Kinematics — Structured & Open-Ended
Speed, velocity, acceleration, distance-time and velocity-time graphs, SUVAT equations.
K1
Structured
A car starts from rest and accelerates uniformly to a velocity of 24 m/s in 8 s. It then travels at constant velocity for 12 s before decelerating uniformly to rest in 4 s.
- (a) Sketch a velocity-time graph for the entire journey. Label key values on both axes. [3]
- (b) Calculate the acceleration during the first 8 s. [2]
- (c) Calculate the total distance travelled. [2]
- (d) Calculate the deceleration during the final 4 s. [1]
✏ Your answer
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Model answers
- (a) Graph must show: straight line from (0,0) to (8,24) [1]; horizontal line from (8,24) to (20,24) [1]; straight line from (20,24) to (24,0) [1]. Axes labelled: velocity (m/s) on y-axis, time (s) on x-axis with key values 0, 8, 20, 24 marked.
- (b) a = (v − u) / t = (24 − 0) / 8 = 3 m/s² [1 for formula/substitution, 1 for correct answer with unit]
- (c) Total distance = area under v-t graph = ½(8)(24) + (12)(24) + ½(4)(24) = 96 + 288 + 48 = 432 m [1 for correct method using areas, 1 for correct total]
- (d) Deceleration = (24 − 0) / 4 = 6 m/s² [1]
Note: deceleration is positive (magnitude of acceleration). Do not penalise if student writes −6 m/s² provided it is clear this means deceleration.
⚠ Common trap: Many students add up time intervals (8+12+4=24 s) and think the car travels 24×24 = 576 m. Total distance is the area under the graph — use three separate areas.
K2
Open-ended
A ball is thrown vertically upward from the ground with initial velocity 20 m/s. Take g = 10 m/s² and ignore air resistance.
- (a) Calculate the maximum height reached by the ball. [2]
- (b) Calculate the time taken to reach maximum height. [1]
- (c) Sketch the velocity-time graph for the ball from the moment it is thrown until it returns to the ground. Mark key values. [3]
✏ Your answer
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Model answers
- (a) Using v² = u² + 2as, at max height v = 0: 0 = 20² − 2(10)s → s = 400/20 = 20 m [1 for correct equation, 1 for answer]
- (b) t = (v − u)/a = (0 − 20)/(−10) = 2 s [1]
- (c) Graph: straight line with negative gradient (−10 m/s²) from (0, +20) crossing zero at t = 2 s, reaching (4 s, −20 m/s) when ball returns to ground [1 for correct shape; 1 for zero at t=2; 1 for returning to −20 at t=4 by symmetry]. Total time of flight = 4 s by symmetry.
Accept upward as positive throughout. The graph is a single straight line from +20 to −20, crossing the time axis at 2 s.
⚠ Common trap: Students often draw a parabola (as if sketching height-time). A velocity-time graph for constant acceleration is always a straight line.
K3
Open-ended
A student records the distance-time data for a cyclist over 50 seconds:
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|
| Distance (m) | 0 | 40 | 80 | 80 | 120 | 160 |
- (a) Describe the motion of the cyclist in each of these intervals: 0–20 s, 20–30 s, 30–50 s. [3]
- (b) Calculate the average speed over the entire 50 s. [1]
- (c) Explain why the average speed is different from the speed during 0–20 s. [1]
✏ Your answer
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Model answers
- (a) 0–20 s: constant speed of 4 m/s (distance increases uniformly by 40 m every 10 s) [1]. 20–30 s: stationary / at rest (distance does not change) [1]. 30–50 s: constant speed of 4 m/s (distance increases uniformly again) [1].
- (b) Average speed = total distance / total time = 160 / 50 = 3.2 m/s [1]
- (c) The average speed is lower (3.2 m/s vs 4 m/s) because it includes the 10 s stationary period when no distance was covered, which reduces the overall average [1].
K4
Open-ended
A stone is dropped from a bridge 45 m above a river. Take g = 10 m/s², ignore air resistance.
- (a) Calculate the time taken for the stone to reach the river. [2]
- (b) Calculate the velocity of the stone just before it hits the water. [2]
✏ Your answer
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- (a) s = ut + ½at². u = 0 (dropped from rest), s = 45, a = 10: 45 = 0 + ½(10)t² → t² = 9 → t = 3 s [1 formula/substitution, 1 answer]
- (b) v = u + at = 0 + 10(3) = 30 m/s [1 formula, 1 answer]. Or: v² = u² + 2as = 0 + 2(10)(45) = 900 → v = 30 m/s.
Forces — Structured & Open-Ended
Newton's laws, F=ma, friction, terminal velocity, free body diagrams.
