Map sapling learning Two skydivers are holding on to each other while falling st

Question

Map sapling learning Two skydivers are holding on to each other while falling straight down at a common terminal speed of 53.50 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): Yu= 5.430 m/s i-3.750 m/s Yu=53.50 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? Number m/s 2.r Number m/s What is the change in kinetic energy of the system? Number Joules

Explanation / Answer

initial total momentum is along z axis.

magnitude of total momentum=mass of first skydiver*his speed+mass of second skydiver*his speed

=94.8*53.5+57.7*53.5=8158.75 kg.m/s

momentum along x and y axis are 0

as there are no external force, total momentum of the system will remain unchanged

hence total momentum along x axis after sepration is 0

hence mass of first skydiver*his speed along x axis+mass of second diver*his speed along x axis=0

==>94.8*5.43+57.7*V2,x=0

==>V2x=-8.92138 m/s

total momentum along y axis after sepration is 0

hence mass of first skydiver*his speed along y axis+mass of second diver*his speed along y axis=0

==>94.8*3.75+57.7*V2,y=0

==>V2x=-6.1611 m/s

initial total kinetic energy=0.5*94.8*53.5^2+0.5*57.7*53.5^2=218246.5625 J

final velocity of first skydiver=sqrt(V1x^2+V1y^2+V1z^2)=sqrt(5.43^2+3.75^2+53.5^2)=53.9054 m/s

final velocity of second skydiver=sqrt(V2x^2+V2y^2+V2z^2)=sqrt(8.92138^2+6.1611^2+53.5^2)=54.5875 m/s

hence final total kinetic energy=0.5*94.8*53.9054^2+0.5*57.7*54.5875^2=223701.6381 J

hence change in kinetic eenrgy=final kinetic energy-initial kinetic energy=5455 J

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