Question

In a circus trapeze act, two acrobats actually fly through the air and grab on to each other, then together grab a swinging bar. One acrobat, with a mass of 70 kg, is moving at 3.9 m/s at an angle of 15° above the horizontal and the other, with a mass of 41 kg, is approaching her with a speed of 3.9 m/s at an angle of 12° above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive x−axis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive x− and y−directions.

Answer 1

Conservation of momentum, v_x = (70 times 3.9 times cos 15 degree) + (41 times 3.9 times cos 12 degree)/(70 + 41) v_x = 3.784 m/s. v_y = (70 times 3.9 times sin 15 degree) + (41 times 3.9 times sin 12 degree)/(70 + 41) v_y = 0.936 m/s. speed 3.898 m/s at 13.89 degree above horizontal.