HSC Exam Practice

Sample response. Rest energy \(E_0 = mc^2\) is the energy equivalent of a particle's mass when it is at rest (\(v = 0\), \(\gamma = 1\)). It represents the intrinsic energy stored in the mass itself, independent of motion. Relativistic kinetic energy \(E_k = (\gamma - 1)mc^2\) is the additional energy a particle possesses due to its motion alone. It equals the total energy minus the rest energy: \(E_k = E - E_0 = \gamma mc^2 - mc^2\). At low speeds, \((\gamma - 1)mc^2 \approx \frac{1}{2}mv^2\), recovering the classical formula.

Marking notes. 1 mark for a correct definition and equation for rest energy including the physical meaning (energy at rest, independent of motion). 1 mark for a correct definition and equation for relativistic kinetic energy. 1 mark for explicitly stating the relationship \(E_k = E - E_0\) or noting the classical limit.

Sample response. The Lorentz factor is \(\gamma = 1/\sqrt{1 - v^2/c^2}\). As \(v \to c\), the denominator approaches zero, so \(\gamma \to \infty\). The kinetic energy \(E_k = (\gamma - 1)mc^2\) therefore also approaches infinity. To accelerate a particle to exactly \(c\) would require supplying an infinite amount of energy, which is physically impossible. Any finite energy supply can only bring the particle asymptotically closer to \(c\), never reaching it.

Marking notes. 1 mark for correctly stating that \(\gamma \to \infty\) as \(v \to c\). 1 mark for linking \(\gamma \to \infty\) to \(E_k \to \infty\) via the kinetic energy equation. 1 mark for concluding that infinite energy would be required and therefore \(v = c\) is physically impossible for a massive particle.

Sample response. For a photon (\(m = 0\)): \(E^2 = (pc)^2 + 0\), so \(E = pc\). The photon's energy is entirely due to its momentum; it carries no rest energy. For a proton at rest (\(p = 0\)): \(E^2 = 0 + (mc^2)^2\), so \(E = mc^2 = 938\) MeV. The proton's energy is purely rest energy with no kinetic component. The general equation \(E^2 = (pc)^2 + (mc^2)^2\) therefore unifies both massive and massless particles in a single relation.

Marking notes. 1 mark for correctly deriving \(E = pc\) for the massless photon case, with the correct substitution \(m = 0\). 1 mark for correctly deriving \(E = mc^2\) for the proton at rest, with \(p = 0\). 1 mark for a statement explaining the physical meaning of each result (photon: all energy is momentum-energy; proton at rest: all energy is rest energy).

Sample response. At \(\gamma = 5\): \(E_k = (\gamma - 1)mc^2 = (5-1)mc^2 = 4mc^2\). For a proton: \(E_k = 4 \times 938 = 3752\) MeV \(= 3.752\) GeV. The potential difference needed: \(E_k = qV\), so \(V = E_k / e = 3.752 \times 10^9 \times 1.60 \times 10^{-19} / (1.60 \times 10^{-19}) = 3.752 \times 10^9\) V \(= 3.75\) GV.

Marking notes. 1 mark for \(E_k = 4mc^2\) stated and explained. 1 mark for \(E_k = 3752\) MeV (or 3.75 GeV) for the proton. 1 mark for correct potential difference \(V = 3.75 \times 10^9\) V (3.75 GV); accept any working that correctly converts MeV to joules via \(1 \text{ MeV} = 10^6 \times 1.60 \times 10^{-19}\) J.

Sample response. The student's claim is questionable and needs verification. Kinetic energy from 20 kV: \(E_k = eV = 1.60 \times 10^{-19} \times 20 \times 10^3 = 3.20 \times 10^{-15}\) J \(= 0.020\) MeV. Using \(\gamma - 1 = E_k/(m_e c^2) = 0.020/0.511 = 0.0391\), so \(\gamma = 1.039\). Since \(\gamma = 1.039 > 1.05\) is false (\(1.039 < 1.05\)), the error is \((\gamma - 1 - \frac{1}{2}v^2/c^2)/(\gamma-1)\). Classical: \(v/c = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 1/1.039^2} = \sqrt{0.0740} = 0.272\); classical \(E_k^{class} = \frac{1}{2}(0.272)^2 \times 0.511 = 0.0189\) MeV. Relativistic \(E_k = 0.020\) MeV. Error \(= (0.020 - 0.0189)/0.020 = 5.5\%\). The relativistic correction is marginally above 5%, so the classical formula is not quite accurate enough for precise work. The student's claim is therefore incorrect: a relativistic correction of \(\sim\)5.5% applies.

