Kinetic energy and momentum are related to each other by the following equation: $$K.E.=\frac{1}{2}\frac{\textbf{P}^2}{m} $$ In inelastic collisions the momentum is conserved but the energy isn't. How can this be correct in the view of previous equation?
Moreover, if I want to rewrite the previous equation in term of change of momentum and change of kinetic energy, is the following true or not?
$$\Delta K.E.=\frac{1}{2}\frac{(\Delta\textbf{P})^2}{m} $$
If that is wrong, what is the true form?
For simplicity's sake, let's restrict ourselves to collisions in 1 dimension, where object 1 collides with object 2. They have momenta $p_1$ and $p_2$ before collision and $p_1'$ and $p_2'$ after collision. We then have $$\begin{align} p_1' + p_2' &= p_1 + p_2,\tag{1}\\ K' = \frac{p_1'^2}{2m_1} + \frac{p_2'^2}{2m_2} &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \Delta K = K + \Delta K,\tag{2} \end{align} $$ where $\Delta K = K' -K$ is the change in kinetic energy. Note that Eq. (1) by itself does not have a unique solution: only the total momentum is conserved, not the individual momenta. Each solution $(p_1',p_2')$ will correspond with a different value $\Delta K$, and there will be only 1 specific solution for which the collision is elastic, i.e. $\Delta K=0$.
It is instructive to write the solutions in terms of the so-called coefficient of restitution $$ C_R = \frac{v_2' - v_1'}{v_1 - v_2} = \frac{m_1p_2' - m_2p_1'}{m_2p_1 - m_1p_2}.\tag{3} $$ Evidently, $v_1 > v_2$, otherwise there would be no collision. Also, $v_2' > v_1'$, because object 1 cannot get passed object 2. So $C_R\geqslant 0$. The value $C_R = 0$ corresponds with a perfectly inelastic collision, where both objects stick together after they collide.
Using Eq. (1) we find $$\begin{align} 1 + C_R &= \frac{m_1(p_2'-p_2) + m_2(p_1-p_1')}{m_2p_1 - m_1p_2}\\ &=\frac{(m_1+m_2)(p_1-p_1')}{m_2p_1 - m_1p_2}\\ &=\frac{(m_1+m_2)(p_2'-p_2)}{m_2p_1 - m_1p_2}, \end{align} $$ so that the possible solutions are of the form $$\begin{align} p_1' &= p_1 - (1 + C_R)\frac{m_2p_1 - m_1p_2}{m_1+m_2},\\ p_2' &= p_2 + (1 + C_R)\frac{m_2p_1 - m_1p_2}{m_1+m_2}. \end{align} $$ Next we derive the relation between $C_R$ and $\Delta K$. First, note that $$\begin{align} 2m_1m_2(m_1 + m_2)K &= (m_1+m_2)(m_2p_1^2 + m_1p_2^2)\\ &= m_1m_2(p_1 + p_2)^2 + (m_2p_1 - m_1p_2)^2, \end{align} $$ and $$\begin{align} 2m_1m_2(m_1 + m_2)K' &= m_1m_2(p_1' + p_2')^2 + (m_1p_2' - m_2p_1')^2\\ &=m_1m_2(p_1 + p_2)^2 + (m_1p_2' - m_2p_1')^2, \end{align} $$ so that $$ 2m_1m_2(m_1 + m_2)\Delta K = (m_1p_2' - m_2p_1')^2 - (m_2p_1 - m_1p_2)^2. $$ We plug this into Eq. (3), and obtain $$\begin{align} C_R &=\sqrt{1+\frac{2m_1m_2(m_1 + m_2)}{(m_2p_1 - m_1p_2)^2}\Delta K}, \end{align} $$ or alternatively, $$ \Delta K = (C_R^2 - 1)\frac{(m_2p_1 - m_1p_2)^2}{2m_1m_2(m_1 + m_2)}. $$ Since $\Delta K\leqslant 0$, we get $0\leqslant C_R\leqslant 1$. For elastic collisions, $\Delta K = 0$ and $C_R= 1$.
To summarize, for each value of $C_R$ between $0$ and $1$ we get a possible solution, each corresponding with a different value of $\Delta K$.
The momentum of one system is conserved if no force acts on it.
Your problem is that you look at one system alone, and when force acts on it, you think that momentum isn't conserved.
If for example a clay ball A of mass $m$ and velocity $v$ collides with an identical ball B moving at opposite direction. They both stop by symmetry.
If you look at system A alone, it's KE went from $\frac{mv^2}{2}$ to $0$, and so did its momentum. But that's to be expected since an outside force acts on it.
Look at systems A and B together. Momentum is vector, so the total momentum of the systems A, B together is 0. since KE is scalar, their total $KE = mv^2$. When the 2 balls collide, the momentum is unchanged and the KE decreases to 0.
