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A 300 g bird flying along at 6.3 m/s sees a 10 g insect heading straight toward it with a speed of 34 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? Express your answer in meters per second.

Respuesta :

tejasvinisinha1623

Explanation:

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

Velocity of Bird = 0.3 x 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 × 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 vf

Initial momentum of the system = Final momentum of the system

1.86 -0.35 = vf ( 0.01 + 0.3 )

vf 0.31 = 4.87 m/s

The Final velocity of Bird = 4.87 m/s

msm555

Answer:

5 m/s

Explanation:

To find the bird's speed immediately after swallowing the insect, we can use the principle of conservation of momentum. According to this principle, the total momentum before the event is equal to the total momentum after the event.

The momentum [tex] p [/tex] of an object is given by the product of its mass [tex] m [/tex] and its velocity [tex] v [/tex]:

[tex] \Large\boxed{\boxed{ p = m \cdot v}} [/tex]

Initially, the total momentum before swallowing is the sum of the momentum of the bird and the momentum of the insect.

Let's denote:

  • [tex] m_b [/tex] as the mass of the bird (300 g = 0.3 kg)
  • [tex] v_b [/tex] as the velocity of the bird before swallowing (6.3 m/s)
  • [tex] m_i [/tex] as the mass of the insect (10 g = 0.01 kg)
  • [tex] v_i [/tex] as the velocity of the insect before swallowing (-34 m/s, since it's moving towards the bird)

The total initial momentum is:

[tex] p_{\textsf{initial}} = m_b \cdot v_b + m_i \cdot v_i [/tex]

[tex] p_{\textsf{initial}} = (0.3 \, \textsf{kg}) \times (6.3 \, \textsf{m/s}) + (0.01 \, \textsf{kg}) \times (-34 \, \textsf{m/s}) [/tex]

[tex] p_{\textsf{initial}} = 1.89 \, \textsf{kg m/s}- 0.34 \, \textsf{kg m/s} [/tex]

[tex] p_{\textsf{initial}} = 1.55 \, \textsf{kg m/s} [/tex]

After the bird swallows the insect, the combined mass is the sum of the masses of the bird and the insect, and the velocity is the velocity of the bird after swallowing, denoted as [tex] v_{\textsf{final}} [/tex].

So, the total momentum after swallowing is:

[tex] p_{\textsf{final}} = (m_b + m_i) \cdot v_{\textsf{final}} [/tex]

[tex] p_{\textsf{final}} = (0.3 \, \textsf{kg} + 0.01 \, \textsf{kg}) \times v_{\textsf{final}} [/tex]

[tex] p_{\textsf{final}} = 0.31 \, \textsf{kg} \times v_{\textsf{final}} [/tex]

Since momentum is conserved:

[tex] p_{\textsf{initial}} = p_{\textsf{final}} [/tex]

[tex] 1.55 \, \textsf{kg m/s} = 0.31 \, \textsf{kg} \times v_{\textsf{final}} [/tex]

[tex] v_{\textsf{final}} = \dfrac{1.55 \, \textsf{kg m/s}}{0.31 \, \textsf{kg}} [/tex]

[tex] v_{\textsf{final}} = 5 \, \textsf{m/s} [/tex]

Therefore, the bird's speed immediately after swallowing the insect is [tex] 5 \, \textsf{m/s} [/tex].

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