Determine the speed at which the volume of an object shrinks to half of its rest value.

Step-by-step solution

Let the rest volume of the object be V₀ = L_x L_y L_z . When the object moves with a speed v along the x-axis, its length in the direction of motion contracts due to length contraction, while the perpendicular dimensions remain unchanged. The new length is L'_x = L_x 1 - v^2 c^2 . The new volume is V = L'_x L_y L_z = V₀ 1 - v^2 c^2 . Given that the volume shrinks to half of its rest value, V = V₀ 2 . V₀ 2 = V₀ 1 - v^2 c^2 1 4 = 1 - v^2 c^2 v^2 c^2 = 3 4 v = 3 ,c 2