An electron with a speed of 0.988c is emitted by a supernova, where c is the speed of light. What is the magnitude of the momentum of this electron?
Thanks for any help you can offer! I've tried so many things!
it's mv/√(1 - v^2/c^2)
see
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html
the momentum is the product (multiply) of the relativistic mass and the velocity
the relativistic mass is the rest mass multiplied by gamma (the relativistic "fudge factor")
gamma = 1 / sqrt[1 - (v² / c²)]
p = gamma * rest mass * v
hyperphysics (web) is a good reference
So this is how I calculated, but can speed be taken for the velocity?
To find the magnitude of the momentum of the electron, we can use the formula for the momentum of an object:
momentum = mass × velocity
In this case, the mass of an electron is constant, so we only need to determine its velocity.
Given that the speed of the electron is 0.988c, where c is the speed of light, we can calculate the velocity using the equation:
velocity = speed / c
Substituting the given values:
velocity = 0.988c / c = 0.988
Now we can calculate the momentum by multiplying the mass of the electron by its velocity. The mass of an electron is approximately 9.11 × 10^(-31) kilograms.
momentum = mass × velocity
= 9.11 × 10^(-31) kg × 0.988
Evaluating this equation gives us the magnitude of the momentum of the electron emitted by the supernova.