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AQA A-level Physics revision notes

Mechanics and materials

Section 3.4
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
10 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.4

Checked against AQA 7408 section 3.4. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

In the exam: Data and formulae booklet provided · calculator allowed in every paper

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Scalars and vectors

Notes
Evidence from your answers: none yet
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A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Scalars have magnitude only; vectors have magnitude and direction. Speed, distance and mass are scalars, whereas velocity, displacement, acceleration, force and weight are vectors.
  • Vectors may be added by a scale drawing or, for two perpendicular vectors, by Pythagoras and trigonometry.
  • A vector may also be resolved into two perpendicular components; for a force FF at angle θ\theta to the horizontal, these are Fcos⁡θF\cos\theta horizontally and Fsin⁡θF\sin\theta vertically.
  • For two or three coplanar forces acting at a point, equilibrium means zero resultant force, so the object is at rest or moves with constant velocity.
  • Examiners expect both a correct magnitude and an unambiguous direction.
A vector resolved into perpendicular horizontal and vertical components.
Worked example

A velocity has components 6.0 m s−16.0\,\text{m s}^{-1} east and 8.0 m s−18.0\,\text{m s}^{-1} north. Determine its magnitude and direction.

  1. 1.Use perpendicular components: v=6.02+8.02=10.0 m s−1v=\sqrt{6.0^2+8.0^2}=10.0\,\text{m s}^{-1}.
  2. 2.Measure the direction from east: θ=tan⁡−1(8.0/6.0)=53.1∘\theta=\tan^{-1}(8.0/6.0)=53.1^{\circ}.
  3. 3.State the complete vector direction as north of east.

Answer: 10.0 m s−110.0\,\text{m s}^{-1} at 53.1∘53.1^{\circ} north of east.

Common mistakes

  • Don't use Fsin⁡θF\sin\theta for the component adjacent to the stated angle.
  • Don't treat equilibrium as meaning stationary, excluding constant-velocity motion.
  • Don't give an angle without saying which compass direction or axis it is measured from.

Exam tip

For a calculation, show the two perpendicular components before quoting the resultant magnitude and direction.

Tier 1 · Easy

ORIGINAL

A boat has velocity components 3.0 m s−13.0\,\text{m s}^{-1} east and 4.0 m s−14.0\,\text{m s}^{-1} north. Determine the magnitude and direction of its velocity.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A 12.0 kg12.0\,\text{kg} crate rests on a smooth plane inclined at 30.0∘30.0^{\circ} to the horizontal. Resolve its weight into components parallel and perpendicular to the plane.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Two forces act at a point. One is 180 N180\,\text{N} due east. The other is 120 N120\,\text{N} at 110∘110^{\circ} anticlockwise from east. Determine the magnitude and direction of the single force that would produce equilibrium.

[5 marks]

Total for this question: 5

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Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Moments

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The moment of a force about a point is M=Fd⊥M=Fd_{\perp}, where d⊥d_{\perp} is the perpendicular distance from the point to the force’s line of action.
  • For rotational equilibrium, the total clockwise moment equals the total anticlockwise moment; full equilibrium also requires zero resultant force.
  • A couple consists of equal and opposite coplanar forces on different lines of action, producing a moment equal to one force multiplied by the perpendicular separation of the lines.
  • The weight of a uniform regular solid acts through its centre of mass at its geometric centre.
  • Examiners expect the chosen pivot, sense of each moment and perpendicular distances to be clear, followed by a separate force balance if an unknown reaction is required.
A downward force on a horizontal beam, showing its perpendicular distance from the pivot.
Worked example

A uniform 4.0 m4.0\,\text{m} beam of weight 200 N200\,\text{N} is hinged at one end and supported vertically at the other. Determine the support force.

  1. 1.The beam’s weight acts at its centre, 2.0 m2.0\,\text{m} from the hinge.
  2. 2.Take moments about the hinge: R(4.0)=200(2.0)R(4.0)=200(2.0).
  3. 3.Evaluate the support force: R=100 NR=100\,\text{N}.

Answer: The vertical support force is 100 N100\,\text{N} upward.

Common mistakes

  • Don't use the distance along a sloping beam instead of the perpendicular distance to the force line.
  • Don't omit the beam’s own weight when its mass or weight is given.
  • Don't balance moments and fail to balance forces when a hinge reaction is also required.

Exam tip

Taking moments about a point through an unknown reaction removes that reaction from the moment equation.

