Mark Scheme

Section A — Module 1

Question 1

(a) (4) — 1 mark for scalar defined as having magnitude only, 1 for vector defined as having both magnitude and direction. 1 mark for two correct scalars (mass, temperature, speed, energy, distance, time), 1 for two correct vectors (force, velocity, acceleration, displacement, momentum, weight). Speed/velocity and distance/displacement offered as a matched pair earn credit; "speed" listed as a vector scores 0 for that mark.

(b) (4)

• Statement (2): the rate of change of momentum of a body is directly proportional to the resultant force acting on it, and takes place in the direction of that force. Both the proportionality and the direction are required.
• Derivation (2): F ∝ Δ(mv)/Δt; for constant mass m, F ∝ m(Δv/Δt) = ma (1); with SI units the constant of proportionality is unity, giving F = ma (1). The final mark requires explicit reference to the constant being 1 in SI units — candidates who simply drop the proportionality sign do not earn it.

(c) (7) Free-body diagram (2): must show four labelled forces on the crate — weight (245 N) vertically down, normal reaction vertically up, tension 150 N at 30.0° above the horizontal, friction 40.0 N horizontally opposing motion. 1 mark for all four present and correctly directed, 1 for correct labelling with values. A diagram omitting the normal reaction, or drawing friction in the direction of motion, scores a maximum of 1. (i) (2) — Tₓ = T cos θ = 150 × cos 30.0° (1) = 130 N (1). Accept 129.9 N or 130. (ii) (2) — Resultant = Tₓ − friction = 129.9 − 40.0 (1) = 89.9 N (1). Accept 90 N. (iii) (1) — a = F/m = 89.9 / 25.0 = 3.60 m s⁻². Error carried forward applies throughout: a candidate who uses sin 30.0° in (i) loses that mark only, and is credited in (ii) and (iii) for correct method on their own figure. Omission of units in the final answer loses 1 mark once across the question.

Question 2

(a) (3) — The total linear momentum of a system remains constant (2), provided no resultant external force acts on the system (1). "In a closed/isolated system" is an acceptable statement of the condition.

(b) (8) (i) (3)

• Momentum before = (2.00 × 3.00) + (4.00 × 0) = 6.00 kg m s⁻¹ (1)
• Momentum after = (2.00 + 4.00)v = 6.00v (1)
• v = 6.00 / 6.00 = 1.00 m s⁻¹ (1) (ii) (5)
• KE before = ½ × 2.00 × 3.00² = 9.00 J (2 — 1 for method, 1 for value)
• KE after = ½ × 6.00 × 1.00² = 3.00 J (1)
• The collision is inelastic (1)
• Reason: kinetic energy is not conserved — 6.00 J has been converted to other forms, principally internal energy (heat), sound and work done in deformation. Momentum, however, is conserved (1). The final mark requires a numerical or explicit statement that KE has decreased. "Inelastic because they stick together" scores 0 for the reason mark — that is the definition of a perfectly inelastic collision, not evidence drawn from the calculation.

(c) (4)

• Vertical motion: s = uₜt + ½gt², with uᵥ = 0, so 45.0 = ½ × 9.81 × t² (1)
• t² = 90.0 / 9.81 = 9.174; t = 3.03 s (1)
• Horizontal motion is at constant velocity: x = uₓt = 12.0 × 3.03 (1)
• x = 36.4 m (1) The physics being tested is the independence of the horizontal and vertical components; a candidate who uses the resultant speed of 12.0 m s⁻¹ in the vertical equation scores 0 for the first two marks. Error carried forward applies to the horizontal distance.

Section B — Module 2

Question 3

(a) (4)

• Motion in which the acceleration is directly proportional to the displacement from a fixed equilibrium position (1) and is always directed towards that position (1).
• Defining equation: a = −ω²x (1).
• The negative sign indicates that acceleration and displacement are always in opposite directions — the restoring force acts back towards equilibrium (1). Both clauses of the definition are required; "proportional to displacement" alone scores 1.

(b) (6) — 2 marks each: 1 for the substituted equation, 1 for the value with units. (i) ω = 2π/T = 2π/0.800 = 7.85 rad s⁻¹ (ii) vₘₐₓ = ωA = 7.85 × 0.0400 = 0.314 m s⁻¹ (iii) aₘₐₓ = ω²A = (7.85)² × 0.0400 = 61.7 × 0.0400 = 2.47 m s⁻² The single commonest error is failure to convert the amplitude from 4.00 cm to 0.0400 m; this loses the value marks in (ii) and (iii) but the method marks are still awarded.

