Calculate recoil speed of a canon given muzzle velocity and barrel length Consider the following scenario which I pose out of curiosity (its not a homework question!)
We have a canon of mass $M$ and barrel length $l$ which uniformly accelerates a projectile of mass $m$ to a muzzle velocity $v_0$. The canon is mounted on wheels and is constrained to move only in the horizontal direction. The canon initially rests on a scale at ground level which measures an average weight of $W$ during the time in which the projectile travels through the barrel.
I would like to know two things: (1) the recoil speed of the canon, (2) how much of the chemical energy stored stored in the canon's explosives is lost to vibrations with the ground.
I'm not sure if there's enough information to solve this, but I'll describe what I did and I'd like to know if its correct.

Denote the canon-projectile system by $S$, the angle formed by the barrel with the ground by $\theta$, and the recoil speed of the canon by $V$. 
The absence of forces acting in the horizontal direction implies conservation of horizontal momentum of $S$: $MV=mv_0\cos(\theta)$. 
The net force acting on $S$ vertically is the normal force $N$ of the ground on the canon plus gravity, so the change in vertical momentum during the time $\Delta t$ during which the projectile travels through the barrel is $[-(m+M)g+W]\Delta t=mv_0\sin(\theta)$. Also 
Finally, kinematics tells us $\delta t=2l/v_0$. Hence $V=\frac{mv_0\cos(\theta)}{M}$ where
$\sin(\theta)=\frac{2l}{mv_0^2}[W-(m+M)g]$
Let's take a real-world example: the M198 howitzer mounted on wheels. We have $M=7000$ kg, $m=40$ kg, $v_0=685$ m/s, $l=6$ m. Let's suppose $\theta=45$ degrees. Then we get a recoil velocity of $2.77$ m/s and the average weight $W$ registered by the scale would be $1.17 *10^6$ N, which is $17$ times greater than the actual weight of the canon. Well something is wrong here - the recoil velocity is much too slow, and the average weight registered seems to high. The explosive content of a howitzer round is roughly $6$ kg of TNT so $2.5 *10^7$ J of energy. Is this capable of making the canon feel $17$ g's?
As for the chemical energy converted to vibrational energy with the ground, I'm not sure how to calculate it. 
 A: Let's look at this a little differently.
First - conservation of momentum. If the projectile of mass $m$ has horizontal velocity $v$, then the recoil velocity of the tank with mass $M$ must be $\frac{mv}{M}$. Putting in numbers, I get the same recoil velocity of $$\frac{40\cdot685\cdot \sin 45}{7000} = 2.8 ~\rm{m/s}$$
If you have constant acceleration in the barrel (unrealistic?), then the time in the barrel is just $t = \frac{2L}{v} = 0.017 ~\rm{s}$. Dividing the vertical momentum by the time gets us the force, so the weight multiplier of the tank is 
$$\begin{align}\mu &= \frac{F}{Mg}+1\\
&=\frac{p_v}{Mgt} +1\\
&=\frac{mv\sin 45}{Mg\frac{2L}{v}} +1\\
&= \frac{m v^2}{MgL\sqrt{8}}+1\\
&=17.1\end{align}$$
Again, the same value that you get.
Note that energy of the explosive has little to do with how much force the tank appears to produce - that is really just a function of the time it takes. The short time in the barrel is what gives you this multiplier... and as you can see, as the velocity goes up, the force goes up quadratically (you give more momentum in less time).
Incidentally you can see that the kinetic energy of the projectile appears in your expression (it is slightly disguised, but it's there).
We can do a sanity check: the kinetic energy of the projectile is 9.4 MJ. - a little less than half of the 25 MJ energy content of the charge.
It looks reasonable.
As for the "energy lost to the ground": that is possible to answer in very general terms only. As you know, work done is force times distance. If the ground is very rigid, the higher weight will not result in any greater displacement, and very little energy will go into vibration (in essence, there is a large impedance match). But clearly since the tank is mounted on a platform that allows (by your description) horizontal recoil, that's where you will get significant energy.
Again, conservation of momentum already told us
$$V_h = v_h \frac{m}{M}$$
So for the kinetic energy of the recoiling tank we can write
$$\begin{align}\frac12 M V_h^2 &= \frac12 M v_h^2 \left(\frac{m}{M}\right)^2\\
&= \frac12 \frac{m^2}{M} v_h^2\end{align}$$
Note that we assumed that $m\ll M$ - obviously, as $m$ gets larger, the recoil velocity will increase and we have to worry about the effect this has on the motion of the projectile in the c.o.m. frame. But I won't worry about that here.
According to the above equation, the energy "lost" to the tank will be only 27 kJ - a small fraction of the energy in the projectile. Given the small vertical motion of the tank, the energy lost to vibration will be even less.
Most of the energy from the explosive that doesn't make it into the projectile is lost through the fact that the gas has not finished expanding / cooling by the time the projectile leaves the barrel. So there is a lot of hot air escaping, and a very loud boom.
