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 55.9 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 94.8 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis):

v1x=4.93 m/s v1y=4.25 m/s v1z=55.9 m/s

What are the x- and y-components of the velocity of the second skydiver, whose mass is 57.7 kg, immediately after separation?

What is the change in kinetic energy of the 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 = 94.8 * 4.93/ 55.9 = 5.87 m/s

My = m1 * Vy - m2 * Vy2 = 0

Vy2 = m1 * Vy / m2 = 94.8 * 6.75/ 55.9 = 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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