Question

Derivation of law of conservation of momentum

Answer 1

Linear momentum before collision=linear momentum after collision
Pb=m2u2(before collision)
Pb^=m2v2(after collision)
Rate of change of l m of B by A ie.F21
=Pb-Pb/t= m2v2-m2u2/t= m2(v2-u2)/t
Rate of change of l m of A by B ie.F12=
Pa-Pa/t=m1v1-m1u1/t=m1(v1-u1)/t
Therefore,F12=-F21
m1(v1-u1)/t= -m2(v2-u2)/t
m1u1+m2u2= - m1v1+m2v2

Answer 2

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Conservation 0f Momentum➫ A system in which the momentum of a system is constant if there are no external forces acting on the system.

Prove: Conservation 0f Momentum:-▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃⤵

➧Let the Body A = m1 & B = m2

➧Let their velocity be u1 & u2
(u1 > u2)

➧Momentum of Body
A = m1u1 & B = m2u2

❱ Initial & Final momentum of System
➾ m1u1 + m2u2

◉ Body A exert force of Action
➾ Fᴀʙ on Body B.

◉ Body B exert force of Reaction
➾ Fʙᴀ on Body A.

▸ Fᴀʙ = Body A
➾ m1 (v1 - u1) / t

▸ Fʙᴀ = Body B
➾ m2 (v2 - u2) / t

▸ Since, Action = - reaction
▸ So, Fᴀʙ = - Fʙᴀ

m1 (v1 - u1) / t = - m2 (v2 - u2) / t

▸ m1v1 - m1u1 = m2v2 + m2u2
▸ m1v1 - m2v2 = m1u1 + m2u2...✔

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Thanks...✊