Answer :
1) The momentum of the airplane is [tex]3.84\cdot 10^5 kg m/s[/tex] east
2) The impulse applied to the airplane is 32,000 kg m/s west
Explanation:
1)
The momentum of an object is defined as
[tex]p=mv[/tex]
where
m is the mass of the object
v is its velocity
The airplane in this problem has
m = 3200 kg is its mass
v = 120 m/s east is its velocity
So, its momentum is
[tex]p=(3200)(120)=3.84\cdot 10^5 kg m/s[/tex]
And since momentum is a vector, it also has a direction, which is the same as the velocity (east).
2)
The impulse applied to an object is equal to the change in momentum of the object:
[tex]I=\Delta p = m(v-u)[/tex]
where
m is the mass
v is the final velocity
u is the initial velocity
For the airplane in this problem,
m = 3200 kg
u = 120 m/s
v = 110 m/s
So, the impulse is
[tex]I=(3200)(110-120)=-32,000 kg m/s[/tex]
And the negative sign means the direction of the impulse is opposite to the airplane motion, so west.
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