In a collinear collision, a particle with an initial speed vo 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 collision, is

As we learnt that

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

From conservation of momentum

$mv_{o}+0=mv_{1}+mv_{2}\; \; \; \; \; \; eq \, 1$

and     $\frac{3}{2}(\frac{1}{2}mv^{2})=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}mv_{2}^{2}$

$or\: v_{1}^{2}+v_{2}^{2}=\frac{3}{4}v_{o}^{2}$                   eq 2

$and \: v_{1}+v_{2}=v_{o}$

From 1 and 2

$v_{1}=\frac{v_{o}}{2}.(1+\sqrt{2})$

$and\: v_{2}=\frac{v_{o}}{2}.(1-\sqrt{2})$

$V_{rel}=\vec{V_{1}}-\vec{V_{2}}=\frac{V_{o}}{2}[1+\sqrt{2}-1+\sqrt{2}]$

$=\sqrt{2}V_{o}$

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