A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

1. Relation of the initial kinetic energy between father and son ${\left({\mathrm{K}}_{\mathrm{father}}\right)}_{\mathrm{i}}=\frac{\left({\mathrm{K}}_{\mathrm{son}}\right)}{2}$.
2. Relation for the kinetic energy is, ${\left({\mathrm{K}}_{\mathrm{father}}\right)}_{\mathrm{f}}=\left({\mathrm{K}}_{\mathrm{son}}\right)$when final velocity is, ${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+1.0\mathrm{m}/\mathrm{s}$where v_i is the initial speed of the father.
3. Mass of both father and son are related as, ${\mathrm{m}}_{\mathrm{son}}=\frac{{\mathrm{m}}_{\mathrm{father}}}{2}$.

Using the formula of kinetic energy, we can write an equation for the kinetic energy of the son and the father. Next, using the given relation between the kinetic energy of father and son, we can find the original speed of father and son.

Formula:

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$

The expression for kinetic energy is,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ (1)

The initial kinetic energy of the father is,

${\left({\mathrm{K}}_{\mathrm{father}}\right)}_{\mathrm{i}}=\frac{\left({\mathrm{K}}_{\mathrm{son}}\right)}{2}$ (2)

The final kinetic energy of the father is,

${\left({\mathrm{K}}_{\mathrm{father}}\right)}_{\mathrm{f}}=\left({\mathrm{K}}_{\mathrm{son}}\right)$

From the above two expressions, we can write,

${\left({\mathrm{K}}_{\mathrm{father}}\right)}_{\mathrm{i}}=\frac{1}{2}{\left({\mathrm{K}}_{\mathrm{father}}\right)}_{\mathrm{f}}$

From equation 1, we can write the above expression as,

$\begin{array}{rcl}\frac{1}{2}{\mathrm{m}}_{\mathrm{father}}{\mathrm{v}}_{\mathrm{i}}^2& =& \frac{{\displaystyle \frac{1}{2}}{\mathrm{m}}_{\mathrm{father}}{\mathrm{v}}_{\mathrm{f}}^2}{2}\\ {\mathrm{v}}_{\mathrm{i}}^2& =& \frac{{\mathrm{v}}_{\mathrm{f}}^2}{2}\end{array}$

Substitute the given data in the above expression, and we get,

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{i}}^2& =& \frac{1}{2}\left[{\left({\mathrm{v}}_{\mathrm{i}}+1.0\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\right]\\ {\mathrm{v}}_{\mathrm{i}}^2& =& \frac{1}{2}\left[\left({\mathrm{v}}_{\mathrm{i}}^2+2{\mathrm{v}}_{\mathrm{i}}+1\right)\right]\\ \frac{1}{2}{\mathrm{v}}_{\mathrm{i}}^2-{\mathrm{v}}_{\mathrm{i}}-\frac{1}{2}& =& 0\end{array}$

This is a quadratic equation; the valid root of the above equation we find is,

${\mathrm{v}}_{\mathrm{i}}=2.4\frac{\mathrm{m}}{\mathrm{s}}$

Therefore, the original speed of the father is 2.4 m/s.

From equations (1) and (2), we can write,

$\frac{1}{2}{\mathrm{m}}_{\mathrm{father}}{\mathrm{v}}_{\mathrm{i}}^2=\frac{{\displaystyle \frac{1}{2}}{\mathrm{m}}_{\mathrm{son}}{\mathrm{v}}_{\mathrm{son}}^2}{2}$

Here v_son is the velocity of the son.

Substitute the given values in the above expression, and we get,

$\begin{array}{rcl}\frac{1}{2}{\mathrm{m}}_{\mathrm{father}}{\mathrm{v}}_{\mathrm{i}}^2& =& \frac{{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\mathrm{m}}_{\mathrm{father}}}{2}}{\mathrm{v}}_{\mathrm{son}}^2}{2}\\ {\mathrm{v}}_{\mathrm{i}}^2& =& \frac{{\displaystyle \frac{1}{2}}{\mathrm{v}}_{\mathrm{son}}^2}{2}\\ {\mathrm{v}}_{\mathrm{son}}^2& =& 4{\mathrm{v}}_{\mathrm{i}}^2\end{array}$

Solving further as,

$\begin{array}{rcl}{\mathrm{v}}_{\mathrm{son}}& =& \sqrt{4{{\mathrm{v}}_{\mathrm{i}}}^2}\\ {\mathrm{v}}_{\mathrm{son}}& =& 2{\mathrm{v}}_{\mathrm{i}}\end{array}$

Substitute the given values in the above expression, and we get,

data-custom-editor="chemistry" $$\begin{gathered} {\mathrm{v}}_{\mathrm{son}}=2\left(2.4\mathrm{m}/\mathrm{s}\right) \\ {\mathrm{v}}_{\mathrm{son}}=4.8\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the original speed of the son is 4.8 m/s.