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Question: State and prove the law of conservation of momentum....

State and prove the law of conservation of momentum.

Explanation

Solution

Here, we will proceed by giving the statement of law of conservation of momentum. Then, we will take a case of two bodies undergoing elastic collision and will use Newton’s second and third laws.
Formula used: p = mv, FAB=FBA{{\text{F}}_{{\text{AB}}}} = - {{\text{F}}_{{\text{BA}}}} and F=dpdt{\text{F}} = \dfrac{{d{\text{p}}}}{{dt}}.

Complete answer:
Law of conservation of momentum states that unless an external force is applied, the two or more objects acting upon each other in an isolated system, the total momentum of the system remains constant. This also means that the total momentum of an isolated system before is equal to the total momentum of the isolated system after.
In order to prove the law of conservation of momentum, let us take two bodies A and B of masses m1{{\text{m}}_1} and m2{{\text{m}}_2} respectively moving with initial velocities u1{{\text{u}}_1} and u2{{\text{u}}_2} respectively in the same direction (right direction) as shown in the figure. Now, elastic collision occurs between these bodies because we have assumed u1>u2{{\text{u}}_1} > {{\text{u}}_2}. After elastic collision let us assume that the final velocities of bodies A and B are v1{{\text{v}}_1} and v2{{\text{v}}_2} respectively and these bodies are moving in the same direction (right direction) even after the collision.

As we know that the momentum ‘p’ of the body of mass m and moving with velocity v is given by
p = mv (1) \to (1)
Using equation (1), we can write
Change in momentum of a body = (Mass of the body)(Final velocity – Initial velocity) (2) \to (2)
Using equation (2), the change in the momentum of body A is given by
Change in momentum of body A = (Mass of body A)(Final velocity of body A – Initial velocity of body A)
\Rightarrow Change in momentum of body A = m1(v1u1){{\text{m}}_1}\left( {{{\text{v}}_1} - {{\text{u}}_1}} \right)
Using equation (2), the change in the momentum of body B is given by
Change in momentum of body B = (Mass of body B)(Final velocity of body B – Initial velocity of body B)
\Rightarrow Change in momentum of body B = m2(v2u2){{\text{m}}_2}\left( {{{\text{v}}_2} - {{\text{u}}_2}} \right)
According to Newton’s third law of motion, we can say that for any action there occurs an equal and opposite reaction.
i.e., Force applied by body B on body A = Force applied by body A on body B
FAB=FBA (3)\Rightarrow {{\text{F}}_{{\text{AB}}}} = - {{\text{F}}_{{\text{BA}}}}{\text{ }} \to {\text{(3)}} where negative sign shows that both forces occur in opposite directions

According to Newton’s second law of motion, force applied F to a body is equal to the rate of change of momentum p of the body with respect to time
F=dpdt F=Δpdt (4)  {\text{F}} = \dfrac{{d{\text{p}}}}{{dt}} \\\ {\text{F}} = \dfrac{{\Delta {\text{p}}}}{{dt}}{\text{ }} \to (4) \\\
Using equation (4), we have
Force applied by body B on body A FAB=Change in momentum of body ATime taken=m1(v1u1)t{{\text{F}}_{{\text{AB}}}} = \dfrac{{{\text{Change in momentum of body A}}}}{{{\text{Time taken}}}} = \dfrac{{{{\text{m}}_1}\left( {{{\text{v}}_1} - {{\text{u}}_1}} \right)}}{{\text{t}}}
Using equation (4), we have
Force applied by body A on body B FBA=Change in momentum of body BTime taken=m2(v2u2)t{{\text{F}}_{{\text{BA}}}} = \dfrac{{{\text{Change in momentum of body B}}}}{{{\text{Time taken}}}} = \dfrac{{{{\text{m}}_2}\left( {{{\text{v}}_2} - {{\text{u}}_2}} \right)}}{{\text{t}}}
Using equation (3), we get
m1(v1u1)t=m2(v2u2)t m1(v1u1)=[m2(v2u2)] m1v1m1u1=(m2v2m2u2) m1v1m1u1=m2v2+m2u2 m1v1+m2v2=m1u1+m2u2  \Rightarrow \dfrac{{{{\text{m}}_1}\left( {{{\text{v}}_1} - {{\text{u}}_1}} \right)}}{{\text{t}}} = - \dfrac{{{{\text{m}}_2}\left( {{{\text{v}}_2} - {{\text{u}}_2}} \right)}}{{\text{t}}} \\\ \Rightarrow {{\text{m}}_1}\left( {{{\text{v}}_1} - {{\text{u}}_1}} \right) = - \left[ {{{\text{m}}_2}\left( {{{\text{v}}_2} - {{\text{u}}_2}} \right)} \right] \\\ \Rightarrow {{\text{m}}_1}{{\text{v}}_1} - {{\text{m}}_1}{{\text{u}}_1} = - \left( {{{\text{m}}_2}{{\text{v}}_2} - {{\text{m}}_2}{{\text{u}}_2}} \right) \\\ \Rightarrow {{\text{m}}_1}{{\text{v}}_1} - {{\text{m}}_1}{{\text{u}}_1} = - {{\text{m}}_2}{{\text{v}}_2} + {{\text{m}}_2}{{\text{u}}_2} \\\ \Rightarrow {{\text{m}}_1}{{\text{v}}_1} + {{\text{m}}_2}{{\text{v}}_2} = {{\text{m}}_1}{{\text{u}}_1} + {{\text{m}}_2}{{\text{u}}_2} \\\
The above equation represents that the total momentum of the system after collision is equal to the total momentum of the system before collision.
Therefore, the law of conservation of momentum is proved.

Note:
In this particular problem, we have considered the right direction as the positive direction. Here, it is very important to take care of directions also because momentum is a vector quantity (as velocity is itself a vector quantity). Here, momentum before collision is equal to the sum of the individual momentum of the bodies A and B before collision and similarly, momentum after collision is equal to the sum of the individual momentum of the bodies A and B after collision.