Secondary 2 Science Semestral Assessment 2 (End of Year) Paper 3

SA2 Practice Paper — Science Secondary 2 (Version 3)

Answer Key & Marking Scheme

Section A: Multiple Choice Questions (10 marks)

Section A Total: 10 marks

Section B: Structured Response (25 marks)

11. State the Principle of Conservation of Energy. [2]

Answer:

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

Marking notes: Award 1 mark for each correct statement. Accept equivalent phrasing such as "total energy before = total energy after." Do not accept vague answers like "energy stays the same" without mentioning conversion.

12. A 0.5 kg stone is dropped from a cliff 20 m high. (Take g = 10 N/kg)

(a) Calculate the gravitational potential energy of the stone at the top of the cliff. [2]

Working: GPE = mgh GPE = 0.5 × 10 × 20 GPE = 100 J [1] for correct working, [1] for correct answer with unit

(b) State the kinetic energy of the stone just before it hits the ground. Assume no air resistance. [1]

Answer: 100 J [1]

Marking notes: By conservation of energy, all GPE is converted to KE. Award the mark for 100 J.

(c) Calculate the speed of the stone just before it hits the ground. [2]

Working: KE = ½mv² 100 = ½ × 0.5 × v² 100 = 0.25v² v² = 400 v = 20 m/s [1] for correct working, [1] for correct answer with unit

Common mistake: Students may forget to take the square root. Award 1 mark for correct substitution even if final answer is wrong.

13. Lever problem.

(a) Calculate the minimum effort required to lift the load. [2]

Working: Using the principle of moments (for equilibrium): Effort × distance from fulcrum = Load × distance from fulcrum Effort × 1.5 = 600 × 0.5 Effort × 1.5 = 300 Effort = 300 / 1.5 Effort = 200 N [1] for correct working, [1] for correct answer with unit

(b) State one way to make it easier to lift the load using this lever. [1]

Answer: Increase the distance from the fulcrum to the effort / Move the effort further from the fulcrum / Apply the effort further away from the fulcrum. [1]

Accept any valid method: e.g., "apply the effort at a greater distance from the fulcrum."

14. A student runs up a flight of stairs. Mass = 55 kg, height = 6.0 m, time = 8.0 s. (Take g = 10 N/kg)

(a) Calculate the gain in gravitational potential energy. [2]

Working: GPE = mgh GPE = 55 × 10 × 6.0 GPE = 3300 J [1] for correct working, [1] for correct answer with unit

(b) Calculate the power developed by the student. [2]

Working: Power = Work done / Time Power = 3300 / 8.0 Power = 412.5 W (or 413 W to 3 s.f.) [1] for correct working, [1] for correct answer with unit

Common mistake: Students may use total distance along the stairs instead of vertical height. Award 0 for that approach.

15. Refraction through a glass block.

(a) Calculate the angle of refraction. Refractive index n = 1.5, angle of incidence = 40°. [3]

Working: Using Snell's law: n₁ sin θ₁ = n₂ sin θ₂ 1.0 × sin 40° = 1.5 × sin θ₂ sin 40° = 0.6428 0.6428 = 1.5 × sin θ₂ sin θ₂ = 0.6428 / 1.5 sin θ₂ = 0.4285 θ₂ = sin⁻¹(0.4285) θ₂ = 25.4° (or 25° to 2 s.f.)

[1] for correct substitution into Snell's law [1] for correct rearrangement [1] for correct final answer (accept 25°–25.4°)

Common mistake: Students may divide 40 by 1.5 directly. This is incorrect — Snell's law uses sine of angles. Award 0 for this approach.

(b) Sketch the refracted ray inside the glass block. [1]

Answer: The refracted ray should bend towards the normal (i.e., angle of refraction < angle of incidence). Award [1] for a ray that bends towards the normal inside the glass.

16. Electric kettle rated at 240 V, 2000 W.

(a) Calculate the current flowing through the kettle. [2]

Working: P = IV 2000 = I × 240 I = 2000 / 240 I = 8.33 A (or 8.3 A to 2 s.f.) [1] for correct working, [1] for correct answer with unit

(b) Calculate the electrical energy consumed in kilowatt-hours. Time = 5 minutes. [2]

Working: Power = 2000 W = 2.0 kW Time = 5 min = 5/60 h = 1/12 h Energy = Power × Time Energy = 2.0 × (5/60) Energy = 2.0 × 0.0833 Energy = 0.167 kWh (or 0.1667 kWh) [1] for correct working, [1] for correct answer with unit

Common mistake: Students may forget to convert minutes to hours or watts to kilowatts. Award 1 mark for correct method with one conversion error.

