Please Help! Thanks a lot!! A particle moves with velocity v\' in the x\'y\' pla

Question

                    Please Help! Thanks a lot!!                 

                    
                

                    
                                     

A particle moves with velocity v' in the x'y' plane of the rocket frame in a direction that makes an angle (fY with the x'-axis. Find the angle (f) that that velocity vector of this particle makes with the x-axis of the laboratory frame. (Hint: Transform space and time displacements rather than velocities.) Why does this angle differ from that found in Exercise L-6 on trans¬formation of angles? Contrast the two results when the relative velocity between the rocket and laboratory frames is very great.

Explanation / Answer

a)

vx = dx/dt

==> vx = (gamma (dx' + v dt'))/dt

==> vx = (gamma dt' (dx'/dt' + v))/dt

==> vx = (gamma (vx' + v)) (dt'/dt)

==> vx = (gamma v' (vx'/v' + v/v')) (dt'/dt)

==> vx = (gamma v' (cos(phi') + v/v')) (dt'/dt)

vy = dy/dt

==> vy = (dy'/dt') (dt'/dt)

==> vy = (vy') (dt'/dt)

==> vy = v' (vy'/v') (dt'/dt)

==> vy = v' (sin(phi')) (dt'/dt)

==> phi = arctan(vy/vx)

==> phi = arctan((sin(phi'))/(gamma (cos(phi') + v/v')))

it is because here we must transform time intervals too.

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