Two skydivers are holding on to each other while falling straight down at a comm
ID: 1606766 • Letter: T
Question
Two skydivers are holding on to each other while falling straight down at a common terminal speed of 60.70 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): v_1, x = 4.430 m/s v_1, y = 3.750 m/s v_1, z = 60.70 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 57.70 kg, immediately after separation? v_2, x = Incorrect. The initial momentum in the x-direction is zero, so both skydivers cannot have a positive momentum in the x-direction after they separate. v_2, y = What is the change in kinetic energy of the system?Explanation / Answer
Apply conservation of momentum in x-direction
final momentum in x-direction = initial momentum in x-direction
m1*v1x + m2*v2x = 0
v2x = -m1*v1x/m2
= -94.8*4.43/57.7
= -7.28 m/s <<<<<<<<<<-----------------Answer
Apply conservation of momentum in y-direction
final momentum in y-direction = initial momentum in y-direction
m1*v1y + m2*v2y = 0
v2y = -m1*v1y/m2
= -94.8*3.75/57.7
= -6.16 m/s <<<<<<<<<<-----------------Answer
change in kinetic energy = kf - ki
= (1/2)*m1*v1^2 + (1/2)*m2*v2^2 - ( (1/2)*m1*v1z^2 + 1/2)*m2*v2z^2 )
= (1/2)*94.8*(4.43^2 + 3.75^2 + 60.7^2) + (1/2)*57.7*(7.28^2 + 6.16^2 + 60.7^2) - ( (1/2)*(94.8 + 57.7)*60.7^2)
= 4220 J <<<<<<<<<<-----------------Answer
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