Why can't I use the law of conservation of energy here? As I'm preparing for my Physics exam next week, I came across 2 questions which look to me like they both require the law of conservation of energy to be worked out, while they actually don't. Here are pictures of both exercises and answers, for clarification. My question continues below.
Puck Collision:

Car Collision:

Now, my question is, I understand why they use the law of conservation of energy to calculate the speed of Puck B (Vb2), but when I try to use to law of conservation of energy to calculate the velocity of the cars after the collision, I get 22 m/s instead of the required 15.6 m/s. Why is the use of the conservation of energy wrong here?
To me it seems like it should work.
My calculation of the velocity after the crash of the Car Exercise:
$$
\begin{aligned}
\text{kinetic-energy(car1) + kinetic-energy(car2)} &= \text{kinetic-energy(cars)}\\
\frac12mv^2 + \frac12mv^2 &= \frac12mv^2\\
(0.5*1500*25^2) + (0.5*2500*20^2) &= 0.5*(1500+2500)*v^2\\
 v &= 22\ \mathrm{m/s}
\end{aligned}
 $$
 A: You are making an implicit assumption that kinetic energy is conserved in all collisions.  In fact, kinetic energy is conserved only in elastic collisions.  If sound is generated, the shape of either object is changed, either object is broken in some way, etc., the collision is not elastic, and the "lost" kinetic energy will be converted into other forms of energy (e.g., heat, crumpled metal, etc.).
The thing that is conserved in all collisions, regardless of whether or not they are elastic, is momentum.  Thus, conservation of momentum is the method of choice for collision problems, as shown in the text sources that you posted.
A: You can use conservation of energy but, in the car problem, you would have to account for the energy lost "by other means" such as sound, deforming the car, heat etc.  Hence, in an inelastic collision,  not all the energy "remains in the 2-car system" during the collision, and precisely how much energy is lost "by other means" is usually unknown (of course if the cars eventually stop, the all energy is eventually lost to the environment).  Hence, energy is conserved in the "2-cars plus environment" system, but that's not enough to help you with the "2-cars"-only part of the problem.
This is to be contrasted with the elastic collision of pucks, where energy is (very nearly) conserved (there is a very small loss to acoustic energy when the sound of the collision is made, but this is very small).  In this case, there is practically no loss to the environment and you can consider the "2-puck" system as closed.
A: It is easiest to analyze collisions by boosting to the center of momentum frame which is frame moving such that the total momentum of the 2 objects is zero. Here we see that in this frame momentum is still zero after the collision as long as the two objects have equal and opposite momenta in magnitude and direction after the collision. The actual value of the 2 momenta depends on whether energy is conserved. If the value is zero they lose all energy in CM frame, and if it is the same momentum magnitude as before then energy is completely conserved. In this case the hockey pucks have the same momentum as before collision and the cars have zero in the CM frame. ie the fact that they are stuck together after the collision means that the energy in the CM frame is completely lost. 
