Two particles of masses $m_{1}, m_{2}$ move with initial velocities $u_{1}$ and $u_{2}$ . On collision, one of the particles get excited to higher level, after absorbing energy $ϵ$ . If final velocities of particles be $v_{1}$ and $v_{2}$ then we must have

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

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

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m_{1}^{2} u_{1} + m_{2}^{2} u_{2} - ϵ = m_{1}^{2} v_{1} + m_{2}^{2} v_{2}

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\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}

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Solution

The correct option is B $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ϵ = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
Using energy conservation principle:
Initial energy is only in form of kinetic energy
$= \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}$ ...(i)
final energy is in form of kinetic energy and potential energy ( excitation energy)
$= \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} + ϵ$ ...(ii)
equation (i) and (ii)
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} + ϵ$
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ϵ = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
hence correct answer is option A.

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Two particles of masses $m_{1}, m_{2}$ moving with initial velocities $u_{1} a n d u_{2}$ . On collision, one of the particles get excited to higher level, after absorbing energy ε . If final velocities of particles be $v_{1} a n d v_{2}$ then we must have :