A rectangular box in a rocket travelling past the Earth at 0.50 c appears to people on the Earth to be a perfect cube,0.50 m on a side.What is the proper volume of the box? A)0.11 m³ B)0.12 m³ C)0.14 m³ D)0.18 m³ E)0.19 m³

Question

Quizplus · Accepted Answer

The proper volume of the box is calculated using the formula for length contraction in special relativity, $L = L_0\sqrt{1 - v^2/c^2}$, where $L$ is the contracted length, $L_0$ is the proper length, $v$ is the velocity of the object, and $c$ is the speed of light. Since the box appears as a cube (0.50 m on a side) to observers on Earth, the length contraction occurs along the direction of motion. The volume of the cube as observed from Earth is $V = 0.50^3 = 0.125 m^3$. However, this is the contracted volume. The proper volume, which is the volume in the box's rest frame, would be larger because one of the dimensions (the length in the direction of motion) is contracted. Given that the contraction only affects one dimension, and the other two dimensions remain unchanged, the proper volume can be calculated by dividing the observed volume by the contraction factor $\sqrt{1 - v^2/c^2}$. Since $v = 0.50c$, the contraction factor is $\sqrt{1 - (0.50)^2} = \sqrt{0.75}$. Therefore, the proper volume is $0.125 / \sqrt{0.75} \approx 0.144 m^3$, which rounds to 0.14 $m^3$, making option C correct.

A Rectangular Box in a Rocket Travelling Past the Earth