An eagle (m1=3.3kg) moving with speed v1=7.8m/ s is on a collision course with a

Question

An eagle (m1=3.3kg) moving with speed v1=7.8m/ s is on a collision course with a second
eagle (m2=4.6kg ) moving at a speed v2=10.2m/ s in a direction perpendicular to the first.
After they collide, they hold on to one another. In what direction, and with what speed are they
moving after the collision. (For simplicity’s sake, choose the initial direction of motion for the
first eagle to be along the positive x-axis and give the direction as an angle between the final
velocity and the positive x-axis).

Explanation / Answer

Given

m1 = 3.3 kg , v1 = 7.8 m/s

m2 = 4.6 kg , v2 = 10.2 m/s

Using law of conservation of momentum along x axis

m1v1 = (m1 +m2)Vx

3.3 kg *7.8 m/s = (3.3 kg + 4.6 kg)Vx

Vx = 3.25 m/s

Using law of conservation of momentum along x axis

m2v2 = (m1 +m2)Vy

4.6 kg *10.2 m/s = (3.3 kg + 4.6 kg) Vy

Vy = 5.93 m/s

Magnitude of velocity v = (3.25 m/s)^2+(5.93 m/s)^2

v = 6.76 m/s

--------------------------------------------------------------------

Angle = tan^-1(vy/vx)

= tan^-1(5.93/3.25) = 61.2^0

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