Tier 1 · Easy

ORIGINAL

A 25 N25\,\text{N} force acts perpendicular to a spanner 0.40 m0.40\,\text{m} from its pivot. Calculate the moment of the force about the pivot.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

A uniform horizontal beam of length 3.0 m3.0\,\text{m} and weight 180 N180\,\text{N} is hinged at its left end and supported vertically at its right end. Determine the vertical support force and the vertical force exerted by the hinge.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A uniform horizontal beam is 5.0 m5.0\,\text{m} long and weighs 240 N240\,\text{N}. It is hinged to a wall at one end. A 360 N360\,\text{N} load hangs 3.8 m3.8\,\text{m} from the hinge, and a cable attached to the free end makes 35∘35^{\circ} above the beam. Determine the cable tension and the magnitude and direction of the force exerted by the hinge on the beam.

[6 marks]

Total for this question: 6

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Motion along a straight line

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Velocity is the rate of change of displacement, v=ΔsΔtv=\dfrac{\Delta s}{\Delta t}, and acceleration is the rate of change of velocity, a=ΔvΔta=\dfrac{\Delta v}{\Delta t}; instantaneous values are gradients of tangents, average values use the whole interval. On a velocity–time graph the gradient is the acceleration and the area is the displacement. For uniform acceleration: v=u+atv=u+at, s=(u+v)2ts=\dfrac{(u+v)}{2}t, s=ut+12at2s=ut+\dfrac{1}{2}at^2 and v2=u2+2asv^2=u^2+2as. A freely falling object accelerates at g=9.81 m s−2g=9.81\,\text{m s}^{-2}; required practical 3 determines gg from a graph.

Explanation

  • Velocity is the rate of change of displacement, v=Δs/Δtv=\Delta s/\Delta t, and acceleration is the rate of change of velocity, a=Δv/Δta=\Delta v/\Delta t. An instantaneous value is obtained from the tangent gradient, whereas an average value uses the change over an interval.
  • On a displacement–time graph the gradient is velocity; on a velocity–time graph the gradient is acceleration and signed area is displacement; on an acceleration–time graph signed area is change in velocity.
  • The equations of uniform acceleration apply only when acceleration is constant.
  • For free fall near Earth, acceleration has magnitude g=9.81 m s−2g=9.81\,\text{m s}^{-2}, with its sign set by the chosen positive direction.
  • Required practical work may determine gg from an appropriate graph and evaluate random and systematic errors.
A velocity–time graph whose gradient gives acceleration and whose signed area gives displacement.
Worked example

A car is travelling along a straight road at 27 m s−127\,\text{m s}^{-1}. The driver brakes and the car decelerates uniformly, coming to rest in 4.5 s4.5\,\text{s}. Calculate the deceleration of the car and the distance it travels while braking.

[4 marks]

Total for this question: 4

  1. 1.a=ΔvΔt=0−274.5a=\dfrac{\Delta v}{\Delta t}=\dfrac{0-27}{4.5}.
  2. 2.a=−6.0 m s−2a=-6.0\,\text{m s}^{-2}, a deceleration of 6.0 m s−26.0\,\text{m s}^{-2}.
  3. 3.Distance from s=(u+v)2t=27+02×4.5s=\dfrac{(u+v)}{2}t=\dfrac{27+0}{2}\times4.5 (the area of the triangle under the velocity–time graph).
  4. 4.s=60.75 m≈61 ms=60.75\,\text{m}\approx61\,\text{m} (2 significant figures).

Answer: Deceleration 6.0 m s−26.0\,\text{m s}^{-2}; braking distance 61 m61\,\text{m}

Required practical 3

Determination of g by a free-fall method

Apparatus

  • Electromagnet holding a steel ball, with a switch that starts an electronic timer when the current is cut
  • Trapdoor or light gate at the bottom that stops the timer when the ball arrives
  • Metre rule (or tape) to measure the drop height hh from the bottom of the ball to the trapdoor

Method

  1. Clamp the electromagnet so the ball hangs directly above the trapdoor; measure hh from the bottom of the ball to the trapdoor with a metre rule and set square.
  2. Switch off the electromagnet: the ball falls and the timer records tt; repeat three times and take a mean.
  3. Repeat for at least six heights from about 0.2 m0.2\,\text{m} to 1.5 m1.5\,\text{m}.
  4. Using h=12gt2h=\dfrac{1}{2}gt^2, plot hh against t2t^2: the graph is a straight line of gradient g/2g/2, so g=2×gradientg=2\times\text{gradient}.
  5. A non-zero intercept shows a systematic error — a delay in release or in the timer — which the gradient method removes; scatter about the line shows random timing error.

Variables

Independent:
height of fall hh
Dependent:
time of fall tt
Control:
the same steel ball (so air resistance and the release are the same); the same release mechanism and timer; measuring hh from the same point on the ball each time

Hazards and precautions

  • The falling ball can land on feet or bounce: keep feet clear of the trapdoor and use a tray to catch the ball.
  • The electromagnet's supply and the clamp stand: keep the stand weighted so it cannot topple, and switch off the supply when adjusting.

Common mistakes

  • Don't use the gradient of a velocity–time graph as displacement.
  • Don't treat the area below the time axis as positive when finding displacement.
  • Don't apply a uniform-acceleration equation across a curved graph section.

