Section A — Short Structured Questions (Q1–5 · 50 marks)
Q1
(a) State the initial vertical velocity of the ball. [1]
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- Initial vertical velocity = 0 m/s [1]. The ball is kicked horizontally, so it has no vertical component of initial velocity.
(b) Calculate the time taken for the ball to reach the sea. [2]
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- Vertical: s = ut + ½at² → 45 = 0 + ½(10)t² [1 for correct equation/substitution]
- t² = 9 → t = 3 s [1 for correct answer with unit]
(c) Calculate the horizontal distance from the cliff base where the ball lands. [2]
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- Horizontal velocity is constant (no air resistance): v_x = 15 m/s [1 for recognising horizontal velocity unchanged]
- Horizontal distance = v_x × t = 15 × 3 = 45 m [1]
💡 Key principle: In projectile motion (no air resistance), horizontal and vertical motions are independent. Horizontal velocity stays constant; vertical velocity increases due to gravity.
(d) Calculate the vertical velocity just before the ball hits the sea. [2]
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- v_y = u + at = 0 + 10 × 3 = 30 m/s (downward) [1 formula/substitution, 1 answer with direction]
Accept v² = u² + 2as = 0 + 2(10)(45) = 900 → v = 30 m/s.
(e) A student claims that a heavier ball kicked at the same speed would land closer to the cliff base. Evaluate this claim. [3]
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- The claim is incorrect. [1]
- The time of flight depends only on the vertical motion: t = √(2h/g). This is independent of mass — both balls would reach the sea in the same time (3 s). [1]
- Since both balls have the same horizontal velocity (15 m/s) and the same time of flight, they land at the same horizontal distance (45 m) from the cliff. Mass does not affect projectile range in the absence of air resistance. [1]
⚠ Trap: Students often confuse this with situations where air resistance matters. Without air resistance, ALL objects have the same acceleration due to gravity regardless of mass (Galileo's principle).
Q2
(a) Calculate the weight of the cyclist. [1]
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- W = mg = 60 × 10 = 600 N [1]
(b) Calculate the gravitational PE gained by the cyclist. [2]
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- GPE = mgh = 60 × 10 × 30 = 18 000 J [1 formula, 1 answer]
Use the vertical height (30 m), NOT the slope length (150 m). GPE depends on vertical height only.
(c) Calculate the work done against the resistive force over the 150 m slope. [2]
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- W = Fd = 80 × 150 = 12 000 J [1 formula, 1 answer]
Use the distance along the slope (150 m) for work done by/against the resistive force, since it acts along the slope. GPE uses vertical height.
(d) Calculate the total useful work done by the cyclist to climb the hill. [2]
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- Total work = GPE gained + work done against resistance [1 for correct principle]
- = 18 000 + 12 000 = 30 000 J [1]
(e) Calculate the minimum power output of the cyclist. [3]
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- Time = distance / speed = 150 / 5 = 30 s [1]
- P = W/t = 30 000 / 30 = 1000 W [1 formula, 1 answer with unit]
Alternative: P = Fv. Total driving force = (Weight component along slope + 80 N). Weight component = 600 × (30/150) = 120 N. Total F = 120 + 80 = 200 N. P = 200 × 5 = 1000 W. ✓
Q3
(a) State which can emits radiation at the greatest rate and explain why. [2]
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- Can A — dull black [1]
- Dull black surfaces are the best emitters of infrared radiation. They emit radiation at the greatest rate for a given temperature. [1]
(b) State which can cools most slowly and explain why. [2]
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- Can B — shiny silver [1]
- Shiny silver surfaces are poor emitters of infrared radiation, so they lose heat by radiation at the lowest rate. The water inside retains heat for longer. [1]
(c) The room temperature rises to 30°C and the cans are refilled with water at 30°C. A student says the radiation detector reading will now be zero. Evaluate this statement. [2]
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- The statement is incorrect. All objects above absolute zero (0 K / −273°C) emit some thermal radiation. [1]
- However, the NET radiation emitted by the cans is zero (they emit and absorb radiation at the same rate since they are at the same temperature as the surroundings). The detector would read background radiation only. [1]
(d) The student heats 0.4 kg of water in can A from 20°C to 100°C, then the heater is turned off. The water cools from 100°C back to 20°C. Calculate the total thermal energy released by the water as it cools. (c_water = 4200 J/(kg°C)) [2]
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- Q = mcΔT = 0.4 × 4200 × (100 − 20) = 0.4 × 4200 × 80 [1]
- = 134 400 J (134.4 kJ) [1]
The energy released on cooling equals the energy absorbed during heating (same mass, same temperature change, same specific heat capacity) — conservation of energy.
