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Date	November 2017	Marks available	4	Reference code	17N.3.HL.TZ0.7
Level	Higher level	Paper	Paper 3	Time zone	0 - no time zone
Command term	Determine	Question number	7	Adapted from	N/A

Question

The ${\Lambda ^0}$ ${\Lambda ^0}$ (Lambda) particle decays spontaneously into a proton and a negatively charged pion of rest mass 140 MeV c^–2. After the decay, the particles are moving in the same direction with a proton momentum of 630 MeV c^–1 and a pion momentum of 270 MeV c^–1.

Determine the rest mass of the ${\Lambda ^0}$ ${\Lambda ^0}$ particle.

[4]

Determine, using your answer to (a), the initial speed of the ${\Lambda ^0}$ ${\Lambda ^0}$ particle.

[2]

Markscheme

$\Lambda$ $\Lambda$ momentum = 900

E_proton = « $\sqrt {p{c^2} + {{\left( {m{c^2}} \right)}^2}} = \sqrt {{{630}^2} + {{938}^2}} =$ $\sqrt {p{c^2} + {{\left( {m{c^2}} \right)}^2}} = \sqrt {{{630}^2} + {{938}^2}} =$ » 1130 «MeV»

E_pion = « $\sqrt {{{270}^2} + {{140}^2}} =$ $\sqrt {{{270}^2} + {{140}^2}} =$ » 304 «MeV»

so rest mass of $\Lambda$ $\Lambda$ = « $\sqrt {{{\left( {1130 + 304} \right)}^2} - {{900}^2}} =$ $\sqrt {{{\left( {1130 + 304} \right)}^2} - {{900}^2}} =$ » 1116 «MeV c^–2»

«E = $\gamma$ $\gamma$ mc² so» $\gamma$ $\gamma$ = « $\frac{{1434}}{{1116}}$ $\frac{{1434}}{{1116}}$ =» 1.28

to give 0.64c

Examiners report

[N/A]

Syllabus sections

Option A: Relativity » Option A: Relativity (Additional higher level option topics) » A.4 – Relativistic mechanics (HL only)

Show 27 related questions

18M.3.HL.TZ1.6b:
State the rest mass of the pion with an appropriate unit.
18M.3.HL.TZ2.6b.ii:
The diagram shows the paths of the incident protons together with the proton and neutron created in the interaction. On the diagram, draw the path of the pion.
18M.3.HL.TZ2.6a:
Calculate the gamma (γ) factor for one of the protons.
18N.3.HL.TZ0.6a:
Show that the momentum of the electron is 1.41 MeV c^–1.
18N.3.HL.TZ0.6b.i: Explain the origin of each equation.
18N.3.HL.TZ0.6b.ii: Calculate, in MeV c–1, p1 and p2.
19M.3.HL.TZ2.8bi: Determine the momentum of the proton.
16N.3.HL.TZ0.8a: Explain, in terms of a conservation law, why two photons need to be created.
17N.3.HL.TZ0.7b:
Determine, using your answer to (a), the initial speed of the ${\Lambda ^0}$ ${\Lambda ^0}$ particle.
16N.3.HL.TZ0.8b:
Determine the speed of the incoming electron.
16N.3.HL.TZ0.8c:
Calculate the energy and the momentum for each photon after the collision.
17M.3.HL.TZ1.5a:
Calculate the potential difference V.
19M.3.HL.TZ2.8bii:
Determine the speed of the proton.
18M.3.HL.TZ2.6b.i:
Determine, in terms of MeV c^–1, the momentum of the pion.
19M.3.HL.TZ1.6a:
Show that energy is conserved in this decay.
19N.3.HL.TZ0.5a(i):
the total energy.
19N.3.HL.TZ0.5a(ii):
the speed.
19N.3.HL.TZ0.5b:
Determine the rest mass of X.
20N.3.HL.TZ0.6c: In relativistic reactions the mass of the products may be less than the mass of the...
20N.3.HL.TZ0.6a: Define rest mass.
17M.3.HL.TZ2.6:
A lambda $\Lambda$ $\Lambda$ ⁰ particle at rest decays into a proton p and a pion ${\pi ^ - }$ ${\pi ^ - }$ according to the reaction

$\Lambda$ $\Lambda$ ⁰ → p + $\pi$ $\pi$ ^–

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

The speed of the pion after the decay is 0.579c. For this speed $\gamma$ $\gamma$ = 1.2265. Calculate the speed of the proton.
18M.3.HL.TZ1.6a.ii:
show that the energy of the pion is about 140 MeV.
20N.3.HL.TZ0.6b:
Calculate the total energy of the deuterium particle in $MeV$ $MeV$ .
19M.3.HL.TZ2.8a: Define total energy.
19M.3.HL.TZ2.8biii:
Calculate the potential difference V.
17M.3.HL.TZ1.5b:
The proton collides with an antiproton moving with the same speed in the opposite direction. As a result both particles are annihilated and two photons of equal energy are produced.

Determine the momentum of one of the photons.
19M.3.HL.TZ1.6b:
Calculate the speed of the pion.

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Option A: Relativity » Option A: Relativity (Additional higher level option topics)

Option A: Relativity

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