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 89.3 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis) : V1, x = 7.43 m/s V1, y = 5.75 m/s V1, z = 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? V2, x = ? m/s V2, y = ? m/s What is the change in kinetic engery of the system? in Joules

Question

Physics

Semaj Simon

11 May, 00:34

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 89.3 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis) : V1, x = 7.43 m/s V1, y = 5.75 m/s V1, z = 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? V2, x = ? m/s V2, y = ? m/s What is the change in kinetic engery of the system? in Joules

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Answers (1)

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Deven Mcclain · Answer 1

10,041.7 J

Explanation:

Using Newton's second law of motion of motion; the linear momentum in each direction will be conserved.

For each direction,

Momentum before collision = momentum after collision

In the x - direction, there was initially no x-component for velocity before collision for the two bodies,

0 = m₁v₁ₓ + m₂v₂ₓ

0 = (89.3) (7.43) + (57.7) (v₂ₓ)

v₂ₓ = (-89.3*7.43) / 57.7

v₂ₓ = - 11.50 m/s (minus indicating that the motion is in the opposite direction the first diver's motion).

In the y - direction, there was initially no y-component for velocity before collision for the two bodies,

0 = m₁v₁ᵧ + m₂v₂ᵧ

0 = (89.3) (5.75) + (57.7) (v₂ᵧ)

v₂ᵧ = (-89.3*5.75) / 57.7

v₂ᵧ = - 8.90 m/s (minus indicating that the motion is in the opposite direction the first diver's motion).

And the velocity in the z-axis for both has been given to be 55.9 m/s

Kinetic energy of the system before leaving each other = (1/2) (89.3+57.7) (55.9²) = 229673.535 J

Kinetic energy of skydiver 1 after collision

To do this, we first find the magnitude of the velocity

v₁ = √ (7.43² + 5.75² + 55.9²) = 56.684 m/s

Kinetic energy = (1/2) (89.3) (56.684²) = 143463.84 J

Kinetic energy of skydiver 2 after collision

To do this, we first find the magnitude of the velocity

v₂ = √[ (-11.50) ² + (-8.90) ² + 55.9²] = 57.76 m/s

Kinetic energy = (1/2) (57.7) (57.76²) = 96251.39 J

Change in kinetic energy of the system = (final kinetic energy) - (initial kinetic energy)

Change in kinetic energy of the system = 143463.84 + 96251.39 - 229673.535 = 10041.695 J = 10041.7 J