MYP 4 Physics 1.1 Types of Forces and Newton's Laws

MYP 4 Physics 1.1 Types of Forces and Newton's Laws - Full notes

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MYP 4 Physics | 1.1 Types of Forces and Newton's Laws

MYP 4 PHYSICS · CHAPTER 1: FORCES AND MOTION

1.1

Types of Forces and Newton's Laws

Full notes for MYP Year 4 · IB-style

Contents

1	What is a force?
2	Types of force in detail
3	Resultant force
4	Newton's first law
5	Newton's second law
6	Newton's third law
7	Newton's laws in action
8	Data-based questions and further practice
9	Common misconceptions and links
10	Exam-style question with markscheme
11	Check your understanding
12	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe the effects of forces and measure force in newtons
distinguish contact and non-contact forces and name the common types
distinguish mass and weight and use W = mg
draw free-body diagrams and find resultant forces
state and apply Newton's first law and explain inertia
use F = ma, including multi-force problems
state Newton's third law and identify force pairs

Guiding questions

How do forces change the motion of objects? How can the motion of an object be predicted from the forces acting on it? Why do forces always come in pairs?

How this subtopic is assessed

Criterion A questions test definitions, F = ma and W = mg calculations, and identifying third-law pairs. Criterion B and C questions use the trolley investigation: variables, method and analysing a-F graphs. Criterion D questions apply Newton's laws to vehicle safety, lifts and sport.

Show every calculation step with units; in explanations, link force balance to changes in motion.

Criterion	Name	What it rewards
A	Knowing and understanding	recalling, explaining and applying scientific knowledge to familiar and unfamiliar situations
B	Inquiring and designing	questions, hypotheses, variables and methods
C	Processing and evaluating	presenting and interpreting data, evaluating conclusions
D	Reflecting on the impacts of science	applications, implications and limitations of science in society

Command term	What the examiner wants
State / Write down	a brief answer with no working or explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate / Find	a numerical answer with working shown
Show that	obtain the given result, showing every step
Sketch	a graph or diagram showing the general shape and key features
Justify	give valid reasons or evidence for an answer
Evaluate	weigh strengths and limitations and reach a judgement
Suggest / Predict	propose a reasonable answer from the information given

How to use these notes

Each section teaches one topic in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error teachers see every year, and how to avoid it
In the real world	where the idea shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 What is a force?

Pushes, pulls and how we measure them.

Think about it

When you lean on a wall, the wall does not move. Is the wall pushing back on you?

A force is a push or a pull on an object caused by an interaction with another object. Forces cannot be seen, but their effects can: they can change an object's speed, its direction of motion or its shape. This booklet builds the ideas you need to describe any force and predict what it will do.

1.1 Forces are interactions

Every force needs two objects: one that exerts the force and one that feels it. When a footballer kicks a ball, the foot pushes the ball and the ball pushes the foot. A force can do three things: start or stop motion, change speed or direction, or change shape (stretch, squash, bend or twist).

•start an object moving, or stop it
•make a moving object speed up or slow down
•change the direction of motion
•change the shape of an object

Quick check

1. Give an example of a force changing the shape of an object.

Answers: 1. squashing a sponge, stretching a spring or denting a car

1.2 Units and measurement

Force is measured in newtons (N), named after Isaac Newton. One newton is roughly the weight of a small apple. Forces are measured with a newton meter (force meter): a spring inside stretches by an amount proportional to the force, and the scale converts the stretch into newtons.

Situation	Typical force
weight of an apple	1 N
weight of a 1 kg bag of sugar	9.8 N
push to open a door	10–30 N
weight of a 60 kg student	590 N
thrust of a car engine	about 3000 N
thrust of a jumbo jet	about 1 000 000 N

Figure 1. A newton meter: the spring stretches in proportion to the force.

