Question

A running man has half the kinetic energy that a boy half his mass has. The man speeds up by 1.0m/s and then has the same energy as the boy. What were the orignal speed of the man and the boy?

Answer 1

Answer is approximately 4.88.

Steps in picture given below

Answer 2

The speed of man and boy is $\bold{2.41 m/s\ and\ 4.82 m/s}$.

Given:

Mass of man = m  

Mass of boy = m/2

Velocity of man $= v_1$

Velocity of boy $= v_2$

Solution:  

The Kinetic energy initially is given below:

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

From question,

⇒ K.E. of man = ½ K.E. of boy

$\Rightarrow\left(\frac{1}{2} m v_{1}^{2}\right)=\frac{1}{2}\left(\frac{1}{2} \times \frac{m}{2} \times v_{2}^{2}\right)$

$v_{1}^{2}=\left(\frac{1}{4}\right) v_{2}^{2}$

$\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{1}{4}$

$\Rightarrow \frac{v_{1}}{v_{2}}=\frac{1}{2} \rightarrow(1)$

From the question, when they have same energy,

⇒ K.E. of man = K.E. of boy

$\Rightarrow \frac{1}{2}\left(m\left(v_{1}+1\right)^{2}\right)=\frac{1}{2}\left(\frac{m}{2} \times v_{2}^{2}\right)$

$\left(v_{1}+1\right)^{2}=\frac{1}{2} \times v_{2}^{2}$

$2\left(v_{1}+1\right)^{2}=v_{2}^{2}$

$v_{2}=\sqrt{2}\left(v_{1}+1\right) \rightarrow(2)$

Thereby, from both case,  

$v_{1}=\frac{v_{2}}{2}$

On substituting equation (2) in equation (1), we get,

$v_{1}=\frac{\sqrt{2}\left(v_{1}+1\right)}{2}$

$2 v_{1}=\sqrt{2}\left(v_{1}+1\right)$

$2 v_{1}=\sqrt{2} v_{1}+\sqrt{2}$

$2 v_{1}-\sqrt{2} v_{1}=\sqrt{2}$

$v_{1}(2-\sqrt{2})=\sqrt{2}$

$v_{1}=\frac{\sqrt{2}}{(2-\sqrt{2})}$

$v_{1}=2.41 \ \mathrm{m} / \mathrm{s}$

On substituting, $v_1$ in equation (1), we can get the velocity of boy,

$v_{2}=2 v_{1}$

$v_{2}=2(2.41)$

$v_{2}=4.82 \ \mathrm{m} / \mathrm{s}$.