DP Physics (last assessment 2024)

Question 23M.3.HL.TZ1.6

[Maximum mark: 5]

23M.3.HL.TZ1.6

A proton is accelerated from rest through a potential difference of 3.40 GV.

(a)

Calculate, in GeV c⁻¹, the momentum of the proton.

[3]

Markscheme

E_T = «0.938 + 3.40 =» 4.34 GeV ✓

attempted use of p²c² = E² − $m_{0}^{2}$ c⁴ «pc = $\sqrt{4 . 34^{2} - 0 . 938^{2}}$ »

p = 4.24 GeV c^−1_✓

Award [2] if omitted rest mass in MP1, the answer is 3.27.

Examiners report

Accelerated proton. The concept of the momentum of an accelerated proton proved to be difficult for HL students.

(b)

Calculate, in terms of c, the speed of the proton.

[2]

Markscheme

$y$ = « $\frac{4.34}{0.938}$ =» 4.63 ✓

v = 0.976c ✓

Allow ECF from (a).