A 340 g bird flying along at 6.0 m/s sees a 14 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?

To determine the bird's speed immediately after swallowing the insect, we can use the principle of conservation of momentum. The total momentum before the insect is swallowed is equal to the total momentum after the insect is swallowed.

Let's denote the bird's mass as Mb, the insect's mass as Mi, the bird's initial velocity as Vb, and the insect's initial velocity as Vi.

Given:
Mb = 340 g = 0.34 kg (bird's mass)
Mi = 14 g = 0.014 kg (insect's mass)
Vb = 6.0 m/s (bird's initial velocity)
Vi = 30 m/s (insect's initial velocity)

The total momentum before swallowing is:
P_initial = Mb * Vb + Mi * Vi

The total momentum after swallowing can be calculated as the final momentum of the bird alone, since the insect becomes part of the bird's mass:
P_final = (Mb + Mi) * V_final

According to the conservation of momentum, P_initial = P_final.

Therefore,
Mb * Vb + Mi * Vi = (Mb + Mi) * V_final

Plugging in the given values:
(0.34 kg) * (6.0 m/s) + (0.014 kg) * (30 m/s) = (0.34 kg + 0.014 kg) * V_final

0.34 kg * 6.0 m/s + 0.014 kg * 30 m/s = (0.34 kg + 0.014 kg) * V_final

2.04 kg·m/s + 0.42 kg·m/s = 0.354 kg * V_final

2.46 kg·m/s = 0.354 kg * V_final

Now, we can solve for V_final, which is the bird's speed immediately after swallowing the insect.

V_final = 2.46 kg·m/s / 0.354 kg

V_final = 6.95 m/s (rounded to two decimal places)

Therefore, the bird's speed immediately after swallowing is approximately 6.95 m/s.