A-Level Mathematics: Mechanics

Edexcel · AQA · OCR A-Level Mathematics · Updated March 2026 · 35 min read

Mechanics applies mathematics to model the physical world. At A-Level, you study the motion of particles, the forces that cause or prevent motion, moments, friction, and projectile motion. This guide covers all Mechanics topics with systematic worked examples.

Exam Board Note

Mechanics is tested in Paper 3 (Statistics and Mechanics) for Edexcel and OCR. For AQA, Mechanics appears in Paper 3 alongside a choice topic. All boards require Newton's three laws, SUVAT equations, and projectile motion. OCR additionally includes moments in Year 1; all boards include it at A-Level. Take $g = 9.8\,\text{m\,s}^{-2}$ unless otherwise stated.

1. Kinematics: Language and Definitions

Key Kinematic Quantities

Displacement $s$ (m): distance in a specified direction from a reference point. A vector quantity.

Velocity $v$ (m s$^{-1}$): rate of change of displacement. $v = \dfrac{ds}{dt}$. A vector quantity.

Speed (m s$^{-1}$): magnitude of velocity. A scalar quantity.

Acceleration $a$ (m s$^{-2}$): rate of change of velocity. $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$. A vector quantity.

Distance (m): total path length travelled. A scalar quantity.

Mechanics models treat objects as particles (no size) unless stated otherwise. This simplifies calculations by allowing all forces to act at a single point. Real-world objects are treated as particles when their size is negligible compared to the distances involved.

2. SUVAT Equations

For motion with constant acceleration $a$, the SUVAT equations relate the five kinematic variables:

Variable	Symbol	Unit
Initial velocity	$u$	m s$^{-1}$
Final velocity	$v$	m s$^{-1}$
Acceleration	$a$	m s$^{-2}$
Displacement	$s$	m
Time	$t$	s

The SUVAT Equations

$$v = u + at$$

$$s = ut + \tfrac{1}{2}at^2$$

$$s = vt - \tfrac{1}{2}at^2$$

$$v^2 = u^2 + 2as$$

$$s = \tfrac{1}{2}(u+v)t$$

Choose the equation that contains the three known quantities and the one unknown. Always define a positive direction.

Worked Example 2.1 — Multi-Stage SUVAT Problem

A car accelerates uniformly from rest to $20\,\text{m\,s}^{-1}$ in $10\,\text{s}$, then travels at constant speed for $30\,\text{s}$, then decelerates uniformly to rest in $5\,\text{s}$. Find the total distance travelled.

Stage 1 $u = 0$, $v = 20$, $t = 10$. $s_1 = \dfrac{1}{2}(0+20)(10) = 100\,\text{m}$.

Stage 2 Constant speed $20\,\text{m\,s}^{-1}$ for $30\,\text{s}$: $s_2 = 20 \times 30 = 600\,\text{m}$.

Stage 3 $u = 20$, $v = 0$, $t = 5$: $s_3 = \dfrac{1}{2}(20+0)(5) = 50\,\text{m}$.

Total $100 + 600 + 50 = \textbf{750}\,\text{m}$.

Worked Example 2.2 — Vertical Motion Under Gravity

A stone is thrown vertically upward from the top of a cliff 45 m high, with initial speed $15\,\text{m\,s}^{-1}$. Taking upward as positive and $g = 9.8\,\text{m\,s}^{-2}$, find:

(a) the maximum height above the base of the cliff, (b) the time taken to hit the sea at the base.

(a) At maximum height, $v = 0$. $u = 15$, $a = -9.8$.

$v^2 = u^2 + 2as \Rightarrow 0 = 225 - 19.6s \Rightarrow s = \dfrac{225}{19.6} \approx 11.5\,\text{m}$ above the top of the cliff.

Maximum height above base $= 45 + 11.5 = 56.5\,\text{m}$.

(b) Taking the top of the cliff as origin ($s = 0$), the base is at $s = -45\,\text{m}$.

