Question

Two skydivers are holding on to each other while falling straight down at a common terminal speed of 57.1 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):

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 the system?

Answer 1

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 = 89.3 * 4.43/ 52.2 = 7.56 m/s

My = m1 * Vy - m2 * Vy2 = 0

Vy2 = m1 * Vy / m2 = 89.3 * 6.75/ 52.2 = 11.54 m/s

(b).

E1 - kinetic energy before separation

E1 = m1*V^2 = 89.3* 57.1^2 / 2 = 145577J

E2 - kinetic energy after separation

E2 = (m1+m2)*V^2/2 = (89.3 + 52.2) * 57.1^2 / 2 = 230674 J

E2 - E1 = 230674 - 145577 = 85097 J