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A Level H2 Physics Energy Power Quiz
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Questions
A-Level Physics H2 Quiz - Energy Power
Name: __________________________
Class: __________
Date: __________
Score: ______ / 42
Duration: 45 minutes
Total Marks: 42
Instructions:
- Answer all questions in the spaces provided.
- Show all working for calculation questions; final answers without working may not receive full marks.
- Use g = 9.81 m s⁻² where necessary, unless otherwise stated.
- 1 u = 1.66 × 10⁻²⁷ kg, 1 u = 931.5 MeV/c², c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C.
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A car of mass 1200 kg is moving at a speed of 25 m s⁻¹. Calculate the kinetic energy of the car.
[2 marks]
-
A stone of mass 0.50 kg is dropped from a height of 20 m. Using the principle of conservation of energy, calculate its kinetic energy just before it hits the ground. Neglect air resistance.
[2 marks]
-
State the principle of conservation of energy.
[1 mark]
-
An electric motor lifts a 50 kg mass vertically upward at a constant speed of 2.0 m s⁻¹. Calculate the minimum power output of the motor.
[2 marks]
-
An electric kettle rated at 2000 W is used to heat 0.80 kg of water from 25 °C to 100 °C in 3.0 minutes.
Specific heat capacity of water = 4200 J kg⁻¹ °C⁻¹.(a) Calculate the useful energy transferred to the water. [1]
(b) Determine the electrical energy supplied by the kettle. [1]
(c) Hence, calculate the efficiency of the kettle. [1]
[3 marks total]
-
Explain what is meant by the binding energy of a nucleus.
[2 marks]
-
The mass of a helium-4 nucleus is 4.002603 u, while the total mass of its separated nucleons is 4.032980 u.
(1 u = 931.5 MeV/c²)(a) Calculate the mass defect of the helium-4 nucleus in u. [1]
(b) Hence, determine the binding energy of the helium-4 nucleus in MeV. [2]
[3 marks total]
-
In a nuclear reaction, the mass defect is 0.0305 u. Calculate the total kinetic energy of the products, expressing your answer in MeV.
[2 marks]
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In an X-ray tube, electrons bombard a metal target. With reference to energy, explain why the electrons must have a minimum energy to produce the characteristic X-ray spectrum of the target metal.
[2 marks]
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The intensity I of radiation from a point source and the distance x from the source are related by a power law x = k I^n, where k and n are constants.
The following data are obtained:
I (W m⁻²) x (m) 100 2.0 25 4.0 Determine the values of n and k.
[3 marks]
-
A car of mass 1500 kg accelerates from rest to a speed of 72 km h⁻¹ in 10 s on a level road. Friction can be neglected.
(a) Show that the final speed is 20 m s⁻¹. [1]
(b) Calculate the gain in kinetic energy of the car. [1]
(c) Hence, determine the average power developed by the engine during this acceleration. [1]
[3 marks total]
-
A ball of mass 0.20 kg is thrown vertically upward with an initial speed of 15 m s⁻¹. Neglecting air resistance, use conservation of energy to find the maximum height reached.
[2 marks]
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A rubber ball is dropped from a certain height and bounces several times. The kinetic energy of the ball just after each bounce is less than just before the bounce. State two reasons for this energy loss.
[2 marks]
-
The binding energy per nucleon of iron-56 (⁵⁶₂₆Fe) is 8.8 MeV.
(1 eV = 1.60 × 10⁻¹⁹ J)(a) Calculate the total binding energy of an iron-56 nucleus in MeV. [1]
(b) Convert this total binding energy into joules. [1]
[2 marks total]
-
A satellite of mass 2000 kg is raised from the Earth’s surface to an orbit where the gravitational potential difference is 2.0 × 10⁷ J kg⁻¹. Calculate the increase in gravitational potential energy of the satellite.
[2 marks]
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A student claims that a machine with an output power of 100 W and an input power of 150 W has an efficiency of 150%. Explain why this claim is impossible.
[2 marks]
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A simple pendulum has a bob of mass 0.10 kg. At the lowest point of its swing, the bob has a speed of 2.0 m s⁻¹.
Calculate the maximum vertical height the bob reaches above the lowest point.[2 marks]
-
A nuclear power plant generates an electrical power output of 1.0 GW at an efficiency of 33%.
(1 GW = 1 × 10⁹ W)(a) Determine the rate at which thermal energy is produced in the reactor core. [1]
(b) Using Einstein’s mass–energy relation, calculate the rate at which mass is converted to energy in the reactor core. [2]
[3 marks total]
-
The stability of a nucleus is related to its binding energy per nucleon. Explain why a larger binding energy per nucleon indicates a more stable nucleus.
