Mechanics (M2) – Forces, Motion & Energy (Cambridge International AS & A Level Mathematics 9709)
1. Syllabus Overview (Paper 4 – Mechanics)
The mechanics component of the 9709 syllabus is divided into the sub‑topics listed below. Each entry indicates the key ideas that must be covered and the typical skills examined.
| Syllabus code | Sub‑topic | Key ideas / required formulae |
|---|
| 4.1 | Forces & equilibrium | Friction (static & kinetic), normal reaction, limiting equilibrium, vector form of forces, equilibrium of rigid bodies (∑ F = 0, ∑ M = 0). |
| 4.2 | Kinematics of a particle | Displacement, velocity, acceleration; constant‑acceleration equations; motion graphs; simple projectile on an inclined plane (optional). |
| 4.3 | Momentum & impulse | Linear momentum, impulse–momentum theorem, conservation of momentum for direct (1‑D) impact (elastic & inelastic collisions are optional). |
| 4.4 | Newton’s laws | Statement of the three laws, application of Σ F = ma to particles and systems (including non‑inertial frames only if required). |
| 4.5 | Energy, work & power | Work‑energy theorem, kinetic & gravitational potential energy, power (P = F v), energy methods for equilibrium (e.g. limiting friction). |
2. Motion of a Body on a Rough Surface
2.1 Fundamental concepts
- Forces are vectors. Write every force as a vector \(\mathbf{F}\) and resolve it into orthogonal components (usually parallel & perpendicular to the surface).
- Normal reaction, \(\mathbf{R}\) – the force exerted by the surface, directed perpendicular to the contact.
- Frictional force, \(\mathbf{F}_f\) – always opposite the direction of relative motion (or impending motion).
- Static (limiting) friction: \(\displaystyle |\mathbf{F}_f| \le \mu_s |\mathbf{R}|\). Equality holds at the point of impending motion.
- Kinetic friction: \(\displaystyle |\mathbf{F}_f| = \mu_k |\mathbf{R}|\) (constant while sliding).
- Coefficients of friction – summarized in Table 1.
| Situation | Coefficient used | Friction formula |
|---|
| Body at rest, no tendency to move | — | \(|\mathbf{F}_f| < \mu_s |\mathbf{R}|\) |
| Body on the verge of moving (limiting equilibrium) | \(\mu_s\) | \(|\mathbf{F}_f| = \mu_s |\mathbf{R}|\) |
| Body sliding (constant speed or accelerating) | \(\mu_k\) | \(|\mathbf{F}_f| = \mu_k |\mathbf{R}|\) |
2.2 Inclined plane – basic relations
For a plane inclined at an angle \(\theta\) to the horizontal:
\[
\begin{aligned}
\mathbf{R} &= mg\cos\theta \;\mathbf{\hat n},\\[4pt]
\mathbf{W}_\parallel &= mg\sin\theta \;\mathbf{\hat t},
\end{aligned}
\]
where \(\mathbf{\hat n}\) is normal to the plane and \(\mathbf{\hat t}\) is directed down the slope.
2.3 Limiting‑equilibrium conditions
- Just about to slide down:
\[
mg\sin\theta = \mu_s\,mg\cos\theta \quad\Longrightarrow\quad \tan\theta = \mu_s .
\]
- Just about to slide up (e.g. an external pull \(P\) acting up‑slope):
\[
P + \mu_s\,mg\cos\theta = mg\sin\theta .
\]
2.4 Worked example – block on a rough incline (constant speed)
Problem: A 5 kg block rests on a rough plane inclined at \(30^{\circ}\). \(\mu_s = 0.40\), \(\mu_k = 0.30\). Determine whether the block moves, and if it does, find its acceleration.
- Normal reaction: \(\displaystyle R = mg\cos30^{\circ}=5\times9.8\times0.866=42.4\text{ N}\).
- Maximum static friction: \(\displaystyle F_{s,\max}= \mu_sR =0.40\times42.4=16.96\text{ N}\).
- Component of weight down the plane: \(\displaystyle W_{\parallel}=mg\sin30^{\circ}=5\times9.8\times0.5=24.5\text{ N}\).
- Since \(W_{\parallel}>F_{s,\max}\) the block will start to slide down.
- Kinetic friction while sliding: \(\displaystyle F_k=\mu_kR=0.30\times42.4=12.72\text{ N}\).
- Apply Newton’s II law parallel to the plane:
\[
mg\sin\theta - F_k = ma\;\Longrightarrow\;24.5-12.72 =5a\;\Longrightarrow\;a=2.36\text{ m s}^{-2}.
