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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    \[KE _{rel}=mc ^{2}(\gamma-1)\] \[KE _{nonrel}=\frac{ 1 }{ 2 }mV ^{2}\]

  2. Astrophysics
    • one year ago
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    Looks like fun, can you show your work please?

  3. Astrophysics
    • one year ago
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    \[\gamma = \frac{ 1 }{ \sqrt{1-\left( \frac{ v }{ c } \right)}^2 }\] just to note the Lorentz factor

  4. anonymous
    • one year ago
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    Sure

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  5. Astrophysics
    • one year ago
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    It's probably best if you do the calculations separately

  6. Astrophysics
    • one year ago
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    \[KE_{rel} = mc^2\left( \frac{ 1 }{ \sqrt{1-\left( \frac{ v }{ c } \right)^2} }-1 \right)\]

  7. Astrophysics
    • one year ago
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    You should find what fraction of the speed of light the given velocities are and then you can find the ratio.

  8. Astrophysics
    • one year ago
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    So \[1.5 \times 10^{-3} m/s = \frac{ 1 }{ 450,000 } c \]

  9. Astrophysics
    • one year ago
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    \[\gamma = \frac{ 1 }{ \sqrt{1-\left( \frac{ (1/450,000)c)^2 }{ c^2 } \right)} }\]

  10. Astrophysics
    • one year ago
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    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...

  11. Michele_Laino
    • one year ago
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    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

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