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A Level H2 Physics Practice Paper 1

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Questions

TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Secondary School (AI)


Subject: Physics H2
Level: A-Level
Paper: PRACTICE Paper 2
Duration: 2 hours
Total Marks: 80

Name: _________________ Class: _________ Date: _________


Instructions to Candidates

  • Answer all questions in the spaces provided
  • Show all working clearly for full credit
  • Give final answers to an appropriate number of significant figures
  • The use of an approved scientific calculator is expected
  • Take g = 9.81 m s⁻² where required
  • Speed of light c = 3.00 × 10⁸ m s⁻¹

Section A [25 marks]

Question 1 [8 marks]

A student investigates the motion of a trolley on an inclined track using the apparatus shown in Fig. 1.1.

(a) State the principle of conservation of energy. [2]



(b) The trolley has mass 0.50 kg and starts from rest. After traveling 1.2 m down the incline, it has velocity 2.4 m s⁻¹.

(i) Calculate the initial potential energy of the trolley relative to its final position. [2]

Working:

Answer: _________________ J

(ii) Calculate the final kinetic energy of the trolley. [2]

Working:

Answer: _________________ J

(iii) Hence determine the energy lost to friction during the motion. [2]

Working:

Answer: _________________ J


Question 2 [9 marks]

Fig. 2.1 shows a circuit containing a battery, resistors, and an ammeter.

(a) Calculate the total resistance of the circuit shown in Fig. 2.1. [3]

Working:

Answer: _________________ Ω

(b) The battery has EMF 12.0 V and internal resistance 0.50 Ω. Calculate the current reading on ammeter A₁. [3]

Working:

Answer: _________________ A

(c) State two precautions that should be taken when setting up this circuit to ensure accurate measurements. [2]

(i) ________________________________________________________________

(ii) ________________________________________________________________

(d) The student replaces one of the 6.0 Ω resistors with a variable resistor. Explain how this modification could be used to investigate the relationship between current and resistance. [1]



Question 3 [8 marks]

A mass spectrometer is used to analyze isotopes of carbon. Fig. 3.1 shows the path of a ¹²C⁺ ion in the magnetic field region.

(a) State the direction of the magnetic field in the region shown. [1]

Answer: _________________

(b) The ¹²C⁺ ion has mass 1.99 × 10⁻²⁶ kg and enters the magnetic field with speed 2.5 × 10⁵ m s⁻¹. The magnetic flux density is 0.15 T.

(i) Calculate the radius of curvature of the ion's path. [3]

Working:

Answer: _________________ m

(ii) A ¹⁴C⁺ ion with the same charge and speed enters the same magnetic field. State, with a reason, how its radius of curvature compares with that of the ¹²C⁺ ion. [2]

Comparison: _____________________________________________________________

Reason: _______________________________________________________________

(c) Explain why the ions must be accelerated to high speeds before entering the magnetic field region. [2]




Section B [30 marks]

Question 4 [15 marks]

A student uses the photoelectric effect to determine Planck's constant. The experimental setup is shown in Fig. 4.1.

(a) State what is meant by the work function of a metal. [2]



(b) Light of wavelength 450 nm is incident on the metal surface. The stopping potential is measured to be 0.85 V.

(i) Calculate the energy of a photon of this wavelength. [2]

Working:

Answer: _________________ J

(ii) Use Einstein's photoelectric equation to calculate the work function of the metal. [3]

Working:

Answer: _________________ J

(c) The intensity of the incident light is now doubled while keeping the wavelength constant.

(i) State what happens to the stopping potential. [1]

Answer: _________________

(ii) Explain your answer to (c)(i). [2]



(d) The student repeats the experiment with different wavelengths and plots a graph of stopping potential against frequency.

