How fast would a rocket have to move past a ground observer if the latter were to observe a 4.0% length shrinkage in the rocket length? (c = 3.00 ´ 10⁸ m/s) A)0.12 x 10⁸ m/s B)0.28 x 10⁸ m/s C)0.84 x 10⁸ m/s D)1.2 x 10⁸ m/s

Question

Quizplus · Accepted Answer

We can use the formula for length contraction:

L' = L * sqrt(1 - v^2/c^2)

where L is the original length, L' is the contracted length, v is the velocity of the rocket, and c is the speed of light.

We are given that L' = 0.96L (since there is a 4.0% length shrinkage). Substituting this and c into the equation, we get:

0.96L = L * sqrt(1 - v^2/c^2)

Squaring both sides and isolating v^2/c^2, we get:

v^2/c^2 = 1 - (0.96)^2

v/c = sqrt(1 - (0.96)^2)

v = sqrt(1 - (0.96)^2) * c

v ≈ 0.84 * 10^8 m/s

Therefore, the answer is (C).

How Fast Would a Rocket Have to Move Past a Ground