Special Relativity: Length Contraction

Classical Mechanics Level 1

Jealous of a Prussian lab's work concerning time dilation \text{ time dilation } an Irish-Sephardic consortium jointly publishes their findings into length contraction \text{ length contraction } of alloy rods. Using the formula: L o b s = L 1 − v 2 c 2 L_{obs} = L \sqrt{1 - \frac{v^2}{c^2}} determine the contracted length observed ( L o b s L_{obs} ) by the consortium when a copper-tin rod of uncontracted length ( L L ) of 67 inches 67 \text{ inches } was accelerated to a velocity ( v v ) of 0.50 c 0.50c .

David's Special Relativity Set


The answer is 58.

David Hontz by David Hontz , 26, USA

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2 solutions

David Hontz
Dec 20, 2016

L o b s = L 1 − v 2 c 2 = 67 1 − ( 0.50 c ) 2 c 2 = 67 1 − 0.25 c 2 c 2 = 67 1 − 0.25 = 67 0.75 = 58 inches L_{obs} = L \sqrt{1 - \frac{v^2}{c^2}} = 67 \sqrt{1 - \frac{(0.50c)^2}{c^2}} = 67 \sqrt{1 - \frac{0.25c^2}{c^2}} = 67 \sqrt{1 - 0.25} = 67 \sqrt{0.75} = \boxed{58 \text{ inches}}

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You should specify the degree of accuracy the answer requires

Joe Freeman - 4 years, 5 months ago

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I change the velocity to include an extra significant figure; therefore it is implied that the answer must be to two significant figures. Thank you for the suggestion.

David Hontz - 4 years, 5 months ago
Navanil Ghosh
Dec 11, 2019

L'=L X SQRT(1- ( 0.5 C ) 2 C 2 \frac{(0.5C)^2}{C^2} = L X SQRT(1-0.25)

THEREFORE, 67=L X SQRT(0.75)

=> L = 67 S Q R T ( 0.75 ) \frac{67}{SQRT(0.75)} = APROXIMATELY 58 INCH

ANS 58
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