A lambda \(\Lambda \)⁰ particle at rest decays into a proton p and a pion \({\pi ^ - }\) according to the reaction

\(\Lambda \)⁰ → p + \(\pi \)^–

where the rest energy of p = 938 MeV and the rest energy of \(\pi \)^– = 140 MeV.

The speed of the pion after the decay is 0.579c. For this speed \(\gamma \) = 1.2265. Calculate the speed of the proton.

pion momentum is \(\gamma mv = 1.2265 \times 140 \times 0.579 = 99.4\) «MeV\(\,\)c^–1»

use of momentum conservation to realize that produced particles have equal and opposite momenta

so for proton \(\gamma v = \frac{{99.4}}{{938}} = 0.106c\)

solving to get v = 0.105c

Accept pion momentum calculation using E ² = p ²c ² +m ²c ⁴.

Award [2 max] for a non-relativistic answer of v = 0.0864c

[4 marks]