F1
Structured
A skydiver of mass 75 kg jumps from a plane. She initially accelerates downward. As she falls, her acceleration decreases. Eventually she reaches terminal velocity before opening her parachute, after which she decelerates to a lower terminal velocity.
- (a) Calculate the weight of the skydiver. (g = 10 N/kg) [1]
- (b) Draw and label a free body diagram showing the forces on the skydiver when she is (i) accelerating downward and (ii) at terminal velocity. [2]
- (c) Explain, using Newton's laws, why her acceleration decreases as she falls faster. [3]
- (d) After she opens her parachute, she decelerates. Explain the forces involved and why she reaches a new, lower terminal velocity. [2]
✏ Your answer
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- (a) Weight = mg = 75 × 10 = 750 N [1]
- (b)(i) Weight arrow (downward, larger) and air resistance arrow (upward, smaller) — net downward force [1]. (ii) Weight and air resistance arrows equal in length [1] — net force = 0.
- (c) As the skydiver falls faster, air resistance increases [1]. The resultant (net) downward force decreases (= weight − air resistance) [1]. By Newton's 2nd law (F = ma), a smaller resultant force produces a smaller acceleration [1].
- (d) Opening the parachute greatly increases air resistance, which now exceeds weight — the resultant force is upward, causing deceleration [1]. As she slows down, air resistance decreases until it again equals weight. At this point the resultant force is zero and she moves at a new, lower constant (terminal) velocity [1].
💡 Exam tip: For all force–motion explanations, follow this structure: (1) state the forces, (2) find the resultant, (3) apply Newton's 2nd law to explain the motion.
F2
Open-ended
A car of mass 1200 kg is driven along a straight road. The engine provides a driving force of 4800 N. The frictional resistance (air resistance + friction) is 1200 N.
- (a) Calculate the resultant force on the car. [1]
- (b) Calculate the acceleration of the car. [2]
- (c) The driver releases the accelerator. The frictional resistance remains 1200 N. Describe and explain the subsequent motion of the car. [3]
✏ Your answer
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- (a) Resultant force = 4800 − 1200 = 3600 N (forward) [1]
- (b) F = ma → a = F/m = 3600/1200 = 3 m/s² [1 formula, 1 answer with unit]
- (c) When the engine is off, the driving force is zero. The only horizontal force is friction (1200 N, backward) [1]. By Newton's 2nd law, there is a net backward force → deceleration: a = 1200/1200 = 1 m/s² [1]. The car decelerates (slows down) at 1 m/s² until it comes to rest (velocity = 0) [1].
F3
Open-ended
A book of mass 0.8 kg rests on a table. A student pushes the book horizontally with a force of 3 N but the book does not move. (g = 10 N/kg)
- (a) State the magnitude and direction of the friction force acting on the book. Explain your answer using Newton's 1st Law. [2]
- (b) The student increases the push force to 6 N, and the book starts to move and accelerates at 1.5 m/s². Calculate the friction force during motion (kinetic friction). [2]
- (c) Explain why the kinetic friction force is less than the maximum static friction force. [1]
✏ Your answer
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- (a) Friction = 3 N, acting horizontally opposite to the push (backward) [1]. Newton's 1st Law: the book is in equilibrium (not moving, no acceleration) → the resultant horizontal force = 0. Therefore friction must equal and oppose the applied force [1].
- (b) F = ma → resultant force = 0.8 × 1.5 = 1.2 N. Resultant = push − friction → friction = 6 − 1.2 = 4.8 N [1 for F=ma, 1 for correct friction value]
- (c) Once the object is moving, the microscopic interlocking of surfaces is reduced (fewer surface contacts during motion), so the friction force is lower than the force needed to initially overcome static friction [1].
F4
Open-ended
A student pushes a 5 kg trolley with 20 N and a friend pulls the same trolley with 15 N in the same direction. A friction force of 8 N opposes motion. Calculate the resultant force and the acceleration of the trolley. [4]
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- Total forward force = 20 + 15 = 35 N [1]
- Resultant force = 35 − 8 = 27 N (forward) [1]
- F = ma → a = 27/5 = 5.4 m/s² [1 formula, 1 answer with unit]
Energy, Work & Power — Structured & Open-Ended
KE, GPE, work done, conservation of energy, power, efficiency.
