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 58.30 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=4.930

v1y=4.250

v1z=58.3

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?

Explanation / Answer

Here,

m1 = 89.30 Kg

m2 = 52.2 Kg

v1x=4.930
v1y=4.250
v1z=58.3

as the momentum is conserved ,

Using conservation of momentum in all directions

v1x * m1 + v2x * m2 = 0

4.930 * 89.3 + v2x * 52.2 = 0

v2x = -8.39 m/s

for the y component

v1y * m1 + v2y * m2 = 0

4.250 * 89.30 + v2y * 52.20 = 0

v2y = -7.27 m/s

the x and y components of the velocity of the second skydiver is -8.39 m/s and -7.27 m/s

-----------------------------------

change in kinetic energy of system = 0.5 * 89.30 * (4.90^2 + 4.250^2) + 0.5 * 52.2 * (7.27^2 + 8.39^2)

change in kinetic energy of system = 5095.2 J

the change in kinetic energy of system is 5095.2 J

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