Question

A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 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?

Answer 1

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}$

Put the value into the formula

$0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}$

$v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}$

$v_{3}=4.98\ m/s$

Hence, The bird's speed immediately after swallowing is 4.98 m/s.