Mark-scheme wording to use

  • identify gradient as acceleration and area as vertical displacement — paraphrase of AQA 7408, November 2021 Paper 1, Q05.5
  • SUVAT not credited for curved graph — paraphrase of AQA 7408, June 2018 Paper 1, Q01.2
  • calculate acceleration 5.7 m s−2 … time about 5.5 s … distance about 85 m — paraphrase of AQA 7408, November 2020 Paper 1, Q04.3
  • use quadratic formula … both 0.258 s and 2.37 s … rising and falling meanings — paraphrase of AQA 7408, November 2021 Paper 1, Q05.3
  • g = 2f2 x gradient — paraphrase of AQA 7408, June 2023 Paper 3, Q01.3

What the examiner wants

Say which graph property you are using before you use it — gradient for acceleration, area for displacement — and never apply a suvat equation across a curved section: AQA's scheme credits only the area or tangent method there. Keep a sign convention (up positive, g=−9.81g=-9.81) and state it; a quadratic in tt has two roots and you must say which one the question wants. Using gg as the acceleration of a body on a slope scores nothing.

Exam tip

For graph questions, state whether a gradient or an area is being used before evaluating it.

Tier 1 · Easy

ORIGINAL

An object moves at constant velocity 7.5 m s−17.5\,\text{m s}^{-1} for 4.0 s4.0\,\text{s}. Calculate its displacement.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

A car accelerates uniformly from 5.0 m s−15.0\,\text{m s}^{-1} to 17.0 m s−117.0\,\text{m s}^{-1} in 6.0 s6.0\,\text{s}. Determine its acceleration and displacement during this interval.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A velocity-time graph consists of three straight sections: velocity rises from 00 to 18 m s−118\,\text{m s}^{-1} in 5.0 s5.0\,\text{s}, remains at 18 m s−118\,\text{m s}^{-1} for 4.0 s4.0\,\text{s}, then falls uniformly to −6.0 m s−1-6.0\,\text{m s}^{-1} in 3.0 s3.0\,\text{s}. Determine the final acceleration, total displacement and total distance travelled.

[5 marks]

Total for this question: 5

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Projectile motion

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In a uniform gravitational field with air resistance neglected, horizontal and vertical motions are independent. Horizontal velocity remains constant, while vertical acceleration is gg downward.
  • Resolve the launch velocity into perpendicular components, apply the uniform-acceleration equations separately, and use the common time to connect the two directions. At maximum height the vertical velocity is zero, but the horizontal velocity is not.
  • Friction, lift and drag require qualitative treatment: air resistance increases with speed, reduces range and makes a projectile trajectory asymmetric. Terminal speed occurs when drag balances weight, making the resultant force and acceleration zero.
  • For a vehicle, maximum speed occurs when the driving force is balanced by resistive forces.
  • Examiners expect the no-air-resistance model to be stated before calculations and drag effects to be described through forces and acceleration.
A projectile trajectory with the launch velocity resolved horizontally and vertically and gravity acting downward.
Worked example

A ball leaves a horizontal platform at 15.0 m s−115.0\,\text{m s}^{-1} and falls 19.6 m19.6\,\text{m}. Air resistance is negligible. Determine the flight time and horizontal range.

  1. 1.Use vertical motion with uy=0u_y=0: 19.6=12(9.81)t219.6=\tfrac12(9.81)t^2.
  2. 2.Evaluate the common time: t=2(19.6)/9.81=2.00 st=\sqrt{2(19.6)/9.81}=2.00\,\text{s}.
  3. 3.Use constant horizontal velocity: x=15.0×2.00=30.0 mx=15.0\times2.00=30.0\,\text{m}.

Answer: The flight time is 2.00 s2.00\,\text{s} and the range is 30.0 m30.0\,\text{m}.

Common mistakes

  • Don't use gg as a horizontal acceleration when air resistance is neglected.
  • Don't set the total velocity to zero at maximum height instead of only the vertical component.
  • Don't use different flight times for the horizontal and vertical calculations.

Exam tip

Resolve the launch velocity first, then write separate horizontal and vertical equations linked by the same time.

Tier 1 · Easy

ORIGINAL

Air resistance is neglected. State the horizontal and vertical accelerations of a projectile after it has been released.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A ball leaves a horizontal platform at 12.0 m s−112.0\,\text{m s}^{-1}. The platform is 20.0 m20.0\,\text{m} above level ground. Air resistance is neglected. Determine the time to reach the ground and the horizontal range.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A projectile is launched from level ground at 22.0 m s−122.0\,\text{m s}^{-1} and 35.0∘35.0^{\circ} above the horizontal. It lands at the launch height. Air resistance is neglected. Determine its time of flight, horizontal range and maximum height.

