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
