In a collinear collision , a particle with an initial speed $v_{o}$ 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 particle after collision, is

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$\frac{v_{o}}{2}$

$\frac{v_{0}}{\sqrt{2}}$

$\frac{v_{\circ}}{4}$

$\sqrt{2} v_{0}$

answer is B.

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Detailed Solution

Momentum is conserved in all type of collisions.

Final Kinetic energy is 50% more than initial kinetic energy

$\Rightarrow \frac{1}{2} m v_{2}^{2} + \frac{1}{2} m v_{1}^{2} = \frac{150}{100} \times \frac{1}{2} m v_{0}^{2}$

$\Rightarrow v_{2}^{2} + v_{1}^{2} = \frac{3}{2} v_{0}^{2} \dots \dots . (i)$

from momentum conservation gives,

$\begin{array}{l} m v_{0} = m v_{1} + m v_{2} \\ v_{0} = v_{2} + v_{1} \dots . (i i) \end{array}$

Squaring Eqs. (ii), we get

$\begin{array}{l} v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} = v_{0}^{2} \\ \Rightarrow 2 v_{1} v_{2} = \frac{- v_{0}^{2}}{2} (from eq (i)) \\ ∴ {(v_{1} - v_{2})}^{2} = {(v_{1} + v_{2})}^{2} - 4 v_{1} v_{2} \\ = 2 v_{0}^{2} \\ or v_{rel} = \sqrt{2} v_{0} \end{array}$