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Question 19M.3.HL.TZ2.8bi

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Date May 2019 Marks available [Maximum mark: 1] Reference code 19M.3.HL.TZ2.8bi
Level HL Paper 3 Time zone TZ2
Command term Determine Question number bi Adapted from N/A
bi.
[Maximum mark: 1]
19M.3.HL.TZ2.8bi

A proton has a total energy 1050 MeV after being accelerated from rest through a potential difference V.

(bi)

Determine the momentum of the proton.

[1]

Markscheme

«p2 c2 = 10502–9382 therefore» p=472«MeVc–1»✔

Examiners report

Most candidates seemed to have the right starting points but mistakes were often made in attempting to convert units. The energy-momentum equation is generally best answered using only ‘MeV’ based units.

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