How is energy conserved in an elastic collision between two unequal masses? 
In an elastic collision between two masses, if one mass is much heavier than the other, then the heavier mass will continue to move with same velocity while the lighter mass doubles its velocity. How is the Law of Conservation of Energy conserved in this?

I concluded that the assumption of neglecting the change in velocity of the heavier mass is responsible for an increase in kinetic energy of the whole system, thus failing to satisfy the Law of Conservation of Energy. Is this a correct explanation?
 A: You should recognize this 

"the heavier mass will continue to move with same velocity" 

as an approximation.
The actual velocity will be reduced, and you can figure out how much from the conservation of momentum. 
$$ \Delta V \approx - 2 V \frac{m}{M}  \tag{for $M \gg m$}\;.$$
(Strictly speaking this is another approximation., but now we're talking about a small correction to a small correction and I'm going to ignore it.)
Now that is a small change in speed but by hypothesis it is connected with a large mass so it results in a non-trivial change of kinetic energy
\begin{align}
\Delta K_M 
&= \frac{1}{2} M \left[ (V + \Delta V)^2 - V^2\right] \\
&= \frac{1}{2} M \left[ (2 V \Delta V + (\Delta V)^2\right] \;.
\end{align}
Now, we factor one power of $\Delta V$ out and notice that it cancels the $M$ to give
\begin{align}
\Delta K_M
&= - m V \left[ 2V + \Delta V \right] \\
&\approx - m V \left[ 2V \right] \\
&= - \frac{1}{2} m (2V)^2 
\end{align}
The large mass losses approximately the same energy as the small one gains.
Keeping all the terms is more algebraically difficult, but follows exactly the same pattern.
