The point of this question is to compare rest energy and kinetic energy at high speeds. An alpha particle (a helium nucleus) is moving at a speed of $\mathbf{0}\mathbf{.}\mathbf{9993}$times the speed of light. Its mass is $\mathbf{6}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathit{k}\mathit{g}$(a) what is its rest energy? (b) Is it okay to calculate its kinetic energy using the expression$\mathit{K}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$?(c) What is its kinetic energy? (d) Which is true? A. the kinetic energy is approximately equal to the rest energy. B. The kinetic energy is much bigger than the rest energy. C. The kinetic energy is much smaller than the rest energy.

The given data can be listed below as,

• The mass of alpha particle is, $m=6.40\times {10}^{-27}kg$.

• The speed of alpha particle is $0.9993$times the speed of light.

The energy equivalent to an object's rest mass is equal to its rest mass multiplied by the square of the speed of light.

The relation of rest energy is expressed as,

$E_0=m{c}^2$

Here $E_0$is the rest energy, $m$is the mass of the baseball and $c$is the speed of light.

Substitute $6.40\times {10}^{-27}\mathrm{kg}$for $\mathrm{m}$and $3\times {10}^8\mathrm{m}/\mathrm{s}$for $c$in the equation (i).

$=5.76\times {10}^{10}J$

Hence the rest energy is, $5.76\times {10}^{-10}J$.

The relation of kinetic energy is expressed as,

$K-\frac{1}{2}m{v}^2$

Here $K$is the kinetic energy, $m$is the mass of the baseball and $V$is the speed of baseball.

Substitute $6.40\times {10}^{27}\mathrm{kg}$for $\mathrm{m}$and $$0.9993\times 3\times {10}^8\mathrm{m}/\mathrm{s}$for $v$in the above equation.

$K=\frac{1}{2}\times 6.40\times {10}^{-27}\mathrm{kg}\times {\left(0.9993\times 3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2$

$=2.876\times {10}^{-10}J$

Hence the kinetic energy is, $2.876\times {10}^{-10}J$.

The kinetic energy of the body is the energy that the body holds due to motion. This is the work done to the body to accelerate from a standstill to a constant speed.

From the result of part (a) and part (b) the rest energy and kinetic energy is expressed as,

$E_0=5.76\times {10}^{-10}J$

$K=2.876\times {10}^{-10}J$

Here$K<E_0$, hence kinetic energy is much smaller than the rest energy.