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Secondary 2 Science Physical Sciences Quiz

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Secondary 2 Science Quiz - Physical Sciences (Answer Key)

Total Marks: 50

Section A: Multiple Choice Questions (10 marks)

  1. C [1]
    Explanation: The Principle of Conservation of Energy states that energy cannot be created or destroyed; it can only be converted from one form to another. The total energy in an isolated system remains constant.

  2. C [1]
    Working:
    GPE at top = mgh=0.5×10×20=100 Jmgh = 0.5 \times 10 \times 20 = 100 \text{ J}
    By conservation of energy, KE at bottom = GPE at top = 100 J.

  3. B [1]
    Working:
    Power = 60 W, Time = 30 min = 1800 s
    Energy = Power × Time = 60×1800=108 000 J60 \times 1800 = 108\,000 \text{ J}

  4. B [1]
    Explanation: At the highest point (A), the pendulum has maximum GPE and zero KE. As it swings down to B, GPE is converted to KE.

  5. C [1]
    Working:
    Work done = Force × Distance = 20×5=100 J20 \times 5 = 100 \text{ J}

  6. B [1]
    Explanation: Power is the rate of doing work or transferring energy (unit: Watt), not a form of energy itself.

  7. B [1]
    Working:
    KE gained = 12mv2=12×1000×202=200 000 J\frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 20^2 = 200\,000 \text{ J}
    Average Power = Work / Time = 200 000/10=20 000 W200\,000 / 10 = 20\,000 \text{ W}

  8. C [1]
    Explanation: A compressed (or stretched) spring stores elastic potential energy.

  9. B [1]
    Working:
    Work against gravity = mgh=50×10×3=1500 Jmgh = 50 \times 10 \times 3 = 1500 \text{ J}
    Power = Work / Time = 1500/4=375 W1500 / 4 = 375 \text{ W}

  10. B [1]
    Explanation: A battery stores chemical energy, which is converted to electrical energy in the circuit, then to light and heat energy in the bulb.

Section B: Structured Questions (24 marks)

  1. (a) Work done is the product of the force applied on an object and the distance moved by the object in the direction of the force. [1]
    Marking note: Must mention "in the direction of the force" for full mark.

    (b) (i) Work done by applied force = F×d=150×8=1200 JF \times d = 150 \times 8 = 1200 \text{ J} [1]

    (ii) Work done against friction = frictional force × distance = 50×8=400 J50 \times 8 = 400 \text{ J} [1]

    (iii) Net work done = Work by applied force – Work against friction = 1200−400=800 J1200 - 400 = 800 \text{ J} [1]

    (iv) Gain in KE = Net work done = 800 J [1]
    Marking note: Work-energy theorem: net work = change in KE.

  2. (a) GPE at top = mgh=500×10×40=200 000 Jmgh = 500 \times 10 \times 40 = 200\,000 \text{ J} [1]

    (b) Total mechanical energy = 200 000 J (conserved, no losses) [1]

    (c) At point B (bottom), GPE = 0, so KE = Total energy = 200 000 J [1]

    (d) KE=12mv2KE = \frac{1}{2}mv^2
    200 000=12×500×v2200\,000 = \frac{1}{2} \times 500 \times v^2
    v2=800v^2 = 800
    v=800=28.3 m/sv = \sqrt{800} = 28.3 \text{ m/s} (or 202 m/s20\sqrt{2} \text{ m/s}) [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

    (e) GPE at C = mgh=500×10×25=125 000 Jmgh = 500 \times 10 \times 25 = 125\,000 \text{ J} [1]

    (f) KE at C = Total energy – GPE at C = 200 000−125 000=75 000 J200\,000 - 125\,000 = 75\,000 \text{ J} [1]

  3. (a) Power is the rate of doing work or the rate of energy conversion. Its SI unit is the watt (W). [2]
    Mark breakdown: 1 mark for definition, 1 mark for unit.

    (b) (i) Work done = Gain in GPE = mgh=200×10×15=30 000 Jmgh = 200 \times 10 \times 15 = 30\,000 \text{ J} [1]

    (ii) Output power = Work / Time = 30 000/20=1500 W30\,000 / 20 = 1500 \text{ W} [1]

    (iii) Efficiency = Output PowerInput Power×100%=15002000×100%=75%\frac{\text{Output Power}}{\text{Input Power}} \times 100\% = \frac{1500}{2000} \times 100\% = 75\% [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with %.

