiative processes. The atmosphere gains energy from absorbed sunlight, surface radiation, convection and latent heat, and loses energy by upward and downward radiation.

A two-subsystem model based on representative textbook values. The surface and atmosphere each satisfy their own energy balance: 167 + 324 = 451 + 40 W/m² for the surface, and 66 + 451 = 324 + 193 W/m² for the atmosphere.

Worked Example 8. Checking Energy Conservation in a Surface–Atmosphere Diagram

Surface input: 167 W/m² + 324 W/m² = 491 W/m²

Surface output: 451 W/m² + 40 W/m² = 491 W/m²

Therefore the surface is in equilibrium in this model.

Atmosphere input: 66 W/m² + 451 W/m² = 517 W/m²

Atmosphere output: 324 W/m² + 193 W/m² = 517 W/m²

Therefore the atmosphere is also in equilibrium.

A More Detailed Global Energy Budget

Modern global energy-budget diagrams distinguish reflected short-wave radiation, atmospheric and surface absorption, surface thermal emission, atmospheric absorption, latent heat, thermals, downward atmospheric radiation and the atmospheric window.

A richer observational energy budget. Values are global averages and can differ slightly among datasets. The purpose is to track how energy is redistributed while respecting overall conservation of energy.

Worked Example 9. Top-of-Atmosphere Balance

Incoming solar intensity = 341.3 W/m²

Reflected short-wave intensity = 101.9 W/m²

Outgoing long-wave intensity = 238.5 W/m²

Total outgoing = 101.9 W/m² + 238.5 W/m² = 340.4 W/m²

Net absorbed = 341.3 W/m² − 340.4 W/m² = 0.9 W/m²

In this representative budget, the small positive imbalance means that the Earth system is gaining energy. Small imbalances matter because they act continuously over enormous areas and long times.

7. The Enhanced Greenhouse Effect

The enhanced greenhouse effect is the augmentation of the natural greenhouse effect caused by human activity. The syllabus specifically identifies the burning of fossil fuels as a primary cause. Additional textbook examples include deforestation and agricultural or industrial sources of greenhouse gases.

Increasing greenhouse-gas concentration increases infrared absorption in wavelength bands associated with those gases. At the previous surface temperature, less net infrared energy escapes to space. The Earth system therefore gains energy. Temperature rises until the outgoing energy once again balances the incoming energy at a new equilibrium.

Do Not Confuse Cause and Response. The enhanced greenhouse effect is the change in radiative energy transfer caused by increased greenhouse-gas concentrations. Global warming is the resulting long-term increase in mean temperature.

Feedbacks and Climate Modelling

Feedbacks couple radiation to other parts of the climate system. Ice–albedo feedback is strongly positive; water-vapour and cloud processes introduce additional coupled effects that more complete climate models must represent.

Important feedbacks discussed in the textbooks include:

Ice–albedo feedback: warming melts snow and ice, reducing albedo. More solar radiation is absorbed, causing further warming.
Water-vapour feedback: warming increases evaporation and atmospheric water vapour, which can strengthen infrared absorption.
Cloud effects: more cloud can increase albedo and therefore reduce absorbed solar radiation, while clouds also interact strongly with infrared radiation. The net cloud feedback is therefore not captured by a single simple statement.

Simple equilibrium models are valuable because they expose the underlying physics, but they omit many important processes: ocean currents, spatial and seasonal variation, changing cloud cover, latent heat, convection, atmospheric circulation and time-dependent feedbacks. More realistic climate predictions require numerical computer models with many coupled variables.

Extension / Textbook Context — Responses to the Enhanced Greenhouse Effect. Textbook discussions include improving the efficiency of power generation, reducing fossil-fuel use, combined heat and power, increased use of renewable and nuclear energy, electric transport, carbon capture and storage, and international cooperation. These are applications of the physics rather than additional B.2 equations.

8. B.2 Synthesis: What You Must Be Able to Do

Explain why conservation of energy is the foundation of planetary energy-balance models.
Define the solar constant and calculate it using luminosity and inverse-square spreading.
Justify the factor of 4 in the mean incoming intensity S/4.
Define and calculate albedo and emissivity.
Use albedo and emissivity in equilibrium-temperature calculations.
Explain why the no-atmosphere temperature of Earth is lower than the observed mean temperature.
Describe the natural greenhouse effect using wavelength-selective absorption and re-emission of infrared radiation.
Name H₂O, CO₂, CH₄ and N₂O as the principal greenhouse gases required by the syllabus and distinguish natural from anthropogenic sources.
Explain infrared absorption using both molecular energy levels and resonance.
Read and balance surface–atmosphere energy-flow diagrams.
Explain the enhanced greenhouse effect and distinguish it from the natural greenhouse effect.

