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 postive z-axis): Yu=5.430 3.750 m/s V,,-57.10 m/s m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.20 kg, immediately after separation? Number Number m/s What is the change in kinetic energy of the system? Number Joules

Explanation / Answer

the momentum needs to be conserved as no external force is acting on them.

in x-direction:

Before separation, velocity along x-axis =0

Momentum before separation = momentum after separation

so, ( 94.80 + 52.20 ) x 0 = 94.80 x 5.430 + 52.20 x V2x

so, V2x = -9.861 m/s

Before separation, velocity along y-axis =0

Momentum before separation = momentum after separation

so, ( 94.80 + 52.20 ) x 0 = 94.80 x 3.75 + 52.20 x V2y

so, V2y = -6.81 m/s

Kinetic energy before separation = 1/2 x total mass x speed2

= 1/2 x ( 94.8 +52.2) x 57.12

= 1/2 x 147 x 57.12

= 239640.135 j

kinetic energy after separation = kinetic energy of 1st sky diver + kinetic energy of 2nd sky diver

= 1/2 x 94.8 x ( 5.432 + 3.752 + 57.12 ) + 1/2 x 52.2 x [ (-9.861)2 + (-6.81)2 + 57.12 ]

= 156607.58 j + 88845.06 j

= 245452.64 j

so change in kinetic energy of the system = kinetic energy after separation - Kinetic energy before separation

= 245452.64 -  239640.135 j

= 5812.50 J

all the best in the course work

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