Secondary 2 Science Practice Paper 2

TuitionGoWhere Practice Paper - Science Secondary 2 (Answer Key)

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

Section A: Multiple Choice Questions [10 marks]

Question 1 [1] — Answer: C

Working:

• Gravitational potential energy at start = $mgh = 2 \times 10 \times 10 = 200 \text{ J}$
• By conservation of energy, this converts entirely to kinetic energy just before impact (ignoring air resistance).
• Kinetic energy = 200 J

Key concept: In free fall without air resistance, loss in GPE = gain in KE.

Question 2 [1] — Answer: A

Explanation: A battery stores chemical energy. When the torch is switched on, chemical energy is converted to electrical energy, which then powers the bulb to produce light energy and heat energy.

Common mistake: Reversing the order of conversions or omitting heat energy.

Question 3 [1] — Answer: C

Working: Work done = Force × Distance moved in direction of force $W = 15 \text{ N} \times 4 \text{ m} = 60 \text{ J}$

Question 4 [1] — Answer: B

Explanation: At position Q (lowest point), gravitational potential energy is minimum, so kinetic energy is maximum (by conservation of mechanical energy). At P and R, the bob momentarily stops, so KE = 0.

Question 5 [1] — Answer: B

Working: Power = 500 W, Time = 3 minutes = 180 seconds Energy = Power × Time = $500 \times 180 = 90\,000 \text{ J}$

Common mistake: Forgetting to convert minutes to seconds (would give 1500 J, option A).

Question 6 [1] — Answer: C

Explanation: The principle of conservation of energy states that energy cannot be created or destroyed; the total energy in a closed (isolated) system remains constant. It can only be converted from one form to another.

Question 7 [1] — Answer: C

Explanation: Work done against gravity depends only on vertical height and weight ($W = mgh$), not on speed. Power = Work/Time, so doing the same work in less time (running) requires more power.

Question 8 [1] — Answer: C

Working: Efficiency = Useful output energy / Input energy × 100% $80\% = 400 / \text{Input} \times 100\%$ Input = $400 / 0.8 = 500 \text{ J}$

Question 9 [1] — Answer: B

Working: $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (2)^2 = 0.25 \times 4 = 1.0 \text{ J}$

Question 10 [1] — Answer: C

Explanation: Natural gas is a fossil fuel formed over millions of years; it is non-renewable. Solar, wind, and hydroelectric are renewable resources.

Section B: Structured Questions [30 marks]

Question 11 [4]

(a) [2]
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.
Marking: 1 mark for "cannot be created or destroyed", 1 mark for "converted from one form to another" or "total energy remains constant".

(b) [1]
$GPE = mgh = 500 \times 10 \times 40 = 200\,000 \text{ J}$ (or 200 kJ)

(c) [1]
At point B, all GPE is converted to KE (negligible friction).
$KE = \frac{1}{2}mv^2 = 200\,000$
$v^2 = \frac{2 \times 200\,000}{500} = 800$
$v = \sqrt{800} \approx 28.3 \text{ m/s}$

Question 12 [5]

(a) [2]
Work done against gravity = Gain in GPE = $mgh = 60 \times 10 \times 8 = 4800 \text{ J}$

(b) [2]
Power = Work done / Time = $4800 / 20 = 240 \text{ W}$

(c) [1]
Some energy is converted to heat energy due to internal body processes (muscle inefficiency, maintaining body temperature, etc.). The human body is not 100% efficient at converting chemical energy to mechanical work.

Question 13 [4]

(a) [1]
Work done by applied force = $F \times d = 20 \times 5 = 100 \text{ J}$

(b) [1]
Work done against friction = $f \times d = 8 \times 5 = 40 \text{ J}$

(c) [1]
Net work done = Work by applied force − Work against friction = $100 - 40 = 60 \text{ J}$
Alternatively: Net force = $20 - 8 = 12 \text{ N}$; Net work = $12 \times 5 = 60 \text{ J}$

(d) [1]
By work-energy theorem: Net work = Change in KE = $\frac{1}{2}mv^2 - 0$
$60 = \frac{1}{2} \times 3 \times v^2$
$v^2 = 40$
$v = \sqrt{40} \approx 6.32 \text{ m/s}$

Question 14 [5]

(a) [1]
Gravitational potential energy of water → Kinetic energy of moving water → Kinetic energy of turbine → Electrical energy (via generator).

