A cosmic-ray proton is moving at such a speed that it can travel from the moon t

Question

A cosmic-ray proton is moving at such a speed that it can travel from the moon to Earth in 2.20 s. (The Earth-moon distance is 3.80 105 km. Ignore Earth's rotation. Use 3. 108 m/s for the speed of light c.)

(a) At what fraction of the speed of light is the proton moving? (Give your answer to three decimal places.)
(b) What is its kinetic energy?
(c) What value would be measured for its mass by an observer in Earth's reference frame?
(d) What percent error is made in the kinetic energy by using the classical relation?

Explanation / Answer

a) speed of proton, v = (3.80 x 10^5 x 10^3 m ) / 2.2

v =1.727 x 10^8 m/s

fraction = v / c = 0.576

b) KE = m0 c^2

     = (1.67 x 10^-27 ) ( 3 x 10^8)^2 / sqrt(1 - 0.576^2)

= 1.838 x 10^-10 J

c) mo = m / sqrt(1 - (v/c)^2)

m0 = (1.67 x 10^-27) / sqrt(1 - 0.576^2)

    = 2.04 x 10^-27 kg

d) KE using classical relation,

Ke = m v^2 /2

= (1.67 x 10^-27) ( 1.727 x 10^8)^2 / 2

= 0.752 x 10^-10 J

%error = (1.838 - 0.752) / (1.838)   x 100   =59.1 %

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