Question

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at 17.1 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 θ = 44.7° below the horizontal, and a speed of 34.9 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

the given collision is an inelastic collision

initial momentum (x axis) (initial) = m * 17.1 + 2m * 34.9cos(-44.7) =P(x)

initial momentum (y axis) (initial) = 2m 34.9 sin(-44.7) = P(y)

After the collision,

the total mass is 3m and

Px(after) = 3m Vx , Py(after) = 3m Vy

Because Px(after) = Px(initial) , we have

3m Vx = m * 17.1 + 2m * 34.9cos(-44.7)

Vx = 22.23 m/s

Likewise:

Vy = 1/3 * (2 * 34.9*sin(-44.7) ) = -16.365m/s

So the resultant speed after catching the prey is

V = sqrt(Vx^2 + Vy^2) = 27.60 m/s

The angle with the horizontal is:

tan(angle) = Vy/Vx = -16.365/22.23 = -36.35 degrees

angle = -36.35 deg