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

TuitionGoWhere Exam Practice (AI)

Secondary 1 Science – SA2 Physical Sciences

Answer Key and Marking Scheme Version 1 of 5

Total Marks: 60

---

Section A: Multiple Choice (Questions 1–8)

2 marks each

---

1. Answer: B

Working/Explanation:

• When pulling a box at constant velocity, the student uses chemical energy from food/muscles.
• Some energy goes into kinetic energy of the moving box.
• Friction between the box and floor converts some energy to thermal energy (heat).
• Since velocity is constant, kinetic energy is constant, but energy must continuously be supplied to overcome friction.

Common mistake: Choosing A (forgets kinetic energy) or C (confuses with falling objects).

Marks: 2

---

2. Answer: A

Working/Explanation:

• Work done = Force × Distance moved in the direction of the force
• $W = F \times d$
• If the object does not move, $d = 0$
• Therefore $W = 15 \times 0 = 0$ J

Key concept: No work is done if there is no displacement, no matter how large the force.

Common mistake: Choosing C (confusing work with power, which involves time).

Marks: 2

---

3. Answer: D

Working/Explanation:

• Definition: Pressure = $\frac{\text{Force}}{\text{Area}}$ (force per unit area)
• Pressure in fluids acts equally in all directions — this is a key property
• Pressure is not a force (it's force divided by area)
• Pressure is a scalar quantity — it does not have a specific direction

Common mistake: Choosing B (confusing pressure with force, which is a vector).

Marks: 2

---

4. Answer: B

Working/Explanation:

• Pascal's principle states that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of the container.
• From $p = \frac{F}{A}$, if pressure is constant: $\frac{F_A}{A_A} = \frac{F_B}{A_B}$
• Since $A_B > A_A$, we get $F_B > F_A$ — force is multiplied.
• With values: $p = \frac{100}{5} = 20$ N/cm², so $F_B = 20 \times 25 = 500$ N

Common mistake: Choosing A (energy is conserved, but the principle specifically explaining force multiplication is Pascal's).

Marks: 2

---

5. Answer: B

Working/Explanation:

• Density = $\frac{\text{Mass}}{\text{Volume}}$
• $\rho = \frac{m}{V} = \frac{270 \text{ g}}{100 \text{ cm}^3}$
• $\rho = 2.7$ g/cm³

Note: This is the density of aluminium, a common reference value.

Common mistake: Unit errors or decimal place errors (0.37 from dividing upside down).

Marks: 2

---

6. Answer: B

Working/Explanation:

• Pressure = $\frac{\text{Weight}}{\text{Area}} = \frac{mg}{A}$
• A: $P = \frac{50 \times 10}{400} = \frac{500}{400} = 1.25$ N/cm²
• B: $P = \frac{50 \times 10}{200} = \frac{500}{200} = 2.50$ N/cm²
• C: $P = \frac{40 \times 10}{300} = \frac{400}{300} = 1.33$ N/cm²
• D: $P = \frac{60 \times 10}{3600} = \frac{600}{3600} = 0.167$ N/cm²

Highest pressure is B (same weight, smallest area).

Key insight: Pressure increases when area decreases, even with same force.

Common mistake: Choosing D (confusing larger mass with larger pressure, ignoring the much larger area).

Marks: 2

---

7. Answer: C

Working/Explanation:

• Deceleration means decreasing speed — the graph shows a negative gradient
• O to P: Speed increases from 0 to 4 m/s (acceleration)
• P to Q: Speed increases from 4 to 10 m/s (acceleration, constant rate)
• Q to (45s): Speed constant at 10 m/s (zero acceleration)
• (45s) to R: Speed decreases from 10 to 0 m/s (deceleration)

The instruction "Between which two points" refers to the labelled points, and between Q and R encompasses the deceleration phase.

Common mistake: Choosing A (confusing initial slowing with the later deceleration, or misreading curved initial acceleration as deceleration).

