Two skydivers are holding on to each other while falling straight down at a comm
ID: 2142430 • Letter: T
Question
Two skydivers are holding on to each other while falling straight down at a common terminal speed of 61.9 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):
V1,x = 7.93 m/s V1,y = 6.25 m/s V1,z = 52.3 m/s
What are the x- and y- components of the velocity of the second skydiver, whose mass is 63.2 kg, immediately after separation?
What is the change in kinetic engery of the system?
Two skydivers are holding on to each other while falling straight down at a common terminal speed of 61.9 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): V1,x = 7.93 m/s V1,y = 6.25 m/s V1,z = 52.3 m/s What are the x- and y- components of the velocity of the second skydiver, whose mass is 63.2 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 * 7.93/52.3 = 5.87 m/s
My = m1 * Vy - m2 * Vy2 = 0
Vy2 = m1 * Vy / m2 = 83.8 * 6.25/52.3 = 8.95 m/s
E1 - kinetic energy before separation
E1 = (m1+m2)*V^2/2 = (89.3 + 52.3) * 63.2^2 / 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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