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Question 18M.3.HL.TZ2.b.i

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Date May 2018 Marks available [Maximum mark: 3] Reference code 18M.3.HL.TZ2.b.i
Level HL Paper 3 Time zone TZ2
Command term Determine Question number b.i Adapted from N/A
b.i.
[Maximum mark: 3]
18M.3.HL.TZ2.b.i

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.

Determine, in terms of MeV c–1, the momentum of the pion.

[3]

Markscheme

energy of pion = (3350 × 2) – 6200 = 500 «MeV»

5002 = p2c2 + 1402

p = 480 «MeV c–1»

 

[3 marks]

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