Problem Solving in Mechanics

Mechanics is the area of A-Level Mathematics where the physical world meets mathematical modelling. Problem solving in mechanics requires you to translate physical situations into mathematical equations, make and evaluate modelling assumptions, and solve multi-step problems that often combine forces, motion, and energy.

The AQA A-Level Mathematics specification (7357) assesses mechanics in Paper 3 (Section A). Many mechanics questions are inherently problem-solving questions because they require you to model a real-world situation, set up equations, and interpret your results physically.

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Modelling Assumptions

One of the most important aspects of mechanics at A-Level is understanding the assumptions that underpin mathematical models. You must know what assumptions are being made and how they affect the validity of your model.

Common Modelling Assumptions

Assumption	Meaning	Effect
Particle	Object has mass but no size	Ignore rotational effects and air resistance due to shape
Light (string/rod)	Has no mass	Tension is the same throughout
Inextensible string	Does not stretch	Connected particles have the same acceleration
Smooth surface	No friction	No frictional force to consider
Rough surface	Friction acts	F = μR at the point of sliding
Rigid body	Does not deform	Shape does not change under forces
Uniform body	Mass is evenly distributed	Centre of mass is at the geometric centre
No air resistance	Ignore drag forces	Only gravity acts on a projectile
Constant g	g = 9.8 m s⁻² everywhere	Valid near Earth's surface
Light smooth pulley	Pulley has no mass and no friction	Tension is the same on both sides of the string

Exam Tip: If a question asks you to "state one modelling assumption," identify which simplification has been made and explain its physical meaning. For example, "The ball is modelled as a particle, so we ignore its size and air resistance."

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Connected Particles

Example 1: Two Particles on a Rough Inclined Plane

Problem: Particles A (mass 3 kg) and B (mass 5 kg) are connected by a light, inextensible string that passes over a smooth, light pulley at the top of a rough inclined plane. A rests on the plane inclined at 30° to the horizontal, and B hangs freely. The coefficient of friction between A and the plane is μ = 0.4. Find the acceleration of the system and the tension in the string.

Solution:

Draw a diagram. Let the acceleration be a m s⁻² and the tension be T N. Let the positive direction be: B moves down, A moves up the plane.

For particle B (hanging freely):

5g − T = 5a     ... (1)

For particle A (on the inclined plane):

Resolving perpendicular to the plane: R = 3g cos 30° = 3(9.8)(√3/2) = 25.46 N

Friction force: F = μR = 0.4 × 25.46 = 10.18 N (acts down the plane, opposing motion up the plane)

Resolving along the plane (up the plane positive for A):

T − 3g sin 30° − F = 3a
T − 3(9.8)(0.5) − 10.18 = 3a
T − 14.7 − 10.18 = 3a
T − 24.88 = 3a     ... (2)

Adding (1) and (2):

5g − 24.88 = 8a
49 − 24.88 = 8a
24.12 = 8a
a = 3.015 m s⁻²

From (1): T = 5g − 5a = 49 − 15.075 = 33.9 N (to 3 s.f.)

Therefore the acceleration is 3.02 m s⁻² and the tension is 33.9 N. ∎

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Projectile Motion

Example 2: Multi-Step Projectile Problem

Problem: A ball is projected from the top of a cliff 40 m above sea level with a speed of 20 m s⁻¹ at an angle of 30° above the horizontal. Find (a) the time taken for the ball to hit the sea, and (b) the horizontal distance from the base of the cliff to the point where the ball hits the sea. Take g = 9.8 m s⁻².

Solution:

Resolve the initial velocity:

Horizontal: uₓ = 20 cos 30° = 10√3 m s⁻¹
Vertical: u_y = 20 sin 30° = 10 m s⁻¹ (upwards)

Taking upwards as positive and the top of the cliff as the origin:

(a) When the ball hits the sea, the vertical displacement is s = −40 m.

