Based on my conservation of momentum calcs (the total mom. of system is zero), assuming that the gun and car form one rigid body, the recoil velocity of the car is 1.41 m/s when a Browning .50 shell is fired from it (mass 700g, ejection velocity 908 m/s used as an example) facing forwards. The kinetic energy of the vehicle is about 450J. To get the car to tip over, you need an energy equivalent to that of lifting the car, if it's tipping from one end. If the gun fires straight and the pivot is back of the cat tracks, PE= mgh gives about 14000J to flip up. Thus it won't. I assumed it would flip end on end because any sideways momentum is too small to bother with!

If turn the gun 90° and fire across the doors of the car, total momentum of the system is still zero, and the recoil will be the same but it's pushing the car sideways instead of backwards. The KE of the car is the same, and you still need to effectively lift the car to tip it over. PE = mgh still applies, but h is now smaller. You need over 6000J of energy to tip the car ...so still not going to happen.