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Consider an electron moving with relativistic speed v and momentum of magnitude p. If the speed of the electron is doubled, the resulting relativistic momentum of the electron will be:a) equal to 2pb) less than 2pc) greater than 2pPlease explain why.

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Answer:

option (b)

Step-by-step explanation:

initial speed = v

initial momentum = p

final speed = 2v

final momentum = p'

The formula for relativistic momentum is given by


p = \frac{mv}{\sqrt{1-(v^(2))/(c^(2))}} .... (1)

The new momentum is given by


p' = \frac{m* 2v}{\sqrt{1-(4v^(2))/(c^(2))}} .... (2)

By comparison, p' < 2p.

Thus, the option is (b) .

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