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A particle moves in the frame K with a velocity v at an angle theta to the x axis. Find the corresponding angle in the frame K^' moving with a velocity V relative to the frame K in the positive direction of its x axis, if the x and x^' axes of the two frames coincide. |
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Answer» Solution :In the frame `K^'` the components of the velocity of the particle are `v_x^'=(vcostheta-V)/(1-(vVcostheta)/(c^2))` `v_y^'=(vsinthetasqrt(1-V^2//c^2))/(1-(vV)/(c^2)costheta)` HENCE, `TANTHETA^'=(v_y^')/(v_x^')=(vsintheta)/(vcostheta-V)SQRT((1-V^2)//c^2)` 
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