O Level Combined Science Practice Paper 2

TuitionGoWhere Practice Paper – Combined Science O-Level

Physical Sciences – ANSWER KEY & MARKING SCHEME

TuitionGoWhere Secondary School (AI) PRACTICE PAPER – Version 2

Total Marks: 65

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Section A: Structured Questions (20 marks)

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1. Pendulum – Energy and Forces

(a) State the principle of conservation of energy. [1]

Answer: Energy cannot be created or destroyed; it can only be converted/transformed from one form to another. The total energy in a closed/isolated system remains constant.

Marking:

• 1 mark for stating energy cannot be created or destroyed AND can be converted/transformed.
• Accept: "Total energy is conserved" or "Energy is neither created nor destroyed, only changed from one form to another."

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(b)(i) Calculate the gravitational potential energy gained. [2]

Answer:

• GPE = mgh = 0.15 × 10 × 0.12
• GPE = 0.18 J

Marking:

• 1 mark for correct formula and substitution (mgh).
• 1 mark for correct answer with unit (0.18 J).
• Accept 0.18 J or 1.8 × 10⁻¹ J.

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(b)(ii) Determine the maximum speed at the lowest position. [2]

Answer:

• By conservation of energy: GPE lost = KE gained
• 0.18 = ½ × 0.15 × v²
• v² = (0.18 × 2) / 0.15 = 2.4
• v = √2.4 = 1.55 m/s (or 1.5 m/s to 2 s.f.)

Marking:

• 1 mark for equating GPE to KE (½mv²).
• 1 mark for correct answer with unit (1.55 m/s or 1.5 m/s).
• Accept answers in range 1.5–1.6 m/s.

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(c) Free-body diagram at lowest position. [2]

Answer:

• Two forces should be shown:
  • Weight (W or mg) acting vertically downwards from the centre of the sphere.
  • Tension (T) acting vertically upwards along the thread from the point of attachment.
• Tension arrow should be longer than weight arrow (since there is a net upward/centripetal force at the lowest point).

Marking:

• 1 mark for correct forces (weight downwards, tension upwards).
• 1 mark for correct relative lengths (tension > weight) and clear labelling.
• Deduct 1 mark if forces are not drawn from the sphere or if arrows are missing.

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2. Heat Conduction

(a) Explain how heat is conducted through the metal rod. [2]

Answer:

• Heat energy causes the particles (atoms/ions) at the heated end to vibrate more vigorously.
• These vibrations are passed to neighbouring particles through collisions, transferring kinetic energy along the rod.
• In metals, free electrons also move and transfer kinetic energy rapidly through the material.

Marking:

• 1 mark for describing particle vibration and energy transfer.
• 1 mark for mentioning free electrons (or delocalised electrons) in metals.
• Accept: "Vibrating particles pass energy to adjacent particles" (1 mark) + "Free electrons carry energy through the metal" (1 mark).

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(b) Why are metals better conductors than non-metals? [1]

Answer: Metals have free/delocalised electrons that can move through the structure and transfer kinetic/heat energy quickly. Non-metals do not have free electrons; heat transfer occurs only through particle vibration, which is slower.

Marking:

• 1 mark for mentioning free electrons in metals OR stating that non-metals lack free electrons.

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(c) Calculate the temperature gradient. [2]

Answer:

• Temperature difference = 180 – 30 = 150 °C
• Length = 0.50 m
• Temperature gradient = 150 / 0.50 = 300 °C/m

Marking:

• 1 mark for correct calculation of temperature difference and division by length.
• 1 mark for correct answer with unit (300 °C/m).
• Accept: °C m⁻¹ or °C per metre.

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3. Sound and Echo

(a) Explain what is meant by an echo. [1]

Answer: An echo is a reflected sound wave that reaches the listener after a time delay, distinguishable from the original sound.

Marking:

• 1 mark for mentioning reflection of sound AND time delay/distinct repetition.
• Accept: "Sound that is reflected from a surface and heard again."

