Secondary 1 Science Semestral Assessment 2 (End of Year) Paper 2

TuitionGoWhere Practice Paper - Science Secondary 1

Answer Key and Marking Scheme (SA2 Version 2)

Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

Question 1 [1]

Answer: C

Working:

• Mass $m = 2 \text{ kg}$
• Height $h = 1.5 \text{ m}$
• $g = 10 \text{ N/kg}$
• Work done against gravity $= mgh = 2 \times 10 \times 1.5 = 30 \text{ J}$

Key concept: Work done against gravity = gain in gravitational potential energy = $mgh$.

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Question 2 [1]

Answer: B

Explanation: In a battery-powered toy car:

1. Chemical energy stored in the battery
2. Converted to electrical energy in the circuit
3. Electrical energy converted to kinetic energy (and some heat/sound) by the motor

Common mistake: Option A omits the electrical energy intermediate step.

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Question 3 [1]

Answer: B

Working:

• Applied force $F = 20 \text{ N}$
• Frictional force $f = 8 \text{ N}$
• Net force $F_{\text{net}} = 20 - 8 = 12 \text{ N}$
• Displacement $s = 4 \text{ m}$
• Net work done $= F_{\text{net}} \times s = 12 \times 4 = 48 \text{ J}$

Alternative method: Work by applied force $= 20 \times 4 = 80 \text{ J}$; Work against friction $= 8 \times 4 = 32 \text{ J}$; Net work $= 80 - 32 = 48 \text{ J}$.

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Question 4 [1]

Answer: A

Explanation: At position X (highest point), the bob has maximum gravitational potential energy and zero kinetic energy (released from rest). At position Y (lowest point), gravitational potential energy is minimum and kinetic energy is maximum. By conservation of energy (ignoring air resistance), GPE at X $\rightarrow$ KE at Y.

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Question 5 [1]

Answer: B

Working:

• Mass $m = 500 \text{ g} = 0.5 \text{ kg}$
• Speed $v = 10 \text{ m/s}$
• Kinetic energy $= \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times 10^2 = 0.25 \times 100 = 25 \text{ J}$

Common mistake: Forgetting to convert grams to kg (would give 25000 J) or forgetting the $\frac{1}{2}$ factor (would give 50 J).

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Question 6 [1]

Answer: C

Explanation: Solar energy is renewable (continuously replenished by the Sun). Coal, natural gas, and uranium for nuclear fission are finite fossil/nuclear fuels — non-renewable.

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Question 7 [1]

Answer: C

Working:

• Efficiency $= \frac{\text{Useful work output}}{\text{Work input}} \times 100\%$
• $80\% = \frac{400}{\text{Work input}} \times 100\%$
• Work input $= \frac{400}{0.8} = 500 \text{ J}$

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Question 8 [1]

Answer: A

Working:

• Hooke's Law: $F = kx$
• $10 = 50 \times x$
• $x = \frac{10}{50} = 0.2 \text{ m}$

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Question 9 [1]

Answer: C

Explanation: When a car brakes, friction between brake pads and discs/drums, and between tyres and road, converts kinetic energy primarily into heat energy. Some sound energy is also produced (squealing brakes, tyre noise).

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Question 10 [1]

Answer: B

Working:

• Power $= \frac{\text{Work done}}{\text{Time taken}} = \frac{200 \text{ J}}{10 \text{ s}} = 20 \text{ W}$

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Section B: Structured Questions [30 marks]

Question 11 [4]

(a) [1] Weight $W = mg = 500 \times 10 = 5000 \text{ N}$ (or $5 \text{ kN}$)

(b) [2] Work done by crane $= \text{Force} \times \text{distance} = \text{Tension} \times \text{height}$ Since constant velocity, Tension $= \text{Weight} = 5000 \text{ N}$ Work done $= 5000 \times 12 = 60000 \text{ J}$ (or $60 \text{ kJ}$)

Mark breakdown: 1 mark for correct tension/force, 1 mark for correct calculation with units.

(c) [1] Chemical energy (in fuel/electricity) $\rightarrow$ Gravitational potential energy (of block) Accept: Electrical energy $\rightarrow$ Gravitational potential energy (if electric crane)

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Question 12 [5]

(a) [1] GPE at A $= mgh_A = 400 \times 10 \times 30 = 120000 \text{ J}$ (or $120 \text{ kJ}$)

(b) [2] At point B (ground level), $h = 0$, so GPE $= 0$. By conservation of energy (frictionless track), total energy at A = total energy at B. GPE at A $=$ KE at B $= 120000 \text{ J}$. Mark breakdown: 1 mark for correct value (120000 J), 1 mark for explanation referencing conservation of energy / frictionless track.

