An attack helicopter is equipped with a 20- mm cannon that fires 87 g shells in the forward direction with a muzzle speed of 853 m/s. the fully loaded helicopter has a mass of 4410 kg. a burst of 176 shells is fired in a 2.93 s interval. what is the resulting average force on the helicopter?

Answer

given,

mass of the shell = 87 g = 0.087 Kg

speed of the muzzle = 853 m/s

mass of the helicopter = 4410 kg

A burst of 176 shell fired in 2.93 s

resulting average force = ?

momentum of the shell = m v

                                       = 0.087 x 853

                                       = 74.21 kgm/s

momentum of 176 shell is = 176 p

                                          = 176 x 74.21

                                          = 13060.96

momentum of helicopter = - 13060.96 kgm/s

amount of speed reduce a =

                                          a=

                                          a = 2.96 m/s²

velocity  = \dfrac{2.96}{2.93}

      v = 1.01 m/s

4457.7 N

Explanation:

87g = 0.087kg

v = 853m/s

t = 2.93 s

total mass hitting = m = 0.087*176 = 15.32 kg

net momentum = mv = 15.32*853 = 13061 kg.m/s

as we know the rate of change of momentum is the average force of impact,

F_avg = = = 4457.7 kg.m/s^2 = 4457.7 N

he kicked it 10 feet straight up into the air? i think this works.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s