$-45 = 15t - \frac{1}{2}(9.8)t^2 = 15t - 4.9t^2$

$4.9t^2 - 15t - 45 = 0$

$t = \dfrac{15 \pm \sqrt{225 + 4 \times 4.9 \times 45}}{2 \times 4.9} = \dfrac{15 \pm \sqrt{225 + 882}}{9.8} = \dfrac{15 \pm \sqrt{1107}}{9.8}$

$t = \dfrac{15 + 33.27}{9.8} \approx \dfrac{48.27}{9.8} \approx 4.93\,\text{s}$ (taking the positive root).

3. Velocity-Time Graphs

A velocity-time graph encodes the complete kinematics of a particle:

Gradient at any point $=$ acceleration at that instant.
Area between graph and time-axis $=$ displacement (area below the axis is negative).
Constant velocity: horizontal line. Constant acceleration: straight line with non-zero gradient.

Worked Example 3.1 — Velocity-Time Graph

A particle's velocity (in m s$^{-1}$) is described by the following $v$-$t$ graph: it starts at rest, increases linearly to $12\,\text{m\,s}^{-1}$ at $t = 4\,\text{s}$, stays constant until $t = 10\,\text{s}$, then decreases linearly to $-4\,\text{m\,s}^{-1}$ at $t = 14\,\text{s}$. Find the total displacement and total distance travelled.

Segment 1 (0 to 4 s): Triangle, area $= \frac{1}{2}(4)(12) = 24\,\text{m}$.

Segment 2 (4 to 10 s): Rectangle, area $= 6 \times 12 = 72\,\text{m}$.

Segment 3 (10 to 14 s): The velocity crosses zero at some time $t^*$. At $t=10$: $v=12$; at $t=14$: $v=-4$. The zero crossing: $12 - 4(t^*-10) = 0 \Rightarrow t^* = 13\,\text{s}$.

Area above axis (10 to 13 s): $\frac{1}{2}(3)(12) = 18\,\text{m}$. Area below axis (13 to 14 s): $\frac{1}{2}(1)(4) = 2\,\text{m}$ (negative displacement).

Total displacement $= 24 + 72 + 18 - 2 = 112\,\text{m}$.

Total distance $= 24 + 72 + 18 + 2 = 116\,\text{m}$.

MECH·1 — Velocity-time graph for Worked Example 3.1. Green shaded areas are positive displacement (above axis); red area is negative displacement (below axis). Gradient of each segment gives acceleration; total area gives displacement.

4. Forces and Newton's Laws

Newton's Three Laws of Motion

First Law: A body remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force.

Second Law: The net force on a body equals the product of its mass and acceleration: $\mathbf{F} = m\mathbf{a}$. In SI units: force in newtons (N), mass in kilograms (kg), acceleration in m s$^{-2}$.

Third Law: For every action there is an equal and opposite reaction. If body A exerts a force on body B, then B exerts an equal and opposite force on A.

Common forces to include in diagrams:

Weight $W = mg$ acting downward.
Normal reaction $R$ acting perpendicular to the surface.
Tension $T$ in a string (always pulls, never pushes).
Friction $F$ opposing motion or tendency of motion.
Applied force $P$ from an engine or push.

Worked Example 4.1 — Connected Particles (Atwood's Machine)

Two particles of mass $3\,\text{kg}$ and $5\,\text{kg}$ are connected by a light inextensible string over a smooth pulley. Find the acceleration of the system and the tension in the string. Take $g = 9.8\,\text{m\,s}^{-2}$.

Step 1 The heavier mass (5 kg) accelerates downward; the lighter (3 kg) accelerates upward. Let acceleration $= a$ and tension $= T$. Take positive in the direction of motion for each particle.

Step 2 Apply $F = ma$ to each particle:

5 kg particle: $5g - T = 5a$ (i)

3 kg particle: $T - 3g = 3a$ (ii)

Step 3 Add (i) and (ii): $2g = 8a$, so $a = \dfrac{g}{4} = \dfrac{9.8}{4} = 2.45\,\text{m\,s}^{-2}$.

