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A Level H2 Physics Energy Power Quiz

Free A Level H2 Physics Energy Power quiz, Gemma31B Exam version, with questions, answers, and A Level-style practice for Singapore students.

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A Level H2 Physics From Real Exams Generated by Gemma 4 31B Updated 2026-08-17

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Questions

A-Level Physics H2 Quiz - Energy Power

Name: ____________________ Class: ____________________ Date: ____________________ Score: / 50

Duration: 60 minutes
Total Marks: 50
Instructions: Answer all questions. Show all working for calculations. Use $g = 9.81 \text{ m s}^{-2}$ and $c = 3.00 \times 10^8 \text{ m s}^{-1}$ where applicable.

Section A: Fundamental Concepts (Questions 1–5)

Define the term binding energy of a nucleus. [2]

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A block of mass $0.50 \text{ kg}$ slides across a rough horizontal surface at $4.0 \text{ m s}^{-1}$ . Calculate its initial kinetic energy. [2]

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State the relationship between the mass defect of a nucleus and its binding energy. [2]

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A constant force of $15 \text{ N}$ acts on a body of mass $2.0 \text{ kg}$ initially at rest. Calculate the kinetic energy of the body after $3.0 \text{ s}$ . [3]

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Explain why a nucleus with a higher binding energy per nucleon is generally more stable. [2]

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Section B: Nuclear Energy & Transitions (Questions 6–12)

In a nuclear decay process, the mass of the parent nucleus is $208.980 \text{ u}$ and the total mass of the decay products is $208.972 \text{ u}$ . Calculate the total kinetic energy of the decay products in MeV. (Take $1 \text{ u} = 931.5 \text{ MeV/c}^2$ ) [3]

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An electron bombards a target metal. To produce a characteristic X-ray spectrum, the electron must have a minimum energy of $69.7 \text{ keV}$ . Explain the physical significance of this minimum energy. [3]

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Calculate the energy released when one nucleus of $^{235}\text{U}$ undergoes fission into two fragments with a total mass defect of $0.20 \text{ u}$ . [3]

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A particle of mass $1.67 \times 10^{-27} \text{ kg}$ moves with a speed of $2.0 \times 10^6 \text{ m s}^{-1}$ . Calculate its kinetic energy in electron-volts (eV). [3]

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Distinguish between the continuous X-ray spectrum and the characteristic X-ray spectrum in terms of the energy loss mechanism of the incident electrons. [3]

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A nucleus undergoes alpha decay. If the total energy released is $5.5 \text{ MeV}$ , explain how this energy is distributed among the products. [2]

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Calculate the wavelength of a photon that carries an energy of $1.2 \text{ keV}$ . [3]

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Section C: Power Laws & Applied Energy (Questions 13–20)

A physical quantity $x$ is related to current $I$ by the power law $x = kI^n$ . If $\log_{10} x$ is plotted against $\log_{10} I$ , what does the gradient of the straight line represent? [1]
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In the power law $x = kI^n$ , the y-intercept of the $\log_{10} x$ vs $\log_{10} I$ graph is $0.477$ . Determine the value of the constant $k$ . [2]

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A motor lifts a load of $100 \text{ kg}$ through a height of $5.0 \text{ m}$ in $10 \text{ s}$ . Calculate the average power output of the motor. [3]

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A variable $y$ depends on $z$ such that $y = kz^n$ . Given that when $z = 2, y = 8$ and when $z = 4, y = 64$ , find the value of $n$ . [3]

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A block of mass $m$ is pushed up a rough incline of angle $\theta$ at a constant speed $v$ . If the coefficient of friction is $\mu$ , derive an expression for the power delivered by the pushing force. [4]

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A sample of radioactive material has a mass defect of $0.015 \text{ u}$ per nucleus. Calculate the energy equivalent of this mass defect in Joules. [3]

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A pump delivers water at a rate of $0.10 \text{ kg s}^{-1}$ to a height of $20 \text{ m}$ . If the pump is $70\%$ efficient, calculate the electrical power input. [4]

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A particle is accelerated from rest by a constant power $P$ . Show that its velocity $v$ as a function of time $t$ is proportional to $t^{1/2}$ . [4]

