Q. When a man increases his speed by $2\ ms^{- 1}$, he finds that his kinetic energy is doubled, the original speed of the man is
A) $2\left( \sqrt{2} - 1 \right)\ ms^{- 1}$
B) $2\left( \sqrt{2} + 1 \right)\ ms^{- 1}$
C) $4.5\ ms^{- 1}$
D) None of these
Explanation
(b) $KE\ = \ \frac{1}{2}mv^{2}$
Given, $v_{2} = (v_{1} + 2)$
$\frac{K_{1}}{K_{2}} = \left( \frac{v_{1}}{v_{2}} \right)^{2}$
$\frac{1}{2} = \frac{v_{1}^{2}}{\left( v_{1} + 2 \right)^{2}}\ \ \ \ \ \ \ \ \ (\therefore k_{2} = 2k_{1})$
$v_{1}^{2} + 4v_{1} + 4 = 2v_{1}^{2}$
$v_{1}^{2} - 4v_{1} - 4 = 0$
$v_{1} = \frac{4 \pm \sqrt{16 + 16}}{2}$
$v_{1} = \frac{4 + \sqrt{32}}{2} = 2\left( \sqrt{2} + 1 \right)ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.