Physics 1 ch 8 problem

Question

The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at vi,2=18.1 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 theta=45.1 degree below horizontal and a speed vi,1=31.9 m/s.

What is the speed of vf of the eagle immediately after it catches its prey?

What is the magnitude /0 of the angle, measured from horizontal , at which the eagle is flying immediately after the strike?

Answer 1

Mass of pigeon = m

Mass of Eagle = 2m

total mass =3m

Initial momentum of pigeon = mv_i,2 = (mv_i,2) i

Initial momentum of eagle = 2mv_i,1 = 2mv_i,1 cos(-θ)i + 2mv_i,1 sin(-θ)j

Total Initial momentum = [mv_i,2 + 2mv_i,1 cos(θ)]i - 2mv_i,1 sin(θ)j

Total final momentum = Total Initial momentum

So, 3mv_f = [mv_i,2 + 2mv_i,1 cos(θ)]i - 2mv_i,1 sin(θ)j

or, v_f = (1/3)[v_i,2 + 2v_i,1 cos(θ)]i - (2/3)v_i,1 sin(θ)j

= (21.045 m/s) i - (15.064 m/s)j

Therefore, v_f =√( (21.045 m/s)² + (-15.064 m/s)² ) = 25.9 m/s (3 sig fig)

θ_f = tan^-1 (-15.064/21.045) = -35.6° (negative means below horizontal)