Roller Coaster — Energy Transformations
A roller coaster car of mass 800 kg starts from rest at point A (height 45 m above the ground). It travels along a track to point B (height 5 m), then up to point C (height 20 m), and finally to point D at ground level (height 0 m). Assume no friction or air resistance unless stated. (g = 10 N/kg)
Data table
| Point | Height (m) | Speed (m/s) | GPE (J) | KE (J) |
|---|---|---|---|---|
| A | 45 | 0 | 360 000 | 0 |
| B | 5 | ? | 40 000 | ? |
| C | 20 | ? | 160 000 | ? |
| D | 0 | ? | 0 | ? |
Total mechanical energy at A = 360 000 J (all GPE, since speed = 0)
(a)
[1]
State the principle used to complete the missing values in the table (assuming no energy losses).
✏ Your answer
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- Conservation of energy: total mechanical energy (KE + GPE) is constant at every point — 360 000 J. [1]
(b)
[4]
Calculate the speed at points B, C, and D. Show working for at least one.
✏ Your answer
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- At B: KE = 360 000 − 40 000 = 320 000 J. ½mv² = 320 000 → v² = 640 000/800 = 800 → v = 28.3 m/s [1]
- At C: KE = 360 000 − 160 000 = 200 000 J. v² = 400 000/800 = 500 → v = 22.4 m/s [1]
- At D: KE = 360 000 − 0 = 360 000 J. v² = 720 000/800 = 900 → v = 30 m/s [1]
- [1 for clear method shown for at least one point using KE = total E − GPE then ½mv²]
(c)
[2]
In practice, the car only reaches 25 m/s at point D instead of 30 m/s. Calculate the energy lost to friction and air resistance over the entire journey.
✏ Your answer
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- Actual KE at D = ½ × 800 × 25² = 400 × 625 = 250 000 J [1]
- Energy lost = 360 000 − 250 000 = 110 000 J [1]
(d)
[3]
The car is brought to rest from 25 m/s in 4 s using brakes. Calculate: (i) the deceleration, (ii) the braking force.
✏ Your answer
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- (i) a = (v−u)/t = (0−25)/4 = −6.25 m/s² (deceleration = 6.25 m/s²) [1]
- (ii) F = ma = 800 × 6.25 = 5000 N [1 formula, 1 answer]
(e)
[4]
A designer proposes a loop section. The loop has a radius of 8 m and the car enters at the bottom with speed 20 m/s. The minimum speed needed at the top of the loop to maintain contact with the track is 9 m/s. (i) Using energy conservation, calculate the speed at the top of the loop (height = 16 m above bottom of loop). (ii) State whether the car maintains contact and justify your answer.
✏ Your answer
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- (i) KE at bottom = ½ × 800 × 20² = 160 000 J. GPE gained going to top = mgh = 800 × 10 × 16 = 128 000 J. KE at top = 160 000 − 128 000 = 32 000 J. v² = 2 × 32 000/800 = 80 → v = 8.9 m/s [1 method, 1 answer] (ignoring friction)
- (ii) The car does NOT maintain contact. [1] The calculated speed at the top (8.9 m/s) is less than the minimum required (9 m/s), so the car loses contact with the track at the top of the loop. [1]
Accept 8.94 m/s. With friction losses, the speed would be even lower — making it more dangerous.
(f)
[4]
Explain, using Newton's laws, what would happen to passengers in the car if it lost contact with the track at the top of the loop.
✏ Your answer
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- At the point of losing contact, the normal force from the track on the car becomes zero. [1]
- The only force acting on the car and passengers is gravity (weight, downward). By Newton's 1st Law, with no centripetal force from the track, the car cannot continue in a circular path. [1]
- The car and passengers follow a parabolic (projectile) path — they continue forward and downward under gravity. [1]
- The passengers would experience free fall momentarily (weightlessness), then collide with the inside of the car or the track structure — extremely dangerous. [1]
Solar Panels — Electrical Power & Efficiency
A house is fitted with 12 solar panels. Each panel has area 1.6 m² and converts solar energy to electrical energy. The solar irradiance (power of sunlight per unit area) is 850 W/m² on a clear day. A monitoring system records the electrical output and usage data below.
Recorded data (average clear day)
| Time period | Solar output (kW) | House consumption (kW) | Grid import/export (kW) |
|---|---|---|---|
| 06:00–09:00 | 1.2 | 2.8 | −1.6 (import) |
| 09:00–15:00 | 8.4 | 3.1 | +5.3 (export) |
| 15:00–18:00 | 3.6 | 2.4 | +1.2 (export) |
| 18:00–22:00 | 0 | 4.2 | −4.2 (import) |
Negative grid value = importing from grid. Positive = exporting to grid.
