Question

${\LARGE{\underline{\underline{\bf{Question : }}}}}$

Why loss of kinetic energy in perfectly inelastic collision is maximum ?

Please answer with proper explanation.


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Answer 1

$\bf \star Question$

Why loss of kinetic energy in perfectly inelastic collision is maximum

$\bf \star Answer$

What is perfectly inelastic collision?

When two object are sticked together and move together then it is called perfectly inelastic collision.

Now,

Coming to Question

When the two object are sticked the kinetic energy loss. It is due to bonding of two objects. Also due to the action of internal friction the kinetic energy in perfectly inelastic collision is maximum. In collision of object which is macroscopic bodies, some kinetic energy is turned into vibrational energy.

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Kinetic Energy as well as the relation of Linear Momentum of the bodies has been used. We see that the moving body is with some initial velocity with which it approaches the target body and after that they travel with same velocity.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\;m\;\times\;(velocity)^{2}}}}$

$\\\;\boxed{\sf{\Delta\;K\;=\;\bf{K_{i}\;-\;K_{f}}}}$

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★ Solution :-

• Perfectly Inelastic Collision :: When the two colliding bodies stick together and move as a single body with a common velocity after the Collision, the collision is perfectly inelastic.

From figure we can see that, the body of mass m₁ is moving with velocity u₁ and collides head - on with another body of mass m₂ which is at rest. After this collision these both bodies move with common velocity v. Then,

» Initial velocity of body with mass m₁ = u₁

» Initial velocity of body with mass m₂ = 0

» Final velocity of body with mass m₁ = v

» Final velocity of body with mass m₂ = v

Now according to the Law of Conservation of Linear Momentum,

$\\\;\sf{\rightarrow\;\;m_{1}u_{1}\;+\;m_{2}\;\times\;0\;=\;\bf{m_{1}v\;+\;m_{2}v}}$

$\\\;\sf{\rightarrow\;\;m_{1}u_{1}\;=\;\bf{(m_{1}v\;+\;m_{2})v}}$

$\\\;\sf{\rightarrow\;\;v\;=\;\bf{\dfrac{m_{1}u_{1}}{(m_{1}v\;+\;m_{2})}}}$

Let this be equation i.)

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~ For the Loss of Kinetic Energy of bodies after Collision ::

We know that,

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{K_{i}\;-\;K_{f}}}$

And by applying the formula of Kinetic Energy,

$\\\;\sf{:\Rightarrow\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\;m\;\times\;(velocity)^{2}}}$

Now combining these, we get,

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\bigg(\dfrac{1}{2}\:m_{1}u_{1} ^{2}\:+\dfrac{1}{2}\:m_{2}u_{2} ^{2}\bigg)\;-\;\bigg(\dfrac{1}{2}\:m_{1}v^{2}\:+\dfrac{1}{2}\:m_{2}v^{2}\bigg)}}$

Since, m₂u₂ = 0

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\bigg(\dfrac{1}{2}\:m_{1}u_{1} ^{2}\bigg)\;-\;\bigg(\dfrac{1}{2}\:(m_{1}\;+\;m_{2})v^{2}\bigg)}}$

Now using equation i.) , we get,

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\dfrac{1}{2}\:m_{1}u_{1} ^{2}\;-\;\dfrac{1}{2}\:\bigg(m_{1}\;+\;m_{2}\bigg)\bigg[\dfrac{m_{1}}{m_{1}\;+\;m_{2}}\:u_{1}\bigg]^{2}}}$

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\dfrac{1}{2}\:m_{1}u_{1} ^{2}\;-\;\dfrac{1}{2}\:\bigg[\dfrac{m_{1} ^{2}}{m_{1}\;+\;m_{2}}\bigg]u_{1} ^{2}}}$

Since, by squaring (m₁ + m₂) will cancel off from numerator and denominator.

Also, m₁m₂u₂ = 0

Now taking the terms common from this, we get,

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\dfrac{1}{2}\:m_{1}u_{1} ^{2}\:\bigg[1\;-\;\dfrac{m_{1}}{m_{1}\;+\;m_{2}}\bigg]}}$

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\dfrac{1}{2}\:m_{1}u_{1} ^{2}\:\bigg[\dfrac{m_{1}\;+\;m_{2}\;-\;m_{1}}{m_{1}\;+\;m_{2}}\bigg]}}$

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\dfrac{1}{2}\:m_{1}u_{1} ^{2}\:\bigg[\dfrac{m_{2}}{m_{1}\;+\;m_{2}}\bigg]}}$

$\\\;\sf{:\Rightarrow\;\;\Delta\;K\;=\;\bf{\dfrac{1}{2}\:\times\:\dfrac{m_{1}m_{2}}{m_{1}\;+\;m_{2}}\:u_{1} ^{2}}}$

Clearly, here we will get a positive answer. This, means,

$\\\;\qquad\qquad\sf{\pink{\rightarrow\;\;K_{f}\;<\;K_{i}}}$

This means that here is Loss of Kinetic Energy in the form of Heat and Sound.

Then,

$\\\;\sf{:\Longrightarrow\;\;\dfrac{K_{f}}{K_{i}}\;=\;\bf{\dfrac{\dfrac{1}{2}\:(m_{1}m_{2})v^{2}}{\dfrac{1}{2}\:m_{1}u_{1} ^{2}}}}$

Cancelling ½, we get,

$\\\;\sf{:\Longrightarrow\;\;\dfrac{K_{f}}{K_{i}}\;=\;\bf{\dfrac{(m_{1}m_{2})}{m_{1}}\;\times\;\dfrac{v^{2}}{u_{1} ^{2}}}}$

Now using equation i.) again, we get,

$\\\;\sf{:\Longrightarrow\;\;\dfrac{K_{f}}{K_{i}}\;=\;\bf{\dfrac{(m_{1}+m_{2})}{m_{1}}\;\times\;\dfrac{\bigg(\dfrac{m_{1}}{m_{1}\;+\;m_{2}}\:u_{1}\bigg)^{2}}{u_{1} ^{2}}}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{K_{f}}{K_{i}}\;=\;\bf{\dfrac{(m_{1}+m_{2})}{m_{1}}\;\times\;\bigg(\dfrac{m_{1}}{m_{1}\;+\;m_{2}}\bigg)^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{K_{f}}{K_{i}}\;=\;\bf{\dfrac{(m_{1}+m_{2})}{m_{1}}\;\times\;\bigg(\dfrac{m_{1}}{m_{1}\;+\;m_{2}}\bigg)\;\times\;\bigg(\dfrac{m_{1}}{m_{1}\;+\;m_{2}}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{K_{f}}{K_{i}}\;=\;\bf{\dfrac{m_{1}}{m_{1}\;+\;m_{2}}}}$

Now we see, that this value is less than one that is < 1. This means that kinetic energy after collision is less that the kinetic energy before the collision.

Now if the target body is very massive that is m₂ >> m₁, then

$\\\;\sf{\green{:\Longrightarrow\;\;\dfrac{K_{f}}{K_{i}}\;\approx\;\bf{0}}}$

This means,

$\\\;\bf{\blue{\leadsto\;\;K_{f}\;<\:<\;K_{i}}}$

Hence the Loss of Kinetic Energy is maximum here because Final Kinetic Energy is much lower than Initial Kinetic Energy. This means nearly all of the Kinetic Energy is lost .