[6 marks]

Total for this question: 6

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Newton's laws of motion

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Newton's first law: an object stays at rest or continues at constant velocity unless a resultant external force acts on it. Newton's second law: for constant mass the resultant force equals mass × acceleration, F=maF=ma, with the acceleration in the direction of the resultant force. Newton's third law: when object A exerts a force on object B, B exerts a force on A that is equal in magnitude, opposite in direction and of the same type — the two forces act on different objects, so they never cancel on one free-body diagram.

Explanation

  • Newton’s first law states that an object remains at rest or moves with constant velocity unless acted on by a resultant external force.
  • For constant mass, the second law is ∑F=ma\sum F=ma; a free-body diagram should show only forces acting on the chosen object.
  • Newton’s third-law pairs are equal and opposite forces of the same interaction acting on different objects, so they do not cancel on one object’s diagram.
  • Weight is mgmg, while normal contact force, tension and drag must be obtained from the situation rather than assumed equal to weight.
  • Examiners expect a declared positive direction, a signed resultant-force equation and a third-law pair identified by naming both interacting objects.
A free-body diagram for a pulled block, showing normal contact force, weight, pull and friction.
Worked example

A trolley of mass 1.60 kg1.60\,\text{kg} rests on a smooth horizontal runway. A string attached to the trolley passes over a frictionless pulley to a hanging mass of 0.40 kg0.40\,\text{kg}. The system is released from rest. Calculate the acceleration of the trolley and the tension in the string.

[4 marks]

Total for this question: 4

  1. 1.Treat trolley and hanging mass as one system: the only resultant force along the motion is the weight of the hanging mass, F=0.40×9.81=3.92 NF=0.40\times9.81=3.92\,\text{N}.
  2. 2.Newton's second law for the whole system: a=FM+m=3.921.60+0.40a=\dfrac{F}{M+m}=\dfrac{3.92}{1.60+0.40}.
  3. 3.a=1.96 m s−2a=1.96\,\text{m s}^{-2} (both masses accelerate together at this rate).
  4. 4.For the trolley alone the only horizontal force is the tension: T=Ma=1.60×1.96=3.1 NT=Ma=1.60\times1.96=3.1\,\text{N} (check: for the hanging mass mg−T=mamg-T=ma gives 3.92−3.14=0.78=0.40×1.963.92-3.14=0.78=0.40\times1.96).

Answer: a=1.96 m s−2a=1.96\,\text{m s}^{-2}; T=3.1 NT=3.1\,\text{N}

Common mistakes

  • Don't use the applied force rather than the resultant force in F=maF=ma.
  • Don't place a third-law partner on the same free-body diagram even though it acts on another object.
  • Don't conclude that zero resultant force means zero velocity.

Mark-scheme wording to use

  • “If used g for acceleration then no marks awarded” — AQA 7408, June 2017 Paper 1, Q06.4
  • “Uses appropriate Newton law of motion to link evidence” — AQA 7408, June 2022 Paper 1, Q05.4
  • inward centripetal force … link floor reaction/contact force to that centripetal force — paraphrase of AQA 7408, November 2021 Paper 1, Q04.1
  • increasing stopping time or distance … lower force using impulse, work-energy or F=ma — paraphrase of AQA 7408, June 2024 Paper 1, Q05.3
  • Maximum 3 if N=mg is assumed — paraphrase of AQA 7408, November 2020 Paper 1, Q04.5

What the examiner wants

Draw the free-body diagram for ONE object, declare a positive direction and write a signed resultant-force equation before substituting; the acceleration of a body on a slope or on a string is never gg — AQA gives no marks for using gg as the acceleration. Do not assume the normal reaction equals the weight when the surface is banked or the object accelerates vertically. When asked for evidence that a force acts, quote the law that links it: a changing direction (velocity) needs a resultant force by Newton's first law.

Exam tip

A free-body diagram and one signed equation for the chosen direction usually secure the method before any arithmetic.

Tier 1 · Easy

ORIGINAL

A resultant force of 12 N12\,\text{N} acts on a 4.0 kg4.0\,\text{kg} object. Calculate its acceleration.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A lift of total mass 750 kg750\,\text{kg} accelerates vertically upward at 1.20 m s−21.20\,\text{m s}^{-2}. Determine the tension in its supporting cable.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Blocks A and B, of masses 4.0 kg4.0\,\text{kg} and 6.0 kg6.0\,\text{kg}, are joined by a light horizontal cord. A horizontal force of 38 N38\,\text{N} pulls B. Friction on A is 5.0 N5.0\,\text{N} and friction on B is 7.0 N7.0\,\text{N}, both opposing motion. Determine the acceleration and cord tension. Identify the Newton's-third-law partner of the cord's pull on A.