(c) (5) — Sketch (3): axes labelled energy (y) against displacement x (x), running from −A to +A. PE drawn as an upward-opening parabola, zero at x = 0 and maximum at ±A. KE drawn as a downward-opening parabola, maximum at x = 0 and zero at ±A. Total energy may be shown as a horizontal line. 1 mark per correct curve, 1 for correct axes and labelling of ±A. Straight lines instead of parabolas score 0 for that curve. Explanation (2):

• PE ∝ x², so it is zero at the equilibrium position and maximum at maximum displacement, where the oscillator is momentarily at rest (1).
• KE is the difference between the constant total energy and the PE, so it is maximum at the centre where speed is greatest and zero at the extremes; the two curves are mirror images and their sum is constant at every displacement (1). Reward candidates who state explicitly that the sum is constant.

Question 4

(a) (4)

• Transverse — oscillations of the particles or fields are perpendicular to the direction of energy propagation (1); example: light, any electromagnetic wave, waves on a string, water surface waves (1).
• Longitudinal — oscillations are parallel to the direction of energy propagation, producing compressions and rarefactions (1); example: sound, ultrasound, P-waves (1). "Up and down" versus "back and forth" scores 0 — the definition must be relative to the direction of propagation.

(b) (5)

• Superposition (2): when two or more waves meet at a point, the resultant displacement at that point is the vector sum of the individual displacements (2; "they add together" alone scores 1).
• Conditions (3, 1 each): the sources must be coherent, i.e. have a constant phase relationship, which requires the same frequency; the waves must have similar amplitudes so that cancellation at minima is appreciable; the waves must be of the same type and, for transverse waves, polarised in the same plane (accept "unpolarised or similarly polarised"). Accept "monochromatic" as part of the coherence mark but not as a separate condition.

(c) (6) (i) (3)

• Third harmonic on a string fixed at both ends: L = 3λ/2, so λ = 2L/3 = 2(1.20)/3 (1) = 0.800 m (1)
• f = v/λ = 240 / 0.800 = 300 Hz (1) (ii) (1) — 4 nodes (including the two fixed ends) and 3 antinodes. Both figures required for the mark. (iii) (2)
• Energy: a progressive wave transfers energy through the medium in the direction of travel; a stationary wave transfers no net energy along its length, energy being stored and interchanged between kinetic and potential within each loop (1).
• Phase: in a progressive wave, adjacent particles have a progressively changing phase; in a stationary wave all particles between adjacent nodes oscillate in phase, and particles in adjacent loops are in antiphase (1). Also acceptable for the second mark: in a progressive wave all particles have the same amplitude, whereas in a stationary wave amplitude varies with position from zero at nodes to maximum at antinodes.

Section C — Module 3

Question 5

(a) (3)

• Heat is energy transferred between two bodies as a result of a temperature difference between them; it is measured in joules (1–2).
• Temperature is a measure of the average kinetic energy of the molecules of a body, and determines the direction of net heat flow — from higher to lower temperature (1–2). Maximum 3. The discriminating idea is that heat is energy in transit while temperature is a property of the body; candidates who define heat as "the energy contained in a body" lose that mark, as that describes internal energy.

(b) (4)

• Specific heat capacity (2): the energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C), with no change of state. Unit J kg⁻¹ K⁻¹.
• Specific latent heat of fusion (2): the energy required to change 1 kg of a substance from solid to liquid at constant temperature. Unit J kg⁻¹. "At constant temperature" and "no change of state" are the respective discriminating phrases; omission loses a mark in each case.

(c) (5)

• E = mcΔθ = 0.400 × 4200 × (100 − 20.0) (1)
• = 0.400 × 4200 × 80.0 = 134 400 J (1)
• t = E/P = 134 400 / 500 (1)
• = 269 s (approximately 4 min 29 s) (1) Reasons, any two for 1 mark total: heat is lost to the surroundings by conduction, convection and radiation; energy is absorbed in heating the container and the heater itself; some water may evaporate before 100 °C, absorbing latent heat; the heater may not deliver its full rated power. Note that Δθ = 80.0 K, not 80.0 °C converted to kelvin — a candidate who adds 273 to the temperature difference loses the first two marks.

