| Date | November 2018 | Marks available | 1 | Reference code | 18N.3.SL.TZ0.5 |
| Level | Standard level | Paper | Paper 3 | Time zone | 0 - no time zone |
| Command term | Label | Question number | 5 | Adapted from | N/A |
Question
The spacetime diagram shows the axes of an inertial reference frame S and the axes of a second inertial reference frame S′ that moves relative to S with speed 0.745c. When clocks in both frames show zero the origins of the two frames coincide.
Event E has coordinates x = 1 m and ct = 0 in frame S. Show that in frame S′ the space coordinate and time coordinate of event E are
A rod at rest in frame S has proper length 1.0 m. At t = 0 the left-hand end of the rod is at x = 0 and the right-hand end is at x = 1.0 m.
x′ = 1.5 m.
ct′ = –1.1 m.
Label, on the diagram, the space coordinate of event E in the S′ frame. Label this event with the letter P.
Label, on the diagram, the event that has coordinates x′ = 1.0 m and ct′ = 0. Label this event with the letter Q.
Using the spacetime diagram, outline without calculation, why observers in frame S′ measure the length of the
rod to be less than 1.0 m.
Using the spacetime diagram, estimate, in m, the length of this rod in the S′ frame.
Markscheme
«» 1.499 ✔
x′ = «» 1.499 × (1.0 − 0) ✔
«x′ = 1.5 m»
t′ = « =» «= »
«ct′ = –1.1 m»
OR
using spacetime interval 0 − 12 = (ct′)2 − 1.52 ⇒ «ct′ = –1.1» ✔
line through event E parallel to ct′ axis meeting x' axis and labelled P ✔
point on x' axis about of the way to P labelled Q ✔
ends of rod must be recorded at the same time in frame S′ ✔
any vertical line from E crossing x’, no label required ✔
right-hand end of rod intersects at R «whose co-ordinate is less than 1.0 m» ✔
0.7 m ✔