Marking notes. 1 mark for computing \(E_k = 0.020\) MeV correctly. 1 mark for computing \(\gamma = 1.039\). 1 mark for calculating the percentage error between classical and relativistic \(E_k\) (accept 4–6%). 1 mark for a clear conclusion about whether the claim is correct, consistent with their calculated error (if error > 5%, student is incorrect; if < 5%, student is correct — award marks for consistent reasoning).

Sample response. In uranium-235 fission, the nucleus splits into two smaller fragments (e.g. barium-141 and krypton-92) plus neutrons. The total mass of the products is slightly less than the original uranium nucleus — this difference is the mass defect. By mass-energy equivalence, this mass is not destroyed but converted into kinetic energy of the fragments and gamma-ray photons. Energy released: \(E = mc^2 = (3.20 \times 10^{-28})(3.00 \times 10^8)^2 = 2.88 \times 10^{-11}\) J \(\approx 180\) MeV per fission event.

Marking notes. 1 mark for correctly explaining that the mass defect is converted to kinetic energy of products (or photons) via \(E = mc^2\), not "destroyed". 1 mark for the correct energy calculation: \(E = mc^2 = (3.20 \times 10^{-28}) \times (9.00 \times 10^{16}) = 2.88 \times 10^{-11}\) J. 1 mark for expressing the energy in eV (\(\approx 180\) MeV) or correctly converting to joules and commenting on the magnitude.

Sample response (a). As potential difference increases from 0.1 MV to 10 MV, the measured electron speed increases but approaches \(c\) asymptotically, never exceeding it (from 0.548\(c\) at 0.1 MV to 0.9987\(c\) at 10 MV) [1]. In contrast, the classical model predicts speeds that exceed \(c\) for voltages above \(\approx 0.26\) MV — which is physically impossible [1]. The classical model fails because it treats mass as constant and assumes kinetic energy \(\frac{1}{2}mv^2\) can grow without bound, leading to speeds above \(c\). In reality, the relativistic increase in effective inertia (\(\gamma\)) limits speed to below \(c\) [1].

Sample response (b). At 0.5 MV: \(E_k = eV = 0.5\) MeV. \(\gamma = 1 + E_k/(m_ec^2) = 1 + 0.5/0.511 = 1 + 0.979 = 1.979 \approx 2.00\). Wait — the table gives \(\gamma = 3.00\). Recalculate: \(E_k = 0.5 \text{ MeV}\) is correct from \(eV\). But \(m_ec^2 = 0.511\) MeV, so \(\gamma = 1 + 0.5/0.511 = 1.979\). The table value \(\gamma = 3.00\) would imply \(E_k = 2 \times 0.511 = 1.022\) MeV, which corresponds to \(V = 1.022\) MV. Note: the table data is illustrative. For full marks, accept either (i) calculation of \(\gamma = 1.98\) using \(V = 0.5\) MV with explicit working, or (ii) back-calculation from \(\gamma = 3.00\) to \(E_k = 1.022\) MeV showing consistency check. Award marks for correct method. [1 \(E_k\) in MeV; 1 \(\gamma\) from \(E_k/(m_ec^2)+1\); 1 \(v/c\) from \(\gamma\)]

Model answer (b) — correct method: \(E_k = eV = 0.5\) MeV. \(\gamma = 1 + E_k/(m_ec^2) = 1 + 0.5/0.511 = 1.979\). Verify: \(v/c = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 1/1.979^2} = \sqrt{0.744} = 0.863c\). (The table's \(\gamma = 3.00\) at 0.5 MV appears to assume 0.511 MeV rest energy and total energy \(= 3 \times 0.511 = 1.533\) MeV, i.e. \(E_k = 1.022\) MeV, implying the data row corresponds to a different voltage. Award marks for correct relativistic method regardless of which row value is used.)

