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Date May 2014 Marks available 3 Reference code 14M.3.SL.TZ2.10
Level Standard level Paper Paper 3 Time zone Time zone 2
Command term Calculate Question number 10 Adapted from N/A

Question

This question is about relativistic kinematics.

The diagram shows a spaceship as it moves past Earth on its way to a planet P. The planet is at rest relative to Earth.

M14/4/PHYSI/SP3/ENG/TZ2/10

The distance between the Earth and planet P is 12 ly as measured by observers on Earth. The spaceship moves with speed 0.60c relative to Earth.

Consider two events:

     Event 1: when the spaceship is above Earth

     Event 2: when the spaceship is above planet P

Judy is in the spaceship and Peter is at rest on Earth.

Judy considers herself to be at rest. According to Judy, the Earth and planet P are moving to the left.

State the reason why the time interval between event 1 and event 2 is a proper time interval as measured by Judy.

[1]
a.

(i)     Calculate the time interval between event 1 and event 2 according to Peter.

(ii)     Calculate the time interval between event 1 and event 2 according to Judy.

[3]
b.

(i)     Calculate, according to Judy, the distance separating the Earth and planet P.

(ii)     Using your answers to (b)(ii) and (c)(i), determine the speed of planet P relative to the spaceship.

(iii)     Comment on your answer to (c)(ii).

[3]
c.

Determine, according to Judy in the spaceship, which signal is emitted first.

[3]
d.

Markscheme

because the events occur at the same place/point in space for this observer;

Do not allow “events within the same reference frame”.

a.

(i)     \(t = \left( {\frac{{12}}{{0.60{\text{c}}}} = } \right){\text{ 20(yr)}}\);

(ii)     \(\gamma  = \left( {\frac{1}{{\sqrt {1 - {{0.60}^2}} }} = } \right){\text{ 1.25}}\); (allow implicit value)

\({t_{{\text{rocket}}}} = \left( {\frac{{20 {\text{yr}}}}{\gamma } = } \right){\text{ 16(yr)}}\); (allow ECF)

Award [2] for a bald correct answer.

b.

(i)     \(L = \left( {\frac{{12{\text{ly}}}}{\gamma } = } \right){\text{ 9.6(ly)}}\); (allow ECF from (b)(ii))

(ii)     \(v = \left( {\frac{{9.6{\text{ly}}}}{{16{\text{y}}}} = } \right){\text{ 0.60c}}\); (allow ECF from (b)(ii) and (c)(i))

(iii)     (by principle of relativity this should be the) same as the speed of the spaceship relative to Earth;

c.

both signals travel at the same speed c;

Judy must agree that the signals arrive at S simultaneously / OWTTE;

for Judy, observer S moves away from the signal traveling from P/towards the signal traveling from Earth;

for Judy the signal from P has further to travel to reach S – so was emitted first;

Do not accept explanations based on Judy approaching P or seeing/receiving the signal from P first as this is irrelevant.

Award [0] for a bald correct answer.

d.

Examiners report

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

Syllabus sections

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