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Date November 2011 Marks available 3 Reference code 11N.3.HL.TZ0.12
Level Higher level Paper Paper 3 Time zone Time zone 0
Command term Calculate and Determine Question number 12 Adapted from N/A

Question

This question is about stellar distances and stellar properties.

According to Vladimir, a clock at rest in the railway carriage will appear to run slower than a clock at rest beside him. However, according to Natasha, Vladimir’s clock will run slower than a clock at rest beside her.

(i) Outline how this time dilation phenomenon leads to the “twin paradox” in which one of the twins embarks on a return journey to a distant star at a speed close to that of light whilst the other twin remains on Earth.

(ii) State the reason behind the resolution of the paradox.

[4]
d.

Evidence for time dilation comes from the decay of muons. A pulse of muons produced by cosmic radiation in the upper atmosphere of Earth travels to Earth with a speed of 0.96c as measured by an observer at rest on the surface of Earth. The half-life of the muons, as measured in the frame of reference in which the muons are at rest, is 3.1×10–6s.

(i) Determine for the muons, the distance that Earth will have travelled towards them after half of the muons in the pulse have decayed.

(ii) Calculate for the Earth observer, the distance that the muon pulse will have travelled towards Earth after half of the muons in the pulse have decayed.

[3]
e.

Suggest how your answers to (e)(i) and (e)(ii) provide evidence that supports the theory of special relativity.

[3]
f.

Markscheme

(i) Look for an argument along these lines:
on the return of the travelling twin according to the twin on Earth the travelling twin will have aged very little compared to himself/herself;
however, since time dilation is symmetric it could be the twin on Earth who has done the least aging;
experiment suggests that it is the travelling twin who ages the least;

(ii) because of the accelerations undergone by the travelling twin the situation is not symmetric / travelling twin is not in the same inertial frame of reference/changes inertial frame of reference;

d.

(i) (0.96×3.0×108×3.1×10-6)=890m;

(ii) \(\gamma  = \frac{1}{{\sqrt {1 - {{\left( {0.96} \right)}^2}} }}\);
distance = (3.57×890=) or (3.57×0.96×3.0×108×3.1×10-6=) 3200m;

e.

using the laboratory half-life, most of the muons would have decayed before reaching Earth;
however many muons are detected at the surface;
showing that the half-life is dilated / to the muons the distance travelled is contracted;

f.

Examiners report

[N/A]
d.
[N/A]
e.
[N/A]
f.

Syllabus sections

Option A: Relativity » Option A: Relativity (Core topics) » A.2 – Lorentz transformations
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