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 16.7 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 34.7° below the horizontal. A). 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? B.) What percentage of the initial kinetic energy of the hawk-dove system is lost during the strike?

Explanation / Answer

let M is the mass hawk and m is the mass of dove

M = 3*m

let x axis is towards north and y axis towards down

let vx and vy are component of hawk after grabbing the dove

Apply conservation of momentum in x-direction

M*46.1*cos(34.7) + m*16.7 = (M + m)*vx

3*m*46.1*cos(34.7) + m*16.7 = (3*m + m)*vx

3*46.1*cos(34.7) + 16.7 = 4*vx

vx = (3*46.1*cos(34.7) + 16.7)/4

= 32.6 m/s

Apply conservation of momentum in y-direction

M*46.1*sin(34.7) = (M + m)*vy

3*m*46.1*sin(34.7) = (3*m + m)*vy

3*46.1*sin(34.7) = 4*vy

vy = 3*46.1*sin(34.7)/4

= 19.7 m/s

so, v = sqrt(vx^2 + vy^2)

= sqrt(32.6^2 + 19.7^2)

= 38.1 m/s <<<<<<<--------Answer

Direction : theta = tan^-1(vy/vx)

= tan^-1(19.7/32.6)

= 31.1 degrees below horizontal <<<<<<<--------Answer

B) Ki = 0.5*m*16.7^2 + 0.5*3*m*46.1^2

= 3327.26*m

Kf = 0.5*(m + 3*m)*38.1^2

= 2903*m

so, (ki - kf)*100/ki = (3327 - 2903)*100/3327

= 12.74 %

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