Question

A pitcher can throw a baseball at about $\mathbf{100}\mathbf{}\mathit{m}\mathit{i}\mathbf{/}\mathit{h}$(about$\mathbf{44}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$). What is the ratio of the kinetic energy to the rest energy role="math" localid="1657706011431" $\mathit{m}{\mathit{c}}^{\mathbf{2}}$?

Short answer

The ratio of kinetic energy to the rest mass energy is$1.076\times {10}^{-14}:1$.

Step-by-step solution

Identification of given data

The given data can be listed below,

The speed of the baseball is,$v=44m/s$

Concept/Significance of rest mass energy

A single atom is perfectly still in intergalactic space (i.e., no kinetic energy or potential energy in the observing frame). It would still possess the energy of some form, and since it is not attributed to the object moving, it is called rest mass-Energy.

Determination of the ratio of the kinetic energy to the rest energy.

The kinetic energy of the ball is given by,

$K.E=\frac{1}{2}m{v}^2$

Here,mis the mass of the baseball andvis the speed of the baseball.

The rest mass energy of the baseball is given by,

$E=m{c}^2$

Here, m is the mass of the baseball and $c$is the speed of light whose value is $3\times {10}^{8}m/s$, the ratio of kinetic energy and rest mass energy is given by,

$\frac{K.E}{E}=\frac{\frac{1}{2}m{v}^2}{m{c}^2}$

$=\frac{v^2}{2c^2}$

Substitute values in the above,

$\frac{K·E}{m{c}^2}-\frac{(44\mathrm{m}/\mathrm{s}{)}^2}{2{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}$

$=1.076\times {10}^{14}:1$

Thus, the ratio of kinetic energy to the rest mass energy is$=1.076\times {10}^{-14}:1$.