O Level Combined Science Practice Paper 5

TuitionGoWhere Practice Paper - Combined Science O-Level

Answer Key and Marking Scheme

Subject: Combined Science (Physics, Chemistry) Level: O-Level Paper: Practice Paper – Physical Sciences Version: 5 of 5 Total Marks: 65

Section A: Structured Questions (40 marks)

Question 1: Measurement and Density (4 marks)

(a) State the SI unit for mass. [1]

• Answer: kilogram / kg
• Marking: 1 mark for correct unit. Accept 'kg' only.

(b) Calculate the density of the stone in g/cm³. [2]

• Answer:
  • Volume of stone = 72 – 50 = 22 cm³ [1 mark for correct volume]
  • Density = mass / volume = 55 / 22 = 2.5 g/cm³ [1 mark for correct answer with unit]
• Marking: Award 1 mark for correct volume calculation. Award 1 mark for correct density with unit. Accept 2.5 g/cm³. Deduct 0.5 marks if unit missing or incorrect.

(c) Explain whether the stone will float or sink in water. [1]

• Answer: The stone will sink because its density (2.5 g/cm³) is greater than the density of water (1.0 g/cm³).
• Marking: 1 mark for correct prediction with valid reason linking density comparison. Accept 'sink' with reference to greater density.

Question 2: Kinematics (4 marks)

(a) Describe the motion of the car between t = 10 s and t = 25 s. [1]

• Answer: The car is travelling at constant velocity / constant speed of 20 m/s.
• Marking: 1 mark for 'constant velocity' or 'constant speed'. Accept 'uniform motion' or 'zero acceleration'.

(b) Calculate the acceleration of the car during the first 10 seconds. [2]

• Answer:
  • a = (v – u) / t = (20 – 0) / 10 = 2.0 m/s² [1 mark for correct substitution, 1 mark for correct answer with unit]
• Marking: Award 1 mark for correct formula/substitution. Award 1 mark for correct answer with unit. Accept 2 m/s².

(c) State what the area under a velocity-time graph represents. [1]

• Answer: Distance travelled / displacement
• Marking: 1 mark for 'distance' or 'displacement'.

Question 3: Forces and Moments (5 marks)

(a) Define the moment of a force. [1]

• Answer: The moment of a force is the product of the force and the perpendicular distance from the pivot to the line of action of the force.
• Marking: 1 mark for definition including 'force × perpendicular distance'. Accept 'turning effect of a force'.

(b) Calculate the moment of the box's weight about the support. [2]

• Answer:
  • Moment = force × perpendicular distance = 150 × 2.0 = 300 N m [1 mark for substitution, 1 mark for correct answer with unit]
• Marking: Award 1 mark for correct formula/substitution. Award 1 mark for correct answer with unit (N m). Accept Nm.

(c) State the principle used to determine the position of this box. [1]

• Answer: Principle of moments (sum of clockwise moments = sum of anticlockwise moments for a body in equilibrium)
• Marking: 1 mark for 'principle of moments' or statement of equilibrium condition.

(d) Calculate the distance from the support at which the 100 N box must be placed. [1]

• Answer:
  • Clockwise moment = Anticlockwise moment
  • 100 × d = 150 × 2.0
  • d = 300 / 100 = 3.0 m
• Marking: 1 mark for correct answer with unit. Accept 3 m.

Question 4: Pressure (4 marks)

(a) Calculate the maximum pressure the block can exert on a horizontal surface. [2]

• Answer:
  • Minimum area = 0.15 × 0.10 = 0.015 m² [1 mark for identifying minimum area]
  • Maximum pressure = force / area = 90 / 0.015 = 6000 Pa [1 mark for correct calculation with unit]
• Marking: Award 1 mark for using the smallest face area. Award 1 mark for correct pressure with unit. Accept 6000 N/m².

(b) Explain why the pressure exerted by the block depends on its orientation. [1]

• Answer: Pressure = force / area. The weight (force) is constant, but the contact area changes with orientation. A smaller area produces a larger pressure.
• Marking: 1 mark for explanation linking area to pressure (inverse relationship).

(c) Calculate the density of the liquid. [1]

• Answer:
  • P = ρgh → ρ = P / (gh) = 4200 / (10 × 0.50) = 840 kg/m³
• Marking: 1 mark for correct answer with unit. Accept 840 kg/m³.

Question 5: Energy and Power (5 marks)

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

• Answer: Energy cannot be created or destroyed; it can only be converted/transferred from one form to another. The total energy in a closed system remains constant.
• Marking: 1 mark for correct statement. Must include 'cannot be created or destroyed' and 'converted/transferred'.

(b) Calculate the work done by the crane in lifting the load. [2]

• Answer:
  • Weight = mg = 500 × 10 = 5000 N [1 mark for correct weight]
  • Work done = force × distance = 5000 × 30 = 150,000 J / 150 kJ [1 mark for correct work]
• Marking: Award 1 mark for correct weight calculation. Award 1 mark for correct work done with unit.