E1
Structured
A 1200 kg car travels at 30 m/s. The driver applies the brakes and the car stops in 60 m. (g = 10 N/kg)
- (a) Calculate the kinetic energy of the car before braking. [2]
- (b) Calculate the braking force, assuming it is the only horizontal force. [2]
- (c) The car travels up a slope (height 15 m) at constant velocity before stopping. Calculate the gravitational PE gained. [2]
- (d) Explain where the kinetic energy goes when a car brakes on a flat road, using conservation of energy. [3]
✏ Your answer
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- (a) KE = ½mv² = ½ × 1200 × 30² = ½ × 1200 × 900 = 540 000 J (540 kJ) [1 formula, 1 answer]
- (b) Work done by braking force = KE lost: W = F × d → F = W/d = 540 000/60 = 9000 N [1 method, 1 answer]
- (c) GPE = mgh = 1200 × 10 × 15 = 180 000 J (180 kJ) [1 formula, 1 answer]
- (d) By conservation of energy, energy cannot be created or destroyed — it is transformed [1]. The kinetic energy is converted into thermal energy (heat) in the brake pads, discs, and tyres due to friction [1], and a small amount into sound energy [1]. The total energy remains constant — it is just no longer in a useful (kinetic) form.
⚠ Common trap: Students write "energy is lost to friction" — this is acceptable but imprecise. Better: "kinetic energy is transferred to thermal (heat) energy in the brakes by friction."
E2
Open-ended
A crane lifts a 500 kg steel beam from the ground to a height of 20 m in 25 s. (g = 10 N/kg)
- (a) Calculate the useful work done on the beam. [2]
- (b) Calculate the minimum power of the crane motor. [2]
- (c) In practice, the crane motor has an efficiency of 75%. Calculate the actual power input to the motor. [2]
✏ Your answer
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- (a) W = Fd = mgh = 500 × 10 × 20 = 100 000 J (100 kJ) [1 formula/method, 1 answer]
- (b) P = W/t = 100 000/25 = 4000 W (4 kW) [1 formula, 1 answer]
- (c) Efficiency = useful output/total input → 0.75 = 4000/P_input → P_input = 4000/0.75 = 5333 W ≈ 5.3 kW [1 rearrangement, 1 answer]
E3
Open-ended
A pendulum bob is released from rest at point A, 0.8 m above its lowest point B. (g = 10 N/kg, mass of bob = 0.2 kg, ignore air resistance)
- (a) Calculate the GPE at point A (relative to B). [2]
- (b) Calculate the speed of the bob at point B. [2]
- (c) Explain why in practice the bob does not rise to exactly 0.8 m on the other side. [1]
✏ Your answer
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- (a) GPE = mgh = 0.2 × 10 × 0.8 = 1.6 J [1 formula, 1 answer]
- (b) At B, all GPE → KE (conservation of energy): ½mv² = 1.6 → v² = 2(1.6)/0.2 = 16 → v = 4 m/s [1 method, 1 answer]
- (c) In practice, air resistance and friction at the pivot convert some kinetic energy to thermal energy, so the bob has less energy at B and cannot rise to the same height on the other side [1].
Thermal Physics — Structured & Open-Ended
Kinetic model, heat transfer (conduction, convection, radiation), Q=mcΔT, latent heat.
T1
Open-ended
A thermal flask (vacuum flask) is designed to keep liquids hot for many hours. It has a double-walled glass container with a vacuum between the walls, and the inner glass surfaces are silvered.
- (a) Explain how the vacuum between the walls reduces heat loss by conduction and convection. [2]
- (b) Explain how the silvered surfaces reduce heat loss by radiation. [2]
- (c) The cap of the flask is made of plastic, not metal. Suggest why, using the concept of thermal conductivity. [1]
- (d) Using Q = mcΔT, calculate how much energy is needed to heat 0.5 kg of water from 20°C to 100°C. (c_water = 4200 J/(kg°C)) [2]
✏ Your answer
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- (a) Conduction requires particles to vibrate and collide to transfer energy. A vacuum has no particles [1], so conduction cannot occur. Convection requires the movement of fluid (liquid or gas) particles — a vacuum has no fluid, so convection is also impossible [1].
- (b) Silvered (shiny) surfaces are poor emitters of infrared radiation [1] and also poor absorbers (good reflectors) — they reflect radiation back into the liquid, reducing heat loss by radiation [1].
- (c) Plastic is a poor thermal conductor (good insulator) [1]. A metal cap would conduct heat from the hot liquid to the outside much faster, increasing heat loss.
- (d) Q = mcΔT = 0.5 × 4200 × (100 − 20) = 0.5 × 4200 × 80 = 168 000 J (168 kJ) [1 substitution, 1 answer with unit]
💡 Exam tip: When explaining how each method of heat transfer is reduced, always first explain how that transfer method works, then explain how the design feature prevents it.
T2
Open-ended
200 g of ice at 0°C is heated until it becomes steam at 100°C. The specific latent heat of fusion of ice is 336 000 J/kg, specific heat capacity of water is 4200 J/(kg°C), and the specific latent heat of vaporisation of water is 2 260 000 J/kg.
