Secondary 2 Science Practice Paper 3

TuitionGoWhere Practice Paper - Science Secondary 2 (Answer Key)

Subject: Science
Level: Secondary 2 (G3)
Paper: Practice Paper – Physical Sciences (Version 3)
Total Marks: 60

Section A: Multiple Choice Questions (10 marks)

1. C
Working:
Gravitational potential energy gained = $mgh = 2.0 \times 10 \times 1.5 = 30 \text{ J}$
Marking: 1 mark for correct answer.

2. B
Explanation: As the ball falls, its height decreases so gravitational potential energy decreases, while its speed increases so kinetic energy increases. Energy is converted from gravitational potential energy to kinetic energy.
Marking: 1 mark for correct answer.

3. C
Working:
Work done = Force × Distance = $25 \times 4 = 100 \text{ J}$
Marking: 1 mark for correct answer.

4. B
Working:
Energy = Power × Time = $500 \text{ W} \times (3 \times 60) \text{ s} = 500 \times 180 = 90\,000 \text{ J}$
Marking: 1 mark for correct answer.

5. C
Explanation: Natural gas is a fossil fuel formed over millions of years and cannot be replenished on a human timescale. Solar, wind, and hydroelectric are renewable.
Marking: 1 mark for correct answer.

6. B
Working:
Kinetic energy gained = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 20^2 = 200\,000 \text{ J}$
Average power = $\frac{\text{Work}}{\text{Time}} = \frac{200\,000}{10} = 20\,000 \text{ W}$
Marking: 1 mark for correct answer.

7. B
Explanation: At the lowest point B, gravitational potential energy is minimum and kinetic energy is maximum (by conservation of energy, ignoring air resistance).
Marking: 1 mark for correct answer.

8. B
Working:
Efficiency = $\frac{\text{Useful energy output}}{\text{Energy input}} \times 100\% = \frac{400}{500} \times 100\% = 80\%$
Marking: 1 mark for correct answer.

9. B
Explanation: Water in a reservoir has gravitational potential energy → converts to kinetic energy as it falls → drives turbines to generate electrical energy.
Marking: 1 mark for correct answer.

10. B
Working:
Kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 0.2 \times 3^2 = 0.1 \times 9 = 0.9 \text{ J}$
Marking: 1 mark for correct answer.

Section B: Structured Questions (30 marks)

11. (a) Principle of conservation of energy: Energy cannot be created or destroyed. It can only be converted from one form to another, and the total amount of energy in a closed system remains constant.
[2]
Marking: 1 mark for "cannot be created or destroyed"; 1 mark for "converted from one form to another" or "total energy remains constant".

(b) GPE at P = $mgh = 500 \times 10 \times 40 = 200\,000 \text{ J}$ (or $2.0 \times 10^5 \text{ J}$)
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(c) By conservation of energy:
Loss in GPE from P to Q = Gain in KE at Q
$mg(h_P - h_Q) = 500 \times 10 \times (40 - 10) = 500 \times 10 \times 30 = 150\,000 \text{ J}$
KE at Q = 150 000 J
[2]
Marking: 1 mark for correct concept (loss in GPE = gain in KE); 1 mark for correct calculation and unit.

(d) $\text{KE} = \frac{1}{2}mv^2$
$150\,000 = \frac{1}{2} \times 500 \times v^2$
$v^2 = \frac{150\,000 \times 2}{500} = 600$
$v = \sqrt{600} \approx 24.5 \text{ m/s}$
[2]
Marking: 1 mark for correct rearrangement/substitution; 1 mark for correct answer with unit (accept 24.5 or 24.49 m/s).

Common mistake: Forgetting to take square root, or using $h = 40$ m instead of height difference.

12. (a) Work done (useful output) = Gain in GPE = $mgh = 0.5 \times 10 \times 1.2 = 6.0 \text{ J}$
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(b) Electrical energy input = $VIt = 6.0 \times 0.8 \times 4.0 = 19.2 \text{ J}$
[2]
Marking: 1 mark for correct formula $E = VIt$; 1 mark for correct answer with unit.

(c) Efficiency = $\frac{\text{Useful output}}{\text{Input}} \times 100\% = \frac{6.0}{19.2} \times 100\% = 31.25\%$
[2]
Marking: 1 mark for correct formula; 1 mark for correct answer (accept 31.3% or 31.25%).

(d) Reason: Energy is lost as heat (thermal energy) in the motor coils due to electrical resistance, and as sound energy from friction in moving parts.
[1]
Marking: 1 mark for any valid reason (heat loss, sound, friction).

13. (a) As the bob swings from A to B, gravitational potential energy is converted to kinetic energy. At A, the bob has maximum GPE and zero KE. At B, GPE is minimum and KE is maximum.
[2]
Marking: 1 mark for identifying GPE → KE conversion; 1 mark for describing the change (max at A to max at B).

