A rocket ship of rest length 100 m is moving at speed 0.8c past a timing device which records the time interval between the passage of the front and back ends of the ship.This time interval is:
A)0.20 $\mu$s
B)0.25 $\mu$s
C)0.33 $\mu$s
D)0.52 $\mu$s
E)0.69 $\mu$s

Question

Quizplus · Accepted Answer

The length of the rocket ship as observed from the timing device's frame of reference is contracted due to Lorentz contraction. The contracted length $L'$ is given by $L' = L\sqrt{1 - v^2/c^2}$, where $L$ is the rest length of the rocket, $v$ is the velocity of the rocket, and $c$ is the speed of light. Substituting $L = 100$ m and $v = 0.8c$, we get $L' = 100\sqrt{1 - 0.8^2} = 100\sqrt{1 - 0.64} = 100\sqrt{0.36} = 60$ m. The time interval $\Delta t$ is the length divided by the speed, so $\Delta t = L'/v = 60\, \text{m} / (0.8c) = 60/(0.8 \times 3 \times 10^8 \, \text{m/s}) = 0.25 \times 10^{-6}$ s, or 0.25 microseconds.

A Rocket Ship of Rest Length 100 M Is Moving μ\muμ

A Rocket Ship of Rest Length 100 M Is Moving $\mu$