Question 18N.3.SL.TZ0.5
| Date | November 2018 | Marks available | [Maximum mark: 9] | Reference code | 18N.3.SL.TZ0.5 |
| Level | SL | Paper | 3 | Time zone | TZ0 |
| Command term | Estimate, Label, Outline, Show that | Question number | 5 | Adapted from | N/A |
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
x′ = 1.5 m.
[2]
«» 1.499 ✔
x′ = «» 1.499 × (1.0 − 0) ✔
«x′ = 1.5 m»
ct′ = –1.1 m.
[1]
t′ = « =» «= »
«ct′ = –1.1 m»
OR
using spacetime interval 0 − 12 = (ct′)2 − 1.52 ⇒ «ct′ = –1.1» ✔
Label, on the diagram, the space coordinate of event E in the S′ frame. Label this event with the letter P.
[1]
line through event E parallel to ct′ axis meeting x' axis and labelled P ✔
Label, on the diagram, the event that has coordinates x′ = 1.0 m and ct′ = 0. Label this event with the letter Q.
[1]
point on x' axis about of the way to P labelled Q ✔
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.
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.
[3]
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» ✔
Using the spacetime diagram, estimate, in m, the length of this rod in the S′ frame.
[1]
0.7 m ✔