In this problem we will observe how the momentum of an electron changes as its v

Question

In this problem we will observe how the momentum of an electron changes as its velocity approaches the speed of light. The mass of an electron is 9.109 times 10^-31 kg. First, what is the momentum of an electron which is moving at the speed of a car on a highway, say 33 m/s, in kilogram meters per second? What is the momentum of an electron moving at 16% of the speed of light, in kilogram meters per second? What is the momentum of an electron moving at 81% of the speed of light, in kilogram meters per second? What is the non-relativistic momentum of an electron moving at this speed (81% of the speed of light), in kilogram meters per second? When the electron is moving at 81% of the speed of light, how many times greater is the relativistic momentum than the non-relativistic momentum? Which of the following facts are true about the ratio of relativistic to non-relativistic momentum of an electron? What is the ratio of the relativistic momentum to the classical momentum of a rocketship moving at 995% of the speed of light, in kg m/s?

Explanation / Answer

a) at low speed, (not comparable to speed of light )

momentum = mass x velcoity

= 9.109 x 10^-31 x 33 = 3 x 10^-29 kg m /s

b) momentum when speed is comparable to the speed of light.

momentum = y m v

where y = 1 / sqrt(1 - (v/c)^2 )

p = ( 9.109 x 10^-31 x (0.16 x 3 x 10^8) ) / [ 1 - (0.16c/c)^2]

p = 4.429 x 10^-23 kg m/s

c) momentum = y m v

where y = 1 / sqrt(1 - (v/c)^2 )

p = ( 9.109 x 10^-31 x (0.81 x 3 x 10^8) ) / [ 1 - (0.81c/c)^2]

p = 3.775 x 10^-22 kg m/s

d) non-relativistic momentum:

p = m v = 9.109 x 10^-31 x 0.16 x 3 x 10^8 = 4.372 x 10^-23 kg m/s

e)
p = m v = 9.109 x 10^-31 x 0.81 x 3 x 10^8 = 2.213 x 10^-23 kg m/s

f) Ratio = (y m v ) / (m v ) = y = 1 / sqrt[1 - (v/c)^2]

g) Ratio = 1 / sqrt [ 1 - (0.995c / c)^2 ]

      = 10.01

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