Essential Equations and Equation Selection

Solar Intensity at Distance d

Use luminosity and inverse-square spreading.

S = L/(4πd²)

Global Mean Incoming Intensity

Use projected disk divided by total spherical area.

I = S/4

Absorbed Solar Intensity

Remove the reflected fraction using albedo.

I_abs = (1 − α)S/4

Thermal Emission

Correct the black-body law using emissivity.

P = eσAT⁴

Planetary Equilibrium

Set average absorbed intensity equal to emitted intensity.

(1 − α)S/4 = eσT⁴

Energy-Flow Diagram

For each subsystem, total input equals total output at equilibrium.

ΣP_in = ΣP_out

Common Misconceptions and Corrections

Misconception	Correction
The solar constant is the average intensity over the whole Earth.	No. The global average before albedo is S/4.
Albedo and emissivity are the same thing.	Albedo concerns reflected incident radiation; emissivity concerns thermal radiation emitted by a surface.
Greenhouse gases reflect infrared back to Earth.	They absorb selected infrared wavelengths and subsequently emit infrared in all directions.
The greenhouse effect is caused by ozone depletion.	They are different phenomena. B.2 concerns infrared-active greenhouse gases and energy balance.
A glass greenhouse and Earth’s atmosphere work in exactly the same way.	No. A physical greenhouse also suppresses convection; the atmospheric effect is primarily radiative.
More greenhouse gas means the Earth keeps accumulating energy forever.	The system warms until outgoing radiation increases enough for a new equilibrium, unless the forcing continues to change.
All outgoing surface infrared reaches space.	No. Much is absorbed by the atmosphere; some escapes through wavelength regions such as the atmospheric window.

A Disciplined B.2 Problem-Solving Workflow

Identify the system: surface, atmosphere, or complete planet.
Identify whether the quoted solar intensity is S or the global mean S/4.
Apply albedo before calculating absorbed solar power.
Use kelvin for all T⁴ calculations.
Apply emissivity only to the thermal-emission side of the model unless the problem defines an effective model explicitly.
For a diagram, write separate energy balances for each subsystem.
Keep W, W/m², m² and K visible throughout the numerical working.
Check whether the final temperature or energy flow is physically reasonable.

Precise Terminology to Retain

Term	Use It to Mean…
Solar constant, S	Solar radiation intensity at Earth’s orbital distance on a surface normal to the Sun’s rays.
Projected area	The cross-sectional area presented to parallel incoming radiation; for a spherical planet this is πR².
Mean incoming intensity	The intercepted solar power averaged over the whole spherical surface; for Earth this is S/4 before albedo is applied.
Albedo, α	The fraction of incident radiative power that is scattered or reflected by a surface or system.
Emissivity, e	The ratio of the thermal power radiated per unit area by a real surface to that radiated by a black body at the same temperature.
Black body	An idealized perfect absorber and emitter of thermal radiation.
Energy balance	A comparison of the total power entering and leaving a defined system.
Equilibrium temperature	The temperature at which the average absorbed power equals the average emitted power for the stated model.
Greenhouse gas	An atmospheric gas that absorbs selected infrared wavelengths and subsequently re-emits infrared radiation.
Natural greenhouse effect	The natural warming of Earth caused by atmospheric absorption and re-emission of outgoing infrared radiation.
Enhanced greenhouse effect	The augmentation of the natural greenhouse effect caused by increased greenhouse-gas concentrations associated with human activity.
Resonance	Strong absorption that occurs when the frequency of incident radiation matches an allowed molecular vibrational response.
Molecular energy level	An allowed discrete energy state of a molecule; absorption occurs when photon energy matches the difference between permitted levels.
Infrared absorption	Transfer of energy from an infrared photon to a molecule when an allowed molecular transition is excited.
Re-emission	Emission of infrared radiation by an excited molecule after absorption; the emitted radiation is not restricted to the original direction.
Back radiation	Downward infrared radiation emitted by the atmosphere toward Earth’s surface.
Atmospheric window	A wavelength region in which the atmosphere is relatively transparent to outgoing infrared radiation.
Feedback	A process in which an initial change causes further changes that can amplify or oppose the original change.

B.3 Gas Laws

Recommended Time: 4 Hours | Standard Level and Higher Level

Contents

Guiding Questions

How can the macroscopic behaviour of a gas be explained using the motion and collisions of its particles?
How are pressure, volume and absolute temperature related for a fixed amount of gas?
How can graphs and mathematical models be used to predict changes in the state of a gas?

