Notes · Further MathematicsUK · A-Levels

Optional application: Mechanics

Mechanics is one of the three optional applied routes in AQA Further Mathematics 7367; a student is examined on it only if their centre enters them for it. It builds on the mechanics of A-Level Mathematics with dimensional analysis, momentum and collisions with restitution, work, energy and power, circular motion, and centres of mass. This note presents the route in full for students who choose it.

5 sections~15 min reading time3 competenciesLevel Standard 2 · Advanced 3

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Exam profile

AO1 · Apply mechanics principles and formulae accuratelyAO2 · Model a physical situation, stating and justifying assumptionsAO3 · Solve extended mechanics problems and interpret the results in context

Operators:findshow thatmodelhencededucecalculate

basic level

As an optional applied route, Mechanics is examined only for students entered for it; AS-level content covers momentum, work-energy and simpler circular motion.

higher level

The full A-Level route adds dimensional analysis, restitution in oblique and successive collisions, motion in a vertical circle, and centres of mass of composite bodies.

Depth

Text

Contents · 5 sections

Optional application: Mechanics

Dimensional analysis#

StandardAQA 7367 Mechanics optionDfE Further Mathematics (optional applied)

Every mechanical quantity is built from three base dimensions: mass

M

, length

L

and time

T

. Writing a quantity in terms of these — its dimensions — gives a powerful check and a way of discovering relationships. Velocity has dimensions

LT^{-1}

, acceleration

LT^{-2}

, force

MLT^{-2}

(from

F = ma

), energy and work

ML^2T^{-2}

, and power

ML^2T^{-3}

. These follow mechanically from the defining equations.

The principle of dimensional homogeneity states that every term in a physically valid equation must have the same dimensions: you cannot add a length to a time, and both sides of an equation must match dimensionally. This gives an immediate error check — if a derived formula is dimensionally inconsistent it is certainly wrong (though a dimensionally consistent formula is not guaranteed correct, since dimensionless constants are invisible to the method).

Dimensional analysis can also suggest the form of a relationship. If a quantity is believed to depend on certain others as a product of powers, equating the dimensions on both sides gives simultaneous equations for the powers. For example, assuming the period

T

of a simple pendulum depends on its length

l

, the gravitational acceleration

g

and possibly its mass

m

, dimensional balance forces the mass power to be zero and yields

T \propto \sqrt{l/g}

— recovering the known result up to a dimensionless constant.

The method's strength and its limitation are the same: it captures the powers of the base dimensions exactly but is blind to dimensionless factors such as

2\pi

. So dimensional analysis tells you that the pendulum period is proportional to

\sqrt{l/g}

but not that the constant is

2\pi

. Used honestly — as a check and as a guide to the structure of a formula, not as a substitute for a full derivation — it is a valuable modelling tool.

[\text{force}] = MLT^{-2}, \quad [\text{energy}] = ML^2T^{-2}, \quad [\text{power}] = ML^2T^{-3}

Dimensions of key quantities

Derived from the defining equations $F = ma$ , $E = Fd$ , $P = E/t$ .

Worked example

The period of a pendulum by dimensions

Assuming the period $T$ of a simple pendulum depends on its length $l$ , mass $m$ and $g$ as $T = k\,l^a m^b g^c$ , find $a$ , $b$ and $c$ .

Write the dimensions
$[T] = \mathsf{T}$ , $[l] = L$ , $[m] = M$ , $[g] = LT^{-2}$ . So $\mathsf{T} = L^a M^b (LT^{-2})^c = M^b L^{a+c} T^{-2c}$ .
Equate the powers
$M$ : $b = 0$ ; $L$ : $a + c = 0$ ; $T$ : $-2c = 1$ , so $c = -\tfrac12$ and $a = \tfrac12$ .
$a = \tfrac12, \quad b = 0, \quad c = -\tfrac12$
Assemble
$T = k\,l^{1/2} g^{-1/2} = k\sqrt{\dfrac{l}{g}}$ ; the mass does not appear.
Comment
This matches the known $T = 2\pi\sqrt{l/g}$ ; dimensional analysis fixes the powers but not the constant $2\pi$ .

Result: $T = k\sqrt{l/g}$ with $a = \tfrac12$ , $b = 0$ , $c = -\tfrac12$ .

Typical mistakes

Believing a dimensionally consistent equation must be correct — dimensionless constants are undetected.
Forgetting that added or subtracted terms must individually share the same dimensions.

Active revision

The drag force $F$ on a sphere is believed to depend on its speed $v$ , radius $r$ and the fluid density $\rho$ as $F = k\rho^a r^b v^c$ . Use dimensional analysis to find $a$ , $b$ and $c$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Momentum, impulse and collisions#

AdvancedAQA 7367 Mechanics optionDfE Further Mathematics (optional applied)

A direct collision of two particles

Fig. 1Before impact,

A

(2 kg) approaches at

5\,\text{m s}^{-1}

and

B

(3 kg) moves at

1\,\text{m s}^{-1}

in the same direction.

