Two skydivers are holding on to each other while falling straight down at a comm
ID: 1284830 • Letter: T
Question
Two skydivers are holding on to each other while falling straight down at a common terminal speed of 52.3 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 = 4.43 m/s v1,y = 4.25 m/s v1, z = 52.3 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 = ?
v2,y = ?
What is the change in kinetic energy of the system? ( Joules )
Explanation / Answer
m1*v1x + m2*v2x = (m1+m2)*ux =0
(83.8*4.43)+(52.2*v2x) = 0
v2x = -7.11 m/s
m1*v1y + m2*v2y = (m1+m2)*uy =0
(83.8*4.25)+(52.2*v2y) = 0
v2y = -6.82 m/s
m1*v1z + m2*v2z = (m1+m2)*uz
(83.8*52.3)+(52.2*v2z) = (83.3+52.2)*52.3
v2z = 51.8 m/s
v1 = sqrt(v1x^2+v1y^2+v1z^2) = 52.6
v2 = sqrt(v2x^2+v2y^2+v2z^2) = 52.7 m/s
K1 = 0.5*(m1+m2)*u^2 = 1859997.2J
K2 = 0.5*m1*v1^2 +0.5*m2*v2^2 = 188414.513 J
change = K2 -K1 = -1671582.687 J
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