(i) Sketch the expected shape of this graph on the axes provided in Fig. 4.2. [2]

(ii) Explain how Planck's constant can be determined from the gradient of this graph. [2]



(e) State one limitation of this experimental method for determining Planck's constant. [1]



Question 5 [15 marks]

A nuclear power station uses the fission of ²³⁵U to generate electricity. The fission reaction can be represented as:

²³⁵U + ¹n → ¹⁴⁴Ba + ⁸⁹Kr + 3¹n

(a) Explain what is meant by nuclear fission. [2]



(b) The atomic masses are:

  • ²³⁵U: 235.044 u
  • ¹n: 1.009 u
  • ¹⁴⁴Ba: 143.923 u
  • ⁸⁹Kr: 88.918 u

(i) Calculate the mass defect for this fission reaction. [3]

Working:

Answer: _________________ u

(ii) Calculate the energy released in this fission reaction. [2]

Working:

Answer: _________________ MeV

(c) In the reactor, a chain reaction occurs. Explain what is meant by a chain reaction and state one condition necessary for a sustained chain reaction. [3]

Explanation: ____________________________________________________________


Condition: _____________________________________________________________

(d) The power station has an electrical output of 1200 MW and operates with an efficiency of 35%.

(i) Calculate the thermal power input required. [2]

Working:

Answer: _________________ MW

(ii) If each fission reaction releases 200 MeV of energy, calculate the number of fission reactions occurring per second in the reactor. [3]

Working:

Answer: _________________ reactions s⁻¹


Section C [25 marks]

Question 6 [25 marks]

A student investigates the relationship between the extension of a spring and the applied force using the apparatus shown in Fig. 6.1.

(a) State Hooke's law. [1]


(b) The student records the following data:

Force / NExtension / cm
0.00.0
2.01.2
4.02.5
6.03.7
8.05.1
10.06.2

(i) Plot a graph of force (y-axis) against extension (x-axis) on the grid provided in Fig. 6.2. [4]

(ii) Draw the best-fit straight line through your plotted points. [1]

(iii) Use your graph to determine the spring constant of the spring. Show your working clearly. [3]

Working:

Answer: _________________ N m⁻¹

(c) The student notices that when the force exceeds 10.0 N, the relationship between force and extension is no longer linear.

(i) State the name given to the point where this occurs. [1]

Answer: _________________

(ii) Explain what happens to the spring beyond this point. [2]



(d) State and explain three precautions the student should take to improve the accuracy of this experiment. [6]

Precaution 1: ___________________________________________________________

Explanation: ___________________________________________________________

Precaution 2: ___________________________________________________________

Explanation: ___________________________________________________________

Precaution 3: ___________________________________________________________

Explanation: ___________________________________________________________

(e) The spring is now used in a simple harmonic oscillator with a mass of 0.25 kg attached.

(i) Calculate the period of oscillation. [3]

Working:

Answer: _________________ s

(ii) If the amplitude of oscillation is 0.08 m, calculate the maximum acceleration of the mass. [2]

Working:

Answer: _________________ m s⁻²

(f) State one assumption made in the analysis of simple harmonic motion that may not be valid in practice. [2]




END OF PAPER

Answers

TuitionGoWhere Practice Paper - Physics H2 A-Level (Mark Scheme)

Total Marks: 80


Section A [25 marks]

Question 1 [8 marks]

(a) State the principle of conservation of energy. [2] Answer: Energy cannot be created or destroyed, only converted from one form to another. / The total energy of an isolated system remains constant. Marking: 1 mark for conservation statement, 1 mark for conversion/constant total energy.

(b)(i) Calculate initial potential energy. [2] Working: Need to find height: From kinematics, v2=u2+2asv^2 = u^2 + 2as (2.4)2=0+2a(1.2)(2.4)^2 = 0 + 2a(1.2), so a=2.4a = 2.4 m s⁻² sinθ=ag=2.49.81=0.245\sin\theta = \frac{a}{g} = \frac{2.4}{9.81} = 0.245 Height h=1.2sinθ=1.2×0.245=0.294h = 1.2 \sin\theta = 1.2 \times 0.245 = 0.294 m PE=mgh=0.50×9.81×0.294=1.44PE = mgh = 0.50 \times 9.81 \times 0.294 = 1.44 J Answer: 1.4 J Marking: 1 mark for method to find height, 1 mark for correct calculation.

