A spaceship passes over an observation station on Earth. Just as the nose of the ship passes the station, a light in the nose of the ship flashes. As the tail of the ship passes the station, a light flashes in the ship's tail. According to an Earth observer, 50.0 ns elapses between the two events. In the astronaut's reference frame, the length of the ship is \(12.0 \mathrm{m} .\) (a) How fast is the ship traveling according to an Earth observer? (b) What is the elapsed time between light flashes in the astronaut's frame of reference?

In this problem, we're given: - Time elapsed between the two events for the Earth observer: \(t_e = 50.0 \,\text{ns}\) - Length of the spaceship according to the astronaut: \(L_0 = 12.0 \,\text{m}\)

According to an Earth observer, the spaceship travels \(L_0\) in the time \(t_e\). We can use the formula for speed: \(v = \frac{d}{t}\) Rearranging for the speed of the spaceship: \(v = \frac{L_0}{t_e}\) Plug in the given values: \(v = \frac{12.0 \,\text{m}}{50.0 \times 10^{-9} \,\text{s}} = 240,000,000 \,\text{m/s}\) The speed of the spaceship according to an Earth observer is \(240,000,000 \,\text{m/s}\).

Now, we need to find the elapsed time between light flashes in the astronaut's frame of reference. We can use the time dilation formula: \(t_0 = \frac{t_e}{\sqrt{1 - \frac{v^2}{c^2}}}\) Where \(t_0\) is the elapsed time in the astronaut's frame of reference, \(t_e\) is the elapsed time in the Earth observer's frame of reference, \(v\) is the speed of the spaceship, and \(c\) is the speed of light (\(3\times10^8 \,\text{m/s}\)). Plug in the given values: \(t_0 = \frac{50.0 \times 10^{-9}\,\text{s}}{\sqrt{1 - \frac{(240,000,000\, \text{m/s})^2}{(3\times10^8\, \text{m/s})^2}}}\) Calculate the value of the time: \(t_0 \approx 86.6 \,\text{ns}\) The elapsed time between light flashes in the astronaut's frame of reference is approximately \(86.6\, \text{ns}\).