(c) Calculate the average power developed by the crane. [1]

• Answer:
  • Power = work / time = 150,000 / 25 = 6000 W / 6.0 kW
• Marking: 1 mark for correct answer with unit.

(d) Suggest one reason why the crane's motor provides more power than the value calculated in (c). [1]

• Answer: Energy is lost due to friction in the moving parts / Energy is converted to heat and sound / The motor must overcome friction and air resistance / Not all electrical energy is converted to useful work.
• Marking: 1 mark for any valid reason relating to energy losses or inefficiency.

Question 6: Thermal Energy Transfer (4 marks)

(a) Describe what the student observes as the rod heats up. [1]

• Answer: The drawing pins fall off one by one, starting from the end nearest the hot water.
• Marking: 1 mark for correct observation (pins fall sequentially from hot end).

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

• Answer:
  • Particles at the hot end gain kinetic energy and vibrate more vigorously. [1 mark]
  • These vibrations are passed to neighbouring particles through collisions. In metals, free electrons also move and transfer kinetic energy rapidly through the rod. [1 mark]
• Marking: Award 1 mark for particle vibration explanation. Award 1 mark for mentioning free electrons in metals.

(c) State and explain how the observation differs with a glass rod. [1]

• Answer: The pins fall off more slowly / take longer to fall. Glass is a poor conductor (insulator) because it lacks free electrons and particle vibrations are transferred more slowly.
• Marking: 1 mark for correct observation with valid explanation.

Question 7: Waves and Light (5 marks)

(a) State what happens to the speed of light as it enters the glass. [1]

• Answer: The speed of light decreases.
• Marking: 1 mark for 'decreases' or 'slows down'.

(b) Calculate the refractive index of the glass. [2]

• Answer:
  • n = sin i / sin r = sin 50° / sin 30° = 0.7660 / 0.5000 = 1.53 [1 mark for correct substitution, 1 mark for correct answer]
• Marking: Award 1 mark for correct formula/substitution. Award 1 mark for correct answer (1.53, accept 1.5).

(c) Draw a labelled diagram. [2]

• Answer: Diagram should show:
  • Ray entering glass and bending towards the normal [1 mark]
  • Ray leaving glass and bending away from the normal (parallel to incident ray)
  • Normal line drawn perpendicular to surface at both entry and exit points
  • Labels: incident ray, refracted ray, emergent ray, normal, angle of incidence (i), angle of refraction (r) [1 mark]
• Marking: Award 1 mark for correct ray paths (bending towards normal on entry, away on exit). Award 1 mark for correct labels including normal and angles.

Question 8: Electricity (5 marks)

(a) Calculate the total resistance of the parallel combination. [2]

• Answer:
  • 1/R_total = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2 [1 mark for correct formula/substitution]
  • R_total = 2 Ω [1 mark for correct answer with unit]
• Marking: Award 1 mark for correct use of parallel resistance formula. Award 1 mark for correct answer with unit.

(b) Calculate the total current drawn from the battery. [1]

• Answer:
  • I = V / R = 12 / 2 = 6.0 A
• Marking: 1 mark for correct answer with unit.

(c) State one advantage of connecting household appliances in parallel rather than in series. [1]

• Answer: Each appliance receives the full mains voltage / Appliances can be switched on/off independently / If one appliance fails, others continue to work.
• Marking: 1 mark for any valid advantage.

(d) State and explain how the total current changes when the 3 Ω resistor is replaced with a 12 Ω resistor. [1]

• Answer: The total current decreases. The total resistance increases (1/R = 1/12 + 1/6 = 1/4, so R = 4 Ω), so for the same voltage, current decreases (I = V/R).
• Marking: 1 mark for correct prediction with valid explanation.

Question 9: Electromagnetic Induction (4 marks)

(a) State what is observed on the ammeter as the magnet moves into the coil. [1]

• Answer: The ammeter needle deflects / shows a reading, indicating a current flows.
• Marking: 1 mark for 'deflection' or 'current flows'.

(b) Explain why this observation occurs. [2]

• Answer:
  • The moving magnet causes a changing magnetic field through the coil. [1 mark]
  • This changing magnetic field induces an electromotive force (e.m.f.) / voltage across the coil, which drives a current through the circuit (electromagnetic induction). [1 mark]
• Marking: Award 1 mark for 'changing magnetic field'. Award 1 mark for 'induced e.m.f./current'.

(c) State one way to increase the magnitude of the induced current. [1]

• Answer: Move the magnet faster / Use a stronger magnet / Use a coil with more turns.
• Marking: 1 mark for any valid method.

Section B: Data-Based and Application Questions (25 marks)

Question 10: Experimental Data Analysis – Pendulum (6 marks)

(a) Calculate the average time for one complete oscillation (the period). [2]

• Answer:
  • Average time for 20 oscillations = (40.2 + 39.8 + 40.0) / 3 = 40.0 s [1 mark]
  • Period = 40.0 / 20 = 2.00 s [1 mark]
• Marking: Award 1 mark for correct average. Award 1 mark for correct period with unit.