(b)(ii) Calculate final kinetic energy. [2] Working: KE=12mv2=12×0.50×(2.4)2=1.44KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.50 \times (2.4)^2 = 1.44 J Answer: 1.4 J Marking: 1 mark for formula, 1 mark for correct calculation.

(b)(iii) Energy lost to friction. [2] Working: Energy lost = Initial PE - Final KE = 1.44 - 1.44 = 0 J Alternative: If using mgh=0.50×9.81×1.2×sinθmgh = 0.50 \times 9.81 \times 1.2 \times \sin\theta where θ\theta found from energy considerations. Answer: 0 J (or small positive value depending on calculation method) Marking: 1 mark for method, 1 mark for calculation.

Question 2 [9 marks]

(a) Calculate total resistance. [3] Working: Two 6.0 Ω resistors in parallel: 1Rp=16.0+16.0=26.0\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0}, so Rp=3.0R_p = 3.0 Ω Total resistance = 3.0+4.0=7.03.0 + 4.0 = 7.0 Ω Answer: 7.0 Ω Marking: 1 mark for parallel combination, 1 mark for series addition, 1 mark for correct answer.

(b) Calculate current in A₁. [3] Working: Total resistance including internal = 7.0+0.50=7.57.0 + 0.50 = 7.5 Ω Current = EMFRtotal=12.07.5=1.6\frac{EMF}{R_{total}} = \frac{12.0}{7.5} = 1.6 A Answer: 1.6 A Marking: 1 mark for including internal resistance, 1 mark for Ohm's law application, 1 mark for correct answer.

(c) State two precautions. [2] Sample answers: (i) Ensure all connections are tight/secure to minimize contact resistance (ii) Use thick connecting wires to minimize wire resistance / Check ammeter is connected in series Marking: 1 mark each for appropriate precautions.

(d) Explain use of variable resistor. [1] Answer: By varying the resistance, the current changes, allowing investigation of the inverse relationship between current and resistance (at constant voltage). Marking: 1 mark for explanation of varying resistance to change current.

Question 3 [8 marks]

(a) Direction of magnetic field. [1] Answer: Into the page / perpendicular to the page, into the plane Marking: 1 mark for correct direction.

(b)(i) Calculate radius of curvature. [3] Working: F=BQv=mv2rF = BQv = \frac{mv^2}{r} r=mvBQ=1.99×1026×2.5×1050.15×1.60×1019=0.207r = \frac{mv}{BQ} = \frac{1.99 \times 10^{-26} \times 2.5 \times 10^5}{0.15 \times 1.60 \times 10^{-19}} = 0.207 m Answer: 0.21 m Marking: 1 mark for formula, 1 mark for substitution, 1 mark for correct answer.

(b)(ii) Compare ¹⁴C⁺ radius. [2] Comparison: The radius will be larger Reason: Radius is proportional to mass (r ∝ m), and ¹⁴C has greater mass than ¹²C Marking: 1 mark for larger radius, 1 mark for mass relationship.

(c) Explain need for high speeds. [2] Answer: High speeds are needed to provide sufficient kinetic energy for the ions to be deflected in a measurable arc / to ensure the magnetic force is large enough to curve the path significantly / to separate different isotopes effectively. Marking: 1 mark for deflection/separation concept, 1 mark for kinetic energy/force explanation.


Section B [30 marks]

Question 4 [15 marks]

(a) Define work function. [2] Answer: The work function is the minimum energy required to remove an electron from the surface of a metal / the minimum photon energy needed to cause photoemission. Marking: 1 mark for minimum energy, 1 mark for electron removal/photoemission.

