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

Question

2)An eagle (m1 = 3.3 kg ) moving with speed v1 = 7.8 m/s is on acollision course with a second eagle (m2 = 4.6 kg) moving at aspeed v2 = 10.2 m/s in a direction perpendicular to thefirst. After they collide, they hold on to one another. Inwhat direction, and with

what speed are they moving after the collision. (Forsimplicity’s sake, choose the initial

direction of motion for the first eagle to be along the positivex-axis and give the

direction as an angle between the final velocity and the positivex-axis).

Explanation / Answer

calculate the x and y (perpendicular) components of momentum if xis going in the x direction (since you want the angle relative toeagle a) and b in going in the y: px = 3.3 kg * 7.8 m/s = 25.74 py = 4.6 kg * 10.2 m/s = 46.92 The angle of the momentum (and velocity) is just = arctan(py/px) (or arctan(vy/vx)) =61.25° To get magnitude, use pathag thm: (25.74^2 + 46.92^2) = 53.52 = total momentum v = momentum/total mass = 53.52/(3.3 + 4.6) = 6.774m/s

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