Question

An airplane flies east at a speed of 120 m/s. The airplane has a mass of 3200 kg.

a. What is the momentum of the airplane? (Remember, momentum is a vector quantity)

b. Air resistance reduces the speed of the plane is 110 m/s. What impulse was applied to the plane by the air resistance?

Answer 1

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

Answer 2

PART A)

Initial momentum of the airplane is product of its initial speed and mass

so here it is

`P = mv`

m = 3200 kg

v = 120 m/s

now we have

`P_1 = 3200 * 120`

`P_1 = 3.84 * 10^5 kg m/s`

Part B)

Speed of the plane is decreases to 110 m/s due to air resistance

so it's final momentum is given as

`P_f = mv_f`

here we have

`P_f = 3200(110)`

`P_f = 3.52 * 10^5 kg m/s`

Now impulse is defined as change in momentum

So impulse on airplane is given as

`I = 3.84 * 10^5 - 3.52 * 10^5`

`I = 32000 kg m/s`