| Date | May 2012 | Marks available | 4 | Reference code | 12M.3.HL.TZ2.15 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Determine | Question number | 15 | Adapted from | N/A |
Question
This question is about mass and energy.
The graph shows the variation with the fraction \(\frac{v}{c}\), of the kinetic energy EK of a particle, where v is the speed of the particle.
Determine, using the value EK =360 MeV from the graph, the rest mass of the particle in MeV c–2.
Determine, using data from the graph, the potential difference required to accelerate the particle in (a) from a speed of 0.63c to a speed of 0.96c. The charge of the particle is +e.
Markscheme
v=0.96c when EK=360(MeV);
calculation of gamma factor \(\gamma = \frac{1}{{\sqrt {1 - {{0.96}^2}} }} = 3.571\);
so that \(m = \frac{{{E_{\rm{K}}}}}{{\left( {\gamma - 1} \right){c^2}}} = \frac{{360}}{{2.571{{\rm{c}}^2}}}\);
m=140MeVc-2;
change in kinetic energy is 360-40 = 320MeV;
so voltage required is 320MV ; (unit needed)
MV needed for second marking point.
Examiners report