[2 marks]
-
A constant force of 20 N acts on an object and moves it through a distance of 5.0 m in the direction of the force. The movement takes 2.0 s.
(a) Calculate the work done by the force. [1]
(b) Calculate the average power. [1]
[2 marks total]
END OF PAPER
Answers
A-Level Physics H2 Quiz - Energy Power – ANSWERS AND MARKING NOTES
Total Marks: 42
Duration: 45 minutes
Question 1 (Kinetic energy)
KE = ½ m v² = 0.5 × 1200 × (25)² = 0.5 × 1200 × 625
= 375 000 J
Answer: 3.75 × 10⁵ J (or 375 kJ)
Mark: 2 marks (1 for formula, 1 for correct answer with unit)
Question 2 (Conservation of energy – free fall)
By conservation of energy: loss in GPE = gain in KE
GPE = m g h = 0.50 × 9.81 × 20 = 98.1 J
Kinetic energy just before hitting ground = 98.1 J ≈ 98 J (or 100 J using g = 10)
Mark: 2 marks (1 for method, 1 for correct value with unit)
Question 3 (Principle of conservation of energy)
Energy cannot be created or destroyed; it can only be transferred from one form to another or transformed from one type to another. The total energy of an isolated system remains constant.
Mark: 1 mark (accept equivalent complete statement)
Question 4 (Power of motor lifting mass)
Minimum power = force × velocity = (weight) × v
Weight = m g = 50 × 9.81 = 490.5 N
Power = 490.5 × 2.0 = 981 W ≈ 980 W
Answer: 981 W
Mark: 2 marks (1 for force/weight, 1 for power calculation with unit)
Question 5 (Kettle efficiency)
(a) Useful energy = m c Δθ = 0.80 × 4200 × (100 – 25) = 0.80 × 4200 × 75
= 252 000 J [1]
(b) Electrical energy supplied = P × t = 2000 × (3.0 × 60) = 2000 × 180 = 360 000 J [1]
(c) Efficiency = (useful energy / electrical energy) × 100%
= (252 000 / 360 000) × 100% = 70%
Answer: 70% [1]
Mark: 3 marks total, as shown above
Question 6 (Binding energy definition)
The binding energy of a nucleus is the minimum energy required to completely separate the nucleus into its individual constituent protons and neutrons (nucleons).
Mark: 2 marks (1 for “separate into nucleons”, 1 for “minimum energy required” or equivalent wording)
Question 7 (Mass defect and binding energy of He-4)
(a) Mass defect Δm = 4.032980 u – 4.002603 u = 0.030377 u [1]
(b) Binding energy = Δm × 931.5 MeV/u = 0.030377 × 931.5
≈ 28.3 MeV [2]
Mark: 3 marks total (1 for mass defect, 2 for correct conversion and answer)
Question 8 (Kinetic energy from mass defect)
Total KE of products = Δm c² = 0.0305 u × 931.5 MeV/u
= 28.4 MeV (accept 28.4–28.5 MeV)
Answer: 28.4 MeV
Mark: 2 marks (1 for conversion factor, 1 for correct value with unit)
Question 9 (Minimum energy for characteristic X-rays)
To produce characteristic X-rays, an incident electron must have enough energy to remove an inner-shell electron (e.g., from the K-shell) of a target atom. The minimum energy required equals the binding energy (ionisation energy) of that inner-shell electron. When such a vacancy is created, an electron from a higher shell drops down, emitting an X‑ray photon of energy equal to the difference between the energy levels; these are the characteristic lines. Hence, without at least the ionisation energy of the relevant shell, characteristic X-rays cannot be produced.
Mark: 2 marks (1 for linking minimum energy to inner-shell ionisation, 1 for linking to emission of characteristic photon or equivalent)
Question 10 (Power law I = k x^n)
Given x = k I^n → taking logarithms: log x = n log I + log k
Using data:
When I = 100 W m⁻², x = 2.0 m: log 2.0 = n log 100 + log k → 0.301 = 2n + log k
When I = 25 W m⁻², x = 4.0 m: log 4.0 = n log 25 + log k → 0.602 = n log 25 + log k
log 25 = log(5²) = 2 log 5 ≈ 2×0.699 = 1.398 (or use exact logs)
Subtract first equation from second: 0.602 – 0.301 = n (log 25 – log 100)
0.301 = n (1.398 – 2.000) = n (–0.602)
So n = –0.500
Substitute n into first: 0.301 = 2(–0.5) + log k → 0.301 = –1.0 + log k → log k = 1.301 → k = 10^1.301 ≈ 20.0
Thus n = –0.5, k = 20 (units: m × (W m⁻²)^0.5).