- (a) Calculate the energy needed to melt all the ice at 0°C. [2]
- (b) Calculate the energy needed to heat the water from 0°C to 100°C. [2]
- (c) Explain why the temperature stays at 0°C while the ice melts, even though energy is being continuously supplied. [2]
✏ Your answer
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- (a) Q = mL = 0.2 × 336 000 = 67 200 J [1 formula, 1 answer]
- (b) Q = mcΔT = 0.2 × 4200 × 100 = 84 000 J [1 formula, 1 answer]
- (c) During melting, the energy supplied is used to break the bonds (intermolecular forces) between water molecules in the solid lattice [1]. This increases the potential energy of the molecules without increasing their kinetic energy. Since temperature is a measure of average kinetic energy, the temperature does not rise while melting occurs [1].
⚠ Common trap: Students often write "the energy is absorbed by the water" — this is too vague. Specify that energy goes into breaking intermolecular bonds (potential energy), not kinetic energy.
T3
Open-ended
Explain, using the kinetic model of matter, why evaporation causes cooling of the remaining liquid. [4 marks]
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- Liquid molecules have a range of kinetic energies — some have much higher KE than average. [1]
- Molecules at the surface with sufficient KE escape (evaporate) from the liquid into the gas phase. [1]
- The molecules that escape are the ones with the highest KE (fastest moving). [1]
- The remaining molecules therefore have a lower average KE → the average temperature of the liquid decreases (it cools). [1]
Waves — Structured & Open-Ended
Wave properties, v=fλ, reflection, refraction, total internal reflection, EM spectrum, sound.
W1
Structured
A ray of light travels from glass (refractive index n = 1.5) into air. The critical angle of the glass is 42°.
- (a) Explain what is meant by the critical angle. [2]
- (b) Describe what happens to a ray hitting the glass–air boundary at exactly 42°, at 30°, and at 55°. [3]
- (c) Optical fibres use total internal reflection. State two conditions that must be satisfied for TIR to occur. [2]
- (d) State one practical application of optical fibres and explain why TIR is essential for it to work. [1]
✏ Your answer
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- (a) The critical angle is the angle of incidence (measured from the normal) in the denser medium at which the refracted ray travels along the boundary (angle of refraction = 90°) [1]. Above this angle, total internal reflection occurs [1].
- (b) At 42° (= critical angle): refracted ray travels along the boundary, angle of refraction = 90° [1]. At 30° (below critical angle): ray is partially refracted into air (bends away from normal) and partially reflected [1]. At 55° (above critical angle): total internal reflection — all light reflects back into the glass, no refraction [1].
- (c) (1) Light must be travelling from a denser medium to a less dense medium (e.g. glass to air) [1]. (2) The angle of incidence must be greater than the critical angle [1].
- (d) Telecommunications / internet data transmission / endoscopy (medical imaging) [accept any valid use] [1]. TIR ensures the light signal bounces along the fibre without escaping through the sides, allowing signals to travel long distances with minimal loss [accept equivalent].
W2
Open-ended
A ship uses sonar (sound waves) to detect the seabed. A pulse of sound is emitted and the echo is received 0.6 s later. The speed of sound in seawater is 1500 m/s.
- (a) Calculate the depth of the seabed below the ship. [2]
- (b) The same ship also uses radar (radio waves, speed = 3 × 10⁸ m/s) to detect another vessel 15 km away. Calculate the time for the radar pulse to return. [2]
- (c) Explain one key difference between sound waves and radio waves in terms of their nature and ability to travel through a vacuum. [2]
✏ Your answer
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- (a) Distance = speed × time = 1500 × 0.6 = 900 m (total path). Depth = 900/2 = 450 m [1 for dividing by 2, 1 for answer]
- (b) Total distance = 2 × 15 000 = 30 000 m. t = d/v = 30 000/(3 × 10⁸) = 1 × 10⁻⁴ s (0.0001 s / 0.1 ms) [1 doubling distance, 1 answer]
- (c) Sound is a longitudinal mechanical wave — it requires a medium (particles) to travel and cannot travel through a vacuum [1]. Radio waves are transverse electromagnetic waves — they do not require a medium and can travel through a vacuum at 3 × 10⁸ m/s [1].
W3
Open-ended
A wave has a frequency of 250 Hz and a wavelength of 1.36 m.
