Edexcel IGCSE Physics · Topic 1

Forces and Motion

Units  ·  Movement and Position  ·  Forces, Shape and Momentum

Core: 1.1, 1.3–1.24 Plus tier: 1.2P, 1.25P–1.33P
EXPAND TO REVEAL: 📖 Full Explanation 📝 Worked Examples 🔬 Practical Notes 🔭 Beyond the Syllabus
1.11.2P 1.31.41.51.6 1.71.81.91.10 1.111.121.131.14 1.151.161.171.18 1.191.201.21 1.221.231.24 1.25P1.26P1.27P1.28P 1.29P1.30P1.31P1.32P1.33P
a

Units

1.1Use the following units: kg, m, m/s, m/s², N, s, N/kg

All physics calculations in this topic must use SI (Système International) base units. Mixing units — for example using km and s together — gives wrong answers.

QuantitySymbolSI UnitAbbrev.
Massmkilogramkg
Distance / Displacementd or smetrem
Speed / Velocityv or umetre per secondm/s
Accelerationametre per second squaredm/s²
Force / WeightF or WnewtonN
Timetseconds
Gravitational field strengthgnewton per kilogramN/kg
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Why SI units matter

Physics is a universal language. For equations like F = ma to work, every scientist — whether in India, Brazil or Norway — must use the same units. The International System of Units (SI) provides this shared standard.

The units you use in this topic are all derived from three base units: the kilogram (mass), the metre (length), and the second (time). The newton, for example, is defined as the force that gives 1 kg an acceleration of 1 m/s² — it's built directly from the base units. This is why the unit of force is kg·m/s².

Converting to SI before calculating

Exam questions frequently give data in non-SI units to test whether you convert first. Common traps:

Conversions to memorise:
1 km = 1000 m  ·  1 g = 0.001 kg  ·  1 minute = 60 s  ·  1 hour = 3600 s  ·  1 km/h = 1 ÷ 3.6 m/s

Always do unit conversions before substituting into a formula, and always write the unit alongside your answer.

Converting units — step by step

Example 1: Convert 72 km/h to m/s
72 km/h × (1000 m ÷ 3600 s) = 72 ÷ 3.6 = 20 m/s
Example 2: Convert 250 g to kg
250 ÷ 1000 = 0.25 kg
Example 3: Convert 3 minutes 20 seconds to seconds
(3 × 60) + 20 = 180 + 20 = 200 s
Q1

A car travels at 90 km/h. Convert this to m/s.

Write your working here…
✓ 90 ÷ 3.6 = 25 m/s
[2 marks: 1 correct method, 1 correct answer + unit]
✗ Common error: Multiplying by 3.6 instead of dividing — always divide to go from km/h to m/s.
1.2PUse the following units: newton metre (Nm), kilogram metre/second (kg m/s)
★ Physics Plus — Higher Tier only

Two additional units are needed for the Plus tier: Nm for moments (turning effects) and kg m/s for momentum — both used in sections 1.30P–1.33P and 1.25P–1.28P respectively.

QuantityUnitAbbrev.
Moment (turning effect of a force)newton metreNm
Momentumkilogram metre per secondkg m/s
b

Movement and Position

1.3Plot and explain distance–time graphs

A distance–time graph shows how far an object has moved from a starting point as time passes. The shape of the line reveals the type of motion: flat = stationary; straight slope = constant speed; steeper slope = faster speed; curve = changing speed (acceleration).

  • Horizontal line → stationary (distance unchanging)
  • Straight slope → constant speed; steeper = faster
  • Curved (steepening) → accelerating
  • Gradient of the line = speed: speed = Δd ÷ Δt
  • A vertical line is physically impossible (infinite speed)
Distance - Time Graph Time (s) Distance (m) slope=0 (stops) Δt Δd gradient = speed instantaneous speed = gradient of tangent same gradient ✓ same gradient ✓ ① Stationary ② Acceleration ③ Constant v ④ Deceleration ⑤ Stationary slope=0: starts from rest ↑ slope=0: comes to rest ↑ Gradient is continuous at every join (no kinks) Quadratic bezier curves for phases 2 and 4 — parabolic shape, slope only ever changes in one direction Acceleration (slope 0 to 0.700) Constant velocity (slope 0.700) Deceleration (slope 0.700 to 0) Stationary (slope = 0)

Fig 1.1 — Annotations placed in clear space only. Phase labels below the axis. Gradient continuity holds exactly at every join.

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What a distance–time graph really tells you

Imagine placing a camera at the starting line and taking a photo every 0.5 seconds of a car driving away. Each photo shows the car at a different position. If you then plot those positions (distances from the start) against the time when each photo was taken, you get a distance–time graph.

When the car moves equal distances in equal time intervals — for example, always 6 metres in every 0.5 seconds — the points fall on a straight line. This straight line tells you instantly that the car is moving at constant speed. The steeper the line, the more distance the car covers each second, and the faster it is moving.

When the car is stationary — perhaps at a traffic light — the distance doesn't change as time passes. Every photo shows the car in the same place. On the graph, this becomes a flat, horizontal line. No matter how much time passes, the distance stays the same.

