Question

$\huge \orange \sf Question$

● 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 1

★ 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 !!

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★ Formula Used :-

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

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★ 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

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~ 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)

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~ 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 :-

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

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~ 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.

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Answer 2

Explanation:

$\Large\underline{\underline{\sf{\purple{Required\ Answer:-}}}}$

★ To explain conservation of momentum, let us take the following example. Consider two balls A and B having masses $\sf {m_1}$ and $\sf {m_2}$ respectively. Let the initial velocity of ball A be $\sf {u_1}$ and that of ball B be $\sf {u_{2}\ (u_1 > u_2).}$ Their collision takes place for a very short interval of time $\sf {t}$ and after that A and B start moving with velocities $\sf {v_1}$ and $\sf {v_{2}\ (now\ v_1 < v_2)}$ respectively as shown in the above figure.

The momentum of ball A before and after the collision is $\sf {m_1\ u_1}$ and $\sf {m_1\ v_1}$ respectively. If there are no external forces acting on the body, then the rate of change of momentum of ball A, during the collision will be,

$\sf \: \: \: \: \: \: \: \: \: {\orange{=\ \dfrac{m_1\ (v_1\ -\ u_1)}{t}}}$

and, similarly the rate of change in momentum of ball B

$\sf \: \: \: \: \: \: \: \: \: {\orange{=\ \dfrac{m_2\ (v_2\ -\ u_2)}{t}}}$

Let $\sf {F_{12}}$ be the force exerted by ball A on B and $\sf {F_{21}}$ be the force exerted by ball B on A. Then, according to Newton's second law of motion,

$\sf \: \: \: \: \: \: \: \: \: {\orange{{F_{12}\ =\ \dfrac{m_1\ (v_1\ -\ u_1)}{t}}\ \: \: \: \: \: {and}\ \: \: \: \: \: {F_{21}\ =\ \dfrac{m_2\ (v_2\ -\ u_2)}{t}}}}$

According to Newton's third law of motion, we have

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\orange{F_{12}\ =\ -F_{21}}}$

$\sf {\orange{{or,}\ \: \: \: \: \: \: \: {\dfrac{m_1\ (v_1\ -\ u_1)}{t}\ =\ -\dfrac {m_2\ (v_2\ -\ u_2)}{t}}}}$

$\bf {\orange{{or,}\ \: \: \: \: \: \: \: \: {m_1\ v_1\ -\ m_1\ u_1\ =\ -m_2\ v_2\ +\ m_2\ u_2\: \: \: \: or\: \: \: \: m_1\ u_1\ +\ m_2\ u_2\ =\ m_1\ v_1\ +\ m_2\ v_2}}}$

$\sf {i.e.,}$ Total momentum before collision = Total momentum after collision

Thus, we find that in a collision between the two balls the total momentum before and after the collision remains unchanged or conserved provided no net force acts on the system. This result is law of conservation of momentum.