Secondary 2 Science Semestral Assessment 2 (End of Year) Paper 1

TuitionGoWhere Practice Paper - Science Secondary 2 (SA2 Version 1) - Answer Key

Total Marks: 60

Section A: Multiple Choice Questions [10 marks]

1. Answer: C [1]

Explanation: As the ball rises, its speed decreases so kinetic energy decreases. Its height increases so gravitational potential energy increases. By conservation of energy (ignoring air resistance), kinetic energy is converted directly to gravitational potential energy. No heat or sound is produced in the ideal case.

Common mistake: Choosing A (includes heat) - this would only be true if air resistance is considered, but the question implies ideal conditions.

2. Answer: C [1]

Working: Work done against gravity = Gain in gravitational potential energy = $mgh$ $= 2 \times 10 \times 5 = 100 \text{ J}$

3. Answer: A [1]

Working: Kinetic energy gained = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 20^2 = 200,000 \text{ J}$ Average power = $\frac{\text{Work done}}{\text{Time}} = \frac{200,000}{10} = 20,000 \text{ W} = 20 \text{ kW}$

4. Answer: D [1]

Explanation: Thermal energy (heat) is the internal energy of particles due to their random motion - it is not a form of potential energy. Gravitational, elastic, and chemical potential energy are all forms of stored energy due to position or configuration.

5. Answer: A [1]

Explanation: At the highest points (A and C), speed is zero so kinetic energy is zero and gravitational potential energy is maximum. At the lowest point (B), height is minimum so potential energy is minimum and kinetic energy is maximum. By conservation of energy, the loss in potential energy from A to B equals the gain in kinetic energy.

6. Answer: C [1]

Working: Energy = Power × Time $= 2000 \text{ W} \times (3 \times 60) \text{ s} = 2000 \times 180 = 360,000 \text{ J}$

Common mistake: Forgetting to convert minutes to seconds (would give 6000 J, option A).

7. Answer: C [1]

Working: Work done = Force × Distance (in direction of force) $= 50 \times 4 = 200 \text{ J}$

8. Answer: B [1]

Explanation: In a hydroelectric power station, water stored at height has gravitational potential energy. As it falls, this converts to kinetic energy. The moving water turns turbines which drive generators to produce electrical energy. Sequence: Gravitational potential → Kinetic → Electrical.

9. Answer: C [1]

Working: Work done = Gain in GPE = $mgh = 50 \times 10 \times 3 = 1500 \text{ J}$ Power = $\frac{\text{Work}}{\text{Time}} = \frac{1500}{4} = 375 \text{ W}$

10. Answer: A [1]

Explanation: When a spring is compressed, it stores elastic potential energy. When released, this energy is converted to kinetic energy of the toy car (plus some heat/sound in reality).

Section B: Structured Questions [30 marks]

11. Roller Coaster Energy Conservation

(a) Principle of Conservation of Energy [2]

• Energy cannot be created or destroyed. [1]
• Energy can be converted from one form to another, and the total amount of energy in a closed system remains constant. [1]

Marking notes: Both points required for full marks. "Energy is conserved" alone is insufficient.

(b) GPE at point A [2] GPE = $mgh = 500 \times 10 \times 40 = 200,000 \text{ J}$ (or 200 kJ) [1 for formula/substitution, 1 for answer with unit]

(c) Speed at point B [2] At B, all GPE is converted to KE (frictionless track). $\frac{1}{2}mv^2 = mgh$ $v^2 = 2gh = 2 \times 10 \times 40 = 800$ $v = \sqrt{800} = 28.3 \text{ m/s}$ [1 for correct method/equation, 1 for answer with unit]

(d) Maximum height with 15% energy loss [2] Energy remaining = 85% of initial = $0.85 \times 200,000 = 170,000 \text{ J}$ At max height, all remaining energy is GPE: $mgh_{\text{max}} = 170,000$ $h_{\text{max}} = \frac{170,000}{500 \times 10} = 34 \text{ m}$ [1 for calculating remaining energy, 1 for height calculation with unit]

12. Crane Lifting Load

(a) Work done [2] Work done = Gain in GPE = $mgh = 800 \times 10 \times 15 = 120,000 \text{ J}$ (or 120 kJ) [1 for formula/substitution, 1 for answer with unit]

(b) Useful power output [2] Power = $\frac{\text{Work}}{\text{Time}} = \frac{120,000}{20} = 6,000 \text{ W} = 6 \text{ kW}$ [1 for formula/substitution, 1 for answer with unit]

(c) Efficiency [2] Efficiency = $\frac{\text{Useful power output}}{\text{Power input}} \times 100\% = \frac{6}{10} \times 100\% = 60\%$ [1 for formula/substitution, 1 for answer with %]

(d) Where remaining energy goes [1] The remaining 40% of energy is converted to heat and sound due to friction in the motor, gears, and cables, and air resistance.

