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 57.10 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=5.430m/s V1y=4.250m/s V1z=57.10m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.20 kg, immediately after separation V2x= m/s Vyx= m/s What is the change in kinetic energy of the system? joules

Explanation / Answer

initial net momentum= (94.8+52.2)*57.1 =8393.7 k^

Apply conservation of momentum,

along x direction, 94.80*v1x + 52.2*v2x= 0 { since initial mementum along x=0}

v2x= -9.861

along y direction, 94.80*v1y + 52.2*v2y= 0 ; v2y= -7.718

along z direction, 94.80*v1z + 52.2*v2z=(94.8+52.2)*57.1 =8393.7

v2z=57.1 ms-1

initial kinetic energy(Ti)= mv2/2 = 2.396 x 105 J

final velocity of the 1st diver(v1); v12=v1x2 + v1y2 + v1z2 ; kinetic energy of the 1st diver(K1)

K1= 94.8*v12/2 =1.568 x 105

final velocity of the 2nd diver(v1); v22=v2x2 + v2y2 + v2z2 ; kinetic energy of the 2nd diver(K2)

K2=8.92 x 104

Total final energy of the system(Tf) = K1+K2

change in kinetic energy=Tf- Ti =-6.4 x 104 J

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