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Answers

Answer Key - A-Level Physics H2 Quiz: Energy Power

Binding Energy: The minimum energy required to completely separate a nucleus into its constituent protons and neutrons. [2]
Calculation: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ J}$ [2]
Relationship: The binding energy is the energy equivalent of the mass defect, given by $E = \Delta m c^2$ . [2]
Calculation: $a = F/m = 15/2.0 = 7.5 \text{ m s}^{-2}$ $v = u + at = 0 + (7.5)(3.0) = 22.5 \text{ m s}^{-1}$ $KE = \frac{1}{2}(2.0)(22.5)^2 = 506.25 \text{ J}$ [3]
Stability: Higher binding energy per nucleon means more energy is required to remove a nucleon from the nucleus, making it more tightly bound and stable. [2]
Calculation: $\Delta m = 208.980 - 208.972 = 0.008 \text{ u}$ $E = 0.008 \times 931.5 = 7.452 \text{ MeV}$ [3]
X-ray Significance: The minimum energy corresponds to the ionization energy of the K-shell electrons of the target metal. Electrons must have at least this energy to eject a K-shell electron to produce characteristic X-rays. [3]
Calculation: $E = 0.20 \times 931.5 = 186.3 \text{ MeV}$ [3]
Calculation: $KE = \frac{1}{2}(1.67 \times 10^{-27})(2.0 \times 10^6)^2 = 3.34 \times 10^{-15} \text{ J}$ $KE \text{ in eV} = (3.34 \times 10^{-15}) / (1.6 \times 10^{-19}) = 20,875 \text{ eV} \approx 20.9 \text{ keV}$ [3]
Comparison:
- Continuous: Produced by Bremsstrahlung (braking radiation) where electrons lose varying amounts of energy as they are decelerated by the nucleus.
- Characteristic: Produced when an incident electron knocks out an inner-shell electron, and an outer-shell electron drops down to fill the vacancy, emitting a photon of a specific energy. [3]
Distribution: The energy is shared as kinetic energy between the alpha particle and the daughter nucleus. Due to conservation of momentum, the lighter alpha particle carries the majority of the kinetic energy. [2]
Calculation: $E = 1.2 \times 10^3 \times 1.6 \times 10^{-19} = 1.92 \times 10^{-16} \text{ J}$ $\lambda = hc/E = (6.63 \times 10^{-34} \times 3 \times 10^8) / 1.92 \times 10^{-16} = 1.036 \times 10^{-9} \text{ m} = 1.04 \text{ nm}$ [3]
Gradient: The gradient represents the exponent $n$ . [1]
Calculation: $\log_{10} k = 0.477$ $k = 10^{0.477} \approx 3.0$ [2]
Calculation: $W = mgh = 100 \times 9.81 \times 5.0 = 4905 \text{ J}$ $P = W/t = 4905 / 10 = 490.5 \text{ W}$ [3]
Calculation: $8 = k(2^n)$ and $64 = k(4^n)$ Divide: $64/8 = (4/2)^n \Rightarrow 8 = 2^n \Rightarrow n = 3$ [3]
Derivation: Force $F = mg \sin\theta + \mu mg \cos\theta$ $P = Fv = (mg \sin\theta + \mu mg \cos\theta)v = mgv(\sin\theta + \mu \cos\theta)$ [4]
Calculation: $\Delta m = 0.015 \times 1.66 \times 10^{-27} \text{ kg} = 2.49 \times 10^{-29} \text{ kg}$ $E = \Delta m c^2 = 2.49 \times 10^{-29} \times (3 \times 10^8)^2 = 2.24 \times 10^{-12} \text{ J}$ [3]
Calculation: $P_{\text{out}} = \frac{dE}{dt} = \frac{dm}{dt}gh = 0.10 \times 9.81 \times 20 = 19.62 \text{ W}$ $P_{\text{in}} = P_{\text{out}} / 0.70 = 19.62 / 0.70 = 28.03 \text{ W}$ [4]
Proof: $P = \frac{dE}{dt} = \frac{d}{dt}(\frac{1}{2}mv^2) = mv \frac{dv}{dt}$ $P dt = mv dv$ Integrate: $Pt = \frac{1}{2}mv^2$ $v^2 = \frac{2Pt}{m} \Rightarrow v = \sqrt{\frac{2P}{m}} t^{1/2}$ Thus $v \propto t^{1/2}$ . [4]