Deduce that few muons would be expected to arrive at the surface of the Earth if non-relativistic mechanics are assumed.

Calculate the average decay time of a muon as observed by an observer on the surface of the Earth.

Explain, with a calculation, why many muons reach the surface of the Earth before they have decayed.

\(t = \left( {\frac{d}{v} = \frac{{{{10}^4}}}{{0.98 \times 3 \times {{10}^8}}} = } \right){\text{ }}3.4 \times {10^{ - 5}}{\text{ s}}\);

which is \(\frac{{3.4 \times {{10}^{ - 5}}}}{{2.2 \times {{10}^{ - 6}}}} \approx 15\) decay times;

(so very few muons will reach Earth)

or

\(d = vt = 0.98 \times 3 \times {10^8} \times 2.2 \times {10^{ - 6}} = 650{\text{ m}}\) in one decay time;

in travelling to Earth, there are \(\frac{{10000}}{{650}} \approx 15\) decay times;

(so very few muons will reach Earth)

\(\gamma  = \sqrt {\frac{1}{{1 - {{0.98}^2}}}}  = 5.0\);

\(t = \gamma {t_0} = 5.0 \times 2.2 \times {10^{ - 6}} = 1.1 \times {10^{ - 5}}{\text{ s}}\);

distance travelled in one half-life is

\(d = vt = 0.98 \times 3 \times {10^8} \times 1.1 \times {10^{ - 5}} = 3200{\text{ m}}\);

for observers on Earth, the muons are able to travel much further compared with those in (a);

(so many more will be able to reach the Earth before they have decayed)

Many well prepared candidates presented good, well structured and clear answers.

Many well prepared candidates presented good, well structured and clear answers.

Many well prepared candidates presented good, well structured and clear answers. Some candidates struggled with (b)(ii).