(a)
[3]
Calculate the total solar power incident on all 12 panels and hence the efficiency of the solar panels during 09:00–15:00.
✏ Your answer
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- Total area = 12 × 1.6 = 19.2 m² [1]
- Total incident solar power = 850 × 19.2 = 16 320 W = 16.32 kW [1]
- Efficiency = useful output / input = 8.4 / 16.32 = 0.515 = 51.5% [1]
(b)
[3]
Calculate the total electrical energy (in kWh) generated by the solar panels over the full day shown in the table.
✏ Your answer
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- 06:00–09:00: 1.2 kW × 3 h = 3.6 kWh [1]
- 09:00–15:00: 8.4 × 6 = 50.4 kWh; 15:00–18:00: 3.6 × 3 = 10.8 kWh; 18:00–22:00: 0 kWh
- Total = 3.6 + 50.4 + 10.8 + 0 = 64.8 kWh [1 correct method, 1 correct total]
(c)
[2]
During 09:00–15:00, the panels export 5.3 kW to the grid at 230 V AC. Calculate the current exported to the grid.
✏ Your answer
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- P = IV → I = P/V = 5300/230 [1]
- = 23.0 A [1]
(d)
[4]
The homeowner installs a battery storage system to store excess solar energy instead of exporting it. The battery can store 20 kWh. (i) Calculate the total energy that could be stored during the export periods. (ii) If the battery costs $6 000 and electricity costs $0.30/kWh, calculate how many days it would take for the battery savings to pay back its cost, assuming the stored energy replaces grid imports at night.
✏ Your answer
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- (i) Export energy: 09:00–15:00: 5.3 × 6 = 31.8 kWh; 15:00–18:00: 1.2 × 3 = 3.6 kWh. Total available = 35.4 kWh. Battery capacity = 20 kWh, so 20 kWh stored (battery is full). [1 correct export calculation, 1 for identifying battery cap as limiting factor]
- (ii) Energy saved per day = 20 kWh (replaces grid import). Daily saving = 20 × $0.30 = $6.00 [1]. Days to payback = $6 000 / $6.00 = 1 000 days ≈ 2.7 years [1]
Accept answers that use actual stored energy (20 kWh limited by battery capacity).
(e)
[4]
The DC output from the solar panels passes through an inverter that converts it to 230 V AC for home use. The inverter has efficiency 94%. (i) If the panels produce 8.4 kW DC, calculate the AC power output of the inverter. (ii) Explain one reason why 100% efficiency is impossible for the inverter.
✏ Your answer
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- (i) AC output = 0.94 × 8400 = 7896 W ≈ 7.9 kW [1 formula, 1 answer]
- (ii) Some electrical energy is converted to thermal energy (heat) in the inverter's components (transistors, transformers, resistors) as current flows through them. By conservation of energy, this wasted heat means not all input energy reaches the output as electrical energy. [1 for identifying heat loss, 1 for correct energy reasoning]
Medical Ultrasound Imaging
Ultrasound is used in hospitals to image internal organs. A probe emits ultrasound pulses at 3.5 MHz (3 500 000 Hz) into the body. The speed of ultrasound in soft tissue is approximately 1540 m/s. When the pulse reaches a boundary between different tissues, some energy is reflected back to the probe and some is transmitted.
Reflection data from a scan
| Reflected pulse | Time delay after emission (μs) | Depth of boundary (cm) |
|---|---|---|
| Echo 1 (skin–muscle boundary) | 2.6 | ? |
| Echo 2 (muscle–organ boundary) | 18.2 | ? |
| Echo 3 (far wall of organ) | 26.0 | ? |
(a)
[4]
Calculate the depth of each boundary. Show working for Echo 1.
✏ Your answer
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- depth = (speed × time) / 2 [1 for dividing by 2 — pulse travels to boundary AND back]
- Echo 1: d = (1540 × 2.6 × 10⁻⁶) / 2 = 0.002002 m ≈ 0.20 cm [1]
- Echo 2: d = (1540 × 18.2 × 10⁻⁶) / 2 = 0.014014 m ≈ 1.40 cm [1]
- Echo 3: d = (1540 × 26.0 × 10⁻⁶) / 2 = 0.02002 m ≈ 2.00 cm [1]
The organ thickness = depth of Echo 3 − depth of Echo 2 = 2.00 − 1.40 = 0.60 cm.
⚠ Always divide by 2 — the pulse travels to the boundary AND back. Forgetting this gives depths twice as large.
(b)
[2]
Calculate the wavelength of the ultrasound in soft tissue.
✏ Your answer
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- λ = v/f = 1540 / (3.5 × 10⁶) [1]
- = 4.4 × 10⁻⁴ m = 0.44 mm [1]
(c)
[3]
Explain why higher-frequency ultrasound gives better image resolution, and state one disadvantage of increasing frequency.
