**The relation between momentum and kinetic energy:**

Momentum is defined as:

$$\vec{p} = m \vec{v}$$

So, we can write velocity as:

$$ \frac{\vec{p}}{m} = \vec{v}$$

Kinetic energy is defined as:

$$ K = \frac{1}{2} mv^2$$

Using the previous equation,

$$ K = \frac{p^2}{2m}$$

So, it's very easy to see that **it can be said that kinetic energy is a function of momentum and mass.**


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In an **inelastic collision,** the **momentum is conserved** but **the kinetic energy is not.** You may think otherwise from the previous equation I wrote because it says that kinetic energy is a direct function of momentum. 

When we have a collision what happens is the total momentum is split up between the different bodies 
such that if we summed momentum of every single body then the total momentum be the same.

The idea is that **in an inelastic collision**, the total momentum redistributes in a way that when you do kinetic energy calculation on it, the difference is negative. 

In an inelastic collision, the energy loss is due to deformations that occur. When the two bodies get stuck to each other, there is an energy called 'elastic potential energy' which changes which increases. As an analogy, you can think of the energy increase of spring when it's compressed.


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**Example:**

Let two bodies $A$ and $B$ with mass $m_a$ and $m_b$ respectively. Let their initial momentums be $\vec{p_a} $ and $ \vec{p_b}$ and after collision in which they stick let their momentum be $\vec{p_{ab}}$ then their loss in kinetic energy.

$$ K_{loss} = \frac{m_a m_b}{2(m_a +m_b)} (\frac{p_a}{m_a} - \frac{p_b}{m_b})^2$$


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**Derivation:**

Considering the same conditions above, the initial kinetic energy is given as:

$$K_{i} = \frac{ (\vec{p_a})^2}{2m_a} + \frac{ (\vec{p_b})^2}{2m_b}$$

The final kinetic energy is given as:

$$ K_{f} = \frac{ (\vec{p_a} +\vec{p_b})^2}{2(m_b +m_a)}$$

Note the momentum in the final must be equal to the initial momentum of $p_a +p_b$ due to the momentum conservation.

Hence,

$$ K_f = \frac{ (\vec{p_a})^2 + ( \vec{p_b})^2 + 2 \vec{p_a} \cdot \vec{p_b}}{2(m_b +m_a)}$$

Now consider **the difference of kinetic energy** between final and initial states:

$$ K_f - K_i = \frac{ \vec{p_a} \cdot \vec{p_b} }{m_a + m_b} - [ \frac{(m_a \vec{p_b})^2 + (m_b \vec{p_a})^2}{2(m_a + m_b)(m_a m_b)}]$$

$$ K_f - K_i =-\bigg[ \frac{(m_a \vec{p_b})^2 + (m_b \vec{p_a})^2 - 2m_a m_b \vec{p_a} \cdot
 \vec{p_b}}{2(m_a + m_b)(m_a m_b)} \bigg]$$

Or,

$$ K_f -K_i = - \bigg[ \frac{|m_a \vec{p_b} - m_b \vec{p_a}|^2}{2(m_a + m_b)(m_a m_b)} \bigg] $$

Since $m_a m_b$ is a **strictly positive quantity**, we can move it into the modulus :

$$ K_f - K_i =\frac{m_a m_b}{2(m_a +m_b)} (|\frac{p_b}{m_b} - \frac{p_a}{m_a}|)^2$$

Since we can switch terms in the square modulus,

$$ K_f -K_i = \frac{m_a m_b}{2(m_a +m_b)} (|\frac{p_a}{m_a} - \frac{p_b}{m_b}|)^2$$

Now, we can say that kinetic energy change has been reduced in the final state. This absolute value of the amount which is lost is taken as $K_{loss}$ and is given as:

$$ K_{loss} =  \frac{m_a m_b}{2(m_a +m_b)} (|\frac{p_a}{m_a} - \frac{p_b}{m_b}|)^2$$

q.e.d


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**Further reading:**

HC-Verma: Concept's of Physics

 [Feynman lectures][1] ( read under section of Momentum and Energy)


  [1]: https://www.feynmanlectures.caltech.edu/I_10.html