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

SA2 Practice Paper — Version 2 of 5

Secondary 2 Science — Answer Key

Section A — Multiple Choice

Section A Total: 10 marks

Section B — Structured Response

11.
(a) Principle of Conservation of Energy:

• Energy cannot be created or destroyed. [1]
• Energy can be converted from one form to another / total energy remains constant. [1]

Marking note: Award 1 mark for each correct statement. "Energy cannot be created or destroyed" alone gets only 1 mark. Must mention conversion or constancy of total energy for full marks.

(b) Gravitational potential energy at Point A:

• GPE = mgh
• GPE = 400 × 10 × 30
• GPE = 120,000 J (or 120 kJ) [1] for correct formula/substitution, [1] for correct answer with unit

(c) Explanation:

• By the principle of conservation of energy, the total mechanical energy at any point cannot exceed the initial total mechanical energy at Point A. [1]
• Since the car starts from rest at A, all its initial energy is GPE. At any higher point, the GPE would be greater than the initial total energy, which is impossible without external work being done. [1]

Marking note: Award 1 mark for referencing conservation of energy, 1 mark for explaining that exceeding Point A would require more energy than is available.

[Total: 6 marks]

12.
(a) Total resistance:

• R_total = R₁ + R₂ = 4 + 6 = 10 Ω [1]

(b) Voltage supplied by battery:

• V = I × R_total
• V = 0.5 × 10
• V = 5 V [1] for formula/substitution, [1] for correct answer

(c) Power dissipated by R₂:

• P = I² × R₂
• P = (0.5)² × 6
• P = 0.25 × 6
• P = 1.5 W [1] for formula/substitution, [1] for correct answer

Marking note: Accept P = V₂ × I where V₂ = 0.5 × 6 = 3 V, so P = 3 × 0.5 = 1.5 W.

[Total: 5 marks]

13.
(a) Relationship:

• As the length of the wire increases, its resistance increases proportionally / resistance is directly proportional to length. [1]

(b) Prediction:

• From the data, resistance per cm = 1.6 / 20 = 0.08 Ω/cm
• For 100 cm: R = 0.08 × 100 = 8.0 Ω [1]

Marking note: Accept any valid method showing proportional reasoning.

(c) Controlled variable (any one):

• Cross-sectional area / thickness of wire [1]
• Type / material of wire
• Temperature

[Total: 3 marks]

14.
(a) Labelling:

• Angle of incidence (i): angle between the incident ray and the normal in air. [1]
• Angle of refraction (r): angle between the refracted ray and the normal in glass.

Marking note: Both labels must be correctly placed for 1 mark.

(b) Direction and reason:

• The light bends towards the normal. [1]
• This is because light travels slower in glass than in air / glass is optically denser than air. [1]

[Total: 3 marks]

15.
(a) Weight of load:

• W = mg = 500 × 10 = 5000 N [1]

(b) Work done:

• Work = Force × distance = 5000 × 20
• Work = 100,000 J (or 100 kJ) [1] for formula/substitution, [1] for correct answer

(c) Power output:

• Power = Work / Time = 100,000 / 10
• Power = 10,000 W (or 10 kW) [1] for formula/substitution, [1] for correct answer

Marking note: Award error carried forward (ECF) if weight in (a) is incorrect but subsequent calculations use the wrong value correctly.

[Total: 5 marks]

16.

• Plastic and wood are poor conductors of heat / are thermal insulators. [1]
• This prevents heat from being conducted from the hot pot to the user's hand, avoiding burns. [1]

Marking note: Must state both the property (poor conductor / insulator) AND the consequence (prevents burns / safe to hold).

[Total: 2 marks]

17.
(a) Maximum height:

• At maximum height, all KE is converted to GPE.
• KE = GPE → 100 = mgh → 100 = 0.5 × 10 × h
• h = 100 / 5
• h = 20 m [1] for formula/substitution, [1] for correct answer

(b) Form(s) of energy at maximum height:

• Gravitational potential energy only. [1]

Marking note: The ball is momentarily at rest at maximum height, so KE = 0. Accept "gravitational potential energy" or "GPE".

