| Date | May 2013 | Marks available | 2 | Reference code | 13M.3.HL.TZ2.18 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Calculate | Question number | 18 | Adapted from | N/A |
Question
This question is about relativistic mechanics.
A proton, after acceleration from rest through a potential difference V, has momentum 1600MeVc–1.
Calculate the speed of the proton after acceleration.
Markscheme
\(\gamma = \frac{{1855}}{{938}} = 1.977\) or \(\gamma - 1 = \frac{{917}}{{938}}\);
\(\frac{v}{c} = \sqrt {1 - \frac{1}{{{\gamma ^2}}}} = 0.86{\rm{c}}\) or \(v = \frac{{1600}}{{938 \times 1.997}}{\rm{c = 0.86c}}\);
Award [2] for a bald correct answer.
Watch for ECF from (a).
Examiners report
in (b) gamma was often obtained correctly to find V - but many more got lost. Surprisingly no useful 'Pythagorean triangles' were seen which make understanding this topic much easier.
Syllabus sections
Option A: Relativity » Option A: Relativity (Additional higher level option topics) » A.4 – Relativistic mechanics (HL only)
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