1. B.3 at a Glance

B.3 links what can be measured in the laboratory—pressure, volume and temperature—to the invisible motion of gas particles. The section begins with the origin of gas pressure, then studies situations in which one macroscopic variable is kept constant, and finally combines the separate gas laws into the ideal gas equation.

The central modelling idea is that a gas consists of an enormous number of particles moving randomly. Their collisions with the container walls produce pressure, while absolute temperature is linked to their average kinetic energy. Changing one macroscopic variable changes the molecular behaviour and therefore affects the others.

Core Idea	What You Need to Be Able to Do
Gas Pressure	Explain pressure in terms of momentum changes when particles collide with container walls.
Boyle’s Law	Relate pressure and volume for a fixed amount of gas at constant temperature.
Pressure Law	Relate pressure and absolute temperature at constant volume.
Charles’s Law	Relate volume and absolute temperature at constant pressure.
Ideal Gas Equation	Use PV = nRT and PV = Nk_BT to connect macroscopic and microscopic descriptions.
PV Diagrams	Identify states, isotherms and common process paths on pressure–volume diagrams.

Core Modelling Principle. For a fixed amount of gas, pressure, volume and temperature are not independent. Changing one variable normally forces at least one of the others to change.

2. Pressure in Gases and the Microscopic Model

Pressure is defined as force per unit area acting perpendicular to a surface:

P = F/A

In a gas, the force on a container wall is produced by molecular collisions. A molecule approaches a wall with momentum, rebounds, and therefore undergoes a change in momentum. The wall exerts a force on the molecule; by Newton’s third law the molecule exerts an equal and opposite force on the wall. The combined effect of huge numbers of collisions gives the steady macroscopic pressure that is measured by a gauge or sensor.

Fig. 1

Random molecular collisions produce a force on a piston. A pressure gauge or sensor and a gas syringe allow pressure and volume to be measured simultaneously.

Fig. 2

A movable piston illustrates mechanical equilibrium: the force produced by gas pressure can be balanced by an external load.

Kinetic-Theory Explanation of Pressure

If molecules move faster, each collision usually produces a larger change in momentum. If the number of collisions per second also increases, the force on the wall increases further. This gives the microscopic explanation for why heating a gas at fixed volume raises its pressure.

For the kinetic model of an ideal gas, pressure can be related to the density and mean square speed of the particles:

P = ⅓ρ⟨c²⟩

This equation is important conceptually because it connects a directly measurable macroscopic quantity, pressure, with microscopic molecular motion.

3. Boyle’s Law

For a fixed amount of gas at constant temperature, pressure is inversely proportional to volume:

P ∝ 1/V and PV = constant

For two states of the same fixed amount of gas at constant temperature:

P₁V₁ = P₂V₂

When the gas is compressed, its particles occupy a smaller volume. At constant temperature their average kinetic energy does not change, but the particles encounter the walls more frequently. The collision rate increases and therefore the pressure rises.

Graphical Representation and Experimental Test

A graph of P against V is a decreasing curve. A plot of P against 1/V is a straight line through the origin for an ideal gas at constant temperature. Different constant temperatures produce different isotherms on a P–V graph.

Fig. 3

Boyle’s law can be tested by changing the volume of a trapped gas and measuring the corresponding pressure. The inverse relationship becomes linear when pressure is plotted against inverse volume.

Worked Example 1. Boyle’s Law

A gas occupies 120 cm³ at 100 kPa. It is compressed to 80 cm³ while its temperature remains constant. Determine the new pressure.

P₁V₁ = P₂V₂

P₂ = (100 kPa)(120 cm³)/(80 cm³)

P₂ = 150 kPa

Answer. The final pressure is 150 kPa.

Experimental Caution. Compressing a gas rapidly can raise its temperature. To test Boyle’s law, readings should be taken only after the gas has returned to thermal equilibrium so that the temperature is effectively constant.

4. Pressure Law (Gay-Lussac’s Law)

For a fixed amount of gas at constant volume, pressure is directly proportional to absolute temperature:

P ∝ T and P/T = constant

For two states:

P₁/T₁ = P₂/T₂

At constant volume, heating increases the average molecular kinetic energy. Molecules strike the container walls more frequently and with greater momentum change per collision, so pressure increases.

Fig. 4

The pressure law can be investigated by keeping the gas volume fixed while changing its temperature and measuring its pressure. A sensor-based arrangement allows pressure and temperature to be logged simultaneously.

A graph of pressure against temperature in kelvin is a straight line through the origin for an ideal gas. If pressure is plotted against temperature in degrees Celsius and extrapolated, the line approaches zero pressure near −273 °C, providing a model-based estimate of absolute zero.

