| Date | November 2011 | Marks available | 3 | Reference code | 11N.3.HL.TZ0.13 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Determine | Question number | 13 | Adapted from | N/A |
Question
This question is about relativistic energy and momentum.
A proton is accelerated from rest through a potential difference V. The proton reaches a speed of 0.970c. Determine the value of V.
[3]
a.
Calculate, after acceleration for the proton in (a), its
(i) mass.
(ii) momentum.
[2]
b.
Markscheme
V=(γ–1)938×106;
\(\gamma = \left( {\frac{1}{{\sqrt {1 - {{\left( {0.970} \right)}^2}} }}} \right) = 4.11\);
V=(3.11×938×106=) 2.92×109V or 2.92GV;
a.
(i) 3.86×103MeVc-2 or 3.86GeVc-2;
(ii) 3.74×103MeVc-1 or 3.74GeVc-1;
b.
Examiners report
[N/A]
a.
[N/A]
b.
Syllabus sections
Option A: Relativity » Option A: Relativity (Additional higher level option topics) » A.4 – Relativistic mechanics (HL only)
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