✏ Your answer
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- Higher frequency → shorter wavelength (since v = fλ and v is fixed in tissue). [1]
- A shorter wavelength can resolve (distinguish) smaller structures — the image detail is finer since features smaller than one wavelength cannot be detected. Higher frequency gives finer resolution. [1]
- Disadvantage: higher frequency ultrasound is absorbed/attenuated more rapidly in tissue — it cannot penetrate as deeply. It is therefore less suitable for imaging deep organs. [1]
(d)
[3]
Compare ultrasound imaging with X-ray imaging. Give one advantage and one disadvantage of ultrasound compared to X-rays for medical imaging.
✏ Your answer
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- Advantage of ultrasound: ultrasound does not use ionising radiation (unlike X-rays), so it carries no risk of radiation damage to cells or DNA. It is therefore safe for use on pregnant women and repeated scans. [1]
- Disadvantage of ultrasound: ultrasound cannot pass through bone or air-filled cavities (it is totally reflected at these boundaries), so it cannot image structures behind bone or lungs effectively. X-rays pass through soft tissue and can image bone and dense structures more clearly. [1]
- [1 for both answers being scientifically complete and clearly comparative — mentioning both modalities in each point]
(e)
[2]
Ultrasound is also used to break up kidney stones. The probe is focused so that the ultrasound converges at the kidney stone. Explain why this application uses much higher intensity ultrasound than medical imaging.
✏ Your answer
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- For imaging, the ultrasound only needs to be reflected by tissue boundaries — very low energy is sufficient. For breaking kidney stones, the ultrasound must transfer enough energy to physically fracture the stone. [1]
- High-intensity focused ultrasound concentrates large amounts of energy at the focal point (the kidney stone), creating mechanical stress and cavitation (bubble formation and collapse) that shatters the stone without damaging surrounding tissue. [1]
Electric Vehicle — Performance & Range
An electric vehicle (EV) has a 75 kWh battery pack. The motor operates at 400 V DC and draws a maximum current of 750 A during rapid acceleration. The car has a mass of 2100 kg. A drag coefficient analysis shows the total resistive force at 30 m/s is 1200 N.
Performance data
| Speed (m/s) | Resistive force (N) | Driving force required for constant speed (N) | Motor power at constant speed (kW) |
|---|---|---|---|
| 10 | 200 | 200 | 2.0 |
| 20 | 600 | 600 | 12.0 |
| 30 | 1200 | 1200 | 36.0 |
| 40 | 2200 | 2200 | 88.0 |
(a)
[2]
Using the data, describe the relationship between speed and resistive force. Support your answer with calculations.
✏ Your answer
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- The resistive force is NOT proportional to speed — it increases more rapidly. [1]
- When speed doubles from 10 to 20 m/s, force increases by factor 3 (200 to 600 N). When speed doubles from 20 to 40 m/s, force increases by factor 3.67 (600 to 2200 N). The relationship appears to be approximately proportional to speed squared (quadratic/exponential growth) — consistent with aerodynamic drag. [1 for supporting with data comparison]
(b)
[3]
During maximum acceleration from rest, the motor draws 750 A at 400 V. (i) Calculate the maximum motor power. (ii) Calculate the resultant force and acceleration when resistive force is 800 N at this moment.
✏ Your answer
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- (i) P = IV = 750 × 400 = 300 000 W = 300 kW [1]
- (ii) At some speed v: driving force = P/v. But we need resultant force. Assuming full power used as driving force ≈ instantaneous: net F = driving − resistive. If driving force = 300 000/v — need v. Alternative approach: if question means at start (v → 0, resistive = 0): net F = driving force = motor force. Motor force ≈ 300 kW / v — insufficient info for exact value. Accept: resultant = motor force − 800 N. For 1 mark: a = F_net/m = (F_motor − 800)/2100. [1 for correct approach, 1 for stating result depends on driving force or speed]
Full credit for: If motor force given as separate value or if student correctly identifies that P=Fv so F=P/v requires v to be known — and states the limitation clearly.
(c)
[3]
At a constant speed of 30 m/s, the motor uses 36 kW. Calculate how long the 75 kWh battery lasts and the range of the car at this speed.
✏ Your answer
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- Time = Energy / Power = 75 000 Wh / 36 000 W = 2.083 h [1]
- Range = speed × time = 30 m/s × (2.083 × 3600 s) = 30 × 7500 = 225 000 m = 225 km [1 method, 1 answer]
(d)
[4]
Regenerative braking captures kinetic energy during deceleration and stores it back in the battery. The car decelerates from 30 m/s to rest. (i) Calculate the kinetic energy available for capture. (ii) If the regenerative system is 80% efficient, calculate the energy returned to the battery. (iii) Explain what happens to the remaining 20% of the kinetic energy.
