After a disastrous descent into enemy territory our hero has

After a disastrous descent into enemy territory, our hero has found himself completely weaponless, yet only slightly disheveled. In a feat of inconceivable (perhaps even implausible) strength, he pries a perfectly intact chain gun from its mountings beneath his chopper\'s wreckage and cradles it in his arms. As enemy troops swarm in: he unleashes a 1,412 spm (shots per minute) barrage that stops them dead in their tracks. If the muzzle velocity is 3,150 ft/s and each bullet weighs 0.22 lb, what average force F is required of our hero to hold the weapon in place? Neglect, among other things, the mass of both the ammunition belt and the discarded casings. Round your answer to the nearest lb.

Solution

change in momentum of one bullet is given by

mv = 0.22*3150 = 693 lb ft/s
no. of bullets per second is,

=1412/60 = 23.5
Force = change in momentum per second = 693*23.5 =16285.5 ft lb/s²

=16285.5/32.17 = 506 lb