Quick check

1. What unit is force measured in?

Answers: 1. newtons (N)

1.3 Forces are vectors

A force has both size (magnitude) and direction, so it is a vector quantity. We draw forces as arrows: the length shows the size (to scale) and the arrowhead shows the direction. Quantities with size only, such as mass, time, speed and temperature, are scalars.

Vector (size and direction)	Scalar (size only)
force	mass
weight	time
velocity	speed
acceleration	distance
displacement	energy

Common mistake

Treating weight as a scalar. Weight is a force, so it always has a direction: towards the centre of the Earth.

Quick check

1. Is mass a vector or a scalar?

Answers: 1. scalar

1.4 Contact and non-contact forces

Contact forces act only when objects touch: normal contact force, friction, tension, air resistance (drag) and upthrust. Non-contact forces act at a distance through a field: gravitational force (weight), magnetic force and electrostatic force.

Figure 2. Contact forces (red arrows show the force named).

Figure 3. Non-contact forces act through fields.

WORKED EXAMPLE 1 Classifying (basic)

Classify each force as contact or non-contact: (a) the pull of a magnet on a paperclip; (b) the drag on a cyclist; (c) the weight of a book.

Solution

(a) non-contact (magnetic)A1

(b) contact (air resistance); (c) non-contact (gravitational)A1

2 Types of force in detail

Naming every force correctly.

2.1 Weight and mass

Mass is the amount of matter in an object, measured in kilograms; it is the same everywhere. Weight is the gravitational force on the object, measured in newtons. On Earth, W = mg where g = 9.8 N kg⁻¹ is the gravitational field strength.

Figure 4. The same 60 kg student weighs different amounts on different worlds.

WORKED EXAMPLE 2 Weight (standard)

Find the weight of a 65 kg student on Earth and on the Moon (g = 1.6 N kg⁻¹).

Solution

Earth: W = 65 × 9.8 = 637 NA1

Moon: W = 65 × 1.6 = 104 N; mass is still 65 kgA1

Common mistake

Saying 'I weigh 50 kg'. Kilograms measure mass; weight is in newtons (about 490 N for 50 kg).

2.2 Normal contact force

When an object presses on a surface, the surface pushes back perpendicular (normal) to itself. The atoms of the surface are squashed very slightly and push back like tiny springs. On a level surface with nothing else acting vertically, the normal force equals the weight.

WORKED EXAMPLE 3 Normal force (standard)

A 2.0 kg box rests on a level table. Find the normal force.

Solution

N = W = 2.0 × 9.8M1

= 19.6 N upwardsA1

Quick check

1. In which direction does a normal force always act?

Answers: 1. perpendicular to the surface, away from it

2.3 Friction

Friction acts along a surface and opposes sliding (or the tendency to slide). It arises because surfaces are rough at a microscopic level. Friction heats surfaces and causes wear, but without it we could not walk, drive or hold objects.

Useful friction	Unwanted friction
tyres gripping the road	wear on engine parts
brakes slowing a bicycle	energy wasted as heat in machines
shoes gripping the floor	drag slowing a cyclist

Quick check

1. Give one way to reduce friction between moving parts.

Answers: 1. lubricate them with oil (or use ball bearings)

2.4 Tension

Tension is the pulling force in a stretched string, rope, cable or chain. It pulls equally on the objects at both ends. A light string over a frictionless pulley has the same tension throughout.

WORKED EXAMPLE 4 Hanging mass (standard)

A 3.0 kg lamp hangs at rest from a cable. Find the tension.

Solution

Forces balance: T = W = 3.0 × 9.8M1

= 29.4 NA1

2.5 Air resistance and upthrust

Air resistance (drag) opposes motion through air and increases with speed. Water resistance does the same in liquids. Upthrust is the upward push of a fluid on an object immersed in it; a floating object has upthrust equal to its weight.