(b) State the position where the bob has maximum kinetic energy. Explain your answer. [2]

• Answer:
  • Maximum kinetic energy at the lowest point / equilibrium position. [1 mark]
  • At this position, the bob is moving fastest. Gravitational potential energy has been converted to kinetic energy. [1 mark]
• Marking: Award 1 mark for correct position. Award 1 mark for valid explanation linking to speed or energy conversion.

(c) Predict how the period changes with a bob of greater mass. Explain. [2]

• Answer:
  • The period remains the same / does not change. [1 mark]
  • The period of a simple pendulum depends only on the length of the thread and acceleration due to gravity, not on the mass of the bob. [1 mark]
• Marking: Award 1 mark for correct prediction. Award 1 mark for correct explanation (period independent of mass).

Question 11: Circuit Analysis (7 marks)

(a) Calculate the potential difference across the 10 Ω resistor. [1]

• Answer:
  • V = IR = 0.40 × 10 = 4.0 V
• Marking: 1 mark for correct answer with unit.

(b) Calculate the potential difference across resistor R. [2]

• Answer:
  • Total voltage = 6.0 V [1 mark for recognising series circuit voltage division]
  • V_R = 6.0 – 4.0 = 2.0 V [1 mark for correct answer]
• Marking: Award 1 mark for method (total voltage minus known voltage). Award 1 mark for correct answer with unit.

(c) Calculate the resistance of R. [2]

• Answer:
  • R = V / I = 2.0 / 0.40 = 5.0 Ω [1 mark for correct formula, 1 mark for correct answer with unit]
• Marking: Award 1 mark for correct formula/substitution. Award 1 mark for correct answer with unit.

(d) Explain whether the resistance of the filament lamp is greater than, less than, or equal to 10 Ω. [2]

• Answer:
  • The resistance of the filament lamp is equal to 10 Ω. [1 mark]
  • Since the current is still 0.40 A and the potential difference across it would still be 4.0 V (to maintain the same circuit conditions), R = V/I = 4.0/0.40 = 10 Ω. The resistance is the same under these conditions. [1 mark]
• Marking: Award 1 mark for correct answer. Award 1 mark for valid reasoning using V=IR.

Question 12: Energy Efficiency in Context (6 marks)

(a) Calculate the electrical energy supplied by the kettle in 150 seconds. [1]

• Answer:
  • E = P × t = 2200 × 150 = 330,000 J / 330 kJ
• Marking: 1 mark for correct answer with unit.

(b) Calculate the useful thermal energy gained by the water. [2]

• Answer:
  • E = mcΔθ = 1.0 × 4200 × (100 – 25) [1 mark for correct substitution]
  • E = 1.0 × 4200 × 75 = 315,000 J / 315 kJ [1 mark for correct answer with unit]
• Marking: Award 1 mark for correct formula and substitution. Award 1 mark for correct answer with unit.

(c) Calculate the efficiency of the kettle. [2]

• Answer:
  • Efficiency = (useful energy output / total energy input) × 100% [1 mark for correct formula]
  • Efficiency = (315,000 / 330,000) × 100% = 95.5% [1 mark for correct answer]
• Marking: Award 1 mark for correct formula. Award 1 mark for correct answer (accept 95% or 95.5%).

(d) Suggest one reason why the efficiency is less than 100%. [1]

• Answer: Some energy is lost as heat to the surroundings / Energy is used to heat the kettle itself / Some energy is lost as sound.
• Marking: 1 mark for any valid reason relating to energy losses.

Question 13: Forces and Motion in Context (6 marks)

(a) State the magnitude of the forward force exerted by the cyclist. Explain. [2]

• Answer:
  • Forward force = 40 N [1 mark]
  • Since the cyclist travels at constant speed, the resultant force is zero (Newton's First Law). The forward force must equal the resistive force of 40 N. [1 mark]
• Marking: Award 1 mark for correct magnitude. Award 1 mark for explanation using balanced forces / Newton's First Law.

(b) Calculate the work done by the cyclist in travelling 500 m. [2]

• Answer:
  • Work done = force × distance = 40 × 500 [1 mark for correct substitution]
  • Work done = 20,000 J / 20 kJ [1 mark for correct answer with unit]
• Marking: Award 1 mark for correct formula/substitution. Award 1 mark for correct answer with unit.

(c) Explain why the speed increases on a downward slope even though no forward force is applied. [2]

• Answer:
  • On the downward slope, the cyclist loses gravitational potential energy. [1 mark]
  • This potential energy is converted to kinetic energy, causing the speed to increase. [1 mark]
• Marking: Award 1 mark for identifying loss of gravitational potential energy. Award 1 mark for linking to conversion to kinetic energy / increase in speed.

END OF ANSWER KEY

This answer key was generated by TuitionGoWhere AI. Mark allocations reflect typical O-Level Combined Science assessment standards. Partial marks should be awarded where working demonstrates correct scientific reasoning.