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.
$$m_1^2{u_1} + m_2^2{u_2} - \varepsilon = m_1^2{v_1} + m_2^2{v_2}$$
B.
$$\frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 - \varepsilon $$
C.
$$\frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 - \varepsilon = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2$$
D.
$$\frac{1}{2}m_1^2u_1^2 + \frac{1}{2}m_2^2u_2^2 + \varepsilon = \frac{1}{2}m_1^2v_1^2 + \frac{1}{2}m_2^2v_2^2$$
Answer :
$$\frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 - \varepsilon = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2$$
Solution :
Total initial energy $$ = \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2$$
Since, after collision one particle absorb energy $$\varepsilon .$$
∴ Total final energy $$ = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 + \varepsilon $$
From conservation of energy,
$$\eqalign{
& \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 + \varepsilon \cr
& \Rightarrow \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 - \varepsilon = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 \cr} $$