Quick check

1. Why does a feather fall more slowly than a stone in air?

Answers: 1. air resistance is large compared with its small weight

2.6 Free-body diagrams

A free-body diagram shows one object (often as a box or dot) with every force acting ON it drawn as an arrow from the object, labelled with its name and size. Forces exerted BY the object on other things are not included.

Figure 5. Free-body diagrams for three common situations.

Key point

Draw only forces acting on the chosen object, start each arrow on the object, and label every arrow.

Common mistake

Adding a 'force of motion' to a moving object. Moving objects do not need a forward force to keep moving (Newton's first law).

3 Resultant force

Combining forces.

3.1 Forces in a straight line

The resultant force is the single force that has the same effect as all the forces acting together. For forces along one line, add those in the same direction and subtract those in opposite directions, keeping track of direction with a sign.

Figure 6. Finding the resultant of forces along one line.

WORKED EXAMPLE 5 Tug of war (standard)

Team A pulls left with 1200 N and team B pulls right with 1350 N. Find the resultant.

Solution

1350 − 1200M1

= 150 N to the rightA1

3.2 Balanced and unbalanced forces

If the resultant force is zero the forces are balanced: the object stays at rest or keeps moving at constant velocity. If the resultant is not zero the forces are unbalanced and the object accelerates in the direction of the resultant.

WORKED EXAMPLE 6 Balanced or not? (standard)

A car's engine provides 2500 N of driving force; drag and friction total 2500 N. Describe its motion.

Solution

Resultant = 0M1

Car moves at constant velocity (or stays at rest)A1

3.3 Forces at right angles

Two forces at right angles are combined with Pythagoras' theorem for the size and trigonometry for the direction.

Figure 7. Two perpendicular forces combined head to tail.

WORKED EXAMPLE 7 Perpendicular forces (exam level)

A boat is pushed 30 N east by its engine and 20 N north by the wind. Find the resultant.

Solution

R = √(30² + 20²) = 36 NM1 A1

tan θ = 20/30: θ = 34° north of eastA1

3.4 Scale drawings

For forces at any angle, draw them to scale head to tail; the resultant goes from the start of the first arrow to the end of the last. Measure its length and angle.

Figure 8. Scale drawing for a 50 N force and a 40 N force at 60°.

WORKED EXAMPLE 8 Check by calculation (exam level)

Use the cosine rule to check the resultant in the figure.

Solution

R² = 50² + 40² + 2(50)(40) cos 60° = 6100M1

R = 78 NA1

Try this

Pull a small object with two newton meters at different angles and compare the reading of a third meter that balances them with your scale drawing.

4 Newton's first law

Inertia: objects keep doing what they are doing.

Think about it

Why do you lurch forward when a bus brakes suddenly?

4.1 Statement

Newton's first law: an object remains at rest, or continues to move at constant velocity (constant speed in a straight line), unless acted on by a resultant force. The tendency to resist changes in motion is called inertia; the greater the mass, the greater the inertia.

Figure 9. Velocity–time graph for an object with zero resultant force.

Key point

Zero resultant force does not mean zero motion: it means no change in motion.

Quick check

1. Name the property of an object that resists a change in its motion.

Answers: 1. inertia

4.2 Inertia in everyday life

When a car stops suddenly, passengers continue moving forward because no force has yet acted on them to stop them; seatbelts provide that force. A tablecloth pulled quickly from under dishes leaves the dishes behind because friction acts only briefly.

Figure 10. Why seatbelts are needed: the passenger tends to keep moving.

In the real world

In space there is almost no friction or air resistance, so probes like Voyager 1 have kept moving for over 45 years without engines.

4.3 Applying the first law

Use the first law in reverse: if an object is at rest or moving at constant velocity, the forces on it must be balanced, so unknown forces can be found.

WORKED EXAMPLE 9 Steady parachutist (standard)

A parachutist of mass 80 kg falls at constant speed. Find the air resistance.