- (a) Calculate the speed of the wave. [1]
- (b) The wave enters a new medium and slows to 204 m/s. Calculate the new wavelength. State whether this medium is denser or less dense, and explain how you know. [3]
✏ Your answer
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- (a) v = fλ = 250 × 1.36 = 340 m/s [1]
- (b) Frequency does not change between media. New wavelength = v/f = 204/250 = 0.816 m [1]. The new medium is denser [1] because the wave slowed down — waves travel more slowly in denser media, and the wavelength decreased while frequency stayed constant [1].
💡 Key rule: Frequency never changes when a wave moves between media. Only speed and wavelength change.
Electricity — Structured & Open-Ended
Current, voltage, resistance, Ohm's Law, series/parallel circuits, power, household electricity.
Elec1
Structured
A circuit contains a 12 V battery and three resistors: R₁ = 4 Ω in series with a parallel combination of R₂ = 6 Ω and R₃ = 12 Ω.
- (a) Calculate the combined resistance of R₂ and R₃ in parallel. [2]
- (b) Calculate the total resistance of the circuit. [1]
- (c) Calculate the current from the battery. [1]
- (d) Calculate the voltage across R₁ and across the parallel combination. [2]
- (e) Calculate the current through R₂ and through R₃. [2]
- (f) Calculate the total power delivered by the battery. [2]
✏ Your answer
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- (a) 1/R_parallel = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4 → R_parallel = 4 Ω [1 method, 1 answer]
- (b) R_total = R₁ + R_parallel = 4 + 4 = 8 Ω [1]
- (c) I = V/R = 12/8 = 1.5 A [1]
- (d) V_R₁ = IR₁ = 1.5 × 4 = 6 V; V_parallel = 12 − 6 = 6 V (or: 1.5 × 4 = 6 V) [1 each]
- (e) I_R₂ = 6/6 = 1 A; I_R₃ = 6/12 = 0.5 A (check: 1 + 0.5 = 1.5 A ✓) [1 each]
- (f) P = IV = 1.5 × 12 = 18 W [1 formula, 1 answer]
Or use P = V²/R = 144/8 = 18 W, or P = I²R = 2.25 × 8 = 18 W.
Elec2
Open-ended
A 2 kW electric kettle is connected to the 230 V mains supply.
- (a) Calculate the current drawn by the kettle. [2]
- (b) Calculate the resistance of the heating element. [2]
- (c) The kettle runs for 3 minutes. Calculate the electrical energy consumed and its cost if electricity is priced at $0.28 per kWh. [2]
✏ Your answer
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- (a) P = IV → I = P/V = 2000/230 = 8.70 A [1 formula, 1 answer to 3 s.f.]
- (b) V = IR → R = V/I = 230/8.70 = 26.4 Ω [1 formula, 1 answer]. Or R = V²/P = 230²/2000 = 26.45 Ω.
- (c) Energy = Power × time = 2 kW × (3/60) h = 2 × 0.05 = 0.1 kWh [1]. Cost = 0.1 × $0.28 = $0.028 (2.8 cents) [1].
Elec3
Open-ended
A student connects two identical bulbs first in series and then in parallel, using the same battery each time.
- (a) Compare the brightness of each bulb in the two arrangements. Explain your reasoning using circuit theory. [3]
- (b) Which arrangement runs down the battery faster? Explain why. [2]
✏ Your answer
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- (a) Parallel arrangement: bulbs are brighter [1]. In series, total resistance doubles, so total current halves; each bulb gets half the battery voltage → much less power (P = V²/R). In parallel, each bulb gets the full battery voltage and draws full current — same power as one bulb alone [1]. Therefore each bulb in parallel glows at full brightness; each bulb in series glows dimly [1].
- (b) Parallel arrangement drains the battery faster [1] because the total current from the battery is twice as large (both bulbs draw full current simultaneously), so the battery transfers energy at twice the rate [1].
Magnetism & Electromagnetism — Structured & Open-Ended
Magnetic fields, motor effect, electromagnetic induction, transformers, Fleming's rules.
M1
Structured
A transformer has a primary coil of 400 turns connected to a 240 V AC supply. The secondary coil has 60 turns and is connected to a 6 Ω resistor.
- (a) Calculate the secondary voltage. [2]
- (b) Calculate the current in the secondary coil. [1]
- (c) Assuming 100% efficiency, calculate the current in the primary coil. [2]
- (d) Explain why transformers only work with alternating current (AC) and not direct current (DC). [3]
✏ Your answer
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- (a) V_s/V_p = N_s/N_p → V_s = 240 × (60/400) = 36 V [1 formula, 1 answer]
- (b) I_s = V_s/R = 36/6 = 6 A [1]
- (c) For 100% efficiency: V_p I_p = V_s I_s → I_p = (36 × 6)/240 = 216/240 = 0.9 A [1 principle, 1 answer]
- (d) Electromagnetic induction (which transformers rely on) requires a changing magnetic flux [1]. AC current produces a continuously changing magnetic field in the primary coil [1]. This changing flux induces an EMF in the secondary coil. DC produces a constant (non-changing) field once established, so no EMF is induced and the transformer does not work [1].