(b)(i) Calculate photon energy. [2] Working: E=hcλ=6.63×1034×3.00×108450×109=4.42×1019E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{450 \times 10^{-9}} = 4.42 \times 10^{-19} J Answer: 4.4 × 10⁻¹⁹ J Marking: 1 mark for formula, 1 mark for correct calculation.

(b)(ii) Calculate work function. [3] Working: Einstein equation: hf=Φ+Ekmaxhf = \Phi + E_k^{max} Ekmax=eVs=1.60×1019×0.85=1.36×1019E_k^{max} = eV_s = 1.60 \times 10^{-19} \times 0.85 = 1.36 \times 10^{-19} J Φ=hfEkmax=4.42×10191.36×1019=3.06×1019\Phi = hf - E_k^{max} = 4.42 \times 10^{-19} - 1.36 \times 10^{-19} = 3.06 \times 10^{-19} J Answer: 3.1 × 10⁻¹⁹ J Marking: 1 mark for Einstein equation, 1 mark for kinetic energy calculation, 1 mark for work function.

(c)(i) Effect on stopping potential. [1] Answer: Remains the same / no change Marking: 1 mark for correct answer.

(c)(ii) Explain answer. [2] Answer: Stopping potential depends only on the frequency/energy of the photons, not the intensity. Intensity affects the number of photons, hence the current, but not the maximum kinetic energy of individual photoelectrons. Marking: 1 mark for frequency dependence, 1 mark for intensity affects current not energy.

(d)(i) Sketch graph. [2] Answer: Straight line with positive gradient, x-intercept at threshold frequency, y-intercept negative. Marking: 1 mark for straight line with positive gradient, 1 mark for correct intercepts.

(d)(ii) Determine Planck's constant from gradient. [2] Answer: From eVs=hfΦeV_s = hf - \Phi, the gradient equals he\frac{h}{e}, so h=gradient×eh = \text{gradient} \times e. Marking: 1 mark for identifying gradient as h/e, 1 mark for h = gradient × e.

(e) State limitation. [1] Sample answers: Difficulty in measuring very small currents / Contact potential differences / Surface contamination affects work function Marking: 1 mark for appropriate limitation.

Question 5 [15 marks]

(a) Explain nuclear fission. [2] Answer: Nuclear fission is the process where a heavy nucleus splits into two or more lighter nuclei, usually triggered by neutron bombardment, releasing energy and additional neutrons. Marking: 1 mark for splitting of heavy nucleus, 1 mark for energy release and neutron production.

(b)(i) Calculate mass defect. [3] Working: Mass before = 235.044 + 1.009 = 236.053 u Mass after = 143.923 + 88.918 + (3 × 1.009) = 235.868 u Mass defect = 236.053 - 235.868 = 0.185 u Answer: 0.185 u Marking: 1 mark for mass before, 1 mark for mass after, 1 mark for correct difference.

(b)(ii) Calculate energy released. [2] Working: E=Δm×931.5=0.185×931.5=172E = \Delta m \times 931.5 = 0.185 \times 931.5 = 172 MeV Answer: 172 MeV Marking: 1 mark for conversion factor, 1 mark for correct calculation.

(c) Explain chain reaction and condition. [3] Explanation: A chain reaction occurs when the neutrons produced by one fission event cause further fission reactions, leading to a self-sustaining process. Condition: The reproduction factor (k) must be ≥ 1 / Critical mass must be achieved / Sufficient fissile material must be present Marking: 1 mark for self-sustaining process, 1 mark for neutron-induced fission, 1 mark for appropriate condition.

(d)(i) Calculate thermal power input. [2] Working: Efficiency = Electrical outputThermal input\frac{\text{Electrical output}}{\text{Thermal input}} Thermal input = 12000.35=3430\frac{1200}{0.35} = 3430 MW Answer: 3430 MW (or 3400 MW) Marking: 1 mark for efficiency formula, 1 mark for correct calculation.