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(b) Calculate the time taken for sound to travel to the building and back. [2]

Answer:

• Total distance = 2 × 85 = 170 m
• Time = distance / speed = 170 / 340 = 0.50 s

Marking:

• 1 mark for using total distance (170 m, i.e., there and back).
• 1 mark for correct answer with unit (0.50 s or 0.5 s).

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(c) State one condition for a distinct echo to be heard. [1]

Answer:

• The reflecting surface must be at least 17 m away from the sound source/listener.
• OR: The time interval between the original sound and the echo must be at least 0.1 s.
• OR: There should be no other reflecting surfaces nearby to cause reverberation.

Marking:

• 1 mark for any valid condition.

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4. Refraction of Light

(a) Draw the path of the ray through the glass block. [2]

Answer:

• Ray bends towards the normal as it enters the glass (from air to glass).
• Ray travels straight through the glass to the curved surface.
• Ray emerges from the curved surface without bending (since it strikes the curved surface along the normal/at 0° angle of incidence).
• Angle of incidence i labelled in air (between incident ray and normal).
• Angle of refraction r labelled inside glass (between refracted ray and normal).

Marking:

• 1 mark for correct bending towards normal on entry and straight path to curved surface.
• 1 mark for correct emergence (no bending at curved surface) and correct labelling of i and r.

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(b) Calculate the angle of refraction inside the glass. [2]

Answer:

• Snell's law: n₁ sin i = n₂ sin r
• 1 × sin 40° = 1.5 × sin r
• sin r = sin 40° / 1.5 = 0.6428 / 1.5 = 0.4285
• r = sin⁻¹(0.4285) = 25.4° (or 25° to 2 s.f.)

Marking:

• 1 mark for correct application of Snell's law.
• 1 mark for correct answer (25.4° or 25°).
• Accept answers in range 25°–26°.

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Section B: Data-Based Questions (25 marks)

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5. Power When Climbing Stairs

(a) Calculate total vertical height in metres. [1]

Answer:

• Height = 25 × 16 cm = 400 cm = 4.0 m

Marking:

• 1 mark for correct answer with unit (4.0 m).
• Accept 4 m.

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(b) Calculate work done against gravity. [2]

Answer:

• Work done = Weight × height = 520 × 4.0 = 2080 J

Marking:

• 1 mark for correct formula (W = F × d or mgh).
• 1 mark for correct answer with unit (2080 J or 2.08 × 10³ J).

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(c) Calculate average power developed. [2]

Answer:

• Power = Work done / Time = 2080 / 8.0 = 260 W

Marking:

• 1 mark for correct formula (P = W/t).
• 1 mark for correct answer with unit (260 W).

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(d) Determine which student develops greater power. [3]

Answer:

• Student 2: Work done = 650 × 4.0 = 2600 J
• Power = 2600 / 10.0 = 260 W
• Both students develop the same power (260 W).

Marking:

• 1 mark for calculating work done by second student (2600 J).
• 1 mark for calculating power of second student (260 W).
• 1 mark for correct conclusion (same power OR both 260 W).

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6. Cooling Experiment

(a) Plot graph of temperature against time. [4]

Answer:

• Correct axes: Temperature / °C on y-axis, Time / min on x-axis.
• Appropriate scales chosen (e.g., y-axis: 0–80 °C, x-axis: 0–20 min).
• All points plotted accurately for both beakers.
• Smooth curves drawn through points (not dot-to-dot).
• Curves clearly labelled (Beaker A and Beaker B).

Marking:

• 1 mark for correct axes with labels and units.
• 1 mark for appropriate scales (more than half the grid used).
• 1 mark for accurate plotting of all points (±½ small square).
• 1 mark for smooth curves and clear labelling.
• Deduct 1 mark if curves are drawn dot-to-dot.

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(b) Determine temperature of Beaker A after 7 minutes. [1]

Answer: Approximately 56–57 °C (read from graph).

Marking:

• 1 mark for correct reading from graph (±1 °C).
• Accept 56 °C or 57 °C.

