| Date | May 2011 | Marks available | 3 | Reference code | 11M.3.HL.TZ1.17 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Calculate | Question number | 17 | Adapted from | N/A |
Question
This question is about relative velocities and energy at relativistic speeds.
Two identical rockets are moving along the same straight line as viewed from Earth. Rocket 1 is moving away from the Earth at speed 0.80 c relative to the Earth and rocket 2 is moving away from rocket 1 at speed 0.60 c relative to rocket 1.
Calculate the velocity of rocket 2 relative to the Earth, using the
(i) Galilean transformation equation.
(ii) relativistic transformation equation.
Comment on your answers in (a).
The rest mass of rocket 1 is 1.0×103kg. Determine the relativistic kinetic energy of rocket 1, as measured by an observer on Earth.
Markscheme
(i) u'x= ux + v = 0.60c + 0.80c = 1.40c;
(ii) \({u_x}^\prime \left( { = \frac{{{u_x} + v}}{{1 + \frac{{{u_x}v}}{{{c^2}}}}}} \right) = \frac{{0.60c + 0.80c}}{{1 + \frac{{0.60c \times 0.80c}}{{{c^2}}}}}\);
\(\left( { = \frac{{1.40c}}{{1.48}}} \right) = 0.95c\);
Award [1] for answers that use v = –0.80 c to get an answer of –0.38 c.
the answer to (a)(i) exceeds c / the answer to (a)(ii) does not exceed c;
hence the Galilean transformation is not valid / the relativistic transformation must be used / OWTTE;
\(\gamma = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} = \frac{1}{{\sqrt {1 = \frac{{{{0.80}^2}{c^2}}}{{{c^2}}}} }} = 1.7\);
\({E_K} = \left[ {\gamma - 1} \right]{m_0}{c^2} = \left[ {1.667 - 1} \right] \times 1.0 \times {10^3} \times {\left[ {3.0 \times {{10}^8}} \right]^2}\);
=6.0×1019J
Award [0] for answers that use Ek = ½ mv2 to get an answer of 2.9×1019 J.
