Determine the magnitude of and direction of the velocity

[azatkgz]

I have problems on understanding Relativity.So can someone check this one.

Homework Statement

Homework Equations

A spacecraft is launched from the surface of the with velocity of 0.600c at an angle of
50^{\circle} above the horizontal positive x axis.Another spacecraft is moving past,with a velocity of 0.700c in the negative x direction..Determine the magnitude of and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft .

The Attempt at a Solution

v_{1x}=0.6c\times cos50^{\circle}
v'_x=\frac{v_{1x}+v_2}{1+\frac{v_{1x}v_2}{c^2}}
v'_y=v_1sin50^{\circle}
v'=\sqrt{v'_x+v'_y}
tan\theta'=\frac{v'_y}{v'_x}

 

[MathematicalPhysicist]

the equation in the y direction is:
v_y'=\frac{v_y}{(1+v_xv/c^2)\gamma(v))}
other than this it looks good to me.

 

[azatkgz]

Like this
v'_y=\frac{v_1sin50\sqrt{1-v_1^2/c^2}}{(1+v_1^2cos50/c^2)}?

 

[azatkgz]

but should be there similar equation for v_{1x} ?

 

[learningphysics]

I think it's best to use these equations:

u_x' = \frac{u_x - v}{1-u_x v/c^2}

u_y' = \frac{u_y}{\gamma (1-u_x v/c^2)}

v is the velocity of the second ship... ie v = -0.700c, and ux' and uy' are the speeds measured by this second ship...

just plug in ux, uy, v and gamma and you'll have the answers.