(b) [2]
Mass of water per second = 100 kg
GPE lost per second = $mgh = 100 \times 10 \times 50 = 50\,000 \text{ J/s} = 50\,000 \text{ W}$

(c) [2]
Electrical power output = Efficiency × Input power
$= 0.85 \times 50\,000 = 42\,500 \text{ W}$ (or 42.5 kW)

Question 15 [4]

(a) [2]
Elastic potential energy = $\frac{1}{2}kx^2 = \frac{1}{2} \times 400 \times (0.05)^2 = 200 \times 0.0025 = 0.5 \text{ J}$

(b) [2]
At max height, EPE = GPE gained
$0.5 = mgh = 0.02 \times 10 \times h$
$h = \frac{0.5}{0.2} = 2.5 \text{ m}$

Question 16 [4]

(a) [1]
Useful work done = Gain in GPE = $mgh = 10 \times 10 \times 6 = 600 \text{ J}$

(b) [2]
Electrical energy = $VIt = 240 \times 2.5 \times 8 = 4800 \text{ J}$

(c) [1]
Efficiency = Useful output / Input × 100% = $600 / 4800 \times 100\% = 12.5\%$

Question 17 [4]

(a) [2]
Energy per appliance per day (kWh) = Power (kW) × Time (h)

• Refrigerator: $0.150 \times 24 = 3.6 \text{ kWh}$
• Air Conditioner: $1.200 \times 6 = 7.2 \text{ kWh}$
• Washing Machine: $0.500 \times 1 = 0.5 \text{ kWh}$
• LED Lights: $0.060 \times 5 = 0.3 \text{ kWh}$
  Total = $3.6 + 7.2 + 0.5 + 0.3 = 11.6 \text{ kWh}$

(b) [1]
Cost = 11.6 \times 0.28 = \3.248 \approx $3.25$

(c) [1]
Any one valid suggestion, e.g.:

• Set the temperature to a higher (less cold) setting
• Ensure door seals are tight and avoid frequent opening
• Place refrigerator away from heat sources (oven, direct sunlight)
• Defrost regularly if not frost-free

Section C: Longer Structured and Data-Based Questions [20 marks]

Question 18 [7]

(a) [3]
Graph requirements:

• Axes labelled with units: $h$ (m) horizontal, $v^2$ (m²/s²) vertical
• Suitable scales covering data range
• All 5 points plotted correctly: (0.10, 1.69), (0.20, 3.61), (0.30, 5.29), (0.40, 7.29), (0.50, 9.00)
• Best-fit straight line passing through origin (or near origin) Marking: 1 mark for axes + scales, 1 mark for correct plotting, 1 mark for best-fit line.

(b) [2]
Theory: $v^2 = 2gh$, so gradient of $v^2$ vs $h$ graph = $2g$
From graph, gradient = $\frac{\Delta v^2}{\Delta h} \approx \frac{9.00 - 0}{0.50 - 0} = 18.0 \text{ m/s}^2$ (using origin and last point)
$2g = 18.0 \Rightarrow g = 9.0 \text{ m/s}^2$
Accept values in range 8.5–9.5 m/s² based on best-fit line.

(c) [1]
Friction between car wheels and ramp / air resistance / rotational kinetic energy of wheels not accounted for / ramp not perfectly rigid.
Any one valid reason other than human measurement error.

(d) [1]
The graph would not change. Speed at bottom depends only on height ($v = \sqrt{2gh}$), not on mass. Both GPE ($mgh$) and KE ($\frac{1}{2}mv^2$) are proportional to mass, so mass cancels out.

Question 19 [7]

(a) [1]
$a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \text{ m/s}^2$

(b) [1]
$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times (25)^2 = 600 \times 625 = 375\,000 \text{ J}$

(c) [2]
Distance travelled during acceleration: $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2.5 \times 10^2 = 125 \text{ m}$
Work against resistive force = $F \times s = 800 \times 125 = 100\,000 \text{ J}$

(d) [2]
Total work by engine = Gain in KE + Work against resistance = $375\,000 + 100\,000 = 475\,000 \text{ J}$
Average power = Total work / Time = $475\,000 / 10 = 47\,500 \text{ W}$ (or 47.5 kW)

(e) [1]
At constant speed up hill: Engine force balances resistive force + component of weight down slope.
Component of weight down slope = $mg \sin\theta = 1200 \times 10 \times \sin 5^\circ \approx 12000 \times 0.0872 = 1046 \text{ N}$
Total force = $800 + 1046 = 1846 \text{ N}$
Power = Force × Velocity = $1846 \times 25 = 46\,150 \text{ W}$ (or 46.2 kW)

Question 20 [6]

(a) [2]
Vertical height raised: $h = L - L\cos\theta = L(1 - \cos\theta)$
$h = 1.0 \times (1 - \cos 30^\circ) = 1.0 \times (1 - 0.8660) = 0.134 \text{ m}$

(b) [2]
Loss in GPE = Gain in KE at lowest point
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.134} = \sqrt{2.68} \approx 1.64 \text{ m/s}$

(c) [2]
Energy conversions:
Gravitational potential energy → Kinetic energy (at bottom) → Kinetic energy + Gravitational potential energy (rising other side) → ... repeated conversions ...
Eventually all mechanical energy is dissipated as heat energy (and sound) due to air resistance and friction at the pivot.
Why it stops: The system is not closed; external resistive forces (air resistance, pivot friction) do negative work, converting mechanical energy to thermal energy until the bob has no kinetic energy left.

Marking: 1 mark for describing conversions (GPE ↔ KE cyclically, eventually to heat/sound), 1 mark for identifying air resistance/pivot friction as cause of stopping.