Answer to Question #88666 in Mechanics | Relativity for Mary Ndukwu

Question

In the distance future, Ben is in a rocket ship. He passes Jerry. Jerry measures the time it takes Ben's ship to pass him and finds that t=4.03μs. Ben measures the length of the ship to be l=203 m.
a) According to Jerry, what speed is the rocket travelling at?
b) According to Jerry what is the length of the ship?
c) Ben measures the time for the ship to pass Jerry. He turns on his clock when the front of the ship passes Jerry and turns it off when the back of the ship passes Jerry. What time does Ben measure?

Answer 1

Because of Lorentz contraction, the length of the moving ship in Jerry's reference system is $\ell = \ell_0 \sqrt{1 - v^2 / c^2}$, where $\ell_0 = 203\, \text{m}$ is its length in Ben's system, $v$ is the velocity of the ship, and $c = 3 \times 10^8\, \text{m/s}$ is the speed of light. If $t = 4.03 \times 10^{-6}\, \text{s}$ is the time measured by Jerry, then $v t = \ell = \ell_0 \sqrt{1 - v^2 / c^2}$. By squaring both sides of this equation, we have

$$v^2 t^2 = \ell_0^2 \left( 1 - \frac{v^2}{c^2} \right) \, ,$$

whence we find

$$v = \frac{\ell_0}{\sqrt{ \ell_0^2/c^2 + t^2}} = 4.968 \times 10^7\, \text{m/s} \, .$$

The length of the ship according to Jerry is then $\ell = v t = 200.2 \, \text{m}$. Ben measures the time $t_0 = \ell_0 / v = 4.086 \times 10^{-6}\, \text{s} = 4.086\, \mu\text{s}$.

Answer: a) $4.968 \times 10^7\, \text{m/s}$; b) 200.2 m; c) $4.086\, \mu\text{s}$.