Two skydivers are holding on to each other while falling straight down at a comm
ID: 1365472 • Letter: T
Question
Two skydivers are holding on to each other while falling straight down at a common terminal speed of 54.70 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 83.80 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): v_1, x = 4.430 m/s v_1, y = 4.250 m/s v_1, z = 54.70 m/s What are the x-and y-components of the velocity of the second skydiver, whose mass is 52.20 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 = 83.8*4.43/52.2=7.069 m/sec
My = m1 * Vy - m2 * Vy2 = 0
Vy2 = m1 * Vy / m2=83.8*4.25/52.2=6.78 m/sec
E1 - kinetic energy before separation
E1 = (m1+m2)*V^2/2=1/2*(83.8+52.2)*54.7^2=202714.0975 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
The answer b) will be E2 - E1
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.