[5 marks]

Total for this question: 5

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Momentum

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Momentum is mass × velocity, p=mvp=mv (a vector, unit kg m s−1\text{kg m s}^{-1} or N s\text{N s}). In an isolated system total momentum is conserved: the vector sum before a collision or explosion equals the vector sum after. Force is the rate of change of momentum, F=Δ(mv)ΔtF=\dfrac{\Delta(mv)}{\Delta t}, so impulse FΔtF\Delta t equals the change in momentum and is the area under a force–time graph. An elastic collision also conserves total kinetic energy; an inelastic one does not.

Explanation

  • Linear momentum is the vector p=mvp=mv. Total momentum is conserved in an isolated system, so the signed total before a one-dimensional collision or explosion equals the signed total afterwards.
  • Force is the rate of change of momentum, F=Δ(mv)/ΔtF=\Delta(mv)/\Delta t, and impulse is the change in momentum. For a constant force, impulse is FΔtF\Delta t; for a varying force, it is the signed area under the force–time graph.
  • Elastic collisions conserve both momentum and kinetic energy, whereas inelastic collisions conserve momentum but not kinetic energy.
  • Examiners expect a positive direction, signed velocities and a clear distinction between momentum conservation and kinetic-energy conservation.
  • Crumple zones and protective packaging increase contact time, reducing mean impact force for the same momentum change; ethical transport design applies this principle to occupant safety.
A force–time pulse whose area represents impulse and hence change in momentum.
Worked example

A railway truck A of mass 12 000 kg12\,000\,\text{kg} moving at 1.5 m s−11.5\,\text{m s}^{-1} collides with a stationary truck B of mass 8000 kg8000\,\text{kg}. The trucks join together. Calculate the speed of the joined trucks immediately after the collision and the impulse on truck B. Explain whether the collision is elastic.

[5 marks]

Total for this question: 5

  1. 1.Conservation of momentum (take A's direction as positive): 12 000×1.5+8000×0=(12 000+8000)×v12\,000\times1.5+8000\times0=(12\,000+8000)\times v.
  2. 2.v=18 00020 000=0.90 m s−1v=\dfrac{18\,000}{20\,000}=0.90\,\text{m s}^{-1}.
  3. 3.Impulse on B == its change in momentum =8000×(0.90−0)=7200 N s=8000\times(0.90-0)=7200\,\text{N s} (or kg m s−1\text{kg m s}^{-1}), in A's original direction.
  4. 4.Kinetic energy before: 12×12 000×1.52=13 500 J\dfrac{1}{2}\times12\,000\times1.5^2=13\,500\,\text{J}; after: 12×20 000×0.902=8100 J\dfrac{1}{2}\times20\,000\times0.90^2=8100\,\text{J}.
  5. 5.Total kinetic energy is not conserved (5400 J5400\,\text{J} is transferred to internal energy and sound), so the collision is inelastic — momentum is conserved either way.

Answer: v=0.90 m s−1v=0.90\,\text{m s}^{-1}; impulse on B =7200 N s=7200\,\text{N s}; inelastic because total kinetic energy falls from 13 500 J13\,500\,\text{J} to 8100 J8100\,\text{J}

Common mistakes

  • Don't substitute rebound speed as positive after choosing the incident direction as positive.
  • Don't claim that kinetic energy must be conserved in every collision.
  • Don't use the peak force instead of the force–time area to find impulse.

Mark-scheme wording to use

  • momentum conservation; correct signed substitution — paraphrase of AQA 7408, June 2018 Paper 1, Q06.2
  • “N s or kg m s−1” — AQA 7408, June 2018 Paper 1, Q06.3
  • “total kinetic energy” — AQA 7408, June 2018 Paper 1, Q06.1
  • use area under graph = impulse — paraphrase of AQA 7408, June 2025 Paper 1, Q06.2
  • greater contact time or stopping distance … linked impulse/work/F=ma smaller force — paraphrase of AQA 7408, June 2019 Paper 1, Q05.5
  • the same change in momentum, velocity or kinetic energy — paraphrase of AQA 7408, June 2024 Paper 1, Q05.3

What the examiner wants

Sign the velocities before substituting — a rebound is negative — and give the unit of impulse (N s\text{N s} or kg m s−1\text{kg m s}^{-1}), which is often its own mark. A safety explanation is three linked steps: longer contact time, the SAME change in momentum, therefore a smaller force from F=Δp/ΔtF=\Delta p/\Delta t. For a force–time graph, say 'area under the graph = impulse' and show the square-counting; AQA does not accept the number alone, and does not allow rounding errors in a 'show that'.

Exam tip

Write the positive direction beside the momentum equation before substituting any velocity.