Acceleration is different. When a car speeds up from a standstill, it covers a little distance in the first second, more in the second, and even more in the third. The points on the graph curve away from the x-axis, getting progressively steeper. The gradient (slope) of the graph at any point equals the speed at that instant — so a steepening curve means increasing speed.

Impossible shape: A vertical line would mean the car moved a distance in zero time — this requires infinite speed, which is physically impossible. If you ever sketch one, rethink.

Reading and calculating from d–t graphs

Example 1 — find the speed: A straight line on a d–t graph goes from (0 s, 0 m) to (5 s, 40 m).
Speed = gradient = Δd ÷ Δt = (40 − 0) ÷ (5 − 0) = 40 ÷ 5 = 8 m/s
Example 2 — identify the motion: A graph shows a horizontal line from t = 2 s to t = 5 s at d = 30 m, then a steeper straight line.
From 2–5 s: gradient = 0, so the object is stationary. After t = 5 s: positive gradient, so it moves at constant speed.
Example 3 — compare speeds: Object A has gradient 12 m/s; Object B has gradient 4 m/s.
Object A is 3× faster than Object B. The steeper line always represents the faster object.

Speed vs velocity — and displacement vs distance

Distance is the total length of path travelled, regardless of direction — it is a scalar. Displacement is the straight-line distance from start to finish, in a specific direction — it is a vector.

Speed = distance ÷ time (scalar). Velocity = displacement ÷ time (vector). A car going around a roundabout at constant speed has changing velocity — its direction changes every moment. At IGCSE, most questions use "speed" and "distance", but at A-Level the distinction becomes essential.

On a displacement–time graph (a d–t graph using displacement), a line sloping downward means the object is returning toward its starting point — the displacement is decreasing. This is different from simply slowing down.

Q2

A d–t graph shows a straight line rising from (0 s, 0 m) to (5 s, 40 m). State the type of motion and calculate the speed.

Write your answer here…
✓ Constant (uniform) speed. Speed = gradient = 40 ÷ 5 = 8 m/s
[2 marks: 1 for "constant speed", 1 for correct calculation + unit]
✗ Error: Saying "the object accelerates" because the line goes upward — a straight line means constant speed. Only a curve means acceleration.
1.4Know and use the relationship between average speed, distance moved and time taken

Average speed summarises an entire journey as a single value — total distance divided by total time. It does not describe what the speed was at any individual moment.

EQ 1.4
average speed = distance moved ÷ time taken
v = d / t
Rearranges to: d = v × t  and  t = d / v
  • Speed is a scalar — magnitude only, no direction
  • Ensure time is in seconds and distance in metres before calculating
  • Instantaneous speed = speed at one specific moment (what a speedometer shows)
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Average vs instantaneous speed

Imagine driving from Birmingham to Manchester, a journey of about 130 km. You pass through town centres (20 km/h), dual carriageways (100 km/h), and motorways (110 km/h). Your speedometer needle jumps around constantly. This is your instantaneous speed — the speed at any single moment.

Now divide the total journey distance by the total time taken. If the journey takes 1.5 hours, your average speed is 130 ÷ 1.5 = 87 km/h. You never actually drove at exactly 87 km/h — it's a mathematical summary of the whole trip, hiding all the variation inside it.

The formula v = d/t is one of the most frequently used in GCSE Physics. Almost every calculation involving time and distance begins here. The key skill is unit conversion: many questions give data in km and minutes, forcing you to convert to metres and seconds before calculating.

Average speed calculations

Example 1: A sprinter runs 200 m in 22 s.
v = 200 ÷ 22 = 9.1 m/s (3 s.f.)
Example 2: A cyclist travels 1800 m in 4 minutes.
Convert: 4 min = 4 × 60 = 240 s
v = 1800 ÷ 240 = 7.5 m/s
Example 3 (rearranged): A train travels at 50 m/s for 3 minutes. How far does it go?
Convert: 3 min = 180 s
d = v × t = 50 × 180 = 9000 m (9 km)
Q3

A cyclist travels 1800 m in 4 minutes. Calculate their average speed in m/s.

Write your working here — convert units first!
✓ t = 4 × 60 = 240 s;   v = 1800 ÷ 240 = 7.5 m/s
[3 marks: 1 unit conversion, 1 substitution, 1 correct answer + unit]
✗ Error: Using t = 4 directly → v = 1800/4 = 450 m/s (faster than a jet fighter — a sure sign of an error).
1.5Practical: investigate the motion of everyday objects such as toy cars or tennis balls

The core practical involves measuring time over known distances to calculate average speed, then plotting a d–t graph. Light gates reduce human timing error compared to a stopwatch.

  • Mark distance intervals (e.g. every 20 cm) along the ramp with tape
  • Time the car over each interval — repeat 3× per height and average to reduce random error
  • Calculate v = d/t for each interval and plot a d–t graph
  • Increase ramp angle → car accelerates more → d–t graph curves more steeply
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Full method

Apparatus: Ramp, toy car, metre rule, tape, stopwatch or light gates and data logger, clamp stand to hold ramp at fixed angle.

Variables: Independent = ramp height h; Dependent = average speed v over each interval; Control = same car, same distances AB.