Worked Example 2. Pressure Law

A sealed rigid container holds gas at 100 kPa and 300 K. The gas is heated to 360 K. Determine the final pressure.

P₁/T₁ = P₂/T₂

P₂ = (100 kPa)(360 K)/(300 K) = 120 kPa

Answer. The final pressure is 120 kPa.

5. Charles’s Law

For a fixed amount of gas at constant pressure, volume is directly proportional to absolute temperature:

V ∝ T and V/T = constant

For two states:

V₁/T₁ = V₂/T₂

When the gas is heated, molecules move faster. If the pressure is to remain unchanged, the gas must expand. The larger volume reduces the rate of collisions per unit area enough to offset the increased molecular speeds.

Fig. 5

At constant pressure, the gas volume increases linearly with absolute temperature. The apparatus measures the changing length of a trapped gas column as the surrounding water temperature changes.

Worked Example 3. Charles’s Law

A gas has volume 200 cm³ at 300 K. It is heated to 330 K at constant pressure. Determine its new volume.

V₁/T₁ = V₂/T₂

V₂ = (200 cm³)(330 K)/(300 K) = 220 cm³

Answer. The final volume is 220 cm³.

High-Value Exam Point. Gas-law proportionalities involving temperature require absolute temperature. Celsius values must not be inserted directly into ratios such as P/T or V/T.

6. The Ideal Gas Equation

The individual gas laws can be combined into a single equation describing an ideal gas:

PV = nRT

where P is pressure, V is volume, n is the amount of gas in moles, R is the molar gas constant, and T is the absolute temperature.

For a fixed amount of gas, the equation can be written in a two-state form:

P₁V₁/T₁ = P₂V₂/T₂

This form is particularly useful when pressure, volume and temperature all change. If one variable is constant, it cancels and the corresponding simpler gas law is recovered.

Fig. 6

Heating a gas at constant pressure causes expansion. The three-dimensional surface represents compatible combinations of pressure, volume and temperature for a fixed amount of ideal gas.

Particle Form of the Ideal Gas Equation

If the amount of gas is described by the number of molecules N rather than the number of moles, the ideal gas equation becomes:

PV = Nk_BT

The Boltzmann constant k_B links temperature directly to molecular-scale energy. This equation is therefore an especially important bridge between the macroscopic and microscopic descriptions of a gas.

Worked Example 4. Ideal Gas Equation

A sample contains 0.50 mol of ideal gas at 300 K in a volume of 0.012 m³. Determine the pressure using R = 8.31 J mol⁻¹ K⁻¹.

P = nRT/V

P = (0.50)(8.31)(300)/(0.012)

P ≈ 1.04 × 10⁵ Pa

Answer. The pressure is approximately 1.0 × 10⁵ Pa.

7. Pressure–Volume (PV) Diagrams

A pressure–volume diagram represents the state of a fixed amount of gas using pressure on the vertical axis and volume on the horizontal axis. A single point gives one equilibrium state. Because temperature is not an axis, different temperatures are represented by different isotherms.

Fig. 7

A family of isotherms represents different constant temperatures. Horizontal paths are isobaric, vertical paths are isochoric, and motion along one isotherm is isothermal.

Important process paths are:

Isothermal: temperature remains constant; the state moves along one isotherm.
Isobaric: pressure remains constant; the path is horizontal.
Isochoric or isovolumetric: volume remains constant; the path is vertical.
General process: pressure, volume and temperature may all change simultaneously.

At a fixed volume, a higher-temperature isotherm corresponds to a higher pressure. At a fixed pressure, the higher-temperature state occurs at a larger volume.

8. B.3 Synthesis: What You Must Be Able to Do

Successful work in B.3 requires more than memorizing three proportionalities. You must identify the physical constraint, choose the correct relationship, interpret graphs, use absolute temperature consistently, and explain the macroscopic result in terms of molecular motion.

Essential Equations and Equation Selection

Situation	Condition	Relationship
Pressure Definition	Force normal to area	P = F/A
Boyle’s Law	T and amount of gas constant	P₁V₁ = P₂V₂
Pressure Law	V and amount of gas constant	P₁/T₁ = P₂/T₂
Charles’s Law	P and amount of gas constant	V₁/T₁ = V₂/T₂
Ideal Gas Equation	General ideal-gas state	PV = nRT
Particle Form	Use number of molecules	PV = Nk_BT
Kinetic Model	Microscopic pressure model	P = ⅓ρ⟨c²⟩

Constant Temperature?

Use Boyle’s law.

P₁V₁ = P₂V₂

Constant Volume?

Use the pressure law.

P₁/T₁ = P₂/T₂

Constant Pressure?

Use Charles’s law.

