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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

LearnYMA01WME01 · Mechanics 1Dynamics of a particle in a straight lineNewton's laws and F = ma

Dynamics of a particle in a straight line · Subtopic 1 of 4

Newton's Laws and F = ma

Add up the forces in the direction of motion to get the resultant, then $F=ma$ gives the acceleration. If the resultant is zero, the object keeps a constant velocity. In lifts, the floor's reaction is more than the weight when accelerating upwards and less when accelerating downwards.

Edexcel YMA01 WME01≈ 35 minRevision 5 · updated 2026-10-27

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Section 1 of 14

Learning objectives

State and use Newton's three laws.
Find a resultant force and use $F=ma$ .
Solve lift and vertical-string problems.
Combine $F=ma$ with the suvat formulae.

Pause and think

Standing on bathroom scales in a lift, when would the reading go up, and when down?

Section 2 of 14

Core idea

Start herePulling a box at an angle

Pull a box with a rope held at an angle. Part of your pull drags it forward; part lifts it, so the floor pushes up less. Splitting forces into components is the key move in every Newton's-law problem.

The big idea

$F=ma$ along the motion; forces balance perpendicular to it. Resolve angled forces first. For connected bodies, use the whole system for $a$ and one part for internal forces.

1. Resolving

A 2 kg box pulled by 15 N at 30°, with 5 N resistance

Horizontal:

15\cos30^\circ-5=2a

, so

a=3.99

m s

^{-2}

. Vertical:

R+15\sin30^\circ=19.6

, so

R=12.1

Force problem method

Drawing diagram…

The same routine solves almost every force question.

F_{\text{resultant}}=ma

$F_{\text{resultant}}$: forward forces − backward forces (N).
$m$: mass (kg).
$a$: acceleration (m s⁻²).

Worked exampleBox pulled at 30°

2 kg box, 15 N pull at $30^\circ$ above horizontal, 5 N resistance. Acceleration and normal reaction?

Goal: $a$ , $R$ .
Strategy: Resolve horizontally ( $F=ma$ ) and vertically (balance).

$15\cos30^\circ-5=2a$
Horizontal component $12.99$ .
$a=3.99\text{ m s}^{-2}$
$7.99/2$ .
$R+15\sin30^\circ=2(9.8)\Rightarrow R=12.1\text{ N}$
Vertical: the pull lifts by $7.5$ N.

Answer

a=3.99

m s

^{-2}

R=12.1

✓ Check

R<mg

because the rope lifts ✓.

Worked exampleCar and trailer

Car 1000 kg, trailer 300 kg; driving force 2500 N; resistances 400 N (car), 200 N (trailer). Find $a$ and the tow-bar tension.

Goal: $a$ , $T$ .
Strategy: System first, then trailer alone.

$2500-600=1300a\Rightarrow a=1.46$
System: tension is internal.
$T-200=300(1.46)\Rightarrow T=638$
Trailer only.

Answer

1.46

m s

^{-2}

;

638

✓ CheckCar:

2500-400-638=1462\approx1000(1.46)

✓.

Common misconceptionR = mg always Only when no other vertical forces act and there's no vertical acceleration.

Pause and think

A 3 kg mass is lifted with $a=2$ upward. Tension?

Summary

Resolve, then $F=ma$ .
System for $a$ , one part for internal forces.

Resultant force: Net force producing acceleration.

HistoryThe Principia. Isaac Newton published his three laws in Philosophiæ Naturalis Principia Mathematica in 1687. Edmond Halley (of comet fame) paid for the printing himself because the Royal Society had spent its budget on a book about fish.

Fun factWeightless in a falling lift. If a lift accelerated downwards at exactly

g

, the reaction would be

m(g-g)=0

: you would float. Astronauts train for weightlessness in aircraft that fly curved paths so they fall freely for about

20

seconds at a time.

Section 3 of 14

Key formulae

Newton's second law

F_{\text{resultant}}=ma

Weight

W=mg

Lift (up positive)

R-mg=ma

TrickResultant in the direction of a. Always write the equation as "(forces in the direction of

a

)

-

(forces against it)

=ma

". It stops sign errors, especially in lifts and pulleys.

Section 4 of 14

Worked examples

Example 1 · Car with resistance0 / 2 steps

1200

kg car has driving force

3000

N and resistance

600

N. Find its acceleration.

Example 2 · Lift going down0 / 2 steps

60

kg person stands in a lift accelerating downwards at

1.8

m s⁻². Find the reaction from the floor.

Section 5 of 14

Explore the graph

Reaction in a lift

Loading the interactive graph…

Try this

The line shows the floor's reaction $R=m(g+a)$ against the lift's upward acceleration $a$ ; the horizontal line is the weight.
Where does the line meet the weight line? What is the lift doing there?
Follow the line to $a=-9.8$ . What is the reaction, and how would that feel?

