1.

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

Answer»

Solution :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.


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