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A gun, the barrel of which weighs 450 kg-wt, is fired in a horizontal direction. The weight of the projectile is 5 kg-wt and the muzzle velocity is 450 m/sec. When fired, the barrel recoils 45 cm. Find the mean value of the force developed by the anti-recoil device of the gun in absorbing the recoil.

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Answer: F=m2v2/2MS = 1280 kg-wt.

Explanation:

The velocity of the barrel u at the instant of firing is determined from the law of conservation of momentum. It is given by

u = mv/M,

where m and M are the masses of the projectile and the gun barrel, respectively. The entire kinetic energy 1/2(Mu2)  acquired by the barrel at the instant of firing will be expended in work done against the braking force F. Hence

1/2(Mu2) = FS.

  

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