Marks: 2

---

8. Answer: A

Working/Explanation:

• Principle of moments: Clockwise moment = Anticlockwise moment (for equilibrium)
• Load × Load arm = Effort × Effort arm
• $800 \times 0.5 = F \times 2.0$
• $400 = 2.0F$
• $F = 200$ N

Key insight: The longer effort arm provides mechanical advantage — less force needed but over greater distance.

Common mistake: Choosing D (multiplying instead of dividing, or swapping arms).

Marks: 2

---

Section A Total: 16 marks

---

Section B: Short Answer and Calculation (Questions 9–16)

---

9. (a) Work done is defined as the product of the force applied on an object and the distance moved by the object in the direction of the force. [1 mark]

(Accept: Force multiplied by distance in the direction of the force. Accept equation form: $W = F \times d$ with both terms defined.)

---

9. (b)

Working:

• Weight of student: $W = mg = 55 \times 10 = 550$ N
• Work done against gravity: $E_p = mgh = 550 \times 3.0$
• $E_p = 1650$ J

Alternative: $E_p = mgh = 55 \times 10 \times 3.0 = 1650$ J

Mark breakdown:

• Correct formula stated: 1 mark
• Correct substitution and answer with unit: 1 mark

Answer: 1650 J (or 1650 J/work done = 1650 J) [2 marks]

Total for Q9: 3 marks

---

10. (a) Two factors affecting pressure of a solid:

1. The magnitude of the force applied (or weight of the object) [1 mark]
2. The area of contact between the object and the surface [1 mark]

(Accept: Mass of object; surface area in contact. Do not accept "size" without specification.)

---

10. (b)

Explanation:

• A snowshoe has a large surface area compared to normal shoes. [1 mark]
• Since pressure = $\frac{\text{force}}{\text{area}}$, increasing the area (for the same weight/force) decreases the pressure on the snow. [1 mark]
• Lower pressure means the snowshoes do not compress the snow as much, so the person stays on top rather than sinking in.

Key concept: Inversely proportional relationship between pressure and area; same force spread over larger area.

Total for Q10: 4 marks

---

11. (a)

The displacement method is suitable because the stone has an irregular shape, so its volume cannot be calculated using regular formulas (like $V = l \times w \times h$ for a rectangular block). The volume of water displaced equals the volume of the stone regardless of its shape. [1 mark]

---

11. (b)

Working:

• Volume of stone = Final volume − Initial volume = $82 - 50 = 32$ cm³ [1 mark]
• Formula: $\rho = \frac{m}{V}$ [1 mark]
• $\rho = \frac{96}{32} = 3.0$ g/cm³ [1 mark]

Mark breakdown:

• Correct volume calculation: 1 mark
• Correct formula stated: 1 mark
• Correct substitution and final answer with unit: 1 mark

Answer: 3.0 g/cm³ (accept 3 g/cm³) [3 marks]

Total for Q11: 4 marks

---

12. (a)

Working:

• Formula: $KE = \frac{1}{2}mv^2$ [1 mark]
• $KE = \frac{1}{2} \times 1200 \times (20)^2$
• $KE = \frac{1}{2} \times 1200 \times 400$
• $KE = 600 \times 400 = 240\,000$ J [1 mark]

Mark breakdown:

• Correct formula: 1 mark
• Correct substitution and answer with unit: 1 mark

Answer: 240 000 J (or 2.4 × 10⁵ J, or 240 kJ) [2 marks]

---

12. (b)

The kinetic energy is converted to thermal energy (heat) in the brakes and surrounding air. [1 mark]

(Accept: Heat energy, internal energy. Do not accept "lost" without stating where it goes.)

---

12. (c)

Explanation:

• When brakes are applied, friction acts between the brake pads and the wheel discs/drums. [1 mark]
• Work is done against friction, and this work converts kinetic energy to thermal energy in the brake materials. [1 mark]
• With repeated braking, more and more kinetic energy is converted to thermal energy, causing the temperature of the brakes to rise significantly.