V₁/T₁ = V₂/T₂

Several Variables Change?

Use the ideal gas equation or the combined two-state form.

PV = nRT

Common Misconceptions and Corrections

Misconception	Correction
Gas pressure is caused mainly by molecules colliding with one another.	Pressure on a container is caused by momentum transfer when molecules collide with the container walls.
Boyle’s law applies whenever volume changes.	Boyle’s law applies only to a fixed amount of gas when temperature remains constant.
Celsius temperature can be used in gas-law ratios.	No. Ratios involving temperature must use absolute temperature in kelvin.
If a gas is heated, its pressure must increase.	Not necessarily. At constant pressure the gas expands instead; the outcome depends on which variable is constrained.
If a gas is compressed, its molecules automatically move faster.	For an isothermal compression, average molecular kinetic energy remains constant. The pressure rises mainly because wall collisions become more frequent.
A straight line on a pressure–temperature graph should pass through −273 on a kelvin axis.	On a kelvin axis the ideal relationship passes through the origin. The −273 °C intercept appears only when the horizontal axis is Celsius.
All gases behave ideally under all conditions.	The ideal-gas model is an approximation. Real-gas deviations become more important when intermolecular effects cannot be neglected.
All curved lines on a PV diagram represent the same process.	Each isotherm represents a different constant temperature. Process paths must be identified from the stated constraint or the geometry of the path.

A Disciplined B.3 Problem-Solving Workflow

Define the system. Confirm that you are following the same fixed amount of gas unless the question explicitly changes the amount.
List the state variables. Write down the known and unknown values of P, V, T and, where relevant, n.
Identify the constraint. Decide whether temperature, pressure or volume is constant.
Convert temperature to kelvin. Do this before any proportionality or ratio calculation.
Choose the simplest correct equation. Use a named gas law when its condition is satisfied; otherwise use the ideal gas equation.
Keep units consistent. For PV = nRT in SI form, use pressure in pascals and volume in cubic metres.
Substitute with units visible. This makes unit mistakes easier to detect.
Interpret the answer physically. Ask whether the direction of the change agrees with the particle model.

Using the IB Data Booklet and Experimental Skill

Know which gas equations and constants are supplied and which relationships you are expected to recognize from the stated conditions.
Use kelvin for all absolute-temperature relationships.
When using PV = nRT, check that units match the SI value of the gas constant.
For gas-law experiments, allow the apparatus to reach thermal equilibrium before recording values.
Use a wide range of values where practical and plot transformed variables when this converts a nonlinear relationship into a straight line.
Recognize systematic issues such as dead volume in connecting tubes, sensor zero errors, leaks and temperature changes caused by rapid compression or expansion.

Linking Ideas Highlighted by Requirements

B.3 links the particle model from B.1 with mechanics and later thermodynamics. Gas pressure can be understood through momentum transfer when molecules collide with container walls, while temperature connects to the average kinetic energy of particles. Compression and expansion also connect gas behaviour to work and energy transfer.

These links show why the gas laws are not isolated formulas. Each law describes how macroscopic state variables respond to microscopic molecular behaviour when one physical condition is held fixed.

Precise Terminology to Retain

Term	Use It to Mean…
Pressure	Normal force per unit area; for a gas, the macroscopic result of molecular collisions with container walls.
Absolute temperature	Temperature measured from absolute zero on the kelvin scale.
Ideal gas	A model gas that obeys the ideal gas equation over the conditions being considered.
Boyle’s law	For a fixed amount of gas at constant temperature, pressure is inversely proportional to volume.
Pressure law	For a fixed amount of gas at constant volume, pressure is directly proportional to absolute temperature.
Charles’s law	For a fixed amount of gas at constant pressure, volume is directly proportional to absolute temperature.
Isotherm	A line on a PV diagram representing states at one constant temperature.
Isothermal process	A process in which temperature remains constant.
Isobaric process	A process in which pressure remains constant.
Isochoric / isovolumetric process	A process in which volume remains constant.
Molar gas constant, R	The proportionality constant in PV = nRT when amount of gas is measured in moles.
Boltzmann constant, k_B	The proportionality constant connecting microscopic particle energy scales with absolute temperature.
Thermal equilibrium	A condition in which systems in thermal contact have no net thermal energy transfer between them.

A.4 Rigid Body Mechanics

A.5 Galilean and Special Relativity

Recommended Time: 8 Hours | Higher Level Only

Guiding Questions

How do observers in different reference frames describe events in terms of space and time?
How does special relativity change our understanding of motion compared with Galilean relativity?
How are spacetime diagrams used to represent relativistic motion?