The linear momentum of a particle is

\mathbf{p} = m\mathbf{v}

, and impulse is the change it produces: a constant force

F

acting for time

t

delivers an impulse

Ft = \Delta(mv)

, more generally

\int F\,\mathrm{d}t

. Impulse and momentum share the same units, and the impulse-momentum principle is simply Newton's second law integrated over time — the tool for problems involving sudden forces such as blows or jerks.

In a collision between two bodies with no external horizontal force, the total momentum is conserved:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

, where

u

and

v

are the velocities before and after. Conservation of momentum alone gives one equation for the two unknown final velocities, so a second relation is needed to solve a collision completely.

That second relation is Newton's experimental law of restitution: the speed of separation is

e

times the speed of approach,

v_2 - v_1 = e(u_1 - u_2)

, where the coefficient of restitution

e

satisfies

0 \leq e \leq 1

. A perfectly elastic collision has

e = 1

(kinetic energy conserved); a perfectly inelastic collision has

e = 0

(the bodies coalesce). Solving the momentum equation together with the restitution equation determines both final velocities.

Care with signs and directions is essential: choose a positive direction, assign signed velocities consistently, and interpret a negative result as motion in the opposite sense. Kinetic energy is generally not conserved (unless

e = 1

), and the energy 'lost' — converted to heat, sound and deformation — can be computed as the difference in total kinetic energy before and after, a frequent final part of a collision question.

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad (\text{conservation of momentum})

Conservation of momentum

Total momentum is unchanged in the absence of external impulse.

v_2 - v_1 = e\,(u_1 - u_2), \qquad 0 \leq e \leq 1

Newton's law of restitution

Speed of separation is $e$ times speed of approach; $e=1$ is perfectly elastic.

Worked example

A direct collision with restitution

A particle $A$ of mass $2\,\text{kg}$ at $5\,\text{m s}^{-1}$ hits $B$ of mass $3\,\text{kg}$ at $1\,\text{m s}^{-1}$ (same direction), with $e = \tfrac12$ . Find the final velocities.

Conservation of momentum
$2(5) + 3(1) = 2v_1 + 3v_2$ , so $13 = 2v_1 + 3v_2$ .
Restitution
Speed of approach $= 5 - 1 = 4$ , so $v_2 - v_1 = \tfrac12(4) = 2$ .
Solve simultaneously
From the second, $v_2 = v_1 + 2$ ; substitute: $13 = 2v_1 + 3(v_1 + 2) = 5v_1 + 6$ , so $v_1 = 1.4$ and $v_2 = 3.4$ .
$v_1 = 1.4\ \text{m s}^{-1}, \quad v_2 = 3.4\ \text{m s}^{-1}$
Check
Both move in the original direction, and $A$ is now slower than $B$ (they have separated), as expected.

Result: $A$ moves at $1.4\,\text{m s}^{-1}$ and $B$ at $3.4\,\text{m s}^{-1}$ .

Typical mistakes

Assuming kinetic energy is conserved in a collision when $e < 1$ — only momentum is always conserved.
Sign errors in the restitution equation, or mixing up the speed of approach and the speed of separation.

Active revision

A particle of mass $2\,\text{kg}$ moving at $5\,\text{m s}^{-1}$ collides directly with a particle of mass $3\,\text{kg}$ moving at $1\,\text{m s}^{-1}$ in the same direction. Given $e = \tfrac12$ , find the velocities after the collision.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Work, energy and power#

StandardAQA 7367 Mechanics optionDfE Further Mathematics (optional applied)

Work is done when a force moves its point of application: for a constant force, work

= F\,d

along the direction of motion, or

Fd\cos\theta

when the force makes an angle

\theta

with the displacement. Its unit is the joule. Kinetic energy

\tfrac12 mv^2

and gravitational potential energy

mgh

measure stored energy of motion and of position respectively, and both have the same units as work.

The work-energy principle states that the total work done by all the forces on a body equals its change in kinetic energy:

W_{\text{total}} = \Delta\big(\tfrac12 mv^2\big)

. When only conservative forces (like gravity) act, mechanical energy is conserved and

\tfrac12 mv^2 + mgh

is constant; when friction or other resistive forces act, the work they do accounts for the mechanical energy lost. This principle often solves a problem far more directly than resolving forces and using the equations of motion.

Power is the rate of doing work,

P = \dfrac{\mathrm{d}W}{\mathrm{d}t}

, measured in watts. For a force driving a body at speed

v

, the instantaneous power is

P = Fv

. This form is the key to vehicle problems: at maximum speed the driving force equals the total resistance, so

P = (\text{resistance})\times v_{\max}

, which relates a vehicle's engine power to its top speed.

