Final answer:
The velocity of the eagle and the pigeon after the collision is 5.5 m/s.
Step-by-step explanation:
In this scenario, the eagle catches the pigeon while the pigeon is flying north at a velocity of 16.5 m/s. Since the eagle is twice the mass of the pigeon, we can assume that its initial velocity is zero. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
We can calculate the velocity of the eagle and the pigeon after the collision using the equation:
m1v1 + m2v2 = (m1 + m2)v
Where m1 and m2 are the masses of the eagle and the pigeon, v1 and v2 are their initial velocities, and v is their final velocity.
Substituting the given values, we have:
(2m)(0 m/s) + (1m)(16.5 m/s) = (2m + 1m)v
Simplifying, we find:
16.5 m/s = 3mv
Dividing both sides by 3m, we get:
v = 5.5 m/s
Therefore, the velocity of the eagle and the pigeon after the collision is 5.5 m/s.