An attack helicopter is equipped with a 20 - mm cannon that fires 88.7 g shells in the forward direction with a muzzle speed of 929 m/s. the fully loaded helicopter has a mass of 4180 kg. a burst of 89.9 shells is fired in a 4.84 s interval. what is the resulting average force on the helicopter? answer in units of n

The impulse imparted to the shells equals the change in the momentum:

Fav * (Delta t) = Delta m*v.

The mass change is

Delta m = n*m = (89.9shells) * (88.7g) = 7.97Kg

So the average force is

F = ((v) * (Delta m)) / t = ((929) * (7.97)) / 4.84=1529.78 N

Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.