\]
2.5 General procedure for rough‑surface problems
- Draw a clear free‑body diagram (FBD) showing all forces (weight, normal reaction, friction, any applied forces).
- Resolve each force into components parallel (\(\parallel\)) and perpendicular (\(\perp\)) to the surface.
- Write the vector equation \(\displaystyle \sum\mathbf{F}=m\mathbf{a}\) (or \(\displaystyle \sum\mathbf{F}=0\) for static cases).
- Replace the friction term with \(\mu_sR\) for limiting equilibrium or \(\mu_kR\) for sliding, as appropriate.
- Solve the resulting scalar equations for the unknown quantities (acceleration, required pull, maximum angle, etc.).
3. Connected Particles (Massless String, Ideal Pulley)
3.1 Core assumptions
- String is mass‑less and inextensible → all attached masses share the same magnitude of acceleration \(a\).
- Ideal pulley: frictionless, massless, and changes only the direction of the tension. Consequently the tension \(T\) is the same throughout a single continuous string.
3.2 Standard two‑mass system
Two masses, \(m_1\) and \(m_2\) (\(m_1>m_2\)), are connected over a frictionless pulley.
\[
\begin{aligned}
m_1g - T &= m_1 a \quad\text{(downward positive for }m_1\text{)},\\[4pt]
T - m_2g &= m_2 a \quad\text{(downward positive for }m_2\text{)}.
\end{aligned}
\]
Solving simultaneously:
\[
a = \frac{(m_1-m_2)g}{m_1+m_2}, \qquad
T = \frac{2m_1m_2\,g}{m_1+m_2}.
\]
3.3 Example – friction on a horizontal table
Problem: A 4 kg block rests on a horizontal table ( \(\mu_k =0.15\) ). It is linked by a light string over a pulley to a hanging 2 kg mass. Find the acceleration and the tension.
- Forces on the block:
- Weight \(4g\) downwards.
- Normal reaction \(R=4g\) upwards.
- Kinetic friction \(F_k = \mu_kR =0.15\times4g =5.88\text{ N}\) opposite the motion.
- Tension \(T\) to the right.
- Forces on the hanging mass: weight \(2g\) downwards, tension \(T\) upwards.
- Take rightward (block) and downward (hanging mass) as positive:
\[
\begin{aligned}
T - F_k &= 4a,\\
2g - T &= 2a.
\end{aligned}
\]
- Add the equations to eliminate \(T\):
\[
2g - F_k = 6a \;\Longrightarrow\; a = \frac{2g-5.88}{6}=1.68\text{ m s}^{-2}.
\]
- Back‑substitute for tension:
\[
T = 4a + F_k = 4(1.68)+5.88 = 12.6\text{ N}.
\]
3.4 Systems with more than one pulley
If the string passes over several ideal pulleys, the same principles apply: the tension is uniform throughout each continuous segment, and the acceleration of every attached mass is the same magnitude. Write a separate \(\sum F = ma\) equation for each mass, then combine them to eliminate the unknown tensions.
3.5 General solution method for connected‑particle problems
- Draw a free‑body diagram for **each** particle, indicating weight, normal reaction, friction, tension(s) and any applied forces.
- Choose a consistent sign convention (e.g., positive in the direction each particle is expected to move).
- Write \(\displaystyle \sum\mathbf{F}=m\mathbf{a}\) for every particle.
- Use the fact that the string is inextensible to set all accelerations equal in magnitude.
- Eliminate the internal tension(s) by adding or subtracting the equations.
- Solve for the common acceleration \(a\); then substitute back to obtain the tension(s) and any other required quantity.
4. Equilibrium of Rigid Bodies
4.1 Fundamental equilibrium conditions (syllabus wording)
\[
\boxed{\displaystyle \sum\mathbf{F}= \mathbf{0}\quad\text{and}\quad\sum\mathbf{M}_O = \mathbf{0}}
\]
- Translational equilibrium: the vector sum of all external forces is zero.
- Rotational equilibrium: the sum of moments (torques) about any point \(O\) is zero.
4.2 Vector and component form for planar problems
Resolve forces into horizontal (\(x\)) and vertical (\(y\)) components:
\[
\sum F_x = 0,\qquad \sum F_y = 0.
\]
The moment about a chosen pivot \(O\) is
\[
\sum M_O = \sum (F_{\perp}\,d) = 0,
\]
where \(F_{\perp}\) is the component of the force perpendicular to the line joining its point of application to \(O\), and \(d\) is the perpendicular distance (lever arm). Adopt a consistent sign convention (e.g., anticlockwise = positive).