13. Spring and Block on Rough Surface

(a) Elastic potential energy stored [2] $EPE = \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 1 \text{ J}$ [1 for formula/substitution, 1 for answer with unit]

(b) Work done against friction over 0.5 m [2] Friction force = $\mu_k \times \text{Normal reaction} = \mu_k mg = 0.2 \times 0.5 \times 10 = 1 \text{ N}$ Work done against friction = Force × Distance = $1 \times 0.5 = 0.5 \text{ J}$ [1 for friction force calculation, 1 for work done with unit]

(c) Kinetic energy after 0.5 m [2] Initial EPE = 1 J Work done against friction = 0.5 J By work-energy theorem: KE = Initial EPE - Work against friction = $1 - 0.5 = 0.5 \text{ J}$ [1 for correct energy balance, 1 for answer with unit]

14. Hydroelectric Dam

(a) GPE lost per second [2] Mass per second = 200 kg GPE lost per second = $mgh = 200 \times 10 \times 50 = 100,000 \text{ J/s} = 100 \text{ kW}$ [1 for formula/substitution, 1 for answer with unit]

Note: This is the power available from the water.

(b) Efficiency [2] Efficiency = $\frac{\text{Electrical power output}}{\text{Power input from water}} \times 100\% = \frac{80 \times 10^6}{100 \times 10^3} \times 100\% = 80\%$ [1 for formula/substitution, 1 for answer with %]

(c) Forms of "lost" energy [1] Heat (thermal energy) and sound energy. [1 for both, ½ each if only one given]

15. Ramp Experiment - Data Analysis

(a) Missing $v^2$ value [1] $v = 2.4 \text{ m/s}$, so $v^2 = 2.4^2 = 5.76 \text{ m}^2/\text{s}^2$ [1]

(b) Graph of $v^2$ against $h$ [2]

• Axes correctly labelled with units: $v^2 / \text{m}^2\text{s}^{-2}$ and $h / \text{m}$ [½]
• Appropriate scale covering at least 50% of grid [½]
• All 5 points plotted accurately (± half a small square) [½]
• Best-fit straight line through origin (or near origin) [½]

(c) Gradient of graph [2] Gradient = $\frac{\Delta v^2}{\Delta h}$ Using points (0,0) and (0.5, 9.61): Gradient = $\frac{9.61}{0.5} = 19.22 \text{ m/s}^2$ [1 for correct method (rise/run), 1 for value with unit] Accept range 19–20 m/s² depending on plotted points.

(d) Calculate $g$ from gradient [2] Theory: $v^2 = 2gh$, so gradient = $2g$ $g = \frac{\text{gradient}}{2} = \frac{19.22}{2} = 9.61 \text{ m/s}^2$ [1 for relationship, 1 for calculated value with unit]

(e) Reason for difference from 10 m/s² [1] Friction between trolley and ramp / air resistance / rotational kinetic energy of wheels not accounted for / measurement errors in height or speed. [Any one valid reason]

Section C: Longer Structured and Data-Based Questions [20 marks]

16. Solar-Powered Water Pump

(a) Total solar energy incident [2] Power incident = Intensity × Area = $800 \times 2.5 = 2000 \text{ W}$ Time = $4 \times 3600 = 14,400 \text{ s}$ Energy = Power × Time = $2000 \times 14,400 = 28,800,000 \text{ J} = 2.88 \times 10^7 \text{ J}$ [1 for power calculation, 1 for energy with unit]

(b) GPE gained by water [2] GPE = $mgh = 2000 \times 10 \times 12 = 240,000 \text{ J} = 2.4 \times 10^5 \text{ J}$ [1 for formula/substitution, 1 for answer with unit]

(c) Overall efficiency [2] Efficiency = $\frac{\text{Useful energy output}}{\text{Energy input}} \times 100\% = \frac{240,000}{28,800,000} \times 100\% = 0.833\%$ [1 for formula/substitution, 1 for answer with %]

(d) Two reasons for <100% efficiency [2]

1. Solar panel efficiency is limited (typically 15-20%) - much solar energy is reflected or converted to heat. [1]
2. Pump motor has friction losses, electrical resistance losses, and water turbulence/friction in pipes converts energy to heat. [1] Other valid reasons: energy to overcome atmospheric pressure, kinetic energy of water not recovered, etc.

