Conservation of Momentum: An Easy Proof (High School Level)
Theorem
If no external force acts, the total momentum before and after the action of force (internal force) remains constant. In simpler terms, when two objects collide, the sum of each object’s momentum before and after the collision remains the same.
$$ m_{1}v_{1}+m_2v_2=m_{1}{v_{1}}^{\prime}+m_2{v_2}^{\prime} $$
Proof (High School Level)
When two objects $A, B$ collide, by the law of action and reaction, the force each exerts on the other is equal in magnitude but opposite in direction. The force $A$ exerts on $B$ is
$$ \begin{equation} F=F_{AB} \end{equation} $$
The force $B$ exerts on $A$ is
$$ \begin{equation} -F=F_{BA} \end{equation} $$
When the two objects collide, the collision time for both is the same. By $(1)$ and $(2)$,
$$ \begin{equation} Ft+(-F)t=0 \end{equation} $$
Multiplying the force by the time it acts gives $(Ft)$ the impulse$(I)$, and the impulse equals the change in momentum. Therefore,
$$ \begin{equation} \begin{cases} Ft=I_{B}=\Delta p_{B}=m_{B}{v_{B}}^{\prime}-m_{B}v_{B} \\ (-F)t=I_{A}=\Delta p_{A}=m_{A}{v_{A}}^{\prime}-m_{A}v_{A} \end{cases} \end{equation} $$
Combining $(3)$ and $(4)$,
$$ m_{B}{v_{B}}^{\prime}-m_{B}v_{B}+m_{A}{v_{A}}^{\prime}-m_{A}v_{A} =0 \\ \implies m_{A}v_{A}+m_{B}v_{B} = m_{A}{v_{A}}^{\prime}+m_{B}{v_{B}}^{\prime} $$
Hence, the sum of the pre- and post-collision momentum of each object is conserved. Mechanical energy is conserved only when the acting force is a conservative force, but momentum is always conserved if no external force acts. Indeed, the conservation of momentum is a more general law than the law of action and reaction, but it seems that it is not treated with such rigor at the high school level.