Derive an expression for the law of conservation of momentum
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Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies
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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...✊
✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✮✭
┗─━─━─━─━∞◆∞━─━─━─━─┛
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...✊
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