Secondary 1 Science Semestral Assessment 2 (End of Year) Paper 5

TuitionGoWhere Practice Paper - Science Secondary 1 (Version 5 of 5)

Answer Key and Marking Scheme

SA2 Physical Sciences Total Marks: 60

Section A: Multiple Choice [10 marks]

Section B: Structured Response [28 marks]

11. Work done with trolley [5 marks]

(a) Work done = force × distance (moved in direction of force) [1]

(b) Working:

• Work done = F × d = 40 N × 15 m [1]
• = 600 J [1]

(c) No work is done because:

• The trolley does not move (no displacement) [1]
• Work done requires both force AND distance moved in the direction of the force; holding stationary involves force but zero displacement [1]
• Note: The man feels tired because his muscles are contracting and using chemical energy, but this energy is converted to heat internally, not work on the trolley.

12. Electric kettle [5 marks]

(a) Power is the rate of doing work / rate of energy transfer; or Power = work done ÷ time [1]

(b) Working:

• Energy = Power × Time = 2000 W × 45 s [1]
• = 90 000 J (or 90 kJ) [1]
• Unit must be stated for full marks

(c) Suggested improvement:

• Reduce mass of water / Use water at higher initial temperature / Use a kettle with higher power rating [1]
• Explanation: Less energy required to raise temperature to boiling point (E = mcΔθ) OR greater power means faster energy transfer for same energy requirement [1]

13. Heat conduction experiment [7 marks]

(a) The material/type of material (of the rod) [1]

(b) Any two from: [2]

• Length of rods (all 15 cm)
• Diameter/thickness of rods (all 1 cm)
• Initial temperature / room temperature
• Same heat source / same bulb temperature
• Same wax material and size
• Same pin material and mass
• Environmental conditions (e.g., no drafts)

(c) The best conductor transfers heat fastest along the rod [1]

• The wax on that rod will melt first, causing the pin to drop first [1]
• Time measurement of pin drop allows comparison

(d) The wax would melt faster [1]

• Greater cross-sectional area means more particles/vibrations can pass heat energy along the rod per unit time [1]
• Or: Thicker rod has more conducting material in parallel, reducing thermal resistance

14. Vacuum flask design [5 marks]

(a) Conduction requires particles/medium to pass energy; convection requires fluid movement [1]

• A vacuum contains no particles/medium, so neither conduction nor convection can occur [1]

(b) Shiny surfaces are poor emitters of thermal radiation [1]

• This reduces heat loss by radiation from the hot liquid to the surroundings [1]
• Also: shiny surfaces reflect radiant heat back towards the liquid

(c) Any one from: [1]

• Heat still escapes through the stopper (conduction through plastic, air gaps)
• Some radiation may escape if silvering is imperfect
• Heat gains from surroundings through the opening/stopper
• Thermal energy transfers through any solid supports connecting inner to outer wall

15. Wind turbine efficiency [5 marks]

(a) Working:

• Efficiency = (useful energy output ÷ total energy input) × 100% [1]
• = (14 400 J ÷ 48 000 J) × 100% = 30% [1]

(b) Efficiency < 100% due to energy losses: [1]

• Wasted energy forms: thermal energy (heat due to friction in bearings/generator), sound energy (noise from blades), kinetic energy of moving air that escapes, strain/elastic energy in blade deformation [2 marks for any two valid forms]

16. Heating curve analysis [6 marks]

(a) 0°C [1]

(b) In Region B, melting occurs: [3]

• Particles gain sufficient kinetic energy to overcome fixed lattice forces/structure [1]
• Particles break free from fixed positions; arrangement changes from ordered/regular (solid) to disordered/able to move past one another (liquid) [1]
• Temperature stays constant because incoming thermal energy is used to overcome interparticle forces (latent heat of fusion), not to increase kinetic energy of particles [1]

(c) Temperature remains constant during boiling (Region D) because: [2]

• All incoming thermal energy is used as latent heat of vaporisation [1]
• Energy overcomes attractions between liquid particles to form gas; no energy increases particle kinetic energy, so temperature stays at boiling point (100°C) [1]

17. Spring extension investigation [8 marks]

(a) The limit of proportionality is the maximum force (or extension) beyond which the spring no longer extends in direct proportion to the applied force / the point where Hooke's Law ceases to apply [1]

(b) Graph plotting [3 marks]:

• Axes correct and labelled with units: x-axis Weight/N, y-axis Extension/cm [1]
• Correct scale chosen to use most of grid; all points plot within ±½ small square [1]
• All six points correctly plotted with smooth line of best fit (initially straight, then curving after 6 N) [1]

(c) From graph (reading at extension = 5.0 cm): [2]

• Method: Find 5.0 cm on y-axis, trace across to line then down to x-axis [1]
• Expected answer: approximately 6.7 N (acceptable range 6.5–7.0 N depending on line drawn) [1]

(d) The graph is a straight line through origin initially, then curves/bends away from the straight line at higher weights [2]

