Question

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at??,2=16.9vi,2=16.9 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle ?=49.9θ=49.9°below the horizontal and a speed of ??,1=37.3vi,1=37.3 m/s. What is the speed of the eagle immediately after it catches its prey?What is the magnitude of the angle, measured from horizontal, at which the eagle is flying immediately after the strike?

Answer 1

Given the mass of the eagle is twice that of a hunted pigeon, let the mass of the eagle be 2 kg and mass of pigeon be 1 kg. The principle of conservation of momentum is involved. Momentum of moving as one unit will equal to the vector sum of individual momentums just before the catch.

The momentum of the pigeon = 16.9 Kg m/sec

Momentum of the eagle =2x37.3=74.6 Kg m/sec

Horizontal component of momentum of the eagle is=cos(49.9)x74.6=48.05 Kg m/sec

Total horizontal momentum as it grabs the pigeon =48.05+16.9 =64.95 Kg m/sec

Vertical component of momentum of the eagle is=sin(49.9)x74.6=57.06 Kg m/sec

Resultant momentum of both the birds = = 86.45Kg m/sec

The speed of the eagle immediately after it catches its prey = 86.45Kg m/sec/3Kg=28.81m/s

The angle, measured from horizontal, at which the eagle is flying immediately after the strike =arctan (64.95/57.06) =48.69 degrees below horizontal.