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

TuitionGoWhere Practice Paper - Science Secondary 1

SA2 Practice Paper Version 2 of 5 — Answer Key

Subject: Science
Level: Secondary 1 (G3)
Total Marks: 60 marks

SECTION A: Multiple Choice

Section A Total: 8 marks

SECTION B: Structured Response

9. [Total: 3 marks]

(a) Distance = 50 − 30 = 20 cm (or 0.20 m) [1]

The perpendicular distance is measured from the pivot (50 cm mark) to the line of action of the 2 N weight at the 30 cm mark.

(b) The principle of moments (or: the sum of clockwise moments equals the sum of anticlockwise moments when an object is in equilibrium) [1]

(c) By principle of moments: Clockwise moment = Anticlockwise moment [1]

$2 \text{ N} \times 20 \text{ cm} = 4 \text{ N} \times d$

$d = \frac{2 \times 20}{4} = 10 \text{ cm}$ from the pivot

Position = 50 + 10 = 60 cm mark

Common error: Placing at 40 cm instead of 60 cm (forgetting which side of pivot).

10. [Total: 4 marks]

(a) Kinetic energy = $\frac{1}{2}mv^2$ [1 for formula + substitution]

$= \frac{1}{2} \times 1200 \text{ kg} \times (20 \text{ m/s})^2$

$= \frac{1}{2} \times 1200 \times 400$

$= \mathbf{240\,000 \text{ J}}$ (or 240 kJ) [1]

(b) Kinetic energy → Thermal energy (heat) in the brakes and surroundings [1]

The friction in the brake pads converts the kinetic energy of the car into heat, dissipating it to the environment.

(c) By work-energy theorem, work done by braking force = change in kinetic energy = 240 000 J (or 240 kJ) [1]

The braking force does negative work, removing all the kinetic energy the car had.

11. [Total: 4 marks]

(a) By principle of moments: [1 for correct equation]

Clockwise moment = Anticlockwise moment

$450 \text{ N} \times 40 \text{ cm} = E \times 120 \text{ cm}$

$E = \frac{450 \times 40}{120} = \frac{18\,000}{120} = \mathbf{150 \text{ N}}$ [1]

(b) Moving load closer to fulcrum (25 cm instead of 40 cm): [1 for identifying change]

The clockwise moment decreases because the perpendicular distance from the pivot decreases ($450 \times 25 = 11\,250$ N cm, compared to $450 \times 40 = 18\,000$ N cm). [1]

Since clockwise moment is smaller, a smaller anticlockwise moment (hence smaller effort) is needed to balance it. The effort needed decreases.

12. [Total: 4 marks]

(a) Useful work done = increase in gravitational potential energy = $mgh$ [1]

$= 50 \text{ kg} \times 10 \text{ N/kg} \times 3 \text{ m}$

$= \mathbf{1500 \text{ J}}$ [1]

(b) Efficiency = $\frac{\text{useful work output}}{\text{total energy input}} \times 100\%$ [1]

$80\% = \frac{1500}{E_{input}} \times 100\%$

$E_{input} = \frac{1500 \times 100}{80} = \mathbf{1875 \text{ J}}$ [1]

Teaching note: The extra 375 J is lost mainly as heat due to friction and motor heating.

13. [Total: 4 marks]

(a) Independent variable: The surface area of the parachute [1]

**Dependent variable:** The time taken to fall a fixed distance [1]

(b) Any two from: [1 for both correct]

• Mass of the parachute/load
• Height from which the parachute is dropped
• Shape of the parachute (when open)
• Material of the parachute

(c) Air resistance (drag) acts upwards, opposing the motion of the parachute. [1]

As the parachute falls, air resistance increases with speed until it balances the weight, causing the parachute to reach a constant (terminal) velocity. A larger surface area creates greater air resistance, leading to a slower descent.

14. [Total: 3 marks]

(a) Horizontal component = $F \cos \theta = 24 \text{ N} \times \cos 30° = 24 \times 0.866 = \mathbf{20.8 \text{ N}}$ (or ~20.8 N, accept 20.4–21.0 N depending on rounding) [1]

(b) Since the block moves at constant velocity, the net force is zero (Newton's first law / equilibrium). [1]

The horizontal component of the pulling force is balanced by friction.

Therefore, frictional force = 20.8 N (same value as part (a)), acting opposite to the direction of motion. [1]

Section B Total: 22 marks

SECTION C: Extended Response

15. [Total: 5 marks]

(a) Gravitational potential energy → Kinetic energy → Electrical energy [2]

The water at height possesses gravitational potential energy. As it falls, this converts to kinetic energy of moving water. The moving water turns turbines, which transfer kinetic energy to generators that produce electrical energy. Accept: including sound/thermal as intermediate losses if mentioned correctly.

(b) Energy is lost to: [2 marks for any two valid points with explanation]

• Friction in the turbines and pipes (converted to thermal energy)
• Sound energy produced by moving machinery
• Heat generated in the generator due to electrical resistance
• Some water may not fall through the optimal path (turbine design limitations)

Marking: 1 mark per distinct energy loss mechanism, 1 mark for explaining it reduces useful electrical output.

(c) Increase the mass/volume of water flowing through the turbines per second (or: increase the flow rate) [1]

Power = energy/time = (mgh)/t. Without changing h, increasing mass flow rate (m/t) increases power output.

16. [Total: 5 marks]

(a) Graph marking points: [2]

• Correct axes with labels and units: Load (N) vs Extension (cm) [0.5]
• Correct scale and origin starting at (0,0) [0.5]
• All data points accurately plotted [0.5]
• Best-fit straight lines showing two distinct regions [0.5]

Expected: straight line through origin to (8, 4), then distinctly different gradient from (8,4) to (13, 6).

