Couldn't you use a spring to increase energy? When I was a kid I had a BB gun and when it was cocked it would compress a spring, then when you pulled the trigger, the spring would be released and fire a BB out at high velocity.
I was thinking, why can't you put a spring with a cocking mechanism against the wall, roll a bowling ball into it, move the bowling ball out of the way, and use the compressed spring to fire a baseball at high speed. I figure the momentum would be the same, but the baseball would have much higher energy, so you've created energy using a spring and a wall. Then, just to take it a step further, why not do the same thing with a lever that has a cocking spring around the hub (fulcrum). You could throw a baseball at the end of the lever, to cock the spring, then put four baseballs in the middle of the lever (the halfway point) and propel them to half the speed of the original baseball.
This is because the moment of inertia is the square of the radius of a lever times mass. How about instead of four, you propel two baseballs from the middle of the lever, at the same speed as the original baseball and double your momentum. Or even one baseball, from the middle of the lever, at twice velocity as the original baseball. If you could do that, you could not only double momentum, you could also double the energy.
I'm wondering why couldn't you do all this? I would like simple answers and relationships, not an equation festival.
 A: The way I understand your question, the final kinetic energy of the baseball (mass $m$) would be the same as the initial kinetic energy of the bowling ball (mass $M$), but the final momentum of the baseball will be less than the initial momentum of the bowling ball by a factor of $\sqrt{ \left(m/M\right)}$. As James points out, the Earth itself absorbs the missing momentum.
I think the confusion here (unless it's me) is that you must consider both the compression ($\Delta x$) of the spring and the time ($\Delta t$) needed for the spring to compress (and relax). The kinetic energy is associated with $\Delta x$, and the change in momentum with $\Delta t$. $\Delta x$ will be the same for the bowling ball and the baseball, but $\Delta t$ will not be the same.
Let's ignore rolling vs. translational motion, and let everything be translational. Of course, let's also ignore friction and air resistance.
A bowling ball of mass $M$ moves toward a spring with spring constant $k$ at speed $V$. The kinetic energy is $\frac{1}{2}MV^2.$ The ball compresses the spring by an amount $\Delta x$, coming to rest for an instant. At that instant, some kind of locking mechanism is activated and locks the spring in place. The bowling is removed and replaced with a baseball of mass $m$. The lock is released, and the baseball flies away at speed $v.$
The stored energy in the compressed spring is given by $\frac{1}{2}k\Delta x^2,$ and is equal to the bowling ball's kinetic energy immediately before it makes contact with the spring and begins to compress it. When the lock is released, all of this stored energy is imparted to the baseball as kinetic energy, in the amount $\frac{1}{2}mv^2.$ Because mechanical energy is conserved, we can write
$$\frac{1}{2}MV^2 = \frac{1}{2}mv^2.$$
Solving for $v$, we get
$$v = V\sqrt{\frac{M}{m}}.$$
Since $m < M$, $v > V$.
So what is happening to the momentum? The bowling ball's initial momentum is $MV$, and baseball's final momentum is $-mv$. Since the bowling ball stops in a resting state and the baseball starts in a resting state, we can write for the changes in momentum:
$$\Delta p_{\textrm{baseball}} = -mv - 0 = -mv,$$
$$\Delta p_{\textrm{bowling ball}} = 0 - MV = -MV.$$
The total $\Delta p$, ignoring the Earth, is
$$\Delta p_{\textrm{total}} = -mv - \left(-MV\right) = -mv + MV.$$
Substituting for $v$, we can write
$$\Delta p_{\textrm{total}} = -mV\sqrt{\frac{M}{m}} - \left(-MV\right) = MV - \sqrt{mM}V = V\left(M - \sqrt{mM}\right) = V\left(1 - \sqrt{\frac{m}{M}}\right).$$
$\Delta p_{\textrm{total}}$ will be zero only if $m = M$. So the issue is the $\sqrt{\frac{m}{M}}$ term.
Let's see if we can get to the same result in a different way. When either ball is in contact with the spring, it is undergoing simple harmonic motion, with the position given by
$$x\left(t\right) = At\sin{\sqrt{\frac{k}{m}}},$$
where $A$ is the maximum compression of the spring. In general, the period of the motion will depend on the spring constant $k$ and the mass $m$ of the oscillating object:
$$T = 2\pi\sqrt{\frac{m}{k}}.$$
Each ball moves through $\frac{1}{4}$ of a period as the spring compresses (for the bowling ball) and relaxes (for the baseball). But since $T$ depends on the mass, the two balls will not compress the spring in the same amount of time. Calling $\Delta t$ the compression time, we can write
$$\frac{\Delta t_{\textrm{baseball}}}{\Delta t_{\textrm{bowling ball}}} = \frac{\frac{\pi}{2}\sqrt{\frac{m}{k}}}{{\frac{\pi}{2}\sqrt{\frac{M}{k}}}} = \sqrt{\frac{m}{M}}.$$
This is the same ratio calculated above. For each ball, $\Delta p$ is related to $\Delta t$ through Newton's second law: $dp = dF/dt.$ Since $F$ depends on $x$ in exactly the same way for each ball, the ratio of the $\Delta p$ values will depend only on the ratio of the $\Delta t$ values. In fact, it will be the same ratio, $\sqrt{\frac{m}{M}}.$
A: When the bowling ball compresses the spring, it comes to a standstill. A lot of energy is now stored in the spring. But where is the momentum stored? How can the bowling ball's momentum just disappear? 
It can't, and in fact it is absorbed by the earth (or whichever else is holding the spring). 

