Question

A 340 g bird flying along at 6.0 m/s sees a 13 g insect heading straight toward it with a speed of 30 m/s. The bird opens its mouth wide and enjoys a nice lunch. What is the bird's speed immediately after swallowing?
_____m/s

Answer 1

Answer:

V = 4.67 m/s

Explanation:

given,

mass of the flying bird (M)= 340 g = 0.34 Kg

Speed of bird(u₁) = 6 m/s

mass of insect(m) = 13 g = 0.013 Kg

speed of insect(u₂) = 30 m/s

Speed of bird after eating the insect(V) =?

Using conservation of momentum

M u₁ + m u₂ = (M+m)V

$V = \dfrac{M u_1+ m u_2}{(M+m)}$

$V = \dfrac{0.34\times 6 - 0.013 \times 30}{0.34+0.013}$

$V = \dfrac{1.65}{0.353}$

V = 4.67 m/s

the bird's speed immediately after swallowing V = 4.67 m/s