Question

A man who is running has half the kinetic energy of the boy of half his mass. The man speeds up by $1m/s$ and then has the same kinetic energy as the boy. The original speed of the man was

• A. $\sqrt{2}m/s$
• B. $(\sqrt{2}-1)m/s$
• C. $2m/s$
• D. $(\sqrt{2}+1)m/s$

Answer 1

The correct option is D $(\sqrt{2}+1)m/s$

Given mass of boy is half that of man, and kinetic energy is twice.
Let $M$ be mass of the man, $\frac{M}{2}$ be mass of the boy
$V_{boy}$ is velocity of the boy.
$V_{man}$ is velocity of the man.

$2K.E_{man}=K.E{.}_{boy}$
$2\times \frac{1}{2}M\times V_{man}^2=\frac{1}{2}.\frac{M}{2}{V}_{boy}^2$
$V_{man}=\frac{V_{boy}}{2}......\left(i\right)$


Now, If velocity of man is increased by $1m/s$, $K.E_{man}=K.E{.}_{boy}$
$\Rightarrow \frac{1}{2}M(V_{man}+1{)}^2=\frac{1}{2}.\frac{M}{2}{V}_{boy}^2$
$\Rightarrow (V_{man}+1{)}^2=\frac{V_{boy}^2}{2}$
from $\left(i\right)$
$\Rightarrow (V_{man}+1{)}^2=\frac{4V_{man}^2}{2}=2V_{man}^2$
$\Rightarrow (V_{man}+1)=\sqrt{2}{V}_{man}$
$\Rightarrow 1=(\sqrt{2}-1)V_{man}$
$\Rightarrow V_{man}=\frac{1}{\sqrt{2}-1}$
$\Rightarrow V_{man}=(\sqrt{2}+1)m/sec$