Question

give the proof of law of conservation of momentum

Answer 1

momentum : momentum of a body is defined as the product of the mass and velocity of the body.
P=MV
law of conservation not momentum: when there is no external force acting on the system,the total momentum of an isolated system of particles is conserved.
explanation : consider two bodies A and B with initial momenta Pa and Pb.
after collision final momentum are PA and PB .
by second law F ab dt=PA- Pa F ba dt=PB - Pb
by third law F ab = - F ba
PA- Pa= -(PB - pb)
PA+PB = Pa+Pb
total final momentum=total initial momentum

Answer 2

$\bf{\red{Proof \:of \:the\: law \:of\: conservation\: of \:momentum}}$

Consider a system consisting of two bodies X and Y of masses $m_1$ and $m_2$ respectively in (refer to pic).

Suppose two bodies are moving with velocities $u_1$ and$u_2$ and say $u_1>u_2.$

Let these bodies collide with each other for a small interval of time $'dt'$.

At the time of collision, body X exerts a force F on body Y, and body Y exerts and equal and opposite force (-F) on body X. This changes the momentum of both the bodies.

Now let $v_1$ and $v_2$ be the velocity of body X and Y respectively after the collision. Then

Initial momentum of body X = $m_1u_1$

Initial momentum of body Y = $m_2u_2$

Final momentum of body X = $m_1v_1$

Final momentum of body Y =$m_2v_2$

Therefore, Total momentum of the system before collision

$m_1u_1+m_2u_2$

Total momentum of the system after collision

$m_1v_1+m_2v_2$

Now,

The rate of change of momentum pf a body X

$=\dfrac{Change \:in \:momentum\: of\: body \:X}{Time \:taken}$

$=\dfrac{Final \:momentum - Initial\: momentum}{Time\: taken}$

$=\dfrac{(m_1v_1-m_1u_1)}{dt}$

Now,

Rate of change of momentum of body Y

$=\dfrac{Change\: in \:momentum \:of\: body\: Y}{Time \:taken}$

$F=\dfrac{(m_2v_2-m_2u_2)}{dt}$

According to Newton's second law of motion,

Force acting on body Y

= Rate of change of momentum of body Y

$-F=\dfrac{(m_2v_2-m_2u_2)}{dt}$

$F=\dfrac{-(m_2v_2-m_2u_2)}{dt}$

Equating (1) and (2),we get

$⟹\dfrac{(m_1v_1-m_1u_1)}{dt}=\dfrac{-(m_2v_2-m_2u_2)}{dt}$

$⟹m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Thus, total momentum of the system before collision is equal to the total momentum of the system after collision. Hence, the law of conservation of momentum is proved.