M2
Open-ended
Describe how you would use Fleming's Left Hand Rule to determine the direction of motion of a current-carrying conductor in a magnetic field. Then apply it: if the magnetic field points into the page and the conventional current flows upward, in which direction does the conductor move? [5 marks]
✏ Your answer
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✓ Mark Scheme
- Hold the left hand with the thuMb, First finger, and seCond finger mutually perpendicular [1].
- First finger points in the direction of the magnetic Field (B) [1].
- seCond finger points in the direction of conventional Current (I) [1].
- thuMb points in the direction of the Motion (force) [1].
- Application: Field into page (away from viewer) → first finger pointing into page. Current upward → second finger pointing upward. ThumbB direction: pointing to the LEFT (or rightward, depending on orientation — accept either with clear justification). Correct answer: conductor moves to the left [1].
Accept a clearly drawn diagram showing the three-finger setup as an alternative to written description.
Pressure & Moments — Structured & Open-Ended
Pressure P=F/A, fluid pressure P=ρgh, moments, principle of moments, centre of gravity.
P1
Structured
A uniform beam of weight 120 N and length 4 m is balanced on a pivot at its centre. A load of 200 N is hung 0.5 m to the left of the pivot.
- (a) Calculate the moment of the 200 N load about the pivot. State its direction. [2]
- (b) Calculate the distance from the pivot at which a 150 N upward force must be applied on the right side to achieve equilibrium. [2]
- (c) A 60 N load is added 1.5 m to the right of the pivot. The 150 N force is removed. Find the resultant moment and state which direction the beam rotates. [3]
- (d) Why does the weight of a uniform beam act at its midpoint? [1]
✏ Your answer
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✓ Mark Scheme
- (a) Moment = F × d = 200 × 0.5 = 100 Nm, anticlockwise [1 calculation, 1 direction]
- (b) For equilibrium: clockwise moment = anticlockwise moment. 150 × d = 100 → d = 100/150 = 0.667 m (2/3 m) to the right [1 principle, 1 answer]
- (c) ACW moment (200 N load) = 200 × 0.5 = 100 Nm. CW moment (60 N load) = 60 × 1.5 = 90 Nm [1 each]. Resultant = 100 − 90 = 10 Nm anticlockwise [1] → beam rotates anticlockwise (left side goes down).
- (d) The beam is uniform — its mass is evenly distributed along its length, so the centre of gravity (where all weight effectively acts) is at the geometric centre (midpoint) [1].
P2
Open-ended
A diver swims to a depth of 30 m in seawater (density = 1025 kg/m³, g = 10 N/kg). Atmospheric pressure at the surface = 100 000 Pa.
- (a) Calculate the pressure due to the seawater at 30 m depth. [2]
- (b) Calculate the total pressure (water + atmosphere) at 30 m. [1]
- (c) Explain why deep-sea submarines need very thick, strong hulls. [2]
✏ Your answer
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✓ Mark Scheme
- (a) P = ρgh = 1025 × 10 × 30 = 307 500 Pa ≈ 3.08 × 10⁵ Pa [1 formula, 1 answer]
- (b) Total pressure = 307 500 + 100 000 = 407 500 Pa ≈ 4.08 × 10⁵ Pa [1]
- (c) Fluid pressure increases with depth (P = ρgh) [1]. At great depths, the water pressure acting on the submarine hull is enormous. The hull must be thick and strong enough to withstand this inward pressure force without collapsing [1].
Radioactivity — Structured & Open-Ended
Alpha, beta, gamma radiation, half-life, nuclear equations, uses and safety.
R1
Structured
A radioactive source has an initial activity of 4800 counts/min. Its half-life is 6 hours.
- (a) Define half-life. [1]
- (b) Calculate the activity after 18 hours. [2]
- (c) Calculate the activity after 30 hours. [2]
- (d) Sketch a decay curve showing activity (y-axis) against time (x-axis) from 0 to 30 hours. Mark at least four key points. [3]
✏ Your answer
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✓ Mark Scheme
- (a) Half-life is the time taken for the activity (or number of undecayed nuclei) of a radioactive substance to decrease to half its initial value [1].