(d) (3)

• The energy supplied does not increase the average kinetic energy of the molecules, so the temperature stays constant (1).
• It is used to do work against the intermolecular forces of attraction, breaking the rigid bonds holding molecules in fixed positions in the lattice (1).
• The potential energy of the molecules therefore increases while their kinetic energy is unchanged, allowing them to move past one another as a liquid (1). The explicit contrast between potential and kinetic energy is the key discriminator; answers referring only to "breaking bonds" score 1.

Question 6

(a) (6) — 2 marks each: 1 for the definition, 1 for the unit.

• Stress σ = force per unit cross-sectional area, F/A. Unit Pa (N m⁻²).
• Strain ε = extension per unit original length, e/L. No unit (dimensionless) — this must be stated explicitly to earn the mark.
• Young modulus E = stress / strain, valid within the limit of proportionality. Unit Pa (N m⁻²). "Original length" is required in the strain definition; "length" alone loses that mark.

(b) (5)

• Stress = F/A = 180 / (1.50 × 10⁻⁶) (1) = 1.20 × 10⁸ Pa (1)
• Strain = e/L = (1.20 × 10⁻³) / 2.00 (1) = 6.00 × 10⁻⁴ (1)
• E = stress / strain = (1.20 × 10⁸) / (6.00 × 10⁻⁴) = 2.00 × 10¹¹ Pa (1) Accept the single-step route E = FL/(Ae) with correct substitution, awarding 2 for the substitution and 3 for the correct value and unit. The commonest error is failure to convert 1.20 mm to 1.20 × 10⁻³ m, which yields 2.00 × 10⁸ Pa; this loses the strain value mark and the final mark but earns 3 by error carried forward. A candidate whose answer is not of the order 10¹¹ Pa should recognise it as physically implausible for a metal and say so — examiners credit such a comment in the level-marked parts.

(c) (4) — Sketch (2): axes labelled stress (y) against strain (x); an initial straight line through the origin, then a slight curve, a marked yield point after which strain increases rapidly for little additional stress, a maximum, and termination at fracture. 1 mark for the correct overall shape, 1 for all four features correctly labelled: limit of proportionality (end of the straight line), elastic limit (just beyond it), yield point, breaking stress. Explanation (2):

• Beyond the elastic limit the material no longer returns to its original length when the load is removed; it has undergone plastic deformation and retains a permanent extension (1).
• At the atomic level, layers of atoms slip past one another along dislocations to new equilibrium positions, so the deformation is not reversed by the interatomic forces (1). Candidates who place the elastic limit before the limit of proportionality lose the labelling mark.

Level descriptors (extended-response parts)

• Level 4 (top band): Accurate physics with correct terminology; every calculation laid out with the equation quoted, values substituted, and the answer given to appropriate significant figures with units; all quantities converted to SI before substitution; sketches correctly shaped, labelled on both axes and referred to in the prose; explanations distinguish cause from consequence.
• Level 3: Sound method with occasional unit or conversion slips; sketches broadly correct but incompletely labelled; explanations state what happens without fully explaining why.
• Level 2: Equations recalled but applied inconsistently; frequent missing units; sketch absent or unlabelled.
• Level 1: Isolated correct formulae or statements with no developed working.

Sample Answers with Examiner Commentary

Question 1(c) — Sample Answers

Grade I (Distinction) answer (extract)

"[Free-body diagram: crate drawn as a box. Weight W = 245 N acting vertically downwards from the centre; normal reaction N acting vertically upwards; tension T = 150 N acting at 30.0° above the horizontal, arrow pointing up and to the right; friction f = 40.0 N acting horizontally to the left, opposing the motion. All four arrows labelled with symbol and value.]

(i) The rope is inclined, so only the component of tension along the direction of motion accelerates the crate. Tₓ = T cos θ = 150 × cos 30.0° = 150 × 0.866 = 130 N

(ii) The vertical component of the tension and the normal reaction and weight all act perpendicular to the motion, so they do not contribute to the horizontal resultant. Taking the direction of motion as positive: Fᵣₑₛ = Tₓ − f = 129.9 − 40.0 = 89.9 N

(iii) a = Fᵣₑₛ / m = 89.9 / 25.0 = 3.60 m s⁻², directed horizontally in the direction of motion.

It is worth noting that the vertical component of the tension, T sin 30.0° = 75.0 N, reduces the normal reaction from 245 N to 170 N. Since friction depends on the normal reaction, pulling at an angle reduces the frictional force in practice — here the friction is given as constant, so this does not affect the calculation, but it explains why a rope angled upwards is often more effective than a horizontal one."