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(c) Calculate rate of temperature decrease for Beaker B between 4 and 10 minutes. [2]

Answer:

• Temperature at 4 min = 70 °C
• Temperature at 10 min = 58 °C
• Temperature decrease = 70 – 58 = 12 °C
• Time interval = 10 – 4 = 6 min
• Rate = 12 / 6 = 2.0 °C/min

Marking:

• 1 mark for correct temperature difference and time interval.
• 1 mark for correct answer with unit (2.0 °C/min or 2 °C/min).

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(d) Explain why Beaker B cools more slowly. [2]

Answer:

• Cotton wool is a better insulator (poor conductor of heat) than aluminium foil.
• Cotton wool traps air, and air is a poor conductor of heat, reducing heat loss by conduction and convection.
• Aluminium foil is a metal and a good conductor, allowing more rapid heat transfer to the surroundings.

Marking:

• 1 mark for identifying cotton wool as a better insulator/poorer conductor.
• 1 mark for explaining that trapped air reduces heat loss OR that aluminium conducts heat away faster.
• Accept reference to reduction of conduction/convection/radiation as appropriate.

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(e) State one way to improve reliability. [1]

Answer:

• Repeat the experiment and calculate average temperatures.
• Use the same initial temperature more precisely.
• Ensure beakers are identical in size and shape.
• Use a data logger for more frequent/accurate readings.

Marking:

• 1 mark for any valid suggestion that improves reliability (consistency/repeatability).

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7. Spring Investigation

(a) Complete the table – calculate weights. [1]

Answer:

• Weight = mass (kg) × 10
• 0 g → 0 N (given)
• 100 g = 0.1 kg → 1.0 N (given)
• 200 g = 0.2 kg → 2.0 N (given)
• 300 g = 0.3 kg → 3.0 N (given)
• 400 g = 0.4 kg → 4.0 N (given)
• 500 g = 0.5 kg → 5.0 N (given)

Marking:

• 1 mark for all weights correctly stated (already provided in table; accept if student confirms values).

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(b) Plot graph of weight against extension. [3]

Answer:

• Weight / N on y-axis, Extension / cm on x-axis.
• Appropriate scales chosen.
• All six points plotted accurately.
• Best-fit straight line drawn through the origin and the first five points (0 to 500 g).
• The point for 600 g (17.0 cm, 6.0 N) should NOT be on the line.

Marking:

• 1 mark for correct axes with labels and units.
• 1 mark for accurate plotting of all points.
• 1 mark for best-fit straight line through the origin and first five points.

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(c) Determine the spring constant. [2]

Answer:

• Spring constant k = gradient of graph
• Using points from the line: e.g., (5.0 cm, 2.0 N) and (10.0 cm, 4.0 N)
• Gradient = (4.0 – 2.0) / (10.0 – 5.0) = 2.0 / 5.0 = 0.40 N/cm
• OR: k = 40 N/m (if converted to metres)

Marking:

• 1 mark for correct method (gradient calculation).
• 1 mark for correct answer with unit (0.40 N/cm or 40 N/m).
• Accept values in range 0.38–0.42 N/cm or 38–42 N/m.

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(d) Suggest why the 600 g value does not lie on the straight line. [1]

Answer:

• The spring has exceeded its elastic limit / limit of proportionality.
• The spring has been permanently deformed/stretched.
• Hooke's law no longer applies beyond this point.

Marking:

• 1 mark for mentioning elastic limit OR limit of proportionality OR permanent deformation.

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Section C: Free-Response Questions (20 marks)

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8. Electrical Circuits

(a) Draw the circuit diagram. [3]

Answer:

• Correct symbols: battery (two cells or battery symbol), switch, ammeter, two lamps.
• Lamps connected in parallel (each lamp on a separate branch).
• Ammeter connected in series with the battery (to measure total current).
• Switch connected in series with the battery.
• All connections shown with straight lines and right angles.

Marking:

• 1 mark for correct symbols (battery, switch, ammeter, lamps).
• 1 mark for correct parallel arrangement of lamps.
• 1 mark for ammeter and switch in correct positions (series with battery).
• Deduct 1 mark if circuit is drawn as series instead of parallel.

