| Date | May 2018 | Marks available | 3 | Reference code | 18M.3.HL.TZ2.6 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Determine | Question number | 6 | Adapted from | N/A |
Question
Two protons, travelling in opposite directions, collide. Each has a total energy of 3.35 GeV.
As a result of the collision, the protons are annihilated and three particles, a proton, a neutron, and a pion are created. The pion has a rest mass of 140 MeV c–2. The total energy of the emitted proton and neutron from the interaction is 6.20 GeV.
Calculate the gamma (γ) factor for one of the protons.
Determine, in terms of MeV c–1, the momentum of the pion.
The diagram shows the paths of the incident protons together with the proton and neutron created in the interaction. On the diagram, draw the path of the pion.
Markscheme
γ «= \(\frac{{3350}}{{938}}\)» = 3.37
[1 mark]
energy of pion = (3350 × 2) – 6200 = 500 «MeV»
5002 = p2c2 + 1402
p = 480 «MeV c–1»
[3 marks]
path of pion constructed in direction around 4–5 o’clock by eye
[1 mark]
