Secondary 4 Combined Science Physics Preliminary Examination Paper 3

TuitionGoWhere Practice Paper - Combined Science Physics Secondary 4

PRELIMINARY EXAMINATION (Version 3) - ANSWER KEY & MARKING SCHEME

TuitionGoWhere Secondary School (AI)

Subject: Combined Science Physics (5086/5087) Level: Secondary 4 Paper: Physics Theory Paper Total Marks: 65

Section A: Multiple Choice (10 marks)

Section B: Structured Questions (30 marks)

Question 11: Cyclist Motion (7 marks)

(a) Describe the motion of the cyclist between 0 s and 10 s. [1 mark]

• Answer: The cyclist accelerates uniformly from rest / The cyclist moves with constant acceleration / Velocity increases at a constant rate.
• Marking: 1 mark for correct description including "accelerates" or "velocity increases."

(b) State the magnitude of the acceleration of the cyclist between 10 s and 20 s. [1 mark]

• Answer: 0 m/s²
• Marking: 1 mark for correct answer with unit. Accept "zero" or "0."

(c) Calculate the total distance travelled by the cyclist in the 50 seconds. [3 marks]

• Answer: Distance = area under velocity-time graph
  • 0-10 s: Area of triangle = ½ × 10 × 12 = 60 m
  • 10-20 s: Area of rectangle = 10 × 12 = 120 m
  • 20-30 s: Area of triangle = ½ × 10 × 12 = 60 m
  • 30-50 s: Area of triangle = ½ × 20 × 12 = 120 m
  • Total distance = 60 + 120 + 60 + 120 = 360 m
• Marking:
  • 1 mark for identifying method (area under graph)
  • 1 mark for correct calculation of at least two sections
  • 1 mark for correct total with unit (360 m)

(d) Calculate the average speed of the cyclist for the entire journey. [2 marks]

• Answer: Average speed = total distance / total time = 360 / 50 = 7.2 m/s
• Marking:
  • 1 mark for correct formula
  • 1 mark for correct answer with unit (7.2 m/s)

Question 12: Wax Heating (8 marks)

(a) State the melting point of the wax. [1 mark]

• Answer: 40°C
• Marking: 1 mark for correct temperature.

(b) Describe the motion and spacing of the wax particles between t = 2 min and t = 6 min. [2 marks]

• Answer: The particles vibrate faster (increased kinetic energy). The spacing between particles increases slightly (thermal expansion).
• Marking:
  • 1 mark for "particles vibrate faster" or "kinetic energy increases"
  • 1 mark for "spacing increases slightly" or "particles move slightly further apart"

(c) Explain why the temperature remains constant between t = 6 min and t = 10 min, even though heating continues. [2 marks]

• Answer: The energy supplied by the heater is used to overcome the intermolecular forces/bonds between particles during melting (phase change). This energy does not increase the kinetic energy of the particles, so the temperature remains constant.
• Marking:
  • 1 mark for "energy used to overcome intermolecular forces/bonds"
  • 1 mark for "kinetic energy does not increase" or "temperature remains constant during phase change"

(d) Calculate the time taken for the wax to melt completely. [3 marks]

• Answer:
  • Energy required for melting: Q = mL = 0.20 × 180,000 = 36,000 J
  • Power of heater = 40 W
  • Time = Energy / Power = 36,000 / 40 = 900 s = 15 minutes
• Marking:
  • 1 mark for correct energy calculation (Q = mL)
  • 1 mark for correct formula (t = E/P)
  • 1 mark for correct answer with unit (900 s or 15 min)

Question 13: Converging Lens (8 marks)

(a) Complete the ray diagram. [3 marks]

• Answer:
  • Ray 1: Parallel to principal axis, refracted through focal point F on the right side
  • Ray 2: Through optical centre O, undeviated (straight line)
  • Image formed where the two refracted rays intersect on the screen (right side, inverted)
• Marking:
  • 1 mark for correct ray 1 (parallel to axis, through F)
  • 1 mark for correct ray 2 (through O, undeviated)
  • 1 mark for correct image position (on screen, inverted)

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

• Answer: The image is real and inverted. (Also accept: diminished/smaller than object, formed on the screen)
• Marking: 1 mark each for any two correct characteristics.

(c) Calculate the distance of the image from the lens. [3 marks]

• Answer:
  • 1/f = 1/u + 1/v
  • 1/15 = 1/25 + 1/v
  • 1/v = 1/15 - 1/25 = (5 - 3)/75 = 2/75
  • v = 75/2 = 37.5 cm
• Marking:
  • 1 mark for correct substitution into lens formula
  • 1 mark for correct rearrangement
  • 1 mark for correct answer with unit (37.5 cm)

Question 14: Kitchen Appliance (7 marks)

(a) Calculate the current flowing through the heating element. [2 marks]

• Answer: I = P/V = 2000/240 = 8.33 A
• Marking:
  • 1 mark for correct formula (I = P/V)
  • 1 mark for correct answer with unit (8.33 A or 8.3 A)

(b) Calculate the total current drawn from the mains supply. [2 marks]

• Answer:
  • Current through lamp: I = P/V = 40/240 = 0.167 A
  • Total current = 8.33 + 0.167 = 8.50 A (or 8.5 A)
• Marking:
  • 1 mark for calculating lamp current
  • 1 mark for correct total current with unit

(c) Discuss whether the 10 A fuse is appropriate. [3 marks]