(c) [2] At point C: Total energy $=$ GPE at C $+$ KE at C $120000 = mgh_C + \text{KE}_C$ $120000 = (400 \times 10 \times 20) + \text{KE}_C$ $120000 = 80000 + \text{KE}_C$ $\text{KE}_C = 40000 \text{ J}$

$\text{KE}_C = \frac{1}{2}mv_C^2$ $40000 = \frac{1}{2} \times 400 \times v_C^2$ $40000 = 200 \times v_C^2$ $v_C^2 = 200$ $v_C = \sqrt{200} = 10\sqrt{2} \approx 14.1 \text{ m/s}$

Mark breakdown: 1 mark for correct KE at C (40000 J), 1 mark for correct speed calculation with units.

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Question 13 [4]

(a) [1] Net force $= \text{Applied force} - \text{Friction} = 30 - 10 = 20 \text{ N}$ (forward)

(b) [1] $F_{\text{net}} = ma$ $20 = 5 \times a$ $a = 4 \text{ m/s}^2$

(c) [1] Work done by applied force $= F \times s = 30 \times 6 = 180 \text{ J}$

(d) [1] Power $= \frac{\text{Work done}}{\text{Time}} = \frac{180}{4} = 45 \text{ W}$

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Question 14 [5]

(a) [1] GPE at P relative to Q $= mgh = 0.5 \times 10 \times 0.2 = 1 \text{ J}$

(b) [1] At Q (lowest point), all GPE is converted to KE (assuming no air resistance). KE at Q $= 1 \text{ J}$. Explanation: By conservation of energy, loss in GPE = gain in KE.

(c) [2] $\text{KE} = \frac{1}{2}mv^2$ $1 = \frac{1}{2} \times 0.5 \times v^2$ $1 = 0.25 v^2$ $v^2 = 4$ $v = 2 \text{ m/s}$

Mark breakdown: 1 mark for correct substitution/formula, 1 mark for correct answer with units.

(d) [1] The initial gravitational potential energy is converted to kinetic energy, then gradually dissipated as heat and sound energy due to air resistance and friction at the pivot, until the bob comes to rest at Q.

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Question 15 [6]

(a) [1] $a = \frac{v - u}{t} = \frac{20 - 0}{10} = 2 \text{ m/s}^2$

(b) [1] $F = ma = 1200 \times 2 = 2400 \text{ N}$

(c) [1] $\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 20^2 = 600 \times 400 = 240000 \text{ J}$ (or $240 \text{ kJ}$)

(d) [1] Work done by resultant force $=$ Gain in kinetic energy $= 240000 \text{ J}$ (Work-energy theorem: net work = change in KE)

(e) [2] Average power $= \frac{\text{Work done}}{\text{Time}} = \frac{240000}{10} = 24000 \text{ W}$ (or $24 \text{ kW}$)

Mark breakdown: 1 mark for correct work done (or recognition that work done = KE gain), 1 mark for correct power calculation with units.

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Question 16 [6]

(a) [2] Mass of water per second $= 2000 \text{ kg}$ Height $h = 50 \text{ m}$ GPE lost per second $= mgh = 2000 \times 10 \times 50 = 1000000 \text{ J/s} = 1 \text{ MW}$

Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with units (J/s or W).

(b) [2] Efficiency $= 85\% = 0.85$ Electrical power output $= 0.85 \times 1000000 = 850000 \text{ W} = 850 \text{ kW}$

Mark breakdown: 1 mark for correct efficiency calculation, 1 mark for correct answer with units.

(c) [2] Gravitational potential energy of water $\rightarrow$ Kinetic energy of falling water $\rightarrow$ Kinetic energy of turbine rotation $\rightarrow$ Electrical energy (via generator)

Mark breakdown: 1 mark for GPE $\rightarrow$ KE of water, 1 mark for KE of turbine $\rightarrow$ Electrical energy (or complete chain with all steps).

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

Question 17 [7]

(a) [2] Marking points for graph:

• Axes correctly labelled with units (Drop height / cm on x-axis, Bounce height / cm on y-axis) [1]
• All 5 points plotted correctly [1]
• Best-fit straight line through origin [1] Total 2 marks (typically 1 for plotting, 1 for line; or 1 for axes+points, 1 for line)

(b) [1] Bounce height is directly proportional to drop height. (Or: Bounce height increases linearly with drop height; the ratio bounce height/drop height is constant at 0.6).

(c) [2] Drop height $= 100 \text{ cm} = 1 \text{ m}$ Bounce height $= 60 \text{ cm} = 0.6 \text{ m}$ Mass $= 50 \text{ g} = 0.05 \text{ kg}$

Initial GPE $= mgh_{\text{drop}} = 0.05 \times 10 \times 1 = 0.5 \text{ J}$ GPE after bounce $= mgh_{\text{bounce}} = 0.05 \times 10 \times 0.6 = 0.3 \text{ J}$

Percentage retained $= \frac{0.3}{0.5} \times 100\% = 60\%$

Alternative (simpler): Since GPE $\propto h$, percentage retained $= \frac{h_{\text{bounce}}}{h_{\text{drop}}} \times 100\% = \frac{60}{100} \times 100\% = 60\%$.

