plz solve me these?

Question

Plz solve me the following, with elaborate explanations if required (b’coz of the lack of diagrams).




1) Determine the proper length of a rod if in the laboratory frame of reference its velocity is v = c/2, the length l = 1m, and the angle between the rod and its direction of motion is θ = 45°.



Ans: l’ = l √ {(1- v²sin² θ/c²) / (1 - v²/c²)} = 1.08m

Answer 1

Only the component along the direction of motion has its length contracted. This amounts to Lcos(45deg)=Lsin(45deg)=L/( 2^(1/2) ). We then can expand that length to find its rest value L'a= L/( 2(1-v^2/c^2 ) )^(1/2). The other component, in the direction perpendicular to the motion is unchanged, Lb=L/(2^(1/2)). By Pythagoras, L'^2=L'a^2+Lb^2, where La and Lb are the two legs of the triangle here. That gives:

L'=Lsin(45)*(4/3+1)^(1/2) = 1.08L