A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose was flying with a speed of 60 km/hr and the aeroplane’s speed was 870 km/hour. Take the length of the goose to be 1.0 m long. (a) What is the change in momentum of the goose during this interaction?

The change in momentum of the goose during this interaction is 33.334 m/s

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

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