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?

Answer 1

before saperation :

m = 89.30 + 57.7 = 147 kg

v_x =0 m/s      v_y =0 m/s     v_z = 63.1 m/s

after saperation :

for first skydiver

m₁ = 89.30 kg

v_1x=4.93 m/s      v_1y=3.75 m/s     v_1z=63.1 m/s

for second skydiver

m₂ = =57.70 kg

v_2x= ?      v_2y=?     v_2z=63.1 m/s

using conservation of momentum in X-direction

m V_x = m₁ V_1x + m₂ V_2x

147 (0) = (89.30) (4.93) + (57.70) V_2x

V_2x = - 7.63 m/s

using conservation of momentum in Y-direction

m V_y = m₁ V_1y + m₂ V_2y

147 (0) = (89.30) (3.75) + (57.70) V_2y

V_2y = - 5.80 m/s

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

V₂ = sqrt (V_2x² + V_2y² + V_2z²) = 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 (V_1x² + V_1y² + V_1z²) = 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