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