Question

State and derive law of conservation of momentum.

Answer 1

Law of conservation of momentum states that in the absence of external forces, total momentum of a system comprising of two or more interacting bodies is constant.
Derivation:
Let us consider two moving balls $A$ and $B$ of masses $m_1$ and $m_2$ and having initial velocities $u_1$ and $u_2$ such that $u_2<u_1$.
Suppose the balls collide at some point and there is no external force acting on this system.
Let their final velocities be $v_1$ and $v_2$ respectively.
According to Newton’s third law of motion,
Force on ball $B$ due to $A$ = -Force on ball $A$ due to $B$.
Or, $F_{AB}=-F_{BA}\dots \dots \dots \dots \dots \dots .\left(i\right)$
Total initial momentum before collision $\left(p_i\right)=m_1{u}_1+m_2{u}_2$.
Total final momentum after collision $\left(p_f\right)=m_1{v}_1+m_2{v}_2$.
According to Newton’s second law,
$F_{BA}=\frac{p_A^{\prime }-p_A}{t}=\frac{m_1{v}_1-m_1{u}_1}{t}...........\left(ii\right)$
$F_{AB}=\frac{p_B^{\prime }-p_B}{t}=\frac{m_2{v}_2-m_2{u}_2}{t}...........\left(iii\right)$

From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$,

$\frac{m_1{v}_1-m_1{u}_1}{t}=-\frac{m_2{v}_2-m_2{u}_2}{t}$
$\Rightarrow m_1{v}_1-m_1{u}_1=-(m_2{v}_2-m_2{u}_2)$
$\Rightarrow m_1{v}_1+m_2{v}_2=m_1{u}_1+m_2{u}_2$
$\Rightarrow$ Final momentum $\left(p_f\right)$ = Initial momentum $\left(p_i\right)$.
Thus, we conclude that during the interaction between the two balls, momentum before collision is equal to total momentum after collision. Thus, the momentum of the system is conserved in the absence of external forces.