Solution

Constant velocity: forces balanceM1

Air resistance = W = 80 × 9.8 = 784 NA1

WORKED EXAMPLE 10 Lorry (exam level)

A lorry travels at a constant 25 m s⁻¹. Drag is 3200 N and rolling friction 800 N. Find the driving force.

Solution

Driving force = total resistanceM1

= 4000 NA1

5 Newton's second law

Resultant force, mass and acceleration.

5.1 F = ma

Newton's second law: the acceleration of an object is proportional to the resultant force on it and inversely proportional to its mass, and is in the direction of the resultant force. F = ma, where F is the resultant force in N, m the mass in kg and a the acceleration in m s⁻². One newton is the force that gives a 1 kg mass an acceleration of 1 m s⁻².

WORKED EXAMPLE 11 Acceleration (standard)

A resultant force of 600 N acts on a 1200 kg car. Find its acceleration.

Solution

a = F/m = 600/1200M1

= 0.50 m s⁻²A1

WORKED EXAMPLE 12 Force needed (standard)

Find the resultant force needed to accelerate a 70 kg sprinter at 4.0 m s⁻².

Solution

F = 70 × 4.0M1

= 280 NA1

Common mistake

Using one of the forces instead of the resultant force in F = ma. Always find the resultant first.

5.2 Investigating F = ma

A trolley on a level runway is pulled by a string over a pulley to a hanging mass. Light gates measure its acceleration. To vary force at constant total mass, move masses from the trolley to the hanger; to vary mass at constant force, add masses to the trolley.

Variable	Force investigation	Mass investigation
independent	resultant force	mass of trolley
dependent	acceleration	acceleration
controlled	total mass	pulling force

Figure 11. Apparatus for investigating force, mass and acceleration.

Try this

Use a phone accelerometer app on a trolley pulled by different numbers of identical elastic bands stretched by the same amount.

5.3 Acceleration and force: results

With the total mass fixed, a graph of acceleration against resultant force is a straight line through the origin: a ∝ F. The gradient is 1/m.

Figure 12. Acceleration against resultant force for a total mass of 0.50 kg.

WORKED EXAMPLE 13 Mass from gradient (exam level)

The gradient of the line is 2.0 m s⁻² N⁻¹. Find the mass being accelerated.

Solution

gradient = 1/mM1

m = 0.50 kgA1

5.4 Acceleration and mass: results

With the force fixed, doubling the mass halves the acceleration: a ∝ 1/m. A graph of a against m is a curve; a graph of a against 1/m is a straight line.

Figure 13. Acceleration against mass for a constant 1.0 N resultant force.

Quick check

1. What graph would give a straight line for the mass investigation?

Answers: 1. acceleration against 1/mass

5.5 Mass, inertia and braking

The same braking force gives a heavier vehicle a smaller deceleration, so it needs a longer distance to stop. This is why lorries keep larger gaps.

Figure 14. Deceleration produced by a 6000 N braking force against car mass.

WORKED EXAMPLE 14 Two cars (standard)

A 6000 N braking force acts on a 1000 kg car and a 2000 kg car. Compare the decelerations.

Solution

6.0 m s⁻² and 3.0 m s⁻²A1

Heavier car decelerates half as muchA1

5.6 Multi-force problems

To use F = ma with several forces: draw a free-body diagram, find the resultant along the direction of motion, then apply F = ma.

WORKED EXAMPLE 15 Car accelerating (exam level)

A 900 kg car has a driving force of 3200 N and total resistance 500 N. Find its acceleration and the speed after 6.0 s from rest.

Solution

Resultant = 3200 − 500 = 2700 N; a = 2700/900 = 3.0 m s⁻²M1 A1

v = at = 3.0 × 6.0 = 18 m s⁻¹A1

WORKED EXAMPLE 16 Rocket launch (exam level)

A 2.0 × 10⁵ kg rocket has engine thrust 3.0 × 10⁶ N. Find its initial acceleration.

