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Date November 2012 Marks available 3 Reference code 12N.3.SL.TZ0.12
Level Standard level Paper Paper 3 Time zone Time zone 0
Command term Calculate, Determine, Explain, and State Question number 12 Adapted from N/A

Question

This question is about relativistic kinematics.

A source of light S and a detector of light D are placed on opposite walls of a box as shown in the diagram.

According to an observer in the box the distance L between S and D is 6.0m. The box moves with speed v= 0.80c relative to the ground.

Consider the following events.

Event 1: a photon is emitted by S towards D
Event 2: the photon arrives at D

In the context of the theory of relativity, state what is meant by an event.

[1]
a.

(i) Calculate the time interval t between event 1 and event 2 according to an observer in the box.

(ii) According to an observer on the ground the time interval between event 1 and event 2 is T. One student claims that \(T = \frac{t}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\) and another that \(T = t\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} \).

Explain why both students are wrong.

[3]
b.

Relative to an observer on the ground,

(i) calculate the distance between S and D.

(ii) state the speed of the photon leaving S.

(iii) state an expression for the distance travelled by detector D in the time interval T (T is the interval in (b)(ii)).

(iv) determine T, using your answers to (c)(i), (ii) and (iii).

[6]
c.

Markscheme

a point in spacetime / something happening at a particular time and a particular point in space;

a.

(i) \(t = \frac{{6.0}}{{3.0 \times {{10}^8}}} = 2.0 \times {10^{ - 8}}{\rm{s}}\);

(ii) for either formula to be used one of the time intervals must be a proper time interval;
the two events occur at different points in space and so neither observer measures a proper time interval;
the proper time interval is that of the photons;

b.

(i) \(\gamma  = \frac{1}{{\sqrt {1 - {{0.80}^2}} }} = \frac{5}{3} = 1.67\);
\(l = \frac{L}{\gamma } = \frac{{6.0}}{{1.67}} = 3.6{\rm{m}}\);
Award [2] for a bald correct answer.

(ii) c;

(iii) vT or 0.80cT;

(iv) cT=0.80cT+3.6;
\(T = \frac{{3.6}}{{0.20 \times 3.0 \times {{10}^8}}} = 6.0 \times {10^{ - 8}}{\rm{s}}\);
Award [2] for a bald correct answer.

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Option A: Relativity » Option A: Relativity (Core topics) » A.2 – Lorentz transformations
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