Dynamics

This lesson covers dynamics at A-Level — the study of how forces affect the motion of objects. While statics deals with objects in equilibrium, dynamics applies Newton's second law to objects that are accelerating. This lesson brings together forces, acceleration, and the equations of motion.

Newton's Second Law in Practice

The fundamental equation of dynamics is:

$F_{\text{net}} = ma$

where $F_{\text{net}}$ is the resultant (net) force, $m$ is the mass, and $a$ is the acceleration. The acceleration is always in the direction of the net force.

Strategy for dynamics problems:

Identify all forces acting on the object.
Draw a clear force diagram.
Choose a positive direction (usually the direction of motion).
Apply $F_{\text{net}} = ma$ in the chosen direction.
If necessary, resolve perpendicular to the direction of motion (often to find the normal reaction).

Motion on a Rough Horizontal Surface

For an object of mass $m$ on a rough horizontal surface with coefficient of friction $\mu$ , subject to a driving force $D$ and resistance $R_{\text{resist}}$ :

$D - F - R_{\text{resist}} = ma$

where $F = \mu mg$ is the friction force.

Example: A 1200 kg car has a driving force of 4000 N. Road resistance is 500 N and $\mu = 0.05$ . Find the acceleration.

$F = \mu mg = 0.05 \times 1200 \times 9.8 = 588$ N

$F_{\text{net}} = 4000 - 500 - 588 = 2912$ N

$a = \frac{2912}{1200} = 2.43$ m/s $^2$

Motion on an Inclined Plane

For an object on a smooth incline at angle $\theta$ :

Down the slope (taking down as positive):

$mg\sin\theta - T = ma \quad \text{(if a tension T acts up the slope)}$

Up the slope (taking up as positive):

$T - mg\sin\theta = ma$

For a rough incline, add friction $F = \mu R = \mu mg\cos\theta$ opposing the direction of motion.

Connected Particles

Particles on a Surface Connected by a String

Two objects A (mass $m_1$ ) and B (mass $m_2$ ) connected by a light inextensible string. A is pulled by a force $P$ across a smooth surface.

For A: $P - T = m_1 a$ For B: $T = m_2 a$

Adding: $P = (m_1 + m_2) a$ , giving $a = \frac{P}{m_1 + m_2}$

Atwood's Machine (Pulley System)

Two particles of masses $m_1$ and $m_2$ ( $m_1 > m_2$ ) connected by a string over a smooth pulley.

For $m_1$ (heavier, moving down): $m_1 g - T = m_1 a$ For $m_2$ (lighter, moving up): $T - m_2 g = m_2 a$

Adding: $(m_1 - m_2)g = (m_1 + m_2)a$

$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$

$T = \frac{2m_1 m_2 g}{m_1 + m_2}$

Particle on a Table Connected to a Hanging Mass

Mass $m_1$ on a smooth table, connected by a string over a pulley at the edge to a hanging mass $m_2$ .

For $m_1$ (horizontal): $T = m_1 a$ For $m_2$ (vertical): $m_2 g - T = m_2 a$

Adding: $m_2 g = (m_1 + m_2) a$

Exam Tip: For connected particle problems, always draw separate force diagrams for each object. The tension is the same in both diagrams (light string), and the acceleration has the same magnitude. Write $F = ma$ for each object and solve simultaneously.

Motion Under Variable Forces

When the force (and hence acceleration) varies with time, velocity, or position, use calculus:

$F = ma = m\frac{dv}{dt}$

This can be rearranged and integrated to find velocity as a function of time.

Summary

Apply $F_{\text{net}} = ma$ to analyse the motion of accelerating objects.
Always draw force diagrams and choose a consistent positive direction.
On inclined planes, resolve weight into components parallel and perpendicular to the slope.
For connected particles, write $F = ma$ separately for each object and solve simultaneously.
The tension in a light inextensible string is the same throughout, and connected objects share the same acceleration magnitude.
For variable forces, use $F = m \frac{dv}{dt}$ and integrate.

Exam Tip: Connected particle problems almost always require two equations (one for each particle) and two unknowns (tension and acceleration). Set up the equations carefully with consistent directions, then add or subtract to eliminate the tension.

A-Level Deep Dive: Dynamics — Connected Particles via a Smooth Pulley

Spec mapping

AQA A-Level Mathematics (7357) Paper 3 — Mechanics, Section U: Forces and Newton's Laws. This sub-section requires students to "apply Newton's second law to systems of connected particles, including those connected by a light inextensible string passing over a smooth pulley." Connected-particle pulley problems are a flagship Paper 3 topic — they synthesise Newton's laws (Section U), modelling assumptions (Section R), and frequently feed into SUVAT kinematics (Section S) once the acceleration is found. The AQA formula booklet provides no specific pulley formulae; candidates must derive everything from $F = ma$ applied separately to each particle. Pulley work also appears synoptically with friction (Section U continued) when one particle rests on a rough surface, and with vectors (Section T) when motion is resolved into components on an inclined plane.

