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

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Date May 2019 Marks available [Maximum mark: 2] Reference code 19M.3.HL.TZ2.8bii
Level HL Paper 3 Time zone TZ2
Command term Determine Question number bii Adapted from N/A
bii.
[Maximum mark: 2]
19M.3.HL.TZ2.8bii

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

(bii)

Determine the speed of the proton.

[2]

Markscheme

γ = 1050 938 = 1.12  ✔

v = 0.45 c

OR

V = 1.35 × 10 8 «ms–1»  ✔

 

Examiners report

An easy calculation, that was generally well answered.

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