Question

A 2.55-m-long rod, as measured in its rest frame, speeds by you longitudinally at 6.11Ã10^7. You measure its length as it passes. By how many millimeters do you determine the rod has contracted?

Answer 1

Answer:

The length contracts by 54 mm.

Explanation:

According to theory of special relativity the contraction in the length of an object travelling at a speed 'v' is given by

$L=L_{o}\sqrt{1-\frac{v^{2}}{c^{2}}}$

where

L = contracted length of the object

$L_{o}$ = original length of object

v = speed of the object

c = speed of light

Applying values we get

$L=2.55\times \sqrt{1-\frac{(6.11\times 10^{7})^{2}}{(3\times 10^{8})^{2}}}$

$L=2.496m$

Thus the change in length equals $\Delta L=2.55-2.496\\\\\therefore \Delta L=0.054m=54mm$