| Date | May 2014 | Marks available | 6 | Reference code | 14M.3.HL.TZ2.16 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Calculate and Show that | Question number | 16 | Adapted from | N/A |
Question
This question is about relativistic mechanics.
A rho meson \((\rho )\) decays at rest in a laboratory into a pion \(({\pi ^ + })\) and an anti-pion \(({\pi ^ - })\) according to
\(\rho \to {\pi ^ + } + {\pi ^ - }\).
The rest masses of the particles involved are:
\({m_{{\pi ^ + }}} = {m_{{\pi ^ - }}} = 140{\text{ MeV}}\,{{\text{c}}^{ - 2}}\)
\({m_\rho } = \) \(770{\text{ MeV}}\,{{\text{c}}^{ - {\text{2}}}}\)
(i) Show that the initial momentum of the pion is \(360{\text{ MeV}}\,{{\text{c}}^{ - {\text{1}}}}\).
(ii) Show that the speed of the pion relative to the laboratory is 0.932c.
(iii) Calculate, in \({\text{MeV}}\,{{\text{c}}^{ - 2}}\), the mass that has been converted into energy in this decay.
The pion \(({\pi ^ + })\) emits a muon in the same direction as the velocity of the pion. The speed of the muon is 0.271c relative to the pion. Calculate the speed of the muon relative to the laboratory.
Markscheme
(i) use of \({E^2} = {({\text{m}}{{\text{c}}^2})^2} + {p^2}{c^2}\);
by conservation of energy, total energy of pion is \(\frac{{770}}{2} = {\text{385 MeV}}\);
\({385^2} = {140^2} + {p^2}{c^2}\); (award [3] immediately if this marking point is seen)
Solving for momentum gives the answer p = 359 \(MeV\,c\)\(^{ - 1}\) \( \approx \) 360 \(MeV\,c\)–1.
Answer is given, marks are for correct working only.
No ECF if wrong energy used.
(ii) \(\gamma = \frac{{385}}{{140}}{\text{ }}( = 2.75)\);
hence \(v = \sqrt {1 - \frac{1}{{{\gamma ^2}}}} c = \sqrt {1 - \frac{1}{{{{2.75}^2}}}} c{\text{ }}( = 0.932c)\);
Answer given, award marks for working only.
Watch for ECF from (a)(i) or first marking point.
(iii) \((770 - 2 \times 140) = 490{\text{ MeV}}\,{{\text{c}}^{ - 2}}\);
\(u = \left( {\frac{{u' + v}}{{1 + \frac{{u'v}}{{{c^2}}}}} = } \right)\frac{{0.932{\text{c}} + 0.271{\text{c}}}}{{1 + \frac{{0.932{\text{c}} \times 0.271{\text{c}}}}{{{c^2}}}}}\);
\(u = 0.960{\text{c}}\);
Award [2] for a bald correct answer.
Allow working which does not mention c.
Examiners report
(i) In any question with units expressed in terms of MeV and c there is enormous potential for confusion. However a reasonable number are able to use the relativistic energy - momentum equation \(\left( {{E^2} = {{(m{c^2})}^2} + {p^2}{c^2}} \right)\) correctly. The most common mistake was to try to make use of the value of 'c' in the calculation instead of just sticking with the values given. In (a)(ii) few could determine gamma. (iii) was much easier.
(b) required use of the relativistic velocity addition formula. Quite a few performed velocity subtraction.
