| 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, Comment, and Determine | 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.
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.
(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.
(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).
Determine, according to Judy in the spaceship, which signal is emitted first.
Markscheme
because the events occur at the same place/point in space for this observer;
Do not allow “events within the same reference frame”.
(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.
(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;
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.