Worked example with full mark scheme

Question (8 marks):

Two particles $A$ and $B$ of masses $3\,\text{kg}$ and $5\,\text{kg}$ respectively are connected by a light inextensible string. The string passes over a smooth fixed pulley, with $A$ and $B$ hanging vertically on either side. The system is released from rest with the string taut.

(a) Find the acceleration of the system and the tension in the string. (6)

(b) State two modelling assumptions you have used, and comment briefly on the effect each would have if it were relaxed. (2)

Take $g = 9.8\,\text{m s}^{-2}$ .

Solution with mark scheme:

(a) Step 1 — set up the equations of motion.

Because the string is inextensible, the magnitudes of the accelerations of $A$ and $B$ are equal; call this common magnitude $a$ . Because the pulley is smooth and the string is light, the tension $T$ is the same throughout the string. Particle $B$ (the heavier) descends; particle $A$ ascends.

For $B$ (taking downward positive):

$5g - T = 5a \quad (1)$

For $A$ (taking upward positive):

$T - 3g = 3a \quad (2)$

M1 — equation of motion written for one particle, with weight and tension shown in the correct directions and $F = ma$ applied. Common error: writing $5g + T = 5a$ for the descending particle, treating tension as accelerating the particle rather than opposing motion.

A1 — both equations correct, with consistent direction conventions.

Step 2 — solve simultaneously.

Add equations $(1)$ and $(2)$ to eliminate $T$ :

$5g - 3g = 5a + 3a$ $2g = 8a \implies a = \dfrac{g}{4} = \dfrac{9.8}{4} = 2.45\,\text{m s}^{-2}$

M1 — valid method of elimination (adding equations, or solving by substitution).

A1 — $a = 2.45\,\text{m s}^{-2}$ to 3 s.f.

Step 3 — back-substitute to find $T$ .

From $(2)$ : $T = 3a + 3g = 3(2.45) + 3(9.8) = 7.35 + 29.4 = 36.75\,\text{N}$ .

M1 — substitution into either original equation.

A1 — $T = 36.75\,\text{N}$ , or 36.8 N to 3 s.f. A useful sanity check: $T$ must lie strictly between the two weights ( $3g = 29.4$ N and $5g = 49$ N) — and 36.75 N does. Stating this check explicitly often picks up the final A mark even when arithmetic has slipped earlier.

(b) Two modelling assumptions and effects:

String is light (massless): allows the tension to be treated as uniform along the string. If the string had mass, tension would vary along its length (greater nearer the heavier particle's side because the string itself would need to be accelerated), and the single value $T$ in the equations would no longer apply.
Pulley is smooth: ensures no friction between string and pulley, so the tension on either side of the pulley is equal. If the pulley were rough, tension on the descending side would exceed tension on the ascending side (the difference accelerating-decelerating the string at the pulley's contact arc), reducing the net accelerating force and hence the system acceleration.

B1 — one assumption stated correctly with effect described.

B1 — second assumption stated correctly with effect described. Either of "string inextensible" (so accelerations are equal in magnitude — relaxing this would let the particles have different accelerations and the simple coupling would fail) or "no air resistance" is also acceptable.

Total: 8 marks (M3 A3 B2).

Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): Particles $P$ and $Q$ of masses $2\,\text{kg}$ and $7\,\text{kg}$ respectively are connected by a light inextensible string passing over a smooth pulley. The system is released from rest with both particles at the same height, $1.5\,\text{m}$ above the floor. Take $g = 9.8\,\text{m s}^{-2}$ .

(a) Find the acceleration of the system. (2)

(b) Find the speed of $Q$ at the instant it hits the floor. (2)

(c) After $Q$ hits the floor, the string becomes slack. Find the further height $P$ rises before momentarily coming to rest. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — applying $F = ma$ to the system overall (or to each particle): net driving force $(7 - 2)g = 5g$ , total mass $9\,\text{kg}$ , so $a = \dfrac{5g}{9}$ .
A1 (AO1.1b) — $a = \dfrac{5 \times 9.8}{9} \approx 5.44\,\text{m s}^{-2}$ .

