PART A)
Initial momentum of the airplane is product of its initial speed and mass
so here it is
[tex]P = mv[/tex]
m = 3200 kg
v = 120 m/s
now we have
[tex]P_1 = 3200 \times 120 [/tex]
[tex]P_1 = 3.84 \times 10^5 kg m/s[/tex]
Part B)
Speed of the plane is decreases to 110 m/s due to air resistance
so it's final momentum is given as
[tex]P_f = mv_f[/tex]
here we have
[tex]P_f = 3200(110)[/tex]
[tex]P_f = 3.52 \times 10^5 kg m/s[/tex]
Now impulse is defined as change in momentum
So impulse on airplane is given as
[tex]I = 3.84 \times 10^5 - 3.52 \times 10^5[/tex]
[tex]I = 32000 kg m/s[/tex]
a) Let's point x-axis to the east. Thus p=mv=384000 kg*m/s
b) Molecules of air applied impulse p.a=mu-mv=3200*110-3200*120=-32000 kg*m/s