(b) By conservation of energy, the total mechanical energy (GPE + KE) remains constant (ignoring air resistance). At C, the bob momentarily comes to rest so KE = 0. All energy is GPE. Since no energy is added, the GPE at C cannot exceed the initial GPE at A, so the height cannot be greater than A.
[2]
Marking: 1 mark for conservation of energy argument; 1 mark for explaining height at C ≤ height at A.

(c) In a real pendulum, air resistance and friction at the pivot do negative work on the bob. The initial GPE at A is gradually converted to thermal energy (heat) in the surrounding air and at the pivot, and sound energy. Eventually, all mechanical energy is dissipated as thermal/sound energy, and the bob comes to rest at B.
[2]
Marking: 1 mark for identifying air resistance/friction; 1 mark for stating energy is dissipated as heat/sound.

14. (a) Initial KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 25^2 = 600 \times 625 = 375\,000 \text{ J}$
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(b) Work done by braking force = Loss in KE = $375\,000 \text{ J}$
Work = Force × Distance
$F \times 50 = 375\,000$
$F = \frac{375\,000}{50} = 7\,500 \text{ N}$
[2]
Marking: 1 mark for equating work to KE loss; 1 mark for correct force with unit.

(c) Braking distance would double.
Explanation: Initial KE = $\frac{1}{2}mv^2$. If mass doubles, initial KE doubles. Work done by braking force = $F \times d$. Since $F$ is constant, $d$ must double to do twice the work.
[2]
Marking: 1 mark for correct prediction (doubles); 1 mark for correct explanation linking KE ∝ m and Work = Fd.

15. (a) Solar power incident = Intensity × Area = $800 \times 2.0 = 1600 \text{ W}$
[2]
Marking: 1 mark for correct formula; 1 mark for correct answer with unit.

(b) Electrical power output = Efficiency × Incident power = $0.18 \times 1600 = 288 \text{ W}$
[2]
Marking: 1 mark for correct use of efficiency; 1 mark for correct answer with unit.

(c) Electrical power to battery = $VI = 12 \times 2.0 = 24 \text{ W}$
Time = $\frac{\text{Energy}}{\text{Power}} = \frac{1.0 \times 10^6}{24} = 41\,666.7 \text{ s}$
Convert to hours: $\frac{41\,666.7}{3600} \approx 11.6 \text{ hours}$ (or 11 h 34 min)
[2]
Marking: 1 mark for correct power calculation; 1 mark for correct time with unit.

Section C: Longer Structured and Data-Based Questions (20 marks)

16. (a) GPE lost per second = Mass flow rate × $g$ × height = $500 \times 10 \times 80 = 400\,000 \text{ J/s} = 400\,000 \text{ W}$
[2]
Marking: 1 mark for correct formula (power = $\dot{m}gh$); 1 mark for correct answer with unit.

(b) Electrical power output = Efficiency × Input power = $0.85 \times 400\,000 = 340\,000 \text{ W} = 340 \text{ kW}$
[2]
Marking: 1 mark for correct use of efficiency; 1 mark for correct answer with unit (accept 340 kW or 340 000 W).

(c) $P = VI$ → $I = \frac{P}{V} = \frac{340\,000}{400\,000} = 0.85 \text{ A}$
[2]
Marking: 1 mark for correct formula/rearrangement; 1 mark for correct answer with unit.

(d) Explanation: Transmitting at high voltage reduces the current for the same power ($P = VI$). Lower current reduces the power loss in the cables ($P_{\text{loss}} = I^2R$), making transmission more efficient.
[2]
Marking: 1 mark for stating current is reduced; 1 mark for linking to reduced $I^2R$ losses / heating in cables.

17. (a) Daily energy per appliance (kWh):

• Refrigerator: $\frac{150}{1000} \times 24 = 3.6 \text{ kWh}$
• Air conditioner: $\frac{1500}{1000} \times 6 = 9.0 \text{ kWh}$
• Washing machine: $\frac{500}{1000} \times 1 = 0.5 \text{ kWh}$
• LED lights: $\frac{60}{1000} \times 5 = 0.3 \text{ kWh}$
• Laptop: $\frac{50}{1000} \times 4 = 0.2 \text{ kWh}$
  Total = $3.6 + 9.0 + 0.5 + 0.3 + 0.2 = 13.6 \text{ kWh}$
  [3]
  Marking: 1 mark for correct method (Power in kW × time); 1 mark for correct individual calculations; 1 mark for correct total with unit.

(b) Cost for 30 days = 13.6 \times 30 \times 0.28 = 408 \times 0.28 = \114.24$[2] **Marking:** 1 mark for correct monthly energy (408 kWh); 1 mark for correct cost with$ sign.

(c) Actual refrigerator energy = $3.6 \times 0.30 = 1.08 \text{ kWh}$
[1]
Marking: 1 mark for correct answer with unit.