Mark: 3 marks (1 for correct log transformation, 1 for finding n, 1 for finding k)
Question 11 (Car acceleration – average power)
(a) 72 km/h = 72×1000 / 3600 = 20 m s⁻¹ [1]
(b) Gain in KE = ½ m v² = 0.5 × 1500 × (20)² = 750 × 400 = 300 000 J [1]
(c) Average power = work done / time = 300 000 / 10 = 30 000 W = 30 kW [1]
Mark: 3 marks total as shown
Question 12 (Maximum height from energy conservation)
Initial KE = ½ m v² = 0.5 × 0.20 × (15)² = 22.5 J
At maximum height, all KE converted to GPE: m g h = 22.5 J
h = 22.5 / (0.20 × 9.81) = 22.5 / 1.962 = 11.5 m (approx 11.5 m)
Answer: 11.5 m
Mark: 2 marks (1 for energy transfer identification, 1 for correct height with working)
Question 13 (Energy loss in bouncing)
- Air resistance / drag causes heat dissipation.
- Internal friction / deformation of the ball (or ground) converts some kinetic energy into thermal energy (and sound).
Mark: 2 marks (1 for each plausible reason, any two from: air resistance, internal damping, plastic deformation, sound)
Question 14 (Binding energy of iron-56)
(a) Total binding energy = number of nucleons × BE per nucleon = 56 × 8.8 MeV = 492.8 MeV ≈ 493 MeV [1]
(b) In joules: 492.8 MeV = 492.8 × 10⁶ eV × 1.60×10⁻¹⁹ J eV⁻¹ = 7.8848 × 10⁻¹¹ J ≈ 7.88 × 10⁻¹¹ J [1]
Mark: 2 marks total
Question 15 (Gravitational potential energy satellite)
Increase in GPE = mass × gravitational potential difference = 2000 × (2.0 × 10⁷) = 4.0 × 10¹⁰ J
Answer: 4.0 × 10¹⁰ J
Mark: 2 marks (1 for formula, 1 for correct answer with unit)
Question 16 (Efficiency > 100% impossible)
Efficiency = (useful output power / input power) × 100% cannot exceed 100% because energy is always conserved; some energy is inevitably transferred to non‑useful forms (e.g., heat, sound) due to friction, resistance, etc. The output energy cannot be greater than the input energy.
Mark: 2 marks (1 for stating that efficiency cannot exceed 100%, 1 for linking to energy conservation/dissipation)
Question 17 (Maximum height of pendulum bob)
At lowest point: KE = ½ m v² = 0.5 × 0.10 × (2.0)² = 0.20 J
At maximum height: KE → GPE → m g h = 0.20 J
h = 0.20 / (0.10 × 9.81) = 0.20 / 0.981 = 0.204 m ≈ 0.20 m
Answer: 0.20 m (2.0 × 10⁻¹ m)
Mark: 2 marks (1 for method, 1 for correct height)
Question 18 (Nuclear power plant – mass conversion)
(a) Thermal power produced = electrical power / efficiency = (1.0 × 10⁹ W) / 0.33 = 3.03 × 10⁹ W [1]
(b) Rate of mass conversion: P = (Δm/Δt) c² → Δm/Δt = P / c² = (3.03 × 10⁹) / (3.00 × 10⁸)² = 3.03×10⁹ / 9.00×10¹⁶ = 3.37 × 10⁻⁸ kg s⁻¹ [2]
Mark: 3 marks total (1 for thermal power, 2 for mass conversion rate with correct units)
Question 19 (Higher binding energy per nucleon = more stable nucleus)
A larger binding energy per nucleon means that more energy per nucleon is needed to break the nucleus apart; thus, the nucleus is held more tightly together. Such nuclei are more stable because they are less likely to undergo spontaneous disintegration (radioactive decay).
Mark: 2 marks (1 for linking to energy required to separate nucleons, 1 for linking to stability or resistance to decay)
Question 20 (Work and power from constant force)
(a) Work done = force × distance = 20 × 5.0 = 100 J [1]
(b) Average power = work / time = 100 / 2.0 = 50 W [1]
Mark: 2 marks total
END OF ANSWERS