Determine the magnitude of and direction of the velocity

  • Thread starter azatkgz
  • Start date
  • #1
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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
[tex]50^{\circle}[/tex] 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


[tex]v_{1x}=0.6c\times cos50^{\circle}[/tex]
[tex]v'_x=\frac{v_{1x}+v_2}{1+\frac{v_{1x}v_2}{c^2}}[/tex]
[tex]v'_y=v_1sin50^{\circle}[/tex]
[tex]v'=\sqrt{v'_x+v'_y}[/tex]
[tex]tan\theta'=\frac{v'_y}{v'_x}[/tex]
 
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Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
4,518
286
the equation in the y direction is:
[tex]v_y'=\frac{v_y}{(1+v_xv/c^2)\gamma(v))}[/tex]
other than this it looks good to me.
 
  • #3
190
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Like this
[tex]v'_y=\frac{v_1sin50\sqrt{1-v_1^2/c^2}}{(1+v_1^2cos50/c^2)}[/tex]?
 
  • #4
190
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but should be there similar equation for [tex]v_{1x}[/tex] ?
 
  • #5
learningphysics
Homework Helper
4,099
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I think it's best to use these equations:

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

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

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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