Question

frame s' moves relative to frame s at 0.6c in the direction

Answer 1

We use the standard inertial frames S and S which are set up such that the x and x axes ... moves in the positive x-direction with speed v.

Answer 2

Answer:  

i. u = 0.844c

ii. u = 0.212c  

Solution:  

I)        According to Einstein’s theory of relativity, the velocity of the particle “v” with respect to S frame relativitically is given by the equation,

$u = u^{'}+v/ 1+vu^{'}/c^2 \rightarrow(1)$

given:

u’ = 0.620c , v = 0.470c

u’+v = 0.620c+0.470c = 1.09c

u’v= (0.470c)(0.620c) = 0.2914$c^2$

u = 1.09c/ 1+[(0.2914$c^2$)/$c^2$]

u = 1.09c/ 1.2914

u = 0.844c

ii)        The velocity of the particle “v”, with respect to S if it moves in the direction of decreasing x’ in the S’ frame will be, here u’ = 0.620c , v = -0.470c

u = u’+v/ 1+vu’/$c^2 \rightarrow$ (2)

u’+v = 0.620c-0.470c = 0.15c

u’v = -0.2914$c^2$

u = 0.15c/ 1+[(-0.2914$c^2$)/$c^2$]

u = 0.15c/ 0.7086

u = 0.212c