(c) Calculate the cost of using the kettle. Cost = $0.25 per kWh. [1]

Working: Cost = 0.167 × 0.25 Cost = **$0.0417** (or 4.2 cents /$0.042) [1]

Section C: Data Interpretation and Application (15 marks)

17. Crater depth experiment.

(a) Identify the independent and dependent variables. [2]

Answer:

• Independent: Height of drop [1]
• Dependent: Depth of crater [1]

(b) Describe the relationship between height of drop and depth of crater. [1]

Answer: As the height of drop increases, the depth of the crater increases. [1] (Accept: "The greater the height, the deeper the crater" or "positive/direct relationship.")

(c) Explain, in terms of energy, why a ball dropped from a greater height makes a deeper crater. [2]

Answer:

• The ball dropped from a greater height has more gravitational potential energy. [1]
• This energy is converted to kinetic energy as it falls, so it hits the ground with greater kinetic energy / speed, doing more work on the sand and creating a deeper crater. [1]

Marking notes: Award 1 mark for mentioning GPE and 1 mark for linking to KE and crater formation. Do not accept answers that only say "it falls faster" without energy explanation.

(d) Predict the depth of the crater if the ball is dropped from 120 cm. [1]

Answer: Approximately 6.9–7.0 cm [1]

Working (for reference): The relationship is approximately linear. From the data:

• Every 20 cm increase in height → ~1.1–1.2 cm increase in crater depth
• From 100 cm to 120 cm: increase of 20 cm → add ~1.1–1.2 cm
• 5.8 + 1.1 = 6.9 cm (or 5.8 + 1.2 = 7.0 cm)

Accept any answer in the range 6.8–7.2 cm based on reasonable extrapolation.

18. Series circuit with R₁ = 4 Ω, R₂ = 8 Ω, V = 12 V.

(a) Calculate the total resistance. [1]

Answer: R_total = R₁ + R₂ = 4 + 8 = 12 Ω [1]

(b) Calculate the current flowing through the circuit. [2]

Working: V = IR 12 = I × 12 I = 12 / 12 I = 1.0 A [1] for correct working, [1] for correct answer with unit

(c) Calculate the potential difference across R₂. [2]

Working: V₂ = IR₂ V₂ = 1.0 × 8 V₂ = 8.0 V [1] for correct working, [1] for correct answer with unit

(d) If R₂ is replaced with a 2 Ω resistor, state and explain what happens to the brightness of the bulb (R₁). [2]

Answer:

• The bulb becomes brighter. [1]
• Explanation: The total resistance decreases (from 12 Ω to 6 Ω), so the current in the circuit increases. Since the current through R₁ (the bulb) increases, more electrical energy is converted to light and heat per second, so the bulb is brighter. [1]

Marking notes: Award 1 mark for "brighter" and 1 mark for a correct explanation involving increased current. Accept equivalent reasoning.

19. Pendulum swinging from A (highest) to B (lowest).

(a) At which point does the pendulum bob have maximum kinetic energy? [1]

Answer: Point B (the lowest point) [1]

(b) At which point does the pendulum bob have maximum gravitational potential energy? [1]

Answer: Point A (the highest point) [1]

(c) Explain why the pendulum bob reaches the same height on the other side. [2]

Answer:

• At the highest point, all energy is gravitational potential energy. [1]
• As it swings down, GPE is converted to KE. As it swings up the other side, KE is converted back to GPE. Since air resistance is negligible (no energy is lost), the total energy is conserved, so it reaches the same height. [1]

Marking notes: Award 1 mark for identifying energy conversion and 1 mark for linking to conservation of energy / no energy loss. Accept equivalent phrasing.

20. Converging lens. Candle at 30 cm, screen at 60 cm.

(a) Calculate the focal length of the lens. [3]

Working: Object distance u = 30 cm Image distance v = 60 cm Using the lens formula: 1/f = 1/u + 1/v 1/f = 1/30 + 1/60 1/f = 2/60 + 1/60 1/f = 3/60 1/f = 1/20 f = 20 cm

[1] for correct identification of u and v [1] for correct substitution into lens formula [1] for correct final answer with unit

Common mistake: Students may simply average the two distances (30+60)/2 = 45 cm. This is incorrect. Award 0 for this approach.

(b) State two characteristics of the image formed. [2]

Answer: The image is real [1] and inverted [1]. (Also accept: magnified/enlarged.)

Marking notes: Award 1 mark each for any two of: real, inverted, magnified. Do not accept "virtual" or "upright."

(c) The candle is moved to 15 cm from the lens. State what happens to the image and explain. [2]

Answer:

• The image can no longer be formed on the screen / the image becomes virtual. [1]
• Explanation: The object is now placed between the focal point (f = 20 cm) and the lens (u = 15 cm < f). When the object is within the focal length of a converging lens, the image formed is virtual, upright, and magnified. A virtual image cannot be projected onto a screen. [1]

Marking notes: Award 1 mark for stating the image becomes virtual / cannot be captured on screen, and 1 mark for correct explanation involving object being within focal length.

End of Answer Key

Total Marks: 50