Solution

Resultant = 3.0 × 10⁶ − 2.0 × 10⁵ × 9.8 = 1.04 × 10⁶ NM1

a = 1.04 × 10⁶/2.0 × 10⁵ = 5.2 m s⁻²A1

6 Newton's third law

Forces come in pairs.

Think about it

If you push on a wall, why don't you move through it, and why might you roll backwards on a skateboard?

6.1 Statement

Newton's third law: when object A exerts a force on object B, object B exerts a force on object A that is equal in size and opposite in direction. The two forces are of the same type and act on different objects, so they never cancel.

Figure 15. Newton's third-law pairs.

Key point

Third-law pairs: same size, opposite direction, same type of force, act on DIFFERENT objects.

Quick check

1. A hammer hits a nail with 500 N. What force does the nail exert on the hammer?

Answers: 1. 500 N in the opposite direction

6.2 Third-law pairs vs balanced forces

A book resting on a table has two balanced forces on it: its weight (Earth pulls book) and the normal force (table pushes book). These are NOT a third-law pair, because both act on the book and are different types of force. The weight's partner is the book pulling the Earth upwards; the normal force's partner is the book pushing down on the table.

Force on book	Third-law partner
Earth pulls book down (gravity)	book pulls Earth up (gravity)
table pushes book up (contact)	book pushes table down (contact)

WORKED EXAMPLE 17 Identify the pair (exam level)

A person stands on a bathroom scale. State the third-law partner of the force of the scale on the person.

Solution

The person pushes down on the scaleR1

Equal size, opposite direction, both contact forcesR1

Common mistake

Calling weight and normal force a third-law pair. They act on the same object, so they are an example of balanced forces, not a pair.

6.3 Why the effects differ

Although third-law forces are equal, their effects can be very different because the objects have different masses. When an apple falls, the Earth accelerates towards it too, but with an immeasurably small acceleration.

WORKED EXAMPLE 18 Earth's acceleration (exam level)

A 0.10 kg apple falls. Find the Earth's acceleration towards the apple (Earth's mass 6.0 × 10²⁴ kg).

Solution

Force on Earth = 0.98 NM1

a = 0.98/6.0 × 10²⁴ = 1.6 × 10⁻²⁵ m s⁻²A1

WORKED EXAMPLE 19 Skaters (standard)

A 60 kg skater pushes a 40 kg skater with 120 N. Find each acceleration.

Solution

60 kg: 120/60 = 2.0 m s⁻²A1

40 kg: 120/40 = 3.0 m s⁻² in the opposite directionA1

6.4 Walking, swimming and jet propulsion

To walk forward, your foot pushes backwards on the ground; friction from the ground pushes you forwards. Swimmers push water backwards; rockets push exhaust gas backwards. In each case the forward force on you is the third-law partner of your backward push.

Quick check

1. Why is it hard to walk on ice?

Answers: 1. little friction: the ground cannot push you forward strongly

7 Newton's laws in action

Lifts, parachutes and vehicles.

7.1 Lifts

In a lift, the cable tension T acts up and weight W acts down. When the lift accelerates upwards, T > W; at constant speed T = W; when it slows while moving up, T < W. You feel heavier or lighter because the floor's normal force on you changes in the same way.

Figure 16. Forces on a lift.

Figure 17. Velocity–time graph for a lift travelling upwards.

WORKED EXAMPLE 20 Accelerating lift (exam level)

A lift and passengers have total mass 800 kg and accelerate upwards at 1.0 m s⁻². Find the cable tension.

Solution

T − mg = ma: T = 800(9.8 + 1.0)M1

= 8640 NA1

WORKED EXAMPLE 21 Slowing lift (exam level)

Find the tension as the same lift slows at 1.0 m s⁻² while moving up.