4.3 Typical configurations and the equations you will need
| Configuration | Key equilibrium equations | Typical unknowns |
|---|
| Simply supported beam |
\(\sum F_y=0,\;\sum M_A=0\) |
Reactions at supports \(A\) and \(B\) |
| Cantilever beam |
\(\sum F_y=0,\;\sum M_{\text{fixed}}=0\) |
Reaction at the fixed end, bending moment |
| Particle on an inclined plane with a rope |
\(\sum F_{\parallel}=0,\;\sum F_{\perp}=0\) |
Tension, required friction for static equilibrium |
| Rigid body with three non‑collinear forces |
\(\sum F_x=0,\;\sum F_y=0,\;\sum M_O=0\) |
Magnitude and direction of the unknown force |
4.4 Step‑by‑step method for solving equilibrium problems
- Free‑body diagram (FBD). Draw the body in isolation, showing every external force (weights, reactions, tensions, friction, applied loads).
- Choose a pivot point. Prefer a point through which an unknown reaction passes; this removes that unknown from the moment equation.
- Resolve forces. Break each force into its \(x\) and \(y\) components (or parallel/perpendicular to a convenient axis).
- Write the three equilibrium equations:
- \(\displaystyle \sum F_x = 0\)
- \(\displaystyle \sum F_y = 0\)
- \(\displaystyle \sum M_O = 0\)
- Solve the simultaneous equations. Usually three equations for three unknown reactions.
- Check limiting friction (if present). Verify that \(|\mathbf{F}_f|\le\mu_s|\mathbf{R}|\); if equality holds the body is in limiting equilibrium.
4.5 Example – simply supported beam with an off‑centre load
Problem: A uniform beam \(AB\) is 6 m long, weight \(W=120\text{ N}\). It is simply supported at \(A\) and \(B\). A downward load \(P=80\text{ N}\) acts 2 m from \(A\). Find the reactions \(R_A\) and \(R_B\).
- Take upward forces as positive. Vertical‑force equilibrium:
\[
R_A + R_B - W - P = 0 \;\Longrightarrow\; R_A+R_B = 200\text{ N}.
\]
- Choose pivot at \(A\) to eliminate \(R_A\) from the moment equation:
\[
\sum M_A = 0:\; R_B(6) - W(3) - P(2) = 0.
\]
(Weight acts at the centre, 3 m from \(A\).)
- Compute:
\[
6R_B = 120\times3 + 80\times2 = 360 + 160 = 520 \;\Longrightarrow\; R_B = 86.7\text{ N}.
\]
- From the vertical‑force equation:
\[
R_A = 200 - 86.7 = 113.3\text{ N}.
\]
5. Kinematics of a Particle (Constant Acceleration)
5.1 Definitions
- Displacement \(\mathbf{s}\) – vector change in position.
- Velocity \(\mathbf{v}\) – \(\displaystyle \mathbf{v}= \frac{d\mathbf{s}}{dt}\).
- Acceleration \(\mathbf{a}\) – \(\displaystyle \mathbf{a}= \frac{d\mathbf{v}}{dt}\).
5.2 Standard constant‑acceleration equations (syllabus‑required)
\[
\begin{aligned}
v &= u + at,\\[4pt]
s &= ut + \tfrac12 a t^{2},\\[4pt]
v^{2} &= u^{2} + 2as,
\end{aligned}
\]
where \(u\) is the initial speed, \(v\) the final speed, \(t\) the elapsed time and \(s\) the distance travelled along the line of motion.
5.3 Motion graphs (qualitative)
- Displacement‑time graph: slope = velocity.
- Velocity‑time graph: slope = acceleration; area under the curve = displacement.
- Acceleration‑time graph: area = change in velocity.
5.4 Example – projectile on a frictionless incline (optional extension)
A particle is projected up a smooth incline of \(20^{\circ}\) with an initial speed \(u=10\text{ m s}^{-1}\). Find the maximum distance up the plane.
- Component of gravitational acceleration down the plane: \(a = g\sin20^{\circ}=9.8\sin20^{\circ}=3.35\text{ m s}^{-2}\) (acting opposite the motion).
- At the highest point \(v=0\). Use \(v^{2}=u^{2}+2as\):
\[
0 = 10^{2} - 2(3.35)s \;\Longrightarrow\; s = \frac{100}{6.70}=14.9\text{ m}.
\]
6. Momentum, Impulse & Conservation
6.1 Linear momentum
\[
\mathbf{p}=m\mathbf{v}.
