Question 19M.3.SL.TZ1.5d.i
Date | May 2019 | Marks available | [Maximum mark: 2] | Reference code | 19M.3.SL.TZ1.5d.i |
Level | SL | Paper | 3 | Time zone | TZ1 |
Command term | Determine | Question number | d.i | Adapted from | N/A |
The diagram shows space and time axes and ct for an observer at rest with respect to a galaxy. A spacecraft moving through the galaxy has space and time axes ′ and ct′.
A rocket is launched towards the right from the spacecraft when it is at the origin of the axes. This is labelled event 1 on the spacetime diagram. Event 2 is an asteroid exploding at = 100 ly and ct = 20 ly.
An observer in the spacecraft measures that events 1 and 2 are a distance of 120 ly apart. Determine, according to the spacecraft observer, the time between events 1 and 2.
[2]
9600 = 1202 − c2t2 ✔
ct = «−» 69.3 «ly» / t = «−» 69.3 «y» ✔
Allow approach with Lorentz transformation.
Most of the candidates well used the formula for invariant spacetime from the data booklet, but only a few strong candidates were able to determine the time between the events according to the spacecraft observer. This implies a lack of understanding of the concept of invariance for different frames of reference.
