Cambridge A‑Level Mathematics 9709 – Mechanics (Paper 4) – Complete Notes
1. Syllabus Overview (Paper 4)
The Mechanics component of the 9709 syllabus is divided into four main sections. These notes cover the Energy, Power and Elasticity sub‑topic in full and also provide concise refreshers for the other required areas, so the resource can be used as a single, self‑contained pack.
| Syllabus Code |
Topic |
Coverage in these notes |
| 4.1 |
Forces & equilibrium |
Section 2 |
| 4.2 |
Kinematics (straight‑line motion) |
Section 3 |
| 4.3 |
Newton’s laws of motion |
Section 4 |
| 4.4 |
Momentum, impulse and collisions |
Section 5 |
| 4.5 |
Work, energy and power (including elastic strings & springs) |
Sections 6–12 |
2. Forces & Equilibrium (AO1, AO2)
- Resultant force: \(\displaystyle\mathbf{R}= \sum\mathbf{F}_i\).
- Static‑equilibrium checklist (exact syllabus wording):
- ∑ \(F_x = 0\) (horizontal components balance)
- ∑ \(F_y = 0\) (vertical components balance)
- ∑ \(M_O = 0\) moments about any point \(O\) balance (optional for 9709 but useful).
- Resolving forces on an inclined plane (incline angle \(\theta\)):
- Component parallel to the plane: \(F_{\parallel}=F\sin\theta\) (if \(F\) acts vertically) or \(F\cos\theta\) (if \(F\) is horizontal).
- Component normal to the plane: \(N = mg\cos\theta \pm\) any additional vertical component.
- Friction:
- Limiting static friction: \(F_{\text{lim}}=\mu_s N\).
- Kinetic friction: \(F_f=\mu_k N\) (opposes relative motion).
Example 2.1 – Static block on an inclined plane
A 5 kg block rests on a \(30^{\circ}\) incline. A rope pulls up the plane with tension \(T\). The coefficient of static friction is \(\mu_s=0.4\). Find the minimum \(T\) to keep the block at rest.
- Weight component down the plane: \(W_{\parallel}=mg\sin30^{\circ}=5\times9.8\times0.5=24.5\;\text{N}\).
- Normal reaction: \(N=mg\cos30^{\circ}=5\times9.8\times0.866=42.5\;\text{N}\).
- Maximum static friction (up the plane): \(F_{\text{lim}}=\mu_s N=0.4\times42.5=17.0\;\text{N}\).
- Equilibrium parallel to the plane: \(T+F_{\text{lim}}=W_{\parallel}\;\Rightarrow\;T=24.5-17.0=7.5\;\text{N}\).
3. Kinematics – Straight‑Line Motion (AO1, AO2)
All symbols follow the Cambridge convention. The equations below apply for constant acceleration \(a\).
| Quantity | Symbol | Formula (constant \(a\)) |
|---|
| Displacement | \(s\) | \(s = ut + \tfrac12 a t^{2}\) |
| Final velocity | \(v\) | \(v = u + a t\) |
| Velocity–displacement relation | \(v,\;u,\;s\) | \(v^{2}=u^{2}+2as\) |
| Average speed | \(\bar v\) | \(\displaystyle\bar v = \frac{s}{t}\) |
- Graph interpretation (syllabus requirement):
- Gradient of a displacement‑time graph = instantaneous velocity.
- Area under a velocity‑time graph = displacement.
- Gradient of a velocity‑time graph = acceleration.
- Area under an acceleration‑time graph = change in velocity.
Example 3.1 – Projectile up an inclined plane
A particle is projected up a frictionless \(20^{\circ}\) incline with speed \(15\;\text{m s}^{-1}\). Find the maximum distance travelled up the plane.
\[
0 = u^{2} - 2(g\sin20^{\circ})\,s
\;\Longrightarrow\;
s = \frac{u^{2}}{2g\sin20^{\circ}}
= \frac{15^{2}}{2\times9.8\times\sin20^{\circ}}
\approx 5.8\;\text{m}.