Mark: 7/7. Examiner commentary: The free-body diagram is complete and correctly labelled — all four forces present, friction correctly opposing motion, and values attached to the arrows, which is what earns the second diagram mark. Each calculation quotes the relationship before substituting, so every method mark is separately visible. The candidate states explicitly why the vertical forces do not enter the horizontal resultant rather than silently ignoring them, and gives the direction of the acceleration, not merely its magnitude. The closing paragraph is beyond the requirements of the question but demonstrates genuine physical understanding of why the geometry matters, and the candidate is careful to note that it does not alter the given data. Full marks.

Grade III (Pass) answer (extract)

"[Diagram: box with an arrow labelled 150 N pointing right and an arrow labelled 40 N pointing left.]

(i) Tₓ = 150 × sin 30 = 75 N (ii) 75 − 40 = 35 (iii) a = 35 / 25 = 1.4"

Mark: 3/7. Examiner commentary: Three marks are earned, all by error carried forward. The diagram shows only two forces — weight and normal reaction are both absent and the tension is drawn horizontally rather than at 30.0°, so it is not a free-body diagram of the situation described and scores 0. In (i) the candidate has used sin θ where the component along the direction of motion requires cos θ; a quick check would have revealed the problem, since a rope at only 30° to the horizontal must have a horizontal component larger than half the tension, not exactly half. The method in (ii) and (iii) is correct and is credited on the candidate's own figures, earning the method marks. Finally, no units appear on any answer, which costs a further mark under the rubric. Drawing the free-body diagram properly would very likely have prevented the trigonometric error as well as earning its own two marks — the diagram is not decoration, it is the working.

Question 2(b)(ii) — Sample Answers

Grade I (Distinction) answer (extract)

"KE before = ½m₁u₁² + ½m₂u₂² = ½(2.00)(3.00)² + 0 = 9.00 J KE after = ½(m₁ + m₂)v² = ½(6.00)(1.00)² = 3.00 J

The collision is inelastic. Kinetic energy has fallen from 9.00 J to 3.00 J, so 6.00 J — two-thirds of the original kinetic energy — has been converted into other forms: internal energy in the deformed trolleys and the coupling, sound, and work done against friction at the point of contact. Momentum is nonetheless conserved at 6.00 kg m s⁻¹ before and after, since no resultant external force acted. Conservation of momentum holds in all collisions; conservation of kinetic energy holds only in elastic ones, and it is the loss of kinetic energy, not the fact that the trolleys stick together, that establishes the collision as inelastic."

Mark: 5/5. Examiner commentary: Both energies are calculated correctly with the equations shown. What secures the reason mark is that the candidate cites the numerical fall in kinetic energy as the evidence and names where the energy has gone. The final sentence draws the distinction the mark scheme is designed to test — that inelasticity is established by the energy calculation rather than by the trolleys coalescing — and the candidate correctly notes that momentum conservation is unaffected. Full marks.

Grade II (Merit) answer (extract)

"KE before = ½ × 2 × 3² = 9 J KE after = ½ × 6 × 1² = 3 J The collision is inelastic because the two trolleys stick together after the collision."

Mark: 4/5. Examiner commentary: Both calculations are correct and the classification is correct, earning four of the five marks. The reason, however, is not creditable. The trolleys coalescing is a description of what happened, and while a perfectly inelastic collision is indeed one in which the bodies coalesce, the question follows a kinetic energy calculation precisely so that the candidate will use it. The mark is awarded for observing that kinetic energy has fallen from 9.00 J to 3.00 J and that the missing 6.00 J has been converted to internal energy and sound. One additional sentence, drawing on figures the candidate has already obtained, would have secured full marks.

How Unit 1 is assessed

Paper 02 carries 90 marks and contributes 50% of the Unit 1 external assessment mark, alongside Paper 01 (multiple choice, 30%) and the Internal Assessment (20%). All six questions are compulsory and the three Modules carry equal weight. Grades are reported Grade I – Grade VII, with Grade I the highest.

Four habits separate the top band. First, draw the free-body diagram before calculating — in mechanics questions it is worth marks in its own right and it prevents the resolution errors that cost far more marks downstream. Second, quote the equation, substitute, then evaluate, on separate lines: this paper awards method marks at every stage, and error carried forward means a single slip costs one mark rather than all of them, but only if the method is visible. Third, convert to SI units before substituting — centimetres to metres, millimetres to metres, grams to kilograms — since this single omission accounts for more lost marks on Unit 1 than any conceptual error. Fourth, give units and a direction where appropriate on every final answer; a numerical answer without units is incomplete and is penalised under the rubric.