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Secondary 1 Science Practice Paper 3
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TuitionGoWhere Practice Paper - Science Secondary 1 (Answer Key)
Subject: Science
Level: Secondary 1 (G3)
Paper: Practice Paper - Physical Sciences (Forces, Energy & Work)
Total Marks: 50
Section A: Multiple Choice Questions (10 marks)
| Question | Answer | Explanation |
|---|---|---|
| 1 | B | The student uses chemical energy from muscles to lift the book, increasing its gravitational potential energy. |
| 2 | C | Work done = Force × Distance = 15 N × 4 m = 60 J. |
| 3 | B | Work done = Force × Distance moved in direction of force. Holding a weight stationary involves no displacement, so no work is done. |
| 4 | C | GPE at top = mgh = 2 × 10 × 10 = 200 J. By conservation of energy, this converts to KE at bottom (ignoring air resistance). |
| 5 | B | Useful work output = Efficiency × Work input = 0.80 × 250 J = 200 J. |
| 6 | A | A compressed spring stores elastic potential energy, which converts to kinetic energy when released. |
| 7 | C | Work done against gravity depends only on vertical height (same in both cases). Power = Work/Time, so less time means greater power. |
| 8 | A | KE = ½mv² = 0.5 × 0.5 kg × (4 m/s)² = 0.5 × 0.5 × 16 = 4 J. |
| 9 | C | Gravitational potential energy increases when an object moves higher (rocket launching upwards). |
| 10 | A | Work done = F × d × cos θ = 20 N × 5 m × cos 60° = 100 × 0.5 = 50 J. |
Section B: Structured Questions (25 marks)
Question 11 [4]
(a) Weight = mg = 120 kg × 10 N/kg = 1200 N [1]
(b) At constant velocity, net force = 0. Minimum force = Weight = 1200 N [1]
(c) Work done = Force × Distance = 1200 N × 6 m = 7200 J [1]
(d) Chemical energy (worker) → Gravitational potential energy (load) [1]
Question 12 [5]
(a) GPE at A = mgh = 500 kg × 10 N/kg × 30 m = 150,000 J [1]
(b) At point B (ground level), all GPE is converted to KE (frictionless track).
KE at B = GPE at A = 150,000 J [2]
Explanation: By conservation of energy, total mechanical energy is conserved. At A, energy is all GPE. At B, height = 0 so GPE = 0, thus all energy is KE.
(c) KE = ½mv² → 150,000 = 0.5 × 500 × v² → v² = 600 → v = √600 ≈ 24.5 m/s [1]
(d) At point C (height 15 m):
GPE at C = mgh = 500 × 10 × 15 = 75,000 J
Total energy = 150,000 J (conserved)
KE at C = Total energy - GPE at C = 150,000 - 75,000 = 75,000 J [1]
Question 13 [4]
(a) Graph plotting:
- Axes labelled correctly with units [1]
- Suitable scales (1 cm = 1 cm extension, 1 cm = 1 N) [1]
- Points plotted accurately: (0,0), (4,2), (8,4), (12,6), (16,8) [1]
- Best-fit straight line through origin [1]
Total 4 marks for graph, but question allocates marks differently - see below
(b) Force is directly proportional to extension (Hooke's Law). The graph is a straight line passing through the origin. [1]
(c) At F = 6 N, Extension = 12 cm = 0.12 m
Elastic PE = ½ × Force × Extension = 0.5 × 6 N × 0.12 m = 0.36 J [2]
(1 mark for correct extension, 1 mark for correct calculation with units)
Question 14 [6]
(a) Work done against gravity = mgh = 60 kg × 10 N/kg × 3.5 m = 2100 J [2]
(1 mark for formula/substitution, 1 mark for answer with unit)
(b) Power = Work done / Time = 2100 J / 4.0 s = 525 W [2]
(1 mark for formula/substitution, 1 mark for answer with unit)
(c) Efficiency = Useful energy output / Chemical energy input
0.25 = 2100 J / Chemical energy input
Chemical energy input = 2100 J / 0.25 = 8400 J [2]
(1 mark for rearranging formula, 1 mark for answer with unit)
Question 15 [6]
(a) Net force = Applied force - Friction = 25 N - 9 N = 16 N (to the right) [1]
(b) Work done by applied force = Force × Distance = 25 N × 6 m = 150 J [1]
(c) Work done against friction = Friction × Distance = 9 N × 6 m = 54 J [1]
(d) Gain in KE = Net work done = Work by applied force - Work against friction
= 150 J - 54 J = 96 J
Alternatively: Gain in KE = Net force × Distance = 16 N × 6 m = 96 J [1]
(e) KE = ½mv² → 96 = 0.5 × 4 × v² → v² = 48 → v = √48 ≈ 6.93 m/s [2]
(1 mark for correct formula/substitution, 1 mark for answer with unit)
Section C: Free Response Questions (15 marks)
Question 16 [8]
(a) GPE lost per second = mgh = 500 kg × 10 N/kg × 80 m = 400,000 J/s (or 400,000 W) [2]
(1 mark for formula/substitution, 1 mark for answer with unit)
(b) Electrical power output = Efficiency × Input power = 0.90 × 400,000 W = 360,000 W (or 360 kW) [2]
(1 mark for formula/substitution, 1 mark for answer with unit)
(c) High voltage reduces the current for the same power (P = VI). Lower current reduces power loss in cables (P_loss = I²R), making transmission more efficient over long distances. [2]
(1 mark for linking high voltage to low current, 1 mark for explaining reduced power loss)
(d) Advantage: Renewable, no greenhouse gas emissions during operation, reliable base-load power.
Disadvantage: Disrupts river ecosystems, displaces communities, methane from reservoirs, high initial cost.
Any valid advantage and disadvantage [2]
(1 mark each)
Question 17 [7]
(a) Height gained = L - L cos θ = 1.2 - 1.2 cos 30° = 1.2 - 1.2(0.866) = 1.2 - 1.0392 = 0.1608 m ≈ 0.161 m [2]
(1 mark for correct formula/substitution, 1 mark for answer with unit)
(b) GPE = mgh = 0.5 kg × 10 N/kg × 0.1608 m = 0.804 J [1]
(c) By conservation of energy (no losses), KE at lowest point = GPE at release point = 0.804 J [1]
(d) KE = ½mv² → 0.804 = 0.5 × 0.5 × v² → v² = 3.216 → v = √3.216 ≈ 1.79 m/s [2]
(1 mark for correct formula/substitution, 1 mark for answer with unit)
(e) Mechanical energy is gradually converted to thermal energy (heat) and sound energy due to air resistance and friction at the pivot. The total energy is conserved but mechanical energy decreases. [1]
END OF ANSWER KEY
Total Marks: 50