Secondary 1 Science Practice Paper 3

TuitionGoWhere Practice Paper - Science Secondary 1

Answer Key and Marking Scheme

Subject: Science | Level: Secondary 1 (G3) | Version: 3 of 5 | Total Marks: 60

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Section A: Multiple Choice (1 mark each)

Question	Answer	Explanation
1	B	At constant velocity, resultant force is zero (Newton's First Law). The push force equals the opposing friction force.
2	A	As the ball rises, it slows down—kinetic energy decreases while gravitational potential energy increases. At the very highest point, this conversion completes (KE → GPE).
3	B	Density = Mass ÷ Volume (ρ = m/V). This is the defining relationship for density.
4	D	GPE = mgh = 50 × 10 × 4 = 2000 J
5	D	Velocity is a vector (has magnitude and direction). Mass, speed, and distance are scalar quantities.
6	B	Total distance = 120 + 80 = 200 km; Total time = 2 + 1.5 = 3.5 h; Average speed = 200/3.5 = 57.1 km/h
7	B	Water displacement (Archimedes' principle) is the standard method for irregular solids.
8	B	Principle of moments: clockwise moment = anticlockwise moment for equilibrium.
9	B	Pressure = Force/Area. Smaller area → greater pressure for same force. Unit is pascal (Pa), not newtons.
10	B	Velocity ratio = effort distance/load distance; 4 = 8/load distance; load distance = 8/4 = 2 m

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Section B: Structured Response

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Question 11 (5 marks)

(a) Volume of cube [1]

Method: For a cube, V = side³

• V = (3.0)³ = 3.0 × 3.0 × 3.0 = 27 cm³

Marking: Correct formula or correct answer with unit [1]

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(b) Density of the metal [2]

Method: ρ = m/V

• ρ = 216 g / 27 cm³ = 8.0 g/cm³

Working [1], correct answer with unit [1]

Common error: Forgetting to cube the side, or using 3 × 3 = 9 instead of 27

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(c) Identification of metal [2]

Answer: The metal is iron [1]

Reasoning: The calculated density (8.0 g/cm³) is much closer to iron (7.9 g/cm³) than to aluminium (2.7 g/cm³). The small difference (≈1.3%) is likely due to experimental measurement uncertainty [1].

Alternative acceptable answer: The density is approximately 7.9 g/cm³, matching iron.

Key concept: Density is a characteristic property used to identify materials.

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Question 12 (5 marks)

Expected visual features from Q12-fig1: Distance-time graph with three sections—A rising steeply (0-10 min, 0-1200 m), B horizontal (10-15 min), C rising less steeply (15-30 min, 1200-1800 m)

(a) Speed during section A [2]

Method: Speed = gradient = Δdistance/Δtime

• Speed = (1200 − 0) m / (10 − 0) min = 1200/10 = 120 m/min

Or converted: 120 m/min = 2 m/s

Formula [1], correct substitution and answer with unit [1]

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(b) Motion during section B [1]

Answer: The cyclist is stationary (at rest) / not moving.

Explanation: The horizontal line shows zero gradient, meaning no change in distance over time. The distance remains constant at 1200 m for 5 minutes.

Key concept: Zero gradient on a distance-time graph = stationary object.

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(c) Comparison of speeds in A and C [2]

Answer: Speed in section A is greater than speed in section C [1]

Explanation: The gradient of section A is steeper than the gradient of section C. Since speed equals the gradient of a distance-time graph, a steeper gradient means greater speed [1].

Numerical check: Speed in C = (1800−1200)/(30−15) = 600/15 = 40 m/min, which is less than 120 m/min

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Question 13 (5 marks)

Expected visual features from Q13-fig1: Horizontal lever with fulcrum, load 60 N at 0.5 m on one side, effort E at 1.5 m on other side, balanced

(a) Moment of the load [2]

Method: Moment = Force × perpendicular distance from pivot

• Moment = 60 N × 0.5 m = 30 Nm (or 30 N·m)

Formula: Moment = F × d [1]; correct substitution and answer with unit [1]

Key concept: Moment (torque) measures the turning effect of a force.