  4. (a) GPE at A relative to B = mgh=0.2×10×0.3=0.6 Jmgh = 0.2 \times 10 \times 0.3 = 0.6 \text{ J} [1]

    (b) By conservation of energy, KE at B = GPE lost = 0.6 J [1]

    (c) KE=12mv2KE = \frac{1}{2}mv^2
    0.6=12×0.2×v20.6 = \frac{1}{2} \times 0.2 \times v^2
    v2=6v^2 = 6
    v=6=2.45 m/sv = \sqrt{6} = 2.45 \text{ m/s} [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

    (d) Some of the mechanical energy (GPE + KE) is converted to heat and sound energy due to air resistance and friction at the pivot. This energy is dissipated to the surroundings, so the total mechanical energy decreases. Therefore, the bob has less KE at B and less GPE at C, so it does not reach the same height. [2]
    Mark breakdown: 1 mark for identifying energy loss to heat/sound, 1 mark for explaining the consequence (lower height at C).

Section C: Longer Structured and Data-Based Questions (16 marks)

  1. (a) Graph plotting [2]
    Marking points:

    • Axes labeled with quantities and units (1 mark)
    • Appropriate scales covering >50% of grid, points plotted correctly, smooth curve/best-fit line (1 mark)
      Expected graph: Curve showing increasing but decreasing gradient (since v∝hv \propto \sqrt{h}).

    (b) As the height increases, the gravitational potential energy (GPE = mgh) of the car at the top increases. By the Principle of Conservation of Energy, this GPE is converted to kinetic energy (KE = ½mv²) at the bottom. Since GPE is larger at greater heights, the KE at the bottom is larger, resulting in a higher speed. [2]
    Mark breakdown: 1 mark for linking height to GPE, 1 mark for linking GPE to KE/speed via conservation of energy.

    (c) Some GPE is converted to heat and sound energy due to friction between the car wheels/axle and air resistance, so not all GPE is converted to KE. The experimental speed is lower than the theoretical value (which assumes no energy losses). [1]
    Acceptable answers: friction, air resistance, rotational KE of wheels not accounted for, ramp not perfectly smooth.

    (d) KE=12mv2=12×0.05×(2.6)2=0.169 JKE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.05 \times (2.6)^2 = 0.169 \text{ J} [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

  2. (a) GPE lost per second = mass rate × g × h = 500×10×80=400 000 J/s=400 kW500 \times 10 \times 80 = 400\,000 \text{ J/s} = 400 \text{ kW} [2]
    Mark breakdown: 1 mark for correct formula (mgh per second), 1 mark for correct answer with unit (kW or J/s).

    (b) Efficiency = Useful output powerInput power×100%=300400×100%=75%\frac{\text{Useful output power}}{\text{Input power}} \times 100\% = \frac{300}{400} \times 100\% = 75\% [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with %.

    (c) Heat energy and sound energy (also acceptable: kinetic energy of water turbulence, internal energy of water/turbines) [2]
    Mark breakdown: 1 mark each for two valid forms.

  3. (a) At Q (ground level), GPE = 0, so KE = Initial GPE at P
    12mv2=mghP\frac{1}{2}mv^2 = mgh_P
    v=2ghP=2×10×50=1000=31.6 m/sv = \sqrt{2gh_P} = \sqrt{2 \times 10 \times 50} = \sqrt{1000} = 31.6 \text{ m/s} [2]
    Mark breakdown: 1 mark for correct use of conservation of energy/formula, 1 mark for correct answer with unit.

    (b) At R (height 20 m), GPE = mg×20mg \times 20
    KE at R = Initial GPE – GPE at R = mg(50−20)=mg×30mg(50 - 20) = mg \times 30
    12mv2=mg×30\frac{1}{2}mv^2 = mg \times 30
    v=2g×30=600=24.5 m/sv = \sqrt{2g \times 30} = \sqrt{600} = 24.5 \text{ m/s} [2]
    Mark breakdown: 1 mark for correct energy conservation setup, 1 mark for correct answer with unit.