(e) Suggest one reason why the actual energy released may be slightly less than the calculated value. [2]
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- Some energy is also absorbed by the metal can itself (the can gains thermal energy, not just the water) [1], so the water does not supply all 134 400 J — some goes into heating the can. [1 for developed explanation]
Accept: some water evaporates during cooling (taking away latent heat) / heat is also lost to the surroundings through conduction and convection, not just radiation.
Q4
(a) Calculate the combined resistance of R₂ and R₃ in parallel. [2]
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- 1/R = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 [1]
- R_parallel = 12/3 = 4 Ω [1]
(b) Calculate the total circuit resistance and the current from the battery. [2]
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- R_total = R₁ + R_parallel = 4 + 4 = 8 Ω [1]
- I = V/R = 24/8 = 3 A [1]
(c) Calculate the voltage across R₁ and across the parallel combination. [2]
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- V_R₁ = IR₁ = 3 × 4 = 12 V [1]
- V_parallel = 24 − 12 = 12 V (or 3 × 4 = 12 V) [1]. Check: voltages add to 24 V ✓
(d) R₃ is now increased. Describe and explain the effect on (i) the total circuit current and (ii) the voltage across R₁. [4]
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- (i) Total current decreases. [1] Increasing R₃ increases the parallel combination resistance (1/R_parallel = 1/12 + 1/R₃ — as R₃ increases, 1/R₃ decreases, so R_parallel increases). Total circuit resistance increases. By I = V/R (V constant from battery), current decreases. [1]
- (ii) Voltage across R₁ decreases. [1] V_R₁ = IR₁. Since I decreases (from above) and R₁ is unchanged, V_R₁ decreases. Consequently, the voltage across the parallel combination increases (V_parallel = 24 − V_R₁). [1]
💡 Strategy: Start with the effect on total resistance → then current → then voltage drops. Follow the chain logically.
Q5
(a) Calculate the critical angle for this glass. (sin c = 1/n) [2]
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- sin c = 1/n = 1/1.52 = 0.658 [1]
- c = sin⁻¹(0.658) = 41.1° (accept 41°) [1]
(b) Will total internal reflection occur at the glass–air interface when the angle of incidence is 50°? Justify your answer. [2]
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- Yes, TIR will occur. [1]
- The angle of incidence (50°) exceeds the critical angle (41.1°), and light is travelling from a denser medium (glass) to a less dense medium (air). Both conditions for TIR are satisfied. [1]
(c) Describe what happens to the ray at the glass–air interface when TIR occurs. Include a statement about the angle of incidence and angle of reflection. [2]
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- The ray is completely reflected back into the glass — no refraction/transmission into air occurs. [1]
- The angle of reflection equals the angle of incidence (50°), both measured from the normal. [1]
(d) Optical fibres use TIR to transmit data as light pulses. State two advantages of using optical fibres over copper wires for long-distance data transmission. [2]
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- Light travels much faster than electrical signals in copper, allowing higher bandwidth / faster data transmission. [1]
- Less signal loss (attenuation) over long distances; optical fibres do not suffer from electrical interference (EMI); lighter and more flexible; higher capacity (many signals simultaneously by different wavelengths). [1 for any one valid advantage]
(e) A light signal travels along an optical fibre at 2.0 × 10⁸ m/s. The fibre is 6000 km long. Calculate the time for the signal to travel from one end to the other. Give your answer in ms. [2]
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- t = d/v = (6000 × 10³) / (2.0 × 10⁸) = 6 × 10⁶ / 2 × 10⁸ = 0.03 s [1]
- = 30 ms [1 for correct unit conversion]
Section B — Longer Structured Questions (Q6–8 · 30 marks)
Q6
(a) State the number of protons and neutrons in a ²³⁵₉₂U nucleus. [2]
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- Protons = atomic number = 92 [1]
- Neutrons = mass number − atomic number = 235 − 92 = 143 [1]
(b) Write the nuclear equation for the beta decay of strontium-90 (⁹⁰₃₈Sr). [2]
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- ⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁e [1 for correct daughter nucleus yttrium-90, 1 for correct beta particle notation]
Mass number: 90 → 90 (unchanged). Atomic number: 38 → 39 (+1, since beta emission converts a neutron to a proton). Element 39 = Yttrium (Y).