Elastic potential energy is stored in a stretched string or spring obeying Hooke's law,

T = \dfrac{\lambda x}{l}

, where

\lambda

is the modulus of elasticity,

l

the natural length and

x

the extension. The energy stored is

\dfrac{\lambda x^2}{2l}

, the area under the tension-extension graph. Including elastic PE in the energy balance lets you analyse systems with springs, and the whole energy method — identify the energies, apply conservation or the work-energy principle — is the unifying idea of this section.

W = Fd\cos\theta, \quad \text{KE} = \tfrac12 mv^2, \quad W_{\text{total}} = \Delta\!\left(\tfrac12 mv^2\right), \quad P = Fv

Work, energy and power

The work-energy principle and $P = Fv$ are the workhorses.

T = \frac{\lambda x}{l}, \qquad \text{elastic PE} = \frac{\lambda x^2}{2l}

Hooke's law and elastic energy

$\lambda$ is the modulus of elasticity, $l$ the natural length, $x$ the extension.

Worked example

Maximum speed from power and resistance

A car of mass $1200\,\text{kg}$ has engine power $30\,\text{kW}$ against a constant resistance $600\,\text{N}$ . Find its maximum speed on level ground.

Condition for maximum speed
At top speed the acceleration is zero, so the driving force $F$ equals the resistance: $F = 600\,\text{N}$ .
Relate power and speed
Power $P = Fv$ , so at maximum speed $30\,000 = 600\,v_{\max}$ .
$30\,000 = 600\,v_{\max}$
Solve
$v_{\max} = \dfrac{30\,000}{600} = 50\,\text{m s}^{-1}$ .
Comment
The mass is irrelevant to the top speed on level ground; it would matter for the acceleration at lower speeds.

Result: $v_{\max} = 50\,\text{m s}^{-1}$ .

Typical mistakes

Forgetting the work done against friction or air resistance when applying energy conservation.
Confusing the modulus of elasticity $\lambda$ with the spring constant, or omitting the natural length $l$ in the elastic-energy formula.

Active revision

A car of mass $1200\,\text{kg}$ has engine power $30\,\text{kW}$ and experiences a constant resistance of $600\,\text{N}$ . Find its maximum speed on a level road.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Circular motion#

AdvancedAQA 7367 Mechanics optionDfE Further Mathematics (optional applied)

Forces on a particle in circular motion

Fig. 2At the side of a vertical circle the tension points towards the centre (centripetal) while gravity acts vertically; their resultant is the net force.

A particle moving in a circle of radius

r

at constant speed

v

has angular speed

\omega = \dfrac{v}{r}

, and although its speed is constant its velocity is always changing direction. This change requires an acceleration directed towards the centre — the centripetal acceleration — of magnitude

a = \dfrac{v^2}{r} = r\omega^2

. By Newton's second law, a net inward force

F = \dfrac{mv^2}{r} = mr\omega^2

is needed to sustain the circular motion.

The essential modelling step is to identify what provides that centripetal force. It is never a new 'centripetal force' added to the diagram; rather, the resultant of the real forces already present (tension, gravity, normal reaction, friction) must point towards the centre and equal

\dfrac{mv^2}{r}

. Drawing a clear force diagram and resolving towards the centre is therefore the heart of every circular-motion problem.

For horizontal circular motion, such as a conical pendulum or a car on a banked track, the vertical forces balance while the horizontal resultant supplies the centripetal force. Resolving vertically and horizontally, then dividing the equations, typically eliminates the tension or reaction and yields the required speed or angle. The geometry of the cone or the bank sets the direction of the forces.

Vertical circular motion is richer because the speed varies: gravity does work as the particle rises and falls, so energy conservation must be combined with the centripetal equation. At the top of a vertical circle the minimum speed for the string to stay taut (or the particle to maintain contact) is found by setting the tension or reaction to zero, giving

v_{\text{top}}^2 = gr

; below this the particle leaves the circular path. Combining the work-energy principle with the centripetal condition at the critical point is the standard, and demanding, exam technique.

a = \frac{v^2}{r} = r\omega^2, \qquad F = \frac{mv^2}{r} = mr\omega^2

Centripetal acceleration and force

Directed towards the centre; supplied by the resultant of the real forces.

\text{at the top of a vertical circle, taut string: } v_{\text{top}}^2 \geq gr

Critical speed at the top

Set tension to zero for the minimum speed to maintain circular motion.

Worked example

Least speed at the top of a vertical circle

A particle of mass $0.5\,\text{kg}$ on a string of length $0.8\,\text{m}$ moves in a vertical circle. Find the least speed at the top for the string to stay taut ( $g = 9.8\,\text{m s}^{-2}$ ).

