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A 300 gg bird flying along at 5.7 m/sm/s sees a 10 gg insect heading straight toward it with a speed of 29 m/sm/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? _____m/s

Respuesta :

Answer:

bird's speed  = 4.58 m/s

Explanation:

given data

mass of bird mb = 300 g

velocity of bird vb = 5.7 m/s

mass of insect mi = 10 g

velocity of insect vi = - 29 m/s

solution

here momentum of system is constant

we apply here conservation of momentum so

mb × v + mi × v = mb × vb + mi × vi   ....................1

v ( mb + mi ) = mb × vb + mi × vi  

put here value and we get

v ( 300 + 10 ) = { 300 × 5.7 }  +  { 10 × (-29) }

v = [tex]\frac{1710-290}{310}[/tex]  

v = 4.58 m/s

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