Step 4 From (ii): $T = 3g + 3a = 3(9.8) + 3(2.45) = 29.4 + 7.35 = 36.75\,\text{N}$.

Worked Example 4.2 — Particle on an Inclined Plane

A particle of mass $4\,\text{kg}$ rests on a smooth plane inclined at $30°$ to the horizontal. A force $P$ acts up the plane. Find $P$ and the normal reaction $R$.

Step 1 Resolve parallel to the plane (taking up as positive):

$P - mg\sin 30° = 0$ (particle is in equilibrium)

$P = 4 \times 9.8 \times 0.5 = 19.6\,\text{N}$

Step 2 Resolve perpendicular to the plane:

$R = mg\cos 30° = 4 \times 9.8 \times \dfrac{\sqrt{3}}{2} = 33.9\,\text{N}$

5. Moments

Moment of a Force

The moment of a force $F$ about a point $O$ is:

$$M = F \times d$$

where $d$ is the perpendicular distance from $O$ to the line of action of $F$. Units: N m.

Principle of Moments: For a body in equilibrium, the sum of clockwise moments equals the sum of anti-clockwise moments about any point.

Worked Example 5.1 — Uniform Beam in Equilibrium

A uniform beam $AB$ of mass $20\,\text{kg}$ and length $6\,\text{m}$ is supported at $A$ and by a support at point $C$, where $AC = 4\,\text{m}$. A load of $30\,\text{kg}$ hangs from $B$. Find the reactions at $A$ and $C$.

Step 1 Forces: Reaction $R_A$ at $A$ (upward), reaction $R_C$ at $C$ (upward), weight of beam $20g$ at midpoint of $AB$ (3 m from $A$), weight $30g$ at $B$ (6 m from $A$).

Step 2 Take moments about $A$ (eliminates $R_A$):

$R_C \times 4 = 20g \times 3 + 30g \times 6$

$4R_C = 60g + 180g = 240g$

$R_C = 60g = 60 \times 9.8 = 588\,\text{N}$

Step 3 Resolve vertically: $R_A + R_C = 20g + 30g = 50 \times 9.8 = 490\,\text{N}$.

$R_A = 490 - 588 = -98\,\text{N}$.

The negative sign means $R_A$ acts downward (the support at $A$ must pull the beam down, i.e. $A$ is a fixed pivot).

6. Friction

Laws of Friction

Friction acts in the direction opposing motion (or tendency of motion).

Limiting friction: The maximum friction force $F_{\max} = \mu R$, where $\mu$ is the coefficient of friction and $R$ is the normal reaction.

When an object is moving: $F = \mu R$ (kinetic friction).

When an object is on the point of moving (limiting equilibrium): $F = \mu R$.

When an object is stationary and not on the point of moving: $F < \mu R$.

Worked Example 6.1 — Block on a Rough Horizontal Surface

A block of mass $8\,\text{kg}$ is pulled along a rough horizontal surface by a horizontal force $P = 40\,\text{N}$. If $\mu = 0.4$, find the acceleration.

Step 1 Resolve vertically: $R = mg = 8 \times 9.8 = 78.4\,\text{N}$.

Step 2 Friction force: $F = \mu R = 0.4 \times 78.4 = 31.36\,\text{N}$ (opposing motion).

Step 3 Apply $F = ma$ horizontally: $40 - 31.36 = 8a$, so $a = \dfrac{8.64}{8} = 1.08\,\text{m\,s}^{-2}$.

Worked Example 6.2 — Finding the Coefficient of Friction

A particle of mass $5\,\text{kg}$ is pushed up a rough plane inclined at $25°$ to the horizontal by a force of $40\,\text{N}$ acting up the plane. The particle moves at constant speed. Find $\mu$.

Step 1 Constant speed means no acceleration; the net force is zero.

Step 2 Resolve perpendicular: $R = mg\cos 25° = 5 \times 9.8 \times 0.9063 = 44.41\,\text{N}$.

