| Date | May 2015 | Marks available | 2 | Reference code | 15M.3.HL.TZ2.16 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Calculate | Question number | 16 | Adapted from | N/A |
Question
This question is about relativistic mechanics.
A proton is accelerated from rest through a potential difference of 1.5 GV.
Calculate, for the accelerated proton, the
total energy.
momentum.
speed.
Markscheme
change in total energy/kinetic energy is 1.5 GeV;
total energy is 1.5+0.938=2.4(GeV) or 3.8×10-10 (J) or 3.9×10-10 (J);
Award [2] for a bald correct answer.
\(pc\left( { = \sqrt {{E^2} - {{\left[ {m{c^2}} \right]}^2}} } \right) = \sqrt {{{2.4}^2} - {{\left[ {0.938} \right]}^2}} \); (allow ECF from (a))
p=2.2 or 2.3(GeVc-1) or 1.2×10-18 (kgms-1);
Award [2] for a bald correct answer.
\(u = \frac{p}{{\gamma {m_0}}} = \frac{{2.25}}{{2.6 \times 0.938}} = 0.92{\rm{c}}\); (allow ECF from (a) and (b))
Award [2] for a bald correct answer.
or
\(\left( {p = \gamma {m_0}u = \gamma {m_0}{c^2}\frac{u}{{{c^2}}} \Rightarrow } \right)u = \frac{{pc}}{E}c\);
\( = \left( {\frac{{2.2}}{{2.4}}c = } \right)0.92c\); (allow ECF from (a) and (b))
Award [2] for a bald correct answer.
Examiners report
In part (a) the KE was usually easily identified and added to the proton rest energy.
Part (b): In any question with units expressed in terms of MeV and c there is enormous potential for confusion. However an increasing number of candidates are able use the relativistic energy – momentum equation (E2 = (mc2)2 + p2c2) correctly as they realise that it becomes a Pythagorean E2 = m2 + p2 when using the simpler units. The commonest mistake was to try to make use of the value of “c” in the calculation instead of just sticking with the values given.
In (c) gamma was frequently found correctly and converted to a speed, but ECF was often necessary.
