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.

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