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(b)(i) Calculate total resistance. [2]

Answer:

• For two identical resistors in parallel: 1/R_total = 1/R + 1/R = 2/R
• R_total = R/2 = 12/2 = 6.0 Ω

Marking:

• 1 mark for correct formula for parallel resistors.
• 1 mark for correct answer with unit (6.0 Ω).

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(b)(ii) Calculate current through the ammeter. [2]

Answer:

• I = V / R_total = 6.0 / 6.0 = 1.0 A

Marking:

• 1 mark for correct formula (I = V/R).
• 1 mark for correct answer with unit (1.0 A).

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(b)(iii) State and explain what happens to brightness of remaining lamp. [2]

Answer:

• The brightness of the remaining lamp stays the same.
• In a parallel circuit, each lamp receives the full battery voltage (6.0 V).
• Removing one lamp does not change the voltage across the other lamp.
• Since voltage and resistance are unchanged, current through the lamp and therefore its power/brightness remain the same.

Marking:

• 1 mark for stating brightness stays the same.
• 1 mark for correct explanation (voltage across lamp unchanged in parallel circuit).

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9. Crane and Efficiency

(a)(i) Calculate weight of the concrete block. [1]

Answer:

• Weight = mg = 200 × 10 = 2000 N

Marking:

• 1 mark for correct answer with unit (2000 N or 2.0 × 10³ N).

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(a)(ii) Calculate work done by the crane. [2]

Answer:

• Work done = Force × distance = Weight × height
• Work done = 2000 × 15 = 30 000 J (or 30 kJ)

Marking:

• 1 mark for correct formula (W = F × d).
• 1 mark for correct answer with unit (30 000 J or 30 kJ).

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(a)(iii) Calculate useful power output. [2]

Answer:

• Power = Work done / Time = 30 000 / 12 = 2500 W (or 2.5 kW)

Marking:

• 1 mark for correct formula (P = W/t).
• 1 mark for correct answer with unit (2500 W or 2.5 kW).

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(b) Calculate efficiency of the crane. [2]

Answer:

• Efficiency = (Useful power output / Input power) × 100%
• Efficiency = (2500 / 4000) × 100% = 62.5%

Marking:

• 1 mark for correct formula and substitution.
• 1 mark for correct answer (62.5% or 63% or 62.5).

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(c) Suggest two reasons why efficiency is less than 100%. [2]

Answer (any two):

• Friction in the moving parts of the crane (pulleys, gears, cables).
• Energy lost as heat in the electric motor.
• Energy lost as sound.
• Work done against air resistance.
• Energy used to lift the cable/hook itself.

Marking:

• 1 mark for each valid reason (max 2 marks).
• Accept any reasonable suggestion related to energy losses.

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10. Reflection of Light

(a) State the laws of reflection. [2]

Answer:

• First law: The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.
• Second law: The angle of incidence is equal to the angle of reflection (i = r).

Marking:

• 1 mark for each law correctly stated.
• Accept: "Angle of incidence equals angle of reflection" (1 mark) + "Incident ray, reflected ray, and normal are in the same plane" (1 mark).

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(b) Draw labelled diagram of reflection. [3]

Answer:

• Plane mirror drawn as a straight line with hatching on the back.
• Incident ray drawn approaching the mirror at 30° to the normal.
• Reflected ray drawn leaving the mirror at 30° to the normal (on the opposite side).
• Normal drawn as a dashed/dotted line perpendicular to the mirror at the point of incidence.
• Labels: incident ray, reflected ray, normal, angle of incidence (i = 30°), angle of reflection (r = 30°).
• Arrows on rays showing direction.

Marking:

• 1 mark for correct mirror, normal, and ray directions.
• 1 mark for correct angles (i = r = 30°).
• 1 mark for clear labelling of all five elements.
• Deduct 1 mark if angles are not marked or are clearly unequal.

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(c) State the new angle of reflection when angle of incidence is 50°. [1]

Answer: 50° (since angle of reflection = angle of incidence).

Marking:

• 1 mark for 50°.

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END OF ANSWER KEY

Total: 65 marks