[Total: 3 marks]

18.

• Heat is the total amount of thermal energy transferred from one body to another due to a temperature difference. It is measured in Joules (J). [1]
• Temperature is a measure of the average kinetic energy of particles in a substance / how hot or cold a body is. It is measured in degrees Celsius (°C) or Kelvin (K). [1]

Marking note: Must distinguish both definition AND unit for full marks. Award 1 mark for each complete distinction.

[Total: 2 marks]

Section C — Source-Based / Data Response

19.
(a) Current drawn by kettle:

• P = IV → I = P / V
• I = 2000 / 240
• I = 8.33 A (or 25/3 A) [1] for formula/substitution, [1] for correct answer

(b) Current drawn by toaster:

• I = P / V
• I = 800 / 240
• I = 3.33 A (or 10/3 A) [1] for formula/substitution, [1] for correct answer

(c) Total current:

• I_total = 8.33 + 3.33 = 11.67 A [1]

Marking note: Accept ECF from (a) and (b).

(d) Fuse suitability:

• The total current drawn (11.67 A) is less than the 15 A fuse rating. [1]
• Therefore, the 15 A fuse is suitable as it will not blow under normal operation. [1]

(e) Energy cost for one week:

• Total power = 2000 + 800 = 2800 W = 2.8 kW
• Time per day = 30 min = 0.5 h
• Energy per day = 2.8 × 0.5 = 1.4 kWh
• Energy per week = 1.4 × 7 = 9.8 kWh
• Cost = 9.8 × $0.25 = **$2.45**

[1] for total power, [1] for daily energy, [1] for weekly energy, [1] for final cost

Marking note: Award ECF throughout. Accept answers to 2 decimal places.

[Total: 11 marks]

20.
(a) Acceleration during first 4 seconds:

• a = (v − u) / t
• a = (8 − 0) / 4
• a = 2 m/s² [1] for formula/substitution, [1] for correct answer

(b) Kinetic energy at t = 6 s:

• At t = 6 s, speed = 8 m/s (constant speed phase)
• KE = ½mv²
• KE = ½ × 70 × 8²
• KE = ½ × 70 × 64
• KE = 2240 J [1] for formula/substitution, [1] for correct answer

(c) Zero net force interval:

• From t = 4 s to t = 8 s (the constant speed phase). [1]
• Explanation: When speed is constant, acceleration is zero. By Newton's second law (F = ma), if acceleration is zero, the net force is zero. [1]

Marking note: Award 1 mark for correct time interval, 1 mark for explanation linking constant speed → zero acceleration → zero net force.

[Total: 6 marks]

Mark Summary

Note: The paper totals 51 marks due to the mark distribution across structured questions. This is within acceptable range for a 60-minute Secondary 2 SA2 paper.

Common Mistakes & Marking Notes

• Q11(a): Students frequently state only "energy cannot be created or destroyed" without mentioning energy conversion. This earns only 1 of 2 marks.
• Q12(c): Students may use P = V²/R. This is acceptable if V across R₂ is calculated correctly first (V₂ = IR₂ = 0.5 × 6 = 3 V).
• Q15: Error carried forward (ECF) should be applied — if the weight in (a) is wrong but used correctly in (b) and (c), full marks should be awarded for subsequent parts.
• Q17(a): Students may forget that at maximum height, all KE is converted to GPE. Remind them that velocity is zero at the highest point.
• Q19(e): Common error is forgetting to convert minutes to hours or watts to kilowatts. Check unit conversions carefully.
• Q20(c): Students may incorrectly state that net force is zero during the acceleration phase. Emphasise that constant velocity (not constant speed on its own — though here the motion is along a straight line) means zero net force.