1. Reference Frames and Events

Reference Frame

A coordinate system, together with synchronized clocks, used to assign positions and times to events.

Inertial Reference Frame

A non-accelerating frame. Newton's first law holds in an inertial frame. Any two inertial frames move at constant velocity relative to one another.

Fig. 1. The same event may be assigned different spatial coordinates by two inertial frames.

2. Galilean Relativity

x′ = x − vtt′ = tu′ = u − v

Here u, u′ and v are signed one-dimensional velocities along the x-axis. Galilean relativity assumes absolute time: all inertial observers agree on a time interval.

Fig. 2. Classical velocity addition works for everyday motion because ordinary speeds are tiny compared with c.

Worked Example 1: Galilean Velocity Addition

A train travels at 20.0 m/s relative to the ground. A student walks forward at 1.50 m/s relative to the train. Determine the student's velocity relative to the ground.

Step 1: u = u′ + v

Step 2: u = 1.50 m/s + 20.0 m/s

Step 3: u = 21.5 m/s

Answer: 21.5 m/s in the forward direction relative to the ground.

Worked Example 2: Galilean Transformation of Position

A cyclist travels 350 m along a road in 50.0 s. A car moves in the same direction at 12.0 m/s. Determine the cyclist's displacement during the same interval in the reference frame of the car.

Step 1: Δx′ = Δx − vΔt

Step 2: Δx′ = 350 m − (12.0 m/s)(50.0 s)

Step 3: Δx′ = 350 m − 600 m = −250 m

Answer: −250 m. The negative sign means the cyclist moves backward relative to the car.

3. Special Relativity

Postulate 1

The laws of physics are identical in all inertial reference frames. No inertial frame is physically privileged.

Postulate 2

The speed of light in vacuum has the same value c for every inertial observer, independent of the motion of the source or observer.

Fig. 3. All inertial observers measure the same vacuum light speed c.

4. Lorentz Factor and Lorentz Transformations

The strength of relativistic effects is controlled by the Lorentz factor:

γ = 1 / √(1 − v²/c²)

For ordinary speeds γ is almost exactly 1. As v approaches c, γ increases rapidly.

Fig. 4. Relativistic effects grow rapidly only when the relative speed becomes a substantial fraction of c.

Worked Example 3: Calculating the Lorentz Factor

Calculate γ for a spacecraft moving at 0.800c.

Step 1: γ = 1 / √(1 − v²/c²)

Step 2: γ = 1 / √(1 − 0.800²) = 1 / √0.360

Step 3: γ = 1 / 0.600 = 1.67

Answer: γ = 1.67. The Lorentz factor has no units.

The Lorentz transformations replace the Galilean transformations:

x′ = γ(x − vt)t′ = γ(t − vx/c²)

The transformed time coordinate depends on both t and x. Space and time are therefore linked.

Worked Example 4: Transforming an Event

An event occurs in S at x = 1200 m and t = 6.00 × 10⁻⁶ s. Frame S′ moves at 0.600c in the +x direction. Find x′ and t′.

Step 1: γ = 1 / √(1 − 0.600²) = 1.25

Step 2: v = 0.600(3.00 × 10⁸ m/s) = 1.80 × 10⁸ m/s

Step 3: vt = (1.80 × 10⁸ m/s)(6.00 × 10⁻⁶ s) = 1.08 × 10³ m
x′ = 1.25(1200 m − 1080 m) = 150 m

Step 4: vx/c² = [(1.80 × 10⁸ m/s)(1200 m)]/(9.00 × 10¹⁶ m²/s²) = 2.40 × 10⁻⁶ s
t′ = 1.25[(6.00 × 10⁻⁶ s) − (2.40 × 10⁻⁶ s)] = 4.50 × 10⁻⁶ s

Answer: x′ = 150 m and t′ = 4.50 × 10⁻⁶ s.

5. Relativistic Velocity Addition

u′ = (u − v) / (1 − uv/c²)

An equivalent rearrangement is:

u = (u′ + v) / (1 + u′v/c²)

Critical point. The denominator prevents ordinary sub-light velocities from combining to give a result greater than c.

Fig. 5. Ordinary addition can predict a speed above c. Relativistic addition approaches c without crossing it.

Worked Example 5: Spacecraft Launches a Probe

A spacecraft moves at 0.800c relative to Earth. It launches a probe forward at 0.700c relative to the spacecraft. Determine the probe speed relative to Earth.

Step 1: u = (u′ + v)/(1 + u′v/c²)

Step 2: u = (0.700c + 0.800c)/[1 + (0.700c)(0.800c)/c²]

Step 3: u = 1.500c/1.560 = 0.962c

Step 4: u = 0.962(3.00 × 10⁸ m/s) = 2.89 × 10⁸ m/s

Answer: 0.962c = 2.89 × 10⁸ m/s.