Solution

T = 800(9.8 − 1.0)M1

= 7040 NA1

7.2 Falling objects and parachutes

A skydiver first accelerates at about 9.8 m s⁻². As speed rises, air resistance grows, so the resultant force and the acceleration fall. When air resistance equals weight, the skydiver reaches terminal velocity. Opening a parachute suddenly increases air resistance: the skydiver decelerates to a new, much lower terminal velocity.

Figure 18. Speed of a skydiver before and after the parachute opens.

WORKED EXAMPLE 22 Acceleration during the fall (exam level)

An 80 kg skydiver has 500 N of air resistance. Find the acceleration.

Solution

Resultant = 784 − 500 = 284 NM1

a = 284/80 = 3.6 m s⁻² downwardsA1

7.3 Free fall: all objects accelerate equally

Without air resistance every object falls with the same acceleration g, whatever its mass. A heavier object has more weight (more force) but also more mass (more inertia), and a = W/m = mg/m = g. On the Moon, where there is no air, a hammer and a feather dropped by an Apollo 15 astronaut landed together.

Figure 19. Speed of an object falling freely on Earth and on the Moon.

WORKED EXAMPLE 23 Speed after falling (standard)

A stone falls freely from rest for 2.5 s. Find its speed.

Solution

v = gt = 9.8 × 2.5M1

= 24.5 m s⁻¹A1

7.4 Forces on a slope

On a slope, weight still acts vertically downwards, but the normal force is perpendicular to the slope and friction acts along it. Part of the weight pulls the object down the slope; if friction is large enough to balance it, the object stays still.

Figure 20. Forces on a block resting on a slope.

Quick check

1. Why can a car park on a gentle hill but slide on a steep icy one?

Answers: 1. the downhill part of the weight grows with steepness; ice gives too little friction to balance it

7.5 Newton's laws in sport

When a footballer kicks a ball, a large force acts for only a few milliseconds and the ball accelerates enormously (second law). The ball pushes back on the foot just as hard (third law). Following through keeps the foot in contact longer, so the ball leaves faster.

Figure 21. Force on a football during a kick (illustrative).

WORKED EXAMPLE 24 Kick (exam level)

A 0.45 kg ball receives an average force of 450 N for 8.0 ms. Find its acceleration and launch speed from rest.

Solution

a = 450/0.45 = 1000 m s⁻²A1

v = at = 1000 × 0.0080 = 8.0 m s⁻¹A1

7.6 Vehicle safety

Crumple zones, airbags and seatbelts all increase the time over which a passenger is brought to rest. A longer stopping time means a smaller deceleration, so a smaller force (F = ma).

WORKED EXAMPLE 25 Airbag (exam level)

A 70 kg passenger travelling at 15 m s⁻¹ is stopped in 0.10 s by a seatbelt, or 0.30 s with an airbag as well. Compare the average forces.

Solution

a = 150 m s⁻² vs 50 m s⁻²M1

F = 10 500 N vs 3500 N: three times smallerA1

In the real world

Formula 1 cars have carbon-fibre survival cells and crushable nose cones that extend collision times, letting drivers survive crashes at over 300 km h⁻¹.

8 Data-based questions and further practice

Applying the laws.

8.1 Data: trolley experiment

A student pulls a trolley of total mass 1.5 kg with different forces and records its acceleration.

Force / N	0.5	1.0	1.5	2.0	2.5
Acceleration / m s⁻²	0.32	0.68	0.99	1.31	1.70

Figure 22. Results of the trolley experiment with the line a = F/1.5.

WORKED EXAMPLE 26 Evaluating (exam level)

Use the graph to evaluate whether the results support F = ma.

Solution

Points lie close to a straight line through the originR1

Gradient ≈ 0.67 = 1/1.5, matching the massR1

WORKED EXAMPLE 27 Friction (exam level)

Suggest why some points lie slightly below the line.

Solution

Friction opposes motionR1

so the resultant force is a little less than the applied forceR1

8.2 Graded practice: forces

WORKED EXAMPLE 28 Resultant (basic)

Forces of 12 N left and 20 N right act on a box. Find the resultant.

