Two particles of masses move with initial velocities and . On collision, one of the particles gets excited to a higher level, after absorbing energy . If final velocities of particles be and , then we must have [2015]

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Solution

Q.
Two particles of masses $m_{1}, m_{2}$ move with initial velocities $u_{1}$ and $u_{2}$ . On collision, one of the particles gets excited to a higher level, after absorbing energy $ε$ . If final velocities of particles be $v_{1}$ and $v_{2}$ , then we must have [2015]

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

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

3 $m_{1}^{2} u_{1} + m_{2}^{2} u_{2} - ε = m_{1}^{2} v_{1} + m_{2}^{2} v_{2}$

4 $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} - ε$

Ans.
(1)

Total initial energy of two particles

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

Total final energy of two particles

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

Using energy conservation principle,

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

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

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