Question

Rod $R_1$ has rest length 1 m and rod $R_2$ has rest length 2 m. $R_1$ and $R_2$ move with velocities $+v\hat{i}$ and $-v\hat{i}$ respectively relative to the lab. If $R_2$ has a length of 1 m in the rest frame of $R_1$, $\frac{v}{c}$ is ................... (Specify answer up to two digits after decimal.)

Hint:

Always use relativistic velocity addition when two moving frames observe each other.

Answer 1

Correct Answer: 0.5

Step 1: Use Lorentz contraction.
Length in $R_1$ frame: $L = L_0\sqrt{1 - u^2/c^2}$ where $u$ is relative velocity between the rods.
Given: $L = 1$ m, $L_0 = 2$ m.

Step 2: Solve for relative velocity $u$.
$1 = 2\sqrt{1 - u^2/c^2}$
$\sqrt{1 - u^2/c^2} = 1/2$
$1 - u^2/c^2 = 1/4$
$u^2/c^2 = 3/4$
$u/c = \sqrt{3}/2 = 0.866$.

Step 3: Relate $u$ to $v$.
Using relativistic velocity addition: $u = \frac{v + v}{1 + v^2/c^2} = \frac{2v}{1 + v^2/c^2}$.

Step 4: Solve for $v/c$.
$\frac{2(v/c)}{1 + (v/c)^2} = 0.866$.
Solving gives $v/c = 0.75$.