Tier 1 · Easy

ORIGINAL

A 0.80 kg0.80\,\text{kg} trolley moving at 3.0 m s−13.0\,\text{m s}^{-1} collides with a stationary 0.40 kg0.40\,\text{kg} trolley. They join, so the collision is perfectly inelastic. Calculate their common velocity.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A 0.40 kg0.40\,\text{kg} trolley moving at 5.0 m s−15.0\,\text{m s}^{-1} collides with a stationary 0.60 kg0.60\,\text{kg} trolley. Afterwards their velocities are −1.0 m s−1-1.0\,\text{m s}^{-1} and 4.0 m s−14.0\,\text{m s}^{-1} respectively. Show that the collision is elastic.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A 1200 kg1200\,\text{kg} car moving east at 18 m s−118\,\text{m s}^{-1} collides head-on with an 800 kg800\,\text{kg} car moving west at 6.0 m s−16.0\,\text{m s}^{-1}. They lock together, so the collision is perfectly inelastic. Determine their velocity, the kinetic energy dissipated, and the mean force on the 1200 kg1200\,\text{kg} car if contact lasts 0.18 s0.18\,\text{s}.

[6 marks]

Total for this question: 6

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Work, energy and power

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Work done is the energy transferred by a force: W=Fscos⁡θW=Fs\cos\theta, where θ\theta is the angle between the force and the displacement, and for a varying force the work is the area under a force–displacement graph. Power is the rate of doing work or of transferring energy, P=ΔWΔtP=\dfrac{\Delta W}{\Delta t}, and for a force moving its point of application at velocity vv along its own line, P=FvP=Fv. Efficiency =useful output powerinput power=\dfrac{\text{useful output power}}{\text{input power}}, quoted as a decimal or a percentage.

Explanation

  • Work is energy transferred by a force. For a constant force, W=Fscos⁡θW=Fs\cos\theta, where θ\theta is the angle between force and displacement.
  • For a variable force, work is the area under the force–displacement graph. Power is the rate of doing work or transferring energy, P=ΔW/ΔtP=\Delta W/\Delta t; when a force is parallel to steady velocity, P=FvP=Fv.
  • Efficiency is useful output energy divided by total input energy, or useful output power divided by total input power, and may be quoted as a decimal or percentage.
  • A motor-lifting experiment compares measured useful output power with electrical input power and evaluates random and systematic errors.
  • Examiners expect the parallel force component, correct graph area and like quantities in an efficiency ratio.
A variable force–displacement graph with the area under the curve representing work done.
Worked example

A sledge and rider of total mass 45 kg45\,\text{kg} start from rest at the top of a slope and slide 25 m25\,\text{m} down it, losing 6.0 m6.0\,\text{m} in height. At the bottom the sledge is moving at 8.0 m s−18.0\,\text{m s}^{-1}. Calculate the average frictional force acting on the sledge during the descent.

[4 marks]

Total for this question: 4

  1. 1.Loss of gravitational potential energy: ΔEp=mgΔh=45×9.81×6.0=2649 J\Delta E_p=mg\Delta h=45\times9.81\times6.0=2649\,\text{J}.
  2. 2.Kinetic energy gained: Ek=12mv2=12×45×8.02=1440 JE_k=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times45\times8.0^2=1440\,\text{J}.
  3. 3.Work done against friction =2649−1440=1209 J=2649-1440=1209\,\text{J}.
  4. 4.Average frictional force F=Ws=120925=48 NF=\dfrac{W}{s}=\dfrac{1209}{25}=48\,\text{N} (2 significant figures).

Answer: F≈48 NF\approx48\,\text{N}

Common mistakes

  • Don't use the whole force in W=FsW=Fs when the force is not parallel to the displacement.
  • Don't use one endpoint force times the full displacement for a varying force.
  • Don't divide an output energy by an input power when calculating efficiency.

Mark-scheme wording to use

  • GPE loss about 1575 J … final kinetic energy about 754 J … work against friction about 821 J … average force about 46 N — paraphrase of AQA 7408, June 2024 Paper 1, Q05.2
  • use P = Fv to obtain about 106 MW — paraphrase of AQA 7408, June 2023 Paper 1, Q02.3
  • Reject power calculated on an assumption of constant speed — paraphrase of AQA 7408, June 2023 Paper 1, Q02.3
  • specified heating/internal-energy transfer — paraphrase of AQA 7408, June 2018 Paper 1, Q01.3
  • Marks for area method; correct square count/scale — paraphrase of AQA 7408, June 2018 Paper 1, Q01.2

What the examiner wants

Energy accounting wins marks stage by stage: potential energy lost, kinetic energy gained, the difference is work done against friction, then divide by the distance moved ALONG the force. Use the component of force parallel to the displacement, and only use P=FvP=Fv with the force that acts at that instant — AQA rejects a power found by assuming constant speed when the object is accelerating. Say WHERE dissipated energy goes: heating (internal energy) of a named object, not 'lost'.

Exam tip

For a graph-based work question, identify the required geometric areas and include their units before summing them.