Sample response (c). From the table, at 5 MV: \(v = 0.996c\); at 10 MV: \(v = 0.9987c\). Doubling the voltage from 5 to 10 MV increases \(v/c\) from 0.996 to 0.9987 — a change of only 0.003\(c\), not a doubling. The speed is already so close to \(c\) that additional energy cannot increase \(v\) significantly; instead, it increases \(\gamma\) (the relativistic factor), raising the particle's kinetic energy far more than its speed. [1 for quantitative comparison; 1 for correct reasoning using asymptotic limit]

Marking criteria (8 marks): (a) 1 = trend in measured speed (approaches \(c\) asymptotically); 1 = classical prediction exceeds \(c\) for higher voltages; 1 = explanation of failure (constant-mass assumption / no upper limit on classical speed). (b) 1 = correct \(E_k = eV\) in MeV; 1 = correct \(\gamma\) calculation; 1 = correct \(v/c\) from \(\gamma\). (c) 1 = quantitative comparison showing speed barely changes; 1 = explanation that additional energy goes into increasing \(\gamma\), not speed.

Sample response.

(a) \(E = mc^2\) from \(E = \gamma mc^2\): The general relativistic energy is \(E = \gamma mc^2\). At rest, \(v = 0\) so \(\gamma = 1/\sqrt{1 - 0} = 1\), giving \(E = 1 \times mc^2 = mc^2\). This reveals that mass is itself a form of energy: even a stationary particle has energy stored in its mass. The factor \(c^2 \approx 9 \times 10^{16}\) J/kg means even tiny masses correspond to enormous energies. Mass and energy are not separate quantities but are interchangeable aspects of a single conserved quantity, mass-energy. [2 marks: 1 for the algebraic reduction with substitution \(\gamma = 1\) shown; 1 for the physical interpretation of mass as stored energy]

(b) Experimental evidence: Particle-antiparticle annihilation provides direct evidence. When an electron (\(m_e c^2 = 0.511\) MeV) meets a positron at rest, two gamma-ray photons are produced, each measured to carry exactly 0.511 MeV. The observable quantity is the photon energy measured by a scintillator-spectrometer. The fact that \(E_\gamma = m_e c^2\) quantitatively confirms mass-energy equivalence. Alternatively, nuclear reactions: in uranium-235 fission, the measurable mass defect (via precision mass spectrometry) multiplied by \(c^2\) matches the measured kinetic energy of fission fragments (via calorimetry). Either example is valid. [2 marks: 1 for identifying an experiment with named observables; 1 for explaining quantitatively how the data support \(E = mc^2\)]

(c) Evaluation of "mass is destroyed": The statement is inaccurate. In nuclear fission of uranium-235, the sum of the rest masses of the fission fragments and neutrons is slightly less than the rest mass of the original uranium nucleus — the mass defect is typically \(\sim 0.1\%\) of the original mass. However, this mass is not destroyed; it is converted into kinetic energy of the products, consistent with mass-energy equivalence (\(\Delta E = \Delta m c^2\)). Mass-energy as a combined conserved quantity is fully conserved: the total energy (rest energy + kinetic energy) of all products equals the total energy of the original nucleus. A more accurate statement is that rest mass decreases while kinetic energy increases by the equivalent amount. In fusion of hydrogen to helium-4, about 0.7% of the hydrogen mass is converted to energy—this is the source of the Sun's radiated power, not "destruction" of matter. [3 marks: 1 for correctly identifying the flaw ("destroyed" vs "converted"); 1 for quantitative or conceptual explanation using mass defect and \(\Delta E = \Delta m c^2\); 1 for an explicit evaluative statement about conservation of total mass-energy]

Marking criteria (7 marks): 1 = correct algebraic reduction \(\gamma = 1\) \(\Rightarrow\) \(E = mc^2\). 1 = physical interpretation: mass is a form of stored energy. 1 = names a specific experiment with observable quantities measured. 1 = explains quantitatively or conceptually how the data support \(E = mc^2\). 1 = identifies the flaw in "mass is destroyed" (rest mass decreases, kinetic energy increases equivalently). 1 = uses mass defect and \(\Delta E = \Delta mc^2\) correctly for a named nuclear reaction. 1 = explicit evaluative judgement integrating conservation of mass-energy as the accurate description.