Question

In: Physics

A cosmic ray proton streaks through the lab with velocity 0.82c at an angle of 52°...

A cosmic ray proton streaks through the lab with velocity 0.82c at an angle of 52° with the +x direction (in the xy plane of the lab). Compute the magnitude and direction of the proton's velocity when viewed from frame S' moving with β = 0.74.

Solutions

Expert Solution

x-component of the speed:

    ux = (0.82c) cos520   ......... (1)

y-component of the speed:

    uy = (0.82c) sin520 ............ (2)

from given problem, we have

β = v/c = 0.74

   v = 0.74 c

factor γ = 1/√[1-(v2/c2)] = 1.4867

from Lorentz transformation,(relativistic velocity transformation),

              ux' = (ux-v)/[1-(vux/c2)] .......... (3)

              uy' = (uy)/(γ)[1-(vux/c2)]   .......... (4)

substitute the eq (1), in eq (3), we get

   ux' = (((0.82c) cos520)-(0.74c))/[1-((0.74c)((0.82c) cos520 )/c2)]

        = -0.3754 c

substitute the eq (2), in eq (4), we get

   uy' = (((0.82c) sin520)/(1.4867)[1-((0.74c)((0.82c) sin520 )/c2)]

        = 0.8329 c

.................................................

therefore velocity u = √( ux')2 + (uy')2

                              = 0.9136 c

............................................................

angle θ = tan-1( uy'/ux')

            = -65.7o

            = 180-65.7

            = 114.3o (along +ve x-direction)


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