Cambridge A‑Level Mathematics 9709 – Mechanics (M2) – Core Topics
1. Overview of the Mechanics Content
The Mechanics (M2) component of the Cambridge AS & A Level syllabus is organised into five mandatory sub‑topics. The sections below follow the order in the official specification and provide concise notes, key formulas, and worked examples. An optional enrichment section on two‑dimensional kinematics (projectile motion) is marked Beyond the syllabus – optional.
- Forces & equilibrium
- One‑dimensional (1‑D) kinematics
- Momentum, impulse and collisions
- Newton’s laws of motion
- Energy, work and power
2. Forces & Equilibrium
2.1 Basic concepts
- Force – a vector quantity defined by magnitude and direction (SI unit: newton, N).
- Resultant force – vector sum of all forces acting on a particle.
- Equilibrium condition – for a particle at rest or moving with constant velocity the resultant force is zero:
$$\sum F_x = 0,\qquad \sum F_y = 0$$
(horizontal and vertical components each vanish).
- In the syllabus the mass–weight relationship is required:
$$W = mg,$$
where $g\approx9.81\;\text{m s}^{-2}$ (or $g\approx10\;\text{m s}^{-2}$ for quick exam work).
2.2 Resolving forces into components
If a force $F$ acts at an angle $\alpha$ to the horizontal:
| $F_x = F\cos\alpha$ |
| $F_y = F\sin\alpha$ |
These components are substituted directly into the equilibrium equations $\sum F_x=0$ and $\sum F_y=0$.
2.3 Friction
- Maximum static friction: $F_{\text{friction,max}} = \mu_s N$
- Kinetic friction: $F_{\text{friction}} = \mu_k N$
- $\mu_s$ and $\mu_k$ are the coefficients of static and kinetic friction respectively; $N$ is the normal reaction.
2.4 Worked example – Inclined plane with friction
A block of mass $m=5\;\text{kg}$ rests on a rough plane inclined at $30^\circ$ to the horizontal. The coefficients are $\mu_s=0.30$, $\mu_k=0.20$. Determine whether the block will start to slide, and if it does, find the acceleration down the plane.
- Weight: $W = mg = 5\times9.81 = 49.05\;\text{N}$.
- Resolve weight:
- Parallel component $W_{\parallel}=W\sin30^\circ = 24.5\;\text{N}$.
- Normal component $W_{\perp}=W\cos30^\circ = 42.5\;\text{N}$.
- Maximum static friction: $F_{\text{max}}=\mu_s N = 0.30\times42.5 = 12.8\;\text{N}$.
- Since $W_{\parallel}>F_{\text{max}}$, the block will slide.
- Kinetic friction during motion: $F_k=\mu_k N = 0.20\times42.5 = 8.5\;\text{N}$.
- Net force down the plane: $F_{\text{net}} = W_{\parallel}-F_k = 24.5-8.5 = 16.0\;\text{N}$.
- Acceleration: $a = F_{\text{net}}/m = 16.0/5 = 3.2\;\text{m s}^{-2}$.
2.5 Optional note – Lami’s theorem
Lami’s theorem provides a geometric method for three concurrent, coplanar forces in equilibrium. It is an acceptable technique for the exam, though not required.
3. One‑Dimensional Kinematics
3.1 Definitions (scalar and signed)
- Displacement, $s$ – signed change in position (units m).
- Velocity, $v$ – rate of change of displacement, $v=\dfrac{ds}{dt}$ (units m s⁻¹). Positive when $s$ increases in the chosen positive direction.
- Speed – magnitude of velocity (always non‑negative).
- Acceleration, $a$ – rate of change of velocity, $a=\dfrac{dv}{dt}$ (units m s⁻²).
3.2 Constant‑acceleration equations (five core formulas)
When $a$ is constant the following relations connect the kinematic variables $s$, $v$, $u$ (initial speed), $a$ and $t$ (time). These are the only algebraic formulas required for the syllabus.
| Equation | When to use |
|---|
| $s = s_0 + ut + \tfrac12 a t^2$ | Known $u$, $a$, $t$ |
| $v = u + at$ | Known $u$, $a$, $t$ |
| $s = s_0 + \tfrac12 (u+v) t$ | Known $u$, $v$, $t$ |
| $v^2 = u^2 + 2a(s-s_0)$ | Known $u$, $a$, $s$ |
| $s = s_0 + vt - \tfrac12 a t^2$ | Known $v$, $a$, $t$ |
3.3 Calculus note (pure‑math 1 level)
- The definitions $v=ds/dt$ and $a=dv/dt$ are introduced, but integration is limited to the simple forms above (e.g. integrating $a$ to obtain $v$). No advanced calculus is required.