(d) Two ways:

1. Switch off appliances at the socket (avoid standby power).
2. Set air conditioner to a higher temperature (e.g., 25°C instead of 22°C).
3. Use natural light instead of LED lights during the day.
4. Wash clothes with full loads only.
   (Any two valid suggestions)
   [2]
   Marking: 1 mark each for two valid, practical suggestions.

18. (a) Average speed = $\frac{\text{Distance}}{\text{Time}} = \frac{1.0}{0.49} \approx 2.04 \text{ m/s}$
[1]
Marking: 1 mark for correct calculation with unit.

(b) Speed at bottom ≈ average speed over last 1.0 m = $2.04 \text{ m/s}$
KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (2.04)^2 = 0.25 \times 4.1616 \approx 1.04 \text{ J}$
[2]
Marking: 1 mark for using average speed as final speed; 1 mark for correct KE calculation with unit.

(c) Graph plotting:
First, calculate KE for each height using $v_{\text{avg}} = 1.0/t$, $KE = \frac{1}{2} \times 0.5 \times (1.0/t)^2 = \frac{0.25}{t^2}$:

Graph requirements:

• Axes labelled with units: "Height of ramp / m" (x-axis), "Kinetic Energy / J" (y-axis)
• Appropriate scales covering all points
• All 5 points plotted accurately
• Best-fit straight line through origin (theoretically) or close to points
  [3]
  Marking: 1 mark for correct axes labels and scales; 1 mark for all 5 points plotted correctly; 1 mark for best-fit line.

(d) Relationship: The kinetic energy at the bottom of the ramp is directly proportional to the height of the ramp. (Or: KE increases linearly with height.)
[1]
Marking: 1 mark for "directly proportional" or "linear increase".

(e) Reason: The graph does not pass through the origin because of friction between the trolley and ramp, and air resistance. Some gravitational potential energy is converted to thermal/sound energy instead of kinetic energy, so even at very small heights, there is a threshold below which the trolley does not move, or the KE is less than predicted by $mgh$.
[1]
Marking: 1 mark for identifying friction/air resistance as cause of energy loss.

19. (a) Two disadvantages:

1. Releases carbon dioxide (CO₂), a greenhouse gas, contributing to global warming and climate change.
2. Releases pollutants (e.g., sulfur dioxide, nitrogen oxides, particulates) causing air pollution, acid rain, and health problems.
   (Also acceptable: finite/non-renewable resource; habitat destruction from mining; oil spills.)
   [2]
   Marking: 1 mark each for two distinct valid disadvantages.

(b) Explanation: Solar energy comes from the Sun, which will continue to radiate energy for billions of years. It is replenished naturally on a human timescale and cannot be depleted by human use.
[1]
Marking: 1 mark for "replenished naturally" or "virtually inexhaustible" or "Sun will shine for billions of years".

(c) Average electrical power output:
Total solar power incident = Irradiance × Area = $200 \times 50\,000 = 10\,000\,000 \text{ W} = 10 \text{ MW}$
Electrical power output = Efficiency × Incident power = $0.20 \times 10 \text{ MW} = 2 \text{ MW}$
[3]
Marking: 1 mark for incident power calculation; 1 mark for applying efficiency; 1 mark for correct answer in MW.

(d) Advantage: No fuel cost / no greenhouse gas emissions during operation / renewable.
Disadvantage: Intermittent (depends on sunlight, weather, day/night) / requires large land area / high initial cost / energy storage needed for continuous supply.
[2]
Marking: 1 mark for valid advantage; 1 mark for valid disadvantage.

20. (a) Work done against gravity = Gain in GPE = $mgh = 60 \times 10 \times 3.5 = 2100 \text{ J}$
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(b) Power = $\frac{\text{Work}}{\text{Time}} = \frac{2100}{4.0} = 525 \text{ W}$
[2]
Marking: 1 mark for correct formula; 1 mark for correct answer with unit.

(c) Chemical energy expended:
Efficiency = $\frac{\text{Useful work output}}{\text{Chemical energy input}}$
$0.25 = \frac{2100}{E_{\text{chem}}}$
$E_{\text{chem}} = \frac{2100}{0.25} = 8400 \text{ J}$
[2]
Marking: 1 mark for correct efficiency formula rearrangement; 1 mark for correct answer with unit.

(d) Explanation:

• Work done by gravity during descent: The force of gravity (weight) acts downwards, and the displacement is downwards. Since force and displacement are in the same direction, work done by gravity is positive ($W = Fd \cos 0^\circ = +Fd$). Gravity transfers energy to the student (increases kinetic energy if uncontrolled).
• Work done by the student against gravity: The student must exert an upward force to control the descent, while displacement is downwards. Force and displacement are in opposite directions, so work done by the student is negative ($W = Fd \cos 180^\circ = -Fd$). The student's muscles absorb energy (chemical energy is still expended to maintain tension).
  [2]
  Marking: 1 mark for explaining positive work by gravity (force and displacement same direction); 1 mark for explaining negative work by student (force opposite to displacement).

End of Answer Key