(d)(ii) Calculate fission reactions per second. [3] Working: Thermal power = 3430 × 10⁶ W = 3430 × 10⁶ J s⁻¹ Energy per fission = 200 MeV = 200 × 1.60 × 10⁻¹³ J = 3.20 × 10⁻¹¹ J Reactions per second = 3430×1063.20×1011=1.07×1020\frac{3430 \times 10^6}{3.20 \times 10^{-11}} = 1.07 \times 10^{20} s⁻¹ Answer: 1.1 × 10²⁰ reactions s⁻¹ Marking: 1 mark for power conversion, 1 mark for energy per fission, 1 mark for correct division.


Section C [25 marks]

Question 6 [25 marks]

(a) State Hooke's law. [1] Answer: The extension of a spring is directly proportional to the applied force, provided the elastic limit is not exceeded. / F = kx Marking: 1 mark for correct statement.

(b)(i) Plot graph. [4] Marking: 1 mark for correct axes and labels, 1 mark for appropriate scale, 2 marks for accurate plotting of all points.

(b)(ii) Draw best-fit line. [1] Marking: 1 mark for straight line through plotted points.

(b)(iii) Determine spring constant. [3] Working: Spring constant = gradient = ΔFΔx\frac{\Delta F}{\Delta x} From graph: gradient = 10.000.0620=161\frac{10.0 - 0}{0.062 - 0} = 161 N m⁻¹ Answer: 160 N m⁻¹ (accept range 150-170 depending on graph) Marking: 1 mark for gradient method, 1 mark for calculation, 1 mark for correct unit.

(c)(i) Name the point. [1] Answer: Elastic limit / Limit of proportionality Marking: 1 mark for correct term.

(c)(ii) Explain what happens beyond this point. [2] Answer: Beyond the elastic limit, the spring undergoes plastic deformation and will not return to its original length when the force is removed. The relationship between force and extension becomes non-linear. Marking: 1 mark for plastic deformation, 1 mark for permanent change/non-linear relationship.

(d) State and explain three precautions. [6] Sample answers: Precaution 1: Ensure the spring is vertical and not twisted Explanation: This ensures the force acts along the axis of the spring and prevents additional forces that would affect the extension

Precaution 2: Add masses gradually and allow the spring to settle before taking readings Explanation: This allows the spring to reach equilibrium and reduces oscillations that could affect the measurement

Precaution 3: Take readings at eye level to avoid parallax error Explanation: This ensures accurate reading of the extension from the scale

Alternative precautions: Use a set square to ensure vertical alignment, repeat readings and take averages, use appropriate mass range to stay within elastic limit Marking: 2 marks per precaution-explanation pair (1 mark each).

(e)(i) Calculate period of oscillation. [3] Working: T=2πmk=2π0.25160=2π0.00156=0.248T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.25}{160}} = 2\pi\sqrt{0.00156} = 0.248 s Answer: 0.25 s Marking: 1 mark for formula, 1 mark for substitution, 1 mark for correct answer.

(e)(ii) Calculate maximum acceleration. [2] Working: ω=2πT=2π0.248=25.3\omega = \frac{2\pi}{T} = \frac{2\pi}{0.248} = 25.3 rad s⁻¹ amax=ω2A=(25.3)2×0.08=51.2a_{\max} = \omega^2 A = (25.3)^2 \times 0.08 = 51.2 m s⁻² Answer: 51 m s⁻² Marking: 1 mark for finding ω, 1 mark for maximum acceleration calculation.

(f) State one assumption. [2] Sample answers:

  • No damping forces act on the system / Air resistance is negligible
  • The spring obeys Hooke's law throughout the motion
  • The mass of the spring is negligible compared to the attached mass Explanation should relate to why this may not be valid in practice Marking: 1 mark for assumption, 1 mark for explanation of why it may not be valid.

Total: 80 marks

Grade Boundaries (Indicative):

  • A: 65-80 marks (81-100%)
  • B: 55-64 marks (69-80%)
  • C: 45-54 marks (56-68%)
  • D: 35-44 marks (44-55%)
  • E: 25-34 marks (31-43%)