Question

a running man has the half the kinetic energy of a body of half his mass. the man speeds up by 1m/s and has the same kinetic energy as the boy. what were the original speed of the man and the boy?

Answer 1

QUESTION:

A running man has the half the kinetic energy of a body of half his mass. the man speeds up by 1m/s and has the same kinetic energy as the boy. What were the original speed of the man and the boy?

SOLUTION:

The kinetic energy of a man is $E_{k,man} = \frac{m_{man} v_{man}^2}{2}$. The kinetic energy of a boy is

$$E_{k,boy} = \frac{m_{boy} v_{boy}^2}{2} = \frac{m_{man}}{2} \cdot \frac{v_{boy}^2}{2} = \frac{m_{man} v_{boy}^2}{4}. \text{ When man speeds up by a } \Delta v = 1 \, \text{m/s}, \text{ his kinetic energy is } E_{k,man}' = \frac{m_{man} (v_{man} + \Delta v)^2}{2}$$

As a man has the half the kinetic energy of a boy:

$$E_{k,man} = \frac{1}{2} E_{k,boy}$$

And after speeding up man has the same kinetic energy as a boy:

$$E_{k,man}' = E_{k,boy}$$

Hence:

$$\left\{ \begin{array}{l} \frac{m_{man} v_{man}^2}{2} = \frac{1}{2} \cdot \frac{m_{man} v_{boy}^2}{4} \\ \frac{m_{man} (v_{man} + \Delta v)^2}{2} = \frac{m_{man} v_{boy}^2}{4} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} v_{man}^2 = \frac{v_{boy}^2}{4} \\ (v_{man} + \Delta v)^2 = \frac{v_{boy}^2}{2} \end{array} \right. \Rightarrow$$ $$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} v_{man} = \frac{v_{boy}}{2} \\ v_{man} + \Delta v = \frac{v_{boy}}{\sqrt{2}} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} v_{man} = \frac{v_{boy}}{2} \\ v_{boy} = \sqrt{2} (v_{man} + \Delta v) \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} v_{man} = \frac{\sqrt{2} (v_{man} + \Delta v)}{2} \\ v_{boy} = \sqrt{2} (v_{man} + \Delta v) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 2 v_{man} = \sqrt{2} (v_{man} + \Delta v) \\ v_{boy} = \sqrt{2} (v_{man} + \Delta v) \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} 2 v_{man} - \sqrt{2} v_{man} = \sqrt{2} \Delta v \\ v_{boy} = \sqrt{2} (v_{man} + \Delta v) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} v_{man} (2 - \sqrt{2}) = \sqrt{2} \Delta v \\ v_{boy} = \sqrt{2} (v_{man} + \Delta v) \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} v_{man} = \frac{\sqrt{2} \Delta v}{2 - \sqrt{2}} \\ v_{boy} = \sqrt{2} (v_{man} + \Delta v) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} v_{man} = \frac{\sqrt{2} \Delta v}{2 - \sqrt{2}} \\ v_{boy} = \sqrt{2} \left(\frac{\sqrt{2} \Delta v}{2 - \sqrt{2}} + \Delta v\right) \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} v_{man} = \frac{\sqrt{2} \Delta v}{2 - \sqrt{2}} \\ v_{boy} = \frac{2 \sqrt{2} \Delta v}{2 - \sqrt{2}} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} v_{man} = 2.414 \, \text{m/s} \\ v_{boy} = 4.828 \, \text{m/s} \end{array} \right. \end{array}$$

ANSWER:

$$v_{man} = 2.414 \, \text{m/s}$$ $$v_{boy} = 4.828 \, \text{m/s}$$