Answer:
The change in momentum of the goose during this interaction is 33.334 m/s
Step-by-step explanation:
Given;
mass of goose, m₁ = 2.0 kg
mass of commercial airliner, m₂ = 160,000 kg
initial velocity of the bird, u₁ = 60 km/hr = 16.667 m/s
initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s
Change in momentum is given as;
ΔP = mv - mu
where;
u is the initial velocity of the bird
v is the final velocity of the bird
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the final velocity of bird and airliner after collision;
(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)
38,666,753.334 = 160,002v
v = 38,666,753.334 / 160,002
v = 241.664 m/s
Thus, the final velocity of the bird is negligible compared to final velocity of the airliner.
ΔP = mv - mu
ΔP = m(v - u)
ΔP = 2(0 - 16.667)
ΔP = -33.334 m/s
The negative sign implies a deceleration of the bird after the impact.
Therefore, the change in momentum of the goose during this interaction is 33.334 m/s