JinaWu
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An attack helicopter is equipped with a 20-mm cannon that fires 174 g shells in the forward direction with a muzzle speed of 453 m/s. The fully loaded helicopter has a
mass of 2600 kg. A burst of 223 shells is fired in a 6.22 s interval. What is the resulting average force on the helicopter? By what amount is its forward speed reduced? Plz i need help!

Respuesta :

 By F = ∆P/∆t 
=>F = n x m x ∆v/∆t 
=>F = [242 x 66.1 x 10^-3 x 488]/3.57 
=>F = 2186.60 N 
By the law of momentum conservation:- 
=>(M+m)V = Mv1 + mv2 
=>(3630 + 66.1 x 10^-3 x 242) x V = 3630 x v1 + 66.1 x 10^-3 x 242 x 488 
=> if we know the initial speed i.e. V we can calculate final speed v1.
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