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nikdorinn [45]
3 years ago
8

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.
Physics
1 answer:
astraxan [27]3 years ago
8 0

Answer:

option (b)

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-\frac{v^{2}}{c^{2}}}}     .... (1)

The new momentum is given by

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

By comparison, p' < 2p.

Thus, the option is (b) .

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Explanation:

We have given glass surface has a diameter of 1.5 mm

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4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}

v = =0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec

   

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