Physics • Year 12 • Module 7 • Lesson 14
Relativistic Mass, Momentum and Energy
Apply relativistic energy and momentum equations to particle accelerator data, interpret graphs, and evaluate classical versus relativistic predictions.
1. Interpret particle accelerator data
A particle physics team accelerated protons ($m_p c^2 = 938$ MeV) to various speeds. The table shows $v/c$ values. Complete the missing columns. 10 marks (1 per cell)
| $v/c$ | $\gamma$ (3 s.f.) | $E_k$ (MeV) | $p$ (MeV/$c$) | $E_{total}$ (MeV) |
|---|---|---|---|---|
| 0.600 | ||||
| 0.800 | ||||
| 0.960 | ||||
| 0.998 | ||||
| 0.9999 |
Use: $\gamma = 1/\sqrt{1-v^2/c^2}$; $E_k = (\gamma-1)m_p c^2$; $p = \gamma m_p v$ expressed in MeV/$c$; $E_{total} = \gamma m_p c^2$.
1.1 By what factor does the kinetic energy increase as $v/c$ goes from 0.600 to 0.9999? Comment on whether this is linear. 2 marks
1.2 At $v = 0.998c$, calculate the classical kinetic energy $E_k^{class} = \frac{1}{2}m_p v^2$ in MeV and find the percentage error compared to the relativistic value. 2 marks
2. Interpret a graph — proton energy at the Large Hadron Collider
The graph below shows the total energy $E_{total}/m_pc^2$ of a proton as a function of $v/c$ in the Large Hadron Collider (LHC). 7 marks
Figure 2. Total proton energy (in units of $m_p c^2$) as a function of $v/c$. LHC protons reach $\gamma \approx 7500$. Curve is illustrative up to $\gamma = 8$; at LHC conditions, $\gamma$ is off the top of this chart.
2.1 Describe the shape of the curve and explain why it rises more steeply as $v \to c$. 2 marks
2.2 Read from the graph: at what value of $v/c$ does $\gamma \approx 2$? At this point, what is the proton's kinetic energy in MeV? Show your reasoning. 2 marks
2.3 LHC protons have $\gamma \approx 7500$. Calculate their total energy in GeV and their speed, expressing the difference $(c - v)$ in m/s. ($m_p c^2 = 938$ MeV) 3 marks
3. Predict and justify — the Sun’s mass loss
The Sun radiates $3.8 \times 10^{26}$ W and loses mass via nuclear fusion. The Sun's total mass is $2.0 \times 10^{30}$ kg. 7 marks
3.1 Calculate the mass the Sun loses per second (in kg). 2 marks
3.2 Calculate the fraction of the Sun's mass lost per year. Express your answer in scientific notation. 2 marks
3.3 A student claims the Sun will "run out" of mass in about 15 billion years at this rate. Assess this claim using your result from 3.2 and one other piece of scientific reasoning. 3 marks
4. Compare classical and relativistic predictions
Complete the table by applying both classical and relativistic formulas to the same electron data. ($m_e c^2 = 0.511$ MeV) 8 marks (1 per cell)
| Electron speed | Classical $E_k = \frac{1}{2}mv^2$ (MeV) | Relativistic $E_k = (\gamma-1)mc^2$ (MeV) | Classical accurate? (<5% error) |
|---|---|---|---|
| $v = 0.1c$ | |||
| $v = 0.5c$ | |||
| $v = 0.9c$ | |||
| $v = 0.99c$ |
Q1 — Particle accelerator data table
Using $\gamma = 1/\sqrt{1-(v/c)^2}$, $E_k = (\gamma-1) \times 938$ MeV, $p = \gamma \times 938 \times (v/c)$ MeV/$c$, $E_{total} = \gamma \times 938$ MeV:
$v/c = 0.600$: $\gamma = 1.25$; $E_k = 235$ MeV; $p = 703$ MeV/$c$; $E_{total} = 1173$ MeV.
$v/c = 0.800$: $\gamma = 1.667$; $E_k = 625$ MeV; $p = 1251$ MeV/$c$; $E_{total} = 1563$ MeV.
$v/c = 0.960$: $\gamma = 3.571$; $E_k = 2410$ MeV; $p = 3204$ MeV/$c$; $E_{total} = 3350$ MeV.
$v/c = 0.998$: $\gamma = 15.82$; $E_k = 13940$ MeV; $p = 14820$ MeV/$c$; $E_{total} = 14850$ MeV.
$v/c = 0.9999$: $\gamma = 70.71$; $E_k = 65330$ MeV; $p = 66370$ MeV/$c$; $E_{total} = 66340$ MeV.
1.1 $E_k$ increases from $\approx 235$ MeV to $\approx 65330$ MeV, a factor of $\approx 278$. The increase is highly non-linear: doubling from 0.6$c$ to near $c$ multiplies $E_k$ by a factor far exceeding 2, because $\gamma$ diverges as $v \to c$.