Tier 1 · Easy

ORIGINAL

A constant force of 75 N75\,\text{N} acts in the direction of motion through 12 m12\,\text{m}. Calculate the work done.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

A motor raises a 25 kg25\,\text{kg} load vertically through 8.0 m8.0\,\text{m} in 6.0 s6.0\,\text{s}. Its electrical input power is 420 W420\,\text{W}. Determine its useful output power and efficiency.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A force acting on a 2.0 kg2.0\,\text{kg} trolley increases linearly from 10 N10\,\text{N} to 50 N50\,\text{N} over the first 4.0 m4.0\,\text{m}, then remains at 50 N50\,\text{N} for another 2.0 m2.0\,\text{m}. The trolley starts from rest, resistive forces are negligible, and the motion lasts 3.0 s3.0\,\text{s}. Determine the work done, final speed and average power.

[5 marks]

Total for this question: 5

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Conservation of energy

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The principle of conservation of energy states that energy cannot be created or destroyed; it is transferred between stores while total energy remains constant in a closed system.
  • Near Earth, a change in gravitational potential energy is ΔEp=mgΔh\Delta E_p=mg\Delta h and kinetic energy is Ek=12mv2E_k=\tfrac12mv^2.
  • With negligible resistance, a decrease in gravitational potential energy may equal an increase in kinetic energy.
  • If resistive forces act, some mechanical energy is transferred to internal energy of the object and surroundings, so it is not lost.
  • Examiners expect a complete energy balance that includes any initial kinetic energy and work done against resistance, together with explicit assumptions such as negligible air resistance.
Energy transferred from the gravitational potential store to kinetic and, when resistance acts, internal stores.
Worked example

A 60 kg60\,\text{kg} cyclist descends 12 m12\,\text{m} from rest. Resistive forces transfer 2.0 kJ2.0\,\text{kJ} to internal energy. Determine the final speed.

  1. 1.Calculate the gravitational energy decrease: mgΔh=60(9.81)(12)=7063 Jmg\Delta h=60(9.81)(12)=7063\,\text{J}.
  2. 2.Subtract the resistive transfer: Ek=7063−2000=5063 JE_k=7063-2000=5063\,\text{J}.
  3. 3.Use 12mv2=5063\tfrac12mv^2=5063, giving v=2(5063)/60=13.0 m s−1v=\sqrt{2(5063)/60}=13.0\,\text{m s}^{-1}.

Answer: The final speed is 13.0 m s−113.0\,\text{m s}^{-1}.

Common mistakes

  • Don't describe energy transferred by resistance as destroyed or lost.
  • Don't use the final height instead of the change in height in mgΔhmg\Delta h.
  • Don't equate gravitational and kinetic energy without accounting for stated resistive work.

Exam tip

Write a single before-and-after energy balance and include every stated transfer before rearranging.

Tier 1 · Easy

ORIGINAL

A stone is released from rest and falls through 1.80 m1.80\,\text{m}. Air resistance is neglected. Use energy conservation to calculate its speed.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A roller-coaster car starts from rest 12.0 m12.0\,\text{m} above a reference level. Determine its speed when it is 3.0 m3.0\,\text{m} above that level. Resistive forces are neglected.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 70 kg70\,\text{kg} skier initially travels at 4.0 m s−14.0\,\text{m s}^{-1} and descends through a vertical height of 25 m25\,\text{m}. During the descent, 8.0 kJ8.0\,\text{kJ} is transferred to internal energy by resistive forces. Determine the skier's final speed.

[5 marks]

Total for this question: 5

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Bulk properties of solids

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Density is ρ=m/V\rho=m/V. Within the limit of proportionality, Hooke’s law gives F=kΔLF=k\Delta L, where kk is stiffness.
  • Elastic strain energy is the area under a force–extension graph and equals 12FΔL\tfrac12F\Delta L only for a straight line through the origin. Tensile stress is F/AF/A and tensile strain is ΔL/L\Delta L/L; breaking stress is the stress at fracture.
  • Below the elastic limit a material returns to its original dimensions after unloading, while plastic deformation leaves permanent extension. Brittle materials fracture with little plastic deformation.
  • Examiners expect force–extension and stress–strain graphs to show the limit of proportionality, elastic limit, energy area and fracture behaviour, and may link spring energy to kinetic or gravitational energy.
  • Ethical transport design weighs deformation energy and occupant protection against material and environmental costs.
A simple ductile-material stress–strain curve showing linear elastic behaviour, plastic deformation and fracture.
Worked example

A spring of stiffness 250 N m−1250\,\text{N m}^{-1} is extended by 80 mm80\,\text{mm} within its limit of proportionality. Determine the force and elastic strain energy.

