Question

Space cruisers A and B are moving parallel to the positive direction of an x axis. Cruiser A is faster, with a relative speed of $\mathbf{v}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{900}\mathbf{c}$, and has a proper length of $\mathit{L}\mathbf{=}\mathbf{200}\mathit{m}$. According to the pilot of A, at the instant (t = 0) the tails of the cruisers are aligned, the noses are also. According to the pilot of B, how much later are the noses aligned?

Short answer

The noses of cruisers aligned for $1.37\times {10}^{-6}\text{s}$.

Step-by-step solution

Identification of given data

The relative speed of Cruiser A is $v=0.900c$

The proper length of Cruiser A is $L=200\text{m}$

The length observed by pilot B and pilot A is found by Length contraction formula. The duration for alignment of noses is found by difference in lengths by both pilots by relative speed.

Determination of Lengths observed by Pilot A and Pilot B

The length of cruiser for pilot B is given as:

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

Here, $c$is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$.

Substitute all the values in the above equation.

$$\begin{gathered} L_B=\frac{200\text{m}}{\sqrt{1-\frac{{\left(0.900c\right)}^2}{c^2}}} \\ {L}_B=458.8\text{m} \end{gathered}$$

The length of cruiser for pilot A is given as:

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

Substitute all the values in the above equation.

$$\begin{gathered} L_A=\left(200\text{m}\right)\left(\sqrt{1-\frac{{\left(0.900c\right)}^2}{c^2}}\right) \\ {L}_A=87.8\text{m} \end{gathered}$$

Determination of duration for alignment of noses of cruisers

The duration for alignment of noses of cruiser is given as:

$t=\frac{L_B-L_A}{v}$

Substitute all the values in the above equation.

$$\begin{gathered} t=\frac{458.8\text{m}-87.8\text{m}}{\left(0.900c\right)\left(\frac{3\times {10}^8\text{m}/\text{s}}{c}\right)} \\ t=1.37\times {10}^{-6}\text{s} \end{gathered}$$

Therefore, the noses of cruisers aligned for $1.37\times {10}^{-6}\text{s}$.