I figure the momentum would be the same, but the baseball would have much higher energy,

No, this doesn't really make realistic sense. The only thing supplying energy to the baseball is the spring, and the spring cannot supply more than it has stored. Surely, the baseball cannot receive more energy than the bowling ball delivered to the spring. It can't possibly reach higher energy. 
When the baseball is put there instead and the spring is released, momentum is supplied to the baseball by shooting the earth further backwards (so the total momentum remains unchanged at zero). At the same time, the baseball absorbs the stored spring energy (minus what the earth absorbs) and now carries it as kinetic energy.
There is no issue in the conservation laws here, and you will not reach higher kinetic energy than you started out with from this method. The piece that was missing in your analysis seems to be the presence of the Earth. 
A: In your example the wall, in order to stay fixed, must be attached to a massive body and serve as a “momentum sink”.  Thus momentum will not be conserved.  If you make the wall light and movable then you will find that the wall recoils when the other end of the spring puts force on it.
A: It does cost energy to put the spring against the wall and it will cost more energy if there is something between. You where talking about a bowling ball. How much energy does it cost to put the bowling ball omn the spring?
A: Using ideas from the other answers, try looking at it like this (with no equations - which you don't like):


*

*Bowling ball hits spring (attached to wall). Spring compresses. Ball stops. Spring latches.

*All kinetic energy now in spring. Ball momentum transferred to wall (hence Planet Earth) and essentially lost (cannot be recovered).

*Replace bowling ball with baseball. Release spring.

*Energy now in baseball which shoots off much faster than bowling ball came in.

*Kinetic energy of baseball now equal to kinetic energy of bowling ball.

*Momentum of baseball much less than original momentum of bowling ball.

*The ratio by which the baseball is faster is the same as the ratio by which the baseball's momentum is lower (see how hard this becomes to explain when you don't like equations...)


So there is no energy amplification.
Oh, shoot - I have to do some equations:
$$E_{1} = {1\over{2}}m_{1}v_{1}^{2},\;\;\;  p_{1} = m_{1}v_{1}$$
$$E_{2} = {1\over{2}}m_{2}v_{2}^{2},\;\;\;  p_{2} = m_{2}v_{2}$$
$$E_{1} = E_{2}$$
$$m_{1}v_{1}^{2} = m_{2}v_{2}^{2}$$
$$v_{2} = \sqrt{m_{1}\over{m_{2}}}v_{1}$$
So if $m_{1} = 16m_{2}$, $v_{2} = 4v_{1}$ and the baseball is four times faster.
But for the momentum, $p_{1} \neq p_{2}$ (Momentum is not conserved in this way). Instead, you have to put the velocity of the baseball (from above) into the equation for $p_{2}$:
$$p_{2} = m_{2}.\sqrt{m_{1}\over{m_{2}}}v_{1}$$
$$= {m_{2}\over{\sqrt{m_{1}m{2}}}}.m_{1}v_{1}$$
$$= \sqrt{m_{2}\over{m_{1}}}.p_{1}$$
Putting in our example of $m_{1} = 16m_{2}$, we get:
$$p_{2} = {1\over{4}}p_{1}$$
so the momentum of the baseball is a quarter of the momentum of the bowling ball.
