● Derive the mathematical formula of conservation of momentum. ▪�

1.	● Derive the mathematical formula of conservation of momentum. ▪︎Note:-Wrong answer will reported on spot..!!!Correct answer with quality explanation will appreciated and mark as brainliest...!!!
Answer» ★ Concept :- Here the concept of LAW of Conservation of MOVEMENTUM has been used. We see that here we have to derive the mathematical FORMULA of Law of Conservation of Momentum. So firstly we can assume two bodies whose CHANGE in moment we can find out. And then using the relationship of Force, we can find out the force exerted on the bodies by each other. And then we can apply the Newton's Third of Motion and then find the answer. Let's do it !! ______________________________________ ★ Formula Used :- $\;\boxed{\sf{\pink{Force\;=\;\bf{\dfrac{Change\;in\; Momentum}{Time}}}}}$ ______________________________________ ★ Solution :- Let the two bodies be 1 and 2 where the body 1 collides with 2 where 1 and 2 both are in motion. Initial velocity of 1 = u₁ Mass of 1 = m₁ Final velocity of 1 = v₁ Initial velocity of 2 = u₂ Mass of 2 = m₂ Final velocity of 2 = v₂ Force exerted by 1 on 2 = F₁₂ Force exerted by 2 on 1 = F₂₁ Time taken = t ______________________________________ ~ For Change in Momentum of both bodies :: → Change in Momentum = Final Momentum - Initial Momentum • For 1 :: ›»› Change in Momentum of 1 = m₁v₁ - m₁u₁ • For 2 :: ›»› Change in Momentum of 2 = m₂v₂ - m₂u₂ (since, momentum = mass × velocity) ______________________________________ ~ For Force exerted by bodies :: We know that, $\;\sf{\rightarrow\;\;Force\;=\;\bf{\dfrac{Change\;in\;Momentum}{Time}}}$ • For F₁₂ :- $\;\sf{\rightarrow\;\;\green{F_{12}\;=\;\bf{\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}}}}$ • For F₂₁ :- $\;\sf{\rightarrow\;\;\blue{F_{21}\;=\;\bf{\dfrac{m_{2}v_{2}\:-\:m_{2}u_{2}}{t}}}}$ ______________________________________ ~ For the Final Derivation :: By Newton's Third Law of Motion, every action has an equal and opposite reaction. This means after collision, the MAGNITUDE of F₁₂ will be equal to F₂₁ but different in magnitude. So, $\;\bf{\mapsto\;\;F_{12}\;=\;-\:F_{21}}$ By applying values, we get $\;\bf{\mapsto\;\;\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}\;=\;-\bigg(\dfrac{m_{2}v_{2}\:-\:m_{2}u_{2}}{t}\bigg)}$ $\;\bf{\mapsto\;\;\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}\;=\;\dfrac{-(m_{2}v_{2}\:-\:m_{2}u_{2})}{t}}$ $\;\bf{\mapsto\;\;\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}\;=\;\dfrac{m_{2}u_{2}\:-\:m_{2}v_{2}}{t}}$ Cancelling t from both sides, we get $\;\bf{\mapsto\;\;m_{1}v_{1}\:-\:m_{1}u_{1}\;=\;m_{2}u_{2}\:-\:m_{2}v_{2}}$ $\;\bf{\mapsto\;\;\red{m_{1}u_{1}\:+\:m_{2}u_{2}\;=\;m_{1}v_{1}\:+\:m_{2}v_{2}}}$ This is the final answer. $\;\bf{\mapsto\;\;\purple{m_{1}u_{1}\:+\:m_{2}u_{2}\;=\;m_{1}v_{1}\:+\:m_{2}v_{2}}}$ Hence, Derived. ________________________________

● Derive the mathematical formula of conservation of momentum. ▪︎Note:-Wrong answer will reported on spot..!!!Correct answer with quality explanation will appreciated and mark as brainliest...!!!

Answer»

★ Concept :-

Here the concept of LAW of Conservation of MOVEMENTUM has been used. We see that here we have to derive the mathematical FORMULA of Law of Conservation of Momentum. So firstly we can assume two bodies whose CHANGE in moment we can find out. And then using the relationship of Force, we can find out the force exerted on the bodies by each other. And then we can apply the Newton's Third of Motion and then find the answer.

Let's do it !!

______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{Force\;=\;\bf{\dfrac{Change\;in\; Momentum}{Time}}}}}$

______________________________________

★ Solution :-

Let the two bodies be 1 and 2 where the body 1 collides with 2 where 1 and 2 both are in motion.

Initial velocity of 1 = u₁

Mass of 1 = m₁

Final velocity of 1 = v₁

Initial velocity of 2 = u₂

Mass of 2 = m₂

Final velocity of 2 = v₂

Force exerted by 1 on 2 = F₁₂

Force exerted by 2 on 1 = F₂₁

Time taken = t

______________________________________

~ For Change in Momentum of both bodies ::

→ Change in Momentum = Final Momentum - Initial Momentum

• For 1 ::

›»› Change in Momentum of 1 = m₁v₁ - m₁u₁

• For 2 ::

›»› Change in Momentum of 2 = m₂v₂ - m₂u₂

(since, momentum = mass × velocity)

______________________________________

~ For Force exerted by bodies ::

We know that,

$\;\sf{\rightarrow\;\;Force\;=\;\bf{\dfrac{Change\;in\;Momentum}{Time}}}$

• For F₁₂ :-

$\;\sf{\rightarrow\;\;\green{F_{12}\;=\;\bf{\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}}}}$

• For F₂₁ :-

$\;\sf{\rightarrow\;\;\blue{F_{21}\;=\;\bf{\dfrac{m_{2}v_{2}\:-\:m_{2}u_{2}}{t}}}}$

______________________________________

~ For the Final Derivation ::

By Newton's Third Law of Motion, every action has an equal and opposite reaction.

This means after collision, the MAGNITUDE of F₁₂ will be equal to F₂₁ but different in magnitude. So,

$\;\bf{\mapsto\;\;F_{12}\;=\;-\:F_{21}}$

By applying values, we get

$\;\bf{\mapsto\;\;\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}\;=\;-\bigg(\dfrac{m_{2}v_{2}\:-\:m_{2}u_{2}}{t}\bigg)}$

$\;\bf{\mapsto\;\;\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}\;=\;\dfrac{-(m_{2}v_{2}\:-\:m_{2}u_{2})}{t}}$

$\;\bf{\mapsto\;\;\dfrac{m_{1}v_{1}\:-\:m_{1}u_{1}}{t}\;=\;\dfrac{m_{2}u_{2}\:-\:m_{2}v_{2}}{t}}$

Cancelling t from both sides, we get

$\;\bf{\mapsto\;\;m_{1}v_{1}\:-\:m_{1}u_{1}\;=\;m_{2}u_{2}\:-\:m_{2}v_{2}}$

$\;\bf{\mapsto\;\;\red{m_{1}u_{1}\:+\:m_{2}u_{2}\;=\;m_{1}v_{1}\:+\:m_{2}v_{2}}}$

This is the final answer.

$\;\bf{\mapsto\;\;\purple{m_{1}u_{1}\:+\:m_{2}u_{2}\;=\;m_{1}v_{1}\:+\:m_{2}v_{2}}}$