A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 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?

Respuesta :

Answer:

The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]

Explanation:

From the question we are told that  

    The mass of the bird  is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]

    The initial speed of the bird is  [tex]u_1 = 6.2 \ m/s[/tex]

     The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]

       The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]

The negative sign is because it is moving in opposite direction  to the bird

According to the principle of linear momentum conservation

       [tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]

substituting values

        [tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]

    [tex]1.51 = 0.31 v_f[/tex]

     [tex]v_f = 4.87 \ m/s[/tex]

The Final velocity of Bird =  4.87 m/s

Mass of the bird = 300 g = 0.3 kg

Velocity of bird = 6.2 m/s

Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 =  1.86 kgm/s

Mass of the insect = 10 g = 0.01 kg

Velocity of insect =   - 35 m/s

Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35  kgm/s

According to the law of conservation of momentum We can write that

In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.

The bird opens the mouth and enjoys the free lunch  hence

Let the final velocity of bird is [tex]v_f[/tex]

Initial momentum of the system = Final momentum of the system

1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )

1.51 =  [tex]v_f[/tex] 0.31

[tex]v_f[/tex] = 4.87 m/s

The Final velocity of Bird =  4.87 m/s

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https://brainly.com/question/18066930