✏ Your answer
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- (i) KE = ½mv² = ½ × 2100 × 900 = 945 000 J = 945 kJ [1]
- (ii) Energy returned = 0.80 × 945 000 = 756 000 J = 756 kJ [1]
- (iii) The remaining 20% (189 kJ) is converted to thermal energy (heat) in the motor windings, braking resistors, and mechanical friction components — energy that cannot be recovered. [1] This is consistent with conservation of energy — energy is not destroyed, but is no longer in a useful (electrical or mechanical) form. [1]
(e)
[4]
The EV battery is charged by a fast charger at 400 V, 125 A. (i) Calculate the charging power. (ii) Calculate the time to charge a depleted 75 kWh battery assuming 90% charging efficiency. (iii) The electricity costs $0.30/kWh. Calculate the cost to fully charge the battery.
✏ Your answer
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- (i) P = IV = 125 × 400 = 50 000 W = 50 kW [1]
- (ii) Energy needed from charger = 75/0.90 = 83.33 kWh. Time = 83.33 kWh / 50 kW = 1.67 h ≈ 100 minutes [1 for accounting for efficiency, 1 for correct time]
- (iii) Cost = 83.33 kWh × $0.30 = $25.00 [1] (uses energy from charger, not just battery capacity, since 10% is wasted)
Nuclear Power Plant — Radiation Shielding & Waste
A nuclear power plant uses uranium-235 fuel. The control room monitors radiation levels and manages radioactive waste. A sample of iodine-131 (a fission product, beta and gamma emitter) has an initial activity of 3200 Bq and a half-life of 8 days. A sample of caesium-137 has a half-life of 30 years.
Shielding effectiveness data
| Radiation type | Stopped by paper | Stopped by 3mm Al | Reduced by 10 cm Pb | Ionising power |
|---|---|---|---|---|
| Alpha (α) | Yes | Yes | Yes | Highest |
| Beta (β) | No | Yes | Yes | Medium |
| Gamma (γ) | No | No (partially) | Significantly reduced | Lowest |
(a)
[3]
Calculate the activity of the iodine-131 sample after 24 days and after 40 days.
✏ Your answer
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- After 24 days: 24/8 = 3 half-lives. Activity = 3200 × (½)³ = 3200/8 = 400 Bq [1]
- After 40 days: 40/8 = 5 half-lives. Activity = 3200 × (½)⁵ = 3200/32 = 100 Bq [1]
- [1 for correct method shown — counting half-lives and applying (½)ⁿ]
(b)
[3]
Using the shielding data, explain what material(s) should be used to safely contain a mixed alpha, beta, and gamma source, and why. Give the minimum adequate shielding.
✏ Your answer
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- Gamma radiation is the most penetrating and requires the thickest shielding. A thick lead (Pb) container significantly reduces gamma radiation. [1]
- Since lead also stops alpha and beta (both are stopped by less dense materials), a thick lead container alone provides adequate shielding for all three types. [1]
- Alternative: thin plastic/paper (alpha), then 3mm aluminium (beta), then thick lead (gamma) — layered approach. The minimum adequate shielding is determined by the gamma component — everything else is automatically handled. [1]
(c)
[3]
Compare iodine-131 (half-life 8 days) and caesium-137 (half-life 30 years) as long-term nuclear waste disposal concerns. Which poses a greater long-term challenge, and why?
✏ Your answer
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- Caesium-137 poses a greater long-term challenge. [1]
- Iodine-131 decays to a negligible level within weeks (after about 10 half-lives ≈ 80 days, activity is <0.1% of initial). It does not require long-term storage. [1]
- Caesium-137 with a 30-year half-life remains significantly radioactive for centuries — after 300 years (10 half-lives), activity is still ~0.1% of initial. It requires secure, monitored, geological-scale storage for hundreds of years, posing enormous engineering and political challenges. [1]
(d)
[3]
A technician working near a gamma source measures 800 μSv/hr at 1 m distance. Using the inverse square law (intensity ∝ 1/d²), calculate the dose rate at 4 m and explain why working at greater distance significantly reduces radiation exposure.
✏ Your answer
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- Intensity ∝ 1/d². At 4 m (4× the distance): intensity reduces by factor 4² = 16. [1]
- Dose rate at 4 m = 800/16 = 50 μSv/hr [1]
- Doubling the distance reduces intensity by a factor of 4 (not 2). Working at 4 m instead of 1 m reduces exposure by a factor of 16 — a very large reduction for a relatively small increase in working distance. This is why distance is one of the most effective and practical radiation protection measures. [1]