Question

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

• A. $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 -\varepsilon =\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$
• B. $\frac{1}{2}m_1^2u_1^2+\frac{1}{2}m_2^2u_2^2 + \varepsilon =\frac{1}{2}m_1^2v_1^2+\frac{1}{2}m_2^2v_2^2$
• C. $m_!^2u_1+m_2^2u_2-\varepsilon =m_1^2 v_1 +m_2^2v_2$
• D. $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 =\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 -\varepsilon $

Answer 1

The Correct Option is A

Total initial energy of two particles
$=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$
Total final energy of two particles
$=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2+ \varepsilon $
Using energy conservation principle,
$\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 =\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 +\varepsilon $
$\therefore \, \, \, $ $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 -\varepsilon =\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$