A muon travels at a speed of 0.95c relative to an observer - AQA - A-Level Physics - Question 4 - 2020 - Paper 7

To find the distance between the two points in the frame of reference of the muon, we need to use the formula for length contraction:

$L = L_0 \sqrt{1 - \frac{v^2}{c^2}}$

where:

• $L_0$ is the proper length (the distance observed in the stationary frame), which is given as $2.5 \times 10^3$ m.
• $v$ is the speed of the muon ($0.95c$).
• $c$ is the speed of light.

Plugging in the values, we have: $L = 2.5 \times 10^3 \sqrt{1 - (0.95)^2}$

Calculating the value inside the square root: $L = 2.5 \times 10^3 \sqrt{1 - 0.9025} = 2.5 \times 10^3 \sqrt{0.0975}$

Now compute the square root and multiply:

$L \approx 2.5 \times 10^3 \times 0.3122 \approx 781$ m.

Thus, the distance in the muon's frame of reference is approximately 781 m.