Using s = ut + ½at²:

−40 = 10t − ½(9.8)t²
−40 = 10t − 4.9t²
4.9t² − 10t − 40 = 0

Using the quadratic formula:

t = (10 ± √(100 + 784)) / 9.8
t = (10 ± √884) / 9.8
t = (10 ± 29.73) / 9.8

Taking the positive root: t = (10 + 29.73)/9.8 = 39.73/9.8 = 4.054 s.

So t = 4.05 s (to 3 s.f.).

(b) Horizontal distance = uₓ × t = 10√3 × 4.054 = 70.2 m (to 3 s.f.). ∎

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Moments and Equilibrium

Example 3: Beam Problem

Problem: A uniform beam AB of length 6 m and mass 20 kg rests horizontally on two supports at C and D, where AC = 1 m and AD = 5 m. A child of mass 30 kg stands at point E, where AE = 2 m. Find the reaction forces at C and D.

Solution:

The beam is uniform, so its weight acts at the midpoint (3 m from A). Let Rc and Rd be the reaction forces at C and D respectively.

Taking moments about C (to eliminate Rc):

Rd × (5 − 1) = 30g × (2 − 1) + 20g × (3 − 1)
4Rd = 30g + 40g
4Rd = 70g
Rd = 17.5g = 171.5 N

Resolving vertically:

Rc + Rd = 30g + 20g = 50g
Rc = 50g − 17.5g = 32.5g = 318.5 N

Therefore Rc = 319 N and Rd = 172 N (to 3 s.f.). ∎

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Energy Methods

Example 4: Conservation of Energy on a Slope

Problem: A block of mass 2 kg slides down a rough slope inclined at 25° to the horizontal. The coefficient of friction is μ = 0.3. The block starts from rest and slides 4 m down the slope. Find the speed of the block at the bottom using energy methods.

Solution:

The vertical height descended: h = 4 sin 25° = 4 × 0.4226 = 1.690 m.

Normal reaction: R = 2g cos 25° = 2(9.8)(0.9063) = 17.76 N.

Friction force: F = μR = 0.3 × 17.76 = 5.329 N.

Work done against friction: W_f = F × d = 5.329 × 4 = 21.32 J.

Using the work-energy theorem (or conservation of energy with friction):

Gain in KE = Loss in PE − Work done against friction
½mv² = mgh − W_f
½(2)v² = 2(9.8)(1.690) − 21.32
v² = 33.12 − 21.32
v² = 11.80
v = 3.44 m s⁻¹

Therefore the speed at the bottom is 3.44 m s⁻¹ (to 3 s.f.). ∎

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Multi-Step Dynamics

Example 5: Variable Force and Constant Acceleration in Sequence

Problem: A car of mass 1200 kg accelerates from rest along a straight horizontal road. For the first 10 seconds, the driving force is 3600 N and the resistance is 600 N. After 10 seconds, the driver removes the driving force. How far does the car travel before coming to rest? Assume resistance remains at 600 N throughout.

Phase 1 (0 ≤ t ≤ 10):

Resultant force = 3600 − 600 = 3000 N.

Acceleration: a₁ = F/m = 3000/1200 = 2.5 m s⁻².

Speed after 10 s: v = u + at = 0 + 2.5(10) = 25 m s⁻¹.

Distance in Phase 1: s₁ = ½ × 2.5 × 10² = 125 m.

Phase 2 (decelerating under resistance only):

Resultant force = −600 N (opposing motion).

Deceleration: a₂ = 600/1200 = 0.5 m s⁻².

Using v² = u² + 2as with v = 0:

0 = 25² − 2(0.5)s₂
s₂ = 625/1 = 625 m

Total distance: s = s₁ + s₂ = 125 + 625 = 750 m. ∎

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Summary

• Mechanics problems require translating physical situations into mathematical models.
• Always state and evaluate modelling assumptions (particle, light string, smooth surface, etc.).
• For connected particles, draw separate force diagrams for each particle and use Newton's second law.
• For projectiles, resolve into horizontal and vertical components; use suvat equations.
• For beams, take moments about a point to eliminate unknown forces.
• Energy methods are powerful for problems involving changes in speed and height.
• Multi-step problems often involve two phases with different forces — find the transition conditions (speed, position) before moving to the next phase.

