What is the velocity of each of the shells? p=mV and momentum is conserved.
So the momentum of the BB, is equal to and opposite the combined momentum of the shells.
You've got a fairly large V (granted, it isn't anything like V2, but still...)
From a Goolag search.
They fired 2,700 pounds (1,225 kg) armor-piercing projectiles at a muzzle velocity of 2,500 ft/s (762 m/s), or 1,900 pounds (862 kg) high-capacity projectiles at 2,690 ft/s (820 m/s), up to 24 miles (21 nmi; 39 km).
(2700 lbs * 2500 ft/s ) shell * 9 shells = 6.075 x 10**7 ft*lb/s
the mass of the ship is 57,000 tonnes * 2204 lb/tonne =1.25628 x 10**8 lbs
Divide the momentum of the one side (shells), by the mass of the other (ship), to get the initial velocity of the ship:
1.25628 x 10 **8
__________________
6.075 x 10** 7
~= 2.0 ft/sec
That being said, there are other variables, such as the angle of the guns and the buoyancy of the ship in the water, together with the fact that water is fairly incompressible and will spring back.
(2700 lbs * 2500 ft/s ) shell * 9 shells = 6.075 x 10**7 ft*lb/s The mass of the ship is 57,000 tonnes * 2204 lb/tonne =1.25628 x 10**8 lbs
Divide the momentum of the one side (shells), by the mass of the other (ship), to get the initial velocity of the ship:
1.25628 x 10 **8
__________________
6.075 x 10** 7
~= 2.0 ft/sec
Two minor math errors that significantly affect the answer.
The combined momentum of the shells is 6.075x10**5 ft*lb/s vice 6.075x10**7 ft*lb/s.
You reversed the division from your statement. It should be:
6.075 x 10**5
__________________
1.25628 x 10**8
~= 4.849x10-3 ft/sec or ~6 hundredths of an inch.
I submit that this is insignificant when referring to a battleship.
WWG1WGA
Garde la Foi, mes amis! Nous nous sommes les sauveurs de la République! Maintenant et Toujours!
(Keep the Faith, my friends! We are the saviors of the Republic! Now and Forever!)
LonePalm, le Républicain du verre cassé (The Broken Glass Republican)