Method:

  1. Clamp the ramp at a measured height h. Measure h perpendicular to the bench.
  2. Mark fixed distance intervals along the ramp (e.g. every 20 cm).
  3. Release the car from the same start mark (A) each time.
  4. Record times at each marker. Repeat 3× and calculate the mean time.
  5. Calculate v = d/t for each interval. Record in a table.
  6. Plot distance vs time. A straight line → constant speed; a curve → acceleration.

Sources of error and improvements

Human reaction time (stopwatch): improves significantly with light gates, which measure to milliseconds automatically.

Inconsistent release point: Always release from the same mark; a fiducial mark improves repeatability.

Friction variation: Keep the same car and ramp surface throughout. Any oil drip from the axle changes friction during the experiment.

Galileo's original experiment

Galileo (1564–1642) used a ramp studded with small bells to measure acceleration. By listening for the bell sounds, he adjusted the bell positions until they rang at equal time intervals — which meant the ball had travelled a specific ratio of distances between them. He discovered that the distances followed the pattern 1 : 3 : 5 : 7 (odd numbers), which means the ball's displacement from rest follows s ∝ t² — a hallmark of uniform acceleration. This was centuries before precision timers existed.

1.6Know and use the relationship between acceleration, change in velocity and time taken

Acceleration measures how quickly velocity changes. A positive value means speeding up; a negative value (deceleration) means slowing down. It is a vector quantity — direction matters.

EQ 1.6
acceleration = change in velocity ÷ time taken
a = (v − u) / t
u = initial velocity (m/s)  ·  v = final velocity (m/s)  ·  a in m/s²
  • u comes before v in the alphabet — u is the under(initial) velocity
  • Negative acceleration = deceleration (slowing down)
  • Uniform acceleration means a is constant (straight line on v–t graph)
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Understanding acceleration deeply

Speed tells you how fast an object moves. Acceleration tells you how fast that speed is changing. A car that goes from 0 to 60 mph in 3 seconds feels very different from one that takes 10 seconds — the first has much greater acceleration.

The formula a = (v − u)/t measures this rate of change. It's the increase in velocity per second. If a = 4 m/s², it means the object gains 4 m/s of speed every second: after 1 s it's 4 m/s faster, after 2 s it's 8 m/s faster, and so on.

Galileo discovered uniform acceleration by rolling balls down ramps and listening to bells. He found that the velocity increased by the same amount each second — the very definition of uniform (constant) acceleration. Most exam problems involve uniform acceleration, though in reality (e.g. a falling object with air resistance) acceleration is rarely perfectly uniform.

A negative acceleration simply means the velocity is decreasing. The minus sign in the calculation appears automatically when v is less than u — you don't need to "add a minus sign" manually.

Acceleration and deceleration calculations

Example 1 (acceleration): A car speeds up from 5 m/s to 25 m/s in 4 s.
a = (25 − 5) ÷ 4 = 20 ÷ 4 = 5 m/s²
Example 2 (deceleration): A bus slows from 20 m/s to 5 m/s in 6 s.
a = (5 − 20) ÷ 6 = −15 ÷ 6 = −2.5 m/s² (deceleration of 2.5 m/s²)
Example 3 (find final velocity): Starting at u = 0, a = 10 m/s², for t = 5 s.
Rearrange: v = u + at = 0 + (10 × 5) = 50 m/s
Q5

A v–t graph rises from (0 s, 0 m/s) to (8 s, 24 m/s), then is horizontal to (14 s, 24 m/s). (a) Find acceleration in first 8 s. (b) Find total distance in 14 s.

Write your working for both parts…
✓ (a) a = 24 ÷ 8 = 3 m/s²
✓ (b) d = (½ × 8 × 24) + (6 × 24) = 96 + 144 = 240 m
[4 marks: 1 each for gradient method, answer (a), area method, answer (b)]
✗ Error: Using only the rectangle (144 m) and omitting the triangle under the acceleration phase.
1.10Use the relationship between final speed, initial speed, acceleration and distance moved

This equation connects speed, acceleration and distance when time is unknown or not needed. It only works for uniform (constant) acceleration.

EQ 1.10
(final speed)² = (initial speed)² + 2 × acceleration × distance
v² = u² + 2as
Use when time is not given. If starting from rest: u = 0, so v² = 2as
  • Only valid for uniform (constant) acceleration
  • All values must be in SI units before substituting
  • Remember to square-root the answer to find v, not v²
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Example 1 (from rest): Vaccine cylinder dropped from 200 m. a = 10 m/s², u = 0.
v² = 0 + 2 × 10 × 200 = 4000 → v = √4000 = 63.2 m/s
Example 2 (find distance): Car brakes from 30 m/s to 0 at a = −6 m/s². Find stopping distance s.
0 = 30² + 2(−6)s → 12s = 900 → s = 75 m
Example 3 (find acceleration): Ball reaches 12 m/s after rolling 18 m from rest.
12² = 0 + 2a × 18 → 144 = 36a → a = 4 m/s²
Q6

A ball starts from rest and accelerates at 5 m/s² over 20 m. What is its final speed?

Write your working here…
✓ v² = 0 + 2 × 5 × 20 = 200 → v = √200 ≈ 14.1 m/s
[3 marks: correct formula, substitution, square root + unit]
✗ Error: Giving v² = 200 as the answer — always take the square root to find v.
c

Forces, Movement, Shape and Momentum

1.11 – 1.14Effects of forces · Types of force · Vectors vs scalars · Force is a vector

A force can change an object's speed, direction, or shape. Forces are vectors — they have both magnitude and direction. Unlike scalars, adding forces requires attention to direction.

