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Secondary 1 Science Practice Paper 1

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TuitionGoWhere Practice Paper - Science Secondary 1 (Answer Key)

Subject: Science
Level: Secondary 1 (G3)
Paper: Practice Paper 1 – Physical Sciences (Version 1)
Total Marks: 50

Section A: Multiple Choice Questions [10 marks]

1

Answer: B
Explanation: When a student lifts a book at constant speed, the chemical energy stored in the student's muscles is converted into gravitational potential energy of the book. The kinetic energy does not change because the speed is constant.
Marks: [1]

2

Answer: C
Explanation: Work done = Force × Distance moved in direction of force = 15 N × 4 m = 60 J.
Marks: [1]

3

Answer: C
Explanation: Gravitational potential energy = mghmgh = 2 kg × 10 N/kg × 5 m = 100 J.
Marks: [1]

4

Answer: C
Explanation:

  • A is incorrect: Work requires displacement in the direction of the force.
  • B is incorrect: Work can be done when force and displacement are at an angle (using the component of force in the direction of displacement).
  • C is correct: When holding an object stationary, there is no displacement, so no work is done by the upward force.
  • D is incorrect: Work done against friction is converted to thermal energy, not kinetic energy.
    Marks: [1]

5

Answer: B
Explanation: As the ball falls, its height decreases so gravitational potential energy decreases, and its speed increases so kinetic energy increases. The conversion is gravitational potential energy → kinetic energy.
Marks: [1]

6

Answer: B
Explanation: Work done = Force × Distance × cos(θ) = 8 N × 3 m × cos(60°) = 8 × 3 × 0.5 = 12 J.
Marks: [1]

7

Answer: C
Explanation: Kinetic energy = 12mv2\frac{1}{2}mv^2 = 12\frac{1}{2} × 0.5 kg × (4 m/s)² = 0.25 × 16 = 4 J.
Marks: [1]

8

Answer: B
Explanation: Work done = Force × Displacement in direction of force. When pushing a wall that does not move, the displacement is zero, so no work is done. In all other options, there is displacement in the direction of (or opposite to) the force.
Marks: [1]

9

Answer: C
Explanation: By conservation of energy: mgh=12mv2mgh = \frac{1}{2}mv^2 → v=2gh=2×10×15=300≈17.3 m/s≈17 m/sv = \sqrt{2gh} = \sqrt{2 × 10 × 15} = \sqrt{300} \approx 17.3 \text{ m/s} \approx 17 \text{ m/s}.
Marks: [1]

10

Answer: B
Explanation: Power is defined as the rate of doing work, or the rate of energy transfer. Unit: Watt (W) = Joule per second (J/s).
Marks: [1]

Section B: Structured Questions [25 marks]

11

(a) Gain in GPE = mghmgh = 50 kg × 10 N/kg × 3.0 m = 1500 J
Marks: [2] (1 for formula/substitution, 1 for correct answer with unit)

(b) Average power = Work done / Time taken = 1500 J / 6.0 s = 250 W
Marks: [2] (1 for formula/substitution, 1 for correct answer with unit)

(c) The actual power output is greater because the student also does work against internal friction in muscles and joints, and energy is converted to thermal energy in the body. Some energy is also used to move body parts (arms, legs) and overcome air resistance.
Marks: [1] (Accept any valid reason: thermal energy losses, moving body parts, air resistance, etc.)

12

(a) Work done by applied force = Force × Distance = 10 N × 5 m = 50 J
Marks: [1]

(b) Work done by friction = –Friction × Distance = –3 N × 5 m = –15 J (negative because friction opposes motion)
Marks: [1]

(c) Net work done = Work by applied force + Work by friction = 50 J + (–15 J) = 35 J
Marks: [1]

(d) Work done against friction is converted to thermal energy (heat) and sound energy.
Marks: [1] (Accept "thermal energy" or "heat" as the main answer)

(e) By the work-energy principle, net work done = change in kinetic energy.
Increase in kinetic energy = 35 J
Marks: [1]

13

(a) GPE = mghmgh = 0.2 kg × 10 N/kg × 0.15 m = 0.3 J
Marks: [1]

(b) By conservation of energy (no air resistance), loss in GPE = gain in KE.
Kinetic energy at lowest point = 0.3 J
Marks: [1]

(c) KE = 12mv2\frac{1}{2}mv^2 → v=2×KEm=2×0.30.2=3≈1.73 m/sv = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 0.3}{0.2}} = \sqrt{3} \approx \textbf{1.73 m/s}
Marks: [2] (1 for correct formula/substitution, 1 for correct answer with unit)