\]
6.2 Impulse–momentum theorem (useful for “quick‑change” problems)
\[
\displaystyle \mathbf{J}= \int \mathbf{F}\,dt = \Delta\mathbf{p}.
\]
6.3 Conservation of linear momentum (direct impact, 1‑D)
When the net external impulse on a system of particles is zero, the total momentum is conserved:
\[
\sum \mathbf{p}_{\text{initial}} = \sum \mathbf{p}_{\text{final}}.
\]
6.4 Example – perfectly inelastic collision
A 3 kg cart moving at \(4\text{ m s}^{-1}\) collides and sticks to a stationary 2 kg cart. Find the common speed after impact.
\[
(3)(4) + (2)(0) = (3+2)v \;\Longrightarrow\; v = \frac{12}{5}=2.4\text{ m s}^{-1}.
\]
Note: The syllabus does not require detailed treatment of the coefficient of restitution; such material may be labelled “beyond syllabus”.
7. Energy, Work & Power
7.1 Work
\[
W = \mathbf{F}\!\cdot\!\mathbf{s}=Fs\cos\theta.
\]
7.2 Kinetic and gravitational potential energy
\[
\begin{aligned}
\text{Kinetic energy }&K = \tfrac12 mv^{2},\\[4pt]
\text{Gravitational PE }&U = mgh\quad(\text{or }U = mg\,\text{(vertical height)}).
\end{aligned}
\]
7.3 Work–energy theorem
\[
\displaystyle W_{\text{net}} = \Delta K.
\]
7.4 Power
\[
P = \frac{W}{t}=Fv\quad(\text{when the force and velocity are parallel}).
\]
7.5 Energy method for limiting friction (optional exam technique)
When a body is on the verge of moving, the work done by the limiting friction equals the loss of mechanical energy. This provides an alternative to the direct \(\tan\theta=\mu_s\) relation on an incline.
7.6 Example – work done against friction
A 10 kg block is pulled 5 m up a horizontal rough surface (\(\mu_k=0.2\)) by a constant horizontal force of 80 N. Find the work done by the pulling force and the increase in kinetic energy.
- Normal reaction \(R = mg = 10\times9.8 = 98\text{ N}\).
- Kinetic friction \(F_f = \mu_kR = 0.2\times98 = 19.6\text{ N}\) opposite the motion.
- Net horizontal force \(F_{\text{net}} = 80 - 19.6 = 60.4\text{ N}\).
- Work done by the pulling force: \(W_{\text{pull}} = 80\times5 = 400\text{ J}\).
- Work done against friction: \(W_f = -19.6\times5 = -98\text{ J}\).
- Net work \(=400-98 =302\text{ J}\) which equals the increase in kinetic energy \(\Delta K\) (by the work‑energy theorem).
8. Summary of Key Formulae
| Topic | Key formulae |
|---|
| Forces on an incline | \(R = mg\cos\theta,\; W_{\parallel}=mg\sin\theta\) |
| Friction | \(|F_f|\le\mu_s R\) (static), \(|F_f|=\mu_k R\) (kinetic) |
| Limiting equilibrium (incline) | \(\tan\theta = \mu_s\) |
| Two‑mass pulley | \(a=\dfrac{(m_1-m_2)g}{m_1+m_2},\; T=\dfrac{2m_1m_2g}{m_1+m_2}\) |
| Equilibrium of a rigid body | \(\sum\mathbf{F}=0,\;\sum\mathbf{M}_O=0\) |
| Kinematics (constant \(a\)) | \(v=u+at,\; s=ut+\tfrac12at^{2},\; v^{2}=u^{2}+2as\) |
| Momentum | \(\mathbf{p}=m\mathbf{v},\; \Delta\mathbf{p}= \int\mathbf{F}dt\) |
| Energy | \(K=\tfrac12mv^{2},\; U=mgh,\; W_{\text{net}}=\Delta K\) |
| Power | \(P=Fv=\dfrac{W}{t}\) |
9. Quick‑check Checklist for Exam Questions
- Read the question carefully – identify which sub‑topic(s) are being tested.
- Draw a clean free‑body diagram (or diagrams) for every distinct body.
- State the relevant principle(s): Newton’s II law, equilibrium conditions, kinematic equations, conservation of momentum, work‑energy theorem.
- Resolve forces into components; write the scalar equations (∑F_x, ∑F_y, ∑M). Use vector notation where the syllabus expects it.
- Apply the appropriate coefficient of friction (static or kinetic) and check limiting conditions.
- Solve the simultaneous equations; keep track of units and the sign convention.
- Check the answer: does it satisfy all equilibrium/motion conditions? Does it respect the limiting‑friction inequality?