Mark breakdown: 1 mark for correct method (ratio of heights or GPE calculation), 1 mark for correct answer (60%).

(d) [1] During the bounce, some energy is converted to heat and sound energy due to deformation of the ball and floor, and air resistance. This energy is not recovered, so the ball has less kinetic energy after the bounce, resulting in a lower maximum height.

(e) [1] The prediction assumes the linear relationship (60% retention) holds at greater heights. However, at higher drop heights:

• Air resistance becomes more significant (proportional to $v^2$)
• The ball may deform more, increasing energy loss
• The percentage retention may decrease So the actual bounce height would likely be less than 120 cm.

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Question 18 [7]

(a) [2] Elastic potential energy $= \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.15)^2 = 100 \times 0.0225 = 2.25 \text{ J}$

Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with units.

(b) [2] Assuming no energy losses: Elastic PE $\rightarrow$ GPE $\frac{1}{2}kx^2 = mgh$ $2.25 = 0.1 \times 10 \times h$ $2.25 = h$ $h = 2.25 \text{ m}$

Mark breakdown: 1 mark for energy conservation equation, 1 mark for correct answer with units.

(c) [2] Vertical height $h = 2.25 \text{ m}$ Incline angle $\theta = 30^\circ$ Distance along ramp $d = \frac{h}{\sin\theta} = \frac{2.25}{\sin 30^\circ} = \frac{2.25}{0.5} = 4.5 \text{ m}$

Mark breakdown: 1 mark for correct trigonometric relationship, 1 mark for correct answer with units.

(d) [1] Friction would do negative work on the block, converting some mechanical energy to heat. This reduces the kinetic energy available for conversion to GPE, so the maximum vertical height reached would be less than 2.25 m.

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Question 19 [6]

(a) [1] Swept area $A = \pi r^2 = \pi \times 25^2 = 625\pi \approx 1963.5 \text{ m}^2$ (accept $1960 \text{ m}^2$ or $625\pi \text{ m}^2$)

(b) [2] $P_{\text{wind}} = \frac{1}{2} \rho A v^3$ $= \frac{1}{2} \times 1.2 \times 1963.5 \times 12^3$ $= 0.6 \times 1963.5 \times 1728$ $= 0.6 \times 3392928$ $= 2035756.8 \text{ W} \approx 2.04 \text{ MW}$

Using $A = 625\pi$: $P_{\text{wind}} = \frac{1}{2} \times 1.2 \times 625\pi \times 1728 = 648000\pi \approx 2.036 \text{ MW}$

Mark breakdown: 1 mark for correct substitution, 1 mark for correct calculation with units (W or MW).

(c) [1] Efficiency $= 40\% = 0.4$ Electrical power output $= 0.4 \times 2.04 \text{ MW} = 0.816 \text{ MW} = 816 \text{ kW}$

(d) [2] Any two of:

• Wind speed (power $\propto v^3$, so small changes in wind speed cause large changes in power)
• Air density (varies with temperature, altitude, humidity)
• Swept area / blade length (larger blades capture more air)
• Turbine efficiency / design (aerodynamic efficiency, generator efficiency)
• Wind direction relative to turbine orientation

Mark breakdown: 1 mark per valid factor (max 2).

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Question 20 [6]

(a) [1] Energy per 100 g $= 2000 \text{ kJ} = 2000000 \text{ J}$ Energy per 50 g $= \frac{50}{100} \times 2000000 = 1000000 \text{ J}$ (or $1000 \text{ kJ}$)

(b) [3] Useful energy for climbing $= 25\% \times 1000000 = 250000 \text{ J}$ This equals gain in GPE: $mgh = 250000$ $50 \times 10 \times h = 250000$ $500h = 250000$ $h = 500 \text{ m}$

Mark breakdown: 1 mark for calculating useful energy (250000 J), 1 mark for equating to $mgh$, 1 mark for correct height with units.

(c) [2] Any two of:

• The human body also uses energy for basal metabolic processes (breathing, heartbeat, maintaining body temperature), not just climbing.
• Muscles are not 100% efficient even within the 25% figure; the 25% is an average/maximum under ideal conditions.
• Energy is lost as heat during muscle contraction.
• The student would need energy to descend as well (controlled lowering requires muscle work).
• Fatigue and physiological limits prevent sustained maximum efficiency.
• Some energy from food is not fully digested/absorbed.

Mark breakdown: 1 mark per valid reason (max 2).

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End of Answer Key