Question 26 of 130

In a collinear collision, a particle with an initial speed ${v}_{0}$ strikes a stationary particle of the same mass. If the final total kinetic energy is $50$ greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after the collision, is

A$\dfrac{{v}_{0}}{\sqrt{2}}$B$\dfrac{{v}_{0}}{4}$C$\sqrt{2}{v}_{0}$D$\dfrac{{v}_{0}}{2}$

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Answer: (C) $\sqrt{2}{v}_{0}$

$K{E}_{2}=\dfrac{3}{2}m{v}_{0}^{2}$. Conservation of momentum, ${v}_{1}+{v}_{2}={v}_{0}$. Conservation of energy, $\dfrac{1}{2}m{v}_{1}^{2}+\dfrac{1}{2}m{v}_{2}^{2}=\dfrac{1}{2}\times \dfrac{3}{2}m{v}_{0}^{2}$
$\Rightarrow {v}_{1}^{2}+{v}_{2}^{2}=\dfrac{3}{2}{v}_{0}^{2}$$\Rightarrow {v}_{1}^{2}+{v}_{2}^{2}+2{v}_{1}{v}_{2}={v}_{0}^{2}$$\Rightarrow \dfrac{3}{2}{v}_{0}^{2}+2{v}_{1}{v}_{2}={v}_{0}^{2}$
$\Rightarrow 2{v}_{1}{v}_{2}=-\dfrac{1}{2}{v}_{0}^{2}$${\Rightarrow ({v}_{1}-{v}_{2})}^{2}=\dfrac{3}{2}{v}_{0}^{2}+\dfrac{1}{2}{v}_{0}^{2}$$=2{v}_{0}^{2}$
$\Rightarrow {v}_{1}-{v}_{2}=\sqrt{2}{v}_{0}$.