Question

A very long stick ruled with meter markings is placed in empty space. A spaceship of rest length L = 100 m runs lengthwise alongside the stick. Two space travelers equipped with knives and synchronized watches station themselves fore and aft. At a prearranged time, each reaches through a porthole and slices through the stick. If the relative velocity of the stick and ship is v = (4/5)c, how many meter marks are on the cut-off portion of the stick? Do the calculation first in the frame of the ship, and then do the calculation over in the frame of the stick. In each case, draw careful pictures.

Answer 1

Answer:

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m

Explanation:

Given data:

Spaceship  length of L = 100 m

Relative velocity between the ship and stick is given as

$v = \frac{4}{5} c$

The observed length observed by the outside observer is

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

putting all value to get observe length

put$v = \frac{4}{5} c$

$L ' = 100\times \sqrt{1 - \frac{(4/5)^2c^2}{c^2}$

L' = 60 m

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m