\]
4. Newton’s Laws of Motion (AO1, AO2)
- First law (inertia): A particle of constant mass remains at rest or moves with uniform straight‑line velocity unless acted on by a net external force.
- Second law: \(\displaystyle\mathbf{R}=m\mathbf{a}\) (the mass \(m\) is constant; the law applies to particles).
- Third law: For every action \(\mathbf{F}_{AB}\) there is an equal and opposite reaction \(\mathbf{F}_{BA}=-\mathbf{F}_{AB}\).
Typical AO2 question – Block pulled up a rough incline
Derive the acceleration of a block of mass \(m\) pulled by a constant horizontal force \(F\) up a rough incline of angle \(\theta\) (coefficient of kinetic friction \(\mu_k\)).
\[
\begin{aligned}
N &= mg\cos\theta - F\sin\theta,\\[4pt]
\text{Parallel forces: }&F\cos\theta - mg\sin\theta - \mu_k N = ma,\\[4pt]
\therefore\; a &= \frac{F\cos\theta - mg\sin\theta - \mu_k\bigl(mg\cos\theta - F\sin\theta\bigr)}{m}.
\end{aligned}
\]
5. Momentum, Impulse and Collisions (AO1, AO2)
- Linear momentum: \(\displaystyle\mathbf{p}=m\mathbf{v}\) (constant mass).
- Impulse: \(\displaystyle\mathbf{J}= \int_{t_1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p}\).
- Conservation of momentum (isolated system):
\(\displaystyle\sum\mathbf{p}_{\text{initial}} = \sum\mathbf{p}_{\text{final}}\).
- Elastic collision: both momentum and kinetic energy are conserved.
- Inelastic collision: only momentum conserved; kinetic energy is lost (converted to other forms).
- Coefficient of restitution \(e\) (useful for partially elastic collisions):
\[
e = \frac{\text{relative speed after}}{\text{relative speed before}}
= \frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}},\qquad 0\le e\le1.
\]
Example 5.1 – Perfectly elastic head‑on collision
Masses: \(m_1=2\;\text{kg}\) (speed \(u_1=4\;\text{m s}^{-1}\)) and \(m_2=3\;\text{kg}\) (at rest). Find the speeds after the collision.
\[
\begin{aligned}
\text{Momentum:}&\; m_1u_1 = m_1v_1 + m_2v_2,\\
\text{Kinetic energy:}&\; \tfrac12 m_1u_1^{2}= \tfrac12 m_1v_1^{2}+ \tfrac12 m_2v_2^{2}.
\end{aligned}
\]
Solving gives
\[
v_1 = -\frac{m_1-m_2}{m_1+m_2}u_1 = 0.8\;\text{m s}^{-1},\qquad
v_2 = \frac{2m_1}{m_1+m_2}u_1 = 3.2\;\text{m s}^{-1}.
\]
6. Work–Energy Theorem (AO1, AO2)
For a particle of constant mass \(m\) moving under a resultant force \(\mathbf{R}\) along a path \(C\):
\[
W_{\text{total}}=\int_{C}\mathbf{R}\!\cdot\! d\mathbf{s}= \Delta E_k
=\frac12 m v_f^{2}-\frac12 m v_i^{2}.
\]
If both conservative (\(W_c\)) and non‑conservative (\(W_{nc}\)) forces act,
\[
W_c+W_{nc}= \Delta E_k.
\]
Introducing gravitational potential energy \(E_g=mgh\) (a conservative force), the theorem can be written as
\[
W_{nc}= \Delta\!\bigl(E_k+E_g\bigr).
\]
Worked Example 6.1 – Block pulled up a rough incline
Given: \(m=2\;\text{kg},\;\theta=30^{\circ},\;d=5\;\text{m},\;F=15\;\text{N}\) (horizontal), \(\mu_k=0.2\), initial speed \(u=0\).
