Two people jump out of an airplane. They hold on to each other while falling str

Question

Two people jump out of an airplane. They hold on to each other while falling straight down at a shared terminal speed of 63.1 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 83.8 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis):    

V1,x = 5.43 m/s                    V1,y = 3.75m/s                V1,z = 63.1 m/s

What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.2 kg, immediately after separation?

V2,x = ??? m/s

V2,y = ??? m/s

What is the change in kinetic energy of the system?

?KE = ??? Joules

Explanation / Answer

applying conservation of momentum principle :

initial velocity of both persons = V = 63.1 m/s along Z direction

so intital momentum = (m1 +m2)* 63.1 k (vector notation)

final momentum = m1*V1 + m2*V2 = 83.8*(5.43i +3.75j + 63.1 k) + 52.2 (V2,x i + V2y j + V2,z k);

Equate both expressions : (83.8 + 52.2)*63.1 k = (83.8*5.43 + 52.2*V2,x) i + (83.8*3.75 + 52.2*V2,y) j +

                                                                                         (83.8*63.1 + 52.2* V2,z) k ;

vectors along x and y directions equal to zero ; V2,x = -(83.8*5.43/52.2) = -8.717 m/s

                                                                                     V2,y = -(83.8*3.75/52.2) = -6.02 m/s

Initial kinetic energy = 0.5 *(83.8 + 52.2) * 63.1^2 = 270749 J

Final KE = 0.5*83.8*(5.43^2 + 3.75^2 + 63.1^2) + 0.5*52.2*(8.717^2 + 6.02^2 + 63.1^2)

                = 168654 + 106849 = 275503 J

Change in KE = 275503-270749 = 4754 J

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