Step 3 Friction acts down the plane (opposing upward motion). Resolve parallel to plane:

$40 - mg\sin 25° - F = 0$

$F = 40 - 5 \times 9.8 \times 0.4226 = 40 - 20.71 = 19.29\,\text{N}$

Step 4 $\mu = \dfrac{F}{R} = \dfrac{19.29}{44.41} \approx 0.43$.

7. Projectile Motion

A projectile experiences constant acceleration $g = 9.8\,\text{m\,s}^{-2}$ downward and zero acceleration horizontally (assuming no air resistance). The horizontal and vertical components of motion are independent.

Projectile Equations

For initial speed $u$ at angle $\alpha$ to the horizontal:

Horizontal: $x = u\cos\alpha \cdot t$ (uniform motion)

Vertical: $y = u\sin\alpha \cdot t - \dfrac{1}{2}gt^2$ (taking upward positive)

Vertical velocity: $v_y = u\sin\alpha - gt$

Speed at time $t$: $|v| = \sqrt{v_x^2 + v_y^2}$ where $v_x = u\cos\alpha$ (constant).

MECH·2 — Projectile trajectory for Worked Example 7.1: initial speed $20\,\text{m\,s}^{-1}$ at $35°$. The parabolic path has Cartesian equation $y = x\tan 35° - \tfrac{g}{2u^2\cos^2\!35°}\,x^2$. Maximum height $\approx 6.71\,\text{m}$ at $x \approx 19.2\,\text{m}$; range $\approx 38.3\,\text{m}$.

Worked Example 7.1 — Full Projectile Analysis

A ball is kicked from the ground at $20\,\text{m\,s}^{-1}$ at $35°$ above the horizontal. Find:

(a) the maximum height, (b) the time of flight, (c) the horizontal range, (d) the speed and direction of motion when it hits the ground.

Initial components: $v_x = 20\cos 35° = 16.38\,\text{m\,s}^{-1}$, $\quad v_y = 20\sin 35° = 11.47\,\text{m\,s}^{-1}$.

(a) Maximum height: At max height, $v_y = 0$. $v_y^2 = u_y^2 - 2gs \Rightarrow s = \dfrac{u_y^2}{2g} = \dfrac{11.47^2}{19.6} = \dfrac{131.6}{19.6} \approx 6.71\,\text{m}$.

(b) Time of flight: Ball returns to $y = 0$. $0 = u_y t - \frac{1}{2}gt^2 = t(u_y - \frac{1}{2}gt)$.

$t = 0$ (launch) or $t = \dfrac{2u_y}{g} = \dfrac{2 \times 11.47}{9.8} \approx 2.34\,\text{s}$.

(d) Speed on landing: By symmetry, $v_y = -11.47\,\text{m\,s}^{-1}$ (downward). $|v| = \sqrt{16.38^2 + 11.47^2} = \sqrt{268.3 + 131.6} = \sqrt{399.9} \approx 20\,\text{m\,s}^{-1}$. Direction: $\theta = \arctan\!\left(\dfrac{11.47}{16.38}\right) = 35°$ below the horizontal.

8. Practice Problems

Problem 1

A particle starts from rest and accelerates at $3\,\text{m\,s}^{-2}$ for $6\,\text{s}$. It then decelerates at $1.5\,\text{m\,s}^{-2}$ until it stops. Find the total distance travelled.

Show Solution

Phase 1: $u=0$, $a=3$, $t=6$. $v = 0 + 3(6) = 18\,\text{m\,s}^{-1}$. $s_1 = \frac{1}{2}(3)(36) = 54\,\text{m}$.

Phase 2: $u=18$, $v=0$, $a=-1.5$. $v^2 = u^2+2as \Rightarrow 0 = 324 - 3s_2 \Rightarrow s_2 = 108\,\text{m}$.

Total distance $= 54 + 108 = 162\,\text{m}$.

Problem 2

A particle of mass $3\,\text{kg}$ is on a smooth horizontal surface. Forces of $12\,\text{N}$ and $5\,\text{N}$ act at $90°$ to each other. Find the resultant acceleration.

Show Solution

Resultant force $= \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\,\text{N}$.

