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Grade: 11
        
One cosmic-ray proton approaches the Earth along its axis with a velocity of 0.787c toward the north pole and another, with velocity 0.612c, toward the south pole. See Fig. 20-24. Find the relative speed of approach of one particle with respect to the other. (Hint: It is useful to consider the Earth and one of the particles as the two inertial reference frames.)
4 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
							To find out the relative speed of approach of one particle with respect to the other, substitute 0.787c for v0, 0.612c for u in the equation v = (v0 + u) / (1 + v0u/c2),
v = (v0 + u) / (1 + v0u/c2)
= (0.787c + 0.612c) / (1+(0.787c) (0.612c)/c2)
= 1.399c / (1+0.4816)
= 0.944c
From the above observation we conclude that, the relative speed of approach of one particle with respect to the other would be 0.944c.
4 years ago
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