Sapling Learning Two skydivers are holding on to each other while falling straig

Question

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 89.30 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): 4.430 m/s "v=4.250 ml /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,x Number m/s What is the change in kinetic energy of the system? Number Joules

Explanation / Answer

Here, the momentum will be conserved before and after they separate.

So, (m1+m2)v= m1v1+m2v2 {where v= initial speed given along z direction i.e. k and directions vectors are Vx=i, Vy=j and Vz=k}

Or, (89.3+57.7)*53.5 k = 89.3(4.43 i+4.25 j+53.5 k) + 57.7(V2x i+V2y j + V2z k)

Or, V2x i+V2y j + V2z k = ( 7864.5 k - 395.6 i - 379.5 j - 4777.55 k ) / 57.7

Or, V2x i+V2y j + V2z k = ( 3087 k - 395.6 i - 379.5 j ) / 57.7

or, V2x i+V2y j + V2z k = ( - 395.6 i - 379.5 j + 3087 k ) / 57.7

or, V2x i+V2y j + V2z k = -6.85 i - 6.57 j + 53.5 k

So, V2x = -6.85 m/s and V2y = -6.57 m/s

Now, initial Kinetic energy of the system= 0.5*(m1+m2)v^2= 0.5*147*53.5^2 = 210.375 kJ

final velocity of m1 is v1= (4.43^2+4.25^2+53.5^2)^1/2 = 53.85 m/s

So final kinetic energy of m1= 0.5*m1*v1^2= 194.477 kJ

similarly v2= (6.85^2+6.57^2+53.5^2)^1/2 = 54.33 m/s

so kinetic energy of m2 = 0.5*m2*v2^2 = 85.157 kJ

So final kinetic energy= 194.477 kJ+ 85.157 kJ = 279.634 kJ

So change in kinetic energy = final - initial kinetic energy =  279.634 kJ - 210.375 kJ = 69.26 kJ

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