I have found the correct angle (Part B). The mass of a particular eagle is twice

Question

I have found the correct angle (Part B).

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at 16.9 m/s, when the eagle swoops down, grabs the pigeon, and fles off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle = 43.1° below the horizontal, and a speed of 37.1 m/s. What is the speed of the eagle immediately after it catches its prey? Number 43.65 m/ s 37.1 m/s What is the magnitude of the angle, measured from horizontal, at which the eagle is flying immediately after the strike? 16.9 m/s Number 35.5

Explanation / Answer

a). Momentum is conserved during this collision.
The initial total momentum has components:

Px(initial) = m * 16.9 + 2m * 37.1 cos(-43.1)
Py(initial) = 2m 37.1 sin(-43.1)

After the collision, the combination has mass 3m and

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

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

3m Vx = m * 16.9 + 2m * 37.1 cos(-43.1)

The mass drops out from both sides and so
Vx = 1/3 * ( 16.9 + 2* 37.1 cos(-43.1)) = 23.69 m/s

Likewise:

Vy = 1/3 * (2 * 37.1*sin(-43.1) ) = -16.899 m/s

So the speed is

V = sqrt(Vx^2 + Vy^2) = 29.09 m/s

b). The angle with the horizontal is:

tan(angle) = Vy/Vx = -16.899/23.69

angle = -35.5 degrees

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