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The velocity components of a particle moving in the xy plane of the reference frame K are equal to v_x and v_y. Find the velocity v^' of this particle in the frame K in the positive direction of its x axis. |
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Answer» Solution :By the VELOCITY addition FORMULA `v_x'=(v_x-V)/(1-(Vv_x)/(c^2)), v_y'=(v_ysqrt(1-V^2//c^2))/(1-(v_xV)/(c^2))` and `v^'=sqrt(v_x'^2+v_y'^2)=(sqrt((v_x-V)^2+v_y^2(1-V^2//c^2)))/(1-(v_xV)/(c^2))` |
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