# Newton's laws of motion

> CIE A-Level Mathematics · 9709 A-Level Maths (Mechanics)
> Source: https://www.owlsprep.com/study/cie-9709-u3-newton-s-laws-of-motion/

This module covers Newton's three laws of motion, the core relationship $F_{net}=ma$, and applications to 1D and 2D dynamic problems common in CIE A-Level Mechanics 1. You will learn to set up correct force equations for accelerating bodies.

**Prerequisites:** [Forces and force resolution](https://www.owlsprep.com/study/cie-9709-u3-forces-and-equilibrium/); [1D Kinematics](https://www.owlsprep.com/study/cie-9709-u3-kinematics-1d/)

## Learning objectives

- State and apply Newton's three laws of motion correctly
- Use $F=ma$ to solve 1D and 2D dynamic problems
- Resolve forces on inclined planes for accelerating bodies
- Solve common exam problems involving connected particles and lifts

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 M1 past papers (2016–2025): **Newton's laws of motion** appears **96 times** in the last 10 years — **21.4%** of all M1 questions (96 of 443).

Most-tested forms: Connected particles (66), Newton's second law (F=ma) (32).

## Newton's Three Laws of Motion

**Newton's Three Laws of Motion** — 1. First Law: A body remains at rest or constant velocity unless acted on by a resultant external force. 2. Second Law: For constant mass, resultant force equals $\vec{F} = m\vec{a}$. 3. Third Law: If A exerts a force on B, B exerts an equal, opposite force of the same type on A.

*Example:* A book on a table: weight and normal reaction are not a third law pair. The third law pair is Earth pulling the book, and the book pulling Earth.

> **warning**
>
> Newton's third law pairs always act on different bodies, never the same body.

**Check your understanding**

Test your understanding:

1. A tennis ball is hit by a racket. Which is a Newton's third law pair?

   - Racket force on ball, ball force on racket
   - Weight of ball, air resistance on ball
   - Racket force on ball, weight of ball

   *Answer:* Racket force on ball, ball force on racket

   *Why:* Correct! These act on different bodies, are equal/opposite, and same contact force type.

## Newton's Second Law in 1 Dimension

For straight line motion, resolve all forces along the direction of motion, take the direction of acceleration as positive, then apply $F_{net}=ma$ to find unknown acceleration or force.

**Worked example:** A 5 kg box is pulled along a smooth horizontal surface by a horizontal rope with tension 20 N. Find the acceleration of the box.

1. No friction for a smooth surface, so only horizontal force is tension:
2. $$F_{net} = 20 \, \text{N}$$
3. Apply Newton's second law:
4. $$20 = 5a \implies a = 4 \, \text{m s}^{-2}$$

**Worked example:** A 1000 kg car moving at 20 m s⁻¹ brakes to rest over 50 m. Find the constant braking force.

1. First find acceleration with kinematics $v^2 = u^2 + 2as$:
2. $$0^2 = 20^2 + 2a(50) \implies a = -4 \, \text{m s}^{-2}$$
3. Negative acceleration means it opposes motion. Resultant force is the braking force $F$:
4. $$-F = ma = 1000(-4) \implies F = 4000 \, \text{N}$$

> **Exam tip:** Always draw a labelled force diagram before writing your force equation. Stick to your sign convention consistently.

## Newton's Second Law in 2 Dimensions

For motion on inclined planes, resolve forces into two perpendicular directions: one parallel to acceleration (along the plane) and one perpendicular, where acceleration is zero.

**Weight components on an inclined plane** — Weight $mg$ splits into two components: parallel to the plane: $mg\sin\theta$, perpendicular to the plane: $mg\cos\theta$.

*Notation:* Angle $\theta$ to the horizontal

**Worked example:** A 4 kg particle slides down a rough 25° incline, with friction coefficient 0.2. Find the acceleration.