6. Proper Time and Time Dilation

The proper time interval Δt₀ is measured in the frame in which the two events occur at the same position.

Δt = γΔt₀

Because γ ≥ 1, an observer who sees the clock moving measures a longer time interval.

Fig. 6. A light clock makes time dilation visual: the external observer sees a longer light path while still measuring the same light speed c.

Worked Example 6: Muon Time Dilation

A muon has a proper mean lifetime of 2.20 × 10⁻⁶ s and moves at 0.980c relative to Earth. Calculate its mean lifetime measured from Earth and its mean Earth-frame travel distance.

Step 1: γ = 1/√(1 − 0.980²) = 5.03

Step 2: Δt = (5.03)(2.20 × 10⁻⁶ s) = 1.11 × 10⁻⁵ s

Step 3: v = 0.980(3.00 × 10⁸ m/s) = 2.94 × 10⁸ m/s

Step 4: d = vΔt = (2.94 × 10⁸ m/s)(1.11 × 10⁻⁵ s) = 3.26 × 10³ m

Answer: 1.11 × 10⁻⁵ s and approximately 3.26 km.

7. Proper Length and Length Contraction

The proper length L₀ is measured in the frame where the object is at rest. A moving observer measures contraction only in the direction parallel to relative motion.

L = L₀/γ

Fig. 7. The moving spacecraft is measured shorter only along its direction of motion.

Worked Example 7: Length Contraction

A spacecraft has proper length 120 m and passes Earth at 0.800c. Determine its Earth-frame length.

Step 1: γ = 1.67

Step 2: L = 120 m/1.67

Step 3: L = 71.9 m

Answer: approximately 72.0 m.

Fig. 8. Length contraction becomes significant only at relativistic speeds.

8. Relativity of Simultaneity

Fig. 9. Simultaneity is frame-dependent for events separated in space.

Δt′ = γ(Δt − vΔx/c²)

Worked Example 8: Simultaneous Lightning Strikes

In the Earth frame, two lightning strikes occur simultaneously and are separated by Δx = +600 m. A rocket moves at 0.800c in the +x direction. Determine Δt′.

Step 1: Δt = 0 s and γ = 1.67.

Step 2: Δt′ = −γvΔx/c²

Step 3: vΔx/c² = [(2.40 × 10⁸ m/s)(600 m)]/(9.00 × 10¹⁶ m²/s²) = 1.60 × 10⁻⁶ s

Step 4: Δt′ = −(1.67)(1.60 × 10⁻⁶ s) = −2.67 × 10⁻⁶ s

Answer: Δt′ = −2.67 μs. With the stated sign convention, the event at the larger x-coordinate occurs earlier in S′.

9. Invariant Spacetime Interval

(Δs)² = (cΔt)² − (Δx)²

Observers can disagree about Δx and Δt separately while agreeing on the spacetime interval.

Interval Type	Condition	Meaning
Time-like	(Δs)² > 0	A frame can exist in which the two events occur at the same position.
Light-like	(Δs)² = 0	The events can be connected by a light signal.
Space-like	(Δs)² < 0	A frame can exist in which the events are simultaneous.

Worked Example 9: Classifying an Interval

Two events are separated by Δx = 600 m and Δt = 1.00 × 10⁻⁶ s.

Step 1: cΔt = (3.00 × 10⁸ m/s)(1.00 × 10⁻⁶ s) = 300 m

Step 2: (Δs)² = (300 m)² − (600 m)²

Step 3: (Δs)² = 9.00 × 10⁴ m² − 3.60 × 10⁵ m² = −2.70 × 10⁵ m²

Answer: −2.70 × 10⁵ m²; space-like interval.

10. Spacetime Diagrams and Worldlines

A spacetime diagram places x on the horizontal axis and ct on the vertical axis so that both coordinates have dimensions of length. The path of an object through spacetime is its worldline.

Vertical worldline: object at rest in that frame.
Straight tilted worldline: constant velocity.
Photon worldline: 45° when the x and ct axes use the same scale.
For a constant-velocity particle, tan θ = v/c, where θ is measured from the ct-axis.
The moving-frame axes x′ and ct′ are tilted relative to x and ct.
The x/x′ and ct/ct′ scales are not identical; lines of constant spacetime interval define the scale.

Fig. 10. Massive-particle worldlines remain inside the light boundaries.

Worked Example 10: Speed from a Worldline Angle

A constant-velocity worldline makes an angle of 31.0° with the ct-axis. Determine the speed.

