# Determine the magnitude of and direction of the velocity

1. Oct 9, 2007

### azatkgz

I have problems on understanding Relativity.So can someone check this one.
1. The problem statement, all variables and given/known data

2. Relevant 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.

3. The attempt at a solution
$$v_{1x}=0.6c\times cos50^{\circle}$$
$$v'_x=\frac{v_{1x}+v_2}{1+\frac{v_{1x}v_2}{c^2}}$$
$$v'_y=v_1sin50^{\circle}$$
$$v'=\sqrt{v'_x+v'_y}$$
$$tan\theta'=\frac{v'_y}{v'_x}$$

Last edited by a moderator: Jan 7, 2014
2. Oct 9, 2007

### 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.

3. Oct 9, 2007

### azatkgz

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

4. Oct 9, 2007

### azatkgz

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

5. Oct 9, 2007

### 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.

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