Two manned satellites approaching one another at a relative speed of 0.100 m/s intend to dock. the first has a mass of 2.50 ? 103 kg, and the second a mass of 7.50 ? 103 kg. assume that the positive direction is directed from the second satellite towards the first satellite.

(a) calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest.
(b) what is the loss of kinetic energy in this inelastic collision?
(c) repeat both parts, in the frame of reference in which the second satellite was originally at rest.
final velocity
explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both. (do this on paper. your instructor may ask you to turn in this work.)

I'll do it rn and then answer

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s