1. Resolve perpendicular to the plane (acceleration = 0, so resultant force = 0):
2. $$R - 4g\cos 25^\circ = 0 \implies R \approx 35.5 \, \text{N}$$
3. Friction $F = \mu R = 0.2 \times 35.5 \approx 7.1 \, \text{N}$, acting up the slope:
4. Resolve parallel, take down slope as positive:
5. $$4g\sin 25^\circ - F = 4a$$
6. $$16.6 - 7.1 = 4a \implies a \approx 2.4 \, \text{m s}^{-2}$$

> **Exam tip:** Check your components: if $\theta = 0^\circ$ (flat plane), parallel component should be zero. If you get $mg$, you swapped sin and cos.

## Common Exam Application: Connected Particles Over a Pulley

For light inextensible strings over smooth pulleys, tension is the same on both sides, and both particles have the same magnitude of acceleration. Write separate force equations for each particle, then eliminate tension to solve.

**Worked example:** Two particles of mass 2 kg and 3 kg are connected by a light string over a smooth pulley. Find acceleration and tension.

1. Let $a$ = acceleration, $T$ = tension. 3 kg accelerates down, 2 kg accelerates up.
2. Newton's second law for 3 kg (down positive):
3. $$3g - T = 3a$$
4. Newton's second law for 2 kg (up positive):
5. $$T - 2g = 2a$$
6. Add equations to eliminate T:
7. $$g = 5a \implies a = 1.96 \, \text{m s}^{-2}$$
8. Substitute back to find T:
9. $$T = 2g + 2a = 23.52 \, \text{N}$$

> **Exam tip:** Adding the individual force equations for connected particles always eliminates tension, saving you time in exams.

## Common pitfalls

- **Wrong:** Treating action-reaction pairs as acting on the same body and cancelling them
  - Why it fails: Newton's third law pairs act on different bodies, so they never cancel for a single body
  - Correct: Only include forces acting on the body you are analysing in your resultant force calculation
- **Wrong:** Swapping sine and cosine components of weight on an inclined plane
  - Why it fails: For angles measured to the horizontal, the parallel component is $mg\sin\theta$, not $mg\cos\theta$
  - Correct: Test for θ=0° (flat plane): if your parallel component is zero, it is correct, otherwise swap them
- **Wrong:** Using mass instead of weight (forgetting to multiply by g) in force equations
  - Why it fails: Weight is a force, $F=ma$ requires force in newtons, mass in kg
  - Correct: Always write weight as $mg$, not just $m$, when calculating resultant force
- **Wrong:** Writing $T - mg = ma$ for a mass accelerating downwards
  - Why it fails: Resultant force must point in the direction of acceleration
  - Correct: Write $mg - T = ma$ for a mass accelerating downwards to get the correct resultant force direction
- **Wrong:** Using $g = 10 \, \text{m s}^{-2}$ when not specified by the question
  - Why it fails: CIE standard requires $g=9.8 \, \text{m s}^{-2}$, wrong g leads to lost marks
  - Correct: Always use $g=9.8$ unless the question explicitly tells you to use 10

## Cheatsheet

| Concept | Key Rule/Formula |
| --- | --- |
| Newton's First Law | No resultant force → constant velocity |
| Newton's Second Law | $F_{net} = ma$ (constant mass) |
| Newton's Third Law | Equal opposite forces on different bodies |
| Weight on incline (θ to horizontal) | Parallel: $mg\sin\theta$, Perpendicular: $mg\cos\theta$ |
| Connected particles over pulley | Add individual equations to eliminate tension |
| CIE required g value | $g = 9.8 \, \text{m s}^{-2}$ |

## What's next

Newton's laws of motion are the foundation of all A-Level mechanics, and mastery of this topic is critical for all further mechanics content. The core principle $F_{net}=ma$ is used in every dynamic problem you will solve, from friction to circular motion to work and energy. After this, you can move on to more complex applications of Newton's laws, including problems involving friction, more complex connected particle systems, and circular motion. Practise setting up force diagrams correctly to avoid common mistakes and pick up full marks in exams.

- [Energy, work and power](https://www.owlsprep.com/study/cie-9709-u3-energy-work-and-power/)
- [Momentum](https://www.owlsprep.com/study/cie-9709-u3-momentum/)
- [Probability & Statistics](https://www.owlsprep.com/study/cie-9709-u4-overview/)

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