(d) Some mechanical energy is converted to thermal energy due to air resistance and friction at the pivot, so the total mechanical energy (KE + GPE) decreases. The bob cannot rise to the same height because it has less energy.
Marks: [1]

14

(a) Acceleration a=v−ut=20−010=2 m/s2a = \frac{v - u}{t} = \frac{20 - 0}{10} = \textbf{2 m/s}^2
Marks: [1]

(b) Resultant force F=ma=1200 kg×2 m/s2=2400 NF = ma = 1200 \text{ kg} × 2 \text{ m/s}^2 = \textbf{2400 N}
Marks: [1]

(c) Distance s=ut+12at2=0+12×2×102=100 ms = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} × 2 × 10^2 = \textbf{100 m}
Marks: [1]

(d) Work done by resultant force = Force × Distance = 2400 N × 100 m = 240,000 J (or 240 kJ)
Marks: [2] (1 for formula/substitution, 1 for correct answer with unit)

(e) The engine does more work because it must also overcome frictional forces (air resistance, rolling friction, internal engine friction) and increase the internal energy of the car (thermal energy). The resultant force only accounts for the net force causing acceleration.
Marks: [1]

15

(a) From 0 to 4 m, force is constant at 20 N.
Work done = Force × Distance = 20 N × 4 m = 80 J
(Alternatively, area of rectangle = 20 × 4 = 80 J)
Marks: [1]

(b) From 4 to 8 m, force decreases linearly from 20 N to 0 N. The area under the graph is a triangle.
Work done = Area of triangle = 12×base×height=12×4 m×20 N=40 J\frac{1}{2} × \text{base} × \text{height} = \frac{1}{2} × 4 \text{ m} × 20 \text{ N} = \textbf{40 J}
Marks: [2] (1 for identifying triangle area, 1 for correct calculation with unit)

(c) Total work done = 80 J + 40 J = 120 J
Marks: [1]

Section C: Longer Structured and Data-Based Questions [15 marks]

16

(a) Mass of water per second = 2000 kg. Height = 50 m.
GPE lost per second = mgh=2000×10×50=1,000,000 J/smgh = 2000 × 10 × 50 = \textbf{1,000,000 J/s} (or 1 MW)
Marks: [2] (1 for formula/substitution, 1 for correct answer with unit)

(b) Input power = 1,000,000 W = 1000 kW. Output power = 800 kW.
Efficiency = Output powerInput power×100%=8001000×100%=80%\frac{\text{Output power}}{\text{Input power}} × 100\% = \frac{800}{1000} × 100\% = \textbf{80\%}
Marks: [2] (1 for correct formula/substitution, 1 for correct answer with %)

(c) The remaining energy is converted to:

  1. Thermal energy (heat) due to friction in turbines and generators
  2. Sound energy from moving water and machinery
    (Also accept: kinetic energy of water leaving the station, electrical resistance losses in cables)
    Marks: [1] (Any two valid forms)

17

(a) Elastic potential energy = 12kx2=12×400×(0.05)2=200×0.0025=0.5 J\frac{1}{2}kx^2 = \frac{1}{2} × 400 × (0.05)^2 = 200 × 0.0025 = \textbf{0.5 J}
Marks: [2] (1 for formula/substitution, 1 for correct answer with unit)

(b) At maximum height, all elastic PE → gravitational PE.
mgh=0.5 Jmgh = 0.5 \text{ J} → h=0.5mg=0.50.01×10=0.50.1=5 mh = \frac{0.5}{mg} = \frac{0.5}{0.01 × 10} = \frac{0.5}{0.1} = \textbf{5 m}
Marks: [2] (1 for energy conservation equation, 1 for correct answer with unit)

(c) In practice, some elastic potential energy is converted to thermal energy due to air resistance and internal friction in the spring, so less energy is available for gravitational potential energy.
Marks: [1]

18

(a) For h=0.30 mh = 0.30 \text{ m}, v=2.4 m/sv = 2.4 \text{ m/s}.
v2=(2.4)2=5.76 m2/s2v^2 = (2.4)^2 = \textbf{5.76 m}^2/\text{s}^2
Marks: [1]

(b) Graph requirements:

