Question

A 320 g bird flying along at 7.0 m/s sees a 12 g insect heading straight toward it with a speed of 30 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. What is the bird's speed immediately after swallowing?

Answer 1

To solve the problem it is necessary to apply the conservation equations of the moment.

The conservation equation of the moment is defined as,

`m_1v_1+m_2v_2 = (m_1+m_2)v_f`

Where,

`m_(1) =` Mass of Bird

`m_2 =` Mass of Insect

`v_(1) =`Velocity Bird before swallowing

`v_2 =`Velocity of insect

`v_f =` final velocity (both)

Replacing with our values we have,

`m_1v_1+m_2v_2 = (m_1+m_2)v_3`

`(0.32)(7)+(0.012)(30)= (0.32+0.012)v_f`

`2.6 = 0.332v_f`

`v_f = 7.83m/s`

Therefore the bird's speed immediately after swallowing is 7.83m/s