• Specifically: Points at 8 N and 10 N do not fall on the straight line through (0,0), (2, 1.5), (4, 3.0), (6, 4.5); the extension increases disproportionately more at higher forces [1]
• This non-linear region indicates the spring has exceeded its limit of proportionality [1]

Section C: Data Analysis and Extended Response [22 marks]

18. Light bulb comparison [10 marks]

(a) Working: [2]

• Energy = Power × Time = 60 W × 1 000 h = 60 000 Wh = 60 kWh [1 for correct method, 1 for correct answer with unit]
• Or: 0.060 kW × 1 000 h = 60 kWh

(b) LED calculation over 25 000 hours: [4]

Bulb costs:

• Lifespan = 25 000 hours, so 1 LED bulb needed
• Bulb cost = $15.00 [1]

Electricity cost:

• Energy = 10 W × 25 000 h = 250 000 Wh = 250 kWh [1]
• Cost = 250 kWh × $0.20/kWh = **$50.00** [1]

Total cost = $15.00 +$50.00 = $65.00 [1]

(c) CFL versus LED over 25 000 hours: [3]

CFL costs:

• Bulbs needed: 25 000 ÷ 8 000 = 3.125 → round up to 4 bulbs (since can't buy partial; or 3 bulbs cover 24 000 h, need partial 4th)

• Or accept: 3 bulbs if assuming 24 000 h acceptable; or exact calculation with 3.125

• Using exact: Bulb cost = 3.125 × $6.00 = **$18.75** (or 4 × $6 =$24 if rounding up)

• Energy = 14 W × 25 000 h = 350 000 Wh = 350 kWh

• Electricity cost = 350 × $0.20 = **$70.00**

• CFL total = $18.75 +$70.00 = **$88.75** (or$94 with 4 bulbs)

Comparison:

• LED ($65.00) is **cheaper** than CFL ($88.75 or $94) [1 for correct comparison]
• Even though LED bulb costs more initially, its much lower power and longer lifespan save electricity and replacement costs [1 for clear reasoning]
• Conclusion: LED is more economical overall [1]

(Accept method marks for correct approach even if arithmetic slips; award conclusion mark only if supported by working)

(d) Any valid reason: [1]

• Reduced carbon dioxide emissions / environmental impact
• Less waste (fewer bulbs discarded)
• Less heat produced (safety/comfort in homes)
• Longer lifespan means less maintenance/access difficulty

19. Roller coaster energy [11 marks]

(a) Working: [2]

• GPE = m × g × h = 800 kg × 10 N/kg × 35 m [1]
• = 280 000 J (or 280 kJ) [1]

(b) Working: [3]

• By conservation: GPE at top = KE at bottom (assuming no friction, no air resistance)
• 280 000 J = ½mv² = ½ × 800 kg × v² [1]
• 280 000 = 400 × v² [1]
• v² = 700; v = √700 = 26.5 m/s (accept 26.4–26.5 m/s) [1]

Alternative working shown clearly also accepted

(c) Explanation: [2]

• Friction does work against motion, converting some mechanical energy to thermal energy / heat [1]
• This means not all GPE from Hill 1 converts to GPE at Hill 3; some is "lost" (dissipated), so the coaster may not have enough energy to reach 15 m [1]
• At Hill 3 height of 15 m, required GPE = 800 × 10 × 15 = 120 000 J. If energy losses exceed 160 000 J, the coaster cannot reach Hill 3.

(d) Two modifications with explanations: [4]

Modification 1: Increase the height of Hill 1 (or use a motor/powered launch)

• Explanation: More initial GPE provides more total mechanical energy to overcome friction losses and still reach Hill 3 [2]

Modification 2: Reduce friction (smoother wheels, smoother track, lubrication, streamline shape)

• Explanation: Less energy dissipated as heat, so more mechanical energy conserved for reaching Hill 3 [2]

Or: Decrease height of Hill 3 (changes requirements but valid design change) Or: Add a powered booster section between hills

20. Solar water heater [6 marks]

(a) Black surfaces are good absorbers of infrared/thermal radiation [1]

• Dark/matte surfaces absorb more radiant heat energy than shiny or light-coloured surfaces, heating water more efficiently [1]

(b) Convection circulation: [3]

• Water heated in panels becomes warmer and less dense than cooler water [1]
• This less dense, heated water rises to the storage tank due to buoyancy [1]
• Cooler, denser water from the tank sinks to replace the heated water, creating a continuous convection current that circulates water without a pump [1]

(c) Variation explanation: [3]

Less than 60% savings when:

• Cloudy/rainy days with less solar radiation intensity [1]
• More hot water used, requiring backup electric heating

More than 60% savings when:

• Very sunny days with high solar radiation intensity exceeding average [1]
• Less hot water demand, so stored solar-heated water sufficient without any electric backup [1]

Need both directions for full marks; accept other valid weather/usage factors

Summary of Mark Allocation

Marking notes: Award method marks (M) for correct physics even if arithmetic slips. Award independent of follow-through where appropriate. Units penalties: deduct ½ mark per question (max once per question) for missing or incorrect units where required.