$a = \dfrac{F}{m} = \dfrac{13}{3} \approx 4.33\,\text{m\,s}^{-2}$.

Direction: $\arctan\!\left(\dfrac{5}{12}\right) \approx 22.6°$ from the 12 N force.

Problem 3

A uniform plank of length 4 m and mass 10 kg is supported at its two ends. A person of mass 70 kg stands 1.5 m from one end. Find the reactions at each support.

Show Solution

Let $R_A$ be the reaction at the near end, $R_B$ at the far end. Weight acts at midpoint (2 m). Person is 1.5 m from $A$.

Take moments about $A$: $R_B(4) = 10g(2) + 70g(1.5) = 20g + 105g = 125g$

$R_B = \dfrac{125g}{4} = \dfrac{125 \times 9.8}{4} = 306.25\,\text{N}$.

Vertical equilibrium: $R_A = (10+70)g - R_B = 784 - 306.25 = 477.75\,\text{N}$.

Problem 4

A ball is thrown horizontally from a height of $20\,\text{m}$ with speed $15\,\text{m\,s}^{-1}$. Find (a) the time to hit the ground, (b) the horizontal distance, (c) the speed on impact.

Show Solution

(a) Vertical: $20 = \frac{1}{2}(9.8)t^2 \Rightarrow t^2 = \dfrac{40}{9.8} \approx 4.082 \Rightarrow t \approx 2.02\,\text{s}$.

(b) $x = 15 \times 2.02 \approx 30.3\,\text{m}$.

(c) $v_y = gt = 9.8 \times 2.02 = 19.8\,\text{m\,s}^{-1}$ (downward). $|v| = \sqrt{15^2 + 19.8^2} = \sqrt{225 + 392} = \sqrt{617} \approx 24.8\,\text{m\,s}^{-1}$.

Problem 5

Two particles of mass $m$ and $3m$ are connected by a string over a smooth pulley. They are released from rest. Show that the acceleration is $\dfrac{g}{2}$ and find the tension in terms of $m$ and $g$.

Show Solution

For $3m$ (moves down): $3mg - T = 3ma$ ... (i)

For $m$ (moves up): $T - mg = ma$ ... (ii)

Add: $2mg = 4ma \Rightarrow a = \dfrac{g}{2}$.

From (ii): $T = mg + m \cdot \dfrac{g}{2} = \dfrac{3mg}{2}$.

Problem 6

A block of mass $6\,\text{kg}$ is in limiting equilibrium on a rough inclined plane at angle $\theta$ where $\tan\theta = \frac{3}{4}$. Find $\mu$.

Show Solution

$\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ (from $\tan\theta = 3/4$ and Pythagoras).

In limiting equilibrium on the verge of sliding: $F = \mu R$.

Perpendicular: $R = mg\cos\theta = 6g \cdot \frac{4}{5} = \frac{24g}{5}$.

Parallel: $F = mg\sin\theta = 6g \cdot \frac{3}{5} = \frac{18g}{5}$.

$\mu = \dfrac{F}{R} = \dfrac{18g/5}{24g/5} = \dfrac{18}{24} = \dfrac{3}{4} = 0.75$.

Problem 7

A stone is projected from the edge of a cliff 30 m above the sea with initial velocity $10\,\text{m\,s}^{-1}$ at $20°$ above horizontal. Find how far out to sea it lands.

Show Solution

$v_x = 10\cos 20° = 9.397\,\text{m\,s}^{-1}$, $\quad u_y = 10\sin 20° = 3.420\,\text{m\,s}^{-1}$.

Vertical (taking upward positive): $-30 = 3.42t - 4.9t^2$

$4.9t^2 - 3.42t - 30 = 0$

$t = \dfrac{3.42 + \sqrt{3.42^2 + 4(4.9)(30)}}{2(4.9)} = \dfrac{3.42 + \sqrt{11.70 + 588}}{9.8} = \dfrac{3.42 + 24.50}{9.8} \approx 2.85\,\text{s}$

Horizontal distance $= 9.397 \times 2.85 \approx 26.8\,\text{m}$.