Question

A 45.0-kg skater is traveling due east at a speed of 2.85 m/s. A 72.5-kg skater is moving due south at a speed of 7.35 m/s. They collide and hold on to each other after the collision, managing to move off at an angle θ south of east, with a speed of vf. Find the following.
(a) the angle θ
°

(b) the speed vf, assuming that friction can be ignored
m/s

Answer 1

south momentum = 75 * 7.35 = 551
east momentum = 45 * 2.85 = 128

new mass = 45 + 75 = 120

new momentum = old momentum (Newton #1 law)

south speed = 551/120 = 4.59
east speed = 128/120 = 1.07

theta = tan^-1 (4.59/1.07) = 76.9 degrees

Vf = sqrt(4.59^2+1.07^2) = 4.71 m/s

Answer 2

Answer