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Date May 2011 Marks available 2 Reference code 11M.3.HL.TZ1.17
Level Higher level Paper Paper 3 Time zone Time zone 1
Command term Comment 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.

[3]
a.

Comment on your answers in (a).

[2]
b.

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.

[2]
c.

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.

a.

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;

b.

\(\gamma  = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} = \frac{1}{{\sqrt {1 = \frac{{{{0.80}^2}{c^2}}}{{{c^2}}}} }} = 1.7\);
\({E_K} = \left[ {\gamma  - 1} \right]{m_0}{c^2} = \left[ {1.667 - 1} \right] \times 1.0 \times {10^3} \times {\left[ {3.0 \times {{10}^8}} \right]^2}\);
=6.0×1019J
Award [0] for answers that use Ek = ½ mv2 to get an answer of 2.9×1019 J.

 

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Option A: Relativity » Option A: Relativity (Core topics) » A.1 – The beginnings of relativity

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