Exam Tip: In AQA mechanics questions, always draw a clear, labelled force diagram. This is not just good practice — it often earns marks. Show all forces (weight, normal reaction, friction, tension, applied forces) and state the positive direction. Multi-step problems require you to identify where one phase ends and another begins. Always check that your answer makes physical sense — if you get a negative speed or an impossibly large distance, re-check your working.

A-Level Deep Dive: Problem-Solving in Mechanics

Spec mapping

AQA 7357 specification, Section M (Mechanics), as examined in Paper 3 Section B: the specification requires students to "model situations in mechanics by making appropriate assumptions, identify the relevant principle (kinematics, Newton's laws, moments, friction), apply the correct technique, and evaluate the model". Mechanics problem-solving is the synoptic culmination of sub-strands M1 (quantities and units), M2 (kinematics in 1D and 2D), M3 (forces and Newton's laws), M4 (moments), and M6 (friction). Although Paper 3 Section A also tests Statistics, Section B is exclusively Mechanics and rewards strategic identification — which tools to use — over isolated technical skill. The AQA formula booklet supplies SUVAT equations and the projectile range formula, but does not supply Newton's laws, moment definitions, or the friction inequality $F \leq \mu R$F≤μR, all of which must be memorised.

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Worked example with full mark scheme

Question (8 marks):

A particle $P$P of mass $3\,\text{kg}$3kg rests on a rough plane inclined at $30°$30° to the horizontal. $P$P is connected by a light inextensible string passing over a smooth pulley at the top of the slope to a particle $Q$Q of mass $5\,\text{kg}$5kg hanging freely. The coefficient of friction between $P$P and the plane is $\mu = 0.2$μ=0.2. The system is released from rest with the string taut. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

(a) State two modelling assumptions implied by the description. (2)

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

Solution with mark scheme:

(a) B1, B1 — any two of:

• The string is light (massless), so tension is uniform along its length.
• The string is inextensible, so $P$P and $Q$Q have equal-magnitude accelerations.
• The pulley is smooth, so tension is unchanged across it.
• $P$P and $Q$Q are particles, so rotational motion and air resistance are ignored.

(b) Step 1 — draw the force diagram and resolve weight on the slope.

For $P$P on the incline, the weight $3g$3g resolves into:

• Component down the slope: $3g \sin(30°) = 3 \cdot 9.8 \cdot 0.5 = 14.7\,\text{N}$3gsin(30°)=3⋅9.8⋅0.5=14.7N.
• Component perpendicular to the slope (into the surface): $3g \cos(30°) = 3 \cdot 9.8 \cdot \dfrac{\sqrt{3}}{2} \approx 25.46\,\text{N}$3gcos(30°)=3⋅9.8⋅23​​≈25.46N.

M1 — resolving the weight on the inclined plane into parallel and perpendicular components.

Step 2 — normal reaction and friction on $P$P.

Perpendicular equilibrium: $R = 3g\cos(30°) \approx 25.46\,\text{N}$R=3gcos(30°)≈25.46N. Maximum friction on $P$P acts down the slope (opposing motion up the slope, since $Q$Q pulls $P$P upward):

$F = \mu R = 0.2 \cdot 25.46 \approx 5.09\,\text{N}$F=μR=0.2⋅25.46≈5.09N

M1 — correct friction magnitude with direction stated.

Step 3 — equation of motion for $P$P along the slope (positive up the slope).

$T - 3g\sin(30°) - F = 3a$T−3gsin(30°)−F=3a $T - 14.7 - 5.09 = 3a \quad (1)$T−14.7−5.09=3a(1)

M1 — Newton's second law applied to $P$P with all three forces correctly signed.

Step 4 — equation of motion for $Q$Q (positive downward).

$5g - T = 5a$5g−T=5a $49 - T = 5a \quad (2)$49−T=5a(2)

M1 — Newton's second law applied to $Q$Q.

Step 5 — solve the simultaneous system.

