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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.)
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.)

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5 years ago

## Answers : (1)

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)= 0.944cFrom the above observation we conclude that, the relative speed of approach of one particle with respect to the other would be 0.944c.
```
5 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions