A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 27 m/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?
To calculate the bird's speed immediately after swallowing, we can apply the principle of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event.
The momentum of an object is given by the product of its mass and velocity. Mathematically, momentum (p) is represented as:
p = m * v
Where:
p = momentum
m = mass
v = velocity
Let's denote:
m1 = mass of the bird (300 g)
v1 = velocity of the bird before swallowing (6.2 m/s)
m2 = mass of the insect (10 g)
v2 = velocity of the insect (27 m/s)
vf = velocity of the bird immediately after swallowing
According to the conservation of momentum, we have:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
Substituting the given values:
(0.300 kg * 6.2 m/s) + (0.010 kg * 27 m/s) = (0.300 kg + 0.010 kg) * vf
Simplifying,
(1.86 kg·m/s) + (0.27 kg·m/s) = (0.31 kg) * vf
Combining the terms, we get:
2.13 kg·m/s = 0.31 kg * vf
Now, solving for vf:
vf = (2.13 kg·m/s) / (0.31 kg)
vf ≈ 6.871 m/s
Therefore, the bird's speed immediately after swallowing the insect is approximately 6.871 m/s.