Date | May 2012 | Marks available | 3 | Reference code | 12M.3.HL.TZ2.14 |
Level | Higher level | Paper | Paper 3 | Time zone | Time zone 2 |
Command term | Calculate | Question number | 14 | Adapted from | N/A |
Question
This question is about relativistic kinematics.
A short pulse containing many nuclei of a radioactive isotope is emitted from a source S in a laboratory. The nuclei have speed v = 0.920c as measured with respect to the laboratory.
The pulse arrives at a detector D. The detector is 250 m away as measured by an observer in the laboratory.
Calculate the time it takes the pulse to travel from S to D, according to
(i) an observer in the laboratory.
(ii) an observer Q moving along with the pulse.
Calculate the distance between the source S and the detector D according to observer Q.
A particular nucleus in the pulse decays by emitting an electron in the same direction as that of the nucleus. The speed of the electron measured in the laboratory is 0.985c.
Calculate the speed of the electron as measured by observer Q.
The laboratory observer and observer Q agree that by the time the pulse arrives at D, half of the nuclei in the pulse have decayed.
Outline, without further calculation, how this is evidence in support of time dilation.
Markscheme
(i) 9.1×10-7 s;
(ii) the gamma factor is \(\gamma = \left( {\frac{1}{{\sqrt {1 - {{0.920}^2}} }} = } \right)2.55\);
and so \(t' = \left( {\frac{{9.1 \times {{10}^{ - 7}}}}{{2.55}} = } \right)3.5 \times {10^{ - 7}}{\rm{s}}\);
98(m);
Allow ECF from gamma factor in (a)(ii).
\(u' = \left( {\frac{{u - v}}{{1 - \frac{{uv}}{{{c^2}}}}} = } \right)\frac{{0.985{\rm{c}} - 0.920{\rm{c}}}}{{1 - \frac{{0.985{\rm{c}} \times 0.920{\rm{c}}}}{{{c^2}}}}}\);
\(u'\)=0.693c;
both observers measure different values for the half-life;
and the two half-lives are related by the gamma factor;
Examiners report