Question

Determine the ratio of the relativistic to the non relativistic kinetic energy when V=1.5x10^-3 m/s and when V = 0.97 m/s.
I have tried it three times and for some reason my answer is always zero

Answer 1

\[KE _{rel}=mc ^{2}(\gamma-1)\]
\[KE _{nonrel}=\frac{ 1 }{ 2 }mV ^{2}\]

Answer 2

Looks like fun, can you show your work please?

Answer 3

\[\gamma = \frac{ 1 }{ \sqrt{1-\left( \frac{ v }{ c } \right)}^2 }\] just to note the Lorentz factor

Answer 4

Sure

Answer 5

It's probably best if you do the calculations separately

Answer 6

\[KE_{rel} = mc^2\left( \frac{ 1 }{ \sqrt{1-\left( \frac{ v }{ c } \right)^2} }-1 \right)\]

Answer 7

You should find what fraction of the speed of light the given velocities are and then you can find the ratio.

Answer 8

So \[1.5 \times 10^{-3} m/s = \frac{ 1 }{ 450,000 } c \]

Answer 9

\[\gamma = \frac{ 1 }{ \sqrt{1-\left( \frac{ (1/450,000)c)^2 }{ c^2 } \right)} }\]

Answer 10

The reason you're getting 0 is, when you subtract by 1 using a calculator it's approximating to 1 as the numbers are so small, we get something insane such as 0.9999999...

Answer 11

for small velocities (v<<c), we can write this (using Taylor expansion):
\[\Large E = \gamma m{c^2} \cong m{c^2}\left( {1 + \frac{{{v^2}}}{{2{c^2}}}} \right) = m{c^2} + m\frac{{{v^2}}}{2}\]
so we can write:
\[\Large K{E_{rel}} = \gamma m{c^2} - m{c^2} \cong m\frac{{{v^2}}}{2} = K{E_{not\;rel}}\]
so, in the case v << c, your ratio is equal to 1