Question

Two skydivers are holding on to each other while falling straight down at a common terminal speed of 57.10 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver, who has a mass my of 83.80 kg, has a velocity v1 with the following components ("straight down" corresponds to the positive z-axis). 01.x = 4.930 m/s 01.y = 4.750 m/s 01,2 = 57.10 m/s What are the x- and y-components of the velocity U2 of the second skydiver, whose mass m2 is 52.20 kg, immediately after separation? Assume that there are no horizontal forces of air resistance acting on the skydivers. U2.x = -7.914444444 m/s 02.y = -7.625478927 m/s What is the change in kinetic energy of the system AK system? AKsystem = AKystem = 81615.29269 81615.29269 joules joules

Answer 1

Here ,

for this we shall use conservation of momentum

in the horizontal direction

m1 * vx1 + m2 * vx2 = 0

83.80 * 4.930 + 52.20 * vx2 = 0

solving for vx2

vx2 = -7.91 m/s

for the vertical direction

Using conservation of momentum

m1 * vy1 + m2 * vy2 = initial y momentum

83.80 * 4.750 + 52.20 * vy2 = 0

v2y = -7.62 m/s

hence , x and y components of the skydiver is vx2 is -7.91 m/s and -7.62 m/s

Now, for the change in kinetic energy

change in kinetic energy = increase in kinetic energy

change in kinetic energy = 0.50 * 83.80 * (4.93^2 + 4.75^2) + 0.50 * 52.20 * (7.91^2 + 7.62^2)

change in kinetic energy = 5112 J

the change in kinetic energy is 5112 J