Effects of a force:

  • Change in speed (accelerate or decelerate)
  • Change in direction of motion
  • Change in shape (stretch, compress, bend)

Types of force — illustrated in context:

Earth Object W

Gravitational (Weight)

Earth pulls object toward its centre: W = mg

+ attract each other

Electrostatic

Opposite charges attract; like charges repel

Block N

Normal contact

Surface pushes perpendicular (90°) to contact point

Block motion F

Friction / Drag

Acts along surface, opposing direction of motion

Rope / cable T T pulls at both ends

Tension

Force in a stretched rope, string or cable

Object U

Upthrust

Fluid pushes submerged object upward

Vectors vs Scalars SCALAR magnitude only Speed Distance Mass Energy Time Temperature Pressure VECTOR magnitude + direction Velocity Displacement Force Acceleration Momentum Weight Electric field
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Why vectors and scalars are different in practice

The distinction between scalar and vector quantities is one of the most important conceptual ideas in physics. A scalar is fully described by a number and a unit. A vector needs a direction too.

Consider two cars each driving at 60 km/h. If one goes north and one goes south, they have the same speed but opposite velocities. If they collide head-on, the physics is completely different from two cars travelling in the same direction at 60 km/h. Direction matters enormously — and this is why velocity, not speed, appears in the momentum equation p = mv.

Force is also a vector. If you push a box east with 30 N and friction acts west with 10 N, the resultant is 20 N east — but only because you're dealing with a vector subtraction along a line. At higher levels, vectors in two dimensions are added using trigonometry or scale diagrams, but at IGCSE you only need to handle vectors along a straight line.

Q7

Classify each as vector or scalar: (a) 30 m/s north, (b) 50 kg, (c) 200 N downwards, (d) 25°C.

Write your answers here…
✓ (a) Vector — velocity; (b) Scalar — mass; (c) Vector — force; (d) Scalar — temperature
[1 mark each] The giveaway is always whether a direction is stated or required.
1.15Calculate the resultant of forces that act along a line

The resultant force is the single force that has the same effect as all forces acting together. Along a straight line: choose a positive direction, then add or subtract force values with appropriate signs.

Resultant Force — Three Scenarios Same direction → ADD Block 20N 15N 35N → 20+15 = 35 N east → accelerates Opposite → SUBTRACT Block 30N 10N← 20N → 30−10 = 20 N east → accelerates Equal opposite = 0 N Block 25N 25N← 0 N 25−25 = 0 N constant velocity or stationary

Fig 1.3 — When resultant = 0 N, the object is in equilibrium (Newton's First Law)

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Why the resultant is the key to all force problems

In the real world, objects are always subject to multiple forces simultaneously. A car has engine thrust, road friction, air resistance, and gravity all acting at once. To predict its acceleration, you don't need to analyse each force separately — you need just one number: the resultant force.

The technique is simple for forces along a line: choose one direction as positive (usually the direction of motion), assign positive values to forces in that direction and negative values to forces opposing it, then add everything up algebraically. The result — positive, negative, or zero — tells you everything: a positive resultant means acceleration in the positive direction; negative means deceleration; zero means the object is in equilibrium.

Newton's First Law is implicit here: when the resultant is exactly zero, the object neither speeds up nor slows down. A stationary object stays still; a moving object continues at exactly the same velocity. This is why a car on a motorway at constant speed still has its engine running — not to accelerate, but to balance the friction and air resistance forces to maintain zero resultant.

Example 1: Engine thrust 3000 N forward, friction 800 N backward, air resistance 500 N backward.
Resultant = 3000 − 800 − 500 = 1700 N forward
Example 2 (equilibrium): A 60 kg box on a table. Weight = 600 N downward. What is the normal reaction?
Resultant = 0 (stationary) → Normal = 600 N upward
Example 3: Forces of 20 N east, 35 N east, 12 N west.
Resultant = 20 + 35 − 12 = 43 N east

Newton's First and Second Laws

First Law: An object remains at rest or in uniform motion unless acted upon by a resultant force. This is why astronauts in space drift at constant velocity without any engine thrust — there's nothing to slow them down.

Second Law: The resultant force on an object equals its mass times its acceleration (F = ma). This is actually a special case of Newton's more general statement: resultant force = rate of change of momentum — which you'll meet in spec 1.28P.

Q8

Forces of 20 N east, 35 N east, and 12 N west act on a box. Calculate the resultant and state its direction.

Show your working — choose a positive direction first…
✓ Taking east as positive: 20 + 35 − 12 = 43 N east
[2 marks: calculation + direction]
✗ Error: Adding all three magnitudes (67 N) without considering the west-acting force.
1.16 – 1.17Friction opposes motion · F = ma

Friction is a contact force that always opposes relative motion, converting kinetic energy to heat. F = ma connects the resultant force on an object to the acceleration it produces.

