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 $\varepsilon$. If final velocities of particles be $v_1$ and $v_2$, then we must have:
(2015)
$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 - \varepsilon$
$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \varepsilon = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$
$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 + \varepsilon$
$m_1^2 u_1 + m_2^2 u_2 - \varepsilon = m_1^2 v_1 + m_2^2 v_2$
Solution:
According to the conservation of energy, the total initial energy equals the total final energy plus the energy absorbed ($\varepsilon$). Thus, initial kinetic energy minus absorbed energy equals final kinetic energy: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \varepsilon = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$.
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