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 63.1 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 89.3 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): v1x= 4.43m/s v1y= 5.25m/s v1z= 63.1 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.2 kg, immediately after separation? What is the change in kinetic energy of system?

Explanation / Answer

We need to use momentum conservation law. Momentum of the system is conserved in any direction. Let's denote m1 mass of the first skydiver , m2 - 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 = 83.8 * 4.43/ 63.2 = 5.87 m/s My = m1 * Vy - m2 * Vy2 = 0 Vy2 = m1 * Vy / m2 = 83.8 * 6.75/ 63.2 = 8.95 m/s E1 - kinetic energy before separation E1 = (m1+m2)*V^2/2 = (83.8 + 63.2) * 52.3^2 / 2 = 201044 J E2 - kinetic energy after separation You can finish youself. You have to calculate their velocities as V1 = sqrt (Vz^2 + Vx^2 + Vy^2); V2 = ... The answer b) will be E2 - E1

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