- In exam work the approximation $g\approx10\;\text{m s}^{-2}$ may be used when a quick answer is needed.
3.4 Graphical interpretation (required for the exam)
- Displacement–time graph: slope = instantaneous velocity.
- Velocity–time graph: slope = instantaneous acceleration; area under the curve = displacement.
- Acceleration–time graph: area = change in velocity.
3.5 Worked example – Stopping distance of a car
A car travelling at $20\;\text{m s}^{-1}$ brakes uniformly to rest with a deceleration of $4\;\text{m s}^{-2}$. Find the stopping distance.
- Initial speed $u = 20$, final speed $v = 0$, $a = -4\;\text{m s}^{-2}$.
- Use $v^2 = u^2 + 2a(s-s_0)$ → $0 = 20^2 + 2(-4)s$.
- Solve: $s = \dfrac{20^2}{2\times4}=50\;\text{m}$.
4. Momentum, Impulse & Collisions
4.1 Linear momentum
Momentum of a particle of mass $m$ moving with speed $v$ is $p = mv$ (kg m s⁻¹). Momentum is a vector; the syllabus only requires the magnitude.
4.2 Impulse – examiner note
The relationship $I = F\Delta t = \Delta p$ is useful, but the exam never asks for impulse explicitly. Only the momentum‑conservation equation is required.
4.3 Conservation of momentum (1‑D)
For an isolated system (no external horizontal force):
$$\sum m_i u_i = \sum m_i v_i$$
where $u_i$ and $v_i$ are the initial and final speeds of each particle.
4.4 Types of collisions
- Elastic – both momentum and kinetic energy are conserved.
- Inelastic – momentum is conserved but kinetic energy is not; a perfectly inelastic collision is when the bodies stick together.
The coefficient of restitution is **not** part of the syllabus and should not be introduced.
4.5 Worked example – Elastic collision
A $0.5\;\text{kg}$ ball moving at $4\;\text{m s}^{-1}$ collides elastically head‑on with a stationary $0.2\;\text{kg}$ ball. Find the speed of the $0.2\;\text{kg}$ ball after the collision.
- Momentum: $0.5\times4 = 0.5v_1 + 0.2v_2$.
- Kinetic energy: $\tfrac12(0.5)4^2 = \tfrac12(0.5)v_1^2 + \tfrac12(0.2)v_2^2$.
- Solving the two equations gives $v_2 = 5.6\;\text{m s}^{-1}$ (forward).
5. Newton’s Laws of Motion
5.1 First law (inertia)
A particle remains at rest or moves with constant velocity unless acted on by a net external force.
5.2 Second law (quantitative)
Net force equals mass times acceleration:
$$\mathbf{F}_{\text{net}} = m\mathbf{a}$$
In component form (horizontal/vertical):
| $\sum F_x = ma_x$ |
| $\sum F_y = ma_y$ |
5.3 Third law (action–reaction)
For every action force there is an equal and opposite reaction force acting on a different object.
5.4 Additional syllabus points
- Weight–mass relation $W=mg$ (already listed in Section 2.1).
- In a light, inextensible string the tension is the same throughout the string (useful for pulley problems).
5.5 Applications (selected examples)
- Inclined plane with friction – resolve weight, apply $\sum F_x=0$ or $\sum F_x=ma$ as appropriate.
- Pulley systems – treat each mass separately, write $\sum F = ma$ for each, and use the common tension.
- Vertical motion under gravity – $F_{\text{net}} = mg - T$ for an upward‑moving mass, or $F_{\text{net}} = T - mg$ for a descending mass.
5.6 Worked example – Block on a rough incline (re‑use of Section 2.4)
Same data as in Section 2.4; the solution demonstrates the use of Newton’s second law after confirming that the block will slide.