Key concept: Friction converts mechanical energy to thermal energy; temperature is a measure of average kinetic energy of particles.

Total for Q12: 5 marks

---

13. (a)

One advantage: The pulley system allows the load to be lifted with less effort force than lifting directly (mechanical advantage > 1), OR it allows the operator to pull downward while the load moves upward (more convenient direction). [1 mark]

---

13. (b)

Working:

• Velocity ratio (VR) = $\frac{\text{Distance moved by effort}}{\text{Distance moved by load}}$
• VR = $\frac{4.0}{2.0}$ [1 mark]
• VR = 2 [1 mark]

Answer: 2 [2 marks]

---

13. (c)

Working:

• Mechanical Advantage (MA) = $\frac{\text{Load}}{\text{Effort}} = \frac{300}{180} = \frac{5}{3} \approx 1.67$ [1 mark]
• Efficiency = $\frac{\text{MA}}{\text{VR}} \times 100\%$ [1 mark]
• Efficiency = $\frac{1.67}{2} \times 100\% = 83.3\%$ or exactly $\frac{5/3}{2} \times 100\% = \frac{5}{6} \times 100\% = 83.3\%$

Or using work: Efficiency = $\frac{\text{Useful work output}}{\text{Total work input}} \times 100\%$

• Useful work = $300 \times 2.0 = 600$ J
• Input work = $180 \times 4.0 = 720$ J
• Efficiency = $\frac{600}{720} \times 100\% = 83.3\%$ [1 mark]

Answer: 83.3% (accept 83%, or exactly 250/3 %, or fraction 5/6)

Mark breakdown:

• Correct MA or work values: 1 mark
• Correct efficiency formula: 1 mark
• Correct final answer: 1 mark

Total for Q13: 6 marks

---

14. (a)

• Melting point: 0°C [1 mark]
• Boiling point: 100°C [1 mark]

(These match the plateau temperatures on the graph.)

---

14. (b)

Explanation:

• During melting, the heat energy supplied is used to overcome the forces of attraction between the particles in the solid. [1 mark]
• This energy (called latent heat of fusion) does not increase the kinetic energy of the particles. [1 mark]
• Since temperature is a measure of the average kinetic energy of the particles, the temperature remains constant during the phase change even though energy is being added. [1 mark]

Key concept: Latent heat changes potential energy of particle arrangement, not kinetic energy; temperature ∝ average kinetic energy.

Total for Q14: 5 marks

---

15.

Method	Significance	Explanation
Conduction	Significant through metal body of kettle	Metal is a good conductor; thermal energy passes through metal walls [1 mark + 1 mark for explanation of metal conduction]
Convection	Not significant (or minimal) in handle	Hot water at bottom heats up, rises, but plastic handle is at top and plastic is poor conductor [1 mark]
Radiation	Minimal/significant from hot surfaces	Infrared radiation from hot metal, but distance and plastic material reduce effect [1 mark]

Expected answer structure:

Heat transfer 1: Conduction

• Thermal energy is conducted through the metal body of the kettle from the hot water. [1 mark]
• Metal is a good conductor of heat, so conduction is efficient through the metal. However, the plastic handle is a poor conductor/insulator, so very little heat reaches your hand by conduction through the handle. [1 mark]

Heat transfer 2: Convection

• Convection currents in the air may carry hot air upward, but this is not significant in heating the handle because: plastic is a poor conductor and does not heat up easily from the air; the handle is positioned away from the main convection current path. [1 mark + explanation]

OR Heat transfer 2: Radiation

• The hot kettle emits infrared radiation, but this is relatively weak compared to conduction at these temperatures, and the plastic handle absorbs little radiation. [1 mark + explanation]

Total for Q15: 4 marks

---

16. (a)

The principle of moments states that for a body in equilibrium, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point. [2 marks]

(Accept: Clockwise moment = Anticlockwise moment for equilibrium. Award 1 mark for partial statement mentioning moments and equilibrium without specifying direction.)

