| Date | November 2017 | Marks available | 1 | Reference code | 17N.3.SL.TZ0.5 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Define | Question number | 5 | Adapted from | N/A |
Question
Two rockets, A and B, are moving towards each other on the same path. From the frame of reference of the Earth, an observer measures the speed of A to be 0.6c and the speed of B to be 0.4c. According to the observer on Earth, the distance between A and B is 6.0 x 108 m.
Define frame of reference.
Calculate, according to the observer on Earth, the time taken for A and B to meet.
Identify the terms in the formula.
u′ = \(\frac{{u - v}}{{1 - \frac{{uv}}{{{c^2}}}}}\)
Determine, according to an observer in A, the velocity of B.
Determine, according to an observer in A, the time taken for B to meet A.
Deduce, without further calculation, how the time taken for A to meet B, according to an observer in B, compares with the time taken for the same event according to an observer in A.
Markscheme
a co-ordinate system in which measurements «of distance and time» can be made
Ignore any mention to inertial reference frame.
closing speed = c
2 «s»
u and v are velocities with respect to the same frame of reference/Earth AND u′ the relative velocity
Accept 0.4c and 0.6c for u and v
\(\frac{{ - 0.4 - 0.6}}{{1 + 0.24}}\)
«–» 0.81c
\(\gamma \) = 1.25
so the time is t = 1.6 «s»
gamma is smaller for B
so time is greater than for A
