Question

Two particles of masses $m_1,m_2$ movie with initial velocities $u_1\text{and}{u}_2$.On collision, one of the particles get excited to higher level, after absorbing energy If final velocities of particles be $v_1$ and $v_2$ then we must have :

• A. $\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2-\epsilon$
• B. $\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2+\epsilon =\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$
• C. $\frac{1}{2}{m}_1^2{u}_1^2+\frac{1}{2}{m}_2^2{u}_2^2-\epsilon =\frac{1}{2}{m}_1^2{v}_1^2+\frac{1}{2}{m}_2^2{v}_2^2$
• D. $m_1^2{u}_1+m_2^2{u}_2-\epsilon =m_1^2{v}_1+m_2^2{v}_2$

Answer 1

The correct option is C $\frac{1}{2}{m}_1^2{u}_1^2+\frac{1}{2}{m}_2^2{u}_2^2-\epsilon =\frac{1}{2}{m}_1^2{v}_1^2+\frac{1}{2}{m}_2^2{v}_2^2$

$\begin{array}{c}Totalinitialenergyoftwoparticles\\ =\frac{1}{2}{m}_1{u_1}^2+\frac{1}{2}{m}_2{u_2}^2\\ Totalfinalenergyoftwoparticles\\ =\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2+\in \\ U\sin genergyconservationprinciple,\\ \frac{1}{2}{m}_1{u_1}^2+\frac{1}{2}{m}_2{u_2}^2=\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2+\in \\ \therefore \frac{1}{2}{m}_1{u_1}^2+\frac{1}{2}{m}_2{u_2}^2-\in =\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2\end{array}$

Hence,

option $\left(C\right)$ is correct answer.