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Relative Velocity

Find the tota...

Question

Find the total kinetic energy $(T)$ of the two particles in the reference frame fixed to their centre of inertia having masses $m_{1}$ and $m_{2}$ and velocities $_{1}$ and $_{2}$ (perpendicular to each other) respectively. Here $μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$

A

T = \frac{1}{2} μ (|_{1} |^{2} + |_{2} |^{2})

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B

T = \frac{1}{2} μ (|_{1} |^{2} + 2 |_{2} |^{2})

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C

T = \frac{1}{3} μ (|_{1} |^{2} + |_{2} |^{2})

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D

T = \frac{1}{2} μ (2 |_{1} |^{2} + |_{2} |^{2})

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Solution

The correct option is A $T = \frac{1}{2} μ (|_{1} |^{2} + |_{2} |^{2})$
Velocity of center of inertia frame $= \frac{m_{1}_{1} + m_{2}_{2}}{m_{1} + m_{2}}$
IN COM reference frame:

Velocity of $m_{1}$ relative to this frame $= {\to v}_{r_{1}} = {\to v}_{1} - \frac{m_{1} {\to v}_{1} + m_{2} {\to v}_{2}}{m_{1} + m_{2}} = \frac{m_{2} ({\to v}_{1} - {\to v}_{2})}{m_{1} + m_{2}}$
Similarly, velocity of $m_{2}$ relative to this frame $= {\to v}_{r_{2}} = {\to v}_{2} - \frac{m_{1} {\to v}_{1} + m_{2} {\to v}_{2}}{m_{1} + m_{2}} = \frac{m_{1} ({\to v}_{2} - {\to v}_{1})}{m_{1} + m_{2}}$
Hence the kinetic energy with respect to this frame $T = \frac{1}{2} m_{1} v_{r_{1}}^{2} + \frac{1}{2} m_{2} v_{r_{2}}^{2}$
$T = \frac{1}{2} m_{1} (\frac{m_{2} ({\to v}_{1} - {\to v}_{2})}{m_{1} + m_{2}})^{2} + \frac{1}{2} m_{2} (\frac{m_{1} ({\to v}_{2} - {\to v}_{1})}{m_{1} + m_{2}})^{2}$
$T = \frac{1}{2} μ | {\to v}_{2} - {\to v}_{1} |^{2}$

As given velocities are perpendicular to each other,
$| {\to v}_{2} - {\to v}_{1} |^{2} = (|_{1} |^{2} + |_{2} |^{2})$

$T = \frac{1}{2} μ (|_{1} |^{2} + |_{2} |^{2})$

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