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 $ϵ$. If final velocities of particles be $v_1$ and $v_2$ then we must have

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

Answer 1

Answer: (A)

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2-ϵ=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

Using energy conservation principle:
Initial energy is only in form of kinetic energy
$=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$...(i)
final energy is in form of kinetic energy and potential energy ( excitation energy)
$=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+ϵ$...(ii)
equation (i) and (ii)
$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+ϵ$
$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2-ϵ=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$
hence correct answer is option A.