| Date | November 2016 | Marks available | 2 | Reference code | 16N.3.SL.TZ0.7 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Suggest | Question number | 7 | Adapted from | N/A |
Question
Identical twins, A and B, are initially on Earth. Twin A remains on Earth while twin B leaves the Earth at a speed of 0.6c for a return journey to a point three light years from Earth.
Calculate the time taken for the journey in the reference frame of twin A as measured on Earth.
Determine the time taken for the journey in the reference frame of twin B.
Draw, for the reference frame of twin A, a spacetime diagram that represents the worldlines for both twins.
Suggest how the twin paradox arises and how it is resolved.
Markscheme
«0.6 ct = 6 ly» so t = 10 «years»
Accept: If the 6 ly are considered to be measured from B, then the answer is 12.5 years.
ALTERNATIVE 1
102 − 62 = t2 − 02
so t is 8 «years»
Accept: If the 6 ly are considered to be measured from B, then the answer is 10 years.
ALTERNATIVE 2
gamma is \(\frac{5}{4}\)
10 × \(\frac{4}{5}\) = 8 «years»
Allow ECF from a
Allow ECF for incorrect γ in mp1
three world lines as shown
Award mark only if axes OR world lines are labelled.
according to both twins, it is the other one who is moving fast therefore clock should run slow
Allow explanation in terms of spacetime diagram.
«it is not considered a paradox as» twin B is not always in the same inertial frame of reference
OR
twin B is actually accelerating «and decelerating»