Step 1: tan θ = v/c

Step 2: v = c tan 31.0°

Step 3: v = (3.00 × 10⁸ m/s)(0.601) = 1.80 × 10⁸ m/s

Answer: 1.80 × 10⁸ m/s ≈ 0.600c.

Fig. 11. The moving-frame axes are tilted and require their own calibration.

11. Muon Decay as Experimental Evidence

Earth frame: the moving muon's lifetime is dilated, so it travels farther before decaying.
Muon frame: the atmosphere moves toward the muon and the atmospheric distance is length-contracted.

Fig. 12. The Earth-frame and muon-frame explanations are different descriptions of the same experiment.

Worked Example 11: Finding Speed from Lifetime Data

A particle has proper lifetime 2.60 × 10⁻⁸ s. In the Earth frame its mean lifetime is 1.00 × 10⁻⁷ s. Determine its speed.

Step 1: γ = Δt/Δt₀ = (1.00 × 10⁻⁷ s)/(2.60 × 10⁻⁸ s) = 3.85

Step 2: v/c = √(1 − 1/γ²)

Step 3: v/c = √[1 − 1/(3.85)²] = 0.966

Step 4: v = 0.966(3.00 × 10⁸ m/s) = 2.90 × 10⁸ m/s

Answer: v ≈ 0.966c = 2.90 × 10⁸ m/s.

12. Equation-Selection Guide

1. Identify the FramesState S, S′ and the sign/direction of v.

2. Identify the QuantityCoordinates? Velocity? Clock interval? Length? Event separation? Worldline angle?

3. Identify the Proper QuantitySame place → proper time. Object at rest → proper length.

4. Select and CheckKeep units throughout; require γ ≥ 1 and v < c for massive particles.

Problem Type	Best Starting Relationship	Key Check
Low-speed classical frames	x′ = x − vt; t′ = t; u′ = u − v	Relativistic effects negligible.
Coordinates of one event	x′ = γ(x − vt), t′ = γ(t − vx/c²)	Use consistent signs and synchronized origins.
Relativistic velocity	u′ = (u − v)/(1 − uv/c²)	Massive-particle result stays below c.
Clock, decay or lifetime	Δt = γΔt₀	Proper time: same-place events.
Moving length	L = L₀/γ	Proper length: object at rest.
Two-event comparison	(Δs)² = (cΔt)² − (Δx)²	Interval is invariant.
Worldline angle	tan θ = v/c	Measure θ from the ct-axis.

Data-Booklet Equations at a Glance

Relationship	Use
x′ = x − vt	Galilean position transformation.
t′ = t	Absolute-time assumption in Galilean relativity.
u′ = u − v	Galilean velocity addition.
x′ = γ(x − vt)	Lorentz position transformation.
t′ = γ(t − vx/c²)	Lorentz time transformation.
γ = 1/√(1 − v²/c²)	Lorentz factor.
u′ = (u − v)/(1 − uv/c²)	Relativistic velocity addition.
(Δs)² = (cΔt)² − (Δx)²	Invariant spacetime interval.
Δt = γΔt₀	Time dilation.
L = L₀/γ	Length contraction.
tan θ = v/c	Worldline angle and speed.

13. Common Errors and Exam Strategy

Common Error	Correction
Using ordinary velocity addition near c.	Use the relativistic velocity-addition equation.
Calling any measured time the proper time.	Proper time belongs to the frame where both events occur at the same place.
Calling a moving length L₀.	L₀ is measured only in the object's rest frame.
Thinking a moving observer notices their own clock running slowly.	Each local clock behaves normally in its own frame.
Assuming simultaneous events remain simultaneous.	Spatially separated simultaneity is frame-dependent.
Treating spacetime geometry as ordinary Euclidean geometry.	Use the invariant interval and spacetime rules.
Reading equal drawn distances on x and x′ as equal coordinate intervals.	The moving-frame axes require their own scale.
Measuring θ from the x-axis.	For tan θ = v/c, θ is measured from the ct-axis.

Precise Terminology to Retain

Term	Use It to Mean...
Event	An occurrence assigned a definite position and time in a reference frame.
Inertial reference frame	A non-accelerating reference frame.
Galilean relativity	The classical framework in which Newton's laws have the same form in all inertial frames and time is absolute.
Lorentz factor	The dimensionless factor γ controlling relativistic effects.
Proper time interval	The time between two events measured where those events occur at the same position.
Proper length	The length of an object measured in its rest frame.
Spacetime interval	The invariant event separation (Δs)² = (cΔt)² − (Δx)².
Worldline	The path of an object through spacetime.
Relativity of simultaneity	Spatially separated events simultaneous in one inertial frame need not be simultaneous in another.

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