- Component of the pull parallel to the plane: \(F_{\parallel}=F\cos30^{\circ}=12.99\;\text{N}\).
- Weight component down the plane: \(W_{\parallel}=mg\sin30^{\circ}=9.80\;\text{N}\).
- Normal reaction: \(N = mg\cos30^{\circ} - F\sin30^{\circ}=8.49\;\text{N}\).
- Work of kinetic friction: \(W_f = -\mu_k N d = -0.2\times8.49\times5 = -8.49\;\text{J}\).
- Work of the pull: \(W_F = F_{\parallel} d = 12.99\times5 = 64.95\;\text{J}\).
- Work done against gravity: \(W_g = -mg\sin30^{\circ} d = -9.80\times5 = -49.0\;\text{J}\).
- Net work: \(W_{\text{net}} = W_F + W_f + W_g = 7.46\;\text{J}\).
- Apply the theorem: \(\tfrac12 m v^{2}=W_{\text{net}}\;\Rightarrow\;
v = \sqrt{\dfrac{2W_{\text{net}}}{m}} \approx 2.73\;\text{m s}^{-1}\).
7. Power (AO1, AO2)
- Instantaneous power: \(P = \dfrac{dW}{dt}= \mathbf{F}\!\cdot\!\mathbf{v}\).
- If the force makes a constant angle \(\theta\) with the velocity: \(P = Fv\cos\theta\).
- Average power over a time interval \(\Delta t\):
\(\displaystyle\bar P = \frac{W}{\Delta t}\).
- Efficiency (AO3): \(\displaystyle\eta = \frac{\text{useful power output}}{\text{power input}}\times100\%.\)
Example 7.1 – Motor lifting a mass
A motor lifts a \(10\;\text{kg}\) load through \(8\;\text{m}\) in \(4\;\text{s}\) (ignore friction). Find the average power output.
\[
W = mgh = 10\times9.8\times8 = 784\;\text{J},\qquad
\bar P = \frac{784}{4}=196\;\text{W}.
\]
8. Elastic Potential Energy (AO1, AO2)
- Hooke’s law (elastic region): \(\displaystyle\mathbf{F} = -k\mathbf{x}\), where \(k\) is the spring constant and \(\mathbf{x}\) is the extension/compression measured from the natural length.
- Elastic (strain) energy stored when stretching/compressing from \(x=0\) to \(x\):
\[
E_{\text{elastic}} = \int_{0}^{x} kx'\,dx' = \frac12 kx^{2}.
\]
- AO3 note – elastic limit: the formula \(\tfrac12 kx^{2}\) is valid only while the material remains within its elastic limit. Beyond that point permanent deformation occurs and the energy stored is no longer given by this expression.
9. Springs in Series and Parallel (AO1, AO2)
- Series combination (same force, extensions add):
\[
\frac{1}{k_{\text{eq}}}= \frac{1}{k_1}+ \frac{1}{k_2}+ \cdots + \frac{1}{k_n}.
\]
The total extension \(x = x_1+x_2+\dots\) and the force in each spring is identical.
- Parallel combination (same extension, forces add):
\[
k_{\text{eq}} = k_1+k_2+\dots+k_n.
\]
The total force is the sum of the individual spring forces.
Example 9.1 – Two identical springs in series
Each spring has \(k=200\;\text{N m}^{-1}\). A block pulls the combined system \(d=0.10\;\text{m}\) from equilibrium. Find the elastic energy stored.
\[
k_{\text{eq}} = \frac{k}{2}=100\;\text{N m}^{-1},\qquad
E_{\text{elastic}} = \frac12 k_{\text{eq}} d^{2}
= \frac12 \times100\times(0.10)^{2}
= 0.50\;\text{J}.
\]
10. Further Applications (AO2)
Typical A‑Level questions combine the ideas above with:
- Blocks on rough inclined planes (including work of friction \(W_f=-\mu_k N d\)).
