Question 11 of 16 Map Sapling Learning macmillan learning Two skydivers are hold

Question

Question 11 of 16 Map Sapling Learning macmillan learning Two skydivers are holding on to each other while falling straight down at a common terminal speed of 63.10 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): = 3.750 m / s 1,y v1.1 = 63.10 m/s v, r = 4.930 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 m/s 2.x Number m/s 2.y What is the change in kinetic energy of the system? Number Joules

Explanation / Answer

Conserving momentum in each direction

94.8*4.9=52.2*v2x

We got v2x=-8.9 m/s

Similarly v2y=-6.8 m/9

For v2z

Initial momentum=-(94.8+52.2)*63.1=94.8*63.1+52.2*v2z

We got v2z=-292.3 m/s

(- means direction is opposite to direction of first person)

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