A hungry hawk is on the prowl when it spots a tasty dove in flight. The dove is

Question

A hungry hawk is on the prowl when it spots a tasty dove in flight. The dove is flying due north at 18.5 m/s when the hawk swoops down, grabs the dove, and flies off. At the instant right before grabbing the dove, the hawk is flying toward the dove at a speed of 46.1 m/s and at an angle of 63.1° below the horizontal. If mass of the hawk is three times that of the dove, what is the velocity (speed and direction) at which the hawk is flying immediately after catching the dove?

Speed of hawk-dove after strike and direction below horizontal

What percentage of the initial kinetic energy of the hawk-dove system is lost during the strike?

Explanation / Answer

here,

initial speed of dove , ud = 18.5 m/s j

speed of hawk , uh = 46.1 m/s

theta = 63.1 degree

let the final speed be v and angle be phi

using conservation of momentum

m*ud + 3 * m * uh * cos(theta) = (3m + m) * ( v * cos(phi))

18.5 + 46.1 * 3 * cos(63.1) = 4 * v * cos(phi) .....(1)

and

3 * uh * sin(theta) = 4 * v * sin(phi)

3 * 46.1 * sin(63.1) = 4 * v * sin(phi) .....(2)

from(1) and (2)

v = 39.05 m/s

phi = 52.15 degree

the final speed is 39.05 m/s at an angle of 52.15 degree

percentage of the initial kinetic energy of the hawk-dove system is lost during the strike , % = (3m * 46.1^2 + m * 18.5^2 - 4 * m * 39.05^2 ) /( 3m * 46.1^2 + m * 18.5^2)

% = (6717.88 - 6099.61) /( 6717.88)

% = 9.2 %

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