Date | May 2015 | Marks available | 9 | Reference code | 15M.3.HL.TZ1.14 |
Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
Command term | Calculate and Determine | Question number | 14 | 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 0.85c but in opposite directions. The observer measures the rate of increase of distance between the spaceships to be 1.7c.
(i) Outline whether this observation contravenes the theory of special relativity.
(ii) Determine, according to an observer in one of the spaceships, the speed of the other spaceship.
The observer on Earth in (a) watches one spaceship as it travels to a distant star at a speed of 0.85c. 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) Outline whether the time interval measured by the observer on Earth is a proper time interval.
(iii) Calculate, according to the observer on Earth, the distance from Earth to the star.
(iv) The observers in the spaceship send a message to Earth halfway through their journey. Determine how long it takes the message to arrive at Earth according to the observers on the spaceship.
Markscheme
(i) 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;
(ii) recognition that v is negative relative to ux;
use of \(\frac{{0.85c + 0.85c}}{{1 + \frac{{{{\left( {0.85c} \right)}^2}}}{{{c^2}}}}}\);
(0.9869c≈) 0.99c;
Accept first marking point implied in the second marking point.
(i) \(\gamma \)=1.89;
interval on Earth = \(\gamma \)×interval on spaceship;
(interval on Earth 1.90×8 years=)15 years;
Award [3] for a bald correct answer.
(ii) time interval measured by observer on Earth is not proper because the time interval between the two events is not measured at same place/not the shortest time;
(iii) observer on Earth thinks spaceship has travelled for 15 years;
so distance is 0.85c×15=12.8≈13 ly;
Award [2] for a bald correct answer.
or
the spaceship observer observes the distance moved by the Earth=0.85c×8.0yr;
proper distance=1.90×0.85c×8.0 yr=12.9≈13ly;
Award [2] for a bald correct answer.
(iv) Earth is at a distance of 4×0.85c=3.4ly when signal is emitted;
signal reaches Earth in time T where cT+3.4=0.85cT;
T=22.7≈23 years;