A 300 g bird flying along at 6.0 m/s sees a 10 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?

6.77m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the velocities before collision

v is the final collision

Given

m1 = 300g  = 0.3kg

u1 = 6.0m/s

m2 = 10g = 0.01kg

u2 = 30m/s

Required

The bird's speed immediately after swallowing v

Substitute the given values into the formula

m1u1 + m2u2 = (m1+m2)v

0.3(6) + 0.01(30) = (0.3+0.01)v

1.8+0.3 = 0.31v

2.1 = 0.31v

v = 2.1/0.31

v = 6.77m/s

Hence the bird's speed immediately after swallowing is 6.77m/s

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