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 51.10 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 83.80 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis):

V1x=4.430m/s                                V1y=4.750m/s                                         V1z=51.10m/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?

Explanation / Answer

Intital momentum of the system is along positive z-axis only as both skydivers are falling straight down. So intital momentum along x and y axis is zero. So final meomentum of the system along x and y axis must be zero.

Let the mass of first skydiver be M1 and mass of second skydiver be M2.

Let the x,y and z-component of the velocity of the second skydiver after separation be V2x, V2y and V2z respectively.

Applying conservation of momentum along x-axis we have:

M1V1x + M2V2x = 0

or (83.8kg)(4.43m/s) = -(52.2kg)V2x

or V2x = -7.11m/s = x-component of the velocity of the second skydiver after separation

Applying conservation of momentum along y-axis we have:

M1V1y + M2V2y = 0

(83.8kg)(4.75m/s) = -(52.2kg)V2y

or V2y = -7.625m/s =  y-component of the velocity of the second skydiver after separation

Applying conservation of momentum along z-axis we have:

(M1 + M2)(51.1m/s) = M1V1z + M2V2z

(83.8kg +M2)(51.1m/s) = (83.8kg)(51.10m/s) + M2V2z

or M2V2z = M2(51.1m/s)

or V2z = 51.1m/s =  z-component of the velocity of the second skydiver after separation

Intital kinetic energy = 1/2(M1+M2)(51.10 m/s)2 = 1.77X105J

Final kinetic energy = 1/2M1(V1x2 + V1y2 + V1z2) + 1/2M2(V2x2 + V2y2 + V2z2)

= 1/2(83.8kg)[(4.43m/s)2 + (4.75m/s)2 + (51.1m/s)2] + 1/2(52.2kg)[(-7.11m/s)2 + (-7.625m/s)2 + (51.1m/s)2]

= 1.11X105J + 0.7X105J = 1.81X105J

So change in KE = Final kinetic energy - Intital kinetic energy = 1.81X105J - 1.77X105J = 4000J approx.

This concludes the answers. If you find anything lacking please let me know.. I will resolve your query without delay....

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