$R=m(g+a)$ : heavier going up, lighter going down.

In the real worldSeat belts. In a crash a car can decelerate at

20g

or more.

F=ma

shows the force needed to stop you is enormous; seat belts and crumple zones spread the stop over a longer time to reduce

a

, and so the force on your body.

Section 6 of 14

Common mistakes

Incorrect

F=ma

with

F

the driving force only

CorrectUse the resultant

Subtract resistances first.

Incorrect

R=mg

in an accelerating lift

Correct

R-mg=ma

Only true at constant velocity.

IncorrectUsing

g=10

Correct

g=9.8

IAL standard.

IncorrectMass in N

CorrectMass in kg

Weight is

mg

newtons.

You push a wall

The wall pushes you back with exactly the same force. Rude, but it's Newton's third law.

Section 7 of 14

Exam tip

Exam tipDraw the force diagram. A labelled diagram with every force and the direction of acceleration is often worth a mark on its own, and makes the equation almost write itself.

Section 8 of 14

Guided practice

SA-IAL-M1-DY-G012 marksL1 FoundationCalculatorF = ma

A resultant force of

12

N acts on a particle of mass

3

kg. Find its acceleration.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-IAL-M1-DY1-0011 markL1 FoundationNon-calculatorF = ma

Find the resultant force needed to give a

2

kg particle an acceleration of

3

m s⁻².

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY1-0021 markL1 FoundationNon-calculatorF = ma

A resultant force of

20

N gives a particle an acceleration of

4

m s⁻². Find its mass.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY-0012 marksL1 FoundationCalculatorF = ma

A car of mass

1200

kg has a driving force of

3000

N and experiences a resistance of

600

N. Find its acceleration.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY1-0062 marksL2 StandardNon-calculatorF = ma

A car of mass

1000

kg experiences a resistance of

400

N. Find the driving force needed for an acceleration of

1.5

m s⁻².

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY1-0041 markL1 FoundationCalculatorLifts & suvat

A person of mass

60

kg stands in a lift moving upwards at a constant

2

m s⁻¹. Find the reaction from the floor.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY-0023 marksL2 StandardCalculatorLifts & suvat

A person of mass

70

kg stands in a lift that is accelerating upwards at

1.2\ \text{m s}^{-2}

. Find the normal reaction from the floor.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY1-0033 marksL2 StandardCalculatorLifts & suvat

A person of mass

60

kg stands in a lift that is accelerating downwards at

1.8

m s⁻². Find the reaction from the floor.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY1-0052 marksL2 StandardCalculatorLifts & suvat

A ball of mass

0.5

kg is pulled vertically upwards by a light string with tension

7

N. Find its acceleration.

Salam Academy Original Practice · Proposed mark scheme

SA-IAL-M1-DY1-0073 marksL2 StandardNon-calculatorLifts & suvat

A particle of mass

2

kg starts from rest on a smooth horizontal surface and is pushed by a horizontal force of

10

N. How far does it travel in

4

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-IAL-M1-DY1-0084 marksL3 ExamNon-calculatorLifts & suvat

A car of mass

1200

kg travelling at

20

m s⁻¹ brakes and stops in

50

m. Assuming a constant braking force, find its size.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

No resultant force: constant velocity.
Resultant $=ma$ , in the direction of $a$ .
Lifts: $R-mg=ma$ .
Find $a$ , then use suvat.

Section 12 of 14

Quick quiz

Q1 · SA-IAL-M1-DY1-0091 markL1 FoundationCalculatorF = ma

A particle moves in a straight line with constant velocity. What can you say about the forces on it?

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: IAL Mathematics, unit WME01 (Mechanics 1)
Where: M1 · 4 Dynamics of a particle moving in a straight line
Expected: Use Newton's laws of motion; F = ma for motion in a straight line, including vertical motion (lifts, strings) and combined with the constant acceleration formulae.
Calculator: Calculators are permitted, but exact answers (surds, fractions, π) are expected where the question asks.
Typical style: "A lift of mass 500 kg carries a man of mass 80 kg… find the tension in the cable and the reaction on the man."
Open in: YMA01 · WME01

Section 14 of 14

Past paper questions

4 questions (35 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

WME01/01 · June 2025Question 47 marks
Two tensions on a barge: an angle, then acceleration against resistance.
Question paper, page 12 Mark scheme
WME01/01 · January 2025Question 18 marks
Acceleration from forces, then a constant that fixes the direction of the resultant.
Question paper, page 2 Mark scheme
WME01/01 · June 2024Question 36 marks
Newton’s second law for particles connected vertically.
Question paper, page 6 Mark scheme
WME01/01 · June 2024Question 514 marks
Motion under gravity then with resistance, and a speed–time graph.
Question paper, page 12 Mark scheme

All YMA01 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YMA01, 2018 specification (official specification).