The mass of a pigeon hawk is twice that of the pigeons it hunts. Suppose a pigeon is gliding north at a speed of Vp = 22.3 m/s when a hawk swoops down, grabs the pigeon, and flies off, as shown in the figure. The hawk was flying north at a speed of Vh = 37.4 m/s, at an angle theta = 45 degrees below the horizontal at the instant of the attack.

What is the birds final speed Vf just after the attack?

What is the angle below the horizontal below the horizontal of the final velocity vector of the birds just after the attack?

You want the final speed and direction of a pigeon hawk and the pigeon it grabs if the hawk was initially diving at 37.4 m/s 45° below horizontal, and the pigeon was initially flying horizontally at 22.3 m/s. The hawk has twice the mass of the pigeon.

Momentum

In (north, up) coordinates, the initial momentum of the pigeon hawk is ...

mv = (2)(37.4∠-45°) = 74.8(cos(-45°), sin(-45°)) ≈ (52.892, -52.892)

Similarly, the initial momentum of the pigeon is ...

mv = (1)(22.3∠0°) = (22.3, 0)

So, the total momentum of the two birds prior to the attack is ...

(52.892, -52,892) +(22.3, 0) = (75.192, -52.892) . . . . (mass)·(m/s)

Final velocity

Converting this momentum to a magnitude and direction, we have ...

|mv| = √(75.192² +(-52.892)²) ≈ 91.931

∠mv = arctan(-52.892/75.192) ≈ -35.123°

The final velocity will be this total momentum, divided by the total mass of the two birds:

91.931/3 ≈ 30.6 . . . . m/s

The birds' final speed is about 30.6 m/s.

The angle of their velocity vector is about 35° below horizontal.