In a circus trapeze act, two acrobats actually fly through the air and grab on t
ID: 1536092 • Letter: I
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 59 kg, is moving at 3.5 m/s at an angle of 11° above the horizontal and the other, with a mass of 49 kg, is approaching her with a speed of 3.1 m/s at an angle of 15° above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive xaxis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive x and ydirections. The final velocity of the acrobats is m/s at ° above the xaxis.
Explanation / Answer
initial momentum of 1st acrobat = 59 x 3.5 x (cos11 i^ + sin11 j^)
initial momentum of 2nd acrobat = 49 x 3.1 x (-cos15 i^ + sin15 j^)
final momentum = (59 + 49)v = 108v
=> 59 x 3.5 x (cos11 i^ + sin11 j^) + 49 x 3.1 x (-cos15 i^ + sin15 j^) = 108v
=> (206.5cos11 - 151.9cos15) i^ + (206.5sin11 + 151.9sin15) j^ = 108v
=> v = 0.51835 i^ + 0.728858 j^
|v| = speed = 0.8944 m/s
angle above x axis = tan-1(0.728858/0.51835) = 54.58 degree in +ve x direction