Forces on a Moving Block Block Push Friction W ↓ N ↑ W acts from centre of mass N and friction act at contact surface

Fig 1.4 — W from centre of mass; N and friction act at the contact surface

EQ 1.17
Force = mass × acceleration
F = m × a
F = resultant force (N)  ·  m in kg  ·  a in m/s²
  • F is the resultant (net) force — always
  • Double force → double acceleration (same mass)
  • Double mass → half acceleration (same force)
  • Rearranges: a = F/m  and  m = F/a
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F = ma — the most useful equation in mechanics

Newton's second law, F = ma, is the engine of all mechanics problems. It tells you that the harder you push something and the lighter it is, the faster it accelerates. This sounds obvious, but it gives precise, quantitative predictions.

The critical word in "F = ma" is the "F": it stands for the resultant (net) force — not just one of the forces acting. If a 1000 kg car has an engine thrust of 4000 N but friction of 1500 N, F = 4000 − 1500 = 2500 N. The acceleration is then a = F/m = 2500/1000 = 2.5 m/s². Many students use just the engine thrust and get the wrong answer.

Friction always opposes the direction of motion. In a car, friction between the tyres and road prevents the wheels from spinning — it actually propels the car forward. The friction in the wheel bearings and air resistance oppose motion. Understanding which friction helps and which hinders is a conceptual skill the exam tests regularly.

Example 1: A 1200 kg car has resultant force 3600 N. Find acceleration.
a = F/m = 3600/1200 = 3 m/s²
Example 2 (with friction): Engine thrust 4000 N, friction 1500 N, mass 1000 kg.
F_net = 4000 − 1500 = 2500 N; a = 2500/1000 = 2.5 m/s²
Example 3 (find force): A 500 g ball accelerates at 4 m/s².
Convert: m = 0.5 kg; F = 0.5 × 4 = 2 N

Why friction exists — at the molecular level

Even apparently smooth surfaces are rough at the nanometre scale. When two surfaces touch, tiny peaks (asperities) interlock and form temporary bonds. Sliding the surfaces apart continuously breaks these bonds — each broken bond releases a tiny amount of energy as heat. The total of millions of these events per second is what we feel as friction.

This explains why: (1) heavier loads create more friction — more weight presses more asperities into contact; (2) rough surfaces create more friction — more asperities per unit area; (3) lubrication (oil) works — it fills the gaps between asperities, allowing surfaces to glide over the oil film rather than over each other.

Q9

A 1200 kg car has engine thrust 5000 N and friction forces totalling 1400 N. Calculate the acceleration.

Write your working here — find resultant force first!
✓ F_net = 5000 − 1400 = 3600 N; a = 3600 ÷ 1200 = 3 m/s²
[3 marks: resultant force, substitution, answer + unit]
✗ Error: Using F = 5000 N directly → a = 4.17 m/s² — always find the resultant force before applying F = ma.
1.18Know and use: weight = mass × gravitational field strength

Weight is the gravitational force acting on a mass. Unlike mass (which is constant), weight depends on the gravitational field strength of the location and is always measured in newtons.

EQ 1.18
weight = mass × gravitational field strength
W = m × g
W in N  ·  m in kg  ·  g = 10 N/kg on Earth (Edexcel standard value)
  • Mass is the amount of matter (kg) — constant everywhere in the universe
  • Weight is a force (N) — changes with gravitational field strength
  • On Earth g = 10 N/kg; on Moon g ≈ 1.6 N/kg; on Mars g ≈ 3.7 N/kg
⚠️Classic error: "I weigh 60 kg." Incorrect. 60 kg is your mass; your weight is 60 × 10 = 600 N (on Earth).
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Mass vs weight — the clearest way to think about it

Imagine you're an astronaut. On Earth, you struggle to lift a 100 kg barbell. On the Moon, the same barbell feels like it's only about 16 kg. What changed? Not the amount of metal in the barbell — that's the mass, and it's constant. What changed is the gravitational pull on it — the weight.

Mass is a fundamental property of matter: it tells you how much substance is there, and how hard it is to accelerate (inertia). Weight is the gravitational force that happens to act on that mass in a particular location. Weight = mass × g, and g varies across the universe: near a black hole it's enormous; in deep space far from any star it's essentially zero.

In everyday language, "weight" and "mass" are used interchangeably, which causes endless confusion in physics. The key discipline is: whenever you see a force acting downward due to gravity, that's weight in newtons. Whenever you see a quantity measuring the amount of matter, that's mass in kilograms.

Example 1 (Earth): Person of mass 65 kg on Earth (g = 10 N/kg).
W = 65 × 10 = 650 N
Example 2 (Moon): Same person on Moon (g = 1.6 N/kg).
W = 65 × 1.6 = 104 N (mass unchanged: still 65 kg)
Example 3 (find mass): An object weighs 360 N on Earth.
m = W/g = 360/10 = 36 kg
Q10

An astronaut has a mass of 80 kg. Calculate their weight (a) on Earth (g = 10 N/kg) and (b) on Mars (g = 3.7 N/kg).

Write your working for both parts…
✓ (a) W = 80 × 10 = 800 N; (b) W = 80 × 3.7 = 296 N
[1 mark each; bonus mark for stating mass remains 80 kg in both]
✗ Error: Changing the mass value in part (b) — mass is constant regardless of location.
1.19 – 1.21Stopping distance · Factors affecting stopping distance · Terminal velocity

Total stopping distance = thinking distance + braking distance. Terminal velocity is reached when air resistance equals weight, giving zero resultant force and constant speed.

