Two skydivers are holding on to each other while falling straight down at a comm

Question

Two skydivers are holding on to each other while falling straight down at a common terminal speed of 63.10 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 89.30 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis):

v1x=4.93 m/s      v1y=3.75 m/s     v1z=63.1 m/s

What are the x- and y-components of the velocity of the second skydiver, whose mass is 57.70 kg, immediately after separation?

What is the change in kinetic energy of the system?

Explanation / Answer

before saperation :

m = 89.30 + 57.7 = 147 kg

vx =0 m/s      vy =0 m/s     vz = 63.1 m/s

after saperation :

for first skydiver

m1 = 89.30 kg

v1x=4.93 m/s      v1y=3.75 m/s     v1z=63.1 m/s

for second skydiver

m2 = =57.70 kg

v2x= ?      v2y=?     v2z=63.1 m/s

using conservation of momentum in X-direction

m Vx = m1 V1x + m2 V2x

147 (0) = (89.30) (4.93) + (57.70) V2x

V2x = - 7.63 m/s

using conservation of momentum in Y-direction

m Vy = m1 V1y + m2 V2y

147 (0) = (89.30) (3.75) + (57.70) V2y

V2y = - 5.80 m/s

initial total KE = (0.5) m V2 = (0.5) (147) (63.1)2 = 292648.34 J

V2 = sqrt (V2x2 + V2y2 + V2z2) = sqrt ((-7.63)2 + (-5.80)2 + 63.12) = 63.82

KE of second skydiver = (0.5) m2 v22 = (0.5) (57.7) (63.82)2 = 117505.83 J

V1 = sqrt (V1x2 + V1y2 + V1z2) = sqrt (4.932 + 3.752 + 63.12) = 63.4

KE of first skydiver = (0.5) m1 v12 = (0.5) (89.30) (63.4)2 = 179473.4 J

change in KE = 292648.34 - (179473.4 + 117505.83) = - 4330.89

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