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An unstable particle moves in the reference frame K^' along its y^' axis with a velocity v^'. In its turn, the frame K^' moves relative to the frame K in the positive direction of its x axis with a velocity V. The x^' and x axes of the two reference frames coincide, the y^' and y axes are parallel. Find the distance which the particle traverses in the frame K, if its proper lifetime is equal to Deltat_0. |
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Answer» SOLUTION :The COMPONENTS of the velocity of the unstable particle in the FRAME K are `(V, v^'SQRT(1-V^2/c^2),0)` so the velocity relative to K is `sqrt(V^2+v^('^2)-(v^('^2V^2)/(c^2))` The life time in this frame dilates to `Deltat_0//sqrt(1-V^2/c^2-v^('^2)/c^2+(v^('^2)V^2)/(c^4))` and the distance traversed is `Deltat_0(sqrt(V^2+v^('^2)-(v^('^2)V^2)//c^2))/(sqrt(1-V^2//c^2)sqrt(1-v^('^2)//c^2))` 
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