- (b) 18 hours = 3 half-lives. Activity = 4800 × (½)³ = 4800/8 = 600 counts/min [1 number of half-lives, 1 answer]
- (c) 30 hours = 5 half-lives. Activity = 4800 × (½)⁵ = 4800/32 = 150 counts/min [1 number of half-lives, 1 answer]
- (d) Smooth curve starting at (0, 4800) through key points: (6h, 2400), (12h, 1200), (18h, 600), (24h, 300), (30h, 150) [1 correct shape; 1 each for at least two correct marked values; award max 3]
⚠ Common trap: Students subtract half-life from the activity (4800−4800=0 after one half-life). Half-life means the activity HALVES — not becomes zero.
R2
Open-ended
A hospital uses three radioactive sources for different medical purposes: (A) an alpha source for measuring paper thickness in manufacturing, (B) a gamma source for cancer treatment, and (C) a beta source for monitoring blood flow.
- (a) For each source A, B, and C, explain why that type of radiation is most suitable for that application. [3]
- (b) Explain two safety precautions a radiographer should take when handling gamma sources. [2]
- (c) Why should radioactive sources used in medical applications have short half-lives rather than long ones? [2]
✏ Your answer
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- (a) Source A (alpha, paper thickness): Alpha particles are stopped by a few cm of air or a sheet of paper. If the paper is too thick, fewer alpha particles reach the detector — the count rate drops and triggers adjustment. Alpha is ideal because it is stopped by paper but not by air gaps [1].
- Source B (gamma, cancer treatment): Gamma rays are highly penetrating and can pass through skin and tissue to reach a deep tumour [1]. Targeted beams can destroy cancer cells while minimising damage to surrounding tissue. Alpha and beta lack the penetration.
- Source C (beta, blood flow monitoring): Beta particles can penetrate tissue (a few mm) and are detected outside the body, allowing blood flow to be traced. Beta is suitable because it is penetrating enough to be detected externally but less damaging than gamma [1].
- (b) Any two of: handle with long tongs or robotic arms to increase distance (inverse square law reduces dose) [1]; wear lead-lined apron/shield (lead absorbs gamma) [1]; minimise time of exposure [1]; work behind lead-glass shielding [1].
- (c) Short half-life means the source becomes non-radioactive quickly [1], reducing the patient's total radiation dose and the time surrounding tissue is exposed to radiation, reducing long-term risk of radiation damage or cancer [1].
R3
Open-ended
Complete and balance the following nuclear equations:
(a) Ra-226 undergoes alpha decay: ²²⁶₈₈Ra → ? + ⁴₂He [2]
(b) C-14 undergoes beta decay: ¹⁴₆C → ? + ⁰₋₁e [2]
✏ Your answer
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- (a) Mass number: 226 − 4 = 222. Atomic number: 88 − 2 = 86. Element 86 = Radon. ²²²₈₆Rn [1 mass number, 1 atomic number / element]
- (b) Mass number: 14 − 0 = 14. Atomic number: 6 − (−1) = 7. Element 7 = Nitrogen. ¹⁴₇N [1 mass number, 1 atomic number / element]
💡 Key rule: Mass numbers and atomic numbers must balance on both sides. Alpha decay: mass −4, atomic number −2. Beta decay: mass unchanged, atomic number +1.
Measurement — Structured & Open-Ended
SI units, scalars and vectors, experimental errors, significant figures, measuring instruments.
Meas1
Open-ended
A student measures the period of a pendulum (time for one complete swing) by timing 20 complete oscillations. Her results were: 28.4 s, 28.6 s, 28.2 s.
- (a) Calculate the mean time for 20 oscillations and the period of the pendulum. [2]
- (b) Explain why timing 20 oscillations rather than 1 gives a more reliable period measurement. [2]
- (c) Identify one source of random error and one source of systematic error in this experiment. [2]
- (d) Explain the difference between accuracy and precision in the context of this experiment. [1]
✏ Your answer
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- (a) Mean = (28.4 + 28.6 + 28.2)/3 = 85.2/3 = 28.4 s. Period T = 28.4/20 = 1.42 s [1 mean, 1 period]
- (b) Timing 20 oscillations reduces the effect of random errors in reaction time (starting and stopping the stopwatch) [1]. Any timing error is spread over 20 oscillations, so the error per oscillation is 1/20th of the total timing error — making the period much more precise [1].
- (c) Random error: inconsistent reaction time when starting/stopping the stopwatch [1]. Systematic error: zero error in the stopwatch (if it doesn't read exactly 0 at start); or the pendulum bob is not a perfect point mass (affecting true period from theory) [1].
- (d) Precision: the three readings (28.4, 28.6, 28.2 s) are close together — the results are precise (consistent) [1]. Accuracy: whether the mean (28.4 s / 1.42 s period) is close to the true theoretical period. We cannot determine accuracy without knowing the true value.
Challenge Questions — Cross-Topic & Unfamiliar Contexts
Higher-order questions combining multiple topics — typical of O-Level Paper 2 Section C.