Stopping Distance Diagram ! Thinking distance Braking distance

Fig 1.5 — Driver sees hazard at the car; thinking distance passes before brakes engage

Thinking distance ↑ when:

  • Higher speed
  • Tiredness, alcohol, drugs
  • Distractions (phone use)

Braking distance ↑ when:

  • Higher speed (most significant)
  • Wet or icy road surface
  • Worn tyres or faulty brakes
  • Greater vehicle mass

Terminal velocity (spec 1.21)

As an object falls, drag increases with speed. When drag = weight, resultant = 0 and the object falls at constant terminal velocity.

Terminal Velocity — Three Stages ① Just released W=800N D≈0N Large resultant ↓ large acceleration ② Speeding up W=800N D≈400N Smaller resultant ↓ still accelerating ③ Terminal velocity W=800N D=800N W = D resultant = 0 → constant v

Fig 1.6 — Arrow lengths ∝ force magnitude. W and D equal at terminal velocity → resultant = 0.

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Terminal velocity — the force balance in detail

When a skydiver jumps from a plane, gravity immediately pulls them downward. At the very instant of jumping (velocity = 0), there is no air resistance at all — the only force is weight, so the acceleration is the full g ≈ 10 m/s².

As the skydiver speeds up, air resistance (drag) builds. Drag depends on speed — roughly proportional to speed squared for high speeds. As drag grows, the resultant downward force (weight − drag) shrinks. A smaller resultant force means a smaller acceleration. The skydiver is still accelerating, but more and more slowly.

Eventually, drag grows to exactly equal weight. At this point, resultant = 0 N, so by Newton's first law, the skydiver stops accelerating. They continue to fall, but at a constant speed: terminal velocity (typically 55–75 m/s for a skydiver in the spread-eagle position).

When the parachute opens, the surface area suddenly increases enormously, causing drag to spike far above weight. The resultant force now acts upward — the skydiver decelerates (but keeps falling downward). As they slow down, drag decreases until a new equilibrium is reached at a much lower terminal velocity (around 5–7 m/s), safe for landing.

Q11

A skydiver falls at terminal velocity then opens her parachute. Describe what happens to the forces and velocity immediately after opening.

Write your explanation using force terms…
✓ Drag increases greatly (larger area); drag now exceeds weight → resultant force acts upward → skydiver decelerates (slows down); eventually drag reduces as speed drops and a new lower terminal velocity is reached where W = D again.
[4 marks: drag increases, drag > weight, decelerates, new terminal velocity]
✗ Error: Saying she "moves upward" — she still falls downward, just at decreasing speed.
1.22 – 1.24Practical: Hooke's Law · Linear force-extension region · Elastic behaviour

Extension is proportional to applied force in the linear region (Hooke's Law). Beyond the elastic limit, the material deforms permanently.

EQ 1.22
Force = spring constant × extension
F = k × e
k = spring constant (N/m)  ·  e = extension (m)  ·  valid only in the linear region
Spring Extension Diagrams Zero load L₀ Natural length L₀ No weight attached Load applied Mass m F = mg ↓ L₀ end e Extended by e; F = ke Extension (m) Force (N) Elastic limit Hooke's Law F ∝ e Non-linear grad = k Force – Extension

Fig 1.7 — Left: natural length vs extended spring (extension e shown). Right: F–e graph; gradient of linear region = spring constant k.

  • In the linear region: double the force → double the extension
  • Beyond the elastic limit → permanent (plastic) deformation
  • Rubber bands do not obey Hooke's Law — their F–e graph is curved from the start
  • k = gradient of the linear section of the F–e graph
Expand:

Hooke's Law — understanding the spring constant

In 1678, Robert Hooke published his observation: "As the extension, so the force." This simple sentence encodes a powerful idea: for a given spring, the more you stretch it, the greater the restoring force — and the relationship is exactly linear.

The spring constant k captures how stiff a spring is. A stiff spring (high k, e.g. 500 N/m) requires a large force to achieve even a small extension. A soft spring (low k, e.g. 5 N/m) stretches easily. The unit N/m tells you directly: it's the force in newtons needed to extend the spring by one metre. Stiff car suspension springs have k values of thousands of N/m; gentle precision balance springs might have k of just a few N/m.

The elastic limit is the boundary of Hooke's Law. Push a spring beyond it and the metal's internal crystal structure is permanently rearranged — the spring doesn't return to its original length. The F–e graph curves beyond this point, then may eventually become steep again as the coils jam together. Engineering materials are almost always designed to operate well below their elastic limits.

Example 1 (find force): Spring with k = 25 N/m, extension = 0.16 m.
F = 25 × 0.16 = 4 N
Example 2 (find k from graph): A spring extends 0.3 m under 9 N.
k = F/e = 9/0.3 = 30 N/m
Example 3 (find extension): k = 40 N/m, F = 6 N.
e = F/k = 6/40 = 0.15 m (15 cm)

Practical method — spring extension investigation

Apparatus: Helical spring, 100 g masses with hanger, metre rule, clamp stand and boss, set square.

Variables: Independent = applied force (F = mg); Dependent = extension (e); Control = same spring, same ruler position.

