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Westkost
1 month ago
14

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 w

as 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?
Physics
1 answer:
kicyunya [3.2K]1 month ago
4 0

Response:

The change in momentum experienced by the goose during this encounter is 33.334 m/s

Explanation:

Provided;

goose's mass, m₁ = 2.0 kg

airliner mass, m₂ = 160,000 kg

goose's initial speed, u₁ = 60 km/hr  = 16.667 m/s

airliner initial speed, u₂ = 870 km/hr = 241.667 m/s

The alteration in momentum is defined as;

ΔP = mv - mu

where;

u being the foremost velocity of the bird

v representing the concluding velocity of the bird

Using linear momentum conservation principles;

Total momentum pre-collision = Total momentum post-collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the bird and airliner speed 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 minimal in comparison to the final velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative value indicates a slowdown in the bird following the collision.

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

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