- Projectile motion with air resistance (non‑conservative work).
- Mass–spring systems undergoing simple harmonic motion (use \(E_k+E_{\text{elastic}}=\text{constant}\)).
- Energy methods for particles on varying‑force inclined planes (e.g. \(F = -kx + c\)).
- Power and efficiency of machines (e.g. lifts, engines, pulleys).
- Collisions with a given coefficient of restitution \(e\) (partial elasticity).
11. Worked Example – Mass‑Spring Oscillator (AO2)
Problem: A mass \(m=0.5\;\text{kg}\) is attached to a horizontal spring of constant \(k=80\;\text{N m}^{-1}\). It is pulled \(x_0=0.15\;\text{m}\) from equilibrium and released from rest. Find the maximum speed of the mass.
- Initial elastic energy: \(E_{\text{elastic}}=\tfrac12 kx_0^{2}= \tfrac12 \times80\times0.15^{2}=0.90\;\text{J}\).
- At the equilibrium position all elastic energy has been converted to kinetic energy:
\(\tfrac12 m v_{\max}^{2}=E_{\text{elastic}}\).
- Maximum speed:
\[
v_{\max}= \sqrt{\frac{2E_{\text{elastic}}}{m}}
= \sqrt{\frac{2\times0.90}{0.5}}
\approx 1.90\;\text{m s}^{-1}.
\]
12. Summary of Key Formulas (Quick Reference)
| Quantity | Formula | Comments |
|---|
| Resultant force | \(\displaystyle\mathbf{R}= \sum\mathbf{F}_i\) | Static equilibrium: \(\sum F_x=0,\;\sum F_y=0\). |
| Work (variable force) | \(\displaystyle W=\int\mathbf{F}\!\cdot\!d\mathbf{s}\) | Positive if force has component along displacement. |
| Kinetic energy | \(\displaystyle E_k=\frac12 mv^{2}\) | Mass is constant. |
| Gravitational potential energy | \(\displaystyle E_g=mgh\) | Reference level arbitrary. |
| Elastic potential energy | \(\displaystyle E_{\text{elastic}}=\frac12 kx^{2}\) | Valid only within elastic limit. |
| Work‑Energy theorem | \(\displaystyle W_{\text{total}}=\Delta(E_k+E_g)\) | Include non‑conservative work if present. |
| Power (instantaneous) | \(\displaystyle P=\mathbf{F}\!\cdot\!\mathbf{v}\) | Average: \(\bar P=W/\Delta t\). |
| Series spring constant | \(\displaystyle \frac1{k_{\text{eq}}}= \sum\frac1{k_i}\) | Force same in each spring. |
| Parallel spring constant | \(\displaystyle k_{\text{eq}}= \sum k_i\) | Extension same for each spring. |
| Momentum | \(\displaystyle \mathbf{p}=m\mathbf{v}\) | Conserved in isolated system. |
| Impulse | \(\displaystyle \mathbf{J}= \int\mathbf{F}\,dt = \Delta\mathbf{p}\) | Change in momentum. |
| Coefficient of restitution | \(\displaystyle e=\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}}\) | 0 (perfectly inelastic) ≤ e ≤ 1 (perfectly elastic). |
13. Checklist for Exam Preparation (AO1 – AO3)
- Identify which forces are conservative (gravity, spring) and which are non‑conservative (friction, air resistance).
- State clearly any assumptions required by the syllabus:
- Mass is constant.
- Spring obeys Hooke’s law throughout the motion.
- Surfaces are rigid and planar unless otherwise specified.
- Air resistance is neglected unless explicitly included.
- When using energy methods, write the full energy equation before substituting numbers:
\[
W_{nc}= \Delta\!\bigl(E_k+E_g+E_{\text{elastic}}\bigr).
\]
- For power questions, decide whether the problem asks for instantaneous, average, or efficiency‑related power.
- Check units at every step – the syllabus expects SI units (N, J, W, s, m).