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A Level H2 Physics Energy Power Quiz

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A Level H2 Physics From Real Exams Generated by Gemma 4 31B Updated 2026-06-03

Questions

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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 m s2g = 9.81 \text{ m s}^{-2} and c=3.00×108 m s1c = 3.00 \times 10^8 \text{ m s}^{-1} where applicable.


Section A: Fundamental Concepts (Questions 1–5)

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

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  2. A block of mass 0.50 kg0.50 \text{ kg} slides across a rough horizontal surface at 4.0 m s14.0 \text{ m s}^{-1}. Calculate its initial kinetic energy. [2]

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

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  4. A constant force of 15 N15 \text{ N} acts on a body of mass 2.0 kg2.0 \text{ kg} initially at rest. Calculate the kinetic energy of the body after 3.0 s3.0 \text{ s}. [3]

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  5. 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)

  1. In a nuclear decay process, the mass of the parent nucleus is 208.980 u208.980 \text{ u} and the total mass of the decay products is 208.972 u208.972 \text{ u}. Calculate the total kinetic energy of the decay products in MeV. (Take 1 u=931.5 MeV/c21 \text{ u} = 931.5 \text{ MeV/c}^2) [3]

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  2. An electron bombards a target metal. To produce a characteristic X-ray spectrum, the electron must have a minimum energy of 69.7 keV69.7 \text{ keV}. Explain the physical significance of this minimum energy. [3]

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  3. Calculate the energy released when one nucleus of 235U^{235}\text{U} undergoes fission into two fragments with a total mass defect of 0.20 u0.20 \text{ u}. [3]

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  4. A particle of mass 1.67×1027 kg1.67 \times 10^{-27} \text{ kg} moves with a speed of 2.0×106 m s12.0 \times 10^6 \text{ m s}^{-1}. Calculate its kinetic energy in electron-volts (eV). [3]

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  5. 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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  6. A nucleus undergoes alpha decay. If the total energy released is 5.5 MeV5.5 \text{ MeV}, explain how this energy is distributed among the products. [2]

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  7. Calculate the wavelength of a photon that carries an energy of 1.2 keV1.2 \text{ keV}. [3]

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

  1. A physical quantity xx is related to current II by the power law x=kInx = kI^n. If log10x\log_{10} x is plotted against log10I\log_{10} I, what does the gradient of the straight line represent? [1]
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  2. In the power law x=kInx = kI^n, the y-intercept of the log10x\log_{10} x vs log10I\log_{10} I graph is 0.4770.477. Determine the value of the constant kk. [2]

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  3. A motor lifts a load of 100 kg100 \text{ kg} through a height of 5.0 m5.0 \text{ m} in 10 s10 \text{ s}. Calculate the average power output of the motor. [3]

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

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

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  6. A sample of radioactive material has a mass defect of 0.015 u0.015 \text{ u} per nucleus. Calculate the energy equivalent of this mass defect in Joules. [3]

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  7. A pump delivers water at a rate of 0.10 kg s10.10 \text{ kg s}^{-1} to a height of 20 m20 \text{ m}. If the pump is 70%70\% efficient, calculate the electrical power input. [4]

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  8. A particle is accelerated from rest by a constant power PP. Show that its velocity vv as a function of time tt is proportional to t1/2t^{1/2}. [4]

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Answers

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

  1. Binding Energy: The minimum energy required to completely separate a nucleus into its constituent protons and neutrons. [2]

  2. Calculation: KE=12mv2=12(0.50)(4.0)2=4.0 JKE = \frac{1}{2}mv^2 = \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ J} [2]

  3. Relationship: The binding energy is the energy equivalent of the mass defect, given by E=Δmc2E = \Delta m c^2. [2]

  4. Calculation: a=F/m=15/2.0=7.5 m s2a = F/m = 15/2.0 = 7.5 \text{ m s}^{-2} v=u+at=0+(7.5)(3.0)=22.5 m s1v = u + at = 0 + (7.5)(3.0) = 22.5 \text{ m s}^{-1} KE=12(2.0)(22.5)2=506.25 JKE = \frac{1}{2}(2.0)(22.5)^2 = 506.25 \text{ J} [3]

