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$E = 3.2\times 10^{20} \times 1.602\times 10^{-19} = 51.264\textrm{ J}$
If we equate this to a hammer head of mass $0.5\textrm{ kg}$, being flung at some speed, we find that
$$v = 14.32\textrm{ m/s} = 52\textrm{ km/h}\;.$$
This is an absolutely vast amount of energy for a single, fundamental particle to have!
If we rearrange $E = \frac{1}{2}mv^2$, using the rest mass of the particle for m, we find that
$v = 2.5\times 10^{14}\textrm{ m/s}$, which is about a million times faster than light. We can therefore assume that the mass of the particle is greater than the rest mass.
The energy per kilo of the photon (in terms of rest mass) is $51.264/(1.67\times 10^{-27}) = 3.07 \times 10^{29}\textrm{ J/kg}\;.$
Given that the mass of the earth is only $6\times 10^{24}\textrm{ kg}$, the ball of iron would contain enough energy to propel the earth to a velocity of about $320\textrm{ m/s}$, or $1151\textrm{ km/h}$, using the formula for kinetic energy.
Perhaps it maybe might just pass straight through the Earth, vaporising everything it touched, leaving the bulk a little shaken, but intact. Even one proton possessing this energy is extremely rare though, thankfully
If you rearrange the given formula to $v = c\sqrt{1 - \left(\frac{m_0c}{E}\right)^2} \approx 3\times 10^{8}\textrm{ m/s}$, i.e. as far as my calculator is concerned, almost the speed of light.
The Ultra Particle
Age 16 to 18
Challenge Level 
$E = 3.2\times 10^{20} \times 1.602\times 10^{-19} = 51.264\textrm{ J}$
If we equate this to a hammer head of mass $0.5\textrm{ kg}$, being flung at some speed, we find that
$$v = 14.32\textrm{ m/s} = 52\textrm{ km/h}\;.$$
This is an absolutely vast amount of energy for a single, fundamental particle to have!
If we rearrange $E = \frac{1}{2}mv^2$, using the rest mass of the particle for m, we find that
$v = 2.5\times 10^{14}\textrm{ m/s}$, which is about a million times faster than light. We can therefore assume that the mass of the particle is greater than the rest mass.
The energy per kilo of the photon (in terms of rest mass) is $51.264/(1.67\times 10^{-27}) = 3.07 \times 10^{29}\textrm{ J/kg}\;.$
Given that the mass of the earth is only $6\times 10^{24}\textrm{ kg}$, the ball of iron would contain enough energy to propel the earth to a velocity of about $320\textrm{ m/s}$, or $1151\textrm{ km/h}$, using the formula for kinetic energy.
Perhaps it maybe might just pass straight through the Earth, vaporising everything it touched, leaving the bulk a little shaken, but intact. Even one proton possessing this energy is extremely rare though, thankfully
If you rearrange the given formula to $v = c\sqrt{1 - \left(\frac{m_0c}{E}\right)^2} \approx 3\times 10^{8}\textrm{ m/s}$, i.e. as far as my calculator is concerned, almost the speed of light.
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