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(b) Effort force E [2]

Method: For equilibrium, clockwise moment = anticlockwise moment

• Moment of load = Moment of effort
• 30 Nm = E × 1.5 m
• E = 30/1.5 = 20 N

Principle of moments stated or applied [1]; correct answer [1]

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(c) Reducing the effort [1]

Answer: Any one of:

• Increase the effort distance (move effort further from fulcrum) [1]
• Decrease the load distance (move load closer to fulcrum) [1]

Key concept: Lever systems trade force for distance using the principle of moments.

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Question 14 (5 marks)

(a) Graph plotting [2]

Expected visual from Q14-fig1: Graph with load (0-6 N) on x-axis, extension (0-60 mm) on y-axis, points plotted at (0,0), (1,8), (2,16), (3,24), (4,32), (5,40), (6,52)

Marking points:

• Correct axes with labels and units [1]
• Correctly plotted points and reasonable straight line through first 5 points, with point at (6,52) noticeably off the straight line [1]

Common error: Drawing a single straight line through all points misses that Hooke's Law fails after 5 N.

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(b) Extension at 2.5 N [1]

Method: From the linear part of the graph, extension is proportional to load.

• At 2 N: 16 mm; at 3 N: 24 mm
• At 2.5 N: (16 + 24)/2 = 20 mm

Or by direct proportion: extension = 8 × load, so 8 × 2.5 = 20 mm

Answer: 20 mm [1]

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(c) Hooke's Law conclusion [2]

Answer: The conclusion is not correct / only partially correct [1]

Explanation: Hooke's Law states that extension is directly proportional to load, meaning the graph should be a straight line through the origin. From 0 to 5 N, the extension increases by 8 mm per Newton (constant ratio), confirming proportionality. However, from 5 N to 6 N, the extension increases by 12 mm (not 8 mm), so the ratio is no longer constant. The spring has exceeded its limit of proportionality [1].

Key concept: Hooke's Law (F = kx) only applies within the elastic limit/limit of proportionality.

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Question 15 (5 marks)

(a) Energy to heat water [2]

Method: Q = mcΔθ

• Q = 1.5 kg × 4200 J/(kg·°C) × (100 − 25)°C
• Q = 1.5 × 4200 × 75
• Q = 472 500 J or 4.725 × 10⁵ J

Formula [1]; correct substitution and answer with unit [1]

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(b) Minimum time [2]

Method: P = E/t, therefore t = E/P

• t = 472 500 J / 2000 W
• t = 236.25 s ≈ 236 s (or 3 minutes 56 seconds)

Rearrangement or correct formula [1]; correct answer [1]

Note: Accept 236 s or 240 s (to 2 s.f.) or exact value.

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(c) Longer actual time [1]

Answer: Any one valid reason:

• Heat energy is lost to the surroundings (air, kettle body) [1]
• Not all electrical energy is converted to heat in the water [1]
• Some energy is needed to heat the kettle itself (metal body) [1]
• Energy losses due to evaporation of some water [1]

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Question 16 (5 marks)

Expected visual from Q16-fig1: Water cylinder with three holes at depths 5 cm, 15 cm, 25 cm; water jets with C longest, B medium, A shortest

(a) Comparing jet distances [1]

Answer: The water jet from C travels furthest, from B travels an intermediate distance, and from A travels shortest [1].

Or equivalent statement ordering the distances correctly.

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(b) Pressure and depth relationship [1]

Answer: Water pressure increases with depth [1].

The deeper the hole, the greater the pressure, the faster the water emerges, the further the jet travels.

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(c) Horizontal then curved motion [2]

Explanation of horizontal start: The water emerges through a horizontal hole, so the initial pressure force acts perpendicular to the container wall, giving the water a horizontal velocity [1].

Explanation of downward curve: Once outside, gravity acts downward on the water, causing it to accelerate downward and follow a parabolic (curved) path [1].

Key concept: Motion has horizontal and vertical components; gravity only affects the vertical component (projectile motion basics).

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(d) Investigating density effect [1]

Answer: Replace the water with a different liquid of known different density (e.g., oil, salt water, alcohol) but keep the depth of the hole the same [1].

Or: Use the same apparatus with water at different temperatures (density changes slightly).

