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Question: Box 1 and box 2 are identical when at rest relative to each other. An astronaut floating in intergal...

Box 1 and box 2 are identical when at rest relative to each other. An astronaut floating in intergalactic space sees the two boxes fly by from the astronaut’s left to his right. Box 1 flies by at 0.8c (80%\% of the speed of light) and box 2 flies by 0.9c. Because of the relativistic effects, what is true from the point of view of the astronaut?
A) Box 2 is longer and taller than box 1.
B) Box 2 is longer and not as tall as box 1.
C) Box 1 is longer and taller than box 2.
D) Box 1 is longer and not as tall as box 2.
E) Box 2 is shorter and the same height as box 1.

Explanation

Solution

We need to understand the dependence of the length and the height of an object moving with relativistic speeds in a directive with respect to the frame of reference which here is the astronaut himself in order to find the solution for the problem.

Complete Step-by-Step Solution:
We are given a situation in which two identical boxes are in motion at relativistic speeds with respect to an astronaut in the direction along the horizontal. We need to know that the relativistic properties are only applicable in the direction of the motion of the object, i.e., in this case, the length of the box is applicable to a variation in size with respect to each other. Therefore, the height of the two boxes will remain the same as observed by the astronaut.

In order to find the length contraction of the two boxes we can use the Lorentz equations which are given by –
L=L01v2c2L={{L}_{0}}\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}

For box 1,

& {{L}_{1}}={{L}_{0}}\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\\ & v=0.8c \\\ & \Rightarrow {{L}_{1}}={{L}_{0}}\sqrt{1-\dfrac{{{(0.8c)}^{2}}}{{{c}^{2}}}} \\\ & \Rightarrow {{L}_{1}}={{L}_{0}}\sqrt{0.36} \\\ & \therefore {{L}_{1}}=0.6{{L}_{0}} \\\ \end{aligned}$$ For box 2, $$\begin{aligned} & {{L}_{2}}={{L}_{0}}\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\\ & v=0.9c \\\ & \Rightarrow {{L}_{2}}={{L}_{0}}\sqrt{1-\dfrac{{{(0.9c)}^{2}}}{{{c}^{2}}}} \\\ & \Rightarrow {{L}_{2}}={{L}_{0}}\sqrt{0.11} \\\ & \therefore {{L}_{2}}=0.44{{L}_{0}} \\\ \end{aligned}$$ From this we can understand that $${{L}_{1}}>{{L}_{2}}$$, so box 1 is longer as observed by the astronaut. **Hence, the correct answer is option E.** **Note:** The apparent length observed by the observer of an object moving with relativistic speed decreases with the increasing relative motion. For an object moving with the speed equal to light will have the length contracted to zero units which is impossible.