| Date | May 2015 | Marks available | 2 | Reference code | 15M.3.HL.TZ2.18 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | Draw | Question number | 18 | Adapted from | N/A |
Question
This question is about general relativity.
A rocket is in outer space far from all masses. It moves along the dotted line according to an inertial observer located outside the rocket.
A ray of light is moving at right angles to the direction of the rocket according to the same inertial observer. The ray of light enters the rocket through a window W. Draw the path of the light ray according to an observer at rest inside the rocket,
(i) when the rocket is moving at constant speed v.
(ii) when the rocket is moving at constant positive acceleration a.
The acceleration of the rocket in (a)(ii) is 12ms–2. A gamma ray is emitted from the base of the rocket. The frequency at the base is fbase= 3.4×1018Hz. A detector in the rocket is at a distance of 25m above the base. The frequency measured by the detector is fdetector. Determine the frequency shift fdetector–fbase.
Markscheme
(i) any straight line with negative slope;
(ii) a downward curve (projectile like);
\(\Delta f = \left( {{f_{\detector}} - {f_{base}} = \frac{{fgh}}{{{c^2}}} = } \right)\frac{{3.4 \times {{10}^{18}} \times 12 \times 25}}{{9.0 \times {{10}^{16}}}}\);
=1.1×104 (Hz);
negative sign/red-shifted;
Award [3] for a bald correct answer of -1.1×10 4 (Hz) .
Examiners report
In part (a) most candidates drew a projectile path for (ii) but thought that light would travel horizontally for the rocket observer in (i).
Part (b) was an easy substitution into the gravitational frequency shift formula. Many forgot to square the speed of light or failed to give a negative value for ∆f.