(b) From graph: load of 3.5 N falls in linear region. Extension ≈ 7.0 cm [1]

(Accept 6.8–7.2 cm depending on graph reading; proportional: 3.5/1 × 2.0 = 7.0 cm)

(c) Beyond 4 N (extension 8.0 cm), the spring shows non-linear behaviour [1]

The load-extension graph is no longer a straight line through the origin, indicating the spring has exceeded its limit of proportionality / elastic limit. [1]

For a reliable measuring device, we need a linear (proportional) relationship so that scale markings are evenly spaced. Above 4 N, equal changes in load do not produce equal changes in extension.

17. [Total: 5 marks]

(a) With 3 supporting strands, the load is supported by 3 sections of rope sharing the tension. [1]

The effort only needs to pull with 1/3 of the load force, but in return, the effort must move through 3 times the distance the load rises (distance sacrificed). Work input = Work output for ideal case, so smaller force requires greater distance. [1]

(b)(i) Ideal effort = $\frac{\text{Load}}{n} = \frac{600}{3} = \mathbf{200 \text{ N}}$ [1]

(b)(ii) Efficiency = $\frac{\text{Mechanical Advantage}}{\text{Velocity Ratio}} \times 100\%$ or: $\frac{\text{useful work}}{\text{total work}}$

Actual effort = $\frac{\text{Ideal effort}}{\text{efficiency}} = \frac{200}{0.75} = \mathbf{266.7 \text{ N}}$ (or 267 N) [1]

(c) Efficiency < 100% because: [2 marks for any two valid points]

• Friction in the pulley bearings requires extra effort work
• Weight of the moving pulley block must also be lifted
• Rope stiffness/stretch absorbs some energy
• Air resistance on moving parts (minor effect)

18. [Total: 5 marks]

(Scale diagram method or calculation method both acceptable)

(a) Scale diagram method: [3]

• Choose scale (e.g., 1 cm = 1000 N or 1 cm = 500 N) [0.5]
• Draw vectors to scale at correct angles: 40° east of north and 30° west of north [1]
• Complete parallelogram and draw diagonal, OR use head-to-tail method [1]
• Measure resultant: approximately 12,400 N (or 12 000–12 800 N acceptable due to measurement variation) [0.5]

Calculation method (resolving):

Component north from A: $8000 \cos 40° = 8000 \times 0.766 = 6128$ N Component east from A: $8000 \sin 40° = 8000 \times 0.643 = 5142$ N

Component north from B: $6000 \cos 30° = 6000 \times 0.866 = 5196$ N Component west from B: $6000 \sin 30° = 6000 \times 0.5 = 3000$ N

Total north: $6128 + 5196 = 11,324$ N Net east: $5142 - 3000 = 2142$ N (east)

Resultant = $\sqrt{11,324^2 + 2142^2} = \sqrt{1.283 \times 10^8 + 4.59 \times 10^6} = \sqrt{1.329 \times 10^8} \approx \mathbf{11,530 \text{ N}}$

Accept answers in range 11,000–13,000 N depending on method and rounding.

(b) Direction: $\tan \theta = \frac{2142}{11,324} = 0.189$, so $\theta \approx \mathbf{10.7°}$ east of north (accept 10°–12°) [1]

(c) Assumption: The two forces act in the same vertical plane / simultaneously / from the same point [1]

Alternative acceptable answers: No other forces act on the barge; water resistance is negligible; the barge moves slowly so acceleration effects are negligible.

19. [Total: 5 marks]

(a) GPE = $mgh$ [1 for formula + substitution]

$= 70 \text{ kg} \times 10 \text{ N/kg} \times 4 \text{ m}$

$= \mathbf{2800 \text{ J}}$ [1]

(b) Work against component of weight = force × distance [1]

The component of weight parallel to slope = $mg \sin \theta = 70 \times 10 \times \frac{4}{50} = 70 \times 10 \times 0.08 = 56$ N

OR: Work = GPE gained = 2800 J (since no acceleration, and assuming no friction) — but careful: this equals total work against gravity, not just the pedalling against the component.

Actually: if friction exists, work done against weight component = $56 \times 50 = 2800$ J [1]

However, this equals the GPE gain. The cyclist does 2800 J of work just to overcome gravity.

(c) The total work done by the cyclist is greater because: [1]

The cyclist must also do work against frictional forces (air resistance, rolling resistance) in addition to overcoming the component of weight. The pedalling force (120 N over 50 m = 6000 J) exceeds the minimum needed for gravity alone, indicating significant resistance forces.

20. [Total: 5 marks]

(a) Solar energy is renewable because: [1]

The Sun's energy is continuously replenished and will last for billions of years. It is not depleted by human use, unlike fossil fuels which take millions of years to form.

(b) Two design features for efficiency: [2]

• Variable speed compressors that adjust cooling output to match demand, rather than running at full power continuously [1]
• Heat recovery systems that reuse waste heat for water heating [1]
• Better insulation of cooled spaces to reduce heat gain from outside [1]
• Smart thermostats that optimize cooling schedules [1]

(Any two valid, distinct points accepted)

(c) Energy saved per day: [2]

Old: $E = Pt = 500 \text{ W} \times 8 \text{ h} = 4000 \text{ Wh} = 4 \text{ kWh}$

New: $E = 300 \text{ W} \times 8 \text{ h} = 2400 \text{ Wh} = 2.4 \text{ kWh}$

Daily saving: $4 - 2.4 = 1.6 \text{ kWh}$ [1]

Annual saving: $1.6 \times 365 = \mathbf{584 \text{ kWh}}$ [1]

Alternative: Power difference = 200 W = 0.2 kW. Annual saving = 0.2 kW × 8 h × 365 = 584 kWh.

Section C Total: 30 marks

GRAND TOTAL: 60 MARKS