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 54.70 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 94.80 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis):

V1x = 5.430 m/s                                           V1y= 4.250 m/s                                    V1z = 54.70 m/s

What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.20 kg, immediately after separation?

WHAT IS THE CHANGE IN KINETIC ENERGY OF THE SYSTEM IN JOULES?

Explanation / Answer

V2x = m1 V1x / m2 = 52.2 * 5.43 / 94.8 = 2.99 m/s   conservation of momentum
V2y = m1 V1y / m2 = 2.34 m/s
V1^2 = 5.43^2 + 4.25^2 + 54.7^2 = 3040 m^2/s^2
V2^2 = 2.99^2 + 2.34^2 + 54.7^2 = 3007 m^2/s^2
E = 1/2 m1 V1^2 + 1/2 m2 V2^2      after separation
E = 144100 + 78480 = 2.23 * 10E5 J
E (before separation) = 1/2 (m1 + m2) Vy^2 = 2.20 * 10E5 J
Energy is not conserved because of energy expended in the separation
Difference is about 3000 J

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