17. Athlete Running Up Stairs

(a) Total vertical height [1] Height = $80 \times 0.18 = 14.4 \text{ m}$ [1]

(b) Work done against gravity [2] Work = $mgh = 60 \times 10 \times 14.4 = 8,640 \text{ J}$ [1 for formula/substitution, 1 for answer with unit]

(c) Average power [2] Power = $\frac{\text{Work}}{\text{Time}} = \frac{8,640}{12} = 720 \text{ W}$ [1 for formula/substitution, 1 for answer with unit]

(c) Chemical energy from food [2] Efficiency = 25% = 0.25 Useful energy output = Work done = 8,640 J Chemical energy input = $\frac{\text{Useful output}}{\text{Efficiency}} = \frac{8,640}{0.25} = 34,560 \text{ J}$ [1 for correct rearrangement, 1 for answer with unit]

(e) Why athlete feels hot [2]

• Only 25% of chemical energy is converted to useful work (GPE gain). [1]
• The remaining 75% is converted to thermal energy (heat) in the body, raising body temperature. [1]
• The body sweats to cool down, but the heat generated makes the athlete feel hot.

18. Toy Gun Spring Launch

(a) Elastic potential energy in spring [2] $EPE = \frac{1}{2}kx^2 = \frac{1}{2} \times 150 \times (0.08)^2 = 0.48 \text{ J}$ [1 for formula/substitution, 1 for answer with unit]

(b) Maximum height reached [2] At max height, all EPE → GPE (no losses assumed) $mgh = \frac{1}{2}kx^2$ $h = \frac{kx^2}{2mg} = \frac{150 \times (0.08)^2}{2 \times 0.01 \times 10} = \frac{0.96}{0.2} = 4.8 \text{ m}$ [1 for energy equivalence, 1 for answer with unit]

(c) Why actual height is lower [2]

• Air resistance acts on the pellet, converting some kinetic energy to heat and sound. [1]
• Friction between pellet and gun barrel / internal friction in spring converts some energy to heat. [1]
• Sound energy is produced during launch. Any two valid energy loss mechanisms.

19. Household Electrical Energy

(a) Total daily energy in kWh [3]

[1 for correct unit conversion (W to kW), 1 for correct individual calculations, 1 for correct total]

(b) Daily cost [1] Cost = 11.80 \times 0.28 = \3.304 \approx $3.30$ [1]

(c) Two changes to reduce bill by 20% [2] Target reduction: 20% of 11.80 = 2.36 kWh/day

Any two valid suggestions with justification, e.g.:

1. Reduce air conditioner usage from 6 h to 4 h (saves 2.4 kWh) - largest single consumer. [1]
2. Replace refrigerator with a more energy-efficient model (e.g., 100 W instead of 150 W saves 1.2 kWh/day). [1] Other valid: use fans instead of AC, wash full loads only, switch off lights when not in use, use laptop power-saving mode.

20. Bungee Jump Energy Analysis

(a) GPE lost after falling 20 m [2] GPE lost = $mgh = 70 \times 10 \times 20 = 14,000 \text{ J}$ [1 for formula/substitution, 1 for answer with unit]

(b) Elastic potential energy at lowest point [2] Cord stretch = 15 m $EPE = \frac{1}{2}kx^2 = \frac{1}{2} \times 40 \times (15)^2 = 4,500 \text{ J}$ [1 for formula/substitution, 1 for answer with unit]

(c) Speed when cord becomes taut [2] After falling 20 m, all GPE lost → KE (cord not yet stretched) $\frac{1}{2}mv^2 = mgh$ $v^2 = 2gh = 2 \times 10 \times 20 = 400$ $v = 20 \text{ m/s}$ [1 for energy equivalence, 1 for answer with unit]

(d) Why jumper doesn't hit water [2]

• Initially, GPE converts to KE during free fall (first 20 m). [½]
• As cord stretches, KE converts to elastic potential energy in the cord. [½]
• At lowest point, all KE is stored as EPE in the stretched cord (plus some GPE lost during stretch). [½]
• The cord then recoils, converting EPE back to KE and GPE, pulling jumper up before reaching water. [½] Key: Energy conversions prevent the jumper from reaching the water if the cord is properly designed (total energy at start = GPE at platform = 35,000 J; max EPE at 15 m stretch = 4,500 J; GPE lost during 35 m total fall = 24,500 J; total = 29,000 J < 35,000 J, so jumper stops above water).

End of Answer Key