Question

Y push away from each other. immediately after separation, the first skydiver, who has a mass ð‘š1 of 89.30 kg, has a velocity ð‘£1 with the following components ("straight down" corresponds to the positive ð‘§- axis). ð‘£1,ð‘¥=4.930 m/sð‘£1,ð‘¦=3.750 m/sð‘£1,ð‘§=61.90 m/s what are the ð‘¥- and ð‘¦- components of the velocity ð‘£2 of the second skydiver, whose mass ð‘š2 is 57.70 kg, immediately after separation? assume that there are no horizontal forces of air resistance acting on the skydivers. ð‘£2,ð‘¥= -7.63 m/s ð‘£2,ð‘¦= -5.8 m/s what is the change in kinetic energy of the system Δð¾system?

Answer 1

The resulting velocity components are v'2 = (-7.63 m/s)i + (-5.8 m/s)j. To find the velocity components of the second skydiver after separation, we can use the principle of conservation of momentum and solve for the velocity components using the given masses and velocities.

Step-by-step explanation:

To find the velocity components of the second skydiver after separation, we can use the principle of conservation of momentum. Since there are no horizontal forces of air resistance acting on the skydivers, the momentum in the horizontal direction is conserved.

The total momentum before separation is equal to the total momentum after separation. Therefore, we can write the equation as:

(m1 × v1) + (m2 × v2) = (m1 × v'1) + (m2 × v'2)

Given that the first skydiver has a mass (m1) of 89.30 kg and a velocity of v1 = (4.930 m/s)i + (3.750 m/s)j + (61.90 m/s)k, and the second skydiver has a mass (m2) of 57.70 kg, we can substitute these values into the equation and solve for the velocity components of the second skydiver. The resulting velocity components are v'2 = (-7.63 m/s)i + (-5.8 m/s)j.