  1. Measure and record the natural length of the unstretched spring.
  2. Add masses one at a time. After each mass is added, wait for oscillations to stop, then measure the new spring length. Extension = new length − natural length.
  3. Record force (F = mass × 10) and extension in a table.
  4. Continue until the spring has clearly exceeded its elastic limit (extension grows more than expected for a given force).
  5. Plot force (y-axis) vs extension (x-axis). Draw line of best fit.
  6. Calculate k = gradient of the linear section.

Key sources of error

Parallax error: Reading the ruler at an angle. Use a set square held at 90° to the ruler to read the bottom of the spring.

Oscillation: The spring bounces when masses are added. Always wait for it to settle.

Zero error: Ensure ruler reads 0 at the natural length reference point, or subtract consistently.

Q12

A spring has spring constant 25 N/m and is stretched by 0.16 m. (a) What force was applied? (b) Is this likely in the linear region?

Write both answers here…
✓ (a) F = ke = 25 × 0.16 = 4 N
✓ (b) Likely yes — if F = ke gives a consistent answer, the spring is in the linear region. Beyond the elastic limit, F = ke no longer applies and the calculated k would not be a constant.
[3 marks: substitution, answer + unit, explanation about linear region]
✗ Error: Using e = 16 cm instead of 0.16 m → F = 25 × 16 = 400 N — always convert to SI units.
1.25P – 1.28PMomentum · Safety features · Conservation of momentum · F = Δp/t
★ Physics Plus — Higher Tier only

Momentum (p = mv) is conserved in all collisions and explosions when no external force acts. Extending the time of a collision reduces the force experienced.

EQ 1.25P
momentum = mass × velocity
p = m × v
p in kg m/s  ·  vector quantity
EQ 1.28P
force = change in momentum ÷ time
F = (mv − mu) / t
F in N  ·  Δp in kg m/s  ·  t in s
Conservation of Momentum BEFORE 4 kg 6 m/s→ 2 kg at rest p = (4×6)+(2×0) = 24 kg m/s 💥 collide AFTER 6 kg together v = ? m/s → 24 = 6×v → v = 4 m/s ✓ Total momentum before = Total momentum after (no external forces) ✓

Fig 1.8 — Conservation of momentum in a perfectly inelastic collision

💡Safety link (1.26P): Crumple zones, airbags, and seatbelts all increase the time of a collision. Since F = Δp/t, a longer time → smaller force, even though the total momentum change is identical.
Expand:

Why momentum is conserved — the deep reason

Conservation of momentum is one of the most powerful and universal laws in physics. It doesn't just apply to trolleys on tracks — it governs billiard balls, galaxies colliding over billions of years, and subatomic particle interactions. Why is it so universal?

The reason lies in Newton's Third Law. When two objects collide — say trolley A and trolley B — A exerts a force on B, and (Third Law) B exerts an equal and opposite force on A. Crucially, both forces act for the same time — the contact time. So the impulse (F × t) on A equals the impulse on B, but in opposite directions. This means A loses exactly the momentum that B gains. The total remains constant.

This only works when there are no external forces on the system. In practice, friction between trolleys and the track is an external force, which is why lab experiments use air tracks (to minimise friction) and still show slight discrepancies. In real collisions — car crashes, nuclear reactions — we always account for any external forces that might affect the total momentum.

Example 1 (perfectly inelastic): 4 kg at 6 m/s hits stationary 2 kg; they stick.
p_before = 4×6 + 2×0 = 24 kg m/s
p_after = (4+2)×v = 6v = 24 → v = 4 m/s
Example 2 (explosion): 5 kg trolley at rest fires a 0.5 kg projectile at 20 m/s. Find recoil v of trolley.
p_before = 0; so 0.5×20 + 5×v = 0 → 10 + 5v = 0 → v = −2 m/s (recoils)
Example 3 (F = Δp/t): 50 kg person decelerates from 20 m/s to 0 in 0.02 s (no airbag).
Δp = 50×20 = 1000 kg m/s; F = 1000/0.02 = 50 000 N
With airbag (t = 0.5 s): F = 1000/0.5 = 2000 N — 25× smaller force

Impulse — the area under a force–time graph

Rearranging F = Δp/t gives: F × t = Δp. The quantity F × t is called the impulse, measured in N·s (equivalent to kg m/s). On a force–time graph, the area under the curve = impulse = change in momentum.

This explains why professional boxers "roll with a punch": by moving their head in the direction of the punch, they increase the contact time, reducing the peak force on their skull even though the change in momentum is the same. Cricket batters "give" with the ball for the same reason.

Q13

A 3 kg trolley at 4 m/s collides with a stationary 1 kg trolley. They stick together. Find their combined velocity.

Write your working — use conservation of momentum…
✓ p_before = 3×4 + 1×0 = 12 kg m/s; 12 = (3+1)×v; v = 12/4 = 3 m/s
[4 marks: p_before, conservation equation, substitution, answer + unit + direction]
✗ Error: (3+1)×(4+0) = 16 — you cannot add velocities; conserve momentum correctly.
1.29PDemonstrate an understanding of Newton's third law
★ Physics Plus — Higher Tier only

Newton's Third Law: if A exerts a force on B, then B exerts an equal and opposite force on A. Third Law pairs always act on different objects — they can never cancel each other.

