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Oblique Collision

In a collinea...

Question

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

A

\frac{v_{0}}{2}

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B

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

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C

\frac{v_{0}}{4}

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D

\sqrt{2} v_{0}

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Solution

The correct option is D $\sqrt{2} v_{0}$
By conservation of linear momentum
${m v}_{0} + 0 = {m v}_{1} + {m v}_{2}$
$v_{0} = v_{1} + v_{2} . . . . . . . . . . (1)$
$\frac{3}{2} [\frac{1}{2} {m v}_{0}^{2}] = \frac{1}{2} {m v}_{1}^{2} + \frac{1}{2} {m v}_{2}^{2}$
$\frac{3}{2} v_{0}^{2} = v_{1}^{2} + v_{2}^{2} \dots \dots \dots \cdot (2)$

Solving equation (1) and (2)

$v_{1} = \frac{v_{0}}{2} (1 + \sqrt{2})$
$v_{2} = \frac{v_{0}}{2} (1 - \sqrt{2})$
${¯ ¯ ¯ v}_{r e l} = {¯ ¯ ¯ v}_{1} - {¯ ¯ ¯ v}_{2}$
$\frac{v_{0}}{2} [1 + \sqrt{2} - 1 + \sqrt{2}]$
$= \frac{v_{0}}{2} \times 2 \sqrt{2}$
$= \sqrt{2} v_{0}$

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