nope they do not move sideways; we had a huge post on MW about this
Here is a cut and paste from the techincal board
Do Battleships move sideways when they fire?
By R. A. Landgraff and Greg Locock 
Updated 02 August 2000 
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I get asked about once a month if the Iowa (or Bismarck, Yamato, etc.) move sideways when they fire a full broadside.  To save myself some time, I've plagiarized Dick Landgraff's very good answer to this same question (what are friends for?).  In addition, Greg Locock kindly pointed out an error in my math when I ineptly tried to calculate a real number for the motion. 
Tony DiGiulian 
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What looks like a side-ways wake is just the water being broiled up by the muzzle blasts. The ship doesn't move an inch or even heel from a broadside. 
The guns have a recoil slide of up to 48 inches and the shock is distributed evenly through the turret foundation and the hull structure. The mass of a 57,000 ton ship is just too great for the recoil of the guns to move it. Well, theoretically, a fraction of a millimeter. 
But because of the expansive range of the overpressure (muzzle blast), a lot of the rapidly displaced air presses against the bulkheads and decks. Those structures that are not armored actually flex inwards just a bit, thus displacing air quickly inside the ship and causing loose items to fly around. Sort of like having your house sealed up with all windows and vents closed and when you slam the front door quickly the displaced air pops open the kitchen cabinets. 
R. A. Landgraff 
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To calculate the velocity of the USS New Jersey moving sideways, what you need to consider is conservation of momentum.  A 16" Mark 8 APC shell weighs 2,700 lbs. and the muzzle velocity when fired is 2,500 feet per second (new gun).  The USS New Jersey weighs about 58,000 tons fully loaded (for ships, a ton is 2,240 lbs.).  All weights must be divided by 32.17 to convert them to mass. 
If the battleship were standing on ice, then: 
Mass of broadside * Velocity of broadside = Mass of ship * Velocity of ship 
9 * (2,700 / 32.17) * 2,500 = 58,000 * (2,240 / 32.17) * Velocity of ship 
Solving for the ship's velocity: 
Velocity of ship = [9 * (2,700 / 32.17) * 2,500] / [58,000 * (2,240 / 32.17)] = 0.46 feet per second 
So, ship's velocity would be about 6 inches per second, ON ICE. 
This analysis excludes effects such as (1) roll of the ship, (2) elevation of the guns (3) offset of the line of action of the shell from the centre of gravity of the ship and (4) forces imposed by the water on the ship. These are variously significant, and will all tend to reduce the velocity calculated above. 
Greg Locock 
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I need to point out that in Greg's masterful analysis he assumes that the guns are at zero degrees elevation, that is, the guns are pointed directly at the horizon.  In actuality, they are almost never fired at this elevation as it would mean that the shells would only go a short distance before they struck the water.  At a higher, more realistic elevation, the force of the broadside would also have to be multiplied by the cosign of the angle of elevation.  This means that the horizontal velocity imparted to the ship would be even less than the numbers calculated above. 
Tony DiGiulian 
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I just want to add the forces needed to push 800+ ft of ship through the water sideways is just not in the recoil forces; like it says; on ice it might move