Question

When a man increases his speed by $2m{s}^{-1}$, he finds that his kinetic energy is doubled, the original speed of the man is

• A. $2(\sqrt{2}-1)m{s}^{-1}$
• B. $2(\sqrt{2}+1)m{s}^{-1}$
• C. $4.5m{s}^{-1}$
• D. None of these

Answer 1

Answer: (B)

$KE=\frac{1}{2}m{v}^2$
Given, $v_2=\left(v_1+2\right)$
$\therefore \frac{K_1}{K_2}={\left(\frac{v_1}{v_2}\right)}^2$
$\Rightarrow \frac{1}{2}=\frac{v_1^2}{{\left(v_1+2\right)}^2}$
$\left(\therefore K_2=2K_1\right)$
$v_1^2+4v_1+4=2v_1^2$
$v_1^2-4v_1-4=0$
$v_1=\frac{4\pm \sqrt{16+16}}{2}$
$v_1=\frac{4+\sqrt{32}}{2}$
$=2(\sqrt{2}+1)m{s}^{-1}$