A 300 g bird flying along at 5.5 m/sm/s sees a 10 g insect heading straight toward it with a speed of 26 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch. Required:
What is the bird's speed immediately after swallowing?

the bird's speed immediately after swallowing is 4.48 m/s

Explanation:

Given;

mass of the bird, m₁ = 300 g = 0.3 kg

initial velocity of the bird, v₁ = 5.5 m/s

mass of the insect, m₂ = 10 g = 0.01 kg

initial velocity of the insect, v₂ = - 26 m/s (negative because it is moving in opposite direction to the bird)

let the final velocity of the bird-insect system = v

Apply the principle of conservation of linear momentum for inelastic collision to determine the final velocity of the bird after swallowing.

m₁v₁ + m₂v₂ = v(m₁ + m₂)

(0.3 x 5.5) + (0.01 x - 26) = v(0.3 + 0.01)

1.65 - 0.26 = v(0.31)

1.39 = v(0.31)

v = 1.39 / 0.31

v = 4.48 m/s

Therefore, the bird's speed immediately after swallowing is 4.48 m/s

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