Forces at the top
At the top, both the weight $mg$ and the tension $T$ act downwards, towards the centre, so $T + mg = \dfrac{mv^2}{r}$ .
Condition for taut string
The string is on the point of going slack when $T = 0$ , giving $mg = \dfrac{mv^2}{r}$ , so $v^2 = gr$ .
$v_{\text{top}}^2 = gr = 9.8 \times 0.8 = 7.84$
Solve
$v_{\text{top}} = \sqrt{7.84} = 2.8\,\text{m s}^{-1}$ .

Result: The least speed at the top is $2.8\,\text{m s}^{-1}$ .

Typical mistakes

Adding a fictitious outward 'centrifugal' force or a separate inward centripetal force to the diagram.
Forgetting that the speed changes in a vertical circle, so energy conservation is needed alongside the force equation.

Active revision

A particle of mass $0.5\,\text{kg}$ is attached to a string of length $0.8\,\text{m}$ and moves in a vertical circle. Find the least speed at the top of the circle for the string to remain taut.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Centres of mass#

AdvancedAQA 7367 Mechanics optionDfE Further Mathematics (optional applied)

Centre of mass of an L-shaped lamina

Fig. 3Treating the L-shape as a

3\times 1

and a

1\times 2

rectangle gives the centre of mass at

(1.1, 1.1)

The centre of mass of a system of particles is the mass-weighted average of their positions:

\bar x = \dfrac{\sum m_i x_i}{\sum m_i}

, and similarly for

\bar y

. It is the point at which the whole mass may be considered to act, and around which the system balances. For a discrete set of particles the calculation is a direct weighted mean; the coordinate axes can be chosen to simplify the arithmetic.

For a uniform lamina the centre of mass lies on any axis of symmetry, which often locates it immediately. A composite lamina — one built from, or with a piece removed from, standard shapes — is handled by treating each part as a particle at its own centre of mass, with mass proportional to area. Summing the moments of the parts and dividing by the total area gives the centre of mass of the whole, and a removed piece is included with a negative area.

The centre of mass governs equilibrium and toppling. A body resting on a surface is in equilibrium only while the vertical line through its centre of mass falls within the base of support; when that line passes outside the base, the body topples. For a suspended body, the centre of mass hangs directly below the point of suspension, which is how a plumb-line experiment locates it and how hanging problems are solved.

These ideas connect to moments and equilibrium more broadly: the weight of an extended body acts at its centre of mass, so replacing a body by a particle there is exactly what makes moment calculations valid. Problems combine the centre-of-mass calculation with a moments or equilibrium condition — for instance finding the angle at which a suspended lamina hangs, or the greatest incline on which a body will not topple — synthesising the whole route.

\bar x = \frac{\sum m_i x_i}{\sum m_i}, \qquad \bar y = \frac{\sum m_i y_i}{\sum m_i}

Centre of mass

Mass-weighted average of positions; use areas for a uniform lamina.

Worked example

Centre of mass of an L-shape

An L-shaped uniform lamina is a $3\times 1$ rectangle (base) with a $1\times 2$ rectangle standing on its left. Find its centre of mass.

Split into rectangles
Rectangle 1: $3\times 1$ , area $3$ , centre $(1.5, 0.5)$ . Rectangle 2: $1\times 2$ (from $y=1$ to $3$ , $x=0$ to $1$ ), area $2$ , centre $(0.5, 2)$ .
Total area and x̄
Total area $5$ . $\bar x = \dfrac{3(1.5) + 2(0.5)}{5} = \dfrac{4.5 + 1}{5} = 1.1$ .
$\bar x = \frac{4.5 + 1}{5} = 1.1$
Find ȳ
$\bar y = \dfrac{3(0.5) + 2(2)}{5} = \dfrac{1.5 + 4}{5} = 1.1$ .
State
The centre of mass is at $(1.1, 1.1)$ , which lies inside the L-shape as expected.

Result: The centre of mass is at $(1.1, 1.1)$ .

Typical mistakes

Forgetting to weight each part by its mass or area, taking a plain average of the part centroids instead.
Getting the sign wrong for a removed piece, or misjudging where the vertical through the centre of mass falls relative to the base.

Active revision

A uniform lamina is an L-shape formed from a $3 \times 1$ rectangle and a $1 \times 2$ rectangle. Taking axes along the outer edges, find the coordinates of its centre of mass.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

References & sources

Sources

AQA

AQA A-level Further Mathematics 7367 specification

Optional application: Mechanics

Key points

The period of a pendulum by dimensions

Key points

A direct collision with restitution

Key points

Maximum speed from power and resistance

Key points

Least speed at the top of a vertical circle

Key points

Centre of mass of an L-shape

Sources