The Relativity of Length — Question 7

Question 7

Two passing rods use contracted lengths and a laboratory closing rate

Two rods each have proper length $L_0=100\,\mathrm m$. In one laboratory, one moves right at $0.60c$ and the other left at $0.80c$. At $t=0$ their leading ends meet. Ignore interactions. Use $\gamma=(1-v^2/c^2)^{-1/2}$ and $c=3.00\times10^8\,\mathrm{m/s}$.

1. Find their laboratory lengths.

2. Find the time until their trailing ends meet, and explain why using $1.40c$ in this laboratory calculation is legitimate.

Question 7 - Solution

Two passing rods use contracted lengths and a laboratory closing rate

(a) Method and calculation

$$L_1=L_0\sqrt{1-0.60^2}=\boxed{80.0\,\mathrm m}.$$$$L_2=L_0\sqrt{1-0.80^2}=\boxed{60.0\,\mathrm m}.$$

(b) Interpretation and further calculation

$$(0.60c+0.80c)\Delta t=L_1+L_2.$$$$\boxed{\Delta t=100/c=0.333333\,\mu\mathrm s}.$$

Physical check

The sum is a separation-change rate between two laboratory positions, not a relative speed measured in either rod rest frame.