A common of mass `m_(1) = 12000 kg` locatede on a smooth horizontal platform fires a shell of mass `m _(2) = 300 kg ` in horizontal direction with a velocity `v _(2)=400 ms//s.` Find the velocity of the cannon after it is shot.

Explanation

Since the pressure of the powder gases in the bore of the cannon is an iternal force the net external force acting on cannon during the firing is zero.
Let `v _(1)` be the velocity of the cannon after shot. The initial momentum of system is zero.
The final momentum of the system `=m _(1) v_(1) + m _(2) v_(2)`
From the conservation of linear momentum, We get,
`m _(1) v_(1) +m_(2) v_(2) =0`
`m_(1)v_(1) =-m_(2)v_(2)`
`v _(1) =-m_(2) v_(2) //m_(1)`
Substituting the given values in the above equation, we get
`v _(1) =- ((300kg) x (400m//s))/(112 000 kg)`
`=-10 m//s.`
Thus the velocity of cannon is 10 m/s after the slot,
Here .-. sign indicates that the canon moves in a direction opposite to the motion of the bullet.