| Date | May 2014 | Marks available | 4 | Reference code | 14M.3.HL.TZ1.16 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Determine | Question number | 16 | Adapted from | N/A |
Question
This question is about relativistic dynamics.
A proton is accelerated from rest through a potential difference of 2.5 GV. Determine the momentum of the proton after acceleration.
Markscheme
total energy of proton = eV + rest mass;
([2500+938]MeV=)3438;
\({p^2}{c^2} = \left( {{E_{{\rm{tot}}}}^2 = {m_0}^2{c^4} = } \right){3438^2} - {938^2}\);
\(p = 3.3\left( {{\rm{GeV}}{{\rm{c}}^{ - 1}}} \right)\) or 1.76×10-18(kgms-1);
NOTE: The question paper stated the units of potential difference in GeV. Watch for answers stating that the unit of potential difference is V, not eV. For such answers without calculation, award [1].
Award [4] for correct use of potential difference (2.5GeV) divided by e.
ie \(\frac{{2.5 \times {{10}^9}\left( {eV} \right)}}{{1.6 \times {{10}^{19}}\left( {\rm{C}} \right)}} = 1.56 \times {10^{28}}\left( V \right)\).
Examiners report
The units of potential difference were incorrectly stated as GeV in this question. The markscheme was adjusted to ensure no candidate was disadvantaged and all examiners were asked to identify any candidates who appeared to have been thrown by the error. On the whole, candidates had interpreted the question with the correct unit of GV.
Correct answers were given by those who worked in logical manner and who clearly stated that total energy of the proton is the sum of kinetic energy and rest mass energy. The derivation of momentum from the formula was not easy for candidates. Candidates with basic arithmetic and algebra failed here. The more able candidates found the momentum in a clear, straightforward way. Looking through the formulas in data-booklet without understanding is not appropriate here. Many candidates confused kinetic energy with total energy.