(c) A nuclear waste container holds 640 g of strontium-90. Calculate the mass remaining after 87 years. [3]
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- Number of half-lives = 87 / 29 = 3 [1]
- Mass remaining = 640 × (½)³ = 640 / 8 = 80 g [1 for correct method, 1 for answer]
(d) Explain why strontium-90 is considered dangerous to human health even though it emits beta particles (which are less penetrating than gamma rays). [3]
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- Strontium-90 is chemically similar to calcium and can be incorporated into bone tissue if ingested (through contaminated food or water). [1]
- Once inside the body, the beta radiation directly irradiates bone marrow and surrounding cells from within — the short range of beta particles (which would stop at the skin externally) is not a protection when the source is internal. [1]
- The long half-life (29 years) means it remains in the body for decades, causing prolonged radiation exposure that can lead to bone cancer, leukaemia, and other radiation-induced diseases. [1]
Q7
(a) A step-up transformer increases the voltage from 22 kV to 400 kV. If the primary coil has 550 turns, calculate the number of secondary turns. [2]
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- Vs/Vp = Ns/Np → Ns = Np × Vs/Vp = 550 × 400/22 [1]
- = 550 × 18.18 = 10 000 turns [1]
(b) A power station generates 50 MW at 22 kV. Calculate the current in the transmission lines (at 400 kV). Assume 100% transformer efficiency. [3]
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- P = IV → I = P/V = 50 × 10⁶ / 400 × 10³ [1 for correct substitution]
- = 50 000 000 / 400 000 = 125 A [1 for answer with unit]
Compare: at 22 kV, current would be 50×10⁶/22×10³ ≈ 2273 A — much larger. Transmission at high voltage greatly reduces current.
(c) The transmission lines have a total resistance of 8 Ω. Calculate the power wasted in the transmission lines. [2]
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- P_waste = I²R = 125² × 8 = 15 625 × 8 [1]
- = 125 000 W = 125 kW [1]
This is only 125/50 000 = 0.25% of the generated power — very efficient. At 22 kV (2273 A), P_waste = 2273² × 8 = 41.3 MW — 82.6% wasted! This demonstrates why high-voltage transmission is essential.
(d) Explain why transformers only work with alternating current (AC) and would not work with direct current (DC). [3]
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- A transformer works by electromagnetic induction — an EMF is induced in the secondary coil when there is a changing magnetic flux linking the two coils. [1]
- AC in the primary coil creates a continuously changing magnetic field in the iron core. This changing flux induces an alternating EMF in the secondary coil. [1]
- DC produces a constant (non-changing) magnetic field once established. A constant magnetic flux induces no EMF — by Faraday's law, EMF is proportional to the RATE OF CHANGE of flux. So DC produces no output from the secondary coil. [1]
Q8
(a) Calculate the weight of the submarine. [1]
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- W = mg = 8 × 10⁶ × 10 = 8 × 10⁷ N (80 MN) [1]
(b) Calculate the upthrust on the fully submerged submarine. [2]
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- Upthrust = weight of seawater displaced = ρVg = 1025 × 9000 × 10 [1]
- = 92 250 000 N ≈ 9.23 × 10⁷ N (92.25 MN) [1]
(c) Determine the resultant force on the submarine when fully submerged and state its direction. [2]
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- Resultant = Upthrust − Weight = 92 250 000 − 80 000 000 = 12 250 000 N ≈ 12.25 MN upward [1 calculation, 1 direction]
The upthrust exceeds the weight → net upward force → submarine would rise unless the crew adjusts ballast.
(d) To remain stationary at a fixed depth, the crew floods ballast tanks with seawater to increase the submarine's mass. Calculate the total mass of seawater needed to flood the tanks to achieve neutral buoyancy (upthrust = weight). [3]
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- For neutral buoyancy: total weight = upthrust → total mass × g = 9.225 × 10⁷ [1]
- Total mass needed = 9.225 × 10⁷ / 10 = 9 225 000 kg [1]
- Additional seawater mass = 9 225 000 − 8 000 000 = 1 225 000 kg = 1225 tonnes [1]
(e) The submarine descends to 500 m. Calculate the pressure due to the seawater at this depth. [2]
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- P = ρgh = 1025 × 10 × 500 [1]
- = 5 125 000 Pa ≈ 5.1 × 10⁶ Pa (5.1 MPa) [1]
This is approximately 50 times atmospheric pressure — explaining why submarine hulls must be extremely thick and strong.
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