A hockey player with a mass of 53.5 kg is traveling due east with a speed of 3.2

Question

A hockey player with a mass of 53.5 kg is traveling due east with a speed of 3.20 m/s. A second hockey player with a mass of 71.5 kg is moving due south with a speed of 6.85 m/s. They collide and hold on to each other after the collision, managing to move off at an angle ? south of east, with a speed of vf. Assume friction may be ignored and determine the following. (a) the angle ?

Explanation / Answer

applying conservation of momentum (m1+m2)vf cos x = m1 * v1 = 53.5*3.2=171.2 vf * cos x = 1.37 ------------ (1) (m1+m2)vf * sinx = m2*v2 = 71.5*6.85 =489.77 vf * sinx = 3.91 ---------------- (2) (1)^2 + (2)^2 -----> vf^2 = 1.37^2 + 3.91 ^2 = 17.165 vf = 4.14 m/s put vf in (1) cosx = 1.37/4.14 = 0.331 x = 70.67 degrees south of east

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