---

16. (b)

Working:

• Let the 0.30 kg mass be placed at distance $d$ from the fulcrum (50 cm mark)
• Clockwise moment (from 0.40 kg at 20 cm): The mass is 30 cm left of fulcrum
  • Moment = $0.40 \times 10 \times (50-20) = 4.0 \times 30 = 120$ Ncm (anticlockwise, actually)

Actually, taking moments properly:

• 0.40 kg at 20 cm mark: distance from fulcrum = 30 cm to the left

• This creates an anticlockwise moment (tends to lift right side)

• Need clockwise moment from 0.30 kg to balance

• Anticlockwise moment = $0.40 \times 10 \times 30 = 120$ Ncm [1 mark]

• For equilibrium: Clockwise moment = Anticlockwise moment

• $0.30 \times 10 \times d = 120$ [1 mark]

• $3.0 \times d = 120$

• $d = 40$ cm [1 mark]

• Position from 0 cm mark: $50 + 40 = 90$ cm mark

Answer: 90 cm mark (or 40 cm to the right of the fulcrum) [3 marks]

Mark breakdown:

• Correct moment calculation for known mass: 1 mark
• Setting up equation with unknown distance: 1 mark
• Correct answer with position stated: 1 mark

Total for Q16: 5 marks

---

Section B Total: 32 marks

---

Section C: Structured Response (Questions 17–20)

---

17. (a)

Variable	Answer
Independent variable	Surface area of the parachute canopy / Size of parachute (A, B, or C) [1 mark]
Dependent variable	Time of fall / Time taken to reach the ground [1 mark]
Controlled variable (any one)	Height of drop; mass of load; material of parachute; length of suspension lines; same drop position/environmental conditions (no wind) [1 mark]

---

17. (b)

Explanation:

• A larger parachute has a greater surface area exposed to the air. [1 mark]
• This creates more air resistance (drag force) opposing the downward motion, which slows the descent. [1 mark]
• With greater air resistance, the parachute reaches a lower terminal velocity, so it takes longer to fall the same distance.

Total for Q17: 5 marks

---

18. (a)

Working:

• Gravitational potential energy lost: $GPE = mgh$ [1 mark]
• $GPE = 5000 \times 10 \times 80$
• $GPE = 5000 \times 800$
• $GPE = 4\,000\,000$ J or $4.0 \times 10^{6}$ J [1 mark]

Answer: 4 000 000 J (or 4.0 × 10⁶ J, or 4000 kJ, or 4 MJ) [2 marks]

---

18. (b)

Working:

• Power = $\frac{\text{Energy converted}}{\text{Time}}$ = $\frac{4\,000\,000}{1} = 4\,000\,000$ W [1 mark]

(Per second means time = 1 s)

Answer: 4 000 000 W (or 4.0 × 10⁶ W, or 4000 kW, or 4 MW) [1 mark]

---

18. (c)

• Form of energy: Thermal energy / Heat energy / Sound energy [1 mark]
• Explanation: Friction in the turbine bearings and water turbulence convert some mechanical energy to heat. OR Some water splashes and does not hit the turbine blades effectively. OR Energy is lost as sound from moving machinery. [1 mark]

Total for Q18: 5 marks

---

19. (a)

Working:

• Total mass = mass of bricks + mass of wheelbarrow = $50 + 10 = 60$ kg [1 mark]
• Total weight (load force) = $mg = 60 \times 10 = 600$ N [1 mark]

Answer: 600 N [2 marks]

---

19. (b)

Working:

• Taking moments about the wheel (fulcrum):
• The centre of gravity of the wheelbarrow is typically at its geometric centre. Assuming uniform wheelbarrow, CG of empty wheelbarrow is at 0.60 m (midpoint of 1.20 m, but this needs clarification).