Add (1) and (2):

$49 - 14.7 - 5.09 = 8a \implies 29.21 = 8a \implies a \approx 3.65\,\text{m s}^{-2}$49−14.7−5.09=8a⟹29.21=8a⟹a≈3.65m s−2

Substitute into (2): $T = 49 - 5(3.65) = 49 - 18.25 = 30.75\,\text{N}$T=49−5(3.65)=49−18.25=30.75N.

A1 — $a \approx 3.65\,\text{m s}^{-2}$a≈3.65m s−2.

A1 — $T \approx 30.8\,\text{N}$T≈30.8N (3 s.f.).

Total: 8 marks (B2 + M4 A2).

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Specimen question modelled on the AQA 7357 Paper 3 Section B format

Question (6 marks): A uniform rod $AB$AB of length $4\,\text{m}$4m and mass $8\,\text{kg}$8kg rests horizontally on two supports at $A$A and at point $C$C, where $AC = 3\,\text{m}$AC=3m. A child of mass $25\,\text{kg}$25kg stands on the rod at point $D$D where $AD = x\,\text{m}$AD=xm. Find the largest value of $x$x for which the rod remains in equilibrium without tipping about $C$C. Take $g = 9.8\,\text{m s}^{-2}$g=9.8m s−2.

Mark scheme decomposition by AO:

• M1 (AO3.1b) — recognising that the rod is on the point of tipping about $C$C when the reaction at $A$A becomes zero. Setting $R_A = 0$RA​=0 is the modelling step that reduces the problem to a single moment equation.
• M1 (AO1.1a) — taking moments about $C$C. Clockwise moments (caused by the child at $D$D, acting beyond $C$C) must equal anticlockwise moments (caused by the rod's weight at the centre, on the $A$A side of $C$C).
• A1 (AO1.1b) — correct moment equation: $25g(x - 3) = 8g(3 - 2)$25g(x−3)=8g(3−2), i.e. $25(x - 3) = 8$25(x−3)=8.
• M1 (AO1.1b) — solving: $x - 3 = \dfrac{8}{25} = 0.32$x−3=258​=0.32.
• A1 (AO1.1b) — $x = 3.32\,\text{m}$x=3.32m.
• A1 (AO3.5a) — interpretation: "The child can stand at most $0.32\,\text{m}$0.32m beyond support $C$C before the rod tips. For $x > 3.32\,\text{m}$x>3.32m the model predicts equilibrium fails."

Total: 6 marks split AO1 = 3, AO3 = 3. This question is AO3-heavy because the strategic step ($R_A = 0$RA​=0 at the tipping point) is non-procedural — it requires the student to reframe equilibrium as a limiting case.

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Synoptic links

Connects to:

1. Kinematics (M2): every Newton's-second-law problem produces an acceleration, which then feeds SUVAT to find velocity at a later time, displacement covered, or time to reach a given speed. The connected-particle question above is incomplete unless paired with "find the speed of $Q$Q after it has descended $2\,\text{m}$2m" — $v^2 = u^2 + 2as = 0 + 2(3.65)(2)$v2=u2+2as=0+2(3.65)(2), so $v \approx 3.82\,\text{m s}^{-1}$v≈3.82m s−1.

2. Forces and Newton's laws (M3): Newton's third law — equal and opposite reactions — is the reason tension on $P$P and tension on $Q$Q have the same magnitude. The pulley acts as a force redirector, not a force generator. Confusing tension with weight is the most common synoptic slip.

3. Moments (M4): the rod-on-supports specimen above is a moments problem. Moments and Newton's laws unite under the principle "equilibrium requires zero net force and zero net moment about any point". Paper 3 Section B routinely demands both conditions.

4. Friction (M6): the friction inequality $F \leq \mu R$F≤μR is strict — $F = \mu R$F=μR holds only at the limit of static friction or while sliding. Below the limit, friction takes whatever value is required to maintain equilibrium. This subtlety distinguishes "find the minimum force to start the block moving" from "find the friction force when a 10 N push is applied".

5. Modelling (M1): every mechanics question begins with assumptions — light string, smooth pulley, particle (no rotation or air resistance), uniform rod (centre of mass at geometric centre). Stating these assumptions explicitly earns B-marks and signals examiner-aware practice.