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Q.
In an isolated system, (i.e. a system with no external force) total momentum of interacting particles is conserved. It follows
Laws of Motion
Solution:
Case I Consider two particles of a system having masses $m_{1}$ and $m_{2}$ are moving with velocities $v _{1}$ and $v _{2}$, respectively. Then, total linear momentum of the system
$p =m_{1} v _{1}+m_{2} v _{2}= p _{1}+ p _{2}$
If $F$ is the external force acting on the system, then according to Newton's second law of motion,
$F =\frac{d p }{d t}$
For an isolated system, $F =0 \Rightarrow \frac{d p }{d t}=0$
$\Rightarrow p =$ constant i.e. $p _{1}+ p _{2}=$ constant
Case II Consider two particles of masses $m_{1}$ and $m_{2}$ moving along a straight line in opposite direction collide to each other, if $\Delta p_{1}$ and $\Delta p_{2}$ be the changes in momenta produced in time $\Delta t$, then according to the law of conservation of momentum, if no external force (for isolated system) is applied on the system
$\Rightarrow \Delta p_{1}+\Delta p_{2} =0$
$\Rightarrow \Delta p_{2} =-\Delta p_{1}$
$\Rightarrow \frac{\Delta p_{2}}{\Delta t} =-\frac{\Delta p_{1}}{\Delta t}$
$\Rightarrow$ Force on $m_{2} =-$ Force on $m_{1}$
$\left[\because F=\frac{d p}{d t}\right]$
$F_{2} =-F_{1}$
So, in an isolated system, when total momentum of interacting particles is conserved, it follow both, Newton's second and third laws.
Hence, options (b) and (c) are correct.