    (c) Centripetal acceleration ac=v2r=60010=60 m/s2a_c = \frac{v^2}{r} = \frac{600}{10} = 60 \text{ m/s}^2 [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

    (d) Yes, the car will maintain contact with the track.
    At the top of the loop, the centripetal force required is provided by the weight plus the normal reaction force: mg+N=macmg + N = ma_c.
    Here ac=60 m/s2>g(10 m/s2)a_c = 60 \text{ m/s}^2 > g (10 \text{ m/s}^2), so the required centripetal acceleration is greater than g. This means the weight alone is insufficient to provide the centripetal force; the track must push down on the car (N > 0). Since the normal force is positive, the car stays in contact with the track. [2]
    Mark breakdown: 1 mark for correct conclusion with reasoning (comparing aca_c to gg or N>0N > 0), 1 mark for clear explanation of forces at the top of the loop.

  4. (a) Thermal energy gained = mcΔθ=1.0×4200×(100−25)=315 000 Jmc\Delta\theta = 1.0 \times 4200 \times (100 - 25) = 315\,000 \text{ J} [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

    (b) Electrical energy supplied = Power × Time = 1500×210=315 000 J1500 \times 210 = 315\,000 \text{ J} [2]
    Mark breakdown: 1 mark for correct formula/substitution (power in watts), 1 mark for correct answer with unit.

    (c) Efficiency = Useful energy outputEnergy input×100%=315 000315 000×100%=100%\frac{\text{Useful energy output}}{\text{Energy input}} \times 100\% = \frac{315\,000}{315\,000} \times 100\% = 100\% [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer.
    Note: In this idealised calculation, efficiency is 100%. Part (d) addresses why real kettles are less efficient.

    (d) Some electrical energy is converted to heat energy that is lost to the surroundings (e.g., heating the kettle body, heating the air around the kettle, sound energy) rather than being transferred to the water. [1]

  5. (a) Elastic PE = 12kx2=12×400×(0.05)2=0.5 J\frac{1}{2}kx^2 = \frac{1}{2} \times 400 \times (0.05)^2 = 0.5 \text{ J} [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

    (b) By conservation of energy, Elastic PE at start = GPE at max height
    0.5=mgh=0.02×10×h0.5 = mgh = 0.02 \times 10 \times h
    h=0.50.2=2.5 mh = \frac{0.5}{0.2} = 2.5 \text{ m} [2]
    Mark breakdown: 1 mark for correct energy conservation equation, 1 mark for correct answer with unit.

    (c) At point of release, Elastic PE = KE
    0.5=12mv2=12×0.02×v20.5 = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.02 \times v^2
    v2=50v^2 = 50
    v=50=7.07 m/sv = \sqrt{50} = 7.07 \text{ m/s} [2]
    Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

  6. (a) Energy per day = ∑Power (W)×Time (h)1000\sum \frac{\text{Power (W)} \times \text{Time (h)}}{1000}
    Refrigerator: 150×241000=3.6 kWh\frac{150 \times 24}{1000} = 3.6 \text{ kWh}
    Air Conditioner: 1200×61000=7.2 kWh\frac{1200 \times 6}{1000} = 7.2 \text{ kWh}
    Washing Machine: 500×11000=0.5 kWh\frac{500 \times 1}{1000} = 0.5 \text{ kWh}
    LED TV: 80×41000=0.32 kWh\frac{80 \times 4}{1000} = 0.32 \text{ kWh}
    Total = 3.6+7.2+0.5+0.32=11.62 kWh3.6 + 7.2 + 0.5 + 0.32 = 11.62 \text{ kWh} [2]
    Mark breakdown: 1 mark for correct individual calculations/method, 1 mark for correct total with unit.

    (b) Daily cost = 11.62 \times 0.28 = \3.253630−daycost= 30-day cost =3.2536 \times 30 = $97.61 (or \97.60) [2]
    Mark breakdown: 1 mark for correct daily cost calculation, 1 mark for correct 30-day cost with $ sign.

    (c) Original daily energy (AC) = 7.2 kWh
    New daily energy (AC) = 900×61000=5.4 kWh\frac{900 \times 6}{1000} = 5.4 \text{ kWh}
    Reduction = 7.2−5.4=1.8 kWh7.2 - 5.4 = 1.8 \text{ kWh}
    Percentage reduction = 1.87.2×100%=25%\frac{1.8}{7.2} \times 100\% = 25\% [2]
    Mark breakdown: 1 mark for correct new energy/reduction calculation, 1 mark for correct percentage with % sign.

    (d) Switch off appliances at the mains when not in use (avoid standby mode) / Use natural ventilation instead of air conditioning / Use energy-saving settings on appliances / Reduce usage duration. [1]
    Any one valid, practical suggestion.

End of Answer Key