Let us say I have a clay ball of mass $1 \ \text{kg}$. I throw it at a door of mass $10 \ \text{kg}$ with a speed of $10 \ \text{m/s}$. Let us say the ball sticks to the door on contact. I am trying to find the final velocity $v$ of the door and clay.

Using conservation of momentum:

$$1 \cdot 10+10 \cdot 0=11 \cdot v$$

$$v= \frac{10}{11}\approx 0.9$$

Using conservation of energy:

$$0.5 \cdot 1 \cdot 100+0.5 \cdot 10 \cdot 0=0.5 \cdot11 \cdot v \cdot v$$

$$v=\sqrt{\frac{100}{11}} \approx \ 3$$

So, as per google, inelastic collisions do not conserve energy, only momentum is conserved. Therefore, $v=10/11$ is the correct answer to choose.

Now, calculating energy lost for other stuff not involved in moving the door:

$$ \text{Lost energy} =0.5 \cdot 1 \cdot 10 \cdot 10-0.5 \cdot 11 \cdot \left(\frac{10}{11}\right)^2 \approx 50-4.5=45.5$$

So, looks like $90 \%$ of initial energy is lost in the process of clay ball sticking to the door! Is my calculation correct? What is the intuitive explanation for this loss of energy?