| Date | May 2015 | Marks available | 2 | Reference code | 15M.3.SL.TZ1.11 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Outline | Question number | 11 | Adapted from | N/A |
Question
This question is about relativistic kinematics.
An observer at rest relative to Earth observes two spaceships. Each spaceship is moving with a speed of 0.85 c but in opposite directions. The observer measures the rate of increase of distance between the spaceships to be 1.7 c. Outline whether this observation contravenes the theory of special relativity.
The observer on Earth in (a) watches one spaceship as it travels to a distant star at a speed of 0.85 c. According to observers on the spaceship, this journey takes 8.0 years.
(i) Calculate, according to the observer on Earth, the time taken for the journey to the star.
(ii) Calculate, according to the observer on Earth, the distance from Earth to the star.
(iii) At the instant when the spaceship passes the star, the observer on the spaceship sends a radio message to Earth. The spaceship continues to move at a speed of 0.85 c. Determine, according to the spaceship observer, the time taken for the message to arrive on Earth.
Markscheme
theory suggests that no object can travel faster than light;
the 1.7c is not the speed of a physical object;
so is not in violation of the theory;
(i) y \( = \) 1.90;
interval on Earth \( = \) y\( \times \)interval on spaceship;
(interval on Earth 1.90\( \times \)8 years \( = \) )15 years;
Award [3] for a bald correct answer.
(ii) observer on Earth thinks spaceship has travelled for 15 years;
so distance is 0.85c\( \times \)15 \( = \) 12.8≈13ly;
Award [2] for a bald correct answer
or
the spaceship observer observes the distance moved by the Earth \( = \) 0.85c×8.0 yr;
proper distance \( = \) 1.90\( \times \)0.85c\( \times \)8.0yr \( = \) 12.9≈13ly ;
Award [2] for a bald correct answer
(iii) (take time for message to arrive at Earth in spaceship frame to be T)
distance moved by Earth in spaceship frame before message arrives = 0.85cT;
distance of Earth from spaceship when message sent=0.85c\( \times \)8.0=6.8(ly);
\(\left( {cT = 0.85cT + 6.8} \right){\text{so }}T = \frac{{6.8}}{{0.15}} = 45.3{\text{ years}}\)
