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

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(a)A cosmic ray proton streaks through the lab with velocity0.82c at an angle of 52Â° with the +x direction (in the xy plane ofthe lab). Compute the magnitude and direction of the proton'svelocity when viewed from frame S' moving with Î² = 0.74.
(b) Suzanne observes two light pulses to be emitted from thesame location, but separated in time by 3.50 Âµs. Mark sees theemission of the same two pulses separated in time by 8.00Âµs.
How fast is Mark moving relativeto Suzanne?c
According to Mark, what is theseparation in space of the two pulses?
m

Answer & Explanation Solved by verified expert

3.9 Ratings (715 Votes)

The x-component of the speed is,

Â Â Â u_x = (0.82c) cos52^0Â Â Â Â Â Â Â Â Â Â Â (1)

The y-component of the speed is,

Â Â Â u_y = (0.82c) sin52⁰ Â Â Â Â Â Â Â Â Â Â Â Â Â Â (2)

From given problem, we have

Î² = v/c

Â Â = 0.74

Â Â Â Â Â Â Â Â Â Â Â v = 0.74 c

Factor

Î³ = 1/âˆš[1-(v²/c²)]

Â Â Â Â = 1.49

From Lorentz transformation,(relativistic velocity transformation),

Â Â Â Â Â Â Â Â Â Â Â Â Â u_x' = (u_x-v)/[1-(vu_x/c²)] .......... (3)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â u_y' = (u_y)/(Î³)[1-(vu_x/c²)]Â Â .......... (4)

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

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

Â Â Â Â Â Â Â = -0.273 c

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

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

Â Â u_y' = (((0.82c) sin52⁰)/(1.49)[1-((0.74c)((0.82c) sin52⁰ )/c²)]

Â Â Â Â Â Â Â = 0.831 c

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

Therefore velocity u = âˆš( u_x')² + (u_y')²

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.875 c

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

Angle Î¸ = tan^-1( u_y'/u_x')

Â Â Â Â Â Â Â Â Â Â Â = -71.81 0

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

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Physics

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