1.2 At $v = 0.998c$: classical $E_k = \frac{1}{2}(0.998)^2 m_p c^2 = 0.498 \times 938 = 467$ MeV. Relativistic $E_k = 13940$ MeV. Percentage error $= (13940-467)/13940 \times 100\% \approx 97\%$. The classical formula underestimates by 97% at this speed.
Marking criteria: 1 mark per correctly completed cell (5 rows × 4 computed columns = 20 cells, but only 10 marks awarded: award for $\gamma$ and $E_k$ columns; accept $\pm5\%$ rounding). 1 mark for correct ratio with comment on non-linearity. 1 mark for classical $E_k$ calculation; 1 mark for correct percentage error.
Q2 — Graph interpretation
2.1 The curve is almost flat (close to $\gamma = 1$) for small $v/c$, then rises increasingly steeply as $v \to c$. It rises steeply because the denominator $\sqrt{1-v^2/c^2}$ approaches zero as $v \to c$, making $\gamma = 1/\sqrt{1-v^2/c^2}$ diverge to infinity. [1 for shape description, 1 for mathematical reason]
2.2 From the graph, $\gamma = 2$ occurs at approximately $v/c \approx 0.87$ (i.e. $v = c\sqrt{1 - 1/4} = c\sqrt{0.75} \approx 0.866c$). At $\gamma = 2$: $E_k = (\gamma-1)m_pc^2 = 1 \times 938 = 938$ MeV. [1 for reading $v/c \approx 0.866$; 1 for correct $E_k = 938$ MeV]
2.3 Total energy $= \gamma m_p c^2 = 7500 \times 938 \text{ MeV} = 7035 \text{ GeV} \approx 7.0$ TeV. Speed: $v = c\sqrt{1 - 1/\gamma^2}$. For large $\gamma$, $c - v \approx c/(2\gamma^2) = (3.00 \times 10^8)/(2 \times 7500^2) \approx 2.67$ m/s. So LHC protons travel only $\approx 3$ m/s slower than light. [1 mark for total energy; 1 mark for $v$ calculation; 1 mark for $c - v$ result and comment]
Q3 — Sun's mass loss
3.1 $\Delta m/\Delta t = P/c^2 = (3.8 \times 10^{26})/(3.00 \times 10^8)^2 = 3.8 \times 10^{26} / 9.0 \times 10^{16} = 4.2 \times 10^9$ kg/s. [1 mark for method; 1 mark for correct answer]
3.2 Mass lost per year $= 4.2 \times 10^9 \times 3.156 \times 10^7 = 1.33 \times 10^{17}$ kg. Fraction $= 1.33 \times 10^{17} / 2.0 \times 10^{30} = 6.6 \times 10^{-14}$ per year. [1 mark for mass per year; 1 mark for fraction in scientific notation]
3.3 At a fraction of $6.6 \times 10^{-14}$ per year, it would take $1/(6.6 \times 10^{-14}) \approx 1.5 \times 10^{13}$ years ($\approx 15$ trillion years) to deplete all mass via radiation — so the student's "15 billion years" underestimates by 1000×. However, the claim is still misleading: the Sun's actual lifetime limit is not mass loss but hydrogen fuel depletion. The Sun has only $\approx 0.7\%$ of its hydrogen available for fusion, giving a total main-sequence lifetime of $\approx 10$ billion years (consistent with the known age). [2 marks for quantitative assessment using the fraction; 1 mark for the additional reasoning about fuel vs mass]
Q4 — Classical vs relativistic comparison
Using $m_e c^2 = 0.511$ MeV and $E_k^{class} = \frac{1}{2}(v/c)^2 m_e c^2$:
$v = 0.1c$: Classical $= 0.5 \times 0.01 \times 0.511 = 0.00256$ MeV; $\gamma = 1.00504$; Relativistic $= 0.00504 \times 0.511 = 0.00258$ MeV; error $\approx 0.8\%$; Yes, classical is accurate.
$v = 0.5c$: Classical $= 0.5 \times 0.25 \times 0.511 = 0.0639$ MeV; $\gamma = 1.155$; Relativistic $= 0.155 \times 0.511 = 0.0792$ MeV; error $\approx 19\%$; No.
$v = 0.9c$: Classical $= 0.5 \times 0.81 \times 0.511 = 0.207$ MeV; $\gamma = 2.294$; Relativistic $= 1.294 \times 0.511 = 0.661$ MeV; error $\approx 69\%$; No.
$v = 0.99c$: Classical $= 0.5 \times 0.9801 \times 0.511 = 0.250$ MeV; $\gamma = 7.089$; Relativistic $= 6.089 \times 0.511 = 3.11$ MeV; error $\approx 92\%$; No.
Marking criteria: 1 mark per row for correct relativistic and classical values (to 2 s.f.); 1 mark per row for correct accuracy assessment (accept any answer consistent with a <5% threshold applied correctly).