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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.

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

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3.4.1.1

Scalars and vectors

Notes
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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

  • 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 FcosθF\cos\theta horizontally and FsinθF\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 6.0m s16.0\,\text{m s}^{-1} east and 8.0m s18.0\,\text{m s}^{-1} north. Determine its magnitude and direction.

  1. 1.Use perpendicular components: v=6.02+8.02=10.0m s1v=\sqrt{6.0^2+8.0^2}=10.0\,\text{m s}^{-1}.
  2. 2.Measure the direction from east: θ=tan1(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: 10.0m s110.0\,\text{m s}^{-1} at 53.153.1^{\circ} north of east.

Common mistakes

  • Don't use FsinθF\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 3.0m s13.0\,\text{m s}^{-1} east and 4.0m s14.0\,\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 12.0kg12.0\,\text{kg} crate rests on a smooth plane inclined at 30.030.0^{\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 180N180\,\text{N} due east. The other is 120N120\,\text{N} at 110110^{\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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3.4.1.2

Moments

Notes
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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=FdM=Fd_{\perp}, where dd_{\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 4.0m4.0\,\text{m} beam of weight 200N200\,\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, 2.0m2.0\,\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=100NR=100\,\text{N}.

Answer: The vertical support force is 100N100\,\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 25N25\,\text{N} force acts perpendicular to a spanner 0.40m0.40\,\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 3.0m3.0\,\text{m} and weight 180N180\,\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 5.0m5.0\,\text{m} long and weighs 240N240\,\text{N}. It is hinged to a wall at one end. A 360N360\,\text{N} load hangs 3.8m3.8\,\text{m} from the hinge, and a cable attached to the free end makes 3535^{\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

3.4.1.3

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.

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=9.81m s2g=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 accelerates uniformly from 4.0m s14.0\,\text{m s}^{-1} to 16.0m s116.0\,\text{m s}^{-1} in 6.0s6.0\,\text{s}. Determine its acceleration and displacement.

  1. 1.Calculate the gradient: a=(16.04.0)/6.0=2.0m s2a=(16.0-4.0)/6.0=2.0\,\text{m s}^{-2}.
  2. 2.For uniform acceleration, the mean velocity is (4.0+16.0)/2=10.0m s1(4.0+16.0)/2=10.0\,\text{m s}^{-1}.
  3. 3.Use the area under the velocity–time graph: s=10.0×6.0=60ms=10.0\times6.0=60\,\text{m}.

Answer: a=2.0m s2a=2.0\,\text{m s}^{-2} and s=60ms=60\,\text{m}.

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.

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 7.5m s17.5\,\text{m s}^{-1} for 4.0s4.0\,\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 5.0m s15.0\,\text{m s}^{-1} to 17.0m s117.0\,\text{m s}^{-1} in 6.0s6.0\,\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 18m s118\,\text{m s}^{-1} in 5.0s5.0\,\text{s}, remains at 18m s118\,\text{m s}^{-1} for 4.0s4.0\,\text{s}, then falls uniformly to 6.0m s1-6.0\,\text{m s}^{-1} in 3.0s3.0\,\text{s}. Determine the final acceleration, total displacement and total distance travelled.

[5 marks]

Total for this question: 5

3.4.1.4

Projectile motion

Notes
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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 15.0m s115.0\,\text{m s}^{-1} and falls 19.6m19.6\,\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=2.00st=\sqrt{2(19.6)/9.81}=2.00\,\text{s}.
  3. 3.Use constant horizontal velocity: x=15.0×2.00=30.0mx=15.0\times2.00=30.0\,\text{m}.

Answer: The flight time is 2.00s2.00\,\text{s} and the range is 30.0m30.0\,\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 12.0m s112.0\,\text{m s}^{-1}. The platform is 20.0m20.0\,\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 22.0m s122.0\,\text{m s}^{-1} and 35.035.0^{\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

3.4.1.5

Newton's laws of motion

Notes
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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

  • 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 5.0kg5.0\,\text{kg} block is pulled horizontally by 24N24\,\text{N} while friction is 9.0N9.0\,\text{N}. Determine its acceleration.

  1. 1.Choose the pull direction as positive.
  2. 2.Find the resultant force: F=249.0=15NF=24-9.0=15\,\text{N}.
  3. 3.Apply F=maF=ma: a=15/5.0=3.0m s2a=15/5.0=3.0\,\text{m s}^{-2}.

Answer: 3.0m s23.0\,\text{m s}^{-2} in the direction of the pull.

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.

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

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A lift of total mass 750kg750\,\text{kg} accelerates vertically upward at 1.20m s21.20\,\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 4.0kg4.0\,\text{kg} and 6.0kg6.0\,\text{kg}, are joined by a light horizontal cord. A horizontal force of 38N38\,\text{N} pulls B. Friction on A is 5.0N5.0\,\text{N} and friction on B is 7.0N7.0\,\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

3.4.1.6

Momentum

Notes
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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

  • 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 0.20kg0.20\,\text{kg} ball travels at 12m s112\,\text{m s}^{-1} east and rebounds at 8.0m s18.0\,\text{m s}^{-1} west. Contact lasts 0.050s0.050\,\text{s}. Determine the mean force on the ball.

  1. 1.Take east as positive, so u=+12m s1u=+12\,\text{m s}^{-1} and v=8.0m s1v=-8.0\,\text{m s}^{-1}.
  2. 2.Calculate the momentum change: Δp=0.20(8.012)=4.0kg m s1\Delta p=0.20(-8.0-12)=-4.0\,\text{kg m s}^{-1}.
  3. 3.Use F=Δp/Δt=4.0/0.050=80NF=\Delta p/\Delta t=-4.0/0.050=-80\,\text{N}.

