A body ‘A’ experiences a perfectly elastic collision with a stationary body ‘B’. If after the collision the bodies fly apart in the opposite direction with equal velocities, the mass ratio of ‘A’ and ‘B’ is
$\begin{align}
  & a)\dfrac{1}{2} \\
 & b)\dfrac{1}{3} \\
 & c)\dfrac{1}{4} \\
 & d)\dfrac{1}{5} \\
\end{align}$

Hint: It is given in the question that the two bodies collide elastically. Therefore we can conclude that the momentum as well as the kinetic energy is conserved. The velocities of the bodies A i.e. ${{v}_{1}}$ and B i.e. ${{v}_{2}}$ after collision is said to be equal but in opposite direction, therefore we can say that $v={{v}_{1}}=-{{v}_{2}}$ Hence equating the initial and the final momentum and the kinetic energy of the two bodies we can obtain the required ratio.

Complete step by step answer:

In the above figure we can see the bodies A and B travel in opposite directions after the collision. The velocities of the two bodies are the same i.e. $v={{v}_{1}}=-{{v}_{2}}$. Let us say the velocity of the body A before collision is V and therefore the velocity of the body A after collision will be ‘–v’ and that of B will be ‘v’. Let us say the mass of A is ${{\text{m}}_{\text{A}}}$and mass of B as ${{\text{m}}_{\text{B}}}$.Therefore by using the law of conservation of momentum we get i.e. momentum of the bodies before collision equals to the momentum of the bodies after collision. Hence we can represent this mathematically as,
$\begin{align}
  & {{\text{m}}_{\text{A}}}V=-{{m}_{A}}v+{{m}_{B}}v \\
 & \Rightarrow V=\dfrac{\left( {{m}_{B}}-{{m}_{A}} \right)v}{{{\text{m}}_{\text{A}}}}...(1) \\
\end{align}$
Similarly the kinetic energy (K.E) is also conserved. Therefore we can write,
$\begin{align}
  & Final(K.E)=Initial(K.E) \\
 & \dfrac{1}{2}{{m}_{A}}{{V}^{2}}=\dfrac{1}{2}{{m}_{A}}{{v}^{2}}+\dfrac{1}{2}{{m}_{B}}{{v}^{2}} \\
 & {{m}_{A}}{{V}^{2}}={{m}_{A}}{{v}^{2}}+{{m}_{B}}{{v}^{2}} \\
\end{align}$
Further on substitution of ‘V’ we get,
$\begin{align}
  & {{m}_{A}}{{\left( \dfrac{\left( {{m}_{B}}-{{m}_{A}} \right)v}{{{\text{m}}_{\text{A}}}} \right)}^{2}}={{m}_{A}}{{v}^{2}}+{{m}_{B}}{{v}^{2}} \\
 & \Rightarrow {{m}_{A}}\left( \dfrac{{{\left( {{m}_{B}}-{{m}_{A}} \right)}^{2}}{{v}^{2}}}{{{\text{m}}_{\text{A}}}^{2}} \right)={{m}_{A}}{v^2}+{{m}_{B}}{{v}^{2}} \\
 & \dfrac{{{\left( {{m}_{B}}-{{m}_{A}} \right)}^{2}}{{v}^{2}}}{{{\text{m}}_{\text{A}}}}={{m}_{A}}{{v}^{2}}+{{m}_{B}}{{v}^{2}}\text{, }\because {{\left( a-b \right)}^{2}}=a+b-2ab \\
 & \dfrac{{{m}_{A}}^{2}+{{m}_{B}}^{2}-2{{m}_{A}}{{m}_{B}}}{{{\text{m}}_{\text{A}}}}={{m}_{A}}+{{m}_{B}} \\
 & \Rightarrow {{m}_{A}}^{2}+{{m}_{B}}^{2}-2{{m}_{A}}{{m}_{B}}={{m}_{A}}^{2}+{{m}_{B}}{{\text{m}}_{\text{A}}} \\
 & \Rightarrow {{m}_{B}}^{2}-2{{m}_{A}}{{m}_{B}}={{m}_{B}}{{\text{m}}_{\text{A}}} \\
 & \Rightarrow {{m}_{B}}^{2}=3{{m}_{B}}{{\text{m}}_{\text{A}}} \\
 & \Rightarrow {{m}_{B}}=3{{\text{m}}_{\text{A}}}\text{, hence} \\
 & \Rightarrow \dfrac{{{\text{m}}_{\text{A}}}}{{{m}_{B}}}=\dfrac{1}{3} \\
\end{align}$
Therefore we can conclude that the correct answer is option b.

Note:
The velocity of the body ‘A’ after collision is considered to be negative. This is because the convention of the velocity V is taken as positive and the direction opposite to it has to be negative. It is also to be noted that in elastic collision both the momentum and the kinetic energy are conserved, But in case of inelastic collision, the momentum is conserved but the kinetic energy is not conserved.