Question

Derive conservation of momentum

Answer 1

suppose a body of mass m1 moves with another body of mass m2 with velocity u1 and u2, after time 't' they collide and then again begin to move with velocity v1 and v2 respectively.

newton's third law says each and every action has equal and opposite reaction.
so,

let force applied be body of mass m1 = F1

and the force applied be body of mass m2 be
F2 .

now,
According to newton's third law of motion
F1 = - F2 ( it's in opposite direction)

[as force = mass × acceleration]

m1 × {(v1- u1)/t} = -[m2 ×{(v2 -u2)/t}]

m1 (v1 - u1) = -{m2 (v2 -u2)}
(as 't' cancels both sides)

m1v1 - m1u1 = -m2 (v2 -u2)
m1v1 -m1u1 = -m2v2 + m2u2
m1v1 + m2v2 = m2u2 + m1u1
m1v1 + m2v2 = m1u1 + m2u2.




This means
the total momentum after collision = total momentum bedorr collision

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...✔

_________
Thanks...✊