Actually, using the diagram values: load includes both bricks and wheelbarrow weight. We need to find the effective centre of gravity or treat separately.

Better approach:

• Moment of bricks (clockwise about wheel): $50 \times 10 \times 0.40 = 500 \times 0.40 = 200$ Nm
• Moment of wheelbarrow weight (clockwise about wheel): $10 \times 10 \times 0.60 = 100 \times 0.60 = 60$ Nm (assuming CG at 0.60 m, the midpoint)
• Total clockwise moment = $200 + 60 = 260$ Nm

But the question simplifies: "total load force" in (a) suggests treating as combined 600 N. If the load acts at effective distance:

• For combined load: need effective position, or separate moments.

Using separate moments (more accurate):

• Total clockwise moment = $(50 \times 10 \times 0.40) + (10 \times 10 \times 0.60) = 200 + 60 = 260$ Nm [1 mark]
• Anticlockwise moment from effort: $F \times 1.20$ [1 mark]
• For equilibrium: $F \times 1.20 = 260$
• $F = \frac{260}{1.20} = 216.67$ N ≈ 217 N or about 220 N

However, if we simplify and assume the question intends the 600 N to act at the brick position (0.40 m) for approximate calculation:

• $600 \times 0.40 = F \times 1.20$
• $240 = 1.20F$
• $F = 200$ N

Given the marking scheme flexibility, both approaches accepted with appropriate working. The intended simpler answer is likely 200 N treating combined load at brick position, or 217 N with separate moments.

Mark breakdown (for 200 N approach):

• Correct moment equation set up: 1 mark
• Correct substitution: 1 mark
• Correct answer: 1 mark

Mark breakdown (for accurate 217 N):

• Correct identification of both moments: 1 mark
• Correct equation and substitution: 1 mark
• Correct answer: 1 mark

Answer: 200 N (simplified) or 217 N/220 N (with separate moments) [3 marks]

Total for Q19: 5 marks

---

20. (a)

Two possible observations:

1. Measure the time taken for each wax ball to melt and fall off — the metal with the shortest time has the highest thermal conductivity. [1 mark]
2. Measure the temperature at a fixed point along each rod (e.g., 10 cm from heat source) after a fixed time — the rod with the highest temperature has the highest thermal conductivity. [1 mark]

(Accept: Observe which wax ball drops first; use a timer to measure melting time; use temperature sensors/clinical thermometers with dab of wax at equal positions.)

---

20. (b)

Order of thermal conductivity (highest to lowest): Copper > Aluminium > Iron [1 mark]

---

20. (c)

Importance of same diameter:

• The cross-sectional area affects the rate of heat flow. If diameters differ, the thicker rod conducts more heat simply due to larger area, regardless of material conductivity. Same diameter ensures fair comparison of material properties only. [1 mark]

Importance of same starting temperature:

• If rods start at different temperatures, one may reach melting point faster simply because it started warmer, not because it conducts better. Same starting temperature ensures that any difference in melting time is due to thermal conductivity, not initial conditions. [1 mark]

Total for Q20: 5 marks

---

Section C Total: 12 marks

---

GRAND TOTAL: 60 marks

---

Summary of Marks Distribution

Section	Question Range	Marks	Question Types
A	1–8	16	Multiple choice
B	9–16	32	Short answer, calculation, explanation
C	17–20	12	Extended response, experimental design, application

---

Cognitive Demand Analysis

Level	Questions	Approximate Proportion
Recall/Knowledge	Q9a, Q10a, Q14a, Q16a, Q20b	~20%
Application/Calculation	Q5, Q9b, Q11b, Q12a, Q13b,c, Q16b, Q18a,b, Q19	~45%
Analysis/Explanation	Q10b, Q12b,c, Q14b, Q15, Q17b, Q18c, Q19b, Q20c	~30%
Synthesis/Experimental	Q17a, Q20a	~5%