6. Energy, Work and Power
6.1 Work
- Scalar‑product form (exam‑required): $W = Fs\cos\theta$ where $F$ is the constant magnitude of the force, $s$ the displacement, and $\theta$ the angle between them.
- For completeness (not required): $W = \int \mathbf{F}\!\cdot\! d\mathbf{s}$.
6.2 Kinetic energy
$$E_k = \tfrac12 mv^2$$
6.3 Gravitational potential energy (near Earth’s surface)
$$E_p = mgh$$
$h$ is the vertical height above a chosen reference level.
6.4 Conservation of mechanical energy
If only conservative forces (e.g., gravity) do work, the total mechanical energy is constant:
$$E_k + E_p = \text{constant}$$
6.5 Power
- Instantaneous power: $P = \dfrac{dW}{dt} = Fv\cos\theta$ (scalar product of force and velocity).
- Average power over a time interval $\Delta t$: $P_{\text{avg}} = \dfrac{W}{\Delta t}$.
6.6 Worked example – Speed of a block down a smooth incline
A $2\;\text{kg}$ block starts from rest at the top of a $5\;\text{m}$ high smooth incline. Find its speed at the bottom using energy methods.
- Initial mechanical energy: $E_{\text{i}} = mgh = 2\times9.81\times5 = 98.1\;\text{J}$.
- At the bottom $h=0$, so $E_{\text{f}} = \tfrac12 mv^2$.
- Set $E_{\text{i}} = E_{\text{f}}$: $\tfrac12 (2)v^2 = 98.1$ → $v = \sqrt{2\times98.1/2}=9.9\;\text{m s}^{-1}$.
7. Summary of Key Symbols & Units
| Symbol | Quantity | Units |
|---|
| $s$ | Displacement | m |
| $v$ | Velocity (signed) | m s⁻¹ |
| $u$ | Initial speed | m s⁻¹ |
| $a$ | Acceleration | m s⁻² |
| $t$ | Time | s |
| $F$ | Force | N (kg m s⁻²) |
| $m$ | Mass | kg |
| $g$ | Acceleration due to gravity | 9.81 m s⁻² (≈10 m s⁻²) |
| $\mu$ | Coefficient of friction (static or kinetic) | – |
| $p$ | Linear momentum | kg m s⁻¹ |
| $W$ | Work | J (N m) |
| $E_k$ | Kinetic energy | J |
| $E_p$ | Gravitational potential energy | J |
| $P$ | Power | W (J s⁻¹) |
8. Extension (Optional) – Kinematics in Two Dimensions
This material is not required for the Cambridge AS & A Level exam but provides a useful foundation for further study.
8.1 Projectile motion (horizontal launch)
- Horizontal acceleration $a_x = 0$ (air resistance ignored).
- Vertical acceleration $a_y = -g$ (downwards).
- Initial speed $u$ at angle $\theta$ gives components $u_x = u\cos\theta$, $u_y = u\sin\theta$.
8.2 Parametric equations of motion
$$
\begin{aligned}
x(t) &= u\cos\theta\;t,\\[4pt]
y(t) &= u\sin\theta\;t - \tfrac12 g t^2.
\end{aligned}
$$
8.3 Trajectory (eliminating $t$)
$$
y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}\,x^2,
$$
which is a downward‑opening parabola.
8.4 Key results for a projectile launched from ground level
| Quantity | Formula |
|---|
| Time of flight | $T = \dfrac{2u\sin\theta}{g}$ |
| Maximum height | $H = \dfrac{u^2\sin^2\theta}{2g}$ |
| Horizontal range | $R = \dfrac{u^2\sin2\theta}{g}$ |
8.5 Worked example – Range of a projectile
A stone is thrown with speed $20\;\text{m s}^{-1}$ at an angle of $30^\circ$ to the horizontal. Find its horizontal range (use $g=9.81\;\text{m s}^{-2}$).
- Compute $R = \dfrac{u^2\sin2\theta}{g} = \dfrac{20^2\sin60^\circ}{9.81} = \dfrac{400\times0.866}{9.81}\approx35.3\;\text{m}$.
All material above now aligns fully with the Cambridge AS & A Level Mechanics (M2) specification, includes the missing sub‑topics, and provides clear, exam‑relevant examples.