Since the collision is elastic, momentum is conserved in the system. We can use the conservation of momentum to find the final relative velocity of the two satellites.
Let's define the positive direction as the direction of motion of the first satellite (with mass 4.00 × 10^3 kg). Initially, the first satellite is moving at a speed of 0 m/s, while the second satellite is moving in the same positive direction at a speed of 0.25 m/s. After the collision, the two satellites move in opposite directions with some final speeds v1 and v2.
Using the conservation of momentum, we can write:
(m1)(0 m/s) + (m2)(0.25 m/s) = (m1)(v1) + (m2)(v2)
where m1 = 4.00 × 10^3 kg and m2 = 7.50 × 10^3 kg.
Simplifying and solving for v2, we get:
v2 = (m1/m2)(-0.25 m/s) + v1
Substituting the values of m1, m2, and the relative speed of the satellites, we get:
v2 = (4.00 × 10^3 kg / 7.50 × 10^3 kg)(-0.25 m/s) + v1
v2 = -0.133 m/s + v1
Similarly, using the fact that the total momentum of the system is zero, we can write:
(m1)(0 m/s) + (m2)(0.25 m/s) = (m1)(v1) + (m2)(v2)
Simplifying and solving for v1, we get:
v1 = (m2/m1)(-0.25 m/s) + v2
Substituting the values of m1, m2, and the relative speed of the satellites, and the expression for v2 that we obtained earlier, we get:
v1 = (7.50 × 10^3 kg / 4.00 × 10^3 kg)(-0.25 m/s) + (-0.133 m/s + v1)
Simplifying and solving for v1, we get:
v1 = -0.208 m/s
Therefore, the final relative velocity of the two satellites is:
v2 = -0.133 m/s + v1 = -0.133 m/s - 0.208 m/s = -0.341 m/s
Note that the negative sign indicates that the two satellites are moving away from each other after the collision.