Two skydivers are holding on to each other while falling straight dawn at a comm

Question

Two skydivers are holding on to each other while falling straight dawn at a common terminal speed of 6 Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass 89.30 kg) has the following velocity components (with "štraight down" corresponding to the positive z-axis) 1.90 m/s. of x-4930 m/s 1, -4250 m/s 61.90 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 63.20 kg, immediately after separation? Number x6.96 m/s Number 6.005 m/s What is the change in kinetic energy of the systen? Number 57185.71 Joules

Explanation / Answer

We must use the law of conservation of linear momentum Momentum of the system is conserved in any direction

Let us denote m1 mass of the first skydiver and m2 the second

When they fall together Mx = 0, My = 0
When they push away from each other Mx = (m1 * Vx) - (m2 * Vx2) = 0
Vx2 = (m1 * Vx) / m2 = (89.30 * 4.93)/ 63.2 = 6.96 m/s
My = (m1 * Vy) - (m2 * Vy2) = 0
Vy2 = (m1 * Vy) / m2 = (89.3 * 4.250)/ 63.2 = 6.00 m/s
E1 - kinetic energy before separation
E1 = (1/2) (m1 + m2)V2

= (1/2) (89.3 + 63.2) (61.9)2

= 292160 J
E2 - kinetic energy after separation
to get the velocites
V1 = sqrt (Vz2 + Vx2 + Vy2);

and also v2
the change in kinetic energy will be E2 - E 1

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