Solution

20 − 12M1

= 8 N rightA1

WORKED EXAMPLE 29 Weight on Mars (standard)

A rover has mass 900 kg. Find its weight on Mars (g = 3.7 N kg⁻¹).

Solution

W = 900 × 3.7M1

= 3330 NA1

WORKED EXAMPLE 30 Three forces (exam level)

A sledge is pulled with 150 N forwards; friction is 40 N and air resistance 10 N. Its mass is 50 kg. Find its acceleration.

Solution

Resultant = 100 NM1

a = 2.0 m s⁻²A1

8.3 Graded practice: Newton's laws

WORKED EXAMPLE 31 Time to stop (exam level)

A 1500 kg car at 20 m s⁻¹ brakes with a 7500 N force. Find the deceleration and stopping time.

Solution

a = 7500/1500 = 5.0 m s⁻²A1

t = 20/5.0 = 4.0 sA1

WORKED EXAMPLE 32 Unknown mass (standard)

A 24 N resultant force gives an object an acceleration of 3.0 m s⁻². Find its mass.

Solution

m = F/aM1

= 8.0 kgA1

WORKED EXAMPLE 33 Third law (standard)

A gun fires a bullet with a force of 2000 N. State the force on the gun and explain why the gun's acceleration is much smaller.

Solution

2000 N backwards on the gunA1

Gun's mass is far larger, so a = F/m is much smallerR1

8.4 Exam technique

•Always draw a free-body diagram before using F = ma.
•Find the resultant force first, then apply F = ma.
•Use W = mg with g = 9.8 N kg⁻¹ unless told otherwise; weight is in N, mass in kg.
•For third-law questions, name both objects and both directions: 'A pushes B ... B pushes A'.
•In 'explain' questions about motion, link the balance of forces to the change (or no change) in velocity.

8.5 Multi-step problem: lift and passenger

WORKED EXAMPLE 34 Bathroom scale in a lift (exam level)

A 50 kg student stands on scales in a lift. Find the scale reading when the lift (a) is at rest; (b) accelerates upwards at 2.0 m s⁻²; (c) accelerates downwards at 2.0 m s⁻².

Solution

(a) N = mg = 490 NA1

(b) N = m(g + a) = 590 NM1 A1

8.6 Key terms

Term	Meaning
force	a push or pull resulting from an interaction (N)
resultant force	single force with the same effect as all forces acting
weight	gravitational force on an object, W = mg
mass	amount of matter (kg); same everywhere
inertia	tendency of an object to resist changes in motion
free-body diagram	diagram of all forces acting on one object
Newton's third law pair	equal, opposite forces of the same type on two different objects

8.7 Self-test

Try these without notes, then check the answers.

Quick check

1. State Newton's first law.

2. Find the acceleration when 30 N acts on 6.0 kg.

3. Find the weight of 2.5 kg on Earth.

4. Name two non-contact forces.

5. Why do weight and normal force on a book not form a third-law pair?

6. What happens to a skydiver's acceleration as speed increases?

Answers: 1. an object stays at rest or moves at constant velocity unless a resultant force acts 2. 5.0 m s⁻² 3. 24.5 N 4. gravitational, magnetic, electrostatic 5. they act on the same object and are different types of force 6. it decreases (air resistance grows)

8.8 Common exam errors

•Giving weight in kilograms.
•Using a single force instead of the resultant in F = ma.
•Drawing forces exerted by the object on a free-body diagram.
•Pairing weight with normal force as a third-law pair.
•Saying a moving object needs a forward force to keep moving at constant velocity.

