Question

A 300 g bird is flying along at 6.0 m/s and sees a 10 g insect heading straight towards it with a speed of 30 m/s. The bird opens its mouth wide and swallows the insect. a. What is the birds speed immediately after swallowing the insect? b. What is the impulse on the bird? c. If the impact lasts 0.015 s, what is the force between the bird and the insect?

Answer 1

Answer:

(a): The bird speed after swallowing the insect is V= 4.83 m/s

(b): The impulse on the bird is I= 0.3 kg m/s

(c): The force between the bird and the insect is F= 20 N

Explanation:

ma= 0.3 kg

va= 6 m/s

mb= 0.01kg

vb= 30 m/s

(ma*va - mb*vb) / (ma+mb) = V

V= 4.83 m/s (a)

I= mb * vb

I= 0.3 kg m/s  (b)

F*t= I

F= I/t

F= 20 N (c)

Answer 2

The answers to your questions are as listed below

A) The birds speed after swallowing the insect is = 4.84 m/s

B) The impulse on the bird is = 0.348 N-s

C) The force between the bird and the insect when the impact last 0.015s = 23.2 N

Given data :

mass of bird ( M₁ )  = 300 g

speed of bird ( U₁ ) = 6 m/s

mass of insect ( M₂ ) = 10 g

speed of insect ( U₂ )  = 30 m/s

Determine the birds speed and the impulse on the bird

A) Birds speed after swallowing the insect

we will apply the principle of momentum conservation

M₁U₁  +  M₂U₂ = ( M₁ +M₂ )V

= 300*6 - 10*30 = ( 300 + 10 ) V

= 1800 - 300 = 310 V

therefore ; V =  1500 / 310 = 4.84 m/s

B) Impulse on the bird

Impulse = m * V - U₁

             = 0.3 * ( 4.84 - 6 )

             = -0.348 N-s

therefore the magnitude of the impulse = 0.348 N-s

C) The force between the the bird and insect if the impacts lasts for 0.015 s

Force = impulse / time

          = 0.348 / 0.015

          = 23.2 N

Hence we can conclude that the answers to the question are as listed above.

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