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Secondary 3 Combined Science Practice Paper 1
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TuitionGoWhere Practice Paper - Combined Science Secondary 3 (Marking Scheme)
Total Marks: 80
Section A: Multiple Choice Questions [20 marks]
| Question | Answer | Explanation |
|---|---|---|
| 1 | B | Conservation of energy states energy cannot be created or destroyed, only converted |
| 2 | C | PE = mgh = 2×10×10 = 200 J; by conservation, KE = 200 J |
| 3 | D | Velocity has both magnitude and direction (vector); others are scalars |
| 4 | B | Newton's First Law - object continues at constant velocity unless unbalanced force acts |
| 5 | C | Parallel: 1/R = 1/6 + 1/6 = 2/6, so R = 3 Ω |
| 6 | B | Fuse breaks circuit when current exceeds safe limit |
| 7 | A | a = (v-u)/t = (30-0)/6 = 5 m/s² |
| 8 | C | Power is measured in Watts (W) |
| 9 | C | Compressed spring stores elastic potential energy |
| 10 | B | Ohm's Law: V = IR |
Marking: 2 marks per correct answer
Section B: Structured Questions [60 marks]
Question 11: Energy Transformations [12 marks]
(a) Calculate the gravitational potential energy [3 marks]
Answer:
- PE = mgh [1]
- PE = 500 × 10 × 20 [1]
- PE = 100,000 J (or 1.0 × 10⁵ J) [1]
(b) Calculate speed at bottom of hill [4 marks]
Answer:
- By conservation of energy: PE = KE [1]
- 100,000 = ½mv² [1]
- 100,000 = ½ × 500 × v² [1]
- v = 20 m/s [1]
Alternative method: v² = u² + 2as, where a = g = 10 m/s², s = 20 m
(c) Calculate energy lost to friction [3 marks]
Answer:
- Actual KE = ½mv² = ½ × 500 × 15² = 56,250 J [2]
- Energy lost = 100,000 - 56,250 = 43,750 J [1]
(d) Two ways energy could be lost [2 marks]
Answer:
- Air resistance/drag [1]
- Heat generated by friction with track [1]
- Sound energy [1]
- Deformation of wheels/track [1]
Marking: Accept any two reasonable answers
Question 12: Forces and Motion [15 marks]
(a) Calculate acceleration [3 marks]
Answer:
- v² = u² + 2as [1]
- 25² = 0² + 2a(200) [1]
- a = 625/400 = 1.56 m/s² (or 25²/400) [1]
(b) Calculate net force [2 marks]
Answer:
- F = ma = 2000 × 1.56 [1]
- F = 3125 N (or 3120 N) [1]
(c) Calculate friction force [2 marks]
Answer:
- Net force = Driving force - Friction force [1]
- Friction force = 4000 - 3125 = 875 N [1]
(d) Explain constant speed motion [3 marks]
Answer:
- At constant speed, acceleration = 0 [1]
- Therefore net force = 0 (Newton's First Law) [1]
- Driving force equals friction force [1]
(e) Newton's Second Law and application [5 marks]
Answer:
- Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass [2]
- OR F = ma [2]
- Application: The net force of 3125 N causes the 2000 kg truck to accelerate at 1.56 m/s² [2]
- The larger the force, the greater the acceleration for the same mass [1]
Question 13: Electrical Circuits [18 marks]
(a) Combined resistance of parallel resistors [3 marks]
Answer:
- 1/R = 1/4 + 1/8 [1]
- 1/R = 2/8 + 1/8 = 3/8 [1]
- R = 8/3 = 2.67 Ω [1]
(b) Total circuit resistance [2 marks]
Answer:
- Total R = 2.67 + 6 [1]
- Total R = 8.67 Ω [1]
(c) Total current from battery [2 marks]
Answer:
- I = V/R = 12/8.67 [1]
- I = 1.38 A [1]
(d) Current through 4 Ω resistor [3 marks]
Answer:
- Voltage across parallel section = 12 - (1.38 × 6) = 3.72 V [1]
- Current through 4 Ω = V/R = 3.72/4 [1]
- I = 0.93 A [1]
(e) Power dissipated by 6 Ω resistor [3 marks]
Answer:
- Current through 6 Ω = 1.38 A [1]
- P = I²R = (1.38)² × 6 [1]
- P = 11.4 W [1]
(f) Fuse purpose and rating [3 marks]
Answer:
- Purpose: To break the circuit if current exceeds safe limit, preventing overheating and fire [1]
- Suggested rating: 2 A or 3 A [1]
- Justification: Should be slightly higher than normal operating current (1.38 A) but low enough to provide protection [1]
(g) Two other safety features [2 marks]
Answer:
- Earth wire/earthing [1]
- Insulation on wires [1]
- Circuit breakers [1]
- RCD (Residual Current Device) [1]
- Double insulation [1]
Marking: Accept any two appropriate safety features
Question 14: Waves and Energy Transfer [15 marks]
(a) Complete energy efficiency table [4 marks]
Answer:
- Working for 1.0 m: Efficiency = (bounce height/drop height) × 100% = (0.6/1.0) × 100% = 60% [2]
- Completed table:
- 1.0 m: 60% [given in working]
- 1.5 m: 60% [1]
- 2.0 m: 60% [1]
- 2.5 m: 60% [1]
(b) Plot graph [4 marks]
Answer:
- Correct axes labels with units [1]
- Appropriate scale [1]
- All points plotted correctly [1]
- Straight line through points [1]
(c) Describe relationship [2 marks]
Answer:
- Linear/directly proportional relationship [1]
- As drop height increases, bounce height increases proportionally [1]
(d) Predict bounce height for 3.0 m drop [1 mark]
Answer:
- 1.8 m [1]
Marking: Accept values between 1.7-1.9 m based on graph reading
(e) Explain why ball doesn't return to original height [2 marks]
Answer:
- Energy is lost during impact [1]
- Energy converted to heat, sound, and deformation [1]
(f) Two factors affecting energy efficiency [2 marks]
Answer:
- Surface material (hardness) [1]
- Ball material/construction [1]
- Air resistance [1]
- Temperature [1]
- Ball pressure/inflation [1]
Marking: Accept any two reasonable factors
Total: 80 marks
Grade Boundaries (Suggested):
- A: 72-80 marks (90-100%)
- B: 64-71 marks (80-89%)
- C: 56-63 marks (70-79%)
- D: 48-55 marks (60-69%)
- E: 40-47 marks (50-59%)
- F: Below 40 marks (<50%)