(b)

M1 (AO3.1a) — applying SUVAT $v^2 = u^2 + 2as$ with $u = 0$ , $a = 5.44$ , $s = 1.5$ .
A1 (AO1.1b) — $v = \sqrt{2 \times 5.44 \times 1.5} = \sqrt{16.33} \approx 4.04\,\text{m s}^{-1}$ .

(c)

M1 (AO3.1b) — recognising that, once slack, $P$ is in free flight under gravity alone with initial upward speed $4.04\,\text{m s}^{-1}$ . Apply $v^2 = u^2 - 2gs$ with $v = 0$ .
A1 (AO2.5) — $s = \dfrac{u^2}{2g} = \dfrac{16.33}{19.6} \approx 0.833\,\text{m}$ .

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. Notice how parts (b) and (c) push the question into AO3 territory — applying mechanics across two distinct regimes (string taut, then slack) is exactly the synoptic skill the AQA Mechanics paper rewards.

Synoptic links

Connects to:

Section U — Newton's laws (foundation): every connected-particle problem rests on $F = ma$ applied separately to each particle, with tension as the coupling force. The technique generalises directly to three-particle systems, particles on inclined planes, and systems with multiple pulleys.
Section U — Friction: when one particle rests on a rough horizontal surface (or rough inclined plane) and is connected by a string over a smooth pulley to a hanging particle, the friction force $F = \mu R$ enters one of the equations of motion. The tension equation for the surface particle becomes $T - \mu m g = m a$ (horizontal surface) or $T - mg\sin\theta - \mu mg\cos\theta = ma$ (rough incline, sliding up). Setting $\mu = 0$ recovers the smooth case.
Section R — Modelling assumptions: "light, inextensible, smooth" is a triple of assumptions whose individual roles must be understood: light gives uniform tension along the string; inextensible gives equal acceleration magnitudes; smooth (of the pulley) gives equal tension on either side of the pulley. AQA explicitly tests "state and justify a modelling assumption" as a discrete sub-question.
Section S — SUVAT kinematics: once $a$ is found by Newton's laws, the motion of either particle becomes constant-acceleration. Subsequent sub-parts routinely ask for the speed at which one particle hits the ground ( $v^2 = u^2 + 2as$ ), the time taken ( $s = ut + \tfrac{1}{2}at^2$ ), or the further distance travelled after the string becomes slack.
Section T — Vectors / resolving forces: when the pulley sits at the top of an inclined plane, weights must be resolved along and perpendicular to the slope. The component along the slope ( $mg\sin\theta$ ) drives the motion; the component perpendicular ( $mg\cos\theta$ ) determines the normal reaction and hence any friction force. Pulley problems with inclines are the standard A* synthesis.

Mark-scheme literacy

Connected-particle pulley questions on 7357 Paper 3 split AO marks distinctively:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Setting up correct equations of motion, applying $F = ma$ with correct signs, performing the simultaneous solution
AO2 (reasoning / interpretation)	20–30%	Justifying that tension is uniform (citing light/smooth), justifying that accelerations have equal magnitude (citing inextensible), commenting on modelling assumptions
AO3 (problem-solving)	15–25%	Multi-stage problems (taut then slack, particle hits floor then rebounds, finding when particles next meet)

Mark-scheme-aligned phrasing earns method marks: "applying $F = ma$ to particle $A$ "; "since the string is inextensible, the accelerations have equal magnitude"; "since the pulley is smooth and the string is light, the tension is the same on both sides". Phrases that lose marks: writing $T_1$ and $T_2$ with no justification for using a single $T$ later; mixing direction conventions between the two particles' equations (a positive-up choice for one and positive-down for the other introduces a sign error if not handled with explicit care).

A specific AQA pattern: "comment on the modelling assumption that the pulley is smooth". A bare "this means tension is the same on both sides" earns one mark; the second mark requires saying "if the pulley were rough, the tensions would differ and the system acceleration would be smaller". Effects must be explicit, not implied.

Grade-band model answers

3-mark question

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

Grade C response (~120 words):

For the 3 kg particle (descending): $3g - T = 3a$ .

For the 2 kg particle (ascending): $T - 2g = 2a$ .

Add the equations: $3g - 2g = 5a$ , so $g = 5a$ and $a = g/5 = 1.96\,\text{m s}^{-2}$ .

Dynamics

Dynamics

Newton's Second Law in Practice

Motion on a Rough Horizontal Surface

Motion on an Inclined Plane

Connected Particles

Particles on a Surface Connected by a String

Atwood's Machine (Pulley System)

Particle on a Table Connected to a Hanging Mass

Motion Under Variable Forces

Summary

A-Level Deep Dive: Dynamics — Connected Particles via a Smooth Pulley

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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