Question

Derive third law of motion.​

Answer 1

We know,

Distance Travelled = (Average velocity)(Time)

Mathematically,

$s = ( \frac{v - u}{2} )t$

Also, from the eqⁿ : $v = u + at$,

$t = \frac{v + u}{t}$

Substituting 't':-

$s = ( \frac{v - u}{2} )t$

$\implies s = ( \frac{v - u}{2} )(\frac{v + u}{a})$

$\implies s = \frac{v^2 - u^2}{2a}$

$\implies 2as = v^2 - u^2$

Hence, derived.

Answer 2

Answer :-

$\:\\\large{\underbrace{\underline{\sf{Let's\;understand\;the\;Question\;:-}}}}$

Here the concept of Second Law of Motion has been used. We see the Third Equation of Motion has been given already. But we can derive it only using the relation given by Second Equation of Motion.

Third equation of motion states that, 'To every action, there is always an equal and opposite reaction'. This means when a force is applied, the body will face a force back with equal magnitude but in opposite direction. Using this, let's do it !

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★ Formula Used :-

$\;\\\large{\boxed{\sf{F\;\;=\;\;\bf{m\;\times\;a}}}}$

$\;\\\large{\boxed{\sf{Rate\;of\;change\;of\;Momentum\;=\;\bf{\dfrac{\Delta\:p}{\Delta\:t}\;\:=\;\;\dfrac{dp}{dt}}}}}$

$\;\\\large{\boxed{\sf{Rate\;of\;change\;of\;momentum\;\;\propto\;\;\bf{Force\;Applied}}}}$

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★ Solution :-

Let us consider two bodies 1 and 2 on which there is no external force applied.

That means, External Force = 0

(Note : the first attachment shows this when bodies are not collided)

• Now when the bodies collide with each other then,

$\:\\\sf{\mapsto\;\;\:Force\;acting\;on\;body\;2\;by\;body\;1\;=\;\bf{F_{12}}}$

$\:\\\sf{\mapsto\;\;\:Force\;acting\;on\;body\;1\;by\;body\;2\;=\;\bf{F_{21}}}$

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~ For rate of change of momentum of both bodies in differentiation form :-

Rate of change of momentum is the change in magnitude of momentum by difference in time taken.

$\;\\\sf{:\rightarrow\;\;\: Rate\;of\;change\;of\;Momentum\;=\;\bf{\dfrac{\Delta\:p}{\Delta\:t}\;\:=\;\;\dfrac{dp}{dt}}}$

Since, change in momentum is equal to product of mass and change in velocity by time (i.e., acceleration).

» Rate of change of momentum ∝ (m × a)

$\:\\\large{\sf{:\rightarrow\;\;\:Rate\;of\;change\;of\;momentum\;of\;first\;body,\;p_{1}\;=\;\bf{m\;\times\;a\;\;=\;\; \dfrac{dp_{1}}{dt}}}}$

$\:\\\large{\sf{:\rightarrow\;\;\:Rate\;of\;change\;of\;momentum\;of\;second\;body,\;p_{2}\;=\;\bf{m\;\times\;a\;\;=\;\; \dfrac{dp_{2}}{dt}}}}$

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~ For Force exerted by bodies on each other.

Now since there is rate of change of momentum in bodies, that means they will collide with each other. And after colliding they will exert force also on each other.

From Newton's Second Law of motion, we get

$\:\\\sf{:\rightarrow\;\;\: Rate\;of\;change\;of\;momentum\;\;\propto\;\;\bf{Force\;Applied}}$

$\:\\\large{\sf{:\rightarrow\;\;\: F_{12}\;=\;\bf{\dfrac{dp_{2}}{dt}}}}$

This is force exerted on body 2 by body 1.

$\:\\\large{\sf{:\rightarrow\;\;\: F_{21}\;=\;\bf{\dfrac{dp_{1}}{dt}}}}$

This is force exerted on body 1 by body 2.

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~ For deriving the Third Law of Motion :-

Now since we got the values of both the forces. We can add them.

$\:\\\large{\sf{:\Longrightarrow\;\;\:F_{12}\;+\;F_{21}\;=\;\bf{\dfrac{dp_{2}}{dt}\;+\;\dfrac{dp_{1}}{dt}}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\:F_{12}\;+\;F_{21}\;=\;\bf{\dfrac{(dp_{2}\;+\;dp_{1})}{dt}}}}$

Since, already we assumed that there is no external force applied on the bodies.

So, the rate of change of momentum will also be 0.

(from the above formula)

Then,

$\:\\\large{\sf{:\longrightarrow\;\;\:\dfrac{(dp_{2}\;+\;dp_{1})}{dt}\;=\;\bf{0}}}$

Now applying this value into the above equation, we get,

$\:\\\large{\sf{:\Longrightarrow\;\;\:F_{12}\;+\;F_{21}\;=\;\bf{0}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\:F_{12}\;=\;\bf{-\;F_{21}}}}$

Clearly, we that forces exerted by both bodies on each other are equal in magnitude but since there is negative sign in one so they both are in opposite direction to each other.

$\:\\\qquad\large{\underline{\boxed{\bf{F_{12}\;=\;\bf{-\;F_{21}}}}}}$

Now if we assume one force as F. Then,

$\:\\\qquad\large{\underline{\boxed{\bf{F\;=\;\bf{-\;F}}}}}$

This means that,

Force exerted by body 1 on body 2 = - (Force exerted by body 2 on body 1)

Thus, negative sign shows, that force is in opposite direction.

This proves our Third Law of Motion and hence shows its derivation.

$\;\\\qquad\;\large{\boxed{\boxed{\rm{Thus,\;Derived}}}}$

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★ More formulas to know :-

$\;\\\tt{\leadsto\;\;\; \vec{p}\;=\;m \vec{v}}$

$\;\\\tt{\leadsto\;\;\;\vec{W}\;=\; \vec{F}\:\times\:\vec{s}}$

$\;\\\tt{\leadsto\;\;\; T \; =\; mg\; \cos\:\theta\;+\;\dfrac{mv^{2}}{r}}$

$\;\\\tt{\leadsto\;\;\; v\;+\;u \;=\;at}$

$\;\\\tt{\leadsto\;\;\; v^{2}\;-\;u^{2} \;=\;2as}$

$\;\\\tt{\leadsto\;\;\; s \;=\;ut\;+\;\dfrac{1}{2}\:at^{2}}$

$\;\\\tt{\leadsto\;\;\; s_{n_{(th)}}\;=\;u\;+\;\dfrac{a}{2}\:(2n\;-\;1)}$

$\;\\\tt{\leadsto\;\;\; s\;=\;\dfrac{u\;+\;v}{2}\;\times\;t}$