Answer: The mean force is 80N80\,\text{N} west.

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.

Exam tip

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

Tier 1 · Easy

ORIGINAL

A 0.80kg0.80\,\text{kg} trolley moving at 3.0m s13.0\,\text{m s}^{-1} collides with a stationary 0.40kg0.40\,\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 0.40kg0.40\,\text{kg} trolley moving at 5.0m s15.0\,\text{m s}^{-1} collides with a stationary 0.60kg0.60\,\text{kg} trolley. Afterwards their velocities are 1.0m s1-1.0\,\text{m s}^{-1} and 4.0m s14.0\,\text{m s}^{-1} respectively. Show that the collision is elastic.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A 1200kg1200\,\text{kg} car moving east at 18m s118\,\text{m s}^{-1} collides head-on with an 800kg800\,\text{kg} car moving west at 6.0m s16.0\,\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 1200kg1200\,\text{kg} car if contact lasts 0.18s0.18\,\text{s}.

[6 marks]

Total for this question: 6

3.4.1.7

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.

Explanation

  • Work is energy transferred by a force. For a constant force, W=FscosθW=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 motor lifts a 30kg30\,\text{kg} load through 5.0m5.0\,\text{m} in 4.0s4.0\,\text{s} while taking 450W450\,\text{W} input power. Determine its useful power and efficiency.

  1. 1.Find useful energy: ΔEp=mgh=30(9.81)(5.0)=1471.5J\Delta E_p=mgh=30(9.81)(5.0)=1471.5\,\text{J}.
  2. 2.Find useful power: P=1471.5/4.0=368WP=1471.5/4.0=368\,\text{W}.
  3. 3.Calculate efficiency: 368/450=0.818368/450=0.818, or 81.8%81.8\%.

Answer: The useful power is 368W368\,\text{W} and the efficiency is 81.8%81.8\%.

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.

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 75N75\,\text{N} acts in the direction of motion through 12m12\,\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 25kg25\,\text{kg} load vertically through 8.0m8.0\,\text{m} in 6.0s6.0\,\text{s}. Its electrical input power is 420W420\,\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 2.0kg2.0\,\text{kg} trolley increases linearly from 10N10\,\text{N} to 50N50\,\text{N} over the first 4.0m4.0\,\text{m}, then remains at 50N50\,\text{N} for another 2.0m2.0\,\text{m}. The trolley starts from rest, resistive forces are negligible, and the motion lasts 3.0s3.0\,\text{s}. Determine the work done, final speed and average power.

[5 marks]

Total for this question: 5

3.4.1.8

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

  1. 1.Calculate the gravitational energy decrease: mgΔh=60(9.81)(12)=7063Jmg\Delta h=60(9.81)(12)=7063\,\text{J}.
  2. 2.Subtract the resistive transfer: Ek=70632000=5063JE_k=7063-2000=5063\,\text{J}.
  3. 3.Use 12mv2=5063\tfrac12mv^2=5063, giving v=2(5063)/60=13.0m s1v=\sqrt{2(5063)/60}=13.0\,\text{m s}^{-1}.

Answer: The final speed is 13.0m s113.0\,\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 1.80m1.80\,\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 12.0m12.0\,\text{m} above a reference level. Determine its speed when it is 3.0m3.0\,\text{m} above that level. Resistive forces are neglected.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 70kg70\,\text{kg} skier initially travels at 4.0m s14.0\,\text{m s}^{-1} and descends through a vertical height of 25m25\,\text{m}. During the descent, 8.0kJ8.0\,\text{kJ} is transferred to internal energy by resistive forces. Determine the skier's final speed.

[5 marks]

Total for this question: 5

3.4.2.1

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 250N m1250\,\text{N m}^{-1} is extended by 80mm80\,\text{mm} within its limit of proportionality. Determine the force and elastic strain energy.

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

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

Common mistakes

  • Don't use millimetres directly in F=kΔLF=k\Delta L when stiffness is in N m1\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 2.16kg2.16\,\text{kg} and volume 8.00×104m38.00\times10^{-4}\,\text{m}^3. Calculate its density.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A spring of stiffness 320N m1320\,\text{N m}^{-1} is compressed by 75mm75\,\text{mm}. It launches a 0.150kg0.150\,\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 720N720\,\text{N} and 0.015m0.015\,\text{m}. The force then rises linearly to 900N900\,\text{N} at fracture, where the extension is 0.027m0.027\,\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

3.4.2.2

The Young modulus

Notes
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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 2.0m2.0\,\text{m}, area 4.0×107m24.0\times10^{-7}\,\text{m}^2 and extends 1.5mm1.5\,\text{mm} under a 30N30\,\text{N} load. Determine its Young modulus.

  1. 1.Convert extension: ΔL=1.5×103m\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)/[(4.0×107)(1.5×103)]=1.0×1011PaE=30(2.0)/[(4.0\times10^{-7})(1.5\times10^{-3})]=1.0\times10^{11}\,\text{Pa}.

Answer: E=1.0×1011PaE=1.0\times10^{11}\,\text{Pa}, or 100GPa100\,\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 120N120\,\text{N} acts on a wire of cross-sectional area 2.0mm22.0\,\text{mm}^2. Calculate the tensile stress in pascals.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A wire is 2.00m2.00\,\text{m} long and 0.800mm0.800\,\text{mm} in diameter. A tensile force of 55.3N55.3\,\text{N} produces an extension of 2.00mm2.00\,\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 2.50m2.50\,\text{m} and mean diameter 0.420mm0.420\,\text{mm}. Below the limit of proportionality, a force-extension graph has gradient 6.20×103N m16.20\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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