  5. 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]

  6. Calculation: Δm=208.980208.972=0.008 u\Delta m = 208.980 - 208.972 = 0.008 \text{ u} E=0.008×931.5=7.452 MeVE = 0.008 \times 931.5 = 7.452 \text{ MeV} [3]

  7. 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]

  8. Calculation: E=0.20×931.5=186.3 MeVE = 0.20 \times 931.5 = 186.3 \text{ MeV} [3]

  9. Calculation: KE=12(1.67×1027)(2.0×106)2=3.34×1015 JKE = \frac{1}{2}(1.67 \times 10^{-27})(2.0 \times 10^6)^2 = 3.34 \times 10^{-15} \text{ J} KE in eV=(3.34×1015)/(1.6×1019)=20,875 eV20.9 keVKE \text{ in eV} = (3.34 \times 10^{-15}) / (1.6 \times 10^{-19}) = 20,875 \text{ eV} \approx 20.9 \text{ keV} [3]

  10. 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]
  11. 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]

  12. Calculation: E=1.2×103×1.6×1019=1.92×1016 JE = 1.2 \times 10^3 \times 1.6 \times 10^{-19} = 1.92 \times 10^{-16} \text{ J} λ=hc/E=(6.63×1034×3×108)/1.92×1016=1.036×109 m=1.04 nm\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]

  13. Gradient: The gradient represents the exponent nn. [1]

  14. Calculation: log10k=0.477\log_{10} k = 0.477 k=100.4773.0k = 10^{0.477} \approx 3.0 [2]

  15. Calculation: W=mgh=100×9.81×5.0=4905 JW = mgh = 100 \times 9.81 \times 5.0 = 4905 \text{ J} P=W/t=4905/10=490.5 WP = W/t = 4905 / 10 = 490.5 \text{ W} [3]

  16. Calculation: 8=k(2n)8 = k(2^n) and 64=k(4n)64 = k(4^n) Divide: 64/8=(4/2)n8=2nn=364/8 = (4/2)^n \Rightarrow 8 = 2^n \Rightarrow n = 3 [3]

  17. Derivation: Force F=mgsinθ+μmgcosθF = mg \sin\theta + \mu mg \cos\theta P=Fv=(mgsinθ+μmgcosθ)v=mgv(sinθ+μcosθ)P = Fv = (mg \sin\theta + \mu mg \cos\theta)v = mgv(\sin\theta + \mu \cos\theta) [4]

  18. Calculation: Δm=0.015×1.66×1027 kg=2.49×1029 kg\Delta m = 0.015 \times 1.66 \times 10^{-27} \text{ kg} = 2.49 \times 10^{-29} \text{ kg} E=Δmc2=2.49×1029×(3×108)2=2.24×1012 JE = \Delta m c^2 = 2.49 \times 10^{-29} \times (3 \times 10^8)^2 = 2.24 \times 10^{-12} \text{ J} [3]

  19. Calculation: Pout=dEdt=dmdtgh=0.10×9.81×20=19.62 WP_{\text{out}} = \frac{dE}{dt} = \frac{dm}{dt}gh = 0.10 \times 9.81 \times 20 = 19.62 \text{ W} Pin=Pout/0.70=19.62/0.70=28.03 WP_{\text{in}} = P_{\text{out}} / 0.70 = 19.62 / 0.70 = 28.03 \text{ W} [4]

  20. Proof: P=dEdt=ddt(12mv2)=mvdvdtP = \frac{dE}{dt} = \frac{d}{dt}(\frac{1}{2}mv^2) = mv \frac{dv}{dt} Pdt=mvdvP dt = mv dv Integrate: Pt=12mv2Pt = \frac{1}{2}mv^2 v2=2Ptmv=2Pmt1/2v^2 = \frac{2Pt}{m} \Rightarrow v = \sqrt{\frac{2P}{m}} t^{1/2} Thus vt1/2v \propto t^{1/2}. [4]