Question

0.013*30A 340 g bird flying along at 6.0 m/s sees a 13 g insect heading straight toward it with a speed of 30 m/s. The bird opens its mouth wide and enjoys a nice lunch. What is the bird's speed immediately after swallowing?

Answer 1

Answer:

The speed of the bird immediately after swallow = 4.67 m/s

Explanation:

According to the law of momentum,

initial momentum before collision(swallow) = final momentum after collision (swallow)

Note: The collision between the bird and the insect is inelastic, as both moves with the same velocity after collision (swallow)

m₁u₁ - m₂u₂ =V (m₁ + m₂).................................. Equation 1

V = (m₁u₁-m₂u₂)/(m₁+m₂)................................... Equation 2

Note: The bird and the insect moves in opposite direction

where m₁ = mass of the bird, m₂ = mass of the insect, u₁ = initial velocity of the bird, u₂ = initial velocity of the insect, V = common velocity of the bird and the insect.

Given: m₁ = 340 g = (340/1000) kg = 0.34 kg, m₂ = 13 g = 0.013 kg, u₁ = 6 m/s, u₂ = 30 m/s

Substituting these values into equation 2

V = [0.34(6) -0.013(30)]/(0.34+0.013)

V = (2.04-0.39)/0.353

V = 1.65/0.353

V = 4.67 m/s

Thus the speed of the bird immediately after swallow = 4.67 m/s