9 Common misconceptions and links

Students often think…	Actually…
moving objects need a force to keep moving	constant velocity needs zero resultant force
weight is measured in kg	weight is a force in newtons
weight and normal force are a third-law pair	they act on the same object
heavier objects fall faster	without air resistance all objects fall with the same acceleration
third-law forces cancel out	they act on different objects, so they cannot cancel

How 1.1 connects to the rest of the course

Topic	Connection
1.2 Balanced vs unbalanced forces	resultant force and motion
1.3 Mass, weight and gravity	W = mg
1.4 Friction and terminal velocity	air resistance
2.2 Work, power and efficiency	work done by forces

10 Exam-style question with markscheme

Try this under exam conditions (about 20 minutes) before reading the markscheme.

A car of mass 1200 kg accelerates from rest along a straight, level road. The driving force is 3000 N and the total resistive force is 600 N.

(a) Draw a free-body diagram showing the horizontal forces on the car. [2] (Criterion A)

(b) Calculate the resultant force on the car. [1] (Criterion A)

(d) Explain, using Newton's laws, why the car's acceleration decreases as its speed increases. [3] (Criterion A)

(e) State the third-law partner of the force of the road on the car's tyres. [1] (Criterion A)

(f) Suggest one way a car designer could increase the car's acceleration and evaluate a disadvantage of your change. [2] (Criterion D)

Markscheme

Part	Answer	Marks
(a)	driving force forwards and resistance backwards, labelled	A1
	forward arrow longer	A1
(b)	3000 − 600 = 2400 N	A1
(c)	a = F/m = 2400/1200	M1
	2.0 m s⁻²	A1
(d)	air resistance increases with speed	R1
	resultant force decreases	R1
	a = F/m so acceleration decreases	R1
(e)	the tyres push backwards on the road	A1
(f)	reduce mass / increase engine force / streamline	A1
	valid disadvantage, e.g. lighter materials cost more or are less safe; bigger engine uses more fuel	R1

Examiner's comments. In (d), many answers said the driving force decreases. The driving force can stay the same: it is the growing air resistance that reduces the resultant.

11 Check your understanding

Section A: multiple choice

1. The unit of force is the

A. newton B. kilogram C. joule D. watt

2. Which is a non-contact force?

A. friction B. tension C. weight D. normal force

3. A 4.0 kg mass has a resultant force of 12 N. Its acceleration is

A. 0.33 m s⁻² B. 3.0 m s⁻² C. 16 m s⁻² D. 48 m s⁻²

4. An object moving at constant velocity has

A. a constant forward resultant force B. no forces acting on it C. increasing weight D. zero resultant force

5. Newton's third-law pairs act on

A. the same object B. different objects C. only moving objects D. only objects at rest

6. Weight on the Moon compared with Earth is

A. smaller, but mass is the same B. the same C. larger D. zero

7. Inertia depends on

A. speed B. weight only C. volume D. mass

8. Two forces of 3 N and 4 N at right angles have a resultant of

A. 1 N B. 7 N C. 5 N D. 12 N

Section B: structured

9. State Newton's second law in words and as an equation. [2]

10. Explain why passengers jerk forward when a bus brakes. [2]

11. Calculate the acceleration of a 2.0 kg trolley with 8.0 N forwards and 2.0 N friction. [2]

12. Identify the third-law partner of the weight of a person. [1]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Acceleration is proportional to resultant force and inversely proportional to mass; F = ma.

10. Passengers keep moving (inertia) until a force (seatbelt) acts on them.

11. Resultant 6.0 N; a = 3.0 m s⁻².

12. The person's gravitational pull on the Earth (upwards).

12 Summary and topic map

Idea	Remember
Force	push or pull; vector; newtons
Types	contact and non-contact
Weight	W = mg
First law	zero resultant: constant velocity
Second law	F = ma
Third law	equal and opposite, different objects

Before the exam, make sure you…

•draw a free-body diagram first
•find the resultant before F = ma
•use newtons for weight, kg for mass
•name both objects in third-law pairs

Types of Forces and Newton's Laws on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 23. Topic map for Types of Forces and Newton's Laws.

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