Question 18M.3.SL.TZ2.4b
| Date | May 2018 | Marks available | [Maximum mark: 3] | Reference code | 18M.3.SL.TZ2.4b |
| Level | SL | Paper | 3 | Time zone | TZ2 |
| Command term | Determine | Question number | b | Adapted from | N/A |
When a spaceship passes the Earth, an observer on the Earth and an observer on the spaceship both start clocks. The initial time on both clocks is 12 midnight. The spaceship is travelling at a constant velocity with γ = 1.25. A space station is stationary relative to the Earth and carries clocks that also read Earth time.
The spaceship passes the space station 90 minutes later as measured by the spaceship clock. Determine, for the reference frame of the Earth, the distance between the Earth and the space station.
[3]
ALTERNATIVE 1
time interval in the Earth frame = 90 × γ = 112.5 minutes
«in Earth frame it takes 112.5 minutes for ship to reach station»
so distance = 112.5 × 60 × 0.60c
1.2 × 10m12 «m»
ALTERNATIVE 2
Distance travelled according in the spaceship frame = 90 × 60 × 0.6c
= 9.72 × 1011 «m»
Distance in the Earth frame «= 9.72 × 1011 × 1.25» = 1.2 × 1012 «m»
[3 marks]