C1
Challenge
A hydroelectric power station converts the gravitational potential energy of water into electrical energy. Water falls through a vertical height of 80 m at a flow rate of 500 kg/s. (g = 10 N/kg)
- (a) Calculate the gravitational PE lost by the water per second. [2]
- (b) The generator has an efficiency of 90%. Calculate the electrical power output. [2]
- (c) The electrical output is at 11 000 V. It is stepped up by a transformer to 400 000 V for transmission. Calculate the ratio N_s : N_p. [2]
- (d) Explain why electricity is transmitted at very high voltage (400 000 V) rather than at the generation voltage of 11 000 V. Include a calculation to support your answer. [3]
✏ Your answer
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- (a) GPE/s = mgh/t = (500 kg/s) × 10 × 80 = 400 000 W = 400 kW [1 method, 1 answer]
- (b) Electrical output = 0.90 × 400 000 = 360 000 W = 360 kW [1 efficiency, 1 answer]
- (c) N_s/N_p = V_s/V_p = 400 000/11 000 ≈ 36.4 : 1 (or 3640 : 100, accept equivalent ratio) [1 formula, 1 answer]
- (d) Transmitting at high voltage means lower current (since P = IV, same power at higher V → lower I) [1]. Power lost as heat in cables = I²R. A lower current means much less power wasted: e.g. if I is reduced by factor 36.4, power loss (I²R) is reduced by factor 36.4² ≈ 1325 [1 calculation or reasoning]. This makes long-distance transmission much more efficient and economical [1].
💡 This question links: Energy (GPE, efficiency) + Electricity (power, P=IV) + Magnetism (transformers, V_s/V_p = N_s/N_p). Cross-topic questions like this often appear in O-Level Paper 2.
C2
Challenge
A student wants to investigate how the resistance of a nichrome wire depends on its length. She has a power supply, ammeter, voltmeter, rheostat, connecting wires, and nichrome wire of different lengths.
- (a) Draw a labelled circuit diagram for this investigation. [2]
- (b) State the independent variable (IV), dependent variable (DV), and two controlled variables (CV). [2]
- (c) Describe the procedure to measure resistance for each length of wire. [2]
- (d) Predict and explain the expected relationship between length and resistance. [2]
✏ Your answer
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- (a) Ammeter in series with the nichrome wire; voltmeter in parallel across the nichrome wire; rheostat (variable resistor) in series to control current; power supply [1 for correct ammeter position (series); 1 for correct voltmeter position (parallel)].
- (b) IV: length of nichrome wire (measured with ruler) [1]. DV: resistance of the wire (calculated from R = V/I) [1]. CVs (any two): cross-sectional area of wire; material of wire; temperature of wire; voltage applied [1 for any two correct CVs].
- (c) Set the length of wire using a ruler; connect in circuit; adjust rheostat to set a suitable current; record ammeter reading (I) and voltmeter reading (V) [1]; calculate R = V/I [1]. Repeat for each length of wire. Take multiple readings and average to improve reliability.
- (d) Expected: resistance is directly proportional to length (R ∝ L) — as length doubles, resistance doubles [1]. Explanation: a longer wire has more collisions between electrons and metal ions per unit length, providing greater opposition to current flow (higher resistance) [1].
C3
Challenge
In a car crash test, a dummy of mass 75 kg travels at 15 m/s and is brought to rest by an airbag in 0.15 s, or by a rigid steel dashboard in 0.008 s. (g = 10 N/kg)
- (a) Calculate the change in momentum of the dummy in both cases. [2]
- (b) Calculate the average force on the dummy in each case. [2]
- (c) Use Newton's laws to explain why the airbag reduces injury compared to hitting the dashboard. [3]
✏ Your answer
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- (a) Change in momentum = mΔv = 75 × 15 = 1125 kg m/s in both cases (same initial and final velocity — change in momentum is the same regardless of stopping method) [1 formula, 1 answer with note that both are equal]
- (b) F = Δp/t. Airbag: F = 1125/0.15 = 7500 N [1]. Dashboard: F = 1125/0.008 = 140 625 N ≈ 141 kN [1].
- (c) The impulse (change in momentum) is the same in both cases (same Δp = 1125 kg m/s) [1]. The airbag increases the time over which the momentum change occurs [1]. By Newton's 2nd Law (F = Δp/t), a longer stopping time produces a much smaller average force on the dummy — reducing the force on the human body and therefore reducing injury [1].
💡 Key physics: Impulse = Ft = Δp. The impulse is fixed by the crash conditions. Longer time → smaller force. This is the physics of all safety features: seat belts, airbags, crumple zones, helmets.
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