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 

Answers (1)
G Gaurav

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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