  • Third law pairs are always equal in magnitude and opposite in direction
  • They are always the same type of force (both contact, both gravitational, etc.)
  • Balanced forces act on the same object; 3rd Law pairs act on different objects
Expand:

The most misunderstood law in mechanics

Newton's Third Law is often stated glibly as "every action has an equal and opposite reaction" — and then misapplied constantly. The key question is always: reaction on what?

When you stand on the floor, your weight (a downward force from Earth's gravity on you) acts on you. The Third Law pair to this is the gravitational force you exert on the Earth — pulling the Earth upward, with exactly the same magnitude. The Earth barely moves because of its enormous mass, but the force is real and equal.

The floor pushes you up with a normal contact force (your weight). You push the floor down with an equal normal force. These are also a Third Law pair — but they're a different Third Law pair from the gravity pair. This is confusing, but the rule always works: identify the force, identify what is exerting it and what it's being exerted on, then the Third Law pair is the same type of force with source and target swapped.

Rockets work in empty space using the Third Law. The engine pushes exhaust gases backward; the exhaust gases push the rocket forward. There's nothing to "push against" — the push comes from the gases expelled, not from the ground or air.

Example 1 — identify the 3rd Law pair:
A horse pulls a cart with force 800 N forward.
3rd Law pair: the cart pulls the horse backward with force 800 N.
Example 2 — why does the cart accelerate?
The cart has two horizontal forces: horse pulls it forward (800 N); road friction acts backward (200 N). Resultant = 600 N forward → cart accelerates. The "equal and opposite" force is on the horse, not on the cart — so it doesn't cancel the 800 N forward.
Example 3 — rocket:
Exhaust gases expelled downward at high speed → 3rd Law pair: rocket propelled upward with equal force. No air or ground needed.
1.30P – 1.33PMoment of a force · Centre of gravity · Principle of moments · Beam on two supports
★ Physics Plus — Higher Tier only

The moment of a force is its turning effect. For equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about any pivot.

EQ 1.30P
moment = force × perpendicular distance from pivot
M = F × d
M in Nm  ·  d must be measured perpendicular to the force
Principle of Moments — Balanced Beam F₁=40N F₂=20N d₁=1.4m d₂=2.8m Anticlockwise: 40×1.4=56 Nm  =  Clockwise: 20×2.8=56 Nm ✓ Balanced

Fig 1.9 — For equilibrium: Σ clockwise moments = Σ anticlockwise moments about any pivot

  • Centre of gravity (1.31P): point through which the entire weight appears to act
  • Principle of moments (1.32P): for equilibrium, Σ CW moments = Σ ACW moments about any pivot
  • Beam on supports (1.33P): the closer support always carries a greater fraction of the load
Expand:

Why moments depend on distance — not just force

Try pushing open a heavy door near its hinge — it barely moves. Push at the handle, far from the hinge — it swings easily. The force is identical; the turning effect is vastly different. This is the essence of a moment.

The moment of a force is its turning effect about a pivot. It equals the force multiplied by the perpendicular distance from the pivot to the line of action of the force. A large force close to the pivot can be balanced by a small force far from the pivot — this is the principle behind levers, wheelbarrows, crowbars, and see-saws.

The perpendicular distance is crucial. If a force acts along a line that passes through the pivot, the perpendicular distance is zero — the force has no turning effect at all, regardless of its magnitude. This is why you must always measure the distance at 90° to the direction of the force.

Example 1 — calculate a moment:
Force of 30 N acts 0.6 m from pivot.
M = 30 × 0.6 = 18 Nm
Example 2 — find unknown force (principle of moments):
Child A, weight 200 N, sits 1.5 m from pivot. Child B sits 1.0 m from pivot. Seesaw balanced.
ACW: 200 × 1.5 = 300 Nm; CW: F × 1.0 = 300 → F = 300 N
Example 3 — find unknown distance:
300 N weight 2 m left of pivot balanced by 150 N weight on right.
300 × 2 = 150 × d → 600 = 150d → d = 4 m from pivot

Stability and the centre of gravity

An object topples when its centre of gravity moves outside its base of support. A car with a low centre of gravity and wide wheelbase (like a Formula 1 car) is extremely stable — even extreme cornering doesn't topple it. A double-decker bus keeps its lower deck heavy to keep the centre of gravity low; a bus with a full upper deck and empty lower deck is more prone to toppling on sharp bends.

This is why racing motorcycles lean deeply into corners: the lean allows the combined centre of gravity of rider and bike to stay above the tyres' contact patch, preventing toppling. Engineers use moment calculations to design stable structures — cranes, bridges, and tall buildings all require careful analysis of where forces act and what moments they produce.

Q15

A child of weight 300 N sits 1.5 m left of a seesaw pivot. A second child of weight 450 N sits on the right. Where must the second child sit for the seesaw to balance?

Use the principle of moments…
✓ ACW: 300 × 1.5 = 450 Nm; CW: 450 × d = 450 Nm → d = 1.0 m from pivot
[3 marks: ACW moment, balance equation, correct answer + unit]
✗ Error: Adding the weights (750 N) instead of using moments — moments = force × distance, not just force.