Question 19M.3.SL.TZ2.7
| Date | May 2019 | Marks available | [Maximum mark: 7] | Reference code | 19M.3.SL.TZ2.7 |
| Level | SL | Paper | 3 | Time zone | TZ2 |
| Command term | Determine, Label, Show, Suggest | Question number | 7 | Adapted from | N/A |
A spaceship moves away from the Earth in the direction of a nearby planet. An observer on the Earth determines the planet is 4 ly from the Earth. The spacetime diagram for the Earth’s reference frame shows the worldline of the spaceship. Assume the clock on the Earth, the clock on the planet, and the clock on the spaceship were all synchronized when ct = 0.
Show, using the spacetime diagram, that the speed of the spaceship relative to the Earth is 0.80c.
[1]
Evidence of finding 1/gradient such as:
use of any correct coordinate pair to find v - eg or .
OR
measures tan of angle between ct and ct’ as about 39° AND tan 39 ≈ 0.8 ✔
Answer 0.8c given, so check coordinate values carefully.
Most candidates could show that the velocity of the spacecraft was 0.8c.
Label, with the letter E, the event of the spaceship going past the planet.
[1]
E labelled at = 4, ct = 5 ✔
Check that E is placed on the worldline of S.
Event E was usually correctly labelled on the space-time diagram.
Determine, according to an observer on the spaceship as the spaceship passes the planet, the time shown by the clock on the spaceship.
[2]
OR
Allow solutions involving the use of Lorentz equations.
A very simple time dilation question which most candidates got wrong at SL but the question was better answered at HL.
Determine, according to an observer on the spaceship as the spaceship passes the planet, the time shown by the clock on the planet.
[1]
t = 5 years OR ct = 5 ly ✔
Many candidates tried to use time dilation again without realising that the clock on P must also read 5 years at event E because that is the time on the Earth clock in P’s frame for the event.
On passing the planet a probe containing the spaceship’s clock and an astronaut is sent back to Earth at a speed of 0.80c relative to Earth. Suggest, for this situation, how the twin paradox arises and how it is resolved.
[2]
On return to Earth the astronaut will have aged less than Earthlings «by 4 years»
OR
time passed on Earth is greater than time passed for the astronaut «by 4 years» ✔
astronaut accelerated/changed frames but Earth did not
OR
for astronaut the Earth clock jumps forward at turn-around ✔
OWTTE
Treat as neutral any mention of both the Earth and astronaut seeing each other’s clock as running slow.
The twin paradox is now well understood and there were some good quality answers. Some candidates even knew that the Earth clock jumps forward when the Astronaut turns around.