• Answer: The total current drawn is 8.5 A. A 10 A fuse is appropriate because:
  • The normal operating current (8.5 A) is below the fuse rating, so the fuse will not blow during normal operation.
  • The fuse rating (10 A) is only slightly above the normal current, so if a fault causes excessive current, the fuse will blow quickly to protect the appliance and wiring.
  • A fuse rated much higher (e.g., 13 A) would not provide adequate protection.
• Marking:
  • 1 mark for comparing calculated current with fuse rating
  • 1 mark for explaining that fuse should be slightly above normal current
  • 1 mark for safety reasoning (fuse protects against excessive current)

Section C: Free Response Questions (25 marks)

Question 15: Energy Conservation (7 marks)

(a) Calculate the gravitational potential energy of the ball before it is dropped. [2 marks]

• Answer: GPE = mgh = 0.50 × 10 × 3.0 = 15 J
• Marking:
  • 1 mark for correct formula
  • 1 mark for correct answer with unit (15 J)

(b) Calculate the kinetic energy of the ball just before it hits the ground. [2 marks]

• Answer: KE = ½mv² = ½ × 0.50 × (7.0)² = 0.25 × 49 = 12.25 J ≈ 12.3 J
• Marking:
  • 1 mark for correct formula
  • 1 mark for correct answer with unit (12.25 J or 12.3 J)

(c) Explain why the loss in GPE is different from the gain in KE and state how the law of conservation of energy applies. [3 marks]

• Answer: The loss in GPE (15 J) is greater than the gain in KE (12.3 J) because some energy is dissipated as heat and sound due to air resistance acting on the ball as it falls. The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. Therefore: GPE lost = KE gained + energy dissipated to surroundings (heat and sound). Total energy is conserved.
• Marking:
  • 1 mark for identifying air resistance/friction as cause of energy difference
  • 1 mark for stating that energy is dissipated as heat/sound
  • 1 mark for applying conservation of energy (total energy constant, GPE = KE + dissipated energy)

Question 16: Reflection and Refraction (5 marks)

(a) State the angle of reflection. [1 mark]

• Answer: 35°
• Marking: 1 mark for correct answer.

(b) State the two laws of reflection. [2 marks]

• Answer:
  1. The incident ray, reflected ray, and normal all lie in the same plane.
  2. The angle of incidence equals the angle of reflection (i = r).
• Marking: 1 mark for each correct law.

(c) Explain what happens to the light ray at the water-air boundary. [2 marks]

• Answer: The angle of incidence (50°) is greater than the critical angle (49°). Therefore, total internal reflection occurs. The light ray is completely reflected back into the water; no light is refracted into the air.
• Marking:
  • 1 mark for stating that angle of incidence > critical angle
  • 1 mark for stating total internal reflection occurs (light reflected back into water)

Question 17: Electric Iron (8 marks)

(a) Calculate the resistance of the heating element. [3 marks]

• Answer:
  • P = V²/R, so R = V²/P
  • R = (240)² / 1200 = 57,600 / 1200 = 48 Ω
• Marking:
  • 1 mark for correct formula (R = V²/P or P = V²/R)
  • 1 mark for correct substitution
  • 1 mark for correct answer with unit (48 Ω)

(b) Calculate the cost of using the iron for 30 days. [3 marks]

• Answer:
  • Daily energy consumption: E = Pt = 1.2 kW × 2 h = 2.4 kWh
  • Monthly energy consumption: 2.4 × 30 = 72 kWh
  • Cost = 72 × $0.25 =$18.00
• Marking:
  • 1 mark for correct daily energy calculation
  • 1 mark for correct monthly energy calculation
  • 1 mark for correct cost with unit ($18.00)

(c) Explain why the metal body of the iron must be earthed. [2 marks]

• Answer: If a fault occurs and the live wire touches the metal body, the earth wire provides a low-resistance path for the current to flow to the ground. This causes a large current to flow, which blows the fuse or trips the circuit breaker, disconnecting the appliance. This prevents the user from receiving an electric shock if they touch the metal body.
• Marking:
  • 1 mark for "provides low-resistance path to ground"
  • 1 mark for "prevents electric shock" or "blows fuse to disconnect appliance"

Question 18: Pressure (6 marks)

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

• Answer: W = mg = 2.4 × 10 = 24 N
• Marking: 1 mark for correct answer with unit.

(b) Calculate the minimum pressure the block can exert on the table. [3 marks]

• Answer:
  • Minimum pressure occurs with maximum contact area
  • Maximum area = 0.20 × 0.10 = 0.020 m²
  • Pressure = Force / Area = 24 / 0.020 = 1200 Pa
• Marking:
  • 1 mark for identifying maximum area (0.20 × 0.10)
  • 1 mark for correct formula (P = F/A)
  • 1 mark for correct answer with unit (1200 Pa or 1.2 kPa)

(c) Explain how the pressure exerted by the block changes when it is placed on a different face. [2 marks]

• Answer: Pressure depends on the contact area. When the block is placed on a smaller face, the contact area decreases. Since pressure = force/area and the weight (force) remains constant, the pressure increases. Conversely, placing the block on a larger face decreases the pressure.
• Marking:
  • 1 mark for stating that pressure changes with contact area
  • 1 mark for explaining inverse relationship (smaller area → larger pressure, or vice versa)

END OF ANSWER KEY

Marking Scheme Summary

Grade Boundaries (Guideline):

• A: 52-65 (80-100%)
• B: 42-51 (65-79%)
• C: 33-41 (50-64%)
• D: 26-32 (40-49%)
• F: Below 26 (Below 40%)