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Section C: Application and Analysis

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Question 17 (5 marks)

(a) Work done [2]

Method: Work done = Force × distance = weight × height = mgh

• Weight = 500 kg × 10 N/kg = 5000 N
• Work = 5000 N × 3.0 m = 15 000 J (or 1.5 × 10⁴ J)

Formula W = Fd or W = mgh [1]; correct substitution and answer with unit [1]

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(b) Power output [2]

Method: P = Work/time = W/t

• P = 15 000 J / 5.0 s
• P = 3000 W (or 3.0 kW)

Formula [1]; correct answer with unit [1]

Common error: Using P = Fv also works: v = 3.0/5.0 = 0.6 m/s; P = 5000 × 0.6 = 3000 W

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(c) Meaning of mechanical advantage [1]

Answer: A mechanical advantage of 4.0 means the load force is 4 times greater than the effort force, or equivalently, the effort force needed is only 1/4 of the load force [1].

Key concept: Mechanical advantage = Load/Effort; values > 1 mean force multiplication.

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Question 18 (5 marks)

Expected visual from Q18-fig1: Block and tackle with 4 supporting rope segments, load 800 N, effort rope pulled upward

(a) Velocity ratio [2]

Reasoning: The velocity ratio equals the number of supporting rope segments that share the load. In this system, there are 4 rope segments directly supporting the movable block [1].

Answer: Velocity ratio = 4 [1]

Key concept: For simple block and tackle, VR = number of supporting segments.

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(b) Effort force (ideal) [2]

Method: For ideal machine, Mechanical Advantage = Velocity Ratio = 4

• MA = Load/Effort
• 4 = 800 N / E
• E = 800/4 = 200 N

Or: The 800 N load is shared by 4 rope segments, so each segment tension = 800/4 = 200 N [1]; correct answer with reasoning [1]

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(c) Efficiency [1]

Method: Efficiency = (MA/VR) × 100% = (Actual MA / Theoretical MA) × 100%

Actual MA = Load/Effort = 800/250 = 3.2

Efficiency = (3.2/4.0) × 100% = 80%

Or: Efficiency = (Useful work output / Total work input) × 100%

Correct answer [1]

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Question 19 (5 marks)

(a) Variables [2]

Variable	Answer
Independent variable	The type of insulating material used (or different materials) [1]
Dependent variable	The temperature of the water (after fixed time intervals / rate of temperature drop) [1]

Note: Both must be clearly stated with the quantity, not just "material" or "temperature."

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(b) Controlled variables [2]

Any two valid:

• Initial temperature of the hot water [1]
• Volume/mass of water in each bottle [1]
• Type/shape of hot water bottle used [1]
• Ambient/surrounding temperature [1]
• Time intervals between temperature readings [1]
• Thickness of insulating material wrapped [1]

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(c) Reliability with longer timing [1]

Answer: A 30-minute period allows a greater temperature drop to be measured, making differences between materials more noticeable and easier to distinguish [1]. Short periods may show very small changes that are within measurement uncertainty of the thermometer [1].

Or: Longer time reduces percentage error in timing and temperature measurement.

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Question 20 (5 marks)

Expected visual from Q20-fig1: Ramp 3.0 m long, 0.8 m high, trolley pushed by 25 N parallel to ramp at constant speed, weight 60 N

(a) Why useful work is less than work by effort [2]

Answer: The effort force (25 N) acts over the full 3.0 m length of the ramp [1].

However, only the vertical component of this motion overcomes gravity. The useful work (against gravity) = weight × vertical height = 60 × 0.8 = 48 J, while total work input = 25 × 3.0 = 75 J.

The difference (75 − 48 = 27 J) is work done against friction between the trolley and the ramp surface [1].

Key concept: Simple machines are not 100% efficient; friction dissipates energy as heat.

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(b) Useful work done [2]

Method: Useful work = weight × vertical height = mgh

• Useful work = 60 N × 0.8 m = 48 J

Or using the ramp relationship: The vertical height gain represents the actual gravitational potential energy increase.

Formula [1]; correct answer with unit [1]

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(c) Efficiency [1]

Method: Efficiency = (Useful work output / Total work input) × 100%

• Total work input = 25 N × 3.0 m = 75 J
• Efficiency = (48/75) × 100% = 64%

Or using MA/VR: VR = 3.0/0.8 = 3.75; MA = 60/25 = 2.4; Efficiency = 2.4/3.75 = 0.64 = 64%

Answer: 64% or 0.64 [1]

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END OF ANSWER KEY