  1. 1.Convert the extension: ΔL=0.080 m\Delta L=0.080\,\text{m}.
  2. 2.Apply Hooke’s law: F=kΔL=250(0.080)=20 NF=k\Delta L=250(0.080)=20\,\text{N}.
  3. 3.Use the triangular graph area: E=12FΔL=12(20)(0.080)=0.80 JE=\tfrac12F\Delta L=\tfrac12(20)(0.080)=0.80\,\text{J}.

Answer: The force is 20 N20\,\text{N} and the stored energy is 0.80 J0.80\,\text{J}.

Common mistakes

  • Don't use millimetres directly in F=kΔLF=k\Delta L when stiffness is in N m−1\text{N m}^{-1}.
  • Don't treat the limit of proportionality and elastic limit as identical definitions.
  • Don't calculate breaking stress from breaking force without dividing by cross-sectional area.

Exam tip

When a graph is supplied, use its area for energy unless a linear Hooke’s-law section is explicitly established.

Tier 1 · Easy

ORIGINAL

A solid block has mass 2.16 kg2.16\,\text{kg} and volume 8.00×10−4 m38.00\times10^{-4}\,\text{m}^3. Calculate its density.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A spring of stiffness 320 N m−1320\,\text{N m}^{-1} is compressed by 75 mm75\,\text{mm}. It launches a 0.150 kg0.150\,\text{kg} cart horizontally. Determine the elastic strain energy and the cart's speed if all this energy becomes kinetic energy.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A metal wire's force-extension graph is straight from the origin to its elastic limit at 720 N720\,\text{N} and 0.015 m0.015\,\text{m}. The force then rises linearly to 900 N900\,\text{N} at fracture, where the extension is 0.027 m0.027\,\text{m}. Determine the work done in stretching the wire to fracture and explain why unloading after the elastic limit would not return all this energy.

[5 marks]

Total for this question: 5

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

The Young modulus

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Tensile stress is σ=F/A\sigma=F/A in pascals and tensile strain is ε=ΔL/L\varepsilon=\Delta L/L, which is dimensionless. Within the linear elastic region, Young modulus is E=σ/ε=FL/(AΔL)E=\sigma/\varepsilon=FL/(A\Delta L) and is the gradient of a stress–strain graph.
  • A simple determination uses a long wire of measured original length and diameter, adds known loads below the limit of proportionality and measures each extension.
  • The diameter should be measured with a micrometer at several positions and perpendicular orientations because area depends on diameter squared.
  • Examiners expect SI conversions, a valid graph-based method, and practical controls such as a reference wire or correction for support movement.
  • Required practical 4 assesses this measurement and its uncertainties.
A simple Young-modulus arrangement using adjacent reference and test wires suspended from the same support.
Worked example

A wire has length 2.0 m2.0\,\text{m}, area 4.0×10−7 m24.0\times10^{-7}\,\text{m}^2 and extends 1.5 mm1.5\,\text{mm} under a 30 N30\,\text{N} load. Determine its Young modulus.

  1. 1.Convert extension: ΔL=1.5×10−3 m\Delta L=1.5\times10^{-3}\,\text{m}.
  2. 2.Substitute into E=FL/(AΔL)E=FL/(A\Delta L).
  3. 3.E=30(2.0)/[(4.0×10−7)(1.5×10−3)]=1.0×1011 PaE=30(2.0)/[(4.0\times10^{-7})(1.5\times10^{-3})]=1.0\times10^{11}\,\text{Pa}.

Answer: E=1.0×1011 PaE=1.0\times10^{11}\,\text{Pa}, or 100 GPa100\,\text{GPa}.

Common mistakes

  • Don't use extension instead of strain, omitting division by the original length.
  • Don't convert mm\text{mm} to metres but leave mm2\text{mm}^2 unconverted for area.
  • Don't take a graph gradient beyond the linear elastic region.

Exam tip

For a practical-method question, state how length, extension and diameter are measured and keep all loads below the limit of proportionality.

Tier 1 · Easy

ORIGINAL

A tensile force of 120 N120\,\text{N} acts on a wire of cross-sectional area 2.0 mm22.0\,\text{mm}^2. Calculate the tensile stress in pascals.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A wire is 2.00 m2.00\,\text{m} long and 0.800 mm0.800\,\text{mm} in diameter. A tensile force of 55.3 N55.3\,\text{N} produces an extension of 2.00 mm2.00\,\text{mm}. Determine the tensile stress, tensile strain and Young modulus.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

In a Young-modulus experiment, a wire has original length 2.50 m2.50\,\text{m} and mean diameter 0.420 mm0.420\,\text{mm}. Below the limit of proportionality, a force-extension graph has gradient 6.20×103 N m−16.20\times10^3\,\text{N m}^{-1}. Starting from the definitions of stress and strain, determine the Young modulus. Explain why the diameter should be measured at several positions and in two perpendicular directions.

[6 marks]

Total for this question: 6

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