  • Axes labelled: h/mh / \text{m} (x-axis), v2/m2/s2v^2 / \text{m}^2/\text{s}^2 (y-axis)
  • Suitable scales (e.g., 2 cm = 0.1 m on x-axis, 2 cm = 2 m²/s² on y-axis)
  • All 5 points plotted correctly: (0.10, 1.96), (0.20, 4.00), (0.30, 5.76), (0.40, 7.84), (0.50, 9.61)
  • Best-fit straight line passing through origin (0,0)
    Marks: [2] (1 for correct plotting of points, 1 for best-fit line through origin with labelled axes)

(c) Gradient of graph = Δv2Δh=2g\frac{\Delta v^2}{\Delta h} = 2g
Using points from best-fit line (e.g., (0,0) and (0.50, 9.61)):
Gradient = 9.61−00.50−0=19.22\frac{9.61 - 0}{0.50 - 0} = 19.22
2g=19.222g = 19.22 → g=19.222=9.61 m/s2g = \frac{19.22}{2} = \textbf{9.61 m/s}^2
(Accept values in range 9.5–9.8 m/s² depending on best-fit line)
Marks: [2] (1 for correct gradient calculation, 1 for correct gg value with unit)

(d) The value is less than 10 m/s² because some gravitational potential energy is converted to thermal energy due to friction (between trolley wheels and axle, and between trolley and ramp) and air resistance, so the kinetic energy at the bottom is less than the theoretical mghmgh.
Marks: [1] (Accept: friction, air resistance, rotational kinetic energy of wheels not accounted for)

19

(a) Vertical height gained = Distance along slope × sin(5°) = 200 m × 0.087 = 17.4 m
Marks: [1]

(b) Total mass = 60 kg + 10 kg = 70 kg.
Gain in GPE = mgh=70×10×17.4=12,180 Jmgh = 70 × 10 × 17.4 = \textbf{12,180 J} (or 12.18 kJ)
Marks: [2] (1 for correct mass and height, 1 for correct answer with unit)

(c) Time taken = Distance / Speed = 200 m / 4 m/s = 50 s.
Power against gravity = Work done / Time = 12,180 J / 50 s = 243.6 W (≈ 244 W)
Marks: [2] (1 for time calculation, 1 for power calculation with unit)

(d) The cyclist also does work against:

  1. Air resistance (drag)
  2. Rolling friction between tyres and road
    (Also accept: internal friction in bearings, friction in chain/gears)
    Marks: [1] (Any two valid answers)

20

(a) KE at bottom = 12mv2=12×500×(25)2=250×625=156,250 J\frac{1}{2}mv^2 = \frac{1}{2} × 500 × (25)^2 = 250 × 625 = \textbf{156,250 J} (or 156.25 kJ)
Marks: [1]

(b) Height at top of loop = diameter = 2 × 10 m = 20 m.
By conservation of energy:
KE_bottom = KE_top + GPE_top
12mvb2=12mvt2+mgh\frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + mgh
Divide by mm: 12vb2=12vt2+gh\frac{1}{2}v_b^2 = \frac{1}{2}v_t^2 + gh
12(25)2=12vt2+10×20\frac{1}{2}(25)^2 = \frac{1}{2}v_t^2 + 10 × 20
312.5=12vt2+200312.5 = \frac{1}{2}v_t^2 + 200
12vt2=112.5\frac{1}{2}v_t^2 = 112.5
vt2=225v_t^2 = 225
vt=15 m/sv_t = \textbf{15 m/s}
Marks: [3] (1 for energy conservation equation, 1 for correct substitution, 1 for correct answer with unit)

(c) Centripetal force at top = mvt2r=500×(15)210=500×22510=11,250 N\frac{mv_t^2}{r} = \frac{500 × (15)^2}{10} = \frac{500 × 225}{10} = \textbf{11,250 N}
Marks: [2] (1 for formula/substitution, 1 for correct answer with unit)

(d) At the top of the loop, forces acting on the car: weight (mgmg) downwards, normal reaction (NN) downwards.
Net downward force = mg+N=500×10+12,500=5,000+12,500=17,500 Nmg + N = 500 × 10 + 12,500 = 5,000 + 12,500 = 17,500 \text{ N}
Required centripetal force = 11,250 N (from part c).
Since net downward force (17,500 N) > required centripetal force (11,250 N), the car presses firmly on the track and maintains contact.
(Alternatively: For contact to be maintained, N≥0N \ge 0. Here N=12,500 N>0N = 12,500 \text{ N} > 0, so contact is maintained.)
Marks: [2] (1 for correct force analysis/condition, 1 for correct conclusion with reasoning)

Total Marks: 50
End of Answer Key