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3 years ago

The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

Physics

1 answer:

FrozenT [24]3 years ago

6 0

Answer:

<em>Speed of the electron is 2.46 x 10^8 m/s</em>

Explanation:

momentum of the electron before relativistic effect = $M_{0} V$

where $M_{0}$ is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = $M_{0}/ \sqrt{1 - \beta ^{2} }$

where

$\beta = V/c$

c is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> $M_{0}V/ \sqrt{1 - \beta ^{2} }$

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75 $M_{0} V$ = $M_{0}V/ \sqrt{1 - \beta ^{2} }$

the equation reduces to

1.75 = $1/ \sqrt{1 - \beta ^{2} }$

square both sides of the equation, we have

3.0625 = 1/ $(1 - \beta ^{2} )$

3.0625 - 3.0625 $\beta ^{2}$ = 1

2.0625 = 3.0625 $\beta ^{2}$

$\beta ^{2}$ = 0.67

β = 0.819

substitute for $\beta = V/c$

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = <em>2.46 x 10^8 m/s</em>

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d) The work done by the applied force on up along the ramp is 500 J

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The work done by a force is given